数理解析研究所講究録 1475 巻 2006 年 121-138

121

Global Existence for a Quasilinear System
Arising in Shape Memory Alloys
東北大学大学院理学研究科

吉川 周二 (Shuji Yoshikawa)

Mathematical Institute, Tohoku University,
ポーランド科学アカデミー Irena Pawiow
Systems Research Institute, Polish Academy of Sciences,
ポーランド科学アカデミー Wojciech M. Zajaczkowski
Institute of Mathematics, Polish Academy of Sciences

Introduction

1

This paper is based on the result of [12]. We consider the following initial-boundary
value problem in quasi-linear thermoelasticity $(TE)_{n}$ :
(1.1)
,
(1.2)
,
.
(1.3)
on ,
$\mathrm{u}_{tt}+\triangle^{2}\mathrm{u}-\nu\triangle \mathrm{u}_{t}=\nabla\cdot(G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})+\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u}))$$[1-\theta G’(\theta)H(\nabla \mathrm{u})]\theta_{t}-\triangle\theta=\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})+\nu|\nabla \mathrm{u}_{t}|^{2}$$\mathrm{u}=\triangle \mathrm{u}=\nabla\theta$$\mathrm{u}(0_{7}\cdot)=\mathrm{u}_{0}$where$\Omega\subseteq \mathbb{R}^{n}$,$\mathrm{i}\mathrm{r}\}$$\Omega_{T}$$S_{T}$$\mathrm{n}=0$$\mathrm{u}_{t}(0, \cdot)=\mathrm{u}_{1}$,in$\theta(0, \cdot)=\theta_{0}\geq 0$$\Omega$$(n=2, 3)$ is a bounded domain with a smooth boundary. Let, and is unit outward normal to, $S_{\Gamma},=[0, T)$andtemperature
absolute
vector,0
thedisplacement
denote theelastic energy density.
We use the following notation$(0, T)>\mathrm{i}\Omega$$\mathrm{x}$$\partial\Omega$$\partial\Omega$$\mathrm{n}$,$f_{j}= \frac{\partial f}{\partial x_{j}})$$\nabla \mathrm{u}=(u_{i,g})$,$\partial\Omega$,$\Omega_{T}:=$$\mathrm{u}=(u_{i})\in \mathbb{R}^{n}$$F\in \mathbb{R}$$f_{t}= \frac{\partial f}{\partial t}$

(1.4)

,

is called thle

$F_{\nabla \mathrm{u}},=( \frac{\partial F}{\partial \mathrm{t}xi,j},)$.
where
In this article, we consider the following structure of the elastic energy density:
satisfy the following conditions.
and
,
(A)
$u_{i,j}= \frac{\partial}{\partial}x_{j}\mathrm{r}u$$G(\theta)$(i)$\mathrm{H}(\mathrm{V}\mathrm{u})$$G\in \mathrm{C}^{3}(\mathbb{R}, \mathbb{R})$$\overline{H}(\nabla \mathrm{u})$is as follows:$C_{1}\theta$$G(\theta)=\{$$\varphi(\theta)$$C_{2}\theta^{f}$

if
if
if

$0\in[0, \theta_{1}]$$\theta\in[\theta_{1}, \theta_{2}]$$\theta\in[\theta_{2\}}\infty)$

,

122

where
fixed ,
function.

,are positive constants for someandandsatisfying $0<\theta_{1}<\theta_{2}<\infty$ . We extend defined on as an odd$\varphi’\leq 0$$\varphi\in \mathrm{C}^{3}(\mathbb{R}_{7}\mathbb{R})$$\theta_{1}$(ii)$\theta_{2}$$G$satisfies thatsymmetric second order tensors in$H\in \mathrm{C}^{3}(\mathbb{R}^{n^{2}}, \mathbb{R})$(iii)$\overline{H}\in \mathrm{C}^{3}(\mathbb{R}^{n^{2}}, \mathbb{R})$(iv)$H(\nabla \mathrm{u})$and$\overline{H}$$H(\nabla \mathrm{u})$satisfies that$\mathbb{R}^{d}$, where$\mathbb{R}$$\mathbb{R}^{n^{2}}$denotes the set of.

, where$C_{3}$

are some real number,(Vu) satisfy the following growth conditions:,$|H_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{1}-2}$$|\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{2}-1}$,,$|\overline{H}_{\nabla\}\mathrm{u}\nabla \mathrm{u}}(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{2}-2}$$|H_{\nabla \mathrm{u}\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{1}-3}$$|\nabla \mathrm{u}|$$\geq 0$$\overline{H}(\nabla \mathrm{u})\geq-C_{3}$$|H_{\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{1}-1}$

for large$C_{2}$$C_{1}$,,$|\overline{H}_{\nabla \mathrm{u}\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C|\nabla \mathrm{u}|^{K_{2}-3}$.

Here we note that the regularity assumption forthere exists a positive constantsuch that$H(\nabla \mathrm{u})$and$\overline{H}(\nabla \mathrm{u})$assures that$\lambda/I$$|H_{\nabla \mathrm{u}},(\nabla \mathrm{u})|+|H_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|+|H_{\nabla \mathrm{u}\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|$$+|\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})|+|\overline{H}_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|+|\overline{H}_{\nabla \mathrm{u}\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq M$.
for small
For the related results, we refer to [11] and [12]. Our main result of this paper
is as follow .
$|\nabla \mathrm{u}|$$\mathrm{s}$Theorem 1.1. (i) Let $5<p\leq q<\infty$ . The exponents r,
following conditions$0 \leq r<\frac{5}{6}$Then, for any $T>0$ andexists at least one solution$(\mathrm{u},$$\ )$Moreover, if we assume,$0\leq K_{1}$, $K_{2}<6$ ,$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in B_{p,p}^{4-2/p}\mathrm{x}$$(\mathrm{u}, \theta)$$K_{1}$

and$K_{2}$

satisfy the$6r+K_{1}<6$ .

$B_{p,p}^{2-2/p}\mathrm{x}$$B_{q,q}^{2-2/q}=:U(p, q)$$(1.\mathrm{t}1)r$, thereto (1.1)-(1.4) satisfying$\in W_{p}^{4,2}(\Omega_{T})\mathrm{x}$$W_{q}^{2,1}(\Omega_{\lrcorner}\tau)=:V_{T}(p, q)$$\min_{\Omega}\theta_{0}=\theta_{*}>0$.

then there exists a positive constant$\omega$such that$\theta\geq\theta_{*}$$\exp(-\omega t)$in$\Omega_{T}$.

(ii) Let $4<p\leq q<\infty$ and assume thai$0\leq r<1$ ,$0\leq K_{17}K_{2}<\infty$

.

(1.6)

Then for the two-dimensional system $(TE)_{2}$ the same conclusion as in (i) holds

123Here, we have used and will be used the following function spaces.
$\bullet$$\bullet$is the standard Lebesgue space. We often
for some interval .

