Asymptotic Stability of the Magnetohydrodynamic Flows with Temperature-Dependent Transport Coefficients

Abstract

The objective of this paper is to analyze the asymptotic stability of global strong solutions to the boundary value problem of the compressible magnetohydrodynamic (MHD) equations for the ideal polytropic gas in which the viscosity $\lambda$ and heat conductivity $\kappa$ depend on temperature, i.e., $\lambda ={\theta }^{\alpha }$ and $\kappa ={\theta }^{\beta }$ with $\alpha ,\beta \in [0,+\infty )$. Both the global-in-time existence and uniqueness of strong solutions are obtained under certain assumptions on the parameter $\alpha$ and initial data. Moreover, based on accurate uniform-in-time estimates, we show that the global large solutions decay exponentially in time to the equilibrium states. Compared with the existing results, the initial data could be large if $\alpha$ is small and the growth exponent $\beta$ can be arbitrarily large.

1. Introduction

The motion of a conducting fluid in an electromagnetic field is governed by the compressible MHD equations in Lagrange coordinates (see [1,2])

$$\left\{\begin{array}{c}{\tau }_t-u_x=0,\hfill \\ u_t+{\left[p+{\displaystyle \frac{1}{2}}{b}^2\right]}_x={\left[{\displaystyle \frac{\lambda u_x}{\tau }}\right]}_x,\hfill \\ {\left(\tau b\right)}_t={\left[{\displaystyle \frac{\nu b_x}{\tau }}\right]}_x,\hfill \\ E_t+{\left[u\left(p+{\displaystyle \frac{1}{2}}{b}^2\right)\right]}_x={\left[{\displaystyle \frac{\lambda u{u}_x+\nu b{b}_x+\kappa {\theta }_x}{\tau }}\right]}_x.\hfill \end{array}\right.$$

(1)

Here, $\tau$ denotes the specific volume; $u\in \mathbf{R}$, the velocity; $b\in \mathbf{R}$, the magnetic field; $\theta$ the temperature; $p=p(\tau ,\theta )$, the pressure; $\lambda =\lambda \left(\theta \right)$ is the bulk viscosity coefficient, $\nu$ is the magnetic diffusivity acting as a magnetic diffusion coefficient of the magnetic field, $\kappa =\kappa \left(\theta \right)$ is the heat conduction coefficient and the total energy E is given by

$$E=e+{\displaystyle \frac{1}{2}}{u}^2+{\displaystyle \frac{1}{2}}\tau b^2,$$

where $e=e(\tau ,\theta )$ is the internal energy, ${\displaystyle \frac{1}{2}}{u}^2$ is the kinetic energy, and ${\displaystyle \frac{1}{2}}\tau b^2$ is the magnetic energy. The pressure p and the initial energy e satisfy the following forms:

$$p(\tau ,\theta )={\displaystyle \frac{R\theta }{\tau }},e(\tau ,\theta )=c_v\theta ,$$

(2)

where $R>0$ is the perfect gas constant, and $c_v>0$ is the specific heat at constant volume.

It is well known that the transport coefficients of the fluids generally vary with temperature and density in very–high–temperature and very–high–density environments. It also follows from the kinetic theory that the viscosity and heat-conductivity coefficients are temperature-dependent, when the fluid dynamics is derived from the Boltzmann system with slab symmetry for monatomic gases by using the Chapman-Enskog expansion [3]. Motivated by these facts, and according to the constitutive relations of the Stefan-Boltzmann model [4], we consider the following case when the transport coefficients $\lambda$ and $\kappa$ depend on the temperature $\theta$:

$$\lambda =\tilde{\lambda }{\theta }^{\alpha },\kappa =\tilde{\kappa }{\theta }^{\beta },$$

(3)

where $\tilde{\lambda },\tilde{\kappa }>0$ are constants and $\alpha ,\beta ⩾0$. As initial and boundary conditions, we consider

$$(\tau ,u,b,\theta ){|}_{t=0}=({\tau }_0,u_0,b_0,{\theta }_0)\left(x\right),x\in \overline{\mathrm{\Omega }}=[0,1],$$

(4)

$$(u,b,{\theta }_x){|}_{\partial \mathrm{\Omega }}=0.$$

(5)

Before stating our main results, we first recall some related studies. If $b=0$, the system (1) reduces to the Navier–Stokes model which has been extensively studied by many authors [5,6,7,8,9,10,11]. Antontsev–Kazhikhov–Monakhov in [5] investigated the boundary value problems in the mechanics of the Navier–Stokes equation. Kazhikhov–Shelukhin [8] and Jiang [6] considered the initial boundary value problems for a viscous heat-conducting one-dimensional real gas, respectively, and obtained the well-posedness of global strong solutions. Kawohl [7] obtained the global existence of large solutions to initial boundary value problems for the equations of the one-dimensional motion of viscous polytropic gases. Refs. [9,10,11] also considered the initial the well-posedness theory and asymptotic behavior of solutions. In the cases where the transport coefficients $\lambda$ and $\kappa$ are both positive constants, Kazhikhov [12,13] and Kawashima–Nishida [14] considered the one-dimensional ideal gas respectively, and also established the existence of global smooth solutions based on the positive upper and lower bounds of a specific volume $\tau$ and temperature $\theta$. This method can also be used to deal with the case that the transport coefficients are both depend on $\tau$ and $\theta$. Wang–Zhao in [15] considered the following case:

$$\lambda =\tilde{\lambda }g\left(\tau \right){\theta }^{\alpha },\kappa =\tilde{\kappa }g\left(\tau \right){\theta }^{\alpha }.$$

(6)

Here, $\tilde{\lambda },\tilde{\kappa }>0$ are constants, and $g(·)\in C^3(0,\infty )$ satisfies

$${\tau }^{r_1}+{\tau }^{-r_2}⩽Cg\left(\tau \right),g^{\prime }{\left(\tau \right)}^2\tau ⩽Cg{\left(\tau \right)}^3,\forall \tau \in (0,\infty ),$$

(7)

for some positive numbers $r_1,r_2⩾1$ and $C>0$. They proved the global existence by using the Kanel’s ideas [16] to derive the upper and lower bound of a specific volume $\tau$. However, they excluded the case when $r_1=r_2=0$, i.e., $g(·)=constant$. Recently, Sun–Zhang–Zhao in [17] generalized the results of [15], and considered 1D viscous and heat-conducting ideal polytropic fluids under the hight temperatures, and they also obtained the global existence, uniqueness and long-time behavior of strong solutions. If $b\ne 0$, for the compressible MHD Equation (1), there are also lots of studies on the well-posedness theory and asymptotic behavior of solutions (see [18,19,20,21,22] and references therein). Chen and Wang [23] investigated a free boundary problem with general large initial data, and they obtained the global existence of continuous of solutions. Fan–Jiang–Nakamura in [24] considered the problem of a vanishing shear viscosity limit in 1D MHD equations under the following condition that $\kappa (\rho ,\theta )$ satisfies:

$$C^{-1}(1+{\theta }^q)⩽\kappa (\rho ,\theta )⩽C(1+{\theta }^q)\mathrm{for}q⩾1,\mathrm{or}\kappa (\rho ,\theta )⩾{\displaystyle \frac{C}{\rho }},$$

(8)

and proved that the global weak solutions converge to a solution of the original equations with zero shear viscosity when the shear viscosity goes to zero. Lately, in [25], Zhang-Xie studied the influence of radiation on the dynamics at high-temperature regimes for 3D compressible nonlinear MHD equations with the growth conditions:

$${\kappa }_1(1+{\theta }^q)⩽\kappa (\rho ,\theta ),\left|{\kappa }_{\rho }(\rho ,\theta )\right|⩽{\kappa }_2(1+{\theta }^q)\mathrm{for}\mathrm{any}q>{\displaystyle \frac{5}{2}}$$

(9)

where the positive constants ${\kappa }_1$ and ${\kappa }_2$ statify ${\kappa }_1⩽{\kappa }_2$. They proved the global existence of a unique classical solution with large initial data to the initial boundary value problem. For radiative fluids, Qin-Liu-Yang [26] proved the global existence and exponential stability of solutions in $H_{+}^i$$(i=1,2,4)$ for 1D compressible and radiative MHD equations in a bounded domain under the assumptions that the initial total volume is small and the following case is hold:

$${\kappa }_1(1+{\theta }^q)⩽\kappa (\tau ,\theta )⩽{\kappa }_2(1+{\theta }^q),|{\kappa }_{\tau }+{\kappa }_{\tau \tau }|⩽{\kappa }_2(1+{\theta }^q)\mathrm{for}\mathrm{any}q>2$$

(10)

with positive constants ${\kappa }_1⩽{\kappa }_2$.

Recently, Hou–Liu–Wang–Xu [27] constructed a family of global-in-time solutions of the three-dimensional full compressible Navier-Stokes equations with temperature-dependent transport coefficients, and obtained that at arbitrary times and arbitrary strength, this family of solutions converges to a planar rarefaction wave, which is connected to the vacuum state, as the viscosity vanishes in the sense of $L^{\infty }\left({\mathbb{R}}^3\right)$. Baranovskii-Brizitskii-Saritskaia [28] considered the solvability of optimal control problems and obtained on both weak and strong solutions of a boundary value problem for the nonlinear reaction-diffusion-convection equation with variable coefficients.

As it was pointed out in (8)–(10), the assumptions on $\kappa (\tau ,\theta )$ have been technically used for the positive upper and lower bounds of a specific volume $\lambda$ and a temperature $\theta$ in which the global existence and long-time behavior can be shown to hold. However, the assumptions on $\kappa (\tau ,\theta )$ exclude the model (3) and the extension of results to the case that $q=0$, i.e., $\kappa$ is a positive constant, is unknown. The long-time behavior of the global solutions is also one of the most fundamental and interesting topics in mathematical fluid dynamics. There have been numerous researches in this direction for the case of bounded or unbounded domains. However, because of the strong coupling between the fluid motion and the magnetic field, the uniform-in-time upper and lower bounds of the specific volume and temperature cannot be achieved in a similar manner as that used for 1D compressible Navier–Stokes equations, and thus, the long-time behavior of 1D compressible MHD equations with large initial data are more difficult and completely different. Note that although the growth condition on heat-conductivity is physically reasonable when the radiative effects at high temperature are involved, it is indeed a regularizing condition from the mathematical point of view, since it leads to some additional estimates of the temperature from the starting energy-entropy estimates.

The works mentioned above only considered the case that $\lambda$ and $\nu$ are either positive constants or functions of a specific volume and $\kappa$ may vary with temperature. The methodology developed in [12,13,16] can not be applied directly to the case of temperature-dependent viscosity, and the existing results about such a more physically important case are very few. In fact, regardless of whether $\lambda$ is either a positive constant or a function of a specific volume, the first and the second equation of (1) can be written as:

$${\left({\displaystyle \frac{\lambda \left(\tau \right){\tau }_x}{\tau }}\right)}_t=u_t+p_x+b{b}_x,$$

(11)

and thus, the ideas in [12,13,16] can be adopted to derive the desired $t-$ independent bounds of the specific volume. Alternately, when the viscosity $\lambda$ is temperature-dependent (e.g., $\lambda =\lambda \left(\theta \right)$), it holds that

$${\left({\displaystyle \frac{\lambda \left(\theta \right){\tau }_x}{\tau }}\right)}_t=u_t+p_x+b{b}_x+{\displaystyle \frac{{\lambda }_{\theta }\left(\theta \right)}{\tau }}({\theta }_t{\tau }_x-u_x{\theta }_x),$$

(12)

where the last term on the right-hand side is highly nonlinear and will cause some serious difficulties in the mathematical analysis.

For the 1D compressible Navier-Stokes equations with temperature-dependent viscosity and heat-conductivity, Liu-Yang-Zhao-Zou [29] proved a Nishida-Smoller’s type global stability result under the assumptions that the adiabatic exponent $\gamma$ is close enough to 1 (i.e., $0<\gamma -1\ll 1$) and that the oscillation of temperature is sufficiently small (i.e., $\parallel {\theta }_0{-1\parallel }_{L^{\infty }}\ll 1$). Assume that both the viscosity and the heat-conductivity are functions of density and temperature, i.e.,

$$\lambda =\tilde{\lambda }h\left(\tau \right){\theta }^{\alpha },\mathrm{and}\kappa =\tilde{\kappa }h\left(\tau \right){\theta }^{\alpha },$$

(13)

where $\lambda$ and $\kappa$ are positive constants, and $h(·)\in C^3(0,\infty )$ satisfies

$${\tau }^{l_1}+{\tau }^{-l_2}⩽Ch\left(\tau \right),h^{\prime }{\left(\tau \right)}^2\tau ⩽Ch{\left(\tau \right)}^3,\forall \tau \in (0,\infty ),$$

(14)

for some positive numbers $l_1,l_2⩾1$ and $C>0$, Wang-Zhao [15] showed that if $\alpha ⩾0$ satisfies certain smallness restriction related to the initial data, then the Cauchy problem of 1D compressible Navier-Stokes equations admits a unique global solution on $\mathbf{R}\times [0,\infty )$. The factor $h(·)$, satisfying (14) with $l_1,l_2⩾1$, is mathematically technical and enables the authors to adopt the Kanel’s ideas [16] to derive the pointwise bounds of the specific volume. However, the assumption in (14) of $h(·)$ excluded the physically important model (1.3) and the extension of the result in [15] to the case that $l_1=l_2=0$ (i.e., $h(·)=const$) is nontrivial. Recently, by modifying the arguments of Kazhikhov–Shelukhin [8] and Wang–Zhao [15] in a non-standard way, Sun–Zhang–Zhao [17] proved the nonlinearly exponential stability for an initial-boundary value problem of 1D compressible Navier-Stokes system with temperature-dependent viscosity and heat-conductivity as in (3), provided that the growth exponent $\alpha ⩾0$ satisfies some smallness conditions similar to those in [15]. We emphasize here that the smallness of $\alpha$ plays a technical role in controlling the nonlinear terms on the right-hand side of (12).

Our main aim is to establish the global existence and non-linear exponential stability of solutions to the initial and boundary value problem of (1)–(5). The main difficulty arises from the higher order nonlinearities of the temperature $\theta$ in $\lambda \left(\theta \right)$ and $\kappa \left(\theta \right)$, which makes the upper bound for $\theta$ become more complicated. In order to overcome this problem, we take full advantage of Lemma 2 to reduce the higher order of $\theta$, and the assumption on $\alpha$ plays an important role. Another difficulty is that we have to obtain the uniform estimates independent of time t in order to study the non-linear behavior. This will be done in Section 2 by a careful analysis. Without loss of generality, we assume that $\nu =\tilde{\lambda }=\tilde{\kappa }=R=c_v=1$. Our main results in the present paper now reads as follows:

Definition 1.

A pair of functions $(\tau ,u,\theta ,b)$ is called a strong solution to the problem (1)–(5) in $[0,1]\times [0,\infty )$, if for some positive constant $C>0$,

$$0<C^{-1}⩽\tau (x,t),\theta (x,t)⩽C,\mathrm{for}\mathrm{all}(x,t)\in [0,1]\times [0,\infty ),$$

$$(\tau ,u,b,\theta )\in C([0,\infty );H^2),{\tau }_x\in L^2(0,\infty ;H^1),(u_x,b_x,{\theta }_x)\in L^2(0,\infty ;H^2),$$

and $(\tau ,u,\theta ,b)$ satisfies the Equation (1) almost everywhere in $[0,1]\times [0,\infty )$.

Theorem 1.

Assume that there are two positive constants $N_0$ and $M_0$, such that if

$$\underset{x\in [0,1]}{inf}{\tau }_0\left(x\right)⩾N_0,\underset{x\in [0,1]}{inf}{\theta }_0⩾N_0,{\parallel ({\tau }_0,u_0,b_0,{\theta }_0)\parallel }_{H^2}⩽M_0.$$

(15)

Then, there exists a positive constant $\epsilon >0$, which depends only on $N_0,M_0,\beta$, such that the initial boundary value problem (1)–(5) with $\left|\alpha \right|⩽\epsilon$ and $\beta ⩾0$ possesses a unique global strong solution $(\tau ,u,b,\theta )$ on $[0,1]\times [0,\infty )$, satisfying

$$\underset{(x,t)\in [0,1]\times [0,\infty )}{inf}\left\{\tau (x,t),\theta (x,t)\right\}>0,\underset{(x,t)\in [0,1]\times [0,\infty )}{sup}\left\{\tau (x,t),\theta (x,t)\right\}<\infty ,$$

(16)

and

$$(\tau ,u,b,\theta )\in C([0,\infty );H^2),{\tau }_x\in L^2(0,\infty ;H^1),(u_x,b_x,{\theta }_x)\in L^2(0,\infty ;H^2).$$

(17)

Moreover, for any $t⩾0$, it holds that

$$\parallel (\tau -m_0,u,b,\theta -{\pi }_0){\parallel }_{H^1}⩽C{e}^{-\sigma t}$$

(18)

with positive constants C and σ. $m_0$ and ${\pi }_0$ are the initial total volume and the initial total energy, respectively, given by

$$m_0\triangleq {\int }_0^1{\tau }_{0}dx,{\pi }_0\triangleq {\int }_0^1\left({\theta }_0+{\displaystyle \frac{1}{2}}{u}_0^2+{\displaystyle \frac{1}{2}}{\tau }_0{b}_0^2\right)dx.$$

Remark 1.

