Stability of Viscous Shock Wave for One-Dimensional Compressible Navier-Stokes System

Author NameAffiliation
Lin ChangSchool of Mathematical Sciences, Beihang University, Beijing 100191, China

Abstract:

Key words:initial value problemviscous shockasymptotic stability
DOI：10.11916/j.issn.1005-9113.21052
Clc Number:O1
Fund:

Lin Chang. Stability of Viscous Shock Wave for One-Dimensional Compressible Navier-Stokes System[J]. Journal of Harbin Institute of Technology (New Series), 2022, 29(4): 49-57. DOI: 10.11916/j.issn.1005-9113.21052
Corresponding author Lin Chang, Ph.D.Student.E-mail: changlin23@buaa.edu.cn Article history Received: 2021-10-16



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Stability of Viscous Shock Wave for One-Dimensional Compressible Navier-Stokes System
Lin Chang
School of Mathematical Sciences, Beihang University, Beijing 100191, China
Received: 2021-10-16; Available online: 2022-03-11
Corresponding author: Lin Chang, Ph.D.Student.E-mail: changlin23@buaa.edu.cn.

Abstract: In this article, the nonlinear stability of viscous shock wave for 1-D compressible Navier-Stokes system is studied. By the standard local existence method, it is found that the solution exists on a finite time interval [0, T] (T < ∞). However, this method is not available for global existence since the solution may blow up as time t tends to infinity. Thus a priori estimate needs to be established, which can reduce the upper bound of the solution on the time interval [0, T]. Moreover, the bound of the solution at time t=T is made equal to the bound at the initial time. By the same method, it is known the solution exists on [T, 2T], [2T, 3T], ….Thus the global existence of the solution is obtained. During the process of obtaining a priori estimate by the standard method, some additional conditions are proposed. To weaken those conditions, two suitable weighted functions were chosen, a double side weighted energy method was used, and a priori estimate was obtained under some weaker conditions. Thus when the adiabatic exponent γ satisfies 1 < γ < 1.5, the solution not only exists globally but also tends to a viscous shock wave as time goes to infinity.
Keywords: initial value problemviscous shockasymptotic stability
0 Introduction In this paper, the initial value problem of one-dimensional compressible non-isentropic Navier-Stokes system for ideal gas is investigated, which reads in the Lagrangian coordinate as^[1]
$\left\{\begin{array}{l}v_t-u_x=0, \\u_t+p_x=\left(\mu \frac{u_x}{v}\right)_x, (t, x) \in \mathbb{R}^{+} \times \mathbb{R} \\\left(e+\frac{u^2}{2}\right)_t+(p u)_x=\left(\kappa \frac{\theta_x}{v}\right)_x+\left(\mu \frac{u u_x}{v}\right)_x\end{array}\right.$ (1)
The unknown function v(x, t)>0 represents the specific volume, the unknown function u(x, t) is the velocity, the unknown function p(x, t) represents the pressure, the unknown function e(x, t) stands for the internal energy, and the unknown function θ(x, t)>0 is the absolute temperature. The positive coefficient μ stands for the heat-conductivity coefficient. The coefficient κ>0 stands for the viscosity coefficient. For ideal gas, there is^[2]
$\begin{aligned}p=R \frac{\theta}{v} &=B v^{-\gamma} \exp \left\{\frac{\gamma-1}{R} S\right\} \\e &=\frac{R}{\gamma-1} \theta+C\end{aligned}$
where B, C are two positive constants, R>0, S, γ>1 are the gas constant, specific entropy, and the adiabatic exponent, respectively.
The initial value problem was investigated for expression (1) equipped with the initial data
$(v, u, \theta)(0, x)=\left(v_0, u_0, \theta_0\right)(x), x \in \mathbb{R}$ (2)
and the far field condition:
$(v, u, \theta)(t, x) \rightarrow\left\{\begin{array}{l}\left(\bar{v}_r, \bar{u}_l, \bar{\theta}_l\right), x \rightarrow-\infty \\\left(\bar{v}_r, \bar{u}_r, \bar{\theta}_r\right), x \rightarrow \infty\end{array}\right.$
where $\bar{v}_r>0, \bar{v}_l>0, \bar{u}_r, \bar{u}_l, \bar{\theta}_r>0, \bar{\theta}_r>0$, are six given constants. The Navier-Stokes expression (1) were established to describe the motion of general fluid, such as water, air, and so on. A lot of attention has been paid to this model from the beginning due to its importance. One of the most important problems is the stability of the viscous shock wave. The focus is on the compressible case and readers may refer to Refs. [3-12] for the general conservation laws.
Consider the case where entropy S is a constant. So the pressure becomes p=p(v)=a(v)^-γ for constant a. Thus expression (1) become
$\left\{\begin{array}{l}v_t-u_x=0 \\u_t+p_x=\left(\mu \frac{u_x}{v}\right)_x\end{array}\right.$ (3)
and the corresponding initial condition is
$(v, u)(t, x) \rightarrow\left\{\begin{array}{l}\left(\bar{v}_l, \bar{u}_l\right), x \rightarrow-\infty \\\left(\bar{v}_r, \bar{u}_r\right), x \rightarrow \infty\end{array}\right.$ (4)
When μ is a constant, Matsumura and Nishihara^[13] first studied the stability of viscous shock to Cauchy problem as shown in expressions (3) and (4), provided that $(\gamma-1)\left|\bar{v}_r-\bar{v}_l\right|$ is sufficiently small. It means that $\left|\bar{v}_r-\bar{v}_l\right|$ could be large if the adiabatic exponent γ tends to 1. The condition was improved to be $\left|\bar{v}_r-\bar{v}_l\right| < C(\gamma-1)^{-2}$ as in Ref. [14] later. For the case μ=μ(v)=μ₀v^-α, it is shown in Ref. [15] that the strong shock wave (the strength of shock wave $\left|\bar{v}_r-\bar{v}_l\right|$ need not be small) is asymptotically stable by a clever weighted energy method as $\alpha>\frac{\gamma-1}{2}$. Vasseur and Yao^[16] removed the condition $\alpha>\frac{\gamma-1}{2}$ by using an effective velocity. The result of Ref. [16] is extended by He and Huang^[17] to be more general p(v) and μ(v).
For the non-isentropic case, as shown in expressions (1) and (2), the situation becomes more complex than the isentropic case as shown in expressions (3) and (4). If $(\gamma-1)\left|\bar{v}_r-\bar{v}_l\right|$ is sufficiently small, Kawashima and Matsumura^[1] proved that the shock wave for the initial value problem in expressions (1) and (2) is non-linearly stable when μ and κ are constants. Afterward, if the strengths of two shock waves are "small with same order, " Huang and Matsumura^[18] obtained the stability of the composite of two shocks.
When γ∈(1, 1.5) (including the air) and $2 \gamma p_v\left(\bar{v}_r\right)>p_v\left(\bar{v}_l\right)$, the nonlinear stability of viscous shock wave was obtained for the initial value problem as shown in expressions (1) and (2), provided that the strength of shock satisfies that "(γ-1)³· $\left|\bar{v}_r-\bar{v}_l\right|^2$ is sufficiently small", which partially improves the stability condition that "$(\gamma-1)\left|\bar{v}_r-\bar{v}_l\right|$ is sufficiently small" in Ref. [1]. Due to the involvement of θ(t, x), the effective velocity was introduced in Refs. [16] and [17], which does not help to simplify the system but make it more complicated. Enlightened by Wang and Matsumura^[15], a double side weighted energy method was used to solve this problem by choosing two suitable weight functions. Now, μ=κ=R=1 is assumed for simple calculation.
The rest of the article is arranged as follows. In Section 1, some properties of the viscous shock wave are proposed and the main result is described. In Section 2, the original problem is reformulated by the anti-derivative method. In Section 3, the a prior estimation is obtained by a weighted energy estimate.
