Research ArticleVariational and Numerical Analysis for Frictional ContactProblem with Normal Compliance inThermo-Electro-Viscoelasticity

Mustapha Bouallala El Hassan Essoufi and Mohamed Alaoui

Univ Hassan 1 Laboratory MISI 26000 Settat Morocco

Correspondence should be addressed to Mustapha Bouallala bouallalamustaphaangmailcom

Received 4 September 2018 Accepted 17 July 2019 Published 2 September 2019

Academic Editor Gaston Mandata Nrsquoguerekata

Copyright copy 2019 Mustapha Bouallala et al -is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In this paper we consider a mathematical model of a contact problem in thermo-electro-viscoelasticity with the normalcompliance conditions and Trescarsquos friction law We present a variational formulation of the problem and we prove the existenceand uniqueness of the weak solution We also study the numerical approach using spatially semidiscrete and fully discrete finiteelement schemes with Eulerrsquos backward scheme Finally we derive error estimates on the approximate solutions

1 Introduction

In the recent years piezoelectric contact problems havebeen of great interest to modern engineering Generalmodels of electroelastic characteristics of piezoelectricmaterials can be found in [1 2] -e problems of piezo-viscoelastic materials have been studied with differentcontact conditions within linearized elasticity in [3ndash5] andwithin nonlinear viscoelasticity in [6ndash8] -e modeling ofthese problems does not take into account the thermiceffect Mindlin [9] was the first to propose the thermo-piezoelectric model -e mathematical model which de-scribes the frictional contact between a thermo-piezo-electric body and a conductive foundation is alreadyaddressed in the static case in [10 11]

Sofonea et al considered in [12] the modeling of qua-sistatic viscoelastic problem with normal compliance fric-tion and damage they proved the existence and uniquenessof the weak solution and they derived error estimates on theapproximate solutions

In the article [13] we find the recent result of a newquasistatic mathematical model which describes the chosenthermo-electro-viscoelastic body behavior and the contactby Signorini condition with nonfrictional and non-conductive foundation also the variational formulation of

this problem is derived and its unique weak solvability isestablished

In this paper we consider a quasi-static contact problemwith Trescarsquos friction between a thermo-electro viscoelasticbody and an electrically and thermally conductive rigidfoundation -e novelty in this model which can be con-sidered as the generalization of the model presented in [13]lies in the use of the penalized normal compliance contactcondition

σ] u] minus g( 1113857 minus1ε

u] minus g1113858 1113859+ εgt 0 (1)

-is means that we allow a weak interpenetration be-tween the body and the foundation On the contact zone weconsider the following regularized electrical and thermalconditions

D middot ] ψ u] minus g( 1113857ϕL φ minus φF( 1113857

zq

z] kc u] minus g( 1113857ϕL θ minus θF( 1113857

(2)

which describe both the thermal and electrical conductivitiesof the foundation -is leads to nonlinear coupling between

HindawiInternational Journal of Differential EquationsVolume 2019 Article ID 6972742 14 pageshttpsdoiorg10115520196972742

the mechanical displacement and thermal and electricalfields and hence more complexities on the model

Since the friction conditions are inequalities we derive aquasivariational formulation of this problem and we provethe existence and uniqueness of the weak solution based onarguments variational inequalities Galerkin method com-pactness method and Banach fixed point theorem Wederive error estimates for the numerical approximationsbased on discrete schemes

-e paper is structured as follows In Section 2 wepresent the model of equilibrium process of the thermo-electro-viscoelastic body in frictional contact with a con-ductive rigid foundation and we introduce the notationsand assumptions on the problem data We also derive thevariational formulation of the problem We state the mainresults concerning the existence and the uniqueness of aweak solution We present a spatially semidiscrete schemeand a fully discrete scheme to approximate the contactproblem We then use the finite element method to dis-cretize the domain Ω and Eulerrsquos forward scheme to dis-cretize the time derivatives Finally the proofs areestablished in Section 3

2 Formulation and Main Results

21 Problem Setting We consider a body of a piezoelectricmaterial which occupies the domainΩ sub Rd(d 2 3) in thereference configuration which will be supposedly boundedwith a smooth boundary zΩ Γ -is boundary is dividedinto three open disjoint parts ΓD ΓN and ΓC on one handand a partition of ΓD cup ΓN into two open parts Γa and Γb onthe other hand such that meas(ΓD)gt 0 and meas(Γa)gt 0Let [0 T] be the time interval where Tgt 0

-e body is submitted to the action of body forces ofdensity f0 a volume electric charge of density q0 and a heatsource of constant strength q1 It is also submitted to me-chanical electrical and thermal constants on the boundaryIndeed the body is assumed to be clamped in ΓD andtherefore the displacement field vanishes there Moreoverwe assume that a density of traction forces denoted by f2acts on the boundary part ΓN We also assume that theelectrical potential vanishes on Γa and a surface electricalcharge of density q2 is prescribed on Γb We consider that thetemperature θ0 is prescribed on the surface ΓD cup ΓN

In the reference configuration the body may come incontact over ΓC with an electrically thermally conductivefoundation Assume that its potential and temperature aremaintained at φF and θF -e contact is frictional and theremay be electrical charges and heat transfer on the contactsurface -e normalized gap between ΓC and the rigidfoundation is denoted by g

In the following sections we use Sd to denote the spaceof second-order symmetric tensors on Rd while ldquordquo and ||

will represent the inner product and the Euclidean norm onSd and Rd that is

u middot v uivi

|v| (v middot v)12

forallu v isin Rd

σ middot τ σijτij

|τ| (τ middot τ)12

forallσ τ isin Sd

(3)

We denote u Ω times [0 T]⟶ Rd as the displacementfield σ Ω⟶ Sd and σ (σij) the stress tensor θ Ω times

[0 T]⟶ Rd the temperature q Ω⟶ Rd and q (qi)

the heat flux vector and D Ω⟶ Rd and D (Di) theelectric displacement field We also denote E(φ) (Ei(φ))

as the electric vector field where φ Ω times [0 T]⟶ R is theelectric potential Moreover let ε(u) (εij(u)) denote thelinearized strain tensor given by εij(u) 12(uij + uji) andldquoDivrdquo and ldquodivrdquo denote the divergence operators for tensorand vector valued functions respectively ie Div σ (σijj)

and div ξ (ξjj) We shall adopt the usual notation fornormal and tangential components of displacement vectorand stress υn υ middot n υτ υ minus υnn σn (σn) middot n andστ σn minus σnn where n denotes the outward normal vectoron Γ

Problem (P) Find a displacement field u Ω times ]0

T[⟶ Rd an electric potential φ Ω times ]0 T[⟶ R and atemperature field θ Ω times ]0 T[⟶ R such that

σ Iε(u) minus εlowastE(φ) minus θM

+ Cε( _u) inΩ times(0 T)

(4)

D Eε(u) + βE(φ)

minus θ minus θlowast( 1113857P inΩ times(0 T)(5)

Div σ + f0 0 inΩ times(0 T) (6)

divD q0 inΩ times(0 T) (7)

_θ + div q q1 inΩ times(0 T) (8)

u 0 on ΓD times(0 T) (9)

σ] f2 on ΓN times(0 T) (10)

φ 0 on Γa times(0 T) (11)

D middot ] qb on Γb times(0 T) (12)

θ 0 on ΓD cup ΓN( 1113857 times(0 T) (13)

u(0 x) u0 inΩ (14)

2 International Journal of Differential Equations

θ(0 x) θ0 inΩ (15)

σ] u] minus g( 1113857 minus1ε

u] minus g1113858 1113859+ εgt 0 on ΓC times(0 T) (16)

στ

le S on ΓC times(0 T) (17)

στ

lt S⟹ _uτ 0 on ΓC times(0 T) (18)

στ

S⟹existλne 0such that στ minus λ _uτ on ΓC times(0 T)

(19)

D middot ] ψ u] minus g( 1113857ϕL φ minus φF( 1113857 on ΓC times(0 T) (20)

zq

z] kc u] minus g( 1113857ϕL θ minus θF( 1113857 on ΓC times(0 T) (21)

Here the equations (4) and (5) represent the thermo-electro-viscoelastic constitutive law of the material in whichσ (σij) denotes the stress tensor ε(u) is the linearizedstrain tensor E(φ) is the electric field I (fijkl)E (eijk) M (mij) β (βij) P (pi) and C (cijkl)

are respectively elastic piezoelectric thermal expansionelectric permittivity pyroelectric tensor and (fourth-order)viscosity tensor Elowast is the transpose of E given by

Elowast

elowastijk1113872 1113873 e

lowastijk ekij (22)

Eσυ σElowastυ

forallσ isin Sd

υ isin Rd

(23)

-e constant θlowast represents the reference temperatureFourierrsquos law of heat conduction is given by

q minus Knablaθ inΩ times(0 T) (24)

where K (kij) denotes the thermal conductivity tensorEquations (6)ndash(8) represent the equilibrium equations

for the stress Relations (9) and (10) (11) and (12) and (13)represent the mechanical the electrical and the thermalboundary conditions -e unilateral boundary condition(16) represents the normal compliance condition and(17)ndash(19) represent Trescarsquos friction law in which S is thecoefficient of friction

Following [14] the contact conditions (20) and (21) onΓC are obtained as follows

When there is no contact at a point on the surface thereis no free electrical charges on the surface and no thermaltransfer that is

u] ltg⟹D middot ] 0

q middot ] 01113896 (25)

If the contact holds ie u] geg the normal component ofthe electric displacement field or the free charge (respthermal transfer) is assumed to be proportional to the

difference between the potential of foundation and thebodyrsquos surface potential (resp to the difference between thetemperature of foundation and the bodyrsquos surface temper-ature) -us

u] geg⟹D middot ] kφ φ minus φF( 1113857

q middot ] kθ θ minus θF( 1113857

⎧⎪⎨

⎪⎩(26)

We combine (25) and (26) to obtain

D middot ] kφχ[0+infin) u] minus g( 1113857 φ minus φF( 1113857

q middot ] kθχ[0+infin) u] minus g( 1113857 θ minus θF( 1113857

⎧⎪⎨

⎪⎩(27)

where χ[0+infin) is the characteristic function of the interval[0 +infin) defined by

χ[0+infin)(s) 0 if slt 0

1 if sge 01113896 (28)

Equation (27) represents the regularization electricalcontact condition and the heat flux condition on ΓCwhere

ϕL(s)

minus L if slt minus L

s if minus Lle sleL

L if sgt L

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

ψ(r)

0 if rlt 0

cδr if 0le rle1δ

c if rgt1δ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

and where c kφ kθ and L is a large positive constant δ gt 0is a small parameter kc r⟶ kc(r) is supposed to be zerofor rlt 0 and positive otherwise nondecreasing and Lipschitzcontinuous

Remark 1 We note that when ψ equiv 0 equality (20) becomes

D middot ] 0 on ΓC times(0 T) (30)

which models the case when the foundation is a perfectelectric insulator

Similarly in equality (21) we have

q middot ] 0 on ΓC times(0 T) (31)

22 Weak Formulation and Uniqueness Result To obtain avariational formulation of Problem (P) we need addi-tional notations and need to recall some definitions in thesequel

International Journal of Differential Equations 3

We use the following functional Hilbert spaces

L2(Ω) L

2(Ω)

d

H1(Ω) H

1(Ω)

d

H σ isin Sd σ σij σij σji isin L

2(Ω)1113966 1113967

(32)

W D (D)i isin H1(Ω) Di isin L

2(Ω) divD isin L

2(Ω)1113966 1113967

(33)

endowed with the canonical inner product given by

(u v)L2(Ω) 1113946Ω

uividx

(σ τ)H 1113946Ωσiτidx

(u v)H1(Ω) (u v)L2(Ω) +(ε(u) ε(v))H

(D middot E) (D middot E)L2(Ω) +(divD middot divE)L2(Ω)

(34)

and the associated norms L2(Ω) H1(Ω) H and Wrespectively

LetVW andQ be the closed subspaces ofH1(Ω) given by

V v isin H1(Ω) v 0 on ΓD1113966 1113967

W ξ isin H1(Ω) ξ 0 on Γa1113966 1113967

Q η isin H1(Ω) η 0 on ΓD cup ΓN1113966 1113967

(35)

and the set of admissible displacements

Vad v isin V v] minus gle 0 on ΓC1113864 1113865 (36)

It is known that VW and Q are real Hilbert spaces withthe inner products (u v)V (ε(u) ε(v))H (φ ξ)W

(nablaφnablaξ)L2(Ω) and (θ η)Q (nablaθnablaη)L2(Ω) respectivelyMoreover the associated norm ||v||V ||ε(v)||H is

equivalent on V to the usual norm H1(Ω) and ||ξ||W

||nablaξ||L2(Ω) and ||η||Q ||nablaη||L2(Ω) are equivalent on W andQ respectively with the usual norms H1(Ω)

By Sobolevrsquos trace theorem there exists three positiveconstants Cs1 Cs2 and Cs3 depending on Ω ΓC ΓN ΓD Γaand Γb

||v||L2(Γ)d le Cs1||v||V forallv isin V (37)

