数理解析研究所講究録
1495 巻 2006 年 8-25

8Global Solution to the One-Dimensional Equations
for a Self-Gravitating Viscous Radiativeand Reactive Gas
梅原守道 (Morimichi UMEHARA), 谷温之 (Atusi TANI)
慶慮義塾大学理工学部
Department of Mathematics, Keio University

1

Introduction.

We consider the one-dimensional motion of a compressible, viscous and heat conductive gas driven by the self-gravitation in the free-boundary case. In addition to thissituation, we take into account the energy producing process inside the medium, thatis, the gas consists of a reacting mixture and the combustion process is current at thehigh temperature stage.
The motion mentioned above is described by the following four equations in the
Euler coordinate system corresponding to the conservation laws of mass, momentum
and energy, and an equation of reaction-diffusion type:$\{\rho(v_{t}+vv_{y})\rho(e_{t}+ve_{y})\rho(z_{t}+vz_{y})\rho_{t}+v\rho_{y}$(1.1)$====-\rho v_{y}(-p+’\mu v_{y})_{y}+\rho f(\kappa\theta_{y})_{y}+(-p+\mu v_{y}’)v_{y}+\lambda\phi z(d\rho z_{y})_{y}-\phi z$’

in$\bigcup_{t>0}(\Omega(t)\cross\{t\})$, where $\Omega(t):=\{y\in \mathbb{R}|y_{1}(t)\leq y\leq y_{2}(t)\}$ and$y_{i}(\cdot)$

for $i=1,2$ arefluctuating boundary functions. Here the density $\rho=\rho(y, t)$ , the velocity $v=v(y, t)$ ,and the mass fraction of the reactant $z=z(y, t)$
the absolute temperature
and are the coefficients ofare the unknown functions, and positive constants
viscosity, the species diffusion and the difference in heat between the reactant and the
product.
The external force per unit mass $f=f(y, t)$ is given by $f=-U_{y}$ , where is thesolution of the boundary value problem$\theta=\theta(y, t)$$\mu,$$d$$\lambda$$U$$U_{yy}=G\rho$in$U|_{y=\nu 1(t)}=U|_{y=y_{2}(t)}=0$

for $t>0$ .

$\bigcup_{t>0}(\Omega(t)\cross\{t\})$

,

(1.2)

$\{$

Here is the Newtonian gravitational constant. The rate function
by the Arrhenius law$G$$\phi(\theta)=\theta^{\beta}\mathrm{e}^{-T}A$,$\phi=\phi(\theta)$

is defined
(1.3)

9

where is the activation energy (a positive constant) and is a non-negative number.
At high tempereture regimes, pressure $p=p(\rho, \theta)$ and internal energy $e=e(\rho, \theta)$ aregiven by $p=p_{G}+p_{R}$ and$A$$\beta$$e=C_{\mathrm{v}} \theta+a\frac{\theta^{4}}{\rho}$with the specific heat at constant volume (positive constant) , the Stefan-Boltzmann
constant $a>0$ , respectively. Hereis the gaseous (elastic and thermal) pressure, which is described for example, by Benedict-Webb-Rubin equation ofstate [21]$C_{\mathrm{v}}$$p_{G}=p_{G}(\rho, \theta)$$p_{G}=R\rho\theta+(RB_{0}\theta-A_{0}-C_{0}\theta^{-2})\rho^{2}$$+(Rb_{2}\theta-b_{1})\rho^{3}+b_{1}\alpha\rho^{6}+b_{3}\theta^{-2}\rho^{3}(1+\gamma\rho^{2})\mathrm{e}^{-\gamma\rho^{2}}$depenwith the perfect gas constant and positive constants
dent on the concrete media, andis the radiative pressure given by Stefan
law [14]$R$$A_{0},$ $B_{0},$ $C_{0},$$b_{1},$$b_{2},$$b_{3},$$\alpha,$$\gamma$$p_{R}=p_{R}(\rho, \theta)$$p_{R}= \frac{a}{3}\theta^{4}$.

For technical reason, we neglect the terms except the first in the right hand side of. We also assume the conductivity
has the following form (see forexample, [1], [8] :$\kappa=\kappa(\rho, \theta)$$p_{G}$$)$

tc$=\kappa_{1}+\kappa_{2^{\frac{\theta^{q}}{\rho}}}$,and are positive constants.
where
We impose the dynamical and kinematic boundary conditions for $i=1,2$$\kappa_{1},$$\kappa_{2}$$q$$(-p+\mu v_{x})|_{y=y:(t)}=-p_{\mathrm{e}}$for $t>0$ ,$\frac{\mathrm{d}y_{i}(t)}{\mathrm{d}t}=v(y:(t), t)$

for $t>0$ ,$\{$where the positive constant is the external pressure, and the thermal and chemicalboundary conditions for $i=1,2$$p_{e}$$(\kappa\theta_{y})|_{y=y_{*}(t)}=0$

for $t>0$ ,$(d\rho z_{y})|_{y=y:(t)}=0$

for $t>0$ ,$\{$and the initial conditon$(\rho, v, \theta, z)|_{\mathrm{t}=0}=(\rho_{0}(y), v_{0}(y),$$\theta_{0}(y),$$z_{0}(y))$

for$y\in\overline{\Omega(0)}$.

10

We introduce the Lagrangian transformation. For arbitrary fixed point $(y, t)\in$, we consider the solution curveof the Cauchy problem$Y_{y_{)}t}(\tau)$$\bigcup_{t\geq 0}(\overline{\Omega(t)}\cross\{t\})$$\frac{\mathrm{d}\mathrm{Y}_{y,i}(\tau)}{\mathrm{d}\tau}=v(\mathrm{Y}_{y,t}(\tau), \tau)$

for $0<\tau<t$ ,$\{$$\mathrm{Y}_{y,t}(t)=y$.

The unique existence of such a solution curve is guaranteed from the fundamental
existence theorem of an ordinary differential equation as long as is suitably smooth.
. Then this is uniquely solvable in ,Let$v$$\mathrm{Y}_{y,t}(0)=\xi$$y$$y= \mathrm{Y}_{\xi,0}(t)=\xi+\int_{0}^{t}v(\mathrm{Y}_{\xi,0}(\tau), \tau)\mathrm{d}\tau$.

