on the sign-stability of numerical solutions of one-dimensional parabolic problems

9
On the sign-stability of numerical solutions of one-dimensional parabolic problems q Ro ´ bert Horva ´th * Institute of Mathematics and Statistics, University of West-Hungary, Erzse ´bet u. 9, H-9400 Sopron, Hungary Received 1 December 2006; received in revised form 1 March 2007; accepted 10 April 2007 Available online 6 July 2007 Abstract The preservation of the qualitative properties of physical phenomena in numerical models of these phenomena is an important requirement in scientific computations. In this paper, the numerical solutions of a one-dimensional linear par- abolic problem are analysed. The problem can be considered as a altitudinal part of a split air pollution transport model or a heat conduction equation with a linear source term. The paper is focussed on the so-called sign-stability property, which reflects the fact that the number of the spatial sign changes of the solution does not grow in time. We give sufficient con- ditions that guarantee the sign-stability both for the finite difference and the finite element methods. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Parabolic problems; Numerical solution; Qualitative properties; Sign-stability 1. Introduction The concentration of r air-pollutants can be modelled and forecasted by the so-called air pollution trans- port model (e.g., [1]), which has the form ov l ot ¼ rðuv l Þ þ rðkrv l Þþ R l þ E þ rv l ðl ¼ 1; ... ; rÞ: ð1Þ Here the unknown function v l = v l (x, y, z, t) is the concentration of the lth pollutant, the function u = u(x, y, z, t) describes the wind velocity, k = k(x, y, z, t) is the diffusion coefficient, R l = R l (x, y, z, t, v 1 , ... , v l ) describes the chemical reactions between the investigated pollutants, E = E(x, y, z, t) is the emission function and r = r(x, y, z, t) describes the deposition process. Because of its complexity, system (1) is generally solved applying the so-called operator splitting technique. The system is split into several subproblems according to the physical and chemical processes involved in the model: advection, diffusion, chemical reaction, emission and deposition. These subproblems are solved 0307-904X/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2007.04.016 q The author of the paper was supported by the National Scientific Research Fund (OTKA) No. T043765. * Tel./fax: +36 99 518 423. E-mail address: [email protected] Available online at www.sciencedirect.com Applied Mathematical Modelling 32 (2008) 1570–1578 www.elsevier.com/locate/apm

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Page 1: On the sign-stability of numerical solutions of one-dimensional parabolic problems

Available online at www.sciencedirect.com

Applied Mathematical Modelling 32 (2008) 1570–1578

www.elsevier.com/locate/apm

On the sign-stability of numerical solutionsof one-dimensional parabolic problems q

Robert Horvath *

Institute of Mathematics and Statistics, University of West-Hungary, Erzsebet u. 9, H-9400 Sopron, Hungary

Received 1 December 2006; received in revised form 1 March 2007; accepted 10 April 2007Available online 6 July 2007

Abstract

The preservation of the qualitative properties of physical phenomena in numerical models of these phenomena is animportant requirement in scientific computations. In this paper, the numerical solutions of a one-dimensional linear par-abolic problem are analysed. The problem can be considered as a altitudinal part of a split air pollution transport model ora heat conduction equation with a linear source term. The paper is focussed on the so-called sign-stability property, whichreflects the fact that the number of the spatial sign changes of the solution does not grow in time. We give sufficient con-ditions that guarantee the sign-stability both for the finite difference and the finite element methods.� 2007 Elsevier Inc. All rights reserved.

Keywords: Parabolic problems; Numerical solution; Qualitative properties; Sign-stability

1. Introduction

The concentration of r air-pollutants can be modelled and forecasted by the so-called air pollution trans-port model (e.g., [1]), which has the form

0307-9

doi:10

q Th* Tel

E-m

ovl

ot¼ �rðuvlÞ þ rðkrvlÞ þ Rl þ E þ rvl ðl ¼ 1; . . . ; rÞ: ð1Þ

Here the unknown function vl = vl(x,y,z, t) is the concentration of the lth pollutant, the function u = u(x,y,z, t) describes the wind velocity, k = k(x,y,z, t) is the diffusion coefficient, Rl = Rl(x,y,z, t,v1, . . . ,vl) describesthe chemical reactions between the investigated pollutants, E = E(x,y,z, t) is the emission function andr = r(x,y,z, t) describes the deposition process.

