Electronic Journal of Differential Equations, Vol. 2007(2007), No. 92, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
CONTINUOUS DEPENDENCE FOR THE BRINKMAN
EQUATIONS OF FLOW IN DOUBLE-DIFFUSIVE CONVECTION
HONGLIANG TU, CHANGHAO LIN
Abstract. This paper concerns the structural stability for convective motion
in a fluid-saturated porous medium under the Brinkman scheme. Continuous
dependence for the solutions on the gravity coefficients and the Soret coefficient
are proved. First of all, an a priori bound in L2 norm is derived whereby
we show the solution depends continuously in L2 norm on changes in the
gravity coefficients and the Soret coefficient. This estimate also implies that
the solutions decay exponentially.
1. Introduction
Within the context of fluid flow in porous media, or simply within theory of fluid
flow, there has been substantial recent interest in deriving stability estimates where
changes in coefficient are allowed, or even the model (the equations themselves)
changes. This type of stability has earned the name structural stability, and is
different from continuous dependence on the initial data. The concept of structural
stability in which the study of continuous dependence (or stability) is on changes in
the model itself rather than the initial data. Thus structural stability constitutes
a class of stability problem every bit important. Structural stability is focus of
attention now, and for the relevant results, the reader is referred to [1, 2, 3, 5, 6, 7,
8, 9, 10].
The Brinkman model is believed accurate when the flow velocity is too large
for Darcy’s law to be valid, and additionally the porosity is not too small. In
this article, we are concerned with structural stability for the Brinkman equations
modeling the double diffusive convection. The temperature field is non-constant
and a solute is also diffused throughout the porous body.
The Brinkman equations governing the flow of fluid in double-diffusive convection
are
ui,t = ν∆ui − aui − p,i + gi T + hi C
T,t + ui T,i = ∆T
(1.1)
C,t + ui C,i = ∆C + ρ∆T in Ω × t > 0
ui,i = 0
2000 Mathematics Subject Classification. 35B30, 35K55, 35Q35.
Key words and phrases. Continuous dependence; structural stability;
gravity coefficients; Soret coefficient; Brinkman equations.
c
2007
Texas State University - San Marcos.
Submitted April 9, 2007. Published June 16, 2007.
1
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H. TU, C. LIN
EJDE-2007/92
where ui , T , C and p represent fluid velocity, temperature, salt concentration and
pressure, respectively. The quantities gi (x) and hi (x) are gravity vector terms, and
positive constants ρ, a are known the Soret coefficient and the Darcy coefficient,
respectively. Ω is a bounded domain of of R3 , and with sufficiently smooth boundary
∂Ω. ∆ is the Laplace operator, k · k and hu, vi denote the norm and inner product
on L2 (Ω) .
Associated with (1.1), we impose the boundary data
ui = 0;
T = 0; C = 0
on ∂Ω
(1.2)
and the initial data
ui (x, 0) = zi (x); T (x, 0) = T0 (x); C(x, 0) = C0 (x); x ∈ Ω
(1.3)
In (1.1) and in the equations throughout, a comma is used to denote partial differ∂ui
i
entiation: For example ui,i denotes ∂u
∂xi and ui,t denotes ∂t , and we employ the
convention of summing over repeated indices from 1 to 3 .
2. A priori bounds
Multiplying (1.1)2 by T and integrating over Ω, we have
1 d
kT k2 = −k∇T k2
(2.1)
2 dt
Multiplying (1.1)3 by C and integrating over Ω, then using arithmetic-geometric
mean inequality, we obtain
1
ρ2
1 d
kCk2 + k∇Ck2 ≤ k∇T k2
(2.2)
2 dt
2
2
Multiplying (1.1)1 by ui and integrating over Ω, furthermore using CauchySchwarz, arithmetic-geometric mean, then using Poincare’s inequality, we find
a
1
1 d
kuk2 + νk∇uk2 + kuk2 ≤ [g 2 kT k2 + h2 kCk2 ]
2 dt
2
a
(2.3)
1 2
[g k∇T k2 + h2 k∇Ck2 ]
≤
aλ1
where
g 2 = max gi gi ; h2 = max hi hi .
