# Global Solutions to the Spherically Symmetric Compressible Navier-Stokes Equations with Density-Dependent Viscosity

Authors: Ruxu Lian; Lan Huang; Jian Liu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

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 ast→+∞.


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## Body

## 1. Introduction

The isentropic compressible Navier-Stokes equations with density-dependent viscosity coefficients read as follows:(1.1)ρt+div⁡(ρU)=0,(ρU)t+div⁡(ρU⊗U)+∇P(ρ)-div⁡(μ(ρ)D(U))-∇(λ(ρ)div⁡U)=0,
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
(1.2)D(U)=∇(U)+t∇(U)2
is the strain tensor and μ(ρ), λ(ρ) are the Lamé viscosity coefficients satisfying
(1.3)μ(ρ)>0,μ(ρ)+Nλ(ρ)≥0.There are many important progress achieved recently on the compressible Navier-Stokes 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. Asγ>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 Section2, 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 takeμ(ρ)=ρ and λ(ρ)=0 and D(U)=∇U in (1.1). The isentropic compressible Navier-Stokes equations become
(2.1)ρt+div⁡(ρU)=0,(ρU)t+div⁡(ρU⊗U)+∇P(ρ)-div⁡(ρ∇U)=0.
Firstly, we consider the exterior problem, namely, the initial data and boundary conditions of (2.1) are imposed as follows:
(2.2)(ρ,U)(x,0)=(ρ0,U0)(x),x∈Ω,U=0,on⁡∂Ω,lim⁡|x|→+∞(ρ,u)(x,t)=(ρ¯,0),t∈[0,T],
where Ω∶=R3/Ωr-, Ωr- is a ball of radius r- centered at the origin in R3, and ρ¯>0 is a constant.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
(2.3)|x|=r,ρ(x,t)=ρ(r,t),U(x,t)=u(r,t)xr,
which leads to the following system of equations for r>0,
(2.4)ρt+(ρu)r+2ρur=0,(ρu)t+(ρu2+ργ)r+2ρu2r-(ρur)r-ρ(2ur)r=0,
with the initial data and boundary conditions
(2.5)(ρ,u)(r,0)=(ρ0,u0)(r),r∈[r-,+∞),u(r-,t)=0,lim⁡r→+∞(ρ,u)(r,t)=(ρ¯,0)t∈[0,T],
and the initial data satisfies for some constants ρ¯ and ρ̲>0(2.6)ρ0-ρ¯∈L1([r-,+∞))∩L2([r-,+∞)),inf⁡x∈[r-,+∞)ρ0>ρ̲>0,rρ0r∈L2([r-,+∞)),r2u0∈H2([r-,+∞)).
Then, we define that
(2.7)E0∶=12∫r-+∞ρ0u02r2dr+∫r-+∞ρ0(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))r2dr,E1∶=12∫r-+∞ρ0(u0+ρ0-1(ρ0)r)2r2dr+∫r-+∞ρ0(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))r2dr,
and give the main results as follows.Theorem 2.1.
Letγ>1. Assume that the initial data satisfies (2.6) and E0+E1<νr-2ρ¯γ/2+1/2, 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
(2.8)0<ρ*≤ρ(r,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)(2.9)‖(ρ-ρ¯,u)(⋅,t)‖L∞([r-,+∞))⟶0,t⟶+∞.Then investigates the initial boundary value problem, and the initial data and boundary conditions of (2.1) are assumed as follow:
(2.10)(ρ,U)(x,0)=(ρ0,U0)(x),x∈Ω,U(x,t)=0,x∈∂Ω,t∈[0,T].
By (2.3), one considers (2.4) with the initial data and boundary conditions
(2.11)(ρ,u)(r,0)=(ρ0,u0)(r),r∈[r-,r+],u(r-,t)=u(r+,t)=0,t∈[0,T],
and the initial data satisfies for some constant ρ̲>0(2.12)ρ0∈L1([r-,r+])∩W1,∞([r-,r+]),inf⁡x∈[r-,r+]ρ0>ρ̲>0,r2u0∈H2([r-,r+]).Then, can give the main results as follows.Theorem 2.2.
Letγ>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
(2.13)0<ρ*≤ρ(r,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+])).
Furthermore, the solution (ρ,u) tends to the equilibrium state exponentially
(2.14)‖(ρ-ρ̃,u)(⋅,t)‖L∞([r-,r+])≤C0e-C1t,
where C0 and C1 are positive constants independent of time and ρ̃=∫r-r+ρ(r,t)r2dr.Remark 2.3.
Theorems2.1 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(3.1)x=∫r-rρ(r,t)r2dr,τ=t,
which map (r,t)∈[r-,+∞)×R+ into (x,τ)∈[0,+∞)×R+. The relation between Lagrangian and Eulerian coordinates are satisfied as
(3.2)∂x∂r=ρr2,∂x∂t=-ρur2.
The exterior problem (2.4) and (2.5) is reformulated to
(3.3)ρτ+ρ2(r2u)x=0,r-2uτ+(ργ)x=(ρ2(r2u)x)x-2ρxur,(ρ,u)(x,0)=(ρ0,u0)(x),x∈[0,+∞),u(0,t)=0,lim⁡x→+∞(ρ,u)=(ρ¯,0),τ∈[0,+∞),
where the initial data satisfies
(3.4)ρ0-ρ¯∈L1([0,+∞))∩L2([0,+∞)),inf⁡x∈[0,+∞)ρ0>ρ->0,r2ρ0x∈L2([0,+∞)),1r2ρ0(r2u0)∈L2([0,+∞)),r2ρ0(r2u0)x∈L2([0,+∞)),1r2ρ0(r2ρ(r2ρ(r2u0)x)x)∈L2([0,+∞)).
First, we will establish the a-priori estimates for the solution (ρ,u) to the exterior problem (3.3).Lemma 3.1.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.5)12∫0+∞u2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0τ∫0+∞(2u2r2+ρ2ux2r4)dxds≤E0,τ∈[0,T].Proof.
Multiplying (3.3) 2 by r2u and integrating the result with respect to x over [0,+∞), making use of (3.3) 1 and (3.4), we have
(3.6)12ddτ∫0+∞u2dx+ddτ∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0+∞ρ2(r2u)x2dx=2∫0+∞ρ(ru2)xdx,
integrating (3.6) with respect to τ, we obtain
(3.7)12∫0+∞u2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0τ∫0+∞(2u2r2+ρ2ux2r4)dxds=12∫0+∞u02dx+∫0+∞(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))dx.