1 Viscous Shock Wave and Main Result Set ζ=x-σt, where σ is a constant indicating the speed of the shock, which will be determined later. The travel wave solution (v^s, u^s, θ^s)(ζ) of expression (1) is considered to be
$\left\{\begin{array}{l}-\sigma\left(v^s\right)^{\prime}-\left(u^s\right)^{\prime}=0 \\-\sigma\left(u^s\right)^{\prime}+\left(p^s\right)^{\prime}=\left(\frac{\left(u^s\right)^{\prime}}{v^s}\right)^{\prime} \\-\sigma\left(e^s+\frac{\left(u^s\right)^2}{2}\right)^{\prime}+\left(u^s\right)^{\prime}=\left(\frac{\left(\theta^s\right)^{\prime}}{v^s}\right)^{\prime}+\left(\frac{u^s\left(u^s\right)^{\prime}}{v^s}\right)^{\prime}\end{array}\right.$ (5)
And
$ \left(v^s, u^s, \theta^s\right)(\zeta) \longrightarrow\left\{\begin{array}{l}\left(\bar{v}_l, \bar{u}_l, \bar{\theta}_l\right), \zeta \rightarrow-\infty \\\left(\bar{v}_r, \bar{u}_r, \bar{\theta}_r\right), \zeta \rightarrow \infty\end{array}\right. $ (6)
where
$\begin{gathered}\prime=\frac{\mathrm{d}}{\mathrm{d} \zeta}, p^s=\frac{\theta^s}{v^s} \\e^s=\frac{\theta^s}{\gamma-1}+\text { constant }\end{gathered}$
Integrating expression (5) concerning ζ with (ζ, ∞) or (ζ, -∞), there is
$\left(\begin{array}{l}-\sigma v^s-u^s=-\sigma \bar{v}_{r, l}-\bar{u}_{r, l} \\-\sigma u^s+p^s-\frac{\left(u^s\right)^{\prime}}{v^s}=-\sigma \bar{u}_{r, l}+\bar{p}_{r, l} \\-\sigma\left(\frac{\theta^s}{\gamma-1}+\frac{\left(u^s\right)^2}{2}\right)+p^s u^s-\frac{\left(\theta^s\right)^{\prime}}{v^s}-\frac{u^s\left(u^s\right)^{\prime}}{v^s}= \\\quad-\sigma\left(\frac{\bar{\theta}_{r, l}}{\gamma-1}+\frac{\bar{u}_{r, l}^2}{2}\right)+\bar{p}_{r, l} \bar{u}_{r, l}\end{array}\right.$ (7)
where $\bar{p}_r=\frac{\bar{\theta}_r}{\bar{v}_r}, \bar{p}_l=\frac{\bar{\theta}_l}{\bar{v}_l}$. From expression (7), it is known that the constants $\bar{p}_r, \bar{\theta}_r, \bar{v}_r, \bar{p}_l, \bar{\theta}_l, \bar{v}_l$ and σ should satisfy R-H condition^[1]:
$\left\{\begin{array}{l}-\sigma\left(\bar{v}_r-\bar{v}_l\right)-\left(\bar{u}_r-\bar{u}_l\right)=0 \\-\sigma\left(\bar{u}_r-\bar{u}_l\right)+\left(\frac{\bar{\theta}_r}{\bar{v}_r}-\frac{\bar{\theta}_l}{\bar{v}_l}\right)=0 \\-\sigma\left\{\left(\frac{\bar{\theta}_r}{\gamma-1}+\frac{\bar{u}_r^2}{2}\right)-\left(\frac{\bar{\theta}_l}{\gamma-1}+\frac{\bar{u}_l^2}{2}\right)\right\}+ \\ \;\;\;\;\;\;\left(\bar{p}_r \bar{u}_r-\bar{p}_l \bar{u}_l\right)=0\end{array}\right.$ (8)
For simplicity, the focus of this paper is on the case σ>0 (third shock). Furthermore, it is assumed that the shock speed σ satisfies the following Lax's shock condition:
$\frac{\sqrt{\gamma \bar{\theta}_r}}{\bar{v}_r} <\sigma<\frac{\sqrt{\gamma \bar{\theta}_l}}{\bar{v}_l}$ (9)
Lemma 1.1^{[1, 18]} ? Suppose the constants $\left(\bar{v}_r, \bar{u}_r\right.$$\left.\bar{\theta}_r, \bar{v}_l, \bar{u}_l, \bar{\theta}_l\right)$ and σ satisfy expressions (8) and (9). Then the ordinary differential expressions (5) and (6) have a unique solution (v^s, u^s, θ^s)(ζ) for any ζ∈R satisfying
$\left\{\begin{array}{l}\left|\left(v^s\right)^{\prime \prime}\right|, \left|\left(\theta^s\right)^{\prime \prime}\right| \leqslant C\left|\bar{v}_r-\bar{v}_l\right| \\\left|\frac{\left(\theta^s\right)^{\prime}}{\left(v^s\right)^{\prime}}\right| \leqslant C(\gamma-1) \\\left|\left(v^s\right)^{\prime}\right| \leqslant C\left|\bar{v}_r-\bar{v}_l\right|^2 \\\sigma\left(v^s\right)^{\prime}=-\left(u^s\right)^{\prime}>0, \sigma\left(\theta^s\right)^{\prime} <0\end{array}\right.$ (10)
Next, the main result of this study is to be described. Suppose that
$\left(v_0-v^s, u_0-u^s, \theta_0-\theta^s\right) \in H^1$ (11)
and for ?x∈R, the integrals Ψ₀(x), Φ₀(x), Z₀(x) exist, as follows,
$\begin{array}{l}{\mathit{\Psi }_0}(x) = \int_{ - \infty }^x {\left( {{v_0} - {v^s}} \right)} (y){\rm{d}}y\\{\mathit{\Phi }_0}(x) = \int_{ - \infty }^x {\left( {{u_0} - {u^s}} \right)} (y){\rm{d}}y\\{{\bar Z}_0}(x) = \int_{ - \infty }^x {\left( {\frac{{{\theta _0}}}{{\gamma - 1}} - \frac{{{\theta ^s}}}{{\gamma - 1}} + \frac{{u_0^2}}{2} - \frac{{{{\left( {{u^s}} \right)}^2}}}{2}} \right)} (y){\rm{d}}y\end{array}$
Furthermore, Define ${Z_0}: = (\gamma - 1)\left({{{\bar Z}_0} - {u^s}{\mathit{\Phi }_0}} \right)$, and assume (Ψ₀, Φ₀, Z₀) satisfies
$\left( {{\mathit{\Psi }_0}, {\mathit{\Phi }_0}, {Z_0}} \right) \in {L^2}$ (12)
According to Eq.(6) and Lemma 1.1, it is known that |u^s| is bounded. Combining with (Ψ₀, Z₀)∈L², and by the Minkowski inequality, Z₀∈L² is obtained. Thus expression(12) becomes
$\left( {{\mathit{\Psi }_0}, {\mathit{\Phi }_0}, {{\bar Z}_0}} \right)( + \infty ) = 0$
So
$\left\{\begin{array}{l}\int_{-\infty}^{+\infty}\left(v_0-v^s\right)(x) \mathrm{d} x=0 \\\int_{-\infty}^{+\infty}\left(u_0-u^s\right)(x) \mathrm{d} x=0 \\\int_{-\infty}^{+\infty}\left(\frac{\theta_0}{\gamma-1}-\frac{\theta^s}{\gamma-1}+\frac{u_0^2}{2}-\frac{\left(u^s\right)^2}{2}\right)(x) \mathrm{d} x=0\end{array}\right.$
Let v_- < v₊, u₊and θ_- < θ₊ be any fixed positive constants. Then it is assumed that the six constants $\bar{v}_l, \bar{u}_l \bar{v}_r, \bar{u}_r, \bar{\theta}_r$ and ${\bar \theta _l}$ in Eq.(8) satisfy the following inequalities:
$\begin{gathered}\bar{v}_l, \bar{v}_r \in\left[v_{-}, v_{+}\right], \left|\bar{u}_{r, } \bar{u}_l\right| \in\left[0, u_{+}\right], \\\bar{\theta}_{r, } \bar{\theta}_l \in\left[\theta_{-}, \theta_{+}\right]\end{gathered}$ (13)
Define
$\begin{aligned}N(0):=&\left\|{\mathit{\Psi}}_0, {\mathit{\Phi}}_0, Z_0(\gamma-1)^{-\frac{1}{2}}\right\|+\\&\left\|v_0-v^s, u_0-u^s, \left(\theta_0-\theta^s\right)(\gamma-1)^{-\frac{1}{2}}\right\|_1\end{aligned}$
Then the result can be stated as:
Theorem 1.1 ? Assume ($\bar{v}_r, \bar{u}_r, \bar{\theta}_r, \bar{v}_l, \bar{u}_l, \bar{\theta}_l$) and σ>0 are seven given constants that satisfy expressions (8), (9), and (13), and (v^s, u^s, θ^s)(ζ) is the viscous shock wave given in Lemma 1.1. The initial data (v₀(x), u₀(x), θ₀(x)) is assumed to satisfy expressions (11) and (12). Then there are two positive constants δ₀ and ε₀, which are independent of γ and ($\bar{v}_r, \bar{u}_r, \bar{\theta}_r, \bar{v}_l, \bar{u}_l, \bar{\theta}_l$), such that if γ∈(1, 1.5), N(0) < δ₀, and
