on the monotonicity conservation in numerical solutions of the heat equation

Applied Numerical Mathematics 42 (2002) 189–199www.elsevier.com/locate/apnum

On the monotonicity conservation in numerical solutionsof the heat equation

Róbert Horváth

University of Technology Eindhoven, P.O. Box 513, HG 8.25, 5600 MB Eindhoven, The Netherlands

Abstract

It is important to choose such numerical methods in practice that mirror the characteristic properties of thedescribed process beyond the stability and convergence. The investigated qualitative property in this paper is theconservation of the monotonicity in space of the initial heat distribution. We prove some statements about themonotonicity conservation and total monotonicity of one-step vector-iterations. Then, applying these results, weconsider the numerical solutions of the one-dimensional heat equation. Our main theorem formulates the necessaryand sufficient condition of the uniform monotonicity conservation. The sharpness of the conditions is demonstratedby numerical examples. 2001 IMACS. Published by Elsevier Science B.V. All rights reserved.

Keywords:Numerical solution; Qualitative properties; Heat equation; Monotonicity conservation

1. Introduction

The temperature changes of a homogeneous, isotropic rod can be described by the one-dimensionalheat equation, which has the form

∂u

∂t= ∂2u

∂x2, x ∈ (0,1), t > 0, (1)

u(0, t)= µ1, t 0, (2)

u(1, t)= µ2, t 0, (3)

u(x,0)= u0(x), x ∈ [0,1], (4)

where we have chosen constant boundary conditions and have used non-dimensional variables [6]. Thesufficiently smooth functionu0 : [0,1] → R describes the initial temperature of the rod (u0(0) = µ1,u0(1)= µ2) and the functionu= u(x, t) denotes the temperature in the pointx ∈ [0,1] and at the pointof time t 0.

E-mail address:[email protected] (R. Horváth).

0168-9274/01/$22.00 2001 IMACS. Published by Elsevier Science B.V. All rights reserved.PII: S0168-9274(01)00150-7

190 R. Horváth / Applied Numerical Mathematics 42 (2002) 189–199

To solve the above problem numerically we define a uniform mesh on the setΩ = [0,1]×[0,∞) withthe parametersτ > 0 andh= 1/(n+ 1) (n ∈ N) as follows

Ωh,τ := (xi, tj ) ∈Ω | xi = ih (i = 0, . . . , n+ 1), tj = jτ (j ∈ N)

. (5)

We denote the approximation to the exact valueu(ih, jτ) by y(j)i and we sety(j) = (y(j)

1 , . . . , y(j)n ) ∈ R

n.We assume throughout the paper the conditionn > 1 because the casen= 1 is of no practical importance.In this paper the approximating values are generated by the so-called(σ, θ)-method (see paper [7] andfor a detailed analysis of the method consult [2]), that is we generate the valuesy

(j)

i by the iteration

My(j+1) − y(j)

τ= −θ 1

h2Qy(j+1) − (1− θ)

1

h2Qy(j) + 1

h2

(µ1e1 +µ2en

), j ∈ N. (6)

Heree1 = (1,0, . . . ,0), en = (0, . . . ,1) ∈ Rn, Q = tridiag[−1,2,−1] ∈ R

n×n andM = I−σQ ∈ Rn×n

are tridiagonal matrices (I is the unit matrix),σ ∈ [0,1/4) andθ ∈ [0,1] are given parameters. The vectory(0) is a suitable approximation to the initial functionu0.

The(σ, θ)-method unites a few remarkable numerical methods. For example, the(0, θ)-method givesthe classical finite differenceθ -method (the(0,1/2)-method is the well-known Crank–Nicolson method)and the(1/6, θ)-method results in a finite element method with linear elements. In this sense the(σ, θ)-method can be considered as a generalization of the classical methods.