ξL2 Γc( ) le Cs2ξW forallξ isinW (38)

ηL2 Γc( ) le Cs3ηQ forallη isin Q (39)

Since meas(ΓD)gt 0 and Kornrsquos inequality hold

||ε(v)||H ge CK||v||H1(Ω) forallv isin V (40)

where CK in a nonnegative constant depending only on Ωand ΓD Notice also that since meas(Γa)gt 0 the followingFriedrichsndashPoincare inequalities hold

||nablaξ||W geCF1||ξ||W forallξ isinW (41)

nablaηL2(Ω) geCF2ηQ forallη isin Q (42)

where CF1 and CF2 are the positive constants which dependonly on Ω Γa ΓD and ΓN

For a real Banach space X and 1lepleinfin we consider theBanach spaces C(0 T X) and C1(0 T X) of continuous andcontinuously differentiable functions from [0 T] to X withthe norms

uC(0TX) suptisin[0T]

u(t)X

uC1(0TX) suptisin[0T]

u(t)X

+ suptisin[0T]

_u(t)X

(43)

To simplify the writing we denote by a V times V⟶ Rb W times W⟶ R c V times V⟶ R and d Q times Q⟶ R

the following bilinear and symmetric applications

a(u v) ≔ (Iε(u) ε(v))H

b(φ ξ) ≔ (βnablaφnablaξ)L2(Ω)

c(u v) ≔ (Cε(u) ε(v))H

d(θ η) ≔ (Knablaθnablaη)L2(Ω)

(44)

and by e V times W⟶ R m Q times V⟶ R andp Q times W⟶ R the following bilinear applications

e(v ξ) ≔ (Eε(v)nablaξ)L2(Ω) Elowastnablaξ ε(v)( 1113857V

m(θ v) ≔ (Mθ ε(v))Q

p(θ ξ) ≔ (Pnablaθnablaξ)L2(Ω)

(45)

In the study of mechanical Problem (P) we make thefollowing assumptions

HP1 -e elasticity operator I Ω times Sd⟶ Sd theelectric permittivity tensor β (βij) ⟶Ω times Rd

⟶ Rd the viscosity tensor c Ω times Sd⟶ Sd andthe thermal conductivity tensor K (kij) Ω times Rd

⟶ Rd satisfy the usual properties of symmetryboundedness and ellipticity

fijkl fjikl flkij isin Linfin

(Ω)

βij βji isin Linfin

(Ω)

cijkl cjikl clkij isin Linfin

(Ω)

kij kji isin Linfin

(Ω)

(46)

and there exists that mI mβ mc mK gt 0 such that

fijkl(x)ξkξl ge mI||ξ||2 forallξ isin Sd

forallx isin Ω

cijkl(x)ξkξl ge mc||ξ||2 forallξ isin Sd

forallx isin Ω

βijζ iζj ge mβ||ζ||2 kijζ iζj ge mK||ζ||

2 forallζ isin Rd

(47)

4 International Journal of Differential Equations

HP2 -e piezoelectric tensor Ε (eijk) Ω times Sd

⟶ R the thermal expansion tensor M (mij)

Ω times R⟶ R and the pyroelectric tensor P (pi)

Ω⟶ Rd satisfy

eijk eikj isin Linfin

(Ω)

mij mji isin Linfin

(Ω)

pi isin Linfin

(Ω)

(48)

and there exist the positive constantsMI Mβ Mc MK ME MM and MP such that

|a(u v)|leMIuVvV

|b(φ ξ)|leMβφWξW

|c(u v)|leMcuVvV

|d(θ η)|leMKθQηQ

|e(v ξ)|leMΕvVξW

|m(θ v)|leMMθQvV

|p(θ ξ)|leMPθQξW

(49)

HP3 -e surface electrical conductivityψ ΓC times R⟶ R+ and the thermal conductance kc

ΓC times R⟶ R+ satisfy the following hypothesis for(π ψ kc) existMπ gt 0 such that |π(x u)|leMπ foralluisin R ae x isin ΓC and x⟶ π(x u) is measurable onΓC for all u isin R and is zero for all ule 0-e function u⟶ π(x u) is a Lipschitz function onRfor all x isin Γ3

|π(x u1) minus π(x u2)|leLπ|u1 minus u2| forallu1 u2 isin R where Lπis a positive constant

HP4 -e forces the traction the volume the surfacescharge densities and the strength of the heat source areas follows

f0 isin C 0 T L2(Ω)

d1113872 1113873

f2 isin C 0 T L2 ΓN( 1113857

d1113872 1113873

qb isin L2 0 T L

2 Γb( 11138571113872 1113873

q1 isin L2 0 T L

2(Ω)1113872 1113873

q0 isin L2 0 T L

2(Ω)1113872 1113873

(50)

-e potential and the temperature satisfy

φF isin L2 0 T L

2 ΓC( 11138571113872 1113873

θF isin L2 0 T L

2 ΓC( 11138571113872 1113873(51)

-e initial conditions the friction-bounded functionand the gap function satisfy

u0 isin K

θ0 isin Q

S isin Linfin ΓC( 1113857

Sge 0

g isin L2 ΓC( 1113857

gge 0

(52)

Using Rieszrsquos representation theorem we definef [0 T]⟶ V qe [0 T]⟶W and qth [0 T]⟶ Q

by the following

(f(t) v)V 1113946Ω

f0(t) middot v dx + 1113946ΓN

f2(t) middot v da forallv isin V

(53)

qe(t) ξ( 1113857W 1113946Ω

q0(t) middot ξ dx minus 1113946Γb

qb(t) middot ξ da forallξ isin V

(54)

qth(t) η1113872 1113873Q

1113946Ω

q1(t) middot η dx forallη isin Q (55)

We define the mappings j V⟶ R 9 V times V⟶ Rℓ V times W2⟶ R and χ V times Q2⟶ R ae t isin ]0 T[ by

j(v) 1113946ΓC

S vτ

da forallv isin V (56)

9(u v) 1113946ΓC

u]1113858 1113859+

minus g( 1113857v]da lang u]1113858 1113859+

minus g v]rangΓC

forallu v isin V

(57)

ℓ(u(t) φ(t) ξ) 1113946ΓCψ u](t) minus g( 1113857ϕL φ(t) minus φF( 1113857ξ da

forallu isin V forallφ ξ isinW

(58)

χ(u(t) θ(t) η) 1113946ΓC

kc u](t) minus g( 1113857ϕL θ(t) minus θF( 1113857η da

forallu isin V forallθ η isin Q

(59)

respectivelyNow by a standard variational technique it is

straightforward to see that if (uφ θ) satisfies the conditions(4)ndash(21) ae t isin ]0 T[ then

(σ(t) ε(v) minus ε( _u(t)))H +1ε

9(u(t) v minus _u(t)) + j(v)

minus j( _u(t))ge (f(t) v minus _u(t))V forallv isin Va d

(60)

(D(t)nablaξ)L2(Ω) ℓ(u(t) φ(t) ξ) minus qe(t) ξ( 1113857W

forallξ isinW(61)

International Journal of Differential Equations 5

(q(t)nablaη)L2(Ω) ( _θ(t) η)Q + χ(u(t) θ(t) η) minus qth(t) η1113872 1113873Q

forallη isin Q

(62)

We assume that the initial conditions u0 and θ0 satisfy thefollowing compatibility condition there existsφ0 isinW such that

b φ0 ξ( 1113857 minus e u0 ξ( 1113857 minus p θ0 ξ( 1113857 + ℓ u0φ0 ξ( 1113857 qe0 ξ1113872 1113873

W

forallξ isinW

(63)

-is nonlinear problem has a unique solution φ0 byusing the fixed point theorem

Using all of these assumptions notations and E minus nablaφwe obtain the following variational formulation of theProblem (P)

Problem (PV) find a displacement field u ]0

T[⟶ Vad an electric potential φ ]0 T[⟶W and atemperature field θ ]0 T[⟶ Q ae t isin ]0 T[ such that

c( _u(t) v minus _u(t)) + a(u(t) v minus _u(t)) + e(v minus _u(t)

φ(t)) minus m(θ(t) v minus _u(t)) +1ϵ

9(u(t) v minus _u(t))

+ j(v) minus j( _u(t)) ge (f(t) v minus _u(t))V forallv isin Vad

(64)

b(φ(t) ξ) minus e(u(t) ξ) minus p(θ(t) ξ)

+ ℓ(u(t) φ(t) ξ) qe(t) ξ( 1113857W forallξ isinW(65)

d(θ(t) η) +( _θ(t) η)Q + χ(u(t) θ(t) η) qth(t) η1113872 1113873Q

forallη isin Q

(66)

u(0 x) u0(x)

θ(0 x) θ0(x)(67)

We present now the existence and the uniqueness ofsolution to Problem (PV)

Theorem 1 Assume that the assumptions (HP1) minus (HP4)(37)ndash(42) for εgt 0and the conditions

mβ gtMψc21

mK ltc2 Mkc

c2 + LkkLc01113872 1113873

2

(68)

hold Ben Problem (PV) has a unique solution as follows

u isin C1(0 T V)

φ isin L2(0 T W)

θ isin L2(0 T Q)

(69)

23 Spatially Semidiscrete Approximation In this para-graph we consider a semidiscrete approximation of theProblem (PV) by discretizing the spatial domain usingthe finite element method Let Th be a regular finite el-ement partition of the domain Ω compatible with theboundary partition Γ ΓC cup ΓD cup ΓN We then define afinite element space Vh sub V and Vh

a d Va d capVh for theapproximates of the displacement field u Wh subW for theelectric potential φ and Qh sub Q for the temperature θdefined by

Vh

vh isin [C(Ω)]

d v

h|Tr isin P1(Tr)1113858 1113859

dforallTr isin Th v

h 0 on ΓC1113966 1113967

Wh

ξh isin C(Ω) ξh|Tr isin P1(Tr)forallTr isin Th

ξh on Γa1113966 1113967

Qh

ηh isin C(Ω) ηh|Tr isin P1(Tr)forallTr isin Th

ηh 0 on ΓD cup ΓN1113966 1113967

(70)

A spatially semidiscrete scheme can be formed as thefollowing problem

Problem (PVh)Find uh ]0 T[⟶ Vh φh ]0

T[⟶Wh and θh ]0 T[⟶ Qh ae t isin ]0 T[ such thatfor vh isin Vh

a d ξh isinWh and ηh isin Qh

c _uh(t) v

hminus _u

h(t)1113872 1113873 + a u

h(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θh(t) v

hminus _u

h1113872 1113873

+1ε

9 uh(t) v

hminus _u

h(t)1113872 1113873 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873

ge f(t) vh

minus _uh(t)1113872 1113873

(71)

b φh(t) ξh

1113872 1113873 minus e uh(t) ξh

1113872 1113873 minus p θh(t) ξh

1113872 1113873

+ ℓ uh(t)φh

(t) ξh1113872 1113873 qe(t) ξh

1113872 1113873(72)

d θh(t) ηh

1113872 1113873 + _θh(t) ηh

1113874 1113875 + χ uh(t) ηh

(t) ηh1113872 1113873

qth(t) ηh1113872 1113873

(73)

uh(0) u

h0

φh(0) φh

0

θh(0) θh

0

(74)

Here uh0 isin Vh φh

0 isinWh and θh0 isin Qh are appropriate

approximations of u0 φ0 and θ0 respectivelyUsing the same argument presented in Section 2 it can

be shown that Problem (PVh) has a unique solutionuh isin C1(0 T Vh) φh isin L2(0 T Wh) and θh isin L2(0 T Qh)

In this paragraph we are interested in obtaining esti-mates for the errors (u minus uh) (φ minus φh) and (θ minus θh)

Using the initial value condition we have

6 International Journal of Differential Equations

u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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Submit your manuscripts atwwwhindawicom

θ(0 x) θ0 inΩ (15)

σ] u] minus g( 1113857 minus1ε

u] minus g1113858 1113859+ εgt 0 on ΓC times(0 T) (16)

στ

le S on ΓC times(0 T) (17)

στ

lt S⟹ _uτ 0 on ΓC times(0 T) (18)

στ

S⟹existλne 0such that στ minus λ _uτ on ΓC times(0 T)

(19)

D middot ] ψ u] minus g( 1113857ϕL φ minus φF( 1113857 on ΓC times(0 T) (20)

zq

z] kc u] minus g( 1113857ϕL θ minus θF( 1113857 on ΓC times(0 T) (21)

Here the equations (4) and (5) represent the thermo-electro-viscoelastic constitutive law of the material in whichσ (σij) denotes the stress tensor ε(u) is the linearizedstrain tensor E(φ) is the electric field I (fijkl)E (eijk) M (mij) β (βij) P (pi) and C (cijkl)

are respectively elastic piezoelectric thermal expansionelectric permittivity pyroelectric tensor and (fourth-order)viscosity tensor Elowast is the transpose of E given by

Elowast

elowastijk1113872 1113873 e

lowastijk ekij (22)

Eσυ σElowastυ

forallσ isin Sd

υ isin Rd

(23)

