Global Strong Solutions to the Vacuum Free Boundary Problem for 1D Liquid Crystal Flow with Degenerate Viscosity

Abstract

In this paper, we consider the one-dimensional liquid crystal flow with a vacuum free boundary and a degenerate viscosity coefficient. The global existence and long-time dynamics of strong solutions are established under a smallness condition on the initial energy at the basic level. The main challenges come from the degeneracy near the moving boundary and the strong nonlinear coupling between the velocity and the director field. To overcome these, we obtain uniform-in-time and space point-wise bounds of the deformation variable, and we construct uniform-in-time weighted energy estimates via singular multiplier techniques. Unlike previous works, the density is allowed to vanish and the viscosity coefficient is taken to be density-dependent rather than constant.

1. Introduction

The behaviors of nematic liquid crystal flows, with gravitational force and free boundary, in one dimension are described by the following equations [1]:

$$\left\{\begin{array}{c}{\rho }_t+{\left(\rho u\right)}_x=0,\hfill \\ {\left(\rho u\right)}_t+{\left(\rho u^2\right)}_x+p{\left(\rho \right)}_x+\alpha \left(\right|{\mathbf{n}}_x{{|}^2)}_x={\left(\lambda \left(\rho \right)u_x\right)}_x-\rho g,\hfill \\ {\mathbf{n}}_t+u{\mathbf{n}}_x=\beta ({\mathbf{n}}_{xx}+|{\mathbf{n}}_x{|}^2\mathbf{n}),\hfill \\ u(0,t)={\mathbf{n}}_x(0,t)=0,\hfill \\ \rho (\Gamma \left(t\right),t)={\mathbf{n}}_x(\Gamma \left(t\right),t)=0,\hfill \\ \left[p\left(\rho \right)-\lambda \left(\rho \right)u_x\right](\Gamma \left(t\right),t)=0,\hfill \\ (\rho ,u,\mathbf{n})(x,0)=({\rho }_0,u_0,{\mathbf{n}}_0),\hfill \end{array}\right.$$

(1)

where spatial and temporal variables $(x,t)\in (0,\Gamma (t\left)\right)\times (0,+\infty )$, $\Gamma \left(t\right)$ is the moving boundary, and unknown functions $(\rho ,u,\mathbf{n})\in \mathbb{R}\times \mathbb{R}\times {\mathbb{R}}^3$ represent the density, the velocity and the director field, respectively. The pressure p satisfies

$$p\left(\rho \right)=A{\rho }^{\gamma },$$

where $\gamma >1$ is the adiabatic component; the viscosity coefficient

$$\lambda \left(\rho \right)=\mu {\rho }^{\theta },$$

where $\theta >0$ is a constant.

We make several remarks on the system (1) from a modeling viewpoint. The director $\mathbf{n}$ describes the macroscopic average of the orientation field of the rod-like molecules of the nematic liquid crystal, and it is a unit vector, i.e., $\left|\mathbf{n}\right|=1$. The constant $\alpha >0$ measures the competition between the kinetic and the elastic energy, and the constant $\beta >0$ stands for the relaxation time of the director field (see [1]). The moving boundary $\Gamma \left(t\right)$ is the interface between the liquid crystal material and the vacuum. The density-dependent viscosity $\lambda \left(\rho \right)=\mu {\rho }^{\theta }$ degenerates at the vacuum boundary; such a degeneracy has been widely used in the literature on compressible flows with vacuum; see [2,3] and the references therein.

We assume that the initial density satisfies the following conditions:

$${\rho }_0>0\mathrm{on}[0,l_0),{\rho }_0\left(l_0\right)=0,-\infty <{\left({\rho }_0^{\gamma -1}\right)}_x\left(l_0\right)<0\mathrm{and}{\int }_0^{l_0}{\rho }_0\left(x\right)dx=M.$$

(2)

So, ${\rho }_0^{\gamma -1}\sim l_0-x$ as x close to $l_0$, which means the initial sound speed is $C^{1/2}$-Hölder continuous across the vacuum boundary. This corresponds to the so-called physical vacuum singularity. The conservation of mass, that is, Equation ${\left(1\right)}_1$, gives that

$${\int }_0^{\Gamma \left(t\right)}\rho (x,t)dx={\int }_0^{l_0}{\rho }_0\left(x\right)dx=M.$$

The steady solution $(\rho ,u,\mathbf{n})=(\overline{\rho },0,\overline{\mathbf{n}})$ of (1) with finite mass M is defined uniquely by

$${\left({\overline{\rho }}^{\gamma }\right)}_x=-\overline{\rho }g\mathrm{and}{\int }_0^{\infty }\overline{\rho }\left(x\right)dx=M.$$

Clearly,

$${\overline{\rho }}^{\gamma -1}\left(x\right)={\displaystyle \frac{g(\gamma -1)}{\gamma }}(\overline{l}-x)\mathrm{for}0\le x\le \overline{l},\mathrm{where}\overline{l}={\displaystyle \frac{\gamma }{(\gamma -1)g}}{\left(Mg\right)}^{{\displaystyle \frac{\gamma -1}{\gamma }}}.$$

(3)

The mathematical analysis of nematic liquid crystal flows, governed by the Ericksen–Leslie system, is essential for understanding the complex coupling between fluid motion and molecular orientation. Early mathematical results focused on the incompressible case with Lin and Liu [4] proving the global existence of weak solutions for a simplified model. The transition to the compressible regime posed greater difficulties due to the strong coupling between density, velocity, and the director field. The first breakthrough for the compressible NLC system with vacuum was achieved by Ding, Huang, and Xia [5], who proved the global existence of weak solutions for the Cauchy problem in ${\mathbb{R}}^3$. For more related studies with different boundary conditions in three dimensions, we refer to [6,7].

Owing to the significant applications, the theory governing free boundary problems in compressible fluids has been extensively studied by mathematicians and physicists in areas such as gas dynamics, nematic liquid crystal flows, and so on.

When the density of fluid always admits a positive lower bound, fruitful important progress is achieved; for instance, see [8,9,10,11,12,13,14,15,16] for the results of isentropic or non-isentropic Navier–Stokes equations over the last few decades.

For one-dimensional nematic liquid crystal equations in an isentropic regime, the global classical solution of the free boundary problem was established by Ding, Huang and Xia [17] with the assumption that the initial density is away from a vacuum.

In this paper, our work focuses on the scenario where the fluid connects to the vacuum continuously, leading to a strong singularity near the free boundary. While numerous local-in-time results exist for smooth solutions to compressible hydrodynamic equations (cf. [18,19,20]), the study of global-in-time behavior—particularly the existence of global weak or strong solutions for the Navier–Stokes equations and related models—has been addressed in works such as [2,3,21,22,23,24,25,26,27,28,29] and references therein.

For the Ericksen–Leslie equations with a vacuum free boundary, the first rigorous result on the global existence of weak solutions was established by Ding and Huang [1]. Subsequently, in their later work [30], they studied the non-isothermal case and proved the existence of global classical solutions for large initial data. In both papers, however, the viscosity coefficients are taken as constants rather than depending on the density. The global well-posedness of strong solutions for the Navier–Stokes equations with degenerate viscosity has been investigated by Ou and Zeng [3]. Nevertheless, when the director field is included, the analysis becomes considerably more difficult.

To our best knowledge, the global existence of strong solutions for vacuum free boundary problems of Ericksen–Leslie equations with degenerate viscosity, in particular with long-time convergence, is still open due to the difficulties arising from the effects of the degeneracy of the system near the vacuum free boundary and the strong coupling of the director field and the velocity field.

In this paper, we obtained the global existence, uniqueness and long-time stability of strong solutions to the free boundary problems of Ericksen–Leslie equations with density-dependent viscosity coefficients when the fluid always connects to vacuum continuously. Furthermore, we proved the time decay behavior of the solution toward the steady solution as the time approaches infinity. The main difficulties are that the system is degenerate near the free boundary, and the director field is nonlinearly coupled with the velocity field. To overcome these troubles, we established point-wise estimates and weighted estimates by singular multipliers. Previous results on the global solutions of free boundary problems of Ericksen–Leslie equations only care about the constant viscosity; in addition, they either require that the density of fluid admits a positive lower bound [17], or they lack the regularities and the long-time asymptotics [1].

In the study of free boundary problems, an important issue is to trace the free boundary, which requires at least that the velocity field is Lipschitz continuous in spatial variables. Motivated by [2,3], we take a method of flow trajectory to establish global-in-time weighted estimates so that the particle path, including the free boundary, is well defined in the classical sense. However, the system is degenerate near a free boundary; thus, the classical theory for parabolic equations cannot be applied directly to this problem. Moreover, the director field is strongly coupled with the velocity field and can be viewed as a source term which prevents the long-time stability of the solutions. To overcome the trouble, we first derive the point-wise upper and lower bounds for the deformation variable ${\eta }_x(x,t)$ based on the estimate of basic energy. It indeed dominates the evolution of the free boundary. Another ingredient is that we apply singular multipliers and Hardy’s inequality to establish the weighted estimates in the Lagrangian coordinates and then use interpolation inequalities to remove the weight and give the convergence of the strong solutions toward the steady state. Moreover, the higher-order estimates of solutions can be obtained by the lower order estimates and the Hardy inequality.

The structure of the paper is outlined below. Section 2 reformulates the free boundary problem (1) in Lagrangian coordinates, converting it into a fixed-boundary problem, and presents the main theorems. In Section 3, we introduce necessary preliminary estimates and technical tools. Section 4 is devoted to establishing uniform weighted energy estimates and decay rates of the solution, which leads to the global existence of strong solutions.

2. Lagrangian Reformulation and Main Results

We fix the free boundary by using Lagrangian variable $\eta (x,t)$, which is defined as follows:

$${\eta }_t(x,t)=u(\eta (x,t),t)\mathrm{for}t>0\mathrm{and}\eta (x,0)={\eta }_0\left(x\right),x\in (0,\overline{l}).$$

(4)

Here, ${\eta }_0:(0,\overline{l})\to (0,l_0)$ is a diffeomorphism defined by

$${\int }_0^{{\eta }_0\left(x\right)}{\rho }_0\left(y\right)dy={\int }_0^x\overline{\rho }\left(y\right)dy=M-g^{{\displaystyle \frac{1}{\gamma -1}}}{\left[{\displaystyle \frac{\gamma -1}{\gamma }}(\overline{l}-x)\right]}^{{\displaystyle \frac{\gamma }{\gamma -1}}},$$

which is

$${\rho }_0\left({\eta }_0\left(x\right)\right){\eta }_0^{\prime }\left(x\right)=\overline{\rho }\left(x\right)={\left[{\displaystyle \frac{g(\gamma -1)}{\gamma }}(\overline{l}-x)\right]}^{{\displaystyle \frac{1}{\gamma -1}}},$$

(5)

thus, ${\eta }_0$ is one to one and continuous due to (2) and (3); therefore, ${\eta }_0$ is well defined.