Ω
Ω
and λ1 is the first eigenvalue of the problem
∆φ + λφ = 0
in Ω
φ = 0 on ∂Ω
Multiplying (1.1)1 by ui,t and integrating over Ω , furthermore using CauchySchwarz and arithmetic-geometric mean inequality, then using Poincare’s Poincare’s
inequality, we find
a d
ν d
1
kuk2 +
k∇uk2 ≤ [g 2 kT k2 + h2 kCk2 ]
2 dt
2 dt
2
(2.4)
1 2
≤
[g k∇T k2 + h2 k∇Ck2 ]
2λ1
Multiplying (2.1) by Γ11 , (2.2) by Γ12 and (2.3) by Γ13 , then adding all results
to (2.4) leads to
d
Q1 (t) + G(t) ≤ 0
(2.5)
dt
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CONTINUOUS DEPENDENCE
3
where Γ1i , (i = 1, 2, 3) are positive constants at our disposal,
Γ11
Γ12
(Γ13 + a)
ν
kT k2 +
kCk2 +
kuk2 + k∇uk2
2
2
2
2
(2.6)
(2Γ11 − ρ2 Γ12 )
1
1 Γ13 )
aΓ13
− g2 ( +
]k∇T k2 +
kuk2
2
λ1
2
a
2
Γ12
1 2 1 Γ13 )
+[
−
h ( +
]k∇Ck2 + Γ13 νk∇uk2
2
λ1 1
2
a
(2.7)
Q1 (t) =
and
G(t) = [
We can select Γ1i , (i = 1, 2, 3) to secure that all the all the coefficients of (2.7)
are positive, such as
Γ13 =
a
;
2
Γ12 =
4h2
;
λ1
Γ11 >
(2h2 ρ2 + g 2 )
.
λ1
Thus, with the help of Poincare’s inequality, we can show that
1 Γ13 )
(2Γ11 − ρ2 Γ12 )
− g2 ( +
]kT k2 + Γ13 νk∇uk2
2
2
a
2λ1 Γ12
1 Γ13
a
+[
− h2 ( +
]kCk2 + Γ13 kuk2
2
2
a
2
G(t) ≥ [λ1
(2.8)
We can easily show that
κ1 Q1 (t) ≤ G(t)
(2.9)
where κ1 is a positive constant represented by
κ1 = min
1
2Γ13 )
],
[λ1 (2Γ11 − ρ2 Γ12 ) − g 2 (1 +
Γ11
a
1
aΓ13
2Γ13 )
],
, 2Γ13
[Γ12 λ1 − h2 (1 +
Γ12
a
Γ13 + a
Thus, from (2.5), we can derive that
d
Q1 (t) + κ1 Q1 (t) ≤ 0
dt
(2.10)
Q1 (t) ≤ Q1 (0)e−κ1 t
(2.11)
Furthermore, we obtain
Therefore, recalling the definition of Q1 (t) and combining (2.5), we can obtain
Z t
2
2
2
2
kT k ≤ M1 ; kCk ≤ M1 ; kuk ≤ M1 ; k∇uk ≤ M1 ;
kuk2 dη ≤ M1 ;
0
Z t
Z t
Z t
k∇uk2 dη ≤ M1 ;
k∇T k2 dη ≤ M1 ;
k∇Ck2 dη ≤ M1 .
0
0
0
(2.12)
where M1 is a generic positive constant depending on the coefficients of (1.1) and
the initial data terms in (1.3).
It also follows from inequality (2.11) that k∇uk2 , kT k2 and kCk2 decay exponentially as t tends to ∞.
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H. TU, C. LIN
EJDE-2007/92
3. Continuous dependence on the gravity coefficients
Let (ui , T, C, P ) and (u∗i , T ∗ , C ∗ , P ∗ ) be solutions to (1.1) with the same boundary and initial data (1.2), (1.3), but with different gravity coefficients (gi , hi ) and
(gi∗ , h∗i ), respectively.
Namely,
ui,t = ν∆ui − aui − p,i + gi T + hi C
T,t + ui T,i = ∆T
C,t + ui C,i = ∆C + ρ∆T in Ω × t > 0
ui,i = 0
and
u∗i,t = ν∆u∗i − au∗i − p∗,i + gi∗ T ∗ + h∗i C ∗
T,t∗ + u∗i T,i∗ = ∆T ∗
C,t∗ + u∗i C,i∗ = ∆C ∗ + ρ∆T ∗
in Ω × t > 0
u∗i,i = 0 .
