Boni.dvi


CUBO A Mathematical Journal
Vol.12, No¯ 01, (23–40). March 2010

Quenching for Discretizations of a Nonlocal Parabolic

Problem with Neumann Boundary Condition

Théodore K. Boni

Institut National Polytechnique Houphouët-Boigny de Yamoussoukro,

BP 1093 Yamoussoukro, (Côte d’Ivoire)

email : theokboni@yahoo.fr

and

Diabaté Nabongo

Université d’Abobo-Adjamé, UFR-SFA,

Département de Mathématiques et Informatiques,

16 BP 372 Abidjan 16, (Côte d’Ivoire)

email : nabongo diabate@yahoo.fr

ABSTRACT

In this paper, under some conditions, we show that the solution of a discrete form of a nonlocal

parabolic problem quenches in a finite time and estimate its numerical quenching time. We

also prove that the numerical quenching time converges to the real one when the mesh size

goes to zero. Finally, we give some computational results to illustrate our analysis.

RESUMEN

En este art́ıculo mostramos, bajo algunas condiciones, que la solución de una forma discreta

de un problema parabólico no local se sofoca en tiempo finito y estimamos su tiempo de

sofocamiento numérico. Probamos también que el tiempo de sofocamiento numérico converge

par un real cuando el tamaño de la malla tiende a cero. Finalmente damos algunos resultados

computacionales para ilustrar nuestros análisis.



24 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Key words and phrases: Nonlocal diffusion, quenching, numerical quenching time.

Math. Subj. Class.: 5B40, 45A07, 45G10, 65M06.

1 Introduction

Consider the following initial value problem

ut =

∫

Ω

J(x − y)(u(y, t) − u(x, t))dy + (M − u)−p in Ω × (0, T ), (1)

u(x, 0) = u0(x) ≥ 0 in Ω, (2)

where Ω = (−1, 1)N , M = const > 0, p = const > 0, J: RN → R is a C1 nonnegative function.

In addition, J is symmetric (J(z) = J(−z)), and
∫

RN
J(z)dz = 1. The initial data u0 ∈ C

0(Ω),

0 ≤ u0(x) < M , x ∈ Ω.

Here, (0, T ) is the maximal time interval on which the solution u exists. The time T may be finite

or infinite. When T is infinite, then we say that the solution u exists globally. When T is finite,

then the solution u develops a singularity in a finite time, namely,

lim
t→T

‖u(·, t)‖∞ = M,

where ‖u(·, t)‖∞ = supx∈Ω |u(x, t)|. In this last case, we say that the solution u quenches in a finite

time, and the time T is called the quenching time of the solution u. Recently, nonlocal diffusion

has been the subject of investigation of many authors (see, [1]-[7], [10]-[12], [14]-[18], [20], [24], [27],

and the references cited therein). Nonlocal evolution equations of the form

ut =

∫

RN

J(x − y)(u(y, t) − u(x, t))dy,

and variations of it, have been used by several authors to model diffusion processes (see, [3], [4],

[10], [17], [18]). The solution u(x, t) can be interpreted as the density of a single population at

the point x, at the time t, and J(x − y) as the probability distribution of jumping from location

y to location x. Then, the convolution (J ∗ u)(x, t) =
∫

RN
J(x − y)u(y, t)dy is the rate at which

individuals are arriving to position x from all other places, and −u(x, t) = −
∫

RN
J(x − y)u(x, t)dy

is the rate at which they are leaving location x to travel to any other site (see, [17]). Let us notice

that the reaction term (M − u)−p in the equation (1) can be rewritten as follows

(M − u(x, t))
−p

=

∫

RN

J(x − y) (M − u(x, t))
−p

dy.

Therefore, in view of the above equality, the reaction term (M − u)−p can be interpreted as a force

that decreases the rate at which individuals are leaving location x to travel to any other site. Due

to the presence of the term (M − u)−p, we shall see later that the phenomenon of quenching occurs



CUBO
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Quenching for Discretizations ... 25

for the density u(x, t). On the other hand, the integral in (1) is taken over Ω. Thus, there is no

individuals that enter or leave the domain Ω. It is the reason why in the title of the paper, we

have added Neumann boundary condition. In the current paper, we are interested in the numerical

study of the phenomenon of quenching using a discrete form of (1)-(2). Let us notice that, setting

v = M − u, the problem (1)-(2) is equivalent to

vt(x, t) =

∫

Ω

J(x − y)(v(y, t) − v(x, t))dy − v−p in Ω × (0, T ), (3)

v(x, 0) = v0(x) > 0 in Ω, (4)

where v0(x) = M − u0(x). Consequently, the solution u of (1)-(2) quenches at the time T if and

only if the solution v of (3)-(4) quenches at the time T , that is,

lim
t→T

vmin(t) = 0,

where vmin(t) = minΩ v(x, t). Thus, by convenience, we shall often utilize the problem (3)-(4)

instead of (1)-(2). We start by the construction of an explicit adaptive scheme as follows. Let I

be a positive integer, and let h = 2/I. Define the grid xi = −1 + ih, 0 ≤ i ≤ I, and approximate

the solution v of (3)-(4) by the solution U
(n)
h

= (U
(n)
K

)K∈Γ of the following discrete equations

δtU
(n)
K

=
∑

L∈Γ∗

hN J(xK − xL)
(
U

(n)
L

− U
(n)
K

)
−
(
U

(n)
K

)−p
, K ∈ Γ, n ≥ 0, (5)