It is well known that the kinematic boundary condition implies that for each $t\geq 0$ thisonto
. We put $y_{1}(0)=0$
mapping $(y, t)rightarrow(\xi, t)$ is one-to-one fromand $y_{2}(0)=L$ . Futhermore, we introduce the mass transformation$\overline{\Omega(0)}\cross\{t\}$$\overline{\Omega(t)}\cross\{t\}$$\xirightarrow x=\int_{0}^{\xi}\rho_{0}(s)\mathrm{d}s$.

Then problem (1.2) is reduced to$(\tilde{\rho}U_{x})_{x}=G$

in $(0, M)\mathrm{x}(0, \infty)$ ,$\tilde{U}|_{x=0,M}=0$

for $t>0$ ,

$\{$

where

and the tilde ” means transformed functions. Through the
we can get the explicit formula$”\sim$$M= \int_{0}^{L}\rho_{0}(\xi)\mathrm{d}\xi$relations$\tilde{f}=-\tilde{\rho}\tilde{U}_{x}$$\tilde{f}(x, t)=-G(x-\frac{\int_{0}^{M}\eta\tilde{\rho}(\eta,t)^{-1}\mathrm{d}\eta}{\int_{0}^{M}\tilde{\rho}(\eta,t)^{-1}\mathrm{d}\eta})$

.

(1.4)

Consequently, by putting $v(x, t):=1/\tilde{\rho}(x, t),$ $u(x, t):=\tilde{v}(x, t)$ and normalizing $M=1$
our problem becomes$v_{t}$$=u_{x}$,$u_{t}$$=$$(-p+ \frac{\mu}{v}u_{x})_{x}-G(x-\frac{\int_{0}^{1}\eta v(\eta,t)\mathrm{d}\eta}{\int_{0}^{1}v(\eta,t)\mathrm{d}\eta})$$e_{1}$$=$$( \frac{\kappa}{v}\theta_{x})_{x}+(-p+\frac{\mu}{v}u_{x})u_{x}+\lambda\phi z$$z_{t}$$=$$( \frac{d}{v^{2}}z_{x})_{x}-\phi z$

$\{$

,

,
(1.5)

11

in $(0,1)\cross(0, \infty)$ with the boundary conditions$(-p+ \frac{\mu}{v}u_{x},$$\frac{\kappa}{v}\theta_{x},$for $t>0$ ,(1.6)

for $x\in[0,1]$ .

(1.7)$\frac{d}{v^{2}}z_{x})|_{x=0,1}=(-p_{e}, 0,0)$

and the initial condition$(v, u, \theta, z)|_{t=0}=(v_{0}(x), u_{0}(x),$ $\theta_{0}(x),$ $z_{0}(x))$

One-dimensional problems have been studied under various conditions. For theviscous polytropic ideal gas a pioneering work of global in time existence with large
initial data was due to Kazhikhov and Shelukhin [11] under Dirichlet boundary condition with respect to the velocity. In the free-boundary case, Nagasawa [15] discussed
the global existence problem and the asymptotic behavior for the polytropic ideal gas
with the external pressure depending on time. Dafermos and Hsiao [3], Kawohl [10]and Jiang [9] considered problems for some general fluids. Also Chen [2] studied amodel equations for a reacting mixture. All works mentioned above were not taken
into account the influence of an external force.
Ducomet $[4]-[7]$ treated a one-dimensional self-gravitating gaseous model as somelarge-scale structure of the universe, called “pancakes” in the astrophysical literature(see [12], [17]). Following the spirit of [17], he adopted as the self-gravitational term$\tilde{f}(x, t)=-G(x-\frac{1}{2}M)$

not the exact form (1.4), and also assumed that the initial data and the solution are
symmetric.
Now, by integration ofwith respect to over $[0,1]$ we get$(1.5)^{2}$$x$$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}u\mathrm{d}x=-G(\frac{1}{2}-\frac{\int_{0}^{1}\eta v(\eta,t)\mathrm{d}\eta}{\int_{0}^{1}v(\eta,t)\mathrm{d}\eta})$Denoting$u- \int_{0}^{1}u\mathrm{d}x$

by

$u$

.

(1.8)

again, we obtain the final form:

in $(0,1)\cross(0, \infty)$ with the same initial-boundary conditions (1.6) and (1.7). For thisfor the(which corresponds tosystem it is natural that initial function$u_{0}$$u_{0}- \int_{0}^{1}u_{0}\mathrm{d}x$12original system (1.5) satisfies$)$$\int_{0}^{1}u_{0}\mathrm{d}x=0$.

(1.10)In this paper we construct the unique global classical solution of system (1.9), (1.6),(1.7) with the equations of state$p=R \frac{\theta}{v}+\frac{a}{3}\theta^{4}$,(1.11)$e=C_{\mathrm{v}}\theta+av\theta^{4}$and the conductivity$\kappa=\kappa_{1}+\kappa_{2}v\theta^{q}$,(1.12)

without the symmetric assumption to the initial data and the solution. From (1.8)
it is easily seen that this solution leads to the one for the original problem $(1.5)-(1.7)$
describing the exact one-dimensional self-gravitating fluid model, not the approximatedone: “pancakes” which has been considered by Ducomet. The difficulty of our problem
is mainly caused by radiative components of equations of state and -dependency ofthe conductivity. We can solve the problem only for the case of some $q\geq 4$ , which is
physically valid [22]. Similar result has been obtained in [7], but the proof in it is notclear for the authors.
Let $\Omega:=(0,1),$ $m$ be a nonnegative integer and $0<\sigma<1$ . is a positive constantand $Q\tau:=\Omega\cross(0, T)$ . We denote$\theta$$T$$|u|^{(0)}:= \sup_{(x,t)\in Q_{T}}|u(x, t)|$and use the familiar notations
spaces (see for example, [13]).
Our main result is$C^{m+\sigma}(\Omega),$$C_{x,t}^{\sigma,\sigma/2}(Q_{T}),$‘ $1+\sigma/2(Q_{T})$$C_{x,l}^{2+\sigma}$

for the H\"older

Theorem 1 (Global Solution) Let $a\in(0,1),$ $4\leq q\leq 16$ and $0\leq\beta\leq 13/2$ .
Assume that
(1.13)$(v_{0}, u_{0}, \theta_{0}, z_{0})\in C^{1+\alpha}(\Omega)\cross(C^{2+\alpha}(\Omega))^{3}$

satisfies the compatibility conditions, (1.10) and$v_{0}(x),$$\theta_{0}(x)>0$,$0\leq z_{0}(x)\leq 1$

for$x\in$

Sl.

