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Journal of Applied Mathematics Volume 2008, Article ID 753518,29pages doi:10.1155/2008/753518

Research Article

Numerical Blow-Up Time for a Semilinear Parabolic Equation with Nonlinear Boundary Conditions

Louis A. Assal ´e,1Th ´eodore K. Boni,1and Diabate Nabongo2

1Institut National Polytechnique Houphou¨et-Boigny de Yamoussoukro, BP 1093, Yamoussoukro, Cote D’Ivoire

2D´epartement de Math´ematiques et Informatiques, Universit´e d’Abobo-Adjam´e, UFR-SFA, 16 BP 372 Abidjan 16, Cote D’Ivoire

Correspondence should be addressed to Diabate Nabongo,nabongo diabate@yahoo.fr Received 29 April 2008; Revised 15 December 2008; Accepted 29 December 2008 Recommended by Jacek Rokicki

We obtain some conditions under which the positive solution for semidiscretizations of the semilinear equationut uxxax, tfu, 0 < x < 1, t ∈ 0, T, with boundary conditions ux0, t 0,ux1, t btgu1, t, blows up in a finite time and estimate its semidiscrete blow-up time. We also establish the convergence of the semidiscrete blow-up time and obtain some results about numerical blow-up rate and set. Finally, we get an analogous result taking a discrete form of the above problem and give some computational results to illustrate some points of our analysis.

Copyrightq2008 Louis A. Assal´e et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

In this paper, we consider the following boundary value problem:

utuxx −ax, tfu, 0< x <1, t∈0, T, ux0, t 0, ux1, t btg

u1, t

, t∈0, T, ux,0 u0x≥0, 0≤x≤1,

1.1

wheref : 0,∞ → 0,∞is aC1 function,f0 0, g : 0,∞ → 0,∞is aC1 convex function,g0 0,aC00,1×R,ax, t ≥ 0 in0,1×R,atx, t ≤ 0 in0,1×R, bC1R,bt> 0 inR,bt≥0 inR. The initial datau0C20,1,u00 0,u01 b1gu01.

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Here0, Tis the maximal time interval on which the solutionuof1.1exists. The timeT may be finite or infinite. WhereTis infinite, we say that the solutionuexists globally.

WhenT is finite, the solutionudevelops a singularity in a finite time, namely

t→limTu·, t

∞, 1.2

whereu·, tmax0≤x≤1|ux, t|.

In this last case, we say that the solutionublows up in a finite time and the timeT is called the blow-up time of the solutionu.

In good number of physical devices, the boundary conditions play a primordial role in the progress of the studied processes. It is the case of the problem described in 1.1 which can be viewed as a heat conduction problem where u stands for the temperature, and the heat sources are prescribed on the boundaries. At the boundary x 0, the heat source has a constant flux whereas at the boundaryx 1, the heat source has a nonlinear radition haw. Intensification of the heat source at the boundary x 1 is provided by the functionb. The functiongalso gives a dominant strength of the heat source at the boundary x1.

The theoretical study of blow-up of solutions for semilinear parabolic equations with nonlinear boundary conditions has been the subject of investigations of many authorssee 1–7, and the references cited therein.

The authors have proved that under some assumptions, the solution of1.1blows up in a finite time and the blow-up time is estimated. It is also proved that under some conditions, the blow-up occurs at the point 1. In this paper, we are interested in the numerical study. We give some assumptions under which the solution of a semidiscrete form of1.1 blows up in a finite time and estimate its semidiscrete blow-up time. We also show that the semidiscrete blow-up time converges to the theoretical one when the mesh size goes to zero.