The condition of the convergence of the method can be found in [2]. Moreover, it is known that itis not enough to construct a convergent numerical method in practice, the method must be qualitativelyadequate too. This means that we have to require among others the nonnegativity and shape conservationof the initial function, the sign-stability and maximum norm contractivity of the numerical solution (see,e.g., [1,3–5,7–9]). Let us execute the(σ, θ)-method with the parametersσ = 0, θ = 1/2 (this is theCrank–Nicolson method, which is unconditionally stable),µ1 = 0, µ2 = 1, τ = 20/36 and with theinitial vectory(0) = (0,1/2,1/2,1/2,1) (that ish= 1/6). This method is stable, but it does not suit thequalitative requirements. Namely, the initial vector describes a monotonically increasing heat distributionin space while the first iteratey(1) = (0.32,0.17,0.5,0.82,0.68) does not. In this case we say that thenumerical method does not conserve the monotonicity. If the monotonicity is not conserved, then theconsidered method does not describe a real physical process, because this phenomenon contradicts thesecond law of Thermodynamics.

In this paper we determine the conditions of the monotonicity conservation of the(σ, θ)-method.Introducing the notationsq = τ/h2, z = θq − σ , T1 = I + zQ, T2 = I − (q − z)Q andT = T−1

1 T2 theiteration (6) can be written in the one-step vector-iteration form

y(j+1) = 1 0 0

qT−11 e1 T qT−1

1 en

0 0 1

y(j) = Ty(j), j = 0,1, . . . , (7)

where y(j) = (µ1, (y(j)),µ2) ∈ R

n+2 and T ∈ R(n+2)×(n+2). This is why we investigate the one-step

iterations from the monotonicity conservation point of view in Section 2. In Section 3 we apply the linearalgebraic results of Section 2 for the special iteration (7).

In this paper we call an arbitrary vectorx ∈ Rn nonnegative if its elements are nonnegative, in notation:

x 0. In a similar manner we introduce the notions of positive, nonpositive and negative vectors too. Ifx1 ∈ R

n andx2 ∈ Rn are two vectors, thenx1 x2 (x1 > x2) meansx1 − x2 0 (x1 − x2 > 0). Obviously,

these definitions and notations can be applied to matrices too.

R. Horváth / Applied Numerical Mathematics 42 (2002) 189–199 191

2. Monotonicity conservation of one-step vector-iterations

A vectorx ∈ Rn is said to be monotonically increasing (decreasing) if the relationsx1 x2 · · · xn

(x1 x2 · · · xn) hold. Let us consider the one-step vector-iteration process

x(j+1) = Ax(j), j = 0,1, . . . , (8)

wherex(0) ∈ Rn is an arbitrary initial vector andA ∈ R

n×n is an arbitrary matrix.

Definition 1. We say that the iteration (8) is monotonicity conserving if the monotone increase (monotonedecrease) of any initial vectorx(0) implies the same property for the vectorx(1).

These notions of the monotonicity of a vector and the monotonicity conservation of an iteration canbe handled only with difficulties, therefore we introduce another monotonicity notion, where we canuse a matrix representation form. Let us denote the matrix tridiag[−1,1,0] by D, which is the discreteanalogue of the differential operator. Since the elements of the inverse ofD areD−1

i,j = 1 if i j and

D−1i,j = 0 if i < j , thereforeD is a so-called inverse-positive matrix.

Definition 2. The iteration (8) is called totally monotone if for any vectorx(0) such thatDx(0) 0 (orDx(0) 0) the relationDx(1) 0 (or Dx(1) 0) holds.

Remark 3. Since the relationDx(0) 0 can be satisfied only for nonnegative vectors, therefore atotally monotone iteration produces nonnegative monotonically increasing (decreasing) vectors fromall nonnegative monotonically increasing (decreasing) vectors. However, a monotonicity conservingiteration produces monotonically increasing (decreasing) vectors from all monotonically increasing(decreasing) vectors. Later we will prove (Theorem 11) that in case ofA 0 these properties areequivalent.

Theorem 4. The iteration(8) is totally monotone if and only if both of the conditionsDAD−1 0 andDAD− 0 are fulfilled.