Because of its complexity, system (1) is generally solved applying the so-called operator splitting technique.The system is split into several subproblems according to the physical and chemical processes involved inthe model: advection, diffusion, chemical reaction, emission and deposition. These subproblems are solved

04X/$ - see front matter � 2007 Elsevier Inc. All rights reserved.

.1016/j.apm.2007.04.016

e author of the paper was supported by the National Scientific Research Fund (OTKA) No. T043765../fax: +36 99 518 423.ail address: [email protected]

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R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578 1571

cyclically with some appropriate methods (e.g., [2,3]). Then, the solution of the model can be obtained usingthe solutions of the subproblems. Splitting, however, can be applied not only according to the physical orchemical processes but according to the space coordinates, too. In this case, for instance, the longitudinaland latitudinal space coordinates are handled separately from the altitudinal coordinate.

Naturally, the properties of the solution of an air pollution transport model are determined by the prop-erties of the numerical methods that are applied for the subproblems. In order to get a qualitatively correctnumerical solution of the whole model, the subproblems must be solved with qualitatively adequate numericalmethods, too.

In this paper, we consider the one-dimensional problem

ovot� o

oxj

ovox

� �þ cv ¼ 0; ðx; tÞ 2 ð0; 1Þ � ð0;1Þ; ð2Þ

vðx; 0Þ ¼ v0ðxÞ; x 2 ð0; 1Þ; ð3Þvð0; tÞ ¼ vð1; tÞ ¼ 0; t P 0; ð4Þ

where v0 is a given continuous initial function, the continuous function c : ½0; 1� ! R possesses the property

0 < cmin 6 cðxÞ 6 cmax;

the function j : ½0; 1� ! R fulfills the property

0 < jmin 6 jðxÞ 6 jmax

and it has continuous first derivatives. A function v : ½0; 1� � Rþ0 ! R is called the solution of problem (2)–(4)if it is sufficiently smooth and satisfies equalities (2)–(4). Eq. (2) can be considered as one of the subproblems ofan air pollution transport model, namely the one that describes the altitudinal changes. Eq. (2) is also suitableto describe heat conduction processes. In this case v denotes the temperature, j is the heat conduction coef-ficient and �cv is a linear source term.

The investigation of the number of the sign changes of real functions goes back as far as to 1836 [4]. Then inthe thirties, Polya [5] showed for the case c = 0, j = 1 that the number of the sign-changes of the functionx # v(x, t) (x 2 [0, 1]) does not grow in t. This property is called sign-stability.

A number of qualitative properties of Eq. (2), such as nonnegativity preservation, maximum–minimumprinciple, maximum norm contractivity, etc, are thoroughly investigated in the literature both for finite differ-ence and finite element methods [6–10]. This is not the case for the sign-stability property. There are sufficientconditions given for the finite difference solution of (2), but for the finite element method there are no resultsavailable.

It was shown in [11] that if the relation

s

h26

1

4ð1� hÞ ð5Þ

is satisfied then the finite difference method with the uniform spatial step-size h and with the h time-discreti-zation method with the time-step s is sign-stable for problems where c = 0 and j = 1. If h = 1, then there is noupper bound for the quotient s/h2. In [12], we showed that condition (5) is the necessary and sufficient con-dition of the uniform (independent of h) sign-stability, and we extended the investigation to finite element solu-tions. In [13], a sufficient condition of the sign-stability is given for the explicit finite difference solution of asemi-linear parabolic problem. The proof is based on a six-page-long linear algebraic consideration about thesign-stability of positive tridiagonal matrices. Paper [9] simplifies the proof essentially and gives sufficient con-ditions of the sign-stability of the finite difference methods applied for (2).

2. Sign-stability of tridiagonal matrices

Let n be a fixed natural number. In order to simplify the notations, we introduce the sets N = {1, . . . ,n},Ji = {i � 1, i, i + 1} (i = 2, . . . ,n � 1), J1 = {1,2} and Jn = {n � 1,n}. The elements of a matrix A 2 Rn�n are

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1572 R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578

denoted by ai,j or (A)i,j. Moreover, we extend this notation: if i or j does not belong to N, then ai,j is defined tobe zero.

