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

Authors: Louis A. Assalé; Théodore K. Boni; Diabate Nabongo
Journal: Journal of Applied Mathematics
 (2008)

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## Abstract

We obtain some conditions under which the positive solution for
semidiscretizations of the semilinear equationut=uxx−a(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.


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## Body

## 1. Introduction

In this paper,
we consider the following boundary value problem:(1.1)ut−uxx=−a(x,t)f(u),0<x<1,t∈(0,T),ux(0,t)=0,ux(1,t)=b(t)g(u(1,t)),t∈(0,T),u(x,0)=u0(x)≥0,0≤x≤1,where f:[0,∞)→[0,∞) is a C1 function, f(0)=0,   g:[0,∞)→[0,∞) is a C1 convex
function, g(0)=0, a∈C0([0,1]×ℝ+), a(x,t)≥0 in [0,1]×ℝ+, at(x,t)≤0 in [0,1]×ℝ+, b∈C1(ℝ+), b(t)>0 in ℝ+, b′(t)≥0 in ℝ+. The initial data u0∈C2([0,1]), u0′(0)=0, u0′(1)=b(1)g(u0(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)limt→T∥u(⋅,t)∥∞=+∞,where ∥u(⋅,t)∥∞=max0≤x≤1|u(x,t)|.In this last case, we say that the solutionu 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)=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 [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

LetI 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(t))T of the
following semidiscrete equations(2.1)dUi(t)dt−δ2Ui(t)=−ai(t)f(Ui(t)),0≤i≤I−1,t∈(0,Tbh),(2.2)dUI(t)dt−δ2UI(t)=2hb(t)g(UI(t))−aI(t)f(UI(t)),t∈(0,Tbh),(2.3)Ui(0)=φi≥0,0≤i≤I,where φi+1≥φi, 0≤i≤I−1,(2.4)δ2U0(t)=2U1(t)−2U0(t)h2,δ2UI(t)=2UI−1(t)−2UI(t)h2,δ2Ui(t)=Ui+1(t)−2Ui(t)+Ui−1(t)h2.Here (0,Tbh) is the maximal
time interval on which ∥Uh(t)∥∞ is finite where ∥Uh(t)∥∞=max0≤i≤IUi(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.
Letah(t)∈C0([0,T),ℝI+1) and let Vh(t)∈C1([0,T),ℝI+1) such
that(3.1)dVi(t)dt−δ2Vi(t)+ai(t)Vi(t)≥0,0≤i≤I,t∈(0,T),Vi(0)≥0,0≤i≤I.Then we have Vi(t)≥0, 0≤i≤I, t∈(0,T).Proof.
LetT0<T and define the
vector Zh(t)=eλtVh(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≤T0Zi(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}. It is not hard to see that(3.2)dZi0(t0)dt=limk→0Zi0(t0)−Zi0(t0−k)k≤0,δ2Zi0(t0)=Zi0+1(t0)−2Zi0(t0)+Zi0−1(t0)h2≥0if1≤i0≤I−1,δ2Zi0(t0)=2Z1(t0)−2Z0(t0)h2≥0if⁢i0=0,δ2Zi0(t0)=2ZI−1(t0)−2ZI(t0)h2≥0ifi0=I.A straightforward computation
reveals that(3.3)dZi0(t0)dt−δ2Zi0(t0)+(ai0(t0)−λ)Zi0(t0)≥0.We observe from (3.2) that (ai0(t0)−λ)Zi0(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.
LetVh(t),Uh(t)∈C1([0,T),ℝI+1) and f∈C0(ℝ×ℝ,ℝ) such that for t∈(0,T)(3.4)dVi(t)dt−δ2Vi(t)+f(Vi(t),t)<dUi(t)dt−δ2Ui(t)+f(Ui(t),t),0≤i≤I,(3.5)Vi(0)<Ui(0),0≤i≤I.Then we have Vi(t)<Ui(t), 0≤i≤I, t∈(0,T).Proof.
Define the vectorZh(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(3.6)dZi0(t0)dt=limk→0Zi0(t0)−Zi0(t0−k)k≤0,δ2Zi0(t0)=Zi0+1(t0)−2Zi0(t0)+Zi0−1(t0)h2≥0if1≤i0≤I−1,δ2Zi0(t0)=2Z1(t0)−2Z0(t0)h2≥0ifi0=0,δ2Zi0(t0)=2ZI−1(t0)−2ZI(t0)h2≥0ifi0=I,   which implies that(3.7)dZi0(t0)dt−δ2Zi0(t0)+f(Ui0(t0),t0)−f(Vi0(t0),t0)≤0.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 solutionUh 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 operatorδ2 and the second
one reveals a property of the semidiscrete solution.Lemma 4.1.
LetUh∈ℝI+1 be such that Uh≥0. Then we have(4.1)δ2g(Ui)≥g′(Ui)δ2Uifor0≤i≤I.Proof.