Lemma 3.1 can be obtained.Lemma 3.2.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.8)12∫0+∞(u+r2ρx)2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+γ∫0τ∫0+∞ργ-1ρx2r4dxds≤E1,τ∈[0,T],
and there exist two constants 0<ρ*<ρ* such that
(3.9)0<ρ*≤ρ(x,τ)≤ρ*,(x,τ)∈[0,+∞)×[0,T].Proof.
Differentiating (3.3) 1 with respect to x, we have
(3.10)ρxτ+(ρ2(r2u)x)x=0.
Summing (3.10) and (3.3) 2, we get
(3.11)(r-2u+ρx)τ+(ργ)x=(r-2)τu-2ρxur.
Note that
(3.12)r3(x,τ)=r-3+3∫0x1ρ(z,τ)dz,
and so
(3.13)∂r∂τ=1r2∫0x(1ρ)t(z,t)dz=1r2∫0x(r2u)z(z,t)dz=u(x,τ),
which together with (3.11) yields
(3.14)(r-2u+ρx)τ+(ργ)x=-2r-3u2-2ρxur.
Multiplying (3.14) by (u+r2ρx)r2 and integrating the result with respect to x and τ, we have
(3.15)12∫0+∞(u+r2ρx)2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+γ∫0τ∫0+∞ργ-1ρx2r4dxds=12∫0+∞(u0+r2ρ0x)2dx+∫0+∞(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))dx.
Let(3.16)φ(ρ)∶=1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1),ψ(ρ)∶=∫ρ¯ρφ(η)1/2dη.
It follows from (3.6) and (3.13) that
(3.17)|ψ(ρ)|≤|∫y+∞∂xψ(ρ)dx|≤r--2|∫y+∞φ(ρ)1/2ρxr2dx|≤r--2|∫0+∞φ(ρ)dy∫0+∞(r2ρx)2dx|1/2≤r--2(E0+E1).
We can verify that(1)
Asρ→+∞, it holds for ξ=θρ¯+(1-θ)ρ, where θ∈(0,1)(3.18)lim⁡ρ→+∞ψ(ρ)=lim⁡ρ→+∞∫ρ¯ρ((γ-2)ξγ-3+2ρ¯γξ-3)1/2|η-ρ¯|dη≥lim⁡ρ→+∞((γ-2)ξγ-3+2ρ¯γξ-3)1/2∫ρ¯ρ(η-ρ¯)dη=lim⁡ρ→+∞12((γ-2)(θρ¯+(1-θ)ρ)γ-3+2ρ¯γ(θρ¯+(1-θ)ρ)-3)1/2(ρ-ρ¯)2⟶+∞.(2)
Asρ→0, it holds for ξ=θρ¯+(1-θ)ρ, where θ∈(0,1)(3.19)lim⁡ρ→0ψ(ρ)=lim⁡ρ→0∫ρ¯ρ((γ-2)ξγ-3+2ρ¯γξ-3)1/2|η-ρ¯|dη≤-lim⁡ρ→0((γ-2)ξγ-3+2ρ¯γξ-3)1/2∫ρ¯ρ(η-ρ¯)dη=-lim⁡ρ→012((γ-2)(θρ¯+(1-θ)ρ)γ-3+2ρ¯γ(θρ¯+(1-θ)ρ)-3)1/2(ρ-ρ¯)2=:-νρ¯γ/2+1/2.
Applying (3.17)–(3.19) and E0+E1<νr-2ρ¯γ/2+1/2, where ν is a positive constant, we can prove (3.9).Lemma 3.3.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.20)∫0+∞(r2u)x2dx+∫0+∞(r2u)τ2r-4dx+∫0τ∫0+∞(r2u)s2r-4dxds+∫0τ∫0+∞ρ2(r2u)xs2dxds+∫0τ∫0+∞(r2u)xx2dxds≤C,τ∈[0,T],
where C>0 denotes a constant independent of time.Proof.
Multiplying (3.3)2 by ρ-2(r2u)τ and integrating the result with respect to x over [0,+∞), making use of (3.4), we obtain
(3.21)ddτ∫0+∞(12(r2u)x2-ργ-2(r2u)x)dx+∫0+∞ρ-2(r2u)τ2r-4dx=(γ-2)∫0+∞ργ-1(r2u)x2dx-2∫0+∞ργ-3ρx(r2u)τdx+2∫0+∞ρ-1ρx(r2u)x(r2u)τdx+2∫0+∞ρ-2u2(r2u)τr-3dx-2∫0+∞ρ-2ρxu(r2u)τr-1dx,
which implies
(3.22)∫0+∞(r2u)x2dx+∫0τ∫0+∞(r2u)s2r-4dxds≤C+C∫0+∞(ργ-2-ρ¯γ-2)2dx+C∫0τ∫0+∞(u2r2+ux2r4)dxds+C∫0τ∫0+∞ρx2r4dxds+C∫0τ∫0+∞ρx2(r2u)x2r4dxds+C∫0τ∫0+∞u4r-2dxds+C∫0τ∫0+∞ρx2u2r2dxds≤C+C∫0τ∫0+∞ρx2(r2u)x2r4dxds+Csup⁡τ∈[0,T]‖u‖L∞2.
From (3.3) 2, (3.5), (3.8), and (3.9), we can deduce that for some small ϵ∈(0,1)(3.23)∫0τ∫0+∞ρx2(r2u)x2r4dxds≤ϵ∫0τ∫0+∞ρx2(r2u)x2r4dxds+ϵ∫0τ∫0+∞(r2u)xs2r-4dxds+ϵ∫0τ∫0+∞ρx2dxds+C(ϵ)∫0τ∫0+∞(u2r2+ux2r4)dxds,(3.24)sup⁡τ∈[0,T]‖u‖L∞2≤ϵsup⁡τ∈[0,T]∫0+∞(r2u)x2dx+C(ϵ)sup⁡τ∈[0,T]∫0+∞u2dx,
using (3.22)–(3.24), we can obtain that
(3.25)∫0+∞(r2u)x2dx+∫0τ∫0+∞(r2u)s2r-4dxds≤C+Cϵ∫0τ∫0+∞(r2u)xs2r-4dxds.
Differentiating (3.3) 2 with respect to τ, multiplying the result by (r2u)τ, and integrating the result with respect to x over [0,+∞), we have
(3.26)12ddτ∫0+∞(r2u)τ2r-4dx+∫0+∞ρ2(r2u)xτ2dx=2∫0+∞uuτ(r2u)τr-3dx-12∫0+∞(r-4)τ(r2u)τ2dx+2∫0+∞u(r-1u)τ(r2u)τr-2dx+∫0+∞(ργ)τ(r2u)xτdx-∫0+∞(ρ2)τ(r2u)x(r2u)xτdx-∫0+∞(2ρxur)τ(r2u)τdx+2∫0+∞(r-2)τr-1u2(r2u)τdx.
A complicated computation gives
(3.27)ddτ∫0+∞(r2u)τ2r-4dx+∫0+∞ρ2(r2u)xτ2dx≤C∫0+∞(r2u)τ2r-4dx+Csup⁡τ∈[0,T]‖(r2u)x2‖L∞∫0+∞(r2u)x2dx+C∫0+∞(u2r2+ux2r4)dx(1+sup⁡τ∈[0,T]∫0+∞(r2u)x2dx),
integrating (3.27) with respect to τ, by means of (3.3) 2, (3.5), (3.8), (3.9), and (3.25), it holds that
(3.28)∫0+∞(r2u)τ2r-4dx+∫0τ∫0+∞ρ2(r2u)xs2dxds≤C+C∫0τ∫0+∞(r2u)s2r-4dxds+Csup⁡τ∈[0,T]‖(r2u)x2‖L∞+Csup⁡τ∈[0,T]∫0+∞(r2u)x2dx≤C+C∫0τ∫0+∞(r2u)s2r-4dxds+Csup⁡τ∈[0,T](∫0+∞(r2u)x2dx)1/2(∫0+∞(r2u)xx2dx)1/2+Csup⁡τ∈[0,T]∫0+∞(r2u)x2dx≤C+Cϵ∫0τ∫0+∞ρ2(r2u)xs2dxds+Cϵ∫0+∞(r2u)τ2r-4dx,
choosing the constant ϵ small sufficiently, we can complete the proof of Lemma 3.3.Remark 3.4.
By Lemmas3.1–3.3, the following inequality holds:
(3.29)∫0+∞u2dx+∫0+∞(ρ-ρ¯)2dx+∫0+∞ux2dx+∫0+∞uτ2dx+∫0+∞ρx2dx+∫0τ∫0+∞ρx2dxds+∫0τ∫0+∞ux2dxds+∫0τ∫0+∞us2dxds+∫0τ∫0+∞uxx2dxds+∫0τ∫0+∞uxτ2dxds≤C.Lemma 3.5.
Under the conditions in Theorem2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.30)‖(ρ-ρ¯,u)(⋅,τ)‖L∞([0,+∞))⟶0,τ⟶+∞,
where C>0 denotes a constant independent of time.Proof.
From Lemmas3.1–3.3, we can obtain
(3.31)∫0∞‖(ρ-ρ¯,u)x‖L2([0,+∞))2dτ≤r--4∫0∞‖(ρ-ρ¯,u)xr2‖L2([0,+∞))2dτ≤C,(3.32)∫0∞|ddτ‖(ρ-ρ¯,u)x‖L2([0,+∞))2|dτ=∫0∞|∫0+∞(-4ρρx2(r2u)x-2ρ2ρx(r2u)xx)dx+2∫0+∞uxuxτdx|dτ≤C∫0∞∫0+∞ρx2dxdτ+C∫0∞∫0+∞(u2r2+ux2r4+(r2u)xx2+uxτ2)dxdτ≤C,
which together with (3.31) implies
(3.33)‖(ρ-ρ¯,u)x‖L2([0,+∞))2∈W1,1(R+).
It holds from Gagliardo-Nirenberg-Sobolev inequality that
(3.34)‖(ρ-ρ¯,u)‖L∞([0,+∞))≤‖(ρ-ρ¯,u)‖L2([0,+∞))1/2‖(ρ-ρ¯,u)x‖L2([0,+∞))1/2,
which together with (3.5), (3.9), and (3.33) implies this lemma.