U
(0)
K

= ϕK , K ∈ Γ, (6)

where

Γ = {(j1, · · · , jN ); 0 ≤ j1, · · · , jN ≤ I} ,

Γ∗ = {(j1, · · · , jN ); 0 ≤ j1, · · · , jN ≤ I − 1} ,

xK = (xk1 , · · · , xkN ) , xL = (xl1 , · · · , xlN ) ,

and

δtU
(n)
K

=
U

(n+1)
K

− U
(n)
K

∆tn
.

In order to permit the discrete solution to reproduce the properties of the continuous one when

the time t approaches the quenching time T , we need to adapt the size of the time step so that we

take

∆tn = min

{
h2, τ

(
U

(n)
hmin

)p+1}
,

where U
(n)
hmin

= minK∈Γ U
(n)
K

, τ ∈ (0, 1). Let us notice that the restriction on the time step ensures

the positivity of the discrete solution.

To facilitate our discussion, we need to define the notion of numerical quenching time.



26 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Definition 1.1. We say that the discrete solution U
(n)
h

of (5)-(6) quenches in a finite time if

limn→∞ U
(n)
hmin

= 0, and the series
∑∞

n=0 ∆tn converges. The quantity
∑∞

n=0 ∆tn is called the

numerical quenching time of the discrete solution U
(n)
h

.

In the present paper, under some conditions, we show that the discrete solution quenches in a

finite time and estimate its numerical quenching time. We also show that the numerical quenching

time converges to the real one when the mesh size goes to zero. A similar result has been obtained

by Nabongo and Boni in [25] within the framework of the phenomenon of quenching for local

parabolic problems. One may also consult the papers of the same authors in [22] and [23] for

numerical studies of the phenomenon of quenching where semidiscretizations in space have been

utilized. The remainder of the paper is organized as follows. In the next section, we reveal certain

properties of the continuous problem. In the third section, we exhibit some features of the discrete

scheme. In the fourth section, under some assumptions, we demonstrate that the discrete solution

quenches in a finite time, and estimate its numerical quenching time. In the fifth section, the

convergence of the numerical quenching time is analyzed, and finally, in the last section, we show

some numerical experiments to illustrate our analysis.

2 Local Existence

In this section, we shall establish the existence and uniqueness of solutions of (1)-(2) in Ω × (0, T )

for all small T . Some results about quenching are also given.

Let t0 > 0 be fixed, and define the function space Yt0 = {u; u ∈ C([0, t0], C(Ω))} equipped with the

norm defined by ‖u‖Yt0 = max0≤t≤t0 ‖u(·, t)‖∞ for u ∈ Yt0 . It is easy to see that Yt0 is a Banach

space. Introduce the set

Xt0 =
{
u; u ∈ Yt0 , ‖u‖Yt0 ≤ b0

}
,

where b0 =
‖u0‖∞+M

2
. We observe that Xt0 is a nonempty bounded closed convex subset of Yt0 .

Define the map R as follows

R : Xt0 → Xt0 ,

R(v)(x, t) = u0(x) +

∫ t

0

∫

Ω

J(x − y)(v(y, s) − v(x, s))dyds +

∫ t

0

(M − v(x, s))−pds.

Theorem 2.1. Assume that u0 ∈ C(Ω). Then R maps Xt0 into Xt0 , and R is strictly contractive

if t0 is appropriately small relative to ‖u0‖∞.

Proof. Due to the fact that
∫
Ω

J(x − y)dy ≤
∫

RN
J(x − y)dy = 1, a straightforward computation

reveals that

|R(v)(x, t) − u0(x)| ≤ 2‖v‖Yt0 t +
(
M − ‖v‖Yt0

)−p
t,

which implies that ‖R(v)‖Yt0 ≤ ‖u0‖∞ + 2b0t0 + (M − b0)
−pt0. If

t0 ≤
b0 − ‖u0‖∞

2b0 + (M − b0)−p
, (7)



CUBO
12, 1 (2010)

Quenching for Discretizations ... 27

then ‖R(v)‖Yt0 ≤ b0. Therefore, if (7) holds, then R maps Xt0 into Xt0 . Now, we are going to

prove that the map R is strictly contractive. Letting v, z ∈ Xt0 and setting α = v − z, we discover

that

|(R(v) − R(z))(x, t)| ≤

∣∣∣∣
∫ t

0

∫

Ω

J(x − y)(α(y, s) − α(x, s))dyds

∣∣∣∣

+

∣∣∣∣
∫ t

0

((M − v(x, s))−p − (M − z(x, s))−p)ds

∣∣∣∣ .

