# Asymptotic stability of solutions to a hyperbolic-elliptic coupled system of
the radiating gas on the half line

Shanming Ji  Minyi Zhang  Changjiang Zhu Corresponding author. Email:
<EMAIL_ADDRESS>
###### Abstract

This paper is concerned with the asymptotic stability of the solution to an
initial-boundary value problem on the half line for a hyperbolic-elliptic
coupled system of the radiating gas, where the data on the boundary and at the
far field state are defined as $u_{-}$ and $u_{+}$ satisfying $u_{-}<u_{+}$.
For the scalar viscous conservation law case, it is known by the work of Liu,
Matsumura, and Nishihara (SIAM J. Math. Anal. 29 (1998) 293-308) that the
solution tends toward rarefaction wave or stationary solution or superposition
of these two kind of waves depending on the distribution of $u_{\pm}$.
Motivated by their work, we prove the stability of the above three types of
wave patterns for the hyperbolic-elliptic coupled system of the radiating gas
with small perturbation. A singular phase plane analysis method is introduced
to show the existence and the precise asymptotic behavior of the stationary
solution, especially for the degenerate case: $u_{-}<u_{+}=0$ such that the
system has inevitable singularities. The stability of rarefaction wave,
stationary solution, and their superposition, is proved by applying the
standard $L^{2}$-energy method.

Keywords. Hyperbolic-elliptic coupled system; Initial-boundary value problem;
Asymptotic behavior; $L^{2}$-energy method; Singular phase plane analysis.

AMS subject classifications. 35L65; 35M10; 35B40.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries and Main Theorems
    1. 2.1 Construction and Properties of the Smooth Rarefaction Wave
    2. 2.2 Construction and Properties of the Stationary Solution
    3. 2.3 Preliminary Lemmas
    4. 2.4 Main Theorems
  3. 3 Asymptotics to Rarefaction Wave
    1. 3.1 Reformulation of the Problem in the Case of $u_{+}>u_{-}\geq 0$
    2. 3.2 _A priori_ Estimates
    3. 3.3 Asymptotic Behavior toward the Rarefaction Wave
  4. 4 Asymptotics to Stationary Solution
    1. 4.1 Reformulation of the Problem in the Case of $u_{-}<u_{+}\leq 0$
    2. 4.2 _A priori_ Estimates
    3. 4.3 Asymptotic Behavior toward the Stationary Solution
  5. 5 Asymptotics to Superposition of Nonlinear Waves
    1. 5.1 Reformulation of the Problem in the Case of $u_{-}<0<u_{+}$
    2. 5.2 _A priori_ Estimates
    3. 5.3 Asymptotic Behavior toward the Superposition of Nonlinear Waves

##  1 Introduction

We consider the asymptotic behavior of solutions to a model of hyperbolic-
elliptic coupled system of radiating gas on the half line. The initial-
boundary value problem (IBVP) of the above model on the half-line
$\mathbbm{R}_{+}=(0,+\infty)$ is

$\begin{cases}\displaystyle u_{t}+f(u)_{x}+q_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\\\ -q_{xx}+q+u_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\end{cases}$ (1.1)

with boundary condition

$u(0,t)=u_{-},\quad t>0,$ (1.2)

and initial data

$u(x,0)=u_{0}(x)\ \begin{cases}=u_{-},\ \ \ \ x=0,\\\ \rightarrow u_{+},\ \ \
\ x\rightarrow+\infty,\end{cases}$ (1.3)

where the flux $f(u)$ is a given smooth function of $u$, $u_{\pm}$ are given
constants, $u$ and $q$ are unknown functions of the spacial variable
$x\in\mathbbm{R}_{+}$ and the time variable $t$. Generally, $u$ and $q$
represent the velocity and the heat flux of the gas respectively. Throughout
this paper, we impose the following condition:

$f(0)=f^{\prime}(0)=0,~{}f^{\prime\prime}(u)>0\text{ for
}u\in\mathbbm{R},\quad\text{and }u_{-}<u_{+}.$

Such a hyperbolic-elliptic coupled system appears typically in radiation
hydrodynamics, cf. [37, 43]. The simplified model (1.1) was first recovered by
Hamer in [7], and for the derivation of system (1.1), we refer to [6, 7, 37].
In the _in-flow_ case of $u_{-}>0$, the boundary condition (1.2) is necessary
for the single hyperbolic equation (1.1)1, we additionally impose $q(0,t)=0$
for the well-posedness of the coupled elliptic equation (1.1)2. The problem
with additional condition $q(0,t)=0$ for _in-flow_ case will still be denoted
by (1.1)-(1.3) for the sake of simplicity. On the contrary, in the _out-flow_
case of $u_{-}<0$, the boundary condition (1.2) is enough for the hyperbolic-
elliptic coupled system (1.1), where the single hyperbolic equation (1.1)1 is
over-determined with the boundary condition (1.2) and the single elliptic
equation (1.1)2 is under-determined since no information of $q$ is given at
the boundary.

The system (1.1) has been extensively studied by several authors in different
contexts recently, but most of them are in the case of the whole space. For
the one-dimensional whole space, concerning the large-time behavior of
solutions to the Cauchy problem

$\begin{cases}u_{t}+f(u)_{x}+q_{x}=0,\quad x\in\mathbbm{R},\ \ t>0,\\\
-q_{xx}+q_{x}+u_{x}=0,\quad x\in\mathbbm{R},\ \ t>0,\\\ u(x,0)=u_{0}(x)\to
u_{\pm},\quad x\to\pm\infty,\end{cases}$ (1.4)

Tanaka in [36] proved the stability of the diffusion wave as a self-similar
solution to the viscous Burgers equation for the special case:
$u_{-}=u_{+}=0$. Kawashima and Nishihara in [14] discussed the case of
$u_{-}>u_{+}$ and showed that the solution to the Cauchy problem (1.4)
approaches the travelling wave of shock profile. Kawashima and Tanaka in [15]
investigated the remaining case of $u_{-}<u_{+}$ and proved the stability of
the rarefaction wave for the Cauchy problem (1.4) with the asymptotic
convergence rate. Ruan and Zhang in [31] further studied the case:
$u_{-}<u_{+}$ for general flux $f(u)$. The radiating gas system (1.4) in the
following scalar equation form with convolution

$u_{t}+f(u)_{x}+u-\psi*u=0,\quad x\in\mathbbm{R},\ \ t>0,$ (1.5)

where $\psi(x)$ is the fundamental solution to the elliptic operator
$-\Delta+I$ in $\mathbbm{R}$, was studied in [2, 4, 17, 21, 28, 35, 38]. It
was Schochet and Tadmor in [34] who first proved the $W^{1,\infty}$ regularity
of the solution to (1.5). Then Lattanzio and Marcati in [16] studied the well-
posedness and relaxation limits of weak entropy solutions. Yang and Zhao in
[40] constructed the Lax-Friedrichs’ scheme and obtained the BV estimates. For
the multi-dimensional case, Gao, Ruan and Zhu investigated the asymptotic rate
towards the planar rarefaction waves to the Cauchy problem for a hyperbolic-
elliptic coupled system (see [5, 6, 33]). Di Francesco in [1] studied the
global well-posedness and the relaxation limits of the multi-dimensional
radiating gas system. Duan, Fellner and Zhu in [3] studied the stability and
optimal time decay rates of planar rarefaction waves for a radiating gas model
based on Fourier energy method. The structure of shock waves in the radiating
gas dynamics was also investigated by many authors, see [10, 22, 23, 30, 42].
Moreover, the large time behaviors of the solutions for viscous conservation
laws, and other system were studied by many authors, see [11, 12, 13, 24, 26,
29, 39, 44, 45].

We are concerned with the asymptotic behavior of the solution to (1.1)-(1.3).
For the case of $u_{-}<u_{+}$, the corresponding Riemann problem for the
inviscid Burgers equation

$\begin{cases}u_{t}+f(u)_{x}=0,\quad x\in\mathbbm{R},\ \ t>0,\\\
u(x,0)=u_{0}^{R}(x):=\begin{cases}u_{-},\ \ x<0,\\\ u_{+},\ \
x>0,\end{cases}\end{cases}$

admits a simple rarefaction wave solution

$u^{R}(x/t)=\begin{cases}u_{-},\ \ \ \ x\leq
f^{\prime}(u_{-})t,\\\\[5.69054pt]
\displaystyle(f^{\prime})^{-1}\big{(}\frac{x}{t}\big{)},\ \ \ \
f^{\prime}(u_{-})t\leq x\leq f^{\prime}(u_{+})t,\\\\[5.69054pt] u_{+},\ \ \ \
x\geq f^{\prime}(u_{+})t.\end{cases}$

Additionally, the rarefaction wave $(u,q)$ of the system (1.1) is defined as
$(u,q)=(u^{R}(x/t),-\partial_{x}u^{R}(x/t))$. According to the convex function
$f(u)$ and the arguments used by Liu, Matsumura, and Nishihara in [18], we
have the following five cases (taking the typical form of
$f(u)=\frac{1}{2}u^{2}$ for example) due to the signs of the characteristic
speeds $u_{\pm}$:

$\displaystyle(1)\ \ u_{-}<u_{+}<0\quad(\text{equivalent to
}f^{\prime}(u_{-})<f^{\prime}(u_{+})<0);$ $\displaystyle(2)\ \
u_{-}<u_{+}=0\quad(\text{equivalent to
}f^{\prime}(u_{-})<f^{\prime}(u_{+})=0);$ $\displaystyle(3)\ \
u_{-}<0<u_{+}\quad(\text{equivalent to
}f^{\prime}(u_{-})<0<f^{\prime}(u_{+}));$ $\displaystyle(4)\ \
0=u_{-}<u_{+}\quad(\text{equivalent to
}0=f^{\prime}(u_{-})<f^{\prime}(u_{+}));$ $\displaystyle(5)\ \
0<u_{-}<u_{+}\quad(\text{equivalent to
}0<f^{\prime}(u_{-})<f^{\prime}(u_{+})).$

By using energy method, for the cases: $u_{-}<u_{+}\leq 0$, $0\leq
u_{-}<u_{+}$ and $u_{-}<0<u_{+}$, Liu, Matsumura, and Nishihara in [18] proved
that the initial-boundary value problem on the half line for scalar viscous
generalized Burgers equation admits a unique global solution and it converges
to the stationary solution, the rarefaction wave and the superposition of the
nonlinear waves, respectively, as $t\to\infty$. Since then, the initial
boundary value problem on the half line $\mathbbm{R_{+}}$ for different models
have been studied by many authors, cf. [8, 19, 20, 25, 41], and references
therein. In the case $(4)$, the convergence of the initial-boundary value
problem to a rarefaction wave has been investigated by Ruan and Zhu in [32].
However, there remain four cases to be considered in the previous studies.
Motivated by the classification by Liu, Matsumura, and Nishihara in [18], here
we prove the asymptotic behavior of the solutions to (1.1)-(1.3) for all
remaining cases: $u_{-}<u_{+}\leq 0$, $0<u_{-}<u_{+}$, and $u_{-}<0<u_{+}$.

The main features of the hyperbolic-elliptic coupled system (1.1)-(1.3) on the
half line are different from the previous study on scalar viscous Burgers
equation on the half line in [18], or the Cauchy problem (1.4) (equivalent to
the scalar form (1.5) with convolution), due to the following reasons:

• The hyperbolic-elliptic coupled system (1.1)-(1.3) on the half line cannot
be converted to a scalar equation with convolution, since the information of
the solution $q$ on the boundary ($q(0,t)$ or $q_{x}(0,t)$) is unknown for the
_out-flow_ cases (1) and (2). In fact, the “boundary condition” of the
elliptic problem (1.1)2 is determined in an inverse problem fashion such that
the hyperbolic equation (1.1)1 satisfies the boundary condition
$u(0,t)=u_{-}$, which is unnecessary for a single hyperbolic equation with
characteristic curves running out of the region.

• For the degenerate case (2): $u_{-}<u_{+}=0$, there arise inevitable
singularities in the analysis of the existence and spatial decay rates of the
stationary solution. It should be noted that the spatial decay estimates of
the stationary solution are essential to the energy estimates of the
perturbation problem. We employ a singular phase plane analysis method with a
series of approximated solutions to show the existence (see Lemma 2.5), and
then we utilize the finite series expansion to derive the precise decay
estimates of higher order derivatives (see Lemma 2.8).

• The estimates on the boundary terms are more subtle, especially for
$w_{xx}(0,t)$ of the perturbation $w:=u-\tilde{u}-\hat{u}$ when
$0<u_{-}<u_{+}$, since this case corresponds to the _in-flow_ problem. To
overcome it, we find out the relation (3.25) at boundary $x=0$ to estimate
$w_{xt}(0,t)$, which plays an important role in estimating $w_{xx}(0,t)$.

Additionally, in order to avoid too much tedious estimations in Sections 4 and
5, we introduce Lemma 2.12 to simplify the proof of the asymptotic behavior of
perturbation $z$ after we get the asymptotic behavior of $w$.

This paper is organized as follows. In Section 2, we firstly prepare the basic
properties of the rarefaction wave and stationary solution. Secondly, we give
some inequalities for the maximum norm to elliptic problem, which is essential
in estimating the asymptotic behavior of stationary solution. Finally, we
present our main theorems. In Section 3, we show the asymptotic behavior for
the case (5), which correspond to the rarefaction wave. The cases (1) and (2)
corresponding to the stationary solutions are investigated in Section 4. In
the final Section 5, referring to the results from [32], the combination of
the cases (2) and (4) can help us to consider the case (3) of superposition
waves.

Notations. Hereafter, we denote generic positive constants by $C$ and $c$
unless they need to be distinguished. For function spaces,
$L^{p}=L^{p}(\mathbbm{R}_{+})$ with $1\leq p\leq\infty$ denotes the usual
Lebesgue space on $\mathbbm{R}_{+}$ with the norm $|\cdot|_{p}$. For a non-
negative integer $l$, $H^{l}=H^{l}(\mathbbm{R}_{+})$ denotes the $l$-th order
Sobolev space in the $L^{2}$-sense, equipped with the norm $\|\cdot\|_{l}$. We
note that $H^{0}=L^{2}$ and $\|\cdot\|_{0}=\|\cdot\|$. For simplicity,
$\|f(\cdot,t)\|$ is denoted by $\|f(t)\|$, and $\|f(\cdot,t)\|_{l}$ is denoted
by $\|f(t)\|_{l}$.

##  2 Preliminaries and Main Theorems

Without loss of generality, we may take the typical case of
$f(u)=\frac{1}{2}u^{2}$ to simplify the calculations in the following, since
the proof for general convex function $f(u)$ can be slightly modified.

###  2.1 Construction and Properties of the Smooth Rarefaction Wave

In this subsection we consider the cases (4): $0=u_{-}<u_{+}$ and (5):
$0<u_{-}<u_{+}$. Since the rarefaction wave $u^{R}(x/t)$ is not smooth enough,
we construct the smooth approximation $\tilde{u}=\tilde{u}_{i}(x,t)~{}(i=4,5)$
by employing the ideal of Hattori and Nishihara in [9]. We define
$\tilde{u}_{5}(x,t)$ as a solution of the Cauchy problem

$\begin{cases}\tilde{u}_{t}+\tilde{u}\tilde{u}_{x}=\tilde{u}_{xx},\ \ \ \
x\in\mathbbm{R},\ \ \ t>0\\\ \tilde{u}(x,0)=\tilde{u}_{0}^{R}(x),\ \ \ \
x\in\mathbbm{R},\end{cases}$ (2.1)

with $\tilde{q}=-\tilde{u}_{x}$ and the initial data $\tilde{u}_{0}^{R}(x)$ is
defined by

$\tilde{u}_{0}^{R}(x)=\begin{cases}u_{-},\ \ \ \ x<0,\\\ u_{+},\ \ \ \
x>0,\end{cases}$ (2.2)

for the case $u_{-}>0$. When $u_{-}=0$, $\tilde{u}_{4}(x,t)$ defined as above
does not converge to the corresponding rarefaction wave fast enough near the
boundary $x=0$. Therefore, we need to modify $\tilde{u}_{0}^{R}(x)$ as

$\tilde{u}_{0}^{R}(x)=\begin{cases}-u_{+},\ \ \ \ x<0,\\\ u_{+},\ \ \ \
x>0,\end{cases}$

such that the solution $\tilde{u}_{4}(x,t)$ of (2.1) satisfies
$\tilde{u}_{4}(0,t)=0$. Using the Hopf-Cole transformation, the explicit
formula of $\tilde{u}$ can be obtained. Here, we give some properties of
smooth approximation solutions $\tilde{u}$ in Lemma 2.1. The proof of Lemma
2.1 can be found in [9, 15].

######  Lemma 2.1.

For $1\leq p\leq\infty$ and $t>0$, $\tilde{u}_{i}$ $(i=4,5)$ satisfies the
following estimates:
$\mathrm{(i)}$ $0\leq\tilde{u}_{5}(0,t)-u_{-}\leq C\delta\mathrm{e}^{-c(1+t)}$
for $u_{-}>0$ and $\tilde{u}_{4}(0,t)=0$ for $u_{-}=0$;
$\mathrm{(ii)}$ $|\partial_{x}^{k}\tilde{u}_{5}(0,t)|\leq
C\delta\mathrm{e}^{-c(1+t)}$, $|\partial_{x}^{k}\tilde{u}_{4}(0,t)|\leq
C(1+t)^{-\frac{1}{2}(k+1)}$, $k=1,2,3,4$;
$\mathrm{(iii)}$ $|\tilde{u}_{i}(t)-\tilde{u}^{R}(t)|_{p}\leq
C(1+t)^{-\frac{1}{2}+\frac{1}{2p}}$, $i=4,5$;
$\mathrm{(iv)}$ $|\tilde{u}_{ix}(t)|_{p}\leq
C\delta^{\frac{1}{p}}(1+t)^{-1+\frac{1}{p}}$, $|\tilde{u}_{it}(t)|_{p}\leq
C\delta^{\frac{1}{p}}(1+t)^{-1+\frac{1}{p}}$, $i=4,5$;
$\mathrm{(v)}$ $|\partial_{x}^{k}\partial_{t}^{l}\tilde{u}_{i}(t)|_{p}\leq
C\delta(1+t)^{-\frac{1}{2}(k+l-\frac{1}{p})}$, $k+l=1,2,3,4,\ \ \ \
k,l\in\mathbbm{N},\ \ \ \ i=4,5$;
$\mathrm{(vi)}$ $\tilde{u}_{ix}>0,\ \ \ \ u_{-}\leq\tilde{u}_{i}\leq u_{+},\ \
\ \ x\in\mathbbm{R},\ \ \ \ i=4,5$.

We note that the boundary value $\tilde{u}_{5}(0,t)\not=u_{-}$ if $u_{-}>0$.
In this case, the perturbation $u(x,t)-\tilde{u}(x,t)$ has a “boundary layer”
$u_{-}-\tilde{u}(0,t)$ at $x=0$. To solve this problem, we need to modify
$\tilde{u}$ near the boundary $x=0$. Referring to the method of Nakamura in
[27], our modified smooth approximation $(\varphi(x,t),\psi(x,t))$ is defined
as

$\begin{cases}\varphi(x,t):=\tilde{u}(x,t)-\hat{u}(x,t),\\\
\psi(x,t):=\tilde{q}(x,t)-\hat{q}(x,t),\end{cases}$ (2.3)

where

$\begin{cases}\hat{u}(x,t):=(\tilde{u}(0,t)-u_{-})\mathrm{e}^{-x},\\\
\hat{q}(x,t):=-\tilde{u}_{x}(0,t)\mathrm{e}^{-x}.\end{cases}$ (2.4)

Note that $\hat{u}(0,t)=0$ if $u_{-}=0$. Substituting (2.3) into (2.1), we get
the equation of $(\varphi(x,t),\psi(x,t))$:

$\begin{cases}\varphi_{t}+\varphi\varphi_{x}=\varphi_{xx}-\hat{u}_{t}-\varphi_{x}\hat{u}-\varphi\hat{u}_{x}-\hat{u}\hat{u}_{x}+\hat{u}_{xx},\\\
\varphi(0,t)=u_{-},\ \ \ \ \varphi(+\infty,t)=u_{+},\\\
\varphi(x,0)=\varphi_{0}(x)=\tilde{u}_{0}(x)-\hat{u}(x,0),\ \ \ \
x\in\mathbbm{R}_{+},\end{cases}$

and $\psi(x,t)=-\tilde{u}_{x}(x,t)+\tilde{u}_{x}(0,t)\mathrm{e}^{-x}$ which
satisfies $\psi(0,t)=0$.

According to Lemma 2.1, by simple calculations, we can conclude the following
estimates of $\varphi(x,t)$.