### 3.2. Proof of Theorem2.1

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 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.


## 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(3.1)x=∫r-rρ(r,t)r2dr,τ=t,
which map (r,t)∈[r-,+∞)×R+ into (x,τ)∈[0,+∞)×R+. The relation between Lagrangian and Eulerian coordinates are satisfied as
(3.2)∂x∂r=ρr2,∂x∂t=-ρur2.
The exterior problem (2.4) and (2.5) is reformulated to
(3.3)ρτ+ρ2(r2u)x=0,r-2uτ+(ργ)x=(ρ2(r2u)x)x-2ρxur,(ρ,u)(x,0)=(ρ0,u0)(x),x∈[0,+∞),u(0,t)=0,lim⁡x→+∞(ρ,u)=(ρ¯,0),τ∈[0,+∞),
where the initial data satisfies
(3.4)ρ0-ρ¯∈L1([0,+∞))∩L2([0,+∞)),inf⁡x∈[0,+∞)ρ0>ρ->0,r2ρ0x∈L2([0,+∞)),1r2ρ0(r2u0)∈L2([0,+∞)),r2ρ0(r2u0)x∈L2([0,+∞)),1r2ρ0(r2ρ(r2ρ(r2u0)x)x)∈L2([0,+∞)).
First, we will establish the a-priori estimates for the solution (ρ,u) to the exterior problem (3.3).Lemma 3.1.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.5)12∫0+∞u2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0τ∫0+∞(2u2r2+ρ2ux2r4)dxds≤E0,τ∈[0,T].Proof.
Multiplying (3.3) 2 by r2u and integrating the result with respect to x over [0,+∞), making use of (3.3) 1 and (3.4), we have
(3.6)12ddτ∫0+∞u2dx+ddτ∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0+∞ρ2(r2u)x2dx=2∫0+∞ρ(ru2)xdx,
integrating (3.6) with respect to τ, we obtain
(3.7)12∫0+∞u2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+∫0τ∫0+∞(2u2r2+ρ2ux2r4)dxds=12∫0+∞u02dx+∫0+∞(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))dx.
Lemma 3.1 can be obtained.Lemma 3.2.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.8)12∫0+∞(u+r2ρx)2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+γ∫0τ∫0+∞ργ-1ρx2r4dxds≤E1,τ∈[0,T],
and there exist two constants 0<ρ*<ρ* such that
(3.9)0<ρ*≤ρ(x,τ)≤ρ*,(x,τ)∈[0,+∞)×[0,T].Proof.
Differentiating (3.3) 1 with respect to x, we have
(3.10)ρxτ+(ρ2(r2u)x)x=0.
Summing (3.10) and (3.3) 2, we get
(3.11)(r-2u+ρx)τ+(ργ)x=(r-2)τu-2ρxur.
Note that
(3.12)r3(x,τ)=r-3+3∫0x1ρ(z,τ)dz,
and so
(3.13)∂r∂τ=1r2∫0x(1ρ)t(z,t)dz=1r2∫0x(r2u)z(z,t)dz=u(x,τ),
which together with (3.11) yields
(3.14)(r-2u+ρx)τ+(ργ)x=-2r-3u2-2ρxur.
Multiplying (3.14) by (u+r2ρx)r2 and integrating the result with respect to x and τ, we have
(3.15)12∫0+∞(u+r2ρx)2dx+∫0+∞(1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1))dx+γ∫0τ∫0+∞ργ-1ρx2r4dxds=12∫0+∞(u0+r2ρ0x)2dx+∫0+∞(1γ-1(ρ0γ-1-ρ¯γ-1)+ρ¯γ(ρ0-1-ρ¯-1))dx.
Let(3.16)φ(ρ)∶=1γ-1(ργ-1-ρ¯γ-1)+ρ¯γ(ρ-1-ρ¯-1),ψ(ρ)∶=∫ρ¯ρφ(η)1/2dη.
It follows from (3.6) and (3.13) that
(3.17)|ψ(ρ)|≤|∫y+∞∂xψ(ρ)dx|≤r--2|∫y+∞φ(ρ)1/2ρxr2dx|≤r--2|∫0+∞φ(ρ)dy∫0+∞(r2ρx)2dx|1/2≤r--2(E0+E1).
We can verify that(1)
Asρ→+∞, it holds for ξ=θρ¯+(1-θ)ρ, where θ∈(0,1)(3.18)lim⁡ρ→+∞ψ(ρ)=lim⁡ρ→+∞∫ρ¯ρ((γ-2)ξγ-3+2ρ¯γξ-3)1/2|η-ρ¯|dη≥lim⁡ρ→+∞((γ-2)ξγ-3+2ρ¯γξ-3)1/2∫ρ¯ρ(η-ρ¯)dη=lim⁡ρ→+∞12((γ-2)(θρ¯+(1-θ)ρ)γ-3+2ρ¯γ(θρ¯+(1-θ)ρ)-3)1/2(ρ-ρ¯)2⟶+∞.(2)
Asρ→0, it holds for ξ=θρ¯+(1-θ)ρ, where θ∈(0,1)(3.19)lim⁡ρ→0ψ(ρ)=lim⁡ρ→0∫ρ¯ρ((γ-2)ξγ-3+2ρ¯γξ-3)1/2|η-ρ¯|dη≤-lim⁡ρ→0((γ-2)ξγ-3+2ρ¯γξ-3)1/2∫ρ¯ρ(η-ρ¯)dη=-lim⁡ρ→012((γ-2)(θρ¯+(1-θ)ρ)γ-3+2ρ¯γ(θρ¯+(1-θ)ρ)-3)1/2(ρ-ρ¯)2=:-νρ¯γ/2+1/2.
Applying (3.17)–(3.19) and E0+E1<νr-2ρ¯γ/2+1/2, where ν is a positive constant, we can prove (3.9).Lemma 3.3.
