Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2011, Article ID 264795, 18 pages
doi:10.1155/2011/264795
Research Article
Global Existence of Cylinder Symmetric
Solutions for the Nonlinear Compressible
Navier-Stokes Equations
Lan Huang,1 Fengxiao Zhai,2 and Beibei Zhang3
1
College of Mathematics and Information Science, North China University of Water Resources and
Electric Power, Zhengzhou 450011, China
2
Department of Physics, Zhengzhou University of Light Industry, Zhengzhou 450002, China
3
School of Statistics, Capital University of Economics and Business, Beijing 100070, China
Correspondence should be addressed to Lan Huang, huanglan82@hotmail.com
Received 4 September 2011; Accepted 6 October 2011
Academic Editor: Meng Fan
Copyright q 2011 Lan Huang 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 prove the global existence of cylinder symmetric solutions to the compressible Navier-Stokes
equations with external forces and heat source in R3 for any large initial data. Some new ideas and
more delicate estimates are used to prove this result.
1. Introduction
In this paper, we study the global existence of cylinder symmetric solutions to the nonlinear
compressible Navier-Stokes equations with external forces and heat source in a bounded
domain G {r ∈ R , 0 < a < r < b < ∞} of R3 , where r is the radial variable. In the
Eulerian coordinates, the system under consideration are expressed as
ρu
0,
ρt ρu r r
v2
u
ρ ut uur −
P r − ν ur f1 r, t,
r
r r
uv
v
− μ vr f2 r, t,
ρ vt uvr r
r r
1.1
1.2
1.3
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wr
ρwt uwr − μ wrr f3 r, t,
r
θr
u
P ur − Q gr, t,
CV ρθt uθr − κ θrr r
r
1.4
1.5
where
P γρθ,
2 u 2
v 2
u
2
2
Q λ ur ,
μ vr −
wr 2ur 2
r
r
r
1.6
and ρ is the mass density, θ is the absolute temperature, u, v, w are the radial velocity, angular
velocity, and axial velocity, respectively, and λ, μ, ν, γ, CV , κ, λ are the constants satisfying
γ, CV , κ, μ > 0, 3λ 2μ ≥ 0 ν λ 2μ. f1 , f2 , f3 , and g represent external forces and heat
source, respectively. For system 1.1–1.5, we consider the following initial boundary value
problem:
ρr, 0 ρ0 r,
u, v, wr, 0 u0 , v0 , w0 r,
θr, 0 θ0 r,
θr a, t θr b, t 0,
u, v, wa, t u, v, wb, t 0,
r ∈ G,
t ≥ 0.
1.7
1.8
To show the global existence, it is convenient to transform the system 1.1–1.5 to that
in the Lagrangian coordinates. The Eulerian coordinates r, t are connected to the Lagrangian
coordinates ξ, t by the relation
t
rξ, t r0 ξ 1.9
u
ξ, τdτ,
0
where u
ξ, t urξ, t, t and
−1
r0 ξ η ξ,
ηr r
sρ0 sds,
r ∈ G.
1.10
a
It should be noted that if inf{ρ0 s : s ∈ a, b} > 0, then η is invertible. It follows from 1.1,
1.8, and 1.10 that
r
sρs, tds r0
a
sρ0 sds ξ,
1.11
a
and G is transformed into Ω 0, L with
L
b
a
sρs, tds b
a
sρ0 s,
∀t ≥ 0.
1.12
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3
Differentiating 1.11 with respect to ξ yields
∂ξ rξ, t rξ, t−1 ρ−1 rξ, t, t.
1.13
t φrξ, t, t, we easily get
In general, for a function φr, t φξ,
t ∂t φr, t u∂r φr, t,
∂t φξ,
t ∂r φr, t∂ξ rξ, t ∂ξ φξ,
1.14
∂r φr, t −1
ρ r, t.
r
1.15
still by ρ, u, v, w, θ and ξ, t by x, t.