Compared with the results obtained by Wang-Zhao in [15] where the transport coefficients are assumed to satisfy (6) and (7) with $\lambda ,\kappa >0$, the extension of their result to the case that $\lambda =\kappa \ne constant$ did not obtained. Then, Theorem 1 particularly gives an affirm answer to the question raised by Wang-Zhao ([15], Remark 1.3).

Remark 2.

In Theorem 1, the growth exponent $\beta \in [0,\infty )$ can be arbitrarily large, and the choice of $\epsilon >0$ depends only on $N_0,\beta$ and the $H^2$-norm of the initial data. However, the small condition of α in [15] depends on the $H^3$-norm of the initial data.

Remark 3.

It should be pointed out that Theorem 1 permits $\alpha =0$ and $\beta =0$, and thus, the conditions of the viscosity λ and heat-conductivity κ in (5) include the case of $\lambda =\kappa =constant$.

Next, we comment on the analysis of our main results. Compared with the compressible Navier–Stokes system, the coupling of hydrodynamic and electrodynamic effects will bring seriously mathematical difficulties. As usual, it turns out that the key step is to prove the uniform-in-time lower and upper bounds of the specific volume and temperature. To do this, we first modify the arguments developed by Kazhikhov [12] in a non-standard way to derive an explicit representation of the specific volume $\tau$ of (24). Unfortunately, we have to deal with some additional nonlinear terms induced by the temperature-dependent viscosity. A key observation is that these nonlinear terms are involved with ${\mu }_x$ or ${\mu }_t$, and can be successfully controlled by the smallness of the amplitude of $\alpha$. This in turn asks us to seek the solution within a suitable functional space (see Section 2). Based on the representation of the specific volume and the elementary energy-entropy estimates, we can make use of the smallness of $\alpha$ (cf. (28)) to show that $\tau$ has strictly positive lower bound (see (29)). The uniform upper bound of the specific volume is more complicated and the proofs will be split into two cases, $\beta >0$ and $\beta =0$. For the case that $\beta >0$, we can obtain some higher-integrability of the temperature, which, together with the starting energy-entropy estimates, yields a desired bound of ${\parallel b\parallel }_{L^2(0,T;L^{\infty })}$. By virtue of these estimates, we can then deduce the uniform upper bound of the specific volume from the representation Formulas (24). The derivation of the upper bound of the specific volume in the case when $\beta =0$ relies on the t-independent estimate of ${\parallel (ln\tau )}_x{\parallel }_{L^{\infty }(0,T;L^2)}$. To achieve this, we adopt some ideas in [13] to derive a log-type inequality of ${\parallel (ln\tau )}_x{\parallel }_{L^{\infty }(0,T;L^2)}$, which, combined with some elementary analysis and the Gronwall’s inequality, leads to the uniform upper bound of $\tau$. However, we have to deal with the magnetic effects and develop some new estimates of the magnetic field to control the norm of ${\parallel b\parallel }_{L^{\infty }}$, based on the special structure of 1D equations. With the upper and lower bounds of the specific volume at hand, we then proceed to estimate the first-order spatial derivatives of the velocity and magnetic field. Since the growth exponent $\beta ⩾0$ of heat-conductivity can be arbitrarily large, we need to carry out some careful analysis for the different value ranges of $\beta$. The next step is to estimate the $H^1$-norm of the temperature, which particularly leads to the desired upper bound of temperature. The lower bound of temperature will be obtained by combining the long-time behavior and the finite-time boundedness of the temperature. We emphasize here that all these estimates are built upon the key a priori assumption (28), which indicates that the amplitude of $\alpha$ strongly depends on the initial data. So, the $H^2$-norm of the solutions are essentially required to close the a priori assumptions. This will be done by using the standard energy method and the global estimates achieved.

The rest of this paper is organized as follows. In Section 2, we establish the global uniform-in-time estimates of the solutions to the problems (1)–(5). With the help of global (uniform) estimates at hand, and the local existence results stated in Lemma 9, we prove Theorem 1 in Section 3.

2. A Priori Estimates

For positive constants $m_1,m_2,N$ and T, we define

$$\begin{array}{cc}\hfill & X(0,T;m_1,m_2,N)\hfill \\ \hfill & \triangleq \left\{(\tau ,u,b,\theta ):(\tau ,u,b,\theta )\in C([0,T];H^2),{\tau }_t\in C([0,T];H^1),\right.\hfill \\ \hfill & (u_x,b_x,{\theta }_x)\in L^2(0,T;H^2),({\tau }_x,u_t,b_t,{\theta }_t)\in L^2(0,T;H^1),\hfill \\ \hfill & \left.\mathcal{E}(0,T)⩽N^2,\tau (x,t)⩾m_1,\theta (x,t)⩾m_2,\forall (x,t)\in [0,1]\times [0,T]\right\},\hfill \end{array}$$

(19)

where

$$\mathcal{E}(0,T)\triangleq \underset{t\in [0,T]}{sup}\parallel ({\tau }_x,u_x,b_x,{\theta }_x){\parallel }_{H^1}^2+{\int }_0^T{\parallel {\theta }_t\parallel }_{L^2}^{2}dt.$$

(20)

Without loss of generality, we assume

$${\int }_0^1{\tau }_0\left(x\right)dx=1,{\int }_0^1\left({\theta }_0+{\displaystyle \frac{1}{2}}{u}_0^2+{\displaystyle \frac{1}{2}}{\tau }_0{b}_0^2\right)\left(x\right)dx=1,$$

(21)

the Equation (1) yields

$${\int }_0^1{\tau }_0\left(x\right)dx={\int }_0^1\tau (x,t)dx=1,$$

(22)

and

$${\int }_0^1\left(\theta +{\displaystyle \frac{1}{2}}{u}^2+{\displaystyle \frac{1}{2}}\tau b^2\right)(x,t)dx={\int }_0^1\left({\theta }_0+{\displaystyle \frac{1}{2}}{u}_0^2+{\displaystyle \frac{1}{2}}{\tau }_0{b}_0^2\right)\left(x\right)dx=1.$$

(23)

Next, we adapt and modify the idea of Kazhikhov [13] for the polytropic ideal gas to give a representation of $\tau$ which is crucial to get the upper and lower bounds of the specific volume $\tau$. Similar to the proof in [8,30], it is easy to obtain the following lemma, here, we omit its proof as well.

Lemma 1.

For any $t⩾0$, there exists a point ${\eta }_0\left(t\right)\in (0,1)$ such that the specific volume $\tau (x,t)$ of the problems (1)–(5) possesses the following expression

$$\begin{array}{cc}\hfill \tau (x,t)& =D(x,t)B\left(t\right)+{\int }_0^t{\displaystyle \frac{D(x,t)B\left(t\right)}{D(x,s)B\left(s\right)}}\tau (x,s)\left\{{\displaystyle \frac{1}{\lambda }}\left(p+\frac{1}{2}{b}^2\right)\right.\hfill \\ \hfill & \left.+{\int }_0^{x}g\left(y\right)dy-{\int }_0^1\tau (x,s)\left({\int }_0^{x}g\left(y\right)dy\right)dx\right\}(x,s)ds,\hfill \end{array}$$

(24)

where

$$g(x,t)\triangleq -\left[u{\left({\displaystyle \frac{1}{\lambda }}\right)}_t+\left(p+{\displaystyle \frac{1}{2}}{b}^2\right){\left({\displaystyle \frac{1}{\lambda }}\right)}_x+{\displaystyle \frac{{\lambda }_x{u}_x}{\lambda \tau }}\right],$$

(25)

$$B\left(t\right)\triangleq exp\left\{-{\int }_0^t{\int }_0^1{\displaystyle \frac{1}{\lambda }}\left(u^2+\theta +{\displaystyle \frac{1}{2}}\tau b^2\right)(x,s)dxds\right\},$$

(26)

and

$$D(x,t)\triangleq {\tau }_0\left(x\right)exp\left\{{\int }_{{\eta }_0\left(t\right)}^x{\displaystyle \frac{u}{\lambda }}dy-{\int }_0^x{\displaystyle \frac{u_0}{{\lambda }_0}}dy+{\int }_0^1{\tau }_0\left({\int }_0^x{\displaystyle \frac{u_0}{{\lambda }_0}}dy\right)dx\right\}$$

(27)

with ${\lambda }_0\triangleq \lambda \left({\theta }_0\right)$.

Throughout this paper, we denote by C and $C_i(i=1,2,\cdots )$ the generic positive constants depending solely on $N_0,M_0$ and $\beta$. Now, we begin with the following important lemma, which is the basis of our analysis.

Lemma 2.

Suppose that there are two positive constant $C_0$ and ${\epsilon }_0$, depending solely on $N_0$, $M_0$ and β, such that if $(\tau ,u,b,\theta )\in X(0,T;m_1,m_2,N)$ is a solution of the problems (1)–(5) on $(0,T)$, satisfying

$$m_2^{-\alpha }⩽2,N^{\alpha }⩽2,\alpha \mathcal{H}(m_1,m_2,N)⩽{\epsilon }_0$$

(28)

with $\mathcal{H}(m_1,m_2,N)\triangleq {\left(1+m_1^{-1}+m_2^{-1}+N\right)}^6$, then

$$C_0^{-1}⩽\tau (x,t)⩽C_0,\forall (x,t)\in {\overline{\mathrm{\Omega }}}_T\triangleq [0,1]\times [0,T].$$

(29)

Proof. 

Note that the fourth equation of (1) can be rewritten as

$${\theta }_t+{\displaystyle \frac{\theta }{\tau }}{u}_x={\left({\displaystyle \frac{\kappa {\theta }_x}{\tau }}\right)}_x+{\displaystyle \frac{\lambda u_x^2+\nu b_x^2}{\tau }},\mathrm{with}\lambda \left(\theta \right)={\theta }^{\alpha },\kappa \left(\theta \right)={\theta }^{\beta }.$$

(30)

Multiplying the first, the second and the third equation of (1) and (30) by $1-{\tau }^{-1}$, u, b and $1-{\theta }^{-1}$, respectively, and integrating by parts over ${\overline{\mathrm{\Omega }}}_T$, after adding them together, we have

$$\begin{array}{cc}\hfill & \underset{t\in [0,T]}{sup}{\int }_0^1\left({\displaystyle \frac{1}{2}}{u}^2+{\displaystyle \frac{1}{2}}\tau b^2+(\theta -ln\theta -1)+(\tau -ln\tau -1)\right)(x,t)dx\hfill \\ \hfill & +{\int }_0^{T}V\left(t\right)dt⩽2E_0\hfill \end{array}$$

(31)

with

$$V\left(t\right)\triangleq {\int }_0^1\left({\displaystyle \frac{\kappa {\theta }_x^2}{\tau {\theta }^2}}+{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau \theta }}\right)(x,t)dx,$$

$$E_0\triangleq {\int }_0^1\left({\displaystyle \frac{1}{2}}{u}_0^2+{\displaystyle \frac{1}{2}}{\tau }_0{b}_0^2+({\theta }_0-ln{\theta }_0-1)+({\tau }_0-ln{\tau }_0-1)\right)\left(x\right)dx$$

Jensen’s inequality together with (23) and (31) yields that there exists a positive constant $r_0\in (0,1)$, such that

$$\overline{\theta }\left(t\right)={\int }_0^1\theta (x,t)dx\in [r_0,1],\forall t\in [0,T].$$

(32)

Here, we have used the following Jensen’s inequality:

$$f\left({\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}udx\right)⩽{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}f\left(u\right)dx,$$

where $f:\mathbb{R}\to \mathbb{R}$ is convex, $\mathrm{\Omega }\in \mathbb{R}$ is open and bounded domain.

Next, we will prove the lower bound of $\tau$. For ${\eta }_0\left(t\right)\in (0,t)$ in Lemma 4, one has from (23), and (28) that

$$\left|{\int }_{{\eta }_0\left(t\right)}^x{\displaystyle \frac{u}{\lambda }}dy\right|⩽m_2^{-\alpha }{\parallel u\left(t\right)\parallel }_{L^2}⩽2\sqrt{2},$$

thus, (27) gives

$$0<C^{-1}⩽D(x,t)⩽C,\forall (x,t)\in [0,1]\times [0,T].$$

(33)

Applying the mean value theorem of integrals, one obtains

$$\parallel (\theta -\overline{\theta })\left(t\right){\parallel }_{L^{\infty }}⩽{\parallel {\theta }_x\left(t\right)\parallel }_{L^2}⩽N,\forall t\in [0,T],$$

which implies

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{4}}& ⩽{\left(2N\right)}^{-\alpha }{\int }_0^1({\displaystyle \frac{1}{2}}{u}^2+\theta +{\displaystyle \frac{1}{2}}\tau b^2)dx⩽{\int }_0^1{\displaystyle \frac{1}{\lambda }}({\displaystyle \frac{1}{2}}{u}^2+\theta +{\displaystyle \frac{1}{2}}\tau b^2)dx\hfill \\ \hfill & ⩽2m_2^{-\alpha }{\int }_0^1({\displaystyle \frac{1}{2}}{u}^2+\theta +{\displaystyle \frac{1}{2}}\tau b^2)dx⩽4,\hfill \end{array}$$

thus, (26) yields

$$e^{-4t}⩽B\left(t\right)⩽e^{-\frac{t}{4}},e^{-4(t-s)}⩽{\displaystyle \frac{B\left(t\right)}{B\left(s\right)}}⩽e^{-\frac{t-s}{4}}.$$

(34)

We deduce from (25) and (28) that

$$\begin{array}{cc}\hfill \left|\tau {\int }_0^{x}gdy\right|& =\left|\tau {\int }_0^x\left[u{\left({\displaystyle \frac{1}{\lambda }}\right)}_t+\left(p+{\displaystyle \frac{1}{2}}{b}^2\right){\left({\displaystyle \frac{1}{\lambda }}\right)}_x+{\displaystyle \frac{{\lambda }_x{u}_x}{\lambda \tau }}\right]dy\right|\hfill \\ \hfill & ⩽{\alpha \parallel \tau \parallel }_{L^{\infty }}{\int }_0^1\left(|{\theta }^{-(\alpha +1)}{\theta }_{t}u|+|{\theta }^{-\alpha }{\tau }^{-1}{\theta }_x|+|{\theta }^{-(\alpha +1)}{\theta }_x{b}^2|+\left|{\displaystyle \frac{{\theta }_x{u}_x}{\theta \tau }}\right|\right)dx\hfill \\ \hfill & ⩽2\alpha N{m}_2^{-\alpha }\left({\displaystyle \frac{1}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}{\parallel u\parallel }_{L^2}+{\displaystyle \frac{1}{m_1}}\parallel {\theta }_x{\parallel }_{L^2}+{\displaystyle \frac{1}{m_2}}{\parallel b\parallel }_{L^{\infty }}{\parallel {\theta }_x\parallel }_{L^2}\right)\hfill \\ \hfill & +{\displaystyle \frac{2\alpha N}{m_1{m}_2}}\parallel {\theta }_x{\parallel }_{L^2}{\parallel u_x\parallel }_{L^2}\hfill \\ \hfill & ⩽4\alpha N\left({\displaystyle \frac{2}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}+{\displaystyle \frac{N}{m_1}}+{\displaystyle \frac{N}{m_2}}{\parallel b\parallel }_{L^{\infty }}\right)+{\displaystyle \frac{4\alpha N^3}{m_1{m}_2}}\hfill \\ \hfill & ⩽{\displaystyle \frac{\alpha N}{m_2}}(\parallel {\theta }_t{\parallel }_{L^2}^2+{\parallel b\parallel }_{L^{\infty }}^2)+C\alpha \mathcal{H}(m_1,m_2,N),\hfill \end{array}$$

(35)

here, we have used the following facts:

$${\parallel \tau \parallel }_{L^{\infty }}⩽{\parallel \tau \parallel }_{L^1}+\parallel {\tau }_x{\parallel }_{L^2}{,\parallel u\parallel }_{L^2}⩽{\parallel u_x\parallel }_{L^2}⩽N.$$

