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SYSTEM WITH NONLOCAL SOURCEJUN ZHOU, CHUNLAI MU, AND ZHONGPING LI Received 23 January 2006; Revised 3 April 2006; Accepted 7 April 2006 We deal with the blowup properties of the solution

SYSTEM WITH NONLOCAL SOURCE

JUN ZHOU, CHUNLAI MU, AND ZHONGPING LI

Received 23 January 2006; Revised 3 April 2006; Accepted 7 April 2006

We deal with the blowup properties of the solution to the degenerate and singular par-abolic system with nonlocal source and homogeneous Dirichlet boundary conditions The existence of a unique classical nonnegative solution is established and the sufficient conditions for the solution that exists globally or blows up in finite time are obtained Furthermore, under certain conditions it is proved that the blowup set of the solution is the whole domain

Copyright © 2006 Jun Zhou et al This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

In this paper, we consider the following degenerate and singular nonlinear

reaction-diffusion equations with nonlocal source:

x q1 u t −x r1 u x



x =

a

0v p1 dx, (x,t) ∈(0,a) ×(0,T),

x q2v t −x r2v x

x =

a

0 u p2dx, (x,t) ∈(0,a) ×(0,T), u(0,t) = u(a,t) = v(0,t) = v(a,t) =0, t ∈(0,T), u(x,0) = u0(x), v(x,0) = v0(x), x ∈[0,a],

(1.1)

whereu0(x),v0(x) ∈ C2+α(D) for some α ∈(0, 1) are nonnegative nontrivial functions

u0(0)= u0(a) = v0(0)= v0(a) =0,u0(x) ≥0,v0(x) ≥0,u0,v0 satisfy the compatibility condition,T > 0, a > 0, r1,r2∈[0, 1),| q1|+r1=0,| q2|+r2=0, andp1> 1, p2> 1.

LetD =(0,a) and Ω t = D ×(0,t], D and Ω tare their closures, respectively Since| q1|+

r1=0,| q2|+r2=0, the coefficients of ut,u x,u xxandv t,v x,v xx may tend to 0 or∞asx

tends to 0, we can regard the equations as degenerate and singular

Hindawi Publishing Corporation

Boundary Value Problems

Volume 2006, Article ID 21830, Pages 1 19

Floater [9] and Chan and Liu [4] investigated the blowup properties of the following degenerate parabolic problem:

x q u t − u xx = u p, (x,t) ∈(0,a) ×(0,T), u(0,t) = u(a,t) =0, t ∈(0,T), u(x,0) = u0(x), x ∈[0,a],

(1.2)

whereq > 0 and p > 1 Under certain conditions on the initial datum u0(x), Floater [9] proved that the solutionu(x,t) of (1.2) blows up at the boundaryx =0 for the case 1<

p ≤ q + 1 This contrasts with one of the results in [10], which showed that for the case

q =0, the blowup set of solutionu(x,t) of (1.2) is a proper compact subset ofD.

The motivation for studying problem (1.2) comes from Ockendon’s model (see [14]) for the flow in a channel of a fluid whose viscosity depends on temperature

whereu represents the temperature of the fluid In [9] Floater approximatede ubyu pand considered (1.2) Budd et al [2] generalized the results in [9] to the following degenerate quasilinear parabolic equation:

x q u t =u m

with homogeneous Dirichlet conditions in the critical exponentq =(p −1)/m, where q >

0,m ≥1, and p > 1 They pointed out that the general classification of blowup solution

for the degenerate equation (1.4) stays the same for the quasilinear equation (see [2,17])

u t =u m

For the casep > q + 1, in [4] Chan and Liu continued to study problem (1.2) Under certain conditions, they proved thatx =0 is not a blowup point and the blowup set is a proper compact subset ofD.