An analogous study has been also done for a discrete scheme. For the semidiscrete scheme, some results about numerical blow-up rate and set have been also given. A similar study has been undertaken in8,9where the authors have considered semilinear heat equations with Dirichlet boundary conditions. In the same way in10 the numerical extinction has been studied using some discrete and semidiscrete schemesa solutionuextincts in a finite time if it reaches the value zero in a finite time. Concerning the numerical study with nonlinear boundary conditions, some particular cases of the above problem have been treated by several authorssee11–15. Generally, the authors have considered the problem1.1in the case whereax, t 0 and bt 1. For instance in15, the above problem has been considered in the case whereax, t 0 andbt 1. In16, the authors have considered the problem1.1in the case whereax, t λ >0,bt 1,fu up,gu uq. They have shown that the solution of a semidiscrete form of 1.1 blows up in a finite time and they have localized the blow-up set. One may also find in17–22similar studies concerning other parabolic problems.

The paper is organized as follows. In the next section, we present a semidiscrete scheme of1.1. InSection 3, we give some properties concerning our semidiscrete scheme. In Section 4, under some conditions, we prove that the solution of the semidiscrete form of1.1 blows up in a finite time and estimate its semidiscrete blow-up time. InSection 5, we study the convergence of the semidiscrete blow-up time. InSection 6, we give some results on the numerical blow-up rate andSection 7is consecrated to the study of the numerical blow-up

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set. InSection 8, we study a particular discrete form of1.1. Finally, in the last section, taking some discrete forms of1.1, we give some numerical experiments.

2. The semidiscrete problem

Let I be a positive integer and define the grid xi ih, 0iI, where h 1/I. We approximate the solutionuof1.1by the solutionUht U0t, U1t, . . . , UItT of the following semidiscrete equations

dUit

dtδ2Uit −aitf Uit

, 0≤iI−1, t∈ 0, Tbh

, 2.1

dUIt

dtδ2UIt 2 hbtg

UIt

aItf UIt

, t∈ 0, Tbh

, 2.2

Ui0 ϕi≥0, 0≤iI, 2.3

whereϕi1ϕi, 0≤iI−1,

δ2U0t 2U1t−2U0t

h2 , δ2UIt 2UI−1t−2UIt

h2 ,

δ2Uit Ui1t−2Uit Ui−1t

h2 .

2.4

Here 0, Tbh is the maximal time interval on which Uht is finite where Uht max0≤i≤IUit. WhenTbh is finite, we say that the solution Uht blows up in a finite time and the timeTbhis called the blow-up time of the solutionUht.

3. Properties of the semidiscrete scheme

In this section, we give some lemmas which will be used later.

The following lemma is a semidiscrete form of the maximum principle.

Lemma 3.1. Letaht∈C00, T,RI1and letVht∈C10, T,RI1such that dVit

dtδ2Vit aitVit≥0, 0≤iI, t∈0, T, Vi0≥0, 0≤iI.

3.1

Then we haveVit≥0, 0iI,t∈0, T.

Proof. LetT0 < Tand define the vectorZht eλtVhtwhereλis large enough thatait−λ >

0 for t ∈ 0, T0, 0 ≤ iI. Let m min0≤i≤I,0≤t≤T0Zit. Since fori ∈ {0, . . . , I}, Zit is a continuous function, there existst0 ∈0, T0such thatmZi0t0for a certaini0 ∈ {0, . . . , I}.

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It is not hard to see that

dZi0 t0 dt lim

k→0

Zi0 t0

Zi0 t0k

k ≤0,

δ2Zi0 t0

Zi01 t0

−2Zi0

t0

Zi0−1 t0

h2 ≥0 if 1≤i0I−1, δ2Zi0

t0 2Z1

t0

−2Z0

t0

h2 ≥0 ifi00, δ2Zi0

t0

2ZI−1 t0

−2ZI t0

h2 ≥0 ifi0I.

3.2

A straightforward computation reveals that

dZi0

t0

dtδ2Zi0 t0

ai0

t0

λ Zi0

t0

≥0. 3.3

We observe from 3.2 thatai0t0λZi0t0 ≥ 0 which implies thatZi0t0 ≥ 0 because ai0t0λ >0. We deduce thatVht≥0 fort∈0, T0and the proof is complete.

Another form of the maximum principle for semidiscrete equations is the following comparison lemma.