Now set
wi = ui − u∗i ,
Π = p − p∗ , S = T − T ∗ ,
γi = gi − gi∗ ,
Σ = C − C∗
µi = hi − h∗i
(3.1)
Clearly, (wi , Π, S, Σ) satisfies the equations
wi,t = ν∆wi − awi − Π,i + γi T + gi∗ S + µi C + h∗i Σ
S,t + wi T,i + u∗i S,i = ∆S
Σ,t + wi C,i + u∗i Σ,i = ∆Σ + ρ∆S
wi,i = 0
with the boundary-initial data
wi = S = Σ = 0
(3.2)
in Ω × {t > 0}
on ∂Ω × {t > 0}
wi (x, 0) = S(x, 0) = Σ(x, 0) = 0
(3.3)
in Ω
Multiplying (3.2)1 by wi and integrating over Ω, then using Cauchy-Schwarz’s
inequality, and the arithmetic-geometric mean inequality, we obtain
a
1 d
2
νk∇wk2 + kwk2 +
kwk2 ≤ [(g ∗ )2 kSk2 +(h∗ )2 kΣk2 +µ2 kCk2 +γ 2 kT k2 ], (3.4)
2
2 dt
a
where
γ 2 = max γi γi ;
Ω
µ2 = max ui ;
Ω
(g ∗ )2 = max gi∗ gi∗ ;
Ω
(h∗ )2 = max h∗i h∗i .
Ω
Multiplying (3.2)1 by wi,t and integrating over Ω, then using Cauchy-Schwarz
inequality, and the arithmetic-geometric mean inequality, we have
1 d
a d
ν k∇wk2 +
kwk2 ≤ (g ∗ )2 kSk2 + (h∗ )2 kΣk2 + µ2 kCk2 + γ 2 kT k2
(3.5)
2 dt
2 dt
Multiplying (3.2)2 by S and integrating over Ω, then using Cauchy-Schwarz, the
arithmetic-geometric mean inequality, the Sobolev inequality ,which holds for all
ϕ ∈ C01 (ϕ),
kϕk4 ≤ Λkϕ,i k
4
where k · k4 is the norm on L (Ω).
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CONTINUOUS DEPENDENCE
5
we derive
d
kSk2 = −2k∇Sk2 + 2hT,i , wi Si;
dt
therefore
d
kSk2 ≤ −2k∇Sk2 + 2k∇T k · kwk4 kSk4
dt
1
≤ −2k∇Sk2 + 2Λ 2 k∇T k · k∇wk · k∇Sk
Λ
≤ −2k∇Sk2 + 2α11 k∇Sk2 +
k∇wk2 k · k∇T k2
2α11
(3.6)
where α11 is a positive constant at our disposal.
Using similar method as in (3.6), from equation (3.2)3 , we find that
Λ
ρ2
d
kΣk2 ≤ (−2 + α12 + α13 )k∇Σk2 +
k∇wk2 · k∇Ck2 +
k∇Sk2
dt
α13
α12
(3.7)
where α1i , i = 2, 3 are positive constants at our disposal.
Multiplying (3.6) by Γ21 and adding (3.7) leads to
d
(kΣk2 + Γ21 kSk2 )
dt
Γ21
Λ
2
≤ k∇wk2 (
k∇T k2 +
k∇Ck2 )
2
α11
α13
ρ2
+ 2[(α11 − 1) +
]Γ21 k∇Sk2 + (−2 + α12 + α13 )k∇Σk2
2α12 Γ21
(3.8)
where Γ21 is a positive constant at our disposal.
