Journal of Applied Mathematics
Volume 2008 (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

1Institut National Polytechnique Houphouët-Boigny de Yamoussoukro, BP 1093, Yamoussoukro, Cote D'Ivoire
2Département de Mathématiques et Informatiques, Université d'Abobo-Adjamé, UFR-SFA, 16 BP 372 Abidjan 16, Cote D'Ivoire

Received 29 April 2008; Revised 15 December 2008; Accepted 29 December 2008

Academic Editor: Jacek Rokicki

Copyright © 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.

Abstract

We obtain some conditions under which the positive solution for semidiscretizations of the semilinear equation 𝑢 𝑡 = 𝑢 𝑥 𝑥 − 𝑎 ( 𝑥 , 𝑡 ) 𝑓 ( 𝑢 ) , 0 < 𝑥 < 1 , 𝑡 ∈ ( 0 , 𝑇 ) , with boundary conditions 𝑢 𝑥 ( 0 , 𝑡 ) = 0 , 𝑢 𝑥 ( 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: 𝑢 𝑡 − 𝑢 𝑥 𝑥 𝑢 = − 𝑎 ( 𝑥 , 𝑡 ) 𝑓 ( 𝑢 ) , 0 < 𝑥 < 1 , 𝑡 ∈ ( 0 , 𝑇 ) , 𝑥 ( 0 , 𝑡 ) = 0 , 𝑢 𝑥 ( 1 , 𝑡 ) = 𝑏 ( 𝑡 ) 𝑔 𝑢 ( 1 , 𝑡 ) , 𝑡 ∈ ( 0 , 𝑇 ) , 𝑢 ( 𝑥 , 0 ) = 𝑢 0 ( 𝑥 ) ≥ 0 , 0 ≤ 𝑥 ≤ 1 , ( 1 . 1 ) where 𝑓 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) is a 𝐶 1 function, 𝑓 ( 0 ) = 0 ,   𝑔 ∶ [ 0 , ∞ ) → [ 0 , ∞ ) is a 𝐶 1 convex function, 𝑔 ( 0 ) = 0 , 𝑎 ∈ 𝐶 0 ( [ 0 , 1 ] × ℝ + ) , 𝑎 ( 𝑥 , 𝑡 ) ≥ 0 in [ 0 , 1 ] × ℝ + , 𝑎 𝑡 ( 𝑥 , 𝑡 ) ≤ 0 in [ 0 , 1 ] × ℝ + , 𝑏 ∈ 𝐶 1 ( ℝ + ) , 𝑏 ( 𝑡 ) > 0 in ℝ + , 𝑏 ( 𝑡 ) ≥ 0 in ℝ + . The initial data 𝑢 0 ∈ 𝐶 2 ( [ 0 , 1 ] ) , 𝑢 0 ( 0 ) = 0 , 𝑢 0 ( 1 ) = 𝑏 ( 1 ) 𝑔 ( 𝑢 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 l i m 𝑡 → 𝑇 ‖ ‖ ‖ ‖ 𝑢 ( ⋅ , 𝑡 ) ∞ = + ∞ , ( 1 . 2 ) where ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ∞ = m a x 0 ≤ 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 [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 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 ) = 𝑢 𝑝 , 𝑔 ( 𝑢 ) = 𝑢 𝑞 . 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 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 𝑥 𝑖 = 𝑖 ℎ , 0 ≤ 𝑖 ≤ 𝐼 , where ℎ = 1 / 𝐼 . We approximate the solution 𝑢 of (1.1) by the solution 𝑈 ℎ ( 𝑡 ) = ( 𝑈 0 ( 𝑡 ) , 𝑈 1 ( 𝑡 ) , … , 𝑈 𝐼 ( 𝑡 ) ) 𝑇 of the following semidiscrete equations 𝑑 𝑈 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑈 𝑖 ( 𝑡 ) = − 𝑎 𝑖 𝑈 ( 𝑡 ) 𝑓 𝑖 ( 𝑡 ) , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 , ( 2 . 1 ) 𝑑 𝑈 𝐼 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑈 𝐼 2 ( 𝑡 ) = ℎ 𝑈 𝑏 ( 𝑡 ) 𝑔 𝐼 ( 𝑡 ) − 𝑎 𝐼 𝑈 ( 𝑡 ) 𝑓 𝐼 ( 𝑡 ) , 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 𝑈 , ( 2 . 2 ) 𝑖 ( 0 ) = 𝜑 𝑖 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 , ( 2 . 3 ) 𝜑 𝑖 + 1 ≥ 𝜑 𝑖 0 ≤ 𝑖 ≤ 𝐼 − 1 where 𝛿 2 𝑈 0 ( 𝑡 ) = 2 𝑈 1 ( 𝑡 ) − 2 𝑈 0 ( 𝑡 ) ℎ 2 , 𝛿 2 𝑈 𝐼 ( 𝑡 ) = 2 𝑈 𝐼 − 1 ( 𝑡 ) − 2 𝑈 𝐼 ( 𝑡 ) ℎ 2 , 𝛿 2 𝑈 𝑖 𝑈 ( 𝑡 ) = 𝑖 + 1 ( 𝑡 ) − 2 𝑈 𝑖 ( 𝑡 ) + 𝑈 𝑖 − 1 ( 𝑡 ) ℎ 2 . ( 2 . 4 ) , ( 0 , 𝑇 ℎ 𝑏 ) , ‖ 𝑈 ℎ ( 𝑡 ) ‖ ∞ Here ‖ 𝑈 ℎ ( 𝑡 ) ‖ ∞ = m a x 0 ≤ 𝑖 ≤ 𝐼 𝑈 𝑖 ( 𝑡 ) is the maximal time interval on which 𝑇 ℎ 𝑏 is finite where 𝑈 ℎ ( 𝑡 ) . When 𝑇 ℎ 𝑏 is finite, we say that the solution 𝑈 ℎ ( 𝑡 ) blows up in a finite time and the time 𝑎 ℎ ( 𝑡 ) ∈ 𝐶 0 ( [ 0 , 𝑇 ) , ℝ 𝐼 + 1 ) is called the blow-up time of the solution 𝑉 ℎ ( 𝑡 ) ∈ 𝐶 1 ( [ 0 , 𝑇 ) , ℝ 𝐼 + 1 ) .

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 𝑑 𝑉 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑉 𝑖 ( 𝑡 ) + 𝑎 𝑖 ( 𝑡 ) 𝑉 𝑖 𝑉 ( 𝑡 ) ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 , 𝑡 ∈ ( 0 , 𝑇 ) , 𝑖 ( 0 ) ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 . ( 3 . 1 ) and let 𝑉 𝑖 ( 𝑡 ) ≥ 0 such that 0 ≤ 𝑖 ≤ 𝐼 Then we have 𝑡 ∈ ( 0 , 𝑇 ) , 𝑇 0 < 𝑇 , 𝑍 ℎ ( 𝑡 ) = 𝑒 𝜆 𝑡 𝑉 ℎ ( 𝑡 ) .

Proof. Let 𝜆 and define the vector 𝑎 𝑖 ( 𝑡 ) − 𝜆 > 0 where 𝑡 ∈ [ 0 , 𝑇 0 ] is large enough that 0 ≤ 𝑖 ≤ 𝐼 for 𝑚 = m i n 0 ≤ 𝑖 ≤ 𝐼 , 0 ≤ 𝑡 ≤ 𝑇 0 𝑍 𝑖 ( 𝑡 ) , 𝑖 ∈ { 0 , … , 𝐼 } . Let 𝑍 𝑖 ( 𝑡 ) . Since for 𝑡 0 ∈ [ 0 , 𝑇 0 ] , 𝑚 = 𝑍 𝑖 0 ( 𝑡 0 ) is a continuous function, there exists 𝑖 0 ∈ { 0 , … , 𝐼 } such that 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 = l i m 𝑘 → 0 𝑍 𝑖 0 𝑡 0 − 𝑍 𝑖 0 𝑡 0 − 𝑘 𝑘 𝛿 ≤ 0 , 2 𝑍 𝑖 0 𝑡 0 = 𝑍 𝑖 0 + 1 𝑡 0 − 2 𝑍 𝑖 0 𝑡 0 + 𝑍 𝑖 0 − 1 𝑡 0 ℎ 2 ≥ 0 i f 1 ≤ 𝑖 0 𝛿 ≤ 𝐼 − 1 , 2 𝑍 𝑖 0 𝑡 0 = 2 𝑍 1 𝑡 0 − 2 𝑍 0 𝑡 0 ℎ 2 ≥ 0 i f 𝑖 0 𝛿 = 0 , 2 𝑍 𝑖 0 𝑡 0 = 2 𝑍 𝐼 − 1 𝑡 0 − 2 𝑍 𝐼 𝑡 0 ℎ 2 ≥ 0 i f 𝑖 0 = 𝐼 . ( 3 . 2 ) for a certain 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 − 𝛿 2 𝑍 𝑖 0 𝑡 0 + 𝑎 𝑖 0 𝑡 0 𝑍 − 𝜆 𝑖 0 𝑡 0 ≥ 0 . ( 3 . 3 ) . It is not hard to see that ( 𝑎 𝑖 0 ( 𝑡 0 ) − 𝜆 ) 𝑍 𝑖 0 ( 𝑡 0 ) ≥ 0 A straightforward computation reveals that 𝑍 𝑖 0 ( 𝑡 0 ) ≥ 0 We observe from (3.2) that 𝑎 𝑖 0 ( 𝑡 0 ) − 𝜆 > 0 which implies that 𝑉 ℎ ( 𝑡 ) ≥ 0 because 𝑡 ∈ [ 0 , 𝑇 0 ] . We deduce that 𝑉 ℎ ( 𝑡 ) for 𝑈 ℎ ( 𝑡 ) ∈ 𝐶 1 ( [ 0 , 𝑇 ) , ℝ 𝐼 + 1 ) and the proof is complete.

Another form of the maximum principle for semidiscrete equations is the following comparison lemma.Lemma 3.2. Let 𝑓 ∈ 𝐶 0 ( ℝ × ℝ , ℝ ) , 𝑡 ∈ ( 0 , 𝑇 ) and 𝑑 𝑉 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑉 𝑖 𝑉 ( 𝑡 ) + 𝑓 𝑖 < ( 𝑡 ) , 𝑡 𝑑 𝑈 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑈 𝑖 𝑈 ( 𝑡 ) + 𝑓 𝑖 𝑉 ( 𝑡 ) , 𝑡 , 0 ≤ 𝑖 ≤ 𝐼 , ( 3 . 4 ) 𝑖 ( 0 ) < 𝑈 𝑖 ( 0 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 3 . 5 ) such that for 𝑉 𝑖 ( 𝑡 ) < 𝑈 𝑖 ( 𝑡 ) 0 ≤ 𝑖 ≤ 𝐼 𝑡 ∈ ( 0 , 𝑇 ) . Then we have 𝑍 ℎ ( 𝑡 ) = 𝑈 ℎ ( 𝑡 ) − 𝑉 ℎ ( 𝑡 ) , 𝑡 0 , 𝑡 ∈ ( 0 , 𝑇 )

Proof. Define the vector 𝑍 𝑖 ( 𝑡 ) > 0 . Let 𝑡 ∈ [ 0 , 𝑡 0 ) be the first 0 ≤ 𝑖 ≤ 𝐼 such that 𝑍 𝑖 0 ( 𝑡 0 ) = 0 for 𝑖 0 ∈ { 0 , … , 𝐼 } , 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 = l i m 𝑘 → 0 𝑍 𝑖 0 𝑡 0 − 𝑍 𝑖 0 𝑡 0 − 𝑘 𝑘 𝛿 ≤ 0 , 2 𝑍 𝑖 0 𝑡 0 = 𝑍 𝑖 0 + 1 𝑡 0 − 2 𝑍 𝑖 0 𝑡 0 + 𝑍 𝑖 0 − 1 𝑡 0 ℎ 2 ≥ 0 i f 1 ≤ 𝑖 0 𝛿 ≤ 𝐼 − 1 , 2 𝑍 𝑖 0 𝑡 0 = 2 𝑍 1 𝑡 0 − 2 𝑍 0 𝑡 0 ℎ 2 ≥ 0 i f 𝑖 0 𝛿 = 0 , 2 𝑍 𝑖 0 𝑡 0 = 2 𝑍 𝐼 − 1 𝑡 0 − 2 𝑍 𝐼 𝑡 0 ℎ 2 ≥ 0 i f 𝑖 0 = 𝐼 , ( 3 . 6 ) , but 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 − 𝛿 2 𝑍 𝑖 0 𝑡 0 𝑈 + 𝑓 𝑖 0 𝑡 0 , 𝑡 0 𝑉 − 𝑓 𝑖 0 𝑡 0 , 𝑡 0 ≤ 0 . ( 3 . 7 ) for a certain 𝑈 ℎ . We observe that 𝛿 2 which implies that 𝑈 ℎ ∈ ℝ 𝐼 + 1 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 𝑈 ℎ ≥ 0 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 𝛿 2 𝑔 𝑈 𝑖 ≥ 𝑔 𝑈 𝑖 𝛿 2 𝑈 𝑖 f o r 0 ≤ 𝑖 ≤ 𝐼 . ( 4 . 1 ) and the second one reveals a property of the semidiscrete solution.Lemma 4.1. Let 𝑔 𝑈 1 𝑈 = 𝑔 0 + 𝑈 1 − 𝑈 0 𝑔 𝑈 0 + 𝑈 1 − 𝑈 0 2 2 𝑔 𝜂 0 , 𝑔 𝑈 𝑖 + 1 𝑈 = 𝑔 𝑖 + 𝑈 𝑖 + 1 − 𝑈 𝑖 𝑔 𝑈 𝑖 + 𝑈 𝑖 + 1 − 𝑈 𝑖 2 2 𝑔 𝜃 𝑖 𝑔 𝑈 , 1 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑖 − 1 𝑈 = 𝑔 𝑖 + 𝑈 𝑖 − 1 − 𝑈 𝑖 𝑔 𝑈 𝑖 + 𝑈 𝑖 − 1 − 𝑈 𝑖 2 2 𝑔 𝜂 𝑖 𝑔 𝑈 , 1 ≤ 𝑖 ≤ 𝐼 − 1 , 𝐼 − 1 𝑈 = 𝑔 𝐼 + 𝑈 𝐼 − 1 − 𝑈 𝐼 𝑔 𝑈 𝐼 + 𝑈 𝐼 − 1 − 𝑈 𝐼 2 2 𝑔 𝜂 𝐼 , ( 4 . 2 ) be such that 𝜃 𝑖 . Then we have 𝑈 𝑖