Without danger of confusion, we denote ρ,
u
, v , w,
θ
We set τ : 1/ρ to denote the specific volume. Therefore, by virtue of 1.13–1.15, system
1.1–1.8 in the new variables x, t read
r 2 θx
CV θt κ
τ
τt rux ,
νrux − γθ
v2
ut r
f1 rx, t, t,
τ
r
x
rvx
uv
f2 rx, t, t,
vt μr
−
τ
r
x
rwx
τw
wt μr
μ 2 f3 rx, t, t,
τ
r
x
x
1.16
1.17
1.18
1.19
rvx 2
r 2 wx2
1
μ
− 2μ u2 v2 grx, t, t,
νrux − γθ rux μ
x
τ
τ
τ
1.20
together with
τx, 0 τ0 x,
u, v, wx, 0 u0 , v0 , w0 x,
u, v, w0, t u, v, wL, t 0,
θx, 0 θ0 x,
θx 0, t θx L, t 0,
x ∈ 0, L,
1.21
t ≥ 0.
1.22
By 1.9 and 1.13, we have
rx, t r0 x t
ux, sds,
0
rt x, t ux, t,
r0 x a 2
2
x
τ0 y dy
0
1/2
,
1.23
rx, trx x, t τx, t.
Now let us first recall the related results in the literature. When there were no external
forces and heat source, in two or three dimensions, the global existence and large time
behavior of smooth solutions to the equations of a viscous polytropic ideal gas have been
4
Journal of Applied Mathematics
investigated for general domains only in the case of sufficiently small initial data, see, for
example, 1–3. For any large initial data, the global existence of generalized solutions was
shown in 4–7. Recently, Qin 8 proved the exponential stability in H 1 and H 2 , and Qin and
Jiang 9 studied the global existence and exponential stability in H 4 with smallness of initial
total energy.
When there exist external forces and heat forces, for one-dimensional case, the system
is isentropic compressible Navier-Stokes equations. Mucha 10 obtained the exponential
stability under various boundary conditions, Yanagi 11 established the existence of classical
solutions, and Qin and Zhao 12 proved the global existence and asymptotic behavior of
solutions for pressure P ργ with γ 1. Later on, Zhang and Fang 13 studied the global
existence and uniqueness for γ > 1. For nonisentropic compressible Navier-Stokes equations,
Qin and Yu 14 proved the global existence and asymptotic behavior for perfect gas. In twoor three-dimensional case and the external force and heat source f / 0, g / 0, Qin and Wen 15
proved the global existence of spherically symmetric solutions. In this paper, we will prove
the global existence of cylinder symmetric solutions with external forces and heat source in a
bounded domain in R3 .
The notation in this paper will be as follows: Lp , 1 ≤ p ≤ ∞, W m,p , m ∈ N, H 1 W 1,2 ,
1
H0 W01,2 denote the usual Sobolev spaces on 0, L. In addition, · B denotes the norm
in the space B; we also put · · L2 . We denote by Ck I, B, k ∈ N0 , the space of ktimes continuously differentiable functions from I ⊆ R into a Banach space B, and likewise
by Lp I, B, 1 ≤ p ≤ ∞ the corresponding Lebesgue spaces. Subscripts t and x denote the
partial derivatives with respect to t and x, respectively. We use C1 to denote the generic
positive constant depending on the H 1 -norm of the initial data and time T .
We suppose that fi rx, t, ti 1, 2, 3, gr, t satisfy, for any T > 0,
fi ∈ L1 0, T , L∞ 0, L ∩ L2 0, T , L2 0, L ,
gr, t > 0,
g ∈ L1 0, T , L∞ 0, L ∩ L2 0, T , L2 0, L .
1.24
1.25
We are now in a position to state our main theorems.
Theorem 1.1. Assume that 1.24-1.25 hold; if τ0 , u0 , v0 , w0 , θ0 ∈ H 1 0, L × H01 0, L ×
H01 0, L×H01 0, L×H 1 0, L, τ0 x > 0, θ0 x > 0 on 0, L and the initial data are compatible with
the boundary conditions 1.22, then for problem 1.16–1.22 there exists a unique global solution
τ, u, v, w, θ ∈ C0, T , H 1 0, L × H01 0, L × H01 0, L × H01 0, L × H 1 0, L such that, for any
T > 0,
0 < a ≤ rx, t ≤ b,
0 < C1−1 ≤ τx, t ≤ C1 ,
x, t ∈ 0, L × 0, T ,
x, t ∈ 0, L × 0, T ,
τt2H 1 ut2H 1 vt2H 1 wt2H 1 θt2H 1 rt2H 2
t
τ2H 1 u2H 2 v2H 2 w2H 2 θ2H 2 ut 2
0
vt 2 wt 2 θt 2 τdτ ≤ C1 ,
∀t ∈ 0, T .