LetT>0. Under the conditions in Theorem 2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.20)∫0+∞(r2u)x2dx+∫0+∞(r2u)τ2r-4dx+∫0τ∫0+∞(r2u)s2r-4dxds+∫0τ∫0+∞ρ2(r2u)xs2dxds+∫0τ∫0+∞(r2u)xx2dxds≤C,τ∈[0,T],
where C>0 denotes a constant independent of time.Proof.
Multiplying (3.3)2 by ρ-2(r2u)τ and integrating the result with respect to x over [0,+∞), making use of (3.4), we obtain
(3.21)ddτ∫0+∞(12(r2u)x2-ργ-2(r2u)x)dx+∫0+∞ρ-2(r2u)τ2r-4dx=(γ-2)∫0+∞ργ-1(r2u)x2dx-2∫0+∞ργ-3ρx(r2u)τdx+2∫0+∞ρ-1ρx(r2u)x(r2u)τdx+2∫0+∞ρ-2u2(r2u)τr-3dx-2∫0+∞ρ-2ρxu(r2u)τr-1dx,
which implies
(3.22)∫0+∞(r2u)x2dx+∫0τ∫0+∞(r2u)s2r-4dxds≤C+C∫0+∞(ργ-2-ρ¯γ-2)2dx+C∫0τ∫0+∞(u2r2+ux2r4)dxds+C∫0τ∫0+∞ρx2r4dxds+C∫0τ∫0+∞ρx2(r2u)x2r4dxds+C∫0τ∫0+∞u4r-2dxds+C∫0τ∫0+∞ρx2u2r2dxds≤C+C∫0τ∫0+∞ρx2(r2u)x2r4dxds+Csup⁡τ∈[0,T]‖u‖L∞2.
From (3.3) 2, (3.5), (3.8), and (3.9), we can deduce that for some small ϵ∈(0,1)(3.23)∫0τ∫0+∞ρx2(r2u)x2r4dxds≤ϵ∫0τ∫0+∞ρx2(r2u)x2r4dxds+ϵ∫0τ∫0+∞(r2u)xs2r-4dxds+ϵ∫0τ∫0+∞ρx2dxds+C(ϵ)∫0τ∫0+∞(u2r2+ux2r4)dxds,(3.24)sup⁡τ∈[0,T]‖u‖L∞2≤ϵsup⁡τ∈[0,T]∫0+∞(r2u)x2dx+C(ϵ)sup⁡τ∈[0,T]∫0+∞u2dx,
using (3.22)–(3.24), we can obtain that
(3.25)∫0+∞(r2u)x2dx+∫0τ∫0+∞(r2u)s2r-4dxds≤C+Cϵ∫0τ∫0+∞(r2u)xs2r-4dxds.
Differentiating (3.3) 2 with respect to τ, multiplying the result by (r2u)τ, and integrating the result with respect to x over [0,+∞), we have
(3.26)12ddτ∫0+∞(r2u)τ2r-4dx+∫0+∞ρ2(r2u)xτ2dx=2∫0+∞uuτ(r2u)τr-3dx-12∫0+∞(r-4)τ(r2u)τ2dx+2∫0+∞u(r-1u)τ(r2u)τr-2dx+∫0+∞(ργ)τ(r2u)xτdx-∫0+∞(ρ2)τ(r2u)x(r2u)xτdx-∫0+∞(2ρxur)τ(r2u)τdx+2∫0+∞(r-2)τr-1u2(r2u)τdx.
A complicated computation gives
(3.27)ddτ∫0+∞(r2u)τ2r-4dx+∫0+∞ρ2(r2u)xτ2dx≤C∫0+∞(r2u)τ2r-4dx+Csup⁡τ∈[0,T]‖(r2u)x2‖L∞∫0+∞(r2u)x2dx+C∫0+∞(u2r2+ux2r4)dx(1+sup⁡τ∈[0,T]∫0+∞(r2u)x2dx),
integrating (3.27) with respect to τ, by means of (3.3) 2, (3.5), (3.8), (3.9), and (3.25), it holds that
(3.28)∫0+∞(r2u)τ2r-4dx+∫0τ∫0+∞ρ2(r2u)xs2dxds≤C+C∫0τ∫0+∞(r2u)s2r-4dxds+Csup⁡τ∈[0,T]‖(r2u)x2‖L∞+Csup⁡τ∈[0,T]∫0+∞(r2u)x2dx≤C+C∫0τ∫0+∞(r2u)s2r-4dxds+Csup⁡τ∈[0,T](∫0+∞(r2u)x2dx)1/2(∫0+∞(r2u)xx2dx)1/2+Csup⁡τ∈[0,T]∫0+∞(r2u)x2dx≤C+Cϵ∫0τ∫0+∞ρ2(r2u)xs2dxds+Cϵ∫0+∞(r2u)τ2r-4dx,
choosing the constant ϵ small sufficiently, we can complete the proof of Lemma 3.3.Remark 3.4.
By Lemmas3.1–3.3, the following inequality holds:
(3.29)∫0+∞u2dx+∫0+∞(ρ-ρ¯)2dx+∫0+∞ux2dx+∫0+∞uτ2dx+∫0+∞ρx2dx+∫0τ∫0+∞ρx2dxds+∫0τ∫0+∞ux2dxds+∫0τ∫0+∞us2dxds+∫0τ∫0+∞uxx2dxds+∫0τ∫0+∞uxτ2dxds≤C.Lemma 3.5.
Under the conditions in Theorem2.1, it holds for any solution (ρ,u) to the exterior problem (3.3) that
(3.30)‖(ρ-ρ¯,u)(⋅,τ)‖L∞([0,+∞))⟶0,τ⟶+∞,
where C>0 denotes a constant independent of time.Proof.
From Lemmas3.1–3.3, we can obtain
(3.31)∫0∞‖(ρ-ρ¯,u)x‖L2([0,+∞))2dτ≤r--4∫0∞‖(ρ-ρ¯,u)xr2‖L2([0,+∞))2dτ≤C,(3.32)∫0∞|ddτ‖(ρ-ρ¯,u)x‖L2([0,+∞))2|dτ=∫0∞|∫0+∞(-4ρρx2(r2u)x-2ρ2ρx(r2u)xx)dx+2∫0+∞uxuxτdx|dτ≤C∫0∞∫0+∞ρx2dxdτ+C∫0∞∫0+∞(u2r2+ux2r4+(r2u)xx2+uxτ2)dxdτ≤C,
which together with (3.31) implies
(3.33)‖(ρ-ρ¯,u)x‖L2([0,+∞))2∈W1,1(R+).
It holds from Gagliardo-Nirenberg-Sobolev inequality that
(3.34)‖(ρ-ρ¯,u)‖L∞([0,+∞))≤‖(ρ-ρ¯,u)‖L2([0,+∞))1/2‖(ρ-ρ¯,u)x‖L2([0,+∞))1/2,
which together with (3.5), (3.9), and (3.33) implies this lemma.