(1.14)

Then there enists a unique solutionof the initial-boundary value problem(1.9), (1.6), (1.7) with (1.3), (1.11), (1.12) such that for $anyT>0$$(v, u, \theta, z)$$(v, v_{x}, v_{t})\in(C_{x,t}^{\alpha,\alpha/2}(Q_{T}))^{3}$$v(x, t),$$\theta(x, t)>0$,,$(u, \theta, z)\in(C_{x,t}^{2+\alpha,1+\alpha/2}(Q_{T}))^{3}$$0\leq z(x, t)\leq 1$

for$(x, t)\in\overline{Q_{T}}$.

,(1.15)(1.16)

13
Proof of Theorem 1 is based on the local existence theorem and a priori estimates.
The fundamental theorem about the existence and the uniqueness of the local in time
solution in three-dimensional case was firstly established by Tani [19], [20] under sufficiently general initial-boundary conditions. For a radiative fluid, Secchi [16] obtained
the corresponding result.
We can easily obtain suitable unique local solution to our problem in the samemanner as these works (see for example, [18]). Therefore, to prove Theorem 1 it issufficient to establish the following a priori boundedness.
Proposition 1 (A priori estimates) Let
be an arbitrary positive constant, $4\leq$$q\leq 16$ and $0\leq\beta\leq 13/2$ .
Assume that the initial data satisfy the hypotheses ofTheorem 1 and problem (1.9), (1.6), (1.7) with (1.3), (1.11), (1.12) has a solution$(v, u, \theta, z)$ such that$T$$(v, v_{x}, v_{t})\in(C_{x,t}^{\alpha,\alpha/2}(Q_{T}))^{3}$,

Then there exists a positive constant $M$ depending on the initial data and,$|v,$$v(x, t),$2$v_{x},$ $v_{t}|_{\alpha,\alpha/2},$$\theta(x, t)\geq 1/M$

,

$|u,$

(i.17)$(u, \theta, z)\in(C_{x,t}^{2+\alpha,1+\alpha/2}(Q_{T}))^{3}$$\theta,$$z|_{2+\alpha,1+\alpha/2}\leq M$

for$0\leq z(x, t)\leq 1$.

$T$such that(1.18)(1.19)$(x, t)\in\overline{Q_{T}}$Proof of Proposition 1.

In proving Proposition 1, we need several lemmas concerning the estimates of thesolution and its derivatives. Our methods are mainly based on the techniques in Kadenotes usual
wohl [10] and Jiang [9]. We use as positive constants, and$C$$L^{2}$$||\cdot||$norm.

Lemma 1 For any $t\in[0, T]$$\int_{0}^{1}(\frac{1}{2}u^{2}+e+\lambda z+f(x)v)\mathrm{d}x$$= \int_{0}^{1}(\frac{1}{2}u_{0^{2}}+e_{0}+\lambda z_{0}+f(x)v_{0})\mathrm{d}x:=E_{0}$,.

$\int_{0}^{1}\frac{1}{2}z^{2}\mathrm{d}x+\int_{0}^{t}\int_{0}^{1}(\frac{d}{v^{2}}z_{x^{2}}+\phi z^{2})\mathrm{d}_{X}\mathrm{d}\tau=\int_{0}^{1}\frac{1}{2}z_{0^{2}}\mathrm{d}x$$e_{0}:=C_{\mathrm{v}}\theta_{0}+av_{0}\theta_{0}^{4},$$f(x):=p_{e}+ \frac{1}{2}Gx(1-x)$

(2.1)
(2.2)

$U(t)+ \int_{0}^{t}V(\tau)\mathrm{d}\tau\leq C$

Here

,

and

$U(t)$$:= \int_{0}^{1}[C_{\mathrm{v}}(\theta-1-\log\theta)+R(v-1-\log v)]\mathrm{d}x$$V(t)$$:= \int_{0}^{1}(\frac{\mu u_{x^{2}}}{v\theta}+\frac{\kappa\theta_{x^{2}}}{v\theta^{2}}+\lambda\frac{\phi}{\theta}z)\mathrm{d}x$

$\{$

.

,

(2.3)

14

Proof. It is easy to see from (1.9) and (1.6)(2.4)$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}(\frac{1}{2}u^{2}+f(x)v)\mathrm{d}x+\int_{0}^{1}\frac{\mu}{v}u_{x^{2}}\mathrm{d}x=\int_{0}^{1}pu_{x}\mathrm{d}x$and$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}(e+\lambda z)\mathrm{d}x=\int_{0}^{1}(-p+\frac{\mu}{v}u_{x})u_{x}\mathrm{d}x$.

Adding these and integrating over $[0, t]$ , we obtain (2.1).
Rewriting $(1.9)^{3}$ as(2.5)$e_{\theta} \theta_{t}+\theta p_{\theta}u_{x}=\frac{\mu}{v}u_{x}+2(\frac{\kappa}{v}\theta_{x})_{x}+\lambda\phi z$

and multiplying this by$\theta^{-1}$, we have.

$\frac{\mathrm{d}}{\mathrm{d}t}(C_{\mathrm{v}}\log\theta+R\log v+\frac{4}{3}av\theta^{3})=\frac{\mu u_{x^{2}}}{v\theta}+\frac{1}{\theta}(\kappa\frac{\theta_{x}}{v})_{x}+\lambda\frac{\phi}{\theta}z$

Integrating this over $(0,1)\cross(0, t)$ yields$U(t)+ \int_{0}^{t}V(\tau)\mathrm{d}\tau\leq C(1+\int_{0}^{1}v\theta^{3}\mathrm{d}x)$.