Proof. Apply Taylor's expansion to obtain 𝑈 𝑖 + 1 where 𝜂 𝑖 is an intermediate between 𝑈 𝑖 − 1 and 𝑈 𝑖 and 𝛿 2 𝑔 𝑈 0 = 𝑔 𝑈 0 𝛿 2 𝑈 0 + 𝑈 1 − 𝑈 0 2 ℎ 2 𝑔 𝜂 0 , 𝛿 2 𝑔 𝑈 𝐼 = 𝑔 𝑈 𝐼 𝛿 2 𝑈 𝐼 + 𝑈 𝐼 − 1 − 𝑈 𝐼 2 ℎ 2 𝑔 𝜂 𝐼 . ( 4 . 3 ) the one between 𝛿 2 𝑔 𝑈 𝑖 = 𝑔 𝑈 𝑖 𝛿 2 𝑈 𝑖 + 𝑈 𝑖 + 1 − 𝑈 𝑖 2 2 ℎ 2 𝑔 𝜃 𝑖 + 𝑈 𝑖 − 1 − 𝑈 𝑖 2 2 ℎ 2 𝑔 𝜂 𝑖 , 1 ≤ 𝑖 ≤ 𝐼 − 1 . ( 4 . 4 ) and 𝑔 ( 𝑠 ) ≥ 0 . The first and last equalities imply that 𝑠 ≥ 0 Combining the second and third equalities, we see that 𝑈 ℎ ≥ 0 Use the fact that 𝑈 ℎ for 𝑈 𝑖 + 1 ( 𝑡 ) > 𝑈 𝑖 ( 𝑡 ) , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 . ( 4 . 5 ) and 𝑡 0 to complete the rest of the proof.

Lemma 4.2. Let 𝑡 > 0 be the solution of (2.1)–(2.3). Then we have 𝑈 𝑖 + 1 ( 𝑡 ) > 𝑈 𝑖 ( 𝑡 )

Proof. Let 0 ≤ 𝑖 ≤ 𝐼 − 1 be the first 𝑈 𝑖 0 + 1 ( 𝑡 0 ) = 𝑈 𝑖 0 ( 𝑡 0 ) such that 𝑖 0 ∈ { 0 , … , 𝐼 − 1 } for 𝑖 0 but 𝑍 𝑖 ( 𝑡 ) = 𝑈 𝑖 + 1 ( 𝑡 ) − 𝑈 𝑖 ( 𝑡 ) for a certain 0 ≤ 𝑖 ≤ 𝐼 − 1 . Without loss of generality, we may suppose that 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 = l i m 𝑘 → 0 𝑍 𝑖 0 𝑡 0 − 𝑍 𝑖 0 𝑡 0 − 𝑘 𝑘 𝛿 ≤ 0 , 2 𝑍 𝑖 0 𝑡 0 = 𝑍 𝑖 0 + 1 𝑡 0 − 2 𝑍 𝑖 0 𝑡 0 + 𝑍 𝑖 0 − 1 𝑡 0 ℎ 2 > 0 i f 1 ≤ 𝑖 0 𝛿 ≤ 𝐼 − 2 , 2 𝑍 𝑖 0 𝑡 0 = 𝛿 2 𝑍 0 𝑡 0 = 𝑍 1 𝑡 0 − 3 𝑍 0 𝑡 0 ℎ 2 > 0 i f 𝑖 0 𝛿 = 0 , 2 𝑍 𝑖 0 𝑡 0 = 𝛿 2 𝑍 𝐼 − 1 𝑡 0 = 𝑍 𝐼 − 2 𝑡 0 − 3 𝑍 𝐼 − 1 𝑡 0 ℎ 2 > 0 i f 𝑖 0 = 𝐼 − 1 , ( 4 . 6 ) is the smallest integer which satisfies the equality. Introduce the functions 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 − 𝛿 2 𝑍 𝑖 0 𝑡 0 − 𝑎 𝑖 0 + 1 𝑡 0 𝑓 𝑈 𝑖 0 + 1 ( 𝑡 0 + 𝑎 𝑖 0 𝑡 0 𝑓 𝑈 𝑖 0 𝑡 0 < 0 i f 0 ≤ 𝑖 0 ≤ 𝐼 − 2 , 𝑑 𝑍 𝑖 0 𝑡 0 𝑑 𝑡 − 𝛿 2 𝑍 𝑖 0 𝑡 0 + 2 ℎ 𝑏 𝑡 0 𝑔 𝑖 0 + 1 𝑡 0 − 𝑎 𝑖 0 + 1 𝑡 0 𝑓 𝑈 𝑖 0 + 1 ( 𝑡 0 + 𝑎 𝑖 0 𝑡 0 𝑓 𝑈 𝑖 0 𝑡 0 < 0 i f 𝑖 0 = 𝐼 − 1 . ( 4 . 7 ) for 𝑥 𝐼 . We get 𝑈 ℎ which implies that 𝛿 2 𝜑 𝑖 − 𝑎 𝑖 𝜑 ( 0 ) 𝑓 𝑖 𝛿 ≥ 0 , 1 ≤ 𝑖 ≤ 𝐼 − 1 , 2 𝜑 𝐼 − 𝑎 𝐼 𝜑 ( 0 ) 𝑓 𝐼 + 𝑏 ( 0 ) 𝑔 𝐼 𝜑 𝐼 𝜑 ≥ 𝐴 𝑔 𝐼 . ( 4 . 8 ) 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 𝑓 ( 𝑠 ) 𝑔 ( 𝑠 ) − 𝑓 ( 𝑠 ) 𝑔 ( 𝑠 ) ≥ 0 f o r 𝑠 ≥ 0 . ( 4 . 9 ) .

Now, we are in a position to state the main result of this section.Theorem 4.3. Let 𝑈 ℎ be the solution of (2.1)–(2.3). Suppose that there exists a positive integer A such that 𝑇 ℎ 𝑏 Assume that 𝑇 ℎ 𝑏 ≤ 1 𝐴 + ∞ ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ 𝑑 𝑠 𝑔 ( 𝑠 ) . ( 4 . 1 0 ) Then the solution ( 0 , 𝑇 ℎ 𝑏 ) blows up in a finite time ‖ 𝑈 ℎ ( 𝑡 ) ‖ ∞ < ∞ and we have the following estimate 𝑇 ℎ 𝑏

Proof. Since 𝐽 ℎ is the maximal time interval on which 𝐽 𝑖 ( 𝑡 ) = 𝑑 𝑈 𝑖 ( 𝑡 ) 𝑑 𝑡 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝐽 𝐼 ( 𝑡 ) = 𝑑 𝑈 𝐼 ( 𝑡 ) 𝑈 𝑑 𝑡 − 𝐴 𝑔 𝐼 ( 𝑡 ) . ( 4 . 1 1 ) , our aim is to show that 𝑑 𝐽 𝑖 𝑑 𝑡 − 𝛿 2 𝐽 𝑖 = 𝑑 𝑑 𝑡 𝑑 𝑈 𝑖 𝑑 𝑡 − 𝛿 2 𝑈 𝑖 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑑 𝐽 𝐼 𝑑 𝑡 − 𝛿 2 𝐽 𝐼 = 𝑑 𝑑 𝑡 𝑑 𝑈 𝐼 𝑑 𝑡 − 𝛿 2 𝑈 𝐼 − 𝐴 𝑔 𝑈 𝐼 𝑑 𝑈 𝐼 𝑑 𝑡 + 𝐴 𝛿 2 𝑔 𝑈 𝐼 . ( 4 . 1 2 ) is finite and satisfies the above inequality. Introduce the vector 𝛿 2 𝑔 ( 𝑈 𝐼 ) ≥ 𝑔 ( 𝑈 𝐼 ) 𝛿 2 𝑈 𝐼 such that 𝑑 𝐽 𝐼 𝑑 𝑡 − 𝛿 2 𝐽 𝐼 ≥ 𝑑 𝑑 𝑡 𝑑 𝑈 𝐼 𝑑 𝑡 − 𝛿 2 𝑈 𝐼 − 𝐴 𝑔 𝑈 𝐼 𝑑 𝑈 𝐼 𝑑 𝑡 − 𝛿 2 𝑈 𝐼 . ( 4 . 1 3 ) A straightforward calculation gives 𝑑 𝐽 𝑖 𝑑 𝑡 − 𝛿 2 𝐽 𝑖 ≥ − 𝑎 𝑖 𝑈 ( 𝑡 ) 𝑓 𝑖 − 𝑎 𝑖 ( 𝑡 ) 𝑓 𝑈 𝑖 𝑑 𝑈 𝑖 𝑑 𝑡 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑑 𝐽 𝐼 𝑑 𝑡 − 𝛿 2 𝐽 𝐼 ≥ − 𝑎 𝐼 𝑈 ( 𝑡 ) 𝑓 𝐼 − 𝑎 𝐼 ( 𝑡 ) 𝑓 𝑈 𝐼 𝑑 𝑈 𝐼 + 2 𝑑 𝑡 ℎ 𝑏 𝑈 ( 𝑡 ) 𝑔 𝐼 + 2 ℎ 𝑏 ( 𝑡 ) 𝑔 𝑈 𝐼 𝑑 𝑈 𝐼 𝑑 𝑡 − 𝐴 𝑔 𝑈 𝐼 − 𝑎 𝐼 𝑈 ( 𝑡 ) 𝑓 𝐼 + 2 ℎ 𝑈 𝑏 ( 𝑡 ) 𝑔 𝐼 . ( 4 . 1 4 ) From Lemma 4.1, we have 𝑎 𝑖 ( 𝑡 ) ≤ 0 which implies that 𝑏 ( 𝑡 ) ≥ 0 Using (2.1), we get 𝑑 𝑈 𝑖 / 𝑑 𝑡 = 𝐽 𝑖 + 𝐴 𝑔 ( 𝑈 𝑖 ) It follows from the fact that 𝑑 𝐽 𝐼 𝑑 𝑡 − 𝛿 2 𝐽 𝐼 ≥ − 𝑎 𝐼 ( 𝑡 ) 𝑓 𝑈 𝐼 + 2 ℎ 𝑏 ( 𝑡 ) 𝑔 𝑈 𝐼 𝐽 𝐼 + 𝐴 𝑎 𝐼 𝑔 ( 𝑡 ) 𝑈 𝐼 𝑓 𝑈 𝐼 − 𝑓 𝑈 𝐼 𝑔 𝑈 𝐼 . ( 4 . 1 5 ) , 𝑑 𝐽 𝑖 𝑑 𝑡 − 𝛿 2 𝐽 𝑖 ≥ − 𝑎 𝑖 ( 𝑡 ) 𝑓 𝑈 𝑖 𝐽 𝑖 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑑 𝐽 𝐼 𝑑 𝑡 − 𝛿 2 𝐽 𝐼 ≥ − 𝑎 𝐼 ( 𝑡 ) 𝑓 𝑈 𝐼 + 2 ℎ 𝑏 ( 𝑡 ) 𝑔 𝑈 𝐼 𝐽 𝐼 . ( 4 . 1 6 ) and 𝐽 𝑖 ( 0 ) = 𝛿 2 𝜑 𝑖 − 𝑎 𝑖 𝜑 ( 0 ) 𝑓 𝑖 𝐽 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝐼 ( 0 ) = 𝛿 2 𝜑 𝐼 − 𝑎 𝐼 𝜑 ( 0 ) 𝑓 𝐼 + 𝑏 ( 0 ) 𝑔 𝐼 𝜑 𝐼 𝜑 − 𝐴 𝑔 𝐼 ≥ 0 . ( 4 . 1 7 ) that 𝐽 𝑖 ( 𝑡 ) ≥ 0 We deduce from (4.9) that 0 ≤ 𝑖 ≤ 𝐼 From (4.8), we observe that 𝑑 𝑈 𝐼 / 𝑑 𝑡 ≥ 𝑔 ( 𝑈 𝐼 ) We deduce from Lemma 3.1 that 0 ≤ 𝑖 ≤ 𝐼 , 𝑑 𝑈 𝐼 𝑔 𝑈 𝐼 ≥ 𝐴 𝑑 𝑡 . ( 4 . 1 8 ) , which implies that ( 𝑡 , 𝑇 ℎ 𝑏 ) , 𝑇 ℎ 𝑏 1 − 𝑡 ≤ 𝐴 𝑈 + ∞ 𝐼 ( 𝑡 ) 𝑑 𝑠 𝑔 ( 𝑠 ) , ( 4 . 1 9 ) . Obviously we have 𝑇 ℎ 𝑏 ≤ 1 𝐴 + ∞ ‖ ‖ 𝑈 ℎ ( 0 ) ‖ ‖ ∞ 𝑑 𝑠 𝑔 ( 𝑠 ) . ( 4 . 2 0 ) Integrating this inequality over 𝑈 ℎ , we arrive at ‖ 𝑈 ℎ ( 0 ) ‖ ∞ = ‖ 𝜑 ℎ ‖ ∞ which implies that 𝑇 ℎ 𝑏 − 𝑡 0 ≤ 1 𝐴 ∫ + ∞ ‖ ‖ 𝑈 ℎ ( 𝑡 0 ) ‖ ‖ ∞ 𝑑 𝑠 𝑔 ( 𝑠 ) i f 0 < 𝑡 0 < 𝑇 ℎ 𝑏 , 𝑈 𝑖 𝐴 𝑇 ( 𝑡 ) ≤ 𝐻 ℎ 𝑏 − 𝑡 , 0 ≤ 𝑖 ≤ 𝐼 , ( 4 . 2 1 ) Since the quantity on the right hand side of the above inequality is finite, we deduce that the solution 𝐻 ( 𝑠 ) blows up in a finite time. Use the fact that ∫ 𝐺 ( 𝑠 ) = 𝑠 + ∞ ( 𝑑 𝑧 / 𝑔 ( 𝑧 ) ) to complete the rest of the proof.