Use Taylor’s expansion to obtain

|(R(v) − R(z))(x, t)| ≤ 2‖α‖Yt0 t + t‖v − z‖Yt0 p
(
M − ‖β‖Yt0

)−p−1
,

where β is a function which is localized between v and z. We deduce that

‖R(v) − R(z)‖Yt0 ≤ 2‖α‖Yt0 t0 + t0‖v − z‖Yt0 p
(
M − ‖β‖Yt0

)−p−1
,

which implies that ‖R(v) − R(z)‖Yt0 ≤ (2t0 + t0p(M − b0)
−p−1)‖v − z‖Yt0 . If

t0 ≤
1

4 + 2p(M − b0)−p−1
, (8)

then ‖R(v) − R(z)‖Yt0 ≤
1
2
‖v − z‖Yt0 . Hence, we see that R(v) is a strict contraction in Yt0 , and

the proof is complete.

It follows from the contraction mapping principle that for appropriately chosen t0, R has a

unique fixed point u ∈ Yt0 which is a solution of (1)-(2). If ‖u‖Yt0 < M , then taking as initial

data u(·, t0) ∈ C(Ω) and arguing as before, it is possible to extend the solution up to some interval

[0, t1) for certain t1 > t0. Hence, we conclude that if the maximal time interval of existence of

the solution, (0, T ), is finite then the solution quenches in a finite time in L∞(Ω) norm, namely,

limt→T ‖u(·, t)‖∞ = M .

Remark 2.1. Let us notice that we can define the map R in the space Yt0 = C
1,2(Ω × [0, t0]).

Besides, if u0 ∈ C
1(Ω), then arguing as in the proof of Theorem 2.1, it is not hard to see that

Theorem 2.1 remains valid. Consequently, if u0 ∈ C
1(Ω), then the solution u of (1)-(2) belongs to

C1,2(Ω × [0, T )) when T is finite.

The following lemma is a version of the maximum principle for nonlocal problems.

Lemma 2.1. Let a ∈ C0(Ω × [0, T )), and let u ∈ C0,1(Ω × [0, T )) satisfy the following inequalities

ut −

∫

Ω

J(x − y)(u(y, t) − u(x, t))dy + a(x, t)u(x, t) ≥ 0 in Ω × (0, T ), (9)

u(x, 0) ≥ 0 in Ω. (10)

Then, we have u(x, t) ≥ 0 in Ω × (0, T ).



28 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Proof. Let T0 be any positive quantity satisfying T0 < T . Since a(x, t) is bounded in Ω × [0, T0],

then there exists λ such that a(x, t) − λ > 0 in Ω × [0, T0]. Define z(x, t) = e
λtu(x, t) and let

m = min
x∈Ω,t∈[0,T0]

z(x, t). Due to the fact that z is continuous in Ω × [0, T0], then it achieves its

minimum in Ω × [0, T0]. Consequently, there exists (x0, t0) ∈ Ω × [0, T0] such that m = z(x0, t0).

We get z(x0, t0) ≤ z(x0, t) for t ≤ t0 and z(x0, t0) ≤ z(y, t0) for y ∈ Ω. This implies that

zt(x0, t0) ≤ 0,

∫

Ω

J(x0 − y)(z(y, t0) − z(x0, t0))dy ≥ 0. (11)

With the aid of the first inequality of the lemma, it is not hard to see that

zt(x0, t0) −

∫

Ω

J(x0 − y)(z(y, t0) − z(x0, t0))dy + (a(x0, t0) − λ)z(x0, t0) ≥ 0.

We deduce from (11) that (a(x0, t0) − λ)z(x0, t0) ≥ 0. Since a(x0, t0) − λ > 0, we get z(x0, t0) ≥ 0.

This implies that u(x, t) ≥ 0 in Ω × [0, T0], and the proof is complete.

An immediate consequence of the above lemma is that the solution u of (1)-(2) is nonnegative

in Ω × (0, T ) because the initial data u0(x) is nonnegative in Ω.

Now, let us give a result about quenching which says that the solution u of (1)-(2) always quenches

in a finite time. This assertion is stated in the theorem below.

Theorem 2.2. The solution u of (1)-(2) quenches in a finite time, and its quenching time T

satisfies the following estimate

T ≤
(M − A)p+1

p + 1
,

where A = 1
|Ω|

∫
Ω

u0(x)dx.

