Global Strong Solutions to the One-Dimensional Isentropic Compressible Liquid Crystal Equations with a Vacuum Free Boundary and Large Initial Data

Abstract

In this paper, we establish the global existence and uniqueness of strong solutions to the one-dimensional vacuum free boundary problem for the isentropic compressible liquid crystal equations under the influence of gravity with large initial data, where the density is allowed to vanish continuously at the free boundary. The main difficulty is the degeneracy of the momentum equation caused by the vanishing of the density, which prevents standard energy methods from giving pointwise control of the velocity gradient. Working in Lagrangian coordinates, we derive a time-uniform pointwise lower bound and a finite-time pointwise upper bound on the Jacobian ${\eta }_x$, together with a finite $L^{\infty }$-bound on the velocity gradient $v_x$ on any finite time interval $[0,T]$, which, in particular, guarantees that the free boundary is a well-defined $C^1$ curve. This appears to be the first global strong solution result for this problem; the earlier work of Huang and Ding establishes global weak solutions, for which the free boundary is not addressed in a pointwise sense.

1. Introduction

Liquid crystal flows belong to a class of complex fluids whose rich mathematical structure has attracted continuous research interest over several decades. The basic continuum theory for nematic liquid crystals was established by Ericksen [1] and Leslie [2], and their constitutive framework still serves as the foundation for most later mathematical treatments. In the incompressible regime, the analytical theory was greatly developed by the work of Lin [3] and a series of papers by Lin and Liu [4,5,6], which proved the existence and partial regularity of suitable weak solutions for flows of variable-length liquid crystals. The extension to the compressible case is physically important and mathematically more challenging. The corresponding compressible models were derived in [7,8], and under the assumption that the Oseen–Frank energy density is simplified to the Dirichlet form, the governing system for compressible liquid crystals in a domain $\Omega \subset {\mathbb{R}}^d$ takes the following form [7,8]:

$$\left\{\begin{array}{c}{\rho }_t+div\left(\rho u\right)=0,\hfill \\ {\left(\rho u\right)}_t+div(\rho u\otimes u)+\nabla \left(P\left(\rho \right)\right)\hfill \\ =\mu \Delta u+(\kappa +\mu )\nabla divu-\lambda div\left(\nabla n\odot \nabla n-{\displaystyle \frac{{|\nabla n|}^2}{2}}{I}_d\right),\hfill \\ n_t+u·\nabla n={\theta (\Delta n+|\nabla n|}^{2}n),\hfill \end{array}\right.$$

(1)

where ⊗ denotes the tensor product, and $\nabla n\odot \nabla n$ is the $d\times d$ matrix whose $(i,j)$-entry is ${\nabla }_{i}n·{\nabla }_{j}n$.

The well-posedness theory for (1) has been studied in various settings. In three spatial dimensions, Huang, Wang, and Wen [9] proved the local-in-time existence and uniqueness of strong solutions to both the Cauchy problem and related boundary value problems, with vacuum allowed in the initial data. For global-in-time results, Li, Xu, and Zhang [10] constructed classical solutions to the three-dimensional Cauchy problem under a smallness condition on the initial energy, while allowing arbitrarily large oscillations and either vacuum or non-vacuum far-field states. More recent progress on multi-dimensional compressible nematic liquid crystal flows includes the work of Zhong and Zhou [11], who established global well-posedness of strong solutions to the two-dimensional Cauchy problem with large initial data and vacuum allowed under a geometric angle condition on the initial direction field, as well as the work of Sun and Zhong [12] and Liu and Zhong [13], who obtained global strong solutions in two- and three-dimensional bounded domains with large oscillations and vacuum allowed under a smallness condition on the initial total energy.

The one-dimensional case allows stronger results and is the main setting of the present paper. We study the isentropic compressible liquid crystal equations on the time-dependent domain $(0,a(t\left)\right)$, with unknowns $(\rho ,u,n):[0,a\left(t\right)]\times [0,+\infty )\to {\mathbb{R}}_{+}\times \mathbb{R}\times S^2$:

$$\left\{\begin{array}{ccc}{\rho }_t+{\left(\rho u\right)}_x=0,\hfill & \mathrm{in}& (0,a(t\left)\right),\\ {\left(\rho u\right)}_t+{\left(\rho u^2\right)}_x+{\left(P\left(\rho \right)\right)}_x=\mu u_{xx}-\lambda \left(\right|n_x{{|}^2)}_x-\rho g,\hfill & \mathrm{in}& (0,a(t\left)\right),\\ n_t+u{n}_x=\theta (n_{xx}+|n_x{|}^{2}n),\hfill & \mathrm{in}& (0,a(t\left)\right),\\ {(\rho ,u,n)|}_{t=0}=({\rho }_0,u_0,n_0),\hfill & \mathrm{in}& (0,a(0\left)\right),\\ u(0,t)=0,n_x(0,t)=0,\hfill & \mathrm{on}& \{x=0\},\\ (P\left(\rho \right)-\mu u_x)(a\left(t\right),t)=0,n_x(a\left(t\right),t)=0,\hfill & \mathrm{on}& \{x=a(t\left)\right\},\end{array}\right.$$

(2)

for $t⩾0$. The derivation of this system can be found in [7,8]. Here $\rho$, u, and n are the fluid density, velocity, and the unit-vector orientation field ($\left|n\right|=1$) of the nematic director, respectively. The positive constants $\mu$, $\lambda$, $\theta$ represent the shear viscosity, the elastic-kinetic coupling coefficient, and the director’s relaxation time. The pressure law is $P=R{\rho }^{\gamma }$ for constants $R>0$ and $\gamma >1$. Compared with the system studied in [14,15,16,17], the present model includes a gravitational term $-\rho g$ with $g>0$ in the momentum equation. This term contributes a lower-order forcing that is straightforwardly handled within the energy framework, but it affects the structure of the steady state and enters explicitly into the pointwise bounds on the Lagrangian Jacobian ${\eta }_x$. The free boundary $a\left(t\right)$ moves according to

$${\displaystyle \frac{da}{dt}}\left(t\right)=u(a\left(t\right),t),t⩾0,a\left(0\right)=a_0>0.$$

(3)

The boundary conditions in (2)₅ give a no-slip condition for the velocity and a Neumann condition for n at the fixed wall $x=0$; those in (2)₆ are the stress-free condition for the fluid and the Neumann condition for n at the moving vacuum interface.

For the problem on a fixed domain, Ding, Lin, Wang, and Wen [16] proved the global existence and uniqueness of classical solutions under the assumption that the initial density satisfies ${\rho }_0⩾c_0>0$, while Ding, Wang, and Wen [15] relaxed this condition to ${\rho }_0⩾0$, but only obtained weak solutions due to the possible vanishing of the density. Both results do not cover the physically important situation where the fluid region is bounded by a moving surface that separates the liquid from the vacuum.

The free boundary problem, where the fluid occupies a time-dependent interval $(0,a(t\left)\right)$ with $a\left(t\right)$ moving according to the fluid velocity, brings in significant additional difficulties. When the density is uniformly positive, Ding, Huang, and Xia [17] obtained global classical solutions by using the fact that the momentum equation remains uniformly parabolic. The degenerate case, where $\rho$ vanishes at the free boundary, was studied by Huang and Ding [14], who obtained global weak solutions and investigated the lifespan of smooth solutions under a geometric growth condition on the density support. However, in [14] the velocity gradient $u_x$ was only controlled through the weighted quantity $\parallel \rho u_x^2{\parallel }_{L^{\infty }}$, which is not enough to conclude that the free boundary is well-defined in a pointwise sense. In particular, no global strong solution result has been obtained for this problem.

For the compressible Navier–Stokes equations, a series of related results gives useful guidance. In one space dimension with constant viscosity, Okada [18] proved local well-posedness of the vacuum free boundary problem, and Luo, Xin, and Yang [19] studied the behavior of the fluid-vacuum interface. Global smoothness of solutions for the isentropic case, including the physical vacuum condition, was later proved by Zeng [20]. For equations with density-dependent degenerate viscosity, Ou and Zeng [21] proved a global strong solution result for the vacuum free boundary problem with gravitational effects. More recently, Li and Guo [22] established the global existence and nonlinear asymptotic stability of axisymmetric strong solutions to the multi-dimensional vacuum free boundary problem for the isentropic compressible Navier–Stokes equations with self-gravitation for small initial data, and captured the precise physical behavior by showing that the sound speed is $C^{1/2}$-Hölder continuous across the vacuum boundary for $\gamma \in (5/4,2)$. In the non-isentropic case with heat conduction, global classical solutions to the one-dimensional vacuum free boundary problem were obtained by Ou [23]. The Lagrangian coordinate framework used throughout this paper follows the approach developed in [24,25,26] for compressible Euler equations in a physical vacuum.

For liquid crystal flows, Mei [27] recently proved global classical solutions to the one-dimensional vacuum free boundary problem for the full non-isothermal compressible liquid crystal system with large initial data, which gives the liquid crystal analogue of the results in [23]. The isentropic and non-isothermal models differ in modeling, and these differences, in turn, induce structurally different analytical features: in [27], the pressure $p=R\rho \theta$ is linear in $\rho$ and the temperature equation contributes a parabolic structure, while in the present setting, the strong nonlinearity $p={\rho }^{\gamma }$ with $\gamma >1$ governs the analysis. A more detailed comparison is given in Remark 3 below. Related results on a fixed domain for the full non-isothermal system with vacuum can be found in [28]. Some functional inequalities used in this paper are taken from [29,30].

To the best of our knowledge, no global strong solution result has been established for the one-dimensional isentropic compressible liquid crystal equations with a vacuum free boundary and large initial data, and the present paper fills this gap. Two features distinguish our result. First, while [15] treated the fixed-domain problem and obtained global weak solutions, we work on the free boundary setting and establish global strong solutions. Second, Ref. [14] handled the free boundary case with control of $u_x$ through the weighted quantity $\parallel \rho u_x^2{\parallel }_{L^{\infty }}$. In contrast, here we establish a finite-time $L^{\infty }$-bound on the ratio $v_x/{\eta }_x$ on $[0,T]$ in Lagrangian coordinates, which, together with the pointwise bounds on ${\eta }_x$, yields an unweighted $L^{\infty }$-bound on $v_x$ on every finite time interval and identifies the free boundary $a\left(t\right)$ as a $C^1$ curve on $[0,T]$ for any $T>0$.

The main difficulty of the problem comes from the degeneracy when $\rho \to 0$ near $x=a\left(t\right)$: the momentum equation (2)₂ loses its parabolic character, and standard energy methods can no longer give pointwise control of $u_x$. To overcome this difficulty, we transform to Lagrangian coordinates (Section 2) and reformulate the free boundary problem (2) as an initial boundary value problem (7) on the fixed domain $I=(0,1)$. The main steps of the proof are as follows. We first derive a basic energy identity and establish pointwise upper and lower bounds on the Lagrangian Jacobian ${\eta }_x$; the uniform lower bound $B_0>0$ is an important ingredient throughout. We then obtain $L^2\left(I\right)$-estimates on ${({\mathbf{N}}_x/{\eta }_x)}_x$ and on $\sqrt{{\rho }_0}{v}_t$. These estimates, combined with the integral identity (11), give $\parallel v_x/{\eta }_x{\parallel }_{L^{\infty }\left(I\right)}⩽B_1\left(T\right)<\infty$ for any $T>0$. Together with the pointwise upper bound on ${\eta }_x$ on $[0,T]$, this yields a finite $L^{\infty }\left(I\right)$-bound on $v_x$ for each $t\in [0,T]$, which, in particular, guarantees that the free boundary $a\left(t\right)=\eta (1,t)$ is a well-defined $C^1$ curve on $[0,T]$. Our approach is motivated by the framework developed by Zeng [20] for the one-dimensional vacuum free boundary problem of the isentropic compressible Navier–Stokes equations. The additional coupling between the velocity field and the director field $\mathbf{N}$ introduces extra nonlinear terms in the energy estimates that are not present in the Navier–Stokes setting, which we handle through the weighted $L^2$-estimate on ${({\mathbf{N}}_x/{\eta }_x)}_x$ above.

Without loss of generality, we normalize

$$a_0=R=g=\mu =\lambda =\theta =1.$$

In particular, the normalization $a_0=1$ means that the initial fluid domain is $(0,1)$, and the Lagrangian fixed domain is $I=(0,1)$.

