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Source-type Solutions of the Heat Equationwith Nonlinear Convection in n-space

Dimensionsby

Guofu Lu

Department of Mathematics

Putian University

Putian, Fujian Province, CHINA

and

Hong-Ming Yin

Department of Mathematics

Washington State University

Pullman, WA 99164, USA.

Abstract

In this paper we study the existence or nonexistence of a source-type solution for the heat

equation with nonlinear convection: ut = ∆u+~b ·∇un, (x, t) ∈ ST = IRN × (0, T ], u(x, 0) =

δ(x), x ∈ IRN , where δ(x) denotes Dirac-delta measure in IRN , N ≥ 2, n ≥ 0 and ~b =

(b1, · · · , bN) ∈ IRN is a constant vector. It is shown that there exists a critical number

pc = N+2N

such that the source-type solution to the above problem exists and is unique if

0 ≤ n < pc, while such solution does not exist if n ≥ pc. Moreover, the asymptotic behavior

of the solution near origin, when it exists, is derived. It is shown that the source-type solution

near the origin has the same behavior as that for the heat equation without convection if

0 ≤ n < N+1N

.

Keywords and Phases: Source-type solution, heat equation with convection, existence

and nonexistence.

Classification: 35K55, 35K65

1

1. IntroductionIn this paper we are concerned with the following Cauchy problem:

ut = ∆u +~b · ∇un, (x, t) ∈ S = IRN × (0,∞), (1.1)

u(x, 0) = δ(x), (1.2)

where δ(x) represents the Dirac-delta measure in IRN ,~b = (b1, · · · , bn) with constant compo-

nent bi ≥ 0, i = 1, 2, · · · , N and |~b| > 0 and n ≥ 0 is a constant.

Equation (1.1) is the classical heat conduction with a nonlinear convection, where various

constants such as density, thermal conductivity are normalized. Without the nonlinear term

in Eq.(1.1) it is well known that a solution to (1.1)-(1.2) is the source-type solution (or,

sometimes, called fundamental solution), which plays an important role in study of parabolic

equations. On the other hand, this type of problem models the heat conduction with the

concentrated source at a single point such as nuclear reaction. A natural question is whether

or not this type of solution exists when a nonlinear convection term is involved in (1.1).

It seems quite challenging to answer this question. However, this problem for the porous

medium equation has been studied by several authors. In [1], S. Kamin proved the existence

of the source-type solution for the porous medium equation in one-space dimension. Brezis-

Friedman [2] and Kamin-Pelteier [3] extended the results into n-dimensional case, which

includes a possible absorption in the equation. Furthermore, A. Friedman and S. Kamin

in [4] studied the asymptotic behavior of the source-type solution for the porous medium

equation in n-space dimensions. We would also like to mention some recent papers with lower

regularity data in the equation or on the boundary (see [19] and the references therein). In

the best of our knowledge it seems that there is no result on source-type of solutions to the

heat equation with a nonlinear convection.

The present work is devoted to the investigation of the source-type solution to Eq.(1.1)-

(1.2). It will be seen that the exponent number n in the nonlinear convection term dramat-

ically affects the existence or nonexistence of a source-type solution (see Theorem A below).

We would like to point out that our method is quite different from those used in ([1], [2], [3],

[4], [10]). Because of the nonlinear convection there is no self-similarity solution for Eq.(1.1).

As a result, the methods presented in the previous papers seem not suitable to deal with

the current problem (1.1)-(1.2). In this paper we will use some ideas from the work [5], [6],

2

[12] and [13] as well as Moser’s iteration technique to prove the existence or nonexistence,

depending on the exponent n, of source-type solutions to (1.1)-(1.2). Various auxiliary func-

tions are constructed in order to derive some uniform estimates. Sard’s theorem plays an

important role in the proof of uniqueness result. We also study the asymptotic behavior

of the source-type solution near origin (0,0). Throughout this paper, we assume that the

dimension N ≥ 2.

Definition 1.1: A function u(x, t) defined in ST = IRN × (0, T ] (T > 0) is called a

source-type solution (or fundamental solution) of the problem (1.1)-(1.2) if:

(i) u(x, t) is nonnegative, continuous in ST\(0, 0) and smooth in ST .

(ii) The equation (1.1) holds in the classical sense.

(iii) As τ → +0,

u(x, τ) → δ(x)

in the sense of distribution,i.e.

limt→+0

∫

IRNu(x, t)η(x)dx = η(0) (1.3)

for any η(x) ∈ C∞0 (IRN).

The main results of this paper are as follows:

Theorem A: There exists a critical exponent pc := N+2N

such that

(a) If 0 ≤ n < pc, then there exists a unique source-type solution to the problem (1.1)-(1.2),

(b) If n ≥ pc, then there exists no source-type solution to (1.1)-(1.2).

Theorem B: Assume that 0 ≤ n < N+1N

and u(x, t) is a source-type solution to the

problem (1.1)-(1.2), then

tN/2|u(x, t) − w(x, t)| → 0

as t → +0 uniformly with respect to x in any domain

(x, t) ∈ S

∣∣∣ |x| ≤ AtN/2

for some constant A > 0, where w(x, t) is source-type solution to the standard heat equation:

w(x, t) =1

(4πt)N/2e−|x|24t

3

Theorem C: If n = N+1N

and ~b = (b1, · · · , bN) ∈ IRN with positive constant component

bi(i = 1, 2, · · · , N), then there exists a unique similarity source-type solution to the equation

(1.1). In particular, if ~b = ~e = (1, 1, · · · , 1), then

u(x, t) = t−N2 e−

|x|24t

(1

Ne− 1

4Nt

N−1Öi=1

x2i·

∫ xN√t

−∞

e−s2

4N ds + C0

),

where the constant C0 is determined by∫

IRNu(x, t)dx = 1.

The rest of the paper is organized as follows: In Section 2, we will derive some uniform

estimates for the solutions of the approximate problem. These estimates are important for

the existence and uniqueness proof. In Section 3, the existence (Theorem A, part (a)) will be

established. In Sections 4, we show the uniqueness result. The nonexistence result (Theorem

A, part (b)) will be established in Section 5. The section 6 is devoted to investigation of

the asymptotic behavior of source-type solution near origin (Theorem B) and a similarity

source-type solution (Theorem C) is constructed in this section.

Remark 1.1: The critical exponent pc in Theorem A is total different from the Fujita type

one (see surveys [20]-[21] for Fujita type of critical exponents).

2. Some Uniform EstimatesLet u0,k(x) be a δ-type sequence in IRN , where u0,k(x) is smooth and nonnegative

function satisfies

supp u0,k ⊂ Bk(0) =

x ∈ IRN

∣∣∣ |x| <1

k

∫

IRNu0,k(x)dx = 1 (2.1)

k = 1, 2, · · · and u0,k → δ(x) as k → ∞ in the sense of distribution. Let uk(x, t) be the

solution to the equation (1.1) corresponding to the initial u0,k(x). According to the standard

theory of parabolic partial differential equations, this Cauchy problem has a unique classical

4

solution which is smooth in ST (see [16], [17]). We now derive some uniform estimates for

uk(x, t).

Theorem 2.1. Let uk(x, t) be the unique solution of (1.1) with the initial datum u0,k(x),

then uk(x, t) ≥ 0 on IRN × [0, T ] and uk(x, t) > 0 in ST . Moreover,

∫

IRNuk(x, t)dx = 1, t ∈ (0, T ].

Proof: The nonnegativity of uk(x, t) follows from the maximum principle since u0,k(x) ≥ 0.

The positivity of uk(x, t) in ST follows from the strong maximum principle. We integrate

the equation (1.1) over a big ball Br(0) = x ∈ RN : |x| < r for large r > 0 to see

∫

Br(0)

uk(x, t)dx =

∫

Br(0)

u0,k(x)dx +

∫

∂Br(0)

[∇uk −~bun] · νds. (2.2)

Note that, for fixed k,uk(x, t), ∂uk(x,t)∂xi

→ 0 as |x| → ∞ for all i, we see from (2.2) as r → ∞

that ∫

RN

uk(x, t)dx =

∫

RN

u0,k(x)dx = 1.

Q.E.D.

Before proving our major estimate (Theorem 2.3), we present an iteration lemma.

Lemma 2.2. Let ϕ(t) be nondecreasing function in [t1, t2] and satisfy

ϕ(s) ≤ θϕ(t) +A1

(t− s)β+ A2, for any s, t : t1 ≤ s < t ≤ t2,

where 0 < θ < 1, A1 ≥ 0, A2 ≥ 0 and β ≥ 0. Then there exists the constant C > 0,

depending only on θ and β such that

ϕ(s) ≤ C

[A1

(t− s)β+ A2

](2.3)

for all t ∈ [t1, t2].

The proof of Lemma 2.2 can be found in [18]. We skip it here.

Theorem 2.3. Assume 0 ≤ n < N+2N

. Then there exists a constant C(T ) such that

||uk(·, t)||L∞(IRN

)≤ C(T )t−

N2 (2.4)

5

for all t ∈ (0, T ] and k = 1, 2, · · · , where constant C(T ) is independent of k and t.

Proof: For simplicity we omit the subscript k of uk(x, t) in the proof.

Let σ ∈ [1, 2]. Set

Stσ = IRN × (

1

2σ, t], 1 ≤ σ ≤ 2.

Let d ≥ 2 and ξ(x, t) be a cut-off function in Bd(0) × ( t2θ, t] in which ξ(x, t) is smooth and

equal to 1 in Bd−1(0) × ( t2σ, t]. Moreover,

|ξt| ≤C

(θ − σ)t, |∇ξ| ≤ C,

where 1 ≤ σ < θ ≤ 1. Without loss of generality, we can suppose ut ∈ L2(SτT ). Otherwise

we take the average of u in time instead of u (see [16]). Taking φ = ξ2uα (α ≥ 1) as a test

function in (1.1) and integrating it over Stθ, then we have

∫∫

Stθ

[u(ξ2uα)t + u∆(ξ2uα) − un ·~b · ∇(ξ2uα)]dxdt =

∫

IRNξ2uα+1(x, s) |τ1

2θ−1t

dx. (2.5)

Integrating by part, we obtain

1

α + 1

∫

IRNξ2uα+1(x, t)dx =

2

α + 1

∫∫

×Stθ

ξξtuα+1dxdt− 2

∫∫

×Stθ

ξuα∇ξ · ∇u dxdt

−4α

(α + 1)2

∫∫

×Stθ

ξ2 | ∇uα+1

2 |2 dxdt−2n

α + n

∫∫

×Stθ

ξuα+n ~b · ∇ξ dxdt (2.6)

Applying Cauchy-Schwarz’s inequality, we find from (2.6) that

1

α + 1

∫

IRNξ2uα+1(x, t)dx +

2α

(α + 1)2

∫∫

×Stθ

| ∇uα+1

2 |2 dxdt ≤

C

1

(α + 1)(θ − σ)t

∫∫

ØStθ

uα+1dxdt +C

α

∫∫

ØStθ

uα+1dxdt +2n|~b|

α + n

∫∫

ØStθ

ξuα+ndxdt

(2.7)

6

Let

Iα = sup12θ−1t≤τ≤t

∫

IRNξ2uα+1(x, τ)dx +

∫∫

ÙStθ

|∇(ξuα+1

2 )|2dxdt

Notice

∫∫

ÙStθ

|∇(ξuα+1

2 )|2dxdt ≤ C

∫∫

ÙStθ

ξ2|∇uα+1

2 |2dxdt +

∫∫

ÙStθ

uα+1|∇ξ|2dxdt

(2.8).

Without loss of generality, we assume

∫∫

ÙStθ

uα+1dxdt ≥ 1 (2.9)

for any α ≥ 1. Combining (2.7) and (2.8) with (2.9), we have

Iα ≤ C(α + 1)

(θ − σ)

[1

t+ ‖ u ‖n−1

L∞(ÙStθ)

] ∫ t

12θ−1t

∫

Bd(0)

uα+1dxdt, (2.10)

where 1 < n < N+2N

, (as for 0 ≤ n ≤ 1, the term ‖ u ‖n−1

L∞(ÙStθ)

in (2.10) can be dropped), and

the constant C depends only on N and |~b|.