Proof. Since (0, T ) is the maximal time interval of existence of the solution u, our aim is to show

that T is finite and satisfies the above inequality. Due to the fact that the initial data u0(x) is

nonnegative in Ω, we know from Lemma 2.1 that the solution u(x, t) of (1)-(2) is nonnegative in

Ω × (0, T ). Integrating both sides of (1) over (0, t), we find that

u(x, t) − u0(x) =

∫ t

0

∫

Ω

J(x − y)(u(y, s) − u(x, s))dyds

+

∫ t

0

(M − u(x, s))−pds for t ∈ (0, T ). (12)

Integrate again in the x variable and apply Fubini’s theorem to obtain

∫

Ω

u(x, t)dx −

∫

Ω

u0(x)dx =

∫ t

0

(∫

Ω

(M − u(x, s))−pdx

)
ds for t ∈ (0, T ). (13)

Set

w(t) =
1

|Ω|

∫

Ω

u(x, t)dx for t ∈ [0, T ).



CUBO
12, 1 (2010)

Quenching for Discretizations ... 29

Taking the derivative of w in t and using (13), we arrive at

w′(t) =

∫

Ω

1

|Ω|
(M − u(x, t))−pdx for t ∈ (0, T ).

It follows from Jensen’s inequality that w′(t) ≥ (M − w(t))−p for t ∈ (0, T ), or equivalently

(M − w)pdw ≥ dt for t ∈ (0, T ). (14)

Integrate the above inequality over (0, T ) to obtain

T ≤
(M − w(0))p+1

p + 1
.

Since the quantity on the right hand side of the above inequality is finite, we deduce that u quenches

in a finite time at the time T which obeys the above inequality. Use the fact that w(0) = A to

complete the rest of the proof.

3 Properties of the Semidiscrete Scheme

In this section, we give some results about the discrete maximum principle of nonlocal problems

for our subsequent use.

The lemma below is a discrete version of the maximum principle for nonlocal parabolic problems.

Lemma 3.1. For n ≥ 0, let U
(n)
h

, a
(n)
h

be two vectors such that

δtU
(n)
K

≥
∑

L∈Γ∗

hN J(xK − xL)
(
U

(n)
L

− U
(n)
K

)
+ a

(n)
K

U
(n)
K

, K ∈ Γ, n ≥ 0,

U
(0)
K

≥ 0, K ∈ Γ.

Then, we have U
(n)
K

≥ 0, K ∈ Γ, n > 0 when ∆tn ≤
1

2N ‖J‖∞+‖a
(n)
h

‖∞
, where ‖a

(n)
h

‖∞ =

supK∈Γ |a
(n)
K

|.

Proof. If U
(n)
h

≥ 0, then a straightforward computation reveals that

U
(n+1)
K

≥ U
(n)
K

(
1 − 2N ‖J‖∞∆tn − ‖a

(n)
h

‖∞∆tn

)
, K ∈ Γ, n ≥ 0. (15)

To obtain the above inequalities, we have used the fact that
∑

L∈Γ∗

hN J(xK − xL) ≥ 0, K ∈ Γ,

and
∑

L∈Γ∗

hN J(xK − xL) ≤ ‖J‖∞

(
I−1∑

l=0

h

)N
= 2N ‖J‖∞, K ∈ Γ.



30 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Making use of (15) and an argument of recursion, we easily check that U
(n+1)
h

≥ 0, n ≥ 0. This

finishes the proof.

An immediate consequence of the above result is the following comparison lemma. Its proof

is straightforward.

Lemma 3.2. For n ≥ 0, let U
(n)
h

, V
(n)
h

and a
(n)
h

be three vectors such that

δtU
(n)
K

−
∑

L∈Γ∗

hN J(xK − xL)
(
U

(n)
L

− U
(n)
K

)
+ a

(n)
K

U
(n)
K

≥ δtV
(n)
K

−
∑

L∈Γ∗

hN J(xK − xL)
(
V

(n)
L

− V
(n)
K

)
+ a

(n)
K

V
(n)
K

, K ∈ Γ, n ≥ 0,

U
(0)
K

≥ V
(0)
K

, K ∈ Γ.

Then, we have U
(n)
K

≥ V
(n)
K

, K ∈ Γ, n > 0 when ∆tn ≤
1

2N ‖J‖∞+‖a
(n)
h

‖∞
.

Remark 3.1. For n ≥ 0, introduce the vector Z
(n)
h

defined as follows Z
(n)
K

= ‖ϕh‖∞ − U
(n)
K

,

K ∈ Γ, where U
(n)
h

is the solution of (5)-(6). A straightforward computation reveals that

δtZ
(n)
K

≥
∑

L∈Γ∗

hN J(xK − xL)
(
Z

(n)
L

− Z
(n)
K

)
, K ∈ Γ, n ≥ 0,

Z
(0)
K

≥ 0, K ∈ Γ.