Remark 4.4. The inequality (4.19) implies that 𝑔 ( 𝑠 ) = 𝑠 𝑞 where 𝐺 ( 𝑠 ) = 𝑠 1 − 𝑞 / ( 𝑞 − 1 ) is the inverse of 𝐻 ( 𝑠 ) = ( ( 𝑞 − 1 ) 𝑠 ) 1 / ( 1 − 𝑞 ) .

Remark 4.5. If [ 0 , 𝑇 ] , then 𝑢 and 𝑈 ℎ ( 𝑡 ) .

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 𝑢 where ℎ is defined, the solution 𝑢 ∈ 𝐶 4 , 1 ( [ 0 , 1 ] × [ 0 , 𝑇 ] ) of (2.1)–(2.3) approximates ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ = 𝑜 ( 1 ) a s ℎ → 0 , ( 5 . 1 ) , when the mesh parameter 𝑢 ℎ ( 𝑡 ) = ( 𝑢 ( 𝑥 0 , 𝑡 ) , … , 𝑢 ( 𝑥 𝐼 , 𝑡 ) ) 𝑇 goes to zero.Theorem 5.1. Assume that (1.1) has a solution 𝑈 ℎ ∈ 𝐶 1 ( [ 0 , 𝑇 ] , ℝ 𝐼 + 1 ) and the initial condition at (2.3) satisfies m a x 0 ≤ 𝑡 ≤ 𝑇 ‖ ‖ 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ ‖ ‖ ( 𝑡 ) ∞ ‖ ‖ 𝜑 = 𝑂 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + ℎ 2 a s ℎ → 0 . ( 5 . 2 ) where 𝛼 > 0 . Then, for h sufficiently small, the problem (2.1)–(2.3) has a unique solution ‖ ‖ ‖ ‖ 𝑢 ( ⋅ , 𝑡 ) ∞ ≤ 𝛼 f o r 𝑡 ∈ [ 0 , 𝑇 ] . ( 5 . 3 ) such that ℎ

Proof. Let 𝑈 ℎ ∈ 𝐶 1 ( [ 0 , 𝑇 ℎ 𝑏 ) , ℝ 𝐼 + 1 ) be such that 𝑡 ( ℎ ) ≤ m i n { 𝑇 , 𝑇 ℎ 𝑏 } The problem (2.1)–(2.3) has for each 𝑡 > 0 , a unique solution ‖ ‖ 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ ‖ ‖ ( 𝑡 ) ∞ < 1 f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) . ( 5 . 4 ) . Let 𝑡 ( ℎ ) > 0 the greatest value of ℎ such that ‖ ‖ 𝑈 ℎ ‖ ‖ ( 𝑡 ) ∞ ≤ ‖ ‖ ‖ ‖ 𝑢 ( ⋅ , 𝑡 ) ∞ + ‖ ‖ 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ ‖ ‖ ( 𝑡 ) ∞ f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) , ( 5 . 5 ) The relation (5.1) implies that ‖ ‖ 𝑈 ℎ ‖ ‖ ( 𝑡 ) ∞ ≤ 1 + 𝛼 f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) . ( 5 . 6 ) for 𝑒 ℎ ( 𝑡 ) = 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ ( 𝑡 ) sufficiently small. By the triangle inequality, we obtain 𝑡 ∈ ( 0 , 𝑡 ( ℎ ) ) which implies that 𝑑 𝑒 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑒 𝑖 ( 𝑡 ) = − 𝑎 𝑖 ( 𝑡 ) 𝑓 𝜉 𝑖 𝑒 ( 𝑡 ) 𝑖 ℎ ( 𝑡 ) + 𝑜 2 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑑 𝑒 𝐼 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑒 𝐼 ( 𝑡 ) = − 𝑎 𝐼 ( 𝑡 ) 𝑓 𝜉 𝐼 𝑒 ( 𝑡 ) 𝐼 2 ( 𝑡 ) + ℎ 𝑏 ( 𝑡 ) 𝑔 𝑈 𝐼 𝑒 ( 𝑡 ) 𝐼 ℎ ( 𝑡 ) + 𝑜 2 , ( 5 . 7 ) Let 𝜃 𝐼 ( 𝑡 ) be the error of discretization. Using Taylor's expansion, we have for 𝑈 𝐼 ( 𝑡 ) , 𝑢 ( 𝑥 𝐼 , 𝑡 ) where 𝜉 𝑖 ( 𝑡 ) is an intermediate value between 𝑈 𝑖 ( 𝑡 ) and 𝑢 ( 𝑥 𝑖 , 𝑡 ) and 𝐾 the one between 𝐿 and 𝑑 𝑒 𝑖 ( 𝑡 ) 𝑑 𝑡 − 𝛿 2 𝑒 𝑖 | | | 𝑒 ( 𝑡 ) ≤ 𝐿 𝑖 | | | ( 𝑡 ) + 𝐾 ℎ 2 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑑 𝑒 𝐼 ( 𝑡 ) − 𝑑 𝑡 2 𝑒 𝐼 − 1 ( 𝑡 ) − 2 𝑒 𝐼 ( 𝑡 ) ℎ 2 ≤ 𝐿 | | | 𝑒 𝐼 | | | ( 𝑡 ) ℎ | | | 𝑒 + 𝐿 𝐼 | | | ( 𝑡 ) + 𝐾 ℎ 2 . ( 5 . 8 ) . Using (5.3) and (5.6), there exist two positive constants 𝑧 ( 𝑥 , 𝑡 ) = 𝑒 ( ( 𝑀 + 1 ) 𝑡 + 𝐶 𝑥 2 ) ( ‖ 𝜑 ℎ − 𝑢 ℎ ( 0 ) ‖ ∞ + 𝑄 ℎ 2 ) and 𝑀 such that 𝐶 Consider the function 𝑄 where 𝑧 𝑡 ( 𝑥 , 𝑡 ) − 𝑧 𝑥 𝑥 ( 𝑥 , 𝑡 ) = ( 𝑀 + 1 − 2 𝐶 − 4 𝐶 2 𝑥 2 𝑧 ) 𝑧 ( 𝑥 , 𝑡 ) , 𝑥 ( 0 , 𝑡 ) = 0 , 𝑧 𝑥 ( 1 , 𝑡 ) = 2 𝐶 𝑧 ( 1 , 𝑡 ) , 𝑧 ( 𝑥 , 0 ) = 𝑒 𝐶 𝑥 2 ( ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 ) . ( 5 . 9 ) , 𝑀 , 𝐶 are constants which will be determined later. We get 𝑄 By a semidiscretization of the above problem, we may choose 𝑑 𝑧 𝑥 𝑑 𝑡 𝑖 , 𝑡 > 𝛿 2 𝑧 𝑥 𝑖 | | | 𝑧 𝑥 , 𝑡 + 𝐿 𝑖 | | | , 𝑡 + 𝐾 ℎ 2 𝑑 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑧 𝑥 𝑑 𝑡 𝐼 , 𝑡 > 𝛿 2 𝑧 𝑥 𝐼 + 𝐿 , 𝑡 ℎ | | | 𝑧 𝑥 𝐼 | | | | | | 𝑧 𝑥 , 𝑡 + 𝐿 𝐼 | | | , 𝑡 + 𝐾 ℎ 2 , 𝑧 𝑥 𝑖 , 0 > 𝑒 𝑖 ( 0 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 5 . 1 0 ) , 𝑧 𝑥 𝑖 , 𝑡 > 𝑒 𝑖 ( 𝑡 ) f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 5 . 1 1 ) , 𝑧 𝑥 𝑖 , 𝑡 > − 𝑒 𝑖 ( 𝑡 ) f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) , 0 ≤ 𝑖 ≤ 𝐼 , ( 5 . 1 2 ) large enough that 𝑧 𝑥 𝑖 > | | | 𝑒 , 𝑡 𝑖 | | | ( 𝑡 ) f o r 𝑡 ∈ 0 , 𝑡 ( ℎ ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 5 . 1 3 ) It follows from Lemma 3.2 that ‖ ‖ 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ ‖ ‖ ( 𝑡 ) ∞ ≤ 𝑒 ( 𝑀 𝑡 + 𝐶 ) ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 , 𝑡 ∈ 0 , 𝑡 ( ℎ ) . ( 5 . 1 4 ) By the same way, we also prove that 𝑡 ( ℎ ) = 𝑇 which implies that 𝑇 > 𝑡 ( ℎ ) We deduce that ‖ ‖ 𝑈 1 = ℎ 𝑡 ( ℎ ) − 𝑢 ℎ ‖ ‖ 𝑡 ( ℎ ) ∞ ≤ 𝑒 ( 𝑀 𝑇 + 𝐶 ) ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 . ( 5 . 1 5 ) Let us show that ℎ . Suppose that 1 ≤ 0 . From (5.4), we obtain 𝑡 ( ℎ ) = 𝑇 Since the term on the right hand side of the above inequality goes to zero as 𝑇 𝑏 tends to zero, we deduce that 𝑢 ∈ 𝐶 4 , 1 ( [ 0 , 1 ] × [ 0 , 𝑇 𝑏 ) ) , which is impossible. Consequently ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ = 𝑜 ( 1 ) a s ℎ → 0 . ( 5 . 1 6 ) , 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 𝑈 ℎ such that 𝑇 ℎ 𝑏 and the initial condition at (2.3) satisfies l i m ℎ → 0 𝑇 ℎ 𝑏 = 𝑇 𝑏 . ( 5 . 1 7 ) Under the assumptions of Theorem 4.3, the problem (2.1)–(2.3) admits a unique solution 𝜀 > 0 which blows up in a finite time 𝑁 and we have the following relation 1 𝐴 𝑥 + ∞ 𝑑 𝑠 ≤ 𝜀 𝑔 ( 𝑠 ) 2 f o r 𝑥 ∈ [ 𝑁 , + ∞ ) . ( 5 . 1 8 )