1.26
Journal of Applied Mathematics
5
2. Proof of Theorem 1.1
In this section we will complete the proof of Theorem 1.1. To this end, we assume that in this
section all assumptions in Theorem 1.1 hold. The proof of Theorem can be divided into the
following several lemmas.
Lemma 2.1. One has
a r0, t ≤ rx, t ≤ rL, t b,
∀x, t ∈ 0, L × 0, ∞.
2.1
Proof. The proof of 2.1 is borrowed from 6, 8; please refer to 2.1 in 6 or Lemma 2.1 in
8 for detail.
Lemma 2.2. The global solution τt, ut, vt, wt, θt to problems 1.16–1.22 satisfies the
following estimates:
L 1 2
u v2 w2 CV θ x, tdx ≤ C1 ,
0 2
L
0
2.2
−1
2
t L 2 2
τr v − rvx
g
κr θx τu2 u2x ru2x wx2
Ux, tdx
x, sdx ds ≤ C1 ,
θ
τθ
τθ
τθ
τθ
θ
τθ2
0 0
2.3
where
Ux, t 1 2
u v2 w2 γ τ − log τ − 1 CV θ − log θ − 1 .
2
2.4
Proof. Multiplying 1.17–1.19 by u, v, and w, respectively, adding up the results, and using
1.16, we have
d 1 2
u v2 w2 CV θ
dt 2
κr 2 θx ruνrux − γθ μrvrvx μr 2 wwx
− 2μ u2 v2
τ
τ
τ
τ
f1 u f2 v f3 w g.
x
2.5
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Journal of Applied Mathematics
Integrating 2.5 with respect to x and t over QT 0, L × 0, t t ∈ 0, T , ∀T > 0, using
boundary condition 1.22, we obtain
L L 1 2
1 2
2
2
2
2
u v w CV θ dx u0 v0 w0 CV θ0 dx
0 2
0 2
t L
f1 u f2 v f3 w g x, sdx ds
0
0
≤ C1 C1
t L
u2 v2 w2 x, sdx ds
0
C1
t L
0
0
0
t
f12 f22 f32 x, sdx ds C1 g L∞ ds
0
2.6
which, by using Gronwall’s inequality and 1.24-1.25, gives 2.2.
By 1.16–1.20, we can easily obtain
κr 2 θx2 νru2x μrv2x μr 2 wx2 2μ u2 v2 x g
−
Ut τθ
θ
θ
τθ2
κθ − 1r 2 θx ruνrux − γθ μrvrvx μr 2 wwx
2
2
− 2μ u v γru
τθ
τ
τ
τ
x
f1 u f2 v f3 w g.
2.7
Note that constants ν λ 2μ and
2
νru2x μrv2x 2μ u2 v2 x 2μ τ 2 r −2 u2 r 2 u2x λru2x μ τr −1 v − rvx
−
τθ
θ
τθ
τθ
2 τu2 u2x λru2x μ τr −1 v − rvx
−1
≥ C1
.
θ
τθ
τθ
2.8
Integrating 2.7 with respect to x and t over QT , using 1.22, 1.24-1.25, and 2.8, we
conclude
2
L
t L 2 2
2
g
κr θx τu2 u2x rux wx2 τr −1 v − rvx
dx ds
Ux, tdx θ
τθ
θ
τθ2
0
0 0
t L
≤ C1 f1 u f2 v f3 w g x, sdx ds
2.9
0 0
t L
t
u2 v2 w2 f12 f22 f32 x, sdx ds C1 g L∞ ds
≤ C1 C1
0
0
≤ C1 .
The proof is complete.
0
Journal of Applied Mathematics
7
Next we adapt and modify an idea of Qin and Wen 15 for one-dimensional case to
give a representation for τ.