In [7], Chen and Xie discussed the following degenerate and singular semilinear para-bolic equation:

u t −x α u x

x =

a

0 f

u(x,t)

dx, (x,t) ∈(0,a) ×(0,T), u(0,t) = u(a,t) =0, t ∈(0,T),

u(x,0) = u0(x), x ∈[0,a],

(1.6)

they established the local existence and uniqueness of a classical solution Under appro-priate hypotheses, they obtained some sufficient conditions for the global existence and blowup of a positive solution

In [6], Chen et al consider the following degenerate nonlinear reaction-diffusion equation with nonlocal source:

x q u t −x γ u x

x =

a

0u p dx, (x,t) ∈(0,a) ×(0,T), u(0,t) = u(a,t) =0, t ∈(0,T),

u(x,0) = u0(x), x ∈[0,a],

(1.7)

they established the local existence and uniqueness of a classical solution Under appro-priate hypotheses, they also got some sufficient conditions for the global existence and blowup of a positive solution Furthermore, under certain conditions, it is proved that the blowup set of the solution is the whole domain

In this paper, we generalize the results of [6] to parabolic system and investigate the effect of the singularity, degeneracy, and nonlocal reaction on the behavior of the solution

of (1.1) The difficulties are the establishment of the corresponding comparison principle and the construction of a supersolution of (1.1) It is different from [4,9] that under certain conditions the blowup set of the solution of (1.1) is the whole domain But this is consistent with the conclusions in [1,18,19]

This paper is organized as follows: in the next section, we show the existence of a unique classical solution InSection 3, we give some criteria for the solution (u(x,t),v(x, t)) to exist globally or blow up in finite time and in the last section, we discuss the blowup

set

2 Local existence

In order to prove the existence of a unique positive solution to (1.1), we start with the following comparison principle

Lemma 2.1 Let b1(x,t) and b2(x,t) be continuous nonnegative functions defined on [0,a] ×

[0,r] for any r ∈(0,T), and let (u(x,t),v(x,t)) ∈(C(Ω r)∩ C2,1(Ωr))2satisfy

x q1 u t −x r1 u x



x ≥

a

0b1(x,t)v(x,t)dx, (x,t) ∈(0,a) ×(0,r],

x q2v t −x r2v x

x ≥

a

0b2(x,t)u(x,t)dx, (x,t) ∈(0,a) ×(0,r], u(0,t) ≥0, u(a,t) ≥0, v(0,t) ≥0, v(a,t) ≥0, t ∈(0,r],

u(x,0) ≥0, v(x,0) ≥0, x ∈[0,a].

(2.1)

Then, u(x,t) ≥0,v(x,t) ≥ 0 on [0, a] ×[0,T).

Proof At first, similar to the proof of Lemma 2.1 in [20], by using [15, Lemma 2.2.1], we can easily obtain the following conclusion

IfW(x,t) and Z(x,t) ∈ C(Ω r)∩ C2,1(Ωr) satisfy

x q1 W t −x r1 W x



x ≥

a

0b1(x,t)Z(x,t)dx, (x,t) ∈(0,a) ×(0,r],

x q2 Z t −x r2 Z x



x ≥

a

0b2(x,t)W(x,t)dx, (x,t) ∈(0,a) ×(0,r], W(0,t) > 0, W(a,t) ≥0, Z(0,t) > 0, Z(a,t) ≥0, t ∈(0,r],

W(x,0) ≥0, Z(x,0) ≥0, x ∈[0,a],

(2.2)

then,W(x,t) > 0, Z(x,t) > 0, (x,t) ∈(0,a) ×(0,r].

Next letr1 ∈(r1, 1),r2 ∈(r2, 1) be positive constants and

W(x,t) = u(x,t) + η

1 +x r1 − r1

e ct, Z(x,t) = v(x,t) + η

1 +x r 2− r2

e ct, (2.3) where η > 0 is sufficiently small and c is a positive constant to be determined Then W(x,t) > 0, Z(x,t) > 0 on the parabolic boundary of Ω r, and in (0,a) ×(0,r], we have

x q1 W t −x r1 W x



x −

a

0b1(x,t)Z(x,t)dx

≥ x q1 η

1 +x r1 − r1

ce ct+



r1 − r1



1− r1



ηe ct

x2− r 1 −

a

0b1(x,t)η

1 +x r 2− r2

e ct dx

≥ ηe ct



cx q1+



r1 − r1 

1− r1



x2− r1 − a

1 +a r 2− r2 

max (x,t) ∈[0,a] ×[0,r] b1(x,t)



,

x q2 Z t −x r2 Z x

x −

a

0b2(x,t)W(x,t)dx

≥ ηe ct



cx q2+



r2 − r2



1− r2



x2− r2 − a

1 +a r 1− r1

max (x,t) ∈[0,a] ×[0,r] b2(x,t)



.

