research article on the harmonic problem with nonlinear...
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Research ArticleOn the Harmonic Problem with NonlinearBoundary Integral Conditions
Saker Hacene
LMA Department of Mathematics Faculty of Sciences University of Badji Mokhtar PO Box 12 23000 Annaba Algeria
Correspondence should be addressed to Saker Hacene h sakeryahoofr
Received 18 November 2013 Revised 7 January 2014 Accepted 8 January 2014 Published 23 February 2014
Academic Editor Shamsul Qamar
Copyright copy 2014 Saker HaceneThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
In the present work we deal with the harmonic problems in a bounded domain of R2 with the nonlinear boundary integralconditions After applying the Boundary integral method a nonlinear boundary integral equation is obtained the existence anduniqueness of the solution will be a consequence of applying theory of monotone operators
1 Introduction
For the harmonic problem the simplest boundary conditionwe can impose specifies 119906 at all points on the boundary Γ andis known as the Dirichlet boundary condition The Dirichletproblem for the Laplace equation can easily be solved usingthe boundary integral equation [1] If the normal derivativeof 119906 that is 120597119906120597119899 where 119899 is the outward normal to theboundary Γ is specified at all points on the boundary Γ thatis the Neumann boundary condition with int
Γ(120597119906120597119899)119889119904 = 0
then given the value of 119906 at one point on Γ enables a uniquesolution to be obtained [1]
In this work we impose more general boundary condi-tions namely the nonlinear integral equation of Urysohntype [2 3]
Much attention has been paid to the resolution of bound-ary value problems for partial differential operators withnonlinear boundary conditions by the method of integralequations in many directions (see eg Atkinson and Chan-dler [4 5] and Ruotsalainen and Wendland [6])
Problems involving nonlinearities form a basis of math-ematical models of various steady-state phenomena andprocesses in mechanics physics and many other areas ofscience Among these is the steady-state heat transfer Alsosome electromagnetic problems contain nonlinearities inthe boundary conditions for instance problems where theelectrical conductivity of the boundary is variable [7] Furtherapplications arise in heat radiation and heat transfer [7 8]
In the present paper we look for the solution of the Lapla-cian equation with nonlinear data of the form
Δ119906 (119909) = 0 119909 isin Ω (1)
120597119906
120597119899
(119909) + int
Γ
119870 (119909 119910 119906 (119910)) 119889119904119910
= 119891 (119909) 119909 isin Γ
(2)
We recall that the nonlinear boundary integral operatordefined by
119860 (119909 119906 (119909)) = int
Γ
119870 (119909 119910 119906 (119910)) 119889119904119910 119909 isin Γ (3)
is the nonlinear integral operator of Urysohn typeIn (1) we assumeΩ is an open bounded region inR2 with
a smooth boundary Γ = 120597Ω and
119891 Γ 997888rarr R 119870 Γ times Γ times R 997888rarr R (4)
are given real value functionsBy the Green representation formula we formulate a
nonlinear integral equation on the boundary Γ of the domainΩ Under some assumptions on the Kernel of the nonlinearintegral equation of Urysohn 119870(119909 119910 119906) we prove the exis-tence and uniqueness of the solution
Hindawi Publishing CorporationInternational Journal of AnalysisVolume 2014 Article ID 976520 5 pageshttpdxdoiorg1011552014976520
2 International Journal of Analysis
11 Definitions and Notations
Definition 1 (see [1 9]) Let 119898 isin N one denotes by 119867119898
(Ω)
the Sobolev space
119867119898
(Ω) = 119906 isin 1198712
(Ω) 119863120572119906 isin 1198712
(Ω) |120572| le 119898 (5)
Definition 2 (see [1 9]) Let 119904 isin R one denotes by119867119904(R119899) the
Sobolev space
119867119904(R119899) = 119906 isin 119871
2(R119899) (1 +
10038161003816100381610038161205851003816100381610038161003816
2
)
1199042
|119865 [119906]| isin 1198712
(R119899)
(6)
and the associated norm
119906119867119904 = (int
R119899(1 +
10038161003816100381610038161205851003816100381610038161003816
2
)
119904
|119865 [119906]|2119889120585)
12
(7)
with 119865[sdot] the Fourier transform
Definition 3 (see [1 9]) Let Ω sub R119899 a bounded domain andΓ = 120597Ω one defined
119867119904(Ω) = 119906 | Ω 119906 isin 119867
119904(R119899) 119904 isin R
119867119904(Γ) =
119906|Γ
119906 isin 119867119904+(12)
(R119899) 119904 gt 0
1198712
(Γ) 119904 = 0
(119867minus119904
(Γ))1015840
(dual space) 119904 lt 0
(8)
Definition 4 (see [1 9]) The Fichera trace spaces 119867119904(Γ) for
0 lt 119904 lt 1 is defined to be the completion of
1198620
119904(Γ) = 120593 isin 119862
0(Γ)
1003817100381710038171003817120593
1003817100381710038171003817119867119904(Γ)
lt infin (9)
with respect to the norm
119906119867119904(Γ)
= 1199062
1198712(Γ)
+ ∬
Γ
1003816100381610038161003816119906 (119909) minus 119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
12
(10)
2 The Boundary Integral Method
21 Representative Formula and Boundary Operator Weintroduce the fundamental solution of the Laplacian operatorin the plane defined by
119864 (119909 119910) =
1
2120587
log 1003816100381610038161003816119909 minus 119910
1003816100381610038161003816 (11)
We first consider some standard boundary integral operatorsFor 119909 isin Ω the single layer potential is
119878Ω