For a given vector x ¼ ½x1; . . . ; xn�> 2 Rn, let us denote the number of sign-changes in the sequencex1,x2, . . . ,xn, where we leave out the occurrent zero values, by SðxÞ. If x is the zero vector, then we setSðxÞ ¼ �1. For example, Sð½�1; 2; 3;�2; 4�>Þ ¼ 3 and Sð½�1; 0; 0;�2; 0; 3�>Þ ¼ 1. Naturally, the trivial rela-tions �1 6 SðxÞ 6 n� 1 and SðxÞ ¼ Sð�xÞ hold.

Definition 2.1. A matrix A 2 Rn�n is said to be sign-stable (resp. sign-unstable) if the condition SðAxÞ 6 SðxÞ(resp. SðAxÞP SðxÞ) is fulfilled for all vectors x 2 Rn.

Remark 2.2. Clearly, the inverse of a regular sign-stable matrix is sign-unstable, and vice versa. Moreover, theproduct of sign-stable matrices are sign-stable and the same is true also for sign-unstable matrices. It is knownthat totally nonnegative matrices (all their minors are nonnegative) or minordefinite matrices (no two minorswith same order that have different sign) are sign-stable [14,15].

The next two theorems give sufficient conditions for the sign-stability and sign-instability of tridiagonalmatrices.

Theorem 2.3. Let A 2 Rn�n be a tridiagonal matrix with the properties

(P1) ai,j > 0 if i 2 N and j 2 Ji,

(P2) ai,i P jai,i�1j + jai,i+1j if i 2 N (weak row-diagonal dominance).

Then the matrix A is sign-stable.

Theorem 2.4. Let A 2 Rn�n be a tridiagonal matrix with the properties

(Q1) ai,i > 0 if i 2 N,(Q2) ai,j < 0 if i 2 N and j 2 Ji, j 5 i,

(Q3) ai,i P jai,i�1j + jai,i+1j if i 2 N (weak row-diagonal dominance).

Then the matrix A is sign-unstable.

The direct proofs of the theorems are based on the LU-decomposition A = LU of A. It can be shown withinduction that both L and U are sign-stable or sign-unstable. This fact trivially implies the statements of thetheorems. Theorem 2.3 can be proven also showing the total nonnegativity of A based on the implication thata tridiagonal matrix A 2 Rn�n is totally nonnegative if and only if ai,i+1, ai+1,i P 0 (i = 1, . . . ,n � 1) and eachmain-minor of the matrix is nonnegative [14].

3. Finite difference and finite element solutions

In this section we construct the finite difference and the finite element solution of problem (2)–(4). Our maingoal is to write these methods into a vector iteration form, for which the results of the previous section can beapplied.

In order to obtain a finite difference spatial approximation, we define a spatial mesh Xh with the grid points0 = x0 < x1 < � � � < xn < xn+1 = 1, dividing the interval [0,1] into n + 1 subintervals. Denoting the semi-discret-ization of the solution v of (2)–(4) at a grid point xi by vi(t), we approximate (2) at the point xi (i 2 N) as

dviðtÞdt� vixxðtÞ þ civiðtÞ ¼ 0; ð6Þ

where

vixxðtÞ ¼2

hiþ þ hi�jiþ1=2

viþ1ðtÞ � viðtÞhiþ

� ji�1=2

viðtÞ � vi�1ðtÞhi�

� �: ð7Þ

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R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578 1573

The distance between the points xi and xi+1 is denoted in two ways: h(i+1)� or hi+. The value ji+1/2 denotes theapproximate value of j on the segment [xi,xi+1] (typically the midpoint value), ci denotes the approximate va-lue of c at the point xi (typically ci = c(xi)). Applying Eq. (6) with i = 1, . . . ,n, we arrive at a Cauchy problemfor the system of ordinary differential equations

MdvðtÞ

dtþ KvðtÞ ¼ 0; vð0Þ ¼ ½v0ðx1Þ; . . . ; v0ðxnÞ�>; ð8Þ

where v(t) = [v1(t), . . . ,vn(t)]>, M 2 Rn�n with

Mij ¼1 if i ¼ j

0 if i 6¼ j;

and K 2 Rn�n is a sparse matrix with the elements

Ki;i�1 ¼�2ji�1=2

hi�ðhi� þ hiþÞ; Ki;iþ1 ¼

�2jiþ1=2

hiþðhi� þ hiþÞ;