By Sobolev’s imbedding theorem (see [16] the inequality (3.2) in Chapter 2), we obtain

∫∫

ÙStθ

ξ2(1+ 2N)u(α+1) 2

N+α+1 dxdt ≤ C sup

12θ−1t<τ≤t

(∫

Bd(0)

ξ2uα+1(x, τ)dx

) 2N

×

∫∫

ÙStθ

|∇(ξuα+1

2 )|2 dxdt ≤ C I1+ 2

Nα ≤

C(α + 1)1+ 2N

(θ − σ)1+ 2N

[1

t+ ‖ u ‖n−1

L∞(ÙStθ)] ×

∫ t

12θ−1t

∫

Bd(0)

uα+1dxdt

1+ 2N

(2.11)

Now we introduce a sequence

tl =σ−1

2t +

θ−1 − σ−1

2l+1t, S(l) = Bd(0) × (tl, t]

and

α + 1 = (A + 1)ql, l = 0, 1, 2, · · · ,

7

where d ≥ 2, A = (n−1)(N+2)2

, q = 1 + 2N

. The estimate (2.11) gives

∫∫

S(l+1)

u(α+1)(1+ 2N

) dxdt ≤ C

[4l(α + 1)

θ − σ

]1+ 2N

d

1

t+ ‖ u ‖n−1

L∞( ÚStθ)e

∫∫

S(l)

uα+1dxdt

1+ 2N

.

That is

∫∫

S(l+1)

u(A+1)ql+1

dxdt ≤ Cd(4(A + 1)q)l

θ − σeq

[1

t+ ‖ u ‖n−1

L∞( ÚStθ)

] ∫∫

S(l)

u(A+1)qldxdt

q

The standard Moser’s iteration (see [7]) yields

‖ u ‖L∞( ÚStσ)≤ Cd

1

θ − σ

(1

t+ ‖ u ‖n−1

L∞( ÚStθ)

)e

q(A+1)(q−1) ×

∫∫

ÚStθ

uA+1dxdt

1A+1

,

where the constant C does not depend on θ, σ, t and d. Thus we have

‖ u ‖L∞( ÚStσ)≤ Cd(θ − σ)t]

−q(A+1)(q−1)

∫∫

ÚStθ

uA+1dxdt

1A+1

+

Cd(θ − σ)−1 ‖ u ‖n−1

L∞( ÚStθ)]

q(A+1)(q−1)

∫∫

ÚStθ

uA+1dxdt

1A+1

= I1 + I2 (2.12)

Since 1 < n < N+2N

(N ≥ 2), combining Theorem 2.1 with the definition of A and using

Young’s inequality, we have

I1 = Cd(θ − σ)t]−(N+2)

(n−1)(N+2)+2

∫∫

ÚStθ

u(n−1)(N+2)

2+1 dxdt

1(n−1)(N+2)

2 +1

≤ Cd(θ − σ)t]−(N+2)

(n−1)(N+2)+2 ‖ u ‖(n−1)(N+2)

(n−1)(N+2)+2

L∞( ÚStϑ

)

∫∫

ÚStθ

u dxdt

1(n−1)(N+2)

2 +1

8

≤ C(θ − σ)−(N+2)

(n−1)(N+2)+2 t−N

(n−1)(N+2)+2 ‖ u ‖(n−1)(N+2)

(n−1)(N+2)+2

L∞( ÛStθ)

≤1

4‖ u ‖L∞( ÛSt

θ) +C(θ − σ)

−(N+2)2 t−

N2 .

and

I2 = Cd(θ − σ)−1 ‖ u ‖n−1

L∞( ÛStθ)]

−(N+2)(n−1)(N+2)+2

∫∫

ÛStθ

u(n−1)(N+2)

2+1dxdt

1(n−1)(N+2)

2 +1

= C(θ − σ)−(N+2)

(n−1)(n+2)+2 ‖ u ‖(n−1)(N+2)

(n−1)(N+2)+2

L∞( ÛStθ)

∫∫

ÛStθ

u(n−1)(N+2)

2+1 dxdt

2(n−1)(N+2)+2

≤ 14‖ u ‖L∞( ÛSt

θ) +C(θ − σ)−

(N+2)2

∫∫

ÛStθ

u(n−1)(N+2)

2+1 dxdt

Substituting I1 and I2 to (2.12), fixing θ = 2, σ = 1, and applying Lemma 2.2, we obtain

‖ u ‖L∞( ÛSt1)≤ C

t−

N2 +

∫∫

ÛSt2

u(n−1)(N+2)

2 dxdt

(2.13),

where the constant C is independent of k, d and t.

Define

g(t) = sup0<τ≤t

τN2 ‖ u(·, τ) ‖

L∞(IRN).

In virtue of (2.13) and Theorem 2.1, we find

tN2 ‖ u(·, t) ‖

L∞(IRN)≤ C

1 +

∫ t

t4

τN2 ‖ u(·, τ) ‖

(n−1)(N+2)2

L∞(IRN)dτ

It follows that

9

g(t) ≤ C

∞∑

k=1

(∫ t

2k−1

t

2k+1

+

∫ t

2k

t

2k+2

)τ

N2

[1−(n−1(N+2)

2] [g(τ)]

(n−1)(N+2)2 dτ + 1

≤ C

∫ t

0

τN2

[1−(n−1(N+2)

2] [g(τ)]

(n−1)(N+2)2 dτ + 1

(2.14)

Notice that n < N+2N

,

N

2(1 − A) =

N

2

[1 −

(n− 1)(N + 2)

2

]> −1.

We consider the following ordinary differential equation

dG(t)

dt= Ct

N2 [1− (n−1)(N+2)

2 ][G(t)](n−1)(N+2)

2 , (2.15)

G(0) = C (2.16),

where the constant C depends only on C of (2.14). The problem 2.15)-(2.16) has the following

explicit solution:

G(t) = C/

1 −

2[(n− 1)(N + 2) − 2]C(n−1)(N+2)

2

(N + 2)[2 −N(n− 1)]t(N+2)[2−N(n−1)]/4

2(n−1)(N+2)−2

(2.17)

for t ∈ d0, T ), where T = ∞ if 1 < n ≤ N+4N+2

, and

T ≤ C−2(n−1)(N+2)

(N+2)[2−N(n−1)] ×

(N + 2)(d2 −N(n− 1)]

4d(n− 1)(N + 2) − 2]

2(n−1)(N+2)−2

if N+4N+2

< n < N+2N

. By choosing suitable positive constant C(T ), we obtain from (2.14) and

(2.17) that

g(t) ≤ G(t) ≤ C(T )

for t ∈ d0, T ∗], where T ∗ ≤ minT, T

. This implies the conclusion of Theorem 2.3 for the

case 1 < n < N+2N

.

As for 0 ≤ n ≤ 1, due to the term ‖ u ‖n−1

L∞( ÜStθ)

dropped in (2.10), we can employ the

similar process to reach the conclusion of Theorem 2.3.

Remark 2.1: The estimate (2.4) in Theorem 2.3 is optimal. When n = 0, the fundamental

solution to the standard heat equation as follows:

10

u(x, t) =1

(4πt)N2

e−|x|24t .

When n = 1,~b = ~e = (1, · · · , 1) ∈ IRN , we can introduce a suitable transformation for (1.1)

and get the formula of fundamental solution to (1.1) as follows:

u(x, t) =1

(4πt)N2

e−|x+~et|2

4t .

When n = N+1N

, we can get the similar fundamental solution u(x, t) to (1.1) in Section 6,

i.e.,

u(x, t) = t−N2 e−|x|2

4t

(1

Ne− 1

4Nt

N−1Ýi=1

x2i

∫ xN√t

−∞

e−s2

4N ds + C0

)−N

where C0 is some constant, such that

∫

IRNu(x, t)dξ = 1.

3. ExistenceWe first show that the singularity of the solution only occurs near the origin.

Theorem 3.1. If 0 ≤ n < N+2N

and 0 < a < d ≤ ∞, then there exists a constant C

independent of k, such that

‖ uk ‖L∞dBd(0)\Ba(0)×[0,T ]≤ C(T, a) (3.1)

for sufficiently large integer k.

Proof. Without loss of generality, we suppose d < +∞. Let ξ(x) ∈ C20(IRN) be a cut-off

function such that

supp ξ ⊂ BD(0)\BA(0),

where 0 < A < a < d < D = d + a− A and

0 ≤ ξ ≤ 1 in IRN , ξ ≡ 1 on Bd(a)\Ba(0).

Taking φ = ξ2uα(α > 0) as a test function in (1.1) and integrating it over STA,D =

(BD(0)\BA(0)

)×

d0, T ], noticing u(x, 0) = 0 for x ∈ IRN\BA(0), we find

11

1

α + 1

∫

IRNξ2uα+1(x, t)dx +

4α

(α + 1)2

∫∫

STA,D

ξ2|∇uα+1

2 |2dxdt

= −2

∫∫

STA,D

ξuα∇ξ · ∇u dxdt−2n

α + n

∫∫

STA,D

ξuα+n~b · ∇ξdxdt (3.2)

By using the same process as in the proof of Theorem 2.3, we have

1

α + 1

∫

IRNξ2uα+1(x, T )dx +

2α

(α + 1)2

∫∫

STA,D

ξ2|∇uα+1

2 |2dxdt

≤ C

∫∫

STA,D

uα+1|∇ξ|2dxdt +2n|~b|

α + n

∫∫

STA,D

ξ|∇ξ|uα+ndxdt

(3.3)

Applying Sobolev’s imbedding theorem, we find from (3.3) that

∫∫

STA,D

ξ2(1+ 2N

)u(α+1) 2N

+α+1dxdt ≤ C sup0≤t≤T

(

∫

BD\BA

ξ2uα+1(x, t)dx)2N×

∫∫

STA,D

|∇(ξuα+1

2 )|2dxdt ≤ C(α + 1)1+ 2N

∫∫

STA,D

uα+1|∇ξ|2dxdt +

∫∫

STA,D

uα+nξ|∇ξ|dxdt

1+ 2N

(3.4)

Set

al = a +A− a

2l, dl = d +

D − d

2l, Q(l) = (Bdl\Bal) × [0, T ]

for l = 0, 1, 2, · · · . Let ξl be a cutoff function in Bdl\Bal with Bdl+1\Bal+1

. Choose α+1 = 2ql

for l = 0, 1, 2, · · ·. It follows from (3.4) that

∫∫

Q(l+1)

u2ql+1

dxdt ≤ Cd(16q)l

(a− A)2(D − d)2]q [(a− A)2 + (D − d)2]q×

(

∫∫

Q(l)

u2qldxdt +

∫∫

Q(l)

u2ql+n−1dxdt)q.

We apply the standard Moser’s iteration technique to arrive

12

‖ u ‖L∞(STa,d

)≤ C[

∫∫

STA,D

u2dxdt +

∫∫

STA,D

un+1dxdt]12 , (3.5)

where the constant C depends only on a− A,D − d and T .

Obviously,

[

∫∫

STA,D

u2dxdt]12 ≤

1

4C‖ u ‖L∞(ST

A,D) +C

∫∫

STA,D

udxdt (3.6)

and

[

∫∫

STA,D

un+1dxdt]12 ≤‖ u ‖

12

L∞(STA,D)

(

∫∫

STA,D

undxdt)12

≤1

4C‖ u ‖L∞(ST

A,D) +C

∫∫

STA,D

undxdt (3.7)

By Theorem 2.3 and the condition 0 ≤ n < N+2N

,

∫∫

STA,D

undxdt ≤ C(T ).

Hence, taking into account for (3.5), (3.6) and (3.7) and by Lemma 2.2, we conclude the esti-

mate (3.1). Since the second term in the right hand-side of (3.7) does not exceed a constant

depending D − d, but independent of D. When d = +∞ we can take any positive integer

m > a and m + 1 instead of d and D respectively. This implies that the case d = +∞ is

permitted in Theorem 3.1.

Theorem 3.2. If 0 < n < 1, then

‖ ∇(u1−nk ) ‖L∞(Sτ

T )≤ C(τ, T ) (3.8),

where C(τ, T ) depends only on known data and τ, T , but not on k.

