Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 395209, 22 pages
doi:10.1155/2012/395209
Research Article
Global Solutions to the Spherically
Symmetric Compressible Navier-Stokes Equations
with Density-Dependent Viscosity
Ruxu Lian,1 Lan Huang,1 and Jian Liu2
1
College of Mathematics and Information Science, North China University of Water Resources and
Electric Power, Zhengzhou 450011, China
2
Department of Mathematics, Capital Normal University, Beijing 100048, China
Correspondence should be addressed to Ruxu Lian, ruxu.lian.math@gmail.com
Received 20 December 2011; Revised 21 March 2012; Accepted 31 March 2012
Academic Editor: Nazim I. Mahmudov
Copyright q 2012 Ruxu Lian et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We consider the exterior problem and the initial boundary value problem for the spherically
symmetric isentropic compressible Navier-Stokes equations with density-dependent viscosity
coefficient in this paper. For regular initial density, we show that there exists a unique global strong
solution to the exterior problem or the initial boundary value problem, respectively. In particular,
the strong solution tends to the equilibrium state as t → ∞.
1. Introduction
The isentropic compressible Navier-Stokes equations with density-dependent viscosity
coefficients read as follows:
ρt div ρU 0,
ρU t div ρU ⊗ U ∇P ρ − div μ ρ DU − ∇ λ ρ div U 0,
1.1
where t ∈ 0, ∞ is the time and x ∈ RN , N is the spatial coordinate, ρ > 0 and u denote the
density and velocity, respectively. Pressure function is taken as P ρ ργ with γ > 1, and
DU ∇Ut ∇U
2
1.2
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Journal of Applied Mathematics
is the strain tensor and μρ, λρ are the Lamé viscosity coefficients satisfying
μ ρ > 0,
μ ρ Nλ ρ ≥ 0.
1.3
There are many important progress achieved recently on the compressible NavierStokes equations with density-dependent viscosity coefficient. For instance, the mathematical
derivations are derived in the simulation of flow surface in shallow region 1–4. The
prototype model is the viscous Saint-Venant. The well posedness of solutions to the free
boundary value problem with initial finite mass and the flow density being connected
with the infinite vacuum either continuously or via jump discontinuity is investigated by
many authors, refer to 5–12 and references therein. Mellet and Vasseur showed the global
existence of strong solutions for α ∈ 0, 1/2 13. The qualitative behaviors of global solutions
and dynamical asymptotics of vacuum states are also made, such as the finite time vanishing
of finite vacuum or asymptotical formation of vacuum in large time, the dynamical behaviors
of vacuum boundary, the large time convergence to rarefaction wave with vacuum, and the
stability of shock profile with large shock strength, refer to 14–17 and references therein.
In this present paper, we consider the exterior problem and the initial boundary
value problem for the spherically symmetric isentropic compressible Navier-Stokes equations
with density-dependent viscosity coefficient and focus on the regularities and dynamical
behaviors of global strong solution, and so forth. As γ > 1, we show that the exterior problem
and the initial boundary value problem with regular initial data both admit the unique global
strong solution. In particular, the strong solution tends to the equilibrium state as t → ∞.
The rest of the paper is arranged as follows. In Section 2, the main results about the
dynamical behaviors of global strong solution for compressible Navier-Stokes equations are
stated. Then, the theorems of the exterior problem and the initial boundary value problem
are proved in Sections 3 and 4, respectively.
2. Notations and Main Results
For simplicity, we will take μρ ρ and λρ 0 and DU ∇U in 1.1. The isentropic
compressible Navier-Stokes equations become
ρt div ρU 0,
ρU t div ρU ⊗ U ∇P ρ − div ρ∇U 0.
2.1
Firstly, we consider the exterior problem, namely, the initial data and boundary conditions of
2.1 are imposed as follows:
ρ, U x, 0 ρ0 , U0 x, x ∈ Ω,
on ∂Ω,
lim ρ, u x, t ρ, 0 ,
U 0,
|x| → ∞
t ∈ 0, T ,
2.2
where Ω : R3 /Ωr− , Ωr− is a ball of radius r− centered at the origin in R3 , and ρ > 0 is a
constant.
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We are concerned with the spherically symmetric solutions of system 2.1 in an
spherically symmetric exterior domain Ω. To this end, we denote that
|x| r,
ρx, t ρr, t,
x
Ux, t ur, t ,
r
2.3
which leads to the following system of equations for r > 0,
2ρu
ρt ρu r 0,
r
2
2ρu2 2u
− ρur r − ρ
ρu t ρu ργ 0,
r
r
r r
2.4
with the initial data and boundary conditions
ρ, u r, 0 ρ0 , u0 r, r ∈ r− , ∞,
ur− , t 0,
lim ρ, u r, t ρ, 0 t ∈ 0, T ,
2.5
r → ∞
and the initial data satisfies for some constants ρ and ρ > 0
ρ0 − ρ ∈ L1 r− , ∞ ∩ L2 r− , ∞,
rρ0r ∈ L2 r− , ∞,
inf
x∈r− ,∞
ρ0 > ρ > 0,
r 2 u0 ∈ H 2 r− , ∞.