-e constant θlowast represents the reference temperatureFourierrsquos law of heat conduction is given by

q minus Knablaθ inΩ times(0 T) (24)

where K (kij) denotes the thermal conductivity tensorEquations (6)ndash(8) represent the equilibrium equations

for the stress Relations (9) and (10) (11) and (12) and (13)represent the mechanical the electrical and the thermalboundary conditions -e unilateral boundary condition(16) represents the normal compliance condition and(17)ndash(19) represent Trescarsquos friction law in which S is thecoefficient of friction

Following [14] the contact conditions (20) and (21) onΓC are obtained as follows

When there is no contact at a point on the surface thereis no free electrical charges on the surface and no thermaltransfer that is

u] ltg⟹D middot ] 0

q middot ] 01113896 (25)

If the contact holds ie u] geg the normal component ofthe electric displacement field or the free charge (respthermal transfer) is assumed to be proportional to the

difference between the potential of foundation and thebodyrsquos surface potential (resp to the difference between thetemperature of foundation and the bodyrsquos surface temper-ature) -us

u] geg⟹D middot ] kφ φ minus φF( 1113857

q middot ] kθ θ minus θF( 1113857

⎧⎪⎨

⎪⎩(26)

We combine (25) and (26) to obtain

D middot ] kφχ[0+infin) u] minus g( 1113857 φ minus φF( 1113857

q middot ] kθχ[0+infin) u] minus g( 1113857 θ minus θF( 1113857

⎧⎪⎨

⎪⎩(27)

where χ[0+infin) is the characteristic function of the interval[0 +infin) defined by

χ[0+infin)(s) 0 if slt 0

1 if sge 01113896 (28)

Equation (27) represents the regularization electricalcontact condition and the heat flux condition on ΓCwhere

ϕL(s)

minus L if slt minus L

s if minus Lle sleL

L if sgt L

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

ψ(r)

0 if rlt 0

cδr if 0le rle1δ

c if rgt1δ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

and where c kφ kθ and L is a large positive constant δ gt 0is a small parameter kc r⟶ kc(r) is supposed to be zerofor rlt 0 and positive otherwise nondecreasing and Lipschitzcontinuous

Remark 1 We note that when ψ equiv 0 equality (20) becomes

D middot ] 0 on ΓC times(0 T) (30)

which models the case when the foundation is a perfectelectric insulator

Similarly in equality (21) we have

q middot ] 0 on ΓC times(0 T) (31)

22 Weak Formulation and Uniqueness Result To obtain avariational formulation of Problem (P) we need addi-tional notations and need to recall some definitions in thesequel

International Journal of Differential Equations 3

We use the following functional Hilbert spaces

L2(Ω) L

2(Ω)

d

H1(Ω) H

1(Ω)

d

H σ isin Sd σ σij σij σji isin L

2(Ω)1113966 1113967

(32)

W D (D)i isin H1(Ω) Di isin L

2(Ω) divD isin L

2(Ω)1113966 1113967

(33)

endowed with the canonical inner product given by

(u v)L2(Ω) 1113946Ω

uividx

(σ τ)H 1113946Ωσiτidx

(u v)H1(Ω) (u v)L2(Ω) +(ε(u) ε(v))H

(D middot E) (D middot E)L2(Ω) +(divD middot divE)L2(Ω)

(34)

and the associated norms L2(Ω) H1(Ω) H and Wrespectively

LetVW andQ be the closed subspaces ofH1(Ω) given by

V v isin H1(Ω) v 0 on ΓD1113966 1113967

W ξ isin H1(Ω) ξ 0 on Γa1113966 1113967

Q η isin H1(Ω) η 0 on ΓD cup ΓN1113966 1113967

(35)

and the set of admissible displacements

Vad v isin V v] minus gle 0 on ΓC1113864 1113865 (36)

It is known that VW and Q are real Hilbert spaces withthe inner products (u v)V (ε(u) ε(v))H (φ ξ)W

(nablaφnablaξ)L2(Ω) and (θ η)Q (nablaθnablaη)L2(Ω) respectivelyMoreover the associated norm ||v||V ||ε(v)||H is

equivalent on V to the usual norm H1(Ω) and ||ξ||W

||nablaξ||L2(Ω) and ||η||Q ||nablaη||L2(Ω) are equivalent on W andQ respectively with the usual norms H1(Ω)

By Sobolevrsquos trace theorem there exists three positiveconstants Cs1 Cs2 and Cs3 depending on Ω ΓC ΓN ΓD Γaand Γb

||v||L2(Γ)d le Cs1||v||V forallv isin V (37)

ξL2 Γc( ) le Cs2ξW forallξ isinW (38)

ηL2 Γc( ) le Cs3ηQ forallη isin Q (39)

Since meas(ΓD)gt 0 and Kornrsquos inequality hold

||ε(v)||H ge CK||v||H1(Ω) forallv isin V (40)

where CK in a nonnegative constant depending only on Ωand ΓD Notice also that since meas(Γa)gt 0 the followingFriedrichsndashPoincare inequalities hold

||nablaξ||W geCF1||ξ||W forallξ isinW (41)

nablaηL2(Ω) geCF2ηQ forallη isin Q (42)

where CF1 and CF2 are the positive constants which dependonly on Ω Γa ΓD and ΓN

For a real Banach space X and 1lepleinfin we consider theBanach spaces C(0 T X) and C1(0 T X) of continuous andcontinuously differentiable functions from [0 T] to X withthe norms

uC(0TX) suptisin[0T]

u(t)X

uC1(0TX) suptisin[0T]

u(t)X

+ suptisin[0T]

_u(t)X

(43)

To simplify the writing we denote by a V times V⟶ Rb W times W⟶ R c V times V⟶ R and d Q times Q⟶ R

the following bilinear and symmetric applications

a(u v) ≔ (Iε(u) ε(v))H

b(φ ξ) ≔ (βnablaφnablaξ)L2(Ω)

c(u v) ≔ (Cε(u) ε(v))H

d(θ η) ≔ (Knablaθnablaη)L2(Ω)

(44)

and by e V times W⟶ R m Q times V⟶ R andp Q times W⟶ R the following bilinear applications

e(v ξ) ≔ (Eε(v)nablaξ)L2(Ω) Elowastnablaξ ε(v)( 1113857V

m(θ v) ≔ (Mθ ε(v))Q

p(θ ξ) ≔ (Pnablaθnablaξ)L2(Ω)

(45)

In the study of mechanical Problem (P) we make thefollowing assumptions

HP1 -e elasticity operator I Ω times Sd⟶ Sd theelectric permittivity tensor β (βij) ⟶Ω times Rd

⟶ Rd the viscosity tensor c Ω times Sd⟶ Sd andthe thermal conductivity tensor K (kij) Ω times Rd

⟶ Rd satisfy the usual properties of symmetryboundedness and ellipticity

fijkl fjikl flkij isin Linfin

(Ω)

βij βji isin Linfin

(Ω)

cijkl cjikl clkij isin Linfin

(Ω)

kij kji isin Linfin

(Ω)

(46)

and there exists that mI mβ mc mK gt 0 such that

fijkl(x)ξkξl ge mI||ξ||2 forallξ isin Sd

forallx isin Ω

cijkl(x)ξkξl ge mc||ξ||2 forallξ isin Sd

forallx isin Ω

βijζ iζj ge mβ||ζ||2 kijζ iζj ge mK||ζ||

2 forallζ isin Rd

(47)

4 International Journal of Differential Equations

HP2 -e piezoelectric tensor Ε (eijk) Ω times Sd

⟶ R the thermal expansion tensor M (mij)

Ω times R⟶ R and the pyroelectric tensor P (pi)

Ω⟶ Rd satisfy

eijk eikj isin Linfin

(Ω)

mij mji isin Linfin

(Ω)

pi isin Linfin

(Ω)

(48)

and there exist the positive constantsMI Mβ Mc MK ME MM and MP such that

|a(u v)|leMIuVvV

|b(φ ξ)|leMβφWξW

|c(u v)|leMcuVvV

|d(θ η)|leMKθQηQ

|e(v ξ)|leMΕvVξW

|m(θ v)|leMMθQvV

|p(θ ξ)|leMPθQξW

(49)

HP3 -e surface electrical conductivityψ ΓC times R⟶ R+ and the thermal conductance kc

ΓC times R⟶ R+ satisfy the following hypothesis for(π ψ kc) existMπ gt 0 such that |π(x u)|leMπ foralluisin R ae x isin ΓC and x⟶ π(x u) is measurable onΓC for all u isin R and is zero for all ule 0-e function u⟶ π(x u) is a Lipschitz function onRfor all x isin Γ3

|π(x u1) minus π(x u2)|leLπ|u1 minus u2| forallu1 u2 isin R where Lπis a positive constant

HP4 -e forces the traction the volume the surfacescharge densities and the strength of the heat source areas follows

f0 isin C 0 T L2(Ω)

d1113872 1113873

f2 isin C 0 T L2 ΓN( 1113857

d1113872 1113873

qb isin L2 0 T L

2 Γb( 11138571113872 1113873

q1 isin L2 0 T L

2(Ω)1113872 1113873

q0 isin L2 0 T L

2(Ω)1113872 1113873

(50)

-e potential and the temperature satisfy

φF isin L2 0 T L

2 ΓC( 11138571113872 1113873

θF isin L2 0 T L

2 ΓC( 11138571113872 1113873(51)

-e initial conditions the friction-bounded functionand the gap function satisfy

u0 isin K

θ0 isin Q

S isin Linfin ΓC( 1113857

Sge 0

g isin L2 ΓC( 1113857

gge 0

(52)

Using Rieszrsquos representation theorem we definef [0 T]⟶ V qe [0 T]⟶W and qth [0 T]⟶ Q

by the following

(f(t) v)V 1113946Ω

f0(t) middot v dx + 1113946ΓN

f2(t) middot v da forallv isin V

(53)

qe(t) ξ( 1113857W 1113946Ω

q0(t) middot ξ dx minus 1113946Γb

qb(t) middot ξ da forallξ isin V

(54)

qth(t) η1113872 1113873Q

1113946Ω

q1(t) middot η dx forallη isin Q (55)

We define the mappings j V⟶ R 9 V times V⟶ Rℓ V times W2⟶ R and χ V times Q2⟶ R ae t isin ]0 T[ by

j(v) 1113946ΓC

S vτ

da forallv isin V (56)

9(u v) 1113946ΓC

u]1113858 1113859+

minus g( 1113857v]da lang u]1113858 1113859+

minus g v]rangΓC

forallu v isin V

(57)

ℓ(u(t) φ(t) ξ) 1113946ΓCψ u](t) minus g( 1113857ϕL φ(t) minus φF( 1113857ξ da

forallu isin V forallφ ξ isinW

(58)

χ(u(t) θ(t) η) 1113946ΓC

kc u](t) minus g( 1113857ϕL θ(t) minus θF( 1113857η da

forallu isin V forallθ η isin Q

(59)

respectivelyNow by a standard variational technique it is

straightforward to see that if (uφ θ) satisfies the conditions(4)ndash(21) ae t isin ]0 T[ then

(σ(t) ε(v) minus ε( _u(t)))H +1ε

9(u(t) v minus _u(t)) + j(v)

minus j( _u(t))ge (f(t) v minus _u(t))V forallv isin Va d

(60)

(D(t)nablaξ)L2(Ω) ℓ(u(t) φ(t) ξ) minus qe(t) ξ( 1113857W

forallξ isinW(61)

International Journal of Differential Equations 5

(q(t)nablaη)L2(Ω) ( _θ(t) η)Q + χ(u(t) θ(t) η) minus qth(t) η1113872 1113873Q

forallη isin Q

(62)

We assume that the initial conditions u0 and θ0 satisfy thefollowing compatibility condition there existsφ0 isinW such that

b φ0 ξ( 1113857 minus e u0 ξ( 1113857 minus p θ0 ξ( 1113857 + ℓ u0φ0 ξ( 1113857 qe0 ξ1113872 1113873

W

forallξ isinW

(63)

-is nonlinear problem has a unique solution φ0 byusing the fixed point theorem

Using all of these assumptions notations and E minus nablaφwe obtain the following variational formulation of theProblem (P)

Problem (PV) find a displacement field u ]0

T[⟶ Vad an electric potential φ ]0 T[⟶W and atemperature field θ ]0 T[⟶ Q ae t isin ]0 T[ such that

c( _u(t) v minus _u(t)) + a(u(t) v minus _u(t)) + e(v minus _u(t)

φ(t)) minus m(θ(t) v minus _u(t)) +1ϵ

9(u(t) v minus _u(t))

+ j(v) minus j( _u(t)) ge (f(t) v minus _u(t))V forallv isin Vad

(64)

b(φ(t) ξ) minus e(u(t) ξ) minus p(θ(t) ξ)

+ ℓ(u(t) φ(t) ξ) qe(t) ξ( 1113857W forallξ isinW(65)

d(θ(t) η) +( _θ(t) η)Q + χ(u(t) θ(t) η) qth(t) η1113872 1113873Q

forallη isin Q

(66)

u(0 x) u0(x)

θ(0 x) θ0(x)(67)