119906 (119909) = minus int
Γ
119864 (119909 119910) 119906 (119910) 119889119904119910 (12)
and the double layer potential is
119863Ω
119906 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910 (13)
Using Greenrsquos identity for harmonic functions we get
119906 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910
minus int
Γ
120597119906 (119910)
120597119899119910
119864 (119909 119910) 119889119904119910
(14)
for 119909 isin Ω which can be written as
119906 (119909) = 119863Ω
119906 (119909) + 119878Ω
120597119906 (119909)
120597119899
for 119909 isin Ω (15)
Sending in (15) 119909 rarr Γ The continuity of the simple layerpotential 119878
Ωand the jump relation of the double layer poten-
tial 119863Ω we can write the integral equation on the boundary
as follows
119906 (119909) minus 119863119906 (119909) = 119878
120597119906 (119909)
120597119899
119909 isin Γ (16)
where
119878
120597119906 (119909)
120597119899
= minus2 int
Γ
119864 (119909 119910)
120597119906 (119910)
120597119899
119889119904119910 119909 isin Γ
119863119906 (119909) = 2 int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910 119909 isin Γ
(17)
Clearly if 119906 isin 1198671(Ω) is the solution of (1) then the Cauchy
data 119906|Γand 120597119906120597119899|
Γsatisfies the integral equation (16)
Then the boundary conditions
120597119906
120597119899
(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (18)
yield
119906 (119909) minus 119863119906 (119909) = minus119878 (119860 (119909 119906 (119909))) + 119878119891 (119909) 119909 isin Γ
(19)
Equation (19) can be written as
(119868 minus 119863) 119906 (119909) + 119878 (119860 (119909 119906 (119909))) = 119878119891 (119909) 119909 isin Γ (20)
Conversely if 119906|Γsolves (20) then the solution of (1) can be
given by the representation formula (15) and will satisfy
120597119906
120597119899
(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (21)
due to (20) For studying the solvability of the nonlinearequation (20) we give some assumptions to be made here
(H1) We assume a diam(Ω) lt 1(H2) The Kernel 119870(sdot sdot sdot) of the Urysohn operator is a
Caratheodory function [3](H3) We assume that 120597119870(119909 119910 119906)120597119906 is measurable satisfy-
ing
0 lt 119886 le
120597119870 (119909 119910 119906)
120597119906
le 119887 lt +infin (22)
for some constants 119886 and 119887
International Journal of Analysis 3
Remark 5 (1) The operator 119878 may have eigenfunctions [1]then (H1) ensures that the integral operator
119878 119867119904(Γ) 997888rarr 119867
119904+1(Γ) (23)
is an isomorphism for every 119904 isin R and
(119878120583 120583) ge 1198881003817100381710038171003817120583
1003817100381710038171003817
2
119867minus12
(24)
for all 120583 isin 119867minus12 with some positive constant 119888 gt 0 [1] By
(sdot sdot) we denote the 1198712(Γ) scalar product
(2) The Kernel 119870(sdot sdot sdot) is a Caratheodory function (H2)that is 119870(sdot sdot 119906) is measurable for all 119906 isin R and 119870(119909 119910 sdot) iscontinuous for almost all 119909 119910 isin Γ
(3) The assumption (H3) implies that the Nemytskiioperator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (25)
is Lipschitz continuous and strongly monotonous such that
(119860119906 minus 119860V 119906 minus V) le 119887 mes (Γ) 119906 minus V20
(119860119906 minus 119860V 119906 minus V) ge 119886 mes (Γ) 119906 minus V20
(26)
for all 119906 V isin 1198712(Γ)
Theorem6 Let assumptions (H1) (H2) and (H3) holdThenfor every 119891 isin 119867
minus12 the nonlinear boundary integral equation(20) has a unique solution in 119867
12(Γ)
Proof The proof follows from the well-known theorem byBrowder and Minty on monotone operators [6 10]
Since the simple layer potential operator on Γ
119878 119867minus12
(Γ) 997888rarr 11986712
(Γ) (27)
is an isomorphism it is sufficient to consider the uniquesolvability of the following equation
119861119906 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ
(28)
We will prove that the operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (29)
is continuous and strongly monotonous
(i) In the first we show that 119861 is continuous
It is clear from the continuity of the mapping propertiesof the simple and double layer operators that
119878minus1
(119868 minus 119863) 11986712
(Γ) 997888rarr 119867minus12
(Γ) (30)
is continuous And from (H3)
119860 11986712
(Γ) 997888rarr 119867minus12
(Γ) (31)
is continuous Hence the boundary integral operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (32)
is continuous
(ii) In the second we show that 119861 is strongly monotoneoperator
The function 120583 isin 119867minus12
(Γ) defined by
120583 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) (33)
for 119906(119909) isin 11986712
(Γ) is the normal derivative of the harmonicfunction
119908 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910
minus int
Γ
120583 (119910) 119864 (119909 119910) 119889119904119910
(34)
for 119909 isin Ω this means that 119908 satisfies the problem
Δ119908 (119909) = 0 119909 isin Ω
119908 (119909) = 119906 (119909) 119909 isin Γ
(35)
Then Greenrsquos theorem yields
(119878minus1
(119868 minus 119863) 119906 119906) = int
Γ
120583119906 119889119904 = int
Γ
120597119908
120597119899
119906119889119904
= int
Γ
120597119908
120597119899
119908 119889119904 = int
Ω
(nabla119908)2119889119909
(36)
Hence for all 119906 V isin 11986712
(Γ)
