maximum norm contractivity in the numerical solution of the one-dimensional heat equation

Applied Numerical Mathematics 31 (1999) 451–462

Maximum norm contractivity in the numerical solution of theone-dimensional heat equation

Róbert Horváth∗University of Sopron, Institute of Mathematics, Ady E. u. 5, Sopron, H-9400, Hungary

Abstract

In this paper we consider the one-dimensional heat conduction equation (Friedmann, 1964). To the numericalsolution of the problem we apply the so-called(σ, θ)-method (Faragó, 1996; Thomée, 1990) which unites a fewnumerical methods. With the choiceσ = 0 we get the finite differenceθ -method and the choiceσ = 1

6 resultsin the finite element method with linear elements. The most important question is the choice of the suitablemesh-parameters. The basic condition arises from the condition of the convergence (Faragó, 1996; Samarskii,1977; Thomée, 1990). Further conditions can be obtained aiming at preserving some qualitative properties of thecontinuous problem. Some of them are the following: non-negativity, convexity, concavity, shape preservation andcontractivity in some norms (Dekker and Verwer, 1984). Now we shall study the maximum norm contractivity.There are some results in the literature for the parameter choices which guarantee this property (Kraaijevanger,1992; Samarskii, 1977; Thomée, 1990). However these papers specialize only on the finite difference methods andgive sufficient conditions. We determine the necessary and sufficient conditions related to the(σ, θ)-method. Weclose the paper with numerical examples. 1999 Elsevier Science B.V. and IMACS. All rights reserved.

Keywords:Numerical solution; Qualitative properties; Heat equation; Maximum norm contractivity

1. Introduction

Let us consider the following parabolic partial differential equation [8], the so-called one-dimensionalheat conduction equation

∂u

∂t= ∂

2u

∂ξ2, ξ ∈ (0,1), t > 0,

u(t,0)= u(t,1)= 0, t > 0,

u(0, ξ )= u0(ξ), ξ ∈ [0,1],(1)

∗ E-mail: [email protected].

0168-9274/99/$20.00 1999 Elsevier Science B.V. and IMACS. All rights reserved.PII: S0168-9274(99)00007-0

452 R. Horváth / Applied Numerical Mathematics 31 (1999) 451–462

in the domainΩ = (0,∞)× (0,1), whereu0 is a suitable smooth function for which there exists a uniquesolution onΩ . The functionu :Ω→ R is called the solution of the problem (1) ifu ∈C1,2(Ω)∩C(Ω)and it satisfies the equations in (1). This problem has some characteristic qualitative properties as thephysical model [3,11,14]. Applying the well-known maximum principle [14] we can show that thefunctiont 7→maxx∈[0,1] |u(t, x)| is monotonically decreasing, that is the absolute value of the temperaturedoes not increase in time. This property is called maximum norm contractivity. It can be shown that thefunction t 7→ u(t, x) tends to 0 for allx ∈ [0,1] if t→∞. (If u0 is a concave or a convex function thenthe convergence is monotone with respect tox.) It is said that the solution composes itself. Moreover,we say that the problem (1) preserves the non-negativity ifu0 > 0 implies thatu> 0 inΩ . Among theabove properties we shall deal with the maximum norm contractivity. For a numerical solution to theproblem (1) we define the following mesh for fixedτ ∈ R+ andh= 1/(n+ 1) (n ∈N) [13,15]:

Ωτ,h := (tj , xi) ∈Ω | xi = ih (i = 0, . . . , n+ 1), tj = jτ (j ∈N).Denoting the approximation to the exact solution at thej th time level byyj ∈Rn and using the so-called(σ, θ)-method [7] we define the following one-step iteration process:

Myj+1− yj

τ=−θ 1

h2Qyj+1− (1− θ) 1

h2Qyj , j = 0,1, . . . . (2)

HereQ = tridiag[−1,2,−1] ∈ Rn×n andM = I − σQ ∈ Rn×n are matrices (the matrixI denotesthe unit matrix),θ ∈ [0,1] and σ ∈ [0, 1

4) are arbitrary parameters and the vectory0 is a suitableapproximation of the initial functionu0. This method unites a few numerical methods. Ifσ = 0 thenwe get the finite difference method and in the case ofσ = 1