######  Lemma 2.2.

For $1\leq p\leq\infty$ and $t\geq 0$, $\varphi(x,t)$ satisfies:
$\mathrm{(i)}$ $\varphi_{x}>0$, $|\varphi(x,t)|<u_{+}$, for $x\in\mathbbm{R}$;
$\mathrm{(ii)}$ $|\varphi(t)-u^{R}(t)|_{p}\leq
C(1+t)^{-\frac{1}{2}+\frac{1}{2p}}$;
$\mathrm{(iii)}$ $|\varphi_{x}(t)|_{p}\leq
C\delta^{\frac{1}{p}}(1+t)^{-1+\frac{1}{p}}$, $|\varphi_{t}(t)|_{p}\leq
C\delta^{\frac{1}{p}}(1+t)^{-1+\frac{1}{p}}$;
$\mathrm{(iv)}$ $|\partial_{x}^{k}\partial_{t}^{l}\varphi(t)|_{p}\leq
C\delta(1+t)^{-\frac{1}{2}(k+l-\frac{1}{p})}$, $k+l=1,2,3,4$,
$k,l\in\mathbbm{N}$;
$\mathrm{(v)}$ $|\partial_{x}^{k}\partial_{t}^{l}R_{1}(t)|_{p}\leq
C\delta\mathrm{e}^{-c(1+t)}$, $k,l=0,1,2$;
$\mathrm{(vi)}$ $|\partial_{x}^{k}\partial_{t}^{l}R_{2}(t)|_{p}\leq
C\delta(1+t)^{-\frac{1}{2}(k+l+3-\frac{1}{p})}$, $k,l=0,1$.
Here,

$R_{1}(x,t)=\hat{u}_{t}+\varphi_{x}\hat{u}+\varphi\hat{u}_{x}+\hat{u}\hat{u}_{x}+\hat{q}_{x},$

and

$R_{2}(x,t)=-\varphi_{xxx}-\hat{u}_{xxx}-\hat{q}_{xx}+\hat{u}_{x}+\hat{q}.$

###  2.2 Construction and Properties of the Stationary Solution

In this subsection, we consider the cases (1): $u_{-}<u_{+}<0$ and (2):
$u_{-}<u_{+}=0$, and we will show that the IBVP (1.1)-(1.3) has a stationary
solution $(\bar{u}(x),\bar{q}(x))$, which satisfies

$\begin{cases}(\frac{1}{2}\bar{u}^{2})_{x}=-\bar{q}_{x},\\\
-\bar{q}_{xx}+\bar{q}+\bar{u}_{x}=0,\\\ \bar{u}(0)=u_{-},\\\
\bar{u}(+\infty)=u_{+},\ \ \ \ \bar{q}(+\infty)=0.\end{cases}$ (2.5)

Integrating the system (2.5)1 with respect to $x$ over $[x,\infty)$, we
conclude that $\frac{1}{2}\bar{u}^{2}=\frac{1}{2}u_{+}^{2}-\bar{q}\geq 0.$
Thus, $\bar{u}$ can be expressed as

$\bar{u}=\pm\sqrt{u_{+}^{2}-2\bar{q}}.$

Under the assumption of $u_{-}<u_{+}\leq 0$, the function $\bar{u}$ connecting
$u_{-}$ and $u_{+}$ should be chosen as

$\bar{u}=-\sqrt{u_{+}^{2}-2\bar{q}},\quad\text{such~{}that}\quad\bar{u}_{x}=\frac{\bar{q}_{x}}{\sqrt{u_{+}^{2}-2\bar{q}}}.$

Substituting $\bar{u}_{x}$ into (2.5)2, the system (2.5) is converted to

$\begin{cases}-\bar{q}_{xx}+\bar{q}+\frac{\bar{q}_{x}}{\sqrt{u_{+}^{2}-2\bar{q}}}=0,\\\
\bar{q}(0)=\bar{q}_{-}<0,\ \ \ \ \bar{q}(+\infty)=0.\end{cases}$ (2.6)

Here, $\bar{q}_{-}$ satisfies
$\bar{q}_{-}=\frac{1}{2}u_{+}^{2}-\frac{1}{2}u_{-}^{2}<0$.

In the cases (1): $u_{-}<u_{+}<0$ and (2): $u_{-}<u_{+}=0$, we prove that the
initial-boundary value problem (2.5) has a stationary solution
$(\bar{u},\bar{q})=(\bar{u}_{i}(x),\bar{q}_{i}(x)),i=1,2$, respectively.

######  Lemma 2.3.

Suppose $u_{-}<u_{+}\leq 0$ and let $\delta:=|u_{-}-u_{+}|$. Then there exists
a solution $\bar{u}_{i},(i=1,2)$ to the stationary problem (2.5), such that
the following estimates hold for some positive constants C and $\lambda$:
$\mathrm{(i)}$ $\bar{u}_{ix}(x)>0,\quad i=1,2$;
$\mathrm{(ii)}$
$\left|\partial_{x}^{k}\left\\{\bar{u}_{1}(x)-u_{+}\right\\}\right|\leq
C\delta\mathrm{e}^{-\lambda x},\quad k=0,1,2,3$;
$\mathrm{(iii)}$ $\displaystyle|\partial_{x}^{k}\bar{u}_{2}(x)|\leq
C\frac{\delta^{k+1}}{(1+\delta x)^{k+1}},\quad k=0,1,2,3,4$;
$\mathrm{(iv)}$
$\displaystyle\left|\frac{\bar{u}_{ixx}^{2}}{\bar{u}_{ix}}\right|\leq
C\delta,\quad\left|\frac{\bar{u}_{ixxx}^{2}}{\bar{u}_{ix}}\right|\leq
C\delta,\ \ \ \ i=1,2$;
$\mathrm{(v)}$
$\displaystyle\left|\frac{\bar{u}_{2xxx}}{\bar{u}_{2x}}\right|\leq
C\delta,\quad\left|\frac{\partial_{x}^{4}\bar{u}_{2}}{\bar{u}_{2x}}\right|\leq
C\delta$.

In the degenerate case (2): $u_{-}<u_{+}=0$, the elliptic problem (2.6) has
singularity near $\bar{q}=0$, and $\bar{q}(+\infty)=0$, which means the
singularity is inevitable. The singularity causes significant difficulty in
the analysis of the existence and asymptotic behavior of the stationary
solution. We will present the detailed proof of Lemma 2.3 for the degenerate
case (2) by applying a generalized singular phase plane analysis method. Then
we sketch the main lines of the proof for case (1).

For the sake of convenience, in this subsection we set

$s(x)=-\bar{q}(x),\quad v(x)=\bar{q}_{x}(x),\quad x>0,$ (2.7)

then $0<s(x)<s_{0}:=\frac{1}{2}u_{-}^{2}-\frac{1}{2}u_{+}^{2}>0$, $v(x)>0$,
are solutions to the following problem

$\begin{cases}s_{x}=-v,\\\ \displaystyle
v_{x}=-s+\frac{v}{\sqrt{u_{+}^{2}+2s}},\\\ 0<s(x)<s_{0},\quad v(x)>0,\quad
x\in(0,+\infty).\end{cases}$ (2.8)

We first focus on the degenerate case of $u_{+}=0$, and the problem has
singularity at where $s=0$

$\begin{cases}s_{x}=-v,\\\ \displaystyle v_{x}=-s+\frac{v}{\sqrt{2s}},\\\
0<s(x)<s_{0},\quad v(x)>0,\quad x\in(0,+\infty).\end{cases}$ (2.9)

######  Lemma 2.4.

For any given $s_{0}>0$, if $\tilde{v}(s)$ solves the following singular
equation

$\begin{cases}\displaystyle\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}=\frac{s}{\tilde{v}}-\frac{1}{\sqrt{2s}},\quad
s\in(0,s_{0}),\\\
\displaystyle\tilde{v}(s)>0\mathrm{~{}for~{}}s\in(0,s_{0}),\quad\lim_{s\rightarrow
0^{+}}\tilde{v}(s)=0,\quad\int_{0}^{s_{0}}\frac{1}{\tilde{v}(s)}\mathrm{d}s=+\infty,\end{cases}$
(2.10)

then the function $s(x)$ defined by

$x=-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau,\qquad
x\in(0,+\infty)$ (2.11)

is a solution of (2.9) with $v(x):=\tilde{v}(s(x))>0$.

Proof. The positivity of $\tilde{v}(s)$ on $s\in(0,s_{0})$ and the singularity
at $s=0$ under the conditions in (2.10) imply that the function $s(x)$ is
well-defined on $x\in(0,+\infty)$ by (2.11) such that $s(0)=s_{0}>0$ and
$s(+\infty)=0$, $s(x)$ is a strictly decreasing function for
$x\in(0,+\infty)$. Differentiating the identity (2.11) with respect to $x$
shows that

$1=-\frac{1}{\tilde{v}(s(x))}s_{x}(x)=-\frac{s_{x}(x)}{v(x)}.$

Further,

$v_{x}(x)=\frac{\mathrm{d}}{\mathrm{d}x}\tilde{v}(s(x))=\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}\cdot\frac{\mathrm{d}s(x)}{\mathrm{d}x}=\Big{(}\frac{s(x)}{\tilde{v}(s(x))}-\frac{1}{\sqrt{2s(x)}}\Big{)}\cdot(-v(x))=-s(x)+\frac{v(x)}{\sqrt{2s(x)}}.$

The proof is completed. $\hfill\Box$

The main feature of the problem (2.10) is that it has two singularities: (i)
for $s$ near zero, the function $-\frac{1}{\sqrt{2s}}$ is un-bounded and not
Lipschitz continuous; (ii) for $\tilde{v}=0$ as the condition
$\lim_{s\rightarrow 0^{+}}\tilde{v}(s)=0$ required, the term
$\frac{s}{\tilde{v}}$ is also un-bounded and not Lipschitz continuous. In
order to handle these two singularities, we consider an approximated problem
for $k\in\mathbbm{N}^{+}$ (without loss of generality we may assume that
$\frac{1}{k^{2}}<s_{0}$, otherwise we only consider large
$k\in\mathbbm{N}^{+}$ with $k\geq k_{0}$, where
$k_{0}:=[\frac{1}{\sqrt{s_{0}}}]+1$)

$\begin{cases}\displaystyle\frac{\mathrm{d}\tilde{v}_{k}}{\mathrm{d}s}=\frac{s}{\tilde{v}_{k}}-\frac{1}{\sqrt{2s}},\quad
s\in\Big{(}\frac{1}{k^{2}},s_{0}\Big{)},\\\\[11.38109pt]
\displaystyle\tilde{v}_{k}(s)>0\mathrm{~{}for~{}}s\in\Big{(}\frac{1}{k^{2}},s_{0}\Big{)},\quad\tilde{v}_{k}\Big{(}\frac{1}{k^{2}}\Big{)}=\frac{1}{k}.\end{cases}$
(2.12)

The above approximated problem is solved by the generalized phase plane
analysis method. See the illustration Figure 1.

Figure 1: The singular phase plane corresponding to system (2.9): the dash-dot
lines correspond to solutions $\tilde{v}_{k}(s)$ for different $k$ in the
proof of Lemma 2.5, the solid line corresponds to the limit function
$\tilde{v}(s)$.

######  Lemma 2.5.

For any $s_{0}>0$ and any $\frac{1}{k^{2}}<s_{0}$ (i.e. $k\geq k_{0}$), the
problem (2.12) admits a solution $\tilde{v}_{k}(s)>0$ on
$s\in(\frac{1}{k^{2}},s_{0})$ such that
$\mathrm{(i)}$ $\tilde{v}_{k}(s)$ is monotone decreasing with respect to $k$,
i.e., $\tilde{v}_{k_{1}}(s)>\tilde{v}_{k_{2}}(s)$ for any $k_{2}>k_{1}\geq
k_{0}$ on their joint interval
$(\frac{1}{k_{1}^{2}},s_{0})\cap(\frac{1}{k_{2}^{2}},s_{0})=(\frac{1}{k_{1}^{2}},s_{0})$;
$\mathrm{(ii)}$ $\tilde{v}_{k}(s)$ has the following upper bound estimate

$\tilde{v}_{k}(s)\leq\overline{\Gamma}_{k}(s):=\begin{cases}\displaystyle\frac{1}{k},~{}&s\in\Big{(}\frac{1}{k^{2}},(\frac{1}{\sqrt{2}k})^{\frac{2}{3}}\Big{]}\\\\[8.53581pt]
\displaystyle\sqrt{2}s^{\frac{3}{2}},~{}&s\in\Big{(}(\frac{1}{\sqrt{2}k})^{\frac{2}{3}},s_{0}\Big{)}\end{cases}=\max\Big{\\{}\frac{1}{k},\sqrt{2}s^{\frac{3}{2}}\Big{\\}},\quad
s\in\Big{(}\frac{1}{k^{2}},s_{0}\Big{)};$

$\mathrm{(iii)}$ $\tilde{v}_{k}(s)$ has the following uniformly lower bound
estimate for any $\mu\in(0,\sqrt{2})$

$\tilde{v}_{k}(s)\geq\Gamma_{\mu}(s):=\begin{cases}\displaystyle(\sqrt{2}-\mu)s^{\frac{3}{2}},~{}&s\in(0,\delta_{\mu}]\\\\[8.53581pt]
\displaystyle(\sqrt{2}-\mu)\delta_{\mu}^{\frac{3}{2}},~{}&s\in(\delta_{\mu},s_{0})\end{cases}=\min\left\\{(\sqrt{2}-\mu)s^{\frac{3}{2}},(\sqrt{2}-\mu)\delta_{\mu}^{\frac{3}{2}}\right\\},$

where $\delta_{\mu}:=\frac{\mu}{3\sqrt{2}}$.

Proof. In the phase plane of (2.9), define an auxiliary function

$\Gamma_{0}(s):=\sqrt{2}s^{\frac{3}{2}},\quad s>0.$

The system (2.9) is locally uniquely solvable at any point
$(s,v)\in\mathbbm{R}^{+}\times\mathbbm{R}^{+}$ and the vector field is denoted
by

$\Phi(s,v):=-v,\qquad\Psi(s,v):=-s+\frac{v}{\sqrt{2s}}.$

There is no stationary point and no closed periodic orbit (since $\Phi(s,v)<0$
or according to Bendixson’s criterion such that
$\mathrm{div}_{(s,v)}(\Phi,\Psi)=\frac{1}{\sqrt{2s}}>0$) within the first
quadrant, then the Poincáre-Bendixson Theorem implies that any trajectory must
runs to the boundary of the first quadrant for $x\rightarrow\pm\infty$.
According to the smoothness and the absence of stationary point of the vector
field $(\Phi,\Psi)$ within the first quadrant, we know that any two
trajectories can not intersect with each other at any point
$(s,v)\in\mathbbm{R}^{+}\times\mathbbm{R}^{+}$. The graph of $\Gamma_{0}$
(also denoted by $\Gamma_{0}$) divides the first quadrant into two parts

$G_{1}:=\\{(s,v)\in\mathbbm{R}^{+}\times\mathbbm{R}^{+};v>\Gamma_{0}(s)\\},\quad
G_{2}:=\\{(s,v)\in\mathbbm{R}^{+}\times\mathbbm{R}^{+};v<\Gamma_{0}(s)\\}.$

For any point $(s,v)\in G_{1}$, locally $\Phi<0$ and $\Psi>0$, which means the
trajectory runs through $(s,v)$ to the left-up direction if the autonomous
independent variable $x$ grows; similarly, for any point $(s,v)\in G_{2}$,
locally $\Phi<0$ and $\Psi<0$.

For any $s_{0}>0$ and any $\frac{1}{k^{2}}<s_{0}$ (i.e. $k\geq k_{0}$),
consider the dynamic system (2.9) with initial condition
$s_{k}(x_{k})=\frac{1}{k^{2}}$ and $v_{k}(x_{k})=\frac{1}{k}$, where
$x_{k}\in(0,+\infty)$ is a constant to be determined. The trajectory
corresponding to this local solution is denoted by
$\Pi_{k}:=\\{(s_{k}(x),v_{k}(x))\\}$. Since the system (2.9) is autonomous, we
will shift $x_{k}$ to a suitable position in the following proof. Within the
first quadrant, $s_{x}(x)=\Phi(s,v)=-v<0$, which means that $s(x)$ is strictly
decreasing with respect to $x$. We can take $x$ as an inverse function of $s$
in the range of $s(x)$ and then regard $v(x)$ as a function of $s$, this is
the local solution $\tilde{v}_{k}(s)$ of (2.12). The choice of $x_{k}$ has no
influence on the function $\tilde{v}_{k}(s)$.

We only consider the trajectory $\Pi_{k}$ in the negative $x$ direction, that
is, we consider the solution $(s_{k}(x),v_{k}(x))$ such that $x<x_{k}$ and $x$
is decreasing. Noticing that $(\frac{1}{k^{2}},\frac{1}{k})\in G_{1}$ (for
$k\geq 2$ without loss of generality), we see that $\Pi_{k}$ runs through
$(\frac{1}{k^{2}},\frac{1}{k})$ in the right-down direction until it reaches
some point at $\Gamma_{0}$. This must happen since $\Gamma_{0}(s)$ is
increasing and $\Gamma_{0}((\frac{1}{\sqrt{2}k})^{\frac{2}{3}})=\frac{1}{k}$.
Therefore, there exists a $\hat{x}_{k}<x_{k}$ such that
$s_{k}(\hat{x}_{k})=:\hat{s}_{k}\in(\frac{1}{k^{2}},(\frac{1}{\sqrt{2}k})^{\frac{2}{3}})$
and $v_{k}(\hat{x}_{k})=:\hat{v}_{k}$, satisfying

$\hat{v}_{k}=\sqrt{2}\hat{s}_{k}^{\frac{3}{2}}\in\Big{(}\Gamma_{0}(\frac{1}{k^{2}}),\Gamma_{0}((\frac{1}{\sqrt{2}k})^{\frac{2}{3}})\Big{)}=\Big{(}\frac{\sqrt{2}}{k^{3}},\frac{1}{k}\Big{)}.$
(2.13)

Locally at the point $(\hat{s}_{k},\hat{v}_{k})\in\Gamma_{0}$, $\Phi<0$ and
$\Psi=0$. Noticing that
$\Gamma_{0}^{\prime}(\hat{s}_{k})=\frac{3\sqrt{2}}{2}\sqrt{\hat{s}_{k}}>0$, we
find that $\Pi_{k}$ runs into $G_{2}$ region as $x$ decreasing from
$\hat{x}_{k}$.

We assert that $\Pi_{k}$ is under $\Gamma_{0}$ for $x<\hat{x}_{k}$ (i.e.,
$\tilde{v}_{k}(s)<\Gamma_{0}(s)$ for $s>\hat{s}_{k}$) and $\tilde{v}_{k}(s)$
is increasing for $s>\hat{s}_{k}$. We prove by contradiction and assume that
there exists a $s^{*}>\hat{s}_{k}$ such that
$\tilde{v}_{k}(s^{*})=\Gamma_{0}(s^{*})$ and $\tilde{v}_{k}(s)<\Gamma_{0}(s)$
for $s\in(\hat{s}_{k},s^{*})$, which means there exists $x^{*}<\hat{x}_{k}$
such that $s_{k}(x^{*})=s^{*}$ and
$v_{k}(x^{*})=\tilde{v}_{k}(s^{*})=\Gamma_{0}(s^{*})=:v^{*}$. Then
$\tilde{v}_{k}^{\prime}(s^{*})\geq\Gamma_{0}^{\prime}(s^{*})$, and
$\Gamma_{0}^{\prime}(s^{*})=\frac{3\sqrt{2}}{2}\sqrt{s^{*}}$, but

$\tilde{v}_{k}^{\prime}(s^{*})=\left[\frac{\frac{\mathrm{d}v_{k}(x)}{\mathrm{d}x}}{\frac{\mathrm{d}s_{k}(x)}{\mathrm{d}x}}\right]_{x=x^{*}}=\frac{-s_{k}+\frac{v_{k}}{\sqrt{2s_{k}}}}{-v_{k}}\Big{|}_{x=x^{*}}=\frac{s^{*}}{v^{*}}-\frac{1}{\sqrt{2s^{*}}}=0$

at this point, which is a contradiction. Now that we have proved
$\tilde{v}_{k}(s)<\Gamma_{0}(s)$ for $s>\hat{s}_{k}$, or equivalently,
$(s_{k}(x),v_{k}(x))\in G_{2}$ for $x<\hat{x}_{k}$. Therefore,

$\tilde{v}_{k}^{\prime}(s)=\frac{s}{\tilde{v}_{k}}-\frac{1}{\sqrt{2s}}>0,\qquad(s,\tilde{v}_{k}(s))\in
G_{2},$ (2.14)

which shows that $\tilde{v}_{k}(s)$ is increasing for $s>\hat{s}_{k}$. To
summarize, we proved that $\tilde{v}_{k}(s)$ is decreasing for
$s\in(\frac{1}{k^{2}},\hat{s}_{k})$ and increasing for $s>\hat{s}_{k}$, which
means $\tilde{v}_{k}(s)\geq\hat{v}_{k}$ with
$\hat{v}_{k}>\frac{\sqrt{2}}{k^{3}}$ satisfying (2.13). Furthermore, we see
that

$s_{k}^{\prime}(x)=\Phi(s_{k},v_{k})=-v_{k}(x)=-\tilde{v}_{k}(s_{k}(x))\leq-\hat{v}_{k}<-\frac{\sqrt{2}}{k^{3}},$

which implies that

$-\frac{k^{3}}{\sqrt{2}}s_{0}\leq-\int_{\frac{1}{k^{2}}}^{s_{0}}\frac{1}{\tilde{v}_{k}(s)}\mathrm{d}s<0$

is finite. We would shift
$x_{k}=\int_{\frac{1}{k^{2}}}^{s_{0}}\frac{1}{\tilde{v}_{k}(s)}\mathrm{d}s$
such that $s_{k}(0)=s_{0}$.