The initial density satisfies

$$0<{\rho }_0\left(x\right)⩽C_0\mathrm{for}x\in [0,1),{\rho }_0\left(1\right)=0,$$

(4)

for some constant $C_0>0$.

Throughout this paper, $B_i$, $C_i$, $E_i$ ($i=0,1,2$) denote positive constants independent of T, and $C\left(T\right)$ denotes a positive constant that may depend on T.

2. Lagrangian Particle Path Reformulation and Main Results

Since the boundary $a\left(t\right)$ is free to move in Eulerian coordinates, it is more convenient to adopt the Lagrangian particle path coordinates so that the free boundary problem (2) can be transformed into an initial boundary value problem on a fixed domain. To fix the boundary, we define the Lagrangian variable (particle path) $\eta (x,t)$ starting at x when $t=0$ by

$${\partial }_t\eta (x,t)=u(\eta (x,t),t)\mathrm{for}t>0\mathrm{and}\eta (x,0)=x,\eta (1,t)=a\left(t\right).$$

(5)

We set the Lagrangian density, velocity, and orientation field, respectively, by

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

Then the Lagrangian version of system (2) can be written as

$$\left\{\begin{array}{ccc}{f}_t+f{\displaystyle \frac{v_x}{{\eta }_x}}=0,\hfill & \mathrm{in}& I\times (0,\infty ),\hfill \\ f{v}_t+{\displaystyle \frac{{\left(f^{\gamma }\right)}_x}{{\eta }_x}}+{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}\right)}_x={\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{v_x}{{\eta }_x}}\right)}_x-f,\hfill & \mathrm{in}& I\times (0,\infty ),\hfill \\ {\mathbf{N}}_t={\displaystyle \frac{1}{{\eta }_x}}\left[{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N}\right],\hfill & \mathrm{in}& I\times (0,\infty ),\hfill \\ (f,v,\mathbf{N})=({\rho }_0,u_0,n_0),\hfill & \mathrm{in}& I\times \{t=0\},\hfill \\ v(0,t)=0,{\mathbf{N}}_x(0,t)=0,\hfill & \mathrm{in}& \{x=0\}\times (0,\infty ),\hfill \\ \left(f^{\gamma }-{\displaystyle \frac{v_x}{{\eta }_x}}\right)(1,t)=0,{\mathbf{N}}_x(1,t)=0,\hfill & \mathrm{in}& \{x=1\}\times (0,\infty ),\hfill \end{array}\right.$$

(6)

where $I:=(0,1)$. Note that ${\eta }_x>0$ for all $(x,t)\in [0,1]\times [0,\infty )$, see (17) for details. It follows from solving (6)₁ that

$$f={\rho }_0/{\eta }_x,$$

so that the system (6) can be rewritten as

$$\left\{\begin{array}{ccc}{\rho }_0{v}_t+{\left({\displaystyle \frac{{\rho }_0^{\gamma }}{{\eta }_x^{\gamma }}}\right)}_x+{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}\right)}_x={\left({\displaystyle \frac{v_x}{{\eta }_x}}\right)}_x-{\rho }_0,\hfill & \mathrm{in}& I\times (0,\infty ),\hfill \\ {\mathbf{N}}_t={\displaystyle \frac{1}{{\eta }_x}}\left[{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N}\right],\hfill & \mathrm{in}& I\times (0,\infty ),\hfill \\ (v,\mathbf{N})(x,0)=(u_0,n_0)\left(x\right),\hfill & \mathrm{in}& I\times \{t=0\},\hfill \\ v(0,t)=0,{\mathbf{N}}_x(0,t)=0,\hfill & \mathrm{in}& \{x=0\}\times (0,\infty ),\hfill \\ v_x(1,t)=0,{\mathbf{N}}_x(1,t)=0,\hfill & \mathrm{in}& \{x=1\}\times (0,\infty ).\hfill \end{array}\right.$$

(7)

For simplicity, we set

$$s={\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}-{\displaystyle \frac{v_x}{{\eta }_x}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}\mathrm{and}q={\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\mathbf{N},$$

where their initial values are denoted, respectively, by

$$s_0={\rho }_0^{\gamma }-u_{0x}+|n_{0x}{|}^2,q_0=n_{0xx}+{\left|n_{0x}\right|}^2{n}_0.$$

We are ready to state the main theorem:

Theorem 1.

Suppose that the initial data $({\rho }_0,u_0,n_0)$ satisfy (4) and

$${∥{\left({\rho }_0^{\gamma }\right)}_x∥}_{L^2\left(I\right)}⩽A_0,{∥u_0∥}_{L^2\left(I\right)}+{∥n_0∥}_{H^2}+{∥{\rho }_0^{-1/2}{s}_{0x}∥}_{L^2\left(I\right)}⩽A_1,$$

(8)

for positive constants $A_0,A_1$, together with the compatibility conditions

$$u_0\left(0\right)=0,n_{0x}\left(0\right)=n_{0x}\left(1\right)=0,\left|n_0\left(x\right)\right|=1\mathit{for}\mathit{all}x\in I.$$

Then the problem (7) admits a unique global strong solution $(\eta ,v,\mathbf{N})$ satisfying for any $T>0$:

$$\begin{array}{cc}\hfill & \eta \in C^1([0,T];H^1\left(I\right))\cap L^{\infty }(0,T;H^2\left(I\right)),\hfill \\ \hfill & v\in L^{\infty }(0,T;H^2\left(I\right))\cap C([0,T];H^1\left(I\right)),\hfill \\ \hfill & \mathbf{N}\in L^{\infty }(0,T;H^2\left(I\right))\cap C([0,T];H^1\left(I\right)),\hfill \\ \hfill & \sqrt{{\rho }_0}{v}_t\in L^{\infty }(0,T;L^2\left(I\right)),{\mathbf{N}}_t\in L^{\infty }(0,T;L^2\left(I\right)),\hfill \end{array}$$

and the estimate

$${∥v(·,t)∥}_{H^2\left(I\right)}+{∥\mathbf{N}(·,t)∥}_{H^2\left(I\right)}+{∥\sqrt{{\rho }_0}{v}_t(·,t)∥}_{L^2\left(I\right)}+{∥{\mathbf{N}}_t(·,t)∥}_{L^2\left(I\right)}⩽C\left(T\right)$$

(9)

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

Remark 1.

The orientation constraint $\left|\mathbf{N}\right(x,t\left)\right|=1$ is preserved for all $(x,t)\in I\times [0,+\infty )$, as can be verified by multiplying (7)₂ by $\mathbf{N}$ and using $|n_0\left(x\right)|=1$.

Remark 2.

The proof of Theorem 1 yields the following additional information on the Lagrangian map $\eta (x,t)$ and the free boundary.

(i) 

(Pointwise bounds on the Jacobian.) For any $T>0$, there exist constants $B_0>0$ independent of T and $C\left(T\right)>0$ (which may grow exponentially in T) such that

$$B_0⩽{\eta }_x(x,t)⩽C\left(T\right)\mathit{for}\mathit{all}(x,t)\in [0,1]\times [0,T],$$

where $B_0$ is given explicitly in (17); the explicit form of $C\left(T\right)$ is given in (16). In particular, ${\eta }_x\in L^{\infty }(I\times [0,T])$, and together with ${\eta }_{xx}\in L^{\infty }(0,T;L^2\left(I\right))$ from (35), this gives $\eta \in L^{\infty }(0,T;H^2\left(I\right))$. Combined with ${\eta }_t=v\in C([0,T];H^1\left(I\right))$ from Theorem 1, this yields the regularity $\eta \in C^1([0,T];H^1\left(I\right))\cap L^{\infty }(0,T;H^2\left(I\right))$ stated above.

(ii) 

(Finite-time bound on $v_x/{\eta }_x$.) For any $T>0$, there exists a constant $B_1=B_1\left(T\right)>0$ such that

$${∥{\displaystyle \frac{v_x}{{\eta }_x}}(·,t)∥}_{L^{\infty }\left(I\right)}⩽B_1\left(T\right)\mathit{for}t\in [0,T].$$

(iii) 

(Regularity of the free boundary.) Since $v\in C([0,T];H^1\left(I\right))$, the trace $v(1,·)$ is continuous in t on $[0,T]$. Combined with the kinematic boundary condition $\dot{a}\left(t\right)=v(1,t)$ in (3), this identifies the free boundary $a\left(t\right)=\eta (1,t)$ as a $C^1$ curve on $[0,T]$ for any $T>0$. A detailed argument is given at the end of Section 4.

Remark 3.

The present isentropic problem and the non-isothermal one studied in [27] differ at the modeling level, and these differences in turn induce a structurally different closure mechanism for the pointwise estimates on ${\eta }_x$. The non-isothermal model of [27] involves the temperature field ϑ governed by a parabolic energy equation, with constitutive law $p=R\rho \vartheta$ linear in ρ; the isentropic model treated here has no temperature field, and the constitutive law is $p=R{\rho }^{\gamma }$ with $\gamma >1$. As a consequence, the pointwise bounds on ${\eta }_x$ in [27] are closed via the $L^{\infty }$-bounds on ϑ furnished by the maximum principle for the energy equation, a route unavailable in the present setting. Instead, the analysis here is closed by direct pointwise control of ${\eta }_x$ in the presence of the nonlinear vacuum degeneracy ${\rho }_0^{\gamma }/{\eta }_x^{\gamma }$ contributed by the pressure law after the Lagrangian transformation; the integral identity (11) is the technical device on which this control is based.

3. Preliminaries

The lemmas in this section will be useful in the proof of the main theorem.

Lemma 1

(see [30]). Assume that ${\rho }_0\left(x\right)⩾0$ for $x\in I=[a,b]$, ${\int }_I{\rho }_{0}dx=M>0$ and ${\int }_I{\rho }_0\left|v\right|dx⩽L$. Then for $q>0$, there exists a constant $C=C(M,L,q)>0$ such that

$$\parallel v^q{\parallel }_{L^{\infty }\left(I\right)}⩽C{\parallel {\left(v^q\right)}_x\parallel }_{L^1\left(I\right)}+C,$$

for any $v^q\in H^1\left(I\right)$. Here $M,L$ and q are positive constants independent of v.

Lemma 2

(see [29]). Assume $X\subset E\subset Y$ are Banach spaces and $X\hookrightarrow E$. Then the following embeddings are compact:

(i)  

$\left\{\phi :\phi \in L^q(0,T;X),{\displaystyle \frac{\partial \phi }{\partial t}}\in L^1(0,T;Y)\right\}\hookrightarrow \hookrightarrow L^q(0,T;E),\mathit{if}1⩽q⩽\infty ;$

(ii) 

$\left\{\phi :\phi \in L^{\infty }(0,T;X),{\displaystyle \frac{\partial \phi }{\partial t}}\in L^r(0,T;Y)\right\}\hookrightarrow \hookrightarrow C([0,T];E),\mathit{if}1<r⩽\infty .$

4. Proof of Theorem 1

We now prove the estimates of Theorem 1. The local-in-time existence and uniqueness of strong solutions to the present problem can be established by a finite difference scheme, adapted from the approach of Okada [18], Luo, Xin and Zeng [31], and Ou, Shi, and Wittwer [32]. In particular, [32] treats the planar magnetohydrodynamic free boundary problem, in which the strongly coupled magnetic field has a structure analogous to the velocity-director coupling of the present system. In this scheme, the spatial interval $I=(0,1)$ is discretized as $x_n=nh$, $h=1/N$, and the free boundary $x=1$ is recovered a posteriori as the image of the fixed endpoint $x_N=1$ under the Lagrangian flow $\eta$. The vacuum degeneracy is built directly into the discrete boundary condition ${\overline{\rho }}_N=0$ at $x_N=1$, so that no regularization of ${\rho }_0$ is needed; on the interior nodes $n<N$, one has ${\overline{\rho }}_n>0$, and the discretized momentum and director equations form a system of ordinary differential equations whose well-posedness on some interval $[0,T_N]$ follows from standard ODE theory. The director coupling enters the scheme through quadratic terms $|{\mathbf{N}}_x{|}^2/{\eta }_x^2$ in the momentum equation and through transport terms in the director equation; both are handled in the same way as the analogous magnetic terms in [32] (Sec. V), and the pointwise constraint $\left|\mathbf{N}\right|=1$ is preserved by the structure of (7)₂. The N-uniform energy estimates of the type derived in this section, together with the Aubin–Lions compactness lemma (Lemma 2), then yield a strong solution over a time interval $[0,T^{★}]$ with $T^{★}$ independent of N by passing to the limit $N\to \infty$. We therefore omit the details of the local existence proof and focus on the global a priori estimates.