2.6
Then, we define that
∞
∞
1 γ−1
ρ0 − ργ−1 ργ ρ0−1 − ρ−1 r 2 dr,
γ −1
r−
r−
∞
∞ 2
1
1 γ−1
ρ0 − ργ−1 ργ ρ0−1 − ρ−1 r 2 dr,
E1 :
ρ0 u0 ρ0−1 ρ0 r r 2 dr ρ0
2 r−
γ −1
r−
2.7
E0 :
1
2
ρ0 u20 r 2 dr ρ0
and give the main results as follows.
Theorem 2.1. Let γ > 1. Assume that the initial data satisfies 2.6 and E0 E1 < νr−2 ργ/21/2 , ν
is a positive constant. Then, there exist two positive constants ρ∗ and ρ∗ and a unique global strong
solution ρ, u to the exterior problem 2.4 and 2.5, namely, satisfying
0 < ρ∗ ≤ ρr, t ≤ ρ∗ , r, t ∈ r− , ∞ × 0, T ,
ρ − ρ ∈ L∞ 0, T ; L2 r− , ∞ , ρr ∈ L∞ 0, T ; L2 r− , ∞ ,
u ∈ L∞ 0, T ; L2 r− , ∞ , ur ∈ L∞ 0, T ; L2 r− , ∞ .
Furthermore, the solution tends to the equilibrium state ρ, 0
2.8
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Journal of Applied Mathematics
ρ − ρ, u ·, t
L∞ r− ,∞
−→ 0,
t −→ ∞.
2.9
Then investigates the initial boundary value problem, and the initial data and
boundary conditions of 2.1 are assumed as follow:
ρ, U x, 0 ρ0 , U0 x,
Ux, t 0,
x ∈ Ω,
2.10
x ∈ ∂Ω, t ∈ 0, T .
By 2.3, one considers 2.4 with the initial data and boundary conditions
ρ, u r, 0 ρ0 , u0 r,
ur− , t ur , t 0,
r ∈ r− , r ,
2.11
t ∈ 0, T ,
and the initial data satisfies for some constant ρ > 0
ρ0 ∈ L1 r− , r ∩ W 1,∞ r− , r ,
inf ρ0 > ρ > 0,
x∈r− ,r r 2 u0 ∈ H 2 r− , r .
2.12
Then, can give the main results as follows.
Theorem 2.2. Let γ > 1. Assume that the initial data satisfies 2.12, there exist two positive
constants ρ∗ and ρ∗ and a unique global strong solution ρ, u to the initial boundary value problem
2.4 and 2.11, namely, satisfying
0 < ρ∗ ≤ ρr, t ≤ ρ∗ , r, t ∈ r− , r × 0, T ,
ρ ∈ L∞ 0, T ; L2 r− , r , ρr ∈ L∞ 0, T ; L2 r− , r ,
u ∈ L∞ 0, T ; L2 r− , r , ur ∈ L∞ 0, T ; L2 r− , r .
2.13
Furthermore, the solution ρ, u tends to the equilibrium state exponentially
ρ − ρ,
u ·, t
L∞ r− ,r ≤ C0 e−C1 t ,
where C0 and C1 are positive constants independent of time and ρ
2.14
r
r−
ρr, tr 2 dr.
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5
Remark 2.3. Theorems 2.1 and 2.2 hold for one-dimensional Saint-Venant’s model for shallow
water, that is, γ 2, α 1.
3. Proof of the Exterior Problem
3.1. The A Priori Estimates
It is convenient to make use of the Lagrangian coordinates so as to establish the uniformly a
priori estimates. Take the Lagrange coordinates transform
x
r
ρr, tr 2 dr,
τ t,
3.1
r−
which map r, t ∈ r− , ∞ × R into x, τ ∈ 0, ∞ × R . The relation between Lagrangian
and Eulerian coordinates are satisfied as
∂x
−ρur 2 .
∂t
∂x
ρr 2 ,
∂r
3.2
The exterior problem 2.4 and 2.5 is reformulated to
ρτ ρ2 r 2 u 0,
x
2ρx u
,
r −2 uτ ργ x ρ2 r 2 u
−
x x
r
ρ, u x, 0 ρ0 , u0 x, x ∈ 0, ∞,
u0, t 0,
lim ρ, u ρ, 0 , τ ∈ 0, ∞,
3.3
x → ∞
where the initial data satisfies
ρ0 − ρ ∈ L1 0, ∞ ∩ L2 0, ∞,
r 2 ρ0x ∈ L2 0, ∞,
r 2 ρ0 r 2 u0 ∈ L2 0, ∞,
x
1
inf ρ0 > ρ− > 0,
x∈0,∞
r 2 u0 ∈ L2 0, ∞,
r 2 ρ0
1 2 2 2 r ρ r ρ r u0
∈ L2 0, ∞.
x x
2
r ρ0
3.4
First, we will establish the a-priori estimates for the solution ρ, u to the exterior problem
3.3.