Proof. Note that uk(x, t) > 0 in ST . Let α = 1 − n and V = uα, then V satisfies the

following equation:

13

Vt = 4V +1 − α

αu−α|∇V |2 + nun−1~b · ∇V. (3.9)

Set

q =

∫ Φ

0

(2Mγτ − sγ)−1ds (3.10)

where Mτ =‖ uα ‖L∞(SτT ), 0 ≤ Φ ≤ Mτ and γ > 0 to be determined. Since q′(Φ) ≥ 1

2Mγτ

, the

inverse function of Φ = Φ(q) is well defined, smooth and monotonic increasing on d0, q(Mτ )].

Let V = Φ(w), then

wt = 4w +

(Φ′′

Φ′+

1 − α

αΦ−1Φ′

)|∇w|2 + nΦ−1~b · ∇w (3.11)

Differentiating (3.11) with respect to xi, then multiplying by wxi and taking the summation

for i = 1, 2, · · · , N , we obtain

N∑i=1

wxi(wxit − ∆wxi) = d(Φ′′

Φ′ )′ − 1−α

αφ−2(Φ′)2 + 1−α

αΦ−1Φ′′]|∇w|2 +

N∑i=1

w2xi

+2(

Φ′′

Φ′ + 1−αα

Φ−1Φ′)∇w

(N∑i=1

wxi∇wxi

)+ n(n−1)

αΦ−2Φ′

N∑i=1

w2xi~b · ∇w

+nΦ−1~b ·

(N∑i=1

wxi · ∇wxi

)

(3.12)

Set p = |∇w|, then the equation (3.12) can be rewrite as follows:

12(p2)t − p4p = d(Φ′′

Φ′ )′ − 1−α

αΦ−2(Φ′)2 + 1−α

αΦ−1Φ′′]p4

+2(Φ′′

Φ′ + 1−αα

Φ−1Φ′)p∇w · ∇p + n(n−1)α

Φ−2Φ′p2~b · ∇w + nΦ−1p ~b · ∇p

+|∇p|2 −N∑i=1

|∇wxi|2

(3.13)

Let ζ be a cut-off function defined in IRN × [0, T ] such that

supp ξ ⊂ BD(0) × (0, T ]

14

0 ≤ ζ ≤ 1 in BD(0) × (0, T ]

and

ζ ≡ 1, on BD−1(0) × [τ, T ]

Set Z = ζ2p2 and suppose Z attains the positive maximum value at (x0, t0) ∈ BD(0)× [τ, T ],

then

∇Z|(x0,t0) = 0, i.e., ζ∇p|(x0,t0) = −p∇ζ|(x0,t0). (3.14)

Moreover,

(Zt − ∆Z) |(x0,t0)≥ 0 (3.15)

Noticing

|∇p|2 =N∑

j=1

(∂p

∂xj

)2 =N∑

j=1

N∑

i=1

WxixjWxi√N∑i=1

W 2xi

≤

N∑

j=1

N∑

i=1

W 2xixj

=N∑

i=1

|∇Wxi |2.

Multiplying (3.13) by ζ and using (3.14), we see at (x0, t0):

ζ2[1

2(p2)t − p4p]

= d(Φ′

Φ′)′ +

1 − α

αΦ−1Φ′′ −

1 − α

αΦ−2(Φ′)2]ζ2p4 − 2

(Φ′′

Φ′+

1 − α

αΦ−1Φ′

)ζp2×

∇w · ∇ζ +n(n− 1)

αφ−2φ′p2ζ2~b · ∇w + p2|∇ζ|2 − ζ2

N∑

i=1

|∇wxi |2 − nφ−1p2ζ~b · ∇ζ

≤ [(Φ′

Φ′)′ +

1 − α

αΦ−1Φ′′ −

1 − α

αΦ−2(Φ′)2]ζ2p4 − 2

(Φ′′

Φ′+

1 − α

αΦ−1Φ′

)ζp2∇w · ∇ζ

+n(n− 1)

αφ−2φ′p2ζ2~b · ∇w + p2|∇ζ|2 − nφ−1p2ζ~b · ∇ζ

≤ [(Φ′

Φ′)′ +

1 − α

αΦ−1Φ′′ −

1 − α

αΦ−2(Φ′)2]ζ2p4 + 2

∣∣∣∣Φ′′

Φ′+

1 − α

αΦ−1Φ′

∣∣∣∣ ζp3|∇ζ|

+n(1 − n)

αφ−2φ′ζ2|~b|p3 + p2|∇ζ|2 + nφ−1p2ζ|∇ζ| (3.16)

By (3.14) and (3.15), we have

15

ζ2d(1

2(p2)t − p4p] ≥ p2(ζ4ζ − 2|∇ζ|2 − ζζt) (3.17)

Combining (3.16) and (3.17), we get

−[(Φ′′

Φ′)′ +

1 − α

αΦ−1Φ′′ −

1 − α

αΦ−2(Φ′)2]ζ2p2 ≤

(2|

Φ′′

Φ′+

1 − α

αΦ−1Φ′|ζ|∇ζ|+

n(1 − n)

αΦ−2Φ′ζ2|~b|

)p + |∇ζ|2 + nΦ−1ζ|∇ζ| − (ζ4ζ − 2|∇ζ|2 − ζζt)

Noticing Φ′′ = −γΦγ−1Φ′ and fixing γ > 1, we find at (x0, t0) that

n

1 − nζ2p2 ≤ [2(

γΦγ+1

(Φ′)2+

n

1 − n

Φ

Φ′)ζ|∇ζ| +

n|~b|

Φ′ζ2]p

nΦ

(Φ′)2ζ|∇ζ| +

Φ2

(Φ′)2(−ζ4ζ + 3|∇ζ|2 − ζζt) (3.18)

It follows from (3.18) and the properties of Φ that

supSτT

|∇w| ≤ C(τ, T )

This implies the conclusion of Theorem 3.2.

Theorem 3.3. If 1 ≤ n < N+2N

, then

‖ ∇un−1k ‖L∞(Sτ

T )≤ C(τ, T ) (3.19)

and

‖ ∇uk ‖L∞(SτT )≤ C(τ, T ) (3.20),

where C(τ, T ) depends only on known data and τ , the upper bound of T , but not on k.

Proof. First we are going to establish the estimate (3.19) for 1 < n < N+2N

. Set

V = un−1

then V (x, t) satisfies the following equation:

Vt = 4V +2 − n

n− 1u1−n|∇V |2 + nun−1~b · ∇V (3.21)

With the similar discussion to the proof of Theorem 3.2, we can derive the estimate (3.19).

16

Now we prove the estimate (3.20).

Set

q =

∫ φ

0

(2Mβτ − sβ)−1ds

where Mτ =‖ u ‖L∞(SτT ), 0 ≤ Φ ≤ Mτ and β > 0 to be determined. Since q

′(φ) > 1

2Mβτ

,

the inverse function φ = φ(q) is well defined , smooth and increasing on d0, q(Mτ )]. Set

u = φ(W ), then W (x, t) satisfies the following equation:

Wt = 4W +φ′′

φ′|∇W |2 + nφn−1~b · ∇W (3.22)

Differentiating (3.22) with respect to xi, multiplying by Wxi and summing up for i =

1, 2, · · · , N , we obtain

N∑

i=1

(Wxi ·Wtxi −Wxi4Wxi) =

(φ′′

φ′

)′ N∑

i−1

W 2xi|∇W |2

+2φ′′

φ′∇W ·(

N∑

i=1

Wxi∇Wxi)+n(n−1)φn−2φ′N∑

i=1

W 2xi·~b·∇W +nφn−1~b·(

N∑

i=1

Wxi∇Wxi) (3.23)

Set p = |∇W |, then

p4p =N∑

i=1

Wxi4Wxi − |∇p|2 +N∑

i=1

|∇Wxi |2 (3.24)

Combining (3.23) and (3.24), we have

1

2(p2)t − p4p = (

φ′′

φ′)′p4 + 2

φ′′

φ′p∇W · ∇p

+n(n− 1)φn−2φ′p2~b · ∇W + nφn−1p~b · ∇p + |∇p|2 −

N∑

i=1

|∇Wxi |2

≤ (φ′′

φ′)′p4 + 2

φ′′

φ′p∇W · ∇p + n(n− 1)φn−2φ′p2~b · ∇W + nφn−1p~b · ∇p (3.25)

Let ζ(x, t) be a cut-off function as defined in the process of proving Theorem 3.2. Set

Z = ζ2p2 and suppose Z attains its positive maximum value at (x0, t0) ∈ BD(0) × dτ, T ],

then

∇Z |(x0,t0)= 0, i.e. ζ∇p = −p∇ζ at (x0, t0) (3.26)

17

and

(Zt −4Z) |(x0,t0)≥ 0 (3.27)

From (3.26) and (3.27), at (x0, t0) we see

ζ2[1

2(p2) − p4p] ≥ p2(ζ4ζ − 2|∇ζ|2 − ζζt) (3.28)

Multiplying (3.25) by ζ2, using (3.26) and noticing φ′′

= −βφβ−1φ′, we have at (x0, t0):

ζ2

[1

2(p2)t − p∆p

]

= (φ′′

φ′)′p4ζ2 − 2

φ′′

φ′p2ζ∇W · ∇ζ + n(n− 1)φn−2p2ζ2~b · ∇W − nφn−1p2ζ~b · ∇ζ

≤ (φ′′

φ′)′p4ζ2 − 2

[φ′′

φ′ζ · |∇ζ| − n(n− 1)φn−2φ′ζ2|~b|

]p3 − nφn−1p2ζ~b · ∇ζ (3.29)

Combining (3.28) and (3.29), we have

−(φ′′

φ′)′p2ζ2 ≤ −d2

φ′′

φ′ζ|∇ζ| − n(n− 1)φn−2φ

′ζ2|~b|]p

−(nφn−1ζ~b · ∇ζ + 2|∇ζ|2 − ζ4ζ + ζζt) (3.30)

Noticing

(φ′′

φ′)′ = β(1 − β)φβ−2φ′,

we pick β > 1 such that n− 14< β < n. Thus, at (x0, t0),

p2ζ2 ≤ [2

β − 1

φ

φ′ζ|∇ζ| +

n(n− 1)

β(β − 1)φn−β ζ2| ~b| ]p

+nφ1−β+n

β(β − 1)φ′ζ |∇ζ| |~b| +

φ2−β

β(β − 1)φ′(|∇ζ|2 − ζ4ζ + ζζt) (3.31)

From the definition of p(x, t) and the properties of φ, we conclude the estimate (3.20).

Remark 3.1: We can prove that the constant C(τ, T ) in (3.19) satisfies the following

estimate:

C(τ, T ) ≤ C(T )τ−ω (3.32)

18

where ω = (n−1)N2

< 1 as 1 ≤ n < N+2N

.

Since the proof is quite similar to that of Theorem 3.2, we leave the detail in Appendix.

Proof of the part (a) in Theorem A: Let uk(x, t) be the solution of the Cauchy’s problem

(1.1) with initial datum u0,k(x). By the estimates of Theorems 3.2 and 3.3 and standard

parabolic partial equations, uk(x, t) are locally uniform Holder continuous in ST with respect

to k. Thus, there exists a subsequence ukj(x, t) and a continuous function u(x, t) such that

limj→∞

ukj(x, t) = u(x, t)

for (x, t) ∈ ST , the convergence is locally uniformly in ST . Due to Theorem 3.1 and the

regularity theory of parabolic equations, u(x, t) ∈ C(ST\(0, 0))⋂C2,1(ST ) for any T > 0.

Now we prove that u(x, t) satisfies the equation (1.1) and (1.3). Let φ(x, t) ∈ C2,10 (ST ) with

φ(x, T ) = 0, then

∫ ∞

τ

∫

IRN(ukjφt + ukj∆φ− un

kj~b · ∇φ)dxdt +

∫

IRNukj(x, τ) φ(x, τ)dx = 0. (3.33)

After taking limit as k → ∞, we obtain

∫ ∞

τ

∫

IRN(uφt + u∆φ− un ~b · ∇φ)dxdt +

∫

IRNu(x, τ) φ(x, τ)dx = 0.

The regularity theory for parabolic equations (see [16] and [17]) implies that u(x, t) ∈

C2,1(ST ) and satisfies the equation (1.1) in the classical sense.

Now we prove that u(x, t) takes δ(x) as the initial value.