Let
νrux − γθ
σx, t :
τ
x
0
x
v2 − u2 f1 r y, t , t
dy,
r
r2
u0
dy r0
hx, t :
0
t
2.10
2.11
σx, sdx.
0
Then, we infer from 1.16 and 1.17 that
hx u
,
r
ht σ.
2.12
By 1.16 and 2.12, we have
hτt ruhx − u νrux − γθ τ
2
x
0
v2 − u2 f1
r
r2
dx.
2.13
Integrating 2.13 with respect to x and t over QT , we obtain
L
0
t L x 2
v − u2 f1
u γθ dx ds dy dx ds
hτdx h0 τ0 dx −
τ
r
r2
0
0 0
0 0
0
L
t L
t L
x 2
v − u2 f1
2
u γθ dx ds dy dx ds
h0 τ0 dx −
rrx r
r2
0
0 0
0 0
0
L
t L
b2 t L v2 − u2 f1
2
u γθ dx ds h0 τ0 dx −
dx ds
2 0 0
r
r2
0
0 0
t L 2
v − u2 rf1
dx ds,
−
2
2
0 0
2.14
L
t L
2
where h0 x : hx, 0. It follows from integration of 1.16 over QT and use of 1.22 that
L
τx, tdx L
0
τ0 xdx τ ∗ .
2.15
0
If we apply the mean value theorem to 2.14 and use 2.15, we conclude there is an x0 t ∈
0, L such that
1
hx0 t, t ∗
τ
L
hx, tτx, tdx.
0
2.16
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Therefore, we derive from 2.11, 2.14, and 2.16 that
t
σx0 t, sds hx0 t, t −
x0 t
0
0
1
− ∗
τ
1
∗
τ
t L
0
L
0
u0
dx
r0
rf1
v 2 u2
γθ 2
2
h0 xτ0 xdx −
0
x0 t
0
b2
dx ds ∗
2τ
t L
0
0
v2 − u2 f1
r
r2
u0
dx.
r0
dx ds
2.17
Using 2.17, we will show the representation of specific volume τ.
Lemma 2.3. One has the following representation:
γ t θx, sBx, s
Dx, t
1
ds ,
τx, t Bx, t
ν 0
Dx, s
x ∈ 0, L,
2.18
where
L
x
x
u0
u
1 1
Dx, t τ0 x exp
dy
τ0 h0 dx −
dy ν τ∗ 0
r
0 0
x0 t r
f1
1 t L rf1
b2 t L f1
−
dy ds − ∗
dx ds ∗
dx ds ,
τ 0 0 2
2τ 0 0 r
0 0 r
t L
1 1
u2 v 2
b2 t L v2 − u2
Bx, t exp
γθ
dx
ds
−
dx ds
ν τ∗ 0 0
2
2τ ∗ 0 0
r2
t x 2
v − u2
dx ds .
r2
0 0
2.19
t x
Proof. By 1.16 and 1.17, we have
v2 − u2 rf1
u
θ
σx ν log τ tx − γ
.
r t
τ x
r2
2.20
Journal of Applied Mathematics
9
Integrating 2.20 over x0 t, x × 0, t and using 2.17, we derive
t
θx, s
ds
0 τx, s
x t
t x 2
v − u2 rf1
u u0
−
ν log τ0 x σx0 t, sds −
dy
r0
r2
0
0 0
x0 t r
rf1
1 t L v 2 u2
b2 t L v2 − u2 f1
γθ ν log τ0 x − ∗
dx ds ∗
dx ds
τ 0 0
2
2
2τ 0 0
r
r2
ν log τx, t − γ
1
∗
τ
L
0
h0 xτ0 xdx −
x
0
u0
dx −
r0
t x
0
0
v2 − u2 rf1
dy ds r2
x
x0 t
u
dy
r
2.21
which, when the exponentials are taken, turns into
Bx, t
1
exp
Dx, t τx, t
t
γ
θx, s
ds .
ν 0 τx, s
2.22
Multiplying 2.22 by γθ/ν and integrating the resulting equation with respect to t, we arrive
at
t
γ
γ t θx, sBx, s
θx, s
exp
ds 1 ds.
2.23
ν 0 τx, s
ν 0
Dx, s
Substituting this into 2.22, we obtain 2.18. The proof is complete.
