# THE STABILITY OF STRONG VISCOUS CONTACT DISCONTINUITY TO A FREE BOUNDARY
PROBLEM FOR COMPRESSIBLE NAVIER-STOKES EQUATIONS

Tingting Zheng111Corresponding author, e-mail<EMAIL_ADDRESS>
Computer and Message Science College, Fujian Agriculture and Forestry
University,
Fuzhou 350001, P. R. China

Abstract. This paper is concerned with nonlinear stability of strong viscous
contact discontinuity to a free boundary problem for the one-dimensional full
compressible Navier-Stokes equations in half space $[0,\infty)$. For the case
when the local stability of the contact discontinuities was first studied by
[1],later generalized by [2], local stability of weak viscous contact
discontinuity is well-established by [3, 4, 5, 6, 7], but for the global
stability of the impermeable gas , fewer strong nonlinear wave stability
results have been obtained excluding zero dissipation [8] or $\gamma\to 1$ gas
see [9]. Our main purpose is to deduce the corresponding nonlinear stability
result by exploiting the elementary energy method. We will show in this paper
that with a certain class of big perturbation the global stability result of
strong viscous contact discontinuity to Navier-Stokes equations can be
obtained.
AMS Subject Classifications (2000). 35B40, 35B45, 76N10,76N17
Keywords: Viscous contact discontinuity, Free boundary problem, Navier-Stokes
equations

##  1 Introduction

This paper is concerned with a free boundary problem for a one-dimensional
compressible viscous heat-conducting flow in the half space
$\mathbb{R}_{+}=[0,\infty)$, which is governed by the following initial-
boundary value problem in Eulerian coordinate $(\tilde{x},t)$:

$\left\\{\begin{array}[]{lll}\tilde{\rho}_{t}+(\tilde{\rho}\tilde{u})_{\tilde{x}}=0,\quad(\tilde{x},t)\in\mathbb{R}_{+}\times\mathbb{R}_{+},\\\\[5.69054pt]
(\tilde{\rho}\tilde{u})_{t}+(\tilde{\rho}\tilde{u}^{2}+\tilde{p})_{\tilde{x}}=\mu\tilde{u}_{\tilde{x}\tilde{x}},\\\\[5.69054pt]
\left(\tilde{\rho}\left(\tilde{e}+\displaystyle\frac{\tilde{u}^{2}}{2}\right)\right)_{t}+\left(\tilde{\rho}\tilde{u}\left(\tilde{e}+\displaystyle\frac{\tilde{u}^{2}}{2}\right)+\tilde{p}\tilde{u}\right)_{\tilde{x}}=\kappa\tilde{\theta}_{\tilde{x}\tilde{x}}+(\mu\tilde{u}\tilde{u}_{\tilde{x}})_{\tilde{x}},\end{array}\right.$
(1.1)

where $\tilde{\rho}$, $\tilde{u}$ and $\tilde{\theta}$ are the density, the
velocity and the absolute temperature, respectively, while $\mu>0$ is the
viscosity coefficient and $\kappa>0$ is the heat-conductivity coefficients,
respectively. The pressure $p=\tilde{p}(\tilde{\rho},\tilde{\theta})$ are
related by the second law of thermodynamics. To simplify our problem, we focus
our attention on the perfect gas. In this situation

$\displaystyle\tilde{p}(\tilde{\rho},\tilde{\theta})=R\tilde{\theta}\tilde{\rho},$
$\displaystyle\tilde{e}(\tilde{\rho},\tilde{\theta})=\displaystyle\frac{R}{\gamma-1}\tilde{\theta}+const,$

where $R>0$ is the gas constant and $\gamma>1$ is the adiabatic exponent. We
consider the system (1.1) in the part $\tilde{x}>\tilde{x}(t)$, where
$\tilde{x}=\tilde{x}(t)$ is a free boundary, with the following boundary
condition

$\displaystyle\frac{d\tilde{x}(t)}{dt}=\tilde{u}(\tilde{x}(t),t),\
\tilde{x}(0)=0,\ \tilde{\theta}(\tilde{x}(t),t)=\theta_{-}>0,$ (1.2)

and

$(\tilde{p}-\mu\tilde{u}_{\tilde{x}})|_{\tilde{x}=\tilde{x}(t)}=p_{-},$ (1.3)

which means the gas is attached at the free boundary $\tilde{x}=\tilde{x}(t)$
to the atmosphere with pressure $p_{-}$ and the initial data

$(\tilde{\rho},\tilde{u},\tilde{\theta})(\tilde{x},0)=({\tilde{\rho}}_{0},{\tilde{u}}_{0},{\tilde{\theta}}_{0})(\tilde{x}),\
\displaystyle\lim_{\tilde{x}\to+\infty}({\tilde{\rho}}_{0},{\tilde{u}}_{0},{\tilde{\theta}}_{0})(\tilde{x})=(\rho_{+},0,\theta_{+}),$
(1.4)

where $\rho_{+},$ $\theta_{+}$ are positive constants and
$\theta_{0}(0)=\theta_{-}$. Because here we only consider the case of a single
contact discontinuity, we require

$p_{-}=p_{+}=R\theta_{+}\rho_{+}.$ (1.5)

Since the boundary condition (1.3) means the particles always stay on the free
boundary $\tilde{x}=\tilde{x}(t)$, if we use $Lagrangian$ coordinates, then
the free boundary becomes a fixed boundary. That is

$\left\\{\begin{array}[]{ll}v_{t}-u_{x}=0,\quad(x,t)\in\mathbb{R}_{+}\times\mathbb{R}_{+},\\\\[5.69054pt]
u_{t}+\left(\displaystyle\frac{R\theta}{v}\right)_{x}=\mu\left(\displaystyle\frac{u_{x}}{v}\right)_{x},\\\\[11.38109pt]
\displaystyle\frac{R}{\gamma-1}\theta_{t}+R\displaystyle\frac{\theta}{v}u_{x}=\kappa\left(\displaystyle\frac{\theta_{x}}{v}\right)_{x}+\mu\displaystyle\frac{u_{x}^{2}}{v},\\\\[11.38109pt]
\theta|_{x=0}=\theta_{-},\quad t>0,\\\\[5.69054pt]
\left(\displaystyle\frac{R\theta_{-}}{v}-\mu\displaystyle\frac{u_{x}}{v}\right)(0,t)=p_{+},\quad
t>0,\\\\[5.69054pt]
(v,u,\theta)|_{t=0}=(v_{0},u_{0},\theta_{0})\to(v_{+},0,\theta_{+})\quad\mbox{as}\quad
x\to+\infty,\end{array}\right.$ (1.6)

where $v_{+}$ and $\theta_{\pm}$ are given positive constants, and $v_{0},\
\theta_{0}>0$. In fact $v=1/\rho(x,t),\ u=u(x,t),\ \theta=\theta(x,t)$ and
$R\theta/v=p(v,\theta)$ are the specific volume, velocity , temperature and
pressure as in (1.1).

In terms of various boundary values, Matsumura [10] classified all possible
large-time behaviors of the solutions for the one-dimensional
(isentropic)compressible Navier-Stokes equations. In the case that $u(0,t)=0$
(resp. $u(0,t)<0$), the problem is called the impermeable wall (resp. outflow)
problem in which the boundary condition of density can’t be imposed. There
have been a lot of works on the asymptotic behaviors of solutions to the
initial-boundary value (or Cauchy) problem for the Navier-Stokes equations
toward these basic waves or their viscous versions, see, for example, [3–25]
and the references therein.

On the other hand, the problem of stability of contact discontinuities are
associated with linear degenerate fields and are less stable than the
nonlinear waves for the inviscid system (Euler equations). It was observed in
[1, 2], where the metastability of contact waves was studied for viscous
conservation laws with artificial viscosity, that the contact discontinuity
cannot be the asymptotic state for the viscous system, and a diffusive wave,
which approximated the contact discontinuity on any finite time interval,
actually dominates the large-time behavior of solutions. The nonlinear
stability of contact discontinuity for the (full) compressible Navier-Stokes
equations was then investigated in [3, 6] for the free boundary value problem
and [4, 5]for the Cauchy problem.

As it is shown in the references, we construct viscous contact wave with
artificial viscousity by the corresponding Euler system of (1.6) with Riemann
initial data which reads as follows:

$\left\\{\begin{array}[]{ll}v_{t}-u_{x}=0,\\\\[5.69054pt]
u_{t}+p(v,\theta)_{x}=0,\\\\[5.69054pt]
\displaystyle\frac{R}{\gamma-1}\theta_{t}+R\displaystyle\frac{\theta}{v}u_{x}=0,\\\\[11.38109pt]
(v,u,\theta)(x,0)=(v_{-},0,\theta_{-})\quad\mbox{if}\quad x<0,\\\\[5.69054pt]
(v,u,\theta)(x,0)=(v_{+},0,\theta_{+})\quad\mbox{if}\quad
x>0.\end{array}\right.$ (1.7)

Because the corresponding Euler equations (1.7) with the Riemann initial data
has the following soluitons

$\left(\overline{V},\overline{U},\overline{\Theta}\right)=\left\\{\begin{array}[]{ll}(v_{-},0,\theta_{-}),\quad
x<0,\\\\[5.69054pt] (v_{+},0,\theta_{+}),\quad x>0,\end{array}\right.$ (1.8)

provided that

$p_{-}=R\displaystyle\frac{\theta_{-}}{v_{-}}=p_{+}=R\displaystyle\frac{\theta_{+}}{v_{+}},$
(1.9)

as that in [3] we conjecture that the asymptotic limit $(V,U,\Theta)$ of (1.6)
is as follows

$P(V,\Theta)=R\displaystyle\frac{\Theta}{V}=p_{+},\;\;\;U(x,t)=\displaystyle\frac{\kappa(\gamma-1)\Theta_{x}}{\gamma
R\Theta},$ (1.10)

and $\Theta$ is the solution of the following problem

$\left\\{\begin{array}[]{ll}\Theta_{t}=a(\ln\Theta)_{xx},\quad
a=\displaystyle\frac{\kappa p_{+}(\gamma-1)}{\gamma R^{2}}>0,\\\\[5.69054pt]
\Theta(0,t)=\theta_{-},\\\\[5.69054pt] \Theta(x,0)=\Theta_{0}\to\theta_{+},\ \
\mathrm{as}\ x\to+\infty\end{array}\right.$ (1.11)

with $\Theta_{0}=\theta_{+}-(\theta_{+}-\theta_{-})\exp\\{1-(1+\alpha
x)^{\delta_{0}}\\}$. It is easy to check that there exist positive constant
$M_{0}$ which is independent of $\delta_{0}$ and $\alpha$ such that

$\displaystyle\|\Theta_{0}-\theta_{+}\|_{L^{1}}\leq
M_{0}\alpha^{-1}\displaystyle\sum_{n=0}^{[\displaystyle\frac{1}{\delta_{0}}]-1}\prod_{i=0}^{n}(\displaystyle\frac{1}{\delta_{0}}-i),$
$\displaystyle|\Theta_{0x}|\leq M_{0}\alpha\delta_{0},$
$\displaystyle|\Theta_{0xx}|\leq M_{0}\alpha^{2}\delta_{0},$
$\displaystyle\|\Theta_{0x}\|^{2}\leq M_{0}\alpha\delta_{0},$
$\displaystyle\|\Theta_{0x}\|_{L^{1}(\mathbb{R}_{+})}\leq M_{0},$
$\displaystyle\|\Theta_{0xx}\|^{2}+\|(\ln\Theta_{0})_{xx}\|^{2}\leq
M_{0}\alpha^{3}\delta_{0}^{2},$
$\displaystyle\|\Theta_{0xxx}\|^{2}+\|(\ln\Theta_{0})_{xxx}\|^{2}\leq
M_{0}\alpha^{5},$
$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}\Theta_{0x}^{2}(1+\alpha
x)dx\leq M_{0}\alpha\delta_{0},$ (1.12)

here and following $\alpha$ is a positive constant which will be determined in
Lemma 2.3 due to the artificial viscosity and $\delta_{0}$ is a small positive
constant which is independent of $\theta_{\pm}$. To sum up, we have
constructed a pair of functions $(V,U,\Theta)$ such that