4.1. Basic Energy Estimate

Multiplying Equation (7)₁ by v and integrating the product with respect to the spatial variable, we obtain

$${\displaystyle \frac{d}{dt}}{\int }_0^1\left({\displaystyle \frac{1}{2}}{\rho }_0{v}^2+{\rho }_0\eta \right)dx+{\int }_0^1{\left({\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}-{\displaystyle \frac{v_x}{{\eta }_x}}\right)}_{x}vdx+{\int }_0^1{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}\right)}_{x}vdx=0.$$

Integration by parts and the boundary condition (7)_4,5 then imply that

$${\displaystyle \frac{d}{dt}}{\int }_0^1\left({\displaystyle \frac{1}{2}}{\rho }_0{v}^2+{\rho }_0\eta +{\displaystyle \frac{1}{\gamma -1}}{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma -1}}}\right)dx+{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx-{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}{v}_{x}dx=0.$$

Multiplying (7)₂ by ${\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x}}$ and integrating with respect to the spatial variable, we obtain

$${\int }_0^1{\mathbf{N}}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx={\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x}}\right|}^{2}dx={\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx,$$

since ${\left|\mathbf{N}\right|}^2=1$. For the term on the left-hand side of the above equality, we use integration by parts and the boundary condition (7)_4,5 to get

$$\begin{array}{cc}\hfill {\int }_0^1{\mathbf{N}}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx& =-{\int }_0^1{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}}{{\eta }_x}}dx\hfill \\ \hfill & =-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\left(\right|{\mathbf{N}}_x{{|}^2)}_t}{{\eta }_x}}dx\hfill \\ \hfill & =-{\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}{v}_{x}dx.\hfill \end{array}$$

It follows from the above equalities that

$${\displaystyle \frac{d}{dt}}{\int }_0^1\left({\displaystyle \frac{1}{2}}{\rho }_0{v}^2+{\rho }_0\eta +{\displaystyle \frac{1}{\gamma -1}}{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma -1}}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\right)dx+{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+2{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx=0.$$

Then, integrating the equation with respect to the time variable, we obtain

$${\int }_0^1\left({\displaystyle \frac{1}{2}}{\rho }_0{v}^2+{\rho }_0\eta +{\displaystyle \frac{1}{\gamma -1}}{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma -1}}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}\right)dx+{\int }_0^t{\int }_0^1\left({\displaystyle \frac{v_x^2}{{\eta }_x}}+2{\left|{\mathbf{N}}_t\right|}^2{\eta }_x\right)dxds=E_0,$$

(10)

where the initial total energy $E_0={\int }_0^1\left({\displaystyle \frac{1}{2}}{\rho }_0{u}_0^2+{\rho }_{0}x+{\displaystyle \frac{1}{\gamma -1}}{\rho }_0^{\gamma }+{\left|n_{0x}\right|}^2\right)dx$ is finite by (4) and (8).

4.2. Pointwise Estimates on ${\eta }_x$

Integrating Equation (7)₁ over $(x,1)$ with respect to the spatial variable, noting that all boundary terms at $y=1$ vanish by the conditions ${\rho }_0\left(1\right)=0$, ${\mathbf{N}}_x(1,t)=0$ and $v_x(1,t)/{\eta }_x(1,t)=0$ from (4) and (7)₅, we have that

$${\left({\int }_x^1{\rho }_{0}vdy\right)}_t+{\int }_x^1{\rho }_{0}dy+{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}={\displaystyle \frac{v_x}{{\eta }_x}}={(ln{\eta }_x)}_t,$$

(11)

due to ${\eta }_t=v$. Integrating the equation above with respect to the time variable gives

$${\int }_x^1{\rho }_{0}vdy{|}_0^t+t{\int }_x^1{\rho }_{0}dy+{\int }_0^t{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}ds+{\int }_0^t{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}ds=ln{\eta }_x,$$

since $\eta (x,0)=x$, from which we can get an implicit expression for ${\eta }_x$, namely

$${\eta }_x=exp\left\{{\int }_x^1{\rho }_{0}vdy{|}_0^t+t{\int }_x^1{\rho }_{0}dy\right\}exp\left\{A(x,t)\right\}exp\left\{{\int }_0^t{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}ds\right\},$$

(12)

where

$$A(x,t):={\int }_0^t{\displaystyle \frac{{\rho }_0^{\gamma }}{{\eta }_x^{\gamma }}}ds.$$

Next, we show how to obtain an explicit expression for ${\eta }_x$. For each $x\in [0,1]$, consider the following ordinary differential equation satisfied by $A(x,t)$:

$$\begin{array}{cc}\hfill {\displaystyle \frac{dA(x,t)}{dt}}={\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}=& {\rho }_0^{\gamma }exp\left\{-\gamma {\int }_x^1{\rho }_{0}vdy{|}_0^t-\gamma t{\int }_x^1{\rho }_{0}dy\right\}\hfill \\ \hfill & \times exp\{-\gamma A(x,t)\}exp\left\{-\gamma {\int }_0^t{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}ds\right\},\hfill \end{array}$$

where the last equality comes from (12). Using separation of variables, one has that for each $x\in [0,1]$,

$$\begin{array}{cc}\hfill & exp\left\{\gamma A\right(x,t\left)\right\}\hfill \\ \hfill =& 1+\gamma {\int }_0^t\left({\rho }_0^{\gamma }exp\left\{-\gamma {\int }_x^1{\rho }_{0}vdy{|}_0^s-\gamma t{\int }_x^1{\rho }_{0}dy\right\}exp\left\{-\gamma {\int }_0^s{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}d\tau \right\}\right)ds.\hfill \end{array}$$

It then follows from (12) that for $(x,t)\in [0,1]\times [0,\infty )$,

$$\begin{array}{cc}\hfill {\eta }_x& =exp\left\{{\int }_x^1{\rho }_{0}vdy{|}_0^t+t{\int }_x^1{\rho }_{0}dy\right\}exp\left\{{\int }_0^t{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}ds\right\}\hfill \\ \hfill & \times {\left[1+\gamma {\int }_0^t\left({\rho }_0^{\gamma }exp\left\{-\gamma {\int }_x^1{\rho }_{0}vdy{|}_0^s-\gamma t{\int }_x^1{\rho }_{0}dy\right\}exp\left\{-\gamma {\int }_0^s{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}d\tau \right\}\right)ds\right]}^{{\displaystyle \frac{1}{\gamma }}}.\hfill \end{array}$$

(13)

We now bound ${\eta }_x$. In view of (13), one can see that ${\eta }_x(x,t)$ is nonnegative for all $x\in [0,1]$ and $t⩾0$. So, it follows from the basic energy estimates (10) that

$${\int }_0^1{\rho }_0{v}^{2}dx⩽2E_0,t⩾0.$$

Therefore, we have for any $x\in [0,1]$ and $t⩾0$,

$$\left|{\int }_x^1{\rho }_{0}vdy\right|⩽{\left({\int }_0^1{\rho }_{0}dy\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{\rho }_0{v}^{2}dy\right)}^{{\displaystyle \frac{1}{2}}}⩽\sqrt{2m_0{E}_0},$$

(14)

where the initial total mass $m_0={\int }_0^1{\rho }_{0}dy$ is bounded by $C_0$.

Using (10), Hölder’s inequality and Cauchy’s inequality, we obtain

$$\begin{array}{cc}\hfill {\int }_0^t{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}ds& ={\int }_0^t{\int }_0^x{\left({\displaystyle \frac{|{\mathbf{N}}_y{|}^2}{{{\eta }_y}^2}}\right)}_{y}dyds=2{\int }_0^t{\int }_0^x{\mathbf{N}}_y·{\mathbf{N}}_{t}dyds\hfill \\ \hfill & ⩽2{\int }_0^t{\left({\int }_0^x{\eta }_y{\left|{\mathbf{N}}_t\right|}^{2}dy\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^x{\displaystyle \frac{|{\mathbf{N}}_y{|}^2}{{\eta }_y}}dy\right)}^{{\displaystyle \frac{1}{2}}}ds\hfill \\ \hfill & ⩽{\int }_0^t{\int }_0^1{\eta }_y{\left|{\mathbf{N}}_t\right|}^{2}dyds+{\int }_0^t{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_y{|}^2}{{\eta }_y}}dyds⩽{\displaystyle \frac{E_0}{2}}+E_{0}t.\hfill \end{array}$$

(15)

Substituting the results of (14) and (15) into (13) yields the upper and lower bounds for ${\eta }_x$ as follows:

$$\begin{array}{cc}\hfill {\eta }_x(x,t)& ⩽exp\left\{2\sqrt{2m_0{E}_0}+m_{0}t\right\}exp\left\{{\displaystyle \frac{E_0}{2}}+E_{0}t\right\}\hfill \\ \hfill & \times {\left[1+{\displaystyle \frac{C_0^{\gamma }}{E_0}}exp\{2\gamma \sqrt{2m_0{E}_0}+{\displaystyle \frac{E_0}{2}}\gamma \}(e^{\gamma E_{0}t}-1)\right]}^{{\displaystyle \frac{1}{\gamma }}},\hfill \end{array}$$

(16)

$${\eta }_x(x,t)⩾exp\{-2\sqrt{2m_0{E}_0}\}:=B_0>0.$$

(17)

4.3. $L^2$-Estimates on ${\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x$

First, differentiating (7)₂ with respect to t, multiplying the result by ${\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x$ and integrating with respect to the spatial variable, we get

$${\int }_0^1{\mathbf{N}}_{tt}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx={\int }_0^1{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx+{\int }_0^1{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{{\eta }_x}^2}}\right)}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx.$$

It follows from (7)₂, (5) and the fact ${\left|\mathbf{N}\right|}^2=1$ that

$$\begin{array}{cc}\hfill {\int }_0^1{\mathbf{N}}_{tt}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx& ={\displaystyle \frac{d}{dt}}{\int }_0^1{\mathbf{N}}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx-{\int }_0^1{\mathbf{N}}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{xt}dx\hfill \\ \hfill & ={\displaystyle \frac{d}{dt}}{\int }_0^1\left[{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x^2}}\right]·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx+{\int }_0^1{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_t{\mathbf{N}}_{tx}dx\hfill \\ \hfill & ={\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx-{\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^3}}dx\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx-{\int }_0^1{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}}{{\eta }_x^2}}{v}_{x}dx.\hfill \end{array}$$

On the other hand, one has

$$\begin{array}{cc}\hfill & {\int }_0^1{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx\hfill \\ \hfill =& {\int }_0^1{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}_t·{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x{\eta }_{x}dx\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{v_x}{{\eta }_x^2}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx.\hfill \end{array}$$

Combining the above equalities together, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill =& {\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^3}}dx+{\int }_0^1{\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{tx}}{{\eta }_x^2}}{v}_{x}dx\hfill \\ \hfill & +{\int }_0^1{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x^2}}\right)}_t·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{v_x}{{\eta }_x^2}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx=:{\displaystyle \sum _{i=1}^4}{I}_i.\hfill \end{array}$$

It remains to estimate $I_2,I_3,I_4$. By the Sobolev embedding $W^{1,1}\left(I\right)\hookrightarrow L^{\infty }\left(I\right)$ in one dimension, Hölder’s inequality, and (10), one obtains

$$\begin{array}{cc}\hfill {∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2& ⩽c_1{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{W^{1,1}\left(I\right)}^2\hfill \\ \hfill ⩽& c_1\left({∥{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}^2∥}_{L^1}+2{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^1}\right)\hfill \\ \hfill ⩽& c_1\left({∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^2\left(I\right)}^2+2{∥{\displaystyle \frac{{\mathbf{N}}_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}\right)\hfill \\ \hfill ⩽& C(c_1,E_0,B_0)\left(1+{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}\right),\hfill \end{array}$$

(18)

where $c_1$ is a Sobolev constant. We estimate $I_2$ by Cauchy’s inequality with parameter $\epsilon >0$, Hölder’s inequality (to extract $\parallel {\mathbf{N}}_x/{\eta }_x{\parallel }_{L^{\infty }\left(I\right)}^2$), and (18) to close the bound:

$$\begin{array}{cc}\hfill I_2& ⩽\epsilon {\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx+{\displaystyle \frac{1}{4\epsilon }}{\int }_0^1{\displaystyle \frac{v_x^2{\left|{\mathbf{N}}_x\right|}^2}{{{\eta }_x}^3}}dx\hfill \\ \hfill & ⩽\epsilon {\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx+{\displaystyle \frac{1}{4\epsilon }}{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx\hfill \\ \hfill & ⩽\epsilon {\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx+C_1{\epsilon }^{-1}\left({\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+1\right){\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx,\hfill \end{array}$$