Lemma 2.4. There are positive constants τ and τ, such that, for any T > 0,
τ ≤ τx, t ≤ τ,
x, t ∈ 0, L × 0, T .
2.24
Proof. Recalling the definition Dx, t, we have by 1.24, Cauchy-Schwarz’s inequality, and
Lemma 2.1 that
t x
t
1 t L rf
f1
b2 t L f1
1
dx ds − ∗
dx ds −
dy ds ≤ C1 f1 L∞ ≤ C1
2.25
∗
τ 0 0 2
2τ 0 0 r
0 0 r
0
which, along with Lemma 2.2, gives
0 < C1−1 ≤ Dx, t ≤ C1 ,
x, t ∈ 0, L × 0, T .
2.26
By Lemmas 2.1 and 2.2, we easily obtain, for any 0 ≤ s ≤ t,
Bx, t ≤ C1 ,
Bx, s
≤ exp{−C1 t − s}.
Bx, t
2.27
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Journal of Applied Mathematics
Therefore, we derive from 2.2, 2.18, and 2.26-2.27 that
τx, t ≥
τx, t ≤ C1 1 Dx, t
≥ τ,
Bx, t
2.28
t
θe
−C1 t−s
ds
0
⎡
1/2 L
1/2 L
1/2 ⎤2
t L
r 2 θx2
τθ
⎦ e−C1 t−s ds
≤ C1 ⎣
θdx
dx
dx
2
2
τθ
r
0
0
0
0
≤ C1 C1 1 t
L
max τ·, s
0 0,L
0
2.29
r 2 θx2
dx ds ds
τθ2
which, by using Gronwall’s inequality and 2.28, gives 2.24. The proof is complete.
Remark 1. If the initial data or initial energy are small enough, we can obtain the uniform
estimate independent of time t about specific volume τ under assumptions of external forces.
Moreover, we can prove the large-time behavior of solutions.
Lemma 2.5. Under the assumptions of Theorem 1.1, one has, for any T > 0 and for all t ∈ 0, T ,
L
0
t L
θ2 u4 v4 w4 dx θx2 u2 u2x v2 vx2 w2 wx2 x, sdx ds ≤ C1 .
0
2.30
0
Proof. Multiplying 2.5 by 1/2u2 v2 w2 CV θ and then integrating the result over QT ,
we have
1
2
L 0
1 2
u v2 w2 CV θ
2
CV κ
≤ C1 −
2
C1
t L
0
0
t L
0
t L
0
0
0
2
dx
r 2 θx2
dx ds
τ
r 2 u2 u2x r 2 v2 vx2 r 2 w2 wx2
4
4
2
2 2
u v θ θ u dx ds
τ
1 2
2
2
u v w CV θ dx ds,
f1 u f2 v f3 w g
2
2.31
Journal of Applied Mathematics
11
where
t L 1 2
2
2
u v w CV θ dx ds
g
2
0 0
t
L 1 2
u v2 w2 CV θ dx ds ≤ C1 ,
≤ g L∞
2
0
0
t L
1 2
2
2
u v w CV θ dx ds
f1 u f2 v f3 w
2
0 0
t
L 2
f1 ∞ f2 ∞ f3 ∞
u u4 v2 v4 w2 w4 dx ds
≤ C1
L
L
L
0
t L
0
2.32
0
θ2 dx ds.
0
Multiplying 1.17 by u3 and then integrating the result over QT , we get
1
4
L
0
t L
ν t 1 2 2 2
u4 θ2 u2 x, sdx ds
u dx ≤ C1 −
r u ux dx ds C1
τ 0 0
0 0
t L 2
v
f1 u3 dx ds
r
0 0
4
t L
ν t 1 2 2 2
u4 θ2 u2 x, sdx ds
r u ux dx ds C1
≤ C1 −
τ 0 0
0 0
t
L
f1 ∞ u2 ∞
v4 u2 u4 dx ds.