Ki;i ¼ ci þ2jiþ1=2

hiþðhi� þ hiþÞþ 2ji�1=2

hi�ðhi� þ hiþÞ; i 2 N :

ð9Þ

Let us turn to the spatial semi-discretization in the case of the finite element method. Let us consider the grid-points 0 = x0 < x1 < � � � < xn < xn+1 = 1 again and let /1, . . . ,/n be basis functions defined as follows: each /i

is required to be continuous on the interval [0,1] and piecewise linear over the segments [xi,xi+1] (i = 0, . . . ,n)such that /i(xj) = dij (i = 1, . . . ,n; j = 0, . . . ,n + 1), where dij is the Kronecker’s symbol. Then the semi-discreteproblem for (2)–(4) reads as follows: Find a function vhðx; tÞ ¼

Pni¼1viðtÞ/iðxÞ such that

vhðx; 0Þ ¼Xn

i¼1

v0ðxiÞ/iðxÞ ð10Þ

and

Z 1

0

ovhðx; tÞot

/jðxÞdxþZ 1

0

jðxÞ ovhðx; tÞox

o/jðxÞox

þ cðxÞvhðx; tÞ/jðxÞ� �

dx ¼ 0 ð11Þ

for all basis functions /j. In order to obtain the unknown functions vi(t), we have to solve the same Cauchyproblem like in (8), but now we have

Mij ¼Z 1

0

/iðxÞ/jðxÞdx ð12Þ

and

Kij ¼Z 1

0

jðxÞ o/iðxÞox

o/jðxÞox

þ cðxÞ/iðxÞ/jðxÞ� �

dx: ð13Þ

The M and K matrices are called mass and stiffness matrices, respectively.In order to get fully discrete numerical schemes, we discretize the system of ordinary differential Eq. (8) also

in time. We choose a time-step s and denote the approximation to v(ks) by vk (k = 0,1, . . .). Applying the so-called h-method (h 2 [0, 1] is an arbitrary parameter) we arrive at the following sequence of the systems of lin-ear algebraic equations:

Mvkþ1 � vk

sþ hKvkþ1 þ ð1� hÞKvk ¼ 0 ð14Þ

(k = 0,1, . . .). Let us introduce the matrices

A1 ¼Mþ hsK; A2 ¼M� ð1� hÞsK:

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1574 R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578

It can be shown that the matrix A1 is regular, namely it is a nonsingular M-matrix in the finite difference case,and a Gram-matrix in the finite element case. Thus the matrix A ¼ A�1

1 A2 is well-defined and iteration (14) canbe rewritten as

vkþ1 ¼ Avk ðk ¼ 0; 1; . . .Þ: ð15Þ

4. Sign-stability of the finite difference and the finite element solutions

For the sake of simplicity, let us introduce the notation

Xh;s ¼ fðx; tÞ 2 ½0; 1� � Rþ0 jx ¼ xi ði ¼ 1; . . . ; nÞ; t ¼ ksðk ¼ 0; 1; . . .Þg

for the discretization mesh. Moreover let hmin be the minimal mesh-size in the spatial discretization, that ishmin = min{h1�,h1+,h2+, . . . ,hn+}. Then, naturally presents itself the following definition.

Definition 4.1. We say that the finite difference or the finite element method is sign-stable on a fixed mesh Xh,s

for problem (2)–(4) if the matrix A 2 Rn�n in (15) is a sign-stable matrix.

4.1. Sign-stability of the finite difference solution

Theorem 4.2. If the condition

s 61

ð1� hÞðcmax þ 4jmax=h2minÞ

ð16Þ

is satisfied, then the finite difference method for problem (2)–(4) is sign-stable on a fixed mesh Xh,s. If h = 1, thenthere is no upper bound for the time-step.