For $I_3$, expanding the time derivative inside the integrand gives

$$\begin{array}{c}\hfill I_3={\int }_0^1\left({\displaystyle \frac{2({\mathbf{N}}_x·{\mathbf{N}}_{xt})\mathbf{N}}{{\eta }_x^2}}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\mathbf{N}}_t}{{\eta }_x^2}}-{\displaystyle \frac{2|{\mathbf{N}}_x{|}^2{v}_x\mathbf{N}}{{\eta }_x^3}}\right)·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{x}dx,\end{array}$$

where ${(1/{\eta }_x^2)}_t=-2v_x/{\eta }_x^3$ follows from ${\eta }_t=v$. Differentiating the constraint ${\left|\mathbf{N}\right|}^2=1$ in x gives $\mathbf{N}·{\mathbf{N}}_x=0$ and $\mathbf{N}·{\mathbf{N}}_{xx}=-{\left|{\mathbf{N}}_x\right|}^2$. A direct computation then yields

$$\mathbf{N}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x={\displaystyle \frac{\mathbf{N}·{\mathbf{N}}_{xx}}{{\eta }_x}}-{\displaystyle \frac{(\mathbf{N}·{\mathbf{N}}_x){\eta }_{xx}}{{\eta }_x^2}}=-{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}.$$

(19)

Applying (19) to the first and third terms (those carrying the factor $\mathbf{N}$), together with the Cauchy–Schwarz inequality $|{\mathbf{N}}_x·{\mathbf{N}}_{xt}|⩽|{\mathbf{N}}_x\left|\right|{\mathbf{N}}_{xt}|$ applied to the first term, we obtain

$$\begin{array}{cc}\hfill I_3⩽& C_1{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^4}}{v}_{x}dx+C_1{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^3}{{\eta }_x^3}}\left|{\mathbf{N}}_{xt}\right|dx\hfill \\ \hfill & +C_1{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\left|{\mathbf{N}}_t\sqrt{{\eta }_x}\right|\left|{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|dx=:C_1(J_1+J_2+J_3).\hfill \end{array}$$

We estimate each $J_i$ via Hölder’s inequality, factoring the integrand into a pointwise part and two weighted $L^2$-parts:

$$\begin{array}{cc}\hfill & {\displaystyle \frac{|{\mathbf{N}}_x{|}^4{v}_x}{{\eta }_x^4}}={\displaystyle \frac{|{\mathbf{N}}_x{|}^3}{{\eta }_x^3}}·{\displaystyle \frac{|{\mathbf{N}}_x|}{\sqrt{{\eta }_x}}}·{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}},\hfill \\ \hfill & {\displaystyle \frac{|{\mathbf{N}}_x{|}^3\left|{\mathbf{N}}_{xt}\right|}{{\eta }_x^3}}={\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}·{\displaystyle \frac{|{\mathbf{N}}_x|}{\sqrt{{\eta }_x}}}·{\displaystyle \frac{|{\mathbf{N}}_{xt}|}{\sqrt{{\eta }_x}}},\hfill \\ \hfill & {\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}|{\mathbf{N}}_t\sqrt{{\eta }_x}|·{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|={\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}·\left|{\mathbf{N}}_t\right|\sqrt{{\eta }_x}·{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|.\hfill \end{array}$$

This gives

$$\begin{array}{cc}\hfill J_1& ⩽{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^3{∥{\displaystyle \frac{{\mathbf{N}}_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)},\hfill \\ \hfill J_2& ⩽{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{∥{\displaystyle \frac{{\mathbf{N}}_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{{\mathbf{N}}_{xt}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)},\hfill \\ \hfill J_3& ⩽{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{∥{\mathbf{N}}_t\sqrt{{\eta }_x}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}.\hfill \end{array}$$

Applying (18) to bound the $L^{\infty }$-factors in $J_1,J_2,J_3$, using (10) to absorb $\parallel {\mathbf{N}}_x/\sqrt{{\eta }_x}{\parallel }_{L^2}$, and using Young’s inequality (to handle the fractional power $\parallel {\mathbf{N}}_x/{\eta }_x{\parallel }_{L^{\infty }}^3$ arising in $J_1$) together with Cauchy’s inequality with parameter $\epsilon >0$ (to absorb the $\parallel {\mathbf{N}}_{xt}/\sqrt{{\eta }_x}{\parallel }_{L^2}$ factor in $J_2$), we conclude

$$\begin{array}{cc}\hfill I_3⩽& \epsilon {∥{\displaystyle \frac{{\mathbf{N}}_{xt}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2\hfill \\ \hfill & +C_1{\epsilon }^{-1}\left({∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+{∥{\mathbf{N}}_t\sqrt{{\eta }_x}∥}_{L^2\left(I\right)}^2+1\right)\left({∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^2+1\right).\hfill \end{array}$$

For $I_4$, we factor the integrand as

$${\displaystyle \frac{v_x}{{\eta }_x^2}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^2={\displaystyle \frac{1}{{\eta }_x}}\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|·{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}·{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x,$$

and apply Hölder’s inequality with the first factor in $L^{\infty }$ and the remaining two in weighted $L^2$:

$$|I_4|⩽C_1{∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^{\infty }\left(I\right)}{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}.$$

Applying the Sobolev embedding $W^{1,1}\left(I\right)\hookrightarrow L^{\infty }\left(I\right)$ in one dimension to the squared function gives

$${∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^{\infty }\left(I\right)}⩽{∥{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}^2∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}}+C_1{∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x·{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}_x∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}}.$$

To handle the third-order quantity ${\left({\displaystyle \frac{1}{{\eta }_x}}{({\mathbf{N}}_x/{\eta }_x)}_x\right)}_x$ in the second term, rewrite the director equation (7)₂ as ${\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x={\mathbf{N}}_t-{\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x^2}}$ and differentiate in x:

$${\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}_x={\mathbf{N}}_{tx}-{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x^2}}\right)}_x.$$

(20)

A direct computation using the product rule gives

$${\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2\mathbf{N}}{{\eta }_x^2}}\right)}_x=2\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)\mathbf{N}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}·{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}.$$

(21)

Substituting (20) and (21) into the bound for $|I_4|$ above gives the three-term decomposition

$$|I_4|⩽I_{41}+I_{42}+I_{43},$$

where

$$\begin{array}{cc}\hfill I_{41}& :=C_1{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}{∥{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}^2∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill I_{42}& :=C_1{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x·{\mathbf{N}}_{tx}∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill I_{43}& :=C_1{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}\hfill \\ \hfill & \times {∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x·\left[2\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)\mathbf{N}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}·{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right]∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

We estimate $I_{41}$ using ${\eta }_x⩾B_0$ from (17):

$${∥{\left({\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)}^2∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}}={∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}⩽{\displaystyle \frac{1}{\sqrt{B_0}}}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)},$$

which gives

$$I_{41}⩽C_1{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^2.$$

For $I_{42}$, by Cauchy–Schwarz inequality,

$${∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x·{\mathbf{N}}_{tx}∥}_{L^1\left(I\right)}⩽{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{{\mathbf{N}}_{tx}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)},$$

which gives

$$I_{42}⩽C_1{∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^{{\displaystyle \frac{3}{2}}}{∥{\displaystyle \frac{{\mathbf{N}}_{tx}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^{{\displaystyle \frac{1}{2}}}.$$

We apply Young’s inequality with exponents $(4,4/3)$ to extract the $\parallel {\mathbf{N}}_{tx}/\sqrt{{\eta }_x}{\parallel }_{L^2}^2$-factor for absorption, and a second application of Young converts $\parallel v_x/\sqrt{{\eta }_x}{\parallel }_{L^2}^{4/3}$ into $\parallel v_x/\sqrt{{\eta }_x}{\parallel }_{L^2}^2+1$, yielding

$$I_{42}⩽\epsilon {∥{\displaystyle \frac{{\mathbf{N}}_{tx}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+C_1{\epsilon }^{-1}\left({∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+1\right){∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^2.$$

For $I_{43}$, the two terms inside the square bracket are estimated by the same Hölder three-factor decomposition: factoring the integrand of the $L^1$-norm as

$${\displaystyle \frac{|{\mathbf{N}}_x{|}^3}{{\eta }_x^3}}\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|={\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}·{\displaystyle \frac{|{\mathbf{N}}_x|}{\sqrt{{\eta }_x}}}·{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|,$$

Hölder’s inequality with the first factor in $L^{\infty }$ and the other two in $L^2$, together with (10), yields

$$\begin{array}{cc}\hfill & {∥{\displaystyle \frac{1}{{\eta }_x}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x·\left[2\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}·{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right)\mathbf{N}+{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}·{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right]∥}_{L^1\left(I\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill ⩽& C_1{∥{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}^2∥}_{L^{\infty }\left(I\right)}^{{\displaystyle \frac{1}{2}}}{∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Substituting this bound and applying (18) to the $L^{\infty }$-factor, followed by Young’s inequality, we conclude

$$I_{43}⩽C_1\left({∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+1\right){∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^2.$$

Combining the bounds on $I_{41},I_{42},I_{43}$,

$$|I_4|⩽\epsilon {∥{\displaystyle \frac{{\mathbf{N}}_{tx}}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+C_1{\epsilon }^{-1}\left({∥{\displaystyle \frac{v_x}{\sqrt{{\eta }_x}}}∥}_{L^2\left(I\right)}^2+1\right){∥{\displaystyle \frac{1}{\sqrt{{\eta }_x}}}{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x∥}_{L^2\left(I\right)}^2.$$

By choosing $\epsilon$ small enough, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill ⩽& {\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^3}}dx+C\left({\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+1\right)\left({\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+1\right).\hfill \end{array}$$

Integrating the above inequality with respect to the time variable, we have

$$\begin{array}{cc}\hfill & {\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+{\int }_0^t{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dxds\hfill \\ \hfill ⩽& C_1+{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^4}{{\eta }_x^3}}dx\hfill \\ \hfill & +C_1{\int }_0^t\left({\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+1\right)\left({\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\right)ds\hfill \\ \hfill ⩽& C_1+{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }}^2{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}dx\hfill \\ \hfill & +C_1{\int }_0^t\left({\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+1\right)\left({\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\right)ds\hfill \\ \hfill ⩽& C_1+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\hfill \\ \hfill & +C_1{\int }_0^t\left({\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+1\right)\left({\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\right)ds.\hfill \end{array}$$

Then it follows from Gronwall’s inequality and (10) that

$${\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx⩽C_{1}exp\left\{{\int }_0^t\left({\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+1\right)ds\right\}.$$

Thus, we have

$${\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx+{\int }_0^t{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dxds⩽C_1,$$

(22)

for some positive $C_1$ depending on $E_0,B_0,T,c_1,\gamma$ and

$$\parallel {\mathbf{N}}_{xx}{(x,t=0)\parallel }_{L^2\left(I\right)}={\parallel n_{0xx}\left(x\right)\parallel }_{L^2\left(I\right)}.$$

Combining (18) and (22), we have

$${∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}⩽\sqrt{C(1+C_1)}:=C_2,$$

(23)

which implies—using (16)—that

$$\begin{array}{cc}\hfill {∥{\mathbf{N}}_x∥}_{L^{\infty }\left(I\right)}& ⩽{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}{∥{\eta }_x∥}_{L^{\infty }\left(I\right)}\hfill \\ \hfill & ⩽C_{2}exp\left\{2\sqrt{2m_0{E}_0}\right\}exp\left\{{\displaystyle \frac{E_0}{2}}+E_{0}t\right\}\hfill \\ \hfill & \times {\left[1+{\displaystyle \frac{C_0^{\gamma }}{E_0}}exp\{2\gamma \sqrt{2m_0{E}_0}+{\displaystyle \frac{E_0}{2}}\gamma \}\left(e^{\gamma E_{0}t}-1\right)\right]}^{{\displaystyle \frac{1}{\gamma }}}.\hfill \end{array}$$

(24)