Proof. Let 𝑢 . There exists a positive constant 𝑇 𝑏 such that 𝑇 1 ∈ ( 𝑇 𝑏 − 𝜀 / 2 , 𝑇 𝑏 ) Since the solution ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ∞ ≥ 2 𝑁 blows up at the time 𝑡 ∈ [ 𝑇 1 , 𝑇 𝑏 ) , then there exists 𝑇 2 = ( 𝑇 1 + 𝑇 𝑏 ) / 2 such that s u p 𝑡 ∈ [ 0 , 𝑇 2 ] | 𝑢 ( 𝑥 , 𝑡 ) | < ∞ for s u p 𝑡 ∈ [ 0 , 𝑇 2 ] | | | 𝑈 ℎ ( 𝑡 ) − 𝑢 ℎ | | | ( 𝑡 ) ∞ ≤ 𝑁 . ( 5 . 1 9 ) . Setting ‖ ‖ 𝑈 ℎ 𝑇 2 ‖ ‖ ∞ ≥ ‖ ‖ 𝑢 ℎ 𝑇 2 ‖ ‖ ∞ − ‖ ‖ 𝑈 ℎ 𝑇 2 − 𝑢 ℎ 𝑇 2 ‖ ‖ ∞ , ( 5 . 2 0 ) , then we have ‖ 𝑈 ℎ ( 𝑇 2 ) ‖ ∞ ≥ 𝑁 . It follows from Theorem 5.1 that 𝑈 ℎ ( 𝑡 ) Applying the triangle inequality, we get 𝑇 ℎ 𝑏 which leads to | | | 𝑇 𝑏 − 𝑇 ℎ 𝑏 | | | ≤ | | | 𝑇 𝑏 − 𝑇 2 | | | + | | | 𝑇 ℎ 𝑏 − 𝑇 2 | | | ≤ 𝜀 2 + 1 𝐴 + ∞ ‖ ‖ 𝑈 ℎ ( 𝑇 2 ) ‖ ‖ ∞ 𝑑 𝑠 𝑔 ( 𝑠 ) ≤ 𝜀 , ( 5 . 2 1 ) . From Theorem 4.3, 𝑈 ℎ blows up at the time 𝑏 ( 𝑡 ) = 1 . We deduce from Remark 4.4 and (5.18) that 𝑈 ℎ ( 𝑡 ) and the proof is complete.

6. Numerical Blow-Up Rate

In this section, we determine the blow-up rate of the solution 𝑈 ℎ ( 𝑡 ) of (2.1)–(2.3) in the case where 𝑇 ℎ 𝑏 . Our result is the following.Theorem 6.1. Let 𝐶 1 , 𝐶 2 be the solution of (2.1)–(2.3). Under the assumptions of Theorem 4.3, 𝐻 𝐶 1 𝑇 ℎ 𝑏 − 𝑡 ≤ 𝑈 𝐼 𝐶 ( 𝑡 ) ≤ 𝐻 2 𝑇 ℎ 𝑏 − 𝑡 , f o r 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 , ( 6 . 1 ) blows up in a finite time 𝐻 ( 𝑠 ) and there exist two positive constants ∫ 𝐺 ( 𝑠 ) = 𝑠 + ∞ ( 𝑑 𝜎 / 𝑔 ( 𝜎 ) ) such that 𝑈 ℎ ( 𝑡 ) where 𝑇 ℎ 𝑏 is the inverse of the function 𝐶 2 > 0 .

Proof. From Theorem 4.3 and Remark 4.4, 𝑈 𝐼 𝐶 ( 𝑡 ) ≤ 𝐻 2 𝑇 ℎ 𝑏 − 𝑡 f o r 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 . ( 6 . 2 ) blows up in a finite time 𝑈 𝐼 − 1 < 𝑈 𝐼 and there exists a constant 𝑑 𝑈 𝐼 / 𝑑 𝑡 ≤ ( 2 / ℎ ) 𝑏 ( 𝑡 ) 𝑔 ( 𝑈 𝐼 ) − 𝑎 𝐼 ( 𝑡 ) 𝑓 ( 𝑈 𝐼 ) such that 𝑑 𝑈 𝐼 / 𝑑 𝑡 ≤ ( 2 𝑏 ( 𝑡 ) / ℎ ) 𝑔 ( 𝑈 𝐼 ) From Lemma 4.2, ( 𝑡 , 𝑇 ℎ 𝑏 ) . Then using (2.2), we deduce that 𝐶 1 , which implies that 𝑈 𝐼 𝐶 ( 𝑡 ) ≥ 𝐻 1 𝑇 ℎ 𝑏 − 𝑡 f o r 𝑡 ∈ 0 , 𝑇 ℎ 𝑏 , ( 6 . 3 ) . Integration this inequality over 𝐶 0 , there exists a positive constant 𝑠 𝐹 ( 𝑠 ) ≤ 𝐶 0 such that 𝑑 𝑈 𝑑 𝑡 𝑖 − 𝛿 2 𝑈 𝑖 ≤ 0 , 0 ≤ 𝑖 ≤ 𝐼 − 1 . ( 7 . 1 ) 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 𝐶 such that 𝑈 𝑖 ≤ 𝐻 𝐶 ( 𝑇 − 𝑡 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 7 . 2 ) and 𝐵 = { 1 } Assume that there exists a positive constant 𝑣 ( 𝑥 ) = 1 − 𝑥 2 such 𝑊 ( 𝑥 , 𝑡 ) = 𝐻 𝛿 𝑣 ( 𝑥 ) + 𝛿 𝐵 ( 𝑇 − 𝑡 ) f o r 0 ≤ 𝑥 ≤ 1 , 𝑡 ≥ 𝑡 0 , ( 7 . 3 ) Then the numerical blow-up set is 𝛿 .

Proof. Let 𝑊 𝑥 ( 0 , 𝑡 ) = 0 , 𝑊 ( 1 , 𝑡 ) = 𝐻 𝛿 𝐵 ( 𝑇 − 𝑡 ) ≥ 𝑢 ( 1 , 𝑡 ) , ( 7 . 4 ) and define 𝑡 ≥ 𝑡 0 where 𝑊 ( 𝑥 , 𝑡 0 ) = 𝐻 𝛿 𝑣 ( 𝑥 ) + 𝛿 ≥ 𝐻 ( 2 𝛿 ) = 𝐻 2 𝛿 𝐵 𝑇 − 𝑡 0 𝐶 ≥ 𝐻 𝑇 − 𝑡 0 ≥ 𝑢 𝑥 , 𝑡 0 . ( 7 . 5 ) is small enough. We have 𝑊 𝑡 ( 𝑥 , 𝑡 ) − 𝑊 𝑥 𝑥 ( 𝑥 , 𝑡 ) = 𝛿 𝐹 𝐻 ( 𝜏 ) 𝐵 − 2 − 4 𝑥 𝐹 𝐻 ( 𝜏 ) ≥ 𝛿 𝐹 𝐻 ( 𝜏 ) 𝐵 − 2 − 4 𝛿 𝐶 0 . ( 7 . 6 ) and for 𝛼 > 0 , we get 𝑊 𝑡 ( 𝑥 , 𝑡 ) − 𝑊 𝑥 𝑥 ( 𝑥 , 𝑡 ) ≥ 𝛼 𝐹 𝐻 ( 𝛿 + 𝛿 𝐵 𝑇 ) . ( 7 . 7 ) A straightforward computation yields 𝐾 > 0 This implies that there exists 𝑑 𝑊 𝑥 𝑑 𝑡 𝑖 , 𝑡 − 𝛿 2 𝑊 𝑥 𝑖 , 𝑡 ≥ 𝛼 𝐹 𝐻 ( 𝛿 + 𝛿 𝐵 𝑇 ) − 𝐾 ℎ 2 , 0 ≤ 𝑖 ≤ 𝐼 , ( 7 . 8 ) such that 𝑥 𝑑 𝑊 𝑖 , 𝑡 𝑑 𝑡 − 𝛿 2 𝑊 𝑥 𝑖 , 𝑡 ≥ 0 . ( 7 . 9 ) Using Taylor's expansion, there exists a constant 𝑈 𝑖 ( 𝑡 ) ≤ 𝐻 𝛿 𝑣 ( 𝑥 ) + 𝛿 𝐵 𝑇 − 𝑡 0 f o r 𝑡 ≥ 𝑡 0 , 0 ≤ 𝑖 ≤ 𝐼 . ( 7 . 1 0 ) such that 𝑈 𝑖 ( 𝑡 ) ≤ 𝐻 𝛿 𝑣 ( 𝑥 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 7 . 1 1 ) which implies that 𝑈 𝑖 ( 𝑇 ) < + ∞ , 0 ≤ 𝑖 ≤ 𝐼 − 1 The maximum principle implies that 𝑎 ( 𝑥 , 𝑡 ) = 1 Hence, we get 𝑏 ( 𝑡 ) = 1 Therefore 𝑓 ( 𝑢 ) = 𝑢 𝑝 , and we have the desired result.

8. Full Discretization

In this section, we consider the problem (1.1) in the case where 𝑔 ( 𝑢 ) = 𝑢 𝑝 , 𝑝 = , > 1 , 𝑢 𝑡 ( 𝑥 , 𝑡 ) = 𝑢 𝑥 𝑥 ( 𝑥 , 𝑡 ) − 𝑢 𝑝 𝑢 ( 𝑥 , 𝑡 ) , 0 < 𝑥 < 1 , 𝑡 ∈ ( 0 , 𝑇 ) , 𝑥 ( 0 , 𝑡 ) = 0 , 𝑢 𝑥 ( 1 , 𝑡 ) = 𝑢 𝑝 ( 1 , 𝑡 ) , 𝑡 ∈ ( 0 , 𝑇 ) , 𝑢 ( 𝑥 , 0 ) = 𝑢 0 ( 𝑥 ) > 0 , 0 ≤ 𝑥 ≤ 1 , ( 8 . 1 ) with 𝑝 > 1 const 𝑢 0 ∈ 𝐶 1 ( [ 0 , 1 ] ) . Thus our problem is equivalent to 𝑢 0 ( 0 ) = 0 where 𝑢 0 ( 1 ) = 𝑢 𝑝 0 ( 1 ) , 𝐼 , ℎ = 1 / 𝐼 and 𝑥 𝑖 = 𝑖 ℎ .

We start this section by the construction of an adaptive scheme as follows. Let 0 ≤ 𝑖 ≤ 𝐼 be a positive integer and let 𝑢 ( 𝑥 , 𝑡 ) . Define the grid 𝑈 ℎ ( 𝑛 ) = ( 𝑈 0 ( 𝑛 ) , 𝑈 1 ( 𝑛 ) , … , 𝑈 𝐼 ( 𝑛 ) ) 𝑇 , 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) = 𝛿 2 𝑈 𝑖 ( 𝑛 ) − 𝑈 𝑖 ( 𝑛 ) 𝑝 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , ( 8 . 2 ) 𝑡 𝑈 𝐼 ( 𝑛 ) = 𝛿 2 𝑈 𝐼 ( 𝑛 ) − 𝑈 𝐼 ( 𝑛 ) 𝑝 + 2 ℎ 𝑈 𝐼 ( 𝑛 ) 𝑝 𝑈 , ( 8 . 3 ) 𝑖 ( 0 ) = 𝜑 𝑖 , 0 ≤ 𝑖 ≤ 𝐼 , ( 8 . 4 ) and approximate the solution 𝑛 ≥ 0 of the problem (8.1) by the solution 𝜑 𝑖 + 1 ≥ 𝜑 𝑖 of the following discrete equations 0 ≤ 𝑖 ≤ 𝐼 − 1 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) = 𝑈 𝑖 ( 𝑛 + 1 ) − 𝑈 𝑖 ( 𝑛 ) Δ 𝑡 𝑛 , 𝛿 2 𝑈 𝑖 ( 𝑛 ) = 𝑈 ( 𝑛 ) 𝑖 + 1 − 2 𝑈 𝑖 ( 𝑛 ) + 𝑈 ( 𝑛 ) 𝑖 − 1 ℎ 2 𝛿 , 1 ≤ 𝑖 ≤ 𝐼 − 1 , 2 𝑈 0 ( 𝑛 ) = 2 𝑈 1 ( 𝑛 ) − 2 𝑈 0 ( 𝑛 ) ℎ 2 , 𝛿 2 𝑈 𝐼 ( 𝑛 ) = 2 𝑈 ( 𝑛 ) 𝐼 − 1 − 2 𝑈 𝐼 ( 𝑛 ) ℎ 2 . ( 8 . 5 ) 𝑡 where Δ 𝑡 𝑛 = m i n { ( 1 − 𝑝 𝜏 ) ℎ 2 / 3 , 𝜏 / ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ 𝑝 − 1 } , 0 < 𝜏 < 1 / 𝑝 , 𝑈 ℎ ( 𝑛 ) , 𝑈 ( 𝑛 ) 𝑖 + 1 ≥ 𝑈 𝑖 ( 𝑛 ) , 0 ≤ 𝑖 ≤ 𝐼 − 1 . ( 8 . 6 ) In order to permit the discrete solution to reproduce the property of the continuous one when the time 𝑍 𝑖 ( 𝑛 ) = 𝑈 ( 𝑛 ) 𝑖 + 1 − 𝑈 𝑖 ( 𝑛 ) approaches the blow-up time, we need to adapt the size of the time step so that we take 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑍 0 ( 𝑛 + 1 ) − 𝑍 0 ( 𝑛 ) Δ 𝑡 𝑛 = 𝑍 1 ( 𝑛 ) − 3 𝑍 0 ( 𝑛 ) ℎ 2 − 𝑈 1 ( 𝑛 ) 𝑝 − 𝑈 0 ( 𝑛 ) 𝑝 , 𝑍 𝑖 ( 𝑛 + 1 ) − 𝑍 𝑖 ( 𝑛 ) Δ 𝑡 𝑛 = 𝑍 ( 𝑛 ) 𝑖 + 1 − 2 𝑍 𝑖 ( 𝑛 ) + 𝑍 ( 𝑛 ) 𝑖 − 1 ℎ 2 − 𝑈 ( 𝑛 ) 𝑖 + 1 𝑝 − 𝑈 𝑖 ( 𝑛 ) 𝑝 𝑍 , 1 ≤ 𝑖 ≤ 𝐼 − 2 , ( 𝑛 + 1 ) 𝐼 − 1 − 𝑍 ( 𝑛 ) 𝐼 − 1 Δ 𝑡 𝑛 = 𝑍 ( 𝑛 ) 𝐼 − 2 − 3 𝑍 ( 𝑛 ) 𝐼 − 1 ℎ 2 − 𝑈 𝐼 ( 𝑛 ) 𝑝 − 𝑈 ( 𝑛 ) 𝐼 − 1 𝑝 + 2 ℎ 𝑈 𝐼 ( 𝑛 ) 𝑝 . ( 8 . 7 ) .