6 the method (2) results in the finite elementmethod with linear elements. First of all we reformulate our problem to a problem of linear algebra. Letus introduce the notationsq = τ/h2, z = θq − σ . If z = 0 then the method (2) is called explicit. Let usdenote the matrixI − qQ by X. If z 6= 0 then we call the method implicit and introduce the notationss = q− z, x = 2+1/z,X1=Q+ (x−2)I andX2= I − sQ. We notice thats > 0 ass = q(1− θ)+ σand|x|> 2 becausez >−1

4. In this caseX denotes the matrix(1/z)X−11 X2. The matrixX is well-defined

because the matrixX1 is invertible for all values|x| > 2 [12]. Thus the iteration (2) can be rewritten inthe form

yj+1=Xyj , j = 0,1, . . . . (3)

It is an interesting question to examine the behaviour of the numerical solution defined by the iteration (3).We seek conditions for both the method and the mesh under which the qualitative properties of thecontinuous problem are preserved in the numerical solution [3–7,9,10]. We require the iteration (3) to becomposing, too. This means that the vector-sequenceyj (j = 0,1, . . .) has to tend to the zero vector forall initial vectorsy0, that is the matrixX must be a so-called convergent matrix. In this case we call themethod (2) weak convergent method. We get the following condition for the weak convergence.

Lemma 1 [7]. Assume thatσ ∈ [0, 14) and θ ∈ [0,0.5). Then the(σ, θ)-method is weak convergent for

fixedn if and only if the condition

q 6 1

2(1− 2θ)

1− 4σ cos2(π/(2(n+ 1)))

cos2(π/(2(n+ 1)))(4)

is satisfied. Moreover, it is weak convergent for alln if and only if the condition

q 6 1− 4σ

2(1− 2θ)(5)

R. Horváth / Applied Numerical Mathematics 31 (1999) 451–462 453

holds. For the parametersσ ∈ [0, 14) andθ ∈ [0.5,1] the method is always weak convergent.

Naturally in this paper we consider schemes satisfying (4) and we analyze the properties of the discretemaximum norm contractivity.

Definition 2. We call the numerical process (2) contractive in the maximum norm if the inequalities∥∥yj+1∥∥∞ 6 ∥∥yj∥∥∞are fulfilled for eachj ∈N.

We note that the maximum norm contractivity is not required for neither the stability nor theconvergence of the method, it is desirable only from qualitative point of view. Obviously the method (3)is contractive in the maximum norm if and only if the maximum norm of the matrixX is not greaterthan 1, that is if the condition‖X‖∞ 6 1 is fulfilled. Our aim is to give conditions for the values of themesh-parameters by a fixed(σ, θ)-method under which the numerical solution is contractive in maximumnorm.

2. The condition of the contractivity

In this section we give the condition of the maximum norm contractivity both by fixedn and for allvaluesn. Let us consider the case ofz = 0. It follows from the form of the matrixX, that if n = 1 orn = 2, then the necessary and sufficient condition of the contractivity isq 6 1 or q 6 2

3, respectively.If n > 3 then the condition isq 6 1

2. Now let us pass over the implicit methods. LetG ∈ Rn×n be asymmetric, one-pair matrix defined as follows:(G)i,j = (Γ )i,j if x > 2 and(G)i,j = (−1)i+j−1(Γ )i,j ifx <−2, where

(Γ )i,j =γi,j , if i 6 j ,γj,i , if i > j , γi,j = sh(iα) sh((n+ 1− j)α)

shα sh((n+ 1)α), α = arch

|x|2. (6)

For our purpose we rewrite the matrixX to an explicit form. The lemma in question can be stated asfollows.