Again, since uk(x, t) is a solution of (1.1) with initial datum u0,k(x), we have

∫ ∞

0

∫

IRN(ukjφt + ukj∆φ− un

kj~b · ∇φ)dxdt +

∫

IRNukj(x, 0) φ(x, 0)dx = 0. (3.34)

Note that

|

∫ τ

0

∫

IRN(ukjφt + ukj ∆φ − un

kj~b · ∇φ) dxdt|

≤ C supST

|φt| + |4φ| + |∇φ|

∫ τ

0

∫

IRN(ukj + un

kj) dxdt. (3.35)

19

By Theorems 2.1 and 2.3,∫

IRNukj(x, t)dx ≤ 1

∫

IRNunkjdx ≤ ||ukj ||

n−1L∞(RN )

∫

IRNukj(x, t)dx ≤ Ctn−1.

Since 1 < n < N+2N

(for the case 0 ≤ n ≤ 1, we use the first inequality), it follows that the

right hand side of (3.35) goes to zero uniformly with respect to kj as τ → 0+. Therefore,

for any ε > 0, there exists τ0 > 0 such that

|

∫ ∞

τ

∫

IRN(ukjφt + ukj∆φ − un

kj~b · ∇φ) dxdt +

∫

IRNukj(x, 0)φ(x, 0)dx |< ε (3.36)

for 0 < τ < τ0, where τ0 is independent of kj. Taking the limit in (3.36) with respect to

j → ∞, we see

|

∫ ∞

τ

∫

IRN(uφt + u∆φ− un~b · ∇φ) dxdt + φ(0, 0) |≤ ε (3.37)

for 0 < τ ≤ τ0. Combining (3.36) and (3.37), we conclude

|

∫

IRNu(x, τ)φ(x, τ)dx− φ(0, 0) |≤ ε (3.38)

for 0 < τ ≤ τ0, which yields the limit (1.3).

Q.E.D.

4. UniquenessIn this section we are going to establish the uniqueness result of the source-type solution

to the problem (1.1)-(1.2). Let ukj be a solution of the equation (1.1) with the initial datum

u0,k(x) + 1j

for j = 1, 2, · · ·, where u0,k(x) is defined the same as in Section 2. ukj converges

to uk(x, t) in ST and uniformly in any open bounded set G ⊂⊂ ST . By the comparison

theorem and the regularity theory we know that ukj is smooth in ST ,

ukj(x, t) ≥1

j

and ukj is locally uniform Holder continuous in ST for j = 1, 2, · · ·. Let u be a source-type

solution to (1.1)-(1.2), then for any ball Bα(0) ⊂ IRN , φ ∈ C2(ST ) such that φ |∂Bα= 0, the

following integral identity holds

20

∫

Bα(0)

[ukj(x, t) − u(x, T )]φ(x, T ) dx

=

∫ T

τ

∫

Bα(0)

(ukj − u)φt + (ukj − u)4φ− (unkj− un) ~b · ∇φ dxdt

−

∫ T

τ

∫

∂Bα

(ukj − u)~b · ∇φ~νdsdt +

∫

Bα(0)

d(ukj(x, τ) − u(x, τ)eφ(x, τ)dx (4.1)

where ~ν is outward normal derivative for ∂Bα(0) × [0, T ].

We define

d(x, t) =

∫ 1

0

n[s ukj + (1 − s) u]n−1 ds, dρ = d ∗ Jρ (4.2)

where ∗ means the integral convolution in RN and Jρ ∈ C∞(IRN×IR+) is a standard mollifier

with the properties:

suppJρ ⊂ (x, t) | |x| < ρ, t < ρ.

and

∫

IRNJρ(x, t)dx = 1.

Clearly, if 0 < n ≤ 1, then

0 < dρ < Cj1−n (4.3)

and if 1 < n < N+2N

, due to the estimate (3.32),

0 < dρ < C(T ) · t−(n−1)N

2 (4.4)

Furthermore, if 1 < n < N+2N

and N ≥ 2, then

‖ ∇dρ ‖L∞(SτT ) ≤ ‖ ∇

(∫ 1

0

n · ds ukj + (1 − s)u]n−1ds ∗ Jρ

)‖L∞(Sτ

T )

≤ C ‖ n(n− 1)

∫ 1

0

ds ukj + (1 − s)u]n−2 · [s∇ukj + (1 − s)∇u]ds ‖L∞(SτT )

21

≤ C n(n− 1)

∫ 1

0

ds ukj + (1 − s)u]n−2ds · max‖ ∇ukj ‖L∞(SτT ), ‖ ∇u ‖L∞(Sτ

T )

≤ C max‖ ∇un−1kj

‖L∞(SτT ), ‖ ∇un−1 ‖L∞(Sτ

T )

Noticing if n > 1, we can take uk instead of ukj in the identity (4.1) and the expression of

dρ. Hence it follows from Theorem 3.3 and Remark 2 that

‖ ∇dρ ‖L∞(SτT )≤ Ct−ω (4.5)

where 0 < ω = N(n−1)2

< 1.

Choose χ ∈ C∞0 (IRN), 0 ≤ χ ≤ 1 and supp χ ⊂ BL(0) ⊂⊂ Bα−1(0), and consider the

following conjugate problem:

Φt + 4Φ − dρ~b · ∇Φ = 0, (x, t) ∈ Bα(0) × (τ, T ) = SτT,α (4.6)

Φ(x, T ) = χ(x), x ∈ Bα(0) (4.7)

Φ |∂Bα= 0, t ∈ (τ, T ) (4.8)

It is well-known from [16] that the Dirichlet problems (4.6) - (4.8) possesses a unique classical

solution Φ(x, t) on SτT,α for any α > 1 and 0 < τ < T . By the maximum principle, we have

0 ≤ Φ ≤ 1, (x, t) ∈ SτT,α

We are going to establish several estimates for Φ(x, t).

Lemma 4.1. Let Φ be a classical solution of the problem (4.6)-(4.8), then

sup(τ,T ]

∫

Bα(0)

|∇Φ|2dx ≤ K1 (4.9)

and

∫ T

τ

∫

Bα(0)

|∆Φ|2dxdt ≤ K2, (4.10)

22

where the constants K1 and K2 depend only on j, the upper bound of T if 0 < n < 1 and

on the upper bound of T if n ≥ 1, respectively.

Proof. Multiplying (4.6) by 4Φ and integrating the result over SτT,α, then

∫ T

τ

∫

Bα(0)

Φt4Φdxdt +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt =

∫ T

τ

∫

Bα(0)

dρ4Φ~b · ∇Φdxdt.

Integrating by part, we have

1

2

∫

Bα(0)

|∇Φ(x, τ)|2dx +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt =

∫ T

τ

∫

Bα(0)

dρ4Φ~b · ∇Φdxdt

+1

2

∫

Bα(0)

|∇χ|2dx

(4.11)

Applying Cauchy-Schwarz’s inequality, it follows from (4.11) that

∫

Bα(0)

|∇Φ(x, τ)|2dx +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt ≤ C

∫ T

τ

∫

Bα(0)

d2ρ|∇Φ|2dxdt

+

∫

Bα(0)

|∇χ|2dx

(4.12)

If 0 < n < 1, (4.3) and (4.12) give

∫

Bα(0)

|∇Φ(x, τ)|2dx +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt ≤ C(j)

∫ T

τ

∫

Bα(0)

|∇Φ|2dxdt + C (4.13)

If 1 ≤ n < N+1N

, (4.4) and (4.12) give

∫

Bα(0)

|∇Φ(x, τ)|2dx +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt ≤ C(T )

∫ T

τ

∫

Bα(0)

t−(n−1)N |∇Φ|2dxdt + C.

(4.14)

23

Applying Gronwall’s ’s inequality for (4.13), we reach the conclusion of the lemma for the

case 0 < n < 1. Noticing (n− 1)N < 1 and applying Gronwall’s inequality with singularity,

we also reach the conclusion of the lemma for the case 1 ≤ n < N+1N

. As for the case

N+1N

≤ n < N+2N

, it follows from (4.11) that

1

2

∫

Bα(0)

|∇Φ(x, τ)|2dx +

∫ T

τ

∫

Bα(0)

|4Φ|2dxdt ≤ C

∫ T

τ

∫

Bα(0)

|∇dρ| |∇Φ|2dxdt

+ C

∫ T

τ

∫

∂Bα(0)

dρ|∇Φ|2dsdt +1

2

∫

Bα(0)

|∇χ|2dx.

(4.15)

By Theorem 2.2 and Lemma 4.2 below, the second integral of the right hand side of (4.15)

goes to zero as α tends to infinity. Combining (4.15) and (4.5) and using Gronwall’s inequality

with singularity, we have the estimates (4.9) and (4.10) for the case N+1N

≤ n < N+2N

.

Now we derive an estimate for the boundary value of |∇Φ|.

Lemma 4.2. Let Φ be a classical solution of (4.6) - (4.8), then there exists the constants

K3 and K4 such that

Φ(x, t) ≤ K3e−|x|, (x, t) ∈ Sτ

T,α (4.16)

and

|∇Φ(x, t)| ≤ K4e−α, (x, t) ∈ ∂Bα(0) × [τ, T ] (4.17)

where the constant K3 depends only on j as 0 < n < 1 and on T as n ≥ 1 respectively. K4

depends only on j as 0 < n < 1, and on τ and T as n > 1. When n = 1, K4 does not depend

on τ .

Proof. Define an auxiliary function as follows:

v(x, t) := Ke−|x|+β(T−t) − φ(x, t).

Then,

Lv ≡ Vt + 4V − dρ~b · ∇v

24

for (x, t) ∈ SτT,α. After computing

Lv = Ke−|x|+β(T−t)[− β + 1 −

N − 1

|x|+

N∑i=1

bixi

|x|dρ

]

≤ Ke−|x|+β(T−t)(−β + 1 + |~b| dρ) (4.18)

If 0 < n < 1, choosing β > supSτT

1 + dρ = Cj1−n and K > eL+1, we have

Lv < 0, (x, t) ∈ SτT,α

v(x, T ) = ke−|x| − χ(x) ≥ 0, x ∈ Bα(0)

and

v(x, t) = Ke−α+β(T−t) ≥ 0, (x, t) ∈ ∂Bα(0) × [τ, T ].

By the maximum principle, v(x, t) must be nonnegative on Sτ

T,α. Thus, we conclude the

estimate (4.16) of the lemma for the case 0 < n < 1. As for the case 1 ≤ n < N+2N

, we set

tk =T

2k, Stk

tk−1,α= Sk

and

vk(x, t) = K(k)e−|x|+βk(tk−1−t) − Φ(x, t)

for (x, t) ∈ Sk and k = 1, 2, · · ·. Due to Theorem 2.3.

|dρ(x, t)| ≤ Ct−(n−1)N

2 ≤ C(T )2(n−1)N

2k,

for (x, t) ∈ Sk. Now choose

βk ≥ supSk

1 + dρ = C(T )2(n−1)Nk

2 (4.19)

for k = 1, 2, · · ·. This gives

25

Lvk < 0, (x, t) ∈ Sk.

Set K(1) = eL+1 and

K(k+1) = K(k) · eβkT

2k ≤ K(k) · eC(T )·2(n−1)N−2

2 k

(4.20)

for k = 1, 2, · · · and by the maximum principle, we have

vk(x, t) ≥ 0

(x, t) ∈ Sk. Thus,

0 ≤ Φ(x, t) ≤ K(k+1)e−|x| (4.21)

for (x, t) ∈ Sk and k = 1, 2, · · ·. Due to (4.20) and by the induction method,

K(k+1) ≤ K(1) · eC(T )

∞Þk=1

2−ωk

(4.22)

for any integer k = 1, 2, · · ·, where ω = 2−(n−1)N2

> 0. From (4.21) and (4.22) we see the

estimate (4.16) for the case n ≥ 1. Now we derive the estimate (4.17). We define the function

w(x, t) = Ke−α+(|x|−α) + Φ(x, t).