The Lagrangian version of the unknown functions are

$$\left(f,v,\mathbf{N}\right)(x,t)=\left(\rho ,u,\mathbf{n}\right)(\eta (x,t),t).$$

For simplicity, we suppose the constants $A=\mu =g=1$, and set $\overline{l}=1$ and $I:=(0,1)$ for convenience of presentation. Then, the Lagrangian version of system (1) with the initial-boundary conditions on the fixed domain is

$$\left\{\begin{array}{ccc}{f}_t+f{\displaystyle \frac{v_x}{{\eta }_x}}=0\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\hfill \\ f{v}_t+{\displaystyle \frac{{\left(f^{\gamma }\right)}_x}{{\eta }_x}}+{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_x={\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{f^{\theta }{v}_x}{{\eta }_x}}\right)}_x-f\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\hfill \\ {\mathbf{N}}_t={\displaystyle \frac{1}{{\eta }_x}}\left[{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N}\right]\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\hfill \\ f>0\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\\ v(0,t)=0,(f^{\theta }{v}_x-f^{\gamma })(1,t)=0\hfill & \hfill \mathrm{for}& t\ge 0,\hfill \\ {\mathbf{N}}_x(0,t)={\mathbf{N}}_x(1,t)=0\hfill & \hfill \mathrm{for}& t\ge 0,\hfill \\ (f,v,\mathbf{N})(x,0)=({\rho }_0,u_0,{\mathbf{n}}_0)\left(x\right)\hfill & \hfill \mathrm{on}& I,\hfill \end{array}\right.$$

(6)

where we still use the same notations for the initial data. We should note that ${\eta }_x>0$ for all $(x,t)\in I\times [0,T]$, which can be verified in the later estimate. Solving ${\left(6\right)}_1$, it follows that

$$f(x,t)={\displaystyle \frac{\overline{\rho }\left(x\right)}{{\eta }_x(x,t)}},(x,t)\in I\times [0,T].$$

(7)

Therefore, the system (6) can be rewritten as

$$\left\{\begin{array}{ccc}\overline{\rho }{v}_t+{\left[{\left({\displaystyle \frac{\overline{\rho }}{{\eta }_x}}\right)}^{\gamma }-{\overline{\rho }}^{\gamma }\right]}_x+{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_x={\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x}{{\eta }_x^{1+\theta }}}\right)}_x\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\hfill \\ {\eta }_x{\mathbf{N}}_t={\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N}\hfill & \hfill \mathrm{in}& I\times (0,\infty ),\hfill \\ v(0,t)={\mathbf{N}}_x(0,t)={\mathbf{N}}_x(1,t)=0\hfill & \hfill \mathrm{on}& [0,\infty ),\hfill \\ (\eta ,v,\mathbf{N})(x,0)=({\eta }_0,u_0,{\mathbf{n}}_0)\left(x\right)\hfill & \hfill \mathrm{on}& I,\hfill \end{array}\right.$$

(8)

since $\overline{\rho }\left(x\right)$ is the steady-state solution satisfying $\overline{\rho }\left(1\right)=0$. The boundary condition $({\overline{\rho }}^{\theta }{\eta }_x^{-\theta }{v}_x-{\overline{\rho }}^{\gamma }{\eta }_x^{-\gamma })(1,t)=0$ is satisfied automatically provided that $v_x$ is bounded.

Now, we introduce the definition of a strong solution to $\left(8\right)$.

Definition 1.

We say $\left(\eta \right(x,t),v(x,t),\mathbf{N}(x,t\left)\right)$ is a strong solution of $\left(8\right)$ in $I\times [0,T]$ provided that

(1) 

(8) holds for $(x,t)\in I\times [0,T]$, a.e.;

(2) 

$\eta (x,t)\in C^1([0,T];L^{\infty }\left(I\right))$, $v(x,t)\in C([0,T];L^{\infty }\left(I\right))$, $\mathbf{N}(x,t)\in C([0,T];H^1\left(I\right))$

${\eta }_x(x,t),v_x(x,t),{\mathbf{N}}_x(x,t)\in L^{\infty }(0,T;L^{\infty }\left(I\right));$

(3) 

$\sqrt{\overline{\rho }}{v}_t$, ${\mathbf{N}}_t$, ${\left({\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)\right)}_x$, ${\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x$, ${\left({\eta }_x^{-1}{\mathbf{N}}_x\right)}_x$∈$L^{\infty }(0,T;L^2\left(I\right))$;

(4) 

$v(0,t)=0$ holds in the sense of $W^{1,\infty }$-trace and ${\mathbf{N}}_x(x,t){|}_{x=0,1}=0$ hold in the sense that $\underset{ϵ\to 1}{lim}\parallel {\mathbf{N}}_x{(·,t)\parallel }_{L^{\infty }(ϵ,1)}=\underset{ϵ\to 0}{lim}{\parallel {\mathbf{N}}_x(·,t)\parallel }_{L^{\infty }(0,ϵ)}=0$ a.e., for $t\in [0,T]$;

(5) 

${\eta }_t(x,t)=v(x,t)$, a.e. $(x,t)\in I\times [0,T]$.

Notation 1.

We denote

$$\begin{array}{cc}\hfill {\mathcal{E}}_L\left(t\right):=& {\int }_I\left[{\displaystyle \frac{1}{2}}\overline{\rho }{v}^2+{\overline{\rho }}^{\gamma }\left({\displaystyle \frac{1}{\gamma -1}}{\eta }_x^{1-\gamma }+{\eta }_x-{\displaystyle \frac{\gamma }{\gamma -1}}\right)+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\right]dx\left(t\right),\hfill \\ \hfill {\mathcal{E}}_H\left(t\right):=& \parallel {\eta }_x{(·,t)-1\parallel }_{L^{\infty }\left(I\right)}^2+\parallel v_x{(·,t)\parallel }_{L^{\infty }\left(I\right)}^2+{\parallel {\mathbf{N}}_x(·,t)\parallel }_{L^{\infty }\left(I\right)}^2+{\int }_I\left(\overline{\rho }{v}_t^2+{\left|{\mathbf{N}}_t\right|}^2\right)(x,t)dx\hfill \\ & +{\int }_I\left[{\overline{\rho }}^{2\gamma -1}{\eta }_{xx}^2+|{\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x{|}^2+|{\left({\eta }_x^{-1}{\mathbf{N}}_x\right)}_x{|}^2\right](x,t)dx,\hfill \end{array}$$

thus

$$\mathcal{E}\left(t\right):={\mathcal{E}}_L\left(t\right)+{\mathcal{E}}_H\left(t\right).$$

The quantity ${\mathcal{E}}_L\left(t\right)$ is the basic energy of the flow. The three terms in the integrand represent the kinetic energy, the internal energy (which is non-negative by Taylor’s expansion, see Remark 2), and the Oseen–Frank elastic energy of the director field, respectively. The quantity ${\mathcal{E}}_H\left(t\right)$ collects higher-order Sobolev norms, which are needed to capture the regularity of the strong solution near the vacuum boundary.

Now, we are ready to state the main results.

Theorem 1.

Let $\overline{\rho }$ be the stationary solution defined by (5) and let $\gamma ,\theta$ satisfy

$$\gamma >1,0<\theta \le min\{{\displaystyle \frac{\gamma }{2}},\gamma -1\}.$$

(9)

Assume that

(1) 

The initial density ${\rho }_0$ satisfies (2);

(2) 

The compatibilities $u_0\left(0\right)=0$, ${\mathbf{n}}_{0x}\left(0\right)={\mathbf{n}}_{0x}\left(1\right)=0$ hold;

(3) 

$\mathcal{E}\left(0\right)$ is finite;

(4) 

The initial data satisfy

$$\begin{array}{cc}\hfill \parallel {\eta }_{0x}^{-\theta }{\left(x\right)-1\parallel }_{L^{\infty }\left(I\right)}& +2\theta {\left(Mg\right)}^{-\theta /\gamma }\sqrt{M{\mathcal{E}}_L\left(0\right)}\hfill \\ \hfill +& 4{\left({\displaystyle \frac{\gamma }{g(\gamma -1)}}\right)}^{{\displaystyle \frac{\theta }{\gamma -1}}}(1+2^{1/\theta }){\mathcal{E}}_L\left(0\right)\le 1/3;\hfill \end{array}$$

(10)

Then, the problem (8) admits a unique strong solution in $I\times [0,\infty )$ satisfying that for any $\iota \in (0,(\gamma -1)/4)$,

$$\begin{array}{cc}\hfill \mathcal{E}\left(t\right)& +{\int }_I\left[{\overline{\rho }}^{\theta -\kappa +\iota }+(1+t){\overline{\rho }}^{\theta }\right]{({\eta }_x-1)}^2(x,t)dx+{\int }_I\left[{\overline{\rho }}^{2\gamma -1}+\left(1+t\right){\overline{\rho }}^{3\gamma -\theta -1}\right]{\eta }_{xx}^2(x,t)dx\hfill \\ & +(1+t){\int }_I\left(\overline{\rho }{v}_t^2+v_x^2+v^2+{\left|{\mathbf{N}}_t\right|}^2\right)(x,t)dx+{\int }_0^t(1+s){\int }_I\left({\overline{\rho }}^{\theta }{v}_{tx}^2+{\left|{\mathbf{N}}_{tx}\right|}^2\right)(x,s)dxds\hfill \\ & +{\int }_0^t{\int }_I\left[{\overline{\rho }}^{\gamma -\kappa +\iota }+(1+s){\overline{\rho }}^{\gamma }\right]{({\eta }_x-1)}^2(x,s)dxds\hfill \\ & +{\int }_0^t{\int }_I\left[{\overline{\rho }}^{3\gamma -\theta -1}+(1+s){\overline{\rho }}^{4\gamma -2\theta -1}\right]{\eta }_{xx}^2(x,t)dxds\hfill \\ & +{(1+t)}^{{\displaystyle \frac{1}{\gamma -\theta }}}{\int }_I|{\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x{|}^2(x,t)dx+(1+t){\int }_I|{\left({\eta }_x^{-1}{\mathbf{N}}_x\right)}_x{|}^2(x,t)dx\hfill \\ & +{(1+t)}^{{\displaystyle \frac{\kappa -\theta -\iota }{\gamma +\kappa -\theta -\iota }}}{\int }_I{({\eta }_x-1)}^{2}dx+{(1+t)}^{{\displaystyle \frac{min\{2\gamma -2,\gamma \}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}{\int }_I{(\eta -x)}^{2}dx\hfill \\ & +{(1+t)}^{{\displaystyle \frac{min\{\gamma -1,\gamma /2\}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}{\parallel (\eta -x)(·,t)\parallel }_{L^{\infty }\left(I\right)}^2+(1+t){\parallel {\mathbf{N}}_x(·,t)\parallel }_{L^{\infty }\left(I\right)}^2\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^4\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

(11)

where $\kappa =min\{\gamma ,\theta +(\gamma -1\left)\right\}$.