Lemma 3. Assume thatz 6= 0. For the matrixX defined by(1/z)X−11 X2 we have the relation

X = 1

z

[q

zG− sI

]. (7)

Proof. The matrixG is the inverse of the matrixQ+ (x − 2)I [12]. Multiplying the matrixX by thematrix z · (Q+ (x − 2)I ) from the left we obtain the matrixI − sQ. 2

To find the maximum norm of the matrixX we have to examine the structure of the matrix and to sumthe absolute values of the elements in each row. Using the notationν = [(n+ 1)/2] (the integer part of(n+ 1)/2) we can state the following:

Lemma 4. For the diagonal elements of the matrixG are valid the following relations:

γi,i = γn+1−i,n+1−i , i = 1, . . . , ν, γ1,1< γ2,2< · · ·< γν,ν. (8)


Proof. To prove this lemma it is sufficient to analyze the continuous functionω 7→ sh(ωα)sh((n+ 1−ω)α) on the interval[1, n]. 2

Introducing the notationGi =∑nj=1 |(G)i,j | =

∑nj=1(Γ )i,j we have

Lemma 5. The sum of the modulus of the elements in theith row of the matrixG can be expressed asfollows:

Gi = 1

|x| − 2(1− γ1,i − γi,n), i = 1, . . . , n. (9)

Proof. Using the notationz= 1/(|x| − 2) we have the identity(Q+ (|x| − 2)I

) · (z, . . . , z)T = (1, . . . ,1)T + (z,0, . . . ,0, z)T. (10)

Multiplying (10) by the matrixΓ we obtain the statement.2Because of Lemma 4 for the sign ofz and for the sign distribution of diagonal elements of the matrix

X the possible cases are the following:Case I. z > 0 and(X)i,i > 0, i = 1, . . . , n.Case II. z > 0 and(X)i,i < 0, i = 1, . . . , n.Case III. z > 0 and there existsk ∈ 1, . . . , ν−1 such that(X)1,1, . . . , (X)k,k < 0 and(X)k+1,k+1, . . . ,

(X)ν,ν > 0.Case IV. z < 0 and(X)i,i < 0, i = 1, . . . , n.Case V. z < 0 and(X)i,i > 0, i = 1, . . . , n.Case VI. z < 0 and there existsl ∈ 1, . . . , ν− 1 such that(X)1,1, . . . , (X)l,l > 0 and(X)l+1,l+1, . . . ,

(X)ν,ν < 0.For the different cases we examine the condition of the maximum norm contractivity. Our results can besummarized as follows.

Lemma 6. Assume that the method(2) is weak convergent. Then in the caseV the maximum normcontractivity is not fulfilled. In casesI, III, IV andVI the contractivity is valid without further restrictionsfor the parameters. Finally, in the caseII we get the inequality

q

z

(1− sh((n+ 1− ν)α)+ sh(να)+ 2(|x|−2)

shα sh((n+ 1− ν)α)sh(να)

sh((n+ 1)α)

)+ sz6 1 (11)

by fixedn and the condition

q 6 8σ (θ − 1)+ 2− θ +√(8σ (θ − 1)+ 2− θ)2− 16(1− θ)2σ (4σ − 1)

8(1− θ)2 (12)

for all n ∈N which guarantee the contractivity. In case ofθ = 1 no restrictions arise.

Proof. Because of the simple but long calculations of the proof, the steps are not detailed here. For thedetailed proof we refer to Appendix.2

It is remains to examine the six cases above, by which conditions for the parameters are realized.Comparing the bounds, we can give a necessary and sufficient condition of the maximum norm


contractivity. We are not able to give the conditions by fixedn because most of our inequalities areimplicit. To see some numerical results by fixedn, we refer to Section 3. We can formulate the mostimportant result of our paper as follows:

Theorem. Letθ ∈ (0,1) andσ ∈ [0, 14) be fixed. The method(2) is both weak convergent and contractive

in maximum norm for all values ofn if and only if θ > 2σ and the parameterq satisfies the followingcondition:

σ

θ6 q 6 8σ (θ − 1)+ 2− θ +√(8σ (1− θ)− 2+ θ)2− 16(1− θ)2σ (4σ − 1)

8(1− θ)2 . (13)

In case ofθ = 1 the condition isq > σ and in the caseθ = 0 (consequentlyσ = 0) the condition is06 q 6 1

2 .

Proof. The proof is based on Lemma 6. First let us consider the casez < 0. According to Lemma 6the weak convergent method (2) is contractive if and only if there exists a negative element in the maindiagonal of the matrixX (in cases IV and VI). From this condition we get the inequality(X)ν,ν < 0. Thisis fulfilled for all n if it is fulfilled for n = 1. It yields the conditionq > (1− 2σ )/(2(1− θ)). We haveto compare this bound with the boundsz < 0 and (5). From the first comparison we get the necessaryconditionθ < 2σ . On the other hand, in this case

1− 2σ

2(1− θ) >1− 4σ

2(1− 2θ)

so the conditions

q >1− 2σ

2(1− θ) and q 6 1− 4σ

2(1− 2θ)

cannot be fulfilled simultaneously. Thus in the casez < 0 there is no parameter choiceσ, θ, q whichyields both weak convergence and maximum norm contractivity for alln.