After performing routine calculations, we have

Lw = Ke−α+β(|x|−α)[β2 + β(N−1|x|

−dρ

NÞi=1

bjxi

|x|)]

≥ Ke−α+β(|x|−α)(β2 − β|~b|dρ)

(4.23)

for (x, t) ∈ [Bα(0)\Bα−1(0)] × [τ, T ] and α > 1. When 0 < n < 1, we pick β = Cj1−n >

supSτT

dρ, 2, and K = 4eC(T )j1−n= 4K3. By (4.16) for the case 0 < n < 1, we find

w(x, t)|∂Bα−1(0) = Ke−α−β + Φ(x, t)|∂Bα−1(0)

< e−αdKe−β + 3K3(j)] < Ke−α

and

26

w(x, t)|∂Bα(0) = Ke−α

for t ∈ dt, T ], and

w(x, T ) = Ke−α+β(|x|−α) + Φ(x, T ) ≤ Ke−α

for x ∈ Bα(0)\Bα−1(0). By the maximum principle, w(x, t) attains its maximum value on

∈ ∂Bα(0) × [τ, T ]. This implies

∂Φ

∂ν|∂Bα(0) ≥ −Kβe−α

N∑

i−1

xi cosαi

|x|(4.24)

where ν = (cosα1, · · · cosαN) is outward normal and 0 < ( xi, cosαi) <π

2for x =

(x1, . . . , xN) ∈ ∂B(0). On the other hand, we have

∂Φ

∂ν|∂Bα(0) ≤ 0 (4.25)

The inequalities (4.24) and (4.25) give

|∇Φ| ≤ C(j)

for (x, t) ∈ ∂Bα(0)×dt, T ], where the constant C(j) ≤ Cj1−neC(T )j(1−n). As for 1 < n < N+2

N,

we select β = C(T )τ−(n−1)N

2 > supSτT

dρ, 2 and K = 4K3. By the similar discussion as

mentioned above, we can obtain the estimate (4.17) for the case 1 < n < N+2N

. Here we omit

the details. Clearly we can pick β = C(T ), when n = 1. Therefore K4 is independent of τ

for this case.

The proof of uniqueness result in Theorem A. We take the solution Φ of the

problems (4.6)-(4.8) as a test function in (4.1) and set wkj = ukj − u,

∫

Bα(0)

wkj(x, T )χ(x)dx =

∫ T

τ

∫

Bα(0)

(dρ − b)wkj~b · ∇Φdxdt−

∫ T

τ

∫

∂Bα(0)

wkj

∂Φ

∂νdsdt

+

∫

Bα(0)

wkj(x, τ)Φ(x, τ)dx = I1 + I2 + I3

(4.26)

27

Obviously

|I1| ≤ C

∫ T

τ

∫

Bα(0)

(dρ − b)2w2kjdxdt

12 ·

∫ T

τ

∫

Bα(0)

|∇Φ|2dxdt12 (4.27)

and

|I2| ≤ CT sup∂Bα(0)×[τ,T ]

|∇Φ| supSτT

|wkj | (4.28)

When 1 ≤ n < N+2N

, we can take uk instead of ukj in the expression of wkj and use the

estimates (4.9) and (4.17) of Lemma 4.1 for I1 and I2 respectively. If we take limit in (4.26)

first with respect to ρ(ρ → +0), then with respect to α(α → +∞) and finally, with respect

to kj(kj → ∞), we obtain

∫

IRNdu(x, T ) − u(x, T )]χ(x)dx ≤

∫

IRNdu(x, t) − u(x, t)]Φ(x, t)dx (4.29)

for any 0 < τ < T . Since u and u are the fundamental solution of (1.1), for any ε > 0, σ > 0,

there exists τ0 > 0 such that if 0 < τ ≤ τ0, then

|

∫

|x|≥σ

du(x, τ) − u(x, τ)] [Φ(x, τ) − Φ(0, τ)]dx | <ε

4(4.30)

Due to the estimate (4.9) of Lemma 4.1 for the case 1 ≤ n < N+2N

, we have

|∫|x|≤σ

du(x, τ) − u(x, τ)] dΦ(x, τ) − Φ(0, τ)]dx |

≤

∫

|x|≤σ

du(x, τ) + u(x, τ)] |

∫ 1

0

d

dsΦ(sx, τ )ds|dx

≤

∫

|x|≤σ

du(x, τ) + u(x, τ)]|N∑

i=1

∫ xi

0

∂Φ( ξixix, τ)

∂ξidξi|dx

≤ 2|x|12

(∫

IRN|∇Φ|2dx

) 12

< Cσ12

(4.31)

Therefore

28

|

∫

IRNdu(x, τ) − u(x, τ)] Φ(x, τ)dx| = |

∫

IRNdu(x, τ) − u(x, τ)] dΦ(x, τ) − Φ(0, τ)]dx|

≤ |

∫

|x|>σ

du(x, τ) − u(x, τ)] dΦ(x, τ) − Φ(0, τ)]dx | +|

∫

|x|≤σ

du(x, τ) − u(x, τ)]

×dΦ(x, τ) − Φ(0, τ)]dx|(4.32)

Combining (4.30), (4.31) and (4.32), we conclude that, if 1 ≤ n < N+2N

,

limτ→+0

∫

IRN[u(x, τ) − u(x, τ)]Φ(x, τ)dx = 0

Combining (4.29) yields u(x, t) ≤ u(x, t) for all (x, t) ∈ ST . By emplying the analogous

method, we can get the reverse inequality u(x, t) ≥ u(x, t) for all (x, t) ∈ ST . Therefore we

conclude the uniqueness. As for 0 < n < 1, we take the limit in (4.26) first with respect to

ρ(ρ → 0), then

∫

Bα(0)

wkj(x, T )χ(x)dx = −

∫ T

τ

∫

∂Bα(0)

wk,j∂Φ

∂νdsdt +

∫

Bα(0)

wkj(x, τ)Φ(x, τ)dx

= I2 + I3 (4.33)

By Lemma 4.2, if 0 < n < 1,

|∇Φ| ≤ K4(j)e−α

for (x, t) ∈ ∂Bα(0) × dτ, T ], where K4(j) = Cj1−neC(T )j1−nfor j ≥ 1. Thus we claim

|I2| ≤ CT K4(j)e−α supSτT

|wkj | (4.34)

According to (4.34), we pick α > C(T )j and take the limit in (4.33) with respect to j(j →

+∞), we have proved (4.29) is true in the case 0 < n < 1. Now we turn to estimate for

the right hand side of (4.29) when τ is small enough. Let σj = j−1 and introduce the

transformation for the equation (4.6) as follows:

29

x′ =x− x0

σj

, t′ =t0 − t

σ2j

,

(x0, t0) ∈ SτT and (x, t) ∈ Qj = (x, t) | |x− x0| < σj, t0 − σj

2 < t < t0. Denoting

Φ(x′, t′) = Φ(x0 + σjx′, t0 − σj

2t′), dρ(x′, t′) = dρ(x0 + σjx

′, t0 − σj2t′).

then Φ(x′, t′) satisfies the following equation

Φt′− M Φ + σj · dρ~b · ∇Φ = 0

for (x′, t′) ∈ Q′1 = (x′, t′) | |x′| ≤ 1, −1 < t′ < 0. In virtue of the estimate (4.3), we have

|σj dρ| ≤ C

for j = 1, 2, · · ·, where the constant C is independent of j and τ . By the standard Nash’s

interior estimates, there exist the constants C1 and γ ∈ (0, 1) such that

|Φ(x′, t′) − Φ(0, 0)| ≤ C1(|x′|γ + |t′|γ2 )

for any (x′, t′) ∈ Q′12

= (x′, t′) | |x′| < 12, −1

2< t′ < 0. Set

∑j(x0, t0) = (x, t) | |x− x0| ≤ σ2

j , t0 −σ4j

4< t < t0

Now if (x, t) ∈∑

j(x0, t0), we have

|Φ(x, t) − Φ(x0, t0) |≤ C1(|x− x0|γ2 + |t− t0|

γ4 ) (4.35)

for any (x0, t0) ∈ SτT and sufficiently large positive integer j. For any ε > 0, σ =

σ2j

4, we fix

j and take τ0 such that

σj < ε,

and (4.30) follows for 0 < τ ≤ τ0. Therefore (4.35) gives

|

∫

|x|≤σ

[u(x, τ) − u(x, τ)] dΦ(x, τ) − Φ(0, τ)]dx|

< Cσγ2

30

So we have

limτ→+0

∫

IRN[u(x, τ) − u(x, τ)]Φ(x, τ)dx = 0

for 0 < n < 1. This and (4.29) implies

u(x, t) ≤ u(x, t)

for (x, t) ∈ ST . Employing the analogous method, we can derive the reverse inequality.

Q.E.D.

5. NonexistenceIn this section we prove the nonexistence result when n ≥ N+2

N(Theorem B). We first

derive a decay estimate which will be needed in the proof of the main result.

Theorem 5.1. Let n ≥ N+2N

and u(x, t) be a solution of (1.1)-(1.2). Then there exist

the constants M1 and M2 such that if 0 <√

|x|2 + t < M1, then

0 ≤ u(x, t) ≤ M2(|x|2 + t)−1

(n−1) (5.1)

In order to prove Theorem 5.1, we need to prove the following technical lemmas.

Lemma 5.1. Let u(x, t) be a source-type solution of (1.1)-(1.2), (x0, t0) ∈ ST , u(x0, t0) =

ε > 0, where ε is a noncritical value of u. Then there exists a curve

x = β(t) = (β(t), · · · , βN(t)) ∈ C1(0, t0]

such that

x0 = β(t0), u(β(t), t) = ε, |∇u(β(t), t)| 6= 0 t ∈ (0, t0].

Furthermore except for at most a countable number of value ε, limt→+0

β(t) exists.

The proof of Lemma 5.1 can be seen in [8] or [9]. Here we omit the details.

Lemma 5.2. Let u(x, t) be a source-type solution of (1.1)-(1.2), g(s) = supΓs

u(x, t), Γs =

(x, t) | |x|2 + t = s2, t ≥ 0, then g(s) is a nonnegative and decreasing function in (0, ρ]

for any ρ > 0.

31

Proof. Let (x0, t0) ∈ Γs0 , s0 ∈ (0, ρ] and supΓs0

u(x, t) = u(x0, t0) = ε is a noncritical

value. By Lemma 5.1, there exists an equipotential curve x = β(t) ∈ C1(0, t0] such that

x0 = β(t0), u(β(t), t) = ε

and

∇u(β(t), t) 6= 0

for any t ∈ (0, t0]. Noticing u(x, 0) = 0 if x 6= 0, we can choose a sequence tl satisfying tl → 0

and β(tl) → 0 as l → ∞. Thus,

(β(t), t) | 0 < t ≤ t0⋂

Γs 6= ø

This gives

u(x0, t0) = ε ∈ u(Γs) = u(x, t)|(x, t) ∈ Γs

and

g(s0) ≤ g(s)

for any s : 0 < s ≤ s0. If ε is a critical value, by Sard’s lemma and the continuity of u, we

can choose

(xk, tk) ∈ Γs0 , u(xk, tk) = εk,

where εk is a noncritical value for all k = 1, 2, · · · and εk → ε as k → ∞. Since εk ∈ u(Γs0),

then

g(s) ≥ εk

for any positive integer k. After taking limit as k → ∞ we conclude the desired result of the

lemma.

Q.E.D.

32

Proof of Theorem 5.1: Let

Qσρ = (x, t)|σ−1ρ ≤√

|x|2 + t ≤ σρ, t ≥ 0,

for 1 ≤ σ < θ ≤ 2. Let ξ = ξ(|x|2 + t) be a cutoff function in Qσρ such that

0 ≤ ξ ≤ in Qθρ

ξ ≡ 1 on Qσρ

and

|∇ξ| ≤C(N)

(θ − σ)ρ, |ξt| ≤

C(N)

(θ − σ)ρ2(5.2)

As usual we can suppose ξ has zero extension outside Qθρ. Taking ξ2uα(α ≥ 0) as a test

function and integrating over Sτ = IRN × (0, τ ], we obtain

∫∫

Sτ

u(ξ2uα)t −∇u · ∇(ξ2uα) − un ·~b · ∇(ξ2uα)dxdt =

∫

IRNξ2uα+1(x, t)|τ0dx (5.3)

Noticing u(x, 0) = 0 as x 6= 0, by the properties of ξ and integrating by parts, we get

2

α + 1

∫∫

Qτθρ

uα+1ξξt dxdt−4α

(α + 1)2

∫∫

Qτθρ

ξ2|∇uα+1

2 |2 dxdt− 2

∫∫

Qτθρ

ξuα∇ξ · ∇u dx

−2n

α + n

∫∫

Qτθρ

ξ uα+n ~b · ∇ξ dxdt =1

α + 1

∫

IRNξ2uα+1(x, τ) dx

(5.4)

where Qτθρ = Qθρ

⋂0 < t ≤ τ and 0 < τ < (θρ)2. Applying Cauchy-Schwarz’s inequality,

we have

∫∫

Qτθρ

ξuα∇ξ · ∇u dx =2

α + 1

∫∫

Qτθρ

ξuα+1

2 ∇ξ · ∇uα+1

2 dxdt

≤ ε

∫∫

Qτθρ

ξq|∇uα+1

2 |2dxdt +1

(α + 1)2 · ε

∫∫

Qτθρ

|∇ξ|2uα+1dxdt.