$\left\\{\begin{array}[]{ll}R\displaystyle\frac{\Theta}{V}=p_{+},\\\\[11.38109pt]
V_{t}=U_{x},\\\\[5.69054pt]
U_{t}+P(V,\Theta)_{x}=\mu\left(\displaystyle\frac{U_{x}}{V}\right)_{x}+F,\\\\[11.38109pt]
\displaystyle\frac{R}{\gamma-1}\Theta_{t}+R\displaystyle\frac{\Theta}{V}U_{x}=\kappa\left(\displaystyle\frac{\Theta_{x}}{V}\right)_{x}+\mu\displaystyle\frac{U_{x}^{2}}{V}+G,\\\\[11.38109pt]
(V,U,\Theta)(0,t)=(v_{-},\displaystyle\frac{\kappa(\gamma-1)}{\gamma
R}\displaystyle\frac{\Theta_{x}}{\Theta}|_{x=0},\theta_{-}),\\\\[11.38109pt]
(V,U,\Theta)(x,0)=(V_{0},U_{0},\Theta_{0})=(\displaystyle\frac{R}{p_{+}}\Theta_{0},\displaystyle\frac{\kappa(\gamma-1)}{\gamma
R}\displaystyle\frac{\Theta_{0x}}{\Theta_{0}},\Theta_{0})\to(v_{+},0,\theta_{+}),\
as\ \ x\to+\infty,\end{array}\right.$ (1.13)

where

$\displaystyle G$ $\displaystyle=$
$\displaystyle-\mu\displaystyle\frac{U_{x}^{2}}{V}=O((\ln\Theta)^{2}_{xx}),$
$\displaystyle F(x,t)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\kappa(\gamma-1)}{\gamma
R}\left\\{(\ln\Theta)_{xt}-\mu\left(\displaystyle\frac{(\ln\Theta)_{xx}}{V}\right)_{x}\right\\}$
(1.14) $\displaystyle=$ $\displaystyle\displaystyle\frac{\kappa
a(\gamma-1)-\mu
p_{+}\gamma}{R\gamma}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right)_{x}.$

We shall show in the next section that $(V,U,\Theta)$ approximates
$(\overline{V},\overline{U},\overline{\Theta})$ in $L^{p}$ norm with $p\geq 1$
on any finite time interval as the heat conductivity $\kappa$ goes to zero.
So, we call $(V,U,\Theta)$ the viscous contact wave for the Navier-Stokes
system (1.6). The definition can be more precise according to whether
$H(\mathbb{R}_{+})$–norm of the initial perturbation
$(\varphi_{0}(x),\psi_{0}(x),\zeta_{0}(x))$ and (or) $|\theta_{+}-\theta_{-}|$
big or not, the stability results are classified into global (or local)
stability of strong (or weak ) viscous contact wave.

Our main purpose is to justify that the solution $(v,u,\theta)$ of the Navier-
Stokes system (1.3) asymptotically tends to the strong viscous contact
discontinuity $(V,U,\Theta)$. Roughly speaking, the main result is :“if the
oscillation of temperature and density are not small, the viscous contact
discontinuity is asymptotic stable”. To deduce the desired nonlinear stability
result by the elementary energy method as in [3, 4, 5, 6, 7, 9, 11, 12], it is
sufficient to deduce certain uniform (with respect to the time variable $t$)
energy type estimates on the solution $(v(x,t),u(x,t),\theta(x,t))$ and how to
establish the Poincar$\acute{e}$ type inequality in Lemma 4.1 without the
smallness of $|\theta_{+}-\theta_{-}|$ which the arguments employed in [3, 4,
5, 6, 7, 11, 12] is to use both smallness $|\theta_{+}-\theta_{-}|$ and
$N(t)=\displaystyle\sup_{0\leq\tau\leq t}\|(\varphi,\psi,\zeta)\|_{H^{1}}$ to
overcome such difficulties. One of the key points in such an argument is that,
based on the a priori assumption that $\displaystyle\sup_{0\leq\tau\leq
t}\|(\varphi,\psi,\zeta)\|_{H^{1}}(\tau)$ is sufficiently small, one can
deduce a uniform lower and upper positive bounds on the specific volume
$v(x,t)$ and temperature $\theta(x,t)$. With such a bound on $v$ and $\theta$
in hand, one can deduce a priori $H(\mathbb{R}_{+})$ energy type estimates on
$(\varphi,\psi,\zeta)$ in terms of the initial perturbation
$(\varphi_{0},\psi_{0},\zeta_{0})$ provided that $|\theta_{+}-\theta_{-}|$
suitably small, so the stability of weak contact discontinuity can be obtained
. In fact if $N(t)$ not small and the perturbation of
$\|(\varphi_{0x},\psi_{0x},\zeta_{0x})\|_{L^{2}(\mathbb{R}_{+})}$ not small
(see [9]), the combination of the analysis similar as above with the standard
continuation argument, it also can obtain the upper and lower bounds of
$(v,\theta)$, then that yields the global stability of strong viscous contact
discontinuity for the one-dimensional compressible Navier-Stokes equations in
the condition of $\gamma\to 1$. In all, after researching the references
carefully we find it is important to get the uniform time estimates of the
viscous contact discontinuity we constructed, then we can obtain the energy
estimates we expected, then the upper and lower bounds of $(v,\theta)$ come
out . So the global stability result can be obtained. It is easy to see that
in such a result, for all $t\in\mathbb{R}_{+}$, Osc
$\theta(t):=\displaystyle\sup_{x\in\mathbb{R}_{+}}\theta(x,t)-\displaystyle\inf_{x\in\mathbb{R}_{+}}\theta(x,t)\geq|\theta_{+}-\theta_{-}|$,
the oscillation of the temperature $\theta(x,t)$ should not be sufficiently
small when $|\theta_{+}-\theta_{-}|$ is not small. Similarly, we can also
obtain the oscillation of the density $\rho(x,t)$ should not be sufficiently
small too .

Following the above analysis, the rest of this paper is out lined as follows.
In section 2 we study the properties of the viscous continuity $(V,U,\Theta)$
in (1.6). In section 3, we reformulate the problem and give the precise
statement of our main theorem. Finally, we complete the proof of the main
result by the global a priori estimates established in section 1.

Throughout this paper, we shall denote $H^{l}(\mathbb{R}_{+})$ the usual
$l-th$ order Sobolev space with the norm

$\|f\|_{l}=\big{(}\displaystyle\sum_{j=0}^{l}\|\partial_{x}^{j}f\|^{2}\big{)}^{1/2},\
\ \|\cdot\|:=\|\cdot\|_{L^{2}(\mathbb{R}_{+})}.$

For simplicity, we also use $C$ or $C_{i}$ ($i=1,2,3.....$) to denote the
various positive generic constants. $C(z)$ stands for constant about $z$ and
$\displaystyle\lim_{z\to 0}C(z)=0$. $\epsilon$ and
$\epsilon_{i}$($i=1,2,3.....$) stand for a small positive constant in Cauchy-
Schwarz inequality and
$\partial_{x}^{i}=\displaystyle\frac{\partial^{i}}{\partial x^{i}}$.

##  2 Preliminaries

This section is devoted to study of the viscous contact discontinuity
$(V,U,\Theta)$ in (1.13). To finish it, we construct a parabolic equation
about $\theta_{2}$ which play an important role in the time estimates of
$\partial_{x}^{i}\Theta$ $(i=1,2,3)$ , it is shown as follows.

###### Lemma 2.1

If $\delta_{0}$ and $\Theta_{0}$ satisfying the condition in Theorem 3.1 and

$\displaystyle\theta_{2}(x,t)=\displaystyle\int_{0}^{+\infty}(4\pi
at)^{-1/2}(\Theta_{0}(h)-\theta_{-})\left\\{\exp\\{-\displaystyle\frac{(h-x)^{2}}{4at}\\}-\exp\\{-\displaystyle\frac{(h+x)^{2}}{4at}\\}\right\\}\
dh+\theta_{-},$

we can get

$\displaystyle\theta_{2t}=a\theta_{2xx};$
$\displaystyle\theta_{2}(0,t)=\theta_{-};$
$\displaystyle\theta_{2}(x,0)=\theta_{20}(x)=\left\\{\begin{array}[]{ll}\Theta_{0}(x)\to\theta_{+},&\hbox{$x>0$;}\\\
-\Theta_{0}(-x)+2\theta_{-}\to 2\theta_{-}-\theta_{+},&\hbox{$x\leq
0$,}\end{array}\right.$ (2.17)

and

$\displaystyle\displaystyle\int_{0}^{t}\|\theta_{2x}\|^{2}dt\leq
C(1+t)^{1/2},$ (2.18)

Proof. Because $\theta_{2}(x,t)$ can be rewrite to

$\theta_{2}(x,t)=\displaystyle\int_{-\infty}^{+\infty}(4\pi
at)^{-1/2}\theta_{20}(h)\exp\\{-\displaystyle\frac{(x-h)^{2}}{4at}\\}dh,$

and $\theta_{20}(x)\in C^{1}(\mathbb{R})$, we find that $\theta_{2}(x,t)$ is a
fundamental solution of (2.17), it is easy to check $\displaystyle\lim_{t\to
0}\theta_{2}(x,t)=\theta_{20}(x)$, so we finish (2.17).

Because

$\displaystyle\theta_{2x}$ $\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{+\infty}(4\pi
at)^{-1/2}\left(\Theta_{0}(z)-\theta_{-}\right)\exp\\{-\displaystyle\frac{(z-x)^{2}}{4at}\\}\displaystyle\frac{z-x}{2at}dz$
(2.19) $\displaystyle\quad+\displaystyle\int_{0}^{+\infty}(4\pi
at)^{-1/2}\left(\Theta_{0}(z)-\theta_{-}\right)\exp\\{-\displaystyle\frac{(z+x)^{2}}{4at}\\}\displaystyle\frac{z+x}{2at}dz$
$\displaystyle=$ $\displaystyle\displaystyle\int_{0}^{\infty}(4\pi
at)^{-1/2}\Theta_{0z}(z)\exp\\{\displaystyle\frac{-(z-x)^{2}}{4at}\\}dz$
$\displaystyle\quad-\displaystyle\int_{0}^{\infty}(4\pi
at)^{-1/2}\Theta_{0z}(z)\exp\\{\displaystyle\frac{-(z+x)^{2}}{4at}\\}dz,$

we use H$\ddot{o}$lder inequality and Fubini Theorem and
$|\theta_{+}-\theta_{-}|\leq\|\Theta_{0z}\|_{L^{1}(\mathbb{R}_{+})}<C$ to
(2.19), then we can get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\theta_{2x}^{2}dxdt$
$\displaystyle\leq$ $\displaystyle
C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}(4\pi
at)^{-1}\left(\displaystyle\int_{0}^{\infty}\Theta_{0z}\left(\exp\\{-\displaystyle\frac{(z-x)^{2}}{4at}\\}-\exp\\{-\displaystyle\frac{(z+x)^{2}}{4at}\\}\right)dz\right)^{2}dxdt$
$\displaystyle\leq$ $\displaystyle
C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}(4\pi
at)^{-1}\displaystyle\int_{0}^{\infty}|\Theta_{0z}|\exp\\{-\displaystyle\frac{(z-x)^{2}}{4at}\\}dzdx\displaystyle\int_{0}^{\infty}|\Theta_{0z}|dzdt$
$\displaystyle\quad+C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}(4\pi
at)^{-1}\displaystyle\int_{0}^{\infty}|\Theta_{0z}|\exp\\{-\displaystyle\frac{(z+x)^{2}}{4at}\\}dzdx\displaystyle\int_{0}^{\infty}|\Theta_{0z}|dzdt$
$\displaystyle\leq$ $\displaystyle C\sqrt{1+t}.$

So we finish this lemma.$\Box$

Now let’s consider the time estimates of $\partial_{x}^{i}\Theta$ $(i=1,2,3)$
of (1.11), we have the following results.