From H\"older’s inequality$\int_{0}^{1}v\theta^{f}\mathrm{d}x\leq(\int_{0}^{1}v\theta^{4}\mathrm{d}x)^{f/4}(\int_{0}^{1}v\mathrm{d}x)^{(4-f)/4}$and (2.1), (2.2) follows.
is easily obtained by integrating(1.6).

$(1.9)^{4}$$\mathrm{E}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\square (2.3)$

for $0\leq r\leq 4$(2.6)

over $(0,1)\cross(0, t)$ and using

Next lemma concerning the pointwise estimate of is obtained in the same manneras Chen [2]. We omit the proof.
$z$Lemma 2 For any$(x, t)\in\overline{Q_{T}}$$0\leq z(x, t)\leq 1$.

Kazhikhov and Shelukhin firstly derived the useful representation formula forthe present case, we can obtain the following similar form (see [11]).

(2.7)
$v$

. In

Lemma 3 The identity$v(x, t)= \frac{1}{B(x,t)\mathrm{Y}(x,t)D(x,t)}$$\cross(v_{0}+\int_{0}^{t}\frac{R}{\mu}\theta(x, \tau)B(x, \tau)Y(x, \tau)D(x, \tau)\mathrm{d}\tau)$

(2.8)

15

holds, where$B(x, t):= \exp[\frac{1}{\mu}\int_{0}^{x}(u_{0}(\xi)-u(\xi, t))\mathrm{d}\xi]$,$Y(x, t):= \exp(\frac{1}{\mu}f(x)t)$$D(x,t):= \exp(-\frac{a}{3\mu}\int_{0}^{t}\theta(x, \tau)^{4}\mathrm{d}\tau)$Since$( \mu\frac{u_{x}}{v})_{x}=\mu(\log v)_{xt}$follows from$(1.9)^{1}$,.

, integrating $(1.9)^{2}$ over$(0, x)\cross(\mathrm{O}, t)$yields.

$\frac{1}{\mu}\int_{0}^{x}(u-u_{0})\mathrm{d}\xi=\log v-\log v_{0}-\frac{1}{\mu}\int_{0}^{t}p\mathrm{d}\tau+\frac{1}{\mu}f(x)t$

From this one can easily find the lower bound of v:$(x,t) \in \mathrm{m}\mathrm{i}_{\frac{\mathrm{n}}{Q_{T}}}v(x, t)\geq \mathrm{m}\mathrm{i}_{\frac{\mathrm{n}}{\Omega}}v_{0}(x)\exp x\in\{-\frac{1}{\mu}[4E_{0}^{1/2}+(p_{e}+\frac{G}{8})t]\}$

.

(2.9)

This together with (2.6) leads to$\int_{0}^{1}\theta^{r}\mathrm{d}x\leq C$

for $0\leq r\leq 4$ .

(2.10)

Lemma 4 For any $t\in[0, T]$ and $q\geq 0$,$\int_{0}^{t}\max_{0\leq x\leq 1}\theta(x, \tau)^{r}\mathrm{d}\tau\leq C$

Proof. For each $t\in[0, T]$ , there exists$0\leq r\leq q+4$$x^{*}(t)\in[0,1]$.

(2.11)

such that$\theta(x^{*}(t),t)=\int_{0}^{1}\theta \mathrm{d}x$,and therefore, for any $r\geq 0$ and $(x, t)\in Q_{T}$ we have$\theta(x, t)^{r/2}=(\int_{0}^{1}\theta \mathrm{d}x)^{f/2}+\frac{r}{2}\int_{x(t)}^{x}.\theta(\xi, t)^{t/2-1}\theta_{\xi}(\xi, t)\mathrm{d}\xi$$\leq C(1+\int_{0}^{1}\frac{\kappa^{1/2}|\theta_{x}|}{v^{1/2}\theta}\cdot\frac{v^{1/2}\theta^{r/2}}{\kappa^{1/2}}\mathrm{d}x)$$\leq C[1+(\int_{0}^{1}\frac{v\theta^{f}}{1+v\theta^{q}}\mathrm{d}x)^{1/2}V(t)^{1/2}]$

Since$\theta^{f}\leq C(1+\theta^{q+4})$.

holds for $0\leq r\leq q+4$ , we have from (2.10)$\int_{0}^{1}\frac{v\theta^{r}}{1+v\theta^{q}}\mathrm{d}x\leq C\int_{0}^{1}(v+\theta^{4})\mathrm{d}x\leq C$,(2.12)

16

which yields from (2.12) and (2.2)$\int_{0}^{t}\max_{0\leq x\leq 1}\theta(x, \tau)^{r}\mathrm{d}\tau\leq C\int_{0}^{t}(1+V(\tau))\mathrm{d}\tau\leq C.$

Lemma 5 For any$\square$$(x, t)\in\overline{Q_{T}}$.

(2.13)$C^{-1}\leq v(x, t)\leq C$

Proof. The lower bound of is already obtained in (2.9). For the upper bound of we
$v$

$v$

use representaiton (2.8).

From (2.2) and Jensen’s inequality we have$\int_{0}^{1}\theta \mathrm{d}x-\log\int_{0}^{1}\theta \mathrm{d}x-1\leq\frac{C}{C_{\mathrm{v}}}$.

(2.14)

Then, by applying the mean value theorem there exists a point $x^{**}(x, t)\in[0,1]$ foreach fixed $t\in[0, T]$ such that$\theta(x^{**}(t),t)-\log\theta(x^{**}(t),t)-1\leq\frac{C}{C_{\mathrm{v}}}$

Here we define$\alpha_{0}$and.

(2.15)as two positive roots of the equation$\beta_{0}(0<\alpha_{0}<\beta_{0})$y–log$y-1= \frac{C}{C_{\mathrm{v}}}$,

which are independent of . We get from (2.14) and (2.15)$t$$\alpha_{0}\leq\int_{0}^{1}\theta \mathrm{d}x\leq\beta_{0}$,From (2.1) it is easily seen that for any$\alpha_{0}\leq\theta(x^{**}(t), t)\leq\beta_{0}$.

$(x, t)\in\overline{Q_{T}}$$C^{-1}\leq B(x, t)\leq C$.