Lemma 3.2. LetVht,Uht∈C10, T,RI1andfC0R×R,Rsuch that fort∈0, T dVit

dtδ2Vit f Vit, t

< dUit

dtδ2Uit f

Uit, t

, 0≤iI, 3.4

Vi0< Ui0, 0≤iI. 3.5

Then we haveVit< Uit, 0≤iI,t∈0, T.

Proof. Define the vectorZht Uht−Vht. Lett0be the firstt∈0, Tsuch thatZit>0 fort∈0, t0, 0≤iI, butZi0t0 0 for a certaini0 ∈ {0, . . . , I}. We observe that

dZi0 t0 dt lim

k→0

Zi0 t0

Zi0 t0k

k ≤0,

δ2Zi0 t0

Zi01 t0

−2Zi0

t0

Zi0−1 t0

h2 ≥0 if 1≤i0I−1, δ2Zi0

t0 2Z1

t0

−2Z0

t0

h2 ≥0 ifi00, δ2Zi0

t0

2ZI−1 t0

−2ZI t0

h2 ≥0 ifi0I,

3.6

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which implies that dZi0

t0 dtδ2Zi0

t0

f Ui0

t0

, t0

f Vi0

t0

, t0

≤0. 3.7

But this inequality contradicts3.4and the proof is complete.

4. Semidiscrete blow-up solutions

In this section under some assumptions, we show that the solutionUh of2.1–2.3blows up in a finite time and estimate its semidiscrete blow-up time.

Before starting, we need the following two lemmas. The first lemma gives a property of the operatorδ2and the second one reveals a property of the semidiscrete solution.

Lemma 4.1. LetUh∈RI1be such thatUh0. Then we have δ2g

Ui

g Ui

δ2Ui for 0iI. 4.1

Proof. Apply Taylor’s expansion to obtain

g U1

g U0

U1U0 g

U0

U1U0

2

2 g

η0 , g

Ui1 g

Ui

Ui1Ui g

Ui

Ui1Ui

2

2 g

θi

, 1≤iI−1, g

Ui−1 g

Ui

Ui−1Ui g

Ui

Ui−1Ui

2

2 g

ηi

, 1≤iI−1, g

UI−1 g

UI

UI−1UI g

UI

UI−1UI2

2 g

ηI ,

4.2

whereθi is an intermediate betweenUi andUi1 andηi the one betweenUi−1 andUi. The first and last equalities imply that

δ2g U0

g U0

δ2U0

U1U02

h2 g η0

,

δ2g UI

g UI

δ2UI

UI−1UI2

h2 g ηI

.

4.3

Combining the second and third equalities, we see that

δ2g Ui

g Ui

δ2Ui

Ui1Ui2 2h2 g

θi

Ui−1Ui2 2h2 g

ηi

, 1≤iI−1. 4.4

Use the fact thatgs≥0 fors≥0 andUh≥0 to complete the rest of the proof.

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Lemma 4.2. LetUhbe the solution of2.1–2.3. Then we have Ui1t> Uit, 0≤iI−1, t∈

0, Tbh

. 4.5

Proof. Lett0be the firstt >0 such thatUi1t> Uitfor 0≤iI−1 butUi01t0 Ui0t0 for a certain i0 ∈ {0, . . . , I −1}. Without loss of generality, we may suppose that i0 is the smallest integer which satisfies the equality. Introduce the functionsZit Ui1t−Uitfor 0≤iI−1. We get

dZi0 t0 dt lim

k→0

Zi0 t0

Zi0 t0k

k ≤0,

δ2Zi0 t0

Zi01 t0

−2Zi0

t0

Zi0−1 t0

h2 >0 if 1≤i0I−2, δ2Zi0

t0

δ2Z0 t0

Z1

t0

−3Z0

t0

h2 >0 ifi0 0, δ2Zi0

t0

δ2ZI−1 t0

ZI−2 t0

−3ZI−1 t0

h2 >0 ifi0I−1,

4.6

which implies that dZi0

t0 dtδ2Zi0

t0

ai01 t0

f Ui01t0

ai0

t0 f

Ui0 t0

<0 if 0≤i0I−2, dZi0

t0

dtδ2Zi0 t0

2 hb

t0 gi01

t0

ai01 t0

f Ui01t0

ai0

t0 f

Ui0 t0

<0 ifi0I−1.