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Journal of Applied Mathematics
Lemma 3.1. Let T > 0. Under the conditions in Theorem 2.1, it holds for any solution ρ, u to the
exterior problem 3.3 that
1
2
∞
0
u dx 2
∞ 0
τ ∞ 0
0
1 γ−1
ρ − ργ−1 ργ ρ−1 − ρ−1 dx
γ −1
2u2
ρ2 u2x r 4 dx ds ≤ E0 ,
r2
3.5
τ ∈ 0, T .
Proof. Multiplying 3.32 by r 2 u and integrating the result with respect to x over 0, ∞,
making use of 3.31 and 3.4, we have
1 d
2 dτ
∞ 2
d ∞
1 γ−1
γ−1
γ
−1
−1
ρ ρ −ρ
dx ρ −ρ
u dx ρ2 r 2 u dx
x
dτ
γ
−
1
0
0
0
3.6
∞ 2
2
ρ ru dx,
∞
2
x
0
integrating 3.6 with respect to τ, we obtain
1
2
τ ∞ 2
2u
1 γ−1
γ−1
γ
−1
−1
2 2 4
ρ ρ −ρ
dx ρ −ρ
u dx ρ ux r dx ds
γ −1
r2
0
0 0
∞ 1 ∞ 2
1 γ−1
ρ0 − ργ−1 ργ ρ0−1 − ρ−1 dx.
u0 dx 2 0
γ −1
0
3.7
∞
0
∞ 2
Lemma 3.1 can be obtained.
Lemma 3.2. Let T > 0. Under the conditions in Theorem 2.1, it holds for any solution ρ, u to the
exterior problem 3.3 that
1
2
∞ 0
∞ 2
1 γ−1
ρ − ργ−1 ργ ρ−1 − ρ−1 dx
u r 2 ρx dx γ −1
0
τ ∞
γ
ργ−1 ρx2 r 4 dx ds ≤ E1 , τ ∈ 0, T ,
0
3.8
0
and there exist two constants 0 < ρ∗ < ρ∗ such that
0 < ρ∗ ≤ ρx, τ ≤ ρ∗ ,
x, τ ∈ 0, ∞ × 0, T .
3.9
Proof. Differentiating 3.31 with respect to x, we have
0.
ρxτ ρ2 r 2 u
x x
3.10
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7
Summing 3.10 and 3.32 , we get
r −2 u ρx
τ
2ρx u
.
ργ x r −2 u −
τ
r
3.11
Note that
r 3 x, τ r−3 3
x
0
1
dz,
ρz, τ
3.12
and so
∂r
1
∂τ r 2
x 1
1 x 2 r u z, tdz ux, τ,
tdz
z,
z
ρ t
r2 0
0
3.13
which together with 3.11 yields
r −2 u ρx
τ
2ρx u
.
ργ x −2r −3 u2 −
r
3.14
Multiplying 3.14 by u r 2 ρx r 2 and integrating the result with respect to x and τ, we have
1
2
∞ 2
∞ 1 γ−1
γ−1
γ
−1
−1
ρ ρ −ρ
dx
ρ −ρ
γ −1
u r ρx dx 0
τ ∞
γ
ργ−1 ρx2 r 4 dx ds
0 0
∞ 2
1 ∞ 1 γ−1
γ−1
γ
−1
2
−1
ρ −ρ
u0 r ρ0x dx ρ ρ0 − ρ
dx.
2 0
γ −1 0
0
2
0
3.15
Let
ϕ ρ :
1 γ−1
ρ − ργ−1 ργ ρ−1 − ρ−1 ,
γ −1
ρ
1/2
ψ ρ :
ϕ η
dη.
3.16
ρ
It follows from 3.6 and 3.13 that
∞ ∞
1/2
−2 2
ψ ρ ≤ ∂x ψ ρ dx ≤ r− ϕ ρ
ρx r dx
y
y
∞ ∞
2 1/2
2
≤ r −2
r
ϕ
ρ
dy
ρ
dx ≤ r−−2 E0 E1 .
x
− 0
We can verify that
0
3.17
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Journal of Applied Mathematics
1 As ρ → ∞, it holds for ξ θρ 1 − θρ, where θ ∈ 0, 1
ρ lim ψ ρ lim
ρ → ∞
ρ → ∞
ρ
≥ lim
1/2 η − ρdη
γ − 2 ξγ−3 2ργ ξ−3
ρ → ∞
1/2
γ − 2 ξγ−3 2ργ ξ−3
ρ
η − ρ dη
ρ
γ−3
−3 1/2 2
1 lim
2ργ θρ 1 − θρ
ρ−ρ
γ − 2 θρ 1 − θρ
ρ → ∞ 2
−→ ∞.
3.18
2 As ρ → 0, it holds for ξ θρ 1 − θρ, where θ ∈ 0, 1
lim ψ ρ lim
ρ→0
ρ ρ→0
ρ
≤ −lim
ρ→0
1/2 η − ρdη
γ − 2 ξγ−3 2ργ ξ−3
1/2
γ − 2 ξγ−3 2ργ ξ−3
ρ
η − ρ dη
ρ
γ−3
−3 1/2 2
1 −lim
2ργ θρ 1 − θρ
ρ−ρ
γ − 2 θρ 1 − θρ
ρ→02
3.19
: −νργ/21/2 .