## 3.2. Proof of Theorem2.1

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 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(4.1)x=∫r-rρ(r,t)r2dr,τ=t.
By (4.1) and the conservation of mass for (ρ,u)(4.2)∫r-r+ρ(r,t)r2dr=∫r-r+ρ0(r)r2dr=:1,
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(4.3)∂x∂r=ρr2,∂x∂t=-ρur2,
and the initial boundary value problem’s (2.4) and (2.11) are reformulated to
(4.4)ρτ+ρ2(r2u)x=0,r-2uτ+(ργ)x=(ρ2(r2u)x)x-2ρxur,(ρ,u)(x,0)=(ρ0,u0)(x),x∈[0,1],u(0,t)=u(1,t)=0,τ∈[0,+∞),
where the initial data satisfies
(4.5)ρ0∈L1([0,1])∩W1,∞([0,1]),inf⁡x∈[0,1]ρ0>ρ̲>0,1r2ρ0(r2u0)∈L2([0,1]),r2ρ0(r2u0)x∈L2([0,1]),1r2ρ0(r2ρ(r2ρ(r2u0)x)x)∈L2([0,1]).
Then, we will establish the a-priori estimates for the solution (ρ,u) to the initial boundary value problem (4.4).Lemma 4.1.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.6)∫01(12u2+1γ-1ργ-1)dx+∫0τ∫01(2u2r2+ρ2ux2r4)dxds=∫01(12u02+1γ-1ρ0γ-1)dx,τ∈[0,T].Proof.
Multiplying (4.4)2 by r2u and integrating the result with respect to x over [0,1], using (4.4)1 and (4.5), we obtain
(4.7)ddτ∫01(12u2+1γ-1ργ-1)dx+∫01ρ2(r2u)x2dx=2∫01ρ(ru2)xdx,
and integrating (4.7) with respect to τ, we obtain
(4.8)∫01(12u2+1γ-1ργ-1)dx+∫0τ∫01(2u2r2+ρ2ux2r4)dxds=∫01(12u02+1γ-1ρ0γ-1)dx.
Lemma 4.1 can be obtained.Lemma 4.2.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.9)12∫01(u+r2ρx)2dx+1γ-1∫01ργ-1dx+γ∫0τ∫01ργ-1ρx2r4dxds=12∫01(u0+r2ρ0x)2dx+1γ-1∫01ρ0γ-1dx,τ∈[0,T],
where C is a positive constant independent of time.Proof.
Differentiating (4.4)1 with respect to x, we have
(4.10)ρxτ+(ρ2(r2u)x)x=0,
which together with (4.4)2 and ∂r/∂τ=u gives
(4.11)(r-2u+ρx)τ+(ργ)x=-2r-3u2-2ρxur.
Multiplying (4.11) by (u+r2ρx)r2, and integrating the result with respect to x and τ, it holds that
(4.12)12∫01(u+r2ρx)2dx+1γ-1∫01ργ-1dx+γ∫0τ∫01ργ-1ρx2r4dxds=12∫01(u0+r2ρ0x)2dx+1γ-1∫01ρ0γ-1dx.
The proof of (4.9) is completed.Lemma 4.3.
LetT>0. Under the conditions in Theorem 2.2, there exists a constant ρ*>0 such that
(4.13)0<ρ(x,τ)≤ρ*,(x,τ)∈[0,1]×[0,T].Proof.
It follows from (4.6) and (4.9) that
(4.14)ρ(x,τ)=ρ(r,t)≤∫r-r+ρ(r,t)dr+∫r-r+|ρr(r,t)|dr≤r--2∫r-r+ρ(r,t)r2dr+r--2∫r-r+ρ|ρr(r,t)|ρr2dr≤C+C∫r-r+1ρ|ρr(r,t)|2r2dr≤C+C∫01|ρx(x,τ)|2r4dx≤C=:ρ*.Lemma 4.4.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.15)∫01u2ndx+∫0τ∫01(u2nr2+ρ2u2n-2ux2r4)dxds≤C(T),
for any positive integer n∈N, where C(T) is a positive constant dependent of time.Proof.
Multiplying (4.4)2 with u2n-1, integrating by parts over [0,1], we have
(4.16)12nddτ∫01u2ndx+∫01ρ2(r2u)x(r2u2n-1)xdx=∫01(ργ(r2u2n-1)x+ρ(2ru2n)x)dx.
Since it holds that
(4.17)(r2u)x(r2u2n-1)x=(2uρr+r2ux)(2u2n-1ρr+(2n-1)r2u2n-2ux)=4u2nρ2r2+(2n-1)u2n-2ux2r4+4nu2n-1uxrρ,
it follows from (4.16) that
(4.18)ddτ∫01u2n2ndx+2∫01u2nr2dx+(2n-1)∫01ρ2u2n-2ux2r4dx=2∫01ργ-1u2n-1rdx+(2n-1)∫01ργu2n-2uxr2dx≤∫01u2nr2dx+∫01ρ2u2n-2ux2r4dx+C‖ρ‖L∞2(γ-1)∫01u2n-2dx,
which together with (4.13) and Young’s inequality yields
(4.19)ddτ∫01u2ndx+∫01u2nr2dx+∫01ρ2u2n-2ux2r4dx≤C+∫01u2ndx,
and by applying the Gronwall’s inequality to (4.19), we can obtain (4.15).Lemma 4.5.
LetT>0, for n∈N, and n>1/2(γ-1). Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.20)∫0τ‖ρ2n(γ-1)u2n‖L∞([0,1])ds≤C(T),(4.21)∫0τ‖|(ργ)xr2|2n‖L∞([0,1])ds≤C(T),τ∈[0,T],
where C(T) is a positive constant dependent of time.Proof.
By means of Sobolev imbedding theorem and Cauchy-Schwarz inequality, applying (4.6), (4.13), and (4.15), we get
(4.22)∫0τ‖ρ2n(γ-1)u2n‖L∞([0,1])ds≤∫0τ∫01|ρ2n(γ-1)u2n|dxds+∫0τ∫01|(ρ2n(γ-1)u2n)x|dxds≤C(T)+C∫0τ∫01ρ2n(γ-1)-1|ρx|u2ndxds+C∫0τ∫01ρ2n(γ-1)|u2n-1ux|dxds≤C(T)+C∫0τ∫01|ρx|2r4dxds+C∫0τ∫01ρ2(2n(γ-1)-1)u4nr-4dxds+∫0τ∫01ρ2(2n(γ-1)-1)u2nr-4dxds+C∫0τ∫01ρ2u2n-2ux2r4dxds≤C(T).
Next, we find that
(4.23)ρxr2=ρ0xr02+u0-u-∫0τ(ργ)xr2ds,
which together with (4.13), (4.15), and (4.22) gives
(4.24)∫0τ‖|(ργ)xr2|2n‖L∞ds=γ2n∫0τ‖ρ2n(γ-1)|ρxr2|2n‖L∞ds=γ2n∫0τ‖ρ2n(γ-1)|ρ0xr02+u0-u-∫0s(ργ)xr2dl|2n‖L∞ds≤C∫0τ‖ρ2n(γ-1)(ρ0x2n+u02n+u2n)‖L∞ds+C∫0τ‖ρ2n(γ-1)(∫0s(ργ)xr2dl)2n‖L∞ds≤C(T)+C∫0τ‖ρ2n(γ-1)u2n‖L∞ds+C(T)∫0τ∫0s‖|(ργ)xr2|2n‖L∞dlds,