Proof. It is enough to show that the matrix A1 is sign-unstable and A2 is sign-stable for the finite differencemethod (Remark 2.2). The sign-instability of A1 follows from the form (9) of the elements of the matrix K

and Theorem 2.4. The matrix A2 is a tridiagonal matrix. If h = 1, then A2 = M and the unit matrix is triviallysign-stable. Thus the iteration matrix A is sign-stable without any restriction. If h 5 1, then it is sufficient toshow that the conditions (P1)–(P2) are valid under the condition (16). The super-diagonal and the sub-diag-onal and lower diagonals of A2 are positive. We show the weak row-diagonal dominance, which clearly yieldsthat the diagonal is also positive. This implies the sign-stability of A2 based on Theorem 2.3. Thus, we have

ðA2Þi;i � ðA2Þi;i�1 � ðA2Þi;iþ1 ¼ 1þ ð1� hÞsð�ci þ 2ðKi;i�1 þ Ki;iþ1ÞÞ

P 1� ci � 2ðKi;i�1 þ Ki;iþ1Þcmax þ 4jmax=h2

min

P 0 ði 2 NÞ; ð17Þ

where we make use of the relations j Ki;i�1 j6 jmax=h2min; j Ki;iþ1 j6 jmax=h2

min. This completes the proof. h

Remark 4.3. In the above theorem, the choice h = 0, j = 1 results in the same upper bound for the time-stepas the bound that can be obtained with special choices of the coefficient functions in Theorem 4.5 of [13].

4.2. Sign-stability of the finite element method

We consider now the finite element method. The integrals in (12) and (13) can be given as

Mij ¼Z 1

0

/iðxÞ/jðxÞdx ¼

ðhiþ þ hi�Þ=3 if i ¼ j

hiþ=6 if j ¼ iþ 1

hi�=6 if j ¼ i� 1

0 otherwise

; i; j 2 N ;

8>>><>>>:

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R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578 1575

Kij ¼Z 1

0

o/iðxÞox

o/jðxÞox

dx ¼

1=hiþ þ 1=hi� if i ¼ j

�1=hiþ if j ¼ iþ 1

�1=hi� if j ¼ i� 1

0 otherwise

; i; j 2 N ;

8>>><>>>:

which yields that both A1 and A2 are tridiagonal matrices.We start with the investigation of the matrix A1. Our goal is to give a sufficient condition for the sign-insta-

bility of A1.

Lemma 4.4. On any fixed mesh Xh,s, the row-sums of the matrix A1 are positive.

Proof. Let the index i be from the set {2, . . . ,n � 1}. Then we can compute the sum of the ith row as follows:

ðA1Þi;i�1 þ ðA1Þi;i þ ðA1Þi;iþ1 ¼hi�

6þ hiþ

6þ hi� þ hiþ

3þ hs

Z 1

0

jðxÞ o/iðxÞox

oð/i�1ðxÞ þ /iðxÞ þ /iþ1ðxÞÞox

þ hsZ 1

0

cðxÞ/iðxÞð/i�1ðxÞ þ /iðxÞ þ /iþ1ðxÞÞdx

¼ hi�

6þ hiþ

6þ hi� þ hiþ

3þ hs

Z 1

0

cðxÞ/iðxÞdx:

This value is trivially positive. If i = 1, then

ðA1Þ1;1 þ ðA1Þ1;2 ¼h1þ

6þ h1� þ h1þ

3þ hs

Z 1

0

jðxÞ o/1ðxÞox

oð/1ðxÞ þ /2ðxÞÞox

þ hsZ 1

0

cðxÞ/1ðxÞð/1ðxÞ þ /2ðxÞÞdx

¼ h1þ

6þ h1� þ h1þ

3þ hs

Z x1

0

jðxÞ 1

h21�

dxþ hsZ x1

0

cðxÞ/21ðxÞdxþ

Z x2

x1

cðxÞ/1ðxÞdx� �

> 0:

The case i = n can be proven similarly. h

Theorem 4.5. Let us suppose that the conditions h 5 0 and

h2max <

6jmin

cmax

ð18Þ

are satisfied for the finite element solution of problem (2)–(4). Then the condition

s >1

6h jmin=h2max � cmax=6

� � ð19Þ

implies the sign-instability of the matrix A1.