Applying 3.17–3.19 and E0 E1 < νr−2 ργ/21/2 , where ν is a positive constant, we
can prove 3.9.
Lemma 3.3. Let T > 0. Under the conditions in Theorem 2.1, it holds for any solution ρ, u to the
exterior problem 3.3 that
∞ 0
∞ τ ∞ 2
2
2
r 2 u dx r 2 u r −4 dx r 2 u r −4 dx ds
x
τ
s
0
0 0
τ ∞ τ ∞ 2
2
r 2 u dx ds ≤ C, τ ∈ 0, T ,
ρ2 r 2 u dx ds 0
0
xs
0
0
where C > 0 denotes a constant independent of time.
xx
3.20
Journal of Applied Mathematics
9
Proof. Multiplying 3.32 by ρ−2 r 2 uτ and integrating the result with respect to x over
0, ∞, making use of 3.4, we obtain
d
dτ
∞ ∞
2
1 2 2
r u − ργ−2 r 2 u
ρ−2 r 2 u r −4 dx
dx x
x
τ
2
0
0
∞
∞
∞ γ−1 2 2
γ −2
r u dx − 2
r 2 u dx
ρ
ργ−3 ρx r 2 u dx 2
ρ−1 ρx r 2 u
τ
x
τ
x
0
0
0
∞
∞
2
ρ−2 u2 r 2 u r −3 dx − 2
ρ−2 ρx u r 2 u r −1 dx,
τ
0
τ
0
3.21
which implies
∞ 0
2
2
r u
x
dx τ ∞ 0
2
r 2 u r −4 dx ds
s
0
τ ∞
u2
2 4
dx
ds
C
ρ −ρ
dx C
u
r
ρx2 r 4 dx ds
≤CC
x
2
r
0
0 0
0 0
τ ∞ τ ∞
τ ∞
2
C
ρx2 r 2 u r 4 dx ds C
u4 r −2 dx ds C
ρx2 u2 r 2 dx ds
x
0 0
0 0
0 0
τ ∞ 2
≤CC
ρx2 r 2 u r 4 dx ds C sup u2L∞ .
∞ 0
γ−2
γ−2
τ ∞ 2
x
0
τ∈0,T 3.22
From 3.32 , 3.5, 3.8, and 3.9, we can deduce that for some small ∈ 0, 1
τ ∞
0
0
τ ∞ τ ∞ 2
2
2
r 2 u r −4 dx ds
ρx2 r 2 u r 4 dx ds ≤ ρx2 r 2 u r 4 dx ds x
x
xs
0 0
0 0
τ ∞ 2
τ ∞
u
2
2 4
ρx dx ds C
ux r dx ds,
r2
0 0
0 0
sup u2L∞ ≤ sup
τ∈0,T ∞ τ∈0,T 2
r2u
0
x
∞
dx C sup
τ∈0,T u2 dx,
0
3.23
3.24
using 3.22–3.24, we can obtain that
∞ 0
2
r u
2
x
dx τ ∞ 2
−4
r u r dx ds ≤ C C
0
0
2
s
τ ∞ 2
r2u
0
0
xs
r −4 dx ds.
3.25
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Journal of Applied Mathematics
Differentiating 3.32 with respect to τ, multiplying the result by r 2 uτ , and
integrating the result with respect to x over 0, ∞, we have
∞ ∞
2
ρ2 r 2 u dx
τ
xτ
0
0
∞
∞ ∞ 2
1
2
r −4
r 2 u dx 2
r 2 u r −2 dx
uuτ r 2 u r −3 dx −
u r −1 u
τ
τ
τ
τ
τ
2 0
0
0
3.26
∞
∞ ∞ γ 2 u
2ρ
x
ρ2
r2u
r 2 u dx −
r 2 u dx
ρ τ r u dx −
xτ
τ
x
xτ
τ
r
0
0
0
τ
∞ 2
r −2 r −1 u2 r 2 u dx.
1 d
2 dτ
r2u
2
r −4 dx τ
0
τ
A complicated computation gives
d
dτ
∞ ∞
2
ρ2 r 2 u dx
τ
xτ
0
0
∞ 2
2
2
≤C
r 2 u r −4 dx C sup r u 2
r2u
0
C
r −4 dx ∞ 0
τ
∞ x L∞
τ∈0,T 2
r2u
0
x
dx
3.27
∞ 2
u2
2 4
2
r u dx ,
ux r dx 1 sup
x
r2
τ∈0,T 0
integrating 3.27 with respect to τ, by means of 3.32 , 3.5, 3.8, 3.9, and 3.25, it holds
that
∞ τ ∞
2
r2u
0
≤CC
r −4 dx τ
τ ∞ 0
≤CC
0
2
ρ2 r 2 u dx ds
xs
0
2 2
r 2 u r −4 dx ds C sup r u x
s
0
2
τ∈0,T τ ∞ 2
−4
r u r dx ds C sup
0
0
∞
C sup
s
τ∈0,T 0
τ ∞
≤ C C
2
ρ
0
0
2
r2u
2
x
2
τ∈0,T ∞ L∞
2
τ∈0,T 2
r u
0
C sup
x
dx
∞ 2
r2u
x
0
1/2 ∞ 2
2
r u
0
dx
xx
1/2
dx
dx
r u
2
xs
dx ds C
∞ 2
r2u
0
τ
r −4 dx,
choosing the constant small sufficiently, we can complete the proof of Lemma 3.3.