$\mathrm{L}\mathrm{P}(\mathrm{Q}\mathrm{T})=LVXIP$$=L^{p}(0, T;L^{\mathrm{p}}(\Omega))$use the notation$L^{p}(\Omega_{I})=L_{I}^{p}L^{p}$$W_{p}^{2l,l}(\Omega_{T})$$I$

is the Sobolev space equipped with the norm$||u||_{W_{p}^{2l,l}\{\Omega_{T})}:= \sum_{j=0}^{2l}\sum_{2r+|\alpha|=j}||D_{t}^{r}D_{x}^{\alpha}u||_{L^{\mathrm{p}}(\Omega_{T})}$

where$\bullet$$H^{i}$(St)$D_{t}:= \mathrm{i}\frac{\partial}{\partial t}$,$D_{x}^{\alpha}= \prod D_{k}^{\alpha_{k}}\alpha=\alpha_{1}+\alpha_{2}+\alpha_{3}$$:=W_{2}^{j}(\Omega)$, where$D_{k}.-- \mathrm{i}\frac{\partial}{\partial x_{k}}$for multi index$\alpha=(\alpha_{i})_{i=1}^{n}$.

is the Sobolev space equipped with the norm

$W_{p}^{j}$$||u||_{W_{p}^{\mathrm{j}}(\Omega)}:= \sum_{|\alpha|\leq j}||D_{x}^{\alpha}u||_{L^{p}(\Omega)}$$\bullet$

and

,

.

, whereis the Besov space. Namely,is the real interpolation space. For more details we refer to [1] by
Adams and Fournier.
$B_{p,q}^{s}:=[L^{p}(\Omega), W_{p}^{J}(\Omega)]_{s/j,q}$$B_{p,q}^{\mathit{8}}=B_{p,q}^{s}(\Omega)$$[X, 1^{r}]_{s/j,q}$$\bullet$is the Holder space; the set of all continuous functions in
isfying Holder condition in with exponent a and in with exponent$\mathrm{C}^{\alpha,\alpha/2}(\Omega_{T})$

$t$

$x$

$\Omega_{T}$sat-$\alpha/2$.

For completeness we recall also the uniqueness result which follows by repeating
the arguments of the corresponding result in [9, Section 6]
Theorem 1.2. In addition to assumptions of Theorem 1.1, suppose thatto (1.1)-(1.4) constructed above. Then the solution
is unique.
$F(\nabla \mathrm{u}, \theta)\in$$\mathrm{C}^{4}(\mathbb{R}^{n^{2}}\mathrm{x} \mathbb{R}^{+}, \mathbb{R})$$(\mathrm{u}, \theta)\in V_{T}(p, q)$

We prove Theorem 1.1 by using the Leray-Schauder fixed point principle. The
key estimates are the maximal regularity estimate for (1.1), and the classical energy
estimate and the parabolic De Giorgi method for (1.2). In general, the derivativeof a solution is less regular than the right-hand side of the corresponding equation.
However, for parabolic equations such a loss of regularity does not occur, as in thecase of elliptic equations. The estimate ensuring this regularity is called the maximal
regularity. For more precise information on the maximal regularity, we refer to [2]we refer to [4]. Since theand for more recent topics of the maximal
maximal regularity theory is limited to linear parabolic equations, we cannot use
it directly for the quasilinear equation (1.2). To obtain the higher order a priori
estimates we also use the classical energy methods and the parabolic De Giorgi
method (see [6], [7]). Using these methods we can show the Holder continuity of .
By virtue of such regularity, we arrive at the estimate in higher Sobolev norm$L^{p}\mathrm{I}\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{g}\mathrm{u}1\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}$$\theta$

124

Throughout this paper and A are positive constants independent of timeand depending on time , respectively. In particular, we may use A instead offor some $X$ if there is no danger of confusion.
$C$

$T$

$T$

$\Lambda(||(u_{0}, u_{1}, \theta_{0})||_{X})$Remark. We can obtain the same result for the system replacing A in (1.1) withdefined by(A $+\mu$ ),$Q$$Q\mathrm{u}=\mu\triangle \mathrm{u}+$$\nabla(\nabla\cdot \mathrm{u})$where correspondingly we have to replace Vu on the system with the shear strain(Vu
tensor) $/2$ (see [12]$\epsilon=$

2$+^{T}\nabla \mathrm{u}$

$\}$

,

Preliminaries

In this section, we present some auxiliary results which will be used in the subsequent
sections.
(i) Let $p\in(1, \infty)$ . Denote by

Lemma 2,1 (Maximal Regularity).
tion of the linear problem$\mathrm{u}_{tt}+\triangle^{2}\mathrm{u}-\nu\triangle \mathrm{u}_{t}=\nabla$$\{$ $\mathrm{u}=$

.

in

$f$

Au $=0$$\mathrm{u}(0, \cdot)=u_{0}$,on$\Omega_{T}$,$S_{T}$,$\mathrm{u}$the solu-

in 0.

$\mathrm{u}_{t}(0, \cdot)=u_{1}$Then the following estimates hold$||\mathrm{u}||_{W_{p\acute{\backslash }}^{4,2}\Omega_{T})}\leq C(’||\mathrm{u}_{0}||\mathrm{z}+||\mathrm{u}_{1}||?B_{p,p^{\mathrm{P}}}^{4-}B_{P,\nu^{p}}^{2-=}+||\nabla\cdot f||_{L?(\Omega_{T})})$

for any$(\mathrm{u}_{0}, \mathrm{u}_{1})$$\in B_{p,p}^{4-2/p}\cross$$B_{p,p}^{2-2/p}$

and$\nabla\cdot f\in L^{p}(\Omega_{T})$, and$||\nabla \mathrm{u}||_{W_{p}^{2,1}(\Omega_{T})}\leq C(||\mathrm{u}_{0}||B_{p,\rho}^{3-\frac{2}{p}}+||\mathrm{u}_{1}||B_{p,\mathrm{J}^{\mathrm{J}}}^{1-2}\mathrm{p}-^{\mathrm{I}}\ulcorner||f||_{L^{p}(\Omega_{T})})$

for any$(\mathrm{u}_{0}, \mathrm{u}_{1})$$\in B_{p,p}^{3-2/p}\mathrm{x}$$B_{p,p}^{1-2/\mathrm{p}}$and$f\in L^{p}(\Omega_{T})$$\theta_{t}-p\triangle\theta=g$$\{$ $n\cdot\nabla\theta=0$$\theta(0, x)=\theta_{0}(x)$in$\Omega_{T}$onin$S_{T}$$\Omega$(2.2).

(i) Let $q\in(1, \infty)$ . Assume thatis Holder continuous in
0. Denote by 0 the solution of the linear problem$\rho(x)$(2.1)$\overline{\Omega}$such that info$\rho>$,,.