applying the Gronwall’s inequality to (4.24), we obtain (4.21).Lemma 4.6.
LetT>0. Under the conditions in Theorem 2.2, there exists a constant ρ*>0 such that
(4.25)ρ(x,τ)≥ρ*>0,(x,τ)∈[0,1]×[0,T].Proof.
It is easy to verify that(4.26)ρ(γ+1)/2¯(τ)=:∫01ρ(γ+1)/2(x,τ)dx≥r-2r+2(∫r-r+ρ(r,t)r2dr)1+(γ+1)/2=r-2r+2(∫r-r+ρ0(r)r2dr)1+(γ+1)/2>0,(4.27)‖ρ(γ+1)/2-ρ(γ+1)/2¯(τ)‖L2([0,1])2∈W1,1([0,T]).
Indeed, it holds that
(4.28)∫0τ‖ρ(γ+1)/2-ρ(γ+1)/2¯(s)‖L2([0,1])2ds≤C∫0τ‖(ρ(γ+1)/2)xr2‖L2([0,1])2ds≤C,
which together with (4.4), (4.9), and (4.13) gives
(4.29)∫0τ|dds‖ρ(γ+1)/2-ρ(γ+1)/2¯(s)‖L2([0,1])2|ds≤(γ+1)∫0τ|∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))ρ(γ+1)/2-1ρsdx|ds+2∫0τ|∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))(ρ(γ+1)/2¯(s))sdx|ds≤C∫0τ∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))2dxds+∫0τ∫01ρ2ux2r4dxds≤C.
By Gagliardo-Nirenberg-Sobolev inequality and (4.27), it follows that
(4.30)‖ρ(γ+1)/2-ρ(γ+1)/2¯(τ)‖L∞([0,1])2⟶0asτ⟶+∞.
Thus, there is a T0>0 and a constant ρ1>0 such that
(4.31)ρ(x,τ)≥ρ1,x∈[0,1],τ∈[T0,+∞).
Forτ∈[0,T0], denote
(4.32)v(x,τ)=:1ρ(x,τ)r2(x,τ),
then from (4.4), we can obtain
(4.33)vτ=(r2u)xr2-2vur,
multiplying (4.33) by 4v3 and integrating the result over [0,1]×[0,T0], we have
(4.34)∫01v4dx=∫01v04dx+4∫0τ∫01v3(r2u)xr2dxds-8∫0τ∫01v4urdxds=∫01v04dx+12∫0τ∫01v4u(ρxr2)dxds+24∫0τ∫01v4urdxds=∫01v04dx+12∫0τ∫01v4u(ρ0xr02+u0-u-∫0s(ργ)xr2dl)dxds+24∫0τ∫01v4urdxds.
From (4.21), it holds that
(4.35)∫01v4dx+12∫0τ∫01v4u2dxds=∫01v04dx+12∫0τ∫01v4uρ0xr02dxds+12∫0τ∫01v4uu0dxds-12∫0τ∫01v4u(∫0s(ργ)xr2dl)dxds+24∫0τ∫01v4urdxds≤6∫0τ∫01v4u2dxds+C(T0)∫0τ∫01v4dxds,
which yields to
(4.36)∫01v4dx≤C+C(T0)∫0τ∫01v4dxds,
where C(T0) is a positive constant dependent of time T0. By Gronwall’s inequality, (4.36) leads to
(4.37)∫01v4dx=∫011ρ4r8dx≤C(T0).
It holds for (x,τ)∈([0,1])×[0,T0] that
(4.38)1ρ=∫011ρdx+∫01|(1ρ)x|dx≤C+C∫011ρ4r8dx+C(∫011ρ4r8dx)1/2(∫01ρx2r4dx)1/2≤C(T0),
namely,
(4.39)ρ(x,τ)≥C(T0)=:ρ2,(x,τ)∈([0,1])×[0,T0].
Therefore, we can choose
(4.40)ρ*=min⁡{ρ1,ρ2},
to get
(4.41)ρ≥ρ*,(x,τ)∈([0,1])×[0,+∞).Lemma 4.7.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.42)∫01(r2u)x2dx+∫01(r2u)τ2r-4dx+∫0τ∫01(r2u)s2r-4dxds+∫0τ∫01ρ2(r2u)xs2dxds+∫0τ∫01(r2u)xx2dxds≤C,τ∈[0,T],
where C>0 denotes a constant independent of time.Proof.
The proof is similar to Lemma3.3. We omit here.Remark 4.8.
By Lemmas4.1–4.7, the following inequality holds:
(4.43)∫01u2dx+∫01(ρ-ρ¯)2dx+∫01ux2dx+∫01uτ2dx+∫01ρx2dx+∫0τ∫01ρx2dxds+∫0τ∫01ux2dxds+∫0τ∫01us2dxds+∫0τ∫01uxx2dxds+∫0τ∫01uxτ2dxds≤C.Lemma 4.9.
Under the conditions in Theorem2.2, it holds for any solution (ρ,u) to the initial boundary value problems (2.4) and (2.11) that
(4.44)‖(ρ-ρ̃,u)(⋅,t)‖L∞([r-,r+])≤C0e-C1t,t>0,
where C0 and C1 denote the constants independent of time and ρ̃=∫r-r+ρ(r,t)r2dr.Proof.
In a similar argument to show (4.8) and (4.12) with modification, we can obtain the following
(4.45)ddt∫r-r+(12ρu2+1γ-1(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃)))r2dr+∫r-r+(2ρu2+ρur2r2)dr=0,(4.46)12ddt∫r-r+ρ(u+ρ-1ρr)2r2dr+1γ-1ddt∫r-r+(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃))r2dr+γ∫r-r+ργ-2ρr2r2dr=0.
It holds from (4.9), (4.13), and (4.25) that
(4.47)∫r-r+(ρ-ρ̃)2r2dr≤C∫r-r+ργ-2ρr2r2dr.
Denote
(4.48)E(t)=:∫r-r+ρ(u2+(u+ρ-1ρr)2)r2dr+∫r-r+(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃))r2dr.
By (4.45)–(4.48), a complicated computation gives rise to
(4.49)ddtE(t)+C¯E(t)≤0,
which gives
(4.50)E(t)≤E(0)e-C¯t.
By the fact that
(4.51)E(t)≥c(‖u‖L2([r-,r+])2+‖ρ-ρ̃‖L2([r-,r+])2+‖ρr‖L2([r-,r+])2),
where c>0 is a constant independent of time and Gagliardo-Nirenberg-Sobolev inequality
(4.52)‖(ρ-ρ̃,u)‖L∞([r-,r+])≤C‖(ρ-ρ̃,u)‖L2([r-,r+])1/2‖(ρ-ρ̃,u)r‖L2([r-,r+])1/2,
we obtain (4.44).


### 4.2. Proof of Theorem2.2

Proof.
The proof of Theorem2.2 is similar to Theorem 2.1. We omit the details.