Proof. Let x ∈ Rn be an arbitrary nonnegative vector. In this case for the vectorD−1x the condition

D(D−1x) 0 is valid. Thus, if the iteration is totally monotone, then the relationDAD−1x 0 holdsfor all nonnegative vectorsx. So we get the condition that the matrixDAD−1 must be nonnegative.Conversely, if DAD−1 0, then assumingDx(0) 0 follows (DAD−1)(Dx(0)) = Dx(1) 0. TheconditionDAD− 0 can be obtained considering the vectorsD−x. Remark 5. We would like to shed light on the conditions of Theorem 4. Let us introduce the notationsSri,k = ∑n

j=k Ai,j and Sli,k = ∑kj=1Ai,j . Then the iteration (8) is totally monotone if and only if the

relations

Sri+1,k Sri,k, Sli+1,k Sli,k (k = 1, . . . , n; i = 1, . . . , n− 1) (9)

hold and the numbersSr1,k, Sln,k are nonnegative. Using the above consequences fork = 1 andk = n,

respectively, we get a necessary condition of the total monotonicity: the sum of the elements in the rowsof the matrixA is some fixed nonnegative constant.


Remark 6. Multiplying the inequalitiesDAD−1 0 andDAD− 0 by the nonnegative matricesD−1

and D−, respectively, we get that a necessary condition of the total monotonicity isAD−1 0 andAD− 0.

Remark 7. We notice that the nonnegativity of the matrixA is not necessary to the total monotonicity ofthe iteration. As one can see, with the not nonnegative matrix

A =[1 0 0

1 −1 10 0 1

](10)

the iteration is a totally monotone one.

Remark 8. Let I denote the matrix with the elementsIi,(n−i+1) = 1 (i = 1, . . . , n) andIi,j = 0 otherwise.We call the matrixA doubly symmetric if the relationIAI = A holds. If the matrixA is symmetric andit is also symmetric for the secondary diagonal, then the matrixA is doubly symmetric. If the matrixA is doubly symmetric, then the iteration (8) is totally monotone if and only ifDAD−1 0. To see thiswe have to show that the conditionDAD−1 0 yields the conditionDAD− 0. If DAD−1 0, thenIDAD−1I 0. Hence

0 IDAD−1I = DIAID− = DAD−. (11)

Remark 9. If an iteration is totally monotone with the matricesA1 andA2, then it is that with the matrixA1A2 too.

Theorem 10. If the iteration(8) is totally monotone, then it conserves the monotonicity too.

Proof. For simplicity we can assume thatx(0) is a monotonically increasing vector. Thenx(0) can bewritten in the formx(0) = v − ce, wherev 0 is a monotonically increasing vector,e = (1, . . . ,1) andc is a suitable real number. ThenAx(0) = A(v − ce) = Av − cAe. Since the sum of the elements in therows of the matrixA is constant, therefore the vectorcAe is a constant vector. On the other hand, thevectorAv is monotonically increasing because of the total monotonicity. Thus the vectorAx(0) is alsomonotonically increasing. Theorem 11. Assume thatA 0. Then the iteration(8) is totally monotone if and only if it conserves themonotonicity.

Proof. The necessity follows from the previous theorem. To see that the condition is also sufficient letus suppose that for a nonnegative monotonically increasing vectorx(0) the conditionDx(0) 0 is valid.Obviously, Ax(0) is a nonnegative and monotonically increasing vector. So the relationDAx(0) 0 isvalid. For the case ofDx(0) 0 the proof is similar.

3. Monotonicity conservation of the (σ, θ)-method

Definition 12. We say that the(σ, θ)-method is monotonicity conserving (respectively totally monotone)if the iteration (7) is the same. The(σ, θ)-method is said to be uniformly monotonicity conserving


(respectively uniformly totally monotone) for a fixed valueq if the iteration (7) is monotonicityconserving (respectively totally monotone) for all step-sizesh= 1/(n+ 1).