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 𝑍 0 ( 𝑛 + 1 ) = Δ 𝑡 𝑛 ℎ 2 𝑍 1 ( 𝑛 ) + 1 − 3 Δ 𝑡 𝑛 ℎ 2 𝑍 0 ( 𝑛 ) − Δ 𝑡 𝑛 𝑝 𝜉 0 ( 𝑛 ) 𝑝 − 1 𝑍 0 ( 𝑛 ) , 𝑍 𝑖 ( 𝑛 + 1 ) = Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 + 1 + 1 − 2 Δ 𝑡 𝑛 ℎ 2 𝑍 𝑖 ( 𝑛 ) + Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 − 1 − Δ 𝑡 𝑛 𝑝 𝜉 𝑖 ( 𝑛 ) 𝑝 − 1 𝑍 𝑖 ( 𝑛 ) 𝑍 , 1 ≤ 𝑖 ≤ 𝐼 − 2 , ( 𝑛 + 1 ) 𝐼 − 1 ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝐼 − 2 + 1 − 3 Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝐼 − 1 − Δ 𝑡 𝑛 𝑝 𝜉 ( 𝑛 ) 𝐼 − 1 𝑝 − 1 𝑍 ( 𝑛 ) 𝐼 − 1 , ( 8 . 8 ) be the solution of (8.2)–(8.4). Then we have 𝜉 𝑖 ( 𝑛 )

Proof. Let 𝑈 𝑖 ( 𝑛 ) , 𝑈 ( 𝑛 ) 𝑖 + 1 . We observe that 𝑍 𝑖 ( 𝑛 ) ≤ 0 Using the Taylor's expansion, we find that 0 ≤ 𝑖 ≤ 𝐼 − 1 where 𝑍 0 ( 𝑛 + 1 ) ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 1 ( 𝑛 ) + 1 − 3 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 𝑝 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 𝑍 0 ( 𝑛 ) , 𝑍 𝑖 ( 𝑛 + 1 ) ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 + 1 + 1 − 2 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 𝑝 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 𝑍 𝑖 ( 𝑛 ) + Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 − 1 𝑍 , 1 ≤ 𝑖 ≤ 𝐼 − 2 , ( 𝑛 + 1 ) 𝐼 − 1 ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝐼 − 2 + 1 − 3 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 𝑝 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 𝑍 ( 𝑛 ) 𝐼 − 1 . ( 8 . 9 ) is an intermediate value between Δ 𝑡 𝑛 ≤ 𝜏 / ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ 𝑝 − 1 and 𝑍 0 ( 𝑛 + 1 ) ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 1 ( 𝑛 ) + 1 − 3 Δ 𝑡 𝑛 ℎ 2 𝑍 − 𝑝 𝜏 0 ( 𝑛 ) , 𝑍 𝑖 ( 𝑛 + 1 ) ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 + 1 + 1 − 2 Δ 𝑡 𝑛 ℎ 2 𝑍 − 𝑝 𝜏 𝑖 ( 𝑛 ) + Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝑖 − 1 𝑍 , 1 ≤ 𝑖 ≤ 𝐼 − 2 , ( 𝑛 + 1 ) 𝐼 − 1 ≥ Δ 𝑡 𝑛 ℎ 2 𝑍 ( 𝑛 ) 𝐼 − 2 + 1 − 3 Δ 𝑡 𝑛 ℎ 2 𝑍 − 𝑝 𝜏 ( 𝑛 ) 𝐼 − 1 . ( 8 . 1 0 ) . If 1 − 3 ( Δ 𝑡 𝑛 / ℎ 2 ) − 𝑝 𝜏 , 𝑍 𝑖 ( 𝑛 ) ≤ 0 , we deduce that 0 ≤ 𝑖 ≤ 𝐼 − 1 Using the restriction 𝑎 ℎ ( 𝑛 ) , we find that 𝑉 ℎ ( 𝑛 ) We observe that 𝛿 𝑡 𝑉 𝑖 ( 𝑛 ) − 𝛿 2 𝑉 𝑖 ( 𝑛 ) + 𝑎 𝑖 ( 𝑛 ) 𝑉 𝑖 ( 𝑛 ) 𝑉 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 , 𝑛 ≥ 0 , ( 8 . 1 1 ) 𝑖 ( 0 ) ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 . ( 8 . 1 2 ) is nonnegative and by induction, we deduce that 𝑉 𝑖 ( 𝑛 ) ≥ 0 , 𝑛 ≥ 0 . This ends the proof.

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

Lemma 8.2. Let 0 ≤ 𝑖 ≤ 𝐼 be a bounded vector and let Δ 𝑡 𝑛 ≤ ℎ 2 / ( 2 + ‖ 𝑎 ℎ ( 𝑛 ) ‖ ∞ ℎ 2 ) a sequence such that 𝑉 ℎ ( 𝑛 ) ≥ 0 𝑉 0 ( 𝑛 + 1 ) ≥ 2 Δ 𝑡 𝑛 ℎ 2 𝑉 1 ( 𝑛 ) + 1 − 2 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 ‖ ‖ 𝑎 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑉 0 ( 𝑛 ) , 𝑉 𝑖 ( 𝑛 + 1 ) ≥ Δ 𝑡 𝑛 ℎ 2 𝑉 ( 𝑛 ) 𝑖 + 1 + 1 − 2 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 ‖ ‖ 𝑎 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑉 𝑖 ( 𝑛 ) + Δ 𝑡 𝑛 ℎ 2 𝑉 ( 𝑛 ) 𝑖 − 1 𝑉 , 1 ≤ 𝑖 ≤ 𝐼 − 1 , 𝐼 ( 𝑛 + 1 ) ≥ 2 Δ 𝑡 𝑛 ℎ 2 𝑉 ( 𝑛 ) 𝐼 − 1 + 1 − 2 Δ 𝑡 𝑛 ℎ 2 − Δ 𝑡 𝑛 ‖ ‖ 𝑎 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑉 𝐼 ( 𝑛 ) . ( 8 . 1 3 ) Then Δ 𝑡 𝑛 ≤ ℎ 2 / ( 2 + ‖ 𝑎 ℎ ( 𝑛 ) ‖ ∞ ℎ 2 ) for 1 − 2 ( Δ 𝑡 𝑛 / ℎ 2 ) − Δ 𝑡 𝑛 ‖ 𝑎 ℎ ( 𝑛 ) ‖ ∞ , 𝑉 ℎ ( 𝑛 ) ≥ 0 if 𝑎 ℎ ( 𝑛 ) .

Proof. If 𝑏 ℎ ( 𝑛 ) then a routine computation yields 𝑎 ℎ ( 𝑛 ) Since 𝑉 ℎ ( 𝑛 ) , we see that 𝑊 ℎ ( 𝑛 ) is nonnegative. From (8.12), we deduce by induction that 𝛿 𝑡 𝑉 𝑖 ( 𝑛 ) − 𝛿 2 𝑉 𝑖 ( 𝑛 ) + 𝑎 𝑖 ( 𝑛 ) 𝑉 𝑖 ( 𝑛 ) + 𝑏 𝑖 ( 𝑛 ) ≤ 𝛿 𝑡 𝑊 𝑖 ( 𝑛 ) − 𝛿 2 𝑊 𝑖 ( 𝑛 ) + 𝑎 𝑖 ( 𝑛 ) 𝑊 𝑖 ( 𝑛 ) + 𝑏 𝑖 ( 𝑛 ) 𝑉 , 0 ≤ 𝑖 ≤ 𝐼 , 𝑛 ≥ 0 , 𝑖 ( 0 ) ≤ 𝑊 𝑖 ( 0 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 8 . 1 4 ) 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 𝑉 𝑖 ( 𝑛 ) ≤ 𝑊 𝑖 ( 𝑛 ) and 𝑛 ≥ 0 two vectors such that 0 ≤ 𝑖 ≤ 𝐼 is bounded. Let Δ 𝑡 𝑛 ≤ ℎ 2 / ( 2 + ‖ 𝑎 ℎ ( 𝑛 ) ‖ ∞ ℎ 2 ) and 𝛿 𝑡 two sequences such that 𝑈 ( 𝑛 ) ∈ ℝ Then 𝑈 ( 𝑛 ) ≥ 0 for 𝑛 ≥ 0 , 𝛿 𝑡 𝑈 ( 𝑛 ) 𝑝 𝑈 ≥ 𝑝 ( 𝑛 ) 𝑝 − 1 𝛿 𝑡 𝑈 ( 𝑛 ) , 𝑛 ≥ 0 . ( 8 . 1 5 ) if 𝛿 𝑡 𝑈 ( 𝑛 ) 𝑝 𝑈 = 𝑝 ( 𝑛 ) 𝑝 − 1 𝛿 𝑡 𝑈 ( 𝑛 ) + 𝑝 ( 𝑝 − 1 ) 2 Δ 𝑡 𝑛 𝛿 𝑡 𝑈 ( 𝑛 ) 2 𝜃 ( 𝑛 ) 𝑝 − 2 , ( 8 . 1 6 ) .

Now, let us give a property of the operator 𝜃 ( 𝑛 ) .

Lemma 8.4. Let 𝑈 ( 𝑛 ) be such that 𝑈 ( 𝑛 + 1 ) for 𝑈 ( 𝑛 ) ≥ 0 . Then we have 𝑛 ≥ 0

Proof. From Taylor's expansion, we find that 𝑈 ℎ ( 𝑛 ) where l i m 𝑛 → + ∞ ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ = + ∞ , 𝑇 ℎ Δ 𝑡 = l i m 𝑛 → ∞ 𝑛 − 1 𝑖 = 0 Δ 𝑡 𝑖 < + ∞ . ( 8 . 1 7 ) is an intermediate value between 𝑇 ℎ Δ 𝑡 and 𝑈 ℎ ( 𝑛 ) . Use the fact that 𝑈 ℎ ( 𝑛 ) for 𝑈 ℎ ( 𝑛 ) 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 𝐴 ∈ ( 0 , 1 ] of (8.2)–(8.4) blows up in a finite time if 𝛿 2 𝜑 𝑖 − 𝜑 𝑝 𝑖 𝛿 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 − 1 . 2 𝜑 𝐼 − 𝜑 𝑝 𝐼 + 2 ℎ 𝜑 𝑝 𝐼 ≥ 𝐴 𝜑 𝑝 𝐼 . ( 8 . 1 8 ) The number 𝑈 ℎ ( 𝑛 ) is called the numerical blow-up time of 𝑇 ℎ Δ 𝑡 .

The following theorem reveals that the discrete solution 𝑇 ℎ Δ 𝑡 ≤ 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 ( 1 + 𝜏 ) 𝑝 − 1 ‖ ‖ 𝜑 − 1 ℎ ‖ ‖ ∞ 𝑝 − 1 , ( 8 . 1 9 ) of (8.2)–(8.4) blows up in a finite time under some hypotheses.