In a similar manner

$$\begin{array}{cc}\hfill & \left|\tau {\int }_0^1\tau \left({\int }_0^{x}gdy\right)dx\right|\hfill \\ \hfill & =\left|\tau {\int }_0^1\tau \left({\int }_0^x\left[u{\left({\displaystyle \frac{1}{\lambda }}\right)}_t+\left(p+{\displaystyle \frac{1}{2}}{b}^2\right){\left({\displaystyle \frac{1}{\lambda }}\right)}_x+{\displaystyle \frac{{\lambda }_x{u}_x}{\lambda \tau }}\right]dy\right)dx\right|\hfill \\ \hfill & ⩽{\alpha \parallel \tau \parallel }_{L^{\infty }}{\int }_0^1\left(|{\theta }^{-(\alpha +1)}{\theta }_{t}u|+|{\theta }^{-\alpha }{\tau }^{-1}{\theta }_x|+|{\theta }^{-(\alpha +1)}{\theta }_x{b}^2|+\left|{\displaystyle \frac{{\theta }_x{u}_x}{\theta \tau }}\right|\right)dx\hfill \\ \hfill & ⩽2\alpha N{m}_2^{-\alpha }\left({\displaystyle \frac{1}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}{\parallel u\parallel }_{L^2}+{\displaystyle \frac{1}{m_1}}\parallel {\theta }_x{\parallel }_{L^2}+{\displaystyle \frac{1}{m_2}}{\parallel b\parallel }_{L^{\infty }}{\parallel {\theta }_x\parallel }_{L^2}\right)\hfill \\ \hfill & +{\displaystyle \frac{2\alpha N}{m_1{m}_2}}\parallel {\theta }_x{\parallel }_{L^2}{\parallel u_x\parallel }_{L^2}\hfill \\ \hfill & ⩽4\alpha N\left({\displaystyle \frac{2}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}+{\displaystyle \frac{N}{m_1}}+{\displaystyle \frac{N}{m_2}}{\parallel b\parallel }_{L^{\infty }}\right)+{\displaystyle \frac{4\alpha N^3}{m_1{m}_2}}\hfill \\ \hfill & ⩽{\displaystyle \frac{\alpha N}{m_2}}(\parallel {\theta }_t{\parallel }_{L^2}^2+{\parallel b\parallel }_{L^{\infty }}^2)+C\alpha \mathcal{H}(m_1,m_2,N),\hfill \end{array}$$

(36)

Define $f_{+}=max\{f,0\}$. It follows from (22) and (32) that

$$\begin{array}{cc}\hfill {\left({\overline{\theta }}^{\frac{\beta +1}{2}}\left(t\right)-{\theta }^{\frac{\beta +1}{2}}(x,t)\right)}_{+}& ⩽C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^2}}dx\right)}^{1/2}{\left({\int }_0^1\tau \theta {\chi }_{(\theta ⩽\overline{\theta })}dx\right)}^{1/2}⩽CV{\left(t\right)}^{1/2}\hfill \end{array}$$

(37)

with ${\chi }_{(\theta ⩽\overline{\theta })}=1$ if $\theta ⩽\overline{\theta }$, and ${\chi }_{(\theta ⩽\overline{\theta })}=0$ if $\theta ⩾\overline{\theta }$. Thus

$$\underset{x\in [0,1]}{min}\theta (x,t)⩾C_1-C_{2}V\left(t\right).$$

(38)

Plugging (31), (33)–(36), and (38) into (24), then using (28), one has

$$\begin{array}{cc}\hfill \tau (x,t)& ⩾C^{-1}{\displaystyle \frac{1}{{\parallel \theta \parallel }_{L^{\infty }}^{\alpha }}}{\int }_0^t{e}^{-4(t-s)}\underset{x\in [0,1]}{min}\theta (x,s)ds\hfill \\ \hfill & -C{\int }_0^t{e}^{-\frac{t-s}{4}}\left({\displaystyle \frac{\alpha N}{m_2}}\parallel {\theta }_s{\parallel }_{L^2}^2+{\parallel b\parallel }_{L\infty }^2+\alpha \mathcal{H}(m_1,m_2,N)\right)ds\hfill \\ \hfill & ⩾{\displaystyle \frac{C}{1+N^{\alpha }}}{\int }_0^t{e}^{-4(t-s)}\left(C_1-C_{2}V\left(s\right)\right)ds-{\displaystyle \frac{C\alpha N}{m_2}}{\int }_0^t{\parallel {\theta }_s\parallel }_{L^2}^{2}ds\hfill \\ \hfill & -{\displaystyle \frac{C\alpha N}{m_2}}{\parallel b_x\parallel }_{L^2}^2{\int }_0^t{e}^{-\frac{t-s}{4}}ds-C\alpha \mathcal{H}(m_1,m_2,N)\hfill \\ \hfill & ⩾{\displaystyle \frac{C{C}_1}{12}}\left(1-e^{-4t}\right)-C{C}_2{\int }_0^t{e}^{-4(t-s)}V\left(s\right)ds-C{\epsilon }_0.\hfill \end{array}$$

(39)

We deduce from (31) that

$$\begin{array}{cc}\hfill {\int }_0^t{e}^{-4(t-s)}V\left(s\right)ds& ={\int }_0^{\frac{t}{2}}{e}^{-4(t-s)}V\left(s\right)ds+{\int }_{\frac{t}{2}}^t{e}^{-4(t-s)}V\left(s\right)ds\hfill \\ \hfill & ⩽e^{-2t}{\int }_0^{\frac{t}{2}}V\left(s\right)ds+{\int }_{\frac{t}{2}}^{t}V\left(s\right)ds\to 0,t\to \infty ,\hfill \end{array}$$

thus, there exists a sufficiently large number $\tilde{T}>0$, such that

$${\displaystyle \frac{C{C}_1}{12}}{e}^{-4t}+C{C}_2{\int }_0^t{e}^{-4(t-s)}V\left(s\right)ds⩽{\displaystyle \frac{C{C}_1}{36}},\forall t⩾\tilde{T},$$

then

$$\tau (x,t)⩾{\displaystyle \frac{C{C}_1}{36}},\forall x\in [0,1],t⩾\tilde{T}$$

(40)

with $0<{\epsilon }_0⩽min\{1,\frac{C_1}{36C}\}$.

On the other hand, it follows from (24), (28) and (33)–(36) that for any $(x,t)\in [0,1]\times [0,\tilde{T}]$

$$\begin{array}{cc}\hfill \tau (x,t)& ⩾B\left(t\right)D(x,t)-C\alpha \mathcal{H}(m_1,m_2,N){\int }_0^{\tilde{T}}{e}^{-\frac{t-s}{4}}ds\hfill \\ \hfill & -{\displaystyle \frac{C\alpha N}{m_2}}{\int }_0^{\tilde{T}}\parallel {\theta }_s{\parallel }_{L^2}^{2}ds-{\displaystyle \frac{C\alpha N}{m_2}}{\parallel b_x\parallel }_{L^2}^2{\int }_0^t{e}^{-\frac{t-s}{4}}ds\hfill \\ \hfill & ⩾C_3^{-1}{e}^{-4\tilde{T}}-C_3\alpha \mathcal{H}(m_1,m_2,N)⩾C_3^{-1}{e}^{-4\tilde{T}}-C_3{\epsilon }_0⩾{\displaystyle \frac{e^{-4t}}{2C_3}}\hfill \end{array}$$

(41)

with ${\epsilon }_0$ is suitable small such that ${\epsilon }_0⩽\frac{e^{-4\tilde{T}}}{2C_3^2}$. Combining (40) with (41), one has

$$\tau (x,t)⩾C_0^{-1}\triangleq min\left\{{\displaystyle \frac{C{C}_1}{36}},{\displaystyle \frac{e^{-4\tilde{T}}}{2C_3}}\right\},$$

(42)

provided $\alpha \mathcal{H}(m_1,m_2,N)⩽{\epsilon }_0$ with ${\epsilon }_0⩽min\left\{1,\frac{C_1}{36C},\frac{e^{-4\tilde{T}}}{2C_3^2}\right\}$.

Thanks to $\beta ⩾0$, one has

$$\begin{array}{c}\hfill {\overline{\theta }}^{\frac{\beta +1}{2}}\left(t\right)-{\theta }^{\frac{\beta +1}{2}}(x,t)⩽C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^2}}dx\right)}^{1/2}{\left({\int }_0^1\tau \theta dx\right)}^{1/2}⩽CV{\left(t\right)}^{1/2}\underset{x\in [0,1]}{max}{\tau }^{1/2}(x,t),\end{array}$$

which, together with Cauchy-Schwarz’s inequality, yields

$$\theta (x,t)⩽C+CV\left(t\right)\underset{x\in [0,1]}{max}\tau (x,t).$$

(43)

It follows from (28) and (31) that

$$\begin{array}{c}\hfill {\parallel b\parallel }_{L^{\infty }}^2⩽C{\int }_0^1|b{b}_x|dx⩽C{\left({\int }_0^1{\displaystyle \frac{b_x^2}{\tau \theta }}dx\right)}^{1/2}{\left({\int }_0^1\tau b^{2}dx\right)}^{1/2}{\parallel \theta \parallel }_{L^{\infty }}^{1/2}⩽CV{\left(t\right)}^{1/2}{\parallel \theta \parallel }_{L^{\infty }}^{1/2}.\end{array}$$

(44)

Combining the estimates (31), (33)–(36), (43), (44) with Cauchy-Schwarz inequality, one has

$$\begin{array}{cc}\hfill \tau (x,t)& ⩽C{e}^{-\frac{t}{4}}+C{\int }_0^t{e}^{-\frac{t-s}{4}}\left({\displaystyle \frac{1}{{\theta }^{\alpha }}}(\theta +b^2)+\alpha \mathcal{H}(m_1,m_2,N)+{\displaystyle \frac{\alpha N}{m_2}}(\parallel {\theta }_s{\parallel }_{L^2}^2+{\parallel b\parallel }_{L^{\infty }}^2)\right)ds\hfill \\ \hfill & ⩽C+C{m}_2^{-\alpha }{\int }_0^t{e}^{-\frac{t-s}{4}}\underset{x\in [0,1]}{max}\theta (x,t)ds+C{m}_2^{-\alpha }{\int }_0^t{e}^{-\frac{t-s}{4}}{V}^{1/2}\left(s\right){\parallel \theta \parallel }_{L^{\infty }}^{1/2}ds\hfill \\ \hfill & +C\alpha \mathcal{H}(m_1,m_2,N){\int }_0^t{e}^{-\frac{t-s}{4}}ds+{\displaystyle \frac{C\alpha N}{m_2}}{\int }_0^t\parallel {\theta }_s{\parallel }_{L^2}^{2}ds+{\displaystyle \frac{C\alpha N}{m_2}}{\parallel b_x\parallel }_{L^{\infty }}^2{\int }_0^t{e}^{-\frac{t-s}{4}}ds\hfill \\ \hfill & ⩽C+C\alpha \mathcal{H}(m_1,m_2,N)+C{\int }_0^t{e}^{-\frac{t-s}{4}}V\left(s\right)ds\hfill \\ \hfill & +C{\int }_0^t{e}^{-\frac{t-s}{4}}\left(1+V\left(s\right)\underset{x\in [0,1]}{max}\tau (x,s)\right)ds\hfill \\ \hfill & ⩽C+C{\int }_0^{t}V\left(s\right)\underset{x\in [0,1]}{max}\tau (x,s)ds,\hfill \end{array}$$

thus, applying Gronwall’s inequality, one has

$$\tau (x,t)⩽C_0,\forall (x,t)\in [0,1]\times [0,T],$$

which, together with (42), yields (29). □

Lemma 3.

Under the condition of Lemma 2, then for any $\gamma >0$, the following estimates hold

$${\int }_0^T{\int }_0^1{\displaystyle \frac{\kappa {\theta }_x^2}{{\theta }^{\gamma +1}}}dxdt⩽C,{\int }_0^T\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)dt⩽C,$$

(45)

where the positive constant C may depend on γ.

Proof. 

By virtue of (29) and (31), for $\gamma =1$, we obtain

$${\int }_0^T{\int }_0^1{\displaystyle \frac{\kappa {\theta }_x^2}{{\theta }^2}}dxdt⩽C{\int }_0^T{\int }_0^1{\displaystyle \frac{\kappa {\theta }_x^2}{\tau {\theta }^2}}\tau dxdt⩽C.$$

Next, for $\gamma \ne 1$, we Multiply (30) by ${\theta }^{-\gamma }(\gamma \ne 1)$, and integrate the resulting equations by parts, we have from (29) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{\gamma -1}}{\left({\int }_0^1{\theta }^{1-\gamma }dx\right)}_t+\gamma {\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^{\gamma +1}}}dx+{\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau {\theta }^{\gamma }}}dx\hfill \\ \hfill & ={\int }_0^1{\displaystyle \frac{{\theta }^{1-\gamma }-1}{\tau }}{u}_{x}dx+{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽C\left(\gamma \right)\parallel {\theta }^{1/2}{-1\parallel }_{L^{\infty }}{\left({\int }_0^1{\displaystyle \frac{\tau {\theta }^{1-\gamma }}{\lambda }}dx\right)}^{1/2}{\left({\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau {\theta }^{\gamma }}}dx\right)}^{1/2}\hfill \\ \hfill & +C\left(\gamma \right)\parallel {\theta }^{1/2}{-1\parallel }_{L^{\infty }}{\int }_0^1\left|u_x\right|dx+{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau {\theta }^{\gamma }}}dx+C\left(\gamma \right)\parallel {\theta }^{1/2}{-1\parallel }_{L^{\infty }}^2{\int }_0^1{\theta }^{1-\gamma }dx+C\left(\gamma \right){\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^2\hfill \\ \hfill & +C\left(\gamma \right){\left({\int }_0^1\left|u_x\right|dx\right)}^2+{\left({\int }_0^{1}ln\tau dx\right)}_t,\hfill \end{array}$$

(46)

where we have used the following fact:

$${\int }_0^1{\displaystyle \frac{\tau {\theta }^{1-\gamma }}{\lambda }}dx⩽{\int }_0^1{\theta }^{\alpha }\tau {\theta }^{1-\gamma }dx⩽C_0{m}_2^{-\alpha }{\int }_0^1{\theta }^{1-\gamma }dx⩽C{\int }_0^1{\theta }^{1-\gamma }dx.$$

Hölder inequality together with (23), (29) and (31), leads to

$${\int }_0^1\left|u_x\right|dx⩽C{\left({\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau \theta }}\right)}^{1/2}{\left({\int }_0^1{\displaystyle \frac{\tau \theta }{\lambda }}dx\right)}^{1/2}⩽C{V}^{1/2}\left(t\right),$$

(47)

and

$${\parallel b\parallel }_{L^{\infty }}⩽C{\int }_0^1|b_x|dx⩽C{\left({\int }_0^1{\displaystyle \frac{b_x^2}{\tau \theta }}dx\right)}^{1/2}{\left({\int }_0^1\theta dx\right)}^{1/2}{\parallel \tau \parallel }_{L^{\infty }}^{1/2}⩽C{V}^{1/2}\left(t\right),$$

(48)

thus, we get from (23), (47) and (48) that

$$\begin{array}{cc}\hfill |1-{\overline{\theta }}^{1/2}|& ⩽C|1-\overline{\theta }|=C\left|1-{\int }_0^1\theta dx\right|=C\left|{\displaystyle \frac{1}{2}}{\int }_0^1(u^2+\tau b^2)dx\right|\hfill \\ \hfill & ⩽{C\parallel u\parallel }_{L^2}^2+{C\parallel b\parallel }_{L^{\infty }}{\int }_0^1\left|\tau b\right|dx\hfill \\ \hfill & ⩽{C\parallel u\parallel }_{L^2}{\parallel u\parallel }_{L^{\infty }}+C{\parallel b\parallel }_{L^{\infty }}{\left({\int }_0^1\tau dx\right)}^{1/2}{\left({\int }_0^1\tau b^{2}dx\right)}^{1/2}\hfill \\ \hfill & ⩽\left(\parallel u_x{\parallel }_{L^1}+{\parallel b\parallel }_{L^{\infty }}\right)⩽C{V}^{1/2}\left(t\right).\hfill \end{array}$$

(49)

Similarly

$$\begin{array}{cc}\hfill & |{\theta }^{1/2}-{\overline{\theta }}^{1/2}|⩽C\left|{\theta }^{\frac{\beta +1}{2}}(x,t)-{\overline{\theta }}^{\frac{\beta +1}{2}}\left(t\right)\right|⩽C{\int }_0^1{\theta }^{\frac{\beta -1}{2}}\left|{\theta }_x\right|dx\hfill \\ \hfill & ⩽C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^2}}dx\right)}^{1/2}{\left({\int }_0^1\theta \tau dx\right)}^{1/2}\hfill \\ \hfill & ⩽C{V}^{1/2}\left(t\right).\hfill \end{array}$$

(50)

Thus, using (47), (49) and (50), we obtain

$${\int }_0^t{\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^{2}ds+{\int }_0^t{\left({\int }_0^1\left|u_x\right|dx\right)}^{2}ds⩽C{\int }_0^{t}V\left(s\right)ds⩽C.$$

(51)

On the other hand, we infer from (29) that

$${\int }_0^{1}ln\tau dx⩽C,$$

which, combine with (46), (50) and Gronwall’s inequality, yields the first equation of (45) with $\gamma >1$ holds.