C1
L
L
0
2.33
0
Similarly, multiplying 1.18 and 1.19 by v3 and w3 , respectively, and then integrating over
QT , we have
1
4
L
v4 dx ≤ C1 −
0
≤ C1 −
1
4
L
w4 dx ≤ C1 −
0
≤ C1 −
μ
τ
μ
τ
μ
τ
μ
τ
t 1
0
0
t 1
0
0
t 1
0
0
t 1
0
0
r 2 v2 vx2 dx ds C1
r 2 v2 vx2 dx ds C1
t L
v4 u2 v4 f2 v2 v4 dx ds
0
t
r 2 w2 wx2 dx ds C1
r 2 w2 wx2 dx ds C1
0
L
2
4
f2 ∞ u2 ∞ 1
v
dx ds,
v
L
L
0
0
t L
w4 f3 w2 w4 dx ds
0
t
0
0
f3 ∞ 1
L
L
w2 w4 dx ds.
0
2.34
12
Journal of Applied Mathematics
Multiplying 2.31 and 2.33 by μ/2τC1 and μ/ν, respectively, adding up the resulting
inequalities, and using 2.34 to obtain, with the help of 2.32, the following result:
L
u v w θ
4
4
0
≤ C1 t
0
4
t
2
dx t L
0
f1 L∞
0
u2L∞
θx2 u2 u2x v2 vx2 w2 wx2 x, sdx ds
0
f2 L∞ f3 L∞ 1
L
L
θ2 u4 v4 w4 dx ds
2.35
0
θ2 u4 v4 w4 dx ds.
0
On the other hand, by 2.3 and 2.20,
t
0
u2L∞ ds
≤
t L
0
2
2
|ux | dx
ds ≤
0
t L
0
0
u2x
dx
τθ
L
τθdx ds ≤ C1 .
2.36
0
In view of 1.24-1.25 and 2.36, we apply Gronwall’s inequality to 2.35 to obtain 2.30.
The proof is complete.
Lemma 2.6. Under the assumptions of Theorem 1.1, one has, for any T > 0,
L
0
τx2 x, tdx t L
0
0
θτx2 x, sdx ds ≤ C1 ,
∀t ∈ 0, T .
2.37
Proof. By means of 1.16, we rewrite 1.17 as
u ντx
−
r
τ
t
γθτx − τθx v2 − u2 rf1
.
τ2
r2
2.38
Multiplying 2.38 by u/r − ντx /τ in L2 0, L and using Lemmas 2.1–2.5, we arrive at
L 2
2
θτx
1 d
u − ντx νγ
dx
2
2 dt r
τ
0 τ r
L
γθτx − τθx u γντx θx v2 − u2 rf1 u ντx
dx
−
2.39
r
τ
rτ 2
rτ 2
r2
0
L 2
L
θτx
θx2
1
u ντx 2
2
2
2
4
4
2
≤ νγ
θu θx u dx.
dx C1
u v f1 −
2
2
θ
r
τ
0 τ r
0
Journal of Applied Mathematics
13
Integrating it with respect to t, using Lemmas 2.1-2.5, 1.24, and 2.36, we get
t L
u ντx 2
−
θτx2 dx ds
r
τ 0 0
t L
t u ντx 2
θx2
2
2
2
θu f1 θx ≤ C1 −
dx ds
ds C1
τ θ
0 r
0 0
L
t t
t L u ντx 2
1
2
2
≤ C1 −
ds uL∞
θdx ds θx 1 2 dx ds
τ θ
0 r
0
0
0 0
t 2
u ντx ≤ C1 − τ ds.
0 r
2.40
We exploit the Gronwall inequality to 2.40 to obtain 2.37. The proof is complete.
Lemma 2.7. Under the assumptions of Theorem 1.1, one has, for any T > 0,
L
0
L
0
L
0
u2x x, tdx
vx2 x, tdx wx2 x, tdx t L
0
0
u2t u2xx x, sdx ds ≤ C1 ,
∀t ∈ 0, T ,
2.41
2
vt2 vxx
x, sdx ds ≤ C1 ,
∀t ∈ 0, T ,
2.42
t L
0
0
t L
0
0
2
wt2 wxx
x, sdx ds ≤ C1 ,
∀t ∈ 0, T .