(5.5)

33

Clearly,

∫∫

Qτθρ

|∇(ξuα+1

2 )|2 dxdt ≤ C

∫∫

Qτθρ

ξ2|∇uα+1

2 )|2 dxdt +

∫∫

Qτθρ

|∇ ξ|2 uα+1 dxdt. (5.6)

Choosing suitable positive number ε and taking (5.5) and (5.6) into account for (5.4), we

get

∫

IRNξ2 uα+1(x, τ) dx +

∫∫

Qτθρ

|∇(ξuα+1

2 )|2 dxdt ≤ C(α + 1)×

[

∫∫

Qτθρ

uα+1ξ|ξt|dxdt +

∫∫

Qτθρ

|∇ξ|2uα+1 dxdt +

∫∫

Qτθρ

uα+nξ|∇ ξ| dxdt].(5.7)

Without loss of generality, we suppose

∫∫

Qτθρ

uα+1 dxdt ≥ 1 (5.8)

for any α ≥ 0. Set

Iα = sup0<τ<θρ

∫

IRNξ2 uα+1(x, τ)dx +

∫∫

Qτθρ

|∇(ξuα+1

2 )|2. dxdt

Substituting (5.2) and (5.8) into (5.7), we have

Iα ≤ C(α + 1)

(θ − σ)2[ρ−2 + ρ−1 ‖ u ‖n−1

L∞(Qθρ)]

∫∫

Qτθρ

uα+1 dxdt (5.9)

for 0 < τ ≤ (θρ)2 and the constant C is independent of α, τ and ρ.

Define

ρl =σ−1ρ

2+

(θ−1 − σ−1)ρ

2l+1, ρl = σρ +

(θ − σ)ρ

2l

Q(l) =

(x, t) | ρl ≤√

|x|2 + t ≤ ρl

and choose α such that

α + 1 = (A + 1)ql

34

for l = 0, 1, 2, · · · , where the positive constant A will be determined later and q = 1 + 2N

.

It follows from (5.9) and Sobolev’s imbedding theorem that

∫∫

Q(l+1)

u(A+1)ql+1dxdt ≤ CIqα ≤C(α + 1)q

(θ − σ)2q

(ρ−2 + ρ−1 ‖ u ‖n−1L∞(Qθρ))

∫∫

Q(l)

u(A+1)qldxdt

q

The standard Moser’s iteration yields

‖ u ‖L∞(Qσρ)≤ C

(θ − σ)−2 dρ−2 + ρ−1 ‖ u ‖n−1L∞(Qθρ)]

q(A+1)(q−1)

(

∫∫

Qθρ

uA+1 dxdt)1

A+1 . (5.10)

Suppose the conclusion of Theorem 5.1 is false,then there exists a sequence sk such that

sk → 0 as k → +∞ and

s2

n−1

k supΓsk

u(x, t) ≥ C. (5.11)

Due to Lemma 5.2, it follows from (5.10) that

‖ u ‖L∞(Qσρ)≤ C[(θ − σ)−2 ρ−1 ‖ u ‖n−1

L∞(Qσρ)

] q(A+1)(q−1)

(

∫∫

Qσρ

uA+1 dxdt)1

A+1 (5.12)

Choose A + 1 = (N + 2)(n− 1) + 32

and σ = 1, θ = 2, the estimate (5.12) yields

‖ u ‖L∞(Qρ)≤ Cρ−(N+2)

2(N+2)(n−1)+3 ‖ u ‖(N+2)(n−1)

2(N+2)(n−1)+3

L∞(Q2ρ) ×

(∫ 2ρ

ρ4

∫

|ω|=1

[u(s, ω)](N+2)(n−1)+ 32 sN+1 dωds

) 1

(N+2)(n−1)+ 32

where ω =(x, t)√|x|2 + t

and√|x|2 + t = s.

By Young’s inequality, we have

‖ u ‖L∞(Qρ)≤1

2‖ u ‖L∞(Q2ρ)

+C

ρ−

N+2(N+2)(n−1)+1 +

∫ 2ρ

ρ4

∫

|ω|=1

[u(s, ω)](N+2)(n−1)+ 32 sN+1dωds

.

35

Applying Lemma 2.1, we obtain

‖ u ‖L∞(Qρ)≤ C

ρ−

N+2(N+2)(n−1)+1 +

∫ 2ρ

ρ2

∫

|ω|=1

[u(s, ω)](N+2)(n−1)+ 32 sN+1dωds

(5.13)

Define

h(s) = sN+2

(N+2)(n−1)+1 supΓs

u(s, ω) = sN+2

(N+2)(n−1)+1 g(s).

It follows from (5.13) that

sup12ρ≤θ≤2ρ

h(s) ≤ C [1 +

∫ 2ρ

ρ4

s−α dh(s)](N+2)(n−1)+ 32 ds ]

where α =(N+2)(n−1)−N

2

(N+2)(n−1)+1< 1. Therefore

h(ρ) ≤ Cd1 +∞∑

k=1

∫ ρ

2k−1

ρ

2k+2

s−αhA+1(s)ds]

≤ C d1 +

∫ ρ

0

s−αhA+1(s)ds]

for any 0 < ρ < 1.

Consider the following ordinary differential equation:

dH(ρ)

dρ= C ρ−αHA+1(ρ),

H(0) = C,

where the constant C depends on C. This problem can be solved by

H(ρ) = C/[1 − CA+1Aρ1−α

1 − α]

1A

for 0 < ρ < d 2A1−α

CA+1]−1

1−α .

Choose

36

M1 =(

2A1−α

CA+1)− 1

1−α

=

2d2(N+2)(n−1)+3] d(N+2)(n−1)+1]N+2

· C(N+2)(n−1)+ 32

−[2(N+2)(n−1)+2]N+2

,

M2 = C21A = C · 2

22(N+2)(n−1)+1

and suitable C, we have

h(ρ) ≤ J(ρ) ≤ M2

for 0 ≤ ρ ≤ M1. This leads to a contradiction with (5.11) since N+2(N+2)(n−1)+1

< 2n−1

when

n ≥ N+2N

.

Q.E.D.

Theorem 5.2. Let u(x, t) be the solution to the problem (1.1)-(1.2). Then for any τ1, τ2 :

0 < τ1 < τ2 ≤ T , we have

∫

IRNu(x, τ2) dx ≤

∫

IRNu(x, τ1) dx

Proof. Without loss of generality, we suppose ut ∈ L2(ST ) and u is continuously

differentiable with respect to x. Let ηj(x) be the cutoff function in Bj(0) with Bj+1(0).

Multiplying (1.1) by ηj · ud(ηj · u)2 + δ]−12 for any δ > 0 and integrating the resulting

equation over IRN × [τ1, τ2], we find

∫ τ2

τ1

∫

IRN

ηju ut

[(ηju)2 + δ]12

dxdt +

∫ τ2

τ1

∫

IRN∇u · ∇

ηju

[(ηju)2 + δ]12

dxdt−

∫ τ2

τ1

∫

IRN

~b · ∇un ηju

[(ηju)2 + δ]12

dxdt = 0

After performing integration by parts, we have

37

∫

IRNηj(x)

∫ u(x,τ2)

u(x,τ1)

s

[(ηjs)2 + δ]12

dsdx + δ

∫ τ2

τ1

∫

IRN

∇(ηju) · ∇u

[(ηju)2 + δ]32

dxdt

+ δ

∫ τ2

τ1

∫

IRN

∇un~b · ∇(ηju)

d(ηju)2 + δ]32

dxdt = 0

(5.14)

For any fixed δ > 0, the second and third integrals of (5.14) are bounded uniformly with

respect to j. Using Lebesgue dominate convergence (L.D.C.) theorem and taking the limit

in (5.14) with respect to j → +∞, we have

∫

IRN

∫ u(x,τ2)

0

s

(s2 + δ)12

dsdx + δ

∫ τ2

τ1

∫

IRN

|∇u|2

(u2 + δ)32

dxdt

+ δ

∫ τ2

τ1

∫

IRN

~b · ∇

(∫ u(x,t)

0

sn

(s2 + δ)32

ds

)dxdt =

∫

IRN

∫ u(x,τ1)

0

s

(s2 + δ)12

dsdx

(5.15)

Due to the assumption of the theorem, the third term of the left hand side of (5.15) is

bounded, uniformly with respect to δ. This implies the second term of the left hand side

of (5.15) is bounded uniformly with respect to δ. Therefore employing Fatou’s lemma and

taking the limit in (5.15) with respect to δ(δ → 0), we can reach the conclusion of Theorem

5.2.

Proof of nonexistence result for n ≥ N+2N

(Theorem B):

Case 1. n > N+2N

Let φ(x, t) ∈ C∞0 (IRN × (−T, T )) and η(s) be any smooth and nondecreasing function on

IR such that

38

η(s) =

1, t ≥ 20, t ≤ 1

Set

ηk = η(ks)

and

ξk(x, t) = ηk(|x|2 + t) φ(x, t)

Since for every positive integer k, ξk vanishes in a neighborhood of (0, 0), we have

∫∫

ST

(uξkt + u M ξk − un~b · ∇ξk) dxdt = 0 (5.16)

Denote

Dk = (x, t) |1

k≤ |x|2 + t ≤

2

k, t > 0

Clearly,

meas Dk ≤ Ck−(N+2)

2 meas D1 (5.17)

|ξkt| ≤ C(N)k, |∇ξk| ≤ C(N)k12 (5.18)

and

|4ξk| ≤ C(N)k (5.19)

By Theorem 5.1 and n > N+2N

, we have

∫∫

|x|2+t≤ρ2, t≥0

un dxdt ≤

∫ ρ

0

s−2nn−1

+N+1 ds < ∞ (5.20)

for any 0 < ρ < 1. Applying Holder’s inequality and using the fact (5.17), we see

39

∫∫

Dk

|ξkt| u dxdt ≤ Ck

∫∫

Dk

u dxdt ≤ Ck(

∫∫

Dk

un dxdt)1n (meas Dk)1− 1

n

≤ C k1−(N+2)

2(1− 1

n)(

∫∫

Dk

un dxdt)1n .

Since 1 −1

n>

2

N + 2, we find from (5.20) that

limk→+∞

∫∫

Dk

|ξkt|u dxdt = 0. (5.21)

Similarly, we have

limk→+∞

∫∫

Dk

|∇ξk|u dxdt = 0 (5.22)

Taking the limit in (5.16) with respect to k(k → +∞) and substituting (5.21) and (5.20)

into (5.16), we obtain

limk→+∞

∫∫

ST

un~b · ∇ξk dxdt = 0.

This implies that the integral identity

∫∫

S

(uφt + u∆φ− un~b · ∇φ) dxdt = 0 (5.23)

holds for any φ ∈ C∞0 (S), where S = IRN × (−T, T ).

Let w ∈ C∞(IR) such that

w(s) = 0 for |s| ≥ 1

and

∫ ∞

−∞

w(s)ds = 1.

Set

wh(s) = h−1w(s

h)

40

and

ξh(t) = 1 −

∫ t−τ−2h

−∞

wh(s)ds,

for τ ∈ (0, T ). Obviously,

ξh(t) = 1 for t < τ + h

0 ≤ ξh(t) ≤ 1, for t ∈ IR

and

limh→0

ξh = 0 for t > τ.