When $0<\gamma <1$, we have

$${\int }_0^1{\theta }^{1-\gamma }dx⩽C{\int }_0^1(1+\theta )dx⩽C.$$

(52)

Integrating (46) over $[0,T]$, and using (52), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{\gamma -1}}\underset{t\in [0,T]}{sup}\left({\int }_0^1{\theta }^{1-\gamma }dx\right)-{\displaystyle \frac{1}{\gamma -1}}{\int }_0^1{\theta }_0^{1-\gamma }dx+\gamma {\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^{\gamma +1}}}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau {\theta }^{\gamma }}}dx\hfill \\ \hfill & ⩽C+C\left(\gamma \right){\int }_0^T{\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^2\left(1+{\int }_0^1{\theta }^{1-\gamma }dx\right)dt+C\left(\gamma \right){\int }_0^T{\left({\int }_0^1\left|u_x\right|dx\right)}^{2}dt\hfill \\ \hfill & ⩽C,\hfill \end{array}$$

thus, this inequality yields the first equation of (45) with $0<\gamma <1$ holds directly.

In order to obtain the second equation of (45), multiplying the second equation of (1) by u in $L^2$, and integrating by parts over $[0,1]$, we obtain from (23), (28) and (29) that

$$\begin{array}{cc}\hfill & {\left({\int }_0^1{\displaystyle \frac{u^2}{2}}dx\right)}_t+{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau }}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{b}^2{u}_{x}dx={\int }_0^1{\displaystyle \frac{\theta -1}{\tau }}{u}_{x}dx+{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau }}dx+C_0{m}_2^{-\alpha }{\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^2{\int }_0^1(1+\theta )dx+{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau }}dx+C{\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^2+{\left({\int }_0^{1}ln\tau dx\right)}_t.\hfill \end{array}$$

(53)

Multiplying the third equation of (1) by b in $L^2$, and integrating by parts over $[0,1]$, we obtain

$${\left({\int }_0^1{\displaystyle \frac{\tau b^2}{2}}dx\right)}_t+{\int }_0^1{\displaystyle \frac{b_x^2}{\tau }}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{b}^2{u}_{x}dx=0,$$

which, together with (53), leads to

$${\left({\int }_0^1{\displaystyle \frac{u^2+\tau b^2}{2}}dx\right)}_t+{\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau }}dx⩽C{\parallel {\theta }^{1/2}-1\parallel }_{L^{\infty }}^2+{\left({\int }_0^{1}ln\tau dx\right)}_t,$$

thus, it follows from (28), (29) and (49) that

$$\begin{array}{cc}\hfill & {\parallel (u,b)\left(t\right)\parallel }_{L^2}^2+{\int }_0^t(\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2)ds\hfill \\ \hfill & ⩽{\parallel (u,b)\left(t\right)\parallel }_{L^2}^2+C{\int }_0^t{\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau }}dxds\hfill \\ \hfill & ⩽C{\int }_0^t\parallel {\theta }^{1/2}{-1\parallel }_{L^{\infty }}^{2}ds+{\int }_0^{1}ln\tau dx-{\int }_0^{1}ln{\tau }_{0}dx+{\parallel (u_0,b_0)\parallel }_{L^2}^2\hfill \\ \hfill & ⩽C.\hfill \end{array}$$

The proof of Lemma 3 is completed. □

Lemma 4.

Under the condition of Lemma 2, the following estimate holds

$$\underset{t\in [0,T]}{sup}{\parallel {\tau }_x\left(t\right)\parallel }_{L^2}^2+{\int }_0^T{\int }_0^1\left({\tau }_x^2+\theta {\tau }_x^2+b^2{b}_x^2\right)(x,t)dxdt⩽C_2.$$

(54)

Proof. 

Due to the first and the second equation of (1), it is easy to see that

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}_t& ={\left({\displaystyle \frac{\lambda u_x}{\tau }}\right)}_x+{\displaystyle \frac{{\lambda }_t{\tau }_x-{\lambda }_x{u}_x}{\tau }}\hfill \\ \hfill & =u_t-{\displaystyle \frac{\theta {\tau }_x}{{\tau }^2}}+{\displaystyle \frac{{\theta }_x}{\tau }}+b{b}_x+{\displaystyle \frac{\alpha {\theta }^{\alpha }}{\tau \theta }}\left({\theta }_t{\tau }_x-{\theta }_x{u}_x\right).\hfill \end{array}$$

(55)

Multiplying (55) by $\lambda {\tau }_x/\tau$ in $L^2$ and integrating by parts it over $(0,1)$, we infer from ${u|}_{x=0,1}=0$ that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}^{2}dx+{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx\hfill \\ \hfill & ={\int }_0^1{\displaystyle \frac{\lambda {\tau }_x{u}_t}{\tau }}dx+{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x{\tau }_x}{{\tau }^2}}dx+{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}b{b}_x}{\tau }}dx+{\int }_0^1{\displaystyle \frac{\alpha {\lambda }^2({\theta }_t{\tau }_x-{\theta }_x{u}_x){\tau }_x}{{\tau }^2\theta }}dx\hfill \\ \hfill & ={\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx-{\int }_0^1\left[\lambda u{(ln\tau )}_{xt}+{\lambda }_{t}u{(ln\tau )}_x\right]dx\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x{\tau }_x}{{\tau }^2}}dx+{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}b{b}_x}{\tau }}dx+{\int }_0^1{\displaystyle \frac{\alpha {\lambda }^2({\theta }_t{\tau }_x-{\theta }_x{u}_x){\tau }_x}{{\tau }^2\theta }}dx\hfill \\ \hfill & ={\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx-{\int }_0^1\left[{\left({\displaystyle \frac{\lambda u{u}_x}{\tau }}\right)}_x-{\left(\lambda u\right)}_x{\displaystyle \frac{u_x}{\tau }}+{\displaystyle \frac{{\lambda }_{t}u{\tau }_x}{\tau }}\right]dx\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x{\tau }_x}{{\tau }^2}}dx+{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}b{b}_x}{\tau }}dx+{\int }_0^1{\displaystyle \frac{\alpha {\lambda }^2({\theta }_t{\tau }_x-{\theta }_x{u}_x){\tau }_x}{{\tau }^2\theta }}dx\hfill \\ \hfill & ={\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx+{\int }_0^1\left[{\left(\lambda u\right)}_x{\displaystyle \frac{u_x}{\tau }}-{\displaystyle \frac{{\lambda }_{t}u{\tau }_x}{\tau }}\right]dx\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x{\tau }_x}{{\tau }^2}}dx+{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}b{b}_x}{\tau }}dx+{\int }_0^1{\displaystyle \frac{\alpha {\lambda }^2({\theta }_t{\tau }_x-{\theta }_x{u}_x){\tau }_x}{{\tau }^2\theta }}dx\hfill \\ \hfill & \triangleq {\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx+\sum _{i=1}^4{J}_i.\hfill \end{array}$$

(56)

To deal with the right-hand side of (56), we divided the estimates into two cases:

Case I: $\beta =0$.

The Cauchy-Schwarz’s inequality together with (19), (28), (29) yields

$$\begin{array}{cc}\hfill J_1& ⩽C{\int }_0^1\left[\lambda u_x^2+\alpha {\theta }^{\alpha -1}\left(|{\theta }_x\left|\right|u\left|\right|u_x|+|{\theta }_t\left|\right|u\left|\right|{\tau }_x|\right)\right]dx\hfill \\ \hfill & ⩽{C\parallel \theta \parallel }_{L^{\infty }}^{\alpha }\parallel u_x{\parallel }_{L^2}^2+C{\displaystyle \frac{{\parallel \theta \parallel }_{L^{\infty }}^{\alpha }}{m_2}}{\parallel u\parallel }_{L^{\infty }}\left(\parallel u_x{\parallel }_{L^2}\parallel {\theta }_x{\parallel }_{L^2}+\parallel {\tau }_x{\parallel }_{L^2}{\parallel {\theta }_t\parallel }_{L^2}\right)\hfill \\ \hfill & ⩽C(1+N^{\alpha }){\parallel u_x\parallel }_{L^2}^2+{\displaystyle \frac{C\alpha (1+N^{\alpha })}{m_2}}\left(\parallel u_x{\parallel }_{L^2}^2\parallel {\theta }_x{\parallel }_{L^2}+\parallel u_x{\parallel }_{L^2}\parallel {\tau }_x{\parallel }_{L^2}{\parallel {\theta }_t\parallel }_{L^2}\right)\hfill \\ \hfill & ⩽C\parallel u_x{\parallel }_{L^2}^2+{\displaystyle \frac{C\alpha N^2}{m_2}}\left(\parallel u_x{\parallel }_{L^2}^2+\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\theta }_t\parallel }_{L^2}^2\right)\hfill \\ \hfill & ⩽C\left(1+\alpha \mathcal{H}(m_1,m_2,N)\right)\parallel u_x{\parallel }_{L^2}^2+{\displaystyle \frac{C\alpha N^2}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}^2+C\alpha \mathcal{H}(m_1,m_2,N){\parallel {\theta }_x\parallel }_{L^2}^2\hfill \\ \hfill & ⩽C\parallel u_x{\parallel }_{L^2}^2+{\displaystyle \frac{C\alpha N^2}{m_2}}\parallel {\theta }_t{\parallel }_{L^2}^2+C{\epsilon }_0{\parallel {\theta }_x\parallel }_{L^2}^2.\hfill \end{array}$$

(57)

Next, it follows from (29) and (31) that for any $\delta \in (0,1)$

$$\begin{array}{cc}\hfill J_2& ⩽C{\left({\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx\right)}^{1/2}{\left({\int }_0^1{\displaystyle \frac{\lambda {\theta }_x^2}{\theta }}dx\right)}^{1/2}\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+C{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x^2}{\theta }}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+C{\parallel \theta \parallel }_{L^{\infty }}^{\alpha }{\int }_0^1\left(\delta +C\left(\delta \right){\theta }^{-2}\right){\theta }_x^{2}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+C\delta (1+N^{\alpha }){\parallel {\theta }_x\parallel }_{L^2}^2\hfill \\ \hfill & +C\left(\delta \right)(1+N^{\alpha }){\int }_0^1{\displaystyle \frac{{\theta }_x^2}{{\theta }^2}}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+C\delta {\parallel {\theta }_x\parallel }_{L^2}^2+C\left(\delta \right)V\left(t\right),\hfill \end{array}$$

(58)

and

$$\begin{array}{cc}\hfill J_3& ⩽{C\parallel \theta \parallel }_{L^{\infty }}^{\alpha }{\parallel b\parallel }_{L^{\infty }}\parallel {\tau }_x{\parallel }_{L^2}{\parallel b_x\parallel }_{L^2}\hfill \\ \hfill & ⩽C(1+N^{\alpha })\parallel b_x{\parallel }_{L^2}^2{\parallel {\tau }_x\parallel }_{L^2}\hfill \\ \hfill & ⩽C\parallel b_x{\parallel }_{L^2}^2+C\parallel b_x{\parallel }_{L^2}^2{\parallel {\tau }_x\parallel }_{L^2}^2.\hfill \end{array}$$

(59)

Similarly

$$\begin{array}{cc}\hfill J_4& ⩽{\displaystyle \frac{C\alpha (N^{3\alpha /2}+1)}{m_2}}{\int }_0^1{\lambda }^{1/2}\left({\tau }_x^2|{\theta }_t|+|u_x\left|\right|{\tau }_x\left|\right|{\theta }_x|\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+{\displaystyle \frac{C{\alpha }^2}{m_2^2}}\left(\parallel {\tau }_x{\parallel }_{L^{\infty }}^2+{\parallel u_x\parallel }_{L^{\infty }}^2\right){\int }_0^1{\displaystyle \frac{{\tau }^3}{\theta }}\left({\theta }_t^2+{\theta }_x^2\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^3}}\left(\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\theta }_t\parallel }_{L^2}^2\right)\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx+{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^3}}\parallel {\theta }_t{\parallel }_{L^2}^2+C{\left[\alpha \mathcal{H}(m_1,m_2,N)\right]}^2{\parallel {\theta }_x\parallel }_{L^2}^2\hfill \end{array}$$

(60)

Putting (57)–(59) into (56), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}^{2}dx+{\int }_0^1{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}dx\hfill \\ \hfill & ⩽{\left({\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx\right)}_t+{\tilde{C}}_1({\epsilon }_0+\delta )\parallel {\theta }_x{\parallel }_{L^2}^2+C\parallel b_x{\parallel }_{L^2}^2{\parallel {\tau }_x\parallel }_{L^2}^2\hfill \\ \hfill & +C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2+V\left(t\right)\right)+C\left({\displaystyle \frac{C\alpha N^2}{m_2}}+{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^3}}\right){\parallel {\theta }_t\parallel }_{L^2}^2.\hfill \end{array}$$

(61)

Multiplying (1)₄ by $(\theta +\frac{u^2}{2}+\frac{\tau b^2}{2})$, and integrating the resulting equations over $[0,1]$, after using the Cauchy-Schwarz’s inequality, we get from (29) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\left(\theta +{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{\tau b^2}{2}}\right)}^{2}dx+{\int }_0^1\left({\displaystyle \frac{\lambda u^2{u}_x^2+{\theta }_x^2}{\tau }}+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽{\int }_0^1\left(\lambda |u{u}_{x}b{b}_x|+\lambda |u{u}_x{\tau }_x{b}^2|+\lambda |u{u}_x{\theta }_x|+|b{b}_{x}u{u}_x|+|b^3{b}_x{\tau }_x|+|b{b}_x{\theta }_x|\right)dx\hfill \\ \hfill & +{\int }_0^1\left(|{\theta }_{x}u{u}_x|+|{\theta }_x{\tau }_x{b}^2|+|u\theta {\theta }_x|+|u^2\theta u_x|+|u\theta {\tau }_x{b}^2|\right)dx\hfill \\ \hfill & +{\int }_0^1\left(|u\theta b{b}_x|+|u{b}^2{\theta }_x|+|u^2{b}^2{u}_x|+|u{\tau }_x{b}^4|+|u{b}^3{b}_x|\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1\left({\displaystyle \frac{\lambda u^2{u}_x^2+{\theta }_x^2}{\tau }}+b^2{b}_x^2\right)dx+C{\int }_0^1{\displaystyle \frac{{\lambda }^2{u}^2{u}_x^2+u^2{u}_x^2}{\tau }}dx+C{\int }_0^1{\displaystyle \frac{b^2{b}_x^2}{\tau }}dx\hfill \\ \hfill & +C(1+N^{\alpha }){\parallel b\parallel }_{L^{\infty }}^4\parallel {\tau }_x{\parallel }_{L^2}^2+{C\parallel u\parallel }_{L^{\infty }}^2{\parallel \theta \parallel }_{L^2}^2+{C\parallel b\parallel }_{L^{\infty }}^4{\parallel u\parallel }_{L^2}^2,\hfill \end{array}$$

thus

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\left(\theta +{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{\tau b^2}{2}}\right)}^{2}dx+{\int }_0^1\left({\displaystyle \frac{\lambda u^2{u}_x^2+{\theta }_x^2}{\tau }}+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽{\tilde{C}}_2{\int }_0^1{\displaystyle \frac{\lambda u^2{u}_x^2}{\tau }}dx+{\tilde{C}}_3{\int }_0^1{\displaystyle \frac{b^2{b}_x^2}{\tau }}dx\hfill \\ \hfill & +C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)\left(\parallel {\tau }_x{\parallel }_{L^2}^2+{\parallel \theta \parallel }_{L^2}^2\right)+C{\parallel b_x\parallel }_{L^2}^2,\hfill \end{array}$$

(62)

where we have used the Poincaré inequality and the following facts:

$${\parallel b\parallel }_{L^{\infty }}⩽{C\parallel b\parallel }_{L^2}^{1/2}\parallel b_x{\parallel }_{L^2}^{1/2}⩽C\parallel b_x{\parallel }_{L^2}^{1/2}{,\parallel u\parallel }_{L^{\infty }}⩽{C\parallel u\parallel }_{L^2}^{1/2}\parallel u_x{\parallel }_{L^2}^{1/2}⩽C{\parallel u_x\parallel }_{L^2}.$$

Next, multiplying the second equation of (1) by $u^3$, and integrating it over $[0,1]$, we obtain from (28) and (29) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{4}}{\displaystyle \frac{d}{dt}}{\int }_0^1{u}^{4}dx+3{\int }_0^1{\displaystyle \frac{\lambda u^2{u}_x^2}{\tau }}dx=3{\int }_0^1{\displaystyle \frac{u^2\theta u_x}{\tau }}dx+{\displaystyle \frac{3}{2}}{\int }_0^1{u}^2{b}^2{u}_{x}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u^2{u}_x^2}{\tau }}dx+C{m}_2^{-\alpha }{\parallel u\parallel }_{L^{\infty }}^2{\parallel \theta \parallel }_{L^2}^2+C{m}_2^{-\alpha }{\parallel b\parallel }_{L^{\infty }}^4{\parallel u\parallel }_{L^2}^2,\hfill \end{array}$$

then

$${\displaystyle \frac{d}{dt}}{\int }_0^1{u}^{4}dx+{\int }_0^1{\displaystyle \frac{\lambda u^2{u}_x^2}{\tau }}dx⩽C\parallel u_x{\parallel }_{L^2}^2{\parallel \theta \parallel }_{L^2}^2+C{\parallel b_x\parallel }_{L^2}^4.$$

(63)

Similarly, multiplying the third equation of (1) by $b^3$, yields

$$\begin{array}{cc}\hfill & {\displaystyle \frac{3}{4}}{\displaystyle \frac{d}{dt}}{\int }_0^1\tau b^{4}dx+3{\int }_0^1{\displaystyle \frac{b^2{b}_x^2}{\tau }}dx={\displaystyle \frac{3}{4}}{\int }_0^1{u}_x{b}^{4}dx\hfill \\ \hfill & ⩽C\parallel u_x{\parallel }_{L^2}^2{\parallel b\parallel }_{L^4}^4+{\parallel b_x\parallel }_{L^2}^2,\hfill \end{array}$$

(64)

Thus, multiplying (63) and (64) by $({\tilde{C}}_2+1)$ and $({\tilde{C}}_3+1)$, respectively, and then adding the results into (62), we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1\left[{\left(\theta +{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{\tau b^2}{2}}\right)}^2+u^4+\tau b^4\right]dx+{\int }_0^1\left({\displaystyle \frac{\lambda u^2{u}_x^2+{\theta }_x^2}{\tau }}+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)\left(\parallel {\tau }_x{\parallel }_{L^2}^2+{\parallel \theta \parallel }_{L^2}^2+{\parallel b\parallel }_{L^4}^4\right)+C{\parallel b_x\parallel }_{L^2}^2,\hfill \end{array}$$

which, together with (61), and choosing yields

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1\left[{\left(\theta +{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{\tau b^2}{2}}\right)}^2+{\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}^2+u^4+\tau b^4\right]dx\hfill \\ \hfill & +{\int }_0^1\left({\displaystyle \frac{\lambda u^2{u}_x^2+{\theta }_x^2}{\tau }}+{\displaystyle \frac{\lambda \theta {\tau }_x^2}{{\tau }^3}}+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽{\left({\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx\right)}_t+C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)\left(\parallel {\tau }_x{\parallel }_{L^2}^2+{\parallel \theta \parallel }_{L^2}^2+{\parallel b\parallel }_{L^4}^4\right)\hfill \\ \hfill & +C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2+V\left(t\right)\right)+C\left({\displaystyle \frac{C\alpha N^2}{m_2}}+{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^3}}\right){\parallel {\theta }_t\parallel }_{L^2}^2,\hfill \end{array}$$

(65)

provided that $\delta =1/\left(2{\tilde{C}}_1{C}_0\right)$ and ${\epsilon }_0>0$ is chosen to be small enough such that ${\epsilon }_0⩽1/\left(2{\tilde{C}}_1{C}_0\right)$.