$\left\{\begin{array}{l}2 \gamma p_v\left(\bar{v}_r\right)>p_v\left(\bar{v}_l\right) \\(\gamma-1)^3\left|\bar{v}_r-\bar{v}_l\right|^2 <\varepsilon_0\end{array}\right.$ (14)
The Cauchy problem as shown expressions (1) and (2) has a unique global solution (v, u, θ)(x, t) such that
$\left\{ {\begin{array}{*{20}{l}}{v - {v^s} \in {C^0}\left( {0, \infty ;{H^1}} \right) \cap {L^2}\left( {0, \infty ;{H^1}} \right)}\\{\left( {u - {u^s}, \theta - {\theta ^s}} \right) \in {C^0}\left( {0, \infty ;{H^1}} \right) \cap {L^2}\left( {0, \infty ;{H^2}} \right)}\\{\mathop {\inf }\limits_{(x, t) \in \mathbb{R} \times \mathbb{R}^{+}} \{ v(t, x), \theta (t, x)\} > 0}\end{array}} \right.$
and the long time behavior
$\begin{aligned}&\sup\limits_{x \in \mathbb{R}}\left|(v, u, \theta)(t, x)-\left(v^s, u^s, \theta^s\right)(x-\sigma t)\right| \rightarrow 0 \\&\text { as } t \rightarrow \infty.\end{aligned}$
Remark 1.1 ? When $\bar{v}_l$ is suitably large, expression (14) is weaker than the stability condition $(\gamma-1)\left|\bar{v}_r-\bar{v}_l\right| \ll 1$ in Ref. [1]. In this sense, Theorem 1.1 partially improves the result of Ref. [1] when γ∈(1, 1.5).
2 Reformulation of Problem The perturbation is defined as
$\begin{gathered}(\chi, \phi, z)(t, x)=(v, u, \theta)(t, x)-\left(v^s, u^s, \right. \\\left.\theta^s\right)(\zeta), \zeta=x-\sigma t\end{gathered}$
With the aid of expressions (1) and (5), it is known the perturbation (χ, ?, z) satisfies
$\left\{ {\begin{array}{*{20}{l}}{{\chi _t} - {\phi _x} = 0}\\\begin{array}{l}{\phi _t} - \frac{\xi }{{{v^s}}}{\chi _x} + \frac{{{z_x}}}{{{v^s}}} - {\left( {\frac{{{\phi _x}}}{{{v^s}}}} \right)_x} - {\left( {\frac{\xi }{{{v^s}}}} \right)_x}\chi + \\\;\;\;\;{\left( {\frac{1}{{{v^s}}}} \right)_x}z = {f_1}\end{array}\\{\frac{{{z_t}}}{{\gamma - 1}} + \xi {\phi _x} - {{\left( {\frac{{{z_x}}}{{{v^s}}}} \right)}_x} + {{\left( {\frac{{{{\left( {{\theta ^s}} \right)}^{\prime} }\chi }}{{{{\left( {{v^s}} \right)}^2}}}} \right)}_x} - }\\{\frac{{{{\left( {{u^s}} \right)}^{\prime} }}}{{{v^s}}}\left( {\xi - z + {\phi _x}} \right) = {f_2}}\end{array}} \right.$ (15)
where
$\begin{aligned}f_1=&{\left( {\frac{{p - {p^s}}}{{{v^s}}}\chi - \frac{{{\phi _x}\chi }}{{v{v^s}}} + \frac{{{{\left( {{u^s}} \right)}^{\prime} }}}{{v{{\left( {{v^s}} \right)}^2}}}{\chi ^2}} \right)_x}=\\& O(1)\left(|\chi, z|^2+|\chi, z|\left|\chi_x, \phi_x, z_x\right|+\right.\\&\left.\left|\chi_x \phi_x\right|+|\chi|\left|\phi_{x x}\right|\right)\end{aligned}$ (16)
and
$\begin{aligned}f_2=& \frac{1}{v}\left(\xi-z+\phi_x\right)\left(\phi_x-\frac{\left(u^s\right)^{\prime} \chi}{v^s}\right)+\\& {\left[\frac{\chi}{v V}\left(z_x-\frac{\left(\theta^s\right)^{\prime} \chi}{v^s}\right)\right]_x=O(1)\left(|\chi, z|^2+\right.} \\&|\chi, z|\left|\chi_x, \phi_x, z_x\right|+\left|\chi_x, \phi_x\right|\left|\phi_x, z_x\right|+\\&\left.|\chi|\left|z_{x x}\right|\right)\end{aligned}$ (17)
with
$\begin{gathered}\left(\chi_0, \phi_0, z_0\right)(x):=(\chi, \phi, z)(0, x)= \\\left(v_0-v^s, u_0-u^s, \theta_0-\theta^s\right)(x)\end{gathered}$
Motivated by Ref. [1], $\bar{z}: =\frac{z}{\gamma-1}+\frac{\phi^2}{2}+u^s \phi$ is set. The anti-derivative variables are introduced as
$(\mathit{\Psi }, \mathit{\Phi }, \bar{Z})(t, x):=\int_{-\infty}^x(\chi, \phi, \bar{z})(t, y) \mathrm{d} y$
By introducing Z∶ =(γ-1)(Z -u^sΦ), ξ=p^s- $\frac{\left(u^s\right)^{\prime}}{v^s}$, expression (15) can be rewritten as
$\left\{ {\begin{array}{*{20}{l}}{{\mathit{\Psi }_t} - {\mathit{\Phi }_x} = 0}\\{{\mathit{\Phi }_t} - \frac{\xi }{{{v^s}}}{\mathit{\Psi }_x} + \frac{{{Z_x}}}{{{v^s}}} + \frac{{\gamma - 1}}{{{v^s}}}{{\left( {{u^s}} \right)}^{\prime} }\mathit{\Phi } = \frac{{{\mathit{\Phi }_{xx}}}}{{{v^s}}} + {F_1}}\\{\frac{1}{{\gamma - 1}}{Z_t} + \xi {\mathit{\Phi }_x} + u_t^s\mathit{\Phi } - \frac{{\gamma - 1}}{{{v^s}}}{{\left( {{{\left( {{u^s}} \right)}^{\prime} }\mathit{\Phi }} \right)}_x} + }\\{\;\;\;\;\;\;\;\;\;\frac{{{{\left( {{\theta ^s}} \right)}^{\prime} }}}{{{{\left( {{v^s}} \right)}^2}}}{\mathit{\Psi }_x} = \frac{{{Z_{xx}}}}{{{v^s}}} + {F_2}}\end{array}} \right.$ (18)
where
$\begin{array}{l}{F_1} = \frac{{p - {p^s}}}{{{v^s}}}{\mathit{\Psi }_x} + \frac{{\gamma - 1}}{{2{v^s}}}\mathit{\Phi }_x^2 - \frac{{{\mathit{\Phi }_{xx}}{\mathit{\Psi }_x}}}{{v{v^s}}} + \\\;\;\;\frac{{{{\left( {{u^s}} \right)}^{\prime} }\mathit{\Psi }_x^2}}{{v{{\left( {{v^s}} \right)}^2}}} = O(1){\left| {\left( {\chi , \phi , z, {\phi _x}} \right)} \right|^2}\end{array}$ (19)
and
$\begin{aligned}F_2 &=-\left(p-p^s\right) \mathit{\Phi }_x-\frac{\gamma-1}{v^s} \mathit{\Phi }_x \mathit{\Phi }_{x x}-\frac{\mathit{\Psi }_x z_x}{v v^s}+\\& \frac{\left(\theta^s\right)^{\prime}}{v\left(v^s\right)^2} \mathit{\Psi }_x^2+\left(\frac{u_x}{v}-\frac{\left(u^s\right)^{\prime}}{v^s}\right) \mathit{\Phi }_x=\\& O(1)\left|\left(\chi, \phi z, \phi_x z_x\right)\right|^2\end{aligned}$ (20)
with
$\left(\mathit{\Psi }_0, \mathit{\Phi }_0, Z_0\right)(x):=(\mathit{\Psi }, \mathit{\Phi }, Z)(0, x)$ (21)
The solution space of the Cauchy problem as shown in expressions (18) and (21) is defined as
$\begin{gathered}\mathit{\Psi }_x \in L^2\left(0, T ; H^1\right) \\\left(\mathit{\Phi }_x, Z_x\right) \in L^2\left(0, T ; H^2\right) \\N(t):=\sup\limits_{0 <\tau \leqslant t}\left\{\left\|\mathit{\Psi }, \mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}(\tau)\right\|+\right. \\\left.\left.\left\|\chi, \phi, z(\gamma-1)^{-\frac{1}{2}}(\tau)\right\|_1\right\}<\delta\right\}\end{gathered}$
Moreover, the perturbation $\delta < \frac{1}{2} \min \left\{v_{-}, \theta_{-}\right\} \ll 1$ is small. Since the local existence of the solution of expressions (18) and (21) can be proven classically, only the following priori estimate needs to be established.