(119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
= int
Ω
(nabla (1199081
minus 1199082))2
119889119909 =10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
(37)
where (1199081
minus 1199082) denotes the harmonic function correspond-
ing to the Cauchy data 119906 minus V and 119878minus1
(119868 minus 119863)(119906 minus V)On the other hand we note that there exists (]
1minus ]2) isin
119867minus12
(Γ) such that
119878 (]1
minus ]2) = 119906 minus V (38)
on Γ [1] Hence for all 119909 isin Ω we have
119878Ω
(]1
minus ]2) = 1199081
minus 1199082 (39)
The simple layer potential
119878Ω
119867119904(Γ) 997888rarr 119867
119904+(32)(Ω) (40)
is continuous for all 119904 isin R [1] Hence for 119904 = minus32 we find10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
le 1198881
1003817100381710038171003817]1
minus ]2
1003817100381710038171003817119867minus32(Γ)
le 1198882
119906 minus V119867minus12(Γ)
le 1198883
119906 minus V0(41)
for some positive constants 1198881 1198882 and 119888
3
Hence we have
119906 minus V0 ge
1
1198883
10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
(42)
4 International Journal of Analysis
Then with (28) and (37) we get
(119861119906 minus 119861V 119906 minus V) = (119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
+ (119860119906 minus 119860V 119906 minus V)
=10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ (119860119906 minus 119860V 119906 minus V)
(43)
and with (26) we get the inequality
(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ 119886mes (Γ) 119906 minus V20
(44)
hence with (42) we have
(119861119906 minus 119861V 119906 minus V)
ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
ge min1
119886mes (Γ)
1198882
3
times (10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
)
ge min1
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198671(Ω)
ge 1198884
119906 minus V211986712(Γ)
(45)
by the trace theorem [1 9] which completes the proof
Now we prove the regularity of the solution of thenonlinear boundary integral equation (20)
Theorem 7 For all 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 the unique
solution of the nonlinear boundary integral equation (20)belongs to the space 119867
119904(Γ)
In the proof of this theorem we will need the followinglemma
Lemma 8 For every 119906 isin 119867119904(Γ) 0 le 119904 le 1 one has 119860119906 isin
119867119904(Γ) and the mapping 119860 119867
119904(Γ) rarr 119867
119904(Γ) is bounded
Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871
2(Γ) has already been proved
For 119904 = 1 119906 isin 1198671(Γ) 119906 is an absolutely continuous
function By assumption (H3) the function119860119906(119909) is Lipschitzcontinuous Hence 119860119906(119909) is also absolutely continuous func-tion
It remains to prove the case 0 lt 119904 lt 1 by the assumption(H3) and due to the definition of the Sobolev space inDefinition 4 we have
∬
Γ
1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
le 1198872(mes (Γ))
21199062
119867119904(Γ)
(46)
which completes the proof of Lemma 8
Proof of Theorem 7 Let 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 be given
By Theorem 6 there exists a unique solution 119906 isin 11986712
(Γ) ofthe nonlinear boundary integral equation
(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)
Lemma 8 ensure that
119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)
therefore
(119868 minus 119863) 119906 isin 119867119904(Γ) (49)
This implies together with the Fredholm property of thedouble layer potential operator that 119906 isin 119867
119904(Γ) 12 le 119904 le 32
Example 9 Here we give an example to illustrate thetheoretical results We consider the harmonic problems
Δ119906 (119909) = 0 119909 isin Ω
120597119906
120597119899
(119909) + int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910
= 119891 (119909) 119909 isin Γ
(50)
where the nonlinear boundary integral equation of Urysohntype is defined by
119860119906 (119909) = int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ
(51)
and the domain is
Ω = 119909 = (1199091 1199092) 1199092
1+ 1199092
2lt 1199032
lt
1
4
(52)
Clearly the nonlinearity satisfies our assumptions (1198671) (1198672)
and (1198673) such that
diam (Ω) = 2119903 lt 1 (53)
The Kernel (2119906(119910) + sin 119906(119910)) of the nonlinear boundaryintegral equation of Urysohn type is a Caratheodory func-tion And
120597119870 (119909 119910 119906)
120597119906
= 2 + cos 119906 (119910) (54)
is measurable satisfying
1 le
120597 (2119906 (119910) + sin 119906 (119910))
120597119906
le 3 lt +infin (55)
implying that the Nemytskii operator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (56)
is Lipschitz continuous and strongly monotonous such that
2120587119903119906 minus V20
le (119860119906 minus 119860V 119906 minus V) le 6120587119903119906 minus V20
(57)
for all 119906 V isin 1198712(Γ)
International Journal of Analysis 5
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper
References
[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008
[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013
[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964
[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990
[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997
[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988
[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981
[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984
