Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2008, Article ID 753518, 29 pages
doi:10.1155/2008/753518
Research Article
Numerical Blow-Up Time for a Semilinear Parabolic
Equation with Nonlinear Boundary Conditions
Louis A. Assalé,1 Théodore K. Boni,1 and Diabate Nabongo2
1
Institut National Polytechnique Houphouët-Boigny de Yamoussoukro, BP 1093,
Yamoussoukro, Cote D’Ivoire
2
Département de Mathématiques et Informatiques, Université d’Abobo-Adjamé, 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 equation ut uxx − ax, tfu, 0 < x < 1, t ∈ 0, T , with boundary conditions
ux 0, t 0, ux 1, 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.
Copyright q 2008 Louis A. Assalé 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:
ut − uxx −ax, tfu,
ux 0, t 0,
0 < x < 1, t ∈ 0, T ,
ux 1, t btg u1, t , t ∈ 0, T ,
ux, 0 u0 x ≥ 0,
1.1
0 ≤ x ≤ 1,
where f : 0, ∞ → 0, ∞ is a C1 function, f0 0, g : 0, ∞ → 0, ∞ is a C1 convex
function, g0 0, a ∈ C0 0, 1 × R , ax, t ≥ 0 in 0, 1 × R , at x, t ≤ 0 in 0, 1 × R ,
b ∈ C1 R , bt > 0 in R , b t ≥ 0 in R . The initial data u0 ∈ C2 0, 1, u0 0 0, u0 1 b1gu0 1.
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Here 0, T is the maximal time interval on which the solution u of 1.1 exists. The
time T may be finite or infinite. Where T is infinite, we say that the solution u exists globally.
When T is finite, the solution u develops a singularity in a finite time, namely
lim u·, t∞ ∞,
t→T
1.2
where u·, t∞ max0≤x≤1 |ux, t|.
In this last case, we say that the solution u blows up in a finite time and the time T is
called the blow-up time of the solution u.
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 boundary x 1, the heat source has a nonlinear
radition haw. Intensification of the heat source at the boundary x 1 is provided by the
function b. The function g also gives a dominant strength of the heat source at the boundary
x 1.
The theoretical study of blow-up of solutions for semilinear parabolic equations with
nonlinear boundary conditions has been the subject of investigations of many authors see
1–7, and the references cited therein.
The authors have proved that under some assumptions, the solution of 1.1 blows
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 of 1.1
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 in 8, 9 where the authors have considered semilinear heat equations
with Dirichlet boundary conditions. In the same way in 10 the numerical extinction has
been studied using some discrete and semidiscrete schemes a solution u extincts 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 authors see 11–15. Generally, the authors have considered the problem 1.1 in
the case where ax, t 0 and bt 1. For instance in 15, the above problem has been
considered in the case where ax, t 0 and bt 1. In 16, the authors have considered
the problem 1.1 in the case where ax, 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 in 17–22 similar studies concerning other
parabolic problems.
The paper is organized as follows. In the next section, we present a semidiscrete
scheme of 1.1. In Section 3, we give some properties concerning our semidiscrete scheme. In
Section 4, under some conditions, we prove that the solution of the semidiscrete form of 1.1
blows up in a finite time and estimate its semidiscrete blow-up time. In Section 5, we study
the convergence of the semidiscrete blow-up time. In Section 6, we give some results on the
numerical blow-up rate and Section 7 is consecrated to the study of the numerical blow-up
Louis A. Assalé et al.
3
set. In Section 8, we study a particular discrete form of 1.1. Finally, in the last section, taking
some discrete forms of 1.1, we give some numerical experiments.
2. The semidiscrete problem
Let I be a positive integer and define the grid xi ih, 0 ≤ i ≤ I, where h 1/I. We
approximate the solution u of 1.1 by the solution Uh t U0 t, U1 t, . . . , UI tT of the
following semidiscrete equations
dUi t
− δ2 Ui t −ai tf Ui t , 0 ≤ i ≤ I − 1, t ∈ 0, Tbh ,
dt
dUI t
2
− δ2 UI t btg UI t − aI tf UI t , t ∈ 0, Tbh ,
dt
h
Ui 0 ϕi ≥ 0, 0 ≤ i ≤ I,
2.1
2.2
2.3
where ϕi1 ≥ ϕi , 0 ≤ i ≤ I − 1,
δ2 U0 t 2U1 t − 2U0 t
2UI−1 t − 2UI t
,
δ2 UI t ,
2
h
h2
Ui1 t − 2Ui t Ui−1 t
δ2 Ui t .
h2
2.4
Here 0, Tbh is the maximal time interval on which Uh t∞ is finite where Uh t∞ max0≤i≤I Ui t. When Tbh is finite, we say that the solution Uh t blows up in a finite time
and the time Tbh is called the blow-up time of the solution Uh t.
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. Let ah t ∈ C0 0, T , RI1 and let Vh t ∈ C1 0, T , RI1 such that
dVi t
− δ2 Vi t ai tVi t ≥ 0, 0 ≤ i ≤ I, t ∈ 0, T ,
dt
Vi 0 ≥ 0, 0 ≤ i ≤ I.
3.1
Then we have Vi t ≥ 0, 0 ≤ i ≤ I, t ∈ 0, T .
Proof. Let T0 < T and define the vector Zh t eλt Vh t where λ is large enough that ai t−λ >
0 for t ∈ 0, T0 , 0 ≤ i ≤ I. Let m min0≤i≤I, 0≤t≤T0 Zi t. Since for i ∈ {0, . . . , I}, Zi t is a
continuous function, there exists t0 ∈ 0, T0 such that m Zi0 t0 for a certain i0 ∈ {0, . . . , I}.
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It is not hard to see that
Zi0 t0 − Zi0 t0 − k
dZi0 t0
lim
≤ 0,
k→0
dt
k
Zi0 1 t0 − 2Zi0 t0 Zi0 −1 t0
2
δ Zi0 t0 ≥ 0 if 1 ≤ i0 ≤ I − 1,
h2
2Z1 t0 − 2Z0 t0
δ2 Zi0 t0 ≥ 0 if i0 0,
h2
2ZI−1 t0 − 2ZI t0
δ2 Zi0 t0 ≥ 0 if i0 I.
h2
3.2
A straightforward computation reveals that
dZi0 t0
− δ2 Zi0 t0 ai0 t0 − λ Zi0 t0 ≥ 0.
dt
3.3
We observe from 3.2 that ai0 t0 − λZi0 t0 ≥ 0 which implies that Zi0 t0 ≥ 0 because
ai0 t0 − λ > 0. We deduce that Vh t ≥ 0 for t ∈ 0, T0 and the proof is complete.