Proposition 2.1 ? (a priori estimate) It is assumed that $(\mathit{\Psi }, \mathit{\Phi }, Z) \in \mathbb{B}_\delta(0, T)$ is the solution of the problem as shown in expressions (18) and (21) for some T>0. Then there are three positive constants ε₁, δ₁, and C₁, which are independent of T, γ, and $\left(\bar{v}_{r, l}, \bar{u}_{r, l}, \bar{\theta}_{r, l}\right), (\gamma-1)^3\left|\bar{v}_r-\bar{v}_l\right|^2 \leqslant \varepsilon_1, 2 \gamma p_v\left(\bar{v}_r\right)>$ $p_v\left(\bar{v}_l\right)$, and δ < δ₁, it follows that
$\begin{aligned}&\left\|\left(\mathit{\Psi }, \mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2(t)+ \\&\quad\left\|\left(\chi, \phi, z(\gamma-1)^{-\frac{1}{2}}\right)\right\|_1^2(t)+ \\&\quad \int_0^t\left(\left\|\sqrt{\left(v^s\right)^{\prime}}\left(\mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2+\|\chi\|_1^2+\right.\\&\left.\|(\phi, z)\|_2^2\right)(\tau) \mathrm{d} \tau \leqslant C_1\left(\|\left(\mathit{\Psi }_0, \mathit{\Phi }_0, Z_0(\gamma-\right.\right. \\&\left.\left.1)^{-\frac{1}{2}}\right)\left\|^2+\right\|\left(\chi_0, \phi_0, z_0(\gamma-1)^{-\frac{1}{2}}\right) \|_1^2\right), \\&\forall t \in[0, T]\end{aligned}$ (22)
The local solution (Ψ, Φ, Z) can be extended to T=+∞, provided that Proposition 2.1 is obtained.
Theorem 2.1 ? It is assumed that the initial data $\left(\mathit{\Psi }_0, \mathit{\Phi }_0, Z_0\right) \in H^2$. Then there are three positive constants ε₀, δ₀, and C₀ independent of γ, $\left({{{\bar v}_r}, {{\bar u}_r}, {{\bar \theta }_r}} \right., \left. {{{\bar v}_l}, {{\bar u}_l}, {{\bar \theta }_l}} \right) $, such that if $(\gamma-1)^3\left|\bar{v}_r-\bar{v}_l\right|^2 \leqslant \varepsilon_0$, $2 \gamma p_v\left(\bar{v}_r\right)>p_v\left(\bar{v}_l\right)$, and N(0) < δ₀, expressions (18) and (19) have a global solution (Ψ, Φ, Z)∈$\mathbb{B}_{\delta_0}(0, +\infty)$. It satisfies
$\begin{gathered}\sup\limits_{t \geqslant 0}\left(\left\|\left(\mathit{\Psi }, \mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2+\right. \\\left.\left\|\left(\chi, \phi, z(\gamma-1)^{-\frac{1}{2}}\right)\right\|_1^2\right)(t)+ \\\int_0^{\infty}\left(\| \sqrt{\left(v^s\right)^{\prime}}\left(\mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)^2+\right. \\\left.\|\chi\|_1^2+\|(\phi, z)\|_2^2\right)(\tau) \mathrm{d} \tau \leqslant \\C_0\left(\left\|\left(\mathit{\Psi }_0, \mathit{\Phi }_0, Z_0(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2+\right. \\\left.\left\|\left(\chi_0, \phi_0, z_0(\gamma-1)^{-\frac{1}{2}}\right)\right\|_1^2\right)\end{gathered}$
With Theorem 2.1, Theorem 1.1 can be proved directly. So our primary mission in the next section is to prove Proposition 2.1.
3 Proof of Proposition 2.1 In this section, the proof of priori estimation is discussed emphatically. For T>0, it is supposed that expressions (18)-(20) have a solution (Ψ, Φ, Z)∈$\mathbb{B}_{\delta_0}(0, +\infty)$, then there is
$\begin{aligned}\|z\|_1=&(\gamma-1)^{\frac{1}{2}}\left\|z(\gamma-1)^{-\frac{1}{2}}\right\|_1 <\\&\left\|z(\gamma-1)^{-\frac{1}{2}}\right\|_1 \leqslant \delta\end{aligned}$ (23)
Based on expression (23), there is
$\sup \limits_{t \in[0, T]}\|(\chi, z)\|_1 \leqslant \delta$
Note that $\delta < \frac{1}{2} \min \left\{v_{-}, \theta_{-}\right\}$, then using the Sobolev's inequality, for $(x, t) \in \mathbb{R} \times[0, T]$, there is
$\left\{\begin{array}{l}v=\left(v^s+\chi\right) \geqslant v_{-}-\|\chi\|_1 \geqslant \frac{1}{2} v_{-}>0 \\\theta=\left(\theta^s+z\right) \geqslant \theta_{-}-\|z\|_1 \geqslant \frac{1}{2} \theta_{-}>0\end{array}\right.$ (24)
Expression (24) ensures v and θ have positive lower bounds on [0, T].