It follows from Lemma 3.1 that ‖ϕh‖∞ ≥ U
(n)
K

, K ∈ Γ, n > 0 when h is small enough.

4 The Numerical Quenching Time

In this section, under some assumptions, we show that the discrete solution quenches in a finite

time and estimate its numerical quenching time.

Our result concerning the numerical quenching time is stated in the following theorem.

Theorem 4.1. Assume that the initial data at (6) satisfies 2N ‖J‖∞‖ϕh‖
p+1
∞ < 1. Then, the

discrete solution U
(n)
h

of (5)-(6) quenches in a finite time, and its quenching time T ∆th obeys the

following estimate

T ∆th ≤
τ ϕ

p+1
hmin

1 − (1 − τ ′)p+1
,

where τ ′ = A min{h2ϕ
−p−1
hmin

, τ} and A = 1 − 2N ‖J‖∞‖ϕh‖
p+1
∞ .



CUBO
12, 1 (2010)

Quenching for Discretizations ... 31

Proof. We know from Remark 3.1 that ‖U
(n)
h

‖∞ ≤ ‖ϕh‖∞. Since

∑

L∈Γ∗

hN J(xK − xL) ≤ 2
N ‖J‖∞, K ∈ Γ,

exploiting (5), we see that

δtU
(n)
K

≤ 2N ‖J‖∞‖ϕh‖∞ −
(
U

(n)
K

)−p
, K ∈ Γ, n ≥ 0,

or equivalently

δtU
(n)
K

≤ −
(
U

(n)
K

)−p (
1 − 2N ‖J‖∞‖ϕh‖∞

(
U

(n)
K

)p)
, K ∈ Γ, n ≥ 0.

Use the fact that ‖U
(n)
h

‖∞ ≤ ‖ϕh‖∞, n ≥ 0 to arrive at

δtU
(n)
K

≤ −
(
U

(n)
K

)−p (
1 − 2N ‖J‖∞‖ϕh‖

p+1
∞

)
, K ∈ Γ, n ≥ 0.

These estimates may be rewritten as follows

U
(n+1)
K

≤ U
(n)
K

− A∆tn

(
U

(n)
K

)−p
, K ∈ Γ, n ≥ 0. (16)

Let K0 ∈ Γ be such that U
(n)
K0

= U
(n)
hmin

. Replacing K by K0 in (16), we note that

U
(n+1)
K0

≤ U
(n)
hmin

− A∆tn

(
U

(n)
hmin

)−p
, n ≥ 0,

which implies that

U
(n+1)
hmin

≤ U
(n)
hmin

− A∆tn

(
U

(n)
hmin

)−p
, n ≥ 0, (17)

because U
(n+1)
K0

≥ U
(n+1)
hmin

. We observe that

A∆tn(U
(n)
hmin

)−p−1 = A min
{

h2(U
(n)
hmin

)−p−1, τ
}

. (18)

Exploiting (17), we see that U
(n+1)
hmin

≤ U
(n)
hmin

, n ≥ 0, and by induction, we note that U
(n)
hmin

≤

U
(0)
hmin

= ϕhmin. In view of (18), we discover that

A∆tn

(
U

(n)
hmin

)−p−1
≥ A min

{
h2ϕ

−p−1
hmin

, τ
}

= τ ′. (19)

Therefore, employing (17), we get

U
(n+1)
hmin

≤ U
(n)
hmin

(1 − τ ′) , n ≥ 0. (20)

Using an argument of recursion, we find that

U
(n)
hmin

≤ U
(0)
hmin

(1 − τ ′)
n

= ϕhmin (1 − τ
′)

n
, n ≥ 0. (21)



32 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

This implies that U
(n)
hmin

goes to zero as n approaches infinity. Now, let us estimate the discrete

quenching time. The restriction on the time step and (21) lead us to

∞∑

n=0

∆tn ≤ τ ϕ
p+1
hmin

∞∑

n=0

(
(1 − τ ′)

p+1
)n

.

Use the fact that the series on the right hand side of the above inequality converges towards
1

1−(1−τ ′)p+1
to complete the rest of the proof.

Remark 4.1. Due to (20), an argument of recursion reveals that

U
(n)
hmin

≤ U
(q)
hmin

(1 − τ ′)
n−q

, n ≥ q.

In view of the above estimates, the restriction on the time step allows us to write

∞∑

n=q

∆tn ≤ τ
(
U

(q)
hmin

)p+1 ∞∑

n=q

(
(1 − τ ′)

p+1
)n−q

.