3.28
Journal of Applied Mathematics
11
Remark 3.4. By Lemmas 3.1–3.3, the following inequality holds:
∞
0
∞
∞
∞
∞
2
u2 dx u2x dx u2τ dx ρx2 dx
ρ − ρ dx 0
0
0
0
τ ∞
τ ∞
τ ∞
ρx2 dx ds u2x dx ds u2s dx ds
0 0
0 0
0 0
τ ∞
τ ∞
u2xx dx ds u2xτ dx ds ≤ C.
0
0
0
3.29
0
Lemma 3.5. Under the conditions in Theorem 2.1, it holds for any solution ρ, u to the exterior
problem 3.3 that
ρ − ρ, u ·, τ
L∞ 0,∞
−→ 0,
τ −→ ∞,
3.30
where C > 0 denotes a constant independent of time.
Proof. From Lemmas 3.1–3.3, we can obtain
∞
ρ − ρ, u 2 2
x L 0,∞
0
dτ ≤
r−−4
∞ 2
ρ − ρ, u x r 2 2
L 0,∞
0
dτ ≤ C,
∞
d ρ − ρ, u 2 2
dτ
dτ
x L 0,∞ 0
∞ ∞ ∞
2
2
2
2
dτ
−4ρρ
r
r
dx
2
u
−
2ρ
ρ
u
u
u
dx
x
x xτ
x
x
xx
0
0
0
∞ ∞
∞ ∞ 2
2
u
2 4
≤C
ρx2 dx dτ C
u
r
r2u
u2xτ dx dτ ≤ C,
x
xx
r2
0
0
0
0
3.31
3.32
which together with 3.31 implies
ρ − ρ, u 2 2
x L 0,∞
∈ W 1,1 R .
3.33
It holds from Gagliardo-Nirenberg-Sobolev inequality that
ρ − ρ, u L∞ 0,∞
1/2
1/2
≤ ρ − ρ, u L2 0,∞ ρ − ρ, u x L2 0,∞ ,
3.34
which together with 3.5, 3.9, and 3.33 implies this lemma.
3.2. Proof of Theorem 2.1
Proof. The global existence of unique strong solution to the exterior problem as 2.4 and
2.5 can be established in terms of the short-time existence carried out as in 6, the uniform
12
Journal of Applied Mathematics
a-priori estimates and the analysis of regularities, which indeed follow from Lemmas 3.1–3.3.
We omit the details. The large time behaviors follow from Lemma 3.5 directly. The proof of
Theorem 2.1 is completed.
4. Proof of the Initial Boundary Value Problem
4.1. The A-Priori Estimates
Take the Lagrange coordinates transform
x
r
τ t.
4.1
ρ0 rr 2 dr : 1,
4.2
ρr, tr 2 dr,
r−
By 4.1 and the conservation of mass for ρ, u
r
ρr, tr 2 dr r−
r
r−
the Lagrange coordinates transform 4.1 map r, t ∈ r− , r × R into x, τ ∈ 0, 1 × R .
The relation between Lagrangian and Eulerian coordinates are satisfied as
∂x
ρr 2 ,
∂r
∂x
−ρur 2 ,
∂t
4.3
and the initial boundary value problem’s 2.4 and 2.11 are reformulated to
ρτ ρ2 r 2 u 0,
x
2ρx u
,
r −2 uτ ργ x ρ2 r 2 u
−
x x
r
ρ, u x, 0 ρ0 , u0 x, x ∈ 0, 1,
u0, t u1, t 0,
4.4
τ ∈ 0, ∞,
where the initial data satisfies
ρ0 ∈ L1 0, 1 ∩ W 1,∞ 0, 1,
1
r 2 ρ0
r 2 u0 ∈ L2 0, 1,
1
r 2 ρ0
inf ρ0 > ρ > 0,
x∈0,1
r 2 ρ0 r 2 u0 ∈ L2 0, 1,
x
4.5
r 2 ρ r 2 ρ r 2 u0
∈ L2 0, 1.
x x
Then, we will establish the a-priori estimates for the solution ρ, u to the initial boundary
value problem 4.4.
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13
Lemma 4.1. Let T > 0. Under the conditions in Theorem 2.2, it holds for any solution ρ, u to the
initial boundary value problem 4.4 that
1
0
τ 1 2
2u
1 γ−1
1 2
2 2 4
u ρ
ρ ux r dx ds
dx 2
γ −1
r2
0 0
1
1 γ−1
1 2
u0 ρ0
dx, τ ∈ 0, T .