Theorem 8.6. Let 𝜏 = 𝐴 m i n { ( 1 − 𝑝 𝜏 ) ℎ 2 ‖ 𝜑 ℎ ‖ − 𝑝 − 1 i n f / 3 , 𝜏 } . be the solution of (8.2)–(8.4). Suppose that there exists a constant 𝐽 ℎ ( 𝑛 ) such that the initial data at (8.4) satisfies 𝐽 𝑖 ( 𝑛 ) = 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) 𝐽 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑛 ≥ 0 , 𝐼 ( 𝑛 ) = 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) 𝑈 − 𝐴 𝐼 ( 𝑛 ) − 𝑝 , 𝑛 ≥ 0 . ( 8 . 2 0 ) Then 𝛿 𝑡 𝐽 𝑖 ( 𝑛 ) − 𝛿 2 𝐽 𝑖 ( 𝑛 ) = 𝛿 𝑡 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) − 𝛿 2 𝑈 𝑖 ( 𝑛 ) 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝐽 𝐼 ( 𝑛 ) − 𝛿 2 𝐽 𝐼 ( 𝑛 ) = 𝛿 𝑡 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) − 𝛿 2 𝑈 𝐼 ( 𝑛 ) − 𝐴 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) 𝑝 + 𝐴 𝛿 2 𝑈 𝐼 ( 𝑛 ) 𝑝 . ( 8 . 2 1 ) blows up in a finite time 𝛿 𝑡 𝐽 𝑖 ( 𝑛 ) − 𝛿 2 𝐽 𝑖 ( 𝑛 ) = − 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) 𝑝 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝐽 𝐼 ( 𝑛 ) − 𝛿 2 𝐽 𝐼 ( 𝑛 ) = 2 ℎ 𝛿 − 1 − 𝐴 𝑡 𝑈 𝐼 ( 𝑛 ) 𝑝 + 𝐴 𝛿 2 𝑈 𝐼 ( 𝑛 ) 𝑝 . ( 8 . 2 2 ) which satisfies the following estimate 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) 𝑝 𝜉 = 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) 𝜉 = 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝐽 𝑖 ( 𝑛 ) , ( 8 . 2 3 ) where 𝜉 𝑖 ( 𝑛 )

Proof. Introduce the vector 𝑈 𝑖 ( 𝑛 ) defined as follows 𝑈 ( 𝑛 ) 𝑖 + 1 A straightforward computation yields 𝛿 𝑡 𝐽 𝑖 ( 𝑛 ) − 𝛿 2 𝐽 𝑖 ( 𝑛 ) 𝜉 = − 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝐽 𝑖 ( 𝑛 ) 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝐽 𝐼 ( 𝑛 ) − 𝛿 2 𝐽 𝐼 ( 𝑛 ) = 2 ℎ 𝑝 𝑈 − 1 − 𝐴 𝐼 ( 𝑛 ) 𝑝 − 1 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) + 𝐴 𝑝 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) 𝑝 − 1 𝛿 2 𝑈 𝐼 ( 𝑛 ) . ( 8 . 2 4 ) Using (8.2), we arrive at 𝛿 𝑡 𝐽 𝐼 ( 𝑛 ) − 𝛿 2 𝐽 𝐼 ( 𝑛 ) = 2 ℎ 𝑝 𝑈 − 1 𝐼 ( 𝑛 ) 𝑝 − 1 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) − 𝐴 𝑝 𝛿 𝑡 𝑈 𝐼 ( 𝑛 ) 𝑝 − 1 2 ℎ 𝑈 − 1 𝐼 ( 𝑛 ) 𝑝 , ( 8 . 2 5 ) Due to the mean value theorem, we get 𝛿 𝑡 𝐽 𝑖 ( 𝑛 ) − 𝛿 2 𝐽 𝑖 ( 𝑛 ) 𝜉 = − 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝐽 𝑖 ( 𝑛 ) 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝐽 𝐼 ( 𝑛 ) − 𝛿 2 𝐽 𝐼 ( 𝑛 ) = 2 ℎ 𝑝 𝑈 − 1 𝐼 ( 𝑛 ) 𝑝 − 1 𝐽 𝐼 ( 𝑛 ) . ( 8 . 2 6 ) where 𝐽 ℎ ( 0 ) ≥ 0 is an intermediate value between 𝐽 ℎ ( 𝑛 ) ≥ 0 and 𝑈 𝐼 ( 𝑛 + 1 ) ≥ 𝑈 𝐼 ( 𝑛 ) 1 + 𝐴 Δ 𝑡 𝑛 𝑈 𝐼 ( 𝑛 ) 𝑝 − 1 . ( 8 . 2 7 ) . On the other hand, from Lemmas 2.4 and 2.5, we deduce that 𝑈 𝐼 ( 𝑛 ) = ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ It follows from (8.3) that ‖ ‖ 𝑈 ℎ ( 𝑛 + 1 ) ‖ ‖ ∞ ≥ ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 1 + 𝐴 Δ 𝑡 𝑛 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 . ( 8 . 2 8 ) which implies that 𝐴 Δ 𝑡 𝑛 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 = 𝐴 m i n ( 1 − 𝑝 𝜏 ) ℎ 2 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 3 , 𝜏 . ( 8 . 2 9 ) From (8.18), we observe that ‖ 𝑈 ℎ ( 𝑛 + 1 ) ‖ ∞ ≥ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ . It follows from Lemma 8.2 that ‖ 𝑈 ℎ ( 𝑛 + 1 ) ‖ ∞ ≥ ‖ 𝑈 ℎ ( 0 ) ‖ ∞ = ‖ 𝜑 ℎ ‖ ∞ which implies that ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ 𝑝 − 1 ≥ ‖ 𝜑 ℎ ‖ ∞ 𝑝 − 1 From Lemma 8.1, we see that 𝐴 Δ 𝑡 𝑛 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 ≥ 𝐴 m i n ( 1 − 𝑝 𝜏 ) ℎ 2 ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ 𝑝 − 1 3 , 𝜏 = 𝜏 . ( 8 . 3 0 ) which implies that ‖ ‖ 𝑈 ℎ ( 𝑛 + 1 ) ‖ ‖ ∞ ≥ ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ ( 1 + 𝜏 ) ( 8 . 3 1 ) It is not hard to see that ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ ≥ ‖ ‖ 𝑈 ℎ ( 0 ) ‖ ‖ ∞ ( 1 + 𝜏 ) 𝑛 = ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ ( 1 + 𝜏 ) 𝑛 , 𝑛 ≥ 0 . ( 8 . 3 2 ) From (8.28), we get 𝑛 . By induction, we arrive at ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ , which implies that 𝑛 . Therefore, we find that Σ + ∞ 𝑛 = 0 Δ 𝑡 𝑛 ≤ Σ + ∞ 𝑛 = 0 𝜏 ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ 𝑝 − 1 ≤ 𝜏 ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ 𝑝 − 1 Σ + ∞ 𝑛 = 0 1 ( 1 + 𝜏 ) 𝑝 − 1 𝑛 . ( 8 . 3 3 ) Consequently, we arrive at 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 / ( ( 1 + 𝜏 ) 𝑝 − 1 − 1 ) and by induction, we get Σ + ∞ 𝑛 = 0 Δ 𝑡 𝑛 ≤ Σ + ∞ 𝑛 = 0 ( 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 / ( ( 1 + 𝜏 ) 𝑝 − 1 − 1 ) ‖ 𝜑 ℎ ‖ ∞ 𝑝 − 1 ) Since the term on the right hand side of the above equality tends to infinity as ( 1 + 𝜏 ) 𝑝 − 1 = 1 − ( 𝑝 − 1 ) 𝜏 + 𝑜 ( 𝜏 ) approaches infinity, we conclude that 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 = 𝜏 − 1 𝜏 1 ≤ 𝑝 − 1 + 𝑜 ( 1 ) 2 𝜏 𝜏 ( 𝑝 − 1 ) . ( 8 . 3 4 ) tends to infinity as 𝜏 = ℎ 2 approaches infinity. Now, let us estimate the numerical blow-up time. Due to (8.32), the restriction on the time step ensures that 𝜏 𝜏 = 𝐴 m i n 1 − 𝑝 ℎ 2 3 ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ 𝑝 − 1 1 , 1 ≥ 𝐴 m i n 4 ‖ ‖ 𝜑 ℎ ‖ ‖ ∞ 𝑝 − 1 , 1 . ( 8 . 3 5 ) Using the fact that the series on the right hand side of the above inequality converges towards 𝜏 / 𝜏 , we deduce that 𝜏 / ( ( 1 + 𝜏 ) 𝑝 − 1 − 1 ) and the proof is complete.

Remark 8.7. Apply Taylor's expansion to obtain ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ ≥ ‖ ‖ 𝑈 ℎ ( 𝑞 ) ‖ ‖ ∞ ( 1 + 𝜏 ) 𝑛 − 𝑞 f o r 𝑛 ≥ 𝑞 ( 8 . 3 6 ) , which implies that Σ + ∞ 𝑛 = 𝑞 Δ 𝑡 𝑛 ≤ 𝜏 ‖ ‖ 𝑈 ℎ ( 𝑞 ) ‖ ‖ ∞ 𝑝 − 1 Σ + ∞ 𝑛 = 𝑞 1 ( 1 + 𝜏 ) 𝑝 − 1 𝑛 − 𝑞 . ( 8 . 3 7 ) If we take 𝑇 ℎ Δ 𝑡 − 𝑡 𝑞 ≤ 𝜏 ‖ ‖ 𝑈 ℎ ( 𝑞 ) ‖ ‖ ∞ 𝑝 − 1 ( 1 + 𝜏 ) 𝑝 − 1 ( 1 + 𝜏 ) 𝑝 − 1 − 1 . ( 8 . 3 8 ) , we see that 𝜏 = ℎ 2 We deduce that 𝑢 ∈ 𝐶 4 , 2 ( [ 0 , 1 ] × [ 0 , 𝑇 ] ) is bounded from above. We conclude that ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ = 𝑜 ( 1 ) a s ℎ → 0 . ( 9 . 1 ) is bounded from above.

Remark 8.8. From (8.31), we get 𝑈 ℎ ( 𝑛 ) which implies that 0 ≤ 𝑛 ≤ 𝐽 We deduce that m a x 0 ≤ 𝑛 ≤ 𝐽 ‖ ‖ 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ ( 𝑡 𝑛 ) ‖ ‖ ∞ ‖ ‖ 𝜑 = 𝑂 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + ℎ 2 + Δ 𝑡 𝑛 a s ℎ → 0 , ( 9 . 2 )

In the sequel, we take 𝐽 .

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 ∑ 𝐽 − 1 𝑛 = 0 Δ 𝑡 𝑛 ≤ 𝑇 . Assume that the initial data at (8.4) satisfies 𝑡 𝑛 = ∑ 𝑛 − 1 𝑗 = 0 Δ 𝑡 𝑗 Then the problem (8.2)–(8.4) has a solution ℎ for h sufficiently small, 𝑈 ℎ ( 𝑛 ) and we have the following relation 𝑁 ≤ 𝐽 where 𝑛 is such that ‖ ‖ 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ ( 𝑡 𝑛 ) ‖ ‖ ∞ < 1 f o r 𝑛 < 𝑁 . ( 9 . 3 ) and 𝑁 ≥ 1 .