The above arguments imply that $\tilde{v}_{k}(s)\leq\overline{\Gamma}_{k}(s)$
for $s\in(\frac{1}{k^{2}},s_{0})$. Next, we show the monotone dependence of
$\tilde{v}_{k}(s)$ with respect to $k$. For any $k_{2}>k_{1}\geq k_{0}$, the
trajectory $\Pi_{k_{2}}$ runs through $(\frac{1}{k_{2}^{2}},\frac{1}{k_{2}})$
in the right-down direction (for $x$ decreasing) until it reaches some point
$(\frac{1}{k_{1}^{2}},v_{*})\in G_{1}$ with
$v_{*}<\frac{1}{k_{2}}<\frac{1}{k_{1}}$ or some point $(s_{*},v_{*})$ on
$\Gamma_{0}$ with $s_{*}\in(\frac{1}{k_{2}^{2}},\frac{1}{k_{1}^{2}}]$. In the
latter case, $\Pi_{k_{2}}$ runs across $\Gamma_{0}$ into $G_{2}$, and as shown
by the above arguments (i.e., (2.13) and (2.14)),
$\tilde{v}_{k_{2}}(s)<\Gamma_{0}(s)$ and $\tilde{v}_{k_{2}}(s)$ is increasing
for $s>s_{*}$. Therefore,
$\tilde{v}_{k_{2}}(\frac{1}{k_{1}^{2}})\in(v_{*},\Gamma_{0}(\frac{1}{k_{1}^{2}}))\subset(v_{*},\frac{1}{k_{1}})$.
In all cases,
$\tilde{v}_{k_{2}}(\frac{1}{k_{1}^{2}})<\frac{1}{k_{1}}=\tilde{v}_{k_{1}}(\frac{1}{k_{1}^{2}})$.
The comparison $\tilde{v}_{k_{2}}(s)<\tilde{v}_{k_{1}}(s)$ for
$s>\frac{1}{k_{1}^{2}}$ follows from the fact that any two trajectories cannot
intersect with each other in the first quadrant.

Lastly we show the uniformly lower bound of $\tilde{v}_{k}(s)$. For any
$\mu\in(0,\sqrt{2})$ and $\delta_{\mu}:=\frac{\mu}{3\sqrt{2}}$, we consider
the special curve $\Gamma_{\mu}$ defined by $v=\Gamma_{\mu}(s)$. For any point
$(\check{s},\check{v})\in\Gamma_{\mu}$ with $\check{s}\in(0,\delta_{\mu})$ and
$\check{v}=\Gamma_{\mu}(\check{s})$, the direction of vector field

$\frac{\Psi(\check{s},\check{v})}{\Phi(\check{s},\check{v})}=\frac{\check{s}}{\check{v}}-\frac{1}{\sqrt{2\check{s}}}=\frac{\check{s}}{(\sqrt{2}-\mu)\check{s}^{\frac{3}{2}}}-\frac{1}{\sqrt{2\check{s}}}=\frac{\mu}{(\sqrt{2}-\mu)\sqrt{2}\sqrt{\check{s}}}>\frac{\mu}{2\sqrt{\check{s}}},$
(2.15)

and the derivative of the curve $\Gamma_{\mu}$

$\Gamma_{\mu}^{\prime}(\check{s})=\frac{3}{2}(\sqrt{2}-\mu)\sqrt{\check{s}}<\frac{3}{2}\sqrt{2}\sqrt{\check{s}}<\frac{\mu}{2\sqrt{\check{s}}}<\frac{\Psi(\check{s},\check{v})}{\Phi(\check{s},\check{v})}$

for $\check{s}<\delta_{\mu}=\frac{\mu}{3\sqrt{2}}$. For any point
$(\check{s},\check{v})\in\Gamma_{\mu}$ with $\check{s}\geq\delta_{\mu}$ and
$\check{v}=\Gamma_{\mu}(\check{s})=(\sqrt{2}-\mu)\delta_{\mu}^{\frac{3}{2}}$,
the direction of vector field

$\frac{\Psi(\check{s},\check{v})}{\Phi(\check{s},\check{v})}=\frac{\check{s}}{\check{v}}-\frac{1}{\sqrt{2\check{s}}}>0,$
(2.16)

since $(\check{s},\check{v})\in G_{2}$ as $\Gamma_{\mu}(s)<\Gamma_{0}(s)$, and
the curve $\Gamma_{\mu}$ is horizontal. Noticing that $\Phi(s,v)=-v<0$, we see
that any trajectory $(s(x),v(x))$ runs rightwards as $x$ decreasing. It
follows from (2.15) and (2.16) that any trajectory starting from a point
$(s,v)$ above $\Gamma_{\mu}$ cannot run through $\Gamma_{\mu}$ as the
independent variable $x$ decreasing. Therefore,
$\tilde{v}_{k}(s)>\Gamma_{\mu}(s)$ for $s\in(\frac{1}{k^{2}},s_{0})$ since
$(\frac{1}{k^{2}},\frac{1}{k})\in G_{1}$ is above $\Gamma_{\mu}$ for $k\geq
2$. The proof is completed. $\hfill\Box$

The solutions $\\{\tilde{v}_{k}(s)\\}$ to the above approximated problem
(2.12) are not defined for all $s>0$, only on $(\frac{1}{k^{2}},s_{0})$. We
define

$\tilde{v}_{k}^{*}(s):=\begin{cases}\tilde{v}_{k}(s),\ \ \ \
s\in\left.\left(\frac{1}{k^{2}},s_{0}\right.\right],\\\\[11.38109pt]
\frac{1}{k},\ \ \ \
s\in\left.\left(0,\frac{1}{k^{2}}\right.\right],\end{cases}$

and

$\tilde{v}(s)=\lim_{k\rightarrow\infty}\tilde{v}_{k}^{*}(s),\ \ \ \
s\in(0,s_{0}).$ (2.17)

######  Lemma 2.6.

The function $\tilde{v}(s)$ is well-defined in (2.17) for $s\in(0,s_{0})$ and
$\tilde{v}(s)$ is a solution to the singular problem (2.10). Moreover, for any
$\mu\in(0,\sqrt{2})$, let $\Gamma_{\mu}$ be the function defined in Lemma 2.5,
then $\Gamma_{\mu}(s)\leq\tilde{v}(s)\leq\sqrt{2}s^{\frac{3}{2}}$ for all
$s\in(0,s_{0})$, which means
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}+o(s^{\frac{3}{2}})$ as $s\rightarrow
0^{+}$.

Proof. For any fixed $s\in(0,s_{0})$, $\tilde{v}_{k}(s)$ is defined on
$(\frac{1}{k^{2}},s_{0})$, which contains $s$ if $\frac{1}{k^{2}}<s$, i.e.,
$k\geq[\frac{1}{\sqrt{s}}]+1$. According to Lemma 2.5,
$\\{\tilde{v}_{k}(s)\\}$ is monotone decreasing with respect to $k$ and is
bounded. Meanwhile, $\\{\tilde{v}_{k}^{*}(s)\\}$ is monotone decreasing with
respect to $k$ and is bounded on $(0,s_{0}]$. Therefore, the limit
$\tilde{v}(s)=\lim_{k\rightarrow\infty}\tilde{v}_{k}^{*}(s)$ exists, and
satisfies $\Gamma_{\mu}(s)\leq\tilde{v}(s)\leq\sqrt{2}s^{\frac{3}{2}}$. It is
easy to check that $\tilde{v}(s)>0$ for $s\in(0,s_{0})$, $\lim_{s\rightarrow
0^{+}}\tilde{v}(s)=0$, and
$\int_{0}^{s_{0}}\frac{1}{\tilde{v}(s)}\mathrm{d}s=+\infty$.

We show that $\tilde{v}(s)$ satisfies the differential equation (2.10).
Locally in a neighbourhood of any $s_{1}\in(0,s_{0})$, say
$I:=(\frac{s_{1}}{2},s_{0})$, we rewrite the differential equation (2.12) (for
large $k$ such that $\frac{1}{k^{2}}<\frac{s_{1}}{2}$) as

$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}s}(\tilde{v}_{k})^{2}=s-\frac{\tilde{v}_{k}}{\sqrt{2s}},\quad
s\in I.$

Integrating from $s_{1}$ shows

$\frac{1}{2}(\tilde{v}_{k}(s))^{2}-\frac{1}{2}(\tilde{v}_{k}(s_{1}))^{2}=\frac{1}{2}s^{2}-\frac{1}{2}s_{1}^{2}-\int_{s_{1}}^{s}\frac{\tilde{v}_{k}(\tau)}{\sqrt{2\tau}}\mathrm{d}\tau,\quad
s\in I.$ (2.18)

Since $\tilde{v}_{k}(s)$ is monotone decreasing with respect to $k$ and
bounded, Lebesgue’s Dominated Convergence Theorem (or Levi’s Theorem) implies
that

$\frac{1}{2}(\tilde{v}(s))^{2}-\frac{1}{2}(\tilde{v}(s_{1}))^{2}=\frac{1}{2}s^{2}-\frac{1}{2}s_{1}^{2}-\int_{s_{1}}^{s}\frac{\tilde{v}(\tau)}{\sqrt{2\tau}}\mathrm{d}\tau,\quad
s\in I.$ (2.19)

Differentiating (2.19) with respect to $s$ near $s_{1}$, we have

$\tilde{v}(s)\cdot\frac{\mathrm{d}\tilde{v}(s)}{\mathrm{d}s}=s-\frac{\tilde{v}(s)}{\sqrt{2s}},\quad
s\in I.$

Therefore, $\tilde{v}(s)$ is a solution to the problem (2.10). $\hfill\Box$

######  Lemma 2.7.

For $s_{0}=\frac{1}{2}u_{-}^{2}-\frac{1}{2}u_{+}^{2}\in(0,\frac{1}{6})$, let
$\tilde{v}(s)$ for $s\in(0,s_{0})$ be the solution to the problem (2.10) as
defined in (2.17), and let $s(x)$ and $v(x)$ be the functions defined by
(2.11) in Lemma 2.4. Then $\bar{q}(x)=-s(x)$ is a stationary solution to the
problem (2.6) for the degenerate case of $u_{+}=0$, and has the following
decay estimates

$-\frac{1}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{2}}\leq\bar{q}(x)\leq-\frac{1}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{2}},\quad
x\in(0,+\infty),$ (2.20)

and

$\frac{\frac{\sqrt{2}}{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{3}}\leq\bar{q}_{x}(x)\leq\frac{\sqrt{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{3}},\quad
x\in(0,+\infty).$ (2.21)

Proof. According to Lemma 2.6,
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}+o(s^{\frac{3}{2}})$ as $s\rightarrow
0^{+}$, and $\Gamma_{\mu}(s)\leq\tilde{v}(s)\leq\sqrt{2}s^{\frac{3}{2}}$ for
all $s\in(0,s_{0})$ and $\mu=\frac{\sqrt{2}}{2}$, then the function $s(x)$
defined by (2.11) satisfies (note that $0<s(x)<s_{0}$)

$x=-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau\geq-\int_{s_{0}}^{s(x)}\frac{1}{\sqrt{2}\tau^{\frac{3}{2}}}\mathrm{d}\tau=\sqrt{2}\Big{(}\frac{1}{\sqrt{s(x)}}-\frac{1}{\sqrt{s_{0}}}\Big{)},\
\ \ \ x\in(0,+\infty),$ (2.22)

and on the other hand,

$\displaystyle x=$
$\displaystyle-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau\leq-\int_{s_{0}}^{s(x)}\frac{1}{\Gamma_{\frac{\sqrt{2}}{2}}(\tau)}\mathrm{d}\tau=-\int_{s_{0}}^{s(x)}\frac{2}{\sqrt{2}\tau^{\frac{3}{2}}}\mathrm{d}\tau$
$\displaystyle=$ $\displaystyle
2\sqrt{2}\Big{(}\frac{1}{\sqrt{s(x)}}-\frac{1}{\sqrt{s_{0}}}\Big{)},\qquad
x\in(0,+\infty),$ (2.23)

since
$\Gamma_{\frac{\sqrt{2}}{2}}(s)=(\sqrt{2}-\frac{\sqrt{2}}{2})s^{\frac{3}{2}}$
for $s\in(0,\delta_{\frac{\sqrt{2}}{2}})$ and
$\delta_{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}/2}{3\sqrt{2}}=\frac{1}{6}>s_{0}$.
Combining the above estimates (2.22) and (2.23) implies

$\frac{1}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{2}}\leq
s(x)\leq\frac{1}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{2}},\quad
x\in(0,+\infty).$ (2.24)

Since $s(x)$ is a solution to the problem (2.9), we have

$s_{x}(x)=-v(x)=-\tilde{v}(s(x))\in[-\sqrt{2}s^{\frac{3}{2}}(x),-\Gamma_{\frac{\sqrt{2}}{2}}(s(x))],\quad
x\in(0,+\infty).$

That is,

$s_{x}(x)\leq-\frac{\sqrt{2}}{2}s^{\frac{3}{2}}(x)\leq-\frac{\frac{\sqrt{2}}{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{3}},$

and

$s_{x}(x)\geq-\sqrt{2}s^{\frac{3}{2}}(x)\geq-\frac{\sqrt{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{3}},$

for $x\in(0,+\infty)$. $\hfill\Box$

######  Remark 2.1.

The restriction of $s_{0}<\frac{1}{6}$ is not essential for the existence and
the decay estimates of the stationary solution in Lemma 2.7. For general
$s_{0}\geq\frac{1}{6}$, we can modify the estimate (2.23) such that

$-\int_{s_{0}}^{s(x)}\frac{1}{\Gamma_{\frac{\sqrt{2}}{2}}(\tau)}\mathrm{d}\tau=\begin{cases}\displaystyle-\int_{\frac{1}{6}}^{s(x)}\frac{2}{\sqrt{2}\tau^{\frac{3}{2}}}\mathrm{d}\tau-\int_{s_{0}}^{\frac{1}{6}}\frac{1}{\Gamma_{\frac{\sqrt{2}}{2}}(\frac{1}{6})}\mathrm{d}\tau,~{}&\text{if~{}}s(x)<\frac{1}{6},\\\\[17.07164pt]
\displaystyle-\int_{s_{0}}^{s(x)}\frac{1}{\Gamma_{\frac{\sqrt{2}}{2}}(\frac{1}{6})}\mathrm{d}\tau,~{}&\text{if~{}}s(x)>\frac{1}{6}.\end{cases}$

The decay estimates follow similarly. Here we take $s_{0}<\frac{1}{6}$ for the
sake of the simplicity of the expressions. This remark is valid for all the
estimates in this subsection, hence we only present the precise estimates for
small $s_{0}$.

The asymptotic behavior
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}+o(s^{\frac{3}{2}})$ as $s\rightarrow
0^{+}$ implies the decay estimates of $\bar{q}(x)$ and $\bar{q}_{x}(x)$. In
order to derive decay estimates of higher order derivatives, we expand
$\tilde{v}(s)$ to higher order. Define sequences $\\{c_{k}\\}$ and
$\\{a_{k}\\}$ as following

$c_{1}=1,\quad c_{k}=\sum_{i+j=k}(2j+1)c_{i}\cdot c_{j},\quad\forall k\geq 2,$
(2.25)

and

$a_{k}=(-1)^{k+1}\sqrt{2}c_{k},\quad\forall k\geq 1.$ (2.26)

For example, $a_{1}=\sqrt{2}$, $a_{2}=-3\sqrt{2}$, $a_{3}=24\sqrt{2}$,
$a_{4}=-285\sqrt{2}$, $a_{5}=4284\sqrt{2}$. The formal series

$\sum_{i=1}^{\infty}a_{i}s^{\frac{2i+1}{2}}$

is not convergent at any point $s>0$, and then the infinite series expansion
method cannot be applied. However, the finite series expansion still gives the
local behavior of the solution $\tilde{v}(s)$, which leads to the precise
decay estimates of the higher order derivatives of $\bar{q}(x)$ and
$\bar{u}(x)$.

######  Lemma 2.8.

For any $k\in\mathbbm{N}^{+}$, there holds

$\tilde{v}(s)=\sum_{i=1}^{k}a_{i}s^{\frac{2i+1}{2}}+o(s^{\frac{2k+1}{2}}),\quad
s\rightarrow 0^{+}.$

Specifically, for odd $k\in\mathbbm{N}^{+}$, there exist $\delta_{k}>0$ and
$M_{k}>0$ such that

$\sum_{i=1}^{k}a_{i}s^{\frac{2i+1}{2}}-M_{k}s^{\frac{2k+3}{2}}\leq\tilde{v}(s)\leq\sum_{i=1}^{k}a_{i}s^{\frac{2i+1}{2}},\quad
s\in(0,\delta_{k});$

while for even $k\in\mathbbm{N}^{+}$, there exist $\delta_{k}>0$ and $M_{k}>0$
such that

$\sum_{i=1}^{k}a_{i}s^{\frac{2i+1}{2}}\leq\tilde{v}(s)\leq\sum_{i=1}^{k}a_{i}s^{\frac{2i+1}{2}}+M_{k}s^{\frac{2k+3}{2}},\quad
s\in(0,\delta_{k}).$

Proof. We prove the cases $k=2$ and $k=3$. Other cases follow similarly. For
$k=2$, let $\beta(s):=\sqrt{2}s^{\frac{3}{2}}-\tilde{v}(s)$, i.e.,
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}-\beta(s)$. The function
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}+o(s^{\frac{3}{2}})$ as $s\rightarrow
0^{+}$ satisfies the differential equation (2.10), and then $\beta(s)$
satisfies

$\displaystyle\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}=\frac{3}{2}\sqrt{2s}-\frac{\mathrm{d}\beta(s)}{\mathrm{d}s}=\frac{s}{\tilde{v}}-\frac{1}{\sqrt{2s}}=\frac{s}{\sqrt{2}s^{\frac{3}{2}}-\beta(s)}-\frac{1}{\sqrt{2s}}.$

That is,

$\frac{\mathrm{d}\beta(s)}{\mathrm{d}s}=\frac{3}{2}\sqrt{2s}+\frac{1}{\sqrt{2s}}-\frac{s}{\sqrt{2}s^{\frac{3}{2}}-\beta(s)}=\frac{3\sqrt{2}s^{\frac{5}{2}}-(1+3s)\cdot\beta(s)}{2s^{2}-\sqrt{2s}\cdot\beta(s)}.$
(2.27)

We analyze the phase plane $(s,\beta)$ corresponding to the singular
differential equation (2.27) in a similar way as we solve the problem (2.10).

Consider two special curves in the phase plane of $(s,\beta)$

$\Gamma^{*}(s):=3\sqrt{2}s^{\frac{5}{2}},\quad\Gamma_{*}(s):=3\sqrt{2}s^{\frac{5}{2}}-M_{2}s^{\frac{7}{2}},\quad
s\in(0,\delta_{2}),$ (2.28)

where $M_{2}>0$ and $\delta_{2}>0$ are constants to be determined.

We use the same symbol $\Gamma^{*}$ (or $\Gamma_{*}$) to denote the curve as
well as the function. At any point $(\hat{s},\hat{\beta})\in\Gamma^{*}$ (i.e.,
$\hat{\beta}=3\sqrt{2}\hat{s}^{\frac{5}{2}}$), there holds

$\Gamma^{*}{}^{\prime}(\hat{s})=\frac{15}{2}\sqrt{2}\hat{s}^{\frac{3}{2}}>0>\frac{3\sqrt{2}\hat{s}^{\frac{5}{2}}-(1+3\hat{s})\cdot\hat{\beta}}{2\hat{s}^{2}-\sqrt{2\hat{s}}\cdot\hat{\beta}}=\frac{\mathrm{d}\beta(s)}{\mathrm{d}s}\Big{|}_{s=\hat{s},\beta=\hat{\beta}}.$
(2.29)

Meanwhile, at any point $(\hat{s},\hat{\beta})\in\Gamma_{*}$ (i.e.,
$\hat{\beta}=3\sqrt{2}\hat{s}^{\frac{5}{2}}-M_{2}\hat{s}^{\frac{7}{2}}$), we
have

$\Gamma_{*}^{\prime}(\hat{s})=\frac{15}{2}\sqrt{2}\hat{s}^{\frac{3}{2}}-\frac{7}{2}M_{2}\hat{s}^{\frac{5}{2}}<\frac{3\sqrt{2}\hat{s}^{\frac{5}{2}}-(1+3\hat{s})\cdot\hat{\beta}}{2\hat{s}^{2}-\sqrt{2\hat{s}}\cdot\hat{\beta}}=\frac{\mathrm{d}\beta(s)}{\mathrm{d}s}\Big{|}_{s=\hat{s},\beta=\hat{\beta}},$
(2.30)

which is equivalent to

$\frac{15}{2}\sqrt{2}\hat{s}^{\frac{3}{2}}-\frac{7}{2}M_{2}\hat{s}^{\frac{5}{2}}<\frac{(M_{2}-9\sqrt{2})\hat{s}^{\frac{7}{2}}+3M_{2}\hat{s}^{\frac{9}{2}}}{2\hat{s}^{2}-6\hat{s}^{3}+M_{2}\sqrt{2}\hat{s}^{4}}.$

It suffices to take $M_{2}=24\sqrt{2}$ and $\delta_{2}=\frac{1}{8}$. The
analysis of the trajectories according to inequalities (2.29) and (2.30) show
that

$3\sqrt{2}s^{\frac{5}{2}}-M_{2}s^{\frac{7}{2}}\leq\beta(s)\leq
3\sqrt{2}s^{\frac{5}{2}},\quad s\in(0,\delta_{2}).$

Therefore,

$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}-\beta(s)~{}\begin{cases}\geq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}},\\\\[5.69054pt]
\leq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+M_{2}s^{\frac{7}{2}},\end{cases}\qquad
s\in(0,\delta_{2}).$ (2.31)

Next, we consider the case of $k=3$. Let
$\alpha(s):=\tilde{v}(s)-\sqrt{2}s^{\frac{3}{2}}+3\sqrt{2}s^{\frac{5}{2}}$,
i.e.,
$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+\alpha(s)$.
According to the differential equation (2.10) of $\tilde{v}(s)$, we see that
$\alpha(s)$ satisfies

$\displaystyle\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}=\frac{3}{2}\sqrt{2s}-\frac{15}{2}\sqrt{2}s^{\frac{3}{2}}+\frac{\mathrm{d}\alpha(s)}{\mathrm{d}s}=\frac{s}{\tilde{v}}-\frac{1}{\sqrt{2s}}=\frac{s}{\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+\alpha(s)}-\frac{1}{\sqrt{2s}}.$

It is equivalent to

$\displaystyle\frac{\mathrm{d}\alpha(s)}{\mathrm{d}s}=$
$\displaystyle\frac{s}{\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+\alpha(s)}-\frac{1}{\sqrt{2s}}-\frac{3}{2}\sqrt{2s}+\frac{15}{2}\sqrt{2}s^{\frac{3}{2}}$
$\displaystyle=$
$\displaystyle\frac{(-1-3s+15s^{2})\cdot\alpha(s)+(24\sqrt{2}s^{\frac{7}{2}}-45\sqrt{2}s^{\frac{9}{2}})}{2s^{2}-6s^{3}+\sqrt{2s}\cdot\alpha(s)}.$
(2.32)

The two special curves that are used to control the trajectories in the phase
plane $(s,\alpha)$ corresponding to (2.32) are

$\bar{\Gamma}^{*}(s):=24\sqrt{2}s^{\frac{7}{2}},\quad\bar{\Gamma}_{*}(s):=24\sqrt{2}s^{\frac{7}{2}}-M_{3}s^{\frac{9}{2}},\quad
s\in(0,\delta_{3}),$ (2.33)

with $M_{3}=285\sqrt{2}$ and
$\delta_{3}=\min\\{\frac{1}{8},\frac{1}{5},\frac{77}{285}\\}=\frac{1}{8}$.
Here we omit the details showing that at any point
$(\hat{s},\hat{\alpha})\in\bar{\Gamma}^{*}$

$\bar{\Gamma}^{*}{}^{\prime}(\hat{s})>0>\frac{\mathrm{d}\alpha(s)}{\mathrm{d}s}\Big{|}_{s=\hat{s},\alpha=\hat{\alpha}},\quad
s\in(0,\delta_{3}),$

and at any point $(\hat{s},\hat{\alpha})\in\bar{\Gamma}_{*}$

$\bar{\Gamma}_{*}^{\prime}(\hat{s})<\frac{\mathrm{d}\alpha(s)}{\mathrm{d}s}\Big{|}_{s=\hat{s},\alpha=\hat{\alpha}},\quad
s\in(0,\delta_{3}).$

It follows that

$24\sqrt{2}s^{\frac{7}{2}}-M_{3}s^{\frac{9}{2}}\leq\alpha(s)\leq
24\sqrt{2}s^{\frac{7}{2}},\quad s\in(0,\delta_{3}),$

and further

$\tilde{v}(s)=\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+\alpha(s)~{}\begin{cases}\geq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-M_{3}s^{\frac{9}{2}},\\\\[5.69054pt]
\leq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}},\end{cases}\
\ \ s\in(0,\delta_{3}).$ (2.34)

Generally, we can take $M_{k}=|a_{k+1}|$. The proof is completed. $\hfill\Box$

We now show the decay estimates of higher order derivatives of $\bar{q}(x)$.