In (3.8), we choose
α11 =
1
;
4
α12 = α13 =
1
;
2
Γ21 = 4ρ2 .
it follows that
d
(kΣk2 + 4ρ2 kSk2 ) + k∇Σk2 + 4ρ2 k∇Sk2 ≤ 2Λk∇wk2 (4ρ2 k∇T k2 + k∇Ck2 ) (3.9)
dt
Multiplying (3.4) by Γ22 and (3.5) by Γ23 , then adding all the results to (3.9),
then using Poincare’s inequality, we have
d Γ23
(Γ22 + aΓ23 )
νk∇wk2 +
kwk2 + (kΣk2 + 4ρ2 kSk2 )
dt 2
2
2Γ22
∗ 2
+ λ1 − (
+ Γ23 )(g ) kΣk2
a
(3.10)
2
2Γ22
aΓ22
+ 4ρ λ1 − (
+ Γ23 )(h∗ )2 kSk2 + Γ22 νk∇wk2 +
kwk2
a
2
2
2
Γ22
2
2
2
2
2
≤ wk 4ρ k∇T k + kC,i k + (
+ Γ23 ) µ kCk + γ kT k2
a
where Γ2i , i = 2, 3 are positive constants at our disposal.
We can choose Γ22 and Γ23 sufficient small to make sure that all the coefficients
in (3.10) are positive, such as
Γ22 = min{
aλ1 ρ2 aλ1
,
};
8(g ∗ )2 2(h∗ )2
Γ23 = min{
λ1
ρ2 λ 1
,
}.
4(g ∗ )2 (h∗ )2
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H. TU, C. LIN
EJDE-2007/92
Let
Q2 (t) = {
Γ23
(Γ22 + aΓ23 )
νk∇wk2 +
kwk2 + kΣk2 + 4ρ2 kSk2 }
2
2
and
κ2 = min
2Γ22 ν
aΓ22
2Γ22
;
; (λ1 − (
+ Γ23 )(g ∗ )2 );
Γ23 ν (Γ22 + aΓ23 )
a
1
2Γ22
[4ρ2 λ1 − (
+ Γ23 )(h∗ )2 ]
2
4ρ
a
(3.11)
It follows from (3.10) that
Γ22
d
Q2 (t)+κ2 Q2 (t) ≤ (
+Γ23 )[µ2 kCk2 +γ 2 kT k2 ]+2Λk∇wk2 [4ρ2 k∇T k2 +k∇Ck2 ]
dt
a
(3.12)
It is easy to show that
2Λk∇wk2 [4ρ2 k∇T k2 + k∇Ck2 ] ≤ Q2 (t)f2 (t)
(3.13)
where
f2 (t) = τ2 (k∇T k2 + k∇Ck2 ),
τ2 =
4Λ
(4ρ2 + 1).
Γ23 ν
Thus, combining (3.12) and (3.13), we obtain
d
Q2 (t) + κ2 Q2 (t) ≤ M2 (µ2 + γ 2 ) + f2 (t)Q2 (t)
dt
(3.14)
where M2 = 2( Γa22 + Γ23 )M1 .
Now, multiplying both sides of (3.14) by eκ2 t and integrating from 0 and t, we
get
Z t
M2 2
(µ + γ 2 ) +
Q2 (t) ≤
f2 (η)Q2 (η)dη
(3.15)
κ2
0
Hence, applying Gronwall’s lemma and using (2.12), we obtain
Q2 (t) ≤
M2 R t f2 (η)dη
M2 2τ2 M1
e0
· (µ2 + γ 2 ) ≤
e
· (µ2 + γ 2 ), ∀t > 0
κ2
κ2
(3.16)
Consequently, from inequality (3.16), we can see that Q2 (t) → 0, as gi → gi∗ and
hi → h∗i . Recalling the definition of Q2 (t), so (wi , Π, S, Σ) → 0 and (ui , T, C) →
(u∗i , T ∗ , C ∗ ). So, continuous dependence on the gravity coefficients is proved.