Remark 1.

The formal “initial data” for $v_t(x,0)$ and ${\mathbf{N}}_t(x,0)$ are defined from the equations as

$$v_t(x,0):={\overline{\rho }}^{-1}\left[{\left({\overline{\rho }}^{\theta }{\eta }_{0x}^{-1-\theta }{u}_{0x}\right)}_x-{\left({\overline{\rho }}^{\gamma }\left({\eta }_{0x}^{-\gamma }-1\right)\right)}_x-{\mathbf{b}}_0·{\mathbf{b}}_{0x}\right],$$

$${\mathbf{N}}_t(x,0):={\eta }_{0x}^{-1}\left[{\left({\eta }_{0x}^{-1}{\mathbf{n}}_{0x}\right)}_x+{\eta }_{0x}^{-1}{\left|{\mathbf{n}}_{0x}\right|}^2{\mathbf{n}}_0\right],$$

therefore, $\mathcal{E}\left(0\right)$ is well defined.

3. Preliminaries

To prove the theorem, the following Hardy-type inequalities will be often used.

Lemma 1.

(cf. [20]). Let $k>-(\gamma -1)$ be a given real number and let $w\left(x\right)$ be a function satisfying $w\left(0\right)=0$. Then, it holds that

$$\begin{array}{c}\hfill {\int }_I{\overline{\rho }}^k{w}^{2}dx\le C{\int }_I{\overline{\rho }}^{k+2(\gamma -1)}{w}_x^{2}dx,\end{array}$$

(12)

where C is a generic constant independent of w.

Lemma 2.

(cf. [3]). Let $\theta >0,\gamma >1$ and $w\left(x\right)$ be a function satisfying $w\left(0\right)=0$. Then, it holds that for any fixed $0\le \beta <min\{\gamma ,\theta +\gamma -1\},$

$$\begin{array}{c}\hfill {\int }_I{\overline{\rho }}^{1-\beta }{w}^{2}dx\le C{\int }_I{\overline{\rho }}^{\theta }{w}_x^{2}dx,\end{array}$$

(13)

where C is a generic constant independent of w.

Notation 2.

(1) Throughout the rest of paper, C will denote a positive constant which depends only on the parameters of the problem, γ, θ, g and M but does not depend on the initial data or time. They are referred to as universal and can change from one inequality to another one. Also, we use $C\left(\delta \right)$ to denote a certain positive constant depending on the quantity δ.

(2) We will employ the notation $a\lesssim b$ to denote $a\le Cb$, where C is the universal constant as defined above.

(3) In the rest of the paper, we will use the notations

$${\parallel ·\parallel }_{L^p}={:\parallel ·\parallel }_{L^p\left(I\right)}{\mathrm{and}\parallel ·\parallel }_{H^k}=:{\parallel ·\parallel }_{H^k\left(I\right)},$$

for any $1\le p\le +\infty$ and $1\le k<+\infty$.

4. Proof of Theorem 1

4.1. Lower Order Estimates

We first outline the strategy of the proof. The main step of the analysis is the point-wise control of the deformation variable ${\eta }_x$, which dominates the motion of the free boundary. To this end, we establish the basic energy identity for ${\mathcal{E}}_L\left(t\right)$ in Lemma 3 and then derive uniform upper and lower bounds for ${\eta }_x$ in Lemma 4. After that, weighted estimates of ${\eta }_x-1$ with time decay are established in Lemmas 5 and 7 by using singular multipliers in the spirit of [3]. The estimates of $v_t$ and ${\mathbf{N}}_t$ are obtained in Lemma 6.

To carry out the argument, we claim the following a priori assumption that

$$0<{\eta }_x(x,t)\le 1+2^{1/\theta },\forall (x,t)\in I\times (0,\infty ).$$

(14)

We will then verify that $|{\eta }_x-1|$ is small for all time by establishing the uniform boundness in time, and therefore the assumption (14) is reasonable.

Lemma 3.

Assume that (14) holds. Then, for any $t\in (0,\infty ),$ we have

$${\mathcal{E}}_L\left(t\right)+{\int }_0^t{\int }_I\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}+2{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)(x,s)dxds={\mathcal{E}}_L\left(0\right).$$

(15)

Proof. 

Integrating $v·{\left(8\right)}_1$ over I, we have

$$\begin{array}{cc}& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_I\overline{\rho }{v}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)v_{x}dx-{\int }_I{v}_x{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}dx+{\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}dx=0,\hfill \end{array}$$

(16)

Notice that

$$\begin{array}{c}\hfill -{\int }_I{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)v_{x}dx={\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{\gamma }\left({\displaystyle \frac{1}{\gamma -1}}\left({\eta }_x^{1-\gamma }-1\right)+{\eta }_x-1\right)dx,\end{array}$$

(17)

and

$$\begin{array}{c}\hfill -\int {\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}{v}_{x}dx=\int {\left|{\mathbf{N}}_x\right|}^2{\left({\displaystyle \frac{1}{{\eta }_x}}\right)}_{t}dx={\displaystyle \frac{d}{dt}}\int {\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx-2\int {\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{xt}}{{\eta }_x}}dx.\end{array}$$

(18)

We take the integration of ${\left(8\right)}_2·{\mathbf{N}}_t$ over I to obtain

$$\int {\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx=-\int {\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{xt}}{{\eta }_x}}dx+\int {\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N}·{\mathbf{N}}_{t}dx,$$

(19)

and since $\mathbf{N}$ is a unit vector, therefore

$$\mathbf{N}·{\mathbf{N}}_t={\displaystyle \frac{1}{2}}{\left(\right|\mathbf{N}|}^2{)}_t=0.$$

Thus, we substitute the above equalities (17)–(19) back to (16) to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I\left[{\displaystyle \frac{1}{2}}\overline{\rho }{v}^2+{\overline{\rho }}^{\gamma }\left({\displaystyle \frac{1}{\gamma -1}}{\eta }_x^{1-\gamma }+{\eta }_x-{\displaystyle \frac{\gamma }{\gamma -1}}\right)+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\right]dx\hfill \\ & +{\int }_I\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}+2{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)dx=0,\hfill \end{array}$$

(20)

which gives (15) immediately. □

Remark 2.

The particular form in the integral of (17) is indeed non-negative by Taylor’s expansion and the fact that ${\eta }_x-1$ is of order $o\left(1\right)$.

The following lemma provides the key results to handling the main nonlinearities of the estimates; it proves that the assumption in (14) is indeed valid. Consequently, we may remove assumption (14) for the rest of the proof. Note that ${\eta }_x$ determines the deformation of the free interval.

Lemma 4.

Assume that (9), (10) and (14) hold; then, for all $x\in I$ and $t>0$

$${(3/4)}^{1/\theta }\le {\eta }_x(x,t)\le {(3/2)}^{1/\theta }.$$

(21)

Proof. 

The key idea is to trace the evolution of ${\eta }_x^{-\theta }-1$, since its smallness is equivalent to the closeness of ${\eta }_x$ to 1. To this end, we rewrite ${\left(8\right)}_1$ as

$$\begin{array}{c}\hfill \overline{\rho }{v}_t+{\left[{\overline{\rho }}^{\gamma }\left({\eta }_x^{-\gamma }-1\right)\right]}_x+{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_x=-{\displaystyle \frac{1}{\theta }}{\left({\overline{\rho }}^{\theta }{\eta }_x^{-\theta }\right)}_{tx}.\end{array}$$

(22)

Integrating the above equation in $(x,1)\times (t_1,t_2)$, one obtains

$$\begin{array}{cc}\hfill {\eta }_x^{-\theta }(x,t_2)-1=& \left[{\eta }_x^{-\theta }(x,t_1)-1\right]+\theta {\overline{\rho }}^{-\theta }\left(x\right){\int }_x^1\overline{\rho }\left(y\right)v(y,s)dy{|}_{s=t_1}^{t_2}\hfill \\ & -\theta {\overline{\rho }}^{-\theta }\left(x\right){\int }_{t_1}^{t_2}{\overline{\rho }}^{\gamma }\left(x\right)\left({\eta }_x^{-\gamma }(x,s)-1\right)ds-\theta {\overline{\rho }}^{-\theta }\left(x\right){\int }_{t_1}^{t_2}{\left|{\mathbf{N}}_x\right|}^2(x,s)/{\eta }_x^2(x,s)ds,\hfill \end{array}$$

(23)

for any $0\le t_1<t_2\le T$ and $x\in I$.