4.7

But this contradicts2.1-2.2and we have the desired result.

The above lemma says that the semidiscrete solution is increasing in space. This property will be used later to show that the semidiscrete solution attains its minimum at the last nodexI.

Now, we are in a position to state the main result of this section.

Theorem 4.3. LetUhbe the solution of 2.1–2.3. Suppose that there exists a positive integer A such that

δ2ϕiai0f ϕi

≥0, 1≤iI−1, δ2ϕIaI0f

ϕI

b0gI

ϕI

Ag ϕI

. 4.8

Assume that

fsgs−fsgs≥0 fors≥0. 4.9

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Then the solutionUhblows up in a finite timeTbhand we have the following estimate

Tbh≤ 1 A

ϕh

ds

gs. 4.10

Proof. Since0, Tbhis the maximal time interval on whichUht <∞, our aim is to show thatTbhis finite and satisfies the above inequality. Introduce the vectorJhsuch that

Jit dUit

dt , 0≤iI−1, JIt dUIt dtAg

UIt

. 4.11

A straightforward calculation gives dJi

dtδ2Ji d dt

dUi

dtδ2Ui

, 0≤iI−1, dJI

dtδ2JI d dt

dUI

dtδ2UI

Ag UI

dUI

dt 2g UI

.

4.12

FromLemma 4.1, we haveδ2gUIgUIδ2UI which implies that dJI

dtδ2JId dt

dUI

dtδ2UI

Ag UI

dUI dtδ2UI

. 4.13

Using2.1, we get dJi

dtδ2Ji ≥ −aitf Ui

aitf

UidUi

dt , 0≤iI−1, dJI

dtδ2JI ≥ −aItf UI

aItf UI

dUI

dt 2 hbtg

UI

2

hbtg

UIdUI

dtAg UI

aItf UI

2 hbtg

UI .

4.14

It follows from the fact thatait≤0,bt≥0 anddUi/dtJiAgUithat dJI

dtδ2JI

aItf UI

2 hbtg

UI

JIAaIt g

UI

f UI

f UI

g UI

. 4.15

We deduce from4.9that dJi

dtδ2Ji≥ −aitf Ui

Ji, 0≤iI−1, dJI

dtδ2JI

aItf UI

2 hbtg

UI JI.

4.16

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From4.8, we observe that

Ji0 δ2ϕiai0f ϕi

≥0, 0≤iI−1, JI0 δ2ϕIaI0f

ϕI

b0gI

ϕI

Ag ϕI

≥0. 4.17

We deduce fromLemma 3.1thatJit ≥ 0, 0 ≤ iI, which implies that dUI/dtgUI, 0≤iI. Obviously we have

dUI g

UIA dt. 4.18

Integrating this inequality overt, Tbh, we arrive at

Tbht≤ 1 A

UIt

ds

gs, 4.19

which implies that

Tbh≤ 1 A

Uh0

ds

gs. 4.20

Since the quantity on the right hand side of the above inequality is finite, we deduce that the solutionUhblows up in a finite time. Use the fact thatUh0ϕhto complete the rest of the proof.

Remark 4.4. The inequality4.19implies that

Tbht0≤ 1 A

Uht0

ds

gs if 0< t0 < Tbh, Uit≤H

A

Tbht

, 0≤iI,

4.21

whereHsis the inverse ofGs

s dz/gz.

Remark 4.5. Ifgs sq, thenGs s1−q/q−1andHs q−1s1/1−q. 5. Convergence of the semidiscrete blow-up time

In this section, we show the convergence of the semidiscrete blow-up time. Now we will show that for each fixed time interval0, Twhereuis defined, the solutionUhtof2.1–

2.3approximatesu, when the mesh parameterhgoes to zero.