Proof. The proof is based on Theorem 2.4. The main diagonal elements of A1 are trivially positive. Thus con-dition (Q1) is satisfied. In order to show that the condition (Q2) is also valid, we estimate the (A1)i,i+1

(i = 1, . . . ,n � 1) element. The negativity of the sub-diagonal elements can be shown similarly. h

ðA1Þi;iþ1 ¼ Mi;iþ1 þ hsKi;iþ1 ¼hiþ

6þ hs

Z 1

0

jo/iðxÞ

oxo/iþ1ðxÞ

oxþ c/iðxÞ/iþ1ðxÞ

� �dx

6hiþ

6þ hs jmin

�Ki;iþ1 þ cmax

hiþ

6

� �¼ hiþ

6� hs jmin

1

hiþ� cmax

hiþ

6

� �

6hmax

6� hs jmin

1

hmax

� cmax

hmax

6

� �:

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1576 R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578

In view of condition (18), the term in parenthesis is positive, and the nonnegativity of (A1)i,i+1 can be guar-anteed by choosing the time-step s to be sufficiently large according to the condition (19). Condition (Q3) fol-lows from Lemma 4.4 directly.

Now we turn to the matrix A2 and search for conditions that guarantee the sign-stability of this matrix.

Lemma 4.6. On any fixed mesh Xh,s with the property (18), the super-diagonal and the sub-diagonal elements of

the matrix A2 are positive.

Proof. We prove only the positivity of the super-diagonal elements (A2)i,i+1 (i = 1, . . . ,n � 1). The positivity ofthe sub-diagonal ones can be shown similarly.

ðA2Þi;iþ1 ¼ Mi;iþ1 � ð1� hÞsKi;iþ1 ¼hiþ

6� ð1� hÞs

Z 1

0

jo/iðxÞ

oxo/iþ1ðxÞ

oxþ c/iðxÞ/iþ1ðxÞ

� �dx

Phiþ

6� ð1� hÞs jminKi;iþ1 þ cmax

hiþ

6

� �¼ hiþ

6þ ð1� hÞs jmin

1

hiþ� cmax

hiþ

6

� �

Phmin

6þ ð1� hÞs jmin

1

hmax

� cmax

hmax

6

� �

>hmin

6þ ð1� hÞs h2

maxcmax

6� 1

hmax

� cmax

hmax

6

� �¼ hmin

6> 0:

Lemma 4.7. On any fixed mesh Xh,s with the property (18), the condition

s 61

3ð1� hÞ 4jmax=h2min þ ð4cmaxhmax � 3cminhminÞ=3

� � ð20Þ

implies the weak row-diagonal dominance of the matrix A2 for the case h 5 1. If h = 1, then the weak row-diag-

onal dominance is satisfied without any restriction on s.

Proof. In view of Lemma 4.6, the super-diagonal and the sub-diagonal elements of the matrix A2 are positive.We will show that, under the condition (20), the relations (A2)i,i P (A2)i,i�1 + (A2)i,i+1 (i = 2, . . . ,n � 1) arevalid. The relations (A2)1,1 P (A2)1,2 and (A2)n,n P (A2)n,n�1 can be shown similarly. Thus, we have

ðA2Þi;i � ðA2Þi;i�1 � ðA2Þi;iþ1 ¼hiþ þ hi�

3� ð1� hÞsKi;i �

hi�

6þ ð1� hÞsKi;i�1 �

hiþ

6þ ð1� hÞsKi;iþ1

¼ hiþ þ hi�

6þ ð1� hÞsðKi;i�1 þ Ki;i þ Ki;iþ1Þ � 2ð1� hÞsKi;i

¼ hiþ þ hi�

6þ ð1� hÞs

Z 1

0

cðxÞ/iðxÞdx

� 2ð1� hÞsZ 1

0

jðxÞ o/iðxÞox

� �2

cðxÞð/iðxÞÞ2

!dx

Phmin

3þ ð1� hÞscminhmin � 2ð1� hÞs jmax

2

hmin

þ cmax

2hmax

3

� �

¼ hmin

1

3þ ð1� hÞscmin � 2ð1� hÞs jmax

2

h2min

þ cmax

2hmax

3hmin

! !

