Volume 2008, Issue 1 753518
Research Article
Open Access

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

Louis A. Assalé

Louis A. Assalé

Institut National Polytechnique Houphouët-Boigny de Yamoussoukro BP 1093, Yamoussoukro, Cote D′Ivoire , inphb.edu.ci

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Théodore K. Boni

Théodore K. Boni

Institut National Polytechnique Houphouët-Boigny de Yamoussoukro BP 1093, Yamoussoukro, Cote D′Ivoire , inphb.edu.ci

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Diabate Nabongo

Corresponding Author

Diabate Nabongo

Département de Mathématiques et Informatiques Université d′Abobo-Adjamé, UFR-SFA 16 BP 372 Abidjan 16, Cote D′Ivoire , uabobo.ci

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First published: 12 April 2009
Citations: 5
Academic Editor: Jacek Rokicki

Abstract

We obtain some conditions under which the positive solution for semidiscretizations of the semilinear equation ut = uxxa(x, t)f(u),   0 < x < 1,   t ∈ (0, T), with boundary conditions ux(0, t) = 0, ux(1, t) = b(t)g(u(1, 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.

1. Introduction

In this paper, we consider the following boundary value problem:
(1.1)
where f : [0, )→[0, ) is a C1 function, f(0) = 0,   g : [0, )→[0, ) is a C1 convex function, g(0) = 0, aC0([0, 1] × +), a(x, t) ≥ 0 in [0, 1] × +, at(x, t) ≤ 0 in [0, 1] × +, bC1(+), b(t) > 0 in +, b(t) ≥ 0 in +. The initial data u0C2([0, 1]), , .
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
(1.2)
where ∥u(⋅, t)∥ = max0≤x≤1  | u(x, 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 [17], 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 [1115]). Generally, the authors have considered the problem (1.1) in the case where a(x, t) = 0 and b(t) = 1. For instance in [15], the above problem has been considered in the case where a(x, t) = 0 and b(t) = 1. In [16], the authors have considered the problem (1.1) in the case where a(x, t) = λ > 0, b(t) = 1, f(u) = up, g(u) = 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 [1722] 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 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 ≤ iI, where h = 1/I. We approximate the solution u of (1.1) by the solution Uh(t) = (U0(t), U1(t), …, UI(t)) T of the following semidiscrete equations
(2.1)
(2.2)
(2.3)
where φi+1φi, 0 ≤ iI − 1,
(2.4)
Here is the maximal time interval on which ∥Uh(t)∥ is finite where ∥Uh(t)∥ = max0≤iIUi(t). When is finite, we say that the solution Uh(t) blows up in a finite time and the time 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), I+1) and let Vh(t) ∈ C1([0, T), I+1) such that

(3.1)
Then we have Vi(t) ≥ 0, 0 ≤ iI, t ∈ (0, T).

Proof. Let T0 < T and define the vector Zh(t) = eλtVh(t) where λ is large enough that ai(t) − λ > 0 for t ∈ [0, T0], 0 ≤ iI. Let . Since for i ∈ {0, …, I}, Zi(t) is a continuous function, there exists t0 ∈ [0, T0] such that for a certain i0 ∈ {0, …, I}. It is not hard to see that

(3.2)
A straightforward computation reveals that
(3.3)
We observe from (3.2) that which implies that because . 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), I+1) and fC0( × , ) such that for t ∈ (0, T)

(3.4)
(3.5)
Then we have Vi(t) < Ui(t), 0 ≤ iI, 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 ≤ iI, but for a certain i0 ∈ {0, …, I}. We observe that

(3.6)
which implies that
(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 UhI+1 be such that Uh ≥ 0. Then we have

(4.1)

Proof. Apply Taylor′s expansion to obtain

(4.2)
where θi is an intermediate between Ui and Ui+1 and ηi the one between Ui−1 and Ui. The first and last equalities imply that
(4.3)
Combining the second and third equalities, we see that
(4.4)
Use the fact that g′′(s) ≥ 0 for s ≥ 0 and Uh ≥ 0 to complete the rest of the proof.