γ −1
0 2
4.6
Proof. Multiplying 4.42 by r 2 u and integrating the result with respect to x over 0, 1, using
4.41 and 4.5, we obtain
d
dτ
1
0
1 1 2
1 γ−1
1 2
2
2
u ρ
dx ρ r u dx 2 ρ ru2 dx,
x
x
2
γ −1
0
0
4.7
and integrating 4.7 with respect to τ, we obtain
1
0
1
τ 1 2
2u
1 γ−1
1 γ−1
1 2
1 2
2 2 4
dx
ds
u ρ
u
ρ
ρ
u
r
dx dx.
x
0
2
γ −1
γ −1 0
r2
0 0
0 2
4.8
Lemma 4.1 can be obtained.
Lemma 4.2. Let T > 0. Under the conditions in Theorem 2.2, it holds for any solution ρ, u to the
initial boundary value problem 4.4 that
1
2
1 1
τ 1
1
ργ−1 dx γ
ργ−1 ρx2 r 4 dx ds
γ −1 0
0 0
1
2
1 1
1
γ−1
u0 r 2 ρ0x dx ρ dx, τ ∈ 0, T ,
2 0
γ −1 0 0
u r 2 ρx
0
2
dx 4.9
where C is a positive constant independent of time.
Proof. Differentiating 4.41 with respect to x, we have
0,
ρxτ ρ2 r 2 u
x x
4.10
which together with 4.42 and ∂r/∂τ u gives
r −2 u ρx
τ
2ρx u
.
ργ x −2r −3 u2 −
r
4.11
14
Journal of Applied Mathematics
Multiplying 4.11 by u r 2 ρx r 2 , and integrating the result with respect to x and τ, it holds
that
1
2
1 1
τ 1
1
γ−1
u r ρx dx ρ dx γ
ργ−1 ρx2 r 4 dx ds
γ −1 0
0
0 0
1
2
1 1
1
γ−1
u0 r 2 ρ0x dx ρ dx.
2 0
γ −1 0 0
2
2
4.12
The proof of 4.9 is completed.
Lemma 4.3. Let T > 0. Under the conditions in Theorem 2.2, there exists a constant ρ∗ > 0 such that
0 < ρx, τ ≤ ρ∗ ,
x, τ ∈ 0, 1 × 0, T .
4.13
Proof. It follows from 4.6 and 4.9 that
ρx, τ ρr, t ≤
≤
r−−2
r
r−
≤CC
r
r−
r−
ρr, tr dr 2
r
r−
≤CC
ρr, tdr 1
r
r−−2
ρr r, tdr
r
r−
ρr r, t 2
r dr
ρ √
ρ
4.14
2
1 ρr r, t r 2 dr
ρ
ρx x, τ2 r 4 dx ≤ C : ρ∗ .
0
Lemma 4.4. Let T > 0. Under the conditions in Theorem 2.2, it holds for any solution ρ, u to the
initial boundary value problem 4.4 that
1
u dx 2n
0
τ 1
0
0
u2n
2 2n−2 2 4
ρ u
ux r dx ds ≤ CT ,
r2
4.15
for any positive integer n ∈ N, where CT is a positive constant dependent of time.
Proof. Multiplying 4.42 with u2n−1 , integrating by parts over 0, 1, we have
1 d
2n dτ
1
0
u dx 2n
1
ρ
0
2
2
r u
x
2 2n−1
r u
x
1 ργ r 2 u2n−1 ρ 2ru2n
dx.
dx 0
x
x
4.16
Journal of Applied Mathematics
15
Since it holds that
2
r u
x
2 2n−1
r u
x
2u
r 2 ux
ρr
2u2n−1
2n − 1r 2 u2n−2 ux
ρr
4u2n
4nu2n−1 ux r
2 2 2n − 1u2n−2 u2x r 4 ,
ρ
ρ r
4.17
it follows from 4.16 that
d
dτ
1
1 2n
1
u2n
u
dx 2
dx
−
1
ρ2 u2n−2 u2x r 4 dx
2n
2
2n
r
0
0
0
1 γ−1 2n−1
1
ρ u
2
dx 2n − 1 ργ u2n−2 ux r 2 dx
r
0
0
1 2n
1
2γ−1 1 2n−2
u
2 2n−2 2 4
ρ
≤
dx
ρ
u
u
r
dx
C
u
dx,
∞
x
L
2
0 r
0
0
4.18
which together with 4.13 and Young’s inequality yields
d
dτ
1
u2n dx 1
0
0
u2n
dx r2
1
ρ2 u2n−2 u2x r 4 dx ≤ C 0
1
u2n dx,
4.19
0
and by applying the Gronwall’s inequality to 4.19, we can obtain 4.15.
Lemma 4.5. Let T > 0, for n ∈ N, and n > 1/2γ − 1. Under the conditions in Theorem 2.2, it holds
for any solution ρ, u to the initial boundary value problem 4.4 that
τ 2nγ−1 2n u ρ
0
L∞ 0,1
τ γ 2 2n ρ r x
0
L∞ 0,1
ds ≤ CT ,
ds ≤ CT ,
where CT is a positive constant dependent of time.