Since the series on the right hand side of the above inequality converges towards 1
1−(1−τ ′)p+1

, we

infer that
∑∞

n=q ∆tn ≤
τ
(

U
(q)
hmin

)
p+1

1−(1−τ ′)p+1
, or equivalently

T ∆th − tq ≤
τ
(
U

(q)
hmin

)p+1

1 − (1 − τ ′)
p+1

.

Apply Taylor’s expansion to obtain (1−τ ′)p+1 = 1−(p+1)τ ′+o(τ ′). This implies that τ
1−(1−τ ′)p+1

=
τ

τ ′((p+1)+o(1))
. Due to the fact that τ ′ = A min{h2ϕ

−p−1
hmin

, τ}, if we choose τ = h2, then we note

that τ
′

τ
= A min{ϕ

−p−1
hmin

, 1}, which implies that τ
τ ′

= O(1) with the choice τ = h2.

In the sequel, we pick τ = h2.

Under the assumption of the above theorem, we have seen that the discrete solution quenches in a

finite time, and an estimation of its numerical quenching time has been given. In the theorem below,

we derive an upper bound of the numerical quenching time taking into account the assumption of

the earlier theorem.

Theorem 4.2. Assume that Bh2 < 1, where B = 1+2N ‖J‖∞‖ϕh‖
p+1
∞ . Then, under the hypothesis

of Theorem 4.1, the discrete solution U
(n)
h

of (5)-(6) quenches in a finite time, and its quenching

time T ∆th satisfies the following estimate

T ∆th ≥
h2 min{1, ϕ

p+1
hmin

}

1 − (1 − Bh2)p+1
.

Proof. We know from Theorem 4.1 that the discrete solution quenches in a finite time. Thus, our

purpose is to establish the above estimate. Employing (5), we note that

δtU
(n)
K

≥ −2N ‖J‖∞‖ϕh‖∞ −
(
U

(n)
K

)−p
, K ∈ Γ, n ≥ 0. (22)



CUBO
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Quenching for Discretizations ... 33

To obtain the above inequalities, we have used the fact that J is nonnegative, ‖U
(n)
h

‖∞ ≤ ‖ϕ‖∞

and
∑

L∈Γ∗
hN J(xK − xL) ≤ 2

N ‖J‖∞, K ∈ Γ. Since 0 ≤ U
(n)
K

≤ ‖ϕh‖∞, K ∈ Γ, the inequalities

(22) become

δtU
(n)
K

≥ −B
(
U

(n)
K

)−p
, K ∈ Γ, n ≥ 0. (23)

Taking into account the restriction on the time step, it is not hard to see that

∆tn

(
U

(n)
K

)−p−1
≤ ∆tn

(
U

(n)
hmin

)−p−1
≤ h2, K ∈ Γ, n ≥ 0. (24)

In view of (23) and (24), we infer that

U
(n+1)
K

≥ U
(n)
K

(
1 − Bh2

)
, K ∈ Γ, n ≥ 0,

which implies that

U
(n+1)
hmin

≥ U
(n)
hmin

(
1 − Bh2

)
, n ≥ 0. (25)

By induction, we realize that

U
(n)
hmin

≥ U
(0)
hmin

(
1 − Bh2

)n
= ϕhmin

(
1 − Bh2

)n
, n ≥ 0. (26)

Now, let us estimate the numerical quenching time. The restriction on the time step and (26)

reveal that
∞∑

n=0

∆tn ≥

∞∑

n=0

min
{
h2, h2ϕ

p+1
hmin

((
1 − Bh2

)p+1)n}
, (27)

which implies that

∞∑

n=0

∆tn ≥ min
{
h2, h2ϕ

p+1
hmin

} ∞∑

n=0

((
1 − Bh2

)p+1)n
.

Use the fact that the series on the right hand side of the above inequality converges towards
1

1−(1−Bh2)p+1
to complete the rest of the proof.

Remark 4.2. Apply Taylor’s expansion to obtain
(
1 − Bh2

)p+1
= 1 − B(p + 1)h2 + o(h2),

which implies that h
2

1−(1−Bh2)p+1
= 1

B(p+1)+o(1)
.

5 Convergence of the Numerical Quenching Time

In this section, we denote by

vh(t) = (v(xK , t))K∈Γ .

Under some hypotheses, we prove that the discrete solution quenches in a finite time, and its

numerical quenching time converges to the real one when the mesh size goes to zero.

We need the following lemma.



34 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Lemma 5.1. Let f ∈ C1(Ω). Then, we have

∫

Ω

f (x)dx =
∑

J∈Γ∗

hN f (xJ ) + O(h)

The proof of the above lemma is based on the fact that, if g ∈ C1([−1, 1]), then
∫ 1
−1

g(x)dx =
∑I−1

j=0 hg(xj ) + O(h). Making use of the above result, and exploiting Fubini’s theorem, one easily

proves the above lemma.