[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975
[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980
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2 International Journal of Analysis
11 Definitions and Notations
Definition 1 (see [1 9]) Let 119898 isin N one denotes by 119867119898
(Ω)
the Sobolev space
119867119898
(Ω) = 119906 isin 1198712
(Ω) 119863120572119906 isin 1198712
(Ω) |120572| le 119898 (5)
Definition 2 (see [1 9]) Let 119904 isin R one denotes by119867119904(R119899) the
Sobolev space
119867119904(R119899) = 119906 isin 119871
2(R119899) (1 +
10038161003816100381610038161205851003816100381610038161003816
2
)
1199042
|119865 [119906]| isin 1198712
(R119899)
(6)
and the associated norm
119906119867119904 = (int
R119899(1 +
10038161003816100381610038161205851003816100381610038161003816
2
)
119904
|119865 [119906]|2119889120585)
12
(7)
with 119865[sdot] the Fourier transform
Definition 3 (see [1 9]) Let Ω sub R119899 a bounded domain andΓ = 120597Ω one defined
119867119904(Ω) = 119906 | Ω 119906 isin 119867
119904(R119899) 119904 isin R
119867119904(Γ) =
119906|Γ
119906 isin 119867119904+(12)
(R119899) 119904 gt 0
1198712
(Γ) 119904 = 0
(119867minus119904
(Γ))1015840
(dual space) 119904 lt 0
(8)
Definition 4 (see [1 9]) The Fichera trace spaces 119867119904(Γ) for
0 lt 119904 lt 1 is defined to be the completion of
1198620
119904(Γ) = 120593 isin 119862
0(Γ)
1003817100381710038171003817120593
1003817100381710038171003817119867119904(Γ)
lt infin (9)
with respect to the norm
119906119867119904(Γ)
= 1199062
1198712(Γ)
+ ∬
Γ
1003816100381610038161003816119906 (119909) minus 119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
12
(10)
2 The Boundary Integral Method
21 Representative Formula and Boundary Operator Weintroduce the fundamental solution of the Laplacian operatorin the plane defined by
119864 (119909 119910) =
1
2120587
log 1003816100381610038161003816119909 minus 119910
1003816100381610038161003816 (11)
We first consider some standard boundary integral operatorsFor 119909 isin Ω the single layer potential is
119878Ω
119906 (119909) = minus int
Γ
119864 (119909 119910) 119906 (119910) 119889119904119910 (12)
and the double layer potential is
119863Ω
119906 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910 (13)
Using Greenrsquos identity for harmonic functions we get
119906 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910
minus int
Γ
120597119906 (119910)
120597119899119910
119864 (119909 119910) 119889119904119910
(14)
for 119909 isin Ω which can be written as
119906 (119909) = 119863Ω
119906 (119909) + 119878Ω
120597119906 (119909)
120597119899
for 119909 isin Ω (15)
Sending in (15) 119909 rarr Γ The continuity of the simple layerpotential 119878
Ωand the jump relation of the double layer poten-
tial 119863Ω we can write the integral equation on the boundary
as follows
119906 (119909) minus 119863119906 (119909) = 119878
120597119906 (119909)
120597119899
119909 isin Γ (16)
where
119878
120597119906 (119909)
120597119899
= minus2 int
Γ
119864 (119909 119910)
120597119906 (119910)
120597119899
119889119904119910 119909 isin Γ
119863119906 (119909) = 2 int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910 119909 isin Γ
(17)
Clearly if 119906 isin 1198671(Ω) is the solution of (1) then the Cauchy
data 119906|Γand 120597119906120597119899|
Γsatisfies the integral equation (16)
Then the boundary conditions
120597119906
120597119899
(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (18)
yield
119906 (119909) minus 119863119906 (119909) = minus119878 (119860 (119909 119906 (119909))) + 119878119891 (119909) 119909 isin Γ
(19)
Equation (19) can be written as
(119868 minus 119863) 119906 (119909) + 119878 (119860 (119909 119906 (119909))) = 119878119891 (119909) 119909 isin Γ (20)
Conversely if 119906|Γsolves (20) then the solution of (1) can be
given by the representation formula (15) and will satisfy
120597119906
120597119899
(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (21)
due to (20) For studying the solvability of the nonlinearequation (20) we give some assumptions to be made here
(H1) We assume a diam(Ω) lt 1(H2) The Kernel 119870(sdot sdot sdot) of the Urysohn operator is a
Caratheodory function [3](H3) We assume that 120597119870(119909 119910 119906)120597119906 is measurable satisfy-
ing
0 lt 119886 le
120597119870 (119909 119910 119906)
120597119906
le 119887 lt +infin (22)
for some constants 119886 and 119887
International Journal of Analysis 3
Remark 5 (1) The operator 119878 may have eigenfunctions [1]then (H1) ensures that the integral operator
119878 119867119904(Γ) 997888rarr 119867
119904+1(Γ) (23)
is an isomorphism for every 119904 isin R and
(119878120583 120583) ge 1198881003817100381710038171003817120583
1003817100381710038171003817
2
119867minus12
(24)
for all 120583 isin 119867minus12 with some positive constant 119888 gt 0 [1] By
(sdot sdot) we denote the 1198712(Γ) scalar product
(2) The Kernel 119870(sdot sdot sdot) is a Caratheodory function (H2)that is 119870(sdot sdot 119906) is measurable for all 119906 isin R and 119870(119909 119910 sdot) iscontinuous for almost all 119909 119910 isin Γ
(3) The assumption (H3) implies that the Nemytskiioperator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (25)
is Lipschitz continuous and strongly monotonous such that
(119860119906 minus 119860V 119906 minus V) le 119887 mes (Γ) 119906 minus V20