Apply Taylor's expansion to obtain(4.2)g(U1)=g(U0)+(U1−U0)g′(U0)+(U1−U0)22g′′(η0),g(Ui+1)=g(Ui)+(Ui+1−Ui)g′(Ui)+(Ui+1−Ui)22g′′(θi),1≤i≤I−1,g(Ui−1)=g(Ui)+(Ui−1−Ui)g′(Ui)+(Ui−1−Ui)22g′′(ηi),1≤i≤I−1,g(UI−1)=g(UI)+(UI−1−UI)g′(UI)+(UI−1−UI)22g′′(ηI),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)δ2g(U0)=g′(U0)δ2U0+(U1−U0)2h2g′′(η0),δ2g(UI)=g′(UI)δ2UI+(UI−1−UI)2h2g′′(ηI).Combining the second and third
equalities, we see that(4.4)δ2g(Ui)=g′(Ui)δ2Ui+(Ui+1−Ui)22h2g′′(θi)+(Ui−1−Ui)22h2g′′(ηi),1≤i≤I−1.Use the fact that g′′(s)≥0 for s≥0 and Uh≥0 to complete the
rest of the proof.Lemma 4.2.
LetUh be the solution
of (2.1)–(2.3). Then we have(4.5)Ui+1(t)>Ui(t),0≤i≤I−1,t∈(0,Tbh).Proof.
Lett0 be the first t>0 such that Ui+1(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)=Ui+1(t)−Ui(t) for 0≤i≤I−1. We get(4.6)dZi0(t0)dt=limk→0Zi0(t0)−Zi0(t0−k)k≤0,δ2Zi0(t0)=Zi0+1(t0)−2Zi0(t0)+Zi0−1(t0)h2>0if1≤i0≤I−2,δ2Zi0(t0)=δ2Z0(t0)=Z1(t0)−3Z0(t0)h2>0ifi0=0,δ2Zi0(t0)=δ2ZI−1(t0)=ZI−2(t0)−3ZI−1(t0)h2>0ifi0=I−1,which implies
that(4.7)dZi0(t0)dt−δ2Zi0(t0)−ai0+1(t0)f(Ui0+1(t0))+ai0(t0)f(Ui0(t0))<0if0≤i0≤I−2,dZi0(t0)dt−δ2Zi0(t0)+2hb(t0)gi0+1(t0)−ai0+1(t0)f(Ui0+1(t0))+ai0(t0)f(Ui0(t0))<0ifi0=I−1.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 nodexI.Now, we are in a position to state the main result of
this section.Theorem 4.3.
LetUh be the solution
of (2.1)–(2.3). Suppose that there exists a positive integer A such
that(4.8)δ2φi−ai(0)f(φi)≥0,1≤i≤I−1,δ2φI−aI(0)f(φI)+b(0)gI(φI)≥Ag(φI).Assume that(4.9)f(s)g′(s)−f′(s)g(s)≥0fors≥0.Then the solution Uh blows up in a
finite time Tbh and we have the
following estimate(4.10)Tbh≤1A∫∥φh∥∞+∞dsg(s).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(4.11)Ji(t)=dUi(t)dt,0≤i≤I−1,JI(t)=dUI(t)dt−Ag(UI(t)).A straightforward calculation
gives(4.12)dJidt−δ2Ji=ddt(dUidt−δ2Ui),0≤i≤I−1,dJIdt−δ2JI=ddt(dUIdt−δ2UI)−Ag′(UI)dUIdt+Aδ2g(UI).From Lemma 4.1, we have δ2g(UI)≥g′(UI)δ2UI which implies
that(4.13)dJIdt−δ2JI≥ddt(dUIdt−δ2UI)−Ag′(UI)(dUIdt−δ2UI).Using (2.1), we
get(4.14)dJidt−δ2Ji≥−ai′(t)f(Ui)−ai(t)f′(Ui)dUidt,0≤i≤I−1,dJIdt−δ2JI≥−aI′(t)f(UI)−aI(t)f′(UI)dUIdt+2hb′(t)g(UI)+2hb(t)g′(UI)dUIdt−Ag′(UI)(−aI(t)f(UI)+2hb(t)g(UI)).It follows from the fact that ai′(t)≤0, b′(t)≥0 and dUi/dt=Ji+Ag(Ui) that(4.15)dJIdt−δ2JI≥(−aI(t)f′(UI)+2hb(t)g′(UI))JI+AaI(t)(g′(UI)f(UI)−f′(UI)g(UI)).We deduce from (4.9)
that(4.16)dJidt−δ2Ji≥−ai(t)f′(Ui)Ji,0≤i≤I−1,dJIdt−δ2JI≥(−aI(t)f′(UI)+2hb(t)g′(UI))JI.From (4.8), we observe
that(4.17)Ji(0)=δ2φi−ai(0)f(φi)≥0,0≤i≤I−1,JI(0)=δ2φI−aI(0)f(φI)+b(0)gI(φI)−Ag(φI)≥0. We deduce from Lemma 3.1 that Ji(t)≥0, 0≤i≤I, which implies that dUI/dt≥g(UI), 0≤i≤I. Obviously we have(4.18)dUIg(UI)≥A⁢dt.Integrating this inequality over (t,Tbh), we arrive at(4.19)Tbh−t≤1A∫UI(t)+∞dsg(s),which implies
that(4.20)Tbh≤1A∫∥Uh(0)∥∞+∞dsg(s).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)Tbh−t0≤1A∫∥Uh(t0)∥∞+∞dsg(s)if0<t0<Tbh,Ui(t)≤H(A(Tbh−t)),0≤i≤I,where H(s) is the inverse
of G(s)=∫s+∞(dz/g(z)).Remark 4.5.