(2.4)

We will prove that the above inequalities are nonnegative in three cases

Case 1 When

max (x,t) ∈[0,a] ×[0,r] b1(x,t) ≤



r1 − r1



1− r1



a3− r1

1 +a r2 − r2,

max (x,t) ∈[0,a] ×[0,r] b2(x,t) ≤



r2 − r2



1− r2



a3− r 2 

1 +a r1 − r1.

(2.5)

It is obvious that

x q1W t −x r1W x

x −

a

0b1(x,t)Z(x,t)dx ≥0,

x q2Z t −x r2Z x

x −

a

b2(x,t)W(x,t)dx ≥0.

(2.6)

Case 2 If

max (x,t) ∈[0,a] ×[0,r] b1(x,t) >



r1 − r1



1− r1



a3− r 1 

1 +a r2 − r2,

max (x,t) ∈[0,a] ×[0,r] b2(x,t) >



r2 − r2



1− r2



a3− r 

2 

1 +a r 

1− r1.

(2.7)

Letx0andy0be the root of the algebraic equations

a

1 +a r 2− r2 

max (x,t) ∈[0,a] ×[0,r] b1(x,t) =



r1 − r1



1− r1



x2− r1 ,

a

1 +a r 1− r1 

max (x,t) ∈[0,a] ×[0,r] b2(x,t) =



r2 − r2



1− r2



y2− r 2 ,

(2.8)

andC1,C2> 0 be sufficient large such that

C1>

⎧

⎪

⎪

⎪

⎪

max (x,t) ∈[0,a] ×[0,r] b1(x,t)

a

1 +a r 2− r2

x0q1

forq1≥0,

max (x,t) ∈[0,a] ×[0,r] b1(x,t)

a

1 +a r 2− r2

a q1 forq1< 0,

C2>

⎧

⎪

⎪

⎪

⎪

max (x,t) ∈[0,a] ×[0,r] b2(x,t)

a

1 +a r 1− r1

y q20

forq2≥0,

max (x,t) ∈[0,a] ×[0,r] b2(x,t)

a

1 +a r 1− r1

a q2 forq2< 0.

(2.9)

Setc =max{ C1,C2}, then we have

x q1 W t −x r1 W x



x −

a

0b1(x,t)Z(x,t)dx

≥

⎧

⎪

⎪

⎪

⎪

ηe ct r



1− r1



1− r1



x2− r 1 − a

1 +a r2 − r2

max (x,t) ∈[0,a] ×[0,r] b1(x,t)



forx ≤ x0,

ηe ct cx q1 − a

1 +a r2 − r2

max (x,t) ∈[0,a] ×[0,r] b1(x,t)



forx > x0,

≥0,

x q2Z t −x r2Z x

x −

a

0b2(x,t)W(x,t)dx

≥

⎧

⎪

⎪

⎪

⎪

ηe ct r2 − r2



1− r2



x2− r 2 − a

1 +a r1 − r1 

max (x,t) ∈[0,a] ×[0,r] b2(x,t)



forx ≤ y0,

ηe ct cx q2− a

1 +a r1 − r1 

max (x,t) ∈[0,a] ×[0,r] b2(x,t)



forx > y0,

≥0.

(2.10)

Case 3 When

max (x,t) ∈[0,a] ×[0,r] b1(x,t) ≤



r1 − r1



1− r1



a3− r1

1 +a r2 − r2 ,

max (x,t) ∈[0,a] ×[0,r] b2(x,t) >



r2 − r2



1− r2



a3− r 2 

1 +a r1 − r1,

(2.11)

or

max (x,t) ∈[0,a] ×[0,r] b2(x,t) ≤



r2 − r2



1− r2



a3− r2

1 +a r1 − r1,

max (x,t) ∈[0,a] ×[0,r] b1(x,t) >



r1 − r1 

1− r1



a3− r 1 

1 +a r2 − r2.

(2.12)

Combining Cases1with2, it is easy to prove

x q1 W t −x r1 W x

x −

a

0b1(x,t)Z(x,t)dx ≥0,

x q2 Z t −x r2 Z x

x −

a

0 b2(x,t)W(x,t)dx ≥0,

(2.13)

so we omit the proof here

From the above three cases, we know thatW(x,t) > 0, Z(x,t) > 0 on [0,a] ×[0,r].

Lettingη →0+, we haveu(x,t) ≥0,v(x,t) ≥0 on [0,a] ×[0,r] By the arbitrariness of

Obviously, (u,v) =(0, 0) is a subsolution of (1.1), we need to construct a supersolu-tion

Lemma 2.2 There exists a positive constant t0 (t0< T) such that the problem ( 1.1 ) has a supersolution (h1(x,t),h2(x,t)) ∈(C(Ω t0)∩ C2,1(Ωt0))2.