On the other hand, since$\theta(x, t)^{2}=\theta(x^{**}(t), t)^{2}+2\int_{x^{*}(t)}^{x}.\theta(\xi, t)\theta_{\xi}(\xi, t)\mathrm{d}\xi$,

we have$\alpha_{0^{2}}-CV(t)^{1/2}\leq\theta(x, t)^{\mathit{2}}\leq\beta 0^{2}+CV(t)^{1/2}$

,

hence
$\frac{1}{2}\alpha_{0^{4}}-CV(t)\leq\theta(x, t)^{4}\leq 2\beta_{0}^{4}+CV(t)$.

(2.16)

17(2.12) with $r=2$ implies$C-CV(t)\leq\theta(x, t)\leq C(1+V(t))$.

(2.17)Let us decompose $v(x, t)=v_{1}(x, t)+v_{2}(x, t)$ , where$v_{1}(x, t)= \frac{v_{0}(x)}{B(x,t)\mathrm{Y}(x,t)D(x,t)}$,.

$v_{2}(x, t)= \frac{R}{\mu}\int_{0}^{t}\frac{B(x,\tau)\mathrm{Y}(x,\tau)D(x,\tau)}{B(x,t)\mathrm{Y}(x,t)D(x,t)}\theta(x, \tau)\mathrm{d}\tau$Using (2.16), we immediately obtain.

(2.18)$C\mathrm{e}^{-(f(x)-\frac{1}{6}a\alpha_{0^{4}})\frac{\ell}{\mu}}\leq v_{1}(x, t)\leq C\mathrm{e}^{-(f(x)-\frac{2}{\mathrm{s}}a\beta_{0^{4}})\frac{t}{\mu}}$

Also (2.16) and (2.17) yields(2.19)$v_{2}(x, t) \leq C\int_{0}^{t}\mathrm{e}^{-\frac{1}{\mu}(f(x)-\frac{2}{s}a\beta_{0^{4}})(t-\tau)}(1+V(\tau))\mathrm{d}\tau$$\leq C\mathrm{e}^{CT}\int_{0}^{t}(1+V(\tau))\mathrm{d}\tau$.

Consequently, from (2.18) and (2.20) the upper bound of(2.20)$v$

follows.

$\square$Lemma 6 If $q\geq 2$ , for any $t\in[0, T]$(2.21)$||v_{x}||^{\mathit{2}}+ \int_{0}^{t}\int_{0}^{1}\theta v_{x}\mathrm{d}2x\mathrm{d}\tau\leq C.$

Proof.

$(1.9)^{2}$

implies$(u- \frac{\mu}{v}v_{x})_{t}=-p_{x}-G(x-\frac{1}{2})$Multiplying this by$u- \frac{\mu}{v}v_{x}$.

and integrating over $[0,1]$ , we have$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}\frac{1}{2}(u-\frac{\mu}{v}v_{x})^{2}\mathrm{d}x+\int_{0}^{1}\frac{\mu R}{v^{3}}\theta v_{x}\mathrm{d}2x$$= \int_{0}^{1}\frac{R}{v^{\mathit{2}}}u\theta v_{x}\mathrm{d}x-\int_{0}^{1}[(\frac{R}{v}+\frac{4}{3}a\theta^{3})\theta_{x}+G(x-\frac{1}{2})](u-\frac{\mu}{v}v_{x})\mathrm{d}x$Firstly, we have for any$\epsilon>0$$| \int_{0}^{1}\frac{R}{v^{\mathit{2}}}u\theta v_{x}\mathrm{d}x|\leq\epsilon\int_{0}^{1}\theta v_{x^{2}}\mathrm{d}x+C_{\epsilon}\max_{0\leq x\leq 1}\theta\cdot\int_{0}^{1}u^{2}\mathrm{d}x$,.

(2.22)

18

is a positive number depending on
whereof (2.22) is estimated as follows.
$C_{\epsilon}$$\epsilon$. The second term of the right hand side$| \int_{0}^{1}[(\frac{R}{v}+\frac{4}{3}a\theta^{3})\theta_{x}+G(x-\frac{1}{2})](u-\frac{\mu}{v}v_{x})\mathrm{d}x|$$\leq C[\int_{0}^{1}\frac{\kappa\theta_{x}^{2}}{v\theta^{\mathit{2}}}\mathrm{d}x+\int_{0}^{1}\frac{\theta^{2}(1+\theta^{3})^{2}}{\kappa}(u-\frac{\mu}{v}v_{x})^{2}\mathrm{d}x]$$\leq C[V(t)+\max_{0\leq x\leq 1}(1+\theta^{2}+\frac{\theta^{8}}{1+\theta^{q}})\cdot\int_{0}^{1}(u-\frac{\mu}{v}v_{x})^{2}\mathrm{d}x]$

.

Since
$\int_{0}^{t}\max_{0\leq x\leq 1}\frac{\theta^{8}}{1+\theta^{q}}\mathrm{d}\tau\leq C$

holds with $q\geq 2$ , by taking$\epsilon$

small, Gronwall’s inequality implies (2.21).

$\square$Lemma 7 For any $t\in[0, T]$$\int_{0}^{t}||u_{x}||^{2}\mathrm{d}\tau\leq C$

Integrate (2.4) with respect toinequality, we have

Proof.

$t$.

(2.23)

and using Lemma 5 and Cauchy-Schwarz$\int_{0}^{t}||u_{x}||^{2}\mathrm{d}\tau\leq C(1+\int_{0}^{t}\int_{0}^{1}(\theta+\theta^{4})^{2}\mathrm{d}x\mathrm{d}\tau)$$\leq C[1+\int_{0}^{t}(\int_{0}^{1}\theta^{2}\mathrm{d}x+\max_{0\leq x\leq 1}\theta^{4}\cdot\int_{0}^{1}\theta^{4}\mathrm{d}x)\mathrm{d}\tau]$

The right hand side is bounded by (2.10) and (2.11).

.

$\square$Lemma 8 If $q\geq 4$ , for any $t\in[0, T]$.