Ifg(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 u∈C4,1([0,1]×[0,T]) and the initial
condition at (2.3) satisfies(5.1)∥φh−uh(0)∥∞=o(1)ash⟶0,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 Uh∈C1([0,T],ℝI+1) such
that(5.2)max0≤t≤T∥Uh(t)−uh(t)∥∞=O(∥φh−uh(0)∥∞+h2)ash⟶0.Proof.
Letα>0 be such
that(5.3)∥u(⋅,t)∥∞≤αfort∈[0,T].The problem (2.1)–(2.3) has for
each h, a unique solution Uh∈C1([0,Tbh),ℝI+1). Let t(h)≤min{T,Tbh} the greatest
value of t>0 such
that(5.4)∥Uh(t)−uh(t)∥∞<1fort∈(0,t(h)).The relation (5.1) implies that t(h)>0 for h sufficiently
small. By the triangle inequality, we obtain(5.5)∥Uh(t)∥∞≤∥u(⋅,t)∥∞+∥Uh(t)−uh(t)∥∞fort∈(0,t(h)),which implies
that(5.6)∥Uh(t)∥∞≤1+αfort∈(0,t(h)).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)dei(t)dt−δ2ei(t)=−ai(t)f′(ξi(t))ei(t)+o(h2),0≤i≤I−1,deI(t)dt−δ2eI(t)=−aI(t)f′(ξI(t))eI(t)+2hb(t)g′(UI(t))eI(t)+o(h2),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)dei(t)dt−δ2ei(t)≤L|ei(t)|+Kh2,0≤i≤I−1,deI(t)dt−(2eI−1(t)−2eI(t))h2≤L|eI(t)|h+L|eI(t)|+Kh2.Consider the function z(x,t)=e((M+1)t+Cx2)(∥φh−uh(0)∥∞+Qh2) where M, C, Q are constants
which will be determined later. We get(5.9)zt(x,t)−zxx(x,t)=(M+1−2C−4C2x2)z(x,t),zx(0,t)=0,zx(1,t)=2Cz(1,t),z(x,0)=eCx2(∥φh−uh(0)∥∞+Qh2).By a semidiscretization of the
above problem, we may choose M, C, Q large enough
that(5.10)ddtz(xi,t)>δ2z(xi,t)+L|z(xi,t)|+Kh2,0≤i≤I−1,ddtz(xI,t)>δ2z(xI,t)+Lh|z(xI,t)|+L|z(xI,t)|+Kh2,z(xi,0)>ei(0),0≤i≤I.It follows from Lemma 3.2
that(5.11)z(xi,t)>ei(t)fort∈(0,t(h)),0≤i≤I.By the same way, we also prove
that(5.12)z(xi,t)>−ei(t)fort∈(0,t(h)),0≤i≤I,which implies
that(5.13)z(xi,t)>|ei(t)|fort∈(0,t(h)),0≤i≤I.We deduce that(5.14)∥Uh(t)−uh(t)∥∞≤e(Mt+C)(∥φh−uh(0)∥∞+Qh2),t∈(0,t(h)).Let us show that t(h)=T. Suppose that T>t(h). From (5.4), we obtain(5.15)1=∥Uh(t(h))−uh(t(h))∥∞≤e(MT+C)(∥φh−uh(0)∥∞+Qh2).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 u∈C4,1([0,1]×[0,Tb)) and the initial
condition at (2.3) satisfies(5.16)∥φh−uh(0)∥∞=o(1)ash⟶0.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(5.17)limh→0Tbh=Tb.Proof.
Letε>0. There exists a positive constant N such
that(5.18)1A∫x+∞dsg(s)≤ε2forx∈[N,+∞).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]|u(x,t)|<∞. It follows from Theorem 5.1 that(5.19)supt∈[0,T2]|Uh(t)−uh(t)|∞≤N.Applying the triangle
inequality, we get(5.20)∥Uh(T2)∥∞≥∥uh(T2)∥∞−∥Uh(T2)−uh(T2)∥∞,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(5.21)|Tb−Tbh|≤|Tb−T2|+|Tbh−T2|≤ε2+1A∫∥Uh(T2)∥∞+∞dsg(s)≤ε,and the proof is complete.


## 6. Numerical Blow-Up Rate

In this
section, we determine the blow-up rate of the solutionUh of (2.1)–(2.3) in
the case where b(t)=1. Our result is the following.Theorem 6.1.