Theorem 5.1. Assume that1.1has a solutionuC4,10,1×0, Tand the initial condition at 2.3satisfies

ϕhuh0

o1 as h−→0, 5.1

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whereuht ux0, t, . . . , uxI, tT. Then, for h sufficiently small, the problem2.1–2.3has a unique solutionUhC10, T,RI1such that

max0≤t≤TUht−uht

huh0

h2

ash−→0. 5.2

Proof. Letα >0 be such that

u·, t

α fort∈0, T. 5.3

The problem2.1–2.3has for eachh, a unique solutionUhC10, Tbh,RI1. Letth ≤ min{T, Tbh}the greatest value oft >0 such that

Uht−uht

<1 for t∈ 0, th

. 5.4

The relation5.1implies thatth>0 forhsufficiently small. By the triangle inequality, we obtain

Uhtu·, tUht−uht fort∈ 0, th

, 5.5

which implies that

Uht≤1α fort∈ 0, th

. 5.6

Leteht Uht−uhtbe the error of discretization. Using Taylor’s expansion, we have for t∈0, th,

deit

dtδ2eit −aitf ξit

eit o h2

, 0≤iI−1, deIt

dtδ2eIt −aItf ξIt

eIt 2 hbtg

UIt

eIt o h2

,

5.7

whereθItis an intermediate value betweenUItand uxI, tand ξit the one between Uitanduxi, t. Using5.3and5.6, there exist two positive constantsKandLsuch that

deit

dtδ2eit≤L eit Kh2, 0≤iI−1, deIt

dt

2eI−1t−2eIt

h2L eIt

h L eIt Kh2.

5.8

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Consider the functionzx, t eM1tCx2ϕhuh0Qh2whereM,C,Qare constants which will be determined later. We get

ztx, t−zxxx, t M1−2C−4C2x2zx, t, zx0, t 0, zx1, t 2Cz1, t, zx,0 eCx2ϕhuh0Qh2.

5.9

By a semidiscretization of the above problem, we may chooseM,C,Qlarge enough that d

dtz xi, t

> δ2z xi, t

L z

xi, t Kh2, 0≤iI−1, d

dtz xI, t

> δ2z xI, t

L h z

xI, t L z

xI, t Kh2, z

xi,0

> ei0, 0≤iI.

5.10

It follows fromLemma 3.2that z

xi, t

> eit fort∈ 0, th

, 0≤iI. 5.11

By the same way, we also prove that z

xi, t

>−eit fort∈ 0, th

, 0≤iI, 5.12

which implies that

z xi, t

> eit fort∈ 0, th

, 0≤iI. 5.13

We deduce that

Uht−uht

eMtCϕhuh0

Qh2 , t

0, th

. 5.14

Let us show thatth T. Suppose thatT > th. From5.4, we obtain 1Uh

th

uh

th

eMTCϕhuh0

Qh2

. 5.15

Since the term on the right hand side of the above inequality goes to zero ashtends to zero, we deduce that 1≤0, which is impossible. Consequentlyth T, and the proof is complete.

Now, we are in a position to prove the main result of this section.

Theorem 5.2. Suppose that the problem1.1has a solution u which blows up in a finite timeTbsuch thatuC4,10,1×0, Tband the initial condition at2.3satisfies

ϕhuh0

o1 as h−→0. 5.16

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Under the assumptions ofTheorem 4.3, the problem2.1–2.3admits a unique solutionUh which blows up in a finite timeTbhand we have the following relation

h→lim0TbhTb. 5.17

Proof. Letε >0. There exists a positive constantNsuch that 1

A

x

ds gsε

2 forx∈N,∞. 5.18

Since the solutionublows up at the timeTb, then there existsT1 ∈ Tbε/2, Tbsuch that u·, t≥2Nfort∈T1, Tb. SettingT2 T1Tb/2, then we have supt∈0,T2|ux, t|<∞.