For any η(x) ∈ C∞0 (IRN), we pick φ = ηξh in (5.23), then

−

∫∫

ST

wh(t− τ − 2h) u η dxdt +

∫∫

ST

(uξh4η − unξh~b · ∇η) dxdt = 0.

Let h → 0, we obtain

∫

IRNu(x, t)η(x)dx =

∫ t

0

∫

IRN(u4η − un ·~b · ∇η)dxdt.

This implies

limt→+0

∫

IRNu(x, t)η(x)dx = 0

for any η ∈ C∞0 (IRN), which leads to a contradiction with (1.3). Now nonexistence result in

Theorem B is proved for the case n > N+2N

.

Case 2. n = N+2N

Suppose u(x, t) be a source-type solution of the equation (1.1) for the case n = N+2N

.

Let u0,j(x) be δ-type sequence as defined in Section 2 and nk be a sequence such that

nk ↓N+2N

as k → ∞. If u(j)k (x, t) is a solution of the following equation

u(j)kt = 4u

(j)k +~b · ∇(u

(j)k )nj , (x, t) ∈ ST (5.24)

u(j)k (x, 0) = u0,j(x), x ∈ IRN (5.25)

for k, j = 1, 2, · · · . According to the discussion of Sections 2 and 3, we know that for any

positive integer k, there exists a nonnegative and continuous function uk(x, t), (x, t) ∈ ST

and a subsequence of u(j)k which we still denote by u

(j)k such that

41

limj→+∞

u(j)k (x, t) = uk(x, t)

(x, t) ∈ ST , and the convergence is uniformly in any domain D ⊂⊂ ST\(0, 0). We also

readily know that uk(x, t) possesses the properties (i) and (ii) of Definition 1.1, but by the

discussion as mentioned above for any η(x) ∈ C∞0 (IRN), we have

limt→+0

limj→∞

∫

IRNu

(j)k (x, t)η(x) dx = lim

t→+0

∫

IRNuk(x, t)η(x)dx = 0

By the definition of u(j)k (x, t), we see

limj→∞

limt→+0

∫

IRNu

(j)k (x, t)η(x) dx = lim

j→∞

∫

IRNu0,j(x)η(x)dx = η(0)

Hence, we have

limt→+0

limj→∞

∫

IRNu

(j)k (x, t)η(x) dx 6= lim

j→∞limt→+0

∫

IRNu

(j)k (x, t)η(x)dx

for η(x) ∈ C∞0 (IRN). This implies that the limit solution u(x, t) is not a source-type solution

to (1.1)-(1.2) when n = N+2N

.

Clearly the positive function sequence uk(x, t) possesses the following properties:

∫

IRNuk(x, t)dx ≤ 1 (5.26)

and

∫

IRN|uk(x + y, t) − uk(x, t)|dx ≤

∫

IRN|u0,j(x + y) − u0,j(x)|dx (5.27)

for sufficiently large integer j. Hence, uk(x, t) is a bounded and equicontinuous sequence

in L(IRN). In order to show uk(x, t) is a precompact set in L(IRN) uniformly with respect

to t, we need to prove

limG→+∞

∫

|x|≥G

|uk(x, t)|dx = 0 (5.28)

uniformly with respect to k = 1, 2, · · ·. Set

42

W (x, t) = BeK2t(|x|2 + 1)−β

for (x, t) ∈ DM1 = ST

⋂(x, t) | |x|2 + t ≥ M2

1, where M1,M2 are as defined in Theorem 5.1

and T ≤ minM21 ,

1K2

, the positive constants B,K2 and β will be determined later. After

routine calculations, it is easy to see

Wt = BK2eK2t (|x|2 + 1)−β

4W = −2BβNeK2t(|x|2 + 1)−β−1 + 4Bβ(β + 1)eK2t(|x|2 + 1)−β−2|x|2

∇W nk = −2BnkeK2nktnkβ(|x|2 + 1)−βnk−1x

and

LkW = Wt −4W −~b · ∇W nk = BeK2t(|x|2 + 1)−β−2

×[K2(|x|2 + 1)2 − 4β(β + 1)|x|2 + 2βN(|x|2 + 1)

+2Bnk−1eK2(nk−1)tnkβ(|x|2 + 1)−(nk−1)β+1~b · x].

(5.29)

Choose a fixed β > N and B > M−11 M2(M2

1 + 1)β max1,M1 and then choose

K2 > max4Bnk−1enk−1 · nkβ|~b|, 8β(β + 1).

From (5.29) we see

LkW ≥ 0 (x, t) ∈ DM1

and

W (x, t) |(∂DM1 ß t>0)∪(x,0)||x|≥M1≤ uk(x, t) |(∂DM1 ß t>0)∪(x,0)||x|≥M1 .

Using the boundedness of uk(x, t) in DM1 and the comparison principle, we have

W (x, t) ≥ uk(x, t) (5.30)

43

for (x, t) ∈ DM1 and k = 1, 2, · · ·. If T > 1K1

, we can use similar procedure step by step.

Consequently, we see that the inequality (5.30) must be valid in ST

⋂(x, t) | |x|2 + t > M2

1

for any T > 0. Due to (5.26) - (5.28), the functions sequence uk(x, t) is precompact

in L(IRN) uniformly with respect to t ∈ (τ, T ] for any τ : 0 < τ < T . According to

the discussing processes in Sections 2 and 3, uk(x, t) is Holder continuous in any SτT , for

0 < τ < T and uniformly with respect to k. Therefore, by the diagonal technique, we

can obtain a subsequence of uk, (still denoted by uk(x, t)) and nonnegative, continuous

function u(x, t) ∈ L(IRN) such that

limk→∞

∫

IRN|uk(x, t) − u(x, t)| dx = 0

and

limk→∞

uk(x, t) = u(x, t)

for (x, t) ∈ ST , the convergence is uniform in SτT for any 0 < τ ≤ T . By L.D.C. Theorem,

we can pass through the limit procedure in the following integral identities

∫ ∞

τ

∫

IRN(ukφt + uk4φ− unk~b · ∇φ) dxdt +

∫

IRNuk(x, τ)φ(x, τ) dx = 0

for any φ(x, t) ∈ C2.10 (ST ) with φ(x, T ) = 0 and 0 < t < T . Thus, u(x, t) must possess the

properties (i) and (ii) of Definition 1.1 when n = N+2N

.

Now we turn to finding a sequence εk such that εk ↓ 0 and

limk→+∞

∫

IRNu(x, εk)η (x) dx = 0

for any η(x) ∈ C∞0 (IRN).

By using the similar method to the case for n > N+2N

, we have

∫

IRNuk(x, t) η(x) dx =

∫ τ

0

∫

IRN(uk4η − unk

k ·~b · ∇η) dxdt (5.31).

Set

Ω+τ = (x, s) | |x|2 + s ≤ t2, s ≥ 0.

Then,

44

∫∫

Ω+τ

unkk dxdt =

∫ τ

0

∫

|ω|=1

unkk (s, ω) · sN+1 dωds.

By the estimate of Theorem 5.1 we have

∫∫

Ω+τ

unk(x, t) dxdt ≤ C

∫ τ

0

s−

2nknk−1

+N+1ds ≤

C(nk − 1)

(N + 1)(nk − 1) − (nk + 1)sN+1−

nk+1

nk−1 (5.32).

Let

αk = (N + 1)(nk − 1) − (nk + 1) > 0

for k = 1, 2, · · · . Clearly limk→∞

αk = 0 as nk →N+2N

. Picking εk = e−1

α2k , τ ≤ εk, we have

∫∫

Ω+τ

unk(x, t) dxdt ≤ C1

αk

e−1

(nk−1)αk (5.33)

Combining (5.31) and (5.33) with the properties of uk(x, t), we obtain

limk→∞

∫

IRNuk(x, εk) η (x) dx = 0

for any η(x) ∈ C∞0 (IRN). By Lemma 5.3

∫

IRNuk(x, εl) dx ≤

∫

IRNuk(x, εk) dx

for any integer l < k. Applying Fatou’s lemma, we have

∫

IRNu(x, εl) dx ≤ lim

k→+∞

∫

IRNuk(x, εk) dx = 0

for l = 1, 2, · · ·. This leads to a contradiction with the property

u(x, t) → δ(x)

in the sense of distribution as t → +0. Now we conclude the nonexistence of fundamental

solution to (1.1) when n = N+2N

.

Q.E.D.

45

6. Asymptotic Expansion and Similarity SolutionThis section is devoted to the investigation of asymptotic expansion of the source

solution to the equation (1.1) near origin (0,0) when n < N+1N

(Theorem B), and a similarity

solution to the equation (1.1) when n = N+1N

(Theorem C).

Proof of Theorem B. Define

uσ(x, t) = σNu(σ x, σ2t)

where σ > 0, and u(x, t) is the fundamental solution to (1.1). the uσ(x, t) satisfies the

integral identities:

∫ ∞

τ

∫

IRN(uσφ + uσ4φ− σ1−(n−1)Nun

σ~b · ∇φ) dxdt +

∫

IRNuσ(x, t)φ(x, t)dx = 0 (6.1)

∫

IRNuσ(x, t)dx =

∫

IRNσN(σ x, σ2t) dx = 1 (6.2)

for any test function φ(x, t) and 0 < t < T . Then

limτ→+0

∫

IRNuσ(x, τ)η(x) dx = lim

τ→+0

∫

IRNσNu(σx, σ2τ)η(x) dx = η(0) (6.3)

for any η(x) ∈ C∞0 (IRN). According to the estimate of Theorem 2.3 and the expression of

uσ(x, t), when n < N+1N

, we have

‖ uσt(·, t) ‖L∞(IRN)≤ C t−

N2 . (6.4)

In view of (6.1) and (6.3), uσ(x, t) is the source-type solution of the following equation:

uσt = 4uσ + σ1−(n−1)N ~b · ∇unσ (6.5)

for all σ > 0. Due to (6.4), we can show uσ(x, t) must be locally Holder continuous in ST ,

uniformly with respect to σ. Thus by the diagonal technique, there exists a subsequence σk

and a continuous nonnegative function u(x, t) such that σk → 0 and

uσk(x, t) → u(x, t), (x, t) ∈ S,

46

as k → ∞. Moreover, the convergence is uniform in any bounded subset of SτT for 0 < τ < T .

Since we have

supSτT

|∇u1−nσ | ≤ C(τ, T )

when 0 < n < 1, and

supSτT

|∇uσ| ≤ C(τ, T )

when n > 1, we can pass the limit process in (6.1) by L.D. C. theorem. This implies that

u(x, t) is the source-type solution of standard heat conduction equation. By the uniqueness

result, we have

uσ(x, t) → w(x, t), (x, t) ∈ S

as σ → 0, where u(x, t) = w(x, t) = 1

(4πt)N2e−|x|24t . Now Setting σ = t

12 and noticing

w(x, 1) = σN w(σx, σ2), we obtain

tN2 u(xt

12 , t) → t

N2 w(xt

12 , t).

as t → +0. The convergence is uniform for |x| ≤ A. This implies the conclusion of Theorem

C.

Proof of Theorem C. When n = N+1N

, we suppose ~b = (b1, · · · bN). Set

ζ = xt−12 , u = t−

N2 f(xt−

12 ) = t−

N2 f(ζ), ζ = (ζ1, · · · , ζN).

Then,

ut = −12t−

N2−1∇(ζf(ζ)),

~bT∇x(u1+ 1N ) = t−

N2−1~bT∇fn

and

4x u = t−N2−14f,

47

where differential operators are defined as follows:

~bT∇x = (b1∂

∂x1

, · · · , bN∂

∂xN

)

and

~bT∇ = (b1∂

∂ζ1

, · · · , bN∂

∂ζN).