On the other hand, it follows from (31) and the Cauchy- Schwarz’s inequality that

$${\int }_0^1{\displaystyle \frac{\lambda {\tau }_{x}u}{\tau }}dx⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}^{2}dx+C{\int }_0^1{u}^{2}dx⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\left({\displaystyle \frac{\lambda {\tau }_x}{\tau }}\right)}^{2}dx+C,$$

which, together with (19), (20), (28), (29), the second equation of (45), (65) and the Gronwall’s inequality, leads to

$$\underset{t\in [0,T]}{sup}\left({\parallel \theta \parallel }_{L^2}^2+\parallel {\tau }_x{\parallel }_{L^2}^2+{\parallel u\parallel }_{L^4}^4+{\parallel b\parallel }_{L^4}^4\right)+{\int }_0^T\left(u^2{u}_x^2+{\theta }_x^2+\theta {\tau }_x^2+b^2{b}_x^2\right)dt⩽C.$$

(66)

The Cauchy-Schwarz’s inequality together with (51) and (66) yields

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel {\tau }_x\parallel }_{L^2}^{2}dt& ⩽C{\int }_0^T{\int }_0^1\left[{\left(1-{\theta }^{1/2}\right)}^2+\theta \right]{\tau }_x^{2}dxdt\hfill \\ \hfill & ⩽C{\int }_0^T\parallel {\theta }^{1/2}{-1\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^{2}dt+C{\int }_0^T{\int }_0^1\theta {\tau }_x^{2}dxdt\hfill \\ \hfill & ⩽C,\hfill \end{array}$$

which, together with (67) leads to (54) with $\beta =0$.

Case II: $\beta >0$.

Choosing $\gamma =\beta$ in the first equation of (45), one has

$${\int }_0^T{\int }_0^1{\displaystyle \frac{\kappa {\theta }_x^2}{{\theta }^{\beta +1}}}dxdt={\int }_0^T{\int }_0^1{\displaystyle \frac{{\theta }_x^2}{\theta }}dxdt⩽C,$$

thus

$${\int }_0^T{\int }_0^1{\theta }_x^{2}dxdt⩽C{\int }_0^T\left({\parallel \theta \parallel }_{L^{\infty }}{\int }_0^1{\displaystyle \frac{{\theta }_x^2}{\theta }}dx\right)dt⩽C(1+N).$$

(67)

Next, we check (57), (58) and (60) as follows. For $J_1$:

$${\displaystyle \frac{\alpha N^2}{m_2}}{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt⩽{\displaystyle \frac{C\alpha N^2}{m_2}}(1+N)⩽C\alpha \mathcal{H}(m_1,m_2,N)⩽C,$$

where we have used (67). For $J_2$, it follows from the first equation of (45) with $\gamma =\beta$ that

$${\int }_0^T{\int }_0^1{\displaystyle \frac{\lambda {\theta }_x^2}{\theta }}dxdt⩽C{\int }_0^T\left({\parallel \theta \parallel }_{L^{\infty }}^{\alpha }{\int }_0^1{\displaystyle \frac{{\theta }_x^2}{\theta }}dx\right)dt⩽C(1+N^{\alpha }){\int }_0^T{\int }_0^1{\displaystyle \frac{{\theta }_x^2}{\theta }}dxdt⩽C,$$

and it follows from (67) that for $J_3$:

$${\displaystyle \frac{{\alpha }^2{N}^2}{m_2^3}}{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt⩽{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^3}}(1+N)⩽C\alpha \mathcal{H}(m_1,m_2,N)⩽C.$$

Hence, (54) holds for $\beta >0$. The proof of Lemma 4 is completed. □

Lemma 5.

Under the condition of Lemma 2, the following estimate holds

$$\underset{t\in [0,T]}{sup}\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)+{\int }_0^T\left(\parallel (u_t,b_t){\parallel }_{L^2}^2+\parallel (u_{xx},b_{xx}){\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right)dt⩽C_3.$$

(68)

Proof. 

We rewrite the second equation of (1) as follows

$$u_t-{\displaystyle \frac{\lambda u_{xx}}{\tau }}={\displaystyle \frac{{\lambda }_x{u}_x}{\tau }}-{\displaystyle \frac{\lambda u_x{\tau }_x}{{\tau }^2}}-{\displaystyle \frac{{\theta }_x}{\tau }}+{\displaystyle \frac{\theta {\tau }_x}{{\tau }^2}}-b{b}_x.$$

(69)

Multiplying (69) by $u_{xx}$ and integrating by parts over $[0,1]$, one has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{u}_x^{2}dx+{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx+C{\int }_0^1\left({\displaystyle \frac{{\lambda }_x^2{u}_x^2}{\tau \lambda }}+{\displaystyle \frac{\lambda u_x^2{\tau }_x^2}{{\tau }^3}}+{\displaystyle \frac{{\theta }^2}{\tau \lambda }}+{\displaystyle \frac{{\theta }^2{\tau }_x^2}{{\tau }^3\lambda }}+{\displaystyle \frac{\tau b^2{b}_x^2}{\lambda }}\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx+C{\int }_0^1{\lambda }^{-1}\left({\lambda }_x^2{u}_x^2+{\lambda }^2{u}_x^2{\tau }_x^2+{\theta }_x^2+{\theta }^2{\tau }_x^2+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx+C{m}_2^{-\alpha }{\int }_0^1\left({\alpha }^2{m}_2^{-\alpha }{N}^2{u}_x^2+(1+N^{2\alpha })u_x^2{\tau }_x^2+{\theta }_x^2+{\theta }^2{\tau }_x^2+b^2{b}_x^2\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx+C{\int }_0^1\left(u_x^2+u_x^2{\tau }_x^2+{\theta }_x^2+{\theta }^2{\tau }_x^2+b^2{b}_x^2\right)dx\hfill \\ \hfill & \triangleq {\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dx+C{\int }_0^1{b}^2{b}_x^{2}dx+\sum _{i=1}^2{K}_i.\hfill \end{array}$$

(70)

Now, we estimate $K_1$ and $K_2$ as follows.

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_0^1\left(u_x^2+u_x^2{\tau }_x^2\right)dxdt⩽C+C{\int }_0^T{\parallel u_x\parallel }_{L^{\infty }}^2{\int }_0^1{\tau }_x^{2}dxdt\hfill \\ \hfill & ⩽C+C{\int }_0^T\parallel u_x{\parallel }_{L^2}\parallel u_{xx}{\parallel }_{L^2}{\parallel {\tau }_x\parallel }_{L^2}^{2}dt\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^T{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dxdt+C{\int }_0^T{\int }_0^1{u}_x^{2}dxdt+C\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^T{\int }_0^1{\displaystyle \frac{\lambda u_{xx}^2}{\tau }}dxdt+C.\hfill \end{array}$$

(71)

Next, it follows from (54) that

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_0^1\left({\theta }_x^2+{\theta }^2{\tau }_x^2\right)dxdt⩽C{\int }_0^T\left(\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel \theta -\overline{\theta }\parallel }_{L^{\infty }}^2{\int }_0^1{\tau }_x^{2}dx+{\int }_0^1{\tau }_x^{2}dx\right)dt\hfill \\ \hfill & ⩽C+C{\int }_0^T\left(\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\tau }_x\parallel }_{L^2}^2\right)dt⩽C+C{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt.\hfill \end{array}$$

(72)

Integrating (70) over $[0,T]$, and using (54), (71) and (72), we obtain

$$\underset{t\in [0,T]}{sup}\parallel u_x{\parallel }_{L^2}^2+{\int }_0^T\parallel u_{xx}{\parallel }_{L^2}^{2}dt⩽C+\widetilde{C_4}{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt.$$

(73)

Now, we estimate the last term on the right hand side of (73). If $\beta >1$, we taking $\gamma =\beta -1$ in the first equation of (45), then

$${\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^2⩽C,$$

which, together with (73), gives

$$\underset{t\in [0,T]}{sup}{\parallel u_x\parallel }_{L^2}^2+{\int }_0^T\left(\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right)dt⩽C.$$

(74)

On the other hand, it follows from the first and the third equation of (1) that

$$|\tau b_t{|}^2+{\displaystyle \frac{b_{xx}^2}{{\tau }^2}}-2b_t{b}_{xx}={\left(u_{x}b+{\displaystyle \frac{b_x{\tau }_x}{{\tau }^2}}\right)}^2.$$

Integrating the above equation over $(0,1)$ with respect to x, leads to

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1{b}_x^{2}dx+{\int }_0^1(b_t^2+b_{xx}^2)dx& ⩽C\parallel u_x{\parallel }_{L^{\infty }}^2{\parallel b\parallel }_{L^2}^2+C\parallel b_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}\parallel b_{xx}{\parallel }_{L^2}^2+C\parallel u_{xx}{\parallel }_{L^2}^2+C\parallel u_x{\parallel }_{L^2}^2+C{\parallel b_x\parallel }_{L^2}^2.\hfill \end{array}$$

(75)

Integrating (75) over $[0,T]$, and using (45) and (74), one has for $\beta >1$:

$$\underset{t\in [0,T]}{sup}\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)+{\int }_0^T\left(\parallel b_t{\parallel }_{L^2}^2+\parallel (u_{xx},b_{xx}){\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right)dt⩽C.$$

(76)

In the case of $0⩽\beta ⩽1$, we multiply (30) by $\theta$, and integrate the results over [0, 1], then it follows from (23), (29), (31) and (51) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel \theta \parallel }_{L^2}^2+{\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau }}dx={\int }_0^1{\displaystyle \frac{\lambda \theta u_x^2+\theta b_x^2}{\tau }}dx-{\int }_0^1{\displaystyle \frac{{\theta }^2{u}_x}{\tau }}dx\hfill \\ \hfill & ⩽{\int }_0^1{\displaystyle \frac{\lambda \theta u_x^2+\theta b_x^2}{\tau }}dx-{\int }_0^1{\displaystyle \frac{({\theta }^2-{\overline{\theta }}^2)u_x}{\tau }}dx+C|1-{\overline{\theta }}^2|\parallel u_x{\parallel }_{L^1}-{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽{C\parallel \theta \parallel }_{L^{\infty }}^{\alpha }\parallel u_x{\parallel }_{L^{\infty }}^2+C\parallel b_x{\parallel }_{L^{\infty }}^2+\parallel {\theta }^2-{\overline{\theta }}^2{\parallel }_{L^{\infty }}{\parallel u_x\parallel }_{L^1}+CV\left(t\right)-{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽C(1+N^{\alpha })\left(\parallel u_x{\parallel }_{L^2}+{\parallel b_x\parallel }_{L^2}\right)\left(\parallel u_{xx}{\parallel }_{L^2}+{\parallel b_{xx}\parallel }_{L^2}\right)+C{V}^{1/2}\left(t\right){\parallel \theta \parallel }_{L^2}{\parallel {\theta }_x\parallel }_{L^2}\hfill \\ \hfill & +CV\left(t\right)-{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \\ \hfill & ⩽\delta \left(\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)+C\left(\delta \right)\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)+C{V}^{1/2}{\left(t\right)\parallel \theta \parallel }_{L^2}{\parallel {\theta }_x\parallel }_{L^2}\hfill \\ \hfill & +CV\left(t\right)-{\left({\int }_0^{1}ln\tau dx\right)}_t\hfill \end{array}$$

(77)

where we have used the following facts:

$${\int }_0^1|u_x|dx⩽C{\left({\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau \theta }}dx\right)}^{1/2}{\left({\int }_0^1{\displaystyle \frac{\tau \theta }{\lambda }}dx\right)}^{1/2}⩽C{V}^{1/2}\left(t\right){\parallel \theta \parallel }_{L^1}⩽C{V}^{1/2}\left(t\right),$$

and

$$\parallel {\theta }^2-{\overline{\theta }}^2{\parallel }_{L^{\infty }}⩽C{\int }_0^1\left|\theta \right||{\theta }_x{|dx⩽C\parallel \theta \parallel }_{L^2}{\parallel {\theta }_x\parallel }_{L^2}.$$

Due to (29) and the fact that ${\theta }^{\beta -2}+{\theta }^{\beta }⩾1$ for $0⩽\beta ⩽1$, one has

$$\begin{array}{cc}\hfill & \parallel {\theta }_x{\parallel }_{L^2}⩽C{\left({\int }_0^1\left({\theta }^{\beta -2}+{\theta }^{\beta }\right){\theta }_x^{2}dx\right)}^{1/2}\hfill \\ \hfill & ⩽C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau {\theta }^2}}dx\right)}^{1/2}+C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau }}dx\right)}^{1/2}⩽C{V}^{1/2}\left(t\right)+C{\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau }}dx\right)}^{1/2}.\hfill \end{array}$$

(78)

Integrating (77) over $[0,T]$, and using (31), (45), (78) and Cauchy-Schwarz’s inequality, we obtain

$$\underset{t\in [0,T]}{sup}{\parallel \theta \parallel }_{L^2}^2+{\int }_0^T{\int }_0^1{\theta }^{\beta }{\theta }_x^{2}dxdt⩽C\left(\delta \right)+\delta {\int }_0^T\left(\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)dt,$$

which, together with (29) and (31), yields that for $0⩽\beta ⩽1$:

$$\begin{array}{cc}\hfill \underset{t\in [0,T]}{sup}{\parallel \theta \parallel }_{L^2}^2+{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt& ⩽\underset{t\in [0,T]}{sup}{\parallel \theta \parallel }_{L^2}^2+{\int }_0^T{\int }_0^1({\theta }^{\beta }+{\theta }^{\beta -2}){\theta }_x^{2}dxdt\hfill \\ \hfill & ⩽C\left(\delta \right)+\delta {\int }_0^T\left(\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)dt.\hfill \end{array}$$

(79)

Next, it follows from the first equation of (75) that

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\int }_0^1{b}_x^{2}dx+{\int }_0^1(b_t^2+b_{xx}^2)dx& ⩽{\displaystyle \frac{1}{2}}\parallel b_{xx}{\parallel }^2+\delta \parallel u_{xx}{\parallel }^2+C\left(\delta \right)\parallel u_x{\parallel }^2+C{\parallel b_x\parallel }^2,\hfill \end{array}$$

thus

$$\underset{t\in [0,1]}{sup}\parallel b_x{\parallel }_{L^2}^2+{\int }_0^T\left(\parallel b_t{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)dt⩽C\left(\delta \right)+\delta {\int }_0^T{\parallel u_{xx}\parallel }^{2}dt.$$

(80)