LetUh(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(6.1)H(C1(Tbh−t))≤UI(t)≤H(C2(Tbh−t)),fort∈(0,Tbh),where H(s) is the inverse
of the function G(s)=∫s+∞(dσ/g(σ)).Proof.
From
Theorem4.3 and Remark 4.4, Uh(t) blows up in a
finite time Tbh and there
exists a constant C2>0 such that(6.2)UI(t)≤H(C2(Tbh−t))fort∈(0,Tbh).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 (t,Tbh), there exists a positive constant C1 such that(6.3)UI(t)≥H(C1(Tbh−t))fort∈(0,Tbh),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 constantC0 such that sF′(s)≤C0 and(7.1)ddtUi−δ2Ui≤0,0≤i≤I−1.Assume that there exists a
positive constant C such(7.2)Ui≤H(C(T−t)),0≤i≤I.Then the numerical blow-up set
is B={1}.Proof.
Letv(x)=1−x2 and define(7.3)W(x,t)=H(δv(x)+δB(T−t))for0≤x≤1,t≥t0,where δ is small
enough. We have(7.4)Wx(0,t)=0,W(1,t)=H(δB(T−t))≥u(1,t),and for t≥t0, we get(7.5)W(x,t0)=H(δv(x)+δ)≥H(2δ)=H(2δB(T−t0))≥H(C(T−t0))≥u(x,t0).A straightforward computation
yields(7.6)Wt(x,t)−Wxx(x,t)=δF(H(τ))(B−2−4xF′(H(τ)))≥δF(H(τ))(B−2−4δC0).This implies that there exists α>0 such
that(7.7)Wt(x,t)−Wxx(x,t)≥αF(H(δ+δBT)).Using Taylor's expansion, there
exists a constant K>0 such
that(7.8)ddtW(xi,t)−δ2W(xi,t)≥αF(H(δ+δBT))−Kh2,0≤i≤I,which implies
that(7.9)dW(xi,t)dt−δ2W(xi,t)≥0.The maximum principle implies
that(7.10)Ui(t)≤H(δv(x)+δB(T−t0))fort≥t0,0≤i≤I.Hence, we get(7.11)Ui(t)≤H(δv(x)),0≤i≤I.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 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)ut(x,t)=uxx(x,t)−up(x,t),0<x<1,t∈(0,T),ux(0,t)=0,ux(1,t)=up(1,t),t∈(0,T),u(x,0)=u0(x)>0,0≤x≤1,where p>1, u0∈C1([0,1]), u0′(0)=0 and u0′(1)=u0p(1).We start this section by the construction of an
adaptive scheme as follows. LetI be a positive
integer and let h=1/I. Define the grid xi=ih, 0≤i≤I and approximate
the solution u(x,t) of the problem
(8.1) by the solution Uh(n)=(U0(n),U1(n),…,UI(n))T of the
following discrete equations(8.2)δtUi(n)=δ2Ui(n)−(Ui(n))p,0≤i≤I−1,(8.3)δtUI(n)=δ2UI(n)−(UI(n))p+2h(UI(n))p,(8.4)Ui(0)=φi,0≤i≤I,where n≥0, φi+1≥φi, 0≤i≤I−1,(8.5)δtUi(n)=Ui(n+1)−Ui(n)Δtn,δ2Ui(n)=Ui+1(n)−2Ui(n)+Ui−1(n)h2,1≤i≤I−1,δ2U0(n)=2U1(n)−2U0(n)h2,δ2UI(n)=2UI−1(n)−2UI(n)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 take Δtn=min{(1−pτ)h2/3,τ/∥Uh(n)∥∞p−1}, 0<τ<1/p.Let us notice that the restriction on the time step
ensures the nonnegativity of the discrete solution. The lemma below shows that
the discrete solution is increasing in space.Lemma 8.1.
LetUh(n) be the solution
of (8.2)–(8.4). Then we have(8.6)Ui+1(n)≥Ui(n),0≤i≤I−1.Proof.