4.4. $L^2$-Estimates on $\sqrt{{\rho }_0}{v}_t$

Using the first time differentiation of Equation (7)₁ produces

$${\rho }_0{v}_{tt}+2{\left({\displaystyle \frac{{\mathbf{N}}_x·{\mathbf{N}}_{xt}}{{{\eta }_x}^2}}-{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{v}_x}{{{\eta }_x}^3}}\right)}_x-\gamma {\left({\displaystyle \frac{{\rho }_0^{\gamma }}{{\eta }_x^{\gamma +1}}}{v}_x\right)}_x={\left({\displaystyle \frac{v_{xt}}{{\eta }_x}}-{\displaystyle \frac{v_x^2}{{{\eta }_x}^2}}\right)}_x.$$

(25)

Multiplying (25) by $v_t$ and integrating the product with respect to the spatial variable, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\rho }_0{v}_t^{2}dx+{\int }_0^1{\displaystyle \frac{v_{tx}^2}{{\eta }_x}}dx\hfill \\ \hfill =& 2{\int }_0^1{\displaystyle \frac{({\mathbf{N}}_x·{\mathbf{N}}_{xt})v_{tx}}{{{\eta }_x}^2}}dx+{\int }_0^1{\displaystyle \frac{v_x^2{v}_{tx}}{{{\eta }_x}^2}}dx-2{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{v}_x{v}_{tx}}{{{\eta }_x}^3}}dx-\gamma {\int }_0^1{\displaystyle \frac{{\rho }_0^{\gamma }{v}_x{v}_{tx}}{{{\eta }_x}^{\gamma +1}}}dx\hfill \\ \hfill ⩽& \left({\delta }_3+{\delta }_4+{\displaystyle \frac{{\delta }_1}{2}}+{\displaystyle \frac{\gamma {\delta }_2}{2}}\right){\int }_0^1{\displaystyle \frac{v_{tx}^2}{{\eta }_x}}dx+{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\delta }_3{\eta }_x}}dx\hfill \\ \hfill & +{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^4{\int }_0^1{\displaystyle \frac{v_x^2}{{\delta }_4{\eta }_x}}dx+{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{v_x^2}{2{\delta }_1{\eta }_x}}dx+\gamma {\int }_0^1{\displaystyle \frac{{\rho }_0^{2\gamma }}{2{\delta }_2{{\eta }_x}^{2\gamma }}}{\displaystyle \frac{v_x^2}{{\eta }_x}}dx,\hfill \end{array}$$

where we have used the integration by parts and the boundary condition (7)_4,5 to obtain the first equality, and used Cauchy’s inequality to derive the last inequality. That means

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\rho }_0{v}_t^{2}dx+{\int }_0^1{\displaystyle \frac{v_{tx}^2}{{\eta }_x}}dx\hfill \\ \hfill ⩽& 16{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}dx+16{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^4{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx\hfill \\ \hfill & +4{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+4{\gamma }^2{\int }_0^1{\displaystyle \frac{{\rho }_0^{2\gamma }}{{{\eta }_x}^{2\gamma }}}{\displaystyle \frac{v_x^2}{{\eta }_x}}dx\hfill \\ \hfill ⩽& 16C_2^2{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}dx+\left(16C_2^4+{\displaystyle \frac{4{\gamma }^2{C}_0^{2\gamma }}{B_0^{2\gamma }}}\right){\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx+4{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dx,\hfill \end{array}$$

(26)

where we have used (4), (17) and (23) to obtain the last inequality. It remains to bound $\parallel v_x/{\eta }_x{\parallel }_{L^{\infty }\left(I\right)}$. Using (11) and (23), one has

$$\begin{array}{cc}\hfill {∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}& ⩽{∥{\int }_x^1{\rho }_0{v}_{t}dy∥}_{L^{\infty }\left(I\right)}+{∥{\int }_x^1{\rho }_{0}dy∥}_{L^{\infty }\left(I\right)}+{∥{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}∥}_{L^{\infty }\left(I\right)}+{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2\hfill \\ \hfill & ⩽\sqrt{m_0}{\left({\int }_0^1{\rho }_0{v}_t^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}+m_0+{∥{\displaystyle \frac{{\rho }_0^{\gamma }}{{{\eta }_x}^{\gamma }}}∥}_{L^{\infty }\left(I\right)}+{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2.\hfill \end{array}$$

Substituting this into (26) and integrating the last inequality of (26) with respect to the time variable, we have

$$\begin{array}{cc}\hfill & {\int }_0^1{\rho }_0{v}_t^{2}dx+{\int }_0^t{\int }_0^1{\displaystyle \frac{v_{tx}^2}{{\eta }_x}}dxds\hfill \\ \hfill ⩽& {\int }_0^1{\rho }_0{v}_t^{2}dx{|}_{t=0}+16C_2^2{\int }_0^t{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{xt}{|}^2}{{\eta }_x}}dxds+\left(16C_2^4+{\displaystyle \frac{4{\gamma }^2{C}_0^{2\gamma }}{B_0^{2\gamma }}}\right){\int }_0^t{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dxds\hfill \\ \hfill & +4{\int }_0^t{\left[\sqrt{m_0}{\left({\int }_0^1{\rho }_0{v}_t^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{C_0^{\gamma }}{B_0^{\gamma }}}+C_2^2\right]}^2{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dxds\hfill \\ \hfill ⩽& {\int }_0^1{\rho }_0{v}_t^{2}dx{|}_{t=0}+16C_1{C}_2^2+\left(16C_2^4+{\displaystyle \frac{4{\gamma }^2{C}_0^{2\gamma }}{B_0^{2\gamma }}}\right)E_0\hfill \\ \hfill & +4{\int }_0^t{\left[\sqrt{m_0}{\left({\int }_0^1{\rho }_0{v}_t^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{C_0^{\gamma }}{B_0^{\gamma }}}+C_2^2\right]}^2{\int }_0^1{\displaystyle \frac{v_x^2}{{\eta }_x}}dxds.\hfill \end{array}$$

Here, the initial data are defined by

$${\int }_0^1{\rho }_0{v}_t^{2}dx{|}_{t=0}:=-{\int }_0^1{\rho }_0^{-1}{s}_x^{2}dx{|}_{t=0}⩽{\parallel {\rho }_0^{-1/2}{s}_{0x}\parallel }_{L^2\left(I\right)}^2⩽A_1^2.$$

The last step of the above inequality is derived from (4), (17), (10), (22), and (23), which implies—using Gronwall’s inequality—that

$${\int }_0^1{\rho }_0{v}_t^{2}dx+{\int }_0^t{\int }_0^1{\displaystyle \frac{v_{sx}^2}{{\eta }_x}}dxds⩽E_1,\forall t\in (0,T],$$

(27)

for some positive $E_1$ depending on $A_1,E_0,B_0,T,C_0,m_0$.

4.5. $L^{\infty }$-Estimates on $v_x$

The $L^{\infty }\left(I\right)$-bound for $v_x$ can be derived from the pointwise estimates of ${\eta }_x$ and $\parallel \sqrt{{\rho }_0}{v}_t{\parallel }_{L^2\left(I\right)}$. One can easily establish the $L^{\infty }\left(I\right)$-bound for $v_x/{\eta }_x$. Using the integral identity (11), along with (10), (17), and (27), together with Cauchy’s inequality, we obtain

$${∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}⩽\sqrt{m_0{E}_1}+m_0+{\displaystyle \frac{C_0^{\gamma }}{B_0^{\gamma }}}+C_2^2=:B_1\left(T\right)<\infty ,\forall t\in [0,T].$$

(28)

This implies—using (16)—that

$$\begin{array}{cc}\hfill {∥v_x∥}_{L^{\infty }\left(I\right)}& ⩽{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}{∥{\eta }_x∥}_{L^{\infty }\left(I\right)}\hfill \\ \hfill & ⩽B_1\left(T\right)exp\left\{2\sqrt{2m_0{E}_0}\right\}exp\left\{{\displaystyle \frac{E_0}{2}}+E_{0}t\right\}\hfill \\ \hfill & \times {\left[1+{\displaystyle \frac{C_0^{\gamma }}{E_0}}exp\{2\gamma \sqrt{2m_0{E}_0}+{\displaystyle \frac{E_0}{2}}\gamma \}\left(e^{\gamma E_{0}t}-1\right)\right]}^{{\displaystyle \frac{1}{\gamma }}}.\hfill \end{array}$$

(29)

4.6. $L^2$-Estimates on ${\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_x$ and ${\mathbf{N}}_t$

Using (23), (22), we have

$$\begin{array}{cc}\hfill {\int }_0^1{\left|{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{{\eta }_x}^2}}\right)}_x\right|}^{2}dx& =4{\int }_0^1{\left|{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right|}^2·{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\hfill \\ \hfill & ⩽4{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2{\parallel {\eta }_x\parallel }_{L^{\infty }\left(I\right)}{\int }_0^1{\displaystyle \frac{1}{{\eta }_x}}{\left|{\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_x\right|}^{2}dx\hfill \\ \hfill & ⩽4C_1{C}_2^2{\parallel {\eta }_x\parallel }_{L^{\infty }\left(I\right)},\hfill \end{array}$$

(30)

where $\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}$ is bounded by using (16). Using the time differentiation of Equation (7)₂ gives

$${\eta }_x{\mathbf{N}}_{tt}+v_x{\mathbf{N}}_t={\left({\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}\right)}_{xt}+{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x}}{\mathbf{N}}_x\right)}_t.$$

(31)

Multiplying the above equation by ${\mathbf{N}}_t$ and integrating the product with respect to the spatial variable, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill =& {\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_x{|}^2{\left|{\mathbf{N}}_t\right|}^2}{{\eta }_x}}dx+{\int }_0^1{\displaystyle \frac{({\mathbf{N}}_{tx}·{\mathbf{N}}_x)v_x}{{\eta }_x}}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{v}_{x}dx\hfill \\ \hfill ⩽& {\int }_0^1{\displaystyle \frac{{\left|{\mathbf{N}}_x\right|}^2{\left|{\mathbf{N}}_t\right|}^2}{{\eta }_x}}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{{\left|{\mathbf{N}}_{tx}\right|}^2}{{\eta }_x}}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{{\left|{\mathbf{N}}_x\right|}^2{v}_x^2}{{\eta }_x}}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{v}_{x}dx,\hfill \end{array}$$

where we have used the integration by parts and the fact ${\mathbf{N}}_x=0$ on the boundary to obtain the first equality, and used Cauchy’s inequality to derive the last inequality. Hence,

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dx\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{B_0}}\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)}^2\parallel v_x{\parallel }_{L^{\infty }\left(I\right)}^2+\left(2{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^2+{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}\right){\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{B_0}}\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)}^2\parallel v_x{\parallel }_{L^{\infty }\left(I\right)}^2+(2C_2^2+B_1){\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx.\hfill \end{array}$$

Integrating the above inequality with respect to the time variable, using (10), (23) and (28), one gets

$$\begin{array}{cc}\hfill & {\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dx+{\int }_0^t{\int }_0^1{\displaystyle \frac{|{\mathbf{N}}_{tx}{|}^2}{{\eta }_x}}dxds\hfill \\ \hfill ⩽& {\int }_0^1|{\mathbf{N}}_t{|}^2{\eta }_{x}dx{|}_{t=0}+{\displaystyle \frac{1}{B_0}}{\int }_0^t\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)}^2\parallel v_x{\parallel }_{L^{\infty }\left(I\right)}^{2}ds+(2C_2^2+B_1){\int }_0^t{\int }_0^1{\left|{\mathbf{N}}_t\right|}^2{\eta }_{x}dxds\hfill \\ \hfill ⩽& {\int }_0^1|{\mathbf{N}}_t{|}^2{\eta }_{x}dx{|}_{t=0}+{\displaystyle \frac{1}{B_0}}{\int }_0^t\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)}^2{\parallel v_x\parallel }_{L^{\infty }\left(I\right)}^{2}ds+(2C_2^2+B_1)E_0,\hfill \end{array}$$

(32)

where $\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)},{\parallel v_x\parallel }_{L^{\infty }\left(I\right)}$ are bounded by (24) and (29). Here, the initial data are given by

$${\int }_0^1|{\mathbf{N}}_t{|}^2{\eta }_{x}dx{|}_{t=0}:={\parallel q_0\parallel }_{L^2\left(I\right)}^2<A_1^2.$$