We present now the existence and the uniqueness ofsolution to Problem (PV)

Theorem 1 Assume that the assumptions (HP1) minus (HP4)(37)ndash(42) for εgt 0and the conditions

mβ gtMψc21

mK ltc2 Mkc

c2 + LkkLc01113872 1113873

2

(68)

hold Ben Problem (PV) has a unique solution as follows

u isin C1(0 T V)

φ isin L2(0 T W)

θ isin L2(0 T Q)

(69)

23 Spatially Semidiscrete Approximation In this para-graph we consider a semidiscrete approximation of theProblem (PV) by discretizing the spatial domain usingthe finite element method Let Th be a regular finite el-ement partition of the domain Ω compatible with theboundary partition Γ ΓC cup ΓD cup ΓN We then define afinite element space Vh sub V and Vh

a d Va d capVh for theapproximates of the displacement field u Wh subW for theelectric potential φ and Qh sub Q for the temperature θdefined by

Vh

vh isin [C(Ω)]

d v

h|Tr isin P1(Tr)1113858 1113859

dforallTr isin Th v

h 0 on ΓC1113966 1113967

Wh

ξh isin C(Ω) ξh|Tr isin P1(Tr)forallTr isin Th

ξh on Γa1113966 1113967

Qh

ηh isin C(Ω) ηh|Tr isin P1(Tr)forallTr isin Th

ηh 0 on ΓD cup ΓN1113966 1113967

(70)

A spatially semidiscrete scheme can be formed as thefollowing problem

Problem (PVh)Find uh ]0 T[⟶ Vh φh ]0

T[⟶Wh and θh ]0 T[⟶ Qh ae t isin ]0 T[ such thatfor vh isin Vh

a d ξh isinWh and ηh isin Qh

c _uh(t) v

hminus _u

h(t)1113872 1113873 + a u

h(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θh(t) v

hminus _u

h1113872 1113873

+1ε

9 uh(t) v

hminus _u

h(t)1113872 1113873 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873

ge f(t) vh

minus _uh(t)1113872 1113873

(71)

b φh(t) ξh

1113872 1113873 minus e uh(t) ξh

1113872 1113873 minus p θh(t) ξh

1113872 1113873

+ ℓ uh(t)φh

(t) ξh1113872 1113873 qe(t) ξh

1113872 1113873(72)

d θh(t) ηh

1113872 1113873 + _θh(t) ηh

1113874 1113875 + χ uh(t) ηh

(t) ηh1113872 1113873

qth(t) ηh1113872 1113873

(73)

uh(0) u

h0

φh(0) φh

0

θh(0) θh

0

(74)

Here uh0 isin Vh φh

0 isinWh and θh0 isin Qh are appropriate

approximations of u0 φ0 and θ0 respectivelyUsing the same argument presented in Section 2 it can

be shown that Problem (PVh) has a unique solutionuh isin C1(0 T Vh) φh isin L2(0 T Wh) and θh isin L2(0 T Qh)

In this paragraph we are interested in obtaining esti-mates for the errors (u minus uh) (φ minus φh) and (θ minus θh)

Using the initial value condition we have

6 International Journal of Differential Equations

u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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AnalysisInternational Journal of

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Submit your manuscripts atwwwhindawicom

We use the following functional Hilbert spaces

L2(Ω) L

2(Ω)

d

H1(Ω) H

1(Ω)

d

H σ isin Sd σ σij σij σji isin L

2(Ω)1113966 1113967

(32)

W D (D)i isin H1(Ω) Di isin L

2(Ω) divD isin L

2(Ω)1113966 1113967

(33)

endowed with the canonical inner product given by

(u v)L2(Ω) 1113946Ω

uividx

(σ τ)H 1113946Ωσiτidx

(u v)H1(Ω) (u v)L2(Ω) +(ε(u) ε(v))H

(D middot E) (D middot E)L2(Ω) +(divD middot divE)L2(Ω)

(34)

and the associated norms L2(Ω) H1(Ω) H and Wrespectively

LetVW andQ be the closed subspaces ofH1(Ω) given by

V v isin H1(Ω) v 0 on ΓD1113966 1113967

W ξ isin H1(Ω) ξ 0 on Γa1113966 1113967

Q η isin H1(Ω) η 0 on ΓD cup ΓN1113966 1113967

(35)

and the set of admissible displacements

Vad v isin V v] minus gle 0 on ΓC1113864 1113865 (36)

It is known that VW and Q are real Hilbert spaces withthe inner products (u v)V (ε(u) ε(v))H (φ ξ)W

(nablaφnablaξ)L2(Ω) and (θ η)Q (nablaθnablaη)L2(Ω) respectivelyMoreover the associated norm ||v||V ||ε(v)||H is

equivalent on V to the usual norm H1(Ω) and ||ξ||W

||nablaξ||L2(Ω) and ||η||Q ||nablaη||L2(Ω) are equivalent on W andQ respectively with the usual norms H1(Ω)

By Sobolevrsquos trace theorem there exists three positiveconstants Cs1 Cs2 and Cs3 depending on Ω ΓC ΓN ΓD Γaand Γb

||v||L2(Γ)d le Cs1||v||V forallv isin V (37)

ξL2 Γc( ) le Cs2ξW forallξ isinW (38)

ηL2 Γc( ) le Cs3ηQ forallη isin Q (39)

Since meas(ΓD)gt 0 and Kornrsquos inequality hold

||ε(v)||H ge CK||v||H1(Ω) forallv isin V (40)

where CK in a nonnegative constant depending only on Ωand ΓD Notice also that since meas(Γa)gt 0 the followingFriedrichsndashPoincare inequalities hold

||nablaξ||W geCF1||ξ||W forallξ isinW (41)

nablaηL2(Ω) geCF2ηQ forallη isin Q (42)

where CF1 and CF2 are the positive constants which dependonly on Ω Γa ΓD and ΓN

For a real Banach space X and 1lepleinfin we consider theBanach spaces C(0 T X) and C1(0 T X) of continuous andcontinuously differentiable functions from [0 T] to X withthe norms

uC(0TX) suptisin[0T]

u(t)X

uC1(0TX) suptisin[0T]

u(t)X

+ suptisin[0T]

_u(t)X

(43)

To simplify the writing we denote by a V times V⟶ Rb W times W⟶ R c V times V⟶ R and d Q times Q⟶ R

the following bilinear and symmetric applications

a(u v) ≔ (Iε(u) ε(v))H

b(φ ξ) ≔ (βnablaφnablaξ)L2(Ω)

c(u v) ≔ (Cε(u) ε(v))H

d(θ η) ≔ (Knablaθnablaη)L2(Ω)

(44)

and by e V times W⟶ R m Q times V⟶ R andp Q times W⟶ R the following bilinear applications

e(v ξ) ≔ (Eε(v)nablaξ)L2(Ω) Elowastnablaξ ε(v)( 1113857V

m(θ v) ≔ (Mθ ε(v))Q

p(θ ξ) ≔ (Pnablaθnablaξ)L2(Ω)

(45)

In the study of mechanical Problem (P) we make thefollowing assumptions

HP1 -e elasticity operator I Ω times Sd⟶ Sd theelectric permittivity tensor β (βij) ⟶Ω times Rd

⟶ Rd the viscosity tensor c Ω times Sd⟶ Sd andthe thermal conductivity tensor K (kij) Ω times Rd

⟶ Rd satisfy the usual properties of symmetryboundedness and ellipticity

fijkl fjikl flkij isin Linfin

(Ω)

βij βji isin Linfin

(Ω)

cijkl cjikl clkij isin Linfin

(Ω)

kij kji isin Linfin

(Ω)

(46)

and there exists that mI mβ mc mK gt 0 such that

fijkl(x)ξkξl ge mI||ξ||2 forallξ isin Sd

forallx isin Ω

cijkl(x)ξkξl ge mc||ξ||2 forallξ isin Sd

forallx isin Ω

βijζ iζj ge mβ||ζ||2 kijζ iζj ge mK||ζ||

2 forallζ isin Rd

(47)

4 International Journal of Differential Equations

HP2 -e piezoelectric tensor Ε (eijk) Ω times Sd

⟶ R the thermal expansion tensor M (mij)

Ω times R⟶ R and the pyroelectric tensor P (pi)

Ω⟶ Rd satisfy

eijk eikj isin Linfin

(Ω)

mij mji isin Linfin

(Ω)

pi isin Linfin

(Ω)

(48)

and there exist the positive constantsMI Mβ Mc MK ME MM and MP such that

|a(u v)|leMIuVvV

|b(φ ξ)|leMβφWξW

|c(u v)|leMcuVvV

|d(θ η)|leMKθQηQ

|e(v ξ)|leMΕvVξW

|m(θ v)|leMMθQvV

|p(θ ξ)|leMPθQξW

(49)

HP3 -e surface electrical conductivityψ ΓC times R⟶ R+ and the thermal conductance kc

ΓC times R⟶ R+ satisfy the following hypothesis for(π ψ kc) existMπ gt 0 such that |π(x u)|leMπ foralluisin R ae x isin ΓC and x⟶ π(x u) is measurable onΓC for all u isin R and is zero for all ule 0-e function u⟶ π(x u) is a Lipschitz function onRfor all x isin Γ3

|π(x u1) minus π(x u2)|leLπ|u1 minus u2| forallu1 u2 isin R where Lπis a positive constant

HP4 -e forces the traction the volume the surfacescharge densities and the strength of the heat source areas follows

f0 isin C 0 T L2(Ω)

d1113872 1113873

f2 isin C 0 T L2 ΓN( 1113857

d1113872 1113873

qb isin L2 0 T L

2 Γb( 11138571113872 1113873

q1 isin L2 0 T L

2(Ω)1113872 1113873

q0 isin L2 0 T L

2(Ω)1113872 1113873

(50)

-e potential and the temperature satisfy

φF isin L2 0 T L

2 ΓC( 11138571113872 1113873

θF isin L2 0 T L

2 ΓC( 11138571113872 1113873(51)

-e initial conditions the friction-bounded functionand the gap function satisfy

u0 isin K

θ0 isin Q

S isin Linfin ΓC( 1113857

Sge 0

g isin L2 ΓC( 1113857

gge 0

(52)

Using Rieszrsquos representation theorem we definef [0 T]⟶ V qe [0 T]⟶W and qth [0 T]⟶ Q

by the following

(f(t) v)V 1113946Ω

f0(t) middot v dx + 1113946ΓN

f2(t) middot v da forallv isin V

(53)

qe(t) ξ( 1113857W 1113946Ω

q0(t) middot ξ dx minus 1113946Γb

qb(t) middot ξ da forallξ isin V

(54)

qth(t) η1113872 1113873Q

1113946Ω

q1(t) middot η dx forallη isin Q (55)

We define the mappings j V⟶ R 9 V times V⟶ Rℓ V times W2⟶ R and χ V times Q2⟶ R ae t isin ]0 T[ by

j(v) 1113946ΓC

S vτ

da forallv isin V (56)

9(u v) 1113946ΓC

u]1113858 1113859+

minus g( 1113857v]da lang u]1113858 1113859+

minus g v]rangΓC

forallu v isin V

(57)

ℓ(u(t) φ(t) ξ) 1113946ΓCψ u](t) minus g( 1113857ϕL φ(t) minus φF( 1113857ξ da

forallu isin V forallφ ξ isinW

(58)

χ(u(t) θ(t) η) 1113946ΓC

kc u](t) minus g( 1113857ϕL θ(t) minus θF( 1113857η da

forallu isin V forallθ η isin Q

(59)

respectivelyNow by a standard variational technique it is

straightforward to see that if (uφ θ) satisfies the conditions(4)ndash(21) ae t isin ]0 T[ then

(σ(t) ε(v) minus ε( _u(t)))H +1ε

9(u(t) v minus _u(t)) + j(v)

minus j( _u(t))ge (f(t) v minus _u(t))V forallv isin Va d

(60)

(D(t)nablaξ)L2(Ω) ℓ(u(t) φ(t) ξ) minus qe(t) ξ( 1113857W

forallξ isinW(61)

International Journal of Differential Equations 5

(q(t)nablaη)L2(Ω) ( _θ(t) η)Q + χ(u(t) θ(t) η) minus qth(t) η1113872 1113873Q

forallη isin Q

(62)

We assume that the initial conditions u0 and θ0 satisfy thefollowing compatibility condition there existsφ0 isinW such that

b φ0 ξ( 1113857 minus e u0 ξ( 1113857 minus p θ0 ξ( 1113857 + ℓ u0φ0 ξ( 1113857 qe0 ξ1113872 1113873

W

forallξ isinW

(63)

-is nonlinear problem has a unique solution φ0 byusing the fixed point theorem

Using all of these assumptions notations and E minus nablaφwe obtain the following variational formulation of theProblem (P)

Problem (PV) find a displacement field u ]0

T[⟶ Vad an electric potential φ ]0 T[⟶W and atemperature field θ ]0 T[⟶ Q ae t isin ]0 T[ such that

c( _u(t) v minus _u(t)) + a(u(t) v minus _u(t)) + e(v minus _u(t)