$\int_{0}^{t}||u_{x}||_{L^{\theta}(\Omega)}^{3}\mathrm{d}\tau\leq C$

Proof. We use a method due to Dafermos and Hsiao [3]. Putting

using (1.10), we get a new system:(2.24)$w= \int_{0}^{x}u\mathrm{d}\xi$

and

19

General theory of linear parabolic equations (see for example, [13]) gives$\int_{0}^{t}||w_{xx}||_{L^{3}(\Omega)}^{3}\mathrm{d}t\leq C(\int_{0}^{t}||-p+f(x)||_{L^{3}(\Omega)}^{3}\mathrm{d}t+||w_{0}||_{W_{3}^{4/3}(\Omega)})$.

Therefore, we have$\int_{0}^{t}||u_{x}||_{L^{3}(\Omega)}^{3}\mathrm{d}\tau\leq C(1+\int_{0}^{t}||p||_{L^{\theta}(\Omega)}^{3}\mathrm{d}\tau)$$\leq C[1+\int_{0}^{t}(\int_{0}^{1}\theta^{3}\mathrm{d}x+_{0}\max_{\leq x\leq 1}\theta^{8}\cdot\int_{0}^{1}\theta^{4}\mathrm{d}x)\mathrm{d}\tau]$

If $q\geq 4$ , the right hand side is bounded.

.

$\square$As [10] and [9], we introduce the quantities$X:= \int_{0}^{t}\int_{0}^{1}(1+\theta^{q+3})\theta_{t}^{2}\mathrm{d}x\mathrm{d}\tau$,$\mathrm{Y}:=0^{\max_{\leq t\leq T}\int_{0}^{1}}(1+\theta^{q+4})\theta_{x}^{\mathit{2}}\mathrm{d}x$$Z:=0^{\max_{\leq t\leq T}\int_{0}^{1}u_{xx^{2}}\mathrm{d}x}$.

Firstly, by Cauchy-Schwarz inequality we have$|\theta|^{(0)}\leq C(1+\mathrm{Y}^{\frac{1}{q+10}})$.

(2.25)

Secondly, it is easily seen that by interpolation$\max_{0\leq t\leq T}||u_{x}||^{2}\leq C(1+Z^{1/2})$(2.26)

holds and for any $t\in[0, T]$$\leq$$\max_{0\leq x\leq 1}u_{x^{2}}$$\leq$$||u_{x}||^{2}+2||u_{x}||||u_{xx}||$$C(1+Z^{3/4})$ ,which gives$|u_{x}|^{(0)}\leq C(1+Z^{3/8})$.

(2.27)In the same manner as in [10] and [9] we introduce the function$K(v, \theta):=\int_{0}^{\theta}\frac{\kappa(v,\xi)}{v}\mathrm{d}\xi$

Multiplying (2.5) by$K_{t}$.

and integrating it over $(0,1)\cross(0, t)$ , we have$\int_{0}^{t}\int_{0}^{1}(e_{\theta}\theta_{t}+\theta p_{\theta}u_{x}-\frac{\mu}{v}u_{x^{2}})K_{t}\mathrm{d}x\mathrm{d}\tau+\int_{0}^{t}\int_{0}^{1}\frac{\kappa}{v}\theta_{x}K_{xt}\mathrm{d}x\mathrm{d}\tau$$= \int_{0}^{t}\int_{0}^{1}\lambda\phi zK_{t}\mathrm{d}x\mathrm{d}\tau$.

(2.28)

20

Here
,

$K_{t}= \frac{\kappa}{v}\theta_{t}+K_{v}u_{x}$

$\{$

,

$K_{xt}=( \frac{\kappa}{v}\theta_{x})_{t}+K_{vv}v_{x}u_{x}+K_{v}u_{xx}+(\frac{\kappa}{v})_{v}v_{x}\theta_{t}$$|K_{v}|,$$|K_{vv}|\leq C\theta$.

Estimating each term in (2.28), we get the following.

Lemma 9 If $4\leq q\leq 16$ and $0\leq\beta\leq 13/2_{f}$ we have
$X+Y\leq C(1+Z^{7/8})$.

(2.29)

Sketch of proof. At first, note that$\int_{0}^{t}\int_{0}^{1}e_{\theta}\theta_{t}(\frac{\kappa}{v}\theta_{t})\mathrm{d}x\mathrm{d}\tau\geq C\int_{0}^{t}\int_{0}^{1}(1+\theta^{3})(1+\theta^{q})\theta_{t}^{2}\mathrm{d}x\mathrm{d}\tau$$\geq CX$,$\int_{0}^{t}\int_{00}^{11}\frac{\kappa}{v}\theta_{x}(\frac{\kappa}{v}\theta_{x})_{t}\mathrm{d}x\mathrm{d}\tau=\frac{1}{2}\int_{\geq C\mathrm{Y}-}(\frac{\kappa}{c^{v}}.\theta_{x})^{2}\mathrm{d}x-\frac{1}{2}\int_{0}^{1}(\frac{\kappa}{v}\theta_{x})^{2}(x, 0)\mathrm{d}x$

Other terms are also estimated from (2.25)-(2.27), Lemmas 4, 6, 7 and 8, for example,$| \int_{0}^{t}\int_{0}^{1}(\theta p_{\theta}u_{x}-\frac{\mu}{v}u_{x}2)\frac{\kappa}{v}\theta_{t}\mathrm{d}x\mathrm{d}\tau|$$\leq\epsilon X+C_{\epsilon}|u_{x^{2}}|^{(0)}\int_{0}^{t}\max_{0\leq x\leq 1}(1+\theta)^{q+1}\int_{0}^{1}(1+\theta)^{4}\mathrm{d}x\mathrm{d}\tau$$+C_{\epsilon}|1+ \theta^{q-3}|^{(0)}|u_{x}|^{(0)}\int_{0}^{t}||u_{x}||_{3^{3}}\mathrm{d}\tau$$\leq\epsilon X+C_{\epsilon}(1+Z^{3/4}+\mathrm{Y}^{\mathrm{L}_{\frac{\mathrm{a}}{10}}^{-}}q+Z^{3/8})$$\leq\epsilon(X+Y)+C_{\epsilon}(1+Z^{3/4})$,$| \int_{0}^{t}\int_{0}^{1}\frac{\kappa}{v}\theta_{x}K_{v}u_{xx}\mathrm{d}x\mathrm{d}\tau|\leq C|\kappa^{1/2}\theta^{2}|^{(0)}Z^{1/2}\int_{0}^{t}V(\tau)^{1/2}\mathrm{d}\tau$$\leq C(1+Y^{\frac{q+4}{2(l+10)}}Z^{1/2})$$\leq\epsilon Y+C_{\epsilon}(1+Z^{7/8})$

Combining these estimates and taking$\epsilon$.

small, we obtain (2.29).