4.7. $L^2$-Estimates on $v_{xx}$ and ${\eta }_{xx}$

Equation (7)₁ can be rewritten as

$$v_{xx}=\left({\displaystyle \frac{v_x}{{\eta }_x}}-\gamma {\displaystyle \frac{{\rho }_0^{\gamma }}{{\eta }_x^{\gamma }}}\right){\eta }_{xx}+{\rho }_0{v}_t{\eta }_x+{\displaystyle \frac{{\left({\rho }_0^{\gamma }\right)}_x}{{\eta }_x^{\gamma -1}}}+{\left({\displaystyle \frac{{\left|{\mathbf{N}}_x\right|}^2}{{\eta }_x^2}}\right)}_x{\eta }_x+{\rho }_0{\eta }_x.$$

(33)

It then follows from (4), (28), (17), and (5) that

$$\begin{array}{cc}\hfill {∥v_{xx}∥}_{L^2\left(I\right)}⩽& \left[{∥{\displaystyle \frac{v_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}+\gamma {∥{\displaystyle \frac{{\rho }_0}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}^{\gamma }\right]{∥{\eta }_{xx}∥}_{L^2\left(I\right)}+\mathcal{Q}\left(t\right)\hfill \\ \hfill ⩽& \mathcal{P}{\int }_0^t{\parallel v_{xx}(·,s)\parallel }_{L^2\left(I\right)}ds+\mathcal{Q}\left(t\right),\hfill \end{array}$$

where $\mathcal{P}$ is a positive constant, given by $\mathcal{P}=B_1\left(T\right)+\gamma {\left({\displaystyle \frac{C_0}{B_0}}\right)}^{\gamma }<\infty$, and

$$\begin{array}{cc}\hfill \mathcal{Q}\left(t\right)=& \parallel \sqrt{{\rho }_0}{v}_t{\parallel }_{L^2\left(I\right)}\parallel \sqrt{{\rho }_0}{\parallel }_{L^{\infty }\left(I\right)}\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}+\parallel {\eta }_x^{1-\gamma }{\parallel }_{L^{\infty }\left(I\right)}{\parallel {\left({\rho }_0^{\gamma }\right)}_x\parallel }_{L^2\left(I\right)}\hfill \\ \hfill & +{∥{\left({\displaystyle \frac{|{\mathbf{N}}_x{|}^2}{{\eta }_x^2}}\right)}_x∥}_{L^2\left(I\right)}\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}+\parallel {\rho }_0{\parallel }_{L^{\infty }\left(I\right)}{\parallel {\eta }_x\parallel }_{L^{\infty }\left(I\right)}\hfill \\ \hfill ⩽& \sqrt{C_0{E}_1}\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}+B_0^{1-\gamma }{A}_0+2\sqrt{C_1}{C}_2\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}^{{\displaystyle \frac{3}{2}}}+C_0{\parallel {\eta }_x\parallel }_{L^{\infty }\left(I\right)},\hfill \end{array}$$

Here, we use (4), (8), (17), (27) and (30). Note the boundedness of $\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)}$ by (16). Then, by Gronwall’s inequality, we get for $\forall t>0$ that

$${∥v_{xx}(·,t)∥}_{L^2\left(I\right)}⩽\mathcal{P}{\int }_0^t{e}^{\mathcal{P}(t-s)}\mathcal{Q}\left(s\right)ds+\mathcal{Q}\left(t\right),$$

(34)

$${∥{\eta }_{xx}(·,t)∥}_{L^2\left(I\right)}⩽{\int }_0^t{e}^{\mathcal{P}(t-s)}\mathcal{Q}\left(s\right)ds.$$

(35)

From Lemma 1, we have

$$\begin{array}{cc}\hfill {∥v∥}_{L^{\infty }\left(I\right)}⩽& {∥v_x∥}_{L^1\left(I\right)}+\left|{\int }_0^1{\rho }_{0}vdx\right|⩽{∥v_x∥}_{L^{\infty }\left(I\right)}+\left|{\int }_0^1{\rho }_{0}vdx\right|\hfill \\ \hfill ⩽& \sqrt{2m_0{E}_0}+B_{1}exp\left\{2\sqrt{2m_0{E}_0}\right\}exp\left\{{\displaystyle \frac{E_0}{2}}+E_{0}t\right\}\hfill \\ \hfill & \times {\left[1+{\displaystyle \frac{C_0^{\gamma }}{E_0}}exp\{2\gamma \sqrt{2m_0{E}_0}+{\displaystyle \frac{E_0}{2}}\gamma \}(e^{\gamma E_{0}t}-1)\right]}^{{\displaystyle \frac{1}{\gamma }}},\hfill \end{array}$$

(36)

where the last inequality follows from (14) and (29).

4.8. $L^2$-Estimates on ${\mathbf{N}}_{xx}$

Equation (7)₂ can be rewritten as

$${\mathbf{N}}_{xx}={\eta }_x^2{\mathbf{N}}_t+{\displaystyle \frac{{\eta }_{xx}{\mathbf{N}}_x}{{\eta }_x}}-{\left|{\mathbf{N}}_x\right|}^2\mathbf{N},$$

(37)

then we have

$$\begin{array}{cc}\hfill {∥{\mathbf{N}}_{xx}∥}_{L^2\left(I\right)}⩽& {∥{\eta }_x∥}_{L^{\infty }\left(I\right)}^{3/2}{∥{\mathbf{N}}_t\sqrt{{\eta }_x}∥}_{L^2\left(I\right)}+{∥{\displaystyle \frac{{\mathbf{N}}_x}{{\eta }_x}}∥}_{L^{\infty }\left(I\right)}{∥{\eta }_{xx}∥}_{L^2\left(I\right)}+{∥{\mathbf{N}}_x∥}_{L^{\infty }\left(I\right)}^2\hfill \\ \hfill ⩽& {∥{\eta }_x∥}_{L^{\infty }\left(I\right)}^{3/2}{\left(C_3+{\displaystyle \frac{1}{B_0}}{\int }_0^t{∥{\mathbf{N}}_x∥}_{L^{\infty }\left(I\right)}^2{∥v_x∥}_{L^{\infty }\left(I\right)}^{2}ds+\left(2C_2^2+B_1\right)E_0\right)}^{1/2}\hfill \\ \hfill & +C_2\parallel {\eta }_{xx}{\parallel }_{L^2\left(I\right)}+{\parallel {\mathbf{N}}_x\parallel }_{L^{\infty }\left(I\right)}^2,\hfill \end{array}$$

(38)

where $\parallel {\eta }_x{\parallel }_{L^{\infty }\left(I\right)},\parallel {\eta }_{xx}{\parallel }_{L^2\left(I\right)},\parallel {\mathbf{N}}_x{\parallel }_{L^{\infty }\left(I\right)},{\parallel v_x\parallel }_{L^{\infty }\left(I\right)}$ are bounded by (16), (24), (29), and (35).

Combining (24), (27), (29), (32), (34), (36), and (38), we obtain the estimate (9) for almost every $t\in [0,T]$, which gives

$$v\in L^{\infty }(0,T;H^2\left(I\right)),\mathbf{N}\in L^{\infty }(0,T;H^2\left(I\right)),$$

and

$$\sqrt{{\rho }_0}{v}_t\in L^{\infty }(0,T;L^2\left(I\right)),{\mathbf{N}}_t\in L^{\infty }(0,T;L^2\left(I\right)).$$

By the Sobolev embedding $H^2\left(I\right)\hookrightarrow C^1\left(\overline{I}\right)$ in one dimension, it follows that

$$v(·,t),\mathbf{N}(·,t)\in C^1\left(\overline{I}\right)\mathrm{for}\mathrm{almost}\mathrm{every}t\in [0,T].$$

Moreover, since $H^2\left(I\right)\hookrightarrow H^1\left(I\right)$ is a compact embedding in one dimension, applying Lemma 2(ii) with $X=H^2\left(I\right)$, $E=H^1\left(I\right)$, $Y=L^2\left(I\right)$ yields

$$v,\mathbf{N}\in C([0,T];H^1\left(I\right))\hookrightarrow C([0,T];C\left(\overline{I}\right)),$$

where the last embedding follows from $H^1\left(I\right)\hookrightarrow C\left(\overline{I}\right)$ in one dimension. In particular, the trace $v(1,t)$ is well-defined and continuous in t on $[0,T]$. Combined with $\dot{a}\left(t\right)=v(1,t)$ from (5), we conclude that the free boundary $a\left(t\right)=\eta (1,t)$ is a $C^1$ curve on $[0,T]$. More explicitly, the kinematic condition $\dot{a}\left(t\right)=v(1,t)$ identifies $\dot{a}$ as the time-continuous function $t\mapsto v(1,t)$ on $[0,T]$, and integration in time gives $a\left(t\right)=a\left(0\right)+{\int }_0^{t}v(1,s)ds$, so that $a\in C^1\left([0,T]\right)$ with $\dot{a}\in C\left([0,T]\right)$. This $C^1$-regularity is strictly stronger than the Lipschitz regularity that would follow from $v_x\in L^{\infty }(I\times [0,T])$ alone (see (29)); the corresponding free boundary in the global weak solution framework of [14] is not addressed in a pointwise sense. This finishes the proof of Theorem 1.

5. Uniqueness

In this section, we establish the uniqueness part of Theorem 1. Local-in-time existence and uniqueness of strong solutions to the present free boundary problem can be obtained by a finite difference scheme of the type developed by Ou, Shi, and Wittwer [32] for the planar magnetohydrodynamic free boundary problem with degenerate viscosity, which features a strong coupling between the velocity field and an auxiliary vector field (the magnetic field) analogous to the velocity-director coupling of the present system. The same type of construction has also been employed by Mei [27] (via Schauder’s fixed-point theorem) for the non-isothermal liquid crystal free boundary problem. Combined with the global a priori estimates established in Section 4 and a standard continuation argument, this would yield both global existence and uniqueness. However, since the global a priori estimates derived in Section 4 already provide all the strong-solution bounds needed in a direct energy argument, we find it more transparent to present a self-contained uniqueness proof based on the difference of two strong solutions. This direct argument also makes explicit how the liquid crystal coupling enters the energy structure in the isentropic setting.

Let $({\eta }^i,v^i,{\mathbf{N}}^i)$, $i=1,2$, be two strong solutions of system (7) on $I\times [0,T]$ with the regularity stated in Theorem 1 and the same initial data $({\rho }_0,u_0,n_0)$. Define the differences

$$\overline{\eta }:={\eta }^1-{\eta }^2,\overline{v}:=v^1-v^2,\overline{\mathbf{N}}:={\mathbf{N}}^1-{\mathbf{N}}^2.$$

(39)

Throughout this section, C denotes a generic positive constant depending on T, $\gamma$, $C_0$, $E_0$, $B_0$, and on the bounds $\parallel {\eta }_x^i{\parallel }_{L^{\infty }}$, $\parallel v_x^i{\parallel }_{L^{\infty }}$, $\parallel {\mathbf{N}}_x^i{\parallel }_{L^{\infty }}$, $\parallel \sqrt{{\rho }_0}{v}_t^i{\parallel }_{L^2}$ that have been established in Section 4 for $i=1,2$, but is independent of the particular pair of solutions.

From ${\eta }_t=v$ and $\eta (x,0)=x$ we immediately obtain

$$\overline{\eta }(x,t)={\int }_0^t\overline{v}(x,s)ds,{\overline{\eta }}_x(x,t)={\int }_0^t{\overline{v}}_x(x,s)ds,$$

(40)

so that pointwise control of $\overline{\eta }$ and ${\overline{\eta }}_x$ reduces to integral control of $\overline{v}$ and ${\overline{v}}_x$. In particular, by Cauchy–Schwarz,

$$\parallel {\overline{\eta }}_x{(·,t)\parallel }_{L^2\left(I\right)}^2\le t{\int }_0^t{\parallel {\overline{v}}_x(·,s)\parallel }_{L^2\left(I\right)}^{2}ds(0\le t\le T).$$

(41)

5.1. The Equation for $\overline{v}$

Subtracting Equation (7)₁ written for the two solutions, we obtain

$${\rho }_0{\overline{v}}_t+{\left({\displaystyle \frac{{\rho }_0^{\gamma }}{{\left({\eta }_x^1\right)}^{\gamma }}}-{\displaystyle \frac{{\rho }_0^{\gamma }}{{\left({\eta }_x^2\right)}^{\gamma }}}\right)}_x+{\left({\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}-{\displaystyle \frac{|{\mathbf{N}}_x^2{|}^2}{{\left({\eta }_x^2\right)}^2}}\right)}_x={\left({\displaystyle \frac{v_x^1}{{\eta }_x^1}}-{\displaystyle \frac{v_x^2}{{\eta }_x^2}}\right)}_x.$$

(42)

We rewrite each nonlinear difference as a product of a known coefficient with one of ${\overline{\eta }}_x$ or ${\overline{\mathbf{N}}}_x$.