Another form of the maximum principle for semidiscrete equations is the following
comparison lemma.
Lemma 3.2. Let Vh t,Uh t ∈ C1 0, T , RI1 and f ∈ C0 R × R, R such that for t ∈ 0, T dUi t
dVi t
− δ2 Vi t f Vi t, t <
− δ2 Ui t f Ui t, t ,
dt
dt
Vi 0 < Ui 0, 0 ≤ i ≤ I.
0 ≤ i ≤ I,
3.4
3.5
Then we have Vi t < Ui t, 0 ≤ i ≤ I, t ∈ 0, T .
Proof. Define the vector Zh t Uh t − Vh t. Let t0 be the first t ∈ 0, T such that Zi t > 0
for t ∈ 0, t0 , 0 ≤ i ≤ I, but Zi0 t0 0 for a certain i0 ∈ {0, . . . , I}. We observe that
dZi0 t0
Zi0 t0 − Zi0 t0 − k
lim
≤ 0,
k→0
dt
k
Zi0 1 t0 − 2Zi0 t0 Zi0 −1 t0
δ2 Zi0 t0 ≥ 0 if 1 ≤ i0 ≤ I − 1,
h2
2Z1 t0 − 2Z0 t0
δ2 Zi0 t0 ≥ 0 if i0 0,
h2
2ZI−1 t0 − 2ZI t0
2
δ Zi0 t0 ≥ 0 if i0 I,
h2
3.6
Louis A. Assalé et al.
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which implies that
dZi0 t0
− δ2 Zi0 t0 f Ui0 t0 , t0 − f Vi0 t0 , t0 ≤ 0.
dt
3.7
But this inequality contradicts 3.4 and the proof is complete.
4. Semidiscrete blow-up solutions
In this section under some assumptions, we show that the solution Uh of 2.1–2.3 blows
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 operator δ2 and the second one reveals a property of the semidiscrete solution.
Lemma 4.1. Let Uh ∈ RI1 be such that Uh ≥ 0. Then we have
δ2 g Ui ≥ g Ui δ2 Ui
for 0 ≤ i ≤ I.
4.1
Proof. Apply Taylor’s expansion to obtain
2
U1 − U0
g η0 ,
g U1 g U0 U1 − U0 g U0 2
2
Ui1 − Ui
g θi , 1 ≤ i ≤ I − 1,
g Ui1 g Ui Ui1 − Ui g Ui 2
2
Ui−1 − Ui
g ηi , 1 ≤ i ≤ I − 1,
g Ui−1 g Ui Ui−1 − Ui g Ui 2
2
UI−1 − UI
g ηI ,
g UI−1 g UI UI−1 − UI g UI 2
4.2
where θi is an intermediate between Ui and Ui1 and ηi the one between Ui−1 and Ui . The
first and last equalities imply that
2
U1 − U0
g η0 ,
δ g U0 g U0 δ U0 2
h
2
UI−1 − UI
δ2 g UI g UI δ2 UI g ηI .
h2
2
2
4.3
Combining the second and third equalities, we see that
2
2
Ui1 − Ui
Ui−1 − Ui
δ2 g Ui g Ui δ2 Ui g
g ηi ,
θ
i
2
2
2h
2h
1 ≤ i ≤ I − 1.
Use the fact that g s ≥ 0 for s ≥ 0 and Uh ≥ 0 to complete the rest of the proof.
4.4
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Lemma 4.2. Let Uh be the solution of 2.1–2.3. Then we have
Ui1 t > Ui t,
0 ≤ i ≤ I − 1, t ∈ 0, Tbh .
4.5
Proof. Let t0 be the first t > 0 such that Ui1 t > Ui t for 0 ≤ i ≤ I − 1 but Ui0 1 t0 Ui0 t0 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 functions Zi t Ui1 t − Ui t for
0 ≤ i ≤ I − 1. We get
Zi0 t0 − Zi0 t0 − k
dZi0 t0
lim
≤ 0,
k→0
dt
k
Zi0 1 t0 − 2Zi0 t0 Zi0 −1 t0
2
δ Zi0 t0 > 0 if 1 ≤ i0 ≤ I − 2,
h2
Z1 t0 − 3Z0 t0
δ2 Zi0 t0 δ2 Z0 t0 > 0 if i0 0,
h2
ZI−2 t0 − 3ZI−1 t0
δ2 Zi0 t0 δ2 ZI−1 t0 > 0 if i0 I − 1,
h2
4.6
which implies that
dZi0 t0
− δ2 Zi0 t0 − ai0 1 t0 f Ui0 1 t0
dt
ai0 t0 f Ui0 t0 < 0 if 0 ≤ i0 ≤ I − 2,
dZi0 t0
2 − δ2 Zi0 t0 b t0 gi0 1 t0 − ai0 1 t0 f Ui0 1 t0
dt
h
ai0 t0 f Ui0 t0 < 0 if i0 I − 1.
4.7
But this contradicts 2.1-2.2 and 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 node xI .
Now, we are in a position to state the main result of this section.
Theorem 4.3. Let Uh be the solution of 2.1–2.3. Suppose that there exists a positive integer A
such that
δ2 ϕi − ai 0f ϕi ≥ 0, 1 ≤ i ≤ I − 1,
δ2 ϕI − aI 0f ϕI b0gI ϕI ≥ Ag ϕI .
4.8
fsg s − f sgs ≥ 0 for s ≥ 0.
4.9
Assume that
Louis A. Assalé et al.
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Then the solution Uh blows up in a finite time Tbh and we have the following estimate
Tbh ≤
1
A
∞
ϕh ∞
ds
.
gs
4.10
Proof. Since 0, Tbh is the maximal time interval on which Uh t∞ < ∞, our aim is to show
that Tbh is finite and satisfies the above inequality. Introduce the vector Jh such that
Ji t dUi t
,
dt
0 ≤ i ≤ I − 1,
JI t dUI t
− Ag UI t .
dt
4.11
A straightforward calculation gives
dJi
d dUi
− δ 2 Ji − δ2 Ui , 0 ≤ i ≤ I − 1,
dt
dt dt
dUI
dJI
d dUI
− δ 2 JI − δ2 UI − Ag UI
Aδ2 g UI .
dt
dt dt
dt
4.12
From Lemma 4.1, we have δ2 gUI ≥ g UI δ2 UI which implies that
dUI
dJI
d dUI
− δ 2 JI ≥
− δ2 UI − Ag UI
− δ2 UI .
dt
dt dt
dt
4.13
Using 2.1, we get
dUi
dJi
− δ2 Ji ≥ −ai tf Ui − ai tf Ui
,
dt
dt
0 ≤ i ≤ I − 1,
dUI 2 dJI
− δ2 JI ≥ −aI tf UI − aI tf UI
b tg UI
dt
dt
h
2
dU
2
I
− Ag UI
btg UI
− aI tf UI btg UI .
h
dt
h
4.14
It follows from the fact that ai t ≤ 0, b t ≥ 0 and dUi /dt Ji AgUi that
dJI
− δ 2 JI ≥
dt
− aI tf UI
2
btg UI JI AaI t g UI f UI − f UI g UI .
h
4.15
We deduce from 4.9 that
dJi
− δ2 Ji ≥ −ai tf Ui Ji , 0 ≤ i ≤ I − 1,
dt
2
dJI
2
− δ JI ≥ − aI tf UI btg UI JI .
dt
h
4.16
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From 4.8, we observe that
Ji 0 δ2 ϕi − ai 0f ϕi ≥ 0, 0 ≤ i ≤ I − 1,
JI 0 δ2 ϕI − aI 0f ϕI b0gI ϕI − Ag ϕI ≥ 0.