Then the following estimate holds$||\theta||_{W_{\mathrm{q}}^{2,\mathrm{I}}(\Omega_{T})}\leq C(||\theta_{0}||\mathrm{z}B_{q.q}^{2-_{q}}+||g||_{L^{q}(\Omega)})$for any$\theta_{0}\in B_{q,q}^{2-2/q}$, where $C$ depends on info .
$p$

(2.3)

125
For the proof of (i) we refer to [10, Lemma 2.1, Proposition 2.4], and (ii) is theparticular case of [5, 3.2 Examples, 2)]. Next, we recall the useful space-timeembedding lemma.
$\mathrm{A})$

Lemma 2.2 (Embedding [6, Lemma II.3.3]). Let f
and multi index , it follows that$\in W_{p}^{2l,l}(\Omega_{T})$. Then, for$\mathit{1}\in \mathbb{Z}^{+}$$\alpha$,(2.4)$||D_{t}^{r}D_{x}^{\alpha}f||_{L^{q}(\Omega_{T})}\leq C\delta^{l-\psi}||f||_{W_{p}^{2l,l}(\Omega_{T})}+C\delta^{-\psi}||f||_{L^{p}(\Omega_{T})}$

provided $q\geq p$ and $\psi:=r$$+ \frac{|\alpha|}{2}+\frac{n+2}{2}(\frac{1}{p}-\frac{1}{q})\leq l$. $lf$ $\varphi:=r+\frac{|\alpha|}{2}+\frac{n+2}{2p}<l$ , then

,

(2.5}

$||D_{t}^{r}D_{x}^{\alpha}f||_{L^{\infty}(\Omega_{T})}\leq C\delta^{l-\varphi}||f||_{W_{\mathrm{p}}^{2t,1}(\Omega_{T})}+C\delta^{-\varphi}||f||_{L^{p}\langle\Omega_{T})}$

moreover,is H\"older continuous. Here, $\in(0,$
the
cone
in
the
statementofof the cone condition satisfied by$\delta$$D_{t}^{r}D_{x}^{\alpha}f$Lemma 2.3. Let

be given in$\varphi$$(A)-(\mathrm{i})$$\min(\mathrm{T}, \zeta^{2})]$. Then the function$\Omega$$\varphi(s)$,$\langle$is the altitude.

satisfies(2.6)$\varphi(s)-s\varphi’(s)\geq 0$

for any
Proof Putting $f(s)=\varphi(s)-s\varphi’(s)$ , we have $f’(s)=-s\varphi’(s)\geq 0$ and$s\in[\theta_{1}, \theta_{2}]$Then$f(s)=\varphi(s)-s\varphi’(s)\geq 0$in$[\theta_{1}, \theta_{2}]$$f(\theta_{1})=0$.

.

$\square$

To show Theorem 1.1 we apply the Leray-Schauder fixed point principle. We
recall it here in one of its equivalent formulations for the reader ’s convenience.
Theorem 2.4 (Leray-Schauder Fixed Point Principle [3]). Let X be a Banach
space. Assume that : [0, 1] x X $arrow X$ is a map with the following properties.
$\Phi$(L1) For any fixed $\tau\in[0,1]$ the map$\Phi(\tau$,$\cdot$$)$: $Xarrow X$ is compact.

(L2) For every bounded subset
of $X$ , the family of maps
is uniformly equicontinuous.
$B$$\Phi(\cdot, \xi)$: $[0, 1]arrow X$ ,$\xi\in B_{f}$(L3)$\Phi(0$,$\cdot$$)$has precisely one fixed point in $X$ .

(14) There is a bounded subset of $X$ such that any fix $ed$ point in $X$ ofcontained in for every $0\leq\tau\leq 1$ .
$B$

$B$

Then

$\Phi$(1, ) has at least one fixed point in $X$ .
$\cdot$$\Phi(\tau$,$\cdot$

$)$

is

12

$\mathrm{e}$3Proof of Theorem 1.1 (Existence)

We only prove the existence theorem in three-dimensional case. We apply Theorem
from $V_{T}(p, q)$ into $V_{T}(p, q)$ ,2.4 to the map$\Phi_{\tau}$$\Phi_{\tau}$:$(\overline{\mathrm{u}},\overline{\theta})\prec(\mathrm{u}, \theta)$,$\tau\in[0, 1]$,

defined by means of the following initial-boundary value problems:$\mathrm{u}_{tt}+\triangle^{2}\mathrm{u}-\nu\triangle \mathrm{u}_{t}=\tau\nabla\cdot[G(\overline{\theta})H_{\nabla \mathrm{u}},(\overline{\nabla}\mathrm{u})+\overline{H}_{\nabla \mathrm{u}},(\overline{\nabla}\mathrm{u})]${$\theta_{t}-\triangle\theta=\tau$$\theta-G’(\overline{\theta})\theta_{t}H$

(Vu)

,

$+\overline{\theta}G’(\overline{\theta})\partial_{t}H(\nabla \mathrm{u})+\nu|\nabla \mathrm{u}_{t}|^{2}$$\mathrm{u}=\triangle \mathrm{u}=\nabla\theta\cdot \mathrm{n}=0$$\mathrm{u}(0, \cdot)=\tau \mathrm{u}_{0}$,$\mathrm{u}_{t}(0, \cdot)=\tau \mathrm{u}_{1}(x)$

,

}

in

$\Omega_{T}$on$S_{T}$

in$\theta(0, \cdot)=\tau\theta_{0}$$\Omega$,,.

in $V_{T}(p, q)$ is the desired solution of the system $(TE)_{3}$ .
A fixed point of
Therefore to prove the existence statement it is sufficient to check that the mapsatisfies assumptions $(L1)-(L4)$ of Theorem 2,4. We can check assumptions (LI),(L2) and (L3) in the same way as that in [8, Section 3]. Then it is sufficient to, to derive a priori bounds for a fixed point of thecheck the assumption (L4), nam. Without loss of generality we may set $\tau=1$ . Hence from now onsolution map
our purpose is to obtain a priori bounds for $(TE)_{3}$ . To this end we prepare several
lemmas.
$\Phi_{\tau}(1, \cdot)$$\Phi_{\mathcal{T}}$$\mathrm{e}\mathrm{l}\mathrm{y}$$\Phi_{\tau}$a.e. in ,Lemma 3.1 (Energy Conservation Law). Assume thatand $6r+K_{1}\leq 6$ . Then for any t $\in[0,$T] a smooth solution of (1.1)-(1.4) satisfies$\theta\geq 0$$\Omega_{T}$$||\theta(t)||_{L^{1}(\Omega)}+||\mathrm{u}_{t}(t)||_{L^{9}\{\Omega)}\sim+||\triangle \mathrm{u}(t)||_{L^{2}(\Omega)}\leq C(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{H^{2}\mathrm{x}L^{2}\mathrm{x}L^{1}}|)$

Proof Multiplying (1.1) by$\mathrm{u}_{t}$to the space variable, we have.

$K_{2}\leq 6$(3.1)

and integrating the resulting equation with respect$\frac{d}{dt}(\frac{1}{2}||\mathrm{u}_{t}||_{L^{2}}^{2}+\frac{1}{2}||\triangle \mathrm{u}||_{L^{2}}^{2}+\int_{\Omega}\overline{H}(\nabla \mathrm{u})dx)+\nu\int_{\Omega}|\nabla \mathrm{u}_{t}|^{2}dx+\int_{\Omega}G(\theta)\partial_{t}H(\nabla \mathrm{u})dx=0$

Integrating (1.2) over$\Omega$, we obtain.