LetZi(n)=Ui+1(n)−Ui(n), 0≤i≤I−1. We observe that(8.7)Z0(n+1)−Z0(n)Δtn=Z1(n)−3Z0(n)h2−((U1(n))p−(U0(n))p),Zi(n+1)−Zi(n)Δtn=Zi+1(n)−2Zi(n)+Zi−1(n)h2−((Ui+1(n))p−(Ui(n))p),1≤i≤I−2,ZI−1(n+1)−ZI−1(n)Δtn=ZI−2(n)−3ZI−1(n)h2−((UI(n))p−(UI−1(n))p)+2h(UI(n))p.Using the Taylor's expansion, we
find that(8.8)Z0(n+1)=Δtnh2Z1(n)+(1−3Δtnh2)Z0(n)−Δtnp(ξ0(n))p−1Z0(n),Zi(n+1)=Δtnh2Zi+1(n)+(1−2Δtnh2)Zi(n)+Δtnh2Zi−1(n)−Δtnp(ξi(n))p−1Zi(n),1≤i≤I−2,ZI−1(n+1)≥Δtnh2ZI−2(n)+(1−3Δtnh2)ZI−1(n)−Δtnp(ξI−1(n))p−1ZI−1(n),where ξi(n) is an
intermediate value between Ui(n) and Ui+1(n). If Zi(n)≤0, 0≤i≤I−1, we deduce that(8.9)Z0(n+1)≥Δtnh2Z1(n)+(1−3Δtnh2−Δtnp∥Uh(n)∥∞p−1)Z0(n),Zi(n+1)≥Δtnh2Zi+1(n)+(1−2Δtnh2−Δtnp∥Uh(n)∥∞p−1)Zi(n)+Δtnh2Zi−1(n),1≤i≤I−2,ZI−1(n+1)≥Δtnh2ZI−2(n)+(1−3Δtnh2−Δtnp∥Uh(n)∥∞p−1)ZI−1(n).Using the restriction Δtn≤τ/∥Uh(n)∥∞p−1, we find that(8.10)Z0(n+1)≥Δtnh2Z1(n)+(1−3Δtnh2−pτ)Z0(n),Zi(n+1)≥Δtnh2Zi+1(n)+(1−2Δtnh2−pτ)Zi(n)+Δtnh2Zi−1(n),1≤i≤I−2,ZI−1(n+1)≥Δtnh2ZI−2(n)+(1−3Δtnh2−pτ)ZI−1(n).We observe that 1−3(Δtn/h2)−pτ is nonnegative
and by induction, we deduce that Zi(n)≤0, 0≤i≤I−1. This ends the proof.The following lemma is a discrete form of the maximum
principle.Lemma 8.2.
Letah(n) be a bounded
vector and let Vh(n) a sequence such
that(8.11)δtVi(n)−δ2Vi(n)+ai(n)Vi(n)≥0,0≤i≤I,n≥0,(8.12)Vi(0)≥0,0≤i≤I.Then Vi(n)≥0 for n≥0, 0≤i≤I if Δtn≤h2/(2+∥ah(n)∥∞h2).Proof.
IfVh(n)≥0 then a routine
computation yields(8.13)V0(n+1)≥2Δtnh2V1(n)+(1−2Δtnh2−Δtn∥ah(n)∥∞)V0(n),Vi(n+1)≥Δtnh2Vi+1(n)+(1−2Δtnh2−Δtn∥ah(n)∥∞)Vi(n)+Δtnh2Vi−1(n),1≤i≤I−1,VI(n+1)≥2Δtnh2VI−1(n)+(1−2Δtnh2−Δtn∥ah(n)∥∞)VI(n).Since Δtn≤h2/(2+∥ah(n)∥∞h2), we see that 1−2(Δtn/h2)−Δtn∥ah(n)∥∞ is nonnegative. From (8.12), we deduce by induction that Vh(n)≥0 which ends the
proof.A direct consequence of the above result is the
following comparison lemma. Its proof is straightforward.Lemma 8.3.
Suppose
thatah(n) and bh(n) two vectors
such that ah(n) is bounded. Let Vh(n) and Wh(n) two sequences
such that(8.14)δtVi(n)−δ2Vi(n)+ai(n)Vi(n)+bi(n)≤δtWi(n)−δ2Wi(n)+ai(n)Wi(n)+bi(n),0≤i≤I,n≥0,Vi(0)≤Wi(0),0≤i≤I.Then Vi(n)≤Wi(n) for n≥0, 0≤i≤I if Δtn≤h2/(2+∥ah(n)∥∞h2).Now, let us give a
property of the operatorδt.Lemma 8.4.
LetU(n)∈ℝ be such that U(n)≥0 for n≥0. Then we have(8.15)δt(U(n))p≥p(U(n))p−1δtU(n),n≥0.Proof.
From Taylor's expansion, we find that(8.16)δt(U(n))p=p(U(n))p−1δtU(n)+p(p−1)2Δtn(δtU(n))2(θ(n))p−2,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 solutionUh(n) of (8.2)–(8.4)
blows up in a finite time if (8.17)limn→+∞∥Uh(n)∥∞=+∞,ThΔt=limn→∞∑i=0n−1Δti<+∞.The number ThΔt is called the
numerical blow-up time of Uh(n).The following
theorem reveals that the discrete solutionUh(n) of (8.2)–(8.4)
blows up in a finite time under some hypotheses.Theorem 8.6.
LetUh(n) 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)δ2φi−φip≥0,0≤i≤I−1.δ2φI−φIp+2hφIp≥AφIp.Then Uh(n) blows up in a
finite time ThΔt which satisfies
the following estimate(8.19)ThΔt≤τ(1+τ′)p−1((1+τ′)p−1−1)∥φh∥∞p−1,where τ′=Amin{(1−pτ)h2∥φh∥inf−p−1/3,τ}.Proof.