###### Lemma 2.2

If $\Theta_{0x}$ satisfying the condition of (1) and a positive constant
$M_{0}$ is independent of $\delta_{0}$ and $\alpha$ , there exist a positive
constant $C$ such that

$\displaystyle\|(\ln\Theta)_{x}\|^{2}+a\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{2}\ dt\leq M_{0}\alpha\delta_{0}.$ (2.20)
${}_{(see(\ref{2.17})-(\ref{2.19}))}$
$\displaystyle\|\Theta-\theta_{2}\|^{2}+\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{x}\|^{2}\ dt\leq C(1+t)^{1/2}.$ (2.21)
${}_{{(see(\ref{2.14})-(\ref{2.15}))}}$
$\displaystyle\|(\ln\Theta)_{x}\|^{2}\leq C(1+t)^{-1/2}.$ (2.22)
${}_{{(see(\ref{2.20})-(\ref{2.23}))}}$
$\displaystyle\|(\ln\Theta)_{xx}\|^{2}\leq C(1+t)^{-3/2}.$ (2.23)
${}_{(see(\ref{2.24})-(\ref{2.28}))}$
$\displaystyle\|(\ln\Theta)_{xx}\|^{2}(1+t)+\displaystyle\int_{0}^{t}\|\partial^{3}_{x}\ln\Theta\|^{2}(1+t)\
dt\leq C\delta_{0}^{2}.$ (2.24) ${}_{(see(\ref{2.29}))}$
$\displaystyle\|\partial^{3}_{x}\ln\Theta\|^{2}\leq C(1+t)^{-5/2}.$ (2.25)
${}_{(see(\ref{2.30})-(\ref{2.32}))}$
$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}\Theta_{x}^{2}xdx\leq
C\delta_{0}.$ (2.26) ${}_{(see(\ref{2.33})-(\ref{2.35}))}$

Proof. From (1.11) we know

$(\ln\Theta)_{t}=a\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta},$ (2.27)

both side of it multiply by $(\ln\Theta)_{xx}$ and integrate in
$\mathbb{R}_{+}\times(0,t)$ we can get

$\displaystyle\|(\ln\Theta)_{x}\|^{2}+a\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{2}\ dt$ $\displaystyle\leq
C\|(\ln\Theta_{0})_{x}\|^{2}+\displaystyle\int_{0}^{t}(\ln\Theta)_{t}(\ln\Theta)_{x}\big{|}_{0}^{\infty}\
dt.$ (2.28)

So we can get

$\|(\ln\Theta)_{x}\|^{2}+a\displaystyle\int_{0}^{t}\ \|(\ln\Theta)_{xx}\|^{2}\
dt\leq M_{0}\alpha\delta_{0}.$ (2.29)

Then if
$\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\left((\ref{1.11})_{1}-(\ref{2.2})_{1}\right)\times(\Theta-\theta_{2})dxdt$
combine with Cauchy-Schwarz inequality we can get

$\displaystyle\|\Theta-\theta_{2}\|^{2}+\displaystyle\int_{0}^{t}\|(\ln\Theta)_{x}\|^{2}\
dt\leq C\displaystyle\int_{0}^{t}\|{\theta_{2}}_{x}\|^{2}dt.$ (2.30)

Use (2.18) to (2.30) we can get

$\|\Theta-\theta_{2}\|^{2}+\displaystyle\int_{0}^{t}\ \|(\ln\Theta)_{x}\|^{2}\
dt\leq C(1+t)^{1/2}.$ (2.31)

That is (2.21).

Next, from

$\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}(\ref{1.11})_{1}\times\Theta^{-1}(\ln\Theta)_{xx}(1+t)dxdt,$

we can get

$\displaystyle\displaystyle\int_{0}^{t}(1+t)\left((\ln\Theta)_{t}(\ln\Theta)_{x}\right)(0,t)\
dt$ $\displaystyle=a\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\
\displaystyle\frac{(\ln\Theta)^{2}_{xx}}{\Theta}(1+t)\
dxdt+\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\
\left((\ln\Theta)^{2}_{x}\right)_{t}(1+t)\ dxdt.$ (2.32)

Because

$\displaystyle\int_{0}^{t}(1+t)(\ln\Theta)_{t}(\ln\Theta)_{x}(0,t)\ dt=0,$
(2.33)

we can get

$\displaystyle(1+t)\|(\ln\Theta)_{x}\|^{2}+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}(1+t)(\ln\Theta)^{2}_{xx}\ dx\ dt$
$\displaystyle\leq C\|\Theta_{0x}\|^{2}+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ (\ln\Theta)^{2}_{x}\ dx\ dt.$ (2.34)

Combine with (2.31) we can get

$\displaystyle(1+t)\|(\ln\Theta)_{x}\|^{2}+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}(1+t)(\ln\Theta)^{2}_{xx}\ dx\ dt$
$\displaystyle\leq C(1+t)^{1/2}.$ (2.35)

That means $\|(\ln\Theta)_{x}\|^{2}\leq C(1+t)^{-1/2}$, which is(2.22).

Again from (1.11)1 we can get

$(\ln\Theta)_{xt}=a\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right)_{x}.$
(2.36)

Both side of (2.36)multiply $\partial^{3}_{x}\ln\Theta$ and get

$\left((\ln\Theta)_{xt}\partial^{2}_{x}(\ln\Theta)\right)_{x}-1/2(\partial^{2}_{x}\ln\Theta)_{t}=a\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right)_{x}\partial^{3}_{x}(\ln\Theta).$
(2.37)

Because

$\displaystyle\left((\ln\Theta)_{xt}\partial^{2}_{x}(\ln\Theta)\right)_{x}(1+t)^{2}$
$\displaystyle=\left((\ln\Theta)_{xt}(\ln\Theta)_{xx}\right)_{x}(1+t)^{2}$
$\displaystyle=a^{-1}\left((\ln\Theta)_{xt}\Theta_{t}\right)_{x}(1+t)^{2}$

then both side of (2.37) multiply $(1+t)^{2}$ then integrate in
$\mathbb{R}_{+}\times(0,t)$ and combine with $\Theta_{t}(0,t)=0$,
$\Theta_{t}(\infty,t)=0$, $\Theta_{x}(\infty,t)=0$ and Cauchy-Schwarz
inequality to get for some small $\epsilon>0$ we have

$\displaystyle 0\geq a\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\
\displaystyle\frac{(\ln\Theta)_{xxx}^{2}}{\Theta}(1+t)^{2}\ dx\ dt$
$\displaystyle\quad-\epsilon\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ (1+t)^{2}(\ln\Theta)_{xxx}^{2}\ dx\
dt-C\epsilon^{-1}a\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
(1+t)^{2}(\ln\Theta)_{xx}^{2}(\ln\Theta)_{x}^{2}\ dx\ dt$
$\displaystyle\quad+1/2\|(\ln\Theta)_{xx}\|^{2}(1+t)^{2}-1/2\|(\ln\Theta_{0})_{xx}\|^{2}-\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{2}(1+t)\ dx$ $\displaystyle\geq
Ca\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
\displaystyle\frac{(\ln\Theta)_{xxx}^{2}}{\Theta}(1+t)^{2}\ dx\ dt$
$\displaystyle\quad-C\epsilon^{-1}a\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\
(1+t)^{2}\|(\ln\Theta)_{xx}\|\|(\ln\Theta)_{xxx}\|(\ln\Theta)_{x}^{2}\ dx\ dt$
$\displaystyle\quad+1/2\|(\ln\Theta)_{xx}\|^{2}(1+t)^{2}-1/2\|(\ln\Theta_{0})_{xx}\|^{2}-\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{2}(1+t)\ dx.$ (2.38)

Take (2.35) into (2.38) we can get

$\displaystyle\|(\ln\Theta)_{xx}\|^{2}(1+t)^{2}+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ (1+t)^{2}(\ln\Theta)_{xxx}^{2}\ dx\ dt$
$\displaystyle\leq C(1+t)^{1/2},$ (2.39)

which also means

$\|(\ln\Theta)_{xx}\|^{2}\leq C(1+t)^{-3/2},$ (2.40)

and finish (2.23).

If both side of (2.37) multiply by $(1+t)$, similar as the proof of (2.39),
when combine with (2.29) we can get

$\|(\ln\Theta)_{xx}\|^{2}(1+t)+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ (1+t)(\partial^{3}_{x}\ln\Theta)^{2}\ dx\
dt\leq C\delta_{0}^{2},$ (2.41)

which means (2.24).

From (2.36) we can get

$\partial_{t}(\ln\Theta)_{xx}=a\partial^{2}_{x}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right).$
(2.42)

Because

$\displaystyle\left((\ln\Theta)_{xxt}(\ln\Theta)_{xxx}\right)_{x}=\left(a^{-1}\Theta_{tt}(\ln\Theta)_{xxx}\right)_{x}$

when both side of (2.42) multiply $\partial^{4}_{x}\ln\Theta(1+\tau)^{3}$ then
integrate in $\mathbb{R}_{+}\times(0,t)$ we can get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(a^{-1}\Theta_{tt}(\ln\Theta)_{xxx}\right)_{x}(1+\tau)^{3}dxd\tau$
$\displaystyle=\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}a\partial^{2}_{x}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right)\partial^{4}_{x}\ln\Theta(1+\tau)^{3}dxd\tau$
$\displaystyle\quad+\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\displaystyle\frac{1}{2}\left((\partial^{3}_{x}\ln\Theta)^{2}\right)_{t}(1+\tau)^{3}dxd\tau.$
(2.43)

So use (2.35) and (2.39) we can get that for a small $\epsilon>0$, (2.43) can
be change to

$\displaystyle\|\partial^{3}_{x}\ln\Theta\|^{2}(1+t)^{3}+C\displaystyle\int_{0}^{t}\
(1+\tau)^{3}\|\partial^{4}_{x}\ln\Theta\|^{2}\ d\tau$ $\displaystyle\leq
C+C\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
(\partial^{3}_{x}\ln\Theta)^{2}(\ln\Theta)_{x}^{2}(1+\tau)^{3}\ dx\
d\tau+C\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
(\ln\Theta)_{xx}^{4}(1+\tau)^{3}\ dx\ d\tau$
$\displaystyle\quad+C\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\
(\partial^{2}_{x}\ln\Theta)^{2}(\ln\Theta)_{x}^{4}(1+\tau)^{3}\ dx\
d\tau+C\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
(\partial_{x}^{3}\ln\Theta)^{2}(1+\tau)^{2}\ dx\ d\tau$ $\displaystyle\leq
C\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{x}\|^{2}\|\partial^{3}_{x}\ln\Theta\|\|\partial^{4}_{x}\ln\Theta\|(1+\tau)^{3}\
d\tau+C\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{3}\|\partial^{3}_{x}\ln\Theta\|(1+\tau)^{3}\ d\tau$
$\displaystyle\quad+\displaystyle\int_{0}^{t}\
\|(\ln\Theta)_{xx}\|^{4}\|(\ln\Theta)_{x}\|^{2}(1+\tau)^{3}\
d\tau+C(1+t)^{1/2}$ $\displaystyle\leq\epsilon\displaystyle\int_{0}^{t}\
\|\partial^{4}_{x}\ln\Theta\|^{2}(1+\tau)^{3}\
d\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\
\|\partial^{3}_{x}\ln\Theta\|^{2}(1+\tau)^{2}\ d\tau$
$\displaystyle\quad+C\epsilon^{-1}\displaystyle\int_{0}^{t}\
\|\partial^{2}_{x}\ln\Theta\|^{2}(1+\tau)\ d\tau+C(1+t)^{1/2}.$

Again using (2.35) and (2.39) we can get

$\|\partial^{3}_{x}\ln\Theta\|^{2}(1+t)^{3}+\displaystyle\int_{0}^{t}\
(1+\tau)^{3}\|\partial^{4}_{x}\ln\Theta\|^{2}\ d\tau\leq C(1+t)^{1/2}.$ (2.44)

This means (2.25) finished.

Now both side of (2.27) multiply by $(\ln\Theta)_{xx}(x-\beta\tau)$
($\beta>0$) and integrate in $[\beta\tau,\infty)\times(0,t)$ we can get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}\left((\ln\Theta)_{\tau}(\ln\Theta)_{x}(x-\beta\tau)\right)_{x}dxd\tau-1/2\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}\left((\ln\Theta)_{x}^{2}(x-\beta\tau)\right)_{\tau}dxd\tau$
$\displaystyle\quad-a\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{xx}(\ln\Theta)_{x}\Theta^{-1}dxd\tau-\displaystyle\frac{\beta}{2}\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{x}^{2}dxd\tau$
$\displaystyle\quad-\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}a(\ln\Theta)_{xx}^{2}(x-\beta\tau)\Theta^{-1}dxd\tau=\displaystyle\sum_{i=1}^{5}K_{i}=0.$
(2.45)

Use Cauchy-Schwarz inequality

$\displaystyle|K_{3}|\leq\displaystyle\frac{\beta}{4}\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{x}^{2}dxd\tau+C\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{xx}^{2}dxd\tau,$

then combine with (2.20) we can get

$|K_{3}|+K_{4}\leq
C\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{xx}^{2}dxd\tau-\displaystyle\frac{\beta}{4}\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{x}^{2}dxd\tau\leq
C\alpha\delta_{0}-\displaystyle\frac{\beta}{4}\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{x}^{2}dxd\tau.$

From $K_{1}$ to $K_{5}$, (2) can be change to

$\displaystyle\displaystyle\int_{\beta t}^{\infty}(\ln\Theta)_{x}^{2}(x-\beta
t)dx+\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{xx}^{2}(x-\beta\tau)dxd\tau$
$\displaystyle\quad+\displaystyle\frac{\beta}{4}\displaystyle\int_{0}^{t}\displaystyle\int_{\beta\tau}^{\infty}(\ln\Theta)_{x}^{2}dxd\tau\leq
C\delta_{0}.$ (2.46)

Because

$\displaystyle\displaystyle\lim_{\beta\to 0}\left|\displaystyle\int_{\beta
t}^{\infty}(\ln\Theta)_{x}^{2}(x-\beta
t)dx-\displaystyle\int_{0}^{\infty}(\ln\Theta)_{x}^{2}xdx\right|$
$\displaystyle=$ $\displaystyle\displaystyle\lim_{\beta\to
0}\left|\displaystyle\int_{\beta t}^{\infty}(\ln\Theta)_{x}^{2}(x-\beta
t-x)dx+\displaystyle\int_{0}^{\beta t}(\ln\Theta)_{x}^{2}xdx\right|=0,$

we can get

$0<\displaystyle\lim_{\beta\to 0}\displaystyle\int_{\beta
t}^{\infty}(\ln\Theta)_{x}^{2}(x-\beta
t)dx=\displaystyle\int_{0}^{\infty}(\ln\Theta)_{x}^{2}xdx\leq C\delta_{0}.$
(2.47)

So we finish this lemma.$\Box$

The next lemma is concerned with the relation ship between the viscous
continuity and the contact discontinuity. We shall show that as the heat
conductivity $\kappa$ goes to zero, $(V,U,\Theta)$ will approximate
$(\overline{V},\overline{U},\overline{\Theta})$ in $L^{p}(\mathbb{R}_{+})$
$(p\geq 1)$ norm on any finite time interval.