$\frac{d}{dt}\int_{\Omega}\theta dx=\iota/\int_{\Omega}|\nabla \mathrm{u}_{t}|^{2}dx+\int_{\Omega}\theta G’(\theta)\frac{\partial}{\partial t}H(\nabla \mathrm{u})dx+\oint_{\Omega}\theta G’(\theta)\theta_{t}H(\nabla \mathrm{u})dx$

Combining these equalities, we deduce$\frac{d}{dt}(\frac{1}{2}||\mathrm{u}_{t}||_{L^{2}}^{2}+\frac{1}{2}||\triangle \mathrm{u}||_{L^{2}}^{2}+\int_{\Omega}\theta dx+\int_{\Omega}\overline{H}(\nabla \mathrm{u})dx)$$= \int_{\Omega}(\theta G’(\theta)\frac{\partial}{\partial t}H(\nabla \mathrm{u})+\theta G’(\theta)\theta_{t}H(\nabla \mathrm{u})-G(\theta)\frac{\partial}{\partial t}H(\nabla \mathrm{u}))dx$$=- \frac{d}{dt}\int_{\Omega}\overline{G}(\theta)H(\nabla \mathrm{u})dx$

,

.

127

where
$\frac{d}{d\mathrm{f}}$$\overline{G}(\theta)=G(\theta)-\theta G’(\theta)$(. Consequently, we have$\frac{1}{2}\mathrm{I}\mathrm{u}_{t}||_{L^{2}}^{2}+\frac{1}{2}||\triangle \mathrm{u}||_{L^{2}}^{2}+\int_{\Omega}\theta dx+\int_{\Omega}\overline{H}(\nabla \mathrm{u})dx+\oint_{\Omega}\overline{G}(\theta)H(\nabla \mathrm{u})dx)=0$

Here we recall thatis as follows:$\theta\geq 0$

and$H(\nabla \mathrm{u})$$\geq 0$. By the structure of$G(\theta)$.

the function$\overline{G}(\theta)$0$\overline{G}(r)=\{$ $\varphi(\theta)-\theta\varphi’(\theta)$$C_{2}(1-r)\theta^{r}$

Since from Lemma 2.3 we havethat$\overline{G}(\theta)\geq 0$

if
if
if

$\theta\in[0, \theta_{1}]$,$?\in[\theta_{1}, \theta_{2}]$,$\mathit{0}\in[\theta_{2}, \infty)$.

. Consequently, it follows from$(\mathrm{A})-(\mathrm{i}\mathrm{i}\mathrm{i})$$\frac{1}{2}||\mathrm{u}_{t}(t)||_{L^{2}}^{2}+\frac{1}{2}||\mathrm{u}(t)||_{H^{2}}^{2}+||\theta(t)||_{L^{1}}\leq\frac{1}{2}||\mathrm{u}_{0}||_{H^{2}}^{2}+\frac{1}{2}||\mathrm{u}_{1}||_{L^{2}}^{2}+||\theta_{0}||_{L^{1}}+C_{3}|\Omega|$$+ \int_{\Omega}|\overline{H}(\nabla \mathrm{u}_{0})|dx$$+$.

$l_{\{\theta_{2}\geq\theta_{0}\geq\theta_{1}\}\cap\Omega}^{[\varphi(\theta_{0})-\theta_{0}\varphi’(\theta_{0})]H(\nabla \mathrm{u}_{0})dx+C_{2}(1-r)l_{\{\theta 0>\theta_{2}\}\cap\Omega}^{\theta_{0}^{r}H(\nabla \mathrm{u}_{0})dx}}$

Since the smooth function$\varphi(s)$is bounded for$– s\varphi’(s)$$s\in[\theta_{1}, \theta_{2}]$, we have$\int_{\{\theta_{2}\geq\theta_{0}\geq\theta_{1}\}\cap\Omega}[\varphi(\theta_{0})-\theta_{0}\varphi’(\theta_{0})]H(\nabla \mathrm{u}_{0})dx\leq CI_{\Omega}|\nabla \mathrm{u}_{0}|^{K_{1}}dx$$\leq C||\mathrm{u}_{0}||_{H^{2}}^{K_{1}}$for$K_{1}\leq 6$,$\int_{\{\theta_{0}>\theta_{2}\}\cap\Omega}\theta_{0}^{r}H(\nabla \mathrm{u}_{0})dx\leq C||\theta_{0}||_{L^{1}}^{r}||\nabla \mathrm{u}_{0}||$$L^{\Gamma-r}K_{1K,[perp]}$$\leq C||\theta_{0}||_{L^{1}}^{r}||\mathrm{u}_{0}||_{H^{2}}^{K_{1}}$for $6r+K_{1}\leq 6$ and$\int_{\Omega}|\overline{H}(\nabla \mathrm{u}_{0})|dx\leq||\mathrm{u}_{0}||_{H^{2}}^{K_{2}}$for$K_{2}\leq 6$. Hence we conclude the assertion.

Lemma 3.2. Assume that$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in B_{16/5,16/5}^{19/8}\mathrm{x}$$s$$\theta\geq 0\mathrm{a}.\mathrm{e}$.

in$\square$and (1.5) holds.
$tte$
solution,$\Omega_{T}$$B_{16/5,16/5}^{3/8}\mathrm{x}L^{2}=:U_{3}$$(\mathrm{u}, \theta)$

atisfies$||\nabla \mathrm{u}||_{W_{16/5}^{2,1}(\Omega_{T})}+||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L^{2}}\leq\Lambda$

where A depends on

$T$

and

$||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}}$,Then for any
to (1.1)-(1.4)(3.2). Moreover eve have$||\nabla \mathrm{u}||_{L^{\infty}(\Omega_{T}\rangle}+||\theta||_{L^{10/3}}(\Omega_{T})\leq$

A.