$\square$21

Lemma 10 If $4\leq q\leq 16$ and $0\leq\beta\leq 13/2$ , we have for any $t\in[0, T]$$||u_{x}||^{\mathit{2}}+|| \theta_{x}||^{2}+||u_{xx}||^{2}+||u_{t}||^{\mathit{2}}+\int_{0}^{t}(||\theta_{t}||^{\mathit{2}}+||u_{xt}||^{2})\mathrm{d}\tau\leq C$,(2.30)

and(2.31)$|u_{x}|^{(0)}+|u|^{(0)}+|\theta|^{(0)}\leq C.$

Proof. The following culculations are formal because the regularity of the solution
is not sufficient. But as usual we can easily derive tfe rigorous results by using thearguments of mollifiers and passing to the limit.
Differentiating $(1.9)^{2}$ with respect to , multiplying it by and integrating it withrespect to , we have$t$$u_{t}$$x$$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}\frac{1}{2}u_{\mathrm{t}^{\mathit{2}}}\mathrm{d}x+\int_{0}^{1}\frac{\mu}{v}u_{xt^{\mathit{2}}}\mathrm{d}x=\int_{0}^{1}(p_{t}u_{xt}+\frac{\mu}{v^{\mathit{2}}}u_{x}u_{xt})2\mathrm{d}x$

,

hence
$||u_{t}||^{\mathit{2}}+ \int_{0}^{t}||u_{xt}||^{2}\mathrm{d}\tau$$\leq C[1+\int_{0}^{t}(||p_{t}||^{\mathit{2}}+||u_{x}||_{L^{4}(\Omega)}^{4})\mathrm{d}\tau]$$\leq C[\int_{0}^{t}\int_{0}^{1}(1+\theta^{6})\theta_{t}^{2}\mathrm{d}x\mathrm{d}\tau$$+|u_{x}|^{(0)}2 \int_{0}^{t}\int_{0}^{1}\theta^{\mathit{2}}\mathrm{d}x\mathrm{d}\tau+|u_{x}|^{(0)}\int_{0}^{t}||u_{x}||_{\iota^{s_{(\Omega)}}}^{3}\mathrm{d}\tau]$$\leq C(1+X+Z^{3/4}+Z^{3/8})$(2.32)$\leq C(1+Z^{7/8})$

by Lemma 9. From $(1.9)^{2}$ again together with the estimate above, it follows for any$t\in[0,T]$$||u_{xx}||^{2} \leq C[1+\int_{0}^{1}(u_{t^{\mathit{2}}}+p_{x^{\mathit{2}}}+u_{x^{22}}v_{x})\mathrm{d}x]$$\leq C[1+||u_{1}||^{2}+\int_{0}^{1}(1+\theta^{6})\theta_{x}^{2}\mathrm{d}x+(|\theta^{2}|^{(0)}+|u_{x^{2}}|^{(0)})||v_{x}||^{2}]$$\leq C(1+\mathrm{Y}^{\frac{2}{q+10}}+\mathrm{Y}+Z^{3/4}+Z^{7/8})$$\leq C(1+Z^{7/8})$.

This means$Z\leq C(1+Z^{7/8})$.

22

Therefore, we conclude that $Z$ is bounded. Then one can see that $X,$,
are also bounded from Lemma 9, (2.25)-(2.27), and (2.32).
andThe boundedness of is easily derived from$|\theta|^{(0)},$$\mathrm{Y},$$|u_{x}|^{(0)},$$||u_{t}||$$||u_{x}||$$\int_{0}^{t}||u_{xt}||^{2}\mathrm{d}\tau$$u$$|u|^{(0)} \leq\max_{0\leq t\leq T}(||u||_{L^{1}(\Omega)}+||u_{x}||)$

Lemma 11 For any$\square$$(x, t)\in\overline{Q_{T}}$$\theta(x, t)\geq C$

Proof. By putting.

$\Theta:=\frac{1}{\theta},$$(2.5)$.

(2.33)

becomes$e_{\theta}_{t}=( \frac{\kappa}{v}\Theta_{x})_{x}+\frac{vp_{\theta^{2}}}{4\mu}-[\frac{2\kappa\ominus_{x^{2}}}{v\Theta}+\frac{\mu^{2}}{v}(u_{x}-\frac{vp_{\theta}}{2\mu\ominus})^{2}+\lambda \mathrm{O}-^{2}\phi z]$

from which by (2.31) there exists a positive constant$_{t}< \frac{1}{e_{\theta}}(\frac{\kappa}{v}_{x})_{x}+C_{1}$$C_{1}$,

such that.

Standard comparison arguments imply$(x,t) \in \mathrm{m}\mathrm{i}_{\frac{\mathrm{n}}{Q_{T}}}(C_{1}t+\max_{0\leq x\leq 1}\frac{1}{\theta_{0}(x)}-\Theta(x,t))\geq 0$.

$\square$Lemma 12 For any $t\in[0, T]$

1
Proof. Multiplying(2.34)$z_{x}||^{\mathit{2}}+||z_{xx}||^{2}+||z_{t}||^{2}+ \int_{0}^{t}||z_{xt}||^{2}\mathrm{d}\tau\leq C.$$(1.9)^{4}$by$z_{xx}$and integrating it over $[0,1]$ , we have.