Multiplying (69) by $(\widetilde{C_4}+1)$, adding the result to (73) and (80), taking $\delta =1/2({\tilde{C}}_4+2)$, we obtain that (76) with $0⩽\beta ⩽1$. Finally, we will estimate $\parallel u_t{\parallel }_{L^2\left({\mathrm{\Omega }}_T\right)}$. By virtue of (54) and (76), we have

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel \theta {\tau }_x\parallel }_{L^2}^{2}dt& ⩽C{\int }_0^T\left(\parallel {\tau }_x{\parallel }_{L^2}^2+\parallel \theta -\overline{\theta }{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2\right)dt\hfill \\ \hfill & ⩽C+C{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt⩽C,\hfill \end{array}$$

which, together with (28), (29), (45), (54) and (76), gives

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel u_t\parallel }_{L^2}^{2}dt& ⩽C{\int }_0^T{\int }_0^1\left({\displaystyle \frac{{\lambda }^2{u}_{xx}^2}{{\tau }^2}}+{\displaystyle \frac{{\lambda }_x^2{u}_x^2}{{\tau }^2}}+{\displaystyle \frac{{\lambda }^2{u}_x^2{\tau }_x^2}{{\tau }^4}}+{\displaystyle \frac{{\theta }_x^2}{{\tau }^2}}+{\displaystyle \frac{{\theta }^2{\tau }_x^2}{{\tau }^4}}+b^2{b}_x^2\right)dxdt\hfill \\ \hfill & ⩽C{\int }_0^T{\int }_0^1\left(u_{xx}^2+{\displaystyle \frac{{\alpha }^2(1+N^{2\alpha })}{m_2^2}}{\theta }_x^2{u}_x^2+{\tau }_x^2{u}_x^2+{\theta }_x^2+{\theta }^2{\tau }_x^2+b^2{b}_x^2\right)dxdt\hfill \\ \hfill & ⩽C+{\displaystyle \frac{C{\alpha }^2}{m_2^2}}{\int }_0^T\parallel u_x{\parallel }_{L^{\infty }}^2\parallel {\theta }_x{\parallel }_{L^2}^{2}dt+C{\int }_0^T\parallel u_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^{2}dt\hfill \\ \hfill & ⩽C+{\displaystyle \frac{C{\alpha }^2{N}^2}{m_2^2}}{\int }_0^T{\parallel {\theta }_x\parallel }_{L^2}^{2}dt+C{\left({\int }_0^T{\parallel u_x\parallel }_{L^2}^{2}dt\right)}^{1/2}{\left({\int }_0^T{\parallel u_{xx}\parallel }_{L^2}^{2}dt\right)}^{1/2}\hfill \\ \hfill & ⩽C+C{\left[\alpha \mathcal{H}(m_1,m_2,N)\right]}^2⩽C,\hfill \end{array}$$

which, together with (76) leads to (68). Thus, we complete the proof of Lemma 5. □

Lemma 6.

Under the condition of Lemma 2, the following estimates hold

$$C_1^{-1}⩽\theta (x,t)⩽C_1,\forall (x,t)\in [0,1]\times [0,T],$$

(81)

and

$$\underset{t\in [0,1]}{sup}{\parallel {\theta }_x\parallel }_{L^2}^2+{\int }_0^T\left(\parallel {\theta }_t{\parallel }_{L^2}^2+{\parallel {\theta }_{xx}\parallel }_{L^2}^2\right)dt⩽C_4.$$

(82)

Proof. 

By virtue of (29), (31), (68) and (77), we obtain that for $\beta ⩾0$

$$\begin{array}{cc}\hfill & \underset{t\in [0,T]}{sup}{\parallel \theta \parallel }_{L^2}^2+2{\int }_0^T{\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{\tau }}dxdt={\parallel {\theta }_0\parallel }_{L^2}^2+2{\int }_0^T{\int }_0^1{\displaystyle \frac{\lambda \theta u_x^2+\theta b_x^2-{\theta }^2{u}_x}{\tau }}dxdt\hfill \\ \hfill & ⩽C+C{\int }_0^T\left(\parallel u_{xx}{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right)dt\hfill \\ \hfill & +C{\int }_0^T\left(\parallel \theta -\overline{\theta }{\parallel }_{L^{\infty }}^2+\parallel \theta -\overline{\theta }{\parallel }_{L^{\infty }}+|1-{\overline{\theta }}^{1/2}|\right){\parallel u_x\parallel }_{L^1}dt+{\int }_0^T{\left({\int }_0^{1}ln\tau dx\right)}_{t}dt\hfill \\ \hfill & ⩽C+C{\int }_0^T\left(\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel u_x\parallel }_{L^2}^2+V\left(t\right)\right)dt⩽C.\hfill \end{array}$$

(83)

Multiplying (30) by ${\theta }^{\beta }{\theta }_t$, and then integrating the resulting equations over $[0,1]$, one has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1{\displaystyle \frac{{\left({\theta }^{\beta }{\theta }_x\right)}^2}{\tau }}dx+{\int }_0^1{\theta }^{\beta }{\theta }_t^{2}dx\hfill \\ \hfill & ={\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2-\theta u_x}{\tau }}{\theta }^{\beta }{\theta }_{t}dx-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{u_x}{{\tau }^2}}{\left({\theta }^{\beta }{\theta }_x\right)}^{2}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\theta }^{\beta }{\theta }_t^{2}dx+C{\int }_0^1{\theta }^{\beta }\left({\lambda }^2{u}_x^4+b_x^4+{\theta }^2{u}_x^2\right)dx\hfill \\ \hfill & +C\parallel u_x{\parallel }_{L^{\infty }}{\parallel \theta \parallel }_{L^{\infty }}^{\beta /2}{\int }_0^1{\displaystyle \frac{{\theta }^{3\beta /2}{\theta }_x^2}{{\tau }^2}}dx\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\int }_0^1{\theta }^{\beta }{\theta }_t^{2}dx+K_1+K_2.\hfill \end{array}$$

(84)

Now, we estimate $K_1$ and $K_2$ in (84). It follows from (28), (68) and (83) that

$$\begin{array}{cc}\hfill K_1& ⩽{C\parallel \theta \parallel }_{L^{\infty }}^{\beta }{\parallel \theta \parallel }_{L^{\infty }}^{2\alpha }\parallel u_x{\parallel }_{L^{\infty }}^2\parallel u_x{\parallel }_{L^2}^2+{C\parallel \theta \parallel }_{L^{\infty }}^{\beta }\parallel b_x{\parallel }_{L^{\infty }}^2{\parallel b_x\parallel }_{L^2}^2\hfill \\ \hfill & +{C\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}{\parallel \theta \parallel }_{L^2}\parallel u_x{\parallel }_{L^{\infty }}{\parallel u_x\parallel }_{L^2}\hfill \\ \hfill & ⩽C(1+N^{2\alpha })\left(1+{\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}\right)\parallel u_x{\parallel }_{L^2}^3\parallel u_{xx}{\parallel }_{L^2}+C\left(1+{\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}\right)\parallel b_x{\parallel }_{L^2}^3{\parallel b_{xx}\parallel }_{L^2}\hfill \\ \hfill & +{C\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}\parallel u_x{\parallel }_{L^2}^{3/2}{\parallel u_{xx}\parallel }_{L^2}^{1/2}\hfill \\ \hfill & ⩽C\left(1+\parallel {\theta }^{\beta }{\theta }_x{\parallel }_{L^2}\right)\parallel u_x{\parallel }_{L^2}\parallel u_{xx}{\parallel }_{L^2}+C\left(1+\parallel {\theta }^{\beta }{\theta }_x{\parallel }_{L^2}\right)\parallel b_x{\parallel }_{L^2}{\parallel b_{xx}\parallel }_{L^2}\hfill \\ \hfill & +C\left(1+\parallel {\theta }^{\beta }{\theta }_x{\parallel }_{L^2}\right){\parallel u_x\parallel }_{L^2}^2\hfill \\ \hfill & ⩽C\left(\parallel u_x{\parallel }_{L^2}^2+\parallel u_{xx}{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)\hfill \\ \hfill & +C\parallel {\theta }^{\beta }{\theta }_x{\parallel }_{L^2}^2\left(\parallel u_x{\parallel }_{L^2}^2+\parallel u_{xx}{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right),\hfill \end{array}$$

(85)

where we have used the following fact:

$${\theta }^{\beta +1}(x,t)-{\overline{\theta }}^{\beta +1}\left(t\right)={\int }_{x_0}^x{\left[{\theta }^{\beta +1}(y,t)\right]}_y^{\prime }dy,$$

thus

$${\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}⩽C+C{\int }_0^1|{\theta }^{\beta }{\theta }_x{|}^{2}dx⩽C+C{\parallel {\theta }^{\beta }{\theta }_x\parallel }_{L^2}.$$

(86)

Similarly

$$\begin{array}{cc}\hfill K_2& ⩽C\parallel u_x{\parallel }_{L^{\infty }}\left(1+{\parallel \theta \parallel }_{L^{\infty }}^{\beta +1}\right){\left({\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_x^2}{{\tau }^2}}dx\right)}^{1/2}{\left({\int }_0^1{\displaystyle \frac{{\theta }^{2\beta }{\theta }_x^2}{{\tau }^2}}dx\right)}^{1/2}\hfill \\ \hfill & ⩽C\parallel u_x{\parallel }_{L^2}^{1/2}\parallel u_{xx}{\parallel }_{L^2}^{1/2}\left(1+\parallel {\theta }^{\beta }{\theta }_x{\parallel }_{L^2}\right)\parallel {\theta }^{\beta /2}{\theta }_x{\parallel }_{L^2}{\parallel {\theta }^{\beta }{\theta }_x\parallel }_{L^2}\hfill \\ \hfill & ⩽C\parallel {\theta }^{\beta /2}{\theta }_x{\parallel }_{L^2}^2+C{\parallel {\theta }^{\beta }{\theta }_x\parallel }_{L^2}^2\left(\parallel u_x{\parallel }_{L^2}^2+\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel {\theta }^{\beta /2}{\theta }_x\parallel }_{L^2}^2\right)\hfill \\ \hfill & +C\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel u_{xx}\parallel }_{L^2}^2\right)\hfill \end{array}$$

(87)

Putting (85) and (87) into(84), and using (45), (68), (83) and the Gronwall’s inequality, we obtain

$$\underset{t\in [0,T]}{sup}{\parallel {\theta }^{\beta }{\theta }_x\parallel }_{L^2}^2+{\int }_0^T{\int }_0^1{\theta }^{\beta }{\theta }_t^{2}dxdt⩽C,$$

(88)

which, together with (86) gives

$$\theta (x,t)⩽C,\forall (x,t)\in [0,1]\times [0,T].$$

(89)

Next, we will estimate the lower bound of $\theta$. (83) together with (89) yields

$$\begin{array}{cc}\hfill {\int }_0^T{\int }_0^1{\left({\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}\right)}^{2}dxdt& ⩽C{\int }_0^T{\int }_0^1{\theta }^{2\beta }{\theta }_x^{2}dxdt\hfill \\ \hfill & ⩽C\underset{t\in [0,T]}{sup}{\parallel \theta \parallel }_{L^{\infty }}^{\beta }{\int }_0^T{\int }_0^1{\theta }_x^{2}dxdt⩽C,\hfill \end{array}$$

(90)

thus, it follows from (88) and (89) that

$$\begin{array}{cc}\hfill & {\int }_0^T\left|{\displaystyle \frac{d}{dt}}{\int }_0^1{\left({\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}\right)}^{2}dx\right|dt⩽C{\int }_0^T{\int }_0^1|{\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}\left|\right|{\theta }^{\beta }{\theta }_t-{\overline{\theta }}^{\beta }{\overline{\theta }}_t|dxdt\hfill \\ \hfill & ⩽C{\int }_0^T{\int }_0^1{|{\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}|}^{2}dxdt+C{\int }_0^T{\int }_0^1\left(\parallel {\theta }^{\beta }{\theta }_t{\parallel }_{L^2}^2+{\overline{\theta }}^{2\beta }{\overline{\theta }}_t^2\right)dxdt\hfill \\ \hfill & ⩽C+C{\int }_0^T{\overline{\theta }}^{2\beta }{\overline{\theta }}_t^{2}dt⩽C,\hfill \end{array}$$

(91)

where we have used the facts that $\overline{\theta }\left(t\right)⩽1$ and

$$\begin{array}{cc}\hfill {\int }_0^T{\overline{\theta }}_t^{2}dt& ={\int }_0^T{\left({\displaystyle \frac{d}{dt}}{\int }_0^1\theta dx\right)}^{2}dt\hfill \\ \hfill & ={\int }_0^T{\left[{\displaystyle \frac{d}{dt}}\left(1-{\int }_0^1{\displaystyle \frac{u^2}{2}}dx-{\int }_0^1{\displaystyle \frac{\tau b^2}{2}}dx\right)\right]}^{2}dt\hfill \\ \hfill & ⩽C{\int }_0^T{\left[{\int }_0^1\left(\left|u\right||u_t|+|u_x\left|\right|b^2|+|\tau \left|\right|b\left|\right|b_t|\right)dx\right]}^{2}dt\hfill \\ \hfill & ⩽C{\int }_0^T\left({\parallel u\parallel }_{L^2}^2\parallel u_t{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^{\infty }}^2{\parallel b\parallel }_{L^2}^4+{\parallel b\parallel }_{L^{\infty }}^2{\parallel b_t\parallel }_{L^2}^2\right)dt\hfill \\ \hfill & ⩽C{\int }_0^T\left(\parallel u_t{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}\parallel u_{xx}{\parallel }_{L^2}+\parallel b_x{\parallel }_{L^2}^2{\parallel b_t\parallel }_{L^2}^2\right)dt\hfill \\ \hfill & ⩽C.\hfill \end{array}$$

Thus, we infer from (88), (90) and (91) that

$$\underset{t\to +\infty }{lim}{\int }_0^1{\left({\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}\right)}^{2}dx=0,$$

which, together with (88), gives

$$\parallel \left({\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}\right){\left(t\right)\parallel }_{L^{\infty }}^2⩽C\parallel {\theta }^{\beta +1}-{\overline{\theta }}^{\beta +1}{\parallel }_{L^2}{\parallel {\theta }^{\beta }{\theta }_x\left(t\right)\parallel }_{L^2}\to 0,\mathrm{as}t\to +\infty .$$

Thus, by virtue of (32), there exists a time $T_0\gg 1$, such that

$$\theta (x,t)⩾{\displaystyle \frac{r_0}{2}},\forall (x,t)\in [0,1]\times [T_0,+\infty ).$$

(92)

For any fixed $T_0$ in (92), we multiply (30) by ${\theta }^{-q}$ with $q>2$, and integrate it over $[0,1]$, one has from (28) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{q-1}}{\left(\parallel {\theta }^{-1}{\left(t\right)\parallel }_{L^{q-1}}^{q-1}\right)}_t+q{\int }_0^1{\displaystyle \frac{\kappa {\theta }_x^2}{\tau {\theta }^{q+1}}}dx+{\int }_0^1{\displaystyle \frac{\lambda u_x^2+b_x^2}{\tau {\theta }^q}}dx=-{\int }_0^1{\displaystyle \frac{u_x}{\tau {\theta }^{q-1}}}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau {\theta }^q}}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{1}{\tau \lambda }}{\displaystyle \frac{1}{{\theta }^{q-2}}}dx\hfill \\ \hfill & ⩽C{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\lambda u_x^2}{\tau {\theta }^q}}dx+{\displaystyle \frac{1}{2}}{m}_2^{-\alpha }{C}_0^{-1}{\parallel {\theta }^{-1}\parallel }_{L^{q-1}}^{q-2},\hfill \end{array}$$

thus

$${\displaystyle \frac{d}{dt}}{\parallel {\theta }^{-1}\left(t\right)\parallel }_{L^{q-1}}⩽C,$$

where the constant $C>0$ does not depend on q. Integrating the above inequality over $[0,T_0]$, and letting $q\to \infty$, one has

$${\theta }^{-1}(x,t)⩽C(T_0+1),$$

thus

$$\theta (x,t)⩾{\displaystyle \frac{1}{C(T_0+1)}},\forall (x,t)\in [0,1]\times [0,T_0],$$

which, together with (89) and (92) leads to (81). On the other hand, by virtue of (29), (30), (54), (68) and (88), one has

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel {\theta }_{xx}\parallel }_{L^2}^{2}dx& ⩽C{\int }_0^T{\int }_0^1\left({\theta }_t^2+u_x^2+{\theta }_x^4+{\theta }_x^2{\tau }_x^2+u_x^4+b_x^4\right)dxdt\hfill \\ \hfill & ⩽C+C{\int }_0^T\left(\parallel {\theta }_x{\parallel }_{L^{\infty }}^2+\parallel u_x{\parallel }_{L^{\infty }}^2+{\parallel b_x\parallel }_{L^{\infty }}^2\right)dt\hfill \\ \hfill & ⩽C+C{\int }_0^T\left(\parallel {\theta }_x{\parallel }_{L^2}\parallel {\theta }_{xx}{\parallel }_{L^2}+\parallel u_x{\parallel }_{L^2}\parallel u_{xx}{\parallel }_{L^2}+\parallel b_x{\parallel }_{L^2}{\parallel b_{xx}\parallel }_{L^2}\right)dt\hfill \\ \hfill & ⩽C+C{\int }_0^T{\parallel {\theta }_{xx}\parallel }_{L^2}^{2}dt,\hfill \end{array}$$

which, together with (81) and (88) leads to (82). Thus, we complete the proof of Lemma 6. □

Lemma 7.