It follows fromTheorem 5.1that sup

t∈0,T2

Uht−uht

N. 5.19

Applying the triangle inequality, we get Uh

T2

uh

T2

Uh

T2

uh

T2

, 5.20

which leads toUhT2N. FromTheorem 4.3,Uhtblows up at the timeTbh. We deduce fromRemark 4.4and5.18that

TbTbhTbT2 TbhT2ε 2 1

A

UhT2

ds

gsε, 5.21

and the proof is complete.

6. Numerical blow-up rate

In this section, we determine the blow-up rate of the solutionUh of2.1–2.3in the case wherebt 1. Our result is the following.

Theorem 6.1. LetUhtbe the solution of2.1–2.3. Under the assumptions ofTheorem 4.3,Uht blows up in a finite timeTbhand there exist two positive constantsC1, C2such that

H C1

Tbht

UIt≤H C2

Tbht

, fort∈ 0, Tbh

, 6.1

whereHsis the inverse of the functionGs

s dσ/gσ.

Proof. FromTheorem 4.3andRemark 4.4,Uhtblows up in a finite timeTbhand there exists a constantC2>0 such that

UIt≤H C2

Tbht

fort∈ 0, Tbh

. 6.2

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FromLemma 4.2,UI−1 < UI. Then using2.2, we deduce thatdUI/dt ≤2/hbtgUIaItfUI, which implies that dUI/dt ≤ 2bt/hgUI. Integration this inequality over t, Tbh, there exists a positive constantC1such that

UIt≥H C1

Tbht

fort∈ 0, Tbh

, 6.3

which leads us to the result.

7. Numerical blow-up set

In this section, we determine the numerical blow-up set of the semidiscrete solution. This is stated in the theorem below.

Theorem 7.1. Suppose that there exists a positive constantC0such thatsFs≤C0and d

dtUiδ2Ui≤0, 0≤iI−1. 7.1 Assume that there exists a positive constantCsuch

UiH

CTt

, 0≤iI. 7.2

Then the numerical blow-up set isB{1}.

Proof. Letvx 1−x2and define Wx, t H

δvx δBTt

for 0≤x≤1, t≥t0, 7.3 whereδis small enough. We have

Wx0, t 0, W1, t H

δBTt

u1, t, 7.4

and fortt0, we get

Wx, t0 H

δvx δ

H2δ H 2δB

Tt0

H C

Tt0

u x, t0

. 7.5

A straightforward computation yields Wtx, t−Wxxx, t δF

B−2−4xF

δF

B−2−4δC0

. 7.6

This implies that there existsα >0 such that

Wtx, t−Wxxx, t≥αF

HδδBT

. 7.7

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Using Taylor’s expansion, there exists a constantK >0 such that d

dtW xi, t

δ2W xi, t

αF

HδδBT

Kh2, 0≤iI, 7.8

which implies that

dW xi, t

dtδ2W xi, t

≥0. 7.9

The maximum principle implies that

Uit≤H

δvx δB

Tt0

fortt0, 0≤iI. 7.10

Hence, we get

Uit≤H δvx

, 0≤iI. 7.11

ThereforeUiT<∞, 0≤iI−1, and we have the desired result.

8. Full discretization

In this section, we consider the problem1.1in the case whereax, t 1,bt 1,fu up, gu upwithpconst>1. Thus our problem is equivalent to

utx, t uxxx, t−upx, t, 0< x <1, t∈0, T, ux0, t 0, ux1, t up1, t, t∈0, T,

ux,0 u0x>0, 0≤x≤1,

8.1

wherep >1,u0C10,1,u00 0 andu01 up01.