$\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{1}\frac{1}{2}z_{x}^{2}\mathrm{d}x+\int_{0}^{1}\frac{d}{v^{\mathit{2}}}z_{xx}\mathrm{d}2x=\int_{0}^{1}(\frac{2d}{v^{2}}v_{x}z_{x}z_{xx}+\phi zz_{xx})\mathrm{d}x$

In a standard manner, we easily obtain the boundedness ofby $(1.9)^{4}$ again.
which implies
Next, we differentiatewith respect to , multiply it by$[0,1]$ . Then we have$||z_{x}||$and$\int_{0}^{t}||z_{xx}||^{\mathit{2}}\mathrm{d}\tau$,$\int_{0}^{t}||z_{t}||^{2}\mathrm{d}\tau\leq C$$(1.9)^{4}$$t$$z_{t}$and integrate over$||z_{t}||^{\mathit{2}}+ \int_{0}^{t}||z_{xt}||^{2}\mathrm{d}\tau\leq C[1+\int_{0}^{t}(||\theta_{t}||^{2}+||z_{x}||^{2}+||z_{t}||^{2})\mathrm{d}\tau]$

whose right hand side is bounded. From$(1.9)^{4}$again we obtain$||z_{xx}||^{2} \leq C\int_{0}^{1}(1+z_{t^{\mathit{2}}}+z_{x^{\mathit{2}2}}v_{x}+\phi^{2})\mathrm{d}x$$\leq\epsilon||z_{xx}||^{2}+C_{\epsilon}$

which completes (2.34).

$\square$

,

,

23

Lemma 13 For any $t\in[0, T]$.

(2.35)$|| \theta_{xx}||^{\mathit{2}}+||\theta_{t}||^{2}+\int_{0}^{t}||\theta_{xt}||^{2}\mathrm{d}\tau\leq C$

Proof. Differentiating (2.5) with respect to , multiplying it by$t$over $[0,1]$ , we have$|| \theta_{t}||^{\mathit{2}}+\int_{0}^{t}$I$e_{\theta}\theta_{t}$and integrating$\theta_{xt}||^{2}\mathrm{d}\tau\leq C[1+\int_{0}^{t}\int_{0}^{1}(v_{x}2+\theta_{x^{\mathit{2}}})\theta_{t}^{2}\mathrm{d}x\mathrm{d}\tau]$.

Then, by virtue of the interpolation inequality$\max_{0\leq x\leq 1}\theta_{t}(x, t)^{2}\leq\epsilon||\theta_{xt}||^{2}+C_{\epsilon}||\theta_{t}||^{2}$

we conclude that$||\theta_{t}||$and$\int_{0}^{t}||\theta_{xt}||^{2}\mathrm{d}\tau$,

are bounded. Also from (2.5), one can obtain$|| \theta_{xx}||^{2}\leq C\int_{0}^{1}\frac{1}{\kappa^{2}}(1+\theta^{2}+u_{x^{\mathit{2}}}+u_{x^{4}}+v_{x}^{2}\theta_{x}^{2}+\theta_{x}^{4})\mathrm{d}x$$\leq C[1+\max_{0\leq x\leq 1}\theta_{x}^{2}\cdot\int_{0}^{1}(v_{x^{2}}+\theta_{x}^{2})\mathrm{d}x]$$\leq\epsilon||\theta_{xx}||^{2}+C_{\epsilon}$.

$\square$Let us begin the H\"older estimate. From (2.30), (2.34), (2.35) and interpolations weobtain the bounds of. Therefore,andare Lipschitz continuous in .
Applying Cauchy-Schwarz and interpolation inequalities, we have$|\theta_{x}|^{(0)}$$|z_{x}|^{(0)}$$u,$$\theta,$

$x$

$z$

$|u(x,t)-u(x,t’)| \leq(\int_{t}^{t},u_{\iota^{2}}\mathrm{d}\tau)^{1/2}|t-t^{l}|^{1/2}$$\leq[\int_{t}^{t},(||u_{t}||^{2}+2||u_{t}||||u_{xt}||)\mathrm{d}\tau]^{1/2}|t-t’|^{1/2}$$|u_{x}(x, t)-u_{x}(x’, t)| \leq(\int_{x}^{x},u_{xx^{2}}\mathrm{d}\xi)^{1/2}|x-x’|^{1/2}$,,

from which together with (2.30) is H\"older continuous in with exponent 1/2 and isalso H\"older continuous in with exponent 1/2. Thus by a standard interpolation (seefor example, [13], chapter II, Lemma 3.1) one can get. We can also,
get same H\"older estimates for. Recalling that $v_{t}=u_{x}$ andwe deduce. Since it follows from (2.8) that
$t$

$u$

$u_{x}$$x$$u_{x}\in C_{x,t}^{1/3,1/6}(Q_{T})$$\theta,$$z$$v|_{t=0}=v_{0}\in C^{1+\alpha}(\Omega)$$v\in C_{x,t}^{1/3,1/6}(Q_{T})$$v_{x}(x,t)= \frac{1}{B(x,t)\mathrm{Y}(x,t)D(x,t)}\{v_{0’}(x)-A(x, t)v_{0}(x)$$+ \int_{0}^{t}\frac{R}{\mu}[\theta_{x}(x, \tau)+\theta(x, \tau)(A(x, \tau)-A(x,t))]B(x, \tau)Y(x,\tau)D(x, \tau)\mathrm{d}\tau\}(2.36)$

24

with
$A(x, t):= \frac{1}{\mu}[u_{0}(x)-u(x, t)-\frac{4}{3}a\int_{0}^{t}\theta(x, \tau)^{3}\theta_{x}(x, \tau)\mathrm{d}\tau-G(x-\frac{1}{2})t]$

we can easily check
Next we consider,

with $\sigma:=\min\{a, 1/3\}$ .
as the linear parabolic equations:and$v_{x}\in C_{x,t}^{\sigma,\sigma/2}(Q_{T})$$(1.9)^{2},$$(1.9)^{3}$$(1.9)^{4}$

whose coefficients and right hand sides are H\"older continuous in with exponent. By the classical Schauder estimates (see for example, [13]and in with exponent
are Lipschitz. This also implies thator [3] we obtain
continuous in and H\"older in with exponent 1/2. Going back to (2.36), we obtain$x$$\sigma$$\sigma/2$$t$$|u,$ $\theta,$ $z|_{2+\sigma,1+\sigma/2}\leq C$

$)$

$x$

$v_{x}\in C_{x,t}^{\alpha,\alpha/2}(Q_{T})$$v,$ $u_{x},$ $\theta_{x},$ $z_{x}$$t$.

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