######  Lemma 2.9.

For $s_{0}<\frac{1}{8}$, let $\bar{q}(x)$ be the stationary solution proved in
Lemma 2.7. Then

$-\frac{3}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{4}}\leq\bar{q}_{xx}(x)\leq
0,\quad x\in(0,+\infty),$ (2.35)

and

$|\bar{q}_{xxx}(x)|\leq\frac{6\sqrt{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{5}}+\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{7}},\quad
x\in(0,+\infty),$ (2.36)

for some positive constant $C>0$. For $s_{0}<\delta_{4}$ ($\delta_{4}$ is the
positive constant in Lemma 2.8), there holds

$|\bar{q}_{xxxx}(x)|\leq\frac{222}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{6}}+\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{8}},$
(2.37)

for some positive constant $C>0$.

Proof. According to the dynamic system (2.9),

$s_{x}(x)=-v(x),\qquad v_{x}(x)=-s(x)+\frac{v(x)}{\sqrt{2s(x)}}.$

We have

$\displaystyle
s_{xx}(x)=-v_{x}(x)=s(x)-\frac{v(x)}{\sqrt{2s(x)}}=s(x)-\frac{\tilde{v}(s(x))}{\sqrt{2s(x)}}.$

Using the expansion (2.31) in Lemma 2.8 and the decay estimate (2.24) in Lemma
2.7, we deduce

$s_{xx}(x)\leq
s(x)-\frac{\sqrt{2}s^{\frac{3}{2}}(x)-3\sqrt{2}s^{\frac{5}{2}}(x)}{\sqrt{2s(x)}}=3s^{2}(x)\leq\frac{3}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{4}},$

and

$\displaystyle s_{xx}(x)\geq$ $\displaystyle
s(x)-\frac{\sqrt{2}s^{\frac{3}{2}}(x)-3\sqrt{2}s^{\frac{5}{2}}(x)+24\sqrt{2}s^{\frac{7}{2}}(x)}{\sqrt{2s(x)}}$
$\displaystyle\geq$ $\displaystyle
3s^{2}(x)-24s^{3}(x)\geq\max\left\\{\frac{3}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{4}}-\frac{24}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{6}},0\right\\}.$

Similarly, we have

$\displaystyle s_{xxx}(x)=$ $\displaystyle-
v_{xx}(x)=s_{x}(x)-\frac{v_{x}(x)}{\sqrt{2s(x)}}+\frac{v(x)}{2\sqrt{2}s^{\frac{3}{2}}(x)}\cdot(-v(x))$
$\displaystyle=$
$\displaystyle-v(x)-\frac{-s(x)+\frac{v(x)}{\sqrt{2s(x)}}}{\sqrt{2s(x)}}-\frac{v^{2}(x)}{2\sqrt{2}s^{\frac{3}{2}}(x)}$
(2.38) $\displaystyle=$
$\displaystyle-\tilde{v}(s(x))+\frac{\sqrt{s(x)}}{\sqrt{2}}-\frac{\tilde{v}(s(x))}{2s(x)}-\frac{(\tilde{v}(s(x)))^{2}}{2\sqrt{2}s^{\frac{3}{2}}(x)}.$

Utilizing the expansion (2.34) in Lemma 2.8, we deduce

$\displaystyle s_{xxx}(x)\leq$
$\displaystyle-(\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-285\sqrt{2}s^{\frac{9}{2}})+\frac{\sqrt{s}}{\sqrt{2}}$
$\displaystyle-\frac{\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-285\sqrt{2}s^{\frac{9}{2}}}{2s}$
$\displaystyle-\frac{(\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-285\sqrt{2}s^{\frac{9}{2}})^{2}}{2\sqrt{2}s^{\frac{3}{2}}}$
$\displaystyle\leq$
$\displaystyle-6\sqrt{2}s^{\frac{5}{2}}(x)+Cs^{\frac{7}{2}}(x),$

and

$\displaystyle s_{xxx}(x)\geq$
$\displaystyle-(\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}})+\frac{\sqrt{s}}{\sqrt{2}}$
$\displaystyle-\frac{\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}}{2s}-\frac{(\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}})^{2}}{2\sqrt{2}s^{\frac{3}{2}}}$
$\displaystyle\geq$
$\displaystyle-6\sqrt{2}s^{\frac{5}{2}}(x)-Cs^{\frac{7}{2}}(x),$

where $C>0$ is a generic positive constant. Therefore,

$|s_{xxx}(x)|\leq
6\sqrt{2}s^{\frac{5}{2}}(x)+Cs^{\frac{7}{2}}(x)\leq\frac{6\sqrt{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{5}}+\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{7}}.$

We now show the estimates of $s_{xxxx}(x)$. According to (2.38),

$\displaystyle s_{xxxx}(x)=$ $\displaystyle-
v_{x}(x)+\Big{(}\frac{\sqrt{s(x)}}{\sqrt{2}}\Big{)}_{x}-\Big{(}\frac{v(x)}{2s(x)}\Big{)}_{x}-\Big{(}\frac{v^{2}(x)}{2\sqrt{2}s^{\frac{3}{2}}(x)}\Big{)}_{x}$
$\displaystyle=$
$\displaystyle-\Big{(}-s(x)+\frac{v(x)}{\sqrt{2s(x)}}\Big{)}-\frac{v(x)}{2\sqrt{2s(x)}}-\frac{-s(x)+\frac{v(x)}{\sqrt{2s(x)}}}{2s(x)}-\frac{v^{2}(x)}{2s^{2}(x)}$
$\displaystyle-\frac{2v(x)\cdot(-s(x)+\frac{v(x)}{\sqrt{2s(x)}})}{2\sqrt{2}s^{\frac{3}{2}}(x)}-\frac{\frac{3}{2}v^{3}(x)}{2\sqrt{2}s^{\frac{5}{2}}(x)}.$
(2.39)

Substituting $v(x)=\tilde{v}(s(x))$ and the expansion of $\tilde{v}(s)$ for
$k=4$ in Lemma 2.8 such that

$\tilde{v}(s)~{}\begin{cases}\geq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-285\sqrt{2}s^{\frac{9}{2}},\\\\[5.69054pt]
\leq\sqrt{2}s^{\frac{3}{2}}-3\sqrt{2}s^{\frac{5}{2}}+24\sqrt{2}s^{\frac{7}{2}}-285\sqrt{2}s^{\frac{9}{2}}+4284\sqrt{2}s^{\frac{11}{2}},\end{cases}\quad
s\in(0,\delta_{4}),$

into (2.39) implies that

$|s_{xxxx}(x)|\leq
222s^{3}(x)+Cs^{4}(x)\leq\frac{222}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{6}}+\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{8}},$

for some positive constant $C>0$. $\hfill\Box$

######  Lemma 2.10.

For
$s_{0}=\frac{1}{2}u_{-}^{2}-\frac{1}{2}u_{+}^{2}=\frac{1}{2}u_{-}^{2}=\frac{1}{2}\delta^{2}<\min\\{\frac{1}{8},\delta_{4}\\}$,
let $\bar{q}(x)$ be the stationary solution proved in Lemma 2.7 and let
$\bar{u}(x):=-\sqrt{-2\bar{q}(x)}=-\sqrt{2s(x)}$. Then we have

$\displaystyle\frac{C_{1}\delta}{1+\delta
x}\leq-\bar{u}(x)\leq\frac{C_{2}\delta}{1+\delta x},\qquad$
$\displaystyle\frac{C_{1}\delta^{2}}{(1+\delta
x)^{2}}\leq\bar{u}_{x}(x)\leq\frac{C_{2}\delta^{2}}{(1+\delta x)^{2}},$
$\displaystyle\Big{|}\partial^{k}_{x}\bar{u}(x)\Big{|}\leq\frac{C_{2}\delta^{k+1}}{(1+\delta
x)^{k+1}},\qquad$
$\displaystyle\Big{|}\frac{\bar{u}_{xx}^{2}(x)}{\bar{u}_{x}(x)}\Big{|}\leq\frac{C_{2}\delta^{4}}{(1+\delta
x)^{4}},$
$\displaystyle\Big{|}\frac{\bar{u}_{xxx}^{2}(x)}{\bar{u}_{x}(x)}\Big{|}\leq\frac{C_{2}\delta^{6}}{(1+\delta
x)^{6}},\qquad\Big{|}\frac{\bar{u}_{xxx}(x)}{\bar{u}_{x}(x)}\Big{|}\leq$
$\displaystyle\frac{C_{2}\delta^{2}}{(1+\delta
x)^{2}},\qquad\Big{|}\frac{\bar{u}_{xxxx}(x)}{\bar{u}_{x}(x)}\Big{|}\leq\frac{C_{2}\delta^{3}}{(1+\delta
x)^{3}},$

for $k=1,2,3,4$ and some positive constants $C_{1}$, $C_{2}$.

Proof. The lower and upper bounds of $\bar{q}(x)$ and $\bar{q}_{x}(x)$ in
(2.20) and (2.21) in Lemma 2.7 show that

$\bar{u}(x)=-\sqrt{-2\bar{q}(x)}~{}\begin{cases}\displaystyle\leq-\frac{\sqrt{2}}{\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}},\\\\[17.07164pt]
\displaystyle\geq-\frac{\sqrt{2}}{\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}},\end{cases}\quad
x\in(0,+\infty),$

and

$\bar{u}_{x}(x)=\frac{\bar{q}_{x}}{\sqrt{-2\bar{q}}}~{}\begin{cases}\displaystyle\leq\frac{\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{3}}\leq\frac{2}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{2}},\\\\[22.76219pt]
\displaystyle\geq\frac{\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}}{2\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{3}}\geq\frac{1}{4\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{2}},\end{cases}\quad
x\in(0,+\infty).$

Utilizing the higher order estimates (2.35), (2.36) and (2.37) in Lemma 2.9,
we have

$\displaystyle|\bar{u}_{xx}(x)|=$
$\displaystyle\left|\frac{\bar{q}_{xx}(x)}{\sqrt{-2\bar{q}(x)}}+\frac{(\bar{q}_{x}(x))^{2}}{(-2\bar{q}(x))^{\frac{3}{2}}}\right|\leq\frac{3(\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}})}{\sqrt{2}\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{4}}+\frac{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{\sqrt{2}}\Big{)}^{3}}{\sqrt{2}\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{6}}$
$\displaystyle\leq$
$\displaystyle\frac{7\sqrt{2}}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{3}},$

and

$|\bar{u}_{xxx}(x)|=\left|\frac{\bar{q}_{xxx}}{\sqrt{-2\bar{q}}}+\frac{3\bar{q}_{x}\bar{q}_{xx}}{(-2\bar{q})^{\frac{3}{2}}}+\frac{3(\bar{q}_{x})^{3}}{(-2\bar{q})^{\frac{5}{2}}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{4}},$
$|\bar{u}_{xxxx}(x)|=\left|\frac{\bar{q}_{xxxx}}{\sqrt{-2\bar{q}}}+\frac{4\bar{q}_{x}\bar{q}_{xxx}+3(\bar{q}_{xx})^{2}}{(-2\bar{q})^{\frac{3}{2}}}+\frac{18(\bar{q}_{x})^{2}\bar{q}_{xx}}{(-2\bar{q})^{\frac{5}{2}}}+\frac{15(\bar{q}_{x})^{4}}{(-2\bar{q})^{\frac{7}{2}}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{5}},$

where $C>0$ is a generic positive constant. Moreover,

$\left|\frac{\bar{u}_{xxx}}{\bar{u}_{x}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{2}},\qquad\left|\frac{\bar{u}_{xxxx}}{\bar{u}_{x}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{3}},$

and

$\left|\frac{\bar{u}_{xxx}^{2}}{\bar{u}_{x}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{6}},\qquad\left|\frac{\bar{u}_{xx}^{2}}{\bar{u}_{x}}\right|\leq\frac{C}{\Big{(}\frac{1}{\sqrt{s_{0}}}+\frac{x}{2\sqrt{2}}\Big{)}^{4}}.$

The proof is completed. $\hfill\Box$

Proof of Lemma 2.3. The degenerate case (2): $u_{-}<u_{+}=0$ is proved through
a singular phase plane analysis method according to Lemma 2.7 and Lemma 2.10.

Next we show that this method is applicable to the case (1) $u_{-}<u_{+}<0$.
Instead of (2.9), we have a non-degenerate dynamical system (2.8) for the case
of $u_{+}<0$. We sketch the main lines of the proof.

(1) For any $s_{0}:=\frac{1}{2}u_{-}^{2}-\frac{1}{2}u_{+}^{2}>0$, if
$\tilde{v}(s)$ solves the following equation

$\begin{cases}\displaystyle\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}=\frac{s}{\tilde{v}}-\frac{1}{\sqrt{u_{+}^{2}+2s}},\quad
s\in(0,s_{0}),\\\\[17.07164pt]
\displaystyle\tilde{v}(s)>0\mathrm{~{}for~{}}s\in(0,s_{0}),\quad\lim_{s\rightarrow
0^{+}}\tilde{v}(s)=0,\quad\int_{0}^{s_{0}}\frac{1}{\tilde{v}(s)}\mathrm{d}s=+\infty,\end{cases}$
(2.40)

then the function $s(x)$ defined by

$x=-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau,\qquad
x\in(0,+\infty)$ (2.41)

is a solution of (2.8) with $v(x):=\tilde{v}(s(x))>0$.

(2) Consider the approximated problem

$\begin{cases}\displaystyle\frac{\mathrm{d}\tilde{v}_{k}}{\mathrm{d}s}=\frac{s}{\tilde{v}_{k}}-\frac{1}{\sqrt{u_{+}^{2}+2s}},\quad
s\in(0,s_{0}),\\\\[17.07164pt]
\displaystyle\tilde{v}_{k}(s)>0\mathrm{~{}for~{}}s\in(0,s_{0}),\quad\tilde{v}_{k}(0)=\frac{1}{k}.\end{cases}$
(2.42)

For any $s_{0}>0$, the problem (2.42) admits a solution $\tilde{v}_{k}(s)>0$
on $s\in(0,s_{0})$ such that
(i) $\tilde{v}_{k}(s)$ is monotone decreasing with respect to $k$, i.e.,
$\tilde{v}_{k_{1}}(s)>\tilde{v}_{k_{2}}(s)$ for any $k_{2}>k_{1}$ on
$(0,s_{0})$;
(ii) $\tilde{v}_{k}(s)$ has the following upper bound estimate

$\tilde{v}_{k}(s)\leq\overline{\Gamma}_{k}(s):=\max\Big{\\{}\frac{1}{k},s\cdot\sqrt{u_{+}^{2}+2s}\Big{\\}},\quad
s\in(0,s_{0});$

(iii) $\tilde{v}_{k}(s)$ has the following uniformly lower bound estimate

$\tilde{v}_{k}(s)\geq\lambda_{0}s,\quad s\in(0,s_{0}),$

where $\lambda_{0}:=\frac{\sqrt{1+4|u_{+}|^{2}}-1}{2|u_{+}|}$.

(3) The limit function

$\tilde{v}(s):=\lim_{k\rightarrow\infty}\tilde{v}_{k}(s),\quad s\in(0,s_{0})$

is well-defined and $\tilde{v}(s)$ is a solution to the problem (2.40).
Moreover,

$\lambda_{0}s\leq\tilde{v}(s)\leq\lambda_{0}s+bs^{2},\quad s\in(0,s_{0}),$
(2.43)

where $\lambda_{0}:=\frac{\sqrt{1+4|u_{+}|^{2}}-1}{2|u_{+}|}$ is the positive
root of $\lambda_{0}=\frac{1}{\lambda_{0}}-\frac{1}{|u_{+}|}$ and
$b=\frac{1}{2|u_{+}|^{3}}$.

The asymptotic expansion (2.43) plays an essential role in the analysis of
asymptotic decay behavior of the stationary solution, thus we present the
following proof. In the phase plane $(s,v)$, at any point $(\hat{s},\hat{v})$
on the curve $\hat{v}=\Gamma_{*}(\hat{s}):=\lambda_{0}\hat{s}$, we have

$\Gamma_{*}^{\prime}(\hat{s})=\lambda_{0}=\frac{1}{\lambda_{0}}-\frac{1}{|u_{+}|}<\frac{\hat{s}}{\hat{v}}-\frac{1}{\sqrt{u_{+}^{2}+2\hat{s}}}=\left.\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}\right|_{(\hat{s},\hat{v})},$

which means the trajectory lies above $\Gamma_{*}$. At any point
$(\hat{s},\hat{v})$ on the curve
$\hat{v}=\Gamma^{*}(\hat{s}):=\lambda_{0}\hat{s}+b\hat{s}^{2}$, we have

$\Gamma^{*}{}^{\prime}(\hat{s})=\lambda_{0}+2b\hat{s}>\frac{\hat{s}}{\hat{v}}-\frac{1}{\sqrt{u_{+}^{2}+2\hat{s}}}=\left.\frac{\mathrm{d}\tilde{v}}{\mathrm{d}s}\right|_{(\hat{s},\hat{v})},$

since the following auxiliary function

$F(\hat{s}):=\lambda_{0}+2b\hat{s}-\frac{1}{\lambda_{0}+b\hat{s}}+\frac{1}{\sqrt{u_{+}^{2}+2\hat{s}}},\quad
s\in(0,s_{0})$

is monotonically increasing as
$F^{\prime}(\hat{s})=2b+\frac{b}{(\lambda_{0}+b\hat{s})^{2}}-\frac{1}{(|u_{+}|^{2}+2\hat{s})^{\frac{3}{2}}}>0$
for $b=\frac{1}{2|u_{+}|^{3}}$. This shows the trajectory lies between
$\Gamma_{*}$ and $\Gamma^{*}$.

(4) Finally we show the asymptotic behavior of the stationary solution.
According to the definition of $s(x)$ and $\bar{q}(x)$ in (2.41) and the
asymptotic expansion (2.43), we have

$\displaystyle
x=-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau\leq\int_{s(x)}^{s_{0}}\frac{1}{\lambda_{0}\tau}\mathrm{d}\tau=\frac{1}{\lambda_{0}}\ln\Big{(}\frac{s_{0}}{s(x)}\Big{)},\quad
x\in(0,+\infty),$

which implies

$s(x)\leq
s_{0}\mathrm{e}^{-\lambda_{0}x},\quad\bar{q}(x)\geq\bar{q}_{-}\mathrm{e}^{-\lambda_{0}x}.$

On the other hand,

$\displaystyle
x=-\int_{s_{0}}^{s(x)}\frac{1}{\tilde{v}(\tau)}\mathrm{d}\tau\geq\int_{s(x)}^{s_{0}}\frac{1}{\lambda_{0}\tau+b\tau^{2}}\mathrm{d}\tau=\left.\frac{1}{\lambda_{0}}\ln\left(\frac{\tau}{1+\frac{b}{\lambda_{0}}\tau}\right)\right|_{s(x)}^{s_{0}},\qquad
x\in(0,+\infty).$

That is,

$s(x)\geq\frac{\frac{s_{0}}{1+\frac{b}{\lambda_{0}}s_{0}}\cdot\mathrm{e}^{-\lambda_{0}x}}{1-\frac{b}{\lambda_{0}}\cdot\frac{s_{0}}{1+\frac{b}{\lambda_{0}}s_{0}}\cdot\mathrm{e}^{-\lambda_{0}x}}\geq\frac{s_{0}}{1+\frac{b}{\lambda_{0}}s_{0}}\cdot\mathrm{e}^{-\lambda_{0}x},$

which shows

$\bar{q}(x)\leq\frac{\bar{q}_{-}}{1-\frac{b}{\lambda_{0}}\bar{q}_{-}}\cdot\mathrm{e}^{-\lambda_{0}x}.$

The decay estimates of $\bar{q}_{x}(x)$ and other higher order derivatives
follow similarly, which are all exponentially decaying, since

$s_{x}(x)=-v(x)=-\tilde{v}(s(x)),$

and

$s_{xx}(x)=-v_{x}(x)=s(x)-\frac{v(x)}{\sqrt{u_{+}^{2}+2s(s)}}=s(x)-\frac{\tilde{v}(s(x))}{\sqrt{u_{+}^{2}+2s(s)}},$

and according to the asymptotic expansion (2.43). The proof is completed.
$\hfill\Box$

###  2.3 Preliminary Lemmas

In order to show the asymptotic behavior of heat flux $q(x,t)$ which satisfies
an elliptic problem, we prove the following optimal Gagliardo-Nirenberg-
Sobolev inequality. This inequality without optimal constant is known as a
special case of Gagliardo-Nirenberg-Sobolev inequality. Here we present a
primary proof based on fundamental calculus.