In order to obtain the result concerning the convergence of the numerical quenching time, we firstly

prove that the discrete solution approaches the real one in any interval Ω×[0, T −τ ] with τ ∈ (0, T ).

This result is stated in the following theorem.

Theorem 5.1. Assume that the problem (3)-(4) admits a solution v ∈ C1,2(Ω × [0, T − τ ]) such

that mint∈[0,T −τ ] vmin(t) = α > 0 with τ ∈ (0, T ). Suppose that the initial data at (6) satisfies

‖ϕh − vh(0)‖∞ = o(1) as h → 0. (28)

Then, the problem (5)-(6) admits a unique solution U
(n)
h

for h small enough, n ≤ R, and the

following relation holds

sup
0≤n≤R

‖U
(n)
h

− vh(tn)‖∞ = O(‖ϕh − vh(0)‖∞ + h) as h → 0,

where R is a positive integer such that
∑R−1

j=0 ∆tj ≤ T − τ , and tn =
∑n−1

j=0 ∆tj .

Proof. The problem (5)-(6) admits for each n ≥ 0, a unique solution U
(n)
h

. Let d ≤ R be the

greatest integer such that

‖U
(n)
h

− vh(tn)‖∞ <
α

2
for n < d. (29)

Making use of (28), we note that d ≥ 1 for h small enough. An application of the triangle inequality

renders

U
(n)
hmin

≥ vhmin(tn) − ‖U
(n)
h

− vh(tn)‖∞ ≥ α −
α

2
=

α

2
for n < d. (30)

Exploit Taylor’s expansion and use Lemma 5.1 to obtain

δtv(xK , tn) = vt(xK , tn) +
∆tn
2

vtt(xK , t̃n), K ∈ Γ, n < d,

∫

Ω

J(xK − y)(v(y, tn) − v(xK , tn))dy

=
∑

L∈Γ∗

hN J(xK − xL)(v(xL, tn) − v(xK , tn)) + O(h), K ∈ Γ, n < d,



CUBO
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Quenching for Discretizations ... 35

which implies that

δtv(xK , tn) =
∑

L∈Γ∗

hN J(xK − xL)(v(xL, tn) − v(xK , tn))

−(v(xK , tn))
−p +

∆tn
2

vtt(xK , t̃n) + O(h), K ∈ Γ, n < d.

Introduce the error e
(n)
h

defined as follows

e
(n)
K

= U
(n)
K

− v(xK , tn), K ∈ Γ, n < d.

Invoking the mean value theorem, it is easy to see that

δte
(n)
K

=
∑

L∈Γ∗

hN J(xK − xL)
(
e
(n)
L

− e
(n)
K

)
+ p

(
ξ
(n)
K

)−p−1
e
(n)
K

−
∆tn
2

vtt(xK , t̃n) + O(h), K ∈ Γ, n < d,

where ξ
(n)
K

is an intermediate value between v(xK , tn) and U
(n)
K

. We infer that there exists a positive

constant Q such that

δte
(n)
K

≤
∑

L∈Γ∗

hN J(xK − xL)
(
e
(n)
L

− e
(n)
K

)
+ p

(
ξ
(n)
K

)−p−1
e
(n)
K

+Qh, K ∈ Γ, n < d, (31)

because v ∈ C1,2, J ∈ C1(RN ), and ∆tn = O(h
2). Introduce the vector Z

(n)
h

defined as follows

Z
(n)
K

= e(L+1)tn (‖ϕh − vh(0)‖∞ + Qh), K ∈ Γ, n < d,

where L = p
(

α
2

)−p−1
. A straightforward computation reveals that

δtZ
(n)
K

≥
∑

L∈Γ∗

hN J(xK − xL)
(
Z

(n)
L

− Z
(n)
K

)
+ p

(
ξ
(n)
K

)−p−1

+Qh, K ∈ Γ, n < d,

Z
(0)
K

≥ e
(0)
K

, K ∈ Γ.

We deduce from Lemma 3.2 that

Z
(n)
K

≥ e
(n)
K

, K ∈ Γ, n < d.



36 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

In the same way, we also show that

Z
(n)
K

≥ −e
(n)
K

, K ∈ Γ, n < d,

which implies that ‖e
(n)
h

‖∞ ≤ ‖Z
(n)
h

‖∞, n < d, or equivalently

‖U
(n)
h

− vh(tn)‖∞ ≤ e
(L+1)tn (‖ϕh − vh(0)‖∞ + Qh), n < d. (32)

Now, let us reveal that d = R. To prove this result, we argue by contradiction. Assume that

d < R. Replacing n by d in (32), and using (29), we discover that

α

2
≤ ‖U

(d)
h

− vh(td)‖∞ ≤ e
(L+1)T (‖ϕh − vh(0)‖∞ + Qh).