Introduce the vectorJh(n) defined as
follows(8.20)Ji(n)=δtUi(n),0≤i≤I−1,n≥0,JI(n)=δtUI(n)−A(UI(n))−p,n≥0. A straightforward computation
yields(8.21)δtJi(n)−δ2Ji(n)=δt(δtUi(n)−δ2Ui(n)),0≤i≤I−1,δtJI(n)−δ2JI(n)=δt(δtUI(n)−δ2UI(n))−Aδt(UI(n))p+Aδ2(UI(n))p.Using (8.2), we arrive
at(8.22)δtJi(n)−δ2Ji(n)=−δt(Ui(n))p,0≤i≤I−1,δtJI(n)−δ2JI(n)=(2h−1−A)δt(UI(n))p+Aδ2(UI(n))p.Due to the mean value theorem,
we get(8.23)δt(Ui(n))p=p(ξi(n))p−1δt(Ui(n))=p(ξi(n))p−1Ji(n),where ξi(n) is an
intermediate value between Ui(n) and Ui+1(n). On the other hand, from Lemmas 2.4 and 2.5, we
deduce that(8.24)δtJi(n)−δ2Ji(n)=−p(ξi(n))p−1Ji(n),0≤i≤I−1,δtJI(n)−δ2JI(n)=(2h−1−A)p(UI(n))p−1δtUI(n)+Apδt(UI(n))p−1δ2UI(n).It follows from (8.3)
that(8.25)δtJI(n)−δ2JI(n)=(2h−1)p(UI(n))p−1δtUI(n)−Apδt(UI(n))p−1((2h−1)(UI(n))p),which implies
that(8.26)δtJi(n)−δ2Ji(n)=−p(ξi(n))p−1Ji(n),0≤i≤I−1,δtJI(n)−δ2JI(n)=(2h−1)p(UI(n))p−1JI(n).From (8.18), we observe that Jh(0)≥0. It follows from Lemma 8.2 that Jh(n)≥0 which implies
that(8.27)UI(n+1)≥UI(n)(1+AΔtn(UI(n))p−1).From Lemma 8.1, we see that UI(n)=∥Uh(n)∥∞ which implies
that(8.28)∥Uh(n+1)∥∞≥∥Uh(n)∥∞(1+AΔtn∥Uh(n)∥∞p−1).It is not hard to see
that(8.29)AΔtn∥Uh(n)∥∞p−1=Amin{(1−pτ)h2∥Uh(n)∥∞p−13,τ}.From (8.28), we get ∥Uh(n+1)∥∞≥∥Uh(n)∥∞. By induction, we arrive at ∥Uh(n+1)∥∞≥∥Uh(0)∥∞=∥φh∥∞, which implies that ∥Uh(n)∥∞p−1≥∥φh∥∞p−1. Therefore, we find that(8.30)AΔtn∥Uh(n)∥∞p−1≥Amin{(1−pτ)h2∥φh∥∞p−13,τ}=τ′.Consequently, we arrive
at(8.31)∥Uh(n+1)∥∞≥∥Uh(n)∥∞(1+τ′)and by induction, we
get(8.32)∥Uh(n)∥∞≥∥Uh(0)∥∞(1+τ′)n=∥φh∥∞(1+τ′)n,n≥0.Since the term on the right hand
side of the above equality tends to infinity as n approaches
infinity, we conclude that ∥Uh(n)∥∞ 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)Σn=0+∞⁢Δtn≤Σn=0+∞⁢τ∥Uh(n)∥∞p−1≤τ∥φh∥∞p−1Σn=0+∞⁢(1(1+τ′)p−1)n.Using the fact that the series
on the right hand side of the above inequality converges towards τ(1+τ′)p−1/((1+τ′)p−1−1), we deduce that Σn=0+∞Δtn≤Σn=0+∞(τ(1+τ′)p−1/((1+τ′)p−1−1)∥φh∥∞p−1) and the proof
is complete.Remark 8.7.
Apply
Taylor's expansion to obtain(1+τ′)p−1=1−(p−1)τ′+o(τ′), which implies that (8.34)τ(1+τ′)p−1−1=ττ′(1p−1+o(1))≤2ττ′(p−1).If we take τ=h2, we see that(8.35)ττ′=Amin{(1−ph2)3∥φh∥∞p−1,1}≥Amin{14∥φh∥∞p−1,1}.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 (8.36)∥Uh(n)∥∞≥∥Uh(q)∥∞(1+τ′)n−qforn≥qwhich implies
that(8.37)Σn=q+∞⁢Δtn≤τ∥Uh(q)∥∞p−1Σn=q+∞[1(1+τ′)p−1]n−q.We deduce that(8.38)ThΔt−tq≤τ∥Uh(q)∥∞p−1(1+τ′)p−1(1+τ′)p−1−1.In the sequel,
we takeτ=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 u∈C4,2([0,1]×[0,T]). Assume that the initial data at (8.4)
satisfies(9.1)∥φh−uh(0)∥∞=o(1)ash⟶0.Then the problem (8.2)–(8.4) has
a solution Uh(n) for h
sufficiently small, 0≤n≤J and we have the
following relation(9.2)max0≤n≤J∥Uh(n)−uh(tn)∥∞=O(∥φh−uh(0)∥∞+h2+Δtn)ash⟶0,where J is such that ∑n=0J−1Δtn≤T and tn=∑j=0n−1Δtj.Proof.