4.17
We deduce from Lemma 3.1 that Ji t ≥ 0, 0 ≤ i ≤ I, which implies that dUI /dt ≥ gUI ,
0 ≤ i ≤ I. Obviously we have
dUI
≥ A dt.
g UI
4.18
Integrating this inequality over t, Tbh , we arrive at
Tbh − t ≤
∞
UI t
ds
,
gs
4.19
Uh 0∞
ds
.
gs
4.20
1
A
which implies that
Tbh ≤
1
A
∞
Since the quantity on the right hand side of the above inequality is finite, we deduce that the
solution Uh blows up in a finite time. Use the fact that Uh 0∞ ϕh ∞ to complete the rest
of the proof.
Remark 4.4. The inequality 4.19 implies that
∞
ds
if 0 < t0 < Tbh ,
gs
Uh t0 ∞
Ui t ≤ H A Tbh − t , 0 ≤ i ≤ I,
Tbh − t0 ≤
1
A
where Hs is the inverse of Gs ∞
s
4.21
dz/gz.
Remark 4.5. If gs sq , then Gs s1−q /q − 1 and Hs 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 interval 0, T where u is defined, the solution Uh t of 2.1–
2.3 approximates u, when the mesh parameter h goes to zero.
Theorem 5.1. Assume that 1.1 has a solution u ∈ C4,1 0, 1 × 0, T and the initial condition at
2.3 satisfies
ϕh − uh 0 o1
∞
as h −→ 0,
5.1
Louis A. Assalé et al.
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where uh t ux0 , t, . . . , uxI , tT . Then, for h sufficiently small, the problem 2.1–2.3 has a
unique solution Uh ∈ C1 0, T , RI1 such that
maxUh t − uh t∞ O ϕh − uh 0∞ h2
0≤t≤T
as h −→ 0.
5.2
Proof. Let α > 0 be such that
u·, t ≤ α for t ∈ 0, T .
∞
5.3
The problem 2.1–2.3 has for each h, a unique solution Uh ∈ C1 0, Tbh , RI1 . Let th ≤
min{T, Tbh } the greatest value of t > 0 such that
Uh t − uh t < 1
∞
for t ∈ 0, th .
5.4
The relation 5.1 implies that th > 0 for h sufficiently small. By the triangle inequality, we
obtain
Uh t ≤ u·, t Uh t − uh t
∞
∞
∞
for t ∈ 0, th ,
5.5
which implies that
Uh t ≤ 1 α
∞
for t ∈ 0, th .
5.6
Let eh t Uh t − uh t be the error of discretization. Using Taylor’s expansion, we have for
t ∈ 0, th,
dei t
− δ2 ei t −ai tf ξi t ei t o h2 ,
dt
0 ≤ i ≤ I − 1,
deI t
2
− δ2 eI t −aI tf ξI t eI t btg UI t eI t o h2 ,
dt
h
5.7
where θI t is an intermediate value between UI t and uxI , t and ξi t the one between
Ui t and uxi , t. Using 5.3 and 5.6, there exist two positive constants K and L such that
dei t
− δ2 ei t ≤ Lei t Kh2 , 0 ≤ i ≤ I − 1,
dt
2eI−1 t − 2eI t
LeI t
deI t
−
LeI t Kh2 .
≤
2
dt
h
h
5.8
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Consider the function zx, t eM1tCx ϕh −uh 0∞ Qh2 where M, C, Q are constants
which will be determined later. We get
2
zt x, t − zxx x, t M 1 − 2C − 4C2 x2 zx, t,
zx 0, t 0,
zx, 0 e
Cx2
zx 1, t 2Cz1, t,
5.9
ϕh − uh 0∞ Qh .
2
By a semidiscretization of the above problem, we may choose M, C, Q large enough that
d z xi , t > δ2 z xi , t Lz xi , t Kh2 , 0 ≤ i ≤ I − 1,
dt
L d z xI , t > δ2 z xI , t z xI , t Lz xI , t Kh2 ,
dt
h
z xi , 0 > ei 0, 0 ≤ i ≤ I.
5.10
It follows from Lemma 3.2 that
z xi , t > ei t for t ∈ 0, th , 0 ≤ i ≤ I.
5.11
By the same way, we also prove that
z xi , t > −ei t
for t ∈ 0, th , 0 ≤ i ≤ I,
5.12
z xi , t > ei t for t ∈ 0, th , 0 ≤ i ≤ I.
5.13
which implies that
We deduce that
Uh t − uh t ≤ eMtC ϕh − uh 0 Qh2 ,
∞
∞
t ∈ 0, th .
5.14
Let us show that th T . Suppose that T > th. From 5.4, we obtain
1 Uh th − uh th ∞ ≤ eMT C ϕh − uh 0∞ Qh2 .
5.15
Since the term on the right hand side of the above inequality goes to zero as h tends to zero, we
deduce that 1 ≤ 0, which is impossible. Consequently th 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 problem 1.1 has a solution u which blows up in a finite time Tb such
that u ∈ C4,1 0, 1 × 0, Tb and the initial condition at 2.3 satisfies
ϕh − uh 0 o1
∞
as h −→ 0.
5.16
Louis A. Assalé et al.
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Under the assumptions of Theorem 4.3, the problem 2.1–2.3 admits a unique solution Uh which
blows up in a finite time Tbh and we have the following relation
Tb .
lim T h
h→0 b
5.17
Proof. Let ε > 0. There exists a positive constant N such that
1
A
∞
x
ds
ε
≤
gs 2
for x ∈ N, ∞.
5.18
Since the solution u blows up at the time Tb , then there exists T1 ∈ Tb − ε/2, Tb such that
u·, t∞ ≥ 2N for t ∈ T1 , Tb . Setting T2 T1 Tb /2, then we have supt∈0,T2 |ux, t| < ∞.