Lemma 4.2. Let Uh be the solution of (2.1)–(2.3). Then we have

(4.5)

Proof. Let t0 be the first t > 0 such that Ui+1(t) > Ui(t) for 0 ≤ iI − 1 but 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) = Ui+1(t) − Ui(t) for 0 ≤ iI − 1. We get

(4.6)
which implies that
(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

(4.8)
Assume that
(4.9)
Then the solution Uh blows up in a finite time and we have the following estimate
(4.10)

Proof. Since is the maximal time interval on which ∥Uh(t)∥ < , our aim is to show that is finite and satisfies the above inequality. Introduce the vector Jh such that

(4.11)
A straightforward calculation gives
(4.12)
From Lemma 4.1, we have δ2g(UI) ≥ g(UI)δ2UI which implies that
(4.13)
Using (2.1), we get
(4.14)
It follows from the fact that , b(t) ≥ 0 and dUi/dt = Ji + Ag(Ui) that
(4.15)
We deduce from (4.9) that
(4.16)
From (4.8), we observe that
(4.17)
We deduce from Lemma 3.1 that Ji(t) ≥ 0, 0 ≤ iI, which implies that dUI/dtg(UI), 0 ≤ iI. Obviously we have
(4.18)
Integrating this inequality over , we arrive at
(4.19)
which implies that
(4.20)
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

(4.21)
where H(s) is the inverse of .

Remark 4.5. If g(s) = sq, then G(s) = s1−q/(q − 1) and H(s) = ((q − 1)s) 1/(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 uC4,1([0, 1]×[0, T]) and the initial condition at (2.3) satisfies

(5.1)
where uh(t) = (u(x0, t), …, u(xI, t)) T. Then, for h sufficiently small, the problem (2.1)–(2.3) has a unique solution UhC1([0, T], I+1) such that
(5.2)

Proof. Let α > 0 be such that

(5.3)
The problem (2.1)–(2.3) has for each h, a unique solution . Let the greatest value of t > 0 such that
(5.4)
The relation (5.1) implies that t(h) > 0 for h sufficiently small. By the triangle inequality, we obtain
(5.5)
which implies that
(5.6)
Let eh(t) = Uh(t) − uh(t) be the error of discretization. Using Taylor′s expansion, we have for t ∈ (0, t(h)),
(5.7)
where θI(t) is an intermediate value between UI(t) and u(xI, t) and ξi(t) the one between Ui(t) and u(xi, t). Using (5.3) and (5.6), there exist two positive constants K and L such that
(5.8)
Consider the function where M, C, Q are constants which will be determined later. We get
(5.9)
By a semidiscretization of the above problem, we may choose M, C, Q large enough that
(5.10)
It follows from Lemma 3.2 that
(5.11)
By the same way, we also prove that
(5.12)
which implies that
(5.13)
We deduce that
(5.14)
Let us show that t(h) = T. Suppose that T > t(h). From (5.4), we obtain
(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 t(h) = 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 uC4,1([0, 1]×[0, Tb)) and the initial condition at (2.3) satisfies

(5.16)
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 and we have the following relation
(5.17)

Proof. Let ε > 0. There exists a positive constant N such that

(5.18)
Since the solution u blows up at the time Tb, then there exists T1 ∈ (Tbε/2, Tb) such that ∥u(⋅, t)∥ ≥ 2N for t ∈ [T1, Tb). Setting T2 = (T1 + Tb)/2, then we have . It follows from Theorem 5.1 that
(5.19)
Applying the triangle inequality, we get
(5.20)
which leads to ∥Uh(T2)∥N. From Theorem 4.3, Uh(t) blows up at the time . We deduce from Remark 4.4 and (5.18) that
(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 b(t) = 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 and there exist two positive constants C1, C2 such that

(6.1)
where H(s) is the inverse of the function .

Proof. From Theorem 4.3 and Remark 4.4, Uh(t) blows up in a finite time and there exists a constant C2 > 0 such that

(6.2)
From Lemma 4.2, UI−1 < UI. Then using (2.2), we deduce that dUI/dt ≤ (2/h)b(t)g(UI) − aI(t)f(UI), which implies that dUI/dt ≤ (2b(t)/h)g(UI). Integration this inequality over , there exists a positive constant C1 such that
(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

(7.1)
Assume that there exists a positive constant C such
(7.2)
Then the numerical blow-up set is B = {1}.

Proof. Let v(x) = 1 − x2 and define

(7.3)
where δ is small enough. We have
(7.4)
and for tt0, we get
(7.5)
A straightforward computation yields
(7.6)
This implies that there exists α > 0 such that
(7.7)
Using Taylor′s expansion, there exists a constant K > 0 such that
(7.8)
which implies that
(7.9)
The maximum principle implies that
(7.10)
Hence, we get
(7.11)
Therefore Ui(T)<+,   0 ≤ iI − 1, and we have the desired result.