Under the condition of Lemma 2, the following estimate holds

$$\underset{t\in [0,T]}{sup}\parallel \left(u_t,u_{xx},b_t,b_{xx},{\theta }_t,{\theta }_{xx}\right){\left(t\right)\parallel }_{L^2}^2+{\int }_0^T{\parallel \left(u_{xt},b_{xt},{\theta }_{xt}\right)\parallel }_{L^2}^{2}dt⩽C_5.$$

(93)

Proof. 

It follows from the second equation of (1) that

$$u_{tt}+{\left({\displaystyle \frac{{\theta }_t\tau -\theta u_x}{{\tau }^2}}+b{b}_t\right)}_x={\left({\left({\displaystyle \frac{\lambda }{\tau }}\right)}_t{u}_x+{\displaystyle \frac{\lambda }{\tau }}{u}_{xt}\right)}_x,$$

We multiply both sides of the above equation by $u_t$, and integrate the result over $[0,1]$, one has

$${\displaystyle \frac{d}{dt}}{\parallel u_t\parallel }_{L^2}^2+{\int }_0^1{\displaystyle \frac{\lambda }{\tau }}{u}_{xt}^{2}dx⩽C\left(1+\parallel u_x{\parallel }_{L^{\infty }}^2+{\parallel b\parallel }_{L^{\infty }}^2\right)\left(\parallel {\theta }_t{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}^2+{\parallel b_t\parallel }_{L^2}^2\right).$$

(94)

Similarly, it follows from (30) that

$${\theta }_{tt}-{\left({\displaystyle \frac{{\theta }^{\beta }{\theta }_x}{\tau }}\right)}_{xt}=-{\displaystyle \frac{{\theta }_t{u}_x}{\tau }}-{\displaystyle \frac{\theta u_{xt}}{\tau }}+{\displaystyle \frac{\theta u_x^2}{{\tau }^2}}+{\left({\displaystyle \frac{\lambda }{\tau }}\right)}_t{u}_x^2+{\displaystyle \frac{2\lambda }{\tau }}{u}_x{u}_{xt}+{\left({\displaystyle \frac{1}{\tau }}\right)}_t{b}_x^2+{\displaystyle \frac{2}{\tau }}{b}_x{b}_{xt}.$$

Multiplying the above equation by ${\theta }_t$, and integrating it over $[0,1]$, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel {\theta }_t\parallel }_{L^2}^2+{\int }_0^1{\displaystyle \frac{{\theta }^{\beta }{\theta }_{xt}^2}{\tau }}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda }{\tau }}{u}_{xt}^{2}dx+{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{1}{\tau }}{b}_{xt}^{2}dx+C\left(1+\parallel u_x{\parallel }_{L^{\infty }}^2+\parallel b_x{\parallel }_{L^{\infty }}^2+{\parallel {\theta }_x\parallel }_{L^{\infty }}^2\right){\parallel {\theta }_t\parallel }_{L^2}^2\hfill \\ \hfill & +C\left(1+\parallel u_x{\parallel }_{L^{\infty }}^2+\parallel b_x{\parallel }_{L^{\infty }}^2+{\parallel {\theta }_x\parallel }_{L^{\infty }}^2\right)\left(\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right).\hfill \end{array}$$

(95)

Differentiating the third equation of (1) with respect to t, one has from the first equation of (1) that

$$\tau b_{tt}-{\left({\displaystyle \frac{b_{xt}}{\tau }}\right)}_x=-u_{xt}b-2u_x{b}_t-{\left({\displaystyle \frac{b_x{\tau }_x}{{\tau }^2}}\right)}_x.$$

Multiplying the above equation by $b_t$, and integrating it over $[0,1]$, one has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel \tau b_t\parallel }_{L^2}^2+{\int }_0^1{\displaystyle \frac{b_{xt}^2}{\tau }}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{4}}{\int }_0^1{\displaystyle \frac{\lambda }{\tau }}{u}_{xt}^{2}dx+C\left(1+\parallel u_x{\parallel }_{L^{\infty }}^2+{\parallel b_x\parallel }_{L^{\infty }}^2\right)\parallel b_t{\parallel }_{L^2}^2+{C\parallel b\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2.\hfill \end{array}$$

(96)

Combining (94), (95) with (96), we obtain form (45), (54), (68) and (82) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}\left(\parallel u_t{\parallel }_{L^2}^2+\parallel \tau b_t{\parallel }_{L^2}^2+{\parallel {\theta }_t\parallel }_{L^2}^2\right)+{\int }_0^1{\displaystyle \frac{\lambda u_{xt}^2+b_{xt}^2+{\theta }^{\beta }{\theta }_{xt}^2}{\tau }}dx\hfill \\ \hfill & ⩽C\left(1+\parallel \left(u_x,b_x,{\theta }_x\right){\parallel }_{L^{\infty }}^2\right)\left(\parallel b_t{\parallel }_{L^2}^2+\parallel {\theta }_t{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right)\hfill \\ \hfill & ⩽C\left(1+\parallel \left(u_x,b_x,{\theta }_x,u_{xx},b_{xx},{\theta }_{xx}\right){\parallel }_{L^2}^2\right)\left(\parallel b_t{\parallel }_{L^2}^2+\parallel {\theta }_t{\parallel }_{L^2}^2+{\parallel \left(u_x,b_x,{\theta }_x\right)\parallel }_{L^2}^2\right),\hfill \end{array}$$

thus

$$\underset{t\in [0,T]}{sup}\left(\parallel u_t{\parallel }_{L^2}^2+\parallel b_t{\parallel }_{L^2}^2+{\parallel {\theta }_t\parallel }_{L^2}^2\right)+{\int }_0^T\left(\parallel u_{xt}{\parallel }_{L^2}^2+\parallel b_{xt}{\parallel }_{L^2}^2+{\parallel {\theta }_{xt}\parallel }_{L^2}^2\right)dt⩽C.$$

(97)

The second equation of (1) gives

$$\begin{array}{cc}\hfill \parallel u_{xx}{\parallel }_{L^2}^2& ⩽C\left(\parallel u_t{\parallel }_{L^2}^2+\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\tau }_x\parallel }_{L^2}^2\right)+{C\parallel b\parallel }_{L^{\infty }}^2{\parallel b_x\parallel }_{L^2}^2\hfill \\ \hfill & +C\parallel u_x{\parallel }_{L^{\infty }}^2\left(\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\tau }_x\parallel }_{L^2}^2\right)\hfill \\ \hfill & ⩽C+C\parallel u_x{\parallel }_{L^2}{\parallel u_{xx}\parallel }_{L^2}\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\parallel u_{xx}\parallel }_{L^2}^2+C,\hfill \end{array}$$

thus

$$\parallel u_{xx}{\left(t\right)\parallel }_{L^2}⩽C,\forall t\in [0,T].$$

(98)

Similarly

$$\begin{array}{cc}\hfill \parallel b_{xx}{\parallel }_{L^2}^2& ⩽C\parallel b_t{\parallel }_{L^2}^2+{C\parallel b\parallel }_{L^{\infty }}^2\parallel u_x{\parallel }_{L^2}^2+C\parallel b_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2\hfill \\ \hfill & ⩽C+C\parallel b_x{\parallel }_{L^2}{\parallel b_{xx}\parallel }_{L^2}\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\parallel b_{xx}\parallel }_{L^2}^2+C,\hfill \end{array}$$

thus

$$\parallel b_{xx}{\left(t\right)\parallel }_{L^2}⩽C,\forall t\in [0,T].$$

(99)

The inequality of (97) also gives

$$\parallel {\theta }_{xx}{\left(t\right)\parallel }_{L^2}⩽C,\forall t\in [0,T],$$

which, together with (97), (98) and (99) leads to (93). □

Lemma 8.

Under the condition of Lemma 2, the following estimate holds

$$\underset{t\in [0,T]}{sup}{\parallel {\tau }_{xx}\parallel }_{L^2}^2+{\int }_0^T\left(\parallel {\tau }_{xx}{\parallel }_{L^2}^2+\parallel {\tau }_{xxt}{\parallel }_{L^2}^2+\parallel u_{xxx}{\parallel }_{L^2}^2+\parallel b_{xxx}{\parallel }_{L^2}^2+{\parallel {\theta }_{xxx}\parallel }_{L^2}^2\right)dt⩽C_6.$$

(100)

Proof. 

It follows from the second equation of (1) that

$$u_{xt}-\lambda {\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{xt}=-{\left({\displaystyle \frac{{\theta }_x\tau -\theta {\tau }_x}{{\tau }^2}}\right)}_x-{\left(b{b}_x\right)}_x+{\lambda }_x{\left({\displaystyle \frac{u_x}{\tau }}\right)}_x+{\left({\lambda }_x{\displaystyle \frac{u_x}{\tau }}\right)}_x.$$

We multiply both sides of the above equation by ${\left(\frac{{\tau }_x}{\tau }\right)}_x$, and integrate the result over $[0,1]$, one has from (29), (81) and the Cauchy-Schwarz’s inequality that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\int }_0^1\lambda {\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx+{\int }_0^1{\displaystyle \frac{\theta }{\tau }}{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx\hfill \\ \hfill & ={\int }_0^1{u}_{xt}{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx+{\displaystyle \frac{1}{2}}{\int }_0^1\alpha {\theta }^{\alpha -1}{\theta }_t{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx+{\int }_0^1{\left({\displaystyle \frac{{\theta }_x}{\tau }}\right)}_x{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx\hfill \\ \hfill & -{\int }_0^1{\left({\displaystyle \frac{\theta }{\tau }}\right)}_x{\displaystyle \frac{{\tau }_x}{\tau }}{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx+{\int }_0^1{\left(b{b}_x\right)}_x{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx-{\int }_0^1{\lambda }_x{\left({\displaystyle \frac{u_x}{\tau }}\right)}_x{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx\hfill \\ \hfill & -{\int }_0^1{\left({\lambda }_x{\displaystyle \frac{u_x}{\tau }}\right)}_x{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_{x}dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\theta }{\tau }}{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx+C\parallel u_{xt}{\parallel }_{L^2}^2+C{\parallel {\theta }_t\parallel }_{L^{\infty }}^2{\int }_0^1{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx\hfill \\ \hfill & +C{\int }_0^1\left({\theta }_{xx}^2+{\theta }_x^2{\tau }_x^2+{\tau }_x^4+b_x^4+b^2{b}_{xx}^2+{\theta }_x^2{u}_{xx}^2+{\theta }_x^2{u}_x^2{\tau }_x^2+{\theta }_{xx}^2{u}_x^2\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{\theta }{\tau }}{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx+C{\parallel {\theta }_t\parallel }_{H^1}^2{\int }_0^1{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx\hfill \\ \hfill & +C\left(\parallel u_{xt}{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{H^1}^2+\parallel u_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2+{\parallel {\tau }_x\parallel }_{L^4}^4\right)\hfill \\ \hfill & +C\left(\parallel u_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2{\parallel u_x\parallel }_{H^1}^2\right)\left(\parallel u_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2+{\parallel {\tau }_x\parallel }_{L^2}^2\right),\hfill \end{array}$$

thus

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1\lambda {\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx+{\int }_0^1{\displaystyle \frac{\theta }{\tau }}{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx⩽C{\parallel {\theta }_t\parallel }_{H^1}^2{\int }_0^1{\left|{\left({\displaystyle \frac{{\tau }_x}{\tau }}\right)}_x\right|}^{2}dx\hfill \\ \hfill & +C\left(\parallel u_{xt}{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{H^1}^2+\parallel u_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2+{\parallel {\tau }_x\parallel }_{L^4}^4\right).\hfill \end{array}$$

Using Gronwall’s inequality, we obtain

$$\begin{array}{cc}\hfill & \underset{t\in [0,T]}{sup}\parallel {\tau }_{xx}{\parallel }_{L^2}^2+{\int }_0^T\parallel {\tau }_{xx}{\parallel }_{L^2}^{2}dt⩽C+C\underset{t\in [0,T]}{sup}\parallel {\tau }_x{\parallel }_{L^4}^4+{\int }_0^T{\parallel {\tau }_x\parallel }_{L^4}^{4}dt\hfill \\ \hfill & ⩽C+C\underset{t\in [0,T]}{sup}\left(\parallel {\tau }_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2\right)+C{\int }_0^T\parallel {\tau }_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^{2}dt\hfill \\ \hfill & ⩽C+{\displaystyle \frac{1}{2}}\underset{t\in [0,T]}{sup}\parallel {\tau }_{xx}{\parallel }_{L^2}^2+C{\int }_0^T\left(1+\parallel {\tau }_{xx}{\parallel }_{L^2}\right){\parallel {\tau }_x\parallel }_{L^2}^{2}dt\hfill \\ \hfill & ⩽C+{\displaystyle \frac{1}{2}}\underset{t\in [0,T]}{sup}\parallel {\tau }_{xx}{\parallel }_{L^2}^2+{\displaystyle \frac{1}{2}}{\int }_0^T{\parallel {\tau }_{xx}\parallel }_{L^2}^{2}dt,\hfill \end{array}$$

thus

$$\underset{t\in [0,T]}{sup}\parallel {\tau }_{xx}{\parallel }_{L^2}^2+{\int }_0^T{\parallel {\tau }_{xx}\parallel }_{L^2}^{2}dt⩽C,$$

(101)

which, together with (30) and (45), (54), (68), (82) and (93) yields

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel u_{xxx}\parallel }_{L^2}^{2}dt& ⩽C{\int }_0^T\left(\parallel u_{xt}{\parallel }_{L^2}^2+\parallel {\theta }_x{\parallel }_{H^1}^2+\parallel {\tau }_x{\parallel }_{H^1}^2+{\parallel b_x\parallel }_{H^1}^2\right)dt\hfill \\ \hfill & +{\int }_0^T\left(\parallel {\tau }_x{\parallel }_{H^1}^2\parallel u_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^2\parallel u_x{\parallel }_{H^1}^2+\parallel b_x{\parallel }_{H^1}^2{\parallel b_x\parallel }_{L^2}^2\right)dt\hfill \\ \hfill & +{\int }_0^T\left(\parallel {\tau }_x{\parallel }_{H^1}^2\parallel {\theta }_x{\parallel }_{H^1}^2+\parallel {\theta }_x{\parallel }_{H^1}^4+{\parallel {\tau }_x\parallel }_{H^1}^4\right)dt\hfill \\ \hfill & ⩽C,\hfill \end{array}$$

(102)

which, combine with the first equation of (1) gives

$${\int }_0^T{\parallel {\tau }_{xxt}\parallel }_{L^2}^{2}dt⩽C,$$

(103)

Similarly

$${\int }_0^T\left(\parallel {\theta }_{xxx}{\parallel }_{L^2}^2+{\parallel b_{xxx}\parallel }_{L^2}^2\right)dt⩽C,$$

which, together with (101)–(103) leads to (100). Therefore, we complete the proof of Lemma 8. □

3. Proof of Theorem 1.1

Based on the fundamental uniform-in-time estimates in Section 2, we are now ready to prove Theorem 1. To this end, we first need the following local existence theorem of the global solutions, which can be shown via the Banach Fixed Point theorem on a small time interval. We can modify the theory of [31] slightly to obtain the following lemma, which we will not to prove in detail here.

Lemma 9.

Under the condition of (15). Then there exists a positive time $T_0=T_0(N_0,N_0,M_0)$, depending only on β, $N_0$ and $M_0$, such that the MHD system (1)–(5) possesses a unique solution $(\tau ,u,b,\theta )\in X(0,T_0;\frac{1}{2}{N}_0,\frac{1}{2}{N}_0,2M_0)$.

Proof. 