φ(t)) minus m(θ(t) v minus _u(t)) +1ϵ

9(u(t) v minus _u(t))

+ j(v) minus j( _u(t)) ge (f(t) v minus _u(t))V forallv isin Vad

(64)

b(φ(t) ξ) minus e(u(t) ξ) minus p(θ(t) ξ)

+ ℓ(u(t) φ(t) ξ) qe(t) ξ( 1113857W forallξ isinW(65)

d(θ(t) η) +( _θ(t) η)Q + χ(u(t) θ(t) η) qth(t) η1113872 1113873Q

forallη isin Q

(66)

u(0 x) u0(x)

θ(0 x) θ0(x)(67)

We present now the existence and the uniqueness ofsolution to Problem (PV)

Theorem 1 Assume that the assumptions (HP1) minus (HP4)(37)ndash(42) for εgt 0and the conditions

mβ gtMψc21

mK ltc2 Mkc

c2 + LkkLc01113872 1113873

2

(68)

hold Ben Problem (PV) has a unique solution as follows

u isin C1(0 T V)

φ isin L2(0 T W)

θ isin L2(0 T Q)

(69)

23 Spatially Semidiscrete Approximation In this para-graph we consider a semidiscrete approximation of theProblem (PV) by discretizing the spatial domain usingthe finite element method Let Th be a regular finite el-ement partition of the domain Ω compatible with theboundary partition Γ ΓC cup ΓD cup ΓN We then define afinite element space Vh sub V and Vh

a d Va d capVh for theapproximates of the displacement field u Wh subW for theelectric potential φ and Qh sub Q for the temperature θdefined by

Vh

vh isin [C(Ω)]

d v

h|Tr isin P1(Tr)1113858 1113859

dforallTr isin Th v

h 0 on ΓC1113966 1113967

Wh

ξh isin C(Ω) ξh|Tr isin P1(Tr)forallTr isin Th

ξh on Γa1113966 1113967

Qh

ηh isin C(Ω) ηh|Tr isin P1(Tr)forallTr isin Th

ηh 0 on ΓD cup ΓN1113966 1113967

(70)

A spatially semidiscrete scheme can be formed as thefollowing problem

Problem (PVh)Find uh ]0 T[⟶ Vh φh ]0

T[⟶Wh and θh ]0 T[⟶ Qh ae t isin ]0 T[ such thatfor vh isin Vh

a d ξh isinWh and ηh isin Qh

c _uh(t) v

hminus _u

h(t)1113872 1113873 + a u

h(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θh(t) v

hminus _u

h1113872 1113873

+1ε

9 uh(t) v

hminus _u

h(t)1113872 1113873 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873

ge f(t) vh

minus _uh(t)1113872 1113873

(71)

b φh(t) ξh

1113872 1113873 minus e uh(t) ξh

1113872 1113873 minus p θh(t) ξh

1113872 1113873

+ ℓ uh(t)φh

(t) ξh1113872 1113873 qe(t) ξh

1113872 1113873(72)

d θh(t) ηh

1113872 1113873 + _θh(t) ηh

1113874 1113875 + χ uh(t) ηh

(t) ηh1113872 1113873

qth(t) ηh1113872 1113873

(73)

uh(0) u

h0

φh(0) φh

0

θh(0) θh

0

(74)

Here uh0 isin Vh φh

0 isinWh and θh0 isin Qh are appropriate

approximations of u0 φ0 and θ0 respectivelyUsing the same argument presented in Section 2 it can

be shown that Problem (PVh) has a unique solutionuh isin C1(0 T Vh) φh isin L2(0 T Wh) and θh isin L2(0 T Qh)

In this paragraph we are interested in obtaining esti-mates for the errors (u minus uh) (φ minus φh) and (θ minus θh)

Using the initial value condition we have

6 International Journal of Differential Equations

u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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Submit your manuscripts atwwwhindawicom

HP2 -e piezoelectric tensor Ε (eijk) Ω times Sd

⟶ R the thermal expansion tensor M (mij)

Ω times R⟶ R and the pyroelectric tensor P (pi)

Ω⟶ Rd satisfy

eijk eikj isin Linfin

(Ω)

mij mji isin Linfin

(Ω)

pi isin Linfin

(Ω)

(48)

and there exist the positive constantsMI Mβ Mc MK ME MM and MP such that

|a(u v)|leMIuVvV

|b(φ ξ)|leMβφWξW

|c(u v)|leMcuVvV

|d(θ η)|leMKθQηQ

|e(v ξ)|leMΕvVξW

|m(θ v)|leMMθQvV

|p(θ ξ)|leMPθQξW

(49)

HP3 -e surface electrical conductivityψ ΓC times R⟶ R+ and the thermal conductance kc

ΓC times R⟶ R+ satisfy the following hypothesis for(π ψ kc) existMπ gt 0 such that |π(x u)|leMπ foralluisin R ae x isin ΓC and x⟶ π(x u) is measurable onΓC for all u isin R and is zero for all ule 0-e function u⟶ π(x u) is a Lipschitz function onRfor all x isin Γ3

|π(x u1) minus π(x u2)|leLπ|u1 minus u2| forallu1 u2 isin R where Lπis a positive constant

HP4 -e forces the traction the volume the surfacescharge densities and the strength of the heat source areas follows

f0 isin C 0 T L2(Ω)

d1113872 1113873

f2 isin C 0 T L2 ΓN( 1113857

d1113872 1113873

qb isin L2 0 T L

2 Γb( 11138571113872 1113873

q1 isin L2 0 T L

2(Ω)1113872 1113873

q0 isin L2 0 T L

2(Ω)1113872 1113873

(50)

-e potential and the temperature satisfy

φF isin L2 0 T L

2 ΓC( 11138571113872 1113873

θF isin L2 0 T L

2 ΓC( 11138571113872 1113873(51)

-e initial conditions the friction-bounded functionand the gap function satisfy

u0 isin K

θ0 isin Q

S isin Linfin ΓC( 1113857

Sge 0

g isin L2 ΓC( 1113857

gge 0

(52)

Using Rieszrsquos representation theorem we definef [0 T]⟶ V qe [0 T]⟶W and qth [0 T]⟶ Q

by the following

(f(t) v)V 1113946Ω

f0(t) middot v dx + 1113946ΓN

f2(t) middot v da forallv isin V

(53)

qe(t) ξ( 1113857W 1113946Ω

q0(t) middot ξ dx minus 1113946Γb

qb(t) middot ξ da forallξ isin V

(54)

qth(t) η1113872 1113873Q

1113946Ω

q1(t) middot η dx forallη isin Q (55)

We define the mappings j V⟶ R 9 V times V⟶ Rℓ V times W2⟶ R and χ V times Q2⟶ R ae t isin ]0 T[ by

j(v) 1113946ΓC

S vτ

da forallv isin V (56)

9(u v) 1113946ΓC

u]1113858 1113859+

minus g( 1113857v]da lang u]1113858 1113859+

minus g v]rangΓC

forallu v isin V

(57)

ℓ(u(t) φ(t) ξ) 1113946ΓCψ u](t) minus g( 1113857ϕL φ(t) minus φF( 1113857ξ da

forallu isin V forallφ ξ isinW

(58)

χ(u(t) θ(t) η) 1113946ΓC

kc u](t) minus g( 1113857ϕL θ(t) minus θF( 1113857η da

forallu isin V forallθ η isin Q

(59)

respectivelyNow by a standard variational technique it is

straightforward to see that if (uφ θ) satisfies the conditions(4)ndash(21) ae t isin ]0 T[ then

(σ(t) ε(v) minus ε( _u(t)))H +1ε

9(u(t) v minus _u(t)) + j(v)

minus j( _u(t))ge (f(t) v minus _u(t))V forallv isin Va d

(60)

(D(t)nablaξ)L2(Ω) ℓ(u(t) φ(t) ξ) minus qe(t) ξ( 1113857W

forallξ isinW(61)

International Journal of Differential Equations 5

(q(t)nablaη)L2(Ω) ( _θ(t) η)Q + χ(u(t) θ(t) η) minus qth(t) η1113872 1113873Q

forallη isin Q

(62)

We assume that the initial conditions u0 and θ0 satisfy thefollowing compatibility condition there existsφ0 isinW such that

b φ0 ξ( 1113857 minus e u0 ξ( 1113857 minus p θ0 ξ( 1113857 + ℓ u0φ0 ξ( 1113857 qe0 ξ1113872 1113873

W

forallξ isinW

(63)

-is nonlinear problem has a unique solution φ0 byusing the fixed point theorem

Using all of these assumptions notations and E minus nablaφwe obtain the following variational formulation of theProblem (P)

Problem (PV) find a displacement field u ]0

T[⟶ Vad an electric potential φ ]0 T[⟶W and atemperature field θ ]0 T[⟶ Q ae t isin ]0 T[ such that

c( _u(t) v minus _u(t)) + a(u(t) v minus _u(t)) + e(v minus _u(t)

φ(t)) minus m(θ(t) v minus _u(t)) +1ϵ

9(u(t) v minus _u(t))

+ j(v) minus j( _u(t)) ge (f(t) v minus _u(t))V forallv isin Vad

(64)

b(φ(t) ξ) minus e(u(t) ξ) minus p(θ(t) ξ)

+ ℓ(u(t) φ(t) ξ) qe(t) ξ( 1113857W forallξ isinW(65)

d(θ(t) η) +( _θ(t) η)Q + χ(u(t) θ(t) η) qth(t) η1113872 1113873Q

forallη isin Q

(66)

u(0 x) u0(x)

θ(0 x) θ0(x)(67)

We present now the existence and the uniqueness ofsolution to Problem (PV)

Theorem 1 Assume that the assumptions (HP1) minus (HP4)(37)ndash(42) for εgt 0and the conditions

mβ gtMψc21

mK ltc2 Mkc

c2 + LkkLc01113872 1113873

2

(68)

hold Ben Problem (PV) has a unique solution as follows

u isin C1(0 T V)

φ isin L2(0 T W)

θ isin L2(0 T Q)

(69)

23 Spatially Semidiscrete Approximation In this para-graph we consider a semidiscrete approximation of theProblem (PV) by discretizing the spatial domain usingthe finite element method Let Th be a regular finite el-ement partition of the domain Ω compatible with theboundary partition Γ ΓC cup ΓD cup ΓN We then define afinite element space Vh sub V and Vh

a d Va d capVh for theapproximates of the displacement field u Wh subW for theelectric potential φ and Qh sub Q for the temperature θdefined by

Vh

vh isin [C(Ω)]

d v

h|Tr isin P1(Tr)1113858 1113859

dforallTr isin Th v

h 0 on ΓC1113966 1113967

Wh

ξh isin C(Ω) ξh|Tr isin P1(Tr)forallTr isin Th

ξh on Γa1113966 1113967

Qh

ηh isin C(Ω) ηh|Tr isin P1(Tr)forallTr isin Th

ηh 0 on ΓD cup ΓN1113966 1113967

(70)

A spatially semidiscrete scheme can be formed as thefollowing problem

Problem (PVh)Find uh ]0 T[⟶ Vh φh ]0

T[⟶Wh and θh ]0 T[⟶ Qh ae t isin ]0 T[ such thatfor vh isin Vh

a d ξh isinWh and ηh isin Qh

c _uh(t) v

hminus _u

h(t)1113872 1113873 + a u

h(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θh(t) v

hminus _u

h1113872 1113873

+1ε

9 uh(t) v

hminus _u

h(t)1113872 1113873 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873

ge f(t) vh

minus _uh(t)1113872 1113873

(71)

b φh(t) ξh

1113872 1113873 minus e uh(t) ξh

1113872 1113873 minus p θh(t) ξh

1113872 1113873

+ ℓ uh(t)φh

(t) ξh1113872 1113873 qe(t) ξh

1113872 1113873(72)

d θh(t) ηh

1113872 1113873 + _θh(t) ηh

1113874 1113875 + χ uh(t) ηh

(t) ηh1113872 1113873

qth(t) ηh1113872 1113873

(73)

uh(0) u

h0

φh(0) φh

0

θh(0) θh

0

(74)

Here uh0 isin Vh φh

0 isinWh and θh0 isin Qh are appropriate

approximations of u0 φ0 and θ0 respectivelyUsing the same argument presented in Section 2 it can

be shown that Problem (PVh) has a unique solutionuh isin C1(0 T Vh) φh isin L2(0 T Wh) and θh isin L2(0 T Qh)

In this paragraph we are interested in obtaining esti-mates for the errors (u minus uh) (φ minus φh) and (θ minus θh)

Using the initial value condition we have

6 International Journal of Differential Equations

u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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(q(t)nablaη)L2(Ω) ( _θ(t) η)Q + χ(u(t) θ(t) η) minus qth(t) η1113872 1113873Q

forallη isin Q

(62)

We assume that the initial conditions u0 and θ0 satisfy thefollowing compatibility condition there existsφ0 isinW such that

b φ0 ξ( 1113857 minus e u0 ξ( 1113857 minus p θ0 ξ( 1113857 + ℓ u0φ0 ξ( 1113857 qe0 ξ1113872 1113873