8. Full Discretization

In this section, we consider the problem (1.1) in the case where a(x, t) = 1, b(t) = 1, f(u) = up, g(u) = up with p = const >1. Thus our problem is equivalent to
(8.1)
where p > 1, u0C1([0, 1]), and .
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 ≤ iI and approximate the solution u(x, t) of the problem (8.1) by the solution of the following discrete equations
(8.2)
(8.3)
(8.4)
where n ≥ 0, φi+1φi, 0 ≤ iI − 1,
(8.5)
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 take , 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. Let be the solution of (8.2)–(8.4). Then we have

(8.6)

Proof. Let , 0 ≤ iI − 1. We observe that

(8.7)
Using the Taylor′s expansion, we find that
(8.8)
where is an intermediate value between and . If , 0 ≤ iI − 1, we deduce that
(8.9)
Using the restriction , we find that
(8.10)
We observe that 1 − 3(Δtn/h2) − pτ is nonnegative and by induction, we deduce that , 0 ≤ iI − 1. This ends the proof.

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

Lemma 8.2. Let be a bounded vector and let a sequence such that

(8.11)
(8.12)
Then for n ≥ 0, 0 ≤ iI if .

Proof. If then a routine computation yields

(8.13)
Since , we see that is nonnegative. From (8.12), we deduce by induction that which ends the proof.

A direct consequence of the above result is the following comparison lemma. Its proof is straightforward.

Lemma 8.3. Suppose that and two vectors such that is bounded. Let and two sequences such that

(8.14)
Then for n ≥ 0, 0 ≤ iI if .

Now, let us give a property of the operator δt.

Lemma 8.4. Let U(n) be such that U(n) ≥ 0 for n ≥ 0. Then we have

(8.15)

Proof. From Taylor′s expansion, we find that

(8.16)
where θ(n) is an intermediate value between U(n) and U(n+1). Use the fact that U(n) ≥ 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.

Definition 8.5. We say that the solution of (8.2)–(8.4) blows up in a finite time if

(8.17)
The number is called the numerical blow-up time of .

The following theorem reveals that the discrete solution of (8.2)–(8.4) blows up in a finite time under some hypotheses.

Theorem 8.6. Let 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

(8.18)
Then blows up in a finite time which satisfies the following estimate
(8.19)
where

Proof. Introduce the vector defined as follows

(8.20)
A straightforward computation yields
(8.21)
Using (8.2), we arrive at
(8.22)
Due to the mean value theorem, we get
(8.23)
where is an intermediate value between and . On the other hand, from Lemmas 2.4 and 2.5, we deduce that
(8.24)
It follows from (8.3) that
(8.25)
which implies that
(8.26)
From (8.18), we observe that . It follows from Lemma 8.2 that which implies that
(8.27)
From Lemma 8.1, we see that which implies that
(8.28)
It is not hard to see that
(8.29)
From (8.28), we get . By induction, we arrive at , which implies that . Therefore, we find that
(8.30)
Consequently, we arrive at
(8.31)
and by induction, we get
(8.32)
Since the term on the right hand side of the above equality tends to infinity as n approaches infinity, we conclude that tends to infinity as n approaches infinity. Now, let us estimate the numerical blow-up time. Due to (8.32), the restriction on the time step ensures that
(8.33)
Using the fact that the series on the right hand side of the above inequality converges towards , we deduce that and the proof is complete.

Remark 8.7. Apply Taylor′s expansion to obtain , which implies that

(8.34)
If we take τ = h2, we see that
(8.35)
We deduce that τ/τ is bounded from above. We conclude that is bounded from above.

Remark 8.8. From (8.31), we get

(8.36)
which implies that
(8.37)
We deduce that
(8.38)

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.

Theorem 9.1. Suppose that the problem (1.1) has a solution uC4,2([0, 1]×[0, T]). Assume that the initial data at (8.4) satisfies

(9.1)
Then the problem (8.2)–(8.4) has a solution for h sufficiently small, 0 ≤ nJ and we have the following relation
(9.2)
where J is such that and .

Proof. For each h, the problem (8.2)–(8.4) has a solution . Let NJ be the greatest value of n such that

(9.3)
We know that N ≥ 1 because of (9.1). Due to the fact that uC4,2, there exists a positive constant K such that ∥uK. Applying the triangle inequality, we have
(9.4)
Since uC4,2, using Taylor′s expansion, we find that
(9.5)
Let be the error of discretization. From the mean value theorem, we get
(9.6)
where is an intermediate value between u(xi, tn) and . Hence, there exist positive constants L and K such that
(9.7)
Consider the function where M, C, Q are positive constants which will be determined later. We get
(9.8)
By a discretization of the above problem, we obtain
(9.9)
We may choose M, C, Q large enough that
(9.10)
It follows from Comparison Lemma 8.3 that
(9.11)
By the same way, we also prove that
(9.12)
which implies that
(9.13)
Let us show that N = J. Suppose that N < J. From (9.3), we obtain
(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.

