Rev. Anal. Num´er. Th´eor. Approx., vol. 41 (2012) no. 2, pp. 130–148 ictp.acad.ro/jnaat
REMARKS ON THE QUENCHING ESTIMATE FOR A NONLOCAL DIFFUSION PROBLEM WITH A REACTION TERM
HALIMA NACHID∗
Abstract. In this paper, we address the following initial value problem ut=R
ΩJ(x−y)(u(y, t)−u(x, t))dy+f(u(x, t)) in Ω×(0, T), u(x,0) =u0(x)≥0 in Ω,
where Ω is a bounded domain inRN with smooth boundary∂Ω,f: (−∞, b)→ (0,∞) is a C1 convex nondecreasing function, lims→b−f(s) = ∞, R∞ dσ
f(σ) <
∞, with b a positive constant, J : RN → R is a kernel which is measurable, nonnegative and bounded in RN. Under some conditions, we show that the solution of a perturbed form of the above problem quenches in a finite time and estimate its quenching time. We also prove the continuity of the quenching time as a function of the initial datum. Finally, we give some numerical results to illustrate our analysis.
MSC 2000. 35B40, 45A07, 45G10, 65M06.
Keywords. Nonlocal diffusion, quenching, continuity, numerical quenching time, reaction-diffusion equation.
1. INTRODUCTION
Let Ω be a bounded domain in RN with smooth boundary ∂Ω. Consider the following initial value problem
ut= Z
Ω
J(x−y)(u(y, t)−u(x, t))dy+f(u(x, t)) in Ω×(0, T), (1)
u(x,0) =u0(x)≥0 in Ω, (2)
wheref : (−∞, b)→(0,∞) is aC1 convex nondecreasing function,R∞ dσ f(σ) <
∞, lims→b−f(s) =∞, withba positive constant,J :RN →Ris a kernel which is measurable, nonnegative and bounded in RN. In addition,J is symmetric (J(z) = J(−z)) and R
RNJ(z)dz = 1. The initial datum u0 ∈ C0(Ω), 0 ≤
∗Universit´e d’Abobo-Adjam´e, UFR-SFA, D´epartement de Math´ematiques et Informa- tiques, 02 BP 801 Abidjan 02, (Cˆote d’Ivoire), International University of Grand-Bassam Route de Bonoua Grand-Bassam BP 564 Grand-Bassam, (Cˆote d’Ivoire) and Labora- toire de Mod´elisation Math´ematique et de Calcul ´Economique LM2CE settat, (Maroc), e-mail: [email protected].
u0(x) < b in Ω. Let us notice that, if f(s) = (b−s)−p with p a positive constant, then f satisfies the above conditions. 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,
t→Tlimku(·, t)k∞=b,
whereku(·, t)k∞= supx∈Ω|u(x, t)|. In this last case, we say that the solution uquenches 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, [2], [8], [11], [13], [15], [17], [19], [20], [25], [29], [32], and the references cited therein). Nonlocal evolution equations of the form
ut= Z
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, [4], [5], [11], [19], [20]). The solutionu(x, t) can be interpreted as the density of a single population at the pointx, at the timet, andJ(x−y) as the probability distribution of jumping from locationy to locationx. Then the convolution (J ∗ u)(x, t) = R
RNJ(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) =
−R
RNJ(x−y)u(y, t)dy is the rate at which they are leaving location x to travel to any other site (see, [19]). For the problem described in (1)–(2), the integral is taken over Ω. Consequently, there is no individuals that arrive or leave the domain Ω. It is the reason why in the title of the paper, we have added Neumann boundary condition. On the other hand, the term of the sourcef(u) can be rewritten as follows
f(u(x, t)) = Z
RN
J(x−y)f(u(x, t))dy.
Therefore, in view of the above equality, the term f(u) can be interpreted as a force that decreases the rate of individuals which are leaving location x to travel to any other site, provoking as we shall see later, the phenomenon of quenching of the solution u. For local diffusions, solutions which quench in a finite time has been the subject of investigation of many authors (see, for instance [10], [18], [23], [24], [26], [28], [30], and the reference cited therein).
Similar results have been obtained in [32], where the authors considered analogous problems within the framework of the phenomenon of blow-up (we say that a solution blows up in a finite time if it reaches the value infinity in a finite time).