Therefore, if ~b = ~e = (1, · · · , 1), f(ζ) satisfies the following elliptic equation:

4f +1

2∇ · (ζf) + ∇ · (~efn) = 0. (6.6)

By the properties of source-type solution, for x 6= 0, then

lim|ζ|→+∞

ζf = limt→+0

x t−12

+N2 u(x, t) = 0, (6.7)

lim|ζ|→+∞

fn = limt→+0

tnN2 un(x, t) = 0 (6.8)

and

lim|ζ|→+∞

∇ f = limt→+0

tN2

+1∇u(x, t) = 0 (6.9)

Combining (6.7) - (6.9) with (6.6), we see f(ζ) satisfying the following differential system:

∇ f + ~efn + 12ζf = 0, (6.10)

lim|x|→+∞

f(ζ) = 0, (6.11)

∫IRN f(ζ)dζ = 1. (6.12)

(6.10) can be rewritten as

∂f

∂ζi+ fn +

1

2ζi f = 0, (6.13)i

i = 1, 2, · · · , N . Set v = f 1−n, (6.13)i gives

∂v

∂ζi=

1

2Nζiv +

1

N, (6.14)i

48

i = 1, 2, · · · , N . Denote C1 = C1(ζ1, ζ2, · · · , ζN), · · · , Ci = Ci(ζi, · · · , ζN). · · · , CN = CN(ζN)

and C0 is independent of ζi for i = 1, 2, · · · , N . Clearly (6.14)1 can be solved by

v = eζ21

4N

(1

N

∫e−

ζ21

4N dζ1 + C2

)

Substituting it into (6.14)2, we have

eζ21

4N∂C2

∂ζ2

=1

2Nζ2e

ζ21

4N

(1

N

∫e−

ζ21

4N dζ1 + C2

)+

1

N

Then,

∂C2

∂ζ2

=1

2Nζ2

(1

N

∫e−ζ2

14N dζ1 + C2

)+

1

Ne−ζ2

14N .

This can be solved by

C2 =−1

N

∫e−

ζ21

4N dζ1 +1

Ne−ζ2

14N

+−ζ2

24N

∫e−

ζ22

4N dζ2 + C3eζ22

4N .

Hence,

v = eζ21+ζ2

24N (

1

Ne−

ζ21

4N

∫e−

ζ22

4N dζ2 + C3).

Suppose

v = e

N−1ài=1

ζ2i /(4N)

1

Ne−

N−2ài=1

ζ2i /(4N)

∫e−

ζ2N−14N dζN−1 + CN. (6.15)

Substituting it into (6.14)N yields

e

N−1ài=1

ζ2i /(4N)∂CN

∂ζN=

1

2NζN e

N−2ài=1

ζ2i /(4N)

×

1

Ne−

N−2ài=1

ζ2i /(4N)

∫e−

ζ2N−14N dζN−1 + CN +

1

N.

Then,

∂CN

∂ζN=

1

2NζN CN +

1

2N2ζN e

−N−2ài=1

ζ2i /(4N)

×

∫e−ζ2

N−1/(4N) dζN−1 +1

Ne−

N−1ài=1

ζ2i /(4N)

.

49

This can be solved by

CN = e−ζ2N/(4N)−

1

Ne−

N−2ái=1

ζ2i /(4N)−

ζ2N

4N

∫e−

ζ2N−14N dζN−1+

1

Ne−

N−1ái=1

ζ2i /(4N)

∫e−

ζ2N

4N dζN + C0.

Substituting it into (6.15) and by the induction, we have

v = e|ζ|24N

1

Ne−

N−2ái=1

ζ2i /(4N)

∫ ζN

−∞

es2N

4N dsN + C0.

Thus,

f = v−N = e−|ζ|2

4 1

Ne−

N−1ái=1

ζ2i /(4N)

∫ ζN

−∞

e−1

4Ns2NdsN + C0

−N .

Consequently, for case n = N+1N

,~b = ~e = (1, . . . , 1), the similarity source-type solutions of

(1.1) can be expressed by

u(x, t) = t−N2 e−

|x|24t

1

Ne−

N−1ái=1

x2i /(4Nt)

∫ xN√t

−∞

e−s2

4N ds + C0−N ,

where the constant C0 is determined by

∫

IRN

e−|ζ|2

4

[ 1Ne−

N−1ái=1

ζ2i /(4N) ∫ ζN

−∞e−

s2

4N ds + C0]N

dζ = 1

Q.E.D.

Appendix

Proof of (3.32): The discussion of Remark 2.1 implies the conclusion of Remark 3.2

for n = 1. It is sufficient to prove the conclusion for the case 1 < n < N+2N

. Differentiating

(3.21) by xi for i = 1, 2, · · · , N, we get

Vxit = ∆Vxi +2 − n

n− 1(u1−n)xi|∇V |2+

2(2 − n)

n− 1(u1−n)∇V · ∇Vxi + n(un−1)xi

~b · ∇V + nun−1~b · ∇Vxi

50

= ∆Vxi +2 − n

n− 1(

1

V)xi|∇V |2 +

2(2 − n)

n− 1

∇V · ∇Vxi

V+ nVxi

~b · ∇V + nun−1~b · ∇Vxi

Multiplying it by Vxi and summing for i = 1, 2, · · · , N, we have

N∑

i=1

(VxiVxit − Vxi∆Vxi) =2 − n

n− 1

N∑

i=1

[−V 2xi|∇V |2

V+

2∇V · (Vxi∇Vxi)

V

]+ n

N∑

i=1

[V 2xi~b · ∇V + V~b · (Vxi∇Vxi)] (A1)

Set P = |∇V |2, then

P∆P =N∑

i=1

Vxi∆Vxi − |∇P |2 +N∑

i=1

|∇Vxi |2

Substituting it into (A1) yields1

2(P 2)t − P∆P =

−(2 − n)

n− 1

P 4

V 2+

2(2 − n)

n− 1

∇V · ∇P 2

V+ nP 2~b · ∇V +

n

2V~b · ∇P 2

Therefore

Pt − P∆P =−(2 − n)

n− 1

P 3

V 2+

4(2 − n)

n− 1

∇V · ∇P

V+ nP~b · ∇V + nV~b · ∇P

Noticing 2−nn−1

≥ 0 as N ≥ 2, we see

Pt − P∆P ≤4(2 − n)

n− 1

∇V · ∇P

V+ nP~b · ∇V + nV~b · ∇P. (A2)

We are going to estimate P (x0, t0) for any (x0, t0) ∈ STτ . For simplicity, we suppose x0 = 0

and t0 = t. Let ξ be a cut-off function in Qθ = Bθρ(0)× ( θ−1

2t, t] with Qσ = Bσρ(0)× (σ−1

2t, t]

for 1 ≤ σ < θ ≤ 2 such that

0 ≤ ξ ≤ 1, (x, t) ∈ Qθ,

ξ ≡ 1, (x, t) ∈ Qσ,

and

|∇ξ| ≤C

(θ − σ)ρ, |ξt| ≤

C

(θ − σ)t

51

Multiplying (A2) by ξ2Pα(α ≥ 1) and integrating it over Qθ, we find

∫∫

Qθ

(Pαξ2Pt − Pαξ2∆P )dxdt

≤

∫∫

Qθ

[4(2 − n)

n− 1

∇V · ∇P

V+ nP~b · ∇V + nV~b · ∇P ]Pαξ2dxdt

1

α + 1

∫∫

Qθ

(Pα+1)tξ2dxdt +

∫∫

Qθ

∇(Pαξ2) · ∇Pdxdt

≤8(2 − n)

(n− 1)(α + 1)

∫∫

Qθ

Pα+3

2 ξ2|∇Pα+1

2 |V −1dxdt + n|~b|

∫∫

Qθ

Pα+2ξ2dxdt

+2n|~b|

α + 1

∫∫

Qθ

V Pα+1

2 ξ2 · |∇Pα+1

2 |dxdt.

Integrating by parts for the left-side of the above inequality yields

1

α + 1

∫

IRNPα+1(x, t)ξ(x, t)dx +

4α

(α + 1)2

∫∫

Qθ

ξ2|∇Pα+1

2 |2dxdt

≤8(2 − n)

(n− 1)(α + 1)

∫∫

Qθ

Pα+3

2 ξ2|∇Pα+1

2 |V −1dxdt + n|~b|

∫∫

Qθ

Pα+2ξ2dxdt+

2n|~b|

α + 1

∫∫

Qθ

V Pα+1

2 ξ2 · |∇Pα+1

2 |dxdt+

2

α + 1

∫∫

Qθ

ξ|ξt|Pα+1dxdt +

4

α + 1

∫∫

Qθ

ξPα+1

2 |∇ξ||∇Pα+1

2 |dxdt.

Using Cauchy’s inequality, we get

1

α + 1

∫

IRNPα+1(x, t)ξ(x, t)dx +

α

(α + 1)2

∫∫

Qθ

ξ2|∇Pα+1

2 |2dxdt

≤16(2 − n)2

(n− 1)2α

∫∫

Qθ

ξ2Pα+3V −2dxdt + n|~b|

∫∫

Qθ

Pα+2ξ2dxdt+

52

n2|~b|2

α

∫∫

Qθ

ξ2Pα+1V 2dxdt +2

α + 1

∫∫

Qθ

ξ|ξt|Pα+1dxdt +

4

α

∫∫

Qθ

Pα+1|∇ξ|2dxdt

By Sobolev’s imbedding theorem,

∫∫

Qθ

ξ2(1+ 2N

)P (α+1) 2N

+α+1dxdt ≤C(α + 1)1+ 2

N

(θ − σ)1+ 2N

[‖ P 2V −2 ‖L∞(Qθ) +

‖ V 2 ‖L∞(Qθ) + ‖ P ‖L∞(Qθ) +ρ−2 + t−1]

∫∫

Qθ

Pα+1dxdt1+ 2N

Without loss of generality, we suppose

‖ P 2V −2 ‖L∞(Qθ)≤ C1

Otherwise, we have

‖ P ‖L∞(Q′)≥ m ‖ V ‖L∞(Q′)

for large positive number m, where Q′ ⊂ Qθ and meas Q′ > 0. Hence, due to Theorem 2.2

and picking ρ ≥ tN(n−1)

2 , we have

∫∫

Qθ

ξ2(1+ 2N

)P (α+1) 2N

+α+1dxdt ≤C(α + 1)1+ 2

N

(θ − σ)1+ 2N

[‖ P ‖2L∞(Qθ) +t−N(n−1)]

∫∫

Qθ

Pα+1dxdt1+ 2N

(A3)

Set

ρk = σρ +θ − σ

2kρ, tk =

1

2σ−1t +

θ−1 − σ−1

2k+1t,

and α + 1 = (A + 1)qk for k = 0, 1, 2, · · · , where q = 1 + 2N

and the positive constant A to

be determined. It follows from (A3) that

∫∫

Q(k+1)

P (A+1)qk+1

dxdt ≤ C(A + 1)(4q)k

(θ − σ)2[‖ P ‖2

L∞(Qθ) +t−N(n−1)]

∫∫

Q(k)

P (A+1)qkdxdtq

for k = 0, 1, 2, · · · . Standard Moser’s iteration yields

‖ P ‖L∞(Qσ)≤ C1

θ − σ[‖ P ‖2

L∞(Qθ) +t−N(n−1)]q

(A+1)(q−1) (

∫∫

Qθ

P (A+1)dxdt)1

A+1 , (A4)

where the constant C is independent of θ, σ and t.

53

Fix θ = 2, σ = 1,, choose A = N +1+γ, 0 ≤ γ < 2N(n−1)

−1. We apply Young’s inequality

for (3.36) to obtain

‖ P ‖L∞(Q1)≤ C[‖ P ‖N+2

N+2+γ

L∞(Q2) +t−N(N+2)(n−1)

2(N+2+γ) ](

∫∫

Q2

PN+2+γdxdt)1

N+2+γ

≤1

2‖ P ‖L∞(Q2) +C[t−

N(n−1)2 +

∫∫

Q2

PN+2+γdxdt].

By Lemma 2.1, we have

‖ P ‖L∞(Q1)≤ C[t−N(n−1)

2 +

∫∫

Q2

PN+2+γdxdt]. (A5)

Define

Φ(t) = sup0<τ≤t

τN(n−2)

2 · ‖ P ‖L∞(Bρ(0))

The estimate (A5) gives

Φ(t) ≤ C1 +∞∑

k=1

∫ t

2k−1

t

2k+1

[Φ(τ)]N+2+γτ−βdτ

= C1 +

∫ t

0

[Φ(τ)]N+2+γτ−βdτ (A6)

where β = (1+γ)N(n−1)2

< 1. From (A6), we can derive the desired estimate (3.32) by employ-

ing the similar method in the proof of Theorem 2.3.

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Guofu Lu, E-mail: gflu@sina.com.cn

Hong-Ming Yin, E-mail: hyin@wsu.edu

56