2.43
Proof. Multiplying 1.17 by ut , integrating the result over QT , using Lemmas 2.1–2.6, and
taking into account that rux rut x /τ 1/2ru2x /τt − rux u2 x /τ ru3x /2τ 2 , we
obtain
t L
0
0
u2t dx ds
−
t L
0
0
νrux
−
rut x
τ
t L 0
0
γθ
τ
rut dx ds t L
0
x
0
v2
f1 ut dx ds
r
t L
r0 u0 2x
ν L ru2x
τθx − θτx
dx γ
dx −
r
ut dx ds
τ
2
τ
τ2
0
0
0
0 0
t L t L 2
ru2x
v
ν t L
2 rux
−ν
f1 ut dx ds
u
dx ds ru
dx ds τ
2 0 0
r
τ2
0 0
0 0
x
ν
2
L
x
≤ C1 −
ν
2
L
0
ru2x
1
dx τ
4
t L
0
0
u2t dx ds C1
t L
0
0
θ2 τx2 θx2 f12 v4 dx ds
14
Journal of Applied Mathematics
v2 f1 r
ut 1 γθ
dx ds
u
−
−ν
r
r τ x
r2
0 0
v2 f1 r
rux ut 1 γθ
ν t L
dx ds
2ru
−
2 0 0
τ
r
r τ x
r2
t L
2
t L
1 t L 2
ν L ru2x
dx θ2 τx2 θx2 f12 v4 u4 dx ds
ut dx ds C1
2 0 τ
2 0 0
0 0
t L
2
rux
C1
u2
x, sdx ds.
τ
0 0
≤ C1 −
2.44
We derive from 2.30 that
θ2L∞
≤
L
θ dx 2
2
L
0
|θθx |dx ≤ C1
0
L
0
θx2
θ
2
dx ≤ C1 C1
L
0
θx2 dx.
2.45
Therefore, using 1.24, 2.30, 2.36, 2.37, and 2.44-2.45, we conclude
L
0
ru2x
dx τ
t
L
0 0
u2t dx ds
≤ C1 C1
t
0
θ2L∞
L
0
τx2 dx ds
C1
t
0
u2L∞
L
0
ru2x
x, sdx ds
τ
2.46
which, by applying the Gronwall inequality, implies
L
0
u2x x, tdx
t L
0
0
u2t x, sdx ds ≤ C1 .
2.47
By 1.17, we have
t L
0
0
u2xx x, sdxds
≤ C1
t L
0
0
≤ C1 C1
u2t u2x u2 τx2 u2x θ2 τx2 θx2 v4 f12 x, sdx ds
t
0
≤ C1 C1
t
θ2L∞ ux 2L∞
L
2
t L
0
1 θx ux uxx ux sds
0
1
≤ C1 2
0
τx2 dx ds
0
u2xx x, sdx ds.
2.48
Journal of Applied Mathematics
15
Therefore,
t
2.49
uxx s2 ds ≤ C1
0
which, along with 2.47, gives 2.41.
Analogously, multiplying 1.18 by vt , integrating the result over QT , and using
assumptions 1.24 and Lemmas 2.1–2.6, we deduce
μ
2
L
0
t
rv2x
dx vt s2 ds
τ
0
L
μ
μ t L
r0 v0 2x
rv2x
dx ru
dx ds
2 0
τ0
2 0 0
τ
t L x
t L
uvvt
dx ds f2 vt dx ds
r
0 0
0 0
0 0
x
1 t
1 t L rvx
uv
vt − f2 dx ds
≤ C1 u
vt s2 ds −
4 0
2 0 0
τ
r
t L vt − f2 uv
2 dx ds
−
uv
r
r
0 0
L
t L
t
rv2x
1 t
u2 v2 f22 dx ds
dx ds C1
≤ C1 vt s2 ds u2L∞
2 0
τ
0
0
0 0
t
t
L
2
rvx
1
≤ C1 v2 dx ds.
vt s2 ds u2L∞
2 0
τ
0
0
−μ
rvx
τ
uvdx ds −
t L
2.50
In view of 2.36, we apply Gronwall’s inequality to 2.50 to obtain
L
0
rv2x
dx τ
t L
0
0
2.51
vt2 x, sdx ds ≤ C1 .