It follows from Theorem 5.1 that
sup Uh t − uh t∞ ≤ N.
5.19
t∈0,T2 Applying the triangle inequality, we get
Uh T2 ≥ uh T2 − Uh T2 − uh T2 ,
∞
∞
∞
5.20
which leads to Uh T2 ∞ ≥ N. From Theorem 4.3, Uh t blows up at the time Tbh . We deduce
from Remark 4.4 and 5.18 that
Tb − T h ≤ Tb − T2 T h − T2 ≤ ε 1
b
b
2 A
∞
Uh T2 ∞
ds
≤ ε,
gs
5.21
and the proof is complete.
6. Numerical blow-up rate
In this section, we determine the blow-up rate of the solution Uh of 2.1–2.3 in the case
where bt 1. Our result is the following.
Theorem 6.1. Let Uh t be the solution of 2.1–2.3. Under the assumptions of Theorem 4.3, Uh t
blows up in a finite time Tbh and there exist two positive constants C1 , C2 such that
H C1 Tbh − t ≤ UI t ≤ H C2 Tbh − t ,
where Hs is the inverse of the function Gs ∞
s
for t ∈ 0, Tbh ,
6.1
dσ/gσ.
Proof. From Theorem 4.3 and Remark 4.4, Uh t blows up in a finite time Tbh and there exists
a constant C2 > 0 such that
UI t ≤ H C2 Tbh − t
for t ∈ 0, Tbh .
6.2
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From Lemma 4.2, UI−1 < UI . Then using 2.2, we deduce that dUI /dt ≤ 2/hbtgUI −
aI tfUI , which implies that dUI /dt ≤ 2bt/hgUI . Integration this inequality over
t, Tbh , there exists a positive constant C1 such that
UI t ≥ H C1 Tbh − t
for t ∈ 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 constant C0 such that sF s ≤ C0 and
d
Ui − δ2 Ui ≤ 0,
dt
0 ≤ i ≤ I − 1.
7.1
Assume that there exists a positive constant C such
Ui ≤ H CT − t ,
0 ≤ i ≤ I.
7.2
Then the numerical blow-up set is B {1}.
Proof. Let vx 1 − x2 and define
Wx, t H δvx δBT − t
for 0 ≤ x ≤ 1, t ≥ t0 ,
7.3
where δ is small enough. We have
Wx 0, t 0,
W1, t H δBT − t ≥ u1, t,
7.4
and for t ≥ t0 , we get
Wx, t0 H δvx δ ≥ H2δ H 2δB T − t0
≥ H C T − t0 ≥ u x, t0 .
7.5
A straightforward computation yields
Wt x, t − Wxx x, t δF Hτ B − 2 − 4xF Hτ
≥ δF Hτ B − 2 − 4δC0 .
7.6
This implies that there exists α > 0 such that
Wt x, t − Wxx x, t ≥ αF Hδ δBT .
7.7
Louis A. Assalé et al.
13
Using Taylor’s expansion, there exists a constant K > 0 such that
d W xi , t − δ2 W xi , t ≥ αF Hδ δBT − Kh2 ,
dt
0 ≤ i ≤ I,
7.8
which implies that
dW xi , t
− δ2 W xi , t ≥ 0.
dt
7.9
The maximum principle implies that
Ui t ≤ H δvx δB T − t0
for t ≥ t0 , 0 ≤ i ≤ I.
7.10
Hence, we get
Ui t ≤ H δvx ,
0 ≤ i ≤ I.
7.11
Therefore Ui T < ∞, 0 ≤ i ≤ I − 1, and we have the desired result.
8. Full discretization
In this section, we consider the problem 1.1 in the case where ax, t 1, bt 1, fu up ,
gu up with p const > 1. Thus our problem is equivalent to
ut x, t uxx x, t − up x, t,
ux 0, t 0,
0 < x < 1, t ∈ 0, T ,
ux 1, t up 1, t,
ux, 0 u0 x > 0,
t ∈ 0, T ,
8.1
0 ≤ x ≤ 1,
p
where p > 1, u0 ∈ C1 0, 1, u0 0 0 and u0 1 u0 1.
We start this section by the construction of an adaptive scheme as follows. Let I be a
positive integer and let h 1/I. Define the grid xi ih, 0 ≤ i ≤ I and approximate the solution
n
n
n
n
ux, t of the problem 8.1 by the solution Uh U0 , U1 , . . . , UI T of the following
discrete equations
n
δt Ui
n
δt UI
n p
− Ui
, 0 ≤ i ≤ I − 1,
n p 2 n p
n
δ2 UI − UI
,
U
h I
0
Ui ϕi , 0 ≤ i ≤ I,
n
δ2 Ui
8.2
8.3
8.4
14
Journal of Applied Mathematics
where n ≥ 0, ϕi1 ≥ ϕi , 0 ≤ i ≤ I − 1,
n
δt Ui
n
δ2 Ui
n
n
n
Ui1 − 2Ui
n
− Ui
,
Δtn
n
Ui−1
h2
n
δ2 U0 n1
Ui
n
2U1 − 2U0
,
h2
,
1 ≤ i ≤ I − 1,
n
n
δ2 UI
8.5
n
2UI−1 − 2UI
h2
.
In order to permit the discrete solution to reproduce the property of the continuous one when
the time t approaches the blow-up time, we need to adapt the size of the time step so that we
n p−1
take Δtn min{1 − pτh2 /3, τ/Uh ∞ }, 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.
n
Lemma 8.1. Let Uh be the solution of 8.2–8.4. Then we have
n
n
Ui1 ≥ Ui ,
n
Proof. Let Zi
n
8.6
n
Ui1 − Ui , 0 ≤ i ≤ I − 1. We observe that
n1
n1
n1
n
n
n
− Zi
Δtn
n
− Z0
Δtn
Z0
Zi
0 ≤ i ≤ I − 1.
n
Zi1 − 2Zi
n
n
Z1 − 3Z0
h2
n
Zi−1
h2
n
−
−
n p
U1
n p
Ui1
n p − U0
,
n p − Ui
,
1 ≤ i ≤ I − 2,
8.7
n
ZI−1 − ZI−1 ZI−2 − 3ZI−1 n p n p 2 n p
U
− UI
− UI−1
.