(a)

The pressure difference.

By the mean-value theorem, there exists $\xi (x,t)$ between ${\eta }_x^1(x,t)$ and ${\eta }_x^2(x,t)$ such that

$${\displaystyle \frac{1}{{\left({\eta }_x^1\right)}^{\gamma }}}-{\displaystyle \frac{1}{{\left({\eta }_x^2\right)}^{\gamma }}}=-\gamma {\xi }^{-\gamma -1}{\overline{\eta }}_x.$$

Since ${\eta }_x^i\ge B_0>0$ uniformly by (17), we have

$${\displaystyle \frac{{\rho }_0^{\gamma }}{{\left({\eta }_x^1\right)}^{\gamma }}}-{\displaystyle \frac{{\rho }_0^{\gamma }}{{\left({\eta }_x^2\right)}^{\gamma }}}=-\gamma {\rho }_0^{\gamma }{\xi }^{-\gamma -1}{\overline{\eta }}_x,$$

(43)

where the coefficient $\gamma {\rho }_0^{\gamma }{\xi }^{-\gamma -1}$ is bounded in $L^{\infty }(I\times [0,T])$ by $\gamma C_0^{\gamma }{B}_0^{-\gamma -1}$.

(b)

The director coupling difference.

A direct computation gives

$${\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}-{\displaystyle \frac{|{\mathbf{N}}_x^2{|}^2}{{\left({\eta }_x^2\right)}^2}}={\displaystyle \frac{({\mathbf{N}}_x^1+{\mathbf{N}}_x^2)·{\overline{\mathbf{N}}}_x}{{\left({\eta }_x^1\right)}^2}}-{\displaystyle \frac{|{\mathbf{N}}_x^2{|}^2({\eta }_x^1+{\eta }_x^2)}{{\left({\eta }_x^1\right)}^2{\left({\eta }_x^2\right)}^2}}{\overline{\eta }}_x.$$

(44)

By (17) and the $L^{\infty }$-bound (24) on ${\mathbf{N}}_x^i$, both coefficients on the right-hand side of (44) are uniformly bounded on $I\times [0,T]$.

(c)

The viscous difference.

$${\displaystyle \frac{v_x^1}{{\eta }_x^1}}-{\displaystyle \frac{v_x^2}{{\eta }_x^2}}={\displaystyle \frac{{\overline{v}}_x}{{\eta }_x^1}}-{\displaystyle \frac{v_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}.$$

(45)

Substituting (43)–(45) into (42) we obtain

$$\begin{array}{cc}\hfill {\rho }_0{\overline{v}}_t-{\left({\displaystyle \frac{{\overline{v}}_x}{{\eta }_x^1}}\right)}_x=& {\left(\gamma {\rho }_0^{\gamma }{\xi }^{-\gamma -1}{\overline{\eta }}_x\right)}_x-{\left({\displaystyle \frac{({\mathbf{N}}_x^1+{\mathbf{N}}_x^2)·{\overline{\mathbf{N}}}_x}{{\left({\eta }_x^1\right)}^2}}\right)}_x\hfill \\ \hfill & +{\left({\displaystyle \frac{|{\mathbf{N}}_x^2{|}^2({\eta }_x^1+{\eta }_x^2)}{{\left({\eta }_x^1\right)}^2{\left({\eta }_x^2\right)}^2}}{\overline{\eta }}_x\right)}_x-{\left({\displaystyle \frac{v_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}\right)}_x.\hfill \end{array}$$

(46)

The boundary conditions for $\overline{v}$ are inherited from (7)_4,5: $\overline{v}(0,t)=0$ and ${\overline{v}}_x(1,t)=0$, with initial datum $\overline{v}(x,0)=0$.

5.2. The Equation for $\overline{\mathbf{N}}$

Subtracting Equation (7)₂ yields, after introducing the bridging term ${\displaystyle \frac{1}{{\eta }_x^1}}{\left({\displaystyle \frac{{\mathbf{N}}_x^2}{{\eta }_x^1}}\right)}_x$ in the diffusion difference and the bridging term ${\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}{\mathbf{N}}^2$ in the lower-order difference,

$${\overline{\mathbf{N}}}_t={\displaystyle \frac{1}{{\eta }_x^1}}{\left({\displaystyle \frac{{\overline{\mathbf{N}}}_x}{{\eta }_x^1}}\right)}_x+{\mathcal{R}}_1+{\mathcal{R}}_2+{\mathcal{R}}_3,$$

(47)

where

$${\mathcal{R}}_1:={\displaystyle \frac{1}{{\eta }_x^1}}{\left({\displaystyle \frac{{\mathbf{N}}_x^2}{{\eta }_x^1}}\right)}_x-{\displaystyle \frac{1}{{\eta }_x^2}}{\left({\displaystyle \frac{{\mathbf{N}}_x^2}{{\eta }_x^2}}\right)}_x,{\mathcal{R}}_2:={\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}\overline{\mathbf{N}},$$

$${\mathcal{R}}_3:=\left({\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}-{\displaystyle \frac{|{\mathbf{N}}_x^2{|}^2}{{\left({\eta }_x^2\right)}^2}}\right){\mathbf{N}}^2.$$

The remainder ${\mathcal{R}}_3$ is rewritten using (44) (replacing ${\overline{\mathbf{N}}}_x,{\overline{\eta }}_x$ by the corresponding expressions on the right of (44) and multiplying by ${\mathbf{N}}^2$). For ${\mathcal{R}}_1$, we further introduce the bridging term ${\displaystyle \frac{1}{{\eta }_x^2}}{\left({\displaystyle \frac{{\mathbf{N}}_x^2}{{\eta }_x^1}}\right)}_x$ to split it into two pieces, the second of which is a divergence-form term:

$${\mathcal{R}}_1=-{\displaystyle \frac{{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}\left[{\displaystyle \frac{{\mathbf{N}}_{xx}^2}{{\eta }_x^1}}-{\displaystyle \frac{{\mathbf{N}}_x^2{\eta }_{xx}^1}{{\left({\eta }_x^1\right)}^2}}\right]-{\displaystyle \frac{1}{{\eta }_x^2}}{\partial }_x\left[{\displaystyle \frac{{\mathbf{N}}_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}\right].$$

(48)

The boundary conditions for $\overline{\mathbf{N}}$ are ${\overline{\mathbf{N}}}_x(0,t)={\overline{\mathbf{N}}}_x(1,t)=0$, inherited from (7)_4,5, and the initial datum is $\overline{\mathbf{N}}(x,0)=0$.

5.3. Energy Estimate for $\overline{v}$

Multiply (46) by $\overline{v}$ and integrate over I. Integration by parts in the diffusive term, using $\overline{v}(0,t)=0$ and ${\overline{v}}_x(1,t)=0$, gives

$$-{\int }_I{\left({\displaystyle \frac{{\overline{v}}_x}{{\eta }_x^1}}\right)}_x\overline{v}dx={\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dx.$$

For each of the four divergence-form terms on the right of (46), integration by parts produces no boundary contributions, since at $x=1$ the coefficients ${\rho }_0^{\gamma }$, ${\mathbf{N}}_x^i$ and $v_x^2$ all vanish (by (4) and (7)_4,5), while $\overline{v}(0,t)=0$ at the other endpoint. Therefore,

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

Each of the four right-hand-side coefficients is bounded in $L^{\infty }(I\times [0,T])$ by (4), (17), (24) and (29). Denote any one of the four right-hand-side integrals by ${\int }_{I}KW{\overline{v}}_{x}dx$, where $K\in L^{\infty }(I\times [0,T])$ is the relevant coefficient and W is either ${\overline{\eta }}_x$ or ${\overline{\mathbf{N}}}_x$. Cauchy’s inequality with parameter $\epsilon >0$ yields

$$\left|{\int }_{I}KW{\overline{v}}_{x}dx\right|\le \epsilon {\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dx+C{\epsilon }^{-1}\left({\int }_I|{\overline{\eta }}_x{|}^{2}dx+{\int }_I{\left|{\overline{\mathbf{N}}}_x\right|}^{2}dx\right).$$

Choosing $\epsilon$ sufficiently small to absorb the $|{\overline{v}}_x{|}^2/{\eta }_x^1$ contributions into the left-hand side, we obtain

$${\displaystyle \frac{d}{dt}}{\int }_I{\rho }_0{\left|\overline{v}\right|}^{2}dx+{\displaystyle \frac{1}{2}}{\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dx\le C\left({\int }_I|{\overline{\eta }}_x{|}^{2}dx+{\int }_I{\left|{\overline{\mathbf{N}}}_x\right|}^{2}dx\right).$$

By (16), we have ${\eta }_x^1\le C\left(T\right)$, hence

$${\int }_I{\left|{\overline{\mathbf{N}}}_x\right|}^{2}dx={\int }_I{\eta }_x^1·{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx\le C\left(T\right){\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx,$$

(49)

which yields

$${\displaystyle \frac{d}{dt}}{\int }_I{\rho }_0{\left|\overline{v}\right|}^{2}dx+{\displaystyle \frac{1}{2}}{\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dx\le C\left({\int }_I{\left|{\overline{\eta }}_x\right|}^{2}dx+{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx\right).$$

(50)

5.4. Energy Estimate for $\overline{\mathbf{N}}$

Multiply (47) by ${\eta }_x^1\overline{\mathbf{N}}$ and integrate over I. For the time-derivative term, since ${\left({\eta }_x^1\right)}_t=v_x^1$,

$${\int }_I{\eta }_x^1{\overline{\mathbf{N}}}_t·\overline{\mathbf{N}}dx={\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x^1|\overline{\mathbf{N}}{|}^{2}dx-{\displaystyle \frac{1}{2}}{\int }_I{v}_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx.$$

For the principal diffusion term, integration by parts using ${\overline{\mathbf{N}}}_x(0,t)={\overline{\mathbf{N}}}_x(1,t)=0$ gives

$${\int }_I\overline{\mathbf{N}}·{\left({\displaystyle \frac{{\overline{\mathbf{N}}}_x}{{\eta }_x^1}}\right)}_{x}dx=-{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx.$$

Hence

$${\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x^1|\overline{\mathbf{N}}{|}^{2}dx-{\displaystyle \frac{1}{2}}{\int }_I{v}_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx+{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx={\int }_I{\eta }_x^1\overline{\mathbf{N}}·({\mathcal{R}}_1+{\mathcal{R}}_2+{\mathcal{R}}_3)dx.$$

We estimate the right-hand side term by term.