Lemma 3.1 ? Under the same assumptions of Proposition 2.1, there are constants C and ε independent of T. If
$(\gamma-1)^3\left|\bar{v}_r-\bar{v}_l\right|^2 \leqslant \varepsilon_0, 2 \gamma\left|p_v\left(\bar{v}_r\right)\right|>\left|p_v\left(\bar{v}_l\right)\right|$
it holds that
$ \begin{aligned}&\left\|\left(\mathit{\Psi }, \mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2(t)+ \\&\int_0^t\left\|\sqrt{\left(v^s\right)^{\prime}}\left(\mathit{\Phi }, Z(\gamma-1)^{-\frac{1}{2}}\right)\right\|^2(\tau) \mathrm{d} \tau+ \\&\int_0^t\left\|\left(\mathit{\Phi }_x, Z_x\right)\right\|^2(\tau) \mathrm{d} \tau \leqslant C N(0)^2+ \\&C \delta \int_0^t\left\|\left(z, z_x, \mathit{\Phi }_{x x}\right)\right\|^2(\tau) \mathrm{d} \tau+ \\&C(\varepsilon+\delta) \int_0^t\left\|\mathit{\Psi }_x\right\|^2(\tau) \mathrm{d} \tau\end{aligned} $
Proof ? Multiply (18)₁ by Ψ, (18)₂ by $\frac{v^s}{\xi} \mathit{\Phi }$, (18)₃ by $\frac{Z}{\xi^2}$ respectively, and sum over the results. There is
$\begin{aligned}&I_1(\mathit{\Psi }, \mathit{\Phi }, Z)_t+I_2\left(\mathit{\Phi }, \mathit{\Phi }_x\right)+I_3\left(Z, Z_x\right)= \\&\quad I_4\left(\mathit{\Phi }, Z, \mathit{\Psi }_x, Z_x\right)+\frac{v^s}{\xi} F_1 \mathit{\Phi }+\frac{F_2 Z}{\xi^2}+(\cdots)_x\end{aligned}$ (25)
where
$\begin{aligned}&I_1(\mathit{\Psi }, \mathit{\Phi }, Z)=\frac{\mathit{\Psi }^2}{2}+\frac{v^s}{2 \xi} \mathit{\Phi }^2+\frac{Z^2}{2(\gamma-1) \xi^2} \\&I_2\left(\mathit{\Phi }, \mathit{\Phi }_x\right)=\left[\frac{\gamma-1}{\xi}\left(u^s\right)^{\prime}-\left(\frac{v^s}{2 \xi}\right)_t\right] \mathit{\Phi }^2+ \\&\left(\frac{1}{\xi}\right)_x \mathit{\Phi } \mathit{\Phi }_x+\frac{\mathit{\Phi }_x^2}{\xi} \\&I_3\left(Z, Z_x\right)=-\left[\frac{1}{2(\gamma-1) \xi^2}\right]_t Z^2+\frac{Z_x^2}{v^s \xi^2}+ \\&\left(\frac{1}{v^s \xi^2}\right)_x Z Z_x \\&I_4\left(\mathit{\Phi }, Z, \mathit{\Psi }_x, Z_x\right)=\frac{\gamma-1}{v^s \xi^2}\left(u^s\right)^{\prime} \mathit{\Phi } Z_x+ \\&{\left[\frac{\gamma-1}{v^s \xi^2}\right]_x\left(u^s\right)^{\prime} \mathit{\Phi } Z+\frac{\left(\theta^s\right)^{\prime} \mathit{\Psi }_x Z}{\left(v^s\right)^2 \xi^2}:=I_4^1+I_4^2+I_4^3}\end{aligned}$
where (…) represents the conservative terms. It will vanish after integration concerning x over $\mathbb{R}$. (a)_i means the i-th row in (a). By (5)₁ and (5)₂, for ξ=$p^s-\frac{\left(u^s\right)^{\prime}}{v^s}$, there is
$\xi=\bar{p}_r+\sigma^2 \bar{v}_r-\sigma^2 v^s=\bar{p}_l+\sigma^2 \bar{v}_l-\sigma^2 v^s$
By the monotonicity of v^s, there is
$\bar{p}_l \leqslant \xi \leqslant \bar{p}_l$ (26)
Furthermore, there is
$\begin{gathered}\xi_x=-\sigma^2\left(v^s\right)^{\prime}=\sigma(U) \\\xi_t=-\sigma^2\left(u^s\right)^{\prime}\end{gathered}$
Here $\mathit{\Phi }: =\sqrt{\xi} \tilde{\mathit{\Phi }}, Z: =\sqrt{v^s} \xi \tilde{Z}$ is introduced. Then the term I₂ can be rewritten as
$\begin{aligned}I_2=& {\left[\frac{\gamma-1}{\xi}\left(u^s\right)^{\prime}-\left(\frac{v^s}{2 \xi}\right)_t\right] \mathit{\Phi }^2+\left(\frac{1}{\xi}\right)_x \mathit{\Phi } \mathit{\Phi }_x+} \\& \frac{\mathit{\Phi }_x^2}{\xi}:=\tilde{\mathit{\Phi }}_x^2+\left(\frac{3}{2}-\gamma\right)\left(-\left(u^s\right)^{\prime}\right) \tilde{\mathit{\Phi }}^2+\\& \frac{\sigma^2 v^s p^s}{2 \xi^2}\left(-\left(u^s\right)^{\prime}\right) \tilde{\mathit{\Phi }}^2+a_{11}\left[\left(u^s\right)^{\prime}\right]^2 \tilde{\mathit{\Phi }}^2\end{aligned}$ (27)
It is easy to see that if $1 < \gamma < \frac{3}{2}$, the second term of expression (27) is positive. Similarly, the terms I₃ and I₄^j, j=1, 2 can be rewritten as
$\begin{aligned}I_3=&-\left[\frac{1}{2(\gamma-1) \xi^2}\right]_t Z^2+\frac{Z_x^2}{v^s \xi^2}+\left(\frac{1}{v^s \xi^2}\right)_x Z Z_x=\\&-\left[\frac{\sigma^2\left(u^s\right)^{\prime} v^s}{\xi(\gamma-1)}\right] \tilde{Z}^2+\left[\tilde{Z}_x^2\right]-\left(\frac{1}{v^s \xi^2}\right)\left[\left(\sqrt{v^s} \xi\right)_x\right]^2 \cdot \\&\tilde{Z}^2=\left[\tilde{Z}_x^2\right]-\left[\frac{\sigma^2\left(u^s\right)^s v^s}{\xi(\gamma-1)}\right] \tilde{Z}^2-\left(\frac{1}{v^s \xi^2}\right)\left[-\sqrt{v^s} \sigma^2+\right.\\&\left.\frac{1}{2} \frac{1}{\sqrt{v^s}} \xi\right]^2\left(\left(v^s\right)^{\prime}\right)^2 \tilde{Z}^2:=a_{22} \tilde{Z}_x^2+\\&\frac{\sigma^2 v^s p^s}{\xi^2(\gamma-1)}\left(-\left(u^s\right)^{\prime}\right) \tilde{Z}^2+a_{33}\left(u^s\right)^{\prime 2} \tilde{Z}^2\end{aligned}$
$\begin{aligned}I_4^1=& \frac{\gamma-1}{v^s \xi^2}\left(u^s\right)^{\prime} \mathit{\Phi } Z_x=\frac{\gamma-1}{v^s \xi^2}\left(u^s\right)^{\prime} \sqrt{\xi} \tilde{\mathit{\Phi }}\left(\xi \sqrt{v^s} \tilde{Z}\right)_x=\\& \frac{\gamma-1}{v^s \xi^2}\left(u^s\right)^{\prime} \sqrt{\xi} \tilde{\mathit{\Phi }}\left(\xi \sqrt{v^s}\right)_x \tilde{Z}+\\& \frac{\gamma-1}{v^s \xi^2}\left(u^s\right)^{\prime} \sqrt{\xi} \tilde{\mathit{\Phi }}\left(\xi \sqrt{v^s}\right) \tilde{Z}_x=\\& \frac{\gamma-1}{\sqrt{v^s \xi}}\left(u^s\right)^{\prime} \tilde{\mathit{\Phi }} \tilde{Z}_x+\frac{\gamma-1}{v^s \xi^{\frac{3}{2}}}\left(u^s\right)^{\prime}\left(\xi \sqrt{v^s}\right)_x \tilde{Z} \tilde{\mathit{\Phi }}:=\\& 2 a_{12}\left(u^s\right)^{\prime} \tilde{\mathit{\Phi }} \tilde{Z}_x+H_2\end{aligned}$
and