By 1.18 and 2.51, we easily deduce
t L
0
0
2.52
2
vxx
x, sdx ds ≤ C1
which, along with 2.51 and Lemmas 2.1–2.4, implies 2.42. The proof of 2.43 iS similar to
that of 2.41 and 2.42. The proof is now complete.
Lemma 2.8. Under the assumptions of Theorem 1.1, one has, for any T > 0,
L
0
θx2 x, tdx
t L
0
0
2
θt2 θxx
x, sdx ds ≤ C1 ,
∀t ∈ 0, T .
2.53
16
Journal of Applied Mathematics
Proof. Multiplying 1.20 by θt over QT , we have
κ
2
L
0
t L
r 2 θx2
dx CV
τ
L
κ
2
0
θt2 x, sdx ds
0
2
r02 θ0x
κ
dx τ0
2
t L
0
0
0
t L 0
0
r2
τ
θx2 dx ds
t
rvx 2
r 2 wx2
1
μ
− 2μ u2 v2
νrux − γθ rux μ
x
τ
τ
τ
2.54
grx, t, t θt x, sdx ds.
Using Lemmas 2.1–2.7, the Cauchy-Schwarz inequality, and the interpolation inequality, we
have
κ t L r2
2
θx dx ds
2 0 0
τ
t
≤ C1
t
0
≤ C1
uL∞ rux L∞
t
ux rux 0
≤ C1
L
0
θx2 dx ds
1/2
ruxx t
0
≤ C1
1/2
0
ux ruxx 2 rux 2/3
t L
0
L
0
≤ C1 C1
θx2 dx ds
0
u2 τx2
0
t
t
0
L
0
u2x
u2L∞ τx 2 u2 uxx 2
0
≤ C1 C1
C1
t L u2L∞
2
uxx L
0
θx2 dx ds
2.55
θx2 dx ds
u 2
L
0
u2xx
L
dx
0
θx2 dx ds
θx2 dx ds
θx2 dx ds,
t L 1
rvx 2
r 2 wx2
2
2
μ
− 2μ u v
gr, t θt dx ds
νrux − γθ rux μ
0 0 τ
x
τ
τ
1
≤
4
t
0
t L
ru4x rv4x θ2 ru2x wx4 u2 u2x v2 vx2 g 2 dx ds
θt ds C1
2
0
0
Journal of Applied Mathematics
≤
1
4
t
θt 2 ds C1
t
0
0
17
rux 2L∞ rux 2 θ2 rvx 2L∞ rvx 2
t L
u2 u2x v2 vx2 g 2 x, sdx ds.
wx 2L∞ wx 2 sds C1
0
0
2.56
Similarly to 2.55, by virtue of 1.25, 2.30, and 2.41–2.43, we arrive at
t L 1
rvx 2
r 2 wx2
μ
− 2μ u2 v2 gr, t θt dx ds
νrux − γθ rux μ
0 0 τ
x
τ
τ
1
≤ C1 4
≤ C1 ≤ C1 1
4
1
4
t
2
θt ds C1
0
t
0
θt 2 ds C1
0
t
t
rux 2L∞ rvx 2L∞ wx 2L∞
t
0
τx 2 ux 2H 1 vx 2H 1 wx 2H 1
θt 2 ds.
0
2.57
Combining 2.54–2.57, we conclude
θx 2 t
θt s2 ds ≤ C1 C1
0
t
0
u2L∞ uxx 2 θx 2 ds.
2.58
In view of 2.36 and 2.41, we apply Gronwall’s inequality to 2.58 to obtain
θx t2 t
θt s2 ds ≤ C1 ,
∀t ∈ 0, T .
2.59
0
Similarly to proof of 2.41, by Lemmas 2.1–2.7, 1.20, 1.25, and 2.59, we obtain
t
θxx s2 ds ≤ C1
2.60
0
which, together with 2.59, implies 2.53. The proof is complete.
Proof of Theorem 1.1. By Lemmas 2.1–2.8, we complete the proof of Theorem 1.1.
Acknowledgments
The work is in part supported by Doctoral Foundation of North China University of Water
Sources and Electric Power no. 201087 and the Natural Science Foundation of Henan
Province of China no. 112300410040.
18
Journal of Applied Mathematics
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