The first paper which deals with blow-up of (1)–(2) that we are aware as that of Perez-LLanos and Rossi in [32], where they considered the problem (1)-(2) in the case where f(u) = up with p = const > 1. They proved that the solution u of (1)–(2) blows up in a finite time and localized the blow-up
set. Some results about blow-up rate are also given. In the same way, in [29], Nabongo and Boni examined the initial value problem
ut=ε Z
Ω
J(x−y)(u(y, t)−u(x, t))dy+f(u) in Ω×(0, T), u= 0 on (R−Ω)×(0, T),
u(x,0) =u0(x)≥0 in Ω,
whereεis a positive parameter. They showed that, ifεis small enough, then the solutionu of the above problem blows up in a finite time, and its blow-up time goes to that of the solution of the following ODE
α0(t) =f(α(t)), t >0, α(0) = maxx∈Ωu0(x),
as εgoes to zero. In this paper, we are interested in the the phenomenon of quenching of the solutionu, and the continuity of the quenching time for the problem describe in (1)-(2). More precisely, consider the following initial value problem
vt= Z
Ω
J(x−y)(v(y, t)−v(x, t))dy+f(v) in Ω×(0, Th), (3)
v(x,0) =uh0(x) in Ω, (4)
where uh0 ∈ C0(Ω), 0≤uh0(x) ≤u0(x) in Ω, limh→0uh0 =u0. Here (0, Th) is the maximal time interval of existence of the solutionv. In the current paper, under some hypotheses, we show that the solutionv of (3)–(4) quenches in a finite time and estimate its quenching time. We demonstrate in passing that, when the norm of the initial datum is large enough, then the solution v of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution of a certain differential equation
α0(t) =f(α(t)), t >0, α(0) =kuh0k∞,
as kuh0k∞ goes to b. Finally, under some hypotheses, we prove that the so- lution v of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution u of (1)–(2) when h goes to zero. The remainder of the paper is organized in the following manner. In the next section, we prove the local existence and uniqueness of solutions. In the third section, under some conditions, we show that the solution v of (3)–(4) quenches in a finite time and estimate its quenching time. We also show that its quenching time goes to that of the solutionu of (1)–(2) whenhgoes to zero, in the last section, we give some computational results to illustrate our analysis.
2. LOCAL EXISTENCE
In this section, we shall establish the existence and uniqueness of nonneg- ative solutions of (1)–(2) in Ω×(0, T) for all small T. We shall also prove some results concerning the maximum principle within the framework of non- local diffusion problems for our subsequent use. Let us notice that results on local existence and uniqueness are known for our problem if one modifies slightly the proof given by Perez-LLanos and Rossi in [32]. However, for the sake of completeness, we outline them. Let t0 be fixed, and define the func- tion space Yt0 ={u;u∈C([0, t0], C(Ω))} equipped with the norm defined by kukYt
0 = max0≤t≤t0ku(·, t)k∞foru∈Yt0. It is easy to see thatYt0 is a Banach space. Introduce the set
Xt0 ={u;u∈Yt0,kukYt
0 ≤b0},
where b0 = ku0k2∞+b. We observe that Xt0 is a nonempty bounded closed convex subset of Yt0. Define the mapR as follows
R:Xt0 →Xt0
R(v)(x, t) =u0(x) + Z t
0
Z
Ω
J(x−y)(v(y, s)−v(x, s))dyds+ Z t
0
f(v(x, s))ds.
Theorem 1. Assume that u0 ∈ C0(Ω), and 0 ≤ u0(x) < b in Ω. Then R maps Xt0 into Xt0, and R is strictly contractive if t0 is approximately small relative to ku0k∞.
Proof. Due to the fact thatR
ΩJ(x−y)dy≤R
RNJ(x−y)dy = 1,a straight- forward computation reveals that
|R(v)(x, t)−u0(x)| ≤2kvkYt
0t+f(kvkYt
0)t, which implies that kR(v)kYt
0 ≤ ku0k∞+ 2b0t0+f(b0)t0.If t0 ≤ b2b0−ku0k∞
0+f(b0), (5)
then
kR(v)kYt
0 ≤b0.
Therefore, if (5) holds, thenRmapsXt0 intoXt0. Now, we are going to prove that the map R is strictly contractive. Let v, z ∈Xt0. Settingα =v−z, we discover that
|(R(v)−R(z))(x, t)| ≤
Z t 0
Z
Ω
J(x−y)(α(y, s)−α(x, s))dyds
+
Z t 0
(f(v(x, s))−f(z(x, s)))ds . Use Taylor’s expansion to obtain
|(R(v)−R(z))(x, t)| ≤2kαkYt
0t+tkv−zkYt
0f0(kβkYt
0),
whereβ is an intermediate function betweenv and z. We deduce that kR(v)−R(z)kYt
0 ≤2kαkYt
0t0+t0kv−zkYt
0f0(kβkYt
0), which implies that kR(v)−R(z)kYt
0 ≤(2t0+t0f0(b0))kv−zkYt
0. If t0 ≤ 4+2f10(b0),
(6)
then kR(v)−R(z)kYt
0 ≤ 12kv −zkYt
0. Hence, we see that R(v) is a strict
contraction inYt0 and the proof is complete.
It follows from the contraction mapping principle that for appropriately chosen t0 > 0, R has a unique fixed point u(x, t) ∈ Yt0 which is a solution of (1)-(2). If kukYt
0 < b, then taking as initial datum u(·, t0) ∈ C0(Ω) and arguing as before, it is possible to extend the solution up to some interval [0, t1) for certain t1 > t0. Now, to end this section, we shall provide some results about the maximum principle tailored to our study. The following lemma is a version of the maximum principle for nonlocal problems.
Lemma 2. Let a ∈ C0(Ω×[0, T)), and let u ∈ C0,1(Ω×[0, T)) satisfying the following inequalities
(7) ut− Z
Ω
J(x−y)(u(y, t)−u(x, t))dy+a(x, t)u(x, t)≥0 in Ω×(0, T), u(x,0)≥0 in Ω.
(8)
Then, we have u(x, t)≥0 in Ω×(0, T).
Proof. Let T0 be any positive quantity satisfying T0 < T. Since a(x, t) is bounded in Ω×[0, T0], then there existsλsuch thata(x, t)−λ >0 in Ω×[0, T].
Define z(x, t) = eλtu(x, t) and letm = minx∈Ω,t∈[0,T
0]z(x, t). Due to the fact thatz 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, Z
Ω
J(x0−y)(z(y, t0)−z(x0, t0))dy≥0.
(9)
With the aid of the first inequality of the lemma, it is not hard to see that zt(x0, t0)−
Z
Ω
J(x0−y)(z(y, t0)−z(x0, t0))dy+a(x0, t0)−λ)z(x0, t0)≥0.
We deduce from (9) that (a(x0, t0)−λ)z(x0, t0 ≥ 0. Since a(x0, t0)−λ >0, we get z(x0, t0)≥0. This implies thatu(x, t)≥0 in Ω×[0, T0], and the proof
is complete.
An immediate consequence of the above lemma is the following comparison lemma. Its proof is straightforward.
Lemma 3. Let a∈C0(Ω×[0, T)) and letu,v∈C0,1(Ω×[0, T)) satisfying the following inequalities
ut− Z
Ω
J(x−y)(u(y, t)−u(x, t))dy+a(x, t)u(x, t)
≥vt− Z
Ω
J(x−y)(v(y, t)−v(x, t))dy+a(x, t)v(x, t) in Ω×(0, T), u(x,0)≥v(x,0) in Ω.
Then, we have u(x, t)≥v(x, t) in Ω×(0, T).
Remark 4. Invoking the mean value theorem and Lemma 2.2, it is not hard to see that v(x, t) ≤u(x, t) as long as all of them are defined. We infer
that Th ≥T.
3. THE QUENCHING TIME
In this section, under some conditions, we show that the solutionvof (3)–(4) quenches in a finite time and estimate its quenching time. We demonstrate in passing that, if the L∞ norm of the initial datum is large enough, then the solution v of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution of a differential equation as kuh0k∞ goes to b.
Finally, we gather some results that we deem useful to prove the continuity of the quenching time time. Our first result says that the solution v of (3)–(4) always quenches in a finite time if the initial datum is nonnegative. It is stated in the following theorem.
Theorem 5. Let v be the solution of (3)–(4). Then v quenches in a finite time, and its quenching time Th obeys the following estimate
Th ≤ Z b
A ds f(s), where A= |Ω|1 R
Ωuh0(x)dx.
Proof. Since (0, Th) is the maximal time interval on which kv(., t)k∞ < b, our aim is to show that Th is finite and satisfies the above inequality. Due to the fact that the initial datum uh0 is nonnegative in Ω, we know from Lemma 2.1 that the solution v is also nonnegative in Ω×(0, Th).
Integrating both sides of (3) over (0, t), we find that v(x, t)−uh0(x) =
Z t 0
Z
Ω
J(x−y)(v(y, s)−v(x, s))dyds
+ Z t
0
f(v(x, s))ds for t∈(0, Th).
Integrate again in the x variable and apply Fubini’s theorem to obtain Z
Ω
v(x, t)dx−
Z
Ω
uh0(x)dx= Z t
0
Z
Ω
f(v(x, s)
ds for t∈(0, Th).
(10) Set
w(t) = |Ω|1 Z
Ω
v(x, t)dx for t∈[0, Th).
Taking the derivative ofw int and using (10), we arrive at w0(t) =
Z
Ω 1
|Ω|f(v(x, t))dx, for t∈(0, Th)
It follows from Jensen’s inequality that w0(t) ≥ f(w(t)) for t ∈ (0, Th), or equivalently
dw
f(w) ≥dt for t∈(0, Th).
Integrate the above inequality over (0, Th) to obtain Th≤
Z b w(0)
ds f(s).
Since the quantity on the right hand side of the above inequality is finite, we deduce that v quenches in a finite time at the timeTh which obeys the above inequality. Use the fact thatw(0) =A to complete the rest of the proof.
The above theorem allows us to obtain an estimate which depends on the L1 norm of the initial datum. This kinds of estimation is not interesting in order to obtain the continuity of the quenching time as a function of the initial datum. Therefore, we shall give another result which reveals an estimate of the of the quenching time that depends on the L∞norm of the initial datum.
This result is stated in the theorem below.
Theorem 6. Let A = Rb 0
dσ
f(σ). If A < 1, then the solution v of (3)–(4) quenches in a finite time, and its quenching time Th obeys the following esti- mate
Th ≤ 1−A1 Z b
kuh0k∞
dσ f(σ).
Proof. Since (0, Th) is the maximal time interval of existence of the solution v, our aim is to show thatTh is finite and satisfies the above inequality. As in the proof of Theorem 3.1, an application of Lemma 2.1 reveals that the solution v is nonnegative in Ω×(0, Th). Due to the fact thatJ(z) is nonnegative for z∈RN, and
Z
Ω
J(x−y)dy≤ Z
RN
J(x−y)dy= 1 for x∈Ω, we note that
vt(x, t)≥ −v(x, t) +f(v(x, t)) in Ω×(0, Th),
which implies that
vt(x, t)≥f(v(x, t))
1−f(v(x,t))v(x,t)
in Ω×(0, Th).
It is not hard to see that Z b
0 dσ
f(σ) ≥ sup
0≤t<b
Z t 0
dσ
f(σ) ≥ sup
0≤t<b t f(t), because f(s) is nondecreasing for s∈[0, b). We infer that
vt(x, t)≥(1−A)f(v(x, t)) in Ω×(0, Th), or equivalently
dv
f(v) ≥(1−A)dt in Ω×(0, Th).
(11)
Integrate the above inequality over (0, Th) to obtain (1−A)Th≤
Z b uh0(x)
dσ
f(σ) in Ω.
It follows that
Th ≤ 1−A1 Z b
kuh0k∞
dσ f(σ).
We conclude that the solution v of (3)–(4) quenches in a finite time, because the quantity on the right hand side of the above inequality is finite. This
finishes the proof.
Remark 7. Let t0 ∈(0, Th). Integrating the inequality (11) over (t0, Th), we find that
(1−A)(Th−t0)≤ Z b
v(x,t0) dσ
f(σ) for x∈Ω, which implies that
Th−t0 ≤ 1−A1 Z b
kv(·,t0)k∞
dσ f(σ).
It is worth noting that the above estimate is crucial to obtain the continuity of the quenching time as a function of the initial datum. Let us notice that the condition A <1 of the above theorem is very restrictive in certain situations.
By the following theorem, we avoid this condition in the case where the L∞ norm of the initial datum is large enough.
Theorem 8. Let v be the solution of (3)–(4), and suppose that the initial datum at (4) obeys the following condition f(kuh0k∞)> b. Then, the solution v quenches in a finite time, and its quenching time Th is estimated as follows
Th ≤ f(kuf(kuhh0k∞)
0k∞)−b
Z b kuh0k∞
dσ f(σ).
Proof. Since (0, Th) is the maximal time interval of existence of the solution v, our aim is to show thatThis finite and satisfies the above inequality. Owing to Lemma 2.1, we know that the solutionvis nonnegative in Ω×(0, Th) because the initial datumuh0 is nonnegative in Ω. We note that
Z
Ω
J(x−y)dy≤ Z
RN
J(x−y)dy= 1 for x∈Ω, which implies that
vt(x, t)≥ −v(x, t) +f(v(x, t)) in Ω×(0, Th).
(12)
Letx0(t)∈Ω be such that U(t) = max
x∈Ω
v(x, t) =v(x0(t), t) for t∈(0, Th).
It is easy to see that
U0(t) = max
x∈Ω
vt(x, t) for t∈(0, Th).
Consequently, replacing x byx0(t) in (12), we have
U0(t)≥vt(x, t)≥ −U(t) +f(U(t)) for t∈(0, Th).
This estimate may be rewritten as follows U0(t)≥f(U(t))
1−f(U(t))U(t)
for t∈(0, Th).
According to the fact that U(t)≤b, the above estimate becomes U0(t)≥f(U(t))
1−f(U(t))b
for t∈(0, Th).
(13)
We note thatU0(0)>0, and we claim thatU0(t)>0 fort∈(0, Th). To prove the claim, we argue by contradiction. Let t0 be the first t∈(0, Th) such that U0(t) > 0 for t ∈ (0, t0), but U0(t0) = 0. This implies that U(t0) ≥ U(0) = kuh0k∞. Therefore, we get
0 =U0(t0)≥f(kuh0k∞)
1−f(kubh
0k∞)
>0,
which is a contradiction, and the claim is proved. In view of the claim, we find thatU(t)≥ kuh0k∞ fort∈(0, Th), and making use of (13), we arrive at
U0(t)≥
1− b
f(kuh0k∞)
f(U(t)) for t∈(0, Th), or equivalently
dU f(U) ≥
1− b
f(kuh0k∞)
dt for t∈(0, Th).
(14)
Integrating the above estimate over (0, Th) we obtain
1−f(kubh
0k∞)
Th ≤
Z b kuh0k∞
dσ f(σ),
which implies that
Th ≤ f(kuh0k∞)
f(kuh0k∞)−b
Z b kuh0k∞
dσ f(σ).
We use the fact that the quantity on the right hand side of the above inequality
is finite to complete the rest of the proof.
Remark9. Lett0∈(0, Th). Integrating the estimate (14) over (t0, Th), we discover that
Th−t0 ≤ f(kuf(kuhh0k∞)
0k∞)−b
Z b kv(·,t0)k∞
dσ f(σ).
Up to now, the results obtained allow us to see some upper bounds of the quenching time. In the theorem below, we derive a lower bound of the quenching time when quenching occurs.
Theorem 10. Suppose that the solution v of (3)–(4) quenches in a finite time Th. Then, we have
Th ≥ Z b
kuh0k∞
dσ f(σ).
Proof. Letα(t) be the solution of the following ordinary differential equation α0(t) =f(α(t)), t∈(0, Te), α(0) =kuh0k∞,
where (0, Te) is the maximal time interval of existence of the solution α(t).
By a routine computation, one easily sees that Te=Rb kuh0k∞
dσ
f(σ). Now, let us introduce the functionz defined as follows
z(x, t) =α(t) in Ω×[0, Te).
A straightforward calculation yields zt(x, t) =
Z
Ω
J(x−y)(z(y, t)−z(x, t))dy+f(z(x, t)) in Ω×(0, Te), z(x,0)≥v(x,0) in Ω.
Set
w(x, t) =z(x, t)−v(x, t) in Ω×[0, T∗),
whereT∗ = min{Th, Te}. Making use of the mean value theorem, we find that wt(x, t)≥
Z
Ω
J(x−y)(w(y, t)−w(x, t))dy +f0(ξ(x, t))w(x, t) in Ω×(0, T∗), w(x,0)≥0 in Ω,
where ξ(x, t) is an intermediate value between v(x, t) and z(x, t). It follows from Lemma 2.1 that
w(x, t)≥0 in Ω×(0, T∗),
or equivalently
v(x, t)≤α(t) in Ω×(0, T∗).
(15)
We claim that Th ≥ Te. To prove the claim, we argue by contradiction.
Suppose that Th < Te. In view of (15), we see that kv(·, Th)k∞ ≤ α(Th) <
b, which contradicts the fact that (0, Th) is the maximum time interval of existence of the solution v. This demonstrates the claim, and the proof is
complete.
Remark 11. Combining Theorems 3.1 and 3.4, we note that, if the initial datum at (4) satisfies uh0 = β = const ≥ 0, then the solution v of (3)–(4) quenches in a finite timeTh=Rb
β dσ
f(σ).
With the aid of Theorems 3.3 and 3.4, we can derive the following interesting result.
Theorem 12. Let v be the solution of (3)–(4), and suppose that the initial datum (4) obeys the following condition f(kuh0k∞) > b. Then, the solution v quenches in a finite time, and its quenching time Th obeys the following estimates
0≤Th−Te≤ bTe
f(kuh0k∞)+o
Te
f(kuh0k∞)
as kuh0k∞→b, where Te=Rb
kuh0k∞
dσ f(σ).
Proof. Since (0, Th) is the maximal time interval of existence of the solution u, our aim is to show thatThis finite and satisfies the above estimates. Making use of Theorems 3.3 and 3.4, we find that Te is finite and obeys the following estimates
Te≤Th ≤ Te
1− b
f(kuh0k∞)
. (16)
Apply Taylor’s expansion to obtain
1
1− b
f(kuh0k∞)
= 1 +f(kubh
0k∞)+o 1
f(kuh0k∞)
as kuh0k∞→b.
Use (16) and the above relation to complete the rest of the proof.
Remark 13. The estimates of Theorem 3.5 can be rewritten as follows 0≤ TTh
e −1≤ b
f(kuh0k∞)+o
b f(kuh0k∞)
as kuh0k∞→b.
We infer that
lim
kuh0k∞→b Th
Te = 1.
4. CONTINUITY OF THE QUENCHING TIME
In this section, under some assumptions, we show that the solution v of (3)–(4) quenches in a finite time, and its quenching time goes to that of the solution u of (1)–(2) when the parameter h goes to zero. In order to obtain the above result, we firstly reveal that the solution v approaches the solution u in any interval Ω×[0, T −τ] where τ in(0, T). This result is stated in the following theorem.
Theorem14. Assume that the problem(1)–(2)has a solutionu∈C0,1(Ω× [0, T))such thatsupt∈[0,T−τ]ku(·, t)k∞≤b−α,whereα∈(0, b)andτ ∈(0, T).
Suppose that the initial datum at uh0 satisfies the following condition kuh0−u0k∞=o(1) as h→0.
(17)
Then, the problem(3)–(4) admits a unique solutionv∈C0,1(Ω×[0, Th)), and the following relation holds
sup
t∈[0,T−τ]
kv(·, t)−u(·, t)k∞=O(kuh0 −u0k∞) as h→0, where τ ∈(0, T).
Proof. The problem (3)–(4) admits a unique solutionv∈C0,1(Ω×[0, Th)).
In Remark 2.1, we have mentioned thatTh ≥T. Lett(h)≤T−τ be the first tsuch that
kv(·, t)−u(·, t)k∞< α2 for t∈(0, t(h)).
(18)
We know from (17) that t(h) > 0 for h small enough. An application of the triangle inequality yields
kv(·, t)k∞≤ ku(·, t)k∞+kv(·, t)−u(·, t)k∞ for t∈(0, t(h)), which implies that
kv(·, t)k∞≤b−α+ α2 ≤b−α2 for t∈(0, t(h)).
(19)
Introduce the error edefined as follows
e(x, t) =v(x, t)−u(x, t) in Ω×[0, t(h)).
Making use of the mean value theorem, we find that et(x, t) =
Z
Ω
J(x−y)(e(y, t)−e(x, t))dy+f0(ξ(x, t))e(x, t) in Ω×(0, t(h)), e(x,0) =uh0(x)−u0(x) in Ω,
whereξ(x, t) is an intermediate value between v(x, t) and u(x, t). Set z(x, t) =e(L+1)tkuh0−u0k∞ in Ω×[0, T],
where L = f0(b− α2). Due the fact that ku(·, t)k∞ ≤ b−α fort ∈ (0, t(h)), having in mind (19), it is not hard to see thatL≥f0(ξ(x, t))∈Ω×(0, t(h)).
A straightforward computation reveals that zt(x, t)≥
Z
Ω
J(x−y)(z(y, t)−z(x, t))dy+f0(ξ(x, t))z(x, t) in Ω×(0, t(h)), z(x,0)≥e(x,0) in Ω.
Invoking Lemma 2.2, we obtain
z(x, t)≥e(x, t) in Ω×(0, t(h)).
In the same way, we also prove that
z(x, t)≥ −e(x, t) in Ω×(0, t(h)), which implies that
kv(·, t)−u(·, t)k∞≤e(L+1)tkuh0 −u0k∞ for t∈(0, t(h)).
(20)
Now, we claim that t(h) =T −τ. To prove the claim, we argue by contradic- tion. Suppose that t(h)< T−τ. In view of (18) and (20), it is easy to check that
α
2 ≤ kv(·, t(h))−u(·, t(h))k∞≤e(L+1)Tkuh0 −u0k∞.
Since the term on the right hand side of the above inequality goes to zero as h goes to zero, infer that α2 ≤0, which is a contradiction. This demonstrates
the claim, and the proof is complete.
At the moment, we are in a position to prove the main result of this section.
Theorem 15. Assume that the problem (1)–(2) has a solution u which quenches in a finite time T such that u ∈C0,1(Ω×[0, T)). Suppose that the initial datum atuh0 satisfies the condition(17). Then, under the assumption of Theorem 3.2, the problem (3)–(4) admits a unique solution v which quenches in a finite time, and the following relation holds
h→0limTh =T.
Proof. Let 0< ε < T /2. There exists a positive constant α ∈ (0, b) such that
1 1−A
Z b b−α
dσ f(σ) < 2ε. (21)
Sinceuquenches at the timeT, then there exists a timeT0 ∈(T−ε/2, T) such thatku(·, t)k∞≥b−α2 fort∈[T0, T). Invoking Theorem 4.1, we note that the problem (3)–(4) admits a unique solution v, and the following estimate holds kv(·, T0)−u(·, T0)k∞≤ α2. Making use of the triangle inequality, we find that
kv(·, T0)k∞≥ ku(·, T0)k∞− kv(·, T0)−u(·, T0)k∞,
which implies that
kv(·, T0)k∞≥b−α2 −α2 =b−α.
In Remark 2.1 of the paper, we have revealed thatTh ≥T. We infer from (21) and Remark 3.1 that
0≤Th−T ≤Th−T0+T0−T ≤ 2ε+ε2 =ε,
and the proof is complete.
Remark 16. If in Theorem 4.2 we replace the assumption of Theorem 3.2 by that of Theorem 3.3, then the result of Theorem 4.2 remains valid.
5. NUMERICAL RESULTS
In this section, we give some computational experiments to confirm the theory given in the previous section. We consider the problem (1)-(2) in the case where Ω = (−1,1), f(u) = (1−u)p with p >1,
J(x) = (3
2x2, if |x|<1, 0, if |x| ≥1, u0(x) = γ(2−ε(sin(πx))2
4 ) with γ > 0, ε ∈ (0,1]. We start by the construction of some adaptive schemes 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 u of (1)-(2) by the solutionUh(n)= (U0(n),· · · , UI(n))T of the following explicit scheme
Ui(n+1)−Ui(n)
∆tn =
I−1
X
j=0
hJ(xi−xj)(Uj(n)−Ui(n)) + (1−Ui(n))−p, 0≤i≤I, Ui(0) =ϕi, 0≤i≤I,
where ϕi = γ(2−ε(sin(πx4 i))2). In order to permit the discrete solution to re- produce the properties of the continuous one when the time tapproaches the quenching time T, we need to adapt the size of the time step so that we take
∆tn=h2(1− kUh(n)k∞)p+1
withkUh(n)k∞= max0≤i≤I|Ui(n)|.Let us notice that the restriction on the time step ensures the nonnegativity of the discrete solution. We also approximate the solutionu of (1)–(2) by the solution Uh(n) of the implicit scheme below
Ui(n+1)−Ui(n)
∆tn =
I−1
X
j=0
hJ(xi−xj)(Uj(n+1)−Ui(n+1)) + (1−Ui(n))−p, 0≤i≤I, Ui(0) =ϕi, 0≤i≤I.
As in the case of the explicit scheme, here, we also choose
∆tn=h2(1− kUh(n)k∞)p+1.
Let us again remark that for the above implicit scheme, existence and non- negativity of the discrete solution are also guaranteed using standard methods (see, for instance [9]). We need the following definition.
Definition17. We say that the discrete solutionUh(n)of the explicit scheme or the implicit scheme quenches in a finite time if limn→∞kUh(n)k∞= 1, and the seriesP∞
n=0∆tnconverges. The quantityP∞
n=0∆tnis called the numerical quenching time of the discrete solution Uh(n).
In the following tables, in rows, we present the numerical quenching times, the numbers of iterations, the CPU times and the orders of the approximations corresponding to meshes of 16, 32, 64, 128. We take for the numerical blow-up timetn=Pn−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−Tlog(2)2h)/(T2h−Th)).