¼ hmin

1

3þ ð1� hÞs 3cminhmin � 4cmaxhmax

3hmin

� 4jmax

h2min

! !:

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R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578 1577

Because the expression in the inner parenthesis is negative, a sufficiently small time-step that is chosen accord-ing to (20) makes the expression (A2)i,i � (A2)i,i�1 � (A2)i,i+1 nonnegative. If h = 1, then (A2)i,i � (A2)i,i�1 �(A2)i,i+1Phmin/3. Thus the weak row-diagonal dominance is satisfied without any restriction. This completesthe proof. h

Combining the previous results we can give a sufficient condition of the sign-stability of the finite elementsolution of problem (2)–(4).

Theorem 4.8. Let us suppose that we solve problem (2)–(4) on a fixed mesh Xh,s with the finite element method.

Furthermore, let us suppose that the mesh sizes are bounded according to the condition (18). Then, if h 2 (0,1) ands fulfills the condition

1

6h jmin=h2max � cmax=6

� � < s 61

3ð1� hÞ 4jmax=h2min þ ð4cmaxhmax � 3cminhminÞ=3

� � ;

then the finite element method is sign-stable. If h = 1, then the condition is

s >1

6jmin=h2max � cmax

:

Proof. Based on Lemmas 4.6 and 4.7, the assumptions of Theorem 2.3 are satisfied. Thus A2 is a sign-stablematrix. In view of Theorem 4.5, the matrix A1 is a sign-unstable matrix. These facts trivially imply the sign-stability of the finite element solution. h

5. Some concluding remarks

Remark 5.1. From Theorems 4.2 and 4.8, one can see that the finite difference method allows larger time-stepsthan the finite element method. Thus, the finite difference method calculates a qualitatively adequate numericalsolution at a fixed time level faster than the finite element method.

Remark 5.2. It can be important to give the obtained conditions for uniform spatial meshes (h) and for prob-lems with c = 0 and j = 1. In this special case, the condition for the finite difference method is

s 6h2

4ð1� hÞ

(if h = 1, then there is no upper bound for the time-step), while for the finite element method we get thecondition

h2

6h< s 6

h2

12ð1� hÞ

(if h = 1, then the condition is s > h2/6). It can be seen from the last inequality that our sufficient condition forthe sign-stability can guarantee the sign-stability only for methods with h > 2/3.

Remark 5.3. We stress that Theorems 4.2 and 4.8 give only sufficient conditions for the sign-stability. In orderto show that these conditions are not necessary, let us consider the finite difference solution on the uniformspatial mesh with h = 1/4 for a problem with c = 0 and j = 1. The parameter h is chosen to be zero. In thiscase the iteration matrix A has the form

A ¼ tridiag½16s; 1� 32s; 16s� 2 R3�3:

It can be checked easily that this matrix is sign-stable if and only if s 6 1/48, which is a less restrictive con-dition than s 6 1/64 obtained from condition (16).

The sign-stability of the above 3 · 3 matrix can be checked directly applying the definition of the sign-stability or verifying the minordefinity of the matrix. For large matrices, however, this is not a way to proceed.This is why it is important to ensure the sign-stability of the numerical method a priori.

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1578 R. Horvath / Applied Mathematical Modelling 32 (2008) 1570–1578

Remark 5.4. The simple example discussed in the previous remark shows also that the sign-stability results ina stronger condition than the preservation of the nonnegativity of the initial function. Let us choose the time-step to be s = 5/192. Then A is a nonnegative matrix, namely A = tridiag[5/12, 1/6,5/12], but it is not sign-sta-ble: for the vector x = [1/2,1, � 1]>, for which SðxÞ ¼ 1, Ax = [1/12, � 1/24, 1/4]> with SðAxÞ ¼ 2.

Remark 5.5. As it is typical for finite element methods, the preservation of the qualitative properties necessi-tates not only upper bounds for the time-step but lower bounds, too. In order to illustrate this, let us considerthe finite element method with the parameter h = 5/6 on a uniform spatial mesh with h = 1/4. One can checkwith a very simple calculation that the (A)1,3 element of the iteration matrix has the form

ðAÞ1;3 ¼72sð80s� 1Þ

ð40sþ 1Þð6400s2 þ 800sþ 7Þ :

Because the nominator of the above quotient is positive, its sign is determined by the term 80s � 1. Thus ifs < 1/80, then A is not a nonnegative matrix, and A is not a sign-stable matrix either. The sufficient conditionin Theorem 4.8 requires s to be greater than 1/5.

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