######  Lemma 2.11 (Optimal Gagliardo-Nirenberg-Sobolev inequality).

For any function $u(x)\in L^{\infty}(\mathbbm{R})$ with $u_{xx}\in
L^{\infty}(\mathbbm{R})$, there holds

$\|u_{x}\|_{L^{\infty}(\mathbbm{R})}\leq\sqrt{2}\
\|u\|_{L^{\infty}(\mathbbm{R})}^{\frac{1}{2}}\cdot\|u_{xx}\|_{L^{\infty}(\mathbbm{R})}^{\frac{1}{2}},$
(2.44)

and the constant $\sqrt{2}$ is optimal. Moreover, for any function $u(x)\in
L^{\infty}(\mathbbm{R}_{+})$ with $u_{xx}\in L^{\infty}(\mathbbm{R}_{+})$,
there holds

$\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}\leq 2\
\|u\|_{L^{\infty}(\mathbbm{R}_{+})}^{\frac{1}{2}}\cdot\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}^{\frac{1}{2}},$
(2.45)

and the constant $2$ is optimal. For multi-dimensional case,

$\||\nabla u|\|_{L^{\infty}(\mathbbm{R}^{n})}\leq\sqrt{2}\
\|u\|_{L^{\infty}(\mathbbm{R}^{n})}^{\frac{1}{2}}\cdot\||D^{2}u|\|_{L^{\infty}(\mathbbm{R}^{n})}^{\frac{1}{2}},\qquad\forall
u\in W^{2,\infty}(\mathbbm{R}^{n}),$ (2.46)

and

$\||\nabla u|\|_{L^{\infty}(\mathbbm{R}_{+}^{n})}\leq 2\
\|u\|_{L^{\infty}(\mathbbm{R}_{+}^{n})}^{\frac{1}{2}}\cdot\||D^{2}u|\|_{L^{\infty}(\mathbbm{R}_{+}^{n})}^{\frac{1}{2}},\qquad\forall
u\in W^{2,\infty}(\mathbbm{R}_{+}^{n}),$ (2.47)

and the constants $\sqrt{2}$ and $2$ are optimal, where $|\nabla u|$ is the
modulus of a vector $\nabla u$ and $|D^{2}u|$ is the spectral norm of a matrix
$D^{2}u$ such that
$|D^{2}u|:=\sup_{p\in\mathbbm{R}^{n};|p|=1}p^{\mathrm{T}}(D^{2}u)p$.

Proof. We prove that the inequality (2.44) holds for any smooth function $u\in
C^{2}(\mathbbm{R})\cap W^{2,\infty}(\mathbbm{R})$ and the constant is optimal
for $u\in W^{2,\infty}(\mathbbm{R})$ with weak derivatives. Then utilizing an
approximation approach, we see that (2.44) holds for $u\in
W^{2,\infty}(\mathbbm{R})$ with the same optimal constant.

The inequality (2.44) is trivial if $\|u\|_{L^{\infty}(\mathbbm{R})}=0$ or
$\|u_{xx}\|_{L^{\infty}(\mathbbm{R})}=0$, since in the latter case
$u(x)=c_{1}x+c_{2}$ and $c_{1}=0$ according to $u\in L^{\infty}(\mathbbm{R})$.
With the observation that the inequality (2.44) is invariant under the scaling
$\tilde{u}(x):=\lambda u(\mu x)$ for any non-zero $\lambda$ and $\mu$, we only
need to prove that $\|u_{x}\|_{L^{\infty}(\mathbbm{R})}\leq 1$ under the
condition $\|u\|_{L^{\infty}(\mathbbm{R})}=\frac{1}{2}$ and
$\|u_{xx}\|_{L^{\infty}(\mathbbm{R})}=1$, and further
$\|u_{x}\|_{L^{\infty}(\mathbbm{R})}\leq 1$ is optimal. In other words, we
show that if $u_{x}(x_{0})=1$ for some $x_{0}\in\mathbbm{R}$ and
$\|u_{xx}\|_{L^{\infty}(\mathbbm{R})}\leq 1$ then
$\|u\|_{L^{\infty}(\mathbbm{R})}\geq\frac{1}{2}$ and $\frac{1}{2}$ is optimal.

According to Taylor expansion near $x_{0}$ for $u\in C^{2}(\mathbbm{R})\cap
W^{2,\infty}(\mathbbm{R})$, we know that

$\displaystyle u(x_{0}+1)$ $\displaystyle\geq u(x_{0})+u_{x}(x_{0})\cdot
1+\frac{1}{2}u_{xx}(\xi_{1})\cdot 1^{2}\geq u(x_{0})+1-\frac{1}{2},$
$\displaystyle u(x_{0}-1)$ $\displaystyle\leq
u(x_{0})+u_{x}(x_{0})\cdot(-1)+\frac{1}{2}u_{xx}(\xi_{2})\cdot(-1)^{2}\leq
u(x_{0})-1+\frac{1}{2},$

with some $\xi_{1}\in(x_{0},x_{0}+1)$ and $\xi_{2}\in(x_{0}-1,x_{0})$.
Therefore,

$u(x_{0}+1)-u(x_{0}-1)\geq 1,$

and then $\|u\|_{L^{\infty}(\mathbbm{R})}\geq\frac{1}{2}$. The constant is
optimal for the following $\hat{u}_{1}\in W^{2,\infty}(\mathbbm{R})$

$\hat{u}_{1}(x):=\begin{cases}\cdots,\\\\[2.84526pt] x(1-\frac{1}{2}x),\ \ \ \
x\in[0,2),\\\\[2.84526pt] (x-2)(2-\frac{1}{2}x),\ \ \ \
x\in[2,4),\\\\[2.84526pt] \cdots\end{cases}$ (2.48)

which is defined by extension as a $4$-periodic function. We can verify that
$\hat{u}_{1}(x)$ satisfies the following differential equation

$-\hat{u}_{1xx}=\mathrm{sign}u_{1}=\begin{cases}1,\ \ \ \
u_{1}>0,\\\\[2.84526pt] -1,\ \ \ \ u_{1}<0.\end{cases}$

The inequality (2.45) for $u\in W^{2,\infty}(\mathbbm{R}_{+})$ is proved by
extension

$\tilde{u}(x):=\begin{cases}u(x),\ \ \ \ x\geq 0,\\\\[5.69054pt] 2u(0)-u(-x),\
\ \ \ x<0,\end{cases}$

such that
$\|\tilde{u}_{x}\|_{L^{\infty}(\mathbbm{R})}=\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}$,
$\|\tilde{u}_{xx}\|_{L^{\infty}(\mathbbm{R})}=\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}$,
and

$\mathrm{osc}(u;\mathbbm{R}_{+})\leq\mathrm{osc}(\tilde{u};\mathbbm{R})\leq
2\mathrm{osc}(u;\mathbbm{R}_{+})\leq 4\|u\|_{L^{\infty}(\mathbbm{R}_{+})}.$

Here $\mathrm{osc}(f;D):=\sup_{x,y\in D}|f(x)-f(y)|$ is the oscillation of a
given function $f$ and $D$ is its domain of definition. Furthermore, applying
(2.44) (according to the proof, we can replace
$\|\tilde{u}\|_{L^{\infty}(\mathbbm{R})}$ by
$\frac{1}{2}\mathrm{osc}(\tilde{u};\mathbbm{R})$)

$\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}=\|\tilde{u}_{x}\|_{L^{\infty}(\mathbbm{R})}\leq(\mathrm{osc}(\tilde{u};\mathbbm{R}))^{\frac{1}{2}}\cdot\|\tilde{u}_{xx}\|_{L^{\infty}(\mathbbm{R})}^{\frac{1}{2}}\leq
2\
\|u\|_{L^{\infty}(\mathbbm{R})}^{\frac{1}{2}}\cdot\|u_{xx}\|_{L^{\infty}(\mathbbm{R})}^{\frac{1}{2}}.$

The constant $2$ is optimal for the following $\hat{u}_{2}(x)$

$\hat{u}_{2}(x):=\begin{cases}x(1-\frac{1}{2}x)-\frac{1}{4},\ \ \ \
x\in[0,1),\\\\[5.69054pt] \frac{1}{4},\ \ \ \ x\in[1,+\infty),\end{cases}$
(2.49)

such that $\|\hat{u}_{2}\|_{L^{\infty}(\mathbbm{R}_{+})}=\frac{1}{4}$,
$\|\hat{u}_{2x}\|_{L^{\infty}(\mathbbm{R}_{+})}=1$ and
$\|\hat{u}_{2xx}\|_{L^{\infty}(\mathbbm{R}_{+})}=1$.

For the multi-dimensional case, we note that the inequality (2.46) is
invariant under the scaling $\tilde{u}(x):=\lambda u(\mu x)$ for any non-zero
$\lambda$ and $\mu$, and is also invariant under the rotation of coordinates.
Therefore, for any function $u\in C^{2}(\mathbbm{R}^{n})\cap
W^{2,\infty}(\mathbbm{R}^{n})$, if $|\nabla u(x_{0})|=1$ for some
$x_{0}\in\mathbbm{R}^{n}$ and $|D^{2}u(x)|\leq 1$ for all
$x\in\mathbbm{R}^{n}$, then Taylor expansion along the direction $\pm
p:=\pm\nabla u(x_{0})$ shows that

$\displaystyle u(x_{0}+\nabla u(x_{0}))$ $\displaystyle\geq u(x_{0})+\nabla
u(x_{0})\cdot\nabla u(x_{0})+\frac{1}{2}(\nabla
u(x_{0}))^{\mathrm{T}}(D^{2}u(\xi_{1}))(\nabla u(x_{0}))\geq
u(x_{0})+1-\frac{1}{2},$ $\displaystyle u(x_{0}-\nabla u(x_{0}))$
$\displaystyle\leq u(x_{0})-\nabla u(x_{0})\cdot\nabla
u(x_{0})+\frac{1}{2}(-\nabla u(x_{0}))^{\mathrm{T}}(D^{2}u(\xi_{2}))(-\nabla
u(x_{0}))\leq u(x_{0})-1+\frac{1}{2},$

for some $\xi_{1}=x_{0}+\theta_{1}\nabla u(x_{0})$ and
$\xi_{2}=x_{0}-\theta_{2}\nabla u(x_{0})$ with
$\theta_{1},\theta_{2}\in(0,1)$. The rest of the proof follows similarly.
$\hfill\Box$

######  Remark 2.2.

Lemma 2.11 can be seen as a special case of Gagliardo-Nirenberg-Sobolev
inequality with the optimal constant and without the restriction of decay at
infinity such that $\lim_{|x|\rightarrow+\infty}u(x)=0$.

######  Lemma 2.12.

Assume that $f\in L^{\infty}(\mathbbm{R}_{+})$, $g\in\mathbbm{R}$, and $u\in
W^{2,\infty}(\mathbbm{R}_{+})$ solves the following elliptic problem

$\begin{cases}-u_{xx}+u=f(x),\quad&x>0,\\\\[2.84526pt] u_{x}(0)=g,\end{cases}$
(2.50)

then

$\|u\|_{L^{\infty}(\mathbbm{R}_{+})}\leq\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|,\quad\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}\leq\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|,\quad\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}\leq
2\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|,$ (2.51)

and all the above coefficients are optimal.

Proof. Let $v(x):=u(x)+g\mathrm{e}^{-x}$. Then $v\in
W^{2,\infty}(\mathbbm{R}_{+})$ satisfies

$\begin{cases}-v_{xx}+v=f(x),\quad&x>0,\\\\[2.84526pt] v_{x}(0)=0.\end{cases}$
(2.52)

Maximum principle shows that
$\|v\|_{L^{\infty}(\mathbbm{R}_{+})}\leq\|f\|_{L^{\infty}(\mathbbm{R}_{+})}$
and then
$\|u\|_{L^{\infty}(\mathbbm{R}_{+})}\leq\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|$.
Further, according to the equation (2.50) we have
$\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}\leq
2\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|$. According to Lemma 2.11, we see
that

$\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}\leq 2\
\|u\|_{L^{\infty}(\mathbbm{R}_{+})}^{\frac{1}{2}}\cdot\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}^{\frac{1}{2}}\leq
2\
(\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|)^{\frac{1}{2}}\cdot(2\|f\|_{L^{\infty}(\mathbbm{R}_{+})}+|g|)^{\frac{1}{2}}.$

Here the Gagliardo-Nirenberg-Sobolev inequality in Lemma 2.11 is optimal for
all $u\in W^{2,\infty}(\mathbbm{R}_{+})$ but not for the solutions of elliptic
problem (2.50). In order to show optimal estimates, we extend the functions
$f$ and $v$ in (2.52) such that

$\tilde{f}(x):=\begin{cases}f(x),\ \ \ \ x\geq 0,\\\ f(-x),\ \ \ \
x<0,\end{cases}\quad\text{~{}and~{}}\quad\tilde{v}(x):=\begin{cases}v(x),\ \ \
\ x\geq 0,\\\ v(-x),\ \ \ \ x<0.\end{cases}$

Then $\tilde{v}$ can be solved as

$\tilde{v}(x)=\frac{1}{2}\int_{\mathbbm{R}}\mathrm{e}^{-|x-y|}\tilde{f}(y)\mathrm{d}y=\frac{1}{2}\int_{0}^{+\infty}\left(\mathrm{e}^{-|x-y|}+\mathrm{e}^{-|x+y|}\right)f(y)\mathrm{d}y,\quad
x\in\mathbbm{R}.$

Therefore, for $x\geq 0$,

$\displaystyle u(x)$
$\displaystyle=\frac{1}{2}\int_{0}^{+\infty}\left(\mathrm{e}^{-|x-y|}+\mathrm{e}^{-|x+y|}\right)f(y)\mathrm{d}y-g\mathrm{e}^{-x},$
(2.53) $\displaystyle u_{x}(x)$
$\displaystyle=\frac{1}{2}\int_{0}^{+\infty}\left(\mathrm{e}^{-|x-y|}(-\mathrm{sign(x-y)})+\mathrm{e}^{-|x+y|}(-1)\right)f(y)\mathrm{d}y+g\mathrm{e}^{-x},$
(2.54)

and

$u_{xx}(x)=\frac{1}{2}\int_{0}^{+\infty}\left(\mathrm{e}^{-|x-y|}+\mathrm{e}^{-|x+y|}\right)f(y)\mathrm{d}y-g\mathrm{e}^{-x}-f(x).$
(2.55)

The above expressions show that the estimates (2.51) are valid.

Now we show that all these coefficients are optimal. The special case of
$f(x)\equiv 0$, $g=-1$ and $u(x)=\mathrm{e}^{-x}$ implies that the
coefficients of $|g|$ in (2.51) are optimal. The case of $g=0$, $f(x)\equiv 1$
and $u(x)\equiv 1$ shows that the coefficient of
$\|f\|_{L^{\infty}(\mathbbm{R}_{+})}$ in the estimate of
$\|u\|_{L^{\infty}(\mathbbm{R}_{+})}$ is optimal. For any large $x_{0}>0$, we
set $f(x)=\mathrm{sign}(x-x_{0})$ and $g=0$, then (2.54) implies

$\displaystyle u_{x}(x_{0})=$
$\displaystyle\frac{1}{2}\int_{0}^{x_{0}}\left(\mathrm{e}^{-|x_{0}-y|}+\mathrm{e}^{-|x_{0}+y|}\right)\mathrm{d}y+\frac{1}{2}\int_{x_{0}}^{+\infty}\left(\mathrm{e}^{-|x_{0}-y|}-\mathrm{e}^{-|x_{0}+y|}\right)\mathrm{d}y$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{2x_{0}}\mathrm{e}^{-y}\mathrm{d}y+\frac{1}{2}\int_{0}^{+\infty}\mathrm{e}^{-y}\mathrm{d}y-\frac{1}{2}\int_{2x_{0}}^{+\infty}\mathrm{e}^{-y}\mathrm{d}y$
$\displaystyle=$ $\displaystyle
1-\int_{2x_{0}}^{+\infty}\mathrm{e}^{-y}\mathrm{d}y\rightarrow
1,\quad\text{as~{}}x_{0}\rightarrow+\infty,$

which shows that the coefficient of $\|f\|_{L^{\infty}(\mathbbm{R}_{+})}$ in
the estimate of $\|u_{x}\|_{L^{\infty}(\mathbbm{R}_{+})}$ is optimal. Lastly,
for $g=0$, small $\varepsilon>0$, and $f(x)=\mathrm{sign}(x-\varepsilon)$,
according to (2.55), we have

$\displaystyle u_{xx}(0)$
$\displaystyle=\int_{0}^{+\infty}\mathrm{e}^{-y}f(y)\mathrm{d}y-f(0)$
$\displaystyle=\int_{0}^{\varepsilon}\mathrm{e}^{-y}(-1)\mathrm{d}y+\int_{\varepsilon}^{+\infty}\mathrm{e}^{-y}(+1)\mathrm{d}y-(-1)$
$\displaystyle=2-2\int_{0}^{\varepsilon}\mathrm{e}^{-y}\mathrm{d}y\rightarrow
2,\quad\text{as~{}}~{}~{}\varepsilon\rightarrow 0,$

and the coefficient of $\|f\|_{L^{\infty}(\mathbbm{R}_{+})}$ in the estimate
of $\|u_{xx}\|_{L^{\infty}(\mathbbm{R}_{+})}$ is optimal. $\hfill\Box$

###  2.4 Main Theorems

We state our main results for the cases: $u_{-}<u_{+}\leq 0$, $0\leq
u_{-}<u_{+}$ and $u_{-}<0<u_{+}$, that the initial-boundary value problem
(1.1)-(1.3) admits a unique global solution and it converges to the stationary
solution, the rarefaction wave and the superposition of the nonlinear waves,
respectively, as $t\to\infty$.