###### Lemma 2.3

For any given $T\in(0,+\infty)$ independent of $\kappa$ such that for any
$p\geq 1$ and $t\in[0,T]$,

$\|(V-\overline{V},U-\overline{U},\Theta-\overline{\Theta})\|_{L^{p}(\mathbb{R}_{+})}\to
0,\ \ \mathrm{as}\ \ \kappa\to 0.$

Proof. By the definition of $\overline{\Theta}$ in (1.8), to estimate
$\|\Theta-\overline{\Theta}\|_{L^{p}(\mathbb{R}_{+})}$, it suffices to prove

$\|\Theta-\theta_{+}\|_{L^{p}(\mathbb{R}_{+})}\to 0,\ \ \mathrm{as}\ \
\kappa\to 0,\ p\geq 1.$

Because if $M_{0}>0$ is a constant independent of $\alpha$,

$\|\Theta-\theta_{+}\|^{p}_{L^{p}(\mathbb{R}_{+})}\leq
M_{0}\|\Theta-\theta_{+}\|_{L^{1}(\mathbb{R}_{+})},$

the only thing we need to proof is

$\displaystyle\lim_{\kappa\to
0}\|\Theta-\theta_{+}\|_{L^{1}(\mathbb{R}_{+})}=0.$

In fact we set
$sgn_{\eta}(s)=\left\\{\begin{array}[]{ll}1,&\hbox{$s>\eta$;}\\\
s/\eta,&\hbox{$-\eta\leq s\leq\eta$;}\\\
-1,&\hbox{$s<-\eta$.}\end{array}\right.$,
$I_{\eta}(s)=\displaystyle\int_{0}^{s}sgn_{\eta}(s)ds$ and $\eta>0$. Both side
of (1.11)1 multiply by $sgn_{\eta}(\Theta-\theta_{+})$ and integrate in
$(0,+\infty)\times(0,t)$ we can get

$\displaystyle\displaystyle\int_{0}^{t}\left(\displaystyle\int_{0}^{+\infty}I_{\eta}(\Theta-\theta_{+})dx\right)_{\tau}d\tau$
$\displaystyle=$
$\displaystyle-a\displaystyle\int_{0}^{t}(\ln\Theta)_{x}(0,t)sgn_{\eta}(\Theta-\theta_{+})(0,t)d\tau$
$\displaystyle\quad-a\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}(\ln\Theta)_{x}^{2}sgn^{\prime}_{\eta}(\Theta-\theta_{+})dxd\tau.$

When $\eta\to 0$ and use (2.20) , (2.24) we can get

$\displaystyle\|\Theta-\theta_{+}\|_{L^{1}(\mathbb{R}_{+})}+a\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}(\ln\Theta)_{x}^{2}sgn^{\prime}_{\eta}(\Theta-\theta_{+})dxd\tau$
$\displaystyle=a\displaystyle\int_{0}^{t}(\ln\Theta)_{x}(0,t)sgn_{\eta}(\Theta-\theta_{-})(0,t)d\tau+\|\Theta_{0}-\theta_{+}\|_{L^{1}(\mathbb{R}_{+})}.$
(2.48)

Similar as (2.39), when we integrate (2.37) in $\mathbb{R}_{+}\times(0,t)$ and
combine with (2.20) we can get that there exist constant $M_{0}>0$ independent
of $\alpha$ such that

$\|(\ln\Theta)_{xx}\|^{2}+a\displaystyle\int_{0}^{t}\|\partial_{x}^{3}(\ln\Theta)\|^{2}d\tau\leq
M_{0}\alpha^{3}+M_{0}a\alpha^{3}+M_{0}a^{-1}\alpha.$ (2.49)

Since $a=\kappa p_{+}(\gamma-1)/(\gamma R^{2})$ and $\kappa\to 0$, we can
choose $\alpha^{-1/2}=\kappa<1$, use (2.20) and (2.49) such that (2) is meant
if constant $M_{0}>0$ is independent of $\alpha$, we have

$\|\Theta-\theta_{+}\|_{L^{1}(\mathbb{R}_{+})}\leq
M_{0}t((a\alpha)^{3/4}+a^{5/4}\alpha)+M_{0}\alpha^{-1}\leq
M_{0}\kappa^{3/8}(t+1),$

so we get $\|(V-\overline{V},\Theta-\overline{\Theta})\|_{L^{p}}\to 0$ as
$\kappa\to 0$ with any $t\in[0,T]$.

It remains to estimate $\|U-\overline{U}\|_{L^{p}}$. To do so, both side of
(2.36) multiply by $sgn_{\eta}((\ln\Theta)_{x})$ then integrate in
$\mathbb{R}_{+}\times(0,t)$ we can get

$\displaystyle\displaystyle\int_{0}^{t}\left(\displaystyle\int_{\mathbb{R}_{+}}I_{\eta}((\ln\Theta)_{x})dx\right)_{\tau}d\tau+a\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\displaystyle\frac{(\ln\Theta)^{2}_{xx}}{\Theta}sgn^{\prime}_{\eta}((\ln\Theta)_{x})dxd\tau$
$\displaystyle=-a\displaystyle\int_{0}^{t}\theta_{-}^{-1}(\ln\Theta)_{xx}(0,\tau)sgn_{\eta}((\ln\Theta)_{x})(0,\tau)d\tau$
$\displaystyle=-\displaystyle\int_{0}^{t}\theta_{-}^{-1}\Theta_{\tau}(0,t)sgn_{\eta}((\ln\Theta)_{x})(0,\tau)d\tau=0.$

Again let $\eta\to 0$ we can get that there exist constant $M_{0}>0$
independent of $\alpha$ such that

$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}|(\ln\Theta)_{x}|dx+a\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\displaystyle\frac{(\ln\Theta)^{2}_{xx}}{\Theta}sgn^{\prime}_{\eta}((\ln\Theta)_{x})dxd\tau$
$\displaystyle\leq M_{0}.$ (2.50)

Use the definition of $U$ in (1.10) and combine with (2.20),(2.49) and (2.50)
we know that

$\displaystyle\|U-\overline{U}\|^{p}_{L^{p}}\leq
M_{0}\kappa^{p}\|(\ln\Theta)_{x}\|_{L^{1}}\|(\ln\Theta)_{x}\|^{(p-1)/2}\|(\ln\Theta)_{xx}\|^{(p-1)/2}$
$\displaystyle\leq M_{0}\kappa^{p}\alpha^{p-1}.$

Remind that $\alpha=\kappa^{-1/2}$, so we can get

$\displaystyle\lim_{\kappa\to 0}\|U-\overline{U}\|_{L^{p}}=0.$

The proof of Lemma 2.3 is therefore complete, which also means $(V,U,\Theta)$
is viscous contact discontinuity.$\Box$

##  3 Reformulation and Main result

Let $(v,u,\theta)$ be the solution of the problem (1.6), and let
$(V,U,\Theta)$ be the viscous contact discontinuity constructed in (1.13).
Denote

$\displaystyle\varphi(x,t)=v(x,t)-V(x,t),$
$\displaystyle\psi(x,t)=u(x,t)-U(x,t),$
$\displaystyle\zeta(x,t)=\theta(x,t)-\Theta(x,t).$ (3.51)

Combining (1.13) and (1.6), the original problem can be reformulated as

$\left\\{\begin{array}[]{lll}\varphi_{t}=\psi_{x},&\\\
\psi_{t}-(\displaystyle\frac{R\Theta}{vV}\varphi)_{x}+(\displaystyle\frac{R\zeta}{v})_{x}=-\mu(\displaystyle\frac{U_{x}}{vV}\varphi)_{x}+\mu(\displaystyle\frac{\psi_{x}}{v})_{x}-F,&\\\
\displaystyle\frac{R}{\gamma-1}\zeta_{t}+\displaystyle\frac{R\theta}{v}(\psi_{x}+U_{x})-\displaystyle\frac{R\Theta}{V}U_{x}=\kappa(\displaystyle\frac{\zeta_{x}}{v})_{x}-\kappa(\displaystyle\frac{\Theta_{x}\varphi}{vV})_{x}+\mu(\displaystyle\frac{{u_{x}}^{2}}{v}-\displaystyle\frac{{U_{x}}^{2}}{V})-G,&\\\
\zeta(0,t)=0,&\\\
\left(\displaystyle\frac{R\theta_{-}}{V+\varphi}-\mu\displaystyle\frac{U_{x}+\psi_{x}}{V+\varphi}\right)(0,t)=p_{+},&\\\
(\varphi,\psi,\zeta)(x,0)=(\varphi_{0},\psi_{0},\zeta_{0})=(v_{0}-V_{0},u_{0}-U_{0},\theta_{0}-\Theta_{0}).&\\\
\end{array}\right.$ (3.52)

From (1.13) it is easy to check that the initial-boundary data in (3.52)
satisfies the compatible condition, and

$(\varphi,\psi,\zeta)(x,0)=(\varphi_{0},\psi_{0},\zeta_{0})\to(0,0,0)\ \
\mathrm{as}\ \ x\to+\infty.$

To state our main result, we assume throughout of this section that

$(\varphi_{0},\zeta_{0})(x)\in H_{0}^{1}(0,\infty),\ \ \psi_{0}(x)\in
H^{1}(0,\infty).$

Moreover, for an interval $I\in[0,\infty)$ , we define the function space

$X(I)=\left\\{(\varphi,\psi,\zeta)\in C(I,H^{1})|\varphi_{x}\in
L^{2}(I;L^{2}),(\psi_{x},\zeta_{x})\in L^{2}(I;H^{1})\right\\}.$

Our main results of this paper now reads as follows.

###### Theorem 3.1

There exist positive constants $C>1$ and $\eta_{0}$ such that if
$\|(v_{0}-V_{0},u_{0}-U_{0},\theta_{0}-\Theta_{0})\|_{L^{2}}\leq\eta_{0},$
$\|(v_{0x}-V_{0x},u_{0x}-U_{0x},\theta_{0x}-\Theta_{0x})\|_{L^{2}}\leq
C,$(3.52) has a unique global solution $(\varphi,\psi,\zeta)$ satisfying
$(\varphi,\psi,\zeta)\in X([0,\infty))$ and

$\displaystyle\sup_{x\in\mathbb{R}_{+}}|(\varphi,\psi,\zeta)|\to 0,\ as\
t\to\infty.$

In this section, to study the asymptotic behavior of the solution to the free
boundary problem (1.6), we will do some preparation lemmas and list some
priori estimates which are important to the proof of Theorem 3.1.

We shall prove Theorem 3.1 by combining the local existence and the global-in-
time priori estimates. Since the local existence of the solution is well known
(see, for example,[3]), we omit it here for brevity. to prove the global
existence part of Theorem 3.1, it is sufficient to establish the following
priori estimates.