(3.3)

128

Proof. Remark that$||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{H^{2}\mathrm{x}L^{2}\mathrm{x}L^{1}}\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1_{7}}\theta_{0})||_{U_{3}}$(see [1]). From the

Gagliardo-Nirenberg inequality and Lemma 3.1 it follows that
$||$

Vu

$||_{L^{6\mathrm{p}}\langle\Omega_{T})}\leq C||||$

Vu$||_{6}^{\frac{4}{L5}}(\Omega)||$Vu$||_{(p^{2}\Omega)}^{\frac{1}{W5}}||_{L_{T}^{5p}}\leq C||$Vu

( 3.4 )$||_{W_{p}^{2,1}(\Omega_{T}\rangle}^{\frac{1}{5}}$and$||\theta||_{L^{8/3}}\langle\Omega_{T}$

)

(3.5)

$\leq\leq\leq C||||\theta||_{1}^{\frac{1}{L4}}(\Omega)||\theta||_{1}^{\frac{3}{H4}}(\Omega)||_{L_{T}^{\infty}}C||\theta||_{\infty}^{\frac{1}{L4}}||\theta||_{2}^{\frac{3}{L4}}\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L^{2}})^{\frac{3}{4}}\tau^{L_{\tau^{H^{1}}}^{1}}$.
It follows from (3.4) that$|| \overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{16/5}}(\Omega_{T})\leq\Lambda||\nabla \mathrm{u}||_{L^{16}(\Omega_{T})}^{K_{2}-1}\leq\Lambda||\nabla \mathrm{u}||_{W_{16}^{2,1}(\Omega_{T})}^{5}\underline{K}_{arrow-}1\leq\frac{1}{4}||\nabla \mathrm{u}||_{W_{16}^{2_{1}1}(\Omega_{T})}+\mathrm{A}$for $K_{2}\in[1, 6)$ , and$||\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{16/5}}(\Omega_{T})\leq M|\Omega_{T}|^{\frac{5}{16}}\leq\Lambda$

for $K_{2}\in[0,1)$ .
We first consider the case of $K_{1}\geq 1$ . Applying the growth condition and theYoung inequality, we have$||G(\theta)H$, Vu$(\nabla \mathrm{u})||_{L^{1\underline{6}}(\Omega_{T})}\mathrm{B}\leq||\theta||^{r}\S(\Omega_{T})||\nabla \mathrm{u}||_{16(K_{1}-1)}^{K_{1}-1}L$$L\overline{5-6r}(\Omega_{T})$$+ \sup|G(\theta)|||\nabla \mathrm{u}||^{K}\theta\in[0,\theta_{2}]L^{\frac{16(K_{1}-111-1}{5}}(\Omega_{T})$$\leq \mathrm{A}||\theta||_{8}^{r}||\nabla \mathrm{u}||_{L^{16}(\Omega_{T})}^{K_{L}-1}’+\Lambda||\nabla \mathrm{u}||_{L^{16}(\Omega_{T})}^{K_{1}-1}L\mathrm{f}\mathrm{f}(\Omega_{T})$for $6r+K_{1}\leq 6$ (and$K_{1}\leq 6$). Then we have$||\theta||_{L^{8/3}}^{r}(\Omega_{T})||\nabla \mathrm{u}||_{L^{16}\langle\Omega_{T})}^{K_{1}-1}+||\nabla \mathrm{u}||_{L^{16}(\Omega_{T}\rangle}^{K_{1}-1}$$\leq$A$(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{3r/4}||\nabla \mathrm{u}||_{\mathrm{W}_{16/5}^{\gamma^{21}}(\Omega_{T})}^{\langle K_{1}-1)/5}‘+\Lambda||\nabla \mathrm{u}||_{W_{16/5}^{2,1}\{\Omega_{T})}^{(K_{1}-1)/5}$$\leq\frac{1}{4}||\nabla \mathrm{u}||_{W_{16/5}^{2.1}(\Omega_{T})}+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{15r}{4(6-K_{1})}}+$

A

for $6r$ % $K_{1}<6$ (and $K_{1}<6$ ). From the maximal regularity (2.2) it follows that$||\nabla \mathrm{u}||_{W_{16/5}^{2,1}(\Omega_{T})}\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}}+C||G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{16/5}}(\Omega_{T})$(3.6)$+C||\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{16/5}}(\Omega_{T})$$\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}}+\Lambda+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{15r}{4(6-K_{1})}}$

.

129

Next, multiplying (1.2) by 0 and integrating over$\Omega$, we have$\frac{1}{2}\frac{d}{dt}||\theta(t)||_{L^{2}}^{2}+||\nabla\theta||_{L^{2}}^{2}$$= \int_{\Omega}\theta^{2}G’(\theta)\theta_{t}H(\nabla \mathrm{u})dx+\oint_{\Omega}\theta^{2}G’(\theta)\partial_{t}H(\nabla \mathrm{u})dx+\nu$$\oint_{\Omega}\theta|\nabla \mathrm{u}_{t}|^{2}dx$(3.7)$= \int_{\Omega}G_{2}’(\theta)\theta_{t}H(\nabla \mathrm{u})dx+\int_{\Omega}G_{2}(\theta)\partial_{t}H(\nabla \mathrm{u})dx$$+2 \int_{\Omega}\overline{G}_{2}(\theta)\partial_{t}H(\nabla \mathrm{u})dx+\nu$$\int_{\Omega}\theta|\nabla \mathrm{u}_{t}|^{2}dx$,$= \frac{d}{dt}\int_{\Omega}G_{2}(\theta)H(\nabla \mathrm{u})dx+2\oint_{\Omega}\overline{G}_{2}(\theta)\partial_{t}H(\nabla \mathrm{u})dx+\iota/\oint_{\Omega}\theta|\nabla \mathrm{u}_{t}|^{2}dx$

where$G_{2}(\theta)=\theta^{2}G’(\theta)-\overline{G}_{2}(\theta)$$G_{2}(\theta)$

and$\overline{G}_{2}(\theta)=2\int_{0}^{\theta}sG’(s)ds$

and$= \frac{C_{2}r(r-1)}{r+1}\theta^{r+1}\leq 0$. Noting that$\overline{G}_{2}(\theta)=\frac{2C_{2}r}{r+1}\theta^{r+1}$

for$\theta\geq\theta_{2}$,and$\sup_{\theta\in[0,\theta_{2}]}|G_{2}(\theta)|+\sup_{\theta\in[0,\theta_{2}]}|\overline{G}_{2}(\theta)|=:M<\infty$,

we have$-\mathrm{J}$$G_{2}(\theta)H(\nabla \mathrm{u})dX=-I_{\Omega\cap\{\theta\geq\theta_{2}\}}^{G_{2}(\theta)H(\nabla \mathrm{u})dx-l_{\Omega\cap\{\theta_{1}\leq\theta\leq\theta_{2}\}}^{G_{2}(\theta)H(\nabla \mathrm{u})dx}}$$\geq-M\int_{\Omega}|H(\nabla \mathrm{u})|dx$.

Hence integrating (3.7) with respect to time variable, we obtain$\frac{1}{2}||\theta||_{L_{T}^{\infty}L^{2}}^{2}+||\nabla\theta||_{L^{2}(\Omega_{T}\}}^{2}\leq\frac{1}{2}||\theta_{0}||_{L^{2}}^{2}+||\overline{G}_{2}(\theta)\partial_{l}H(\nabla \mathrm{u})||_{L^{1}(\Omega_{T})}+\nu||\theta|\nabla \mathrm{u}_{t}|^{2}||_{L^{1}(\Omega_{T})}$.