For eachh, the problem (8.2)–(8.4) has a solution Uh(n). Let N≤J be the greatest
value of n such
that(9.3)∥Uh(n)−uh(tn)∥∞<1forn<N.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(9.4)∥Uh(n)∥∞≤∥uh(tn)∥∞+∥Uh(n)−uh(tn)∥∞≤1+Kforn<N.Since u∈C4,2, using Taylor's expansion, we find that(9.5)δtu(xi,tn)−δ2u(xi,tn)=−up(xi,tn)+O(h2)+O(Δtn),0≤i≤I−1,δtu(xI,tn)−δ2u(xI,tn)=−up(xI,tn)+2hup(xI,tn)+O(h2)+O(Δtn).Let eh(n)=Uh(n)−uh(tn) be the error of
discretization. From the mean value theorem, we get(9.6)δtei(n)−δ2ei(n)=−p(ξi(n))p−1ei(n)+O(h2)+O(Δtn),0≤i≤I−1,δteI(n)−δ2eI(n)=p(2h−1)(ξI(n))p−1eI(n)+O(h2)+O(Δtn), where ξi(n) is an
intermediate value between u(xi,tn) and Ui(n). Hence, there exist positive constants L and K such
that(9.7)δtei(n)−δ2ei(n)≤−p(ξi(n))p−1ei(n)+Lh2+LΔtn,0≤i≤I−1,n<N,δteI(n)−δ2eI(n)≤p(2h−1)(ξI(n))p−1eI(n)+Lh2+LΔtn,n<N.Consider the function Z(x,t)=e((M+1)t+Cx2)(∥φh−uh(0)∥∞+Qh2+QΔtn) where M, C, Q are positive
constants which will be determined later. We get(9.8)Zt(x,t)−Zxx(x,t)=(M+1−2C−4C2x2)Z(x,t),Zx(0,t)=0,Zx(1,t)=2CZ(1,t),Z(x,0)=e(Cx2)(∥φh−uh(0)∥∞+Qh2+QΔtn).By a discretization of the above
problem, we obtain(9.9)δtZ(xi,tn)−δ2Z(xi,tn)=(M+1−2C−4C2xi2)Z(xi,tn)+h212Zxxxx(xi,tn)−Δtn2Ztt(xi,tn),δtZ(xI,tn)−δ2Z(xI,tn)=(M+1−2C−4C2xI2)Z(xI,tn)+4ChZ(xI,tn)+h212Zxxxx(xI,tn)−Δtn2Ztt(xI,tn).We may choose M, C, Q large enough
that(9.10)δtZ(xi,tn)−δ2Z(xi,tn)>−p(ξi(n))p−1Z(xi,tn)+Lh2+LΔtn,0≤i≤I−1,δtZ(xI,tn)−δ2Z(xI,tn)>p(2h−1)(ξI(n))p−1Z(xI,tn)+Lh2+LΔtn,Zi(0)>ei(0),0≤i≤I.It follows from Comparison Lemma
8.3 that(9.11)Z(xi,tn)>ei(n),0≤i≤I,n<N.By the same way, we also prove
that(9.12)Z(xi,tn)>−ei(n),0≤i≤I,n<N,which implies
that(9.13)∥Uh(n)−uh(t)∥∞≤e(Mtn+C)(∥φh−uh(0)∥∞+Qh2+QΔtn),n<N.Let us show that N=J. Suppose that N<J. From (9.3), we obtain(9.14)1≤∥Uh(N)−uh(tN)∥∞≤e(MT+C)(∥φh−uh(0)∥∞+Qh2+QΔtn).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 u∈C4,2([0,1]×[0,T0)). Assume that the initial data at (2.3)
satisfies(9.15)∥φh−uh(0)∥∞=o(1)ash⟶0.Under the assumption of Theorem
8.6, the problem (8.2)–(8.4) has a solution Uh(n) which blows up
in a finite time ThΔt and the
following relation holds(9.16)limh→0ThΔt=T0.Proof.