###### Proposition 3.1

(A priori estimate) Let $(\varphi,\psi,\zeta)\in X([0,t])$ be a solution of
problem (3.52) for some $t>0$. Then there exist positive constants
$C(\delta_{0})<1$ and $C$ which are all independent of $t$ and $(v,\theta)$ ,
such that if $m\leq v,\theta\leq M$ and
$1<N(t)=\displaystyle\sup_{0\leq\tau\leq t}\|(\varphi,\psi,\zeta)\|_{1}\leq
C$, it holds that

$\displaystyle\displaystyle\sup_{0\leq\tau\leq
t}\|(\psi,\varphi,\zeta)\|^{2}(t)+\displaystyle\int_{0}^{t}\|(\psi_{x},\zeta_{x})\|^{2}(\tau)d\tau$
$\displaystyle\leq C\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{1/2}+C(\delta_{0}).$
$\displaystyle\displaystyle\sup_{0\leq\tau\leq
t}\|(\psi_{x},\varphi_{x},\zeta_{x})\|^{2}(t)+\displaystyle\int_{0}^{t}\left(\|\varphi_{x}\|^{2}(\tau)+\|(\psi_{x},\zeta_{x})\|^{2}_{1}(\tau)\right)d\tau$
$\displaystyle\leq C\|(\varphi_{0},\psi_{0},\zeta_{0})\|_{1}+C(\delta_{0}).$
(3.53)

##  4 Proof of Theorem 3.1

Under the preparations in last section, the main task here is to finish (3.1).
This part we also do some preparations. we must use the results

$\displaystyle|V_{x}|\leq C|\Theta_{x}|,$ $\displaystyle|\Theta_{x}|^{2}\leq
C\|(\ln\Theta)_{x}\|\|(\ln\Theta)_{xx}\|,$ $\displaystyle|U_{x}|\leq
C|(\ln\Theta)_{xx}|,$ $\displaystyle|U_{x}|^{2}\leq
C\|(\ln\Theta)_{xx}\|\|(\ln\Theta)_{xxx}\|,$ (4.54)

which follow from (1.10)–(1.13) . Also we set $C(\delta_{0})$ stands for small
constants about $\delta_{0}$, $\|(\varphi_{0},\psi_{0},\zeta_{0})\|$ is asked
suitably small, $C_{v}=\displaystyle\frac{R}{\gamma-1}$ and

$\epsilon_{1}\ll\epsilon_{2}\ll\epsilon_{3}.$

Before establishing (3.1), we first estimate the value of $\varphi(0,t)$ on
the boundary $x=0$ by the boundary condition (3.52). Let
$\varphi(t)=\varphi(0,t)$. Since $U_{x}(0,t)=V_{t}(0,t)=0$, the boundary
condition of (3.52) yields

$\displaystyle\frac{R\theta_{-}}{v_{-}+\varphi(t)}-\mu\displaystyle\frac{\varphi_{t}(t)}{v_{-}+\varphi(t)}=p_{+},\
t>0$ (4.55)

Direct computation gives

$\varphi_{t}(t)=-\displaystyle\frac{p_{+}}{\mu}\varphi(t),\
\varphi(0)=\varphi_{0}(0).$ (4.56)

It follows then that

$\varphi(t)=\varphi_{0}(0)e^{-p_{+}t/\mu}.$ (4.57)

###### Lemma 4.1

If $C(\delta_{0})>0$ is a small constant about $\delta_{0}$

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\Theta_{x}^{2}(\varphi^{2}+\zeta^{2})dxd\tau\leq
C(\delta_{0})\displaystyle\int_{0}^{t}\|(\varphi_{x},\zeta_{x})\|^{2}d\tau+C\varphi_{0}(0).$

Proof. Because if $x>0$

$\displaystyle\displaystyle\frac{\varphi^{2}}{x+1}=\displaystyle\int_{0}^{x}\big{(}\displaystyle\frac{2\varphi\varphi_{x}}{x+1}-\displaystyle\frac{\varphi^{2}}{(x+1)^{2}}\big{)}dx+\varphi^{2}(0,t)$
$\displaystyle=\displaystyle\int_{0}^{x}\left(\varphi_{x}^{2}-(\varphi_{x}-\displaystyle\frac{\varphi}{x+1})^{2}\right)dx+\varphi^{2}(0,t)\leq\displaystyle\int_{0}^{x}\varphi_{x}^{2}dx+C\varphi_{0}(0)e^{-p_{+}t/\mu}\leq\|\varphi_{x}\|^{2}+C\varphi_{0}(0)e^{-p_{+}t/\mu}.$

similar as above we can get

$\displaystyle\displaystyle\frac{\zeta^{2}}{x+1}\leq\displaystyle\int_{0}^{x}\zeta_{x}^{2}dx\leq\|\zeta_{x}\|^{2}.$

As to

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}\Theta_{x}^{2}(\varphi^{2}+\zeta^{2})dxd\tau$
$\displaystyle\leq\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}\Theta_{x}^{2}(x+1)\displaystyle\frac{(\varphi^{2}+\zeta^{2})}{1+x}dxd\tau$
$\displaystyle\leq\displaystyle\int_{0}^{t}\big{(}\displaystyle\int_{0}^{+\infty}\Theta_{x}^{2}(1+x)dx\big{)}\|(\varphi_{x},\zeta_{x})\|^{2}d\tau+C\varphi_{0}(0),$

use (2.20)and (2.26) we can get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\Theta_{x}^{2}(\varphi^{2}+\zeta^{2})dxd\tau\leq
C(\delta_{0})\displaystyle\int_{0}^{t}\|(\varphi_{x},\zeta_{x})\|^{2}d\tau+C\varphi_{0}(0),$

and we finish this lemma.$\Box$

Now, let’s finish (3.1) by the following lemmas.

###### Lemma 4.2

If $\epsilon_{1}>0$ and $C(\delta_{0})>0$ are small constant about
$\delta_{0}$, we can get

$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}(\varphi^{2}+\psi^{2}+\zeta^{2})dx+\displaystyle\int_{0}^{t}\
\left\|(\psi_{x},\zeta_{x})\right\|^{2}\ d\tau$ $\displaystyle\leq
C(\delta_{0})+C\left\\{\epsilon_{1}\displaystyle\int_{0}^{t}\
\|\varphi_{x}\|^{2}d\tau+\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}\right\\}.$

Proof. Set

$\Phi(z)=z-\ln z-1,$ $\Psi(z)=z^{-1}+\ln z-1,$

where $\Phi^{\prime}(1)=\Phi(1)=0$ is a strictly convex function around $z=1$.
Similar to the proof in [3], we deduce from (3.52) that

$\displaystyle\left(\displaystyle\frac{\psi^{2}}{2}+R\Theta\Phi\left(\displaystyle\frac{v}{V}\right)+C_{v}\Theta\Phi\left(\displaystyle\frac{\theta}{\Theta}\right)\right)_{t}$
$\displaystyle\quad+\mu\displaystyle\frac{\Theta\psi_{x}^{2}}{v\theta}+\kappa\displaystyle\frac{\Theta\zeta_{x}^{2}}{v\theta^{2}}+H_{x}+Q=\mu\left(\displaystyle\frac{\psi\psi_{x}}{v}\right)_{x}-F\psi-\displaystyle\frac{\zeta
G}{\theta},$ (4.58)

where

$H=R\displaystyle\frac{\zeta\psi}{v}-R\displaystyle\frac{\Theta\varphi\psi}{vV}+\mu\displaystyle\frac{U_{x}\varphi\psi}{vV}-\kappa\displaystyle\frac{\zeta\zeta_{x}}{v\theta}+\kappa\displaystyle\frac{\Theta_{x}\varphi\zeta}{v\theta
V},$

and

$\displaystyle Q$ $\displaystyle=$ $\displaystyle
p_{+}\Phi\left(\displaystyle\frac{V}{v}\right)U_{x}+\displaystyle\frac{p_{+}}{\gamma-1}\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)U_{x}-\displaystyle\frac{\zeta}{\theta}(p_{+}-p)U_{x}-\mu\displaystyle\frac{U_{x}\varphi\psi_{x}}{vV}$
$\displaystyle\quad-\kappa\displaystyle\frac{\Theta_{x}}{v\theta^{2}}\zeta\zeta_{x}-\kappa\displaystyle\frac{\Theta\Theta_{x}}{v\theta^{2}V}\varphi\zeta_{x}-2\mu\displaystyle\frac{U_{x}}{v\theta}\zeta\psi_{x}+\kappa\displaystyle\frac{\Theta_{x}^{2}}{v\theta^{2}V}\varphi\zeta+\mu\displaystyle\frac{U_{x}^{2}}{v\theta
V}\varphi\zeta$ $\displaystyle=:$
$\displaystyle\displaystyle\sum_{i=1}^{9}Q_{i}.$

Note that $p=R\theta/v$, $p_{+}=R\Theta/V$ and (1.10), use integrate by part
and Cauchy-Schwarz inequality can get

$\displaystyle Q_{1}+Q_{2}$ $\displaystyle=$ $\displaystyle
Ra\left(\Phi\left(\displaystyle\frac{V}{v}\right)(\ln\Theta)_{x}\right)_{x}+\displaystyle\frac{Ra}{\gamma-1}\left(\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)(\ln\Theta)_{x}\right)_{x}$
(4.59) $\displaystyle\quad-
aR(\ln\Theta)_{x}\left(\displaystyle\frac{V\varphi_{x}\varphi-
V_{x}\varphi^{2}}{Vv^{2}}\right)$
$\displaystyle\quad-a\displaystyle\frac{p_{+}}{\gamma-1}(\ln\Theta)_{x}\left(\displaystyle\frac{\Theta\zeta_{x}\zeta-\Theta_{x}\zeta^{2}}{\Theta\theta^{2}}\right)$
$\displaystyle\geq\left(p_{+}\Phi\left(\displaystyle\frac{V}{v}\right)U+\displaystyle\frac{p_{+}}{\gamma-1}\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)U\right)_{x}$
$\displaystyle\quad-\epsilon(\zeta_{x}^{2}+\varphi_{x}^{2})-C\epsilon^{-1}\Theta_{x}^{2}(\zeta^{2}+\varphi^{2}).$

Similarly, using $p-p_{+}=\displaystyle\frac{R\zeta-p_{+}\varphi}{v}$, we can
get

$Q_{3}\geq\displaystyle\frac{R\zeta-
p_{+}\varphi}{v}(\displaystyle\frac{\zeta}{\theta}U_{x})\geq\left(\displaystyle\frac{R\zeta^{2}U}{v\theta}-\displaystyle\frac{p_{+}\zeta\varphi
U}{\theta
v}\right)_{x}-\epsilon(\zeta_{x}^{2}+\varphi_{x}^{2})-C\epsilon^{-1}\Theta_{x}^{2}(\zeta^{2}+\varphi^{2}).$
(4.60)

And

$\displaystyle(Q_{4}+Q_{7})+(Q_{5}+Q_{6}+Q_{8})+Q_{9}$ $\displaystyle\geq$
$\displaystyle-C\epsilon^{-1}(\ln\Theta)_{xx}^{2}-\epsilon\psi_{x}^{2}$ (4.61)
$\displaystyle-\epsilon\zeta_{x}^{2}-C\epsilon^{-1}\Theta_{x}^{2}(\zeta^{2}+\varphi^{2})$
$\displaystyle-C\epsilon^{-1}|(\ln\Theta)_{xx}|^{2}(\zeta^{2}+\varphi^{2}).$

At the end we use the definition of $F$ and $G$ in (1.14) then combine with
the general inequality skills as above to get

$\displaystyle-F\psi-G\displaystyle\frac{\zeta}{\theta}$ $\displaystyle=$
$\displaystyle-\displaystyle\frac{\kappa a(\gamma-1)-\mu
p_{+}\gamma}{R\gamma}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\right)_{x}\psi$
(4.62) $\displaystyle\quad+\displaystyle\frac{\mu
p_{+}}{R\Theta}\left(\displaystyle\frac{\kappa(\gamma-1)}{R\gamma}(\ln\Theta)_{xx}\right)^{2}\displaystyle\frac{\zeta}{\theta}$
$\displaystyle\leq$ $\displaystyle-\displaystyle\frac{\kappa a(\gamma-1)-\mu
p_{+}\gamma}{R\gamma}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\psi\right)_{x}+\displaystyle\frac{\kappa
a(\gamma-1)-\mu
p_{+}\gamma}{R\gamma}\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\psi_{x}$
$\displaystyle\quad+\displaystyle\frac{\mu
p_{+}}{R\Theta}\left(\displaystyle\frac{\kappa(\gamma-1)}{R\gamma}(\ln\Theta)_{xx}\right)^{2}\displaystyle\frac{\zeta}{\theta}$
$\displaystyle\leq$ $\displaystyle-\displaystyle\frac{\kappa a(\gamma-1)-\mu
p_{+}\gamma}{R\gamma}\left(\displaystyle\frac{(\ln\Theta)_{xx}}{\Theta}\psi\right)_{x}+\epsilon\psi_{x}^{2}+C\epsilon^{-1}(\ln\Theta)_{xx}^{2}.$

Integrating (4.59)$-$(4.62) in $\mathbb{R}\times(0,t)$ , using (2.20), (2.24)
and the boundary condition about $(\varphi,\psi,\zeta)$ of (3.52), (4.57) and
$\Theta_{t}(0,t)=a(\ln\Theta)_{xx}(0,t)=0$,
$\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)(0,t)=0$ to estimate the
terms
$\left(p_{+}\Phi\left(\displaystyle\frac{V}{v}\right)U+\displaystyle\frac{p_{+}}{\gamma-1}\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)U\right)_{x}$,
$\mu\left(\displaystyle\frac{\psi\psi_{x}}{v}\right)_{x}$,
$\left(\displaystyle\frac{(\ln\Theta)_{xx}\psi}{\Theta}\right)_{x}$ and
$H_{x}$, we can know that because