$\begin{aligned}&I_4^2+H_2=\left[\frac{\gamma-1}{v^s \xi^2}\right]_x\left(u^s\right)^{\prime} \mathit{\Phi } Z+ \\&\quad \frac{\gamma-1}{v^s \xi^{\frac{3}{2}}}\left(u^s\right)^{\prime}\left(\xi \sqrt{v^s}\right)_x \tilde{Z} \tilde{\mathit{\Phi }}= \\&{\left[\frac{\gamma-1}{v^s \xi^2}\right]_x\left(u^s\right)^{\prime} v^{\frac{1}{2}} \xi^{\frac{3}{2}} \tilde{\mathit{\Phi }} \tilde{Z}+\frac{\gamma-1}{v^s \xi^{\frac{3}{2}}}\left(u^s\right)\left(\xi \sqrt{v^s}\right)_x \tilde{Z} \tilde{\mathit{\Phi }}=}\\&(\gamma-1)\left(u^s\right)^{\prime} \tilde{\mathit{\Phi }} \tilde{Z}\left[\left(\frac{1}{v^s \xi^2}\right)_x \sqrt{v^s} \xi^{\frac{3}{2}}+\frac{\left(\xi \sqrt{v^s}\right)_x}{v^s \xi^{\frac{3}{2}}}\right]=\\&(\gamma-1)\left(u^s\right)^{\prime} \tilde{\mathit{\Phi }} \tilde{Z}\left[\left(\frac{1}{v^s \xi^2}\right)_x \sqrt{v^s} \xi+\frac{\left(\xi \sqrt{v^s}\right)_x}{v^s \xi^2}\right] \sqrt{\xi}:=\\&2 a_{13}\left[\left(u^s\right)^{-}\right]^2 \tilde{Z} \tilde{\mathit{\Phi }}\end{aligned}$ (28)
where
$\begin{aligned}a_{11}=& \frac{\sigma^2}{4 \xi^2}, a_{22}=1, a_{12}=\frac{\gamma-1}{2 \sqrt{v^s \xi}} \\a_{13}=&-\frac{\gamma-1}{2 v^s \xi^{\frac{3}{2}}}\left[\sigma \sqrt{v^s}-\xi \frac{1}{2 \sqrt{v^s} \sigma}\right] \\a_{33}=& \frac{\sigma^2}{\xi^2(\gamma-1)}\left[1-\left(-\sqrt{v^s} \sigma^2+\right.\right.\\&\left.\left.\frac{1}{2} \frac{1}{\sqrt{v^s}} \xi\right)^2 \frac{\gamma-1}{v^s \sigma^4}\right]\end{aligned}$
Define
$\begin{gathered}X: = \left( {{{\left( {{u^s}} \right)}^{\prime}}\mathit{\tilde \Phi },{{\tilde Z}_x},{{\left( {{u^s}} \right)}^{\prime}}\tilde Z} \right)\\\boldsymbol{A}=\left(\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\a_{12} & a_{22} & 0 \\a_{13} & 0 & a_{33}\end{array}\right)\end{gathered}$
To make sure XAX^T>0, i.e., the matrix A is positive, the following two conditions should be satisfied:
$\left\{\begin{array}{l}\beta^2<\frac{\sigma^2 v^s}{\xi} \\\beta^2+\left(\frac{\sigma^2 v^s}{\xi}+\frac{1}{4} \frac{\xi}{\sigma^2 v^s}-1\right) \beta-\frac{\sigma^2 v^s}{\xi} <0\end{array}\right.$ (29)
where β=γ-1. It is clear that (29)₂ contains(29)₁. So if
$\gamma <1+\min \left\{\frac{\frac{\sigma^2 \bar{v}_l}{\bar{p}_l}}{\frac{\sigma^2 \bar{v}_l}{\bar{p}_l}+\frac{\bar{p}_l}{4 \sigma^2 \bar{v}_l}}, 0.5\right\}$ (30)
holds, expression (29) could be true, which uses the fact that function $f(z)=\frac{z}{z+(4 z)^{-1}}$ is monotonically increasing for z>0 and expression (26). Moreover, if $2 \gamma\left|p_v\left(\bar{v}_r\right)\right|>\left|p_v\left(\bar{v}_l\right)\right|$, expression (30) can be simplified to $\gamma < \frac{3}{2}$. Thus c₀>0 can be chosen, such that
$\boldsymbol{X} \boldsymbol{A} \boldsymbol{X}^{\mathrm{T}} \geqslant c_0\left(\left[\left(u^s\right)^{\prime}\right]^2 \tilde{\mathit{\Phi }}^2+\left[\left(u^s\right)^{\prime}\right]^2 \tilde{Z}^2+\tilde{Z}_x^2\right)$
Besides, by expression (10) and Cauchy inequality, I₄³ can be estimated as
$\begin{aligned}I_4^3=& \frac{\left(\theta^s\right)^{\prime} \mathit{\Psi }_x Z}{\left(v^s\right)^2 \xi^2}=\frac{\left(\theta^s\right)^{\prime} \mathit{\Psi }_x \xi \sqrt{v^s} \tilde{Z}}{\left(v^s\right)^2 \xi^2}=\frac{\left(\theta^s\right)^{\prime} \mathit{\Psi }_x \tilde{Z}}{\left(v^s\right)^{\frac{3}{2}} \xi} \leqslant \\&-\frac{p^s}{2}\left[\frac{\sigma^2\left(u^s\right)^{\prime} v^s}{\xi^2(\gamma-1)}\right] \tilde{Z}^2-\frac{\gamma-1}{2\left(v^s\right)^4 p^s \sigma^2\left(u^s\right)^{\prime}} \cdot \\&\left(\theta^s\right)^{\prime 2} \mathit{\Psi }_x^2 \leqslant-\frac{p^s}{2}\left[\frac{\sigma^2\left(u^s\right)^{\prime} v^s}{\xi^2(\gamma-1)}\right] \tilde{Z}^2+\\& C(\gamma-1)\left|\frac{\left(\theta^s\right)^{\prime}}{\left(v^s\right)^{\prime}}\right|\left(\theta^s\right)^{\prime} \mathit{\Psi }_x^2 \leqslant \\& \frac{1}{2}\left[\frac{\sigma^2 v^s p^s}{\xi^2(\gamma-1)}\right]\left(-\left(u^s\right)^{\prime}\right) \tilde{Z}^2+\\& C(\gamma-1)^3\left|\bar{v}_r-\bar{v}_l\right|^2 \mathit{\Psi }_x^2 \leqslant \\& \frac{1}{2}\left[ {\frac{{{\sigma ^2}{v^s}{p^s}}}{{{\xi ^2}(\gamma - 1)}}} \right]\left( { - {{\left( {{u^s}} \right)}^{\prime} }} \right){\tilde Z^2} + C\varepsilon \mathit{\Psi } _x^2\end{aligned}$
Utilizing expressions (19) and (20), the following expressions are obtained:
$\begin{gathered}\left|\frac{v^s}{\xi} \mathit{\Phi } F_1\right|=\left|\frac{v^s}{\xi^{\frac{1}{2}}} \tilde{\mathit{\Phi }} F_1\right| \leqslant C\left(\mathit{\Psi }_x^2+\mathit{\Phi }_x^2+\right. \\\left.z^2+\phi_x^2\right)|\tilde{\mathit{\Phi }}|\end{gathered}$
and
$\begin{gathered}\left|\frac{Z}{\xi^2} F_2\right|=\left|\frac{\sqrt{v^s} \tilde{Z}}{\xi} F_2\right| \leqslant C\left(\mathit{\Psi }_x^2+\mathit{\Phi }_x^2+z^2+\right. \\\left.\phi_x^2+z_x^2\right)|\tilde{Z}|\end{gathered}$ (31)
From expressions (25) and (28)-(31), there is
$\begin{array}{l}{\left\{ {\frac{{{\mathit{\Psi }^2}}}{2} + \frac{{{v^s}}}{{2\xi }}{\mathit{\Phi }^2} + \frac{{{Z^2}}}{{2(\gamma - 1){\xi ^2}}}} \right\}_t} + \mathit{\widetilde \Phi }_x^2 + \left[ {\left( {\frac{3}{2} - \gamma } \right) + } \right.\\\left. {\;\;\;\frac{{{\sigma ^2}{v^s}{p^s}}}{{2{\xi ^2}}}} \right]\left( { - {{\left( {{u^s}} \right)}^\prime }} \right){\mathit{\widetilde \Phi }^2} + {c_0}{\left( {{{\left( {{u^s}} \right)}^\prime }} \right)^2}{\mathit{\widetilde \Phi }^2} + \\\;\;\;{c_0}\tilde Z_x^2 + \frac{{{p^s}}}{2}\left[ {\frac{{{\sigma ^2}{v^s}}}{{{\xi ^2}(\gamma - 1)}}} \right]\left( { - {{\left( {{u^s}} \right)}^\prime }} \right){{\tilde Z}^2} + \\\;\;\;{c_0}{\left[ {{{\left( {{u^s}} \right)}^\prime }} \right]^2}{{\tilde Z}^2} \le C\varepsilon \mathit{\Psi }_x^2 + C\delta \left( {\mathit{\Psi }_x^2 + \mathit{\Phi }_x^2 + } \right.\\\;\;\;\left. {{z^2} + z_x^2 + \mathit{\Phi }_{xx}^2} \right)\end{array}$ (32)
Integrating expression (32) with $(0, t) \times \mathbb{R}$, recalling the transformations $\mathit{\Phi }: =\sqrt{\xi} \tilde{\mathit{\Phi }}, Z: =\sqrt{v^s} \xi \tilde{Z}$, and taking δ, ε that are sufficiently small, Lemma 3.1 is obtained.