Now let us examine the casez > 0. Because of the conditions (5) andz > 0 we get the necessaryconditionθ > 2σ . In cases I and III the contractivity is fulfilled automatically. These cases are realizedif and only if (X)ν,ν > 0. This is fulfilled for all n if it is fulfilled for n = 1. It yields the conditionq 6 (1− 2σ )/(2(1− θ)). Comparing this condition with the conditionθ > 2σ we obtain the inequalityσ/θ < q 6 (1− 2σ )/(2(1− θ)). (If θ = 1 thenq > σ .) If q > (1− 2σ )/(2(1− θ)) then there exists anatural numbern0 that for alln < n0 the elements in the main diagonal of the matrixX are negative. Butin this case we got the following condition forq in Lemma 6:

q 6 8σ (θ − 1)+ 2− θ +√(8σ (1− θ)− 2+ θ)2− 16(1− θ)2σ (4σ − 1)

8(1− θ)2 .

Thus in the case ofz > 0, the conditions of the contractivity areθ > 2σ and

σ

θ< q 6 8σ (θ − 1)+ 2− θ +√(8σ (1− θ)− 2+ θ)2− 16(1− θ)2σ (4σ − 1)

8(1− θ)2 .

(If θ = 1 thenq > σ .) Let us take into consideration the explicit methods as well. In this case we havegot the conditionq 6 1

2. Thus the theorem is proved.2


To analyze the conditions by fixedσ we introduce the following functions:

lσ (θ)= 1− 4σ

2(1− 2θ)

(θ < 1

2

), pσ (θ)= 1− 2σ

2(1− θ) (θ < 1), eσ (θ)= σθ

(θ > 0),

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

8(1− θ)2 (θ < 1).

They denote the barrier functions of the convergence. By the other values ofθ let us define the values ofthe functions as∞. Then the following statements are true:

(a) Let σ 6= 0. Then lσ (2σ ) = cσ (2σ ) = pσ (2σ ) = eσ (2σ ) = 12, if θ > 2σ then lσ (θ) > cσ (θ) >

pσ (θ)> eσ (θ) and if θ 6 2σ thenlσ (θ)6 cσ (θ)6 pσ (θ)6 eσ (θ).(b) Letσ = 0. Thenl0(θ)6 c0(θ)6 p0(θ).

Remark. Let us apply the theorem for two well-known methods. In the case ofσ = 0 we have the finitedifference method. Then the method (2) is contractive in the maximum norm for alln if and only if

06 q 6 2− θ4(1− θ)2 . (14)

The choiceθ = 0 corresponds to the forward Euler method (in this case the condition isq 6 0.5), θ = 0.5corresponds to the Crank–Nicolson method (in this case the condition isq 6 1.5) and the choiceθ = 1corresponds to the backward Euler method (in this caseq is optional). We have got the upper bound whichwas obtained by Kraaijevanger [10] with other tools. The method of that reference does not make possiblethe examination of the contractivity bound depending onn. Our method as it is shown in Section 3 doescarry out this.

Fig. 1.


The other special choice isσ = 16 which is the finite element method with linear elements. Then the

method (2) is contractive in maximum norm for alln if and only if θ > 13 and the condition

1

6θ6 q 6 θ + 2+√9θ2− 12θ + 12

24(1− θ)2 (15)

holds. (Ifθ = 1 thenq > 16.) The dark area shows the possible parameter choices in Fig. 1.

Remark. According to the continuous case, the method (2) is called non-negativity preserving, ifthe vectorsyj (j = 1,2, . . .) are non-negative for all non-negative initial vectorsy0. Considering theiteration (3) we can see that the necessary and sufficient condition of the non-negativity preservation isthe inequalityX > 0. This condition is fulfilled if and only ifz> 0 and(X)i,i > 0 (i = 1,2, . . . , n). It isapparent from the proof of Lemma 6 that the non-negativity preservation is a sufficient condition of themaximum norm contractivity.