In this section we give the necessary and sufficient condition of the uniform monotonicity conservation(respectively uniform total monotonicity) of the numerical solution. Our results follow from theapplication of the theorems of the previous section for the special iteration (7). Moreover, we applythe fact (see [7]) that the matrixT can be expressed in the form

T = 1

z

[q

zG − (q − z)I

]if z = 0, T = I − qQ if z= 0, (12)

where the matrixG ∈ Rn×n is a symmetric matrix defined as follows:Gi,j = Γi,j if z > 0 andGi,j =

(−1)i+j−1Γi,j if z < 0, whereα = arch|1+ 1/(2z)| and

Γi,j =γi,j , if i j ,γj,i, if i > j , γi,j = sh(iα) sh((n+ 1− j)α)

shα sh((n+ 1)α). (13)

Because of Theorem 10, if the iteration (7) is totally monotone, then it conserves the monotonicity too.If (7) conserves the monotonicity, then it is totally monotone, becausey

(j)

1 0 (respectivelyy(j)n+2 0)

implies y(j+1)1 0 (respectivelyy(j+1)

n+2 0) with the matrixT. Taking into consideration Remark 8 andTheorem 4 the iteration (7) is totally monotone if and only if the condition

DTD−1 0 (14)

holds, where

D =[ 1 0 0

−e1 D 00 −(en) 1

]∈ R

(n+2)×(n+2) (15)

andD ∈ Rn×n is the matrix introduced in the previous section. Expressing the condition (14) with the

parametersσ, θ andq we get our main theorem as follows.

Theorem 13. The (σ, θ)-method is uniformly totally monotone and at the same time it is uniformlymonotonicity conserving if and only if the condition

σ

θ q 8σ (θ − 1)+ 2− θ + √

(8σ (1− θ)− 2+ θ)2 − 16(1− θ)2σ (4σ − 1)

8(1− θ)2(16)

(in case ofθ = 1 the condition isq σ , and if θ = 0, thenσ must equal zero and the condition isq 1/2) holds.

Proof. The proof of the theorem can be found in the Appendix of the paper.Remark 14. Let us notice that comparing the left-hand and the right-hand sides of the condition (16) weget the inequalityθ 2σ , which gives a necessary condition for the uniform monotonicity conservation.This condition results in a restriction for the choice of the suitable methods.

Remark 15. Let us observe that the condition (16) corresponds with the condition of the uniformmaximum norm contractivity [7]. That is the(σ, θ)-method is uniformly totally monotone and at the


Table 1Upper bounds forq with variablen

Upper bound forq Upper bound forq

n 2 2.00000 n 7 1.50204n 3 1.70928 n 8 1.50041n 4 1.53518 n 10 1.50005n 5 1.51896 n 15 1.50000n 6 1.50369 n <∞ 1.50000

same time it is uniformly monotonicity conserving if and only if it is uniformly contractive in maximumnorm.

Our matrix formalism enables the calculation of the necessary and sufficient condition of themonotonicity conservation for fixed values ofn too. For instance, considering the Crank–Nicolsonmethod (σ = 0, θ = 1/2) we can obtain the upper bounds listed in Table 1 from the condition (A.15). Thetable shows that the upper bounds for the monotonicity conservation converge very fast to the uniformupper bound. Accordingly, in practice, we must apply the uniform upper bounds in theq value choices.

We demonstrate the sharpness of the bounds for the monotonicity conservation. Let us choose theCrank–Nicolson method again withn = 128 (h = 1/129), and let the monotonically increasing initialvector be

y(0)i =

0, i < 129/2,1, i > 129/2.

(17)

We can see from Table 1 that the necessary and sufficient upper bound forq is q 1.50000. Choosingq to be q = 5 we get the non-monotone vectory(0) depicted in Fig. 1 in the first iteration step.Setting the valueq = 1.50001 we obtain for the first iterate that 0.5000012= y

(1)64 > y

(1)65 = 0.4999988,

that is the heat distribution is not monotonically increasing. If we choose the valueq = 1.5, theny(1)64 = y

(1)65 = 0.5000000 and we obtain the monotonically increasing heat distribution shown in Fig. 2.

In the numerical example in the Introduction the parameterq was too large. For the Crank–Nicolsonmethod the necessary and sufficient condition of the monotonicity conservation (n = 5) is 0 q 1.51896 (see Table 1). With the choiceq = 1.5 we obtain the iterates

y(1) = (0.2308,0.2692,0.5000,0.7308,0.7692) ,y(2) = (0.1420,0.3580,0.5000,0.6420,0.8580) ,y(3) = (0.1761,0.3239,0.5000,0.6761,0.8239) ,

...

y(20) = (0.1666,0.3333,0.5000,0.6666,0.8333) ,...

which are monotonically increasing vectors, indeed.Summarizing our result, we can establish that the requirement of the monotonicity conservation entails

stricter conditions for the step-size choice than the stability bounds. If we would like to use a qualitativelyadequate numerical method to solve the heat equation, then we have to choose the mesh-parameters


Fig. 1. Heat distribution at the first time level,q = 5.