Next, we fix a spatial point $x_0$ to prove that for any $t^{\prime }\in (0,T]$, where T is an arbitrary positive constant, ${\eta }_x(x_0,t^{\prime })$ will be bounded uniformly from above and below. The local existence shows that ${\eta }_x(x_0,t)$ is $C^1$ in time for $t\in [0,T^{\star }]$, where $T^{\star }$ represents a short existing time of the solution, so we firstly consider the case that $t^{\prime }\in (0,T^{\star }]$. Define

$$t_0:=\underset{0<t\le t^{\prime }}{sup}\left\{t:{\eta }_x^{-\theta }(x_0,t)-1=0\right\},$$

(24)

which implies that $t_0$ is the closest point less than $t^{\prime }$ that makes ${\eta }_x^{-\theta }(x_0,t_0)-1=0$; otherwise, if the set is empty, choose $t_0=0$. This definition shows that

$$-|{\eta }_{0x}^{-\theta }\left(x_0\right)-1|\le {\eta }_x^{-\theta }(x_0,t_0)-1\le |{\eta }_{0x}^{-\theta }\left(x_0\right)-1|.$$

If ${\eta }_x^{-\theta }(x_0,t^{\prime })-1\ge 0$, which is equivalent to ${\eta }_x(x_0,t^{\prime })\le 1$, it then follows that

$${\eta }_x^{-\gamma }(x_0,t)-1\ge 0\mathrm{for}t\in [t_0,t^{\prime }].$$

From (23), we have

$$\begin{array}{c}\hfill 0\le {\eta }_x^{-\theta }(x_0,t^{\prime })-1\le {\eta }_x^{-\theta }(x_0,t_0)-1+\theta {\overline{\rho }}^{-\theta }\left(x_0\right){\int }_{x_0}^1\overline{\rho }v(y,s)dy{|}_{s=t_0}^{t^{\prime }}.\end{array}$$

(25)

It shows from (5) and (15) that for all $x\in I$ and $t\ge 0$,

$$\begin{array}{cc}& {\overline{\rho }}^{-\theta }\left(x\right)\left|{\int }_x^1\overline{\rho }\left(y\right)v(y,t)dy\right|\le {\overline{\rho }}^{-\theta }\left(x\right){\left({\int }_x^1\overline{\rho }\left(y\right)dy\right)}^{1/2}{\left({\int }_0^1\overline{\rho }\left(y\right)v^2(y,t)dy\right)}^{1/2}\hfill \\ & \le {\overline{\rho }}^{-\theta }\left(x\right){\left({\int }_x^1\overline{\rho }\left(y\right)dy\right)}^{1/2}\sqrt{{\mathcal{E}}_L\left(0\right)}={\left[\overline{\rho }\left(x\right)\right]}^{\gamma /2-\theta }\sqrt{{\mathcal{E}}_L\left(0\right)}\hfill \\ & \le M^{-\theta /\gamma }\sqrt{M{\mathcal{E}}_L\left(0\right)},\hfill \end{array}$$

(26)

provided that $\theta \le \gamma /2$. Then, (25) and (26) imply that

$$\begin{array}{cc}\hfill 0\le {\eta }_x^{-\theta }(x_0,t^{\prime })-1\le & |{\eta }_{0x}^{-\theta }\left(x_0\right)-1|+2\theta M^{-\theta /\gamma }\sqrt{M{\mathcal{E}}_L\left(0\right)}.\hfill \end{array}$$

(27)

If ${\eta }_x^{-\theta }(x_0,t^{\prime })-1<0$, it is again obvious that

$${\eta }_x^{-\gamma }(x_0,t)-1<0\mathrm{for}t\in [t_0,t^{\prime }].$$

Similar to the derivation of (27), we can obtain

$$\begin{array}{cc}\hfill 0& >{\eta }_x^{-\theta }(x_0,t^{\prime })-1\hfill \\ & \ge {\eta }_x^{-\theta }(x_0,t_0)-1+\theta {\overline{\rho }}^{-\theta }\left(x_0\right){\int }_{x_0}^1\overline{\rho }\left(y\right)v(y,s)dy{|}_{s=t_0}^{t^{\prime }}-\theta {\overline{\rho }}^{-\theta }\left(x_0\right){\int }_{t_0}^{t^{\prime }}{\displaystyle \frac{|{\mathbf{N}}_x{|}^2(x_0,s)}{{\eta }_x^2(x_0,s)}}ds,\hfill \end{array}$$

(28)

We next estimate the last term on the right-hand side of (28), which comes from the director field. The key fact is that $\mathbf{N}·{\mathbf{N}}_x=0$, since $\mathbf{N}$ is a unit vector. Together with the director Equation ${\left(8\right)}_2$, this yields

$${\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x={\mathbf{N}}_x·{\mathbf{N}}_t.$$

Noticing from (5) that

$$\overline{\rho }\left(x\right)={\left[{\displaystyle \frac{\gamma -1}{\gamma }}(1-x)\right]}^{{\displaystyle \frac{1}{\gamma -1}}},$$

we deduce

$$\begin{array}{cc}& \underset{x\in [x_0,1)}{sup}{\displaystyle \frac{|{\mathbf{N}}_x{(x,t)|}^2}{{\eta }_x^2(x,t)}}=\underset{x\in [x_0,1)}{sup}|-{\int }_x^{1}2{\displaystyle \frac{{\mathbf{N}}_y(y,t)}{{\eta }_y(y,t)}}·{\left({\displaystyle \frac{{\mathbf{N}}_y(y,t)}{{\eta }_y(y,t)}}\right)}_{y}dy|\hfill \\ \hfill =& \underset{x\in [x_0,1)}{sup}|{\int }_x^{1}2{\mathbf{N}}_y(y,t)·{\mathbf{N}}_t(y,t)dy|\hfill \\ \hfill \le & \sqrt{2}\underset{x\in [x_0,1)}{sup}{\left({\int }_x^1{\displaystyle \frac{|{\mathbf{N}}_y{(y,t)|}^2}{{\eta }_y(y,t)}}dy\right)}^{1/2}{\left({\int }_x^{1}2{\eta }_y(y,t){\left|{\mathbf{N}}_t(y,t)\right|}^{2}dy\right)}^{1/2}\hfill \\ \hfill \le & \sqrt{2}\underset{x\in [x_0,1)}{sup}{\displaystyle \frac{|{\mathbf{N}}_x(x,t)|}{{\eta }_x(x,t)}}\underset{x\in [x_0,1)}{sup}{\left|{\eta }_x(x,t)\right|}^{1/2}\underset{x\in [x_0,1)}{sup}\left\{{(1-x)}^{1/2}\right\}{\left({\int }_{I}2{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^{1/2},\hfill \end{array}$$

which yields

$$\begin{array}{cc}\hfill \underset{x\in [x_0,1)}{sup}{\displaystyle \frac{|{\mathbf{N}}_x(x,t)|}{{\eta }_x(x,t)}}\le & \sqrt{2}\underset{x\in [x_0,1)}{sup}|{\eta }_x(x,t){|}^{1/2}{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{1/2}{\overline{\rho }}^{(\gamma -1)/2}\left(x_0\right){\left({\int }_{I}2{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^{1/2}\hfill \\ \hfill \le & \sqrt{2}{(1+2^{1/\theta })}^{1/2}{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{{\displaystyle \frac{\theta }{2(\gamma -1)}}}{\overline{\rho }}^{\theta /2}\left(x_0\right){\left({\int }_{I}2{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^{1/2},\hfill \end{array}$$

since $\theta \le \gamma -1$. Hence, for $x_0\in [0,1)$,

$$\begin{array}{cc}\hfill {\overline{\rho }}^{-\theta }\left(x_0\right){\displaystyle \frac{|{\mathbf{N}}_x(x_0,t){|}^2}{{\eta }_x^2(x_0,t)}}& \le {\overline{\rho }}^{-\theta }\left(x_0\right)\underset{x\in [x_0,1)}{sup}{\displaystyle \frac{|{\mathbf{N}}_x{(x,t)|}^2}{{\eta }_x^2(x,t)}}\hfill \\ & \le 2{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{{\displaystyle \frac{\theta }{\gamma -1}}}(1+2^{1/\theta }){\int }_{I}2{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx,\hfill \end{array}$$

(29)

and consequently

$${\int }_{t_0}^{t^{\prime }}{\overline{\rho }}^{-\theta }\left(x_0\right){\displaystyle \frac{|{\mathbf{N}}_x{|}^2(x_0,s)}{{\eta }_x^2(x_0,s)}}ds\le 2{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{{\displaystyle \frac{\theta }{\gamma -1}}}(1+2^{1/\theta }){\mathcal{E}}_L\left(0\right).$$

(30)

Substituting this result into (28), we have

$$\begin{array}{cc}\hfill {\eta }_x^{-\theta }(x_0,t^{\prime })-1\ge & -|{\eta }_{0x}^{-\theta }\left(x_0\right)-1|-2\theta M^{-\theta /\gamma }\sqrt{M{\mathcal{E}}_L\left(0\right)}\hfill \\ & -2{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{{\displaystyle \frac{\theta }{\gamma -1}}}(1+2^{1/\theta }){\mathcal{E}}_L\left(0\right).\hfill \end{array}$$

(31)

Due to the arbitrariness of $x_0$ and $t^{\prime }$, we derive from (27) and (31) to obtain for all $(x,t)\in I\times [0,T^{\star }]$

$$\begin{array}{cc}\hfill |{\eta }_x^{-\theta }(x,t)-1|\le & |{\eta }_{0x}^{-\theta }\left(x\right)-1|+2\theta M^{-\theta /\gamma }\sqrt{M{\mathcal{E}}_L\left(0\right)}\hfill \\ & +2{\left({\displaystyle \frac{\gamma }{\gamma -1}}\right)}^{{\displaystyle \frac{\theta }{\gamma -1}}}(1+2^{1/\theta }){\mathcal{E}}_L\left(0\right)\le 1/3,\hfill \end{array}$$

(32)

which is due to (10).

Therefore, we apply again the local existence results to the case that the state at $T^{\star }$ is chosen to be the initial state, to achieve that for another period of time $\widehat{T}>0$, provided that the strong solution exists for $t\in [0,T^{\star }+\widehat{T}]$. Then, we repeat the process we have shown above to prove that (32) holds for $(x,t)\in I\times [0,T^{\star }+\widehat{T}]$.

Finally, by iterating the argument, we can write

$$2/3\le {\eta }_x^{-\theta }(x,t)\le 4/3,\forall (x,t)\in I\times [0,T^{\star }+m\widehat{T}],$$

where m is a positive integer. By passing the limit of m to infinity, we then finish the proof of this lemma. □

Remark 3.