3. Numerical examples

Let us assume that(σ, θ) are fixed. To define the mesh on which this method is contractive in themaximum norm by fixedn, we have to solve the implicit inequality (11). In our examples we executeit numerically. The necessary and sufficient condition of both the maximum norm contractivity and theweak convergence by fixedn, θ, σ is the following:

(a) Assume thatσ 6= 0. If θ > 2σ thencσ,n(θ)> q > eσ (θ). Conversely, ifθ < 2σ theneσ (θ) > q >pσ,n(θ) andq 6 lσ,n(θ).

(b) In the caseσ = 0 we have the condition 06 q 6 c0,n(θ).Herecσ,n(θ) denotes such a value ofq for which the relation (11) turns into the equality. On the base ofthe proof of Lemma 6 we can verify that byn→∞ cσ,n(θ)→ cσ (θ) andcσ,n(θ) forms a monotonicallydecreasing sequence inn. Thus we can give for all fixedn a greater bound forq thancσ (θ) defined in [10].

Table 1The case ofσ = 0

n θ = 0.00 θ = 0.30 θ = 0.45 θ = 0.50

1 1 2.5 10 ∞3 0.5 1.0763 2.9894 990237

5 0.5 0.8961 1.5035 2.0000

7 0.5 0.8721 1.3334 1.6180

9 0.5 0.8682 1.2952 1.5352

11 0.5 0.8675 1.2850 1.5113

13 0.5 0.8674 1.2822 1.5037

15 0.5 0.8674 1.2813 1.5012

25 0.5 0.8673 1.2810 1.5000

c0(θ) 0.5 0.8673 1.2810 1.5000


Table 2σ = 1

6, θ = 0.1

n p1/6,n(0.1) l1/6,n(0.1)

1 0.3704 0.8333

2 0.2941 0.4167

3 0.2810 0.3156

4 0.2784 0.2743

5 0.2779 0.2533

7 0.2778 0.2331

9 0.2778 0.2240

∀n c1/6(0.1)= 0.2778 l1/6(0.1)= 0.2083

Table 3σ = 1

6, θ = 0.5

n 3 5 7 9 11 15 25 ∀nc1/6,n(0.5) 2.3333 0.9658 0.9034 0.8973 0.8957 0.8954 0.8954 0.8954

Moreover,lσ,n(θ) denotes the bound derived from (4) andpσ,n(θ) is the solution of the equalityXν,ν = 0for q. It can be verified thatpσ,1(θ)= pσ (θ) and byn→∞ bothpσ,n(θ)→ cσ (θ) andlσ,n(θ)→ lσ (θ)

monotonically decreasing. Because ofpσ (θ) > lσ (θ) if θ < 2σ there does not existq which yields bothcontractivity and weak convergence for alln.

First of all let us consider the caseσ = 0. Table 1 shows the computed valuesc0,n(θ) in the casesθ = 0, 0.3, 0.45, 0.50. We remark that these values satisfy the condition of the weak convergence aswell.

Now let us consider the caseσ = 16. We must consider two different cases according toθ 6 2σ or

θ > 2σ . Let θ be for example 0.1< 2σ = 23. Then by fixedn the conditions of the maximum norm

contractivity and the weak convergence are satisfied if and only if53 > q > p1/6,n(0.1) andq 6 l1/6,n(0.1).

Table 2 shows that weak convergence and contractivity are fulfilled only in the casesn= 1,2,3.Now let θ be 0.5> 1

3. Then by fixedn the conditions of the maximum norm contractivity and weakconvergence are satisfied if and only if1

3 6 q 6 c1/6,n(0.5). Table 3 shows the valuesc1/6,n(0.5) computedwith numerical methods.

Appendix

Proof of Lemma 6. Let us denote byXi (i = 1, . . . , n) the sum of the elements in absolute value in theith row of the matrixX. The valuesXi using Lemma 3 can be expressed in the following form:

Xi = q

z2

n∑j=1j 6=i

(Γ )i,j +∣∣(X)i,i∣∣= q

z2

n∑j=1j 6=i

(Γ )i,j +∣∣∣∣ qz2(G)i,i − s

z

∣∣∣∣, i = 1, . . . , n. (A.1)


Our aim is to compute the value of‖X‖∞ and to give the condition‖X‖∞ 6 1. We have established inLemma 6 that there exist only six sign-distributions in the matrixX. We examine these cases. We applythe expression (A.1) and we use Lemmas 3 and 5 in the calculations.