4. Continuous dependence on the Soret coefficient
Let (ui , T, C, P ) and (u∗i , T ∗ , C ∗ , P ∗ ) be solutions to (1.1 ), with the same boundary and initial data (1.2), (1.3), but with different Soret coefficient ρ and ρ∗ , respectively:
ui,t = ν∆ui − aui − p,i + gi T + hi C
T,t + ui T,i = ∆T
C,t + ui C,i = ∆C + ρ∆T in Ω × t > 0
ui,i = 0
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CONTINUOUS DEPENDENCE
7
and
u∗i,t = ν∆u∗i − au∗i − p∗,i + gi T ∗ + hi C ∗
T,t∗ + u∗i T,i∗ = ∆T ∗
C,t∗ + u∗i C,i∗ = ∆C ∗ + ρ∗ ∆T ∗
u∗i,i
in Ω × t > 0
=0
Now set
wi = ui − u∗i ,
Π = p − p∗ ,
S = T − T ∗,
Σ = C − C∗
(4.1)
Clearly, (wi , Π, S, Σ) satisfies the equations
wi,t = ν∆wi − awi − Π,i + +gi S + +hi Σ
S,t + wi T,i + u∗i S,i = ∆S
Σ,t + wi C,i + u∗i Σ,i = ∆Σ + (ρ − ρ∗ )∆T + ρ∗ ∆S
wi,i = 0
in Ω × {t > 0}
(4.2)
with the boundary-initial data
wi = S = Σ = 0
on ∂Ω × {t > 0}
wi (x, 0) = S(x, 0) = Σ(x, 0) = 0
in Ω
(4.3)
Multiplying (4.2)1 by wi and integrating over Ω, furthermore using the CauchySchwarz’s inequality, and the arithmetic-geometric mean inequality, we have
1 d
1
a
kwk2 ≤ [g 2 kSk2 + h2 kΣk2 ]
νk∇wk2 + kwk2 +
2
2 dt
a
(4.4)
where
g 2 = max gi gi , h2 = max hi hi .
Ω
Ω
Multiplying (4.2)1 by wi,t and integrating over Ω, then using the Cauchy-Schwarz
inequality, and the arithmetic-geometric mean inequality, we get
1 d
a d
1
ν k∇wk2 +
kwk2 ≤ [g 2 kSk2 + h2 kΣk2 ]
2 dt
2 dt
2
(4.5)
Multiplying (4.2)2 by S and integrating over Ω, then using Cauchy-Schwarz,
the arithmetic-geometric mean inequality, and using the Sobolev inequality, we can
derive
d
Λ
kSk2 dη ≤ −2k∇Sk2 + 2α21 k∇Sk2 +
k∇wk2 k∇T k2
(4.6)
dt
2α21
for α21 is a positive constant at our disposal.
Using similar method as in (4.6), from equation (4.2)3 , we can find that
d
Λ
kΣk2 ≤ (−2 + α22 + α23 + α24 )k∇Σk2 +
k∇wk2 k∇Ck2
dt
α23
ρ∗2
(ρ − ρ∗ )2
+
k∇Sk2 +
k∇T k2
α22
α24
for α2i , (i = 2, 3, 4) are positive constants at our disposal.
(4.7)
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H. TU, C. LIN
EJDE-2007/92
Multiplying (4.6) by Γ31 and adding (4.7), leads to
d
(kΣk2 + Γ31 kSk2 )
dt
ρ∗2 Γ31 k∇Sk2
2α22 Γ31
Γ31
(ρ − ρ∗ )2
Λ
2
+
k∇T k2 + k∇wk2
k∇T k2 +
k∇Ck2
α24
2
α21
α23
≤ (−2 + α22 + α23 + α24 )k∇Σk2 + 2 (α21 − 1) +
(4.8)
where Γ31 is a positive constant at our disposal.
In (4.8), we choose α21 = 1/2, α22 = 1, α23 = α24 = 1/4, It follows that
d
(kΣk2 + Γ31 kSk2 ) + k∇Σk2 + (Γ31 − ρ∗2 )k∇Sk2
dt
≤ Λk∇wk2 (Γ31 k∇T k2 + 4k∇Ck2 ) + 4(ρ − ρ∗ )2 k∇T k2
(4.9)
Multiplying (4.4) by Γ32 and (4.5) by Γ33 , furthermore adding all the results to
(4.9), we get
d Γ33
(Γ32 + aΓ33 )
[
νk∇wk2 +
kwk2 + (kΣk2 + Γ4 kSk2 )]
dt 2
2
aΓ32
+ Γ32 νk∇wk2 +
kwk2 + k∇Σk2 + (Γ31 − ρ∗2 )k∇Sk2
2
2Γ32
≤(
+ Γ33 )[g 2 kΣk2 + h2 kSk2 ] + 2Λk∇wk2 [Γ31 k∇T k2
a
+ 4k∇Ck2 ] + 4(ρ − ρ∗ )2 k∇T k2
(4.10)
where Γ3i , (i = 2, 3) are positive constants at our disposal.