Δtn
h I
h2
Using the Taylor’s expansion, we find that
n1
Z0
n1
Zi
n1
ZI−1
n p−1 n
Δtn n
Δtn
n
2 Z1 1 − 3 2 Z0 − Δtn p ξ0
Z0 ,
h
h
Δtn n
Δtn
Δtn n
n
2 Zi1 1 − 2 2 Zi 2 Zi−1
h
h
h
n p−1 n
− Δtn p ξi
Zi , 1 ≤ i ≤ I − 2,
n p−1 n
Δtn n
Δtn
n
≥ 2 ZI−2 1 − 3 2 ZI−1 − Δtn p ξI−1
ZI−1 ,
h
h
8.8
Louis A. Assalé et al.
n
where ξi
that
15
n
is an intermediate value between Ui
n1
Z0
n1
Zi
n1
ZI−1
n
n
and Ui1 . If Zi
≤ 0, 0 ≤ i ≤ I − 1, we deduce
n p−1
Δtn n
Δtn
n
U
Z
1
−
3
−
Δt
p
Z0 ,
n
∞
h
h2 1
h2
n p−1
Δtn n
Δtn
n
≥ 2 Zi1 1 − 2 2 − Δtn pUh ∞ Zi
h
h
Δtn n
2 Zi−1 , 1 ≤ i ≤ I − 2,
h
n p−1
Δtn n
Δtn
n
≥ 2 ZI−2 1 − 3 2 − Δtn pUh ∞ ZI−1 .
h
h
≥
8.9
n p−1
Using the restriction Δtn ≤ τ/Uh ∞ , we find that
n1
Z0
n1
Zi
n1
ZI−1
Δtn n
Δtn
n
Z
1
−
3
−
pτ
Z0 ,
h2 1
h2
Δtn n
Δtn
n
≥ 2 Zi1 1 − 2 2 − pτ Zi
h
h
Δtn n
2 Zi−1 , 1 ≤ i ≤ I − 2,
h
Δtn n
Δtn
n
≥ 2 ZI−2 1 − 3 2 − pτ ZI−1 .
h
h
≥
8.10
n
We observe that 1 − 3Δtn /h2 − pτ is nonnegative and by induction, we deduce that Zi
0 ≤ i ≤ I − 1. This ends the proof.
≤ 0,
The following lemma is a discrete form of the maximum principle.
n
n
Lemma 8.2. Let ah be a bounded vector and let Vh
n
δt Vi
n
− δ2 Vi
n
0
Vi
n
Then Vi
n
ai Vi
≥ 0,
a sequence such that
≥ 0,
0 ≤ i ≤ I, n ≥ 0,
0 ≤ i ≤ I.
8.11
8.12
n
≥ 0 for n ≥ 0, 0 ≤ i ≤ I if Δtn ≤ h2 /2 ah ∞ h2 .
n
Proof. If Vh
≥ 0 then a routine computation yields
n1
V0
n1
Vi
n1
VI
n 2Δtn n
Δtn
a V n ,
V
1
−
2
−
Δt
n
0
∞
h
h2 1
h2
n Δtn n
Δtn
n
≥ 2 Vi1 1 − 2 2 − Δtn ah ∞ Vi
h
h
Δtn n
2 Vi−1 , 1 ≤ i ≤ I − 1,
h
n 2Δtn n
Δtn
n
≥
V 1 − 2 2 − Δtn ah ∞ VI .
h2 I−1
h
≥
8.13
16
Journal of Applied Mathematics
n
n
Since Δtn ≤ h2 /2 ah ∞ h2 , we see that 1 − 2Δtn /h2 − Δtn ah ∞ is nonnegative. From
n
8.12, we deduce by induction that Vh
≥ 0 which ends the proof.
A direct consequence of the above result is the following comparison lemma. Its proof
is straightforward.
n
n
n
n
Lemma 8.3. Suppose that ah and bh two vectors such that ah is bounded. Let Vh
sequences such that
n
δt Vi
n
− δ2 Vi
n
n
ai Vi
n
bi
n
≤ δt W i
0
Vi
n
Then Vi
n
≤ Wi
n
− δ 2 Wi
0
≤ Wi ,
n
n
ai Wi
n
bi ,
n
and Wh two
0 ≤ i ≤ I, n ≥ 0,
0 ≤ i ≤ I.
8.14
n
for n ≥ 0, 0 ≤ i ≤ I if Δtn ≤ h2 /2 ah ∞ h2 .
Now, let us give a property of the operator δt .
Lemma 8.4. Let Un ∈ R be such that Un ≥ 0 for n ≥ 0. Then we have
p
p−1
δt Un ≥ p Un
δt Un ,
n ≥ 0.
8.15
Proof. From Taylor’s expansion, we find that
p
p−1
2 p−2
pp − 1
Δtn δt Un θn
δt Un p Un
δt Un ,
2
8.16
where θn is an intermediate value between Un and Un1 . Use the fact that Un ≥ 0 for
n ≥ 0 to complete the rest of the proof.
To handle the phenomenon of blow-up for discrete equations, we need the following
definition.
n
Definition 8.5. We say that the solution Uh of 8.2–8.4 blows up in a finite time if
n lim Uh ∞ ∞,
n → ∞
ThΔt lim
n→∞
n−1
Δti < ∞.
8.17
i0
n
The number ThΔt is called the numerical blow-up time of Uh .
n
The following theorem reveals that the discrete solution Uh of 8.2–8.4 blows up
in a finite time under some hypotheses.
Louis A. Assalé et al.
17
n
Theorem 8.6. Let Uh be the solution of 8.2–8.4. Suppose that there exists a constant A ∈ 0, 1
such that the initial data at 8.4 satisfies
p
δ2 ϕi − ϕi ≥ 0,
p
δ2 ϕI − ϕI 0 ≤ i ≤ I − 1.
2 p
p
ϕ ≥ AϕI .
h I
8.18
n
Then Uh blows up in a finite time ThΔt which satisfies the following estimate
τ1 τ p−1
p−1 ,
1 τ p−1 − 1 ϕh ∞
ThΔt ≤ 8.19
−p−1
where τ A min{1 − pτh2 ϕh inf /3, τ}.
n
Proof. Introduce the vector Jh defined as follows
n
δt Ui ,
n
JI
n
δt UI
n
0 ≤ i ≤ I − 1, n ≥ 0,
n −p
− A UI
, n ≥ 0.
8.20
n
n δt δt Ui − δ2 Ui , 0 ≤ i ≤ I − 1,
n p
n p
n
n δt δt UI − δ2 UI − Aδt UI
Aδ2 UI
.