Case I (z > 0 and(X)i,i > 0, i = 1, . . . , n). In this case we get the the following estimation:

Xi = q

z2

n∑j=1j 6=i

(Γ )i,j + q

z2(Γ )i,i − s

z= q

z2

n∑j=1

(Γ )i,j − sz

= q

z2

(1

|x| − 2(1− γ1,i − γi,n)

)− sz= qz(1− γ1,i − γi,n)− s

z6 q − s

z= zz= 1.

The inequality above shows that the conditionz > 0 and(X)i,i > 0 (i = 1, . . . , n) is a sufficient conditionfor the maximum norm contractivity.

Case II (z > 0 and(X)i,i < 0, i = 1, . . . , n). In this case we have

Xi = q

z2

n∑j=1j 6=i

(Γ )i,j − q

z2(Γ )i,i + s

z= q

z2

n∑j=1

(Γ )i,j − 2q

z2(Γ )i,i + s

z

= q

z2

1

|x| − 2

(1− f (i)

sh((n+ 1)α)

)+ sz,

where

f (i)= sh((n+ 1− i)α)+ sh(iα)+ 2(|x| − 2)

shαsh((n+ 1− i)α)sh(iα), Df = 1,2, . . . , n.

Extending the functionf on the setR we get a differentiable function. With differentiation we can easilyobtain that this function has a local minimum point ati = (n+ 1)/2 and it has local maximum points at

i = n+ 1

2± 1

αarch

(1

2sh(n+ 1

2α

)shα

).

We verify that the valueXi (i = 1, . . . , n) takes its maximum value by the choicei = [(n+ 1)/2] = ν.Thus we have to show thatf (1) > f (ν). We prove this statement for odd values ofn. In this caseν = (n+ 1)/2 and the inequality can be rewritten in the following form:

shα+ sh(nα)+ 2(|x| − 2)

shαsh(nα)shα > 2sh

(n+ 1

2α

)+ 2(|x| − 2)

shαsh2(n+ 1

2α

).

Using the identities of hyperbolic functions we get the inequality

sh(n+ 1

2α

)ch(n− 1

2α

)− sh

(n+ 1

2α

)> (|x| − 2)

shα

(sh2(n+ 1

2α

)shα sh(nα)

).

This is fulfilled by the choicen= 1. Let us analyze both functions standing on the right and on the lefthand sides, respectively. As earlier we assume that they are functions onR. Since

ch(nα)− ch(n+ 1

2α

)> (|x| − 2)

shα

(sh((n+ 1)α

)− 2shα ch(nα))

(A.2)

therefore the derivative of the left hand side is greater than the derivative of the right hand side. Toprove (A.2) it is sufficient to use the well-known identities for the hyperbolic functions.


We have proved that in case II, the condition of the maximum norm contractivity can be defined asfollows:

‖X‖∞ = qz

(1− f (ν)

sh((n+ 1)α)

)+ sz6 1. (A.3)

We seek the condition which guarantees the contractivity for alln. Letn be an odd natural number. Thenthe condition (A.3) yields

q

z2

1

|x| − 2

(1− 1

ch(n+12 α)

− |x| − 2

shαth(n+ 1

2α

))+ sz6 1.

This is satisfied for alln if and only if

q

z2

1

|x| − 2

(1− |x| − 2

shα

)+ sz6 1.

Substituting the values of the parameters we get the condition

4q2(1− θ)2+ q(8σ (1− θ)− 2+ θ)+ 4σ 2− σ 6 0

from which we obtain the condition

q 6 8σ (θ − 1)+ 2− θ +√(8σ (1− θ)− 2+ θ)2− 16(1− θ)2σ (4σ − 1)

8(1− θ)2 . (A.4)

If θ = 1 then no condition arises.Case III (z > 0 and there existsk ∈ 1, . . . , ν−1 such that(X)1,1, . . . , (X)k,k < 0 and(X)k+1,k+1, . . . ,

(X)ν,ν > 0). In this case we consider the following two functions on the set1,2, . . . , n:

g1(i)= q

z2

n∑j=1

(Γ )i,j − sz, g2(i)= q

z2

1

|x| − 2

(1− f (i)

sh((n+ 1)α)

)+ sz.