(119860119906 minus 119860V 119906 minus V) ge 119886 mes (Γ) 119906 minus V20
(26)
for all 119906 V isin 1198712(Γ)
Theorem6 Let assumptions (H1) (H2) and (H3) holdThenfor every 119891 isin 119867
minus12 the nonlinear boundary integral equation(20) has a unique solution in 119867
12(Γ)
Proof The proof follows from the well-known theorem byBrowder and Minty on monotone operators [6 10]
Since the simple layer potential operator on Γ
119878 119867minus12
(Γ) 997888rarr 11986712
(Γ) (27)
is an isomorphism it is sufficient to consider the uniquesolvability of the following equation
119861119906 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ
(28)
We will prove that the operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (29)
is continuous and strongly monotonous
(i) In the first we show that 119861 is continuous
It is clear from the continuity of the mapping propertiesof the simple and double layer operators that
119878minus1
(119868 minus 119863) 11986712
(Γ) 997888rarr 119867minus12
(Γ) (30)
is continuous And from (H3)
119860 11986712
(Γ) 997888rarr 119867minus12
(Γ) (31)
is continuous Hence the boundary integral operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (32)
is continuous
(ii) In the second we show that 119861 is strongly monotoneoperator
The function 120583 isin 119867minus12
(Γ) defined by
120583 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) (33)
for 119906(119909) isin 11986712
(Γ) is the normal derivative of the harmonicfunction
119908 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910
minus int
Γ
120583 (119910) 119864 (119909 119910) 119889119904119910
(34)
for 119909 isin Ω this means that 119908 satisfies the problem
Δ119908 (119909) = 0 119909 isin Ω
119908 (119909) = 119906 (119909) 119909 isin Γ
(35)
Then Greenrsquos theorem yields
(119878minus1
(119868 minus 119863) 119906 119906) = int
Γ
120583119906 119889119904 = int
Γ
120597119908
120597119899
119906119889119904
= int
Γ
120597119908
120597119899
119908 119889119904 = int
Ω
(nabla119908)2119889119909
(36)
Hence for all 119906 V isin 11986712
(Γ)
(119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
= int
Ω
(nabla (1199081
minus 1199082))2
119889119909 =10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
(37)
where (1199081
minus 1199082) denotes the harmonic function correspond-
ing to the Cauchy data 119906 minus V and 119878minus1
(119868 minus 119863)(119906 minus V)On the other hand we note that there exists (]
1minus ]2) isin
119867minus12
(Γ) such that
119878 (]1
minus ]2) = 119906 minus V (38)
on Γ [1] Hence for all 119909 isin Ω we have
119878Ω
(]1
minus ]2) = 1199081
minus 1199082 (39)
The simple layer potential
119878Ω
119867119904(Γ) 997888rarr 119867
119904+(32)(Ω) (40)
is continuous for all 119904 isin R [1] Hence for 119904 = minus32 we find10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
le 1198881
1003817100381710038171003817]1
minus ]2
1003817100381710038171003817119867minus32(Γ)
le 1198882
119906 minus V119867minus12(Γ)
le 1198883
119906 minus V0(41)
for some positive constants 1198881 1198882 and 119888
3
Hence we have
119906 minus V0 ge
1
1198883
10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
(42)
4 International Journal of Analysis
Then with (28) and (37) we get
(119861119906 minus 119861V 119906 minus V) = (119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
+ (119860119906 minus 119860V 119906 minus V)
=10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ (119860119906 minus 119860V 119906 minus V)
(43)
and with (26) we get the inequality
(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ 119886mes (Γ) 119906 minus V20
(44)
hence with (42) we have
(119861119906 minus 119861V 119906 minus V)
ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
ge min1
119886mes (Γ)
1198882
3
times (10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
)
ge min1
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198671(Ω)
ge 1198884
119906 minus V211986712(Γ)
(45)
by the trace theorem [1 9] which completes the proof
Now we prove the regularity of the solution of thenonlinear boundary integral equation (20)
Theorem 7 For all 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 the unique
solution of the nonlinear boundary integral equation (20)belongs to the space 119867
119904(Γ)
In the proof of this theorem we will need the followinglemma
Lemma 8 For every 119906 isin 119867119904(Γ) 0 le 119904 le 1 one has 119860119906 isin
119867119904(Γ) and the mapping 119860 119867
119904(Γ) rarr 119867
119904(Γ) is bounded
Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871
2(Γ) has already been proved
For 119904 = 1 119906 isin 1198671(Γ) 119906 is an absolutely continuous
function By assumption (H3) the function119860119906(119909) is Lipschitzcontinuous Hence 119860119906(119909) is also absolutely continuous func-tion
It remains to prove the case 0 lt 119904 lt 1 by the assumption(H3) and due to the definition of the Sobolev space inDefinition 4 we have
∬
Γ
1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
le 1198872(mes (Γ))
21199062
119867119904(Γ)
(46)
which completes the proof of Lemma 8
Proof of Theorem 7 Let 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 be given
By Theorem 6 there exists a unique solution 119906 isin 11986712
(Γ) ofthe nonlinear boundary integral equation
(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)
Lemma 8 ensure that
119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)
therefore