Proof. For each 𝑢 ∈ 𝐶 4 , 2 , the problem (8.2)–(8.4) has a solution 𝐾 . Let ‖ 𝑢 ‖ ∞ ≤ 𝐾 be the greatest value of ‖ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ‖ ∞ ≤ ‖ ‖ 𝑢 ℎ 𝑡 𝑛 ‖ ‖ ∞ + ‖ ‖ 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ 𝑡 𝑛 ‖ ‖ ∞ ≤ 1 + 𝐾 f o r 𝑛 < 𝑁 . ( 9 . 4 ) such that 𝑢 ∈ 𝐶 4 , 2 We know that 𝛿 𝑡 𝑢 𝑥 𝑖 , 𝑡 𝑛 − 𝛿 2 𝑢 𝑥 𝑖 , 𝑡 𝑛 = − 𝑢 𝑝 𝑥 𝑖 , 𝑡 𝑛 ℎ + 𝑂 2 + 𝑂 Δ 𝑡 𝑛 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝑢 𝑥 𝐼 , 𝑡 𝑛 − 𝛿 2 𝑢 𝑥 𝐼 , 𝑡 𝑛 = − 𝑢 𝑝 𝑥 𝐼 , 𝑡 𝑛 + 2 ℎ 𝑢 𝑝 𝑥 𝐼 , 𝑡 𝑛 ℎ + 𝑂 2 + 𝑂 Δ 𝑡 𝑛 . ( 9 . 5 ) because of (9.1). Due to the fact that 𝑒 ℎ ( 𝑛 ) = 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ ( 𝑡 𝑛 ) , there exists a positive constant 𝛿 𝑡 𝑒 𝑖 ( 𝑛 ) − 𝛿 2 𝑒 𝑖 ( 𝑛 ) 𝜉 = − 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝑒 𝑖 ( 𝑛 ) ℎ + 𝑂 2 + 𝑂 Δ 𝑡 𝑛 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝑒 𝐼 ( 𝑛 ) − 𝛿 2 𝑒 𝐼 ( 𝑛 ) 2 = 𝑝 ℎ 𝜉 − 1 𝐼 ( 𝑛 ) 𝑝 − 1 𝑒 𝐼 ( 𝑛 ) ℎ + 𝑂 2 + 𝑂 Δ 𝑡 𝑛 , ( 9 . 6 ) such that 𝜉 𝑖 ( 𝑛 ) . Applying the triangle inequality, we have 𝑢 ( 𝑥 𝑖 , 𝑡 𝑛 ) Since 𝑈 𝑖 ( 𝑛 ) , using Taylor's expansion, we find that 𝐿 Let 𝐾 be the error of discretization. From the mean value theorem, we get 𝛿 𝑡 𝑒 𝑖 ( 𝑛 ) − 𝛿 2 𝑒 𝑖 ( 𝑛 ) 𝜉 ≤ − 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝑒 𝑖 ( 𝑛 ) + 𝐿 ℎ 2 + 𝐿 Δ 𝑡 𝑛 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑛 < 𝑁 , 𝑡 𝑒 𝐼 ( 𝑛 ) − 𝛿 2 𝑒 𝐼 ( 𝑛 ) 2 ≤ 𝑝 ℎ 𝜉 − 1 𝐼 ( 𝑛 ) 𝑝 − 1 𝑒 𝐼 ( 𝑛 ) + 𝐿 ℎ 2 + 𝐿 Δ 𝑡 𝑛 , 𝑛 < 𝑁 . ( 9 . 7 ) where 𝑍 ( 𝑥 , 𝑡 ) = 𝑒 ( ( 𝑀 + 1 ) 𝑡 + 𝐶 𝑥 2 ) ( ‖ 𝜑 ℎ − 𝑢 ℎ ( 0 ) ‖ ∞ + 𝑄 ℎ 2 + 𝑄 Δ 𝑡 𝑛 ) is an intermediate value between 𝑀 and 𝐶 . Hence, there exist positive constants 𝑄 and 𝑍 𝑡 ( 𝑥 , 𝑡 ) − 𝑍 𝑥 𝑥 ( 𝑥 , 𝑡 ) = 𝑀 + 1 − 2 𝐶 − 4 𝐶 2 𝑥 2 𝑍 𝑍 ( 𝑥 , 𝑡 ) , 𝑥 ( 0 , 𝑡 ) = 0 , 𝑍 𝑥 ( 1 , 𝑡 ) = 2 𝐶 𝑍 ( 1 , 𝑡 ) , 𝑍 ( 𝑥 , 0 ) = 𝑒 ( 𝐶 𝑥 2 ) ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 + 𝑄 Δ 𝑡 𝑛 . ( 9 . 8 ) such that 𝛿 𝑡 𝑍 𝑥 𝑖 , 𝑡 𝑛 − 𝛿 2 𝑍 𝑥 𝑖 , 𝑡 𝑛 = 𝑀 + 1 − 2 𝐶 − 4 𝐶 2 𝑥 2 𝑖 𝑍 𝑥 𝑖 , 𝑡 𝑛 + ℎ 2 𝑍 1 2 𝑥 𝑥 𝑥 𝑥 𝑥 𝑖 , 𝑡 𝑛 − Δ 𝑡 𝑛 2 𝑍 𝑡 𝑡 𝑥 𝑖 , 𝑡 𝑛 , 𝛿 𝑡 𝑍 𝑥 𝐼 , 𝑡 𝑛 − 𝛿 2 𝑍 𝑥 𝐼 , 𝑡 𝑛 = 𝑀 + 1 − 2 𝐶 − 4 𝐶 2 𝑥 2 𝐼 𝑍 𝑥 𝐼 , 𝑡 𝑛 + 4 𝐶 ℎ 𝑍 𝑥 𝐼 , 𝑡 𝑛 + ℎ 2 𝑍 1 2 𝑥 𝑥 𝑥 𝑥 𝑥 𝐼 , 𝑡 𝑛 − Δ 𝑡 𝑛 2 𝑍 𝑡 𝑡 𝑥 𝐼 , 𝑡 𝑛 . ( 9 . 9 ) Consider the function 𝑀 where 𝐶 , 𝑄 , 𝛿 𝑡 𝑍 𝑥 𝑖 , 𝑡 𝑛 − 𝛿 2 𝑍 𝑥 𝑖 , 𝑡 𝑛 𝜉 > − 𝑝 𝑖 ( 𝑛 ) 𝑝 − 1 𝑍 𝑥 𝑖 , 𝑡 𝑛 + 𝐿 ℎ 2 + 𝐿 Δ 𝑡 𝑛 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝑍 𝑥 𝐼 , 𝑡 𝑛 − 𝛿 2 𝑍 𝑥 𝐼 , 𝑡 𝑛 2 > 𝑝 ℎ 𝜉 − 1 𝐼 ( 𝑛 ) 𝑝 − 1 𝑍 𝑥 𝐼 , 𝑡 𝑛 + 𝐿 ℎ 2 + 𝐿 Δ 𝑡 𝑛 , 𝑍 𝑖 ( 0 ) > 𝑒 𝑖 ( 0 ) , 0 ≤ 𝑖 ≤ 𝐼 . ( 9 . 1 0 ) are positive constants which will be determined later. We get 𝑍 𝑥 𝑖 , 𝑡 𝑛 > 𝑒 𝑖 ( 𝑛 ) , 0 ≤ 𝑖 ≤ 𝐼 , 𝑛 < 𝑁 . ( 9 . 1 1 ) By a discretization of the above problem, we obtain 𝑍 𝑥 𝑖 , 𝑡 𝑛 > − 𝑒 𝑖 ( 𝑛 ) , 0 ≤ 𝑖 ≤ 𝐼 , 𝑛 < 𝑁 , ( 9 . 1 2 ) We may choose ‖ ‖ 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ ‖ ‖ ( 𝑡 ) ∞ ≤ 𝑒 ( 𝑀 𝑡 𝑛 + 𝐶 ) ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 + 𝑄 Δ 𝑡 𝑛 , 𝑛 < 𝑁 . ( 9 . 1 3 ) , 𝑁 = 𝐽 , 𝑁 < 𝐽 large enough that ‖ ‖ 𝑈 1 ≤ ℎ ( 𝑁 ) − 𝑢 ℎ 𝑡 𝑁 ‖ ‖ ∞ ≤ 𝑒 ( 𝑀 𝑇 + 𝐶 ) ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ + 𝑄 ℎ 2 + 𝑄 Δ 𝑡 𝑛 . ( 9 . 1 4 ) It follows from Comparison Lemma 8.3 that ℎ By the same way, we also prove that 1 ≤ 0 which implies that 𝑢 Let us show that 𝑇 0 . Suppose that 𝑢 ∈ 𝐶 4 , 2 ( [ 0 , 1 ] × [ 0 , 𝑇 0 ) ) . From (9.3), we obtain ‖ ‖ 𝜑 ℎ − 𝑢 ℎ ‖ ‖ ( 0 ) ∞ = 𝑜 ( 1 ) a s ℎ → 0 . ( 9 . 1 5 ) Since the term on the right hand side of the second inequality goes to zero as 𝑈 ℎ ( 𝑛 ) goes to zero, we deduce that 𝑇 ℎ Δ 𝑡 , 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 l i m ℎ → 0 𝑇 ℎ Δ 𝑡 = 𝑇 0 . ( 9 . 1 6 ) which blows up in a finite time 𝜏 ( 1 + 𝜏 ) / ( ( 1 + 𝜏 ) 𝑝 − 1 − 1 ) and 𝜀 > 0 . Assume that the initial data at (2.3) satisfies 𝑅 > 0 Under the assumption of Theorem 8.6, the problem (8.2)–(8.4) has a solution 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 𝑥 𝑝 − 1 ( 1 + 𝜏 ) 𝑝 − 1 < 𝜀 − 1 2 f o r 𝑥 ∈ [ 𝑅 , ∞ ) . ( 9 . 1 7 ) which blows up in a finite time 𝑢 and the following relation holds 𝑇 0

Proof. We know from Remark 8.7 that 𝑇 1 ∈ ( 𝑇 0 − 𝜀 / 2 , 𝑇 0 ) is bounded. Letting ‖ 𝑢 ( ⋅ , 𝑡 ) ‖ ∞ ≥ 2 𝑅 , there exists a constant 𝑡 ∈ [ 𝑇 1 , 𝑇 0 ) such that 𝑇 2 = ( 𝑇 1 + 𝑇 0 ) / 2 Since 𝑞 blows up at the time 𝑡 𝑞 = ∑ 𝑞 − 1 𝑛 = 0 Δ 𝑡 𝑛 ∈ [ 𝑇 1 , 𝑇 2 ] , there exists ℎ such that 0 < ‖ 𝑢 ℎ ( 𝑡 𝑛 ) ‖ ∞ < ∞ for 𝑛 ≤ 𝑞 . Let 𝑈 ℎ ( 𝑛 ) and let ‖ 𝑈 ℎ ( 𝑛 ) − 𝑢 ℎ ( 𝑡 𝑛 ) ‖ ∞ < 𝑅 be a positive integer such that 𝑛 ≤ 𝑞 for ‖ ‖ 𝑈 ℎ ( 𝑞 ) ‖ ‖ ∞ ≥ ‖ ‖ 𝑢 ℎ 𝑡 𝑞 ‖ ‖ ∞ − ‖ ‖ 𝑈 ℎ ( 𝑞 ) − 𝑢 ℎ 𝑡 𝑞 ‖ ‖ ∞ ≥ 𝑅 . ( 9 . 1 8 ) small enough. We have 𝑈 ℎ ( 𝑛 ) for 𝑇 ℎ Δ 𝑡 . It follows from Theorem 4.3 that the problem (2.1)–(2.3) has a solution | 𝑇 ℎ Δ 𝑡 − 𝑡 𝑞 | ≤ 𝜏 ( 1 + 𝜏 ) 𝑝 − 1 ‖ 𝑈 ℎ ( 𝑞 ) ‖ ∞ 1 − 𝑝 / ( ( 1 + 𝜏 ) 𝑝 − 1 − 1 ) < 𝜀 / 2 which obeys ‖ 𝑈 ℎ ( 𝑞 ) ‖ ∞ ≥ 𝑅 for | 𝑇 0 − 𝑇 ℎ Δ 𝑡 | ≤ | 𝑇 0 − 𝑡 𝑞 | + | 𝑡 𝑞 − 𝑇 ℎ Δ 𝑡 | ≤ 𝜀 / 2 + 𝜀 / 2 ≤ 𝜀 , which implies that 𝑎 ( 𝑥 , 𝑡 ) = 𝜆 > 0 From Theorem 8.6, 𝑓 ( 𝑢 ) = 𝑢 𝑝 blows up at the time 𝑔 ( 𝑢 ) = 𝑢 𝑞 . It follows from Remark 8.8 and (9.17) that 𝑏 ( 𝑡 ) = 1 because 𝑝 = . We deduce that > 1 , 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 𝑞 = , > 1 , 𝑢 , 𝑈 ℎ ( 𝑛 ) with 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) = 𝛿 2 𝑈 𝑖 ( 𝑛 ) 𝑈 − 𝜆 𝑖 ( 𝑛 ) 𝑝 − 1 𝑈 𝑖 ( 𝑛 + 1 ) 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝑈 𝐼 ( 𝑛 ) = 𝛿 2 𝑈 𝐼 ( 𝑛 ) + 2 ℎ 𝑈 𝐼 ( 𝑛 ) 𝑞 𝑈 − 𝜆 𝐼 ( 𝑛 ) 𝑝 − 1 𝑈 𝐼 ( 𝑛 + 1 ) , 𝑈 𝑖 ( 0 ) = 𝜑 𝑖 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 , ( 1 0 . 1 ) const 𝑢 , 𝑈 ℎ ( 𝑛 ) const 𝛿 𝑡 𝑈 𝑖 ( 𝑛 ) = 𝛿 2 𝑈 𝑖 ( 𝑛 + 1 ) 𝑈 − 𝜆 𝑖 ( 𝑛 ) 𝑝 − 1 𝑈 𝑖 ( 𝑛 + 1 ) 𝛿 , 0 ≤ 𝑖 ≤ 𝐼 − 1 , 𝑡 𝑈 𝐼 ( 𝑛 ) = 𝛿 2 𝑈 𝐼 ( 𝑛 + 1 ) + 2 ℎ 𝑈 𝐼 ( 𝑛 ) 𝑞 𝑈 − 𝜆 𝐼 ( 𝑛 ) 𝑝 − 1 𝑈 𝐼 ( 𝑛 + 1 ) , 𝑈 𝑖 ( 0 ) = 𝜑 𝑖 ≥ 0 , 0 ≤ 𝑖 ≤ 𝐼 . ( 1 0 . 2 ) . We approximate the solution 𝑛 ≥ 0 of (1.1) by the solution Δ 𝑡 𝑛 = m i n ( ℎ 2 / 2 , 𝜏 ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ 1 − 𝑝 ) of the following explicit scheme Δ 𝑡 𝑛 = 𝜏 ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ 1 − 𝑝 We also approximate the solution 𝑈 0 ( 𝑛 + 1 ) = 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 1 ( 𝑛 ) + 1 − 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 0 ( 𝑛 ) 1 + 𝜆 Δ 𝑡 𝑛 𝑈 0 ( 𝑛 ) 𝑝 − 1 , 𝑈 𝑖 ( 𝑛 + 1 ) = 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 ( 𝑛 ) 𝑖 + 1 + 1 − 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 𝑖 ( 𝑛 ) + 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 ( 𝑛 ) 𝑖 − 1 1 + 𝜆 Δ 𝑡 𝑛 𝑈 𝑖 ( 𝑛 ) 𝑝 − 1 , 𝑈 𝐼 ( 𝑛 + 1 ) = 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 ( 𝑛 ) 𝐼 − 1 + 1 − 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 𝐼 ( 𝑛 ) + 2 Δ 𝑡 𝑛 / ℎ 2 𝑈 𝐼 ( 𝑛 ) 𝑞 1 + 𝜆 Δ 𝑡 𝑛 𝑈 𝐼 ( 𝑛 ) 𝑝 − 1 . ( 1 0 . 3 ) of (1.1) by the solution Δ 𝑡 𝑛 ≤ ℎ 2 / 2 of the implicit scheme below 𝐴 𝑛 ℎ 𝑈 ℎ ( 𝑛 + 1 ) = 𝐹 𝑛 , ( 1 0 . 4 ) For the time step, we take 𝐴 ℎ ( 𝑛 ) = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 𝑎 0 𝑏 0 𝑐 0 ⋯ 0 1 𝑎 1 𝑏 1 0 ⋯ 0 ⋱ ⋱ ⋱ ⋮ ⋱ ⋱ ⋱ 𝑏 𝐼 − 1 0 ⋯ 0 𝑐 𝐼 𝑎 𝐼 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , 𝑎 𝑖 = 1 + 2 Δ 𝑡 𝑛 ℎ 2 + 𝜆 Δ 𝑡 𝑛 𝑈 𝑖 ( 𝑛 ) 𝑝 − 1 𝑏 , 0 ≤ 𝑖 ≤ 𝐼 , 𝑖 = − 2 Δ 𝑡 𝑛 ℎ 2 𝑐 , 𝑖 = 0 , … , 𝐼 − 1 , 𝑖 = − 2 Δ 𝑡 𝑛 ℎ 2 , 𝑖 = 1 , … , 𝐼 , ( 𝐹 𝑛 ) 𝑖 = 𝑈 𝑖 ( 𝑛 ) , 𝑖 = 0 , … , 𝐼 − 1 , ( 𝐹 𝑛 ) 𝐼 = 𝑈 𝐼 ( 𝑛 ) + 2 ℎ Δ 𝑡 𝑛 𝑈 𝐼 ( 𝑛 ) 𝑞 . ( 1 0 . 5 ) , 𝐴 ℎ ( 𝑛 ) for the explicit scheme and 𝐴 ℎ ( 𝑛 ) 𝑖 𝑖 𝐴 > 0 , ℎ ( 𝑛 ) 𝑖 𝑗 | | | 𝐴 < 0 , i f 𝑖 ≠ 𝑗 , ℎ ( 𝑛 ) 𝑖 𝑖 | | | > ∑ 𝑗 ≠ 𝑖 | | | 𝐴 ℎ ( 𝑛 ) 𝑖 𝑗 | | | . ( 1 0 . 6 ) for the implicit scheme.