Lemmas 3 and 4, along with Taylor’s expansion, imply that

$${\int }_I\left[\overline{\rho }{v}^2+{\left|{\mathbf{N}}_x\right|}^2+{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^2\right](x,t)dx\lesssim {\mathcal{E}}_L\left(0\right),$$

which, together with (15), gives

$$\begin{array}{cc}& {\int }_I\left[\overline{\rho }{v}^2+{\left|{\mathbf{N}}_x\right|}^2+{\overline{\rho }}^{\gamma }{\left({\eta }_x-1\right)}^2\right](x,t)dx+{\int }_0^t{\int }_I\left({\overline{\rho }}^{\theta }{v}_x^2+|{\mathbf{N}}_t{|}^2\right)(x,s)dxds\lesssim {\mathcal{E}}_L\left(0\right).\hfill \end{array}$$

(33)

In order to obtain the time decay estimate of ${\eta }_x-1$ in some proper norm, we introduce the multiplier ${\int }_0^x({\eta }_y^{-\theta }-1)dy$ in the following lemma. This kind of multiplier was first introduced in [2] for the viscous gaseous star problem.

Lemma 5.

Assume (9) holds. Then, for $t\ge 0$,

$$\begin{array}{cc}& {\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^2(x,t)dx+(1+t){\int }_I\left[\overline{\rho }{v}^2+{\left|{\mathbf{N}}_x\right|}^2+{\overline{\rho }}^{\gamma }{\left({\eta }_x-1\right)}^2\right](x,t)dx\hfill \\ & +{\int }_0^t{\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^2(x,s)dxds+{\int }_0^t(1+s){\int }_I\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}+{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)(x,s)dxds\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)+{\int }_I{\overline{\rho }}^{\theta }{({\eta }_{0x}-1)}^2\left(x\right)dx.\hfill \end{array}$$

(34)

Proof. 

Multiplying ${\left(8\right)}_1$ by ${\int }_0^x({\eta }_y^{-\theta }-1)dy$, one obtains

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I\left[{\displaystyle \frac{1}{2\theta }}{\overline{\rho }}^{\theta }{\left({\eta }_x^{-\theta }-1\right)}^2-\overline{\rho }v{\int }_0^x\left({\eta }_y^{-\theta }-1\right)dy\right]dx+{\int }_I{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)({\eta }_x^{-\theta }-1)dx\hfill \\ \hfill =& -{\int }_I\left({\eta }_x^{-\theta }-1\right)\left(|{\mathbf{N}}_x{|}^2/{\eta }_x\right)dx+\theta {\int }_I\overline{\rho }v\left({\int }_0^x{\eta }_y^{-\theta -1}{v}_{y}dy\right)dx.\hfill \end{array}$$

(35)

Recall (21) and Taylor’s expansion to discover

$$({\eta }_x^{-\gamma }-1)({\eta }_x^{-\theta }-1)=\gamma \theta {({\eta }_x-1)}^2+o\left({({\eta }_x-1)}^3\right),$$

and

$${({\eta }_x^{-\theta }-1)}^2={\theta }^2{({\eta }_x-1)}^2+o\left({({\eta }_x-1)}^3\right).$$

Notice that

$$\begin{array}{cc}& \left|{\int }_I\overline{\rho }v\left({\int }_0^x\left({\eta }_y^{-\theta }-1\right)dy\right)dx\right|\hfill \\ \hfill \le & {\int }_I\overline{\rho }\left|v\right|{\left({\int }_0^x{\overline{\rho }}^{-\theta }dy\right)}^{1/2}{\left({\int }_0^x{\overline{\rho }}^{\theta }{\left({\eta }_y^{-\theta }-1\right)}^{2}dy\right)}^{1/2}dx\hfill \\ \hfill \le & C\left({\delta }^{-1}\right){\int }_I{\overline{\rho }}^2{v}^2\left({\int }_0^x{\overline{\rho }}^{-\theta }dy\right)dx+\delta {\int }_I\left({\int }_0^1{\overline{\rho }}^{\theta }{\left({\eta }_y^{-\theta }-1\right)}^{2}dy\right)dx\hfill \\ \hfill \le & C\left({\delta }^{-1}\right){\int }_I\overline{\rho }{v}^{2}dx+\delta {\int }_0^1{\overline{\rho }}^{\theta }{\left({\eta }_y^{-\theta }-1\right)}^{2}dy,\hfill \end{array}$$

for any positive constant $\delta$ and

$$\begin{array}{cc}& \left|{\int }_I\overline{\rho }v\left({\int }_0^x{\eta }_y^{-\theta -1}{v}_{y}dy\right)dx\right|\lesssim {\int }_I\overline{\rho }\left|v\right|{\left({\int }_0^x{\overline{\rho }}^{-\theta }dy\right)}^{1/2}{\left({\int }_0^x{\overline{\rho }}^{\theta }{v}_y^{2}dy\right)}^{1/2}dx\hfill \\ \hfill \lesssim & {\int }_I{\overline{\rho }}^2{v}^2\left({\int }_0^x{\overline{\rho }}^{-\theta }dy\right)dx+{\int }_I\left({\int }_I{\overline{\rho }}^{\theta }{v}_y^{2}dy\right)dx\lesssim {\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx,\hfill \end{array}$$

due to (5), (21), Lemma 2 and $\theta \le \gamma /2$. From (29) and (21), we have

$$\begin{array}{c}\hfill {\int }_I({\eta }_x^{-\theta }-1)|{\mathbf{N}}_x{|}^{2}dx\lesssim \parallel {\eta }_x^{-\theta }{-1\parallel }_{L^{\infty }}{\int }_I{\eta }_x|{\mathbf{N}}_t{|}^{2}dx\lesssim {\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx.\end{array}$$

Then, substitute the above three inequalities into (35) and integrate with respect to time to obtain

$${\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^2(x,t)dx+{\int }_0^t{\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dxds\lesssim {\int }_I\overline{\rho }{({\eta }_{0x}-1)}^{2}dx+{\mathcal{E}}_L\left(0\right).$$

(36)

It follows from (20) that

$${\displaystyle \frac{d}{dt}}\left[(1+t){\mathcal{E}}_L\left(t\right)\right]+(1+t){\int }_I\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}+2{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)dx={\mathcal{E}}_L\left(t\right),$$

together with Lemma 2, (29), (33) and (36), which gives

$$\begin{array}{cc}& (1+t){\int }_I\left[\overline{\rho }{v}^2+{\left|{\mathbf{N}}_x\right|}^2/{\eta }_x+{\overline{\rho }}^{\gamma }{\left({\eta }_x-1\right)}^2\right]dx\hfill \\ & +{\int }_0^t(1+s){\int }_I\left({\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +1}}}+2{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)dxds\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)+{\int }_0^t{\int }_I\left[\overline{\rho }{v}^2+{\left|{\mathbf{N}}_x\right|}^2/{\eta }_x+{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^2\right]dxds\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)+{\int }_0^t{\int }_I\left[{\overline{\rho }}^{\theta }{v}_x^2+{\eta }_x{\left|{\mathbf{N}}_t\right|}^2+{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^2\right]dxds\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)+{\int }_I{\overline{\rho }}^{\theta }{({\eta }_{0x}-1)}^{2}dx.\hfill \end{array}$$

(37)

So, we obtain (34) from (36) and (37). □

Lemma 6.

Assume (9) holds. Then, for $t\ge 0$,

$$\begin{array}{cc}& (1+t){\int }_I\left(\overline{\rho }{v}_t^2+v_x^2+v^2+{\left|{\mathbf{N}}_t\right|}^2\right)(x,t)dx+{\int }_0^t(1+s){\int }_I\left({\overline{\rho }}^{\theta }{v}_{tx}^2+{\left|{\mathbf{N}}_{tx}\right|}^2\right)(x,s)dxds\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(38)

Proof. 

Integrate ${\left(8\right)}_1$ in $(x,1)$ to obtain

$$\begin{array}{cc}\hfill |v_x|=& |{\overline{\rho }}^{\gamma -\theta }{\eta }_x^{\theta +1}({\eta }_x^{-\gamma }-1)-{\overline{\rho }}^{-\theta }{\eta }_x^{\theta +1}{\int }_x^1\overline{\rho }{v}_{t}dy-{\overline{\rho }}^{-\theta }{\left|{\mathbf{N}}_x\right|}^2|\hfill \\ \hfill \lesssim & {\overline{\rho }}^{\gamma -\theta }|{\eta }_x-1|+{\overline{\rho }}^{\gamma /2-\theta }{\left({\int }_I\overline{\rho }{v}_t^{2}dx\right)}^{1/2}+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill \lesssim & {\overline{\rho }}^{\gamma -\theta }|{\eta }_x-1|+{\left({\int }_I\overline{\rho }{v}_t^{2}dx\right)}^{1/2}+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx.\hfill \end{array}$$

(39)

We now turn to the estimate of ${\mathbf{N}}_t$ in $L^2$. Differentiate ${\left(8\right)}_2$ with respect to t to obtain

$$\begin{array}{c}\hfill {\eta }_x{\mathbf{N}}_{tt}+v_x{\mathbf{N}}_t={\left({\displaystyle \frac{{\mathbf{N}}_{tx}}{{\eta }_x}}\right)}_x-{\left({\displaystyle \frac{{\mathbf{N}}_x{v}_x}{{\eta }_x^2}}\right)}_x+2{\displaystyle \frac{{\mathbf{N}}_{tx}·{\mathbf{N}}_x}{{\eta }_x^2}}\mathbf{N}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}{\mathbf{N}}_t-{\displaystyle \frac{2|{\mathbf{N}}_x{|}^2\mathbf{N}{v}_x}{{\eta }_x^3}},\end{array}$$

(40)

multiply the above equation with ${\mathbf{N}}_t$ and integrate the product over I to obtain

$$\begin{array}{cc}& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx+{\int }_I{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill =& {\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_t\right|}^2}{{\eta }_x}}dx-{\int }_I{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}{v}_x}{{\eta }_x}}dx-{\displaystyle \frac{1}{2}}{\int }_I{v}_x{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill :=& I_1+I_2+I_3,\hfill \end{array}$$

(41)

where

$$I_1={\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_t\right|}^2}{{\eta }_x}}dx,I_2=-{\int }_I{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}{v}_x}{{\eta }_x}}dx,I_3=-{\displaystyle \frac{1}{2}}{\int }_I{v}_x{\left|{\mathbf{N}}_t\right|}^{2}dx.$$