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 uC4,2([0, 1]×[0, T0)). Assume that the initial data at (2.3) satisfies

(9.15)
Under the assumption of Theorem 8.6, the problem (8.2)–(8.4) has a solution which blows up in a finite time and the following relation holds
(9.16)

Proof. We know from Remark 8.7 that is bounded. Letting ε > 0, there exists a constant R > 0 such that

(9.17)
Since u blows up at the time T0, there exists T1 ∈ (T0ε/2, T0) such that ∥u(⋅, t)∥ ≥ 2R for t ∈ [T1, T0). Let T2 = (T1 + T0)/2 and let q be a positive integer such that for h small enough. We have 0 < ∥uh(tn)∥ < for nq. It follows from Theorem 4.3 that the problem (2.1)–(2.3) has a solution which obeys for nq, which implies that
(9.18)
From Theorem 8.6, blows up at the time . It follows from Remark 8.8 and (9.17) that because . We deduce that , 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 a(x, t) = λ > 0, f(u) = up, g(u) = uq, b(t) = 1 with p = const >1, q = const >1. We approximate the solution u of (1.1) by the solution of the following explicit scheme
(10.1)
We also approximate the solution u of (1.1) by the solution of the implicit scheme below
(10.2)
For the time step, we take n ≥ 0, for the explicit scheme and for the implicit scheme.
The problem described in (10.1) may be rewritten as follows
(10.3)
Let us notice that the restriction on the time step Δtnh2/2 ensures the nonnegativity of the discrete solution.
The implicit scheme may be rewritten in the following form
(10.4)
where
(10.5)
The matrix satisfies the following properties
(10.6)
It follows that exists for n ≥ 0. In addition, since is nonnegative, is also nonnegative for n ≥ 0. We need the following definition.

Definition 10.1. We say that the discrete solution of the explicit scheme or the implicit scheme blows up in a finite time if and the series converges. The quantity is called the numerical blow-up time of the solution .

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 which is computed at the first time when
(10.7)
The order (s) of the method is computed from
(10.8)
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).
I Tn n CPU time s
16 0.047927 451
32 0.044695 1260 0.5
64 0.043583 4075 5 1.54
128 0.043225 14555 60 1.64
256 0.043115 55061 1816 1.71
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 Tn n CPU time s
16 0.047631 423
32 0.044645 1234 1
64 0.043576 4050 5 1.49
128 0.043224 14533 99 1.61
256 0.043113 55035 2000 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 Tn n CPU time s
16 0.018286 21750 3
32 0.017181 83838 17
64 0.016729 329960 108 1.30
128 0.016412 1298750 1570 0.51
256 0.016324 6447649 27049 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 Tn n CPU time s
16 0.018283 21741 6
32 0.017181 83831 37
64 0.016729 3299953 347 1.30
128 0.016617 1208495 4640 2.01
256 0.016526 6348765 29957 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 Tn n CPU time s
16 0.024197 1649
32 0.022570 6103 2
64 0.021950 23583 8 1.40
128 0.021734 92985 200 1.52
256 0.021712 369250 3243 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 Tn n CPU time s
16 0.024169 1602
32 0.022566 6066 5
64 0.021950 23551 65 1.38
128 0.021734 92985 1140 1.52
256 0.021713 370240 6709 3.37
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 Tn n CPU time s
16 0.054342 422
32 0.050346 1130
64 0.049027 3539 4 1.60
128 0.048615 12020 28 1.68
256 0.048491 46439 937 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 Tn n CPU time s
16 0.054158 364
32 0.050332 1077 0.6
64 0.049030 3491 7 1.56
128 0.048616 12358 79 1.66
256 0.048519 36919 1123 2.10

Case 1. p = 0, q = 2, φi = 10 + 10  * cos (πih), λ = 1.

Case 2. p = 2, q = 4, φi = 10 + 10  * cos (πih), λ = 1.

Case 3. p = 2, q = 3, φi = 10 + 10  * cos (πih), λ = 1.

Case 4. p = 2, q = 2, φi = 10 + 10  * cos (πih), 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, 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.

Details are in the caption following the image
Evolution of the discrete solution, q = 2, p = 2 (explicit scheme).
Details are in the caption following the image
Evolution of the discrete solution, q = 2, p = 2 (implicit scheme).
Details are in the caption following the image
Evolution of the discrete solution, q = 3, p = 2 (explicit scheme).
Details are in the caption following the image
Evolution of the discrete solution, q = 3, p = 2 (implicit scheme).
Details are in the caption following the image
Evolution of the discrete solution, q = 4, p = 2 (explicit scheme).
Details are in the caption following the image
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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