$\displaystyle\displaystyle\int_{0}^{t}\left(Ra\Phi\left(\displaystyle\frac{V}{v}\right)|(\ln\Theta)_{x}|+\mu\left|\displaystyle\frac{\psi\psi_{x}}{v}\right|+\left|\displaystyle\frac{(\ln\Theta)_{xx}\psi}{\Theta}\right|+|H|\right)(0,\tau)d\tau$
$\displaystyle\leq
C\displaystyle\int_{0}^{t}|(\ln\Theta)_{x}|(0,\tau)\left(|\varphi|(\tau)+|\varphi_{\tau}|(\tau)\right)d\tau\leq
C(\delta_{0}),$

it is easy to get

$\displaystyle\left|\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\left(\mu\left(p_{+}\Phi\left(\displaystyle\frac{V}{v}\right)U+\displaystyle\frac{p_{+}}{\gamma-1}\Phi\left(\displaystyle\frac{\Theta}{\theta}\right)U\right)_{x}+\left(\displaystyle\frac{(\ln\Theta)_{xx}\psi}{\Theta}\right)_{x}+H_{x}\right)dxd\tau\right|$
$\displaystyle\leq C(\delta_{0}).$ (4.63)

In the end from combine with the estimates from $Q_{1}$ to $Q_{9}$, (4) and
Lemma 4.1 we have

$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}\left(R\theta\Phi\left(\displaystyle\frac{v}{V}\right)+\displaystyle\frac{1}{2}\psi^{2}+C_{v}\theta\Phi\left(\displaystyle\frac{\theta}{\Theta}\right)\right)dx+\displaystyle\int_{0}^{t}\
\left\|\left(\psi_{x}/(\sqrt{v\theta}),\zeta_{x}/(\theta\sqrt{v})\right)\right\|^{2}\
d\tau$ $\displaystyle\leq
C\epsilon_{1}^{-1}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\Theta_{x}^{2}(\varphi^{2}+\zeta^{2})\
dxd\tau+C\left\\{\epsilon_{1}\displaystyle\int_{0}^{t}\ \|\varphi_{x}\|^{2}\
d\tau+\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}\right\\}+C(\delta_{0})$
$\displaystyle\leq
C(\delta_{0})\displaystyle\int_{0}^{t}\|(\psi_{x},\zeta_{x})\|^{2}d\tau++C\left\\{\epsilon_{1}\displaystyle\int_{0}^{t}\
\|\varphi_{x}\|^{2}\
d\tau+\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}\right\\}+C(\delta_{0}).$

Use the condition of Proposition 3.1 which is $m\leq v,\theta\leq M$, we
finish this lemma.$\Box$

###### Lemma 4.3

For a small $\epsilon_{2}>0$, $C(\delta_{0})>0$ is a small constant about
$\delta_{0}$,we have

$\displaystyle\|(\varphi,\psi,\zeta)\|^{2}+\|(\psi_{x},\zeta_{x})\|^{2}+\displaystyle\int_{0}^{t}\|(\psi_{xx},\zeta_{xx})\|^{2}dt$
$\displaystyle\leq
C\left(\|(\psi_{0x},\zeta_{0x})\|^{2}+\epsilon_{2}^{-1}\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}\right)+C\epsilon_{2}^{-1}\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+C(\delta_{0})+C\varphi_{0}(0).$

.

Proof. First to get the estimate of $\|\psi_{x}(t)\|$ ,multiply both side of
(3.52)2 to $\psi_{xx}$ to get

$\displaystyle\left(\displaystyle\frac{\psi_{x}^{2}}{2}\right)_{t}+\mu\displaystyle\frac{\psi_{xx}^{2}}{v}=\mu\displaystyle\frac{\psi_{x}v_{x}}{v^{2}}\psi_{xx}+\mu\left(\displaystyle\frac{U_{x}\varphi}{vV}\right)_{x}\psi_{xx}$
$\displaystyle\quad-R\left(\displaystyle\frac{\Theta\varphi}{vV}\right)_{x}\psi_{xx}+R\left(\displaystyle\frac{\zeta}{v}\right)_{x}\psi_{xx}+F\psi_{xx}+(\psi_{t}\psi_{x})_{x}:=\displaystyle\sum_{i=1}^{6}I_{i}.$

use last inequality integrate in $\mathbb{R}_{+}\times(0,t)$

$\displaystyle\|\psi_{x}(t)\|^{2}+\displaystyle\int_{0}^{t}\|\psi_{xx}(\tau)\|^{2}d\tau$
$\displaystyle\quad\leq
C\|\psi_{0x}\|^{2}+C\displaystyle\sum_{i=1}^{6}\left|\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}I_{i}dxd\tau\right|.$
(4.64)

Now deal with $\iint|I_{i}|dxd\tau$ in the right side of (4.64). Using
$\epsilon$ small and $v=\varphi+V$ , $R\Theta/V=p_{+}$ and (2.22), (2.23),
(4)to get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}|I_{1}|dxd\tau\leq
C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}|V_{x}||\psi_{x}||\psi_{xx}|dxd\tau+C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}|\varphi_{x}||\psi_{x}||\psi_{xx}|dxd\tau$
$\displaystyle\quad\leq
C\displaystyle\int_{0}^{t}\|V_{x}\|\|\psi_{x}\|_{L^{\infty}}\|\psi_{xx}\|d\tau+C\displaystyle\int_{0}^{t}\|\psi_{x}\|_{L^{\infty}}\|\varphi_{x}\|\|\psi_{xx}\|d\tau$
$\displaystyle\quad\leq\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}\|V_{x}\|^{4}d\tau+C\displaystyle\int_{0}^{t}\|\psi_{x}\|^{1/2}\|\varphi_{x}\|\|\psi_{xx}\|^{3/2}d\tau$
$\displaystyle\quad\leq
C\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C(\delta_{0})\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau$
$\displaystyle\quad\quad+C\epsilon^{-1}\displaystyle\sup_{t}\|\varphi_{x}\|^{4}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau.$
(4.65)

Next we use the definition of $(V,U,\Theta)$ in (1.10),(1.13), Cauchy-Schwarz
inequality and (2.24), (2.25), (4) to get

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}|I_{2}|dxd\tau$
$\displaystyle\quad\leq
C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(|U_{xx}||\varphi|+|U_{x}||\varphi_{x}|+|U_{x}||V_{x}||\varphi|+|U_{x}||\varphi||\varphi_{x}|\right)|\psi_{xx}|dxd\tau$
$\displaystyle\quad\leq\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|\varphi\|_{L^{\infty}}^{2}\|U_{xx}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|U_{x}\|_{L^{\infty}}^{2}\|\varphi_{x}\|^{2}d\tau$
$\displaystyle\qquad+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|\varphi\|_{L^{\infty}}^{2}\|V_{x}\|^{2}\|U_{x}\|_{L^{\infty}}^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|\varphi\|_{L^{\infty}}^{2}\|U_{x}\|_{L^{\infty}}^{2}\|\varphi_{x}\|^{2}d\tau$
$\displaystyle\quad\leq\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C(\delta_{0})+C(\delta_{0})\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau.$
(4.66)

The same as (4.65) and (4.66), we the definition of $F$ in(1.14) and
(2.22)–(2.24),(4)we can get the estimates about $I_{3}$ to $I_{5}$ as
following.

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}(|I_{3}|+|I_{4}|+|I_{5}|)dxd\tau$
$\displaystyle\leq
C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(|\Theta_{x}||\varphi|+|\Theta||\varphi_{x}|+|\Theta||V_{x}||\varphi|+|\Theta||\varphi||\varphi_{x}|\right)|\psi_{xx}|dxd\tau$
$\displaystyle\quad+C\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(|\zeta_{x}|+|\zeta||V_{x}|+|\zeta||\varphi_{x}|\right)|\psi_{xx}|dxd\tau$
$\displaystyle\quad+\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|F\|^{2}d\tau$
$\displaystyle\leq\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}V_{x}^{2}\varphi^{2}dxd\tau$
$\displaystyle\quad+\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(\zeta_{x}^{2}+V_{x}^{2}\zeta^{2}\right)dxd\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\sup_{t}\|(\varphi,\zeta)\|\|(\varphi_{x},\zeta_{x})\|\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau$
$\displaystyle\quad+\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C(\delta_{0})+C\varphi_{0}(0).$
(4.67)

Because Lemma 4.1 and Lemma 4.2, we know $\|(\varphi,\zeta)\|$ is suitably
small when $C(\delta_{0})$ and $\|(\varphi_{0},\psi_{0},\zeta_{0})\|$ small.
So there exist a small constant $\delta$ about
$\|(\varphi_{0},\psi_{0},\zeta_{0})\|$ and $\delta_{0}$ such that

$\displaystyle\frac{C}{\epsilon}\displaystyle\sup_{t}\|(\varphi,\zeta)\|\|(\varphi_{x},\zeta_{x})\|\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau\leq
C\delta\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau,$

here
$\|\varphi_{x}\|^{2}+\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau\leq C$
can be established in Lemma 4.4. Therefore

$\displaystyle\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}(|I_{3}|+|I_{4}|+|I_{5}|)dxd\tau$
$\displaystyle\leq\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+\displaystyle\frac{C}{\epsilon}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\
(\zeta_{x}^{2}+\varphi_{x}^{2})dxd\tau+C(\delta_{0})+C\varphi_{0}(0).$

At last we integrate by part to the term about $I_{6}$. Because
$\psi_{x}=\varphi_{t}$,

$\psi_{t}\psi_{x}(0,t)=\psi_{t}\varphi_{t}(0,t)=(\psi\varphi_{t})_{t}(0,t)-\varphi_{0}(0)\displaystyle\frac{p_{+}^{2}}{\mu^{2}}\psi(0,t)e^{-p_{+}t/\mu}$
(4.68)

and $\psi(0,t)=C(\ln\Theta)_{x}(0,t)$,

we have

$\displaystyle\left|\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}I_{6}dxd\tau\right|=\left|\displaystyle\int_{0}^{t}(\psi_{t}\varphi_{t})(0,\tau)d\tau\right|$
$\displaystyle\leq
C|\psi(0,t)\varphi_{t}(0,t)|+C|\psi(0,0)\varphi_{t}(0,0)|+C\displaystyle\int_{0}^{t}\varphi_{0}(0)|\psi|(0,t)e^{-p_{+}\tau/\mu}d\tau$
$\displaystyle\leq C(\delta_{0}).$ (4.69)

In all we can get

$\displaystyle\displaystyle\int_{0}^{t}\ \displaystyle\int_{0}^{\infty}\
\displaystyle\sum_{i=1}^{6}|I_{i}|\ dx\ d\tau$ $\displaystyle\leq
C\displaystyle\int_{0}^{t}\ \epsilon\|\psi_{xx}\|^{2}\
d\tau+CN^{4}(t)\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau$
$\displaystyle\quad+C\epsilon^{-1}\displaystyle\int_{0}^{t}\
\|(\varphi_{x},\psi_{x},\zeta_{x})\|^{2}\
d\tau+C(\delta_{0})+C\varphi_{0}(0).$ (4.70)

So (4.64) can be change to

$\displaystyle\|\psi_{x}(t)\|^{2}+\displaystyle\int_{0}^{t}\|\psi_{xx}(\tau)\|^{2}d\tau$
$\displaystyle\leq$ $\displaystyle
C(\delta_{0})+C\varphi_{0}(0)+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|(\varphi_{x},\psi_{x},\zeta_{x})\|^{2}d\tau$
(4.71)
$\displaystyle\quad+CN^{4}(t)\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau+C\|\psi_{0x}\|^{2}.$

The estimate about $\|\zeta_{x}\|$ is similar to $\|\psi_{x}\|$, use (3.52)3
multiply $\zeta_{xx}$ then integrate in $Q_{t}=\mathbb{R}_{+}\times(0,t)$ to
get

$\displaystyle\|\zeta_{x}\|^{2}+\displaystyle\int_{0}^{t}\|\zeta_{xx}\|^{2}\
d\tau$ $\displaystyle\leq
C\|\zeta_{0x}\|^{2}+C\epsilon^{-1}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left(\psi_{x}^{2}+\zeta\psi_{x}^{2}+\zeta^{2}U_{x}^{2}+U_{x}^{2}\varphi^{2}\right)dxd\tau$
$\displaystyle\quad+C\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}|\zeta_{x}|(|\varphi_{x}|+|V_{x}|)|\zeta_{xx}|dxd\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{\infty}\left|\left(\displaystyle\frac{\Theta_{x}\varphi}{vV}\right)_{x}\right|^{2}dxd\tau$
$\displaystyle\quad+C\epsilon^{-1}\displaystyle\int_{0}^{\infty}\displaystyle\int_{0}^{\infty}(U_{x}^{4}+\psi_{x}^{4})dxd\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|G\|^{2}d\tau$
$\displaystyle=:C\|\zeta_{0x}\|^{2}+\displaystyle\sum_{i=1}^{5}J_{i}.$ (4.72)

Use the same method as (4.65)–(4.69)