τ ∈ 0, T ,
4.20
4.21
16
Journal of Applied Mathematics
Proof. By means of Sobolev imbedding theorem and Cauchy-Schwarz inequality, applying
4.6, 4.13, and 4.15, we get
τ 2nγ−1 2n u ρ
0
L∞ 0,1
ds
τ 1
τ 1 2nγ−1 2n 2nγ−1 2n ≤
u dx ds u
ρ
ρ
dx ds
0
0
≤ CT C
≤ CT C
τ 1
0
τ 1
0
τ
0
1
0
0
0
x
0
ρ2nγ−1−1 ρx u2n dx ds C
2 4
ρx r dx ds C
τ 1
0
ρ22nγ−1−1 u2n r −4 dx ds C
0
τ 1
0
ρ2nγ−1 u2n−1 ux dx ds
4.22
0
ρ22nγ−1−1 u4n r −4 dx ds
0
τ 1
0
0
ρ2 u2n−2 u2x r 4 dx ds ≤ CT .
Next, we find that
ρx r 2 ρ0x r02 u0 − u −
τ
0
γ 2
ρ x r ds,
4.23
which together with 4.13, 4.15, and 4.22 gives
τ γ 2 2n ρ r ds
∞
x
L
0
τ 2n 2n
2nγ−1 2 γ
ρx r ds
ρ
L∞
0
τ s
γ 2 2n 2nγ−1 2n
2
γ
ρ x r dl ds
ρ
ρ0x r0 u0 − u −
∞
0
0
L τ 2n τ s
γ 2
2nγ−1
2nγ−1 2n
2n
2n ρ0x u0 u ∞ ds C ρ
≤ C ρ
ρ x r dl
ds
∞
L
0
0
0
L
τ τ s 2n
ργ r 2 dl ds,
≤ CT C ρ2nγ−1 u2n ∞ ds CT ∞
x
L
0
0
0
4.24
L
applying the Gronwall’s inequality to 4.24, we obtain 4.21.
Lemma 4.6. Let T > 0. Under the conditions in Theorem 2.2, there exists a constant ρ∗ > 0 such that
ρx, τ ≥ ρ∗ > 0,
x, τ ∈ 0, 1 × 0, T .
4.25
Journal of Applied Mathematics
17
Proof. It is easy to verify that
ργ1/2 τ
1
:
ρ
γ1/2
x, τdx ≥
0
r−2
r2
r
r−2
1γ1/2
r
r2
ρr, tr dr
1γ1/2
ρ0 rr 2 dr
2
r−
4.26
> 0,
r−
2
γ1/2
− ργ1/2 τ 2
ρ
L 0,1
4.27
∈ W 1,1 0, T .
Indeed, it holds that
τ 2
γ1/2
− ργ1/2 s 2
ρ
L 0,1
0
ds ≤ C
τ 2
γ1/2
r2 2
ρ
x
0
L 0,1
ds ≤ C,
4.28
which together with 4.4, 4.9, and 4.13 gives
τ 2
d γ1/2
γ1/2 s
ρ
ds
−
ρ
ds
2
L 0,1 0
τ 1 γ1/2
γ1/2−1
γ1/2
≤ γ 1
ρ
−ρ
ρs dxds
s ρ
0
0
τ 1 2 ργ1/2 − ργ1/2 s ργ1/2 s dxds
s
0
0
τ 1 τ 1
2
≤C
ργ1/2 − ργ1/2 s dx ds ρ2 u2x r 4 dx ds ≤ C.
0
0
0
4.29
0
By Gagliardo-Nirenberg-Sobolev inequality and 4.27, it follows that
2
γ1/2
− ργ1/2 τ ∞
ρ
L 0,1
−→ 0 as τ −→ ∞.
4.30
Thus, there is a T0 > 0 and a constant ρ1 > 0 such that
ρx, τ ≥ ρ1 ,
x ∈ 0, 1, τ ∈ T0 , ∞.
4.31
For τ ∈ 0, T0 , denote
vx, τ :
1
,
ρx, τr 2 x, τ
4.32
18
Journal of Applied Mathematics
then from 4.4, we can obtain
vτ r2u
x
r2
−
2vu
,
r
4.33
multiplying 4.33 by 4v3 and integrating the result over 0, 1 × 0, T0 , we have
1
0
1
τ 1
r2u
τ 1
v4 u
dx ds
r2
0
0 0
0 0 r
1
τ 1
τ 1
u
4
4
2
v0 dx 12
v u ρx r dx ds 24
v4 dx ds
r
0
0 0
0 0
1
τ 1
s
v04 dx 12
v4 u ρ0x r02 u0 − u −
ργ x r 2 dl dx ds
v4 dx v04 dx 4
0
24
τ 1
0
0
0
v3
x
dx ds − 8
0
4.34
0
u
v4 dx ds.