By using the contractivity of the operator defined by the linearization of the problem (1) on a small time interval, one can obtain the result. For completeness, we give a sketch of the proof. In order to prove Lemma 9, we denote the Banach space as follows:

$$\mathcal{B}\triangleq \left\{\tilde{u}|{\parallel \tilde{u}\parallel }_{\mathcal{B}}⩽K\right\}$$

with the form

$$\parallel \tilde{u}{\parallel }_{\mathcal{B}}\triangleq \parallel \tilde{u}{\parallel }_{L^{\infty }(0,T;H^2)}+{\parallel {\tilde{u}}_x\parallel }_{L^2(0,T;H^2)}.$$

Let $\tilde{u},\tilde{\theta },$ and $\tilde{b}$ be given, the linear problem of (1)–(5) can be written as

$$\left\{\begin{array}{c}{\tau }_t-{\tilde{u}}_x=0,\hfill \\ \tau (·,0)={\tau }_0,\hfill \\ {\left(\tau b\right)}_t={\left[{\displaystyle \frac{\nu b_x}{\tau }}\right]}_x,\hfill \\ u_t+{\left[p+{\displaystyle \frac{1}{2}}{b}^2\right]}_x={\left[{\displaystyle \frac{\lambda u_x}{\tau }}\right]}_x,\hfill \\ E_t+{\left[\tilde{u}\left(p+{\displaystyle \frac{1}{2}}{b}^2\right)\right]}_x={\left[{\displaystyle \frac{\lambda \tilde{u}{u}_x+\nu \tilde{b}{b}_x+\kappa {\theta }_x}{\tau }}\right]}_x.\hfill \end{array}\right.$$

(104)

Let $(\tau ,u,\theta ,b)$ be the unique strong solution to the above linear problem, we define the fixed point map:

$$\mathcal{F}:(\tilde{u},\tilde{\theta },\tilde{b})\in \mathcal{B}\times \mathcal{B}\times \mathcal{B}\to (u,\theta ,b)\in \mathcal{B}\times \mathcal{B}\times \mathcal{B}$$

with

$$\tilde{u}(·,0)=u_0,\tilde{\theta }(·,0)={\theta }_0,\tilde{b}(·,0)=b_0,(\tilde{u},{\tilde{\theta }}_x,\tilde{b}){|}_{\partial \mathrm{\Omega }}=(0,0,0).$$

We will prove the map $\mathcal{F}$ mapping $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$ into $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$ for suitable constant K and small T, and $\mathcal{F}$ is a contraction mapping on $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$ and thus $\mathcal{F}$ has a unique fixed point in $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$. In order to do this, for given $\tilde{u}\in \mathcal{B}$, we first prove that for some small $0<T⩽1$, the above linear problem (104) has a unique solution $\rho$ satisfying

$$C^{-1}⩽\tau ⩽C,{\parallel {\tau }_x\parallel }_{L^2(0,T;H^1)}⩽KC,$$

provided $KT⩽1$, here and later on, C will denote a constant. Since the first equation of (104) is linear with regular $\tilde{u}$, the existence and uniqueness are well-known. Thus, after integration by parts, we have from Gronwall inequality that

$$\tau (x,t)={\tau }_0+{\int }_0^t{\tilde{u}}_{x}ds.$$

Since the fourth equation of (104) in linear with regular $\tau$ and $\tilde{u}$, the existence and uniqueness are well-known. Then, for given $\tilde{u},\tilde{b}\in \mathcal{B}$, we will prove that for some small $0<T⩽1$, the third equation of (104) has a unique solution b satisfying

$${\parallel b\parallel }_{L^{\infty }(0,T;H^2)}+{\parallel b_x\parallel }_{L^2(0,T;H^2)}⩽C_1,$$

provided that $K^{2}T⩽1$. Since (104)₄ in linear with regular $\tau$, $\tilde{u}$ and $\tilde{b}$, the existence and uniqueness are well-known. For given $\rho ,\tilde{u},\tilde{b}\in \mathcal{B}$, we will prove that for some small $0<T⩽1$, the problem (104)₄ has a unique solution u satisfying

$${\parallel u\parallel }_{L^{\infty }(0,T;H^2)}+{\parallel u_x\parallel }_{L^2(0,T;H^2)}⩽C_2,$$

provided that $K^{2}T⩽1$ and $T⩽1$. Since (104)₅ in linear with regular $\tau$, $\tilde{u}$, $\tilde{\theta }$ and $\tilde{b}$, the existence and uniqueness are well-known. For given $\rho ,\tilde{u},\tilde{\theta },\tilde{b}\in \mathcal{B}$, we will prove that for some small $0<T⩽1$, the problem (104)₅ has a unique solution $\theta$ satisfying

$${\parallel \theta \parallel }_{L^{\infty }(0,T;H^2)}+{\parallel {\theta }_x\parallel }_{L^2(0,T;H^2)}⩽C_3,$$

provided that $K^{2}T⩽1$ and $T⩽1$.

Due to the above analysis, we can take $K=max\{C_1,C_2,C_3\}$ and thus $\mathcal{F}$ maps $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$ into $\mathcal{B}\times \mathcal{B}\times \mathcal{B}$. Therefore, in this step, we will prove that $\mathcal{F}$ is contracted in the sense of weaker norm, that is: there is a constant $\alpha \in (0,1)$ such that for any $({\tilde{u}}_i,{\tilde{b}}_i,{\tilde{\theta }}_i)$ and some small $0<T⩽1$, the following estimate holds

$$\parallel \mathcal{F}({\tilde{u}}_1,{\tilde{\theta }}_1,{\tilde{b}}_1)-\mathcal{F}({\tilde{u}}_2,{\tilde{\theta }}_2,{\tilde{b}}_2){\parallel }_{L^2(0,T;H^1)}⩽\alpha {\parallel ({\tilde{u}}_1-{\tilde{u}}_2,{\tilde{\theta }}_1-{\tilde{\theta }}_2,{\tilde{b}}_1-{\tilde{b}}_2)\parallel }_{L^2(0,T;H^1)}.$$

Due to the above steps and the Banach fixed point theorem, we finish the proof of Lemma 9. □

Proof of Global Existence. Lemma 9 yields that the system (1)–(5) possesses a unique solution $(\tau ,u,b,\theta )\in X(0,t_1;\frac{1}{2}{N}_0,\frac{1}{2}{N}_0,2M_0)$ with $t_1=T_0(N_0,N_0,M_0)$.

Furthermore, choosing $\alpha ⩽{\alpha }_1$ with ${\alpha }_1$ being small enough, such that

$${\left({\displaystyle \frac{1}{2}}{N}_0\right)}^{-{\alpha }_1}⩽2,{\left(2M_0\right)}^{{\alpha }_1}⩽2,{\alpha }_1\mathcal{H}(\frac{1}{2}{N}_0,\frac{1}{2}{N}_0,2M_0)⩽{\epsilon }_0,$$

(105)

where ${\epsilon }_0>0$ is chosen in Lemma 2. Thus, Lemmas 1–8 yield that the solution $(\tau ,u,b,\theta )$ of systems (1)–(5) satisfies

$$C_0^{-1}⩽\tau (x,t)⩽C_0,C_1^{-1}⩽\theta (x,t)⩽C_1,\forall (x,t)\in [0,1]\times [0,t_1],$$

(106)

and

$$\underset{t\in [0,t_1]}{sup}{\parallel (\tau ,u,b,\theta )\left(t\right)\parallel }_{H^2}^2+{\int }_0^{t_1}{\parallel {\theta }_t\parallel }_{L^2}^{2}dt⩽C_7^2=\sum _{i=2}^6{C}_i,$$

(107)

where the positive constants $C_i(i=2,3,\cdots ,6)$ are the same ones as in Lemmas 4–26.

Then, taking $(\tau ,u,b,\theta )(·,t_1)$ as the initial data, and using Lemma 9, one can extend the local solution $(\tau ,u,b,\theta )$ to the time interval $[t_1,t_1+t_2]$ with $t_2=T_0(C_0^{-1},C_1^{-1},C_7)$. Thus

$$\tau (x,t)⩾{\displaystyle \frac{1}{2}}{C}_0^{-1},\theta (x,t)⩾{\displaystyle \frac{1}{2}}{C}_1^{-1},\forall (x,t)\in [0,1]\times [t_1,t_1+t_2],$$

(108)

and

$$\underset{t\in [t_1,t_1+t_2]}{sup}{\parallel (\tau ,u,b,\theta )\left(t\right)\parallel }_{H^2}^2+{\int }_{t_1}^{t_1+t_2}{\parallel {\theta }_t\parallel }_{L^2}^{2}dt⩽4C_7^2.$$

(109)

Combining (106) with (107)–(109), one has

$$\tau (x,t)⩾{\displaystyle \frac{1}{2}}{C}_0^{-1},\theta (x,t)⩾{\displaystyle \frac{1}{2}}{C}_1^{-1},\forall (x,t)\in [0,1]\times [0,t_1+t_2],$$

and

$$\underset{t\in [0,t_1+t_2]}{sup}{\parallel (\tau ,u,b,\theta )\left(t\right)\parallel }_{H^2}^2+{\int }_0^{t_1+t_2}{\parallel {\theta }_t\parallel }_{L^2}^{2}dt⩽5C_7^2.$$

Taking $\alpha ⩽min\{{\alpha }_1,{\alpha }_2\}$ with ${\alpha }_1$ is the same one as in (105), and ${\alpha }_2$ is chosen as follows

$${\left({\displaystyle \frac{1}{2}}{C}_0^{-1}\right)}^{-{\alpha }_2}⩽2,{\left(\sqrt{5}{C}_7\right)}^{{\alpha }_2}⩽2,{\alpha }_2\mathcal{H}(\frac{1}{2}{C}_0^{-1},\frac{1}{2}{C}_1^{-1},\sqrt{5}{C}_7)⩽{\epsilon }_0,$$

where ${\epsilon }_0>0$ is chosen in Lemma 2. Thus, Lemmas 1–8 again yield that the solution $(\tau ,u,b,\theta )$ of system satisfies (106) and (107) on $[0,t_1+t_2]$.

Therefore, taking $\epsilon \triangleq min\{{\alpha }_1,{\alpha }_2\}$, and repeating the above procedure, we obtain that the system (1)–(5) possesses a unique global solution $(\tau ,u,b,\theta )\in X(0,+\infty ;C_0^{-1},C_1^{-1},C_7)$. Therefore, we can gradually extend the existence time of the solution of system (1) to $[0,+\infty )$, and then obtain the global existence of the solution of the equations. Thus, we complete the proof of the global existence of Theorem 1. The uniqueness of the solutions can be proved by the standard $L^2$-estimates.

Nonlinear exponential stability. Based on the t-independent estimates in (16) and (17), we are now to derive the long-time behaviors of the initial boundary value problems (1)–(5).

Similar to (31), we can get from (16) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1\left({\displaystyle \frac{1}{2}}{u}^2+{\displaystyle \frac{1}{2}}\tau b^2+(\theta -ln\theta -1)+(\tau -ln\tau -1)\right)dx\hfill \\ \hfill & +{\overline{C}}_1{\int }_0^1\left(u_x^2+b_x^2+{\theta }_x^2\right)dx⩽0.\hfill \end{array}$$

(110)

Multiplying (55) by $\left(u-\lambda \frac{{\tau }_x}{\tau }\right)$, integrating the results over $[0,1]$, and using Cauchy-Schwarz’s inequality, one has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\left(u-\lambda {\displaystyle \frac{{\tau }_x}{\tau }}\right)}^{2}dx+{\overline{C}}_2{\int }_0^1{\tau }_x^{2}dx\hfill \\ \hfill & ⩽C{\int }_0^1\left(|{\theta }_{x}u|+|{\theta }_x{\tau }_x|+|u{\tau }_x|+|b{b}_{x}u|+|b{b}_x{\tau }_x|\right.\hfill \\ \hfill & \left.+|{\theta }_t{\tau }_{x}u|+|{\theta }_t{\tau }_x^2|+|{\theta }_x{u}_{x}u|+|{\theta }_x{u}_x{\tau }_x|\right)dx\hfill \\ \hfill & ⩽{\displaystyle \frac{{\overline{C}}_2}{2}}{\int }_0^1{\tau }_x^{2}dx+C\left(\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+\parallel {\theta }_x{\parallel }_{L^2}^2+{\parallel {\theta }_t\parallel }_{L^2}^2\right).\hfill \end{array}$$

(111)

On the other hand, it follows from the Cauchy-Schwarz’s inequality that

$${\int }_0^1|u_x{\tau }_x{|}^{2}dx⩽\parallel u_x{\parallel }_{L^{\infty }}^2{\parallel {\tau }_x\parallel }_{L^2}^2⩽C\left(\parallel u_x{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}{\parallel u_{xx}\parallel }_{L^2}\right),$$

(112)

which, together with (70), gives

$${\displaystyle \frac{d}{dt}}{\int }_0^1{u}_x^{2}dx+{\overline{C}}_3{\int }_0^1{u}_{xx}^{2}dx⩽C\left(\parallel {\tau }_x{\parallel }_{L^2}^2+\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right).$$

(113)

The inequality (84) together with (112) yields for any $\delta \in (0,1)$

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{\left({\displaystyle \frac{{\theta }^{\beta }{\theta }_x}{\tau }}\right)}^{2}dx+{\overline{C}}_4{\int }_0^1{\theta }_t^{2}dx\hfill \\ \hfill & ⩽\delta \left(\parallel u_{xx}{\parallel }_{L^2}^2+{\parallel b_{xx}\parallel }_{L^2}^2\right)+C\left(\delta \right)\left(\parallel u_x{\parallel }_{L^2}^2+\parallel b_x{\parallel }_{L^2}^2+{\parallel {\theta }_x\parallel }_{L^2}^2\right).\hfill \end{array}$$

(114)

It follows from (75) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\int }_0^1{b}_x^{2}dx+{\int }_0^1(b_t^2+b_{xx}^2)dx\hfill \\ \hfill & ⩽C\parallel u_x{\parallel }_{L^{\infty }}^2+C{\parallel b_x\parallel }_{L^{\infty }}^2\hfill \\ \hfill & ⩽{\displaystyle \frac{1}{2}}\parallel b_{xx}{\parallel }_{L^2}^2+\delta {\parallel u_{xx}\parallel }_{L^2}^2+C\left(\delta \right)\left(\parallel u_x{\parallel }_{L^2}^2+{\parallel b_x\parallel }_{L^2}^2\right).\hfill \end{array}$$

(115)

Let $M_1$, $M_2$ and $M_3$ be properly large positive constants, operate (111)×$M_1+$ (113) + (114) ×$M_2+$ (115), and choose $\delta >0$ suitable small, then add the results to (110)×$M_3$, we obtain

$${\mathcal{F}}^{\prime }\left(t\right)+{\displaystyle \frac{1}{2}}{\parallel ({\tau }_x,u_x,b_x,{\theta }_x)\parallel }_{L^2}^2⩽0,$$

(116)

here

$$\begin{array}{cc}\hfill & \mathcal{F}\left(t\right)\triangleq {\int }_0^1{M}_3\left({\displaystyle \frac{1}{2}}{u}^2+{\displaystyle \frac{1}{2}}\tau b^2+(\theta -ln\theta -1)+(\tau -ln\tau -1)\right)dx\hfill \\ \hfill & +{\int }_0^1\left[\overline{C}{\left(u-\lambda {\displaystyle \frac{{\tau }_x}{\tau }}\right)}^2+{\displaystyle \frac{{\overline{C}}_1{M}_3}{2}}(u_x^2+b_x^2+{\theta }_x^2)\right]dx,\hfill \end{array}$$

where $\overline{C}$ is a positive constant independent of $M_3$.

Thanks to the large number $M_3$ and the Cauchy-Schwarz’s inequality, one has

$$\mathcal{F}\left(t\right)\sim {\parallel \tau -1,u,b,\theta -1\parallel }_{H^1}^2.$$

Consequently, we infer from (116) that there exists a positive constant number $\sigma >0$, such that

$$\parallel ({\tau }_x,u_x,b_x,{\theta }_x)\left(t\right){\parallel }_{L^2}^2⩽C{e}^{-\sigma t}\to 0,\mathrm{as}t\to \infty ,$$

(117)

thus

$${\parallel (\tau -1)\left(t\right)\parallel }_{L^2}^2⩽C{\parallel {\tau }_x\left(t\right)\parallel }_{L^2}^2⩽C{e}^{-\sigma t}\to 0,\mathrm{as}t\to \infty ,$$

and the equality (23) and (29) yield that

$$\begin{array}{cc}\hfill {\parallel (\theta -1)\left(t\right)\parallel }_{L^2}^2& ⩽C\left(\parallel (\theta -\overline{\theta })\left(t\right){\parallel }_{L^2}^2+\parallel u\left(t\right){\parallel }_{L^2}^4+{\parallel b\left(t\right)\parallel }_{L^2}^4\right)\hfill \\ \hfill & ⩽C\left(\parallel {\theta }_x{\left(t\right)\parallel }_{L^2}^2+\parallel u_x{\left(t\right)\parallel }_{L^2}^2+{\parallel b_x\left(t\right)\parallel }_{L^2}^2\right)\hfill \\ \hfill & ⩽C{e}^{-\sigma t}\to 0,\mathrm{as}t\to \infty ,\hfill \end{array}$$

which, together with (117) leads to (18). Therefore, we complete the proof of the Theorem 1. □

4. Conclusions

In this paper, we analyzed the asymptotic stability of the global strong solutions to the boundary value problem of the compressible magnetohydrodynamic (MHD) equations for the ideal polytropic gas in which the viscosity $\lambda$ and heat conductivity $\kappa$ depend on temperature, i.e., $\lambda ={\theta }^{\alpha }$, and $\kappa ={\theta }^{\beta }$ with $\alpha ,\beta \in [0,+\infty )$. Both the global-in-time existence and uniqueness of strong solutions were obtained under certain assumptions on the parameter $\alpha$ and initial data. Moreover, based on the accurate uniform-in-time estimates, we obtained that the global large solutions decay exponentially in time to the equilibrium states. Compared with the existing results, the initial data could be large if $\alpha$ was suitably small and the growth exponent $\beta$ can be large.