$J_{1}\leq
C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau+C\epsilon^{-1}N^{2}(t)\displaystyle\int_{0}^{t}\|U_{x}\|^{2}d\tau\leq
C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{2}d\tau+C(\delta_{0}).$

Again use the same method as(4.65)–(4.69)

$\displaystyle J_{2}$ $\displaystyle\leq$ $\displaystyle
C\displaystyle\int_{0}^{t}\|\zeta_{x}\|_{L^{\infty}}\|\varphi_{x}\|\|\zeta_{xx}\|d\tau+C\displaystyle\int_{0}^{t}\|V_{x}\|\|\zeta_{x}\|_{L^{\infty}}\|\zeta_{xx}\|d\tau$
$\displaystyle\leq$ $\displaystyle
C\displaystyle\int_{0}^{t}\|\zeta_{x}\|^{1/2}\|\zeta_{xx}\|^{3/2}\|\varphi_{x}\|d\tau+\epsilon\displaystyle\int_{0}^{t}\|\zeta_{xx}\|^{2}d\tau+C(\delta_{0})\displaystyle\int_{0}^{t}\|\zeta_{x}\|^{2}d\tau$
$\displaystyle\leq$ $\displaystyle
2\epsilon\displaystyle\int_{0}^{t}\|\zeta_{xx}\|^{2}d\tau+C(\delta_{0})\displaystyle\int_{0}^{t}\|\zeta_{x}\|^{2}d\tau+C\epsilon^{-1}\displaystyle\sup_{t}\|\varphi_{x}\|^{4}\displaystyle\int_{0}^{t}\|\zeta_{x}\|^{2}d\tau.$

Because

$\displaystyle\left|\left(\displaystyle\frac{\Theta_{x}\varphi}{vV}\right)_{x}\right|^{2}$
$\displaystyle=|\displaystyle\frac{\Theta_{xx}\varphi}{vV}+\displaystyle\frac{\Theta_{x}\varphi_{x}}{vV}+\displaystyle\frac{\Theta_{x}\varphi}{vV}(-\displaystyle\frac{V_{x}+\varphi_{x}}{v^{2}}-\displaystyle\frac{V_{x}}{V^{2}})|^{2}$
$\displaystyle\leq
C\Theta_{xx}^{2}\varphi^{2}+C\Theta_{x}^{2}\varphi_{x}^{2}+C\Theta_{x}^{2}V_{x}^{2}\varphi^{2}+C\Theta_{x}^{2}\varphi^{2}\varphi_{x}^{2},$

combine with $R\Theta/V=p_{+}$, use the same method as (4.65)–(4.69) to get

$\displaystyle J_{3}$ $\displaystyle\leq$ $\displaystyle
C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\varphi\|_{L^{\infty}}^{2}\|\Theta_{xx}\|^{2}d\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\Theta_{x}\|_{L^{\infty}}^{2}\|\varphi_{x}\|^{2}d\tau$
$\displaystyle\quad+C\epsilon^{-1}\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ \Theta_{x}^{2}V_{x}^{2}\varphi^{2}\ dx\ d\tau$
$\displaystyle\leq$ $\displaystyle
C(\delta_{0})\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+C(\delta_{0})+C\varphi_{0}(0).$

Use the definition $U$ and similar as (4.65) (4.66) that we combine with Lemma
2.2 to get

$\displaystyle J_{4}$ $\displaystyle\leq$ $\displaystyle
C(\delta_{0})+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|_{L^{\infty}}^{2}\|\psi_{x}\|^{2}d\tau$
$\displaystyle\leq$ $\displaystyle
C(\delta_{0})+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\psi_{x}\|^{3}\|\psi_{xx}\|d\tau$
$\displaystyle\leq$ $\displaystyle
C(\delta_{0})+C\displaystyle\int_{0}^{t}\left(\epsilon^{-2}\|\psi_{x}\|^{2}\|\psi_{x}\|^{4}+\epsilon^{2}\|\psi_{xx}\|^{2}\right)d\tau.$

Use the definition $G$ in (1.14) combine with Lemma 2.2

$\displaystyle J_{5}=C\epsilon^{-1}\displaystyle\int_{0}^{t}\|G\|^{2}d\tau\leq
C(\delta_{0}).$

Use the results from $J_{1}$ to $J_{5}$, the inequality (4.72) can be change
to

$\displaystyle\|\zeta_{x}\|^{2}+\displaystyle\int_{0}^{t}\ \|\zeta_{xx}\|^{2}\
d\tau$ $\displaystyle\leq
C\|\zeta_{0x}\|^{2}+C(\epsilon^{-3}+N^{4}(t))\displaystyle\int_{0}^{t}\|(\psi_{x},\zeta_{x})\|^{2}d\tau+C(\delta_{0})\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+C(\delta_{0})$
$\displaystyle\quad+C\epsilon\displaystyle\int_{0}^{t}\|\psi_{xx}\|^{2}d\tau+C\varphi_{0}(0).$
(4.73)

In fact when combine with (4.71) and (4.73), it is easy to get

$\displaystyle\|(\varphi,\psi,\zeta)\|^{2}+\|(\psi_{x},\zeta_{x})\|^{2}+\displaystyle\int_{0}^{t}\|(\psi_{xx},\zeta_{xx})\|^{2}d\tau$
$\displaystyle\leq
C\left(\|(\psi_{0x},\zeta_{0x})\|^{2}+\epsilon^{-3}\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}\right)+C\epsilon^{-1}\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau+C(\delta_{0})+C\varphi_{0}(0).$

$\Box$

###### Lemma 4.4

For a small $\epsilon_{3}>0$ and $C(\delta_{0})>0$ is a small constant about
$\delta_{0}$ , $C(\delta_{0})>0$ is a small constant about $\delta_{0}$, we
can get

$\displaystyle\|\varphi_{x}\|^{2}+\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}d\tau$
$\displaystyle\leq$ $\displaystyle
C\|\varphi_{0x}\|^{2}+C\epsilon_{3}^{-1}\|(\varphi_{0},\psi_{0},\zeta_{0})\|^{2}+C(\delta_{0})+C\varphi_{0}(0).$
(4.74)

Proof. Set $\bar{v}=\displaystyle\frac{v}{V}$ take it into (3.52)1, (3.52)2
($p=R\theta/v$) to get

$\psi_{t}+p_{x}=\mu\left(\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)_{t}-F,$

Both sides of last equation multiply $\bar{v}_{x}/\bar{v}$ to get

$\displaystyle\left(\displaystyle\frac{\mu}{2}\left(\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)^{2}-\psi\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)_{t}+\displaystyle\frac{R\theta}{v}\left(\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)^{2}+\left(\psi\displaystyle\frac{\bar{v}_{t}}{\bar{v}}\right)_{x}$
$\displaystyle\quad=\displaystyle\frac{\psi_{x}^{2}}{v}+U_{x}\left(\displaystyle\frac{1}{v}-\displaystyle\frac{1}{V}\right)\psi_{x}+\displaystyle\frac{R\zeta_{x}}{v}\displaystyle\frac{\bar{v}_{x}}{\bar{v}}-\displaystyle\frac{R\theta}{v}\left(\displaystyle\frac{1}{\Theta}-\displaystyle\frac{1}{\theta}\right)\Theta_{x}\displaystyle\frac{\bar{v}_{x}}{\bar{v}}+F\displaystyle\frac{\bar{v}_{x}}{\bar{v}}.$
(4.75)

On the boundary condition we have

$\psi(0,\tau)\displaystyle\frac{{\bar{v}}_{\tau}(0,\tau)}{{\bar{v}}(0,\tau)}=\psi(0,\tau)\displaystyle\frac{\varphi_{\tau}(0,\tau)}{v(0,\tau)}$

When use the definition $\psi(0,t)$ in (3.52) and $\varphi_{t}(0,t)$ in (4.56)
we can get

$\displaystyle\int_{0}^{t}\left|\psi(0,\tau)\displaystyle\frac{{\bar{v}}_{\tau}(0,\tau)}{{\bar{v}}(0,\tau)}\right|d\tau\leq
C(\delta_{0})$ (4.76)

On the other hand if we integrate (4.75) in $R_{+}\times(0,t)$, (4.75) is
changed to

$\displaystyle\displaystyle\int_{\mathbb{R}_{+}}\left(\displaystyle\frac{\mu}{2}\left(\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)^{2}-\psi\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)dx-\displaystyle\int_{\mathbb{R}_{+}}\left(\displaystyle\frac{\mu}{2}\left(\displaystyle\frac{\bar{v}_{x}(x,0)}{\bar{v}(x,0)}\right)^{2}-\psi_{0}\displaystyle\frac{\bar{v}_{x}(x,0)}{\bar{v}(x,0)}\right)dx$
$\displaystyle\quad+\displaystyle\int_{0}^{t}\displaystyle\int_{\mathbb{R}_{+}}\left(\displaystyle\frac{R\theta}{v}\left(\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\right)^{2}+\left(\psi\displaystyle\frac{\bar{v}_{t}}{\bar{v}}\right)_{x}\right)dxd\tau$
$\displaystyle\leq C\epsilon^{-1}\left(\displaystyle\int_{0}^{t}\
\|(\zeta_{x},\psi_{x})\|^{2}\ d\tau+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ \Theta_{x}^{2}(\varphi^{2}+\zeta^{2})\ dx\
d\tau\right)$
$\displaystyle+C\epsilon^{-1}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}U_{x}^{2}\varphi^{2}dxd\tau+C\epsilon^{-1}\displaystyle\int_{0}^{t}\displaystyle\int_{0}^{+\infty}|F|^{2}dxd\tau+\epsilon\displaystyle\int_{0}^{t}\|\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\|^{2}d\tau+C(\delta_{0}).$
(4.77)

Insert (4.76) into (4) and combine with the definition of $F$ and $U$ and
Lemma 2.2, we can get

$\displaystyle\displaystyle\int_{0}^{t}\|\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\|^{2}\
d\tau+\|\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\|^{2}-C\epsilon^{-1}\|\psi\|^{2}-C\|\psi_{0}\|^{2}-C\displaystyle\int_{0}^{\infty}\displaystyle\frac{\bar{v}_{x}}{\bar{v}}(x,0)^{2}\
dx$ $\displaystyle\leq C\epsilon^{-1}\left(\displaystyle\int_{0}^{t}\
\|(\zeta_{x},\psi_{x})\|^{2}\ d\tau+\displaystyle\int_{0}^{t}\
\displaystyle\int_{0}^{\infty}\ \Theta_{x}^{2}(\varphi^{2}+\zeta^{2})\ dx\
d\tau+C(\delta_{0})\right)$
$\displaystyle\quad+\epsilon\displaystyle\int_{0}^{t}\|\displaystyle\frac{\bar{v}_{x}}{\bar{v}}\|^{2}d\tau+C\|\varphi_{0x}\|^{2}.$
(4.78)

Because
$C_{1}(\varphi_{x}^{2})-C_{2}V_{x}^{2}\leq(\displaystyle\frac{\bar{v}_{x}}{\bar{v}})^{2}\leq
C_{3}\varphi_{x}^{2}+C_{4}V_{x}^{2}$ ($C_{1},C_{2},C_{3},C_{4}$ stands for
constants about $v$), combine with Lemma 4.1– 4.2 we change (4) to

$\displaystyle\displaystyle\int_{0}^{t}\|\varphi_{x}\|^{2}\
d\tau+\|\varphi_{x}\|^{2}$ $\displaystyle\leq$ $\displaystyle
C\|\varphi_{0x}\|^{2}+\displaystyle\int_{0}^{t}C\epsilon_{3}^{-1}\|(\psi_{x},\zeta_{x})\|^{2}d\tau+C(\delta_{0})+C\varphi_{0}(0).$
(4.79)

So we finish this lemma.$\Box$

From Lemma 4.2 to Lemma 4.4 we know when $\delta_{0}$ and
$\|(\varphi_{0},\psi_{0},\zeta_{0})\|$ suitably small there exist a suitably
small constant $\delta$ such that

$\|(\varphi,\psi,\zeta)\|^{2}+\displaystyle\int_{0}^{t}\|(\psi_{x},\zeta_{x})\|^{2}d\tau\leq
C\delta,$

and

$\|(\varphi_{x},\psi_{x},\zeta_{x})\|^{2}+\displaystyle\int_{0}^{t}\|(\psi_{xx},\zeta_{xx})\|^{2}\leq
C.$

Then we can get $C_{5}\leq|v|\leq C_{6}$ and $C_{7}\leq|\theta|\leq C_{8}$
when $\delta$ small, here $C_{5}$, $C_{6},$ $C_{7}$ and $C_{8}$ are constants
independent of $v$ and $\theta$ . So we can get (3.1) in Proposition 2.2 .