Lemma 3.2 ? Under the same assumptions of Proposition 2.1, it follows that
$\left\|\mathit{\Psi }_x\right\|^2(t)+\int_0^t\left\|\mathit{\Psi }_x\right\|^2(\tau) \mathrm{d} \tau \leqslant C N(0)^2+ \\C \int_0^t\left\|\left(\sqrt{\left(v^s\right)^{\prime}} \mathit{\Phi }, \mathit{\Phi }_x, Z_x\right)\right\|^2(\tau) \mathrm{d} \tau+ \\C \delta \int_0^t\left\|\left(z, \phi_x, z_x\right)\right\|^2(\tau) \mathrm{d} \tau+C \sup\limits_{0 \leqslant \tau \leqslant t}\|\mathit{\Phi }\|^2(\tau)$ (33)
Proof ? Multiply (18)₂ by -v^sΨ_x, using (18)₁, the following equation is obtained:
$\begin{array}{l}\left(\frac{\mathit{\Psi }_x^2}{2}-v^s \mathit{\Psi }_x \mathit{\Phi }\right)_t+\xi \mathit{\Psi }_x^2=\left(v^s\right)^{\prime} \mathit{\Phi }_x \mathit{\Phi }+v^s \mathit{\Phi }_x^2+ \\ \;\;\;\;\mathit{\Psi }_x Z_x+(\gamma-2)\left(u^s\right)^{\prime} \mathit{\Psi }_x \mathit{\Phi }-v^s F_1 \mathit{\Psi }_x+ \\ \;\;\;\;(\cdots)_x\end{array}$ (34)
Integrate expression (34) with respect to (x, t) with $\mathbb{R} \times[0, t]$, there is
$\begin{aligned}\int_{-\infty}^{\infty} &\left(\frac{\mathit{\Psi }_x^2}{2}-v^s \mathit{\Psi }_x \mathit{\Phi }\right) \mathrm{d} x+\xi \int_0^t \int_{-\infty}^{\infty} \mathit{\Psi }_x^2 \mathrm{~d} x \mathrm{~d} \tau=\\&\left.\int_{-\infty}^{\infty}\left(\frac{\mathit{\Psi }_x^2}{2}-v^s \mathit{\Psi }_x \mathit{\Phi }\right) \mathrm{d} x\right|_{t=0}+\int_0^t \int_{-\infty}^{\infty}\left(v^s\right)^{\prime} \mathit{\Phi }_x \mathit{\Phi } \mathrm{d} x \mathrm{~d} \tau+\\& \int_0^t \int_{-\infty}^{\infty} v^s \mathit{\Phi }_x^2 \mathrm{~d} x \mathrm{~d} \tau+\int_0^t \int_{-\infty}^{\infty} \mathit{\Psi }_x Z_x \mathrm{~d} x \mathrm{~d} \tau+\\&(\gamma-2) \int_0^t \int_{-\infty}^{\infty}\left(u^s\right)^{\prime} \mathit{\Psi }_x \mathit{\Phi } \mathrm{d} x \mathrm{~d} \tau-\int_0^t \int_{-\infty}^{\infty} v^s F_1 \mathit{\Psi }_x \mathrm{~d} x \mathrm{~d} \tau :=\\&\left.\int_{-\infty}^{\infty}\left(\frac{\mathit{\Psi }_x^2}{2}-v^s \mathit{\Psi }_x \mathit{\Phi }\right) \mathrm{d} x\right|_{t=0}+\sum\limits_{i=i}^5 I_i\end{aligned}$ (35)
Now, with expressions (19) and (26), I₁-I₅ are estimated term by term:
$\begin{aligned}&I_1+I_2 \leqslant \sup\limits_t\left\|\left(v^s\right)^{\prime}\right\|_{L^{\infty}} \int_0^t\left\|\sqrt{\left(v^s\right)^{\prime}} \mathit{\Phi }\right\|^2(\tau) \mathrm{d} \tau+ \\&C \int_0^t\left\|\mathit{\Phi }_x\right\|^2(\tau) \mathrm{d} \tau \leqslant C \int_0^t\left\|\sqrt{\left(v^s\right)^{\prime}} \mathit{\Phi }\right\|^2(\tau)+ \\&\left\|\mathit{\Phi }_x\right\|^2(\tau) \mathrm{d} \tau \\&I_3+I_4 \leqslant \frac{\xi}{4} \int_0^t\left\|\mathit{\Psi }_x\right\|^2(\tau) \mathrm{d} \tau+ \\&C \int_0^t\left(\left\|\sqrt{\left(v^s\right)^{\prime}} \mathit{\Phi }\right\|^2+\left\|Z_x\right\|^2\right)(\tau) \mathrm{d} \tau \\&I_5 \leqslant C \int_0^t \int_{-\infty}^{\infty}\left|\left(\chi, \phi, z, \phi_x\right)\right|^2|\chi| \mathrm{d} x \mathrm{~d} \tau \leqslant \\&C \delta \int_0^t\left\|\left(\mathit{\Psi }_x, \mathit{\Phi }_x, z, \phi_x\right)\right\|^2(\tau) \mathrm{d} \tau\end{aligned}$
If δ is sufficiently small, expression (33) is obtained.