r
From 4.21, it holds that
1
v4 dx 12
0
1
0
0
v04 dx
− 12
≤6
τ 1
0
12
τ 1
v
0
τ 1
4
0
s
v u
0
τ 1
0
v4 u2 dx ds
0
0
4
uρ0x r02 dx ds
12
τ 1
0
γ 2
ρ x r dl dx ds 24
v4 u2 dx ds CT0 0
τ 1
0
v4 uu0 dx ds
0
τ 1
v
0
0
4u
r
4.35
dx ds
v4 dx ds,
0
which yields to
1
v4 dx ≤ C CT0 0
τ 1
0
v4 dx ds,
4.36
0
where CT0 is a positive constant dependent of time T0 . By Gronwall’s inequality, 4.36 leads
to
1
0
v4 dx 1
0
1
dx ≤ CT0 .
ρ4 r 8
4.37
Journal of Applied Mathematics
19
It holds for x, τ ∈ 0, 1 × 0, T0 that
1 1 1
dx
dx ρ
ρ x
0
0
1
1/2 1
1/2
1
1
1
2 4
≤CC
dx C
dx
ρx r dx
≤ CT0 ,
4 8
4 8
0 ρ r
0 ρ r
0
1
ρ
1
4.38
namely,
x, τ ∈ 0, 1 × 0, T0 .
ρx, τ ≥ CT0 : ρ2 ,
4.39
Therefore, we can choose
ρ∗ min ρ1 , ρ2 ,
4.40
x, τ ∈ 0, 1 × 0, ∞.
4.41
to get
ρ ≥ ρ∗ ,
Lemma 4.7. Let T > 0. Under the conditions in Theorem 2.2, it holds for any solution ρ, u to the
initial boundary value problem 4.4 that
1 2
r u
0
2
x
dx τ 1
ρ
0
2
1 0
2
r u
2
2
r u
xs
0
2
τ
−4
r dx τ 1 2
r 2 u r −4 dx ds
0
dx ds s
0
τ 1 2
2
r u
0
0
xx
4.42
dx ds ≤ C,
τ ∈ 0, T ,
where C > 0 denotes a constant independent of time.
Proof. The proof is similar to Lemma 3.3. We omit here.
Remark 4.8. By Lemmas 4.1–4.7, the following inequality holds:
1
u2 dx 0
1
2
ρ − ρ dx 0
τ 1
0
τ
0
1
0
0
1
0
ρx2 dx ds u2xx dx ds
τ 1
0
0
1
0
u2x dx ds τ 1
0
u2x dx 0
u2τ dx τ 1
0
0
u2xτ dx ds ≤ C.
1
0
ρx2 dx
u2s dx ds
4.43
20
Journal of Applied Mathematics
Lemma 4.9. Under the conditions in Theorem 2.2, it holds for any solution ρ, u to the initial
boundary value problems 2.4 and 2.11 that
ρ − ρ,
u ·, t
≤ C0 e−C1 t ,
L∞ r− ,r where C0 and C1 denote the constants independent of time and ρ
t > 0,
r
r−
4.44
ρr, tr 2 dr.
Proof. In a similar argument to show 4.8 and 4.12 with modification, we can obtain the
following
d
dt
r r−
r 2
1 γ
1 2
r dr ρu ρ − ρ
γ − γ ρ
γ−1 ρ − ρ
2ρu2 ρu2r r 2 dr 0,
2
γ −1
r−
1 d
2 dt
r 2
ρ u ρ−1 ρr r 2 dr r−
γ
r
r−
ργ−2 ρr2 r 2 dr
4.45
r ργ − ρ
γ − γ ρ
γ−1 ρ − ρ
r 2 dr
1 d
γ − 1 dt
r−
4.46
0.
It holds from 4.9, 4.13, and 4.25 that
r
2
ρ − ρ
r 2 dr ≤ C
r
r−
r−
ργ−2 ρr2 r 2 dr.
4.47
Denote
Et :
r r 2 ργ − ρ
γ − γ ρ
γ−1 ρ − ρ
r 2 dr.
ρ u2 u ρ−1 ρr
r 2 dr r−
4.48
r−
By 4.45–4.48, a complicated computation gives rise to
d
Et CEt ≤ 0,
dt
4.49
Et ≤ E0e−Ct .
4.50
2
2
Et ≥ c u2L2 r− ,r ρ − ρ
L2 r− ,r ρr L2 r− ,r ,
4.51
which gives
By the fact that
Journal of Applied Mathematics
21
where c > 0 is a constant independent of time and Gagliardo-Nirenberg-Sobolev inequality
ρ − ρ,
u L∞ r− ,r 1/2
1/2
≤ C ρ − ρ,
u L2 r− ,r ρ − ρ,
u r L2 r
− ,r ,
4.52
we obtain 4.44.
4.2. Proof of Theorem 2.2
Proof. The proof of Theorem 2.2 is similar to Theorem 2.1. We omit the details.
Acknowledgments
The research of R. Lian is partially supported by NNSFC no. 11101145. The research of L.
Huang is partially supported by NNSFC no. 11126323.
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