W

forallξ isinW

(63)

-is nonlinear problem has a unique solution φ0 byusing the fixed point theorem

Using all of these assumptions notations and E minus nablaφwe obtain the following variational formulation of theProblem (P)

Problem (PV) find a displacement field u ]0

T[⟶ Vad an electric potential φ ]0 T[⟶W and atemperature field θ ]0 T[⟶ Q ae t isin ]0 T[ such that

c( _u(t) v minus _u(t)) + a(u(t) v minus _u(t)) + e(v minus _u(t)

φ(t)) minus m(θ(t) v minus _u(t)) +1ϵ

9(u(t) v minus _u(t))

+ j(v) minus j( _u(t)) ge (f(t) v minus _u(t))V forallv isin Vad

(64)

b(φ(t) ξ) minus e(u(t) ξ) minus p(θ(t) ξ)

+ ℓ(u(t) φ(t) ξ) qe(t) ξ( 1113857W forallξ isinW(65)

d(θ(t) η) +( _θ(t) η)Q + χ(u(t) θ(t) η) qth(t) η1113872 1113873Q

forallη isin Q

(66)

u(0 x) u0(x)

θ(0 x) θ0(x)(67)

We present now the existence and the uniqueness ofsolution to Problem (PV)

Theorem 1 Assume that the assumptions (HP1) minus (HP4)(37)ndash(42) for εgt 0and the conditions

mβ gtMψc21

mK ltc2 Mkc

c2 + LkkLc01113872 1113873

2

(68)

hold Ben Problem (PV) has a unique solution as follows

u isin C1(0 T V)

φ isin L2(0 T W)

θ isin L2(0 T Q)

(69)

23 Spatially Semidiscrete Approximation In this para-graph we consider a semidiscrete approximation of theProblem (PV) by discretizing the spatial domain usingthe finite element method Let Th be a regular finite el-ement partition of the domain Ω compatible with theboundary partition Γ ΓC cup ΓD cup ΓN We then define afinite element space Vh sub V and Vh

a d Va d capVh for theapproximates of the displacement field u Wh subW for theelectric potential φ and Qh sub Q for the temperature θdefined by

Vh

vh isin [C(Ω)]

d v

h|Tr isin P1(Tr)1113858 1113859

dforallTr isin Th v

h 0 on ΓC1113966 1113967

Wh

ξh isin C(Ω) ξh|Tr isin P1(Tr)forallTr isin Th

ξh on Γa1113966 1113967

Qh

ηh isin C(Ω) ηh|Tr isin P1(Tr)forallTr isin Th

ηh 0 on ΓD cup ΓN1113966 1113967

(70)

A spatially semidiscrete scheme can be formed as thefollowing problem

Problem (PVh)Find uh ]0 T[⟶ Vh φh ]0

T[⟶Wh and θh ]0 T[⟶ Qh ae t isin ]0 T[ such thatfor vh isin Vh

a d ξh isinWh and ηh isin Qh

c _uh(t) v

hminus _u

h(t)1113872 1113873 + a u

h(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θh(t) v

hminus _u

h1113872 1113873

+1ε

9 uh(t) v

hminus _u

h(t)1113872 1113873 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873

ge f(t) vh

minus _uh(t)1113872 1113873

(71)

b φh(t) ξh

1113872 1113873 minus e uh(t) ξh

1113872 1113873 minus p θh(t) ξh

1113872 1113873

+ ℓ uh(t)φh

(t) ξh1113872 1113873 qe(t) ξh

1113872 1113873(72)

d θh(t) ηh

1113872 1113873 + _θh(t) ηh

1113874 1113875 + χ uh(t) ηh

(t) ηh1113872 1113873

qth(t) ηh1113872 1113873

(73)

uh(0) u

h0

φh(0) φh

0

θh(0) θh

0

(74)

Here uh0 isin Vh φh

0 isinWh and θh0 isin Qh are appropriate

approximations of u0 φ0 and θ0 respectivelyUsing the same argument presented in Section 2 it can

be shown that Problem (PVh) has a unique solutionuh isin C1(0 T Vh) φh isin L2(0 T Wh) and θh isin L2(0 T Qh)

In this paragraph we are interested in obtaining esti-mates for the errors (u minus uh) (φ minus φh) and (θ minus θh)

Using the initial value condition we have

6 International Journal of Differential Equations

u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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u(t) 1113946t

0w(s)ds + u0

θ(t) 1113946t

0_θ(s)ds + θ0

(75)

uh(t) 1113946

t

0_uh(s)ds + u

h0

θh(t) 1113946

t

0_θ

h(s)ds + θh

0

(76)

Theorem 2 Assume that the assumptions stated inBeorem1 are hold for εgt 0 Ben under the conditions

u0 minus uh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

θ0 minus θh0

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868V⟶ 0

as h⟶ 0

(77)

the semi-discrete solution of (PVh) converges as follows

u minus uh

L2(0TV)⟶ 0

φ minus φh

L2(0TW)⟶ 0

θ minus θh

L2(0TQ)⟶ 0

as h⟶ 0

(78)

24 Fully Discrete Approximation In this paragraph weconsider a fully discrete approximation of Problem (PV)We use the finite element spaces Vh Wh and Qh introducedin Section 23 We introduce a partition of the time interval[0 T] 0 t0 lt t1 lt lt tN T We denote the step sizeΔt kn tn minus tnminus 1 for n 1 2 N and let k maxnkn bethe maximal step size For a sequence vn1113864 1113865

N

n0 we denoteδw (wn minus wnminus 1)(kn)

-e fully discrete approximation method is based on thebackward Euler scheme and it has the following form

Problem (PVhkn ) Find a displacement field

uhkn1113864 1113865

N

n0 sub Vha d an electric potential φhk

n1113864 1113865N

n0 subWh and atemperature field θhk

n1113966 1113967N

n0 sub Qh for all vh isin Vh ξh isinWh

ηh isin Qh and n 1 N such that

c whkn v

hminus w

hkn1113872 1113873 + a u

hkn v

hminus w

hkn1113872 1113873 + e v

hminus w

hkn φhk

nminus 11113872 1113873

minus m θhknminus 1 v

hminus w

hkn1113872 1113873 +

1ε

9 uhkn v

hminus w

hkn1113872 1113873 + j v

h1113872 1113873

minus j whkn1113872 1113873ge fn v

hminus w

hkn1113872 1113873

V

(79)

b φhkn ξh

1113872 1113873 minus e uhkn ξh

1113872 1113873 minus p θhkn ξh

1113872 1113873 + ℓ uhkn φhk

n ξh1113872 1113873

qen ξh

1113872 1113873W

(80)

d θhkn ηh

1113872 1113873 + δθhkn ηh

1113872 1113873 + χ uhkn θhk

n ηh1113872 1113873 qthn η

h1113872 1113873

Q (81)

uhk0 u

h0

φhk0 φh

0

θhk0 θh

0

(82)

Remark 2 -e choice of θhnminus 1 and φh

nminus 1 instead of θhn and φh

n

is motivated by the fixed point method in the proof of theexistence and uniqueness Otherwise we may get anotherdifferent condition for the uniqueness of the solution of fixediteration problem (79)ndash(82) In addition this choice will behelpful for the application of discrete Gronwallrsquos lemma inthe next

To simplify again the notation we introduce thevelocity

whkn δu

hkn

n 1 N

uhkn 1113944

n

j1kjw

hkj + u

h0

θhkn 1113944

n

j1kjδθ

hkj + θh

0

nge 1

(83)

-is problem has a unique solution and the proof issimilar to that used in -eorem 1

We now derive the following convergence result

Theorem 3 Assuming that the initial values uh0 isin Vh

φh0 isinWh and θh

0 isin Qh are chosen in such a way that

u0 minus uh0

V⟶ 0

φ0 minus φh0

W⟶ 0

θ0 minus θh0

Q⟶ 0

as h⟶ 0

(84)

under the condition stated inBeorem 1 and for εgt 0 the fullydiscrete solution converges ie

max1lenleN

un minus uhkn

V+ wn minus w

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883⟶ 0

as h k⟶ 0

(85)

3 Proof of Main Results

In this section we prove the theorems presented in theprevious section

International Journal of Differential Equations 7

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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Submit your manuscripts atwwwhindawicom

31 Proof ofBeorem 1 -e proof of -eorem 1 is based onfixed point argument Galerkin method and compactnessmethod similar to those used in [14 15] but with a differentchoice of the operators

We turn now the following existence and uniquenessresult

Let F isin C(0 T V) given by

F(t) v minus _uF(t)( 1113857V e v minus _uF(t)φF(t)( 1113857

minus m θF(t) v minus _uF(t)( 1113857(86)

In the first step we consider the intermediate ProblemPVdf

Problem PVdf Find uF isin Va d for ae t isin ]0 T[ suchthat

c _uF(t) v minus _uF(t)( 1113857 + a uF(t) v minus _uF(t)( 1113857

+ F(t) v minus _uF(t)( 1113857V +1ε

9 uF(t) v minus _u(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge f(t) v minus _uF(t)( 1113857

uF(0) u0 forallv isin V

(87)

We have the following result for PVdf

Lemma 1 For all v isin Va d and for ae t isin ]0 T[ theProblem PVdf has a unique solution uF isin C1(0 T V)

Proof Using Rieszrsquos representation theorem we define theoperator A V⟶ V and the element fF(t) isin V by

fF(t) v( 1113857V (f(t) v)V minus (F(t) v)V (88)

A uF(t) v( 1113857 a uF(t) v( 1113857 +1ε

9 uF(t) v( 1113857 (89)

-en Problem PVdf can be written in the followingform

c _uF(t) v minus _uF(t)( 1113857 + A uF(t) v minus _uF(t)( 1113857

+ j(v) minus j _uF(t)( 1113857ge fF(t) v minus _uF(t)( 1113857

uF(0) u0

(90)

For all u v isin V there exists a constant cgt 0 whichdepends only on MI Cs1 and ϵ such that

|A(u v)|le cuVvV (91)

-e assumption (HP4) and F isin C(0 T V) imply thatfF isin C(0 T V) and by (HP1) minus (HP2) the operator c iscontinuous and coercive

We use now the result presented in pp 61ndash65 in [16] andwe conclude Problem PVdf has a unique solutionuF isin C1(0 T V)

Next we use the displacement field uF obtained in thefirst step and we consider the following lemma proved in[15]

Lemma 2 (a) For all η isin Q and ae t isin ]0 T[ the problem

d θF(t) η( 1113857 + _θF(t) η1113872 1113873Q

+ χ uF(t) θF(t) η( 1113857

qth(t) η1113872 1113873Q

forallη isin Q

θF(0) θ0

(92)

has a unique solution θF isin L2(0 T Q)(b) For all ξ isinW and for ae t isin ]0 T[ the problem

b φF(t) ξ( 1113857 minus e uF(t) ξ( 1113857 minus p θF(t) ξ( 1113857 + ℓ uF(t) φF(t) ξ( 1113857

qe(t) ξ( 1113857W

φF(0) φ0

(93)

has a unique solution φF isin L2(0 T W)

In the last step for F isin L2(0 T V) φF and θF are thefunctions obtained in Lemma 2 and we consider the operatorL C(0 T V)⟶ C(0 T V) defined by

(LF(t) v)V e vφF(t)( 1113857 minus m θF(t) v( 1113857 (94)

for all v isin V and for ae t isin ]0 T[For the operator L we have the following result ob-

tained in [15]

Lemma 3 Bere exists a unique 1113957F isin C(0 T V) such thatL 1113957F 1113957F

We now turn to a proof of -eorem 1

Existence Let 1113957F isin C(0 T V) be the fixed point of the operatorL and us denote 1113957x (1113957uF 1113957φF 1113957θF) the solution of variationalProblem (PVF) for 1113957F F the definition ofL and Problem(PVF) prove that 1113957x is a solution of Problem (PV)

Uniqueness-e uniqueness of the solution follows fromthe uniqueness of the fixed point of the operator L

32 Proof of Beorem 2 To prove -eorem 2 we need thefollowing result

Lemma 4 Assume that (HP1) minus (HP3) Ben we have theestimate as follows

u(t) minus uh(t)

L2(0TV)+ φ(t) minus φh

(t)

L2(0TW)

+ θ(t) minus θh(t)

L2(0TQ)

le clowast inf

vhisinL2 0TVh( )

ξhisinL2 0TWh( )

ηhisinL2 0TQh( )

1113882 _u(t) minus vh

L2(0TV)+ φ(t) minus ξh

L2(0TW)

+ θ(t) minus ηh

L2(0TQ)1113883

+ S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

L2(0T)1113883 + u0 minus u

h0

V+ θ0 minus θh

0

Q

(95)

for a positive constant clowast

8 International Journal of Differential Equations

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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AnalysisInternational Journal of

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Submit your manuscripts atwwwhindawicom