The problem described in (10.1) may be rewritten as follows 𝑈 ℎ ( 𝑛 ) Let us notice that the restriction on the time step 𝑛 ≥ 0 ensures the nonnegativity of the discrete solution.

The implicit scheme may be rewritten in the following form 𝑈 ℎ ( 0 ) where 𝑈 ℎ ( 𝑛 ) The matrix 𝑛 ≥ 0 satisfies the following properties 𝑈 ℎ ( 𝑛 ) It follows that l i m 𝑛 → + ∞ ‖ 𝑈 ℎ ( 𝑛 ) ‖ ∞ = + ∞ exists for ∑ + ∞ 𝑛 = 0 Δ 𝑡 𝑛 . In addition, since ∑ + ∞ 𝑛 = 0 Δ 𝑡 𝑛 is nonnegative, 𝑈 ℎ ( 𝑛 ) is also nonnegative for 𝑛 . We need the following definition.

Definition 10.1. We say that the discrete solution 𝑇 𝑛 = ∑ 𝑛 − 1 𝑗 = 0 Δ 𝑡 𝑗 of the explicit scheme or the implicit scheme blows up in a finite time if Δ 𝑡 𝑛 = | | | 𝑇 𝑛 + 1 − 𝑇 𝑛 | | | ≤ 1 0 − 1 6 . ( 1 0 . 7 ) and the series ( 𝑠 ) converges. The quantity 𝑇 𝑠 = l o g 4 ℎ − 𝑇 2 ℎ / 𝑇 2 ℎ − 𝑇 ℎ l o g ( 2 ) . ( 1 0 . 8 ) is called the numerical blow-up time of the solution 𝑇 𝑛 .

In Tables 1, 2, 3, 4, 5, 6, 7, and 8, in rows, we present the numerical blow-up times, values of 𝑇 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 which is computed at the first time when 𝑇 𝑛 The order 𝑇 𝑛 of the method is computed from 𝑇 𝑛

tab1
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).
tab2
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).
tab3
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).
tab4
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).
tab5
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).
tab6
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).
tab7
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).
tab8
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).

Case 1. 𝜑 𝑖 = 1 0 + 1 0 ∗ c o s ( 𝜋 𝑖 ℎ ) , 𝜆 = 1 , 𝑝 = 2 , 𝑞 = 3 .

Case 2. 𝜑 𝑖 = 1 0 + 1 0 ∗ c o s ( 𝜋 𝑖 ℎ ) , 𝜆 = 1 , 𝑝 = 2 , 𝑞 = 2 .

Case 3. 𝜑 𝑖 = 1 0 + 1 0 ∗ c o s ( 𝜋 𝑖 ℎ ) , 𝜆 = 1 , 0 . 0 4 3 , 𝑞 = 2 .

Case 4. 0 . 0 4 8 , 𝑞 = 2 , 𝑝 = 2 , 𝑞 = 2 .

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 𝑝 = 2 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 𝑝 = 2 for 𝑞 = 3 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.

753518.fig.001
Figure 1: Evolution of the discrete solution, 𝑞 = 3 , 𝑝 = 2 (explicit scheme).
753518.fig.002
Figure 2: Evolution of the discrete solution, 𝑞 = 4 , 𝑝 = 2 (implicit scheme).
753518.fig.003
Figure 3: Evolution of the discrete solution, 𝑞 = 4 , 𝑝 = 2 (explicit scheme).
753518.fig.004
Figure 4: Evolution of the discrete solution, 𝑢 𝑡 = 𝑢 𝑥 𝑥 + 𝑢 2 , 𝑝 (implicit scheme).
753518.fig.005
Figure 5: Evolution of the discrete solution, , (explicit scheme).
753518.fig.006
Figure 6: Evolution of the discrete solution, , (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.

References

  1. T. K. Boni, “Sur l'explosion et le comportement asymptotique de la solution d'une équation parabolique semi-linéaire du second ordre,” Comptes Rendus de l'Académie des Sciences. Série I, vol. 326, no. 3, pp. 317–322, 1998. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  2. T. K. Boni, “On blow-up and asymptotic behavior of solutions to a nonlinear parabolic equation of second order with nonlinear boundary conditions,” Commentationes Mathematicae Universitatis Carolinae, vol. 40, no. 3, pp. 457–475, 1999. View at Zentralblatt MATH · View at MathSciNet
  3. M. Chipot, M. Fila, and P. Quittner, “Stationary solutions, blow up and convergence to stationary solutions for semilinear parabolic equations with nonlinear boundary conditions,” Acta Mathematica Universitatis Comenianae, vol. 60, no. 1, pp. 35–103, 1991. View at Zentralblatt MATH · View at MathSciNet
  4. V. A. Galaktionov and J. L. Vázquez, “The problem of blow-up in nonlinear parabolic equations,” Discrete and Continuous Dynamical Systems. Series A, vol. 8, no. 2, pp. 399–433, 2002. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  5. P. Quittner and P. Souplet, Superlinear Parabolic Problems: Blow-up, Global Existence and Steady States, Birkhäuser Advanced Texts. Basler Lehrbücher, Birkhäuser, Basel, Switzerland, 2007. View at Zentralblatt MATH · View at MathSciNet
  6. A. A. Samarskii, V. A. Galaktionov, S. P. Kurdyumov, and A. P. Mikhailov, Blow-up in Problems for Quasilinear Parabolic Equations, Nauka, Moscow, Russia, 1987. View at MathSciNet
  7. A. A. Samarskii, V. A. Galaktionov, S. P. Kurdyumov, and A. P. Mikhailov, Blow-up in Problems for Quasilinear Parabolic Equations, Walter de Gruyter, Berlin, Germany, 1995. View at MathSciNet
  8. L. M. Abia, J. C. López-Marcos, and J. Martínez, “On the blow-up time convergence of semidiscretizations of reaction-diffusion equations,” Applied Numerical Mathematics, vol. 26, no. 4, pp. 399–414, 1998. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  9. T. Nakagawa, “Blowing up of a finite difference solution to ut=uxx+u2,” Applied Mathematics and Optimization, vol. 2, no. 4, pp. 337–350, 1975. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  10. T. K. Boni, “Extinction for discretizations of some semilinear parabolic equations,” Comptes Rendus de l'Académie des Sciences. Série I, vol. 333, no. 8, pp. 795–800, 2001. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  11. G. Acosta, J. Fernández Bonder, P. Groisman, and J. D. Rossi, “Simultaneous vs. non-simultaneous blow-up in numerical approximations of a parabolic system with non-linear boundary conditions,” M2AN, vol. 36, no. 1, pp. 55–68, 2002. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  12. G. Acosta, J. Fernández Bonder, P. Groisman, and J. D. Rossi, “Numerical approximation of a parabolic problem with a nonlinear boundary condition in several space dimensions,” Discrete and Continuous Dynamical Systems. Series B, vol. 2, no. 2, pp. 279–294, 2002. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  13. C. Brändle, P. Groisman, and J. D. Rossi, “Fully discrete adaptive methods for a blow-up problem,” Mathematical Models & Methods in Applied Sciences, vol. 14, no. 10, pp. 1425–1450, 2004. View at Zentralblatt MATH · View at MathSciNet
  14. C. Brändle, F. Quirós, and J. D. Rossi, “An adaptive numerical method to handle blow-up in a parabolic system,” Numerische Mathematik, vol. 102, no. 1, pp. 39–59, 2005. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  15. R. G. Duran, J. I. Etcheverry, and J. D. Rossi, “Numerical approximation of a parabolic problem with a nonlinear boundary condition,” Discrete and Continuous Dynamical Systems, vol. 4, no. 3, pp. 497–506, 1998. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  16. J. Fernández Bonder and J. D. Rossi, “Blow-up vs. spurious steady solutions,” Proceedings of the American Mathematical Society, vol. 129, no. 1, pp. 139–144, 2001. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  17. A. de Pablo, M. Pérez-Llanos, and R. Ferreira, “Numerical blow-up for the p-Laplacian equation with a nonlinear source,” in Proceedings of the 11th International Conference on Differential Equations (Equadiff '05), pp. 363–367, Bratislava, Slovakia, July 2005.
  18. M. N. Le Roux, “Semidiscretization in time of nonlinear parabolic equations with blowup of the solution,” SIAM Journal on Numerical Analysis, vol. 31, no. 1, pp. 170–195, 1994. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  19. M. N. Le Roux, “Semi-discretization in time of a fast diffusion equation,” Journal of Mathematical Analysis and Applications, vol. 137, no. 2, pp. 354–370, 1989. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  20. D. Nabongo and T. K. Boni, “Numerical blow-up and asymptotic behavior for a semilinear parabolic equation with a nonlilnear boundary condition,” Albanian Journal of Mathematics, vol. 2, no. 2, pp. 111–124, 2008. View at Zentralblatt MATH · View at MathSciNet
  21. D. Nabongo and T. K. Boni, “Numerical blow-up solutions of localized semilinear parabolic equations,” Applied Mathematical Sciences, vol. 2, no. 21–24, pp. 1145–1160, 2008. View at MathSciNet
  22. F. K. N'gohissé and T. K. Boni, “Numerical blow-up solution for some semilinear heat equation,” Electronic Transactions on Numerical Analysis, vol. 30, pp. 247–257, 2008.