######  Theorem 2.13 (In the case of $0<u_{-}<u_{+}$).

Suppose that the boundary condition and far field states satisfy
$0<u_{-}<u_{+}$, the initial data $u_{0}$ satisfies
$u_{0}-\tilde{u}_{0}^{R}\in H^{2}(\mathbbm{R}_{+})$, where $\tilde{u}_{0}^{R}$
is defined in (2.2). Also assume that $\delta=|u_{-}-u_{+}|$ is sufficiently
small. Then there is a positive constant $\epsilon_{0}$ such that if
$\|u_{0}-\tilde{u}_{0}^{R}\|_{2}+\delta\leq\epsilon_{0}$, the problem
(1.1)-(1.3) admits a unique solution $(u(x,t),q(x,t))$, which satisfies

$\displaystyle u-u^{R}\in C^{0}([0,\infty);H^{2})\cap
C^{1}([0,\infty);H^{1}),$ $\displaystyle q+\partial_{x}u^{R}\in
C^{0}([0,\infty);H^{3})\cap L^{2}(0,\infty;H^{3}),$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(u(x,t)-u^{R}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,\\\\[5.69054pt]
&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(q(x,t)+\partial_{x}u^{R}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,2.\end{split}$

######  Theorem 2.14 (In the case of $u_{-}<u_{+}\leq 0$).

Suppose that the boundary condition and far field states satisfy
$u_{-}<u_{+}\leq 0$, the initial data $u_{0}$ satisfies $u_{0}-\bar{u}\in
H^{2}(\mathbbm{R}_{+})$, where $\bar{u}=\bar{u}_{i},(i=1,2)$ is a stationary
solution of Lemma 2.3. Also assume that $\delta=|u_{-}-u_{+}|$ is sufficiently
small. Then there is a positive constant $\epsilon_{0}$ such that if
$\|u_{0}-\bar{u}\|_{2}+\delta\leq\epsilon_{0}$, the problem (1.1)-(1.3) admits
a unique solution $(u(x,t),q(x,t))$, which satisfies

$\displaystyle u-\bar{u}\in C^{0}([0,\infty);H^{2}),\ \ u_{x}-\bar{u}_{x}\in
L^{2}(0,\infty;H^{1}),$ $\displaystyle q-\bar{q}\in
C^{0}([0,\infty);H^{3})\cap L^{2}(0,\infty;H^{3}),$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(u(x,t)-\bar{u}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,\\\\[5.69054pt]
&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(q(x,t)-\bar{q}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,2.\end{split}$

######  Theorem 2.15 (In the case of $u_{-}<0<u_{+}$).

Suppose that the boundary condition and far field states satisfy
$u_{-}<0<u_{+}$, the initial data $u_{0}$ satisfies
$u_{0}-\bar{u}_{2}(\cdot)-\tilde{u}_{4}(\cdot,0)\in H^{2}(\mathbbm{R}_{+})$,
where $\bar{u}_{2}$ and $\tilde{u}_{4}$ is a stationary solution and
rarefaction wave for the cases $u_{-}<u_{+}=0$ and $u_{+}>u_{-}=0$,
respectively. Also assume that $\delta=|u_{-}-u_{+}|$ is sufficiently small.
Then there is a positive constant $\epsilon_{0}$ such that if
$\|u_{0}-\bar{u}_{2}(\cdot)-\tilde{u}_{4}(\cdot,0)\|_{2}+\delta\leq\epsilon_{0}$,
the problem (1.1)-(1.3) admits a unique solution $(u(x,t),q(x,t))$, which
satisfies

$\displaystyle u-\bar{u}_{2}-u_{4}^{R}\in C^{0}([0,\infty);H^{2}),\ \
\partial_{x}(u-\bar{u}_{2}-u_{4}^{R})\in L^{2}(0,\infty;H^{1}),$
$\displaystyle q-\bar{q}_{2}+\partial_{x}u_{4}^{R}\in
C^{0}([0,\infty);H^{3})\cap L^{2}(0,\infty;H^{3}),$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(u(x,t)-\bar{u}_{2}(x)-u_{4}^{R}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,\\\\[5.69054pt]
&\sup_{x\in\mathbbm{R}_{+}}\left|\partial_{x}^{k}\left(q(x,t)-\bar{q}_{2}(x)+\partial_{x}u_{4}^{R}(x)\right)\right|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,2.\end{split}$

##  3 Asymptotics to Rarefaction Wave

###  3.1 Reformulation of the Problem in the Case of $u_{+}>u_{-}\geq 0$

The special case: $0=u_{-}<u_{+}$ has been considered by Ruan and Zhu in [32],
so we will focus on the case of $0<u_{-}<u_{+}$. The case $u_{-}>0$ means that
the fluid blows in through the boundary $x=0$. Hence, this initial boundary
problem is called the _in-flow_ problem. It is worth noticing that the
boundary condition $u(0,t)=u_{-}$ is necessary for the well-posedness of the
problem since the characteristic speed of the first hyperbolic equation (1.1)1
is positive at boundary $x=0$. Moreover, for the second elliptic equation
(1.1)2, we need boundary condition on $q(0,t)$ to ensure the well-posedness of
the problem (1.1). From Lemma 2.1 $\mathrm{(ii)}$ with $k=1$, we note that the
boundary value of $q(x,t)$ can be defined as $q(0,t)=0$. Therefore, in the
case of $0<u_{-}<u_{+}$, the problem (1.1)-(1.3) is rewritten as

$\begin{cases}u_{t}+(\frac{1}{2}u^{2})_{x}+q_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ \ t>0,\\\\[2.84526pt] -q_{xx}+q+u_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ \ t>0,\\\\[2.84526pt] u(0,t)=u_{-},\ \ \ \
q(0,t)=0,\ \ \ \ t\geq 0,\\\\[2.84526pt] u(x,0)=u_{0}(x)=\begin{cases}=u_{-},\
\ \ \ x=0,\\\\[5.69054pt] \rightarrow u_{+},\ \ \ \
x\rightarrow+\infty.\end{cases}\end{cases}$ (3.1)

Set

$u(x,t)=\varphi(x,t)+w(x,t),\ \ \ \ q(x,t)=\psi(x,t)+z(x,t).$ (3.2)

We note that $\psi=-\tilde{u}_{x}-\hat{q}=-\varphi_{x}-\hat{u}_{x}-\hat{q}$,
so we can rewritten (3.2) as

$\begin{cases}u(x,t)=\varphi(x,t)+w(x,t),\\\\[2.84526pt]
q(x,t)=z(x,t)-\varphi_{x}(x,t)-\hat{u}_{x}(x,t)-\hat{q}(x,t).\end{cases}$

Then the perturbation $(w,z)$ satisfies

$\begin{cases}w_{t}+ww_{x}+(\varphi w)_{x}+z_{x}=R_{1},\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\\\\[2.84526pt] -z_{xx}+z+w_{x}=R_{2},\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\\\\[2.84526pt] w(0,t)=0,\ \ \ \ z(0,t)=0,\ \ \
t\geq 0,\\\\[2.84526pt]
w(x,0)=u_{0}(x)-\varphi_{0}(x)=u_{0}(x)-\tilde{u}_{0}(x)+\hat{u}(x,0),\ \ \ \
x\in\mathbbm{R}_{+}.\end{cases}$ (3.3)

We define the solution space as

$X_{1}(0,T)=\left\\{\begin{tabular}[]{c|c}\hbox{\multirowsetup({\it w,
z})}&$w\in C^{0}([0,T);H^{2})\cap C^{1}([0,T);H^{1})$\\\\[2.84526pt] &$z\in
C^{0}([0,T);H^{3})\cap L^{2}(0,T;H^{3})$\end{tabular}\right\\},$

with $0<T\leq+\infty$. Then the problem (3.3) can be solved globally in time
as follows.

######  Theorem 3.1.

Suppose that the boundary condition and far field states satisfy
$0<u_{-}<u_{+}$, the initial data $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and the
wavelength $\delta=|u_{-}-u_{+}|$ are sufficiently small. Then there are the
positive constants $\varepsilon_{1}$ and $C=C(\varepsilon_{1})$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$, the problem (3.3) admits a unique
solution $(w(x,t),z(x,t))\in X_{1}(0,+\infty)$ satisfying

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{3}^{2}+\delta),$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}w(x,t)|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,\\\
&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}z(x,t)|\rightarrow 0\ \ \text{as}\
\ t\rightarrow\infty,\ \ k=0,1,2.\end{split}$ (3.4)

The Combination of the following local existence and the _a priori_ estimates
proves Theorem 3.1.

######  Proposition 3.2 (Local existence).

Suppose the boundary condition and far field states satisfy $u_{-}<u_{+}<0$,
the initial data satisfy $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$. Then there are two positive
constants $C=C(\varepsilon_{1})$ and $T_{0}=T_{0}(\varepsilon_{1})$ such that
the problem (3.3) has a unique solution $(w,z)\in X_{1}(0,T_{0})$, which
satisfies

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{2}^{2}+\delta),$

for $t\in[0,T_{0}]$.

######  Proposition 3.3 (_A priori_ estimates).

Let $T$ be a positive constant. Suppose that the problem (3.3) has a unique
solution $(w,z)\in X_{1}(0,T)$. Then there exist two positive constants
$\varepsilon_{2}(\leq\varepsilon_{1})$ and $C=C(\varepsilon_{2})$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{2}$, then we have the estimate

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{2}^{2}+\delta),$

for $t\in[0,T]$.

###  3.2 _A priori_ Estimates

Under the assumptions of Theorem 3.1, to give the proof of _a priori_
estimates in Proposition 3.3, we devote ourselves to the estimates on the
solution $(w,z)\in X_{1}(0,T)$ (for some $T>0$) of (3.3) under the _a priori_
assumption

$|w_{x}(t)|_{\infty}\leq\varepsilon_{0},\ \ \ \
|w_{t}(t)|_{\infty}\leq\varepsilon_{0},$ (3.5)

where $0<\varepsilon_{0}\ll 1$. For simplicity, we divide the proof of the _a
priori_ estimate into several lemmas.

######  Lemma 3.4.

There are the positive constants $\varepsilon_{1}(\leq\varepsilon_{0})$ and
$C=C(\varepsilon_{1})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$,
then

$\|w(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w(\tau)\|^{2}+\|z(\tau)\|_{1}^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|^{2}+\delta)$ (3.6)

holds for $t\in[0,T]$.

Proof. Multiplying (3.3)1 by $w$ and (3.3)2 by $z$, and adding the two
resulting equations up, we obtain

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w^{2}+\frac{1}{2}\varphi_{x}w^{2}+z_{x}^{2}+z^{2}+\left\\{\frac{1}{2}\varphi
w^{2}+\frac{1}{3}w^{3}-z_{x}z+zw\right\\}_{x}=R_{1}w+R_{2}z.$ (3.7)

Integrating (3.7) over $\mathbbm{R}_{+}\times(0,t)$, using $w(0,t)=z(0,t)=0$,
we get

$\|w(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}+\|z(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C\left(\|w_{0}\|^{2}+\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1}w|+|R_{2}z|\,\mathrm{d}x\mathrm{d}\tau\right).$
(3.8)

From Lemma 2.2 $\mathrm{(v)}$ and $\mathrm{(vi)}$ with $k=l=0$, we have

$\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1}w|\,\mathrm{d}x\mathrm{d}\tau\leq\|w(t)\|\int_{0}^{t}\|R_{1}\|\,\mathrm{d}\tau\leq
C\delta(1+\|w(t)\|^{2})\int_{0}^{t}\mathrm{e}^{-c(1+t)}\,\mathrm{d}\tau\leq
C\delta(1+\|w(t)\|^{2}),$ (3.9)

and

$\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{2}z|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{4}\int_{0}^{t}\|z(\tau)\|^{2}\,\mathrm{d}\tau+\int_{0}^{t}\|R_{2}\|^{2}\,\mathrm{d}\tau\leq\frac{1}{4}\int_{0}^{t}\|z(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{3}{2}}.$
(3.10)

Substituting (3.9) and (3.10) into (3.8), we conclude (3.6). $\hfill\Box$

######  Lemma 3.5.

There are two positive constants $\varepsilon_{2}(\leq\varepsilon_{1})$ and
$C=C(\varepsilon_{2})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$,
then

$\|w_{x}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{x}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta)$ (3.11)

holds for $t\in[0,T]$.

Proof. We differentiate (3.3)1 with respect to $x$ and multiply it by $w_{x}$,
and multiply (3.3)2 by $-z_{xx}$. Then, adding these two equations up, we have

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w_{x}^{2}+\frac{3}{2}\varphi_{x}w_{x}^{2}+z_{xx}^{2}+z_{x}^{2}+\left\\{\frac{1}{2}\varphi
w_{x}^{2}+\frac{1}{2}ww_{x}^{2}-z_{x}z\right\\}_{x}=-\varphi_{xx}ww_{x}-\frac{1}{2}w_{x}^{3}+R_{1x}w_{x}-R_{2}z_{xx}.$
(3.12)

Integrating (3.12) over $\mathbbm{R}_{+}\times(0,t)$, combining it with
$\varphi(0,t)=u_{-}>0$ and $w(0,t)=z(0,t)=0$, we get

$\displaystyle\|w_{x}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{x}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau$
(3.13) $\displaystyle\leq
C\left(\|w_{0}\|_{1}^{2}+u_{-}\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau+\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{xx}ww_{x}|+|w_{x}|^{3}+|R_{1x}w_{x}|+|R_{2}z_{xx}|\,\mathrm{d}x\mathrm{d}\tau\right).$

Since the equation (3.3)1 implies

$u_{-}w_{x}(0,t)=-z_{x}(0,t)+R_{1}(0,t),$

we can estimate the integral on the boundary as follows:

$\displaystyle u_{-}\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}\left(z_{x}^{2}(0,\tau)+R_{1}^{2}(0,\tau)\right)\,\mathrm{d}\tau$
(3.14) $\displaystyle\leq
C\int_{0}^{t}\left(|z_{x}|_{\infty}^{2}+|R_{1}|_{\infty}^{2}\right)\,\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|z_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\int_{0}^{t}\|z_{x}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta.$

Next, we estimate the last four terms on the right-hand side of (3.13). Using
Lemma 2.2 $\mathrm{(iv)}$ with $k=2$ and $l=0$, we have

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{xx}ww_{x}|\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{24}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau+C|w(t)|_{\infty}^{2}\int_{0}^{t}\|\varphi_{xx}(\tau)\|^{2}\,\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{24}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{1}{2}},$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}(\tau)|^{3}\,\mathrm{d}x\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau.$

From Lemma 2.2 $\mathrm{(v)}$ with $k=1$ and $\mathrm{(vi)}$ with $k=0,l=0$,
we get

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1x}w_{x}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{24}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{2}z_{xx}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{8}\int_{0}^{t}\|z_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{3}{2}}.$
(3.15)

On the other hand, from (3.3)2, we have

$\|w_{x}(t)\|^{2}\leq 3(\|z_{xx}(t)\|^{2}+\|z(t)\|^{2}+\|R_{2}(t)\|^{2}).$
(3.16)

In deriving the equation $\eqref{xsbwz}$ we have used the fact that for any
$a,b,c\in\mathbbm{R}$,

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac\leq 3(a^{2}+b^{2}+c^{2}).$

Substituting (3.14)-(3.15) into (3.13) and using (3.16), for some small
$\delta$ and $\varepsilon_{0}(<\frac{1}{6})$, we have

$\displaystyle\|w_{x}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{x}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta).$

This completes the proof of Lemma 3.5. $\hfill\Box$

For (3.14) and (3.16), combining the results of Lemma 3.4 and 3.5, we can
easily show the following Corollary 3.6 and Corollary 3.7.

######  Corollary 3.6.

Under the assumptions of Lemma 3.5, there exists a positive constant $C$ such
that

$\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta),\quad\forall t\in[0,T].$

######  Corollary 3.7.

Under the assumptions of Lemma 3.5, there exists a positive constant $C$ such
that

$\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta),\quad\forall t\in[0,T].$

Next, we try to give the estimate for $w_{xx}$. When estimating $w_{xx}$, we
need to deal with the boundary term $w_{x}x(0,t)$ (see $\eqref{xsbwxxjf}$). It
is quite difficult to estimate the boundary term $w_{x}x(0,t)$ directly.
However, we can get the estimate of the boundary term $w_{xt}(0,t)$ owing to
$w(0,t)=w_{t}(0,t)=w_{tt}(0,t)=0$, and then the estimate of $wxx(0,t)$ is
obtained through the equation $\eqref{xsbwxxzxxbj}$. Thus, to give the
estimate for $w_{xx}$, we firstly proceed to the _a priori_ estimate for the
derivatives $w_{t}$ and $w_{xt}$.

######  Lemma 3.8.

Under the assumptions of Lemma 3.5, there is a positive constant $C$ such that

$\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta).$

Proof. With the help of Lemma 3.4, Corollary 3.7 and Lemma 2.2 $\mathrm{(v)}$
with $k=l=0$, we see from (3.3)1 that

$\displaystyle\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}\int_{\mathbbm{R}_{+}}(w^{2}w_{x}^{2}+\varphi_{x}^{2}w^{2}+\varphi^{2}w_{x}^{2}+z_{x}^{2}+R_{1}^{2})\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq
C\left((|w(t)|_{\infty}^{2}\\!+\\!|\varphi(t)|_{\infty}^{2})\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau\\!+\\!|\varphi_{x}(t)|_{\infty}\int_{0}^{t}\|\sqrt{\varphi_{x}}w(\tau)\|^{2}\,\mathrm{d}\tau\\!+\\!\int_{0}^{t}\|z_{x}(\tau)\|^{2}\,\mathrm{d}\tau\\!+\\!\delta\right)$
$\displaystyle\leq C(\|w_{0}\|_{1}^{2}+\delta).$

Thus, the proof of Lemma 3.8 is completed. $\hfill\Box$

######  Lemma 3.9.

Under the assumptions of Lemma 3.5, there is a positive constant $C$ such that

$\|w_{t}(t)\|^{2}+\int_{0}^{t}\left(\|\varphi_{x}w_{t}(\tau)\|^{2}+\|z_{xt}(\tau)\|^{2}+\|z_{t}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta)$ (3.17)

holds for $t\in[0,T]$.

Proof. We differentiate (3.3) with respect to $t$ and multiply the first and
the second resulting equations by $w_{t}$ and $z_{t}$ respectively. Then,
adding these two equations up, we have

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}$ $\displaystyle
w_{t}^{2}+\frac{1}{2}\varphi_{x}w_{t}^{2}+z_{xt}^{2}+z_{t}^{2}+\left\\{\frac{1}{2}\varphi
w_{t}^{2}+\frac{1}{2}ww_{t}^{2}-z_{xt}z_{t}+z_{t}w_{t}\right\\}_{x}$ (3.18)
$\displaystyle=-\varphi_{xt}ww_{t}-\varphi_{t}w_{x}w_{t}-\frac{1}{2}w_{x}w_{t}^{2}+R_{1t}w_{t}+R_{2t}z_{t}.$

We note that $w_{t}(0,t)=z_{t}(0,t)=0$ due to $w(0,t)=z(0,t)=0$. Integrating
(3.18) over $\mathbbm{R}_{+}\times(0,t)$, we have

$\displaystyle\|w_{t}(t)\|^{2}$
$\displaystyle+\int_{0}^{t}\left(\|\varphi_{x}w_{t}(\tau)\|^{2}+\|z_{xt}(\tau)\|^{2}+\|z_{t}(\tau)\|^{2}\right)\,\mathrm{d}\tau$
(3.19)
$\displaystyle\leq\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{xt}ww_{t}|+|\varphi_{t}w_{x}w_{t}|+|w_{x}w_{t}^{2}|+|R_{1t}w_{t}|+|R_{2t}z_{t}|)\,\mathrm{d}x\mathrm{d}\tau.$

Combining Lemma 3.8 and Lemma 2.2, using Cauchy-Schwarz inequality and (3.5),
we can estimate the terms on the right-hand side of (3.19) as follows:

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{xt}ww_{t}|\,\mathrm{d}x\mathrm{d}\tau\leq\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta|w(t)|_{\infty}^{2}(1+t)^{-\frac{1}{2}},$
(3.20)
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{t}w_{x}w_{t}|\,\mathrm{d}x\mathrm{d}\tau\leq\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau+|\varphi_{t}(t)|_{\infty}^{2}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau,$
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}w_{t}^{2}|\,\mathrm{d}x\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau,$
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1t}w_{t}|\,\mathrm{d}x\mathrm{d}\tau\leq\int_{0}^{t}\|w_{t}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{2t}z_{t}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{4}\int_{0}^{t}\|z_{t}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{5}{2}}.$
(3.21)

Substituting (3.20)-(3.21) into (3.19) yields (3.17). This proves Lemma 3.9.
$\hfill\Box$

Next, we show the estimate for $w_{xt}$ in the following Lemma 3.10.

######  Lemma 3.10.

There are two positive constants $\varepsilon_{3}(\leq\varepsilon_{2})$ and
$C=C(\varepsilon_{3})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{3}$,
then

$\|w_{xt}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xt}(\tau)\|^{2}+\|z_{xxt}(\tau)\|^{2}+\|z_{xt}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta)+C(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau$
(3.22)

holds for $t\in[0,T]$.

Proof. Differentiate (3.3)1 with respect to $x$ and $t$, then multiply it by
$w_{xt}$. Differentiate (3.3)2 with respect to $t$ and multiply it by
$-z_{xxt}$. Finally, adding these two equations up, we have

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w_{xt}^{2}+\frac{3}{2}\varphi_{x}w_{xt}^{2}+z_{xxt}^{2}+z_{xt}^{2}+\left\\{\frac{1}{2}\varphi
w_{xt}^{2}+\frac{1}{2}ww_{xt}^{2}-z_{xt}z_{t}\right\\}_{x}$ (3.23)
$\displaystyle=$
$\displaystyle-\varphi_{xxt}ww_{xt}-\varphi_{xx}w_{t}w_{xt}-2\varphi_{xt}w_{x}w_{xt}-\varphi_{t}w_{xx}w_{xt}-w_{t}w_{xx}w_{xt}-\frac{3}{2}w_{x}w_{xt}^{2}+R_{1xt}w_{xt}-R_{2t}z_{xxt}.$

Integrating (3.23) over $\mathbbm{R}_{+}\times(0,t)$, using
$\varphi(0,t)=u_{-}>0$ and $w(0,t)=z_{t}(0,t)=0$, we have

$\displaystyle\|w_{xt}\|^{2}$
$\displaystyle+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xt}(\tau)\|^{2}+\|z_{xxt}(\tau)\|^{2}+\|z_{xt}(\tau)\|^{2}\right)\,\mathrm{d}\tau$
(3.24) $\displaystyle\leq
C\left(\|w_{0}\|_{1}^{2}+\int_{0}^{t}w_{xt}^{2}(0,\tau)\,\mathrm{d}\tau+\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{xxt}ww_{xt}|+|\varphi_{xx}w_{t}w_{xt}|+|\varphi_{xt}w_{x}w_{xt}|)\,\mathrm{d}x\mathrm{d}\tau\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \
+\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{t}w_{xx}w_{xt}|+|w_{t}w_{xx}w_{xt}|+|w_{x}w_{xt}^{2}|+|R_{1xt}w_{xt}|+|R_{2t}z_{xxt}|)\,\mathrm{d}x\mathrm{d}\tau\right).$

Firstly, from (3.3)1, we have the following equation at the boundary $x=0$,

$u_{-}w_{xt}(0,t)=R_{1t}(0,t)-z_{xt}(0,t),$ (3.25)

which plays an important role in estimating boundary terms. In fact, we
differential (3.3)1 with respect to $t$, then we get

$w_{tt}+w_{t}w_{x}+ww_{xt}+\varphi_{xt}w+\varphi_{t}w_{x}+\varphi
w_{xt}+z_{xt}=R_{1t}.$ (3.26)

For $\varphi(0,t)=u_{-}$ and $w(0,t)=0$, the boundary values at $x=0$ of
(3.26) yields (3.25). According to (3.25), using Lemma 3.9 and Lemma 2.2
$\mathrm{(v)}$ with $k=0,l=1$, we can get

$\displaystyle\int_{0}^{t}w_{xt}^{2}(0,\tau)\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}(R_{1t}^{2}(0,\tau)+z_{xt}^{2}(0,\tau))\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}(|R_{1t}|_{\infty}^{2}+|z_{xt}|_{\infty}^{2})\,\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|z_{xxt}(\tau)\|^{2}\,\mathrm{d}\tau+C\int_{0}^{t}\|z_{xt}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta$
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|z_{xxt}(\tau)\|^{2}\,\mathrm{d}\tau+C(\|w_{0}\|_{1}^{2}+\delta).$

Next, the terms on the right-hand side of (3.24) can be estimated as follows:

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{xxt}ww_{xt}|+|\varphi_{xx}w_{t}w_{xt}|+|\varphi_{xt}w_{x}w_{xt}|)\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq$
$\displaystyle\frac{1}{24}\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta|w|_{\infty}^{2}(1+t)^{-\frac{3}{2}}+C\delta(|w_{t}|_{\infty}^{2}+|w_{x}|_{\infty}^{2})(1+t)^{-\frac{1}{2}}.$

Using the _a priori_ assumption (3.5) and Cauchy-Schwarz inequality, we get

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{t}w_{xx}w_{xt}|+|w_{t}w_{xx}w_{xt}|)\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq$
$\displaystyle|\varphi_{t}(t)|_{\infty}\int_{0}^{t}\|w_{xx}(\tau)\|\|w_{xt}(\tau)\|\,\mathrm{d}\tau+|w_{t}(t)|_{\infty}\int_{0}^{t}\|w_{xx}(\tau)\|\|w_{xt}(\tau)\|\,\mathrm{d}\tau$
$\displaystyle\leq$ $\displaystyle
C(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}w_{xt}^{2}|\,\mathrm{d}x\mathrm{d}\tau\leq|w_{x}(t)|_{\infty}\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau.$

From Lemma 2.2, the last two estimates can be given as follows:

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1xt}w_{xt}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{24}\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{2t}z_{xxt}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{8}\int_{0}^{t}\|z_{xxt}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{5}{2}}.$
(3.27)

On the other hand, from (3.3)2, we have

$\|w_{xt}(t)\|^{2}\leq 3(\|z_{xxt}(t)\|^{2}+\|z_{t}(t)\|^{2}+\|R_{2t}\|^{2}).$
(3.28)

Substituting (3.25)-(3.27) into (3.24) and using (3.28), for some small
$\delta$ and $\varepsilon_{0}$, we have

$\displaystyle\|w_{x}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{x}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta)+C(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau.$

This completes the proof of Lemma 3.10. $\hfill\Box$

######  Corollary 3.11.

Under the same assumptions of Lemma 3.10, there is a positive constant $C$
such that

$\int_{0}^{t}w_{xt}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta)+C(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau,\quad\forall
t\in[0,T].$

Finally, combining it with Lemma 3.10, we show the estimate for $w_{xx}$.