Proof Take v _uh(t) in (64) and add the inequality to (71)we have

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

le c _uh(t) v

hminus _u(t)1113872 1113873 + a u(t) _u

h(t) minus _u(t)1113872 1113873

+ a uh(t) v

hminus _u

h(t)1113872 1113873 + e _u

h(t) minus _u(t)φ(t)1113872 1113873

+ e vh

minus _uh(t) φh

(t)1113872 1113873 minus m θ(t) _uh(t) minus _u(t)1113872 1113873

minus m θh(t) v

hminus _u

h(t)1113872 1113873 +

1ε

9 u(t) _uh(t) minus _u(t)1113872 11138731113960

+ 9 uh(t) v

hminus _u

h(t)1113872 11138731113961 + j v

h1113872 1113873 minus j _u

h(t)1113872 1113873 + j _u

h(t)1113872 1113873

minus j( _u(t)) minus f vh

minus _u(t)1113872 1113873 minus f _uh

minus _u(t)1113872 1113873

(96)

-en

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873

leS t vh _u(t)1113872 1113873 + S

hΦ + c _u

h(t) minus _u(t) v

hminus _u(t)1113872 1113873

+ a uh(t) minus u(t) v

hminus _u

h(t)1113872 1113873 + e v

hminus _u

h(t)φh

(t) minus φ(t)1113872 1113873

minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

(97)

where

S t vh _u(t)1113872 1113873 c _u(t) v

hminus _u(t)1113872 1113873 + a u(t) v

hminus _u(t)1113872 1113873

+ e vh

minus _u(t)φ(t)1113872 1113873 minus m θ(t) vh

minus _u(t)1113872 1113873

+ j vh

1113872 1113873 minus e( _u(t)) minus f vh

minus _u(t)1113872 1113873

(98)

Sh9

1ε

9 u(t) _uh(t) minus _u(t)1113872 1113873 + 9 u

h(t) v

hminus _u

h(t)1113872 11138731113960

minus 9 u(t) vh

minus _u(t)1113872 11138731113961

(99)

We take ξ ξh in (65) and η ηh in (66) and by sub-tracting to (72) and to (73) respectively we deduce that

b φ(t) minus φh(t)φ(t) minus φh

(t)1113872 1113873 Shℓ + b φ(t) minus φh

(t)φ(t) minus ξh1113872 1113873

minus e u(t) minus uh(t)φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873

minus p θ(t) minus θh(t)φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t)φh

(t) minus φ(t)1113872 1113873

(100)

d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873 Shχ + d θ(t) minus θh

(t) θ(t) minus ηh1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875

minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(101)

where

Shℓ minus ℓ u(t)φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t)φ(t) φ(t) minus φh(t)1113872 1113873

+ ℓ uh(t)φh

(t) ξhminus φ(t)1113872 1113873 + ℓ u

h(t)φh

(t)φ(t) minus φh(t)1113872 1113873

(102)

Shχ χ u

h(t) θh

(t) ηhminus θ(t)1113872 1113873 + χ u

h(t) θh

(t) θ(t) minus θh(t)1113872 1113873

minus χ u(t) θ(t) ηhminus θ(t)1113872 1113873 minus χ u(t) θ(t) θ(t) minus θh

(t)1113872 1113873

(103)We now add (87) (100) and (101) we obtain

c _u(t) minus _uh(t) _u(t) minus _u

h(t)1113872 1113873 + b φ(t) minus φh

(t)φ(t) minus φh(t)1113872 1113873

+ d θ(t) minus θh(t) θ(t) minus θh

(t)1113872 1113873leS t vh _u(t)1113872 1113873 + S

h9 + Sℓ + Sχ

+ c _uh(t) minus _u(t) v

hminus _u(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(104)

From (HP1) and the previous inequality it follows that

mc _u(t) minus _uh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ mβ φ(t) minus φh

(t)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W+ mK θ(t) minus θh

(t)

2

Q

leS t vh _u(t)1113872 1113873 + S

h9 + S

hℓ + S

hχ + S

h

(105)

where

Sh

c _uh(t) minus _u(t) v

hminus w(t)1113872 1113873 + a u

h(t) minus u(t) v

hminus _u

h(t)1113872 1113873

+ e vh

minus _uh(t)φh

(t) minus φ(t)1113872 1113873 minus m θh(t) minus θ(t) v

hminus _u

h(t)1113872 1113873

b φ(t) minus φh(t) φ(t) minus ξh

1113872 1113873 minus e u(t) minus uh(t) φ(t) minus ξh

1113872 1113873

minus e u(t) minus uh(t)φh

(t) minus φ(t)1113872 1113873 minus p θ(t) minus θh(t) φ(t) minus ξh

1113872 1113873

minus p θ(t) minus θh(t) φh

(t) minus φ(t)1113872 1113873 + d θ(t) minus θh(t) θ(t) minus ηh

1113872 1113873

minus _θ(t) minus _θh(t) ηh

minus θ(t)1113874 1113875 minus _θ(t) minus _θh(t) θ(t) minus θh

(t)1113874 1113875

(106)

Let us estimate each of the terms in (105) (HP1) minus (HP3)and

xyle βx2

+14β

y2 forallβgt 0 (107)

allowing us to obtain

Sh

11138681113868111386811138681113868

11138681113868111386811138681113868le αh1113896 _u(t) minus _uh(t)

2

V+ u(t) minus u

h(t)

2

V+ φ(t) minus φh

(t)

2

W

+ θ(t) minus θh(t)

2

Q+ _θ(t) minus _θ

h(t)

2

Q+ _u

h(t) minus v

h

2

V

+ φh(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(108)

International Journal of Differential Equations 9

Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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Using the assumptions (HP3) (38) and (58) we find

Shℓ

11138681113868111386811138681113868

11138681113868111386811138681113868le111386811138681113868111386811138681113868 minus ℓ u(t) φ(t) ξh

minus φ(t)1113872 1113873 minus ℓ u(t) φ(t)φ(t)(

minus φh(t)) + ℓ u

h(t)φh

(t) ξhminus φ(t)1113872 1113873

+ ℓ uh(t)φh

(t)φ(t) minus φh(t)1113872 1113873

111386811138681113868111386811138681113868

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

le111386811138681113868111386811138681113868111386811138681113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857(ξ minus φ(t))da

+ 1113946ΓCψ u(t)] minus g( 1113857ϕL φ(t) minus φF( 1113857 φ minus φh

(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 ξhminus φ(t)1113872 1113873da

+ 1113946ΓCψ u

h(t)] minus g1113872 1113873ϕL φh

(t) minus φF1113872 1113873 φ minus φh(t)1113872 1113873da

11138681113868111386811138681113868111386811138681113868

leMψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

111386811138681113868111386811138682da

+ LLψ1113946ΓC

φ(t) minus φh(t)

11138681113868111386811138681113868

11138681113868111386811138681113868 u](t) minus uh](t)

11138681113868111386811138681113868

11138681113868111386811138681113868da

leMψCs1 φ(t) minus φh(t)

2

W+ LLψCs1Cs2 u(t) minus u

h(t)

V

φ(t) minus φh(t)

W

+ MψCs2 φ(t) minus φh(t)

Wφ(t) minus ξh

L2 ΓC( )

+ LLψc0 u(t) minus uh(t)

Vφ(t) minus ξh

L2 ΓC( )

le αℓ φ(t) minus φh(t)

2

W+ u(t) minus u

h(t)

2

V+ φ(t) minus ξh

W1113882 1113883

(109)

Similarly we find

Shχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θ(t) minus θh(t)

2

W+ u(t) minus u

h(t)

2

V+ θ(t) minus ηh

2

Q1113882 1113883

(110)

By combining the inequality

[x]+

minus [y]+

11138681113868111386811138681113868111386811138681113868le |x minus y| (111)

with (107) we observe that

Sh9

11138681113868111386811138681113868

11138681113868111386811138681113868le c9 u(t) minus uh(t)

2

V+ _u(t) minus _u

h(t)

2

V+ v

hminus _u(t)

2

V1113882 1113883

(112)

-us by (106) and (108)ndash(110) we have the inequality

_u(t) minus _uh(t)

2

V+ φ(t) minus φh

(t)

2

W+ θ(t) minus θh

(t)

2

Q

le clowast11113896 u(t) minus u

h(t)

2

V+ _θ(t) minus _θ

h(t)

2

Q+ R t v

h _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868

+ _u(t) minus vh

2

V+ φ(t) minus ξh

2

W+ θ(t) minus ηh

2

Q1113897

(113)-en by (75) and (76) we have

u(t) minus uh(t) 1113946

t

0_u(s) minus _u

h(s)1113872 1113873ds + u0 minus u

h0

θ(t) minus θh(t) 1113946

t

0_θ(s) minus _θ

h(s)1113874 1113875ds + θ0 minus θh

0

(114)

and so

u(t) minus uh(t)

2

Vle clowast2 1113946

t

0_u(s) minus _u

h(s)

2

Vds + u0 minus u

h0

2

V1113896 1113897

θ(t) minus θh(t)

2

Qle clowast3 1113946

t

0_θ(s) minus _θ

h(s)

2

Qds + θ0 minus θh

0

2

Q1113896 1113897

(115)

Consequently from the previous inequalities andGronwallrsquos inequality in (113) we find (95)

Proof of Beorem 2 To estimate the error provided by theapproximation of the finite element space Vh Wh

and Qh weuseΠhuΠhφ andΠhθ the standard finite ele-ment interpolation operator of u φ and θ respectively Wethen have the interpolation error estimate [16]

_u minus Πh_u

Vle ch _uL2(0TV)

φ minus Πhφ

Wle ch|φ|L2(0TW)

θ minus Πhθ

Qle ch|θ|L2(0TQ)

(116)

We bound now the term S( vh() _u())Using theproperties of c a e m f and (98) there exists positiveconstant c depending on Mc MI ME MM fL2(0TV)||u||L2(0TV) ||φ||L2(0TV) and ||θ||L2(0TV) such that

S t vh _u(t)1113872 1113873

11138681113868111386811138681113868

11138681113868111386811138681113868le c vh(t) minus _u(t)

V+ j v

h(t)1113872 1113873 minus j( _u(t))

11138681113868111386811138681113868

11138681113868111386811138681113868

(117)

Taking (116) in (95) we have

u minus uh

L2(0TV)+ φ minus φh

L2(0TW)+ θ minus θh

L2(0TQ)

le c + 1113896 _u minus Πh_u

L2(0TV)φ minus Πhφ

L2(0TW)

+ θ minus Πhθ

L2(0TQ)+ _u minus Πh

_u

12

L2(0TV)

+ j Πh_u1113872 1113873 minus j( _u)

12

L2(0TV)+ u0 minus u

h0

V+ θ0 minus θh

0

Q1113897

(118)

10 International Journal of Differential Equations

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

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Submit your manuscripts atwwwhindawicom

From (116) the constant c is independent of h u φ andθ -is implies

_u minus Πh_u

V⟶ 0

φ minus Πhφ

W⟶ 0

θ minus Πhθ

Q

as h⟶ 0

(119)

From (HP4) and (53) it follows that

j Πh_u1113872 1113873 minus j( _u)

L2(0TV)⟶ 0 (120)

Finally we find the result (78)

33 Proof ofBeorem 3 Add (80) with vh vhn and (64) with

v whkn at t tn we find

c whkn v

hn minus w

hkn1113872 1113873 + c wn w

hkn minus wn1113872 1113873 + a u

hkn v

hn minus w

hkn1113872 1113873

+ a un whkn minus wn1113872 1113873 + e v

hn minus w

hkn φhk

nminus 11113872 1113873 + e whkn minus wnφn1113872 1113873

minus m θhknminus 1 v

hn minus w

hkn1113872 1113873 minus m θn w

hkn minus wn1113872 1113873

+1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 11138731113960 1113961 + j v

hn1113872 1113873 minus j w

hkn1113872 1113873

+ j whkn1113872 1113873 minus j wn( 1113857ge fn v

hn minus w

hkn1113872 1113873 + fn w

hkn minus wn1113872 1113873

(121)

-is equality

c wn minus whkn wn minus w

hkn1113872 1113873 c w

hkn w

hkn minus v

hn1113872 1113873 + c wn wn minus v

hn1113872 1113873

minus c whkn wn minus v

hn1113872 1113873 + c wn v

hn minus w

khn1113872 1113873

(122)

allows us to obtain

c wn minus whkn wn minus w

hkn1113872 1113873leR v

hn wn1113872 1113873 + S

hk9

+ c whkn minus wn v

hn minus wn1113872 1113873

+ a uhkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873

minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

(123)

where

Sn vhn wn1113872 1113873 c wn v

hn minus wn1113872 1113873 + a un v

hn minus wn1113872 1113873

+ e vhn minus wnφn1113872 1113873 minus m θn v

hn minus wn1113872 1113873 + j v

hn1113872 1113873

minus j wn( 1113857 minus fn vhn minus wn1113872 1113873

(124)

Shk9

1ε

9 uhkn v

hn minus w

hkn1113872 1113873 + 9 un w

hkn minus wn1113872 1113873 minus 9 un v

hn minus wn1113872 11138731113960 1113961

(125)