We know from Remark8.7 that τ(1+τ′)/((1+τ′)p−1−1) is bounded. Letting ε>0, there exists a constant R>0 such
that(9.17)τ(1+τ′)p−1xp−1((1+τ′)p−1−1)<ε2forx∈[R,∞).Since u blows up at the
time T0, there exists T1∈(T0−ε/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 tq=∑n=0q−1Δtn∈[T1,T2] for h small enough. We have 0<∥uh(tn)∥∞<∞ for n≤q. It follows from Theorem 4.3 that the problem
(2.1)–(2.3) has a solution Uh(n) which obeys ∥Uh(n)−uh(tn)∥∞<R for n≤q, which implies that(9.18)∥Uh(q)∥∞≥∥uh(tq)∥∞−∥Uh(q)−uh(tq)∥∞≥R.From Theorem 8.6, Uh(n) blows up at the
time ThΔt. It follows from Remark 8.8 and (9.17) that |ThΔt−tq|≤τ(1+τ′)p−1∥Uh(q)∥∞1−p/((1+τ′)p−1−1)<ε/2 because ∥Uh(q)∥∞≥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 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 Uh(n) of the
following explicit scheme(10.1)δtUi(n)=δ2Ui(n)−λ(Ui(n))p−1Ui(n+1),0≤i≤I−1,δtUI(n)=δ2UI(n)+2h(UI(n))q−λ(UI(n))p−1UI(n+1),Ui(0)=φi≥0,0≤i≤I,We also approximate the solution u of (1.1) by
the solution Uh(n) of the implicit
scheme below(10.2)δtUi(n)=δ2Ui(n+1)−λ(Ui(n))p−1Ui(n+1),0≤i≤I−1,δtUI(n)=δ2UI(n+1)+2h(UI(n))q−λ(UI(n))p−1UI(n+1),Ui(0)=φi≥0,0≤i≤I.For the time step, we take n≥0, Δtn=min(h2/2,τ∥Uh(n)∥∞1−p) for the
explicit scheme and Δtn=τ∥Uh(n)∥∞1−p for the
implicit scheme.The problem described in (10.1) may be rewritten
as follows(10.3)U0(n+1)=2(Δtn/h2)U1(n)+(1−2(Δtn/h2))U0(n)1+λΔtn(U0(n))p−1,Ui(n+1)=2(Δtn/h2)Ui+1(n)+(1−2(Δtn/h2))Ui(n)+2(Δtn/h2)Ui−1(n)1+λΔtn(Ui(n))p−1,UI(n+1)=2(Δtn/h2)UI−1(n)+(1−2(Δtn/h2))UI(n)+2(Δtn/h2)(UI(n))q1+λΔtn(UI(n))p−1.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(10.4)AhnUh(n+1)=Fn,where(10.5)Ah(n)=(a0b00⋯0c1a1b10⋯0⋱⋱⋱⋮⋱⋱⋱bI−10⋯0cIaI),ai=1+2Δtnh2+λΔtn(Ui(n))p−1,0≤i≤I,bi=−2Δtnh2,i=0,…,I−1,ci=−2Δtnh2,i=1,…,I,(Fn)i=Ui(n),i=0,…,I−1,(Fn)I=UI(n)+2hΔtn(UI(n))q.     The matrix Ah(n) satisfies the
following properties(10.6)(Ah(n))ii>0,(Ah(n))ij<0,ifi≠j,|(Ah(n))ii|>∑j≠i|(Ah(n))ij|. It follows that Uh(n) exists for n≥0. In addition, since Uh(0) is nonnegative, Uh(n) is also
nonnegative for n≥0. We need the following definition.Definition 10.1.
We say that the discrete solutionUh(n) of the explicit
scheme or the implicit scheme blows up in a finite time if limn→+∞∥Uh(n)∥∞=+∞ and the series ∑n=0+∞Δtn converges. The
quantity ∑n=0+∞Δtn is called the
numerical blow-up time of the solution Uh(n).In Tables1, 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 Tn=∑j=0n−1Δtj which is
computed at the first time when(10.7)Δtn=|Tn+1−Tn|≤10−16.The order (s) of the method
is computed from(10.8)s=log((T4h−T2h)/(T2h−Th))log(2).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).
ITnnCPU times160.047927451——320.04469512600.5—640.043583407551.541280.04322514555601.642560.0431155506118161.71Table 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).
ITnnCPU times160.047631423——320.04464512341—640.043576405051.491280.04322414533991.612560.0431135503520001.67Table 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).
ITnnCPU times160.018286217503—320.0171818383817—640.0167293299601081.301280.016412129875015700.512560.0163246447649270491.85Table 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).
ITnnCPU times160.018283217416—320.0171818383137—640.01672932999533471.301280.016617120849546402.012560.0165266348765299570.30Table 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).
ITnnCPU times160.0241971649——320.02257061032—640.0219502358381.401280.021734929852001.522560.02171236925032433.30Table 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).
ITnnCPU times160.0241691602——320.02256660665—640.02195023551651.381280.0217349298511401.522560.02171337024067093.37Table 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).
ITnnCPU times160.054342422——320.0503461130——640.049027353941.601280.04861512020281.682560.048491464399371.74Table 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).
ITnnCPU times160.054158364——320.05033210770.6—640.049030349171.561280.04861612358791.662560.0485193691911232.10Case 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 to0.043 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 Figures1, 2, 3, 4, 5, and 6, we can appreciate that the discrete solution blows up in a
finite time at the last node.Figure 1
Evolution of the discrete
solution,q=2, p=2 (explicit
scheme).Figure 2
Evolution of the discrete
solution,q=2, p=2 (implicit
scheme).Figure 3
Evolution of the discrete
solution,q=3, p=2 (explicit
scheme).Figure 4
Evolution of the discrete
solution,q=3, p=2 (implicit
scheme).Figure 5
Evolution of the discrete
solution,q=4, p=2 (explicit
scheme).Figure 6
Evolution of the discrete
solution,q=4, p=2 (implicit
scheme).



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*Source: 753518-2009-04-12.xml*