(119868 minus 119863) 119906 isin 119867119904(Γ) (49)
This implies together with the Fredholm property of thedouble layer potential operator that 119906 isin 119867
119904(Γ) 12 le 119904 le 32
Example 9 Here we give an example to illustrate thetheoretical results We consider the harmonic problems
Δ119906 (119909) = 0 119909 isin Ω
120597119906
120597119899
(119909) + int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910
= 119891 (119909) 119909 isin Γ
(50)
where the nonlinear boundary integral equation of Urysohntype is defined by
119860119906 (119909) = int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ
(51)
and the domain is
Ω = 119909 = (1199091 1199092) 1199092
1+ 1199092
2lt 1199032
lt
1
4
(52)
Clearly the nonlinearity satisfies our assumptions (1198671) (1198672)
and (1198673) such that
diam (Ω) = 2119903 lt 1 (53)
The Kernel (2119906(119910) + sin 119906(119910)) of the nonlinear boundaryintegral equation of Urysohn type is a Caratheodory func-tion And
120597119870 (119909 119910 119906)
120597119906
= 2 + cos 119906 (119910) (54)
is measurable satisfying
1 le
120597 (2119906 (119910) + sin 119906 (119910))
120597119906
le 3 lt +infin (55)
implying that the Nemytskii operator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (56)
is Lipschitz continuous and strongly monotonous such that
2120587119903119906 minus V20
le (119860119906 minus 119860V 119906 minus V) le 6120587119903119906 minus V20
(57)
for all 119906 V isin 1198712(Γ)
International Journal of Analysis 5
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper
References
[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008
[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013
[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964
[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990
[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997
[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988
[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981
[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984
[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975
[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980
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Stochastic AnalysisInternational Journal of
International Journal of Analysis 3
Remark 5 (1) The operator 119878 may have eigenfunctions [1]then (H1) ensures that the integral operator
119878 119867119904(Γ) 997888rarr 119867
119904+1(Γ) (23)
is an isomorphism for every 119904 isin R and
(119878120583 120583) ge 1198881003817100381710038171003817120583
1003817100381710038171003817
2
119867minus12
(24)
for all 120583 isin 119867minus12 with some positive constant 119888 gt 0 [1] By
(sdot sdot) we denote the 1198712(Γ) scalar product
(2) The Kernel 119870(sdot sdot sdot) is a Caratheodory function (H2)that is 119870(sdot sdot 119906) is measurable for all 119906 isin R and 119870(119909 119910 sdot) iscontinuous for almost all 119909 119910 isin Γ
(3) The assumption (H3) implies that the Nemytskiioperator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (25)
is Lipschitz continuous and strongly monotonous such that
(119860119906 minus 119860V 119906 minus V) le 119887 mes (Γ) 119906 minus V20
(119860119906 minus 119860V 119906 minus V) ge 119886 mes (Γ) 119906 minus V20
(26)
for all 119906 V isin 1198712(Γ)
Theorem6 Let assumptions (H1) (H2) and (H3) holdThenfor every 119891 isin 119867
minus12 the nonlinear boundary integral equation(20) has a unique solution in 119867
12(Γ)
Proof The proof follows from the well-known theorem byBrowder and Minty on monotone operators [6 10]
Since the simple layer potential operator on Γ
119878 119867minus12
(Γ) 997888rarr 11986712
(Γ) (27)
is an isomorphism it is sufficient to consider the uniquesolvability of the following equation
119861119906 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ
(28)
We will prove that the operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (29)
is continuous and strongly monotonous
(i) In the first we show that 119861 is continuous
It is clear from the continuity of the mapping propertiesof the simple and double layer operators that
119878minus1
(119868 minus 119863) 11986712
(Γ) 997888rarr 119867minus12
(Γ) (30)
is continuous And from (H3)
119860 11986712
(Γ) 997888rarr 119867minus12
(Γ) (31)
is continuous Hence the boundary integral operator
119861 11986712
(Γ) 997888rarr 119867minus12
(Γ) (32)
is continuous
(ii) In the second we show that 119861 is strongly monotoneoperator
The function 120583 isin 119867minus12
(Γ) defined by
120583 (119909) = 119878minus1
(119868 minus 119863) 119906 (119909) (33)
for 119906(119909) isin 11986712
(Γ) is the normal derivative of the harmonicfunction
119908 (119909) = int
Γ
119906 (119910)
120597
120597119899119910
119864 (119909 119910) 119889119904119910
minus int
Γ
120583 (119910) 119864 (119909 119910) 119889119904119910
(34)
for 119909 isin Ω this means that 119908 satisfies the problem
Δ119908 (119909) = 0 119909 isin Ω
119908 (119909) = 119906 (119909) 119909 isin Γ
(35)
Then Greenrsquos theorem yields
(119878minus1
(119868 minus 119863) 119906 119906) = int
Γ
120583119906 119889119904 = int
Γ
120597119908
120597119899
119906119889119904
= int
Γ
120597119908
120597119899
119908 119889119904 = int
Ω
(nabla119908)2119889119909
(36)
Hence for all 119906 V isin 11986712
(Γ)
(119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
= int
Ω
(nabla (1199081