8.21
Ji
A straightforward computation yields
n
δt J i
n
n
− δ 2 Ji
n
− δ 2 JI
δt J I
Using 8.2, we arrive at
n p
n
− δ2 Ji −δt Ui
, 0 ≤ i ≤ I − 1,
n p
n p
2
n
n
δt J I − δ 2 J I − 1 − A δt UI
Aδ2 UI
.
h
n
δt Ji
8.22
Due to the mean value theorem, we get
n p−1 n n p
n p−1 n
δt Ui
p ξi
δt Ui
Ji ,
p ξi
n
n
where ξi is an intermediate value between Ui
2.4 and 2.5, we deduce that
8.23
n
and Ui1 . On the other hand, from Lemmas
n p−1 n
n
− δ2 Ji −p ξi
Ji , 0 ≤ i ≤ I − 1,
n p−1
n p−1 2 n
2
n
− 1 − A p UI
δt UI Apδt UI
δ UI .
h
n
δt J i
n
δt JI
n
− δ 2 JI
8.24
18
Journal of Applied Mathematics
It follows from 8.3 that
n
δt J I
n
− δ 2 JI
n p−1
n p
n p−1
2
2
n
− 1 p UI
− 1 UI
δt UI − Apδt UI
,
h
h
8.25
which implies that
n p−1 n
−p ξi
Ji , 0 ≤ i ≤ I − 1,
n p−1 n
2
n
n
δt J I − δ 2 J I − 1 p UI
JI .
h
n
δt Ji
n
− δ 2 Ji
0
8.26
n
From 8.18, we observe that Jh ≥ 0. It follows from Lemma 8.2 that Jh
that
n1
UI
n
From Lemma 8.1, we see that UI
≥ 0 which implies
n ≥ UI
n p−1 1 AΔtn UI
.
8.27
n
Uh ∞ which implies that
n1 ≥ Un 1 AΔtn Un p−1 .
U
∞
∞
∞
h
h
h
8.28
It is not hard to see that
n p−1
AΔtn Uh ∞ A min
n1
From 8.28, we get Uh
n p−1
1 − pτh2 Uh ∞
3
,τ .
n
8.29
n1
∞ ≥ Uh ∞ . By induction, we arrive at Uh
n p−1
p−1
0
∞ ≥ Uh ∞ ϕh ∞ , which implies that Uh ∞ ≥ ϕh ∞ . Therefore, we find that
n p−1
AΔtn Uh ∞ ≥ A min
p−1 1 − pτh2 ϕh ∞
, τ τ .
3
8.30
Consequently, we arrive at
n1 U
≥ Un 1 τ ∞
∞
h
h
8.31
and by induction, we get
n U ≥ U0 1 τ n ϕh 1 τ n ,
∞
∞
∞
h
h
n ≥ 0.
8.32
Since the term on the right hand side of the above equality tends to infinity as n approaches
n
infinity, we conclude that Uh ∞ tends to infinity as n approaches infinity. Now, let us
Louis A. Assalé et al.
19
estimate the numerical blow-up time. Due to 8.32, the restriction on the time step ensures
that
τ
τ
∞
∞
Σ∞
Σ
n0 Δtn ≤ Σn0 n p−1 ≤ ϕh p−1 n0
U ∞
∞
h
1
1 τ p−1
n
.
8.33
Using the fact that the series on the right hand side of the above inequality converges
∞
p−1
/1τ p−1 −
towards τ1 τ p−1 /1 τ p−1 −1, we deduce that Σ∞
n0 Δtn ≤ Σn0 τ1 τ p−1
1ϕh ∞ and the proof is complete.
Remark 8.7. Apply Taylor’s expansion to obtain 1 τ p−1 1−p −1τ oτ , which implies
that
τ
1 τ p−1
−1
τ
τ
1
p − 1 o1
≤
2τ
.
τ p − 1
8.34
If we take τ h2 , we see that
τ
A min
τ
1 − ph2 p−1
ϕh , 1 ≥ A min 1 ϕh p−1 , 1 .
∞
∞
3
4
8.35
We deduce that τ/τ is bounded from above. We conclude that τ/1 τ p−1 − 1 is bounded
from above.
Remark 8.8. From 8.31, we get
n U ≥ Uq 1 τ n−q
∞
∞
h
h
for n ≥ q
8.36
n−q
τ
1
∞
Σ∞
Δt
≤
Σ
.
n
q p−1 nq
nq
U 1 τ p−1
∞
h
8.37
1 τ p−1
τ
.
ThΔt − tq ≤ q p−1
U 1 τ p−1 − 1
∞
h
8.38
which implies that
We deduce that
In the sequel, we take τ h2 .
9. Convergence of the blow-up time
In this section, under some conditions, we show that the discrete solution blows up in a finite
time and its numerical blow-up time goes to the real one when the mesh size goes to zero. To
start, let us prove a result about the convergence of our scheme.
20
Journal of Applied Mathematics
Theorem 9.1. Suppose that the problem 1.1 has a solution u ∈ C4,2 0, 1 × 0, T . Assume that
the initial data at 8.4 satisfies
ϕh − uh 0 o1
∞
as h −→ 0.
9.1
n
Then the problem 8.2–8.4 has a solution Uh for h sufficiently small, 0 ≤ n ≤ J and we have the
following relation
n
max Uh − uh tn ∞ O ϕh − uh 0∞ h2 Δtn
0≤n≤J
where J is such that
J−1
n0 Δtn
≤ T and tn as h −→ 0,
9.2
n−1
j0 Δtj .
n
Proof. For each h, the problem 8.2–8.4 has a solution Uh . Let N ≤ J be the greatest value
of n such that
n
U − uh tn < 1 for n < N.
∞
h
9.3
We know that N ≥ 1 because of 9.1. Due to the fact that u ∈ C4,2 , there exists a positive
constant K such that u∞ ≤ K. Applying the triangle inequality, we have
n U ≤ uh tn Un − uh tn ≤ 1 K
∞
∞
∞
h
h
for n < N.
9.4
Since u ∈ C4,2 , using Taylor’s expansion, we find that
δt u xi , tn − δ2 u xi , tn −up xi , tn O h2 O Δtn , 0 ≤ i ≤ I − 1,
2 δt u xI , tn − δ2 u xI , tn −up xI , tn up xI , tn O h2 O Δtn .
h
n
9.5
n
Let eh Uh − uh tn be the error of discretization. From the mean value theorem, we get
n
δt e i
n
− δ 2 ei
n
n
δt eI − δ2 eI
n
n p−1 n
−p ξi
ei O h2 O Δtn , 0 ≤ i ≤ I − 1,
n p−1 n
2
− 1 ξI
p
eI O h2 O Δtn ,
h
9.6
n
where ξi is an intermediate value between uxi , tn and Ui . Hence, there exist positive
constants L and K such that
n
δt ei
n
n
− δ 2 ei
n
δt e I − δ 2 e I
n p−1 n
≤ −p ξi
ei Lh2 LΔtn , 0 ≤ i ≤ I − 1, n < N,
n p−1 n
2
− 1 ξI
≤p
eI Lh2 LΔtn , n < N.
h
9.7
Louis A. Assalé et al.