Fig. 2. Heat distribution at the first time level,q = 1.5.

according to the bounds (16). Let us calculate these bounds for two well-known methods. In the case ofσ = 0, the finite difference method is uniformly monotonicity conserving if and only if

0 q 2− θ

4(1− θ)2. (18)

The choiceθ = 0 corresponds to the explicit Euler method (in this case the condition isq 0.5), θ = 0.5corresponds to the Crank–Nicolson method (in this case the condition isq 1.5) and the choiceθ = 1corresponds to the implicit Euler method (in this caseq is optional).

The other special choice isσ = 1/6. Then the finite element method with linear elements is uniformlymonotonicity conserving if and only ifθ 1/3 and the condition

1

6θ q θ + 2+ √

9θ2 − 12θ + 12

24(1− θ)2(19)


holds (if θ = 1, thenq 1/6).Our method is based on the explicit form of the matrixT in Eq. (7). This is why the extension of our

method is very complicated, especially for two- or three-dimensional problems. At the same time otherboundary conditions in the one-dimensional case can be investigated using matrix partition technique.

Appendix

Proof of Theorem 13. If z= 0, thenDTD−1 is the matrix

DTD−1 =

1 0 0 0 . . . 0

0 1− q q 0 . . . 0

0 q 1− 2q q 0 . . . 0

0 0 q 1− 2q q 0 . . . 0...

. . .. . .

. . .. . .

. . .. . .

...

0 . . . 0 q 1− 2q q 0

0 . . . 0 q 1− 2q q

0 . . . 0 q 1− q

. (A.1)

This matrix is nonnegative if and only ifq 1/2. Let us suppose now thatz = 0. ThenDTD−1 can bewritten in the form

DTD−1 =1 0 0

0 DTD−1 + qDT−11 ene qDT−1

1 en

0 −(en)(TD−1 + qT−11 ene)+ e 1− q(en)T−1

1 en

(A.2)

(e = (1, . . . ,1)). Applying thatzT−11 = G this matrix is nonnegative if and only if the conditions

(A) (q/z)DGen 0(B) 1− (q/z)(en)Gen 0(C) −(en)(TD−1 + (q/z)Gene)+ e 0 (A.3)(D) DTD−1 + (q/z)DGene 0

are fulfilled. One can see thatz must be positive (in this caseG = Γ ), because for negative values thematrix G would have “chess-board like” sign distribution and condition (A) could not be satisfied. Forthe term in condition (A) the estimation

(q/z)DGen = (q/z)D(γ1,n, . . . , γn,n)

= q

z · sh((n+ 1)α)D

(sh(α),sh(2α), . . . ,sh(nα)

) 0 (A.4)

is valid because 0< sh(α) < sh(2α) < . . . < sh(nα).Condition (B) results in the relation

1− q

zγn,n 0. (A.5)


This is fulfilled for all values ofn if and only if 1− (q/z)e−α 0, because

limn→∞γn,n = lim

n→∞sh(nα)

sh((n+ 1)α)= e−α (A.6)

and the sequence is monotonically increasing. Due to the equalitiese−α = chα − shα and chα =1 + 1/(2z) the condition (A.5) is valid for all values ofn if and only if

√1+ 4z 2(q − z) − 1.