Since the term on the right hand side of the second inequality goes to zero as h tends to zero, we

deduce that α
2
≤ 0, which is impossible. Consequently, d = R, and the proof is complete.

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

Theorem 5.2. Assume that the problem (3)-(4) has a solution v which quenches in a finite time

T such that v ∈ C1,2(Ω × [0, T )). Suppose that the initial data at (6) satisfies the condition (28).

Then, under the hypotheses of Theorem 4.1, the solution U
(n)
h

of (5)-(6) quenches in a finite time,

and its numerical quenching time T ∆th obeys the following relation

lim
h→0

T ∆th = T.

Proof. Let 0 < ε < T /2. In view of Remark 4.1, we know that τ
1−(1−τ

′
)p+1

is bounded. Thus,

there exists a positive constant ρ such that

τ ρp+1

1 − (1 − τ ′)p+1
≤

ε

2
. (33)

Since u quenches at the time T , there exists a time T0 ∈ (T − ε/2, T ) such that 0 < vhmin(t) <
ρ
2

for t ∈ [T0, T ). Let q be a positive integer such that tq =
∑q−1

n=0 ∆tn ∈ [T0, T ). Invoking

Theorem 5.1, we know that the problem (5)-(6) admits a unique solution U
(n)
h

such that ‖U
(q)
h

−

vh(tq)‖∞ ≤
ρ

2
. An application of the triangle inequality gives

U
(q)
hmin

≤ vhmin(tq) + ‖U
(q)
h

− vh(tq)‖∞,

which implies that U
(q)
hmin

≤ ρ
2

+ ρ
2

= ρ. Taking into account Theorem 4.1, we know that the

discrete solution U
(n)
h

quenches in a finite time T ∆th . It follows from Remark 4.1 and (33) that

|T ∆th − T | ≤ |T
∆t
h − tq| + |tq − T | ≤

ε

2
+

ε

2
= ε.

This finishes the proof.



CUBO
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Quenching for Discretizations ... 37

6 Numerical Results

In this section, we give some computational experiments to illustrate the theory given in the

previous section. We consider the problem (3)-(4) in the case where N = 1,

J(x) =

{
15
16

(1 − x2)2 if |x| ≤ 1,

0 if |x| > 1,

v0(x) = 4 + cos(πx). We consider the explicit scheme defined in (5)-(6). We also use the following

implicit scheme

U
(n+1)
i − U

(n)
i

∆tn
=

I−1∑

l=0

hJ(xi − xl)
(
U

(n+1)
l

− U
(n+1)
i

)

−(U
(n)
i )

−p−1U
(n+1)
i , 0 ≤ i ≤ I, U

(0)
i = ϕi, 0 ≤ i ≤ I.

In both cases, we take ϕi = 4 + cos(πxi). As in the case of the explicit scheme, here, we also choose

∆tn = h
2
(
U

(n)
hmin

)p+1
.

Let us again remark that for the above implicit scheme, existence and positivity of the discrete

solution are also guaranteed using standard methods (see, for instance [9]).

In the following tables, in rows, we present the numerical quenching times, the numbers of iter-

ations, the CPU times and the orders of the approximations corresponding to meshes of 16, 32,

64, 128. We take for the numerical quenching time tn =
∑n−1

j=0 ∆tj which is computed at the first

time when

∆tn = |tn+1 − tn| ≤ 10
−16.

The order (s) of the method is computed from

s =
log((T4h − T2h)/(T2h − Th))

log(2)
.

Numerical experiments for p = 1

Table 1: Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of

the approximations obtained with the explicit Euler method

I tn n CPU time s

16 4.843163 1272 8 -

32 4.553751 4849 40 -

64 4.507777 18651 308 2.65

128 4.501537 59860 3008 2.89



38 Théodore K. Boni and Diabaté Nabongo CUBO
12, 1 (2010)

Table 2: Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of

the approximations obtained with the implicit Euler method

I tn n CPU time s

16 4.903291 1071 5.2 -

32 4.563995 4097 43 -

64 4.509770 15674 358 2.64

128 4.501537 59860 9360 2.71

Remark 6.1. We observe from Tables 1-2 that the numerical quenching time of the discrete

solution is approximately equal to 4.5.

In what follows, we also give some plots to illustrate our analysis. In Figures 1-2, we can

appreciate that the discrete solution quenches in a finite time.

0
200

400
600

800
1000

1200
1400

0

5

10

15

20
0

1

2

3

4

5

ni

U
(i
,n

) 

Figure 1: Evolution of the discrete
solution (explicit scheme).

0
200

400
600

800
1000

1200

0

5

10

15

20
0

1

2

3

4

5

ni

U
(i
,n

) 

Figure 2: Evolution of the discrete
solution (implicit scheme).

Received: October, 2008. Revised: October, 2009.

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