$+M \sup_{t\in[0,T]}\oint_{\Omega}|H(\nabla \mathrm{u}(t))|dx+\oint_{\Omega}|G_{2}(\theta_{0})H(\nabla \mathrm{u}_{0})|dx$

By (3.4), (3.5) and the assumptions we have$||\theta^{r+1}\partial_{t}H$(Vu)$||_{L^{1}(\Omega_{T})}\leq$A$||\theta||_{L^{8/3}}^{r+1}\mathfrak{l}^{\Omega_{T})}||\mathrm{u}||_{W_{16/5}^{2,1}(\Omega_{T})}||\nabla \mathrm{u}||_{L^{16}(\Omega_{T})}^{K_{1}-1}$,$\leq\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3(r+1\mathrm{J}}{4}}||\mathrm{u}||_{W_{16/}^{2,1}\tau)}^{1+^{\underline{K}}\frac{-1}{55\zeta\Omega}}$$||\theta|\nabla \mathrm{u}_{t}|^{2}||_{L^{1}(\Omega_{T})}\leq C||\theta||_{8}||\nabla \mathrm{u}_{t}||_{16}^{2}L\mathrm{F}(\Omega_{T})L\mathrm{Y}(\Omega_{T})$$\leq\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3}{4}}||\nabla \mathrm{u}_{t}||_{\frac{216}{5}(\Omega_{T})}$

,

130

$\oint_{\Omega}|H(\nabla \mathrm{u}(t))|dx\leq C||\mathrm{u}(t)||_{H^{2}}^{K_{1}}\leq$Aand$||\theta_{0}^{r+1}H(\nabla \mathrm{u}_{0})||_{L^{1}(\Omega)}\leq C||\theta_{0}||_{L^{2}(\Omega)}^{r+1}||\nabla \mathrm{u}_{0}||$,$L^{\mathrm{T}\vec{-\tau}}(\Omega)K_{1,2K}$$\leq C||\theta_{0}||_{L^{2}(\Omega)}^{r+1}||\mathrm{u}_{0}||_{H^{2}\{\Omega)}^{K_{1}}$.

Consequently we arrive at$||\theta||_{L_{\mathcal{T}}^{\infty}L^{2}}^{2}+||\nabla\theta||_{L^{2}(\Omega_{T})}^{2}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}})$(3.8)$+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3[r+1]}{4}}||\nabla \mathrm{u}||^{\frac{4}{W5}+}\iota\epsilon/’\acute{o}2,1K\vec{5}(\Omega_{T})$$+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3}{4}}||\nabla \mathrm{u}_{t}||^{2}16$

LB$(\Omega_{T})$Substituting (3.6) into (3.8), we have$||\theta||_{L_{T}^{\infty}L^{2}}^{2}+||\nabla\theta||_{L^{2}(\Omega_{T})}^{2}\leq\Lambda(||(\mathrm{u}_{0},\mathrm{u}_{1},\theta_{0})||_{U_{3}})$$+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3(r+1)}{4}}(||(\mathrm{u}_{0},\mathrm{u}_{1},\theta_{0})||_{U_{3}}+||\nabla\theta||^{\frac{16r}{L^{2}(\Omega_{T})4(6-K_{1})}})^{\frac{4}{\mathrm{s}}+_{\vec{6}}^{K}}$$+\Lambda(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L_{2}})^{\frac{3}{4}}(||(\mathrm{u}_{0_{i}}\mathrm{u}_{1},\theta_{0})||_{U_{3}}+||\nabla\theta||^{\frac{15r}{L^{2}(\Omega_{T})4(6-K_{1})}})^{2}$

Here from the assumption $6r+K_{1}<6$ it follows that$\frac{3(r+1)}{4}+\frac{15r}{4(6-K_{1})}(\frac{4}{5}+\frac{q}{5})=\frac{30r+3(6-K_{1})}{4(6-K_{1})}<2$$\frac{3}{4}+\frac{30r}{4(6-K_{1})}<2$,.

Thus we obtain$||\theta||_{L_{T}^{\infty}L^{2}}+||\nabla\theta||_{L^{2}(\Omega_{T})}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}})+\Lambda||\nabla\theta||_{L^{2}(\Omega_{T})}^{1-}$.

Here we use $p-$ to denote a number less than . Hence by the Young inequality we
have
.
$p$

$|| \theta||_{L_{T}^{\infty}L^{2}}+\frac{1}{2}||\theta||_{L^{2}(\Omega_{T})}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}})$

Substituting the above inequality into (3.6), we also obtain the following$||\nabla \mathrm{u}||_{W_{16/6}^{2.1}(\Omega_{T})}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{3}})$.

Next, we consider the case of $0\leq K_{1}\leq 1$ and $0\leq r<5/6$ . In this case it followsthat$|H_{\nabla \mathrm{u}},(\nabla \mathrm{u})|\leq C<\infty$

.

131

Prom an argument similar to the above we have$||\nabla \mathrm{u}||_{W_{16/6}^{2,1}(\Omega_{T})}\leq||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U_{3}}+||G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{16/5}}(\Omega_{T})$(3.3)$\leq||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U_{3}}+C||\theta||^{r}L^{4}\mathrm{F}_{(\Omega_{T})}+C\sup G(\theta)$$\theta\in\zeta 0,\theta_{2}]$$\leq||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U_{3}}+\Lambda||\theta||^{r}\S(\Omega_{T})+CL^{\cdot}$

Noting that,$||\theta^{r+1}\partial_{t}H(\nabla \mathrm{u})||_{L^{1}(\Omega_{T}\}}\leq\Lambda||\theta||_{L^{8/3}(\Omega_{T})}^{r+1}||\mathrm{u}||_{W_{16/5}^{2,1}(\Omega_{T}\rangle}$

we obtain(0r)$||\theta||_{L_{T}^{\infty}L^{2}}^{2}+||\nabla\theta||_{L^{2}(\Omega_{T})}^{2}\leq||\theta_{0}||_{L^{2}}^{2}+||\theta^{r+1}\partial_{t}H(\nabla \mathrm{u})||_{L^{1}(\Omega_{T})}+||\theta|\nabla \mathrm{u}_{t}|^{2}||_{L^{1}}$$+M \sup_{t\in[0,T]}\oint_{\Omega}|H(\nabla \mathrm{u}(t))|dx+\int_{\Omega}|G_{2}(\theta_{0})H(\nabla \mathrm{u}_{0})|dx$$\leq$A$\leq\Lambda$$(||(\mathrm{u}_{0}, \mathrm{u}_{1},\theta_{0})||_{U_{3}})+\Lambda||\theta||_{L^{8/3}}^{r+1}(\Omega_{T})||\mathrm{u}||_{W_{16/6}^{2,1}}(\Omega_{T})+C||\theta||_{L^{8/3}}\mathfrak{l}^{\Omega_{T})}||\mathrm{u}||_{W_{16/5}^{2,1}(\Omega_{T})}^{2}$

( (
$||$

$\mathrm{u}_{0}$,$\mathrm{u}_{1}$,$\theta_{0}$) C3 )

$+$

$||$

A

$(||\nabla\theta||_{L^{2}(\Omega_{T})}+||\theta||_{L_{T}^{\infty}L^{2}})^{3(2r+1)/4}$.

Since $3(2r+1)/4<2$ , we obtain the desired estimate (3.2).
The estimate (3.3) follows with the help of the embeddings
$||$

Vu

$||_{L^{\infty}(\Omega_{T})}\leq$A$||\nabla \mathrm{u}||_{W_{16/6}^{2,1}(\Omega_{T})}$and of the inequality.