Proof Let

ψ(x) =

x a

1− r1

1− x

a

+

x a

(1− r1)/2

1− x

a

1/2

,

ϕ(x) =

x a

1− r2

1− x

a

+

x a

(1− r2)/2

1− x

a

1/2

,

(2.14)

and letK0be a positive constant such thatK0ψ(x) ≥ u0(x), K0ϕ(x) ≥ v0(x).

Denote the positive constant 1

0[s1− r1(1− s) + s(1− r1)/2(1− s)1/2]p2 ds by b20 and

 1

0[s1− r2(1− s) + s(1− r2 )/2(1− s)1/2]p1ds by b10 Let K10 ∈(0, (1− r1)/(2 − r1)), K20 ∈

(0, (1− r2)/(2 − r2)) be positive constants such that

K10≤2p1+1a3− r1 b10K0p1 −1

−2/(1 − r1)

,

K20≤2p2+1a3− r2 b20K p2 −1−2/(1 − r2)

.

(2.15)

Let (K1(t), K2(t)) be the positive solution of the following initial value problem:

K1(t) =

⎧

⎪

⎪

⎪

⎪

⎪

⎪

b10K2p1(t)

a q1 −1K10q1



K10



1− K10

 1− r1

+K101/2



1− K10

 (1− r1)/2, q1≥0,

b10K2p1(t)

a q1−1 

1− K10 q1 

K10 

1− K10  1− r1

+K101/2



1− K10  (1− r1 )/2, q1< 0,

K1(0)= K0,

K2(t) =

⎧

⎪

⎪

⎪

⎪

⎪

⎪

b20K1p2(t)

a q2 −1K20q2



K20



1− K20

 1− r2

+K201/2



1− K20

 (1− r2)/2, q2≥0,

b20K1p2(t)

a q2−1 

1− K20

q2 

K20



1− K20

 1− r2

+K201/2



1− K20

 (1− r2 )/2, q2< 0,

K2(0)= K0.

(2.16)

SinceK1(t), K2(t) are increasing functions, we can choose t0> 0 such that K1(t) ≤2K0,

K2(t) ≤2K0for allt ∈[0,t0] Seth1(x,t) = K1(t)ψ(x), h2(x,t) = K2(t)ϕ(x), then h1(x,t) ≥

0,h2(x,t) ≥0 onΩt0 We would like to show that (h1(x,t),h2(x,t)) is a supersolution of

(1.1) inΩt0 To do this, let us construct two functionsJ1,J2by

J1= x q1 h1t −x r1 h1x



x −

a

0h2p1 dx, (x,t) ∈Ωt0,

J2= x q2 h2t −x r2 h2x



x −

a

0h1p2 dx, (x,t) ∈Ωt0

(2.17)

Then,

J1= x q1h1t −x r1h1x



x −

a

0h2p1 dx

= x q1K1 ψ(x)+



2− r1

a2− r1 +



1− r1

 2

4 x(r1−3)/2(a − x)1/2+1

2x(r1−1)/2(a − x) −1/2

+1

4x(1+r1)/2(a − x) −3/2



a1− r1/2



K1(t) − ab10K2p1(t)

≥ x q1K1(t)ψ(x) + x(r1−1)/2(a − x) −1/2 K1(t)

2a1− r1/2 − ab10K2p1(t),

J2≥ x q2 K2(t)ϕ(x) + x(r2 −1)/2(a − x) −1/2 K2(t)

2a1− r2/2 − ab20K p2

1 (t).

(2.18)

For (x,t) ∈(0,aK10)×(0,t0]∪(a(1 − K10),a) ×(0,t0], by (2.15), we have

J1≥ x(r1 −1)/2(a − x) −1/2 K1(t)

2a1− r1/2 − ab10K2p1(t)

≥



K(r1−1)/2

10

2a2− r1



K1(t) − ab10K2p1



t0



≥



K(r1−1)/2

10

2a2− r1



K0− ab10



2K0

p1

≥0.