Considering the relationz= θq − σ we get the necessary and sufficient condition

0 q 1− 2(1− θ)σ + √1− 4(1− θ)σ

2(1− θ)2(A.7)

(if θ = 1, then there is no restriction forq).The term in the condition (C) can be rewritten in the following manner

−(en

)(TD−1 + (q/z)Gene) + e

= e −(q

z2Γn,1, . . . ,

q

z2Γn,n−1,

q

z2Γn,n − q − z

z

)D−1 − (q/z)(γn,n, . . . , γn,n). (A.8)

Considering the equality (see [7])n∑

j=1

|Gi,j | =n∑

j=1

Γi,j = 1

|(1/z)+ 2| − 2(1− γ1,i − γi,n) (i = 1, . . . , n) (A.9)

this vector is nonnegative if and only if the condition

1− q

z2· z(1− γ1,n − γn,n)+ q − z

z− q

zγn,n = q

zγ1,n 0 (A.10)

is valid, which in turn means that the condition (C) is always satisfied.For the left-hand side of the inequality in the condition (D) are true the following transformations

DTD−1 + (q/z)DGene = 1

zDG

(T2D−1 + qene)

= 1

zDG

((T1 − qQ)D−1 + qene) = I − q

zDGQD−1 + q

zDGene

= I − q

zDG

(QD−1 − ene) = I − q

zDGD. (A.11)

We investigate the validity of the conditionI − (q/z)DGD 0. The offdiagonal elements of the matrixare nonnegative, because the offdiagonal elements of the matrixDGD are negative. To see this let usconsider the following relations for the indicesi < j settingGi,j = 0 if i = 0 or j = 0 (because of thesymmetricity it is sufficient to consider only these index choices)(

DGD)i,j

= Gi,j + Gi−1,j−1 − Gi−1,j − Gi,j−1 = γi,j + γi−1,j−1 − γi−1,j − γi,j−1

= sh(iα)sh((n+ 1− j)α)+ sh((i − 1)α)sh((n+ 1− j + 1)α)

sh(α)sh((n+ 1)α)

+ −sh((i − 1)α)sh((n+ 1− j)α)− sh(iα)sh((n+ 1− j + 1)α)

sh(α)sh((n+ 1)α)

= (sh((n+ 1− j)α)− sh((n+ 1− j + 1)α)) · (sh(iα)− sh((i − 1)α))

sh(α)sh((n+ 1)α)< 0. (A.12)


The diagonal elements of the matrixDGD are(DGD)

i,i= γi,i + γi−1,i−1 − 2γi−1,i , i = 1,2, . . . , n. (A.13)

One can show with a tedious calculation that(DGD)

i,i γ1+n/2,1+n/2 + γn/2,n/2 − 2γn/2,1+n/2, i = 1, . . . , n, (A.14)

and ifn is even, then this estimation cannot be improved. Thus the following relations are true(I − (q/z)DGD)

i,i= 1− (q/z)

(DGD)

i,i

1− (q/z)(γ1+n/2,1+n/2 + γn/2,n/2 − 2γn/2,1+n/2)

= 1− (q/z)2sh(nα/2)(sh((1+ n/2)α)− sh(nα/2))

shα sh((n+ 1)α)

= 1− (q/z)sh(nα/2)

sh((n+ 1)α/2)ch(α/2)

1− (q/z)e−α/2

ch(α/2)= 1− (q/z)

(1− th(α/2)

). (A.15)

Here we applied the fact that the numbers sh(nα/2)/sh((n+1)α/2) tend toe−α/2 if n approaches infinityand the convergence is monotonically increasing. It follows from the estimations (A.15) that the condition(D) is fulfilled for all values ofn if and only if 1− (q/z)(1− th(α/2)) 0. From this condition we obtainthe inequality 1/

√1+ 4z 1− z/q, which is fulfilled if and only if

σ

θ< q 8σ (θ − 1)+ 2− θ + √

(8σ (1− θ)− 2+ θ)2 − 16(1− θ)2σ (4σ − 1)

8(1− θ)2(A.16)

(in case ofθ = 1 the condition isq σ ). We can easily verify the relation

8σ (θ − 1)+ 2− θ + √(8σ (1− θ)− 2+ θ)2 − 16(1− θ)2σ (4σ − 1)

8(1− θ)2

1− 2(1− θ)σ + √1− 4(1− θ)σ

2(1− θ)2. (A.17)

Comparing the conditions in the cases (A), (B), (C) and (D) we obtain the statement of Theorem 13.

Acknowledgement

The author is very thankful to István Faragó for his fruitful suggestions.

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on the monotonicity conservation in numerical solutions of the heat equation

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