$||\theta||_{L^{10/3}}(\Omega_{T}\}\leq C||||\theta||_{L^{2}(\Omega)}^{2/5}||\theta||_{H^{1}(\Omega)}^{3/5}||_{L_{T}^{10/3}}\leq C||\theta||_{L_{T}^{\infty}L^{2}}^{2/5}||\theta||_{L^{2}H^{1}}^{3/5}$

This completes the proof.

$\square$and (1.5) holds.
Lemma 3.3. Assume that $\theta\geq 0a.e$ . in$H^{1}=U_{4}$ the following estimate holds$\Omega_{T}$$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in B_{4,4}^{6/2}\cross$$B_{4,4}^{1/2}\mathrm{x}$$||\nabla \mathrm{u}||_{W_{4}^{2_{\backslash }1}(\Omega_{T})}+||\nabla\theta||_{L_{T}^{\infty}L^{2}}+||\theta||_{W_{2}^{2,1}(\Omega_{T})}\leq\Lambda$

above constant A depends on

$T$

and

$||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{4}}$$U_{4}\prec$$U_{3}$,. Moreover, we have$||\nabla\theta||_{L^{10/3}}(\Omega_{T})+||\theta||_{L^{10}(\Omega_{T})}+||\triangle \mathrm{u}||_{L^{20}(\Omega_{T}\rangle}\leq$

Proof Remark that

Then for any

A.

. Using (3.3) we have$\Lambda||\theta||_{L^{10/3}}^{r}(\Omega_{T})||\nabla \mathrm{u}||_{L^{\infty}(\Omega_{T}\rangle}^{K_{1}-1}\leq$$||G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{4}(\Omega_{T})}\leq\{$$\Lambda\sup|H_{\nabla \mathrm{u}},|||\theta||_{L^{10/3}}^{r}(\Omega_{T})\leq\Lambda$

A

if
if

$K_{1}\geq 1$,$K_{1}\leq 1$,(3.10)

132
for $r\leq 5/6$ . Then from the maximal regularity (2.2) it follows that.

(3.11)$||\nabla \mathrm{u}||_{W_{4}^{2,1}}\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U_{4}}+C||G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{4}}\leq\Lambda$

Multiplying (1.2) by$\theta_{t}$and integrating over$\Omega_{T}$, we get$|| \theta_{t}||_{L^{2}(\Omega_{T})}^{2}+\frac{1}{2}||\nabla\theta||_{L_{T}^{\infty}L^{2}}^{2}\leq\frac{1}{2}||\theta_{0}||_{H^{1}}^{2}+J\int_{\Omega_{T}}\theta_{t}^{2}\theta G’(\theta)H(\nabla \mathrm{u})dxdt$$+f \int_{\Omega_{T}}\theta_{t}\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})dxdt+J\oint_{\Omega_{T}}\theta_{t}|\nabla \mathrm{u}_{t}|^{2}dxd\mathrm{f}$$\leq\frac{1}{2}||\theta_{0}||_{H^{1}}^{2}+C||\theta_{t}||_{L^{2}(\Omega_{T})}||\theta^{r}H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{4}}||\nabla \mathrm{u}_{t}||_{L^{4}}+C||\theta_{t}||_{L^{2}}||\nabla \mathrm{u}_{t}||_{L^{4}}^{2}$$\leq\frac{1}{2}||\theta_{0}||_{H^{1}}^{2}+\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{4}})||\theta_{t}||_{L^{2}(\Omega_{T}\rangle}$,$\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{4}})+\frac{1}{2}||\theta_{t}||_{L^{2}(\Omega_{T})}^{2}$

where we applied (3.10) and (3.11). Therefore we arrive at(3.12).

$||\nabla \mathrm{u}||_{W_{4}^{2_{1}1}(\Omega_{T})}+||\theta_{t}||_{L^{2}\langle\Omega_{T}\}}+||\nabla\theta||_{L_{T}^{\infty}L^{2}}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{4}})$

Next multiplying (1.2) by,$\frac{-\Delta\theta}{1-\theta G\langle\theta)H(\nabla \mathrm{u})},$and integrating over$\Omega$, we have$\frac{1}{2}\frac{d}{dt}||\nabla\theta(t)||_{L^{2}}^{2}+\int_{\Omega}\frac{|\triangle\theta|^{2}}{1-\theta G’(\theta)H(\nabla \mathrm{u})},dx$.

$\leq\int_{\Omega}\frac{\triangle\theta}{1-\theta G’(\theta)H(\nabla \mathrm{u})}(\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})+\nu|\nabla \mathrm{u}_{\ell}|^{2})dx$

Here we recall that,
) $=:M<\infty$ . Then integrating with respect to time

$1\leq 1-\theta G’(\theta)H$

where$0 \leq\sup_{\theta\geq 0}$(–&G$\prime\prime(\theta)$

variable, we conclude that(Vu)$\leq 1+\mathrm{A}f\Lambda$$|| \nabla\theta(t)||_{L^{2}}^{2}+\frac{2}{1+\Lambda M}||\triangle\theta||_{L^{2}(\Omega_{T})}^{2}$$\leq||\nabla\theta_{0}||_{L^{2}}^{2}+\frac{1}{1+\Lambda NI}||\triangle\theta||_{L^{2}(\Omega_{T})}^{2}+(1+\Lambda M)||\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})+|\nabla \mathrm{u}_{t}|^{2}||_{L^{2}(\Omega_{T})}^{2}$$\leq\Lambda+\frac{1}{1+\Lambda M}||\triangle\theta||_{L^{2}(\Omega_{T})}^{2}+\Lambda||\theta^{r}H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{4}(\Omega_{T})}||\nabla \mathrm{u}_{t}||_{L^{4}(\Omega_{T})}+\Lambda||\nabla \mathrm{u}_{t}||_{L^{4}(\Omega_{T})}^{2}$$\leq\Lambda+\frac{1}{2(1+\Lambda M)}||\triangle\theta||_{L^{2}\langle\Omega_{T})}^{2}$

due to (3.10) and (3.11). Consequently we obtain the first assertion.
With the help of Lemma 2.2, we also obtain$||\nabla\theta||_{L^{10/3}}(\Omega_{T})+||\theta||_{L^{10}\langle\Omega_{T})}+||\triangle \mathrm{u}||_{L^{20}(\zeta \mathit{1}_{T}\rangle}\leq\Lambda(||\theta||_{W_{2}^{2,1}(\Omega_{T})}+||\nabla \mathrm{u}||_{W_{4}^{2,1}\langle\Omega_{T})})\leq\Lambda$

which completes the proof.