minus 1199082))2
119889119909 =10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
(37)
where (1199081
minus 1199082) denotes the harmonic function correspond-
ing to the Cauchy data 119906 minus V and 119878minus1
(119868 minus 119863)(119906 minus V)On the other hand we note that there exists (]
1minus ]2) isin
119867minus12
(Γ) such that
119878 (]1
minus ]2) = 119906 minus V (38)
on Γ [1] Hence for all 119909 isin Ω we have
119878Ω
(]1
minus ]2) = 1199081
minus 1199082 (39)
The simple layer potential
119878Ω
119867119904(Γ) 997888rarr 119867
119904+(32)(Ω) (40)
is continuous for all 119904 isin R [1] Hence for 119904 = minus32 we find10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
le 1198881
1003817100381710038171003817]1
minus ]2
1003817100381710038171003817119867minus32(Γ)
le 1198882
119906 minus V119867minus12(Γ)
le 1198883
119906 minus V0(41)
for some positive constants 1198881 1198882 and 119888
3
Hence we have
119906 minus V0 ge
1
1198883
10038171003817100381710038171199081
minus 1199082
10038171003817100381710038171198712(Ω)
(42)
4 International Journal of Analysis
Then with (28) and (37) we get
(119861119906 minus 119861V 119906 minus V) = (119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
+ (119860119906 minus 119860V 119906 minus V)
=10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ (119860119906 minus 119860V 119906 minus V)
(43)
and with (26) we get the inequality
(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ 119886mes (Γ) 119906 minus V20
(44)
hence with (42) we have
(119861119906 minus 119861V 119906 minus V)
ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
ge min1
119886mes (Γ)
1198882
3
times (10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
)
ge min1
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198671(Ω)
ge 1198884
119906 minus V211986712(Γ)
(45)
by the trace theorem [1 9] which completes the proof
Now we prove the regularity of the solution of thenonlinear boundary integral equation (20)
Theorem 7 For all 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 the unique
solution of the nonlinear boundary integral equation (20)belongs to the space 119867
119904(Γ)
In the proof of this theorem we will need the followinglemma
Lemma 8 For every 119906 isin 119867119904(Γ) 0 le 119904 le 1 one has 119860119906 isin
119867119904(Γ) and the mapping 119860 119867
119904(Γ) rarr 119867
119904(Γ) is bounded
Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871
2(Γ) has already been proved
For 119904 = 1 119906 isin 1198671(Γ) 119906 is an absolutely continuous
function By assumption (H3) the function119860119906(119909) is Lipschitzcontinuous Hence 119860119906(119909) is also absolutely continuous func-tion
It remains to prove the case 0 lt 119904 lt 1 by the assumption(H3) and due to the definition of the Sobolev space inDefinition 4 we have
∬
Γ
1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
le 1198872(mes (Γ))
21199062
119867119904(Γ)
(46)
which completes the proof of Lemma 8
Proof of Theorem 7 Let 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 be given
By Theorem 6 there exists a unique solution 119906 isin 11986712
(Γ) ofthe nonlinear boundary integral equation
(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)
Lemma 8 ensure that
119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)
therefore
(119868 minus 119863) 119906 isin 119867119904(Γ) (49)
This implies together with the Fredholm property of thedouble layer potential operator that 119906 isin 119867
119904(Γ) 12 le 119904 le 32
Example 9 Here we give an example to illustrate thetheoretical results We consider the harmonic problems
Δ119906 (119909) = 0 119909 isin Ω
120597119906
120597119899
(119909) + int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910
= 119891 (119909) 119909 isin Γ
(50)
where the nonlinear boundary integral equation of Urysohntype is defined by
119860119906 (119909) = int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ
(51)
and the domain is
Ω = 119909 = (1199091 1199092) 1199092
1+ 1199092
2lt 1199032
lt
1
4
(52)
Clearly the nonlinearity satisfies our assumptions (1198671) (1198672)
and (1198673) such that
diam (Ω) = 2119903 lt 1 (53)
The Kernel (2119906(119910) + sin 119906(119910)) of the nonlinear boundaryintegral equation of Urysohn type is a Caratheodory func-tion And
120597119870 (119909 119910 119906)
120597119906
= 2 + cos 119906 (119910) (54)
is measurable satisfying
1 le
120597 (2119906 (119910) + sin 119906 (119910))
120597119906
le 3 lt +infin (55)
implying that the Nemytskii operator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (56)
is Lipschitz continuous and strongly monotonous such that
2120587119903119906 minus V20
le (119860119906 minus 119860V 119906 minus V) le 6120587119903119906 minus V20
(57)
for all 119906 V isin 1198712(Γ)
International Journal of Analysis 5
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper
References
[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008
[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013
[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964
[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990
[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997
[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988
[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981
[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984
[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975