Noticing (21), (29) and (33), we estimate the above three terms as follows,

$$\begin{array}{cc}\hfill I_1\lesssim & \parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }}^2{\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\lesssim {\left({\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2,\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill |I_2|\lesssim & \delta {\int }_I|{\mathbf{N}}_{tx}{|}^{2}dx+{\int }_I{v}_x^2{\left|{\mathbf{N}}_x\right|}^{2}dx\hfill \\ \hfill \lesssim & \delta {\int }_I|{\mathbf{N}}_{tx}{|}^{2}dx+{\parallel {\overline{\rho }}^{-\theta }{\mathbf{N}}_x\parallel }_{L^{\infty }}^2{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\hfill \\ \hfill \lesssim & \delta {\int }_I{\left|{\mathbf{N}}_{tx}\right|}^{2}dx+\left({\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)\left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\right),\hfill \end{array}$$

(43)

and by integration by parts,

$$\begin{array}{cc}\hfill |I_3|=& |{\int }_{I}v{\mathbf{N}}_t·{\mathbf{N}}_{tx}dx|\lesssim \delta {\int }_I|{\mathbf{N}}_{tx}{|}^{2}dx+{\int }_I{v}^2{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill \lesssim & \delta {\int }_I|{\mathbf{N}}_{tx}{|}^{2}dx+{\parallel v\parallel }_{L^{\infty }}^2{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill \lesssim & \delta {\int }_I{\left|{\mathbf{N}}_{tx}\right|}^{2}dx+\left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\right)\left({\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right).\hfill \end{array}$$

(44)

where, similar to (29), we can obtain

$${\parallel v\left|\right|}_{L^{\infty }}^2\lesssim {\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx.$$

Summarizing the above three terms and choosing $\delta$ small enough to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x|{\mathbf{N}}_t{|}^{2}dx+{\int }_I{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\lesssim \left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right){\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx,\hfill \end{array}$$

(45)

thus due to Gronwall’s inequality, (10) and (33), we obtain

$$\begin{array}{cc}& {\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx+{\int }_0^t{\int }_I{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dxds\hfill \\ \hfill \lesssim & \left[{\int }_0^t{\int }_I\left({\overline{\rho }}^{\theta }{v}_x^2+{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)dxds\right]\left({\int }_I{\eta }_{0x}{\left|{\mathbf{N}}_{0t}\right|}^{2}dx\right)exp\left\{C{\int }_0^t{\int }_I\left({\overline{\rho }}^{\theta }{v}_x^2+{\eta }_x{\left|{\mathbf{N}}_t\right|}^2\right)dxds\right\}\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)·\mathcal{E}\left(0\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}\hfill \\ \hfill \lesssim & \mathcal{E}\left(0\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}\hfill \end{array}$$

(46)

Take the time derivative of ${\left(8\right)}_1$ to have

$$\begin{array}{c}\hfill \overline{\rho }{v}_{tt}-\gamma \left({\overline{\rho }}^{\gamma }{\eta }_x^{-\gamma -1}{v}_x\right)={\left[{\overline{\rho }}^{\theta }\left({\displaystyle \frac{v_{tx}}{{\eta }_x^{1+\theta }}}-(\theta +1){\displaystyle \frac{v_x^2}{{\eta }_x^{\theta +2}}}\right)\right]}_x-{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_{xt},\end{array}$$

(47)

We multiply it by $v_t$, and integrate it with respect to x to obtain

$${\displaystyle \frac{d}{dt}}{\int }_I\overline{\rho }{v}_t^{2}dx+{\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dx=-\gamma {\int }_I{\overline{\rho }}^{\gamma }{\displaystyle \frac{v_x{v}_{tx}}{{\eta }_x^{\gamma +1}}}dx+(\theta +1){\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_x^2{v}_{tx}}{{\eta }_x^{\theta +2}}}dx+{\displaystyle \frac{1}{2}}{\int }_I{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_t{v}_{tx}dx.$$

which, together with (21) and (39), implies

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I\overline{\rho }{v}_t^{2}dx+{\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dx\hfill \\ \hfill \lesssim & \delta {\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dx+{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\left[1+{\left({\int }_I\overline{\rho }{v}_t^{2}dx\right)}^{1/2}+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right]}^2\left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\right)\hfill \\ & +{\int }_I|{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}{|}^2·|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_t{|}^{2}dx,\hfill \end{array}$$

(48)

choosing $\delta$ small enough, and noticing (29) to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I\overline{\rho }{v}_t^{2}dx+{\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dx\hfill \\ \hfill \lesssim & \left[1+{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\left({\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{v}_x^2}{{\eta }_x}}\right)dx\hfill \\ \hfill \lesssim & \left[1+{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\hfill \\ & +{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_{xt}\right|}^2}{{\eta }_x^2}}dx+\left[1+{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^2}}dx\hfill \\ \hfill \lesssim & {\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }}^2{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx+{\parallel {\mathbf{N}}_x\parallel }_{L^{\infty }}^2{\int }_I{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}dx\hfill \\ & +\left[{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]\left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\parallel {\mathbf{N}}_x\parallel }_{L^{\infty }}^2{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx\right)\hfill \\ \hfill \lesssim & \left[{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right){\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx\right]{\int }_I\overline{\rho }{v}_t^{2}dx\hfill \\ & +{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+\left({\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx+{\int }_I{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}dx\right){\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ & +{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^3{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx,\hfill \end{array}$$

(49)

which, by Gronwall’s inequality, (33) and (46), gives

$$\begin{array}{cc}& {\int }_I\overline{\rho }{v}_t^{2}dx+{\int }_0^t{\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dxds\lesssim \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(50)

We substitute (50) back into (39) to obtain

$$\parallel v_x{\parallel }_{L^{\infty }}\left(t\right)\lesssim \mathcal{E}\left(0\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},t\ge 0.$$

(51)

We now derive time decay estimates by introducing the time weight $(1+t)$. Multiply (41) with $(1+t)$ to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}(1+t){\int }_I{\displaystyle \frac{1}{2}}{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx+(1+t){\int }_I{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill =& (1+t){\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_t\right|}^2}{{\eta }_x}}dx-(1+t){\int }_I{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}{v}_x}{{\eta }_x}}dx-{\displaystyle \frac{1+t}{2}}{\int }_I{v}_x{\left|{\mathbf{N}}_t\right|}^{2}dx,\hfill \end{array}$$

integrate the above equality with respect to t, and notice (33), (34), (46) to obtain

$$\begin{array}{cc}& (1+t){\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx+{\int }_0^t(1+s){\int }_I{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dxds\hfill \\ \hfill \lesssim & {\int }_0^t{\int }_I{\eta }_x|{\mathbf{N}}_t{|}^{2}dxds+{\int }_0^t(1+s)\left({\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx+{\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds{\int }_I{\eta }_x{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill \lesssim & {\mathcal{E}}_L\left(0\right)+{\mathcal{E}}^2\left(0\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}\hfill \\ \hfill \lesssim & \left(\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

(52)

And similarly, from (49), we have

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}(1+t){\int }_I\overline{\rho }{v}_t^{2}dx+(1+t){\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dx\hfill \\ \hfill \lesssim & (1+t)\left[1+{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dx\hfill \\ & +(1+t){\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\left({\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{v}_x^2}{{\eta }_x}}\right)dx.\hfill \end{array}$$

Integrating the above inequality over $(0,t)$ together with (34), (46), (50) and (52), we obtain

$$\begin{array}{cc}& (1+t){\int }_I\overline{\rho }{v}_t^{2}dx+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\theta }{\displaystyle \frac{v_{tx}^2}{{\eta }_x^{\theta +1}}}dxds\hfill \\ \hfill \lesssim & {\int }_0^t{\int }_I\overline{\rho }{v}_t^{2}dxds+{\int }_0^t(1+s)\left[1+{\int }_I\overline{\rho }{v}_t^{2}dx+{\left({\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)}^2\right]{\int }_I{\overline{\rho }}^{\theta }{v}_x^{2}dxds\hfill \\ & +{\int }_0^t(1+s){\int }_I{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\left({\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{v}_x^2}{{\eta }_x}}\right)dxds\hfill \\ \hfill \lesssim & \left(\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right)exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(53)

It then follows from (39), (34) and (53) that

$$(1+t){\int }_I{v}_x^{2}dx\lesssim \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.$$

(54)

Combining (52)–(54), we obtain (38). □

Corollary 1.

From (29) and (52), it holds for $t>0$ that

$$\begin{array}{c}\hfill (1+t)\parallel {\mathbf{N}}_x{(·,t)\parallel }_{L^{\infty }\left(I\right)}^2\le \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\end{array}$$

(55)

Lemma 7.

Assume (9) holds. Then, for any $\iota \in (0,(\gamma -1)/4)$ and $t\ge 0$,

$$\begin{array}{cc}& {\int }_I\left[{\overline{\rho }}^{\theta -\kappa +\iota }+(1+t){\overline{\rho }}^{\theta }\right]{({\eta }_x-1)}^2(x,t)dx\hfill \\ & +{\int }_0^t{\int }_I\left[{\overline{\rho }}^{\gamma -\kappa +\iota }+{\overline{\rho }}^{\theta }+(1+s){\overline{\rho }}^{\gamma }\right]{({\eta }_x-1)}^2(x,s)dxds\hfill \\ & +{(1+t)}^{{\displaystyle \frac{\kappa -\theta -\iota }{\gamma +\kappa -\theta -\iota }}}{\int }_I{({\eta }_x-1)}^{2}dx+{(1+t)}^{{\displaystyle \frac{min\{2\gamma -2,\gamma \}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}{\int }_I{(\eta -x)}^{2}dx\hfill \\ & +{(1+t)}^{{\displaystyle \frac{min\{\gamma -1,\gamma /2\}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}{\parallel \eta (x,t)-x\parallel }_{L^{\infty }\left(I\right)}^2\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(56)

where $\kappa =min\{\gamma ,\theta +(\gamma -1\left)\right\}$.

Proof. 