To finish Theorem 3.1 now we will proof
$\displaystyle\sup_{x\in\mathbb{R}_{+}}|(\varphi,\psi,\zeta)|\to 0,\ as\
t\to\infty.$

Because $\displaystyle\int_{0}^{\infty}\partial_{x}$(3.52)${}_{1}\times
2\varphi_{x}\ dx$ equals to

$0=2\displaystyle\int_{0}^{\infty}\varphi_{x}\psi_{xx}dx-\displaystyle\frac{d}{dt}\|\varphi_{x}\|^{2},$
(4.80)

use Cauchy-Schwarz inequality we get

$2\displaystyle\int_{0}^{\infty}\varphi_{x}\psi_{xx}dx\leq
C\left(\|\varphi_{x}\|^{2}+\|\psi_{xx}\|^{2}\right),$

Again using Lemma 4.3–4.4 , we get

$\displaystyle\displaystyle\int_{0}^{\infty}\left|\displaystyle\frac{d}{dt}\|\varphi_{x}(t)\|^{2}\right|dt$
$\displaystyle\leq
C\displaystyle\int_{0}^{\infty}\left(\|\varphi_{x}\|^{2}+\|\psi_{xx}\|^{2}\right)dt$
$\displaystyle\leq
C\left(C(\delta_{0})+\|(\varphi_{0},\psi_{0},\zeta_{0})\|_{1}^{2}+C\varphi_{0}(0)\right).$
(4.81)

Similar as above, from Lemma 4.2$-$4.4 and combine with Sobolev inequality we
get

$\displaystyle\int_{0}^{\infty}\left(\left|\displaystyle\frac{d}{dt}\|\psi_{x}(t)\|^{2}\right|+\left|\displaystyle\frac{d}{dt}\|\zeta_{x}(t)\|^{2}\right|\right)d\tau\leq
C\left(C(\delta_{0})+\|(\varphi_{0},\psi_{0},\zeta_{0})\|_{1}^{2}+C\varphi_{0}(0)\right).$
(4.82)

It means

$\|(\varphi,\psi,\zeta)(t)\|_{L^{\infty}}^{2}\leq
2\|(\varphi,\psi,\zeta)(t)\|\|(\varphi_{x},\psi_{x},\zeta_{x})(t)\|\to
0\quad\mbox{when}\quad t\to\infty.$

So we finish Theorem 3.1.$\Box$

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Transverse Single Spin Asymmetries of charged hadrons at forward and backward
rapidity from $p^{\uparrow}+p$, $p^{\uparrow}+\mathrm{Al}$, and
$p^{\uparrow}+\mathrm{Au}$ collisions in PHENIX

Jeongsu Bok (for the PHENIX collaboration)

Pusan National University, Busan, South Korea
New Mexico State University, Las Cruces, USA

> Transverse Single Spin Asymmetries (TSSAs) in transversely polarized proton-
> proton collisions ($p^{\uparrow}+p$) have been a fruitful source for
> studying the spin structure of the proton. In the 2015 RHIC data taking
> periods, collisions of polarized protons with nuclei ($p^{\uparrow}+A$) were
> made for the first time. The measurements of TSSAs in $p^{\uparrow}+p$ and
> $p^{\uparrow}+A$ collisions can provide a unique opportunity to investigate
> the origin of TSSA in a gluon-rich target nucleus and provide a tool to
> study nuclear effects in $p+A$ collisions. This presentation will report
> PHENIX results of TSSAs for charged hadrons ($h^{\pm}$) at forward and
> backward rapidity ($1.4<|\eta|<2.4$) over the transverse momentum ranges
> $1.25<p_{T}<7.0\mathrm{\ GeV}/c$ and Feynman-$x$ ranges ($-0.2<x_{F}<0.2$)
> from $p^{\uparrow}+p$, $p^{\uparrow}+\mathrm{Al}$, and
> $p^{\uparrow}+\mathrm{Au}$ collisions at $\sqrt{s_{\rm NN}}=200$ GeV.

> PRESENTED AT
>
>
>
>
> DIS2023: XXX International Workshop on Deep-Inelastic Scattering and Related
> Subjects,
> Michigan State University, USA, 27-31 March 2023
>

##  1 Introduction

Since the 1970’s, large Transverse Single Spin Asymmetries (TSSAs) in single
hadron production have been observed at forward rapidity in transversely
polarized proton-proton collisions ($\text{$p^{\uparrow}$+$p$}\rightarrow
h+X$) in a wide range of collision energies. The phenomena have provided an
important motivation to investigate spin-momentum correlations inside a
nucleon. The analyzing power $A_{N}$ is defined as the left-right asymmetry of
the hadrons with respect to the polarization of the proton where the direction
of polarization is perpendicular to the beam direction.

$A_{N}(\phi)=\frac{\sigma^{\uparrow}-\sigma^{\downarrow}}{\sigma^{\uparrow}+\sigma^{\downarrow}}$
(1)

Figure 1:  Transverse single spin asymmetry measurements for $\pi^{0}$ and
$\pi^{\pm}$ at different center-of-mass energies [1] and $K^{\pm}$ at RHIC
energy as a function of Feynman-x [2].

Figure 1 shows the measurements of $A_{N}$ for neutral and charged pions as a
function of Feynman-$x$ ($x_{F}$) at different energies [1]. The magnitude of
$A_{N}$ increases as a function of $x_{F}$ at large positive $x_{F}$ from low
to high energies up to 500 GeV. $A_{N}$ for $\pi^{+}$ has a positive sign at
$x_{F}>0$, and $A_{N}$ for $\pi^{-}$ has comparable size, but opposite sign.
On the right side of Fig. 1, charged kaons result at 62.4 GeV [2] show that
$A_{N}$ for negatively and positively charged kaons have the same sign at
$x_{F}>0$, which is different from pions.

There are possible origins to explain these nonzero asymmetries. As an initial
state effect, the correlation between the proton spin and the transverse
momentum of the parton is called the Sivers mechanism. The final state effect
is called the Collins mechanism and describes the correlation between the spin
of the fragmenting quark and the transverse momentum of the final state
hadron. It is convoluted with the Transversity distribution functions, the
correlation between the spin of a proton and the spin of a quark. For the
single hadron production in proton-proton collisions, the twist-3 collinear
factorization approach is applicable, and the asymmetry arises from a twist-3
multi-parton correlation and a twist-3 fragmentation function.

In addition to the proton-proton collisions, the polarized proton-nucleus
collision ($p^{\uparrow}$+$A$) is proposed to solve this puzzle with an
interplay between small-$x$ physics and spin physics. Since the mechanisms are
expected to have different nuclear target dependences of $A_{N}$, measuring
$A_{N}$ in $p^{\uparrow}$+$A$ collisions can help to clarify the origin of the
$A_{N}$ [3, 4, 5, 6, 7]. Also, nuclear dependence of $A_{N}$ can be a probe
for the saturation scale in the nucleus [3].

##  2 Experimental Setup

Figure 2: Side view of the PHENIX detector in 2015.

The Relativistic Heavy Ion Collider (RHIC) can collide various kinds of ions
and protons in a wide range of energies, up to 510 GeV for proton-proton
collisions and 200 GeV for ion collisions. Also, RHIC is the first and only
collider which can collide polarized protons. In addition, the first polarized
proton-nucleus collisions were made during the 2015 run period. The PHENIX
experiment has two muon arms which cover $-2.2<\eta<-1.2$ and $1.2<\eta<2.4$.
In the muon arms, charged hadrons such as pions and kaons stop in the
intermediate layer of the muon identifier while muons of large momentum over 3
GeV/$c$ penetrate the whole detector. Therefore reconstructed tracks with a
minimum momentum requirement are selected as charged hadrons [8]. The analysis
methods using the muon arm have been shown in previous studies on heavy flavor
and charged hadrons [9, 10].

Based on the PYTHIA and HIJING event generators and GEANT4 simulation, the
particle composition in the measured charged hadron sample was estimated [11,
12]. Fig. 3 shows the estimated $K/\pi$ ratios at the collision vertex
(Generation) and in the reconstructed muon arm tracks from $p+p$ and $p+A$
collisions. The variations from the different physics lists in the GEANT4
simulation are considered. Due to interaction with the detector material, the
composition of the reconstructed particle is modified from the composition at
the collision vertex.

(a) $K/\pi$ ratio in $p+p$ collisions.

(b) $K/\pi$ ratio in $p+A$ collisions.

Figure 3:  Estimated $K/\pi$ ratios at the collision vertex (Generation) and
$K/\pi$ ratios for reconstructed muon arm tracks in in the geant4 simulation
(Reconstruction) for (a) $p+p$ and (b) $p+A$ collisions [13]. The variations
on the reconstructed $K/\pi$ ratios are from the different physics lists in
the GEANT4 simulation.

##  3 Results and Discussion

Fig. 4 shows $A_{N}$ for negatively and positively charged hadrons in
$p^{\uparrow}$+$p$ collisions as a function of $p_{T}$ and $x_{F}$ [13].
$A_{N}$ at $x_{F}<0$ are consistent with zero for both charges. At $x_{F}>0$,
$A_{N}$ for positively-charged hadrons is positive and increases as a function
of $x_{F}$. Positively charged hadrons are mostly $\pi^{+}$ and $K^{+}$ and
they have similar increasing trend as a function of $x_{F}$ at $x_{F}>0$, the
$A_{N}$ is consistent with it. $A_{N}$ for negatively charged hadrons at
$x_{F}>0$ shows small negative asymmetries. A possible explanation is the
partial cancellation between $\pi^{-}$ and $K^{-}$ where the opposite sign of
$A_{N}$ at $x_{F}>0$ was shown in the previous measurements. The bands at
$x_{F}>0.15$ are obtained from twist-3 model calculations [14] with a
variation of $K/\pi$ ratio by $\pm 30\%$ relative to the central value.

Figure 4:  (Left) $A_{N}$ of charged hadrons from $p^{\uparrow}$+$p$
collisions as a function of $p_{T}$ in the backward ($x_{F}<0$) and forward
($x_{F}>0$) region. (Right) $A_{N}$ of charged hadrons from $p^{\uparrow}$+$p$
collisions as a function of $x_{F}$ [13]. Bands are obtained twist-3 model
calculations [14] by varying the $K/\pi$ ratio by $\pm 30\%$ relative to the
central value.

Fig. 5 and 6 show $A_{N}$ for negatively and positively charged hadrons in
$p^{\uparrow}$+$p$ and $p^{\uparrow}$+$A$ collisions as a function of $p_{T}$
and $x_{F}$ [13]. $A_{N}$ for positively charged hadrons at large positive
$x_{F}$ in $p^{\uparrow}$+Al and $p^{\uparrow}$+Au are smaller than the one in
$p^{\uparrow}$+$p$. For negatively charged hadrons at $x_{F}>0$, the deviation
of $A_{N}$ in $p^{\uparrow}$+$p$ and $p^{\uparrow}$+Au is not significant to
state a modification due to large uncertainty.

Figure 5:  $A_{N}$ of negatively (Left) and positively-charged hadrons (Right)
from $p^{\uparrow}$+$p$, $p^{\uparrow}$+Al, and $p^{\uparrow}$+Au collisions
as a function of $p_{T}$ in the backward ($x_{F}<0$) and forward ($x_{F}>0$)
regions [13]. Vertical bars (boxes) represent statistical (systematic)
uncertainties. Figure 6:  $A_{N}$ of negatively (Left) and positively-charged
hadrons (Right) from $p^{\uparrow}$+$p$, $p^{\uparrow}$+Al, and
$p^{\uparrow}$+Au collisions as a function of $x_{F}$, where $x_{F}>0$ is
along the direction of the polarized proton [13]. Vertical bars(boxes
represent statistical (systematic) uncertainties.

##  4 Summary

In this study, the transverse single spin asymmetry ($A_{N}$) of positively
and negatively charged hadrons ($h^{\pm}$) at forward and backward rapidity
($1.4<|\eta|<2.4$) over the range of $1.5<p_{T}<7.0{\rm\ GeV}/c$ and
$0.04<|x_{F}|<0.2$ from transversely polarized proton-proton
($p^{\uparrow}$+$p$) and proton-nucleus ($p^{\uparrow}$+Al, $p^{\uparrow}$+Au)
collisions are presented while the previous study is limited to positively
charged hadrons at $0.1<x_{F}<0.2$ [10]. The results at $x_{F}<0$ are
consistent with zero in all collision systems. At $x_{F}>0$, negative charged
hadron results show small negative to zero $A_{N}$ in $p^{\uparrow}$+$p$ and
$p^{\uparrow}$+$A$ collisions. $A_{N}$ for positively charged hadrons
increases to positive values as $x_{F}$ increases at $x_{F}>0$ in
$p^{\uparrow}$+$p$ collisions. However, $p^{\uparrow}$+Au results show
suppression of $A_{N}$ at $0.1<x_{F}<0.2$ compared to the $p^{\uparrow}$+$p$
result. The results will be useful to understand the origin of $A_{N}$ and
offer a tool to investigate phenomena and nuclear effects in small-$x$.

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