[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Analysis
Then with (28) and (37) we get
(119861119906 minus 119861V 119906 minus V) = (119878minus1
(119868 minus 119863) (119906 minus V) 119906 minus V)
+ (119860119906 minus 119860V 119906 minus V)
=10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ (119860119906 minus 119860V 119906 minus V)
(43)
and with (26) we get the inequality
(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+ 119886mes (Γ) 119906 minus V20
(44)
hence with (42) we have
(119861119906 minus 119861V 119906 minus V)
ge10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
ge min1
119886mes (Γ)
1198882
3
times (10038161003816100381610038161199081
minus 1199082
1003816100381610038161003816
2
1198671(Ω)
+10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198712(Ω)
)
ge min1
119886mes (Γ)
1198882
3
10038171003817100381710038171199081
minus 1199082
1003817100381710038171003817
2
1198671(Ω)
ge 1198884
119906 minus V211986712(Γ)
(45)
by the trace theorem [1 9] which completes the proof
Now we prove the regularity of the solution of thenonlinear boundary integral equation (20)
Theorem 7 For all 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 the unique
solution of the nonlinear boundary integral equation (20)belongs to the space 119867
119904(Γ)
In the proof of this theorem we will need the followinglemma
Lemma 8 For every 119906 isin 119867119904(Γ) 0 le 119904 le 1 one has 119860119906 isin
119867119904(Γ) and the mapping 119860 119867
119904(Γ) rarr 119867
119904(Γ) is bounded
Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871
2(Γ) has already been proved
For 119904 = 1 119906 isin 1198671(Γ) 119906 is an absolutely continuous
function By assumption (H3) the function119860119906(119909) is Lipschitzcontinuous Hence 119860119906(119909) is also absolutely continuous func-tion
It remains to prove the case 0 lt 119904 lt 1 by the assumption(H3) and due to the definition of the Sobolev space inDefinition 4 we have
∬
Γ
1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)
1003816100381610038161003816
2
1003816100381610038161003816119909 minus 119910
1003816100381610038161003816
1+2119904119889119904119909
119889119904119910
le 1198872(mes (Γ))
21199062
119867119904(Γ)
(46)
which completes the proof of Lemma 8
Proof of Theorem 7 Let 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 be given
By Theorem 6 there exists a unique solution 119906 isin 11986712
(Γ) ofthe nonlinear boundary integral equation
(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)
Lemma 8 ensure that
119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)
therefore
(119868 minus 119863) 119906 isin 119867119904(Γ) (49)
This implies together with the Fredholm property of thedouble layer potential operator that 119906 isin 119867
119904(Γ) 12 le 119904 le 32
Example 9 Here we give an example to illustrate thetheoretical results We consider the harmonic problems
Δ119906 (119909) = 0 119909 isin Ω
120597119906
120597119899
(119909) + int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910
= 119891 (119909) 119909 isin Γ
(50)
where the nonlinear boundary integral equation of Urysohntype is defined by
119860119906 (119909) = int
Γ
(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ
(51)
and the domain is
Ω = 119909 = (1199091 1199092) 1199092
1+ 1199092
2lt 1199032
lt
1
4
(52)
Clearly the nonlinearity satisfies our assumptions (1198671) (1198672)
and (1198673) such that
diam (Ω) = 2119903 lt 1 (53)
The Kernel (2119906(119910) + sin 119906(119910)) of the nonlinear boundaryintegral equation of Urysohn type is a Caratheodory func-tion And
120597119870 (119909 119910 119906)
120597119906
= 2 + cos 119906 (119910) (54)
is measurable satisfying
1 le
120597 (2119906 (119910) + sin 119906 (119910))
120597119906
le 3 lt +infin (55)
implying that the Nemytskii operator
119860 1198712
(Γ) 997888rarr 1198712
(Γ) (56)
is Lipschitz continuous and strongly monotonous such that
2120587119903119906 minus V20
le (119860119906 minus 119860V 119906 minus V) le 6120587119903119906 minus V20
(57)
for all 119906 V isin 1198712(Γ)
International Journal of Analysis 5
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper
References
[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008
[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013
[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964
[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990
[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997
[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988
[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981
[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984
[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975
[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 5
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper
References
[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008
[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013
[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964
[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990
[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997
[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988
[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981
[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984
[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975
[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of