21
Consider the function Zx, t eM1tCx ϕh − uh 0∞ Qh2 QΔtn where M, C, Q are
positive constants which will be determined later. We get
2
Zt x, t − Zxx x, t M 1 − 2C − 4C2 x2 Zx, t,
Zx 1, t 2CZ1, t,
Zx 0, t 0,
2 Zx, 0 eCx ϕh − uh 0∞ Qh2 QΔtn .
9.8
By a discretization of the above problem, we obtain
h2
δt Z xi , tn − δ2 Z xi , tn M 1 − 2C − 4C2 xi2 Z xi , tn Zxxxx xi , tn
12
Δtn Ztt xi , tn ,
−
2
4C Z xI , tn
δt Z xI , tn − δ2 Z xI , tn M 1 − 2C − 4C2 xI2 Z xI , tn h
Δtn h2
Zxxxx xI , tn −
Ztt xI , tn .
12
2
9.9
We may choose M, C, Q large enough that
n p−1 δt Z xi , tn − δ2 Z xi , tn > −p ξi
Z xi , tn Lh2 LΔtn , 0 ≤ i ≤ I − 1,
n p−1 2
δt Z xI , tn − δ2 Z xI , tn > p
− 1 ξI
Z xI , tn Lh2 LΔtn ,
h
0
Zi
0
> ei ,
9.10
0 ≤ i ≤ I.
It follows from Comparison Lemma 8.3 that
n
Z xi , tn > ei ,
0 ≤ i ≤ I, n < N.
9.11
By the same way, we also prove that
n
Z xi , tn > −ei ,
0 ≤ i ≤ I, n < N,
9.12
which implies that
n
U − uh t ≤ eMtn C ϕh − uh 0 Qh2 QΔtn ,
∞
∞
h
n < N.
9.13
Let us show that N J. Suppose that N < J. From 9.3, we obtain
N
1 ≤ Uh − uh tN ∞ ≤ eMT C ϕh − uh 0∞ Qh2 QΔtn .
9.14
Since the term on the right hand side of the second inequality goes to zero as h goes to zero,
we deduce that 1 ≤ 0, which is a contradiction and the proof is complete.
22
Journal of Applied Mathematics
Now, we are in a position to state the main theorem of this section.
Theorem 9.2. Suppose that the problem 1.1 has a solution u which blows up in a finite time T0 and
u ∈ C4,2 0, 1 × 0, T0 . Assume that the initial data at 2.3 satisfies
ϕh − uh 0 o1
∞
as h −→ 0.
9.15
n
Under the assumption of Theorem 8.6, the problem 8.2–8.4 has a solution Uh which blows up in
a finite time ThΔt and the following relation holds
lim ThΔt T0 .
9.16
h→0
Proof. We know from Remark 8.7 that τ1 τ /1 τ p−1 − 1 is bounded. Letting ε > 0,
there exists a constant R > 0 such that
τ1 τ p−1
ε
<
p−1
p−1
2
x
1 τ − 1
for x ∈ R, ∞.
9.17
Since u blows up at the time T0 , there exists T1 ∈ T0 − ε/2, T0 such that u·, t∞ ≥ 2R for
q−1
t ∈ T1 , T0 . Let T2 T1 T0 /2 and let q be a positive integer such that tq n0 Δtn ∈ T1 , T2 for h small enough. We have 0 < uh tn ∞ < ∞ for n ≤ q. It follows from Theorem 4.3 that
n
n
the problem 2.1–2.3 has a solution Uh which obeys Uh − uh tn ∞ < R for n ≤ q, which
implies that
q U ≥ uh tq − Uq − uh tq ≥ R.
∞
∞
∞
h
h
9.18
n
From Theorem 8.6, Uh blows up at the time ThΔt . It follows from Remark 8.8 and 9.17 that
q 1−p
q
|ThΔt − tq | ≤ τ1 τ p−1 Uh ∞ /1 τ p−1 − 1 < ε/2 because Uh ∞ ≥ R. We deduce that
|T0 − ThΔt | ≤ |T0 − tq | |tq − ThΔt | ≤ ε/2 ε/2 ≤ ε, which leads us to the result.
10. Numerical experiments
In this section, we present some numerical approximations to the blow-up time of 1.1 in the
case where ax, t λ > 0, fu up , gu uq , bt 1 with p const > 1, q const > 1. We
n
approximate the solution u of 1.1 by the solution Uh of the following explicit scheme
n
δt Ui
n
δt UI
n
δ2 Ui
n p−1 n1
− λ Ui
Ui
,
n
δ2 UI 0
Ui
0 ≤ i ≤ I − 1,
n p−1 n1
2 n q
− λ UI
UI
,
UI
h
ϕi ≥ 0,
0 ≤ i ≤ I,
10.1
Louis A. Assalé et al.
23
n
We also approximate the solution u of 1.1 by the solution Uh of the implicit scheme below
n
δt Ui
n
δt UI
n1
δ2 Ui
n p−1 n1
− λ Ui
Ui
,
n1
δ2 UI
0
Ui
0 ≤ i ≤ I − 1,
n p−1 n1
2 n q
− λ UI
UI
,
U
h I
ϕi ≥ 0,
10.2
0 ≤ i ≤ I.
n 1−p
For the time step, we take n ≥ 0, Δtn minh2 /2, τUh ∞ for the explicit scheme and
Δtn n 1−p
τUh ∞
for the implicit scheme.
The problem described in 10.1 may be rewritten as follows
n1
U0
n1
Ui
n1
UI
n n
2 Δtn /h2 U1 1 − 2 Δtn /h2 U0
,
n p−1
1 λΔtn U0
n n
n
2 Δtn /h2 Ui1 1 − 2 Δtn /h2 Ui 2 Δtn /h2 Ui−1
,
n p−1
1 λΔtn Ui
n n
n q
2 Δtn /h2 UI−1 1 − 2 Δtn /h2 UI 2 Δtn /h2 UI
.
n p−1
1 λΔtn UI
10.3
Let us notice that the restriction on the time step Δtn ≤ h2 /2 ensures the nonnegativity of the
discrete solution.
The implicit scheme may be rewritten in the following form
n1
Anh Uh
Fn,
10.4
where
⎞
b0 0 · · · 0
a1 b1 0 · · · ⎟
⎟
⎟
.. .. ..
⎟
. . .
⎟,
⎟
⎟
.. .. ..
. . . bI−1 ⎠
· · · 0 cI aI
⎛
n
Ah
a0
⎜c
⎜ 1
⎜
⎜
⎜0
⎜.
⎜.
⎝.