(2.19)

For (x,t) ∈(0,aK20)×(0,t0]∪(a(1 − K20),a) ×(0,t0], by (2.15), we have

J2≥



K20(r2 −1)/2

2a2− r2



K0− ab20



2K0

p2

For (x,t) ∈[aK10,a(1 − K10)]×(0,t0] by (2.16), we have

J1≥ x q1 K1(t)ψ(x) − ab10K p1

2 (t)

≥

⎧

⎪

⎪

a q1 K10q1 K1(t)

K10



1− K10

 1− r1+K101/2



1− K10

 (1− r1)/2

− ab10K2p1(t), q1≥0,

a q1

1− K10

q1

K1(t)

K10



1− K10

 1− r1+K101/2



1− K10

 (1− r1)/2

− ab10K2p1(t), q1<0,

≥0,

(2.21) For (x,t) ∈[aK20,a(1 − K20)]×(0,t0] by (2.16), we have

J2≥ x q2 K2(t)ϕ(x) − ab20K1p2(t)

≥

⎧

⎪

⎪

a q2 K20q2 K2(t)

K20



1− K20

 1− r2+K201/2



1− K20

 (1− r2)/2

− ab20K1p2(t), q2≥0,

a q2

1− K20

q2

K2(t)

K20



1− K20

 1− r2+K201/2



1− K20

 (1− r1)/2

− ab20K1p2(t), q2<0,

≥0.

(2.22)

Thus,J1(x,t) ≥0,J2(x,t) ≥0 inΩt0 It follows fromh1(0,t) = h1(a,t) = h2(0,t) = h2(a,t) =0 andh1(x,0) = K0ψ(x) ≥ u0(x), h2(x,0) = K0ϕ(x) ≥ v0(x) that (h1(x,t),h2(x,t)) is a

super-solution of (1.1) inΩt0 The proof ofLemma 2.2is complete 

To show the existence of the classical solution (u(x,t),v(x,t)) of (1.1), let us intro-duce a cutoff function ρ(x) By Dunford and Schwartz [8, page 1640], there exists a

nondecreasingρ(x) ∈ C3(R) such that ρ(x) =0 ifx ≤0 andρ(x) =1 ifx ≥1 Let 0<

δ < min {(1− r1)/(2 − r1)a,(1 − r2)/(2 − r2)a },

ρ δ(x) =

⎧

⎪

⎪

⎪

⎪

ρ

x

δ −1

, δ < x < 2δ,

(2.23)

andu0δ(x) = ρ δ(x)u0(x), v0δ(x) = ρ δ(x)v0(x) We note that

∂u0δ(x)

⎧

⎪

⎪

⎪

⎪

δ2ρ 

x

δ −1

u0(x), δ < x < 2δ,

∂v0δ(x)

⎧

⎪

⎪

⎪

⎪

δ2ρ 

x

δ −1

v0(x), δ < x < 2δ,

(2.24)

Sinceρ is nondecreasing, we have ∂u0δ(x)/∂δ ≤0,∂v0δ(x)/∂δ ≤0 From 0≤ ρ(x) ≤1, we haveu0(x) ≥ u0δ(x), v0(x) ≥ v0δ(x) and lim δ →0u0δ(x) = u0(x), lim δ →0v0δ(x) = v0(x).

LetD δ =(δ,a), let w δ = D δ ×(0,t0], letD δandw δbe their respective closures, and let

S δ = { δ,a } ×(0,t0] We consider the following regularized problem:

x q1 u δt −x r1 u δx

x =

a

δ v δ p1 dx, (x,t) ∈ w δ,

x q2 v δt −x r2 v δx



x =

a

δ u p2

δ dx, (x,t) ∈ w δ,

u δ(δ,t) = u δ(a,t) = v δ(δ,t) = v δ(a,t) =0, t ∈0,t0



,

u δ(x,0) = u0δ(x), v δ(x,0) = v0δ(x), x ∈ D δ

(2.25)

By using Schauder’s fixed point theorem, we have the following

Theorem 2.3 The problem ( 2.25 ) admits a unique nonnegative solution (u δ,v δ)∈

(C2+α,1+α/2(w δ))2 Moreover, 0 ≤ u δ ≤ h1(x,t), 0 ≤ v δ ≤ h2(x,t), (x,t) ∈ w δ , where h1(x,t),

h2(x,t) are given by Lemma 2.2

Proof By the proof ofLemma 2.1, we know that there exists at most one nonnegative solution (u δ,v δ) To prove existence, we use Schauder’s fixed point theorem