For all nge 1 we subtract (65) from (81) at t tn weobtain that

b φn minus φhkn ξh

1113872 1113873 minus e un minus uhkn ξh

1113872 1113873 minus p θn minus θhkn ξh

1113872 1113873

+ ℓ unφn ξh1113872 1113873 minus ℓ u

hkn φhk

n ξh1113872 1113873 0

(126)

We replace ξh by (φn minus φhkn ) and (φn minus ξh

) in (126) wefind that

b φn minus φhkn φn minus φhk

n1113872 1113873leShkℓ + b φn minus φhk

n φn minus ξh1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873

(127)

where

Shkℓ ℓ unφnφn minus ξh

1113872 1113873 minus ℓ uhkn φhk

n φn minus ξh1113872 1113873

minus ℓ unφnφn minus φhkn1113872 1113873 + ℓ u

hkn φhk

n φn minus φhkn1113872 1113873

(128)

For all nge 1 and similar to (127) and (128) we have

d θn minus θhkn θn minus θhk

n1113872 1113873leShkχ + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(129)

where δθn 1kn(θn minus θnminus 1) and δθhkn 1kn(θhk

n minus θhknminus 1)

and

Shkχ χ un θn θn minus ηh

1113872 1113873 minus χ uhkn θhk

n θn minus ηh1113872 1113873

+ χ uhkn θhk

n θn minus θhkn1113872 1113873 minus χ un θn θn minus θhk

n1113872 1113873(130)

Adding (124) (128) and (130) we find

c wn minus whkn wn minus w

hkn1113872 1113873 + b φn minus φhk

n φn minus φhkn1113872 1113873

+ d θn minus θhkn θn minus θhk

n1113872 1113873leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ

+ c whkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + d θn minus θhkn θn minus ηh

1113872 1113873

+ e un minus uhkn ξh

minus φhkn1113872 1113873 + p θn minus θhk

n ξhminus φhk

n1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873

minus δθhkn θn minus ηh

1113872 1113873 + δθhkn θn minus θhk

n1113872 1113873

(131)

-e ellipticity of operators c b d and the previousinequality follows that

International Journal of Differential Equations 11

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

mC wn minus whkn

2

V+ mβ φn minus φhk

n

2

W+ mK θn minus θhk

n

2

V

leSn vhn wn1113872 1113873 + S

hk9 + S

hkℓ + S

hkχ + S

hk

(132)

whereS

hk c w

hkn minus wn v

hn minus wn1113872 1113873 + a u

hkn minus un v

hn minus w

hkn1113872 1113873

+ e vhn minus w

hkn φhk

nminus 1 minus φn1113872 1113873 minus m θhknminus 1 minus θn v

hn minus w

hkn1113872 1113873

+ b φn minus φhkn φn minus ξh

1113872 1113873 + e un minus uhkn ξh

minus φhkn1113872 1113873

+ p θn minus θhkn ξh

minus φhkn1113872 1113873 + d θn minus θhk

n θn minus ηh1113872 1113873

+ δθn θn minus ηh1113872 1113873 minus δθn θn minus θhk

n1113872 1113873 minus δθhkn θn minus ηh

1113872 1113873

+ δθhkn θn minus θhk

n1113872 1113873

(133)Now we estimate each of the terms in (132)Thanks to (HP1) minus (HP3) (107) (117) (111)and

vhn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 le vhn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 + wn minus whkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868 (134)

we get

Shk

11138681113868111386811138681113868

11138681113868111386811138681113868 le αhk1113896 wn minus whkn

2

V+ un minus u

hkn

2

Vφn minus φhk

n

2

W

+ θn minus θhkn

2

Q+ φn minus φhk

nminus 1

2

W+ θn minus θhk

nminus 1

2

Q

+ θnminus 1 minus θhknminus 1

2

Q+ v

hn minus wn

2

V+ ξh

minus φn

2

W

+ θn minus ηh

2

Q1113897

(135)

Shk9

11138681113868111386811138681113868

11138681113868111386811138681113868 le c9prime un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ wn minus w

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ v

hn minus wn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V1113882 1113883

(136)

Shkℓ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αℓ φn minus φhkn

2

W+ un minus u

hkn

2

V+ φn minus ξh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

W1113882 1113883

(137)

Shkχ

11138681113868111386811138681113868

11138681113868111386811138681113868 le αχ θn minus θhkn

2

Q+ un minus u

hkn

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

V+ θn minus ηh

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

111386811138681113868111386811138682

Q1113882 1113883

(138)

Combining (132) and (135)ndash(138) we have the followingestimate

wn minus whkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883 le c φn minus φhk

nminus 1

W1113882

+ θn minus θhknminus 1

Q+ θnminus 1 minus θhk

nminus 1

Q+ Sn v

hnwn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

V+ ξh

minus φn

W+ ηh

minus θn

Q1113883

(139)

We estimate now the terms ||φn minus φhknminus 1|| θn minus θhk

nminus 1 andθnminus 1 minus θhk

nminus 1 for the first term we have

φhknminus 1 minus φn 1113944

nminus 1

j1kjδφ

hkj + φh

0 minus 1113946tn

0δφ(s)ds minus φ0

1113944j1

nminus 1

kj δφhkj minus δφj1113872 1113873 + φh

0 minus φ0

+ 1113944nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889 minus 1113946tn

tnminus 1

δφ(s)ds

(140)

where δφj 1kj(φj minus φjminus 1) and δφhkj 1kj(φhk

j minus φhkjminus 1)

and

1113944

nminus 1

j1δφjkj minus 1113946

tj

tjminus 1

δφ(s)ds1113888 1113889

W

1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)1113872 1113873ds1113888 1113889

W

le 1113944nminus 1

j11113946

tj

tjminus 1

δφj minus δφ(s)11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Wds Ik(δφ)

(141)

1113946tn

tnminus 1

δφ(s)ds

le 1113946

tn

tnminus 1

δφ(s)Wds le kδφC(0TW)

(142)

-us

φhknminus 1 minus φn

Wle 1113944

nminus 1

j1kj δφkh

j minus δφj

W+ φh

0 minus φ0

W

+ Ik(δφ) + kφC(0TW)

(143)

Similar to (143) we deduce that

θhknminus 1 minus θn

Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

Q+ θh

0 minus θ0

Q+ Ik(δθ)

+ kθC1(0TQ)

(144)

θhknminus 1 minus θnminus 1

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Qle 1113944

nminus 1

j1kj δθkh

j minus δθj

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ θh

0 minus θ011138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868

11138681113868111386811138681113868Q+ Ik(δθ)

(145)

To proceed we need the following discrete version of theGronwallrsquos inequality presented in [12]

Lemma 5 Assuming that gn1113864 1113865N

n1 and en1113864 1113865N

n1 are the twosequences of nonnegative numbers satisfying

en le cgn + c 1113944

nminus 1

j1kjej (146)

12 International Journal of Differential Equations

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

-en

en le c gn + 1113944nminus 1

j1kjgj

⎛⎝ ⎞⎠ n 1 N (147)

-erefore

max1lenleN

en le c max1lenleN

gn (148)

Also the following result

Lemma 6 Let us define the map by

Ik(v) 1113944nminus 1

j11113946

tj

tjminus 1

vj minus v(s)Xds (149)

-en Ik(v) converges to zero as k⟶ 0 for allv isin C(0 T X)

Proof Since v isin C(0 T X) t⟼ v(t) is uniformly con-tinuous on [0 T] -us for any εgt 0 there exists a k0 gt 0such that if klt k0 we have

v(t) minus v(s)X ltεTforalls t isin [0 T] |t minus s|lt k (150)

-en we have

Ik(v)le 1113944N

j11113946

tj

tjminus 1

εT

ds ε (151)

Hence Ik(v) converges to zero as k⟶ 0

We now combine the inequalities (139) and (143)ndash(145)and applying the Lemma 5 we obtain the following estimate

max1lenleN

wn minus whkn

V+ un minus u

hkn

V+ φn minus φhk

n

W+ θn minus θhk

n

Q1113882 1113883

le max1lenleN

infvhisinVh

ξhisinWh

ηhisinQh

c Sn vhn wn1113872 1113873

11138681113868111386811138681113868

1113868111386811138681113868111386812

+ vhn minus wn

Q+ ξh

minus φn

W1113882

+ ηhminus θn

Q1113883 + c φ0 minus φh

0

W+ θ0 minus θh

0

Q1113882 1113883 + c1113882Ik(δφ)

+ Ik(δφ) + kφC1(0TW) + kθC1(0TQ)1113883

(152)

Proof of Beorem 3 We take vhn Πhwn ξ

hn Πhφn and

ηhn Πhθn in (152)-e convergence result follows from Lemma 6 together

with an argument similar to the proof of -eorem 2

4 Conclusion

-e paper presents a numerical analysis of the contact processbetween a thermo-electro-viscoelastic body and a conductivefoundation A spatially semidiscrete scheme and a fullydiscrete scheme to approximate the contact problem werederived We can also study this problem with more general

friction laws -e numerical simulation with the same al-gorithm is an interesting direction for future research

Data Availability

No data were used to support this study

Conflicts of Interest

-e authors declare that they have no conflicts of interest

References

[1] R C Batra and J S Yang ldquoSaint-Venantrsquos principle in linearpiezoelectricityrdquo Journal of Elasticity vol 38 no 2pp 209ndash218 1995

[2] F Maceri and P Bisegna ldquo-e unilateral frictionless contactof a piezoelectric body with a rigid supportrdquo Mathematicaland Computer Modelling vol 28 no 4ndash8 pp 19ndash28 1998

[3] L-E Anderson ldquoA quasistatic frictional problem with normalcompliancerdquo Nonlinear Analysis Beory Methods amp Appli-cations vol 16 no 4 pp 347ndash370 1991

[4] M Cocu E Pratt and M Raous ldquoFormulation and ap-proximation of quasistatic frictional contactrdquo InternationalJournal of Engineering Science vol 34 no 7 pp 783ndash7981996

[5] K Klarbring A Mikelic and M Shillor ldquoA global existenceresult for the quasistatic frictional contact problem withnormal compliancerdquo in Unilateral Peoblems in StructuralAnalysis G Del Piero and F Maceri Eds vol 4 pp 85ndash111Birkhauser Boston MA USA 1991

[6] M Rochdi M Shillor and M Sofonea ldquoA quasistatic contactproblem with directional friction and damped responserdquoApplicable Analysis vol 68 no 3-4 pp 409ndash422 1998

[7] M Rochdi M Shillor and M Sofonea ldquoQuasistatic visco-elastic contact with normal compliance and frictionrdquo Journalof Elasticity vol 51 no 2 pp 105ndash126 1998

[8] M Rochdi M Shillor and M Sofonea ldquoAnalysis of a qua-sistatic viscoelastic problem with friction and damagerdquo Ad-vances in Mathematical Sciences and Applications vol 10pp 173ndash189 2000

[9] R D Mindlin ldquoElasticity piezoelectricity and crystal latticedynamicsrdquo Journal of Elasticity vol 2 no 4 pp 217ndash282 1972

[10] B Benaissa E-H Essoufi and R Fakhar ldquoAnalysis of aSignorini problem with nonlocal friction in thermo-piezo-electricityrdquo Glasnik Matematicki vol 51 no 2 pp 391ndash4112016

[11] H Benaissa E-H Essoufi and R Fakhar ldquoExistence resultsfor unilateral contact problem with friction of thermo-electro-elasticityrdquo Applied Mathematics and Mechanics vol 36 no 7pp 911ndash926 2015

[12] W Han M Shillor and M Sofonea ldquoVariational and nu-merical analysis of a quasistatic viscoelastic problem withnormal compliance friction and damagerdquo Journal of Com-putational and Applied Mathematics vol 137 no 2pp 377ndash398 2001

[13] E-H Essoufi M Alaoui and M Bouallala ldquoError estimatesand analysis results for Signorinirsquos problem in thermo-electro-viscoelasticityrdquo General Letters in Mathematics vol 2 no 2pp 25ndash41 2017

[14] Z Lerguet M Shillor and M Sofonea ldquoA frictinal contactproblem for an electro-viscoelastic bodyrdquo Electronic Journal ofDifferential Equation vol 2007 no 170 pp 1ndash16 2007

International Journal of Differential Equations 13

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

[15] E-H Essoufi M Alaoui and M Bouallala ldquoQuasistaticthermo-electro-viscoelastic contact problem with Signoriniand Trescarsquos frictionrdquo Electronic Journal of DifferentialEquations vol 2019 no 5 pp 1ndash21 2019

[16] W Han and M Sofonea Quasistatic Contact Problems inViscoelasticity and Viscoplasticity Studies in Advance Math-ematics Vol 30 American Mathematical Society-In-ternational Press Providence RI USA 2002

[17] G Duvaut ldquoFree boundary problem connected with ther-moelasticity and unilateral contactrdquo in Free BoundaryProblem Vol II Istituto Nazionale Alta Matematica Fran-cesco Severi Rome Italy 1980

[18] W Han and M Sofonea ldquoEvolutionary variational in-equalities arising in viscoelastic contact problemsrdquo SIAMJournal on Numerical Analysis vol 38 no 2 pp 556ndash5792001

14 International Journal of Differential Equations

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

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Journal of

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Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom