a multi-dimensional quenching problem due to a concentrated nonlinear source in

Nonlinear Analysis 69 (2008) 1494–1514www.elsevier.com/locate/na

A multi-dimensional quenching problem due to a concentratednonlinear source in RN

C.Y. Chan∗, P. Tragoonsirisak

Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA 70504-1010, USA

Received 9 May 2007; accepted 6 July 2007

Abstract

Let B be a N -dimensional ball x ∈ RN: |x | < R centered at the origin with a radius R, B be its closure, and ∂B be its

boundary. Also, let ν(x) denote the unit inward normal at x ∈ ∂B, and χB(x) be the characteristic function, which is 1 for x ∈ B,and 0 for x ∈ RN

\B. This article studies the following multi-dimensional semilinear parabolic first initial-boundary value problemwith a concentrated nonlinear source on ∂B:

ut − 4u = α∂χB(x)

∂νf (u) in RN

× (0, T ],

u(x, 0) = 0 for x ∈ RN , u(x, t) → 0 as |x | → ∞ for 0 < t ≤ T,

where α and T are positive numbers, f is a given function such that limu→c− f (u) = ∞ for some positive constant c, and f (u)and its derivatives f ′(u) and f ′′(u) are positive for 0 ≤ u < c. It is shown that the problem has a unique nonnegative continuoussolution before quenching occurs, and if u quenches in a finite time, then it quenches everywhere on ∂B only. It is proved that ualways quenches in a finite time for N ≤ 2. For N ≥ 3, it is shown that there exists a unique number α∗ such that u exists globallyfor α ≤ α∗ and quenches in a finite time for α > α∗. Thus, quenching does not occur in infinite time. A formula for computing α∗

is given. A computational method for finding the quenching time is devised.c© 2007 Elsevier Ltd. All rights reserved.

MSC: primary 35K60; 35K57; 35B35

Keywords: Multi-dimensional quenching; Concentrated nonlinear source; Unbounded domain; Existence; Uniqueness; Quenching; Computationalmethod; Quenching time

1. Introduction

Let H = ∂/∂t − 4, T be a positive real number, x = (x1, x2, . . . , xN ) be a point in the N -dimensional Euclidianspace RN , Ω = RN

× (0, T ], B be a N -dimensional ball x ∈ RN:∣∣x − b

∣∣ < R centered at a given point b with aradius R, ∂B be the boundary of B, ν(x) denote the unit inward normal at x ∈ ∂B, and

χB(x) =

1 for x ∈ B,0 for x ∈ RN

\B,

∗ Corresponding author. Tel.: +1 337 482 5288; fax: +1 337 482 5346.E-mail addresses: [email protected] (C.Y. Chan), [email protected] (P. Tragoonsirisak).

0362-546X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2007.07.001

http://www.elsevier.com/locate/na

mailto:[email protected]

mailto:[email protected]

http://dx.doi.org/10.1016/j.na.2007.07.001

C.Y. Chan, P. Tragoonsirisak / Nonlinear Analysis 69 (2008) 1494–1514 1495

be the characteristic function. We would like to study the following multi-dimensional semilinear parabolic first initial-boundary value problem with a source on the surface of the ball:

Hu = α∂χB(x)

∂νf (u) in Ω ,

u(x, 0) = 0 for x ∈ RN , u(x, t) → 0 as |x | → ∞ for 0 < t ≤ T,

(1.1)

where α is a positive number. Without loss of generality, let b be the origin. This model is motivated by a N -dimensional ball B having a radius R and situated in RN ; on the surface ∂B of the ball, there is a nonlinearheat source of strength α f (u), where u(x, t) in Ω is the unknown temperature to be determined. We assume thatlimu→c− f (u) = ∞ for some positive constant c, and f (u) and its derivatives f ′(u) and f ′′(u) are positive for0 ≤ u < c. A solution u is said to quench if there exists an extended real number tq ∈ (0,∞] such that

sup

u(x, t) : x ∈ RN

→ c− as t → tq .

If tq < ∞, then u is said to quench in a finite time. If tq = ∞, then u quenches in infinite time. We note that whetherquenching for the heat equation with a non-concentrated source in an unbounded domain occurs in a finite time wasstudied by Dai and Gu [1], and Dai and Zeng [2].

In Section 2, we show that the nonlinear integral equation corresponding to the problem (1.1) has a uniquenonnegative continuous solution u, which is a strictly increasing function of t . We then prove that u is the uniquesolution of the problem (1.1). If tq is finite, we show that u quenches everywhere on ∂B only. In Section 3, we provethat for N ≤ 2, u always quenches in a finite time. This behavior is completely different from that for N ≥ 3. InSection 4, we show that for N ≥ 3, there exists a unique number α∗ such that u exists globally for α ≤ α∗ andquenches in a finite time for α > α∗. This rules out the possibility of quenching in infinite time. We also derive aformula for computing α∗. In Section 5, we give a computational method for determining the finite quenching time tq .

2. Existence, uniqueness, and quenching

Green’s function g(x, t; ξ, τ ) (cf. Stakgold [3, p. 198]) corresponding to the problem (1.1) is determined by thefollowing system:

Hg(x, t; ξ, τ ) = δ(x − ξ)δ(t − τ) for x, ξ ∈ RN and t, τ ∈ (−∞,∞),

g(x, t; ξ, τ ) = 0 for x, ξ ∈ RN , and t < τ,

g(x, t; ξ, τ ) → 0 as |x | → ∞ for ξ ∈ RN and t, τ ∈ (−∞,∞).

It is given by

g(x, t; ξ, τ ) =

1

[4π(t − τ)]N/2 exp

(−

|x − ξ |2

4(t − τ)

), t > τ,

0, t < τ.

To derive the integral equation from the problem (1.1), let us consider the adjoint operator (−∂/∂t − 4) of H . UsingGreen’s second identity, we obtain

u(x, t) = α

∫ t

0

∫RN

g(x, t; ξ, τ )∂χB(ξ)

∂νf (u(ξ, τ ))dξdτ.

Let R denote a positive number larger than R. Then,∫RN

g(x, t; ξ, τ )∂χB(ξ)

∂νf (u(ξ, τ ))dξ = lim

R→∞

∫|ξ |<R

g(x, t; ξ, τ ) f (u(ξ, τ )) [ν(ξ) · ∇χB(ξ)] dξ.

Since χB(ξ) = 0 for ξ ∈ RN\B, it follows from an integration by parts that

limR→∞

∫|ξ |<R

g(x, t; ξ, τ ) f (u(ξ, τ )) [ν(ξ) · ∇χB(ξ)] dξ

1496 C.Y. Chan, P. Tragoonsirisak / Nonlinear Analysis 69 (2008) 1494–1514

= limR→∞

[−

∫|ξ |<R

χB(ξ)

N∑i=1

∂

∂ξi(g(x, t; ξ, τ ) f (u(ξ, τ ))νi (ξ)) dξ

]

= limR→∞

[−

∫B

∇ · (g(x, t; ξ, τ ) f (u(ξ, τ ))ν(ξ)) dξ].

By the Divergence Theorem,

limR→∞

[−

∫B

∇ · (g(x, t; ξ, τ ) f (u(ξ, τ ))ν(ξ)) dξ]

= limR→∞

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξ

=

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξ .

Thus,

u(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ. (2.1)

Let Ω denote the closure of Ω .

Lemma 2.1. Let r(x, t) ∈ C(Ω). On Ω ,

∫ t0

∫∂B g(x, t; ξ, τ )r(ξ, τ )dSξdτ is continuous.

Proof. Let η = (ξi − xi ) /(2√

t − τ). Using∫

∞

−∞e−η2

dη =√π , we have∫

∂Bg(x, t; ξ, τ )r(ξ, τ )dSξ ≤

K

2√π(t − τ)1/2

, (2.2)

where K = max(ξ,τ )∈∂B×[0,T ] r(ξ, τ ).Let ωN denote the surface area of a unit sphere. For t > τ , g(x, t; ξ, τ ) is continuous. Thus for any x and x0 in

RN , t > τ and t0 > τ , we have, for any given positive number ε, there exists some positive number δ such that|g(x, t; ξ, τ )− g(x0, t0; ξ, τ )| < ε/(2ωN RN−1t0 K ) whenever |(x, t)− (x0, t0)| < δ. Since∫ t

0

∫∂B

g(x, t; ξ, τ )r(ξ, τ )dSξdτ = limσ→0

∫ t−σ

0

∫∂B

g(x, t; ξ, τ )r(ξ, τ )dSξdτ,

we let

F(x, t) =

∫ t−σ

0

∫∂B

g(x, t; ξ, τ )r(ξ, τ )dSξdτ.

To show that F(x, t) is continuous, we show that for any given ε, there exists some positive number δ1 such that|F(x, t)− F(x0, t0)| < ε whenever |(x, t)− (x0, t0)| < δ1. To achieve this, let us choose δ1 ≤ min

δ, ε2π/(4K 2)

.

Without loss of generality, we assume that t ≥ t0. Then,∣∣∣∣∫ t−σ

0

∫∂B

g(x, t; ξ, τ )r(ξ, τ )dSξdτ −

∫ t0−σ

0

∫∂B

g(x0, t0; ξ, τ )r(ξ, τ )dSξdτ

∣∣∣∣≤

∣∣∣∣∫ t−σ

t0−σ

∫∂B

g(x, t; ξ, τ )r(ξ, τ )dSξdτ

∣∣∣∣+ ∫ t0−σ

0

∫∂B

|g(x, t; ξ, τ )− g(x0, t0; ξ, τ )| |r(ξ, τ )| dSξdτ

≤K

2√π

∫ t−σ

t0−σ

1√

t − τdτ +

K ε

2ωN RN−1t0 K

∫ t0−σ

0

∫∂B

dSξdτ

≤K

√π

(√t − t0 + σ −

√σ)+ε (t0 − σ)

2t0

≤K

√π

(√t − t0 +

√σ −

√σ)+ε (t0 − σ)

2t0

<K

√π

√δ1 +

ε

2≤ ε.


The lemma then follows by letting σ → 0.

We modify the techniques used in proving Theorems 3 and 4 of Chan and Tian [4] for a blow-up problem toestablish the next two results.

Theorem 2.1. There exists some tq such that for 0 ≤ t < tq , the integral equation (2.1) has a unique continuousnonnegative solution u, and u is a strictly increasing function of t . If tq is finite, then u quenches at tq .

Proof. Let us construct a sequence un in Ω by u0(x, t) = 0, and for n = 0, 1, 2, . . . ,

Hun+1 = α∂χB(x)

∂νf (un) in Ω ,

un+1(x, 0) = 0 for x ∈ RN , un+1(x, t) → 0 as |x | → ∞ for 0 < t ≤ T .

From (2.1),

un+1(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (un(ξ, τ ))dSξdτ. (2.3)

Since

H(u1 − u0) = α∂χB(x)

∂νf (0) in Ω ,

(u1 − u0)(x, 0) = 0 for x ∈ RN , (u1 − u0)(x, t) → 0 as |x | → ∞ for 0 < t ≤ T,

f (0) > 0, and g(x, t; ξ, τ ) > 0 in(x, t; ξ, τ ) : x and ξ are in RN , T ≥ t > τ ≥ 0

, it follows from (2.1) that

u1(x, t) > u0(x, t) in Ω . Let us assume that for some positive integer j , 0 < u1 < u2 < u3 < · · · < u j−1 < u j inΩ . We have

H(u j+1 − u j ) = α∂χB(x)

∂ν

(f (u j )− f (u j−1)

)in Ω ,

(u j+1 − u j )(x, 0) = 0 for x ∈ RN , (u j+1 − u j )(x, t) → 0 as |x | → ∞ for 0 < t ≤ T .

Since f is a strictly increasing function and u j > u j−1, we have f (u j ) − f (u j−1) > 0. It follows from (2.1) thatu j+1 > u j . By the principle of mathematical induction, 0 < u1 < u2 < · · · < un−1 < un in Ω for any positiveinteger n.

To show that each un is an increasing function of t , let us construct a sequence wn such that for n = 0, 1,2, . . . , wn(x, t) = un(x, t + h)− un(x, t), where h is any positive number less than T . Then, w0(x, t) = 0. By (2.3),we have

w1(x, t) = α f (0)(∫ t+h

0

∫∂B

g(x, t + h; ξ, τ )dSξdτ −

∫ t

0

∫∂B

g(x, t; ξ, τ )dSξdτ).

Let σ = τ − h. Then,∫ t+h

0

∫∂B

g(x, t + h; ξ, τ )dSξdτ =

∫ h

0

∫∂B

g(x, t + h; ξ, τ )dSξdτ +

∫ t

0

∫∂B

g(x, t + h; ξ, σ + h)dSξdσ

=

∫ h

0

∫∂B

g(x, t + h; ξ, τ )dSξdτ +

∫ t

0

∫∂B

g(x, t; ξ, σ )dSξdσ

since g(x, t + h; ξ, σ + h) = g(x, t; ξ, σ ). Thus,

w1(x, t) = α f (0)∫ h

0

∫∂B

g(x, t + h; ξ, τ )dSξdτ,

which is positive for 0 < t ≤ T − h. Let us assume that for some positive integer j , w j > 0 for 0 < t ≤ T − h. Then,

w j+1(x, t) = α

(∫ t+h

0

∫∂B

g(x, t + h; ξ, τ ) f (u j (ξ, τ ))dSξdτ −

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u j (ξ, τ ))dSξdτ).


Let σ = τ − h. We have∫ t+h

0

∫∂B

g(x, t + h; ξ, τ ) f (u j (ξ, τ ))dSξdτ

=

∫ h

0

∫∂B

g(x, t + h; ξ, τ ) f (u j (ξ, τ ))dSξdτ +

∫ t

0

∫∂B

g(x, t + h; ξ, σ + h) f (u j (ξ, σ + h))dSξdσ

=

∫ h

0

∫∂B


∫ t

0

∫∂B

g(x, t; ξ, σ ) f (u j (ξ, σ + h))dSξdσ

>

∫ h

0

∫∂B


∫ t

0

∫∂B

g(x, t; ξ, σ ) f (u j (ξ, σ ))dSξdσ.

Thus,

w j+1(x, t) > α

∫ h

0

∫∂B

g(x, t + h; ξ, τ ) f (u j (ξ, τ ))dSξdτ > 0.

By the principle of mathematical induction, wn > 0 for 0 < t ≤ T − h and all positive integers n. Thus, each un is anincreasing function of t .

By (2.2),∫∂B g(x, t; ξ, τ )dSξ is integrable with respect to τ . Thus for any given positive constant M (<c), it follows

from (2.3) and un being an increasing function of t that there exists some t1 such that un+1 ≤ M for 0 ≤ t ≤ t1 andn = 0, 1, 2, . . . . In fact, t1 satisfies

un+1(x, t1) ≤ α f (M)∫ t1

0

∫∂B

g(x, t1; ξ, τ )dSξdτ ≤ M.

Let u denote limn→∞ un . From (2.3) and the Monotone Convergence Theorem (cf. Stromberg [5, p. 266]), we have(2.1) for 0 ≤ t ≤ t1.

To prove that u is unique, let us assume that the integral equation (2.1) has two distinct solutions u and u on theinterval [0, t1]. From (2.1),

u(x, t)− u(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) ( f (u(ξ, τ ))− f (u(ξ, τ ))) dSξdτ. (2.4)

Since f ′′(u) > 0 for u ∈ [0, c), it follows from the Mean Value Theorem that | f (u)− f (u)| ≤ f ′(M) |u − u|. From(2.4),

|u(x, t)− u(x, t)| ≤ α f ′(M)∫ t

0

∫∂B

g(x, t; ξ, τ ) |u(ξ, τ )− u(ξ, τ )| dSξdτ.

By Lemma 2.1, there exists some t2 (≤t1) such that

α f ′(M) supRN ×[0,t2]

(∫ t

0

∫∂B

g(x, t; ξ, τ )dSξdτ)< 1. (2.5)

Let Θ = supRN ×[0,t2] |u − u|. Then,

Θ ≤ α f ′(M) supRN ×[0,t2]

(∫ t

0

∫∂B

g(x, t; ξ, τ )dSξdτ)

Θ .

By (2.5), this gives a contradiction. Thus, we have uniqueness of the solution for 0 ≤ t ≤ t2.If t2 < t1, then for t2 ≤ t ≤ t1,

u(x, t) =

∫RN

g(x, t; ξ, t2)u (ξ, t2) dξ + α

∫ t

t2

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ (2.6)

(cf. Chan and Tian [4]). Thus for t2 ≤ t ≤ t1,

u(x, t)− u(x, t) = α

∫ t

t2

∫∂B

g(x, t; ξ, τ ) ( f (u(ξ, τ ))− f (u(ξ, τ ))) dSξdτ.


Let Θ = supRN ×[t2,min2t2,t1] |u − u|. Then,

Θ ≤ α f ′(M) supRN ×[t2,min2t2,t1]

(∫ t

t2

∫∂B


Θ .

For t ∈ [t2,min2t2, t1],

α f ′(M) supRN ×[t2,min2t2,t1]

(∫ t

t2

∫∂B


= α f ′(M) supRN ×[t2,min2t2,t1]

(∫ t−t2

0

∫∂B

g(x, t; ξ, σ + t2)dSξdσ)

= α f ′(M) supRN ×[t2,min2t2,t1]

(∫ t−t2

0

∫∂B

g(x, t − t2; ξ, σ )dSξdσ)< 1 (2.7)

by (2.5). This gives a contradiction. Hence, we have uniqueness of the solution for 0 ≤ t ≤ min2t2, t1. By proceedingin this way, the integral equation (2.1) has a unique solution u for 0 ≤ t ≤ t1.

To prove that u is continuous on RN× [0, t1], we note that f (un(ξ, τ )) is bounded (by f (M)). It follows from

(2.3), Lemma 2.1 and f being continuous that for n = 0, 1, 2, . . . , un+1(x, t) is continuous on RN× [0, t1]. From

(2.3),

un+1(x, t)− un(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) ( f (un(ξ, τ ))− f (un−1(ξ, τ ))) dSξdτ. (2.8)

Using the Mean Value Theorem, we have

f (un)− f (un−1) ≤ f ′(M) (un − un−1) .

Let Λn = supRN ×[0,t2] (un − un−1). From (2.8),

Λn+1 ≤ α f ′(M) supRN ×[0,t2]

(∫ t

0

∫∂B


Λn .

By (2.5), the sequence un(x, t) converges uniformly to u(x, t) for 0 ≤ t ≤ t2, and hence, u is continuous there.If t2 < t1, then from (2.6),

un+1(x, t) =

∫RN

g(x, t; ξ, t2)u (ξ, t2) dξ + α

∫ t

t2

∫∂B

g(x, t; ξ, τ ) f (un(ξ, τ ))dSξdτ.

Let Λn = supRN ×[t2,min2t2,t1] (un − un−1). Then for t2 ≤ t ≤ t1,

Λn+1 ≤ α f ′(M) supRN ×[t2,min2t2,t1]

(∫ t

t2

∫∂B


Λn .

It follows from (2.7) that for t ∈ [t2,min2t2, t1], u is continuous there. By proceeding in this way, the integralequation (2.1) has a unique continuous solution u for 0 ≤ t ≤ t1.

Let tq be the supremum of the intervals for which the integral equation (2.1) has a unique continuous solution u.If tq is finite, and u does not reach c− at tq , then for any positive constant between supRN u(x, tq) and c, a proofsimilar to the above shows that there exists some t3 (>tq) such that the integral equation (2.1) has a unique continuoussolution u for 0 ≤ t ≤ t3. This contradicts the definition of tq . Hence, if tq is finite, then u reaches c− somewhere inRN at tq .

Since un is an increasing function of t , we have for any positive number h such that t + h < tq ,

u(x, t + h)− u(x, t)

= α

(∫ t+h

0

∫∂B

g(x, t + h; ξ, τ ) f (u(ξ, τ ))dSξdτ −

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ).


As before,∫ t+h

0

∫∂B

g(x, t + h; ξ, τ ) f (u(ξ, τ ))dSξdτ

≥

∫ h

0

∫∂B

g(x, t + h; ξ, τ ) f (u(ξ, τ ))dSξdτ +

∫ t

0

∫∂B

g(x, t; ξ, σ ) f (u(ξ, σ ))dSξdσ.

Hence,

u(x, t + h)− u(x, t) ≥ α

∫ h

0

∫∂B

g(x, t + h; ξ, τ ) f (u(ξ, τ ))dSξdτ > 0,

which shows that u is a strictly increasing function of t .

The next result shows that the solution of the integral equation (2.1) is the solution of the problem (1.1).

Theorem 2.2. The problem (1.1) has a unique solution u for 0 ≤ t < tq .

Proof. Since∫ t

0

∫∂B g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ exists for x ∈ RN , and t in any compact subset of [0, tq), it follows

that for any t4 ∈ (0, t),∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ = limm→∞

∫ t−1/m

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ

= limm→∞

∫ t

t4

∂

∂ζ

(∫ ζ−1/m

0

∫∂B

g(x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξdτ

)dζ

+ limm→∞

∫ t4−1/m

0

∫∂B

g(x, t4; ξ, τ ) f (u(ξ, τ ))dSξdτ.

Since ∣∣gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))∣∣ =

∣∣∣∣∣ 1

[4π(ζ − τ)]N/2

(|x − ξ |2

4(ζ − τ)2

)exp

(−

|x − ξ |2

4(ζ − τ)

)

+1

[4π ]N/2

(−

N

2

)(ζ − τ)−N/2−1 exp

(−

|x − ξ |2

4(ζ − τ)

)∣∣∣∣∣ f (u(ξ, τ ))

is bounded for (ξ, τ ) ∈ ∂B × (0, ζ − 1/m), and hence integrable, it follows from the Leibnitz Rule (cf. Stromberg[5, p. 380]) that

∂

∂ζ

∫∂B

g(x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξ =

∫∂B

gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξ ,

which is continuous and bounded for each τ ∈ (0, ζ − 1/m). By the Leibnitz Rule,

∂

∂ζ

(∫ ζ−1/m

0

∫∂B


)

=

∫∂B

g(x, ζ ; ξ, ζ − 1/m) f (u(ξ, ζ − 1/m))dSξ +

∫ ζ−1/m

0

∂

∂ζ

∫∂B


=

∫∂B

g(x, 1/m; ξ, 0) f (u(ξ, ζ − 1/m))dSξ +

∫ ζ−1/m

0

∫∂B

gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξdτ.

Since g(x, 1/m; ξ, 0) is independent of ζ,

limm→∞

∫ t

t4

∫∂B

g

(x,

1m

; ξ, 0)

dSξdζ =

∫ t

t4lim

m→∞

∫∂B

g

(x,

1m

; ξ, 0)

dSξdζ.


For x 6= ξ , g(x, 1/m; ξ, 0) converges to zero uniformly with respect to ξ on ∂B as m → ∞. Thus,

limm→∞

∫∂B

g

(x,

1m

; ξ, 0)

dSξ =

∫∂B

limm→∞

g

(x,

1m

; ξ, 0)

dSξ . (2.9)

For x = ξ , g(x, 1/m; ξ, 0) increases as m increases. By the Monotone Convergence Theorem, (2.9) holds. Thus,

limm→∞

∫ t

t4

∫∂B

g

(x,

1m

; ξ, 0)

dSξdζ =

∫ t

t4

∫∂B

limm→∞

g

(x,

1m

; ξ, 0)

dSξdζ

=

∫ t

t4

∫∂Bδ(x − ξ)dSξdζ.

Since f is bounded, it follows from the Dominated Convergence Theorem (cf. Stromberg [5, p. 268]) that

limm→∞

∫ t

t4

∫∂B

g

(x,

1m

; ξ, 0)

f

(u

(ξ, ζ −

1m

))dSξdζ =

∫ t

t4

∫∂Bδ(x − ξ) f (u(ξ, ζ ))dSξdζ

=

∫ t

t4

∫RNδ(x − ξ)

∂χB(ξ)

∂νf (u(ξ, ζ ))dξdζ

=

∫ t

t4

∂χB(x)

∂νf (u(x, ζ ))dζ.

Hence,∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ = limm→∞

∫ t

t4

∫∂B

g

(x,

1m

; ξ, 0)

f

(u

(ξ, ζ −

1m

))dSξdζ

+ limm→∞

∫ t

t4

∫ ζ−1/m

0

∫∂B

gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξdτdζ

+ limm→∞

∫ t4−1/m

0

∫∂B

g(x, t4; ξ, τ ) f (u(ξ, τ ))dSξdτ

=

∫ t

t4

∂χB(x)

∂νf (u(x, ζ ))dζ + lim

m→∞

∫ t

t4

∫ ζ−1/m

0

∫∂B


+

∫ t4

0

∫∂B


Let

hm(x, ζ ) =

∫ ζ−1/m

0

∫∂B


Without loss of generality, let m > l. We have

hm(x, ζ )− hl(x, ζ ) =

∫ ζ−1/m

ζ−1/ l

∫∂B


The integrand is integrable. It follows from the Fubini Theorem (cf. Stromberg [5, p. 352]) that

hm(x, ζ )− hl(x, ζ ) =

∫∂B

∫ ζ−1/m

ζ−1/ lgζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dτdSξ .

Since f (u(ξ, τ )) is a monotone function of τ , it follows from the Second Mean Value Theorem (cf. Stromberg[5, p. 328]) that there exists some number γ with ζ − γ ∈ [ζ − 1/ l, ζ − 1/m] such that

1502 C.Y. Chan, P. Tragoonsirisak / Nonlinear Analysis 69 (2008) 1494–1514∫ ζ−1/m

ζ−1/ lgζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dτ

= f

(u

(ξ, ζ −

1l

))∫ ζ−γ

ζ−1/ lgζ (x, ζ ; ξ, τ )dτ + f

(u

(ξ, ζ −

1m

))∫ ζ−1/m

ζ−γ

gζ (x, ζ ; ξ, τ )dτ.

Hence,

hm(x, ζ )− hl(x, ζ ) =

∫∂B

f

(u

(ξ, ζ −

1l

))∫ ζ−γ

ζ−1/ lgζ (x, ζ ; ξ, τ )dτdSξ

+

∫∂B

f

(u

(ξ, ζ −

1m

))∫ ζ−1/m

ζ−γ

gζ (x, ζ ; ξ, τ )dτdSξ .

From gζ (x, ζ ; ξ, τ ) = −gτ (x, ζ ; ξ, τ ), we have

hm(x, ζ )− hl(x, ζ ) =

∫∂B

f

(u

(ξ, ζ −

1l

))(g

(x, ζ ; ξ, ζ −

1l

)− g(x, ζ ; ξ, ζ − γ )

)dSξ

+

∫∂B

f

(u

(ξ, ζ −

1m

))(g(x, ζ ; ξ, ζ − γ )− g

(x, ζ ; ξ, ζ −

1m

))dSξ

=

∫∂B

f

(u

(ξ, ζ −

1l

))(g

(x,

1l; ξ, 0

)− g (x, γ ; ξ, 0)

)dSξ

+

∫∂B

f

(u

(ξ, ζ −

1m

))(g(x, γ ; ξ, 0)− g

(x,

1m

; ξ, 0))

dSξ ,

which can be made as small as we wish by choosing l and m sufficiently large since f (u) is bounded, and g(x, ε; ξ, 0)is continuous for any ε > 0. Thus, hm is a Cauchy sequence converging uniformly with respect to ζ in any compactsubset of (0, tq). Hence,

limm→∞

∫ t

t4

∫ ζ−1/m

0

∫∂B

gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξdτdζ =

∫ t

t4

∫ ζ

0

∫∂B

gζ (x, ζ ; ξ, τ ) f (u(ξ, τ ))dSξdτdζ.

Therefore,∫ t

0

∫∂B


=

∫ t

t4

∂χB(x)

∂νf (u(x, ζ ))dζ +

∫ t

t4

∫ ζ

0

∫∂B


+

∫ t4

0

∫∂B


Thus,

∂

∂t

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ =∂χB(x)

∂νf (u(x, t))+

∫ t

0

∫∂B

gt (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ.

Since

gxi (x, t; ξ, τ ) =1

[4π(t − τ)]N/2

[−(xi − ξi )

2(t − τ)

]exp

(−

|x − ξ |2

4(t − τ)

)

is bounded for (ξ, τ ) ∈ ∂B × (0, t − ε), it follows from the Leibniz Rule that for any x in RN and t in any compactsubset of (0, tq),

∂

∂xi

∫ t−ε

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ =

∫ t−ε

0

∫∂B

gxi (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ.


For any X i = (x1, x2, . . . , xi−1, xi , xi+1, . . . , xN ) and Zi = (x1, x2, . . . , xi−1, ςi , xi+1, . . . , xN ), we have

limε→0

∫ t−ε

0

∫∂B


= limε→0

∫ xi

xi

∂

∂ςi

(∫ t−ε

0

∫∂B

g(Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdτ)

dςi

+ limε→0

∫ t−ε

0

∫∂B

g(X i , t; ξ, τ ) f (u(ξ, τ ))dSξdτ

= limε→0

∫ xi

xi

∫ t−ε

0

∫∂B

gςi (Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdτdςi +

∫ t

0

∫∂B

g(X i , t; ξ, τ ) f (u(ξ, τ ))dSξdτ. (2.10)

By the Fubini Theorem,∫ xi

xi

∫ t−ε

0

∫∂B

gςi (Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdτdςi

=

∫ t−ε

0

∫ xi

xi

∫∂B

gςi (Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdςi dτ

=

∫ t−ε

0

∫ xi

xi

∂

∂ςi

∫∂B

g(Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdςi dτ

=

∫ t−ε

0

∫∂B

(g(x, t; ξ, τ )− g(X i , t; ξ, τ )

)f (u(ξ, τ ))dSξdτ.

Thus,

limε→0

∫ xi

xi

∫ t−ε

0

∫∂B


=

∫ t

0

∫∂B

(g(x, t; ξ, τ )− g(X i , t; ξ, τ )

)f (u(ξ, τ ))dSξdτ,

which exists by Lemma 2.1 since f is continuous and bounded. Therefore, by using the Fubini Theorem,

limε→0

∫ xi

xi

∫ t−ε

0

∫∂B


= limε→0

∫ t−ε

0

∫ xi

xi

∫∂B


=

∫ t

0

∫ xi

xi

∫∂B


=

∫ xi

xi

∫ t

0

∫∂B

gςi (Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdτdςi .

From (2.10),∫ t

0

∫∂B


=

∫ xi

xi

∫ t

0

∫∂B

gςi (Zi , t; ξ, τ ) f (u(ξ, τ ))dSξdτdςi +

∫ t

0

∫∂B

g(X i , t; ξ, τ ) f (u(ξ, τ ))dSξdτ,

which gives

∂

∂xi

∫ t

0

∫∂B


∫ t

0

∫∂B

gxi (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ. (2.11)


We note that

gxi xi (x, t; ξ, τ ) =1

[4π(t − τ)]N/2

(−(xi − ξi )

2(t − τ)

)2

exp

(−

|x − ξ |2

4(t − τ)

)

+1

[4π(t − τ)]N/2

(−

12(t − τ)

)exp

(−

|x − ξ |2

4(t − τ)

)is bounded for (ξ, τ ) ∈ ∂B × (0, t − ε). An argument similar to the above in proving (2.11) shows that for any x inRN and t in any compact subset of

(0, tq

),

∂

∂xi

∫ t

0

∫∂B

gxi (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ =

∫ t

0

∫∂B

gxi xi (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ.

From (2.11),

∂2

∂x2i

∫ t

0

∫∂B


∫ t

0

∫∂B

gxi xi (x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ.

From the integral equation (2.1), we have for x ∈ RN and t ∈ (0, tq),

Hu = α∂χB(x)

∂νf (u(x, t))+ α

∫ t

0

∫∂B

Hg(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ

= α∂χB(x)

∂νf (u(x, t))+ α lim

ε→0

∫ t−ε

0

∫∂Bδ(x − ξ)δ(t − τ) f (u(ξ, τ ))dSξdτ

= α∂χB(x)

∂νf (u(x, t)).

From (2.1), limt→0 u(x, t) = 0 for x ∈ RN , and u → 0 as |x | → ∞ since u is continuous. Thus, the nonnegativecontinuous solution of the integral equation (2.1) is a solution of the problem (1.1). Since a solution of the problem(1.1) is a solution of the integral equation (2.1), which has a unique solution before quenching occurs, u is the solutionof the problem (1.1), and the theorem is proved.

Let

χu<c =

1 if u < c,0 if u ≥ c.

What happens after quenching occurs in a finite time? Physically, the source disappears at the point where u reaches cat tq . Thus to study beyond quenching (cf. Chan and Kong [6], and Chan and Yang [7]), we extend the problem (1.1)as follows:

Hu = α∂χB(x)

∂νf (u)χu<c in RN

× (0,∞) ,

u(x, 0) = 0 for x ∈ RN , u(x, t) → 0 as |x | → ∞ for 0 < t < ∞.

(2.12)

With this, we show in our next result that at tq , quenching occurs on the surface of the ball only.

Theorem 2.3. If tq is finite, then at tq , u quenches everywhere on ∂B only.

Proof. By Theorems 2.1 and 2.2, there exists some tq such that for 0 ≤ t < tq , the problem (1.1) has a uniquenonnegative continuous solution u, which is a strictly increasing function of t . Since u(x, t) on ∂B × (0, tq) is known,let us denote it by g(x, t), and rewrite the problem (1.1) as two initial-boundary value problems:

Hu = 0 in B × (0, tq),u(x, 0) = 0 on B, u(x, t) = g(x, t) on ∂B × (0, tq).

(2.13)

Hu = 0 in (RN\B)× (0, tq),

u(x, 0) = 0 on RN\B, u (x, t) = g(x, t) on ∂B × (0, tq),

u(x, t) → 0 as |x | → ∞ for 0 < t < tq .

(2.14)


To show that u cannot attain its maximum outside B for t > 0, let us prove that (x,5u) < 0 for x ∈ RN\B. For

x ∈ RN\B, it follows from (2.11) that

∂u(x, t)

∂xi= α

∫ t

0

∫∂B

1

[4π(t − τ)]N/2

exp

[−

|x − ξ |2

4(t − τ)

]f (u(ξ, τ ))

[−(xi − ξi )

2(t − τ)

]dSξdτ,

which gives

(x,∇u) = −α

∫ t

0

∫∂B

1

2N+1πN/2(t − τ)1+N/2

exp

[−

|x − ξ |2

4(t − τ)

]f (u(ξ, τ ))

(|x |

2− (x, ξ)

)dSξdτ.

By the Cauchy–Schwarz inequality,

|x |2− (x, ξ) ≥ |x |

2− |x | |ξ | = |x | (|x | − R) > 0.

Thus, (x,∇u) < 0 for x ∈ RN\B.

Let us consider the problem (2.13). It follows from the strong maximum principle (cf. Friedman [8, p. 34]) that uattains its maximum on ∂B × (0, tq). Since u is a strictly increasing function of t , we have, for each given ρ ∈

(0, tq

),

that u attains its maximum for 0 ≤ t ≤ ρ somewhere on ∂B × ρ. Suppose that there exists a smallest positive valueof t , say t0, and some y 6∈ ∂B such that u(y, t0) = minx∈∂B u(x, t0). We claim that for x ∈ ∂B, u(x, t0) = u(y, t0). Ifthis is not true, then there exists some x ∈ ∂B such that u(x, t0) > minx∈∂B u(x, t0). Since u is continuous, there existssome point (y, t0) in a neighborhood of (x, t0) such that y 6∈ ∂B and u(y, t0) > minx∈∂B u(x, t0). This contradictsthe definition of t0. Thus, u attains its maximum at (y, t0) for 0 ≤ t ≤ t0. By the strong maximum principle and thecontinuity of u, we have u ≡ u(y, t0) on B × [0, t0]. This contradicts u(x, 0) = 0. Thus for any t > 0,

u(x, t) > u(y, t) for any x ∈ ∂B, and any y 6∈ ∂B. (2.15)

We claim that for each t > 0, u attains the same value for x ∈ ∂B. If this is not true, then for some t > 0, there existssome x ∈ ∂B such that u(x, t) > minx∈∂B u(x, t). By continuity, there exists some point (y, t) in a neighborhood of(x, t) such that y 6∈ ∂B and u(y, t) > minx∈∂B u(x, t). This contradicts (2.15). Hence for any t > 0,

u(x, t) = M(t) for x ∈ ∂B,M(t) > u(y, t) for any y 6∈ ∂B, (2.16)

where M(t) denotes supx∈RN u(x, t).From (2.16), u quenches everywhere on ∂B as t → tq . Since tq is finite, it follows from (2.12) that the problem

(2.13) holds also at tq with g(x, tq) = c for x ∈ ∂B. Thus, if u quenches somewhere at (z, tq) where z ∈ B, then bythe strong maximum principle and the continuity of u, we have u ≡ c on B × [0, tq ]. This gives a contradiction sinceu(x, 0) = 0. Hence, if u quenches, then it quenches on ∂B only. For the problem (2.14), if u quenches somewhere at(Z , tq) for some Z ∈ RN

\B, then let B be a ball centered at the origin such that Z ∈ B. Since u quenches at (Z , tq),it follows that u ≡ c on (B\B)× [0, tq ]. This contradicts u(x, 0) = 0, and hence, u quenches only on ∂B at tq . Thisproves the theorem.

3. N ≤ 2

Let

I (x, t) =

∫∂B

g(x, t; ξ, 0)dSξ .

Lemma 3.1. For t ≥ 1 and for any b ∈ B,

(4π)−N/2e−R2ωN RN−1t−N/2

≤ I (b, t) ≤ (4π)−N/2ωN RN−1t−N/2.

Proof. For t ≥ 1,

e−R2≤ exp

(−

|b − ξ |2

4t

)≤ 1,


where ξ ∈ ∂B. Thus,

e−R2(4π t)−N/2

∫∂B

dSξ ≤ I (b, t) ≤ (4π t)−N/2∫∂B

dSξ .

The surface area ωN of a unit sphere is given by NπN/2/Γ (N/2 + 1) (cf. Evans [9, p. 615]). Using Γ (N/2 + 1)= (N/2)Γ (N/2), we have ωN = 2πN/2/Γ (N/2). Since

∫∂B dSξ = ωN RN−1, the lemma then follows.

Let us denote the (positive) constants (4π)−N/2e−R2ωN RN−1 and (4π)−N/2ωN RN−1 by c1 and c2 respectively.

Then from Lemma 3.1,

c1t−N/2≤ I (b, t) ≤ c2t−N/2.

Theorem 3.1. For N ≤ 2, the solution u of the problem (1.1) (for any α) always quenches everywhere in a finite timeon ∂B only.

Proof. Since u ≥ 0 and f ′(u) > 0, we have f (u) ≥ f (0) > 0. For any t ∈ (1,∞), it follows from (2.1) that

u(x, t) ≥ α f (0)∫ t

0

∫∂B

g(x, t; ξ, τ )dSξdτ

> α f (0)∫ t−1

0

∫∂B

g(x, t; ξ, τ )dSξdτ

= α f (0)∫ t−1

0

∫∂B

g(x, t − τ ; ξ, 0)dSξdτ

= α f (0)∫ t−1

0I (x, t − τ)dτ

= α f (0)∫ t

1I (x, θ)dθ.

For each t (>0), it follows from (2.16) that u attains its (absolute) maximum M(t) at any b ∈ ∂B. Thus for anyt ∈ (1,∞),

u(b, t) > α f (0)∫ t

1I (b, θ)dθ.

By Lemma 3.1,∫ t

1I (b, θ)dθ ≥ c1

∫ t

1θ−N/2dθ =

2c1

(t1/2

− 1)

if N = 1,

c1 ln t if N = 2.

Thus, u(b, t) quenches in a finite time. The theorem then follows from Theorem 2.3.

4. N ≥ 3

In this section, we show that the quenching behavior for N ≥ 3 is completely different from that for N ≤ 2.

Theorem 4.1. (i) For N ≥ 3, the solution u of the problem (1.1) exists globally for α sufficiently small.(ii) For N ≥ 3, the solution u of the problem (1.1) quenches in a finite time for α sufficiently large.

Proof. (i) Since f ′(u) > 0, we have f (u) ≤ f (c/2) for u(x, t) ≤ c/2. Thus for u(x, t) ≤ c/2, it follows from (2.1)that

u(x, t) ≤ α f( c

2

) ∫ t

0

∫∂B

g(x, t; ξ, τ )dSξdτ.

For 0 < t ≤ 1, it follows from (2.2) that

u(x, t) ≤ α f( c

2

) ∫ t

0

dτ

2√π(t − τ)1/2

=α

√π

f( c

2

)t1/2

≤α

√π

f( c

2

).


For t > 1, it follows from (2.2) and (2.16) that for any b ∈ ∂B,

u(x, t) ≤ α f( c

2

)(∫ t−1

0I (b, t − τ)dτ +

∫ t

t−1I (b, t − τ)dτ

)

≤ α f( c

2

)(∫ t

1I (b, θ)dθ +

∫ t

t−1

dτ

2√π(t − τ)1/2

)< α f

( c

2

)(∫ ∞

1I (b, θ)dθ +

1√π

)since I (b, θ) is positive. By Lemma 3.1,∫

∞

1I (b, θ)dθ ≤ c2

∫∞

1θ−N/2dθ =

c2

N/2 − 1< ∞ for N ≥ 3. (4.1)

Thus for α sufficiently small, u(x, t) ≤ c/2 for (x, t) ∈ RN× (0,∞).

(ii) For any b ∈ ∂B and t ∈ (1,∞),

u(b, t) ≥ α f (0)∫ t

0

∫∂B

g(b, t; ξ, τ )dSξdτ > α f (0)∫ t−1

0I (b, t − τ)dτ = α f (0)

∫ t

1I (b, θ)dθ.

By (4.1), (0<)∫ t

1 I (b, θ)dθ < ∞. We can choose α sufficiently large such that u(b, t) > c. This proves thetheorem.

The fundamental solution (cf. Evans [9, pp. 22 and 615]) of the Laplace equation for N ≥ 3 is given by

G(x) =Γ( N

2 + 1)

N (N − 2)πN/2

1

|x |N−2 . (4.2)

Let us modify the proof of Theorem 3 of Chan and Kaper [10] for a one-dimensional problem with a non-concentratedsource in a bounded domain to obtain the following result.

Theorem 4.2. If u(x, t) ≤ C for some constant C ∈ (0, c), then u(x, t) converges from below to a solutionU (x) = limt→∞ u(x, t) of the nonlinear integral equation,

U (x) = α

∫∂B

G(x − ξ) f (U (ξ))dSξ . (4.3)

Proof. Let

F(x, t) =

∫RN

G(x − ξ)u(ξ, t)dξ for (x, t) ∈ RN× (0,∞). (4.4)

For any point x , we can always choose a number R such that |x | < R. Using Green’s second identity, we have∫RN

G(x − ξ)4 u(ξ, t)dξ = limR→∞

∫|ξ |<R

G(x − ξ)4 u(ξ, t)dξ

= limR→∞

∫|ξ |<R

u(ξ, t)4 G(x − ξ)dξ

= limR→∞

∫|ξ |<R

−u(ξ, t)δ(x − ξ)dξ

= −u(x, t).

Since u is the solution of the problem (1.1), F(x, t) may be regarded as a distribution. Thus,


Ft (x, t) =

∫RN

G(x − ξ)ut (ξ, t)dξ

=

∫RN

G(x − ξ)

(4u(ξ, t)+ α

∂χB(ξ)

∂νf (u(ξ, t))

)dξ

= −u(x, t)+ α

∫RN

G(x − ξ)∂χB(ξ)

∂νf (u(ξ, t))dξ.

From Theorem 2.1, u is a strictly increasing function of t for x ∈ RN . Since f is increasing, the integrand of the secondterm on the right-hand side is monotone increasing with respect to t . It follows from the Monotone ConvergenceTheorem and the continuity of f that

limt→∞

Ft (x, t) = − limt→∞

u(x, t)+ α

∫RN

G(x − ξ)∂χB(ξ)

∂νf ( lim

t→∞u(ξ, t))dξ

= − limt→∞

u(x, t)+ α

∫∂B

G(x − ξ) f ( limt→∞

u(ξ, t))dSξ , (4.5)

which exists since u ≤ C . We note from (4.4) that F is a strictly increasing function of t for x ∈ RN . Iflimt→∞ Ft (x, t) > 0 at some point x , then F(x, t) would increase without bound as t tends to infinity. Since Fis bounded, limt→∞ Ft (x, t) = 0. The theorem follows from (4.5).

The next result shows that there exists a critical value for α.

Theorem 4.3. For N ≥ 3, there exists a unique α∗ such that u exists globally for α < α∗, and u quenches in a finitetime for α > α∗.

Proof. Let us consider the sequence un given by u0(x, t) = 0, and for n = 0, 1, 2, . . . ,

un+1(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (un(ξ, τ ))dSξdτ.

Since un is an increasing sequence as n increases, it follows from the Monotone Convergence Theorem that

u(x, t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (u(ξ, τ ))dSξdτ,

where limn→∞ un(x, t) = u(x, t). To show that the larger the α, the larger the solution u, let α > β, and consider thesequence vn given by v0(x, t) = 0, and for n = 0, 1, 2, . . . ,

vn+1(x, t) = β

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (vn(ξ, τ ))dSξdτ.

Similarly,

v(x, t) = limn→∞

vn(x, t) = β

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (v(ξ, τ ))dSξdτ.

Since un > vn for n = 1, 2, 3, . . . , we have u ≥ v. Hence, the solution u is a nondecreasing function of α. It followsfrom Theorem 4.1 that there exists a unique α∗ such that u exists globally for α < α∗, and u quenches in a finite timefor α > α∗.

We note that the critical value α∗ is determined as the supremum of all positive values α for which a solution Uof (4.3) exists. The proof of the next result, showing that the solution u exists globally when α = α∗, is analogous tothat of Theorem 7 of Chan and Jiang [11] for a degenerate one-dimensional problem in a bounded domain.

Theorem 4.4. For N ≥ 3, the solution u does not quench in infinite time.

Proof. From (2.16), U (x) = limt→∞ u(x, t) attains its maximum at b ∈ ∂B. From (4.3),

U (b) = α

∫∂B

G(b − ξ) f (U (b))dSξ .


Thus,

α∗=

(1∫

∂B G(b − ξ)dSξ

)max

0≤U (b)≤c

(U (b)

f (U (b))

). (4.6)

Let us consider the function ψ(s) = s/ f (s). Since ψ(s) > 0 for 0 < s < c, and ψ(0) = 0 = lims→c− ψ(s), a directcomputation shows that ψ(s) attains its maximum when ψ(s) = 1/ f ′(s), where s ∈ (0, c) by the Rolle Theorem.Thus, max0≤U (b)≤c (U (b)/ f (U (b))) occurs when

U (b)

f (U (b))=

1f ′(U (b))

with 0 < U (b) < c. This implies that U (x) exists when α = α∗. Hence for α ≤ α∗, u exists globally and is uniformlybounded away from c. Since u quenches in a finite time for α > α∗, u does not quench in infinite time.

Our next result gives a formula for α∗.

Theorem 4.5. For N ≥ 3,

α∗=

(N − 2)π (N−3)/2

RΓ(

N−12

) N−3∏i=1

(∫ π0 sini ϕdϕ

) max0≤U (b)≤c

(U (b)

f (U (b))

), (4.7)

where for N = 3,∏N−3

i=1

(∫ π0 sini ϕdϕ

)= 1.

Proof. Using cos2 µ+ sin2 µ = 1, we have

(1 − cosϕ1)2+ (sinϕ1 cosϕ2)

2+ (sinϕ1 sinϕ2 cosϕ3)

2+ · · ·

+ (sinϕ1 · · · sinϕN−2 cos θ)2 + (sinϕ1 · · · sinϕN−2 sin θ)2 = 2 (1 − cosϕ1) . (4.8)

Since∫∂B |b − ξ |2−N dSξ is independent of where b is on ∂B, we may let b = (R, 0, 0, 0, . . . , 0) ∈ RN . Using the

spherical coordinates (cf. Stromberg [5, pp. 369-370]), we have

∫∂B

dSξ|b − ξ |N−2 =

RN−1

RN−2

∫ π

−π

∫ π

0. . .

(N−2 times)

∫ π

0

N−2∏i=1

sinN−i−1 ϕi dϕ1dϕ2 . . . dϕN−2dθ

[2 (1 − cosϕ1)](N−2)/2

=2πR

2(N−2)/2

∫ π

0. . .

(N−2 times)

∫ π

0

N−2∏i=1

sinN−i−1 ϕi dϕ1dϕ2 . . . dϕN−2

(1 − cosϕ1)(N−2)/2

.

Since 1 − cosϕ1 = 2 sin2(ϕ1/2), and sinϕ1 = 2 sin(ϕ1/2) cos(ϕ1/2), we have∫∂B

dSξ|b − ξ |N−2 = 2πR

(∫ π

0cosN−2

(ϕ1

2

)dϕ1

) N−2∏i=2

∫ π

0sinN−i−1 ϕi dϕi ,


i=2

∫ π0 sinN−i−1 ϕi dϕi = 1. By using Mathematica version 5.2, we get

∫ π

0cosN−2

(ϕ1

2

)dϕ1 =

√πΓ

(N−1

2

)Γ( N

2

) .

Therefore,∫∂B

dSξ|b − ξ |N−2 = 2π3/2 R

Γ(

N−12

)Γ( N

2

) N−3∏i=1

(∫ π

0sini ϕN−i−1dϕN−i−1

),



i=1

(∫ π0 sini ϕN−i−1dϕN−i−1

)= 1. From (4.2), and Γ ((N/2)+ 1) = (N/2)Γ (N/2),∫

∂BG(b − ξ)dSξ =

Γ( N

2 + 1)

N (N − 2)πN/2

∫∂B

dSξ|b − ξ |N−2

=

RΓ(

N−12

)(N − 2)π (N−3)/2

N−3∏i=1

(∫ π

0sini ϕdϕ

),


i=1

(∫ π0 sini ϕdϕ

)= 1.

From (4.6), we have (4.7).

For illustration, let f (u) = 1/(1−u). A direct computation shows that U (b) (1 − U (b)) attains its maximum whenU (b) = 1/2. Hence,

α∗=

(N − 2)π (N−3)/2

4RΓ(

N−12

) N−3∏i=1

(∫ π0 sini ϕdϕ

) .

5. Quenching time tq

To determine the quenching time tq , it follows from (2.16) that it is sufficient to consider

M(t) = α

∫ t

0

∫∂B

g(b, t; ξ, τ ) f (M(τ ))dSξdτ,

where b is any point on ∂B. Since u is an increasing function of t , we have f (M(τ )) < f (M(t)) for t > τ . To find alower bound tl of tq , we may use

M(t) < α f (M(t))∫ t

0

∫∂B

g(b, t; ξ, τ )dSξdτ =α f (M(t))

(4π)N/2

∫ t

0

1

(t − τ)N/2

∫∂B

exp

[−

|b − ξ |2

4(t − τ)

]dSξdτ.

(5.1)

Without loss of generality, let b = (R, 0) for N = 2 and b = (R, 0, 0, 0, . . . , 0) for N ≥ 3.For N = 2, we have∫

∂Bexp

[−

|b − ξ |2

4(t − τ)

]dSξ = R

∫ π

−π

exp[−(R − R cos θ)2 + (R sin θ)2

4(t − τ)

]dθ

= R∫ π

−π

exp[−

R2 (1 − cos θ)2(t − τ)

]dθ

= R

exp

[−

R2

2(t − τ)

] ∫ π

−π

exp[

R2 cos θ2(t − τ)

]dθ.

From (5.1),

M(t) <αR f (M(t))

4π

∫ t

0

1(t − τ)

exp

[−

R2

2(t − τ)

] ∫ π

−π

exp[

R2 cos θ2(t − τ)

]dθdτ.

Let η = R/√

2(t − τ). We have dη = R−2η3dτ , and

M(t) <αR f (M(t))

2π

∫∞

R/√

2t

exp(−η2

)η

∫ π

−π

exp(η2 cos θ

)dθdη,

which gives∫∞

R/√

2t

exp(−η2

)η

∫ π

−π

exp(η2 cos θ

)dθdη >

2παR

M(t)

f (M(t)).


Thus, a lower bound tl for the quenching time is given by solving∫∞

R/√

2tl

exp(−η2

)η

∫ π

−π

exp(η2 cos θ

)dθdη =

2παR

max0≤M(t)≤c

(M(t)

f (M(t))

). (5.2)

For N ≥ 3 and α > α∗, it follows from (4.8) that∫∂B

exp

[−

|b − ξ |2

4(t − τ)

]dSξ

= RN−1∫ π

−π

∫ π

0. . .

(N−2 times)

∫ π

0

exp

[−

R2 (1 − cosϕ1)

2(t − τ)

] N−2∏i=1

sinN−i−1 ϕi dϕ1dϕ2 . . . dϕN−2dθ

= 2πRN−1

exp[−

R2

2(t − τ)

](N−3∏i=1

∫ π

0sini ϕdϕ

)∫ π

0

exp

[R2 cosϕ1

2(t − τ)

]sinN−2 ϕ1dϕ1.

From (5.1),

M(t) < 2πRN−1α f (M(t))

(4π)N/2

(N−3∏i=1

∫ π

0sini ϕdϕ

)

×

∫ t

0

1

(t − τ)N/2

exp

[−

R2

2(t − τ)

] ∫ π

0

exp

[R2 cosϕ1

2(t − τ)

]sinN−2 ϕ1dϕ1dτ.

Let η = R/√

2(t − τ). We have dη = R−2η3dτ , and

M(t) <αR f (M(t))

(2π)(N−2)/2

(N−3∏i=1

∫ π

0sini ϕdϕ

)

×

∫∞

R/√

2tηN−3

[exp

(−η2

)] ∫ π

0

[exp

(η2 cosϕ1

)]sinN−2 ϕ1dϕ1dη,

which gives∫∞

R/√

2tηN−3

[exp

(−η2

)] ∫ π

0

[exp

(η2 cosϕ1

)]sinN−2 ϕ1dϕ1dη >

(2π)(N−2)/2

αRN−3∏i=1

∫ π0 sini ϕdϕ

M(t)

f (M(t)).

Thus, a lower bound tl for the quenching time is given by solving∫∞

R/√

2tlηN−3

[exp

(−η2

)] ∫ π

0

[exp

(η2 cosϕ1

)]sinN−2 ϕ1dϕ1dη

=(2π)(N−2)/2

αRN−3∏i=1

∫ π0 sini ϕdϕ

max0≤M(t)≤c

(M(t)

f (M(t))

). (5.3)

From (2.3) and (2.16), we have for N = 2, M (0)(t) = 0, M (k) (0) = 0, and for k = 1, 2, 3, . . . ,

M (k)(t) = α

∫ t

0

∫∂B

g(x, t; ξ, τ ) f (M (k−1)(τ ))dSξdτ

=αR

2π

∫∞

R/√

2t

exp(−η2

)η

f

(M (k−1)

(t −

R2

2η2

))∫ π

−π

[exp

(η2 cos θ

)]dθdη. (5.4)


Similarly for N ≥ 3, M (0)(t) = 0, M (k) (0) = 0, and for k = 1, 2, 3, . . . ,

M (k)(t) =αR

(2π)(N−2)/2

(N−3∏i=1

∫ π

0sini ϕdϕ

)

×

∫∞

R/√

2tηN−3

[exp

(−η2

)]f

(M (k−1)

(t −

R2

2η2

))∫ π

0

[exp

(η2 cosϕ1

)]sinN−2 ϕ1dϕ1dη.

(5.5)

To compute the quenching time tq , we use Mathematica version 5.2. We obtain its lower bound tl by solving (5.2)for N = 2 or (5.3) for N ≥ 3. We then let t = 2tl in (5.4) for N = 2 or (5.5) for N ≥ 3, and check the convergenceof the sequence

M (k)(t)

. If it diverges, then we use 2tl to be an upper bound of tq and tl to be a lower bound. If

M (k)(t)

converges, then we check the convergence for t = 3tl . If it diverges, then we use 3tl to be an upper boundand 2tl to be a lower bound. By this iterative process, we obtain an upper bound j tl for some positive integer j forwhich

M (k)(t)

diverges, and a lower bound ( j − 1)tl for which

M (k)(t)

converges. We then use the bisection

procedure to compute tq .We give below the steps to compute tq :

Step 1. We input the values of R, c and α, and the function f (u). Let j = 2.Step 2. We find tl from (5.2) for N = 2, or from (5.3) for N ≥ 3.Step 3. Let t = j ∗ tl .Step 4. We replace t by t in (5.4) for N = 2, or in (5.5) for N ≥ 3, and let M (0)(τ ) = 0.Step 5. Let h = t/m, where m denotes the number of subdivisions of equal length h. For k = 1, 2, 3, . . . , it follows

from (5.4) that for N = 2,

M (k)(rh) =αR

2π

∫∞

R/√

2rh

exp(−η2

)η

f

(M (k−1)

(rh −

R2

2η2

))∫ π

−π

[exp

(η2 cos θ

)]dθdη, (5.6)

and from (5.5) that for N ≥ 3,

M (k)(rh) =αR

(2π)(N−2)/2

(N−3∏i=1

∫ π

0sini ϕdϕ

)∫∞

R/√

2rhηN−3

[exp

(−η2

)]× f

(M (k−1)

(rh −

R2

2η2

))∫ π

0

[exp

(η2 cosϕ1

)]sinN−2 ϕ1dϕ1dη, (5.7)

where r = 1, 2, 3, . . . ,m with M (k)(0) = 0.Step 6. At the kth iteration, if for some r , M (k)(rh) ≥ c, then an upper bound t is found; otherwise for a given

tolerance δ, if maxr=0,1,2,...,m∣∣M (k)(rh)− M (k−1)(rh)

∣∣ < δ, then the sequence

M (k)(t)

converges; in thiscase, let j = j + 1 and repeat Steps 3–6. However, if maxr=0,1,2,...,m

∣∣M (k)(rh)− M (k−1)(rh)∣∣ ≥ δ, then

we use the interpolation

M (k)(τ ) = Interpolation [rh,M (k)(rh)r=0,...,m]

to approximate M (k)(τ ) and continue the iterative process for the (k + 1)th iteration. In this way, we obtainthe upper bound t (to be j ∗ tl ) and the lower bound t (to be ( j − 1) ∗ tl ) for tq .

Step 7. Let the lower and upper bounds t and t from Step 6 be the initial estimates of a lower bound t (0) and an upperbound t (0) of tq respectively.

Step 8. Let t (n) and t (n) be the (n + 1)th estimates of lower and upper bounds of tq respectively, t = (t (n) + t (n))/2,and h = t/m. With M (0)(τ ) = 0, and M (k) (0) = 0, we use (5.6) for N = 2, or (5.7) for N ≥ 3.

Step 9. At the kth iteration, if for some r , M (k)(rh) ≥ c, then t (n+1)= t (n), and t (n+1)

= t . We go to Steps8 and 9. If for the given tolerance δ, maxr=0,1,2,...,m

∣∣M (k)(rh)− M (k−1)(rh)∣∣ < δ, then the sequence

M (k)(t)

converges; in this case, let t (n+1)= t, t (n+1)

= t (n) and repeat Steps 8 and 9. However, ifmaxr=0,1,2,...,m

∣∣M (k)(rh)− M (k−1)(rh)∣∣ ≥ δ, then we use the interpolation to approximate M (k)(τ ) and


continue the iterative process for the (k + 1)th iteration to obtain an upper bound t and a lower bound t , andgo to Steps 8 and 9.

Step 10. After n iterations, if |t (n) − t (n)| < ε for a given tolerance ε, then t (n)q = (t (n) + t (n))/2 is accepted as thefinal estimate of tq .

For illustration, let N = 3, R = 1, and f (u) = 1/(1 − u). We note that max0≤M(t)≤c (M(t)/ f (M(t))) = 1/4.From (4.7), α∗

= 1/4. From (5.3),∫∞

1/√

2tlexp

(−η2

) ∫ π

0

[exp

(η2 cosϕ1

)]sinϕ1dϕ1dη =

(2π)1/2

α

(14

).

Using Mathematica version 5.2 to do the integration, we obtain

√2π +

√2tl −

√2tle−1/tl −

√2π erf(1/

√tl)−

√2π

4α= 0.

For each given α > α∗, we use

Plot

[√

2π +√

2t −√

2te−1/t−

√2πErf[1/

√t] −

√2π

4α, t, 0, 10,PlotRange- > All

]to find an approximate value z for tl . With this, we use

FindRoot

[√

2π +√

2t −√

2te−1/t−

√2πErf[1/

√t] −

√2π

4α== 0, t, z

]to find tl . Using Steps 1–10 with ε = 10−4, δ = 10−6, and m = 40, we obtain the following table for tq (to foursignificant figures):

α tl from (5.3) tq for N = 3

1.5α∗ 2.525 5.8592α∗ .9250 2.0462.5α∗ .5283 1.1203α∗ .3542 .7303

Similarly, we obtain the following results:

α tl from (5.2) tq for N = 2

.3750 1.168 2.414

.5000 .6797 1.365

.6250 .4531 .8964

.7500 .3247 .6389

We note that in either case, the quenching time tq is a decreasing function of α.

References

[1] Q.Y. Dai, Y.G. Gu, A short note on quenching phenomena for semilinear parabolic equations, J. Differential Equations 137 (1997) 240–250.[2] Q.Y. Dai, X.Z. Zeng, The quenching phenomena for the Cauchy problem of semilinear parabolic equations, J. Differential Equations 175

(2001) 163–174.[3] I. Stakgold, Boundary Value Problems of Mathematical Physics, vol. II, Macmillan, New York, NY, 1968, p. 198.[4] C.Y. Chan, H.Y. Tian, Multi-dimensional explosion due to a concentrated nonlinear source, J. Math. Anal. Appl. 295 (2004) 174–190.[5] K.R. Stromberg, An Introduction to Classical Real Analysis, Wadsworth, Belmont, CA, 1981, pp. 266, 268, 328, 352, 369–370, 380.[6] C.Y. Chan, P.C. Kong, Solution profiles beyond quenching for degenerate reaction–diffusion problems, Nonlinear Anal. 24 (1995) 1755–1763.[7] C.Y. Chan, J. Yang, Beyond quenching for degenerate singular semilinear parabolic equations, Appl. Math. Comput. 121 (2001) 185–201.


[8] A. Friedman, Partial Differential Equations of Parabolic Type, Prentice Hall, Englewood Cliffs, NJ, 1964, p. 34.[9] L.C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, vol. 19, American Mathematical Society, Providence, RI, 1998,

pp. 22, 615.[10] C.Y. Chan, H.G. Kaper, Quenching for semilinear singular parabolic problems, SIAM J. Math. Anal. 20 (1989) 558–566.[11] C.Y. Chan, X.O. Jiang, Quenching for a degenerate parabolic problem due to a concentrated nonlinear source, Quart. Appl. Math. 62 (2004)

553–568.

a multi-dimensional quenching problem due to a concentrated nonlinear source in

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