######  Lemma 3.12.

There are two positive constants $\varepsilon_{4}(\leq\varepsilon_{3})$ and
$C=C(\varepsilon_{4})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{4}$,
then

$\|w_{xx}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xx}(\tau)\|^{2}+\|z_{xxx}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta)$ (3.29)

holds for $t\in[0,T]$.

Proof. Differentiate (3.3)1 twice with respect to $x$, then multiply it by
$w_{xx}$. Differentiate (3.3)2 with respect to $x$ and multiply it by
$-z_{xxx}$. In the end, adding these two equations up, we have

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w_{xx}^{2}+\frac{5}{2}\varphi_{x}w_{xx}^{2}+z_{xxx}^{2}+z_{xx}^{2}+\left\\{\frac{1}{2}\varphi
w_{xx}^{2}+\frac{1}{2}ww_{xx}^{2}-z_{xx}z_{x}\right\\}_{x}$ (3.30)
$\displaystyle=$
$\displaystyle-\varphi_{xxx}ww_{xx}-3\varphi_{xx}w_{x}w_{xx}-\frac{5}{2}w_{x}w_{xx}^{2}+R_{1xx}w_{xx}-R_{2x}z_{xxx}.$

Integrating (3.30) over $\mathbbm{R}_{+}\times(0,t)$, using
$\varphi(0,t)=u_{-}>0$ and $w(0,t)=0$, we obtain

$\displaystyle\|w_{xx}(t)\|^{2}$
$\displaystyle+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xx}(\tau)\|^{2}+\|z_{xxx}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}\right)\,\mathrm{d}\tau$
(3.31) $\displaystyle\leq
C\left(\|w_{0}\|_{2}^{2}+\int_{0}^{t}w_{xx}^{2}(0,\tau)\,\mathrm{d}\tau+\int_{0}^{t}|z_{xx}(0,t)||z_{x}(0,t)|\,\mathrm{d}\tau\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \
+\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\varphi_{xxx}ww_{xx}|+|\varphi_{xx}w_{x}w_{xx}|+|w_{x}w_{xx}^{2}|+|R_{1xx}w_{xx}|+|R_{2x}z_{xxx}|)\,\mathrm{d}x\mathrm{d}\tau\right).$

Firstly, from (3.3)1, we have an equation at the boundary $x=0$,

$u_{-}w_{xx}(0,t)=R_{1x}(0,t)-z_{xx}(0,t)-w_{xt}(0,t)-w_{x}^{2}(0,t)-2\varphi_{x}(0,t)w_{x}(0,t),$
(3.32)

which is important to estimate the boundary terms. In fact, we differential
(3.3)1 with respect to $x$, then we get

$w_{xt}+w_{x}^{2}+ww_{xx}+\varphi_{xx}w+2\varphi_{x}w_{x}+\varphi
w_{xx}+z_{xx}=R_{1x}.$ (3.33)

For $\varphi(0,t)=u_{-}$ and $w(0,t)=0$, the boundary value at $x=0$ of (3.33)
yields (3.32). Using (3.32) and Cauchy-Schwarz inequality, combining Corollary
3.7 and Corollary 3.11, we have

$\displaystyle C\int_{0}^{t}w_{xx}^{2}(0,\tau)\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}\left(R_{1x}^{2}(0,\tau)+z_{xx}^{2}(0,\tau)+w_{xt}^{2}(0,\tau)+w_{x}^{4}(0,\tau)+\varphi_{x}^{2}(0,\tau)w_{x}^{2}(0,\tau)\right)\,\mathrm{d}\tau$
(3.34) $\displaystyle\leq
C(\|w_{0}\|_{1}^{2}+\delta)+C(\delta+\varepsilon_{0})\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau++\frac{1}{8}\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau.$

Combining the results of Lemmas 3.4 and 3.5, we can get

$\displaystyle\int_{0}^{t}|z_{xx}(0,\tau)||z_{x}(0,\tau)|\,\mathrm{d}\tau\leq\frac{1}{8}\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau+C\int_{0}^{t}\left(\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau.$

Next, the rest terms on the right-hand side of (3.31) can be estimated as
follows. According to the _a priori_ assumption (3.5),

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}w_{xx}^{2}|\,\mathrm{d}x\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau.$

From Lemma 2.2, using Cauchy-Schwarz inequality, we have

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{xxx}ww_{xx}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{24}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta|w(t)|_{\infty}^{2}(1+t)^{-\frac{3}{2}},$
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\varphi_{xx}w_{x}w_{xx}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{24}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau++C\delta|w_{x}(t)|_{\infty}^{2}(1+t)^{-\frac{1}{2}},$
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{1xx}w_{xx}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{24}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|R_{2x}z_{xxx}|\,\mathrm{d}x\mathrm{d}\tau\leq\frac{1}{8}\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta(1+t)^{-\frac{5}{2}}.$
(3.35)

In the end, from (3.3)2, we have

$\|w_{xx}(t)\|^{2}\leq
3\left(\|z_{xxx}(t)\|^{2}+\|z_{x}(t)\|^{2}+\|R_{2x}(t)\|^{2}\right).$ (3.36)

Substituting (3.34)-(3.35) into (3.31) and using (3.36), for some small
$\delta$ and $\varepsilon_{0}$, we get

$\displaystyle\|w_{xx}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xx}(\tau)\|^{2}+\|z_{xxx}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}\right)\,\mathrm{d}\tau$
$\displaystyle\leq$ $\displaystyle
C(\|w_{0}\|_{2}^{2}+\delta)+\left(\frac{3}{4}+C\varepsilon_{0}+C\delta\right)\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau,$

which yields (3.29). $\hfill\Box$

Substituting (3.29) into (3.34), using Lemma 3.10 and Lemma 3.11, we can get
the following Corollaries 3.13-3.15.

######  Corollary 3.13.

Under the same assumptions of Lemma 3.12, there exists a positive constant $C$
such that

$\displaystyle\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T],$
$\displaystyle\int_{0}^{t}w_{xx}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T].$

######  Corollary 3.14.

Under the same assumptions of Lemma 3.12, for some small $\varepsilon_{0}$,
there exists a positive constant $C$ such that

$\|w_{xt}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\varphi_{x}}w_{xt}(\tau)\|^{2}+\|z_{xxt}(\tau)\|^{2}+\|z_{xt}(\tau)\|^{2}\right)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T].$

######  Corollary 3.15.

Under the same assumptions of Lemma 3.12, for some small $\varepsilon_{0}$,
there exists a positive constant $C$ such that

$\displaystyle\int_{0}^{t}\|w_{xt}(\tau)\|^{2}\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T],$
$\displaystyle\int_{0}^{t}w_{xt}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T].$

In the end, using the above estimates and the relation between $w$ and $z$, we
can easily get the estimate for $\|z(t)\|_{3}$.

######  Lemma 3.16.

Under the assumptions of Lemma 3.12, there holds

$\|z(t)\|_{3}^{2}\leq C(\|w_{0}\|_{2}^{2}+\delta),\quad\forall t\in[0,T].$
(3.37)

Proof. Firstly, from (3.3)1, we have the following equation at the boundary
$x=0$,

$z_{x}(0,t)=-u_{-}(w_{x}(0,t)+R_{1}(0,t)),$

which can help us to estimate the boundary terms. From (3.3)2, we obtain

$z_{xx}^{2}+z^{2}+2z_{x}^{2}=w_{x}^{2}+R_{2}^{2}+2(z_{x}z)_{x}+2R_{2}w_{x}.$
(3.38)

Integrating (3.38) over $\mathbbm{R}_{+}$, using Cauchy-Schwarz inequality and
$z(0,t)=0$, we get

$\displaystyle\|z_{xx}(t)\|^{2}+2\|z_{x}(t)\|^{2}+\|z(t)\|^{2}\leq
2\|w_{x}(t)\|^{2}+2\|R_{2}\|^{2}\leq 2\|w_{x}(t)\|^{2}+C\delta\leq
C(\|w_{0}\|_{2}^{2}+\delta).$ (3.39)

Secondly, differentiating (3.3)2 with respect to $x$ and integrating these
equations over $\mathbbm{R}_{+}$, combining (3.39), Lemma 2.1 and Lemma 3.12,
we get

$\displaystyle\|z_{xxx}(t)\|^{2}\leq
C\left(\|z_{x}(t)\|^{2}+\|w_{xx}(t)\|^{2}+\|R_{2x}(t)\|^{2}\right)\leq
C\left(\|w_{0}\|_{2}^{2}+\delta\right).$ (3.40)

Combining (3.39) with (3.40), we finish the proof of Lemma 3.16. $\hfill\Box$

###  3.3 Asymptotic Behavior toward the Rarefaction Wave

By combining the local existence, Proposition 3.2 and the _a priori_
estimates, we can get the global in time solution

$\begin{cases}w\in C^{0}([0,\infty);H^{2})\cap C^{1}([0,\infty);H^{1}),\ \
w_{t}\in C^{0}([0,\infty);H^{1})\cap L^{2}(0,\infty;H^{1})\\\\[2.84526pt] z\in
C^{0}([0,\infty);H^{3})\cap L^{2}(0,\infty;H^{3}),\ \ z_{t}\in
L^{2}(0,\infty;H^{2}),\end{cases}$

such that

$\begin{cases}\sup\limits_{t\geq
0}(\|w(t)\|_{2}^{2}+\|w_{t}(t)\|_{1}^{2}+\|z(t)\|_{3}^{2})<\infty,\\\\[11.38109pt]
\int_{0}^{\infty}(\|w_{x}(t)\|_{1}^{2}+\|w_{t}(t)\|_{1}^{2}+\|z(t)\|_{3}^{2}+\|z_{t}(t)\|_{2}^{2})\,\mathrm{d}t<\infty.\end{cases}$
(3.41)

In order to show the large-time behavior (3.4) in Theorem 3.1, using the
Sobolev inequality

$\begin{cases}\sup\limits_{x\in\mathbbm{R}_{+}}|f(x,t)|\leq\sqrt{2}\|f(t)\|^{\frac{1}{2}}\|f_{x}(t)\|^{\frac{1}{2}},\\\\[11.38109pt]
\sup\limits_{x\in\mathbbm{R}_{+}}|f_{x}(x,t)|\leq\sqrt{2}\|f_{x}(t)\|^{\frac{1}{2}}\|f_{xx}(t)\|^{\frac{1}{2}},\\\\[11.38109pt]
\sup\limits_{x\in\mathbbm{R}_{+}}|f_{xx}(x,t)|\leq\sqrt{2}\|f_{xx}(t)\|^{\frac{1}{2}}\|f_{xxx}(t)\|^{\frac{1}{2}},\end{cases}$

we just need to prove

$\|w_{x}(t)\|\to 0,\ \ \|z(t)\|\to 0,\ \ \|z_{xx}(t)\|\to 0,\ \ \text{as}\ \
t\to\infty.$ (3.42)

According to (3.41), we only need to show

$\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|w_{x}(t)\|^{2}\right|\,\mathrm{d}t<\infty,\
\ \ \
\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|z(t)\|^{2}\right|\,\mathrm{d}t<\infty,\
\
\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|z_{xx}(t)\|^{2}\right|\,\mathrm{d}t<\infty.$
(3.43)

Here, we give the proof of (3.43) as follows.

Combining with the results of (3.41), we can easily get

$\displaystyle\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|w_{x}(t)\|^{2}\right|\,\mathrm{d}t$
$\displaystyle\leq
2\int_{0}^{\infty}\int_{\mathbbm{R}_{+}}|w_{x}||w_{xt}|\,\mathrm{d}x\mathrm{d}t$
$\displaystyle\leq\int_{0}^{\infty}\|w_{x}(t)\|^{2}\,\mathrm{d}t+\int_{0}^{\infty}\|w_{xt}(t)\|^{2}\,\mathrm{d}t<\infty.$

Similarly, using (3.41), we can obtain

$\displaystyle\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|z(t)\|^{2}\right|\,\mathrm{d}t$
$\displaystyle\leq
2\int_{0}^{\infty}\int_{\mathbbm{R}_{+}}|z(t)||z_{t}(t)|\,\mathrm{d}x\mathrm{d}t$
$\displaystyle\leq\int_{0}^{\infty}\|z(t)\|^{2}\,\mathrm{d}t+\int_{0}^{\infty}\|z_{t}(t)\|^{2}\,\mathrm{d}t<\infty,$

and

$\displaystyle\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|z_{xx}(t)\|^{2}\right|\,\mathrm{d}t$
$\displaystyle\leq
2\int_{0}^{\infty}\int_{\mathbbm{R}_{+}}|z_{xx}(t)||z_{xxt}(t)|\,\mathrm{d}x\mathrm{d}t$
$\displaystyle\leq\int_{0}^{\infty}\|z_{xx}(t)\|^{2}\,\mathrm{d}t+\int_{0}^{\infty}\|z_{xxt}(t)\|^{2}\,\mathrm{d}t<\infty.$

Therefore, we finish the proof of the large-time behavior (3.4). That is, the
proof of Theorem 3.1 is completed.

##  4 Asymptotics to Stationary Solution

###  4.1 Reformulation of the Problem in the Case of $u_{-}<u_{+}\leq 0$

In the cases (1): $u_{-}<u_{+}<0$ and (2): $u_{-}<u_{+}=0$, the IBVP admits a
stationary solution $(\bar{u},\bar{q})=(\bar{u}_{i}(x),\bar{q}_{i}(x)),i=1,2$,
respectively. The stationary solution $(\bar{u},\bar{q})$ satisfies the
following ordinary differential equations

$\begin{cases}\bar{u}\bar{u}_{x}+\bar{q}_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\\\\[2.84526pt] -\bar{q}_{xx}+\bar{q}+\bar{u}_{x}=0,\ \ \
\ x\in\mathbbm{R}_{+},\\\\[2.84526pt] \bar{u}(0)=u_{-},\ \ \ \
\bar{u}(+\infty)=u_{+},\ \ \ \ \bar{q}(+\infty)=0.\end{cases}$

Put

$u(x,t)=\bar{u}(x)+w(x,t),\qquad q(x,t)=\bar{q}(x)+z(x,t).$

The equation (1.1) can be reformulated as

$\begin{cases}w_{t}+ww_{x}+(\bar{u}w)_{x}+z_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\\\\[2.84526pt] -z_{xx}+z+w_{x}=0,\ \ \ \
x\in\mathbbm{R}_{+},\ \ \ t>0,\\\\[2.84526pt] w(0,t)=0,\ \ \ \
w(+\infty,t)=0,\ \ \ t>0,\\\\[2.84526pt] w(x,0)=w_{0}(x)=u_{0}(x)-\bar{u}(x),\
\ \ \ x\in\mathbbm{R}_{+}.\end{cases}$ (4.1)

Define the solution space of (4.1) by

$X_{2}(0,T)=\left\\{w\in C^{0}([0,T);H^{2}),w_{x}\in L^{2}(0,T;H^{1});z\in
C^{0}([0,T);H^{3})\cap L^{2}(0,T;H^{3})\right\\}$

with $0<T\leq+\infty$. Then the problem (4.1) can be solved globally in time
as follows.

######  Theorem 4.1.

Suppose that the boundary condition and far field states satisfy
$u_{-}<u_{+}\leq 0$, the initial data $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and
the wavelength $\delta=|u_{-}-u_{+}|$ in Lemma 2.3 both are sufficiently
small. Then there are positive constants $\varepsilon_{1}$ and
$C=C(\varepsilon_{1})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$,
the problem (4.1) admits a unique solution $(w(x,t),z(x,t))\in
X_{2}(0,+\infty)$ satisfying

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C\|w_{0}\|_{2}^{2},$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}w(x,t)|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ \ k=0,1,\\\\[5.69054pt]
&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}z(x,t)|\rightarrow 0\ \ \text{as}\
\ t\rightarrow\infty,\ \ \ k=0,1,2.\end{split}$ (4.2)

Theorem 4.1 is proved by combining the local existence of the solution
together with the _a priori_ estimates.

######  Proposition 4.2 (Local existence).

Suppose the boundary condition and far field states satisfy $u_{-}<u_{+}\leq
0$, the initial data satisfies $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{2}$. Then there are two positive
constants $C=C(\varepsilon_{2})$ and $T_{0}=T_{0}(\varepsilon_{2})$ such that
the problem (4.1) has a unique solution $(w,z)\in X_{2}(0,T_{0})$, which
satisfies

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C\|w_{0}\|_{2}^{2},\quad\forall t\in[0,T_{0}].$

######  Proposition 4.3 (A priori estimates).

Let $T$ be a positive constant. Suppose that the problem (4.1) has a unique
solution $(w,z)\in X_{2}(0,T)$. Then there exist positive constants
$\varepsilon_{1}(\leq\varepsilon_{2})$ and $C=C(\varepsilon_{1})$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$, then we have the estimate

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C\|w_{0}\|_{2}^{2},\quad\forall t\in[0,T_{0}].$

###  4.2 _A priori_ Estimates

Under the assumptions of Theorem 4.1, we want to give the proof of the _a
priori_ estimate in Proposition 4.3. To do this, we devote ourselves to the
estimates on the solution $(w,z)\in X_{2}(0,T)$ (for some $T>0$) of (4.1)
under the _a priori_ assumption

$|w_{x}(t)|_{\infty}\leq\varepsilon_{0},$ (4.3)

where $0<\varepsilon_{0}\ll 1$. For simplicity, we divide the proof of the _a
priori_ estimate into the following lemmas.

######  Lemma 4.4.

There are positive constants $\varepsilon_{1}(\leq\varepsilon_{0})$ and
$C=C(\varepsilon_{1})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$,
then

$\|w(t)\|_{1}^{2}+\int_{0}^{t}\
\left(\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}+\|\sqrt{\bar{u}_{x}}w_{x}(\tau)\|^{2}+\|w_{x}(\tau)\|^{2}\right)\,\mathrm{d}{\tau}+\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}{\tau}\leq
C\|w_{0}\|_{1}^{2},$ (4.4)

for $t\in[0,T]$.