## 4.1. The A-Priori Estimates

Take the Lagrange coordinates transform(4.1)x=∫r-rρ(r,t)r2dr,τ=t.
By (4.1) and the conservation of mass for (ρ,u)(4.2)∫r-r+ρ(r,t)r2dr=∫r-r+ρ0(r)r2dr=:1,
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(4.3)∂x∂r=ρr2,∂x∂t=-ρur2,
and the initial boundary value problem’s (2.4) and (2.11) are reformulated to
(4.4)ρτ+ρ2(r2u)x=0,r-2uτ+(ργ)x=(ρ2(r2u)x)x-2ρxur,(ρ,u)(x,0)=(ρ0,u0)(x),x∈[0,1],u(0,t)=u(1,t)=0,τ∈[0,+∞),
where the initial data satisfies
(4.5)ρ0∈L1([0,1])∩W1,∞([0,1]),inf⁡x∈[0,1]ρ0>ρ̲>0,1r2ρ0(r2u0)∈L2([0,1]),r2ρ0(r2u0)x∈L2([0,1]),1r2ρ0(r2ρ(r2ρ(r2u0)x)x)∈L2([0,1]).
Then, we will establish the a-priori estimates for the solution (ρ,u) to the initial boundary value problem (4.4).Lemma 4.1.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.6)∫01(12u2+1γ-1ργ-1)dx+∫0τ∫01(2u2r2+ρ2ux2r4)dxds=∫01(12u02+1γ-1ρ0γ-1)dx,τ∈[0,T].Proof.
Multiplying (4.4)2 by r2u and integrating the result with respect to x over [0,1], using (4.4)1 and (4.5), we obtain
(4.7)ddτ∫01(12u2+1γ-1ργ-1)dx+∫01ρ2(r2u)x2dx=2∫01ρ(ru2)xdx,
and integrating (4.7) with respect to τ, we obtain
(4.8)∫01(12u2+1γ-1ργ-1)dx+∫0τ∫01(2u2r2+ρ2ux2r4)dxds=∫01(12u02+1γ-1ρ0γ-1)dx.
Lemma 4.1 can be obtained.Lemma 4.2.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.9)12∫01(u+r2ρx)2dx+1γ-1∫01ργ-1dx+γ∫0τ∫01ργ-1ρx2r4dxds=12∫01(u0+r2ρ0x)2dx+1γ-1∫01ρ0γ-1dx,τ∈[0,T],
where C is a positive constant independent of time.Proof.
Differentiating (4.4)1 with respect to x, we have
(4.10)ρxτ+(ρ2(r2u)x)x=0,
which together with (4.4)2 and ∂r/∂τ=u gives
(4.11)(r-2u+ρx)τ+(ργ)x=-2r-3u2-2ρxur.
Multiplying (4.11) by (u+r2ρx)r2, and integrating the result with respect to x and τ, it holds that
(4.12)12∫01(u+r2ρx)2dx+1γ-1∫01ργ-1dx+γ∫0τ∫01ργ-1ρx2r4dxds=12∫01(u0+r2ρ0x)2dx+1γ-1∫01ρ0γ-1dx.
The proof of (4.9) is completed.Lemma 4.3.
LetT>0. Under the conditions in Theorem 2.2, there exists a constant ρ*>0 such that
(4.13)0<ρ(x,τ)≤ρ*,(x,τ)∈[0,1]×[0,T].Proof.
It follows from (4.6) and (4.9) that
(4.14)ρ(x,τ)=ρ(r,t)≤∫r-r+ρ(r,t)dr+∫r-r+|ρr(r,t)|dr≤r--2∫r-r+ρ(r,t)r2dr+r--2∫r-r+ρ|ρr(r,t)|ρr2dr≤C+C∫r-r+1ρ|ρr(r,t)|2r2dr≤C+C∫01|ρx(x,τ)|2r4dx≤C=:ρ*.Lemma 4.4.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.15)∫01u2ndx+∫0τ∫01(u2nr2+ρ2u2n-2ux2r4)dxds≤C(T),
for any positive integer n∈N, where C(T) is a positive constant dependent of time.Proof.
Multiplying (4.4)2 with u2n-1, integrating by parts over [0,1], we have
(4.16)12nddτ∫01u2ndx+∫01ρ2(r2u)x(r2u2n-1)xdx=∫01(ργ(r2u2n-1)x+ρ(2ru2n)x)dx.
Since it holds that
(4.17)(r2u)x(r2u2n-1)x=(2uρr+r2ux)(2u2n-1ρr+(2n-1)r2u2n-2ux)=4u2nρ2r2+(2n-1)u2n-2ux2r4+4nu2n-1uxrρ,
it follows from (4.16) that
(4.18)ddτ∫01u2n2ndx+2∫01u2nr2dx+(2n-1)∫01ρ2u2n-2ux2r4dx=2∫01ργ-1u2n-1rdx+(2n-1)∫01ργu2n-2uxr2dx≤∫01u2nr2dx+∫01ρ2u2n-2ux2r4dx+C‖ρ‖L∞2(γ-1)∫01u2n-2dx,
which together with (4.13) and Young’s inequality yields
(4.19)ddτ∫01u2ndx+∫01u2nr2dx+∫01ρ2u2n-2ux2r4dx≤C+∫01u2ndx,
and by applying the Gronwall’s inequality to (4.19), we can obtain (4.15).Lemma 4.5.
LetT>0, for n∈N, and n>1/2(γ-1). Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.20)∫0τ‖ρ2n(γ-1)u2n‖L∞([0,1])ds≤C(T),(4.21)∫0τ‖|(ργ)xr2|2n‖L∞([0,1])ds≤C(T),τ∈[0,T],
where C(T) is a positive constant dependent of time.Proof.
By means of Sobolev imbedding theorem and Cauchy-Schwarz inequality, applying (4.6), (4.13), and (4.15), we get
(4.22)∫0τ‖ρ2n(γ-1)u2n‖L∞([0,1])ds≤∫0τ∫01|ρ2n(γ-1)u2n|dxds+∫0τ∫01|(ρ2n(γ-1)u2n)x|dxds≤C(T)+C∫0τ∫01ρ2n(γ-1)-1|ρx|u2ndxds+C∫0τ∫01ρ2n(γ-1)|u2n-1ux|dxds≤C(T)+C∫0τ∫01|ρx|2r4dxds+C∫0τ∫01ρ2(2n(γ-1)-1)u4nr-4dxds+∫0τ∫01ρ2(2n(γ-1)-1)u2nr-4dxds+C∫0τ∫01ρ2u2n-2ux2r4dxds≤C(T).
Next, we find that
(4.23)ρxr2=ρ0xr02+u0-u-∫0τ(ργ)xr2ds,
which together with (4.13), (4.15), and (4.22) gives
(4.24)∫0τ‖|(ργ)xr2|2n‖L∞ds=γ2n∫0τ‖ρ2n(γ-1)|ρxr2|2n‖L∞ds=γ2n∫0τ‖ρ2n(γ-1)|ρ0xr02+u0-u-∫0s(ργ)xr2dl|2n‖L∞ds≤C∫0τ‖ρ2n(γ-1)(ρ0x2n+u02n+u2n)‖L∞ds+C∫0τ‖ρ2n(γ-1)(∫0s(ργ)xr2dl)2n‖L∞ds≤C(T)+C∫0τ‖ρ2n(γ-1)u2n‖L∞ds+C(T)∫0τ∫0s‖|(ργ)xr2|2n‖L∞dlds,