Let

X1=v1∈ C α,α/2

w δ

: 0≤ v1(x,t) ≤ h2(x,t), (x,t) ∈ w δ

,

X2=u1∈ C α,α/2

w δ) : 0≤ u1(x,t) ≤ h1(x,t), (x,t) ∈ w δ

Obviously,X1,X2are closed convex subsets of Banach spaceC α,α/2(w δ) In order to get the conclusion, we have to define another set:X = X1× X2 Obviously (C α,α/2(w δ))2is a Banach space with the norm

v1,u1

α,α/2 =v1

α,α/2+u1

α,α/2, for any

v1,u1



∈C α,α/2

w δ 2 , (2.27)

andX is a closed convex subset of Banach space (C α,α/2(w δ))2 For anyv1∈ X1,u1∈ X2, let us consider the following linearized uniformly parabolic problem:

x q1 W δt −x r1 W δx

x =

a

δ v1p1 dx, (x,t) ∈ w δ,

x q2 Z δt −x r2 Z δx

x =

a

δ u1p2 dx, (x,t) ∈ w δ,

W δ(δ,t) = W δ(a,t) = Z δ(δ,t) = Z δ(a,t) =0, t ∈0,t0



,

W δ(x,0) = u0δ(x), Z δ(x,0) = v0δ(x), x ∈[δ,a].

(2.28)

It is easy to see that (W(x,t),Z(x,t)) =(0, 0) and (W(x,t),Z(x,t)) =(h1(x,t),h2(x,t))

are subsolution and supersolution of problem (2.28) We also note thatx − q1+r1,x − q1 −1+r1,

x − q1, x − q2+r2, x − q2 −1+r2, x − q2 ∈ C α,α/2(w δ), and x − q1a

δ v1p1 dx, x − q2a

δ u1p2 dx ∈

C α,α/2(w δ),u0δ(x),v0δ(x) ∈ C2+α(D δ) It follows from Theorem 4.2.2 of Laddle et al [11, page 143] that the problem (2.28) has a unique solution (W δ(x,t;v1,u1),Z δ(x,t;v1,u1))∈

(C2+α,1+α/2(w δ))2, which satisfies 0≤ W δ(x,t;v1,u1)≤ h1(x,t), 0 ≤ Z δ(x,t;v1,u1)≤ h2(x,t).

Thus, we can define a mappingY from X into (C2+α,1+α/2(w δ)2, such that

Y

v1(x,t),u1(x,t)

=W δ

x,t;v1,u1



,Z δ

x,t;v1,u1



where (W δ(x,t;v1,u1),Z δ(x,t;v1,u1)) denotes the unique solution of (2.28) correspond-ing to (v1(x,t),u1(x,t)) ∈ X To use Schauder’s fixed point theorem, we need to verify the

fact thatY maps X into itself is continuous and compact.

In fact,YX ⊂ X and the embedding operator form Banach space (C2+α,1+α/2(w δ))2to the Banach space (C α,α/2(w δ))2is compact ThereforeY is compact To show Y is

contin-uous inX1let us consider a sequence{ v1n(x,t) }which converges tov1(x,t) uniformly in

the norm · α,α/2 We know thatv1(x,t) ∈ X1 Analogously, inX2we consider a sequence

{ u1n(x,t) }which converges tou1(x,t) uniformly in the norm · α,α/2andu1(x,t) ∈ X2.

So we get a sequence{(v1n(x,t),u1n(x,t)) } ⊂ X, which converges to (v1(x,t),u1(x,t))

uni-formly in the norm(·,·) α,α/2and (v1(x,t),u1(x,t)) ∈ X Let (W δ n(x,t),Z δ n(x,t)) and

(W δ(x,t),Z δ(x,t)) be the solution of problem (2.28) corresponding to (v1n(x,t),u1n(x,t))

and (v1(x,t),u1(x,t)), respectively Without loss of generality, let us assume that

v1n(x,t)

α,α/2 ≤v1(x,t)

α,α/2+ 1, for anyn ≥1,

u1

n(x,t)

α,α/2 ≤u1(x,t)

fact thatY maps X into itself is continuous and compact.

In fact,YX ⊂ X and the embedding operator form Banach space (C2+α,1+α/2(w... ×(0,t0], letD δand< i>w δbe their respective closures, and let

S δ = { δ,a } ×(0,t0]... δ 2 , (2.27)

and< i>X is a closed convex subset of Banach space (C α,α/2(w δ))2 For anyv1∈