The functiong1(i) results in the sum of theith row of the matrixX andg2(i) gives the sum of theithrow, taking the element in the main diagonal with opposite sign. Thus ifi 6 k theng1(i) < g2(i) =Xi

and if k < i 6 ν thenXi = g1(i) > g2(i). As we know from Lemma 4 and case II these functions taketheir maximum values at the pointi = ν. Sog1(ν) > g1(i) andg1(ν) > g2(i) (i = 1, . . . , n). On the otherhand,g1(ν)6 1, that is the condition‖X‖∞ 6 1 is fulfilled in this case as well.

Case IV (z < 0 and(X)i,i < 0, i = 1, . . . , n). In this case we can calculate as follows:

Xi = q

z2

n∑j=1j 6=i

(Γ )i,j + sz− q

z2(G)i,i = q

z2

n∑j=1

(Γ )i,j + sz

= q

z2

(1

|x| − 2

(1− sh((n+ 1− i)α)+ sh(iα)

sh((n+ 1)α)

))+ sz

6 q

z2

1

|x| − 2+ sz= 4s − 1

4z+ 1= 1+ 2(2(s − z)− 1)

4z+ 1.

Because ofz >−14 we have to prove the relation 2(s − z)6 1. Obviously this inequality is fulfilled in

the caseθ > 12. In case ofθ < 1

2 we get the upper bound

q 6 1− 4σ

2(1− 2θ).


This condition is satisfied automatically if the matrixX is convergent.Case V (z < 0 and(X)i,i > 0, i = 1, . . . , n). In this case the expressionXi can be reformulated as

follows:

Xi = q

z2

n∑j=1j 6=i

(Γ )i,j − q

z2(Γ )i,i − s

z= q

z2

n∑j=1

(Γ )i,j − 2q

z2(Γ )i,i − s

z

= q

z2

1

|x| − 2

(1− sh((n+ 1− i)α)+ sh(iα)+ 2(|x|−2)

shα sh((n+ 1− i)α) sh(iα)

sh((n+ 1)α)

)− sz.

Xi takes its maximum value ati = ν on the set1, . . . , n, that is‖X‖∞ =Xν . We have the estimation

‖X‖∞ = q

z2

1

|x| − 2

(1−

(1

ch n+12 α+ |x| − 2

shαth(n+ 1

2α

)))− sz

> q

z2

1

|x| − 2

(1− |x| − 1

chα

)− sz= −qz2|x| −

s

z> 1.

(The last inequality can be rewritten in the form−1< z|x|. This is true becausez|x| = −2z− 1>−1.)Thus the method is not contractive in the maximum norm in this case.

Case VI (z < 0 and there existsl ∈ 1, . . . , ν − 1 such that(X)1,1, . . . , (X)l,l > 0 and(X)l+1,l+1, . . . ,

(X)ν,ν < 0). In this case our procedure is similar to the case III. Let us consider the following twofunctions defined on the set1,2, . . . , n:

g3(i)= q

z2

n∑j=1

(Γ )i,j + sz, g4(i)= q

z2

1

|x| − 2

(1− f (i)

sh((n+ 1)α)

)− sz.

The functiong3(i) results in the sum of theith row of the matrixX in absolute value (without theelement in the main diagonal) plus the element in the main diagonal with the opposite sign andg4(i)

gives the sum of theith row in maximum value (without the element in the main diagonal) plus theithelement of the row. Thus ifi 6 l theng3(i) < g4(i)=Xi and if l < i 6 ν thenXi = g3(i) > g4(i). Aswe know from Lemma 4 and the case V these functions take their maximum values at the pointi = ν.Sog3(ν) > g3(i) andg3(ν) > g4(i) (i = 1, . . . , n). On the other hand,g3(ν)6 1 (see case IV), that is thecondition‖X‖∞ 6 1 is fulfilled in this case as well.2

Acknowledgements

The author would like to thank István Faragó for the useful advice.

References

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maximum norm contractivity in the numerical solution of the one-dimensional heat equation

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