z can be rewritten as
$z=Z_x+\frac{\gamma-1}{R}\left(\left(u^s\right)^{\prime} \mathit{\Phi }-\frac{1}{2} \phi^2\right)$ (36)
Square the left and right ends of expression (36). Integrate the result with respect to (x, t) over $\mathbb{R} \times[{0}, t]$. The following inequality is obtained:
$\int_0^t\|z\|^2(\tau) \mathrm{d} \tau \leqslant C \int_0^t\left(\left\|Z_x\right\|^2+\left\|\left|\left(v^s\right)^{\prime}\right|^{\frac{1}{2}} \mathit{\Phi }\right\|^2+\right. \\ \;\;\;\;\;\left.\delta^2\|\phi\|^2\right)(\tau) \mathrm{d} \tau$ (37)
Lemma 3.3 ? Under the same assumptions of Proposition 2.1, the following inequality holds:
$\begin{aligned}&\left\|\left(\chi, \phi, z(\gamma-1)^{-\frac{1}{2}}\right)(t)\right\|^2+ \\&\quad \int_0^t\left\|\left(\phi_x, z_x\right)\right\|^2(\tau) \mathrm{d} \tau \leqslant C \int_0^t\|(\chi, \phi, z)\|^2(\tau) \mathrm{d} \tau+ \\&C N(0)^2\end{aligned}$ (38)
Proof ? Multiplying (15)₁ by χ, (15)₂ by v^s?/ξ, (15)₃ by z/ξ², and adding all the results, there is
$\begin{array}{l}\left(\frac{\chi^2}{2}+\frac{v^s}{2 \xi} \phi^2+\frac{z^2}{2(\gamma-1) \xi^2}\right)_t+\frac{\phi_x^2}{\xi}+\frac{z_x^2}{v^s \xi^2}= \\ \;\;\left(\frac{v^s}{2 \xi}\right)_t \phi^2+\left(\frac{1}{2(\gamma-1) \xi^2}\right)_t z^2-\frac{1}{v^s}\left(\frac{v^s}{\xi}\right)_x \phi \phi_x+ \\ \;\;\frac{v^s}{\xi}\left(\frac{\xi}{v^s}\right)_x \chi \phi-\frac{z z_x}{v^s}\left(\frac{1}{\xi^2}\right)_x+\left[\left(\frac{1}{\xi}\right)_x-\frac{v^s}{\xi}\left(\frac{1}{v^s}\right)_x\right] \cdot \\ \;\;\phi z+\frac{\left(u^s\right)^{\prime} \chi}{v^s}\left(\xi-z+\phi_x\right) \frac{z}{\xi^2}-\left(\frac{\left(\theta^s\right)^{\prime} \chi}{\left(v^s\right)^2}\right)_x \frac{z}{\xi^2}+ \\ \;\;f_1 \frac{v^s \phi}{\xi}+f_2 \frac{z}{\xi^2}+(\cdots)_x\end{array}$ (39)
Integrate expression (39) with respect to x and t with $\mathbb{R} \times[{0}, t]$. Using the Young's inequality, the Sobolev inequality and expressions (16) and (17), it follows that
$\left.\int_{-\infty}^{\infty}\left(\frac{\chi^2}{2}+\frac{v^s}{2 \xi} \phi^2+\frac{z^2}{2(\gamma-1) \xi^2}\right) \mathrm{d} x\right|_0 ^t+\\ \;\;\;\int_0^t \int_{-\infty}^{\infty}\left(\frac{\phi_x^2}{\xi}+\frac{z_x^2}{v^s \xi^2}\right) \mathrm{d} x \mathrm{~d} \tau \leqslant(\in+\\\;\;\;C \delta) \int_0^t \int_{-\infty}^{\infty}\left(\phi_x^2+z_x^2\right) \mathrm{d} x \mathrm{~d} \tau+\left(C_{\in}+\right.\\ \;\;\;C \delta) \int_0^t \int_{-\infty}^{\infty}\left(\chi^2+\phi^2+z^2\right) \mathrm{d} x \mathrm{~d} \tau$
Choose ∈ and δ with sufficiently small value in the above inequality, there is expression (38).
Lemma 3.4? Under the same assumptions of Proposition 2.1, there is
$\begin{gathered}\left\|\chi_x(t)\right\|^2+\int_0^t\left\|\chi_x\right\|^2(\tau) \mathrm{d} \tau \leqslant C N(0)^2+ \\C \sup\limits_{0 \leqslant \tau \leqslant t}\|\phi\|^2(\tau)+C \int_0^t\|(\chi, \phi, z)\|^2(\tau) \mathrm{d} \tau+\\C \int_0^t\left\|\left(\phi_x, z_x\right)\right\|^2(\tau) \mathrm{d} \tau+C \delta \int_0^t\left\|\phi_{x x}\right\|^2(\tau) \mathrm{d} \tau\end{gathered}$ (40)
Proof ? Multiply (15)₂ by -v^sχ_x, and integrate the result with respect to t and x over $[0, t] \times \mathbb{R}$. With (15)₁, the following equation is obtained:
$\begin{aligned}&\int_{-\infty}^{\infty}\left(\frac{\chi_x^2}{2}-v^s \chi_x \phi\right) \mathrm{d} x+\int_0^t \int_{-\infty}^{\infty} \xi \chi_x^2 \mathrm{~d} x \mathrm{~d} \tau= \\&\left.\int_{-\infty}^{\infty}\left(\frac{\chi_x^2}{2}-v^s \chi_x \phi\right) \mathrm{d} x\right|_{t=0}-\int_0^t \int_{-\infty}^{\infty} v^s f_1 \chi_x \mathrm{~d} x \mathrm{~d} \tau+ \\&\int_0^t \int_{-\infty}^{\infty} v^s \phi_x^2 \mathrm{~d} x \mathrm{~d} \tau-\int_0^t \int_{-\infty}^{\infty} v_t^s \chi_x \phi \mathrm{d} x \mathrm{~d} \tau+ \\&\int_0^t {\int_{ - \infty }^\infty {{\chi _x}} } {z_x}{\rm{d}}x{\rm{d}}\tau - \int_0^t {\int_{ - \infty }^\infty {{v^s}} } {\left( {\frac{\xi }{{{v^s}}}} \right)_x}\chi {\chi _x}{\rm{d}}x{\rm{d}}\tau + \\&\int_0^t \int_{-\infty}^{\infty} v^s\left(\frac{1}{v^s}\right)_x \chi_x z \mathrm{~d} x \mathrm{~d} \tau-\int_0^t \int_{-\infty}^{\infty} v^s\left(\frac{1}{v^s}\right)_x . \\&\chi_x \phi_x \mathrm{~d} x \mathrm{~d} \tau+\int_0^t \int_{-\infty}^{\infty}\left(v^s\right)^{\prime} \phi_x \phi \mathrm{d} x \mathrm{~d} \tau:= \\&\left.\int_{-\infty}^{\infty}\left(\frac{\chi_x^2}{2}-v^s \chi_x \phi\right) \mathrm{d} x\right|_{t=0}+\sum\limits_{i=1}^8 A_i\end{aligned}$ (41)
With expression (17), there is
$\begin{gathered}A_1 \leqslant C \int_0^t \int_{-\infty}^{\infty}\left|f_1 \chi_x\right| \mathrm{d} x \mathrm{~d} \tau \leqslant \frac{1}{8} \int_0^t \int_{-\infty}^{\infty} \xi \chi_x^2 \mathrm{~d} x \mathrm{~d} \tau+ \\C \delta \int_0^t\left\|\left( {\chi , {\chi _x}, z, {z_x}, {\phi _x}, {\phi _{xx}}} \right)\right\|^2(\tau) \mathrm{d} \tau\end{gathered}$
For i=2-8, the Cauchy inequality yields:
$\begin{aligned}A_i \leqslant & \frac{1}{8} \int_0^t \int_{-\infty}^{\infty} \xi \chi_x^2 \mathrm{~d} x \mathrm{~d} \tau+C \int_0^t\left(\|\phi\|^2+\left\|\phi_x\right\|^2+\right.\\&\left.\|\chi\|^2+\|z\|^2+\left\|z_x\right\|^2\right)(\tau) \mathrm{d} \tau\end{aligned}$ (42)
Using the above inequalities, by taking δ with small enough value, there is expression (40).
By the same method, the highest-order estimate is obtained. For the sake of simplicity, the proof is omitted.
Lemma 3.5 ? Under the same assumptions of Proposition 2.1, it holds that
$ \begin{aligned}&\left\|\left(\chi_x, \phi_x z_x(\gamma-1)^{-\frac{1}{2}}\right)(t)\right\|^2+\int_0^t\left\|\left(\phi_{x x} z_{x x}\right)(\tau)\right\|^2 \mathrm{~d} \tau \leqslant \\&C N(0)^2+C \int_0^t\left\|\left(\chi, \phi, z, \chi_x, \phi_x, z_x\right)(\tau)\right\|^2 \mathrm{~d} \tau\end{aligned} $
Combining expression (37), Lemma 3.1-Lemma 3.5, and δ, ε with suitably small value, expression (22) is obtained. Thus the proof of the a priori estimate is accomplished.

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