5.4.1. The ${\mathcal{R}}_2$ Contribution

Since ${\mathcal{R}}_2={\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\left({\eta }_x^1\right)}^2}}\overline{\mathbf{N}}$, by $\parallel {\mathbf{N}}_x^1{\parallel }_{L^{\infty }}\le C$ from (24) and $1/{\eta }_x^1\le 1/B_0$ from (17),

$$\left|{\int }_I{\eta }_x^1\overline{\mathbf{N}}·{\mathcal{R}}_{2}dx\right|=\left|{\int }_I{\displaystyle \frac{|{\mathbf{N}}_x^1{|}^2}{{\eta }_x^1}}{\left|\overline{\mathbf{N}}\right|}^{2}dx\right|\le C{\int }_I{\left|\overline{\mathbf{N}}\right|}^{2}dx.$$

5.4.2. The ${\mathcal{R}}_3$ Contribution

By (44),

$${\mathcal{R}}_3=\left({\mathbf{b}}_1·{\overline{\mathbf{N}}}_x+b_2{\overline{\eta }}_x\right){\mathbf{N}}^2,$$

where $\parallel {\mathbf{b}}_1{\parallel }_{L^{\infty }}+{\parallel b_2\parallel }_{L^{\infty }}\le C$ by (17) and (24), and $|{\mathbf{N}}^2|=1$. Hence

$${\int }_I{\eta }_x^1\overline{\mathbf{N}}·{\mathcal{R}}_{3}dx={\int }_I{\eta }_x^1\overline{\mathbf{N}}·{\mathbf{N}}^2\left({\mathbf{b}}_1·{\overline{\mathbf{N}}}_x+b_2{\overline{\eta }}_x\right)dx,$$

and using ${\eta }_x^1\le C\left(T\right)$ from (16) together with Cauchy’s inequality with parameter $\epsilon >0$,

$$\left|{\int }_I{\eta }_x^1\overline{\mathbf{N}}·{\mathcal{R}}_{3}dx\right|\le \epsilon {\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx+C{\epsilon }^{-1}\left({\int }_I|{\overline{\eta }}_x{|}^{2}dx+{\int }_I{\left|\overline{\mathbf{N}}\right|}^{2}dx\right).$$

5.4.3. The ${\mathcal{R}}_1$ Contribution

For the first piece in (48), the coefficient ${\displaystyle \frac{1}{{\eta }_x^1{\eta }_x^2}}\left[{\displaystyle \frac{{\mathbf{N}}_{xx}^2}{{\eta }_x^1}}-{\displaystyle \frac{{\mathbf{N}}_x^2{\eta }_{xx}^1}{{\left({\eta }_x^1\right)}^2}}\right]$ belongs to $L^{\infty }(0,T;L^2\left(I\right))$ by (17), (24), (35), and (38). Using Hölder’s inequality followed by the one-dimensional Sobolev embedding $\parallel \overline{\mathbf{N}}{\parallel }_{L^{\infty }}\le C(\parallel \overline{\mathbf{N}}{\parallel }_{L^2}+\parallel {\overline{\mathbf{N}}}_x{\parallel }_{L^2})$,

$$\left|{\int }_I{\eta }_x^1\overline{\mathbf{N}}·\left(-{\displaystyle \frac{{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}\right)\left[{\displaystyle \frac{{\mathbf{N}}_{xx}^2}{{\eta }_x^1}}-{\displaystyle \frac{{\mathbf{N}}_x^2{\eta }_{xx}^1}{{\left({\eta }_x^1\right)}^2}}\right]dx\right|\le \epsilon {\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx+C{\epsilon }^{-1}\left({\int }_I|{\overline{\eta }}_x{|}^{2}dx+{\int }_I{\left|\overline{\mathbf{N}}\right|}^{2}dx\right).$$

For the second (divergence-form) piece, integration by parts gives

$$-{\int }_I{\eta }_x^1\overline{\mathbf{N}}·{\displaystyle \frac{1}{{\eta }_x^2}}{\partial }_x\left[{\displaystyle \frac{{\mathbf{N}}_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}\right]dx={\int }_I{\partial }_x\left({\displaystyle \frac{{\eta }_x^1}{{\eta }_x^2}}\overline{\mathbf{N}}\right)·{\displaystyle \frac{{\mathbf{N}}_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}dx,$$

where the boundary terms vanish thanks to ${\mathbf{N}}_x^2(0,t)={\mathbf{N}}_x^2(1,t)=0$. Expanding the derivative ${\partial }_x({\eta }_x^1/{\eta }_x^2\overline{\mathbf{N}})={({\eta }_x^1/{\eta }_x^2)}_x\overline{\mathbf{N}}+({\eta }_x^1/{\eta }_x^2){\overline{\mathbf{N}}}_x$ and proceeding as before (the coefficient ${({\eta }_x^1/{\eta }_x^2)}_x={\eta }_{xx}^1/{\eta }_x^2-{\eta }_x^1{\eta }_{xx}^2/{\left({\eta }_x^2\right)}^2$ lies in $L^{\infty }(0,T;L^2\left(I\right))$ by (17) and (35), while ${\eta }_x^1/{\eta }_x^2\in L^{\infty }(I\times [0,T])$ by (16) and (17)), we obtain

$$\left|{\int }_I{\partial }_x\left({\displaystyle \frac{{\eta }_x^1}{{\eta }_x^2}}\overline{\mathbf{N}}\right)·{\displaystyle \frac{{\mathbf{N}}_x^2{\overline{\eta }}_x}{{\eta }_x^1{\eta }_x^2}}dx\right|\le \epsilon {\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx+C{\epsilon }^{-1}\left({\int }_I|{\overline{\eta }}_x{|}^{2}dx+{\int }_I{\left|\overline{\mathbf{N}}\right|}^{2}dx\right).$$

5.4.4. The $v_x^1{\left|\overline{\mathbf{N}}\right|}^2$ Term

By (29),

$$\left|{\displaystyle \frac{1}{2}}{\int }_I{v}_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx\right|\le C{\int }_I{\left|\overline{\mathbf{N}}\right|}^{2}dx.$$

Summing the four contributions and choosing $\epsilon$ small enough to absorb the $|{\overline{\mathbf{N}}}_x{|}^2/{\eta }_x^1$ terms into the left-hand side, we arrive at

$${\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx+{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx\le C\left({\int }_I|\overline{\mathbf{N}}{|}^{2}dx+{\int }_I{\left|{\overline{\eta }}_x\right|}^{2}dx\right).$$

(51)

5.5. Closing the Grönwall Argument

Define

$$\begin{array}{cc}\hfill \mathcal{E}\left(t\right)& :=\kappa {\int }_I{\rho }_0|\overline{v}{|}^2(x,t)dx+{\int }_I{\eta }_x^1{\left|\overline{\mathbf{N}}\right|}^2(x,t)dx,\hfill \\ \hfill \mathcal{D}\left(t\right)& :=\kappa {\int }_I{\displaystyle \frac{|{\overline{v}}_x{(x,t)|}^2}{{\eta }_x^1(x,t)}}dx+{\displaystyle \frac{1}{2}}{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{(x,t)|}^2}{{\eta }_x^1(x,t)}}dx,\hfill \end{array}$$

where $\kappa >0$ is a small constant, chosen below, depending only on the constant C appearing in (50).

Multiplying (50) by $\kappa >0$ and adding to (51), the left-hand side becomes

$$\kappa {\displaystyle \frac{d}{dt}}{\int }_I{\rho }_0|\overline{v}{|}^{2}dx+{\displaystyle \frac{d}{dt}}{\int }_I{\eta }_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx+{\displaystyle \frac{\kappa }{2}}{\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dx+{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx,$$

while the right-hand side is bounded by

$$\kappa C\left({\int }_I{\left|{\overline{\eta }}_x\right|}^{2}dx+{\int }_I{\displaystyle \frac{|{\overline{\mathbf{N}}}_x{|}^2}{{\eta }_x^1}}dx\right)+C\left({\int }_I|\overline{\mathbf{N}}{|}^{2}dx+{\int }_I{\left|{\overline{\eta }}_x\right|}^{2}dx\right).$$

Choosing $\kappa$ small enough that $\kappa C\le 1/2$, the term $\kappa C{\int }_I{|{\overline{\mathbf{N}}}_x|}^2/{\eta }_x^{1}dx$ on the right is absorbed into half of ${\int }_I{\left|{\overline{\mathbf{N}}}_x\right|}^2/{\eta }_x^{1}dx$ on the left, leading to

$${\displaystyle \frac{d}{dt}}\mathcal{E}\left(t\right)+\mathcal{D}\left(t\right)\le C{\int }_I|\overline{\mathbf{N}}{|}^{2}dx+C{\int }_I{\left|{\overline{\eta }}_x\right|}^{2}dx,$$

where $\mathcal{E}\left(t\right)$ and $\mathcal{D}\left(t\right)$ are as defined above (now with the same $\kappa$). By ${\eta }_x^1\ge B_0$ from (17), ${\int }_I|\overline{\mathbf{N}}{|}^{2}dx\le B_0^{-1}{\int }_I{\eta }_x^1{\left|\overline{\mathbf{N}}\right|}^{2}dx\le B_0^{-1}\mathcal{E}\left(t\right)$. By (41) together with ${\eta }_x^1\le C\left(T\right)$ from (16),

$${\int }_I|{\overline{\eta }}_x{|}^{2}dx\le T{\int }_0^t{\int }_I{\left|{\overline{v}}_x(·,s)\right|}^{2}dxds\le TC\left(T\right){\int }_0^t{\int }_I{\displaystyle \frac{|{\overline{v}}_x{(·,s)|}^2}{{\eta }_x^1(·,s)}}dxds\le C{\int }_0^t\mathcal{D}\left(s\right)ds.$$

Therefore,

$${\displaystyle \frac{d}{dt}}\mathcal{E}\left(t\right)+\mathcal{D}\left(t\right)\le C\mathcal{E}\left(t\right)+C{\int }_0^t\mathcal{D}\left(s\right)ds.$$

(52)

Setting $\mathcal{F}\left(t\right):=\mathcal{E}\left(t\right)+{\int }_0^t\mathcal{D}\left(s\right)ds$, inequality (52) implies

$${\displaystyle \frac{d\mathcal{F}}{dt}}\left(t\right)\le C\mathcal{F}\left(t\right),\mathcal{F}\left(0\right)=0,$$

since $\overline{v}(·,0)\equiv 0$ and $\overline{\mathbf{N}}(·,0)\equiv 0$ by the assumed common initial data. Grönwall’s inequality then yields $\mathcal{F}\left(t\right)\equiv 0$ for $t\in [0,T]$, hence in particular

$${\int }_I{\rho }_0|\overline{v}{|}^2(x,t)dx\equiv 0,{\int }_I{\eta }_x^1{\left|\overline{\mathbf{N}}\right|}^2(x,t)dx\equiv 0,{\int }_0^T{\int }_I{\displaystyle \frac{|{\overline{v}}_x{|}^2}{{\eta }_x^1}}dxds=0.$$

The second identity, together with ${\eta }_x^1\ge B_0>0$, gives $\overline{\mathbf{N}}\equiv 0$ almost everywhere on $I\times [0,T]$. The third identity, together with $1/{\eta }_x^1\ge 1/C\left(T\right)>0$ from (16), gives ${\overline{v}}_x\equiv 0$ almost everywhere on $I\times [0,T]$; combined with $\overline{v}(0,t)=0$ and the regularity $\overline{v}\in C([0,T];H^1\left(I\right))\hookrightarrow C([0,T];C\left(\overline{I}\right))$ inherited from $v^1,v^2$, this implies $\overline{v}\equiv 0$ on $I\times [0,T]$. The same continuity argument upgrades $\overline{\mathbf{N}}\equiv 0$ from a.e. to everywhere. Finally, (40) gives $\overline{\eta }\equiv 0$. This completes the proof of uniqueness. □

6. Conclusions

In this paper, we have established the global existence and uniqueness of strong solutions to the one-dimensional vacuum free boundary problem for the isentropic compressible hydrodynamic flow of nematic liquid crystals under the influence of gravity, where the initial density is allowed to vanish continuously at the free boundary. To the best of our knowledge, this is the first global strong solution result for this problem.

The key novelty lies in establishing, for any $T>0$, an $L^{\infty }$-bound on the ratio $v_x/{\eta }_x$ on $[0,T]$, which, together with the pointwise time-uniform lower bound and pointwise finite-time upper bound for the Jacobian ${\eta }_x$ gives a finite $L^{\infty }$-bound on the velocity gradient $v_x$ on any finite time interval $[0,T]$. This bound is strong enough to guarantee that the free boundary $a\left(t\right)=\eta (1,t)$ is well-defined as a $C^1$ curve, which improves upon the global weak solutions of [14], where the free boundary was not shown to be well-defined in a pointwise sense.

Several questions remain open, including the global existence of classical solutions and the long-time asymptotic behavior of the strong solutions obtained here.

The extension of our analysis to higher spatial dimensions $d⩾2$ remains an interesting open problem. The Lagrangian reformulation of Section 2 relies on the one-dimensional scalar structure of the Jacobian ${\eta }_x$, and the $L^{\infty }$-bound on $v_x/{\eta }_x$ established in Section 4 makes essential use of the one-dimensional integral ${\int }_x^1{\rho }_{0}vdy$; neither of these has an immediate multi-dimensional counterpart. Partial results under spherical or axisymmetric symmetry have been obtained for related models (see [22] for the isentropic Navier–Stokes setting), but the general multi-dimensional vacuum free boundary problem for the compressible liquid crystal system requires substantially new ideas and lies beyond the scope of the present paper.

From a numerical standpoint, simulating the vacuum free boundary problem treated in this paper requires careful handling of the degeneracy at the interface and the unit-sphere constraint on the director. Modern high-resolution schemes for compressible flows (see, e.g., [33]) provide one possible starting point, and the design of a numerical method tailored to the present setting is a natural direction for future work.