applying the Gronwall’s inequality to (4.24), we obtain (4.21).Lemma 4.6.
LetT>0. Under the conditions in Theorem 2.2, there exists a constant ρ*>0 such that
(4.25)ρ(x,τ)≥ρ*>0,(x,τ)∈[0,1]×[0,T].Proof.
It is easy to verify that(4.26)ρ(γ+1)/2¯(τ)=:∫01ρ(γ+1)/2(x,τ)dx≥r-2r+2(∫r-r+ρ(r,t)r2dr)1+(γ+1)/2=r-2r+2(∫r-r+ρ0(r)r2dr)1+(γ+1)/2>0,(4.27)‖ρ(γ+1)/2-ρ(γ+1)/2¯(τ)‖L2([0,1])2∈W1,1([0,T]).
Indeed, it holds that
(4.28)∫0τ‖ρ(γ+1)/2-ρ(γ+1)/2¯(s)‖L2([0,1])2ds≤C∫0τ‖(ρ(γ+1)/2)xr2‖L2([0,1])2ds≤C,
which together with (4.4), (4.9), and (4.13) gives
(4.29)∫0τ|dds‖ρ(γ+1)/2-ρ(γ+1)/2¯(s)‖L2([0,1])2|ds≤(γ+1)∫0τ|∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))ρ(γ+1)/2-1ρsdx|ds+2∫0τ|∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))(ρ(γ+1)/2¯(s))sdx|ds≤C∫0τ∫01(ρ(γ+1)/2-ρ(γ+1)/2¯(s))2dxds+∫0τ∫01ρ2ux2r4dxds≤C.
By Gagliardo-Nirenberg-Sobolev inequality and (4.27), it follows that
(4.30)‖ρ(γ+1)/2-ρ(γ+1)/2¯(τ)‖L∞([0,1])2⟶0asτ⟶+∞.
Thus, there is a T0>0 and a constant ρ1>0 such that
(4.31)ρ(x,τ)≥ρ1,x∈[0,1],τ∈[T0,+∞).
Forτ∈[0,T0], denote
(4.32)v(x,τ)=:1ρ(x,τ)r2(x,τ),
then from (4.4), we can obtain
(4.33)vτ=(r2u)xr2-2vur,
multiplying (4.33) by 4v3 and integrating the result over [0,1]×[0,T0], we have
(4.34)∫01v4dx=∫01v04dx+4∫0τ∫01v3(r2u)xr2dxds-8∫0τ∫01v4urdxds=∫01v04dx+12∫0τ∫01v4u(ρxr2)dxds+24∫0τ∫01v4urdxds=∫01v04dx+12∫0τ∫01v4u(ρ0xr02+u0-u-∫0s(ργ)xr2dl)dxds+24∫0τ∫01v4urdxds.
From (4.21), it holds that
(4.35)∫01v4dx+12∫0τ∫01v4u2dxds=∫01v04dx+12∫0τ∫01v4uρ0xr02dxds+12∫0τ∫01v4uu0dxds-12∫0τ∫01v4u(∫0s(ργ)xr2dl)dxds+24∫0τ∫01v4urdxds≤6∫0τ∫01v4u2dxds+C(T0)∫0τ∫01v4dxds,
which yields to
(4.36)∫01v4dx≤C+C(T0)∫0τ∫01v4dxds,
where C(T0) is a positive constant dependent of time T0. By Gronwall’s inequality, (4.36) leads to
(4.37)∫01v4dx=∫011ρ4r8dx≤C(T0).
It holds for (x,τ)∈([0,1])×[0,T0] that
(4.38)1ρ=∫011ρdx+∫01|(1ρ)x|dx≤C+C∫011ρ4r8dx+C(∫011ρ4r8dx)1/2(∫01ρx2r4dx)1/2≤C(T0),
namely,
(4.39)ρ(x,τ)≥C(T0)=:ρ2,(x,τ)∈([0,1])×[0,T0].
Therefore, we can choose
(4.40)ρ*=min⁡{ρ1,ρ2},
to get
(4.41)ρ≥ρ*,(x,τ)∈([0,1])×[0,+∞).Lemma 4.7.
LetT>0. Under the conditions in Theorem 2.2, it holds for any solution (ρ,u) to the initial boundary value problem (4.4) that
(4.42)∫01(r2u)x2dx+∫01(r2u)τ2r-4dx+∫0τ∫01(r2u)s2r-4dxds+∫0τ∫01ρ2(r2u)xs2dxds+∫0τ∫01(r2u)xx2dxds≤C,τ∈[0,T],
where C>0 denotes a constant independent of time.Proof.
The proof is similar to Lemma3.3. We omit here.Remark 4.8.
By Lemmas4.1–4.7, the following inequality holds:
(4.43)∫01u2dx+∫01(ρ-ρ¯)2dx+∫01ux2dx+∫01uτ2dx+∫01ρx2dx+∫0τ∫01ρx2dxds+∫0τ∫01ux2dxds+∫0τ∫01us2dxds+∫0τ∫01uxx2dxds+∫0τ∫01uxτ2dxds≤C.Lemma 4.9.
Under the conditions in Theorem2.2, it holds for any solution (ρ,u) to the initial boundary value problems (2.4) and (2.11) that
(4.44)‖(ρ-ρ̃,u)(⋅,t)‖L∞([r-,r+])≤C0e-C1t,t>0,
where C0 and C1 denote the constants independent of time and ρ̃=∫r-r+ρ(r,t)r2dr.Proof.
In a similar argument to show (4.8) and (4.12) with modification, we can obtain the following
(4.45)ddt∫r-r+(12ρu2+1γ-1(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃)))r2dr+∫r-r+(2ρu2+ρur2r2)dr=0,(4.46)12ddt∫r-r+ρ(u+ρ-1ρr)2r2dr+1γ-1ddt∫r-r+(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃))r2dr+γ∫r-r+ργ-2ρr2r2dr=0.
It holds from (4.9), (4.13), and (4.25) that
(4.47)∫r-r+(ρ-ρ̃)2r2dr≤C∫r-r+ργ-2ρr2r2dr.
Denote
(4.48)E(t)=:∫r-r+ρ(u2+(u+ρ-1ρr)2)r2dr+∫r-r+(ργ-ρ̃γ-γρ̃γ-1(ρ-ρ̃))r2dr.
By (4.45)–(4.48), a complicated computation gives rise to
(4.49)ddtE(t)+C¯E(t)≤0,
which gives
(4.50)E(t)≤E(0)e-C¯t.
By the fact that
(4.51)E(t)≥c(‖u‖L2([r-,r+])2+‖ρ-ρ̃‖L2([r-,r+])2+‖ρr‖L2([r-,r+])2),
where c>0 is a constant independent of time and Gagliardo-Nirenberg-Sobolev inequality
(4.52)‖(ρ-ρ̃,u)‖L∞([r-,r+])≤C‖(ρ-ρ̃,u)‖L2([r-,r+])1/2‖(ρ-ρ̃,u)r‖L2([r-,r+])1/2,
we obtain (4.44).


## 4.2. Proof of Theorem2.2

Proof.
The proof of Theorem2.2 is similar to Theorem 2.1. We omit the details.



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*Source: 395209-2012-07-05.xml*