We start this section by the construction of an adaptive scheme as follows. LetIbe a positive integer and leth1/I. Define the gridxiih, 0iIand approximate the solution ux, t of the problem 8.1 by the solution Unh Un0 , Un1 , . . . , UInT of the following discrete equations

δtUni δ2UinUinp

, 0≤iI−1, 8.2

δtUnI δ2UnIUnI p

2 h

UnI p

, 8.3

Ui0ϕi, 0≤iI, 8.4

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wheren≥0,ϕi1ϕi, 0≤iI−1,

δtUni Un1iUni Δtn

, δ2Uni Uni1−2Uni Uni−1

h2 , 1≤iI−1, δ2Un0 2U1n−2Un0

h2 , δ2UIn 2UnI−1−2UIn

h2 .

8.5

In order to permit the discrete solution to reproduce the property of the continuous one when the timetapproaches the blow-up time, we need to adapt the size of the time step so that we takeΔtnmin{1−h2/3, τ/Uhnp−1 }, 0< τ <1/p.

Let us notice that the restriction on the time step ensures the nonnegativity of the discrete solution. The lemma below shows that the discrete solution is increasing in space.

Lemma 8.1. LetUnh be the solution of 8.2–8.4. Then we have

Uni1Uni , 0≤iI−1. 8.6

Proof. LetZni Ui1nUni , 0≤iI−1. We observe that

Zn10Zn0

Δtn Z1n−3Z0n

h2

U1np

U0np

, Zin1Zin

Δtn Zni1−2Zni Zi−1n

h2

Uni1pUinp

, 1≤iI−2, Zn1I−1ZnI−1

Δtn ZnI−2−3ZI−1n

h2

UInp

UnI−1p 2

h UInp

.

8.7

Using the Taylor’s expansion, we find that

Z0n1 Δtn

h2 Z1n

1−3Δtn

h2

Zn0 −Δtnp ξn0 p−1

Zn0 , Zin1 Δtn

h2 Zi1n

1−2Δtn

h2

Zni Δtn

h2 Zi−1n

−Δtnp ξni p−1

Zni , 1≤iI−2, ZI−1n1≥ Δtn

h2 ZI−2n

1−3Δtn

h2

ZI−1n −Δtnp ξnI−1p−1

ZI−1n,

8.8

(15)

whereξni is an intermediate value betweenUinandUni1. IfZni ≤0, 0≤iI−1, we deduce that

Z0n1≥ Δtn

h2 Z1n

1−3Δtn

h2 −ΔtnpUnh p−1

Zn0 , Zin1≥ Δtn

h2 Zi1n

1−2Δtn

h2 −ΔtnpUnh p−1

Zni Δtn

h2 Zi−1n, 1≤iI−2, ZI−1n1≥ Δtn

h2 ZI−2n

1−3Δtn

h2 −ΔtnpUhnp−1 ZI−1n.

8.9

Using the restrictionΔtnτ/Uhnp−1 , we find that Z0n1 ≥ Δtn

h2 Zn1

1−3Δtn

h2

Zn0 , Zin1 ≥ Δtn

h2 Zni1

1−2Δtn

h2

Zni Δtn

h2 Zi−1n, 1≤iI−2, ZI−1n1 ≥ Δtn

h2 ZnI−2

1−3Δtn

h2

ZnI−1.

8.10

We observe that 1−3Δtn/h2 is nonnegative and by induction, we deduce thatZin≤0, 0≤iI−1. This ends the proof.

The following lemma is a discrete form of the maximum principle.

Lemma 8.2. Letanh be a bounded vector and letVhna sequence such that

δtVinδ2Vinani Vin≥0, 0≤iI, n≥0, 8.11

Vi0 ≥0, 0≤iI. 8.12

ThenVin0 forn0, 0iIifΔtnh2/2anh h2. Proof. IfVhn≥0 then a routine computation yields

V0n1≥ 2Δtn

h2 V1n

1−2Δtn

h2 −Δtnanh

V0n, Vin1≥ Δtn

h2 Vi1n

1−2Δtn

h2 −Δtnanh

Vin Δtn

h2 Vi−1n, 1≤iI−1, VIn1≥ 2Δtn

h2 VI−1n

1−2Δtn

h2 −Δtnanh

VIn.

8.13

参照

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