Proof. Multiplying (4.1)1 by $w$ and (4.1)2 by $z$, and adding the two
resulting equations up, we obtain

$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w^{2}+\frac{1}{2}\bar{u}_{x}w^{2}+z_{x}^{2}+z^{2}+\left\\{\frac{1}{2}\bar{u}w^{2}+\frac{1}{3}w^{3}-z_{x}z+zw\right\\}_{x}=0.$
(4.5)

We differentiate (4.1)1 with respect to $x$ and multiply it by $w_{x}$, and
multiply (4.1)2 by $-z_{xx}$. Adding these two equations up, we obtain

$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w_{x}^{2}+\frac{3}{2}\bar{u}_{x}w_{x}^{2}+z_{xx}^{2}+z_{x}^{2}+\left\\{\frac{1}{2}\bar{u}w_{x}^{2}+\frac{1}{2}ww_{x}^{2}-z_{x}z\right\\}_{x}=-\frac{1}{2}w_{x}^{3}-\bar{u}_{xx}ww_{x}.$
(4.6)

On the other hand, rewriting (4.1)2 in the form $w_{x}=z_{xx}-z$ and squaring
this equation, we get

$w_{x}^{2}=z_{xx}^{2}+z^{2}+2z_{x}^{2}-2(z_{x}z)_{x}.$ (4.7)

Adding (4.5), (4.6) and (4.7) up, we get

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}(w^{2}+w_{x}^{2})$
$\displaystyle+\frac{1}{2}\bar{u}_{x}w^{2}+\frac{3}{2}\bar{u}_{x}w_{x}^{2}+w_{x}^{2}$
(4.8)
$\displaystyle+\left\\{\frac{1}{2}\bar{u}w^{2}+\frac{1}{3}w^{3}+zw+\frac{1}{2}\bar{u}w_{x}^{2}+\frac{1}{2}ww_{x}^{2}\right\\}_{x}=-\frac{1}{2}w_{x}^{3}-\bar{u}_{xx}ww_{x}.$

Integrating (4.8) over $\mathbbm{R}_{+}\times(0,t)$, combining it with
$u_{-}<0$ and $w(0,t)=0$, we have

$\displaystyle\frac{1}{2}\|w(t)\|_{1}^{2}$
$\displaystyle+\int_{0}^{t}\left(\frac{1}{2}\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}+\frac{3}{2}\|\sqrt{\bar{u}_{x}}w_{x}(\tau)\|^{2}+\|w_{x}(\tau)\|^{2}\right)\,\mathrm{d}\tau-
u_{-}\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau$ (4.9)
$\displaystyle\leq\frac{1}{2}\|w_{0}\|_{1}^{2}+\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|w_{x}|^{3}+|\bar{u}_{xx}ww_{x}|)\,\mathrm{d}x\mathrm{d}\tau.$

From (4.3) and Lemma 2.3, we get

$\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}|^{3}\,\mathrm{d}x\mathrm{d}\tau\leq|w_{x}(t)|_{\infty}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau,$
(4.10)

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\bar{u}_{xx}ww_{x}|\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq\left|\frac{\bar{u}^{2}_{xx}}{\bar{u}_{x}}\right|_{\infty}\int_{0}^{t}\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}\,\mathrm{d}\tau+\frac{1}{8}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau$
(4.11) $\displaystyle\leq
C\delta\int_{0}^{t}\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}\,\mathrm{d}\tau+\frac{1}{8}\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau.$

Substituting (4.10)-(4.11) into (4.9), for some small $\delta$ and
$\varepsilon_{0}(<\frac{1}{8})$, we get

$\displaystyle\|w(t)\|_{1}^{2}+\int_{0}^{t}\left(\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}+\|\sqrt{\bar{u}_{x}}w_{x}(\tau)\|^{2}+\|w_{x}(\tau)\|_{2}^{2}\right)\,\mathrm{d}\tau+\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C\|w_{0}\|_{1}^{2}.$ (4.12)

Hence, Lemma 4.4 is proved. $\hfill\Box$

The combination of Lemma 4.4 and equation (4.7) yields the following Lemma
4.5.

######  Lemma 4.5.

Under the same assumptions of Lemma 4.4, there exists a positive constant C
such that

$\int_{0}^{t}\|z(\tau)\|_{2}^{2}\,\mathrm{d}\tau\leq C\|w_{0}\|_{1}^{2}$

holds for $t\in[0,T]$.

Proof. Integrating (4.7) over $\mathbbm{R}_{+}\times(0,t)$, we can easily get
the following estimate for $z$,

$\int_{0}^{t}(\|z_{xx}(\tau)\|^{2}+2\|z_{x}(\tau)\|^{2}+\|z(\tau)\|^{2})\,\mathrm{d}\tau\leq\int_{0}^{t}(\|w_{x}(\tau)\|^{2}+2|z_{x}(0,\tau)||z(0,\tau)|)\,\mathrm{d}\tau.$
(4.13)

From (4.1)1, we have the equation at the boundary $x=0$,

$-u_{-}w_{x}(0,t)=z_{x}(0,t).$ (4.14)

Using (4.14) and Cauchy-Schwarz inequality, we have

$\displaystyle 2\int_{0}^{t}|z_{x}(0,\tau)||z(0,\tau)|\,\mathrm{d}\tau\leq
C\int_{0}^{t}|w_{x}(0,\tau)|^{2}\,\mathrm{d}\tau+\frac{1}{2}\int_{0}^{t}|z|_{\infty}^{2}\,\mathrm{d}\tau\leq
C\|w_{0}\|_{1}^{2}+\frac{1}{2}\int_{0}^{t}(\|z_{x}(\tau)\|^{2}+\|z(\tau)\|^{2})\,\mathrm{d}\tau.$
(4.15)

Substituting (4.15) into (4.13), we finish the proof of Lemma 4.5.
$\hfill\Box$

######  Lemma 4.6.

There are positive constants $\varepsilon_{2}(\leq\varepsilon_{1})$ and
$C=C(\varepsilon_{2})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{2}$,
then

$\|w_{xx}(t)\|^{2}+\int_{0}^{t}\left(\|\sqrt{\bar{u}_{x}}w_{xx}(\tau)\|^{2}+\|z_{xxx}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2}+w_{xx}^{2}(0,\tau)\right)\,\mathrm{d}\tau\leq
C\|w_{0}\|_{2}^{2},$ (4.16)

for $t\in[0,T]$.

Proof. We differentiate (4.1) twice with respect to $x$ and multiply the first
and the second resulting equations by $w_{xx}$ and $z_{xx}$ respectively.
Then, adding these two equations up, we have

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w_{xx}^{2}$
$\displaystyle+\frac{5}{2}\bar{u}_{x}w_{xx}^{2}+z_{xxx}^{2}+z_{xx}^{2}+\left\\{\frac{1}{2}\bar{u}w_{xx}^{2}+\frac{1}{2}ww_{xx}^{2}+z_{xx}w_{xx}-z_{xx}z_{xxx}\right\\}_{x}$
(4.17)
$\displaystyle=-\bar{u}_{xxx}ww_{xx}-3\bar{u}_{xx}w_{x}w_{xx}-\frac{5}{2}w_{x}w_{xx}^{2}.$

Integrating (4.17) over $\mathbbm{R}_{+}\times(0,t)$, using $u_{-}<0$ and
$z_{xx}w_{xx}-z_{xx}z_{xxx}=-z_{xx}z_{x}$ due to (4.1)2, we have after some
calculations that

$\displaystyle\|w_{xx}(t)\|^{2}$
$\displaystyle+\int_{0}^{t}(\|\sqrt{\bar{u}_{x}}w_{xx}(\tau)\|+\|z_{xxx}(\tau)\|^{2}+\|z_{xx}(\tau)\|^{2})\,\mathrm{d}\tau-
u_{-}\int_{0}^{t}w_{xx}^{2}(0,\tau)\,\mathrm{d}\tau$ (4.18) $\displaystyle\leq
C\left(\|w_{0}\|_{2}^{2}+\int_{0}^{t}|z_{xx}(0,\tau)||z_{x}(0,\tau)|\,\mathrm{d}\tau\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \
+\int_{0}^{t}\int_{\mathbbm{R}_{+}}(|\bar{u}_{xxx}ww_{xx}|+|\bar{u}_{xx}w_{x}w_{xx}|+|w_{x}|w_{xx}^{2})\,\mathrm{d}x\mathrm{d}\tau\right).$

Firstly, from (4.4), we have

$\displaystyle C\int_{0}^{t}|z_{xx}(0,\tau)||z_{x}(0,\tau)|\,\mathrm{d}\tau$
$\displaystyle\leq
C\int_{0}^{t}|z_{xx}|_{\infty}|z_{x}|_{\infty}\,\mathrm{d}\tau$ (4.19)
$\displaystyle\leq\frac{1}{4}\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau+C\int_{0}^{t}(\|z_{xx}(\tau)\|^{2}+\|z_{x}(\tau)\|^{2})\,\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{4}\int_{0}^{t}\|z_{xxx}(\tau)\|^{2}\,\mathrm{d}\tau+C\|w_{0}\|_{1}^{2}.$

From Lemma 2.3 and (4.3), and using Cauchy-Schwarz inequality, we can get the
estimates on the right-hand side of the equation (4.18)

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\bar{u}_{xxx}ww_{xx}|\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+\left|\frac{\bar{u}_{xxx}^{2}}{\bar{u}_{x}}\right|_{\infty}\int_{0}^{t}\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}\,\mathrm{d}\tau$
(4.20)
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\delta\int_{0}^{t}\|\sqrt{\bar{u}_{x}}w(\tau)\|^{2}\,\mathrm{d}\tau,$

and

$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|\bar{u}_{xx}w_{x}w_{xx}|\,\mathrm{d}x\mathrm{d}\tau$
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+|\bar{u}_{xx}|_{\infty}\int_{0}^{t}\int_{\mathbbm{R}_{+}}w_{x}^{2}\,\mathrm{d}x\mathrm{d}\tau$
(4.21)
$\displaystyle\leq\frac{1}{8}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau+C\int_{0}^{t}\|w_{x}(\tau)\|^{2}\,\mathrm{d}\tau,$
$\displaystyle\int_{0}^{t}\int_{\mathbbm{R}_{+}}|w_{x}|w_{xx}^{2}\,\mathrm{d}x\mathrm{d}\tau\leq|w_{x}|_{\infty}\int_{0}^{t}\int_{\mathbbm{R}_{+}}w_{xx}^{2}\,\mathrm{d}x\mathrm{d}\tau\leq\varepsilon_{0}\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau.$
(4.22)

In the end, from (4.1)2, the relation between $w$ and $z$ satisfies

$\|w_{xx}(t)\|^{2}\leq 2(\|z_{xxx}(t)\|^{2}+\|z_{x}(t)\|^{2}).$ (4.23)

Substituting (4.19)-(4.22) and (4.23) into (4.18), combining Lemma 4.4 and
Lemma 4.5, this yields (4.16) for some small $\delta$ and
$\varepsilon_{0}(<\frac{1}{8})$. $\hfill\Box$

Corollary 4.7 is given by the combination of Lemma 4.6 and (4.23).

######  Corollary 4.7.

Under the same assumptions of Lemma 4.6, the estimate

$\int_{0}^{t}\|w_{xx}(\tau)\|^{2}\,\mathrm{d}\tau\leq C\|w_{0}\|_{2}^{2}$

holds for $t\in[0,T]$.

The final lemma we need for Proposition 4.3 is the following one.

######  Lemma 4.8.

Under the same assumptions of Lemma 4.6. there is a positive constant $C$
independent on $\varepsilon_{2}$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{2}$, then

$\|z(t)\|_{3}^{2}\leq C\|w_{0}\|_{2}^{2},\quad\forall t\in[0,T].$

Proof. Rewriting the equation (4.1)2 as $z_{xx}-z=w_{x}$, and squaring this
equation, we have

$z_{xx}^{2}+2z_{x}^{2}+z^{2}=w_{x}^{2}+(z_{x}z)_{x}.$ (4.24)

Integrating (4.24) over $\mathbbm{R}_{+}$, combining it with (4.14), and using
Cauchy-Schwarz inequality, we get

$\displaystyle\|z(t)\|_{2}^{2}$
$\displaystyle\leq\|w_{x}(t)\|^{2}+C|w_{x}(0,t)|^{2}+\frac{1}{2}\|z(t)\|_{\infty}^{2}$
$\displaystyle\leq\|w_{x}(t)\|^{2}+C|w_{x}(t)|_{\infty}^{2}+\frac{1}{2}(\|z(t)\|^{2}+\|z_{x}(t)\|^{2})$
$\displaystyle\leq
C\|w_{x}(t)\|^{2}+C\|w_{xx}(t)\|^{2}+\frac{1}{2}(\|z(t)\|^{2}+\|z_{x}(t)\|^{2}),$

which yields $\|z\|_{2}^{2}\leq C\|w_{0}\|_{2}^{2}$. To get the
$L^{2}$-estimate on $z_{xxx}$, we differentiate (4.1)2 with respect to $x$,
then

$\|z_{xxx}(t)\|^{2}\leq 2(\|w_{xx}(t)\|^{2}+\|z_{x}(t)\|^{2})\leq
C\|w_{0}\|_{2}^{2}.$

This completes the proof of Lemma 4.8. $\hfill\Box$

###  4.3 Asymptotic Behavior toward the Stationary Solution

The global existence of the unique solution for problem (4.1) and its large
time behavior is an immediate consequence of Proposition 4.3. Indeed,
combining the standard theory of the existence and uniqueness of the local
solution with the _a priori_ estimates, one can extend the local solution for
problem (4.1) globally, that is

$\begin{cases}w\in C^{0}([0,\infty);H^{2}),\quad w_{x}\in
L^{2}(0,\infty;H^{1}),\\\\[2.84526pt] z\in C^{0}([0,\infty);H^{3})\cap
L^{2}(0,\infty;H^{3}).\end{cases}$

Then, the _a priori_ estimates again assert that

$\begin{cases}\sup\limits_{t\geq
0}(\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2})<\infty,\\\\[11.38109pt]
\int_{0}^{t}(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,\mathrm{d}\tau<\infty.\end{cases}$
(4.25)

To complete the proof of Theorem 4.1, by using (4.25), we can easily get

$\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|w_{x}\|^{2}\right|\,\mathrm{d}t<\infty.$
(4.26)

Therefore, it follows from (4.25) and (4.26) that

$\|w_{x}\|\rightarrow 0,\ \ {\rm as}\ t\rightarrow 0.$ (4.27)

From the Sobolev inequality, the desired asymptotic behavior in Theorem 4.1
can be obtained as

$\begin{cases}\sup\limits_{x\in\mathbbm{R}_{+}}|w(x,t)|\leq\sqrt{2}\|w\|^{\frac{1}{2}}\|w_{x}\|^{\frac{1}{2}}\rightarrow
0,\ \ {\rm as}\ t\rightarrow 0,\\\\[11.38109pt]
\sup\limits_{x\in\mathbbm{R}_{+}}|w_{x}(x,t)|\leq\sqrt{2}\|w_{x}\|^{\frac{1}{2}}\|w_{xx}\|^{\frac{1}{2}}\rightarrow
0,\ \ {\rm as}\ t\rightarrow 0.\end{cases}$ (4.28)

The combination of (4.27) and (4.28) completes the proof of Theorem 4.1.

Here, we give the proof of (4.26). In fact, from $\eqref{eq2.82}$, combining
$\eqref{wtjxyfjjg}$, we can conclude that

$\displaystyle\int_{0}^{\infty}\left|\frac{\mathrm{d}}{\mathrm{d}t}\|w_{x}\|^{2}\right|\,\mathrm{d}t$
$\displaystyle\leq
C\int_{0}^{\infty}\int_{\mathbbm{R}_{+}}(|w_{x}^{3}|+|\bar{u}_{xx}ww_{x}|)\,\mathrm{d}x\mathrm{d}t$
$\displaystyle\leq
C\int_{0}^{\infty}(\|w_{x}(t)\|^{2}+\|\sqrt{\bar{u}_{x}}w(t)\|^{2})\,\mathrm{d}t$
$\displaystyle<\infty.$

Thus, we finish the proof of (LABEL:wtjdsjxw)1 in Theorem 4.1. Finally,
according to $\eqref{wtjwzbj}$ and $\eqref{eq2.8}_{2}$, we set
$g=-u_{-}w_{x}(0,t)$, $f(x)=-w_{x}(x,t)$ for any fixed $t\in[0,\infty)$ in
Lemma 2.12. Then, by employing $\eqref{wtjLwq}$, we can obtain the asymptotic
behavior of $z$, which completes the proof of Theorem 4.1.

##  5 Asymptotics to Superposition of Nonlinear Waves

###  5.1 Reformulation of the Problem in the Case of $u_{-}<0<u_{+}$

Referring to the preceding sections, we set

$\Phi_{3}(x,t):=\bar{u}_{2}(x,t)+\tilde{u}_{4}(x,t),\ \ \ \
\Psi_{3}(x,t):=\bar{q}_{2}(x,t)+\tilde{q}_{4}(x,t),$

as an asymptotic state as $t\rightarrow\infty$, where
$(\tilde{u}_{4},\tilde{q}_{4})$ and $(\bar{u}_{2},\bar{q}_{2})$ are given in
Lammas 2.1 and 2.3, respectively. For simplicity, $\bar{u}_{2}$ and
$\tilde{u}_{4}$ are denoted by $\bar{u}$ and $\tilde{u}$, respectively. The
perturbation

$\begin{cases}w(x,t)=u(x,t)-\Phi_{3}(x,t)=u(x,t)-\bar{u}(x)-\tilde{u}(x,t),\\\\[2.84526pt]
z(x,t)=q(x,t)-\Psi_{3}(x,t)=q(x,t)-\bar{q}(x)+\tilde{u}_{x}(x,t),\\\
\end{cases}$

satisfies the reformulated problem

$\begin{cases}w_{t}+(\Phi_{3}w)_{x}+ww_{x}+z_{x}=-\bar{u}\tilde{u}_{x}-\tilde{u}\bar{u}_{x},\\\\[2.84526pt]
-z_{xx}+z+w_{x}=-\tilde{u}_{xxx},\\\\[2.84526pt] w(0,t)=0,\\\\[2.84526pt]
w(x,0)=w_{0}(x)=u_{0}(x)-\bar{u}(x)-\tilde{u}(x,0).\end{cases}$ (5.1)

We seek the solutions of (5.1) in the set of functions $X_{3}(0,T)$ defined by

$X_{3}(0,T)=\left\\{(w,z)|w\in C^{0}([0,T);H^{2})\cap C^{1}([0,T);H^{1}),z\in
C^{0}([0,T);H^{3})\cap L^{2}(0,T;H^{3})\right\\}.$

Firstly, we state the global existence and uniform stability result for the
reformulated problem (5.1).

######  Theorem 5.1.

Suppose that the boundary condition and far field states satisfy
$u_{-}<0<u_{+}$, the initial data $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and the
wavelength $\delta=|u_{-}-u_{+}|$ are sufficiently small. Then there are two
positive constants $\varepsilon_{1}$ and $C=C(\varepsilon_{1})$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$, the problem (5.1) admits a unique
solution $(w(x,t),z(x,t))\in X_{3}(0,+\infty)$ satisfying

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{2}^{2}+\delta^{\frac{1}{3}}),$

and the asymptotic behavior

$\begin{split}&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}w(x,t)|\rightarrow
0\ \ \text{as}\ \ t\rightarrow\infty,\ \ k=0,1,\\\\[8.53581pt]
&\sup_{x\in\mathbbm{R}_{+}}|\partial_{x}^{k}z(x,t)|\rightarrow 0\ \ \text{as}\
\ t\rightarrow\infty,\ \ k=0,1,2.\end{split}$ (5.2)

The combination of the local existence and the _a priori_ estimates proves
Theorem 5.1.

######  Proposition 5.2 (Local existence).

Suppose the boundary condition and far field states satisfy $u_{-}<0<u_{+}$,
the initial data satisfies $w_{0}\in H^{2}(\mathbbm{R}_{+})$ and
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$. Then there are two positive
constants $C=C(\varepsilon_{1})$ and $T_{0}=T_{0}(\varepsilon_{1})$ such that
the problem (5.1) has a unique solution $(w,z)\in X_{3}(0,T_{0})$, which
satisfies

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{2}^{2}+\delta^{\frac{1}{3}}),$

for $t\in[0,T_{0}]$.

######  Proposition 5.3 (A priori estimates).

Let $T$ be a positive constant. Suppose that the problem (5.1) has a unique
solution $(w,z)\in X_{3}(0,T)$. Then there exists positive constants
$\varepsilon_{2}(\leq\varepsilon_{1})$ and $C=C(\varepsilon_{2})$ such that if
$\|w_{0}\|_{2}+\delta\leq\varepsilon_{2}$, then we have the estimate

$\|w(t)\|_{2}^{2}+\|z(t)\|_{3}^{2}+\int_{0}^{t}\
(\|w_{x}(\tau)\|_{1}^{2}+\|z(\tau)\|_{3}^{2})\,d{\tau}\leq
C(\|w_{0}\|_{2}^{2}+\delta^{\frac{1}{3}}),$

for $t\in[0,T]$.

###  5.2 _A priori_ Estimates

Under the assumptions of Theorem 5.1, to show the _a priori_ estimate in
Proposition 5.3, we devote ourselves to the estimates on the solution
$(w,z)\in X_{3}(0,T)$ (for some $T>0$) of (5.1) under the _a priori_
assumption

$|w_{x}(t)|_{\infty}\leq\varepsilon_{0},$ (5.3)

where $0<\varepsilon_{0}\ll 1$. For simplicity, we divide the proof of the _a
priori_ estimate into several lemmas.

######  Lemma 5.4.

There are positive constants $\varepsilon_{1}(\leq\varepsilon_{0})$ and
$C=C(\varepsilon_{1})$ such that if $\|w_{0}\|_{2}+\delta\leq\varepsilon_{1}$,
then

$\displaystyle\|w(t)\|_{1}^{2}+$
$\displaystyle\int_{0}^{t}\left(\|\sqrt{\Phi_{3x}}w(\tau)\|^{2}+\|\sqrt{\Phi_{3x}}w_{x}(\tau)\|^{2}+\|z(\tau)\|_{2}^{2}\right)\,\mathrm{d}\tau+\int_{0}^{t}w_{x}^{2}(0,\tau)\,\mathrm{d}\tau\leq
C(\|w_{0}\|_{1}^{2}+\delta^{\frac{1}{3}})$

holds for $t\in[0,T]$.

Proof. Multiplying (5.1)1 by $w$ and (5.1)2 by $z$, and adding the two
resulting equations up, we obtain

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}w^{2}+\frac{1}{2}\Phi_{3x}w^{2}+z_{x}^{2}+z^{2}+\left\\{\frac{1}{2}\Phi_{3}w^{2}+\frac{1}{3}w^{3}-z_{x}z+zw\right\\}_{x}=-\bar{u}\tilde{u}_{x}w-\tilde{u}\bar{u}_{x}w-\tilde{u}_{xxx}z.$
(5.4)