We multiply ${\left(8\right)}_1$ by

$${\int }_0^x{\overline{\rho }}^{-\beta }\left(y\right)({\eta }_y^{-\theta }(y,t)-1)dy\mathrm{with}0\le \beta <\theta +(\gamma -1),$$

(57)

and integrate the product over I to have

$$\begin{array}{cc}& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{\theta -\beta }{\left({\eta }_x^{-\theta }-1\right)}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma -\beta }{\left({\eta }_x-1\right)}^{2}dx\hfill \\ \hfill =& {\int }_I\overline{\rho }{v}_t\left({\int }_0^x{\overline{\rho }}^{-\beta }\left({\eta }_y^{-\theta }-1\right)dy\right)dx+{\int }_I{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\right)}_x\left({\int }_0^x{\overline{\rho }}^{-\beta }\left({\eta }_y^{-\theta }-1\right)dy\right)dx,\hfill \end{array}$$

(58)

Then, we use the Cauchy inequality, (5), (21), (29) and the Taylor expansion to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{\theta -\beta }{\left({\eta }_x^{-\theta }-1\right)}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma -\beta }{({\eta }_x-1)}^{2}dx\hfill \\ \hfill \lesssim & {\int }_I\overline{\rho }{v}_t{\left({\int }_0^x{\overline{\rho }}^{-\gamma -\beta }\left(y\right)dy\right)}^{1/2}{\left[{\int }_I{\overline{\rho }}^{\gamma -\beta }{({\eta }_x-1)}^{2}dx\right]}^{1/2}dx\hfill \\ & +\parallel {\overline{\rho }}^{-\theta }|{\mathbf{N}}_x{|}^2{\parallel }_{L^{\infty }}{\parallel {\eta }_x^{-\theta }-1\parallel }_{L^{\infty }}{\int }_I{\overline{\rho }}^{\theta -\beta }\left(x\right)dx\hfill \\ \hfill \lesssim & {\int }_I{\overline{\rho }}^{1-\beta }{v}_t^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx.\hfill \end{array}$$

(59)

In view of (2), we then obtain for $0\le \beta <min\{\gamma ,\theta +(\gamma -1\left)\right\}$,

$${\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{\theta -\beta }{\left({\eta }_x^{-\theta }-1\right)}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma -\beta }{({\eta }_x-1)}^{2}dx\lesssim {\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx.$$

(60)

Note that for any $0<\iota \le (\gamma -1)/4$,

$$\iota -(\gamma -1)\le \theta -\kappa +\iota <0\mathrm{and}\gamma -\kappa +\iota >0.$$

We now apply (60) with two different choices of $\beta$. The first choice $\beta =\kappa -\iota$ yields a weighted estimate of ${\eta }_x-1$ with the singular weight ${\overline{\rho }}^{\theta -\kappa +\iota }$ but without time decay. The second choice $\beta =0$, combined with the time weight $(1+t)$, gives the time decay estimate.

Choosing $\beta =\kappa -\iota$ in (60), and then using (5) and (38), we obtain for $t\ge 0$,

$$\begin{array}{cc}& {\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^2(x,t)dx+{\int }_0^t{\int }_I{\overline{\rho }}^{\gamma -\kappa +\iota }{({\eta }_x-1)}^{2}dxds\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(61)

In a similar way, we multiply $(1+t)$ with (60) and set $\beta =0$ to obtain

$$\begin{array}{cc}& (1+t){\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^2(x,t)dx+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dxds\hfill \\ \hfill \lesssim & \mathcal{E}\left(0\right)+{\int }_0^t(1+s)\left({\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds\hfill \\ & +{\int }_0^t{\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^{2}dxds.\hfill \end{array}$$

(62)

To deal with the second term of the above inequality, we notice that

$${\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^{2}dx\le {\left({\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\gamma -\theta }{\gamma +\kappa -\theta -\iota }}}{\left({\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\kappa -\iota }{\gamma +\kappa -\theta -\iota }}},$$

then

$$\begin{array}{cc}& {\int }_0^t{\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^{2}dxds\hfill \\ \hfill \lesssim & {\int }_0^t{(1+s)}^{-1}{\left((1+s){\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\gamma -\theta }{\gamma +\kappa -\theta -\iota }}}\hfill \\ & \times {\left((1+s){\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\kappa -\iota }{\gamma +\kappa -\theta -\iota }}}ds\hfill \\ \hfill \lesssim & \delta {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dxds\hfill \\ & +{\int }_0^t{(1+s)}^{-{\displaystyle \frac{\gamma +\kappa -\theta -\iota }{\gamma -\theta }}+1}{\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dxds.\hfill \end{array}$$

(63)

We substitute (63) into (62) and choose $\delta$ small enough to obtain

$$\begin{array}{cc}& (1+t){\int }_I{\overline{\rho }}^{\theta }{({\eta }_x-1)}^2(x,t)dx+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dxds\hfill \\ \hfill \lesssim & \mathcal{E}\left(0\right)+{\int }_0^t(1+s)\left({\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds\hfill \\ & +{\int }_0^t{(1+s)}^{1-{\displaystyle \frac{\gamma +\kappa -\theta -\iota }{\gamma -\theta }}}{\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dxds\hfill \\ \hfill \lesssim & \mathcal{E}\left(0\right)+{\int }_0^t(1+s)\left({\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds\hfill \\ & +\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}{\int }_0^t{(1+s)}^{1-{\displaystyle \frac{\gamma +\kappa -\theta -\iota }{\gamma -\theta }}}ds\hfill \\ \hfill \lesssim & {\int }_0^t(1+s)\left({\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds+\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

(64)

since (61) and

$$1-{\displaystyle \frac{\gamma +\kappa -\theta -\iota }{\gamma -\theta }}=-{\displaystyle \frac{\kappa -\iota }{\gamma -\theta }}<-1.$$

therefore, by (37), (38) and (64), (56) is proved. Due to (34), we notice that

$$\begin{array}{cc}\hfill {\int }_I{({\eta }_x-1)}^{2}dx\le & {\left({\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\gamma }{\gamma +\kappa -\theta -\iota }}}{\left({\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{\kappa -\theta -\iota }{\gamma +\kappa -\theta -\iota }}}\hfill \\ \hfill \le & {(1+t)}^{-{\displaystyle \frac{\kappa -\theta -\iota }{\gamma +\kappa -\theta -\iota }}}\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}\hfill \end{array}$$

then with (1), we have

$$\begin{array}{cc}\hfill {\int }_I{(\eta -x)}^{2}dx\lesssim & {\int }_I{\overline{\rho }}^{2\gamma -2}{({\eta }_x-1)}^{2}dx\hfill \\ \hfill \lesssim & {(1+t)}^{-{\displaystyle \frac{min\{2\gamma -2,\gamma \}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

Since if $2\gamma -2\ge \gamma$,

$${\int }_I{(\eta -x)}^{2}dx\lesssim {\int }_I{\overline{\rho }}^{2\gamma -2}{({\eta }_x-1)}^{2}dx\lesssim {\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dx,$$

and if $2\gamma -2<\gamma$,

$$\begin{array}{cc}\hfill & {\int }_I{(\eta -x)}^{2}dx\lesssim {\int }_I{\overline{\rho }}^{2\gamma -2}{({\eta }_x-1)}^{2}dx\hfill \\ \hfill \lesssim & {\left({\int }_I{\overline{\rho }}^{\theta -\kappa +\iota }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{2-\gamma }{\gamma -\theta +\kappa -\iota }}}{\left({\int }_I{\overline{\rho }}^{\gamma }{({\eta }_x-1)}^{2}dx\right)}^{{\displaystyle \frac{min\{2\gamma -2\}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}.\hfill \end{array}$$

Then, it yields from the Sobolev embedding, and $\eta (0,t)=0$, that for any $t\ge 0$

$$\begin{array}{cc}\hfill {\parallel \eta (x,t)-x\parallel }_{L^{\infty }\left(I\right)}^2& \le {\parallel \eta (x,t)-x\parallel }_{L^2\left(I\right)}{\parallel {\eta }_x(x,t)-1\parallel }_{L^2\left(I\right)}\hfill \\ \hfill & \lesssim {(1+t)}^{-{\displaystyle \frac{min\{\gamma -1,\gamma /2\}-\theta +\kappa -\iota }{\gamma -\theta +\kappa -\iota }}}\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

□

4.2. Higher-Order Estimates

Based on the lower order estimates established in Section 4.1, we now derive higher-order estimates. The main object is the second-order derivative ${\eta }_{xx}$, which is essential for the regularity of the solution and the well-definedness of the particle path.

Lemma 8.

Suppose the assumptions in Lemma 7 hold. Then, for any $t\ge 0$,

$$\begin{array}{cc}& {\int }_I\left[{\overline{\rho }}^{2\gamma -1}+\left(1+t\right){\overline{\rho }}^{3\gamma -\theta -1}\right]{\eta }_{xx}^2(x,t)dx\hfill \\ & +{\int }_0^t{\int }_I\left[{\overline{\rho }}^{3\gamma -\theta -1}+(1+s){\overline{\rho }}^{4\gamma -2\theta -1}\right]{\eta }_{xx}^2(x,t)dxds\hfill \\ & +{(1+t)}^{{\displaystyle \frac{1}{\gamma -\theta }}}{\int }_I|{\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x{|}^2(x,t)dx+(1+t){\int }_I|{\left({\eta }_x^{-1}{\mathbf{N}}_x\right)}_x{|}^2(x,t)dx\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^4\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(65)

Proof. 