Tables 1–8 show the numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the explicit and implicit Euler method.
I tn n CPU time s
16 0.1267303 74 1.1 -
32 0.1254584 295 1.8 -
64 0.1251387 1179 6.26 1.992 128 0.1250586 4715 78 1.996 Table 1. Explicit Euler method,p= 1,γ= 1,ε= 1/100
I tn n CPU time s
16 0.1267353 75 1.8 -
32 0.1254588 295 2.1 -
64 0.1251389 1179 6.26 1.995 128 0.1250587 4715 78 1.996
Table 2. The implicit Euler method,p= 1,γ= 1,ε= 1/100
I tn n CPU time s
16 0.0326408 35 2.2 -
32 0.0319626 139 15.6 -
64 0.0317912 554 74 1.98
128 0.0317480 2215 79 2.01
Table 3. The explicit Euler method,p= 1,γ= 1.5,ε= 1
I tn n CPU time s
16 0.0326418 35 2.4 -
32 0.0319629 139 16.1 -
64 0.0317915 554 75.4 1.981 128 0.0317481 2215 79 1.985 Table 4. The implicit Euler method,p= 1,γ= 1.5,ε= 1
I tn n CPU time s
16 0.0321300 35 1.4 -
32 0.0314844 137 6.1 -
64 0.0313305 548 62 2.89
128 0.0312703 2189 96 1.35
Table 5. The explicit Euler method,p= 1,γ= 1.5,ε= 1/100
I tn n CPU time s
16 0.0321305 35 1.6 -
32 0.0314846 137 6.1 -
64 0.0313325 548 63 2.08
128 0.0312713 2189 98 1.31
Table 6. The implicit Euler method,p= 1,γ= 1.5,ε= 1/100
I tn n CPU time s
16 3.684595 e-4 4 1.5 -
32 3.371334 e-4 13 9.6 -
64 3.190056 e-4 52 78 2.02
128 3.145623 e-4 207 499 0.78
Table 7. The explicit Euler method,p= 1,γ= 1.95,ε= 1
I tn n CPU time s
16 3.684596 e-4 4 2.5 -
32 3.371394 e-4 13 10.6 -
64 3.190086 e-4 52 79 2.02
128 3.145624 e-4 207 499 0.78
Table 8. The implicit Euler method,p= 1,γ= 1.95,ε= 1
Remark 18. If we consider the problem (1)–(2) in the case wheref(u) = (1−u)−1andu0(x) = 1/2, then we know from Remark 3.3 that the quenching
time of the solutionu equals 0.125. We observe from Tables 1 to 2 that when εis small enough, then the numerical quenching time is approximately equal to 0.125. This fact confirms the result established within the framework of the continuity. On the other hand, the quenching time Te of the solution of the following differential equation α0(t) = (1−α(t))−1, t > 0, α(0) = γ2 is given explicitly by Te= (2−γ)2/8,and
Te=
3.145e−4 when γ = 1.95 .
We note from Tables 3 to 4 that, when γ = 1.95, then numerical quenching time of the discrete solution is approximately equal that toTe. These results
illustrate the idea of Theorem 3.5.
For other illustrations, in what follows, we shall give some plots. In the following figures, we can appreciate that the discrete solution quenches in a finite time.
Fig. 1. Evolution of the discrete solu- tion,γ= 1,ε=1001 (explicit scheme).
Fig. 2. Evolution of the discrete solu- tion,γ= 1,ε= 1001 (implicit scheme).
Fig. 3. Evolution of the discrete solu- tion,γ= 1.5,ε= 1 (explicit scheme).
Fig. 4. Evolution of the discrete solu- tion,γ= 1.5,ε= 1 (implicit scheme).
0 50 100 150 200 250 0
50 100 1500.4
0.6 0.8 1 1.2 1.4
i n
U(n,i)
Fig. 5. Evolution of the discrete solu- tion,γ= 1.95,ε= 1 (explicit scheme).
0 50 100 150 200 250
0 50 100 1500.4
0.6 0.8 1 1.2 1.4
i n
U(n,i)
Fig. 6. Evolution of the discrete solu- tion,γ= 1.95,ε= 1 (implicit scheme).
Acknowledgement. The authors would like to thank the referees for the throughout reading of the manuscript and several suggestions that helped us improve the presentation of the paper.
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Received by the editors: May 1st, 2012.