0
n p−1
Δtn
λΔtn Ui
, 0 ≤ i ≤ I,
2
h
Δtn
bi −2 2 , i 0, . . . , I − 1,
h
Δtn
ci −2 2 , i 1, . . . , I,
h
ai 1 2
n
F n i Ui ,
n
F n I UI i 0, . . . , I − 1,
n q
2
Δtn UI
.
h
10.5
24
Journal of Applied Mathematics
Table 1: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the explicit Euler method defined in 10.1.
Tn
0.047927
0.044695
0.043583
0.043225
0.043115
I
16
32
64
128
256
n
451
1260
4075
14555
55061
s
—
—
1.54
1.64
1.71
CPU time
—
0.5
5
60
1816
n
The matrix Ah satisfies the following properties
n > 0,
Ah ij < 0, if i /
j,
n n A
>
A
.
ii
ij
h
h
n Ah
ii
10.6
j/
i
n
0
It follows that Uh exists for n ≥ 0. In addition, since Uh
nonnegative for n ≥ 0. We need the following definition.
n
is nonnegative, Uh
is also
n
Definition 10.1. We say that the discrete solution Uh of the explicit scheme or the implicit
n
scheme blows up in a finite time if limn → ∞ Uh ∞ ∞ and the series ∞
n0 Δtn converges.
∞
n
The quantity n0 Δtn is called the numerical blow-up time of the solution Uh .
In Tables 1, 2, 3, 4, 5, 6, 7, and 8, in rows, we present the numerical blow-up times,
values of n, the CPU times and the orders of the approximations corresponding to meshes of
16, 32, 64, 128, 256. For the numerical blow-up time we take T n n−1
j0 Δtj which is computed
at the first time when
Δtn T n1 − T n ≤ 10−16 .
10.7
The order s of the method is computed from
s
log
T4h − T2h / T2h − Th
.
log2
10.8
Case 1. p 0, q 2, ϕi 10 10 ∗ cosπih, λ 1.
Case 2. p 2, q 4, ϕi 10 10 ∗ cosπih, λ 1.
Case 3. p 2, q 3, ϕi 10 10 ∗ cosπih, λ 1.
Case 4. p 2, q 2, ϕi 10 10 ∗ cosπih, λ 1.
Remark 10.2. The different cases of our numerical results show that there is a relationship
between the flow on the boundary and the absorption in the interior of the domain. Indeed,
Louis A. Assalé et al.
25
Table 2: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the implicit Euler method defined in 10.2.
I
16
32
64
128
256
Tn
0.047631
0.044645
0.043576
0.043224
0.043113
n
423
1234
4050
14533
55035
CPU time
—
1
5
99
2000
s
—
—
1.49
1.61
1.67
Table 3: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the explicit Euler method defined in 10.1.
I
16
32
64
128
256
Tn
0.018286
0.017181
0.016729
0.016412
0.016324
n
21750
83838
329960
1298750
6447649
CPU time
3
17
108
1570
27049
s
—
—
1.30
0.51
1.85
Table 4: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the implicit Euler method defined in 10.2.
I
16
32
64
128
256
Tn
0.018283
0.017181
0.016729
0.016617
0.016526
n
21741
83831
3299953
1208495
6348765
CPU time
6
37
347
4640
29957
s
—
—
1.30
2.01
0.30
Table 5: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the explicit Euler method defined in 10.1.
I
16
32
64
128
256
Tn
0.024197
0.022570
0.021950
0.021734
0.021712
n
1649
6103
23583
92985
369250
CPU time
—
2
8
200
3243
s
—
—
1.40
1.52
3.30
Table 6: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the implicit Euler method defined in 10.2.
I
16
32
64
128
256
Tn
0.024169
0.022566
0.021950
0.021734
0.021713
n
1602
6066
23551
92985
370240
CPU time
—
5
65
1140
6709
s
—
—
1.38
1.52
3.37
26
Journal of Applied Mathematics
×1013
2
Ui, n
1.5
1
0.5
0
20
15
10
i
5
0
0
100
200
300
400
500
n
Figure 1: Evolution of the discrete solution, q 2, p 2 explicit scheme.
Table 7: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the explicit Euler method defined in 8.2–8.4.
I
16
32
64
128
256
Tn
0.054342
0.050346
0.049027
0.048615
0.048491
n
422
1130
3539
12020
46439
CPU time
—
—
4
28
937
s
—
—
1.60
1.68
1.74
Table 8: Numerical blow-up times, numbers of iterations, CPU times seconds and orders of the
approximations obtained with the implicit Euler method defined in 10.2.
I
16
32
64
128
256
Tn
0.054158
0.050332
0.049030
0.048616
0.048519
n
364
1077
3491
12358
36919
CPU time
—
0.6
7
79
1123
s
—
—
1.56
1.66
2.10
when there is not an absorption on the interior of the domain, we see that the blow-up time is
slightly equal to 0.043 for q 2 whereas if there is an absorption in the interior of the domain,
we observe that the blow-up time is slightly equal to 0.048 for q 2 and p 2. We see that
there is a diminution of the blow-up time. We also remark that if the power of flow on the
boundary increases then the blow-up time diminishes. Thus the flow on the boundary make
blow-up occurs whereas the absorption in the interior of domain prevents the blow-up. This
phenomenon is well known in a theoretical point of view.
For other illustrations, in what follows, we give some plots to illustrate our analysis.
In Figures 1, 2, 3, 4, 5, and, 6, we can appreciate that the discrete solution blows up in a finite
time at the last node.
Louis A. Assalé et al.
27
×1013
2
Ui, n
1.5
1
0.5
0
20
15
10
5
i
0
0
200
100
300
400
500
n
Figure 2: Evolution of the discrete solution, q 2, p 2 implicit scheme.
×1021
7
6
Ui, n
5
4
3
2
1
0
20
15
10
5
i
0
0
20
40
80
60
100
120
n
Figure 3: Evolution of the discrete solution, q 3, p 2 explicit scheme.
×1024
2.5
Ui, n
2
1.5
1
0.5
0
20
15
10
i
5
0
0
20
40
80
60
100
120
n
Figure 4: Evolution of the discrete solution, q 3, p 2 implicit scheme.
28
Journal of Applied Mathematics
×1015
6
Ui, n
5
4
3
2
1
0
20
15
10
5
i
0
0
20
40
60
80
100
n
Figure 5: Evolution of the discrete solution, q 4, p 2 explicit scheme.
×1033
2.5
Ui, n
2
1.5
1
0.5
0
20
15
10
i
5
0
0
20
40
60
80
100
n
Figure 6: Evolution of the discrete solution, q 4, p 2 implicit scheme.
Acknowledgments
We want to thank the anonymous referee for the throughout reading of the manuscript and
several suggestions that help us to improve the presentation of the paper.
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