Multiplying ${\left(8\right)}_1$ by ${\overline{\rho }}^{2\gamma -2\theta -1}{\left[{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right]}_x$ and integrating with respect to spatial variable, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2\theta }}{\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dx-{\int }_I{\left({\overline{\rho }}^{2\gamma -2\theta }{v}_t\right)}_x{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)dx& \\ \hfill +{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}{\left[{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right]}_x{\left[{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)-{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right]}_{x}dx& =0.\hfill \end{array}$$

(66)

By Cauchy’s inequality, (5), (21) and (9), we obtain

$$\begin{array}{cc}& {\left[{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right]}_x{\left[{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)\right]}_x\ge {\displaystyle \frac{\theta \gamma }{2}}{\overline{\rho }}^{\theta +\gamma }{\eta }_x^{-\theta -\gamma -2}{\eta }_{xx}^2-C{\overline{\rho }}^{\theta -\gamma +2}{|{\eta }_x-1|}^2,\hfill \\ & {\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^2\ge {\displaystyle \frac{{\theta }^2}{2}}{\overline{\rho }}^{2\theta }{\eta }_x^{-2\theta -2}{\eta }_{xx}^2-C{\overline{\rho }}^{2\theta -2\gamma +2}{|{\eta }_x-1|}^2,\hfill \end{array}$$

(67)

and

$$\begin{array}{cc}\hfill \left|{\left({\overline{\rho }}^{2\gamma -2\theta }{v}_t\right)}_x{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right|\lesssim & \left({\overline{\rho }}^{2\gamma -2\theta }|v_{tx}|+{\overline{\rho }}^{\gamma -2\theta +1}\left|v_t\right|\right){\overline{\rho }}^{\theta }|{\eta }_x-1|\hfill \\ \hfill \lesssim & {\overline{\rho }}^{\theta }(v_t^2+v_{tx}^2)+{\overline{\rho }}^{\gamma -\theta +1}{({\eta }_x-1)}^2,\hfill \end{array}$$

and, noticing that equation ${\left(8\right)}_2$ implies

$${\mathbf{N}}_x/{\eta }_x·{({\mathbf{N}}_x/{\eta }_x)}_x={\mathbf{N}}_x·{\mathbf{N}}_t,$$

then

$$\begin{array}{cc}& 2|{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}{\left[{\overline{\rho }}^{\theta }\left({\eta }_x^{-\theta }-1\right)\right]}_x{\mathbf{N}}_x/{\eta }_x·{({\mathbf{N}}_x/{\eta }_x)}_{x}dx|\hfill \\ \hfill \lesssim & {\int }_I|{\mathbf{N}}_t{|}^{2}dx+{\parallel {\mathbf{N}}_x\parallel }_{L^{\infty }}^2{\int }_I{\overline{\rho }}^{4\gamma -4\theta -2}|{\left[{\overline{\rho }}^{\theta }\left({\eta }_x^{-\theta }-1\right)\right]}_x{|}^{2}dx\hfill \\ \hfill \lesssim & {\int }_I|{\mathbf{N}}_t{|}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}|{\left[{\overline{\rho }}^{\theta }\left({\eta }_x^{-\theta }-1\right)\right]}_x{|}^{2}dx,\hfill \end{array}$$

(68)

where we use the fact that $\theta \le \gamma /2$. Then, it yields from (66) and the above inequalities that

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dx+{\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\eta }_{xx}^{2}dx\hfill \\ \hfill \lesssim & {\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx{\int }_I{\overline{\rho }}^{2\gamma -2\theta -1}|{\left[{\overline{\rho }}^{\theta }\left({\eta }_x^{-\theta }-1\right)\right]}_x{|}^{2}dx\hfill \\ & +{\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma -\theta +1}{({\eta }_x-1)}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx,\hfill \end{array}$$

(69)

which, by (33), (38), (61) and Gronwall’s inequality, gives

$$\begin{array}{cc}& {\int }_I{\overline{\rho }}^{2\gamma -1}{\eta }_{xx}^2(x,t)dx+{\int }_0^t{\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\eta }_{xx}^{2}dxds\hfill \\ \hfill \lesssim & \mathcal{E}\left(0\right)exp\left\{{\int }_0^t{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dxdx\right\}\hfill \\ & ·\left[1+{\int }_0^t\left({\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dx+{\int }_I{\overline{\rho }}^{\gamma -\theta +1}{({\eta }_x-1)}^{2}dx+{\int }_I{\left|{\mathbf{N}}_t\right|}^{2}dx\right)ds\right]\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(70)

To obtain the time decay estimate, we apply a multiplier with a higher power of $\overline{\rho }$ combined with the time weight $(1+t)$. Similarly, we multiply ${\left(8\right)}_1$ by ${\overline{\rho }}^{3\gamma -3\theta -1}{\left[{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right]}_x$ and integrate over I to obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2\theta }}{\displaystyle \frac{d}{dt}}{\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dx-{\int }_I{\left({\overline{\rho }}^{3\gamma -3\theta }{v}_t\right)}_x{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)dx& \\ \hfill +{\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left[{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right]}_x{\left[{\overline{\rho }}^{\gamma }({\eta }_x^{-\gamma }-1)-{\left|{\mathbf{N}}_x\right|}^2/{\eta }_x^2\right]}_{x}dx& =0.\hfill \end{array}$$

(71)

If $\mu \le 1$, we multiply (71) by $(1+t)$ to obtain

$$\begin{array}{cc}& {\displaystyle \frac{d}{dt}}(1+t){\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dx+(1+t){\int }_I{\overline{\rho }}^{4\gamma -2\theta -1}{\eta }_{xx}^{2}dx\hfill \\ \hfill \lesssim & {\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dx+(1+t){\int }_I{\left({\overline{\rho }}^{3\gamma -3\theta }{v}_t\right)}_x{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)dx\hfill \\ & +(1+t){\int }_I{\overline{\rho }}^{2\gamma -2\theta +1}|{\eta }_x{-1|}^{2}dx+(1+t){\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ & +(1+t)\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }}^2{\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left[{\overline{\rho }}^{\theta }\left({\eta }_x^{-\theta }-1\right)\right]}_x^{2}dx,\hfill \end{array}$$

(72)

we estimate the time integral of every term on the right-hand side of (72), respectively, as following,

$$\begin{array}{cc}& {\int }_0^t{\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left[{\left({\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)\right)}_x\right]}^{2}dxds\hfill \\ \hfill \lesssim & {\int }_0^t{\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\eta }_{xx}^{2}dxds+{\int }_0^t{\int }_I{\overline{\rho }}^{\gamma -\theta +1}{|{\eta }_x-1|}^{2}dxds\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

(73)

due to (56) and (70). Next,

$$\begin{array}{cc}& {\int }_0^t(1+s){\int }_I{\left({\overline{\rho }}^{3\gamma -3\theta }{v}_t\right)}_x{\overline{\rho }}^{\theta }({\eta }_x^{-\theta }-1)dxds\hfill \\ \hfill \lesssim & {\int }_0^t(1+s){\int }_I\left({\overline{\rho }}^{3\gamma -3\theta }|v_{tx}|+{\overline{\rho }}^{2\gamma -3\theta +1}\left|v_t\right|\right){\overline{\rho }}^{\theta }|{\eta }_x-1|dxds\hfill \\ \hfill \lesssim & {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\theta }\left(v_{tx}^2+v_t^2\right)dxds+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{4\gamma -5\theta +2}{|{\eta }_x-1|}^{2}dxds\hfill \\ \hfill \lesssim & {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\theta }{v}_{tx}^{2}dxds+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\gamma }{|{\eta }_x-1|}^{2}dxds\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

and

$$\begin{array}{cc}& {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{2\gamma -2\theta +1}{|{\eta }_x-1|}^{2}dxds\hfill \\ \hfill \lesssim & {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{\gamma }{|{\eta }_x-1|}^{2}dxds\lesssim \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

we then recall (34) and $\theta \le \gamma /2$ to have

$$\begin{array}{cc}& {\int }_0^t(1+s){\int }_I{\overline{\rho }}^{3\gamma -3\theta -1}{\left|{\mathbf{N}}_t\right|}^{2}dxds\lesssim \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^2\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

We apply Gronwall’s inequality to (72) and use (33), (67) and all together the individual estimates above to obtain

$$\begin{array}{cc}& (1+t){\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\eta }_{xx}^{2}dx+{\int }_0^t(1+s){\int }_I{\overline{\rho }}^{4\gamma -2\theta -1}{\eta }_{xx}^{2}dxds\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^4\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

(74)

We deduce from ${\left(8\right)}_1$ and notice (70) and (56) to obtain

$$\begin{array}{cc}\hfill {\int }_I|{\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x{|}^{2}dx\lesssim & {\int }_I{\overline{\rho }}^2{v}_t^{2}dx+{\int }_I{\overline{\rho }}^{2\gamma }{\eta }_{xx}^{2}dx+{\int }_I{\overline{\rho }}^2{({\eta }_x-1)}^{2}dx+{\int }_I|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_t\right|}^{2}dx\hfill \\ \hfill \lesssim & {(1+t)}^{-1}\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}+{\int }_I{\overline{\rho }}^{2\gamma }{\eta }_{xx}^{2}dx,\hfill \end{array}$$

then we apply an interpolation inequality and notice (70) and (74) to obtain

$$\begin{array}{cc}\hfill {\int }_I{\overline{\rho }}^{2\gamma }{\eta }_{xx}^{2}dx\lesssim & {\left({\int }_I{\overline{\rho }}^{2\gamma -1}{\eta }_{xx}^{2}dx\right)}^{{\displaystyle \frac{\gamma -\theta -1}{\gamma -\theta }}}{\left({\int }_I{\overline{\rho }}^{3\gamma -\theta -1}{\eta }_{xx}^{2}dx\right)}^{{\displaystyle \frac{1}{\gamma -\theta }}}\hfill \\ \hfill \lesssim & {(1+t)}^{-{\displaystyle \frac{1}{\gamma -\theta }}}\left[\mathcal{E}\left(0\right)+{\mathcal{E}}^4\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\},\hfill \end{array}$$

thus,

$$\begin{array}{c}\hfill {(1+t)}^{{\displaystyle \frac{1}{\gamma -\theta }}}{\int }_I|{\left({\overline{\rho }}^{\theta }{\eta }_x^{-1-\theta }{v}_x\right)}_x{|}^{2}dx\lesssim \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^4\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\end{array}$$

From ${\left(8\right)}_3$, (34) and (38), we obtain

$$\begin{array}{cc}\hfill (1+t){\int }_I|{\left({\eta }_x^{-1}{\mathbf{N}}_x\right)}_x{|}^{2}dx\lesssim & (1+t)\left({\int }_I|{\mathbf{N}}_t{|}^{2}dx+{\int }_I{\left|{\mathbf{N}}_x\right|}^{4}dx\right)\hfill \\ \hfill \lesssim & \left[\mathcal{E}\left(0\right)+{\mathcal{E}}^3\left(0\right)\right]exp\left\{C{\mathcal{E}}_L\left(0\right)\right\}.\hfill \end{array}$$

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5. Conclusions

In this paper, we have established the global existence, uniqueness and long-time stability of strong solutions to the one-dimensional liquid crystal flow with vacuum free boundary and density-dependent degenerate viscosity, under a smallness condition on the initial energy at the basic level. The strong solution converges to the steady state with explicit decay rates as $t\to \infty$. The main ingredients of the proof are the point-wise upper and lower bounds for the deformation variable ${\eta }_x$, and the weighted estimates with time decay established by singular multipliers and Hardy-type inequalities. The framework developed here also leaves room for further extensions, for instance, to non-isentropic flows or to higher dimensions with symmetry.