We can choose Γ31 to make sure that Γ31 > ρ∗2 , therefore, by the Poincare’s
inequality from (4.10), we have
(Γ32 + aΓ33 )
d Γ33
νk∇wk2 +
kwk2 + (kΣk2 + Γ31 kSk2 )
dt 2
2
aΓ32
Γ32
+ Γ33 )h2 kSk2 +
kwk2
+ λ1 (Γ31 − ρ∗2 ) − (2
a
2
2Γ32
+ Γ32 νk∇wk2 + λ1 − (
+ Γ33 )g 2 kΣk2
a
≤ Λk∇wk2 [Γ31 k∇T k2 + 4k∇Ck2 ] + 4(ρ − ρ∗ )2 k∇T k2
(4.11)
We can choose Γ32 and Γ33 sufficient small to make sure that all the coefficients
in (4.11) are positive, such as we may choose
Γ32 = min{
aλ1 aλ1 (Γ31 − ρ∗2 )
,
};
8g 2
8h2
Γ33 = min{
λ1 λ1 (Γ31 − ρ∗2 )
,
}.
4g 2
4h2
Let
Q3 (t) = {
Γ33
(Γ32 + aΓ33 )
νk∇wk2 +
kwk2 + kΣk2 + Γ31 kSk2 }
2
2
and
κ3 = min{
2Γ32
aΓ32
2Γ32
;
; (λ1 − (
+ Γ33 )g 2 );
Γ33 (Γ32 + aΓ33 )
a
1
2Γ32
[λ1 (Γ31 − ρ∗2 ) − (
+ Γ33 )h2 ]}
Γ31
a
(4.12)
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CONTINUOUS DEPENDENCE
Then, it follows from (4.11) that
d
Q3 (t) + κ3 Q3 (t) ≤ Λkwk2 [Γ31 k∇T k2 + 4kC,i k2 ] + 4(ρ − ρ∗ )2 k∇T k2
dt
Also, it can be easily shown that
Λk∇wk2 [Γ31 k∇T k2 + 4k∇Ck2 ] ≤ f3 (t)Q3 (t)
where
f3 (t) = τ3 (k∇T k2 + k∇Ck2 ), τ3 =
9
(4.13)
(4.14)
2Λ
(Γ31 + 4).
Γ33 ν
Thus, combing (4.13) and (4.14), we obtain
d
Q3 (t) + κ3 Q3 (t) ≤ 4(ρ − ρ∗ )2 k∇T k2 + f3 (t)Q3 (t)
(4.15)
dt
Now, multiplying both sides of (4.15) by eκ3 t and integrating over [0, t], we obtain
Q3 (t) ≤
M3
(ρ − ρ∗ )2 +
κ3
Z
t
f3 (η)Q3 (η)dη
(4.16)
0
where M3 = 4M1 .
Hence, applying Gronwall’s lemma and using (2.12), we obtain
M3 2τ3 M1
M3 R t f3 (η)dη
· (ρ − ρ∗ )2 ≤
Q3 (t) ≤
e0
e
· (ρ − ρ∗ )2 , ∀t > 0
(4.17)
κ3
κ3
As a result, from inequality (4.17), we can see that Q3 (t) → 0, as ρ → ρ∗ .
Recalling the definition of Q3 (t), so (wi , Π, S, Σ) → 0 and (ui , T, C) → (u∗i , T ∗ , C ∗ ).
Consequently, continuous dependence on the Soret coefficient is proved.
Acknowledgments. The authors would like to express their gratitude to Professor
Yan Liu for his help in writing parts of this article.
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H. TU, C. LIN
EJDE-2007/92
Hongliang Tu
Department of Applied Mathematics, Zhuhai college of Jilin University, 519041, China
E-mail address: tuhongliang00@126.com
Changhao Lin
School of Mathematical Sciences, South China Normal University, 510631, China
E-mail address: linchh@scnu.edu.cn