,$\square$133
artd (1.5) holds.
Lemma 3.4. Let $p\in[20/9,10/3]$ and assume that $\theta\geq 0a.e$ . in$H^{1}=:U_{5}(p)$ , the solution
toThen for any(1.1)-(1.4) satisfies$\Omega_{T}$$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in B_{p,p}^{4-2/p}\mathrm{x}$$B_{p,p}^{2-2/p}\mathrm{x}$$(\mathrm{u}, \theta)$$\mathrm{A}$$||\mathrm{u}||_{W_{p}^{4,2}(\Omega_{T})}\leq$

where A depends on

and

$T$

$||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{5}(p)}$Proof. Since the embedding$B_{p,p}^{4-\frac{2}{p}}\{arrow B_{4}^{\frac{5}{42}}$,

we have,.

holds for any$\frac{20}{9}\leq p$, by the Lemma 3.3$||\nabla \mathrm{u}||_{W_{4}^{2,1}(\Omega_{T})}+||\theta||_{W_{2}^{2,1}(\Omega_{T})}\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{B_{4,4}^{5/2}}\cross B_{4,4}^{1/2_{\chi H^{1}}})$$\leq$For any$p \leq\frac{10}{3}$

A ( ( ug ,$||$$\mathrm{u}_{1}$,$\theta_{0}$)$||_{B_{p,p}^{4-2/P}\cross B_{p,p}^{2-2/p}\mathrm{x}H^{1}}$).

we have$||\nabla\cdot(G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u}))||_{L^{p}(\Omega_{T})}\leq \mathrm{A}||\nabla\theta||_{L^{10/3}}(\Omega_{T})||G’(\theta)||_{L^{\infty}(\Omega_{T})}||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}(\Omega_{T})}$

$+$

A ||&

$||_{L^{10}(\Omega_{T})}^{r}||$Au$||_{L^{20}(\Omega_{T})}||H_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}(\Omega_{T})}$$\leq\Lambda$

and

,(Vu)

$||\nabla\cdot\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\mathrm{p}}(\Omega_{T})}\leq\Lambda||\triangle \mathrm{u}||_{L^{20}(\Omega_{T})}||\overline{H}_{\nabla \mathrm{u}\nabla \mathrm{u}}$$||_{L^{\infty}(\Omega_{T})}\leq\Lambda$,

thanks to Lemmas 3.2 and 3.3. Then by the maximal regularity (2.1) we have$||\mathrm{u}||_{W_{p}^{4_{\backslash }2}\langle\Omega_{T})}\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U\mathrm{s}(p)}+C||\nabla\cdot(G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u}))||_{L^{p}\langle\Omega_{T})}$$+C||\nabla$$\leq$.

,$\overline{H}_{\nabla \mathrm{u}}$(Vu)$||_{L^{\mathrm{p}}(\Omega_{T}\rangle})$A.

This completes the proof.

$\square$andLemma 3.5. Let $l>2$ be integer and $p\in(1, \infty)$ . Assume that $\theta\geq 0a.e$ . in$(L^{l}\cap H^{1})=:U_{6}(l)$ ,(1.5) holds. Then for any
to (1.1)-(1.4) satisfies
the solution$\Omega_{T}$$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in B_{10/3,10/3}^{17/5}\cross$$B_{10/3,10/3}^{7/5}\mathrm{x}$$(\mathrm{u}, \theta)$$||\theta||_{L_{T}^{\infty}L_{x}^{p}}\leq\Lambda$

where A$=\Lambda(T, ||(\mathrm{u}_{0}, u_{1}, \theta_{0})||_{U_{6}(l)})$,. $Moreov$ er, $lf$$||\theta||_{L^{\infty}(\Omega_{T})}\leq\Lambda$$w$

here A$=\Lambda(T, ||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{6}(\infty)})$$(B_{p,p}^{1-2/p}\cap B_{10/3,10/3}^{7/5})\mathrm{x}$$(L^{\infty}\cap H^{1})=$$||$

where A, and for

we have,$(\mathrm{u}0, \mathrm{u}1, \theta 0)\in(B^{3-2/p}p,p\cap B_{10/3,10/3}^{17/5})\mathrm{x}$: U7(p) it holds that

Vu$=\Lambda(T, ||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U_{7}(p\rangle})$$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in U_{6}(\infty)$$||_{W_{p}^{2.1}(\Omega_{T})}\leq$

A,

134

Proof. We can deduce that$\frac{1}{l}\frac{d}{dt}||\hat{\theta}||_{L^{f}}^{l}+$ $(/_{-}\mathrm{i})$where we set$\int_{\Omega}\theta^{l-2}|\nabla\theta|^{2}dx=\int_{\Omega}\overline{G}_{l}$$G_{l}(\theta)=\theta^{l}G’(\theta)-\overline{G}_{l}(\theta)$

,

(0 )

$\partial_{t}H$(Vu) $dx+\nu$$\overline{G}_{l}(t)=l\int_{0}^{\theta}s^{l-1}G^{i}(s)ds$$||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L}\infty(\Omega_{T})=\Lambda<\infty$$\theta^{l-1}|\nabla \mathrm{u}_{t}|^{2}dx$, (3.13)

and(3.14).

$\hat{\theta}=\theta(1-\frac{lG_{l}(\theta)H(\nabla \mathrm{u})}{\theta^{l}})^{1/l}\geq\theta$

Since

’

from (3.3), we have$| \int_{\Omega}\overline{G}_{l}(\theta)\partial_{t}H(\nabla \mathrm{u})dx|\leq C||\theta^{l-1}||_{L^{1}(\Omega)}||\theta||_{L\infty(\Omega)}||\nabla \mathrm{u}_{t}||_{L^{\infty}\langle\Omega\rangle}||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}(\Omega)}$$\leq\Lambda||\theta||_{L^{l}(\Omega)}^{l-1}||\theta||_{H^{2}(\Omega\}}||\nabla \mathrm{u}_{t}||_{L^{\infty}(\Omega)}$.

Therefore, we conclude from (3.13) that$\frac{1}{l}\frac{d}{dt}||\theta^{\mathrm{A}}||_{L^{l}(\Omega\rangle}^{l}\leq\Lambda||\nabla \mathrm{u}_{t}||_{L^{\infty}(\Omega)}||\theta||_{H^{2}(\Omega)}||\theta||_{L^{\overline{\iota}}(\Omega)}^{l1}+C||\nabla \mathrm{u}_{t}||_{L(\Omega)}^{2}\infty||\theta||_{L^{p}(\Omega)}^{l-1}$

Here note thatand Lemma 3.4$\partial_{t}||\hat{\theta}||_{L^{l}(\Omega)}^{l}=l||\hat{\theta}||_{L^{l}(\Omega)}^{l-1}\partial_{t}||\hat{\theta}||_{L^{l}(\Omega)}$.

(3.15)and that from the Sobolev embeddin$||\nabla \mathrm{u}_{t}||_{L_{T}^{2}L^{\infty}}\leq\Lambda||\nabla \mathrm{u}_{t}||_{L_{T}^{2}W_{10/3}^{1}}\leq\Lambda||u||_{W_{\iota 0/\mathrm{s}}^{4,2}(\Omega_{T})}\leq\Lambda$$||\theta||_{L_{T}^{2}H^{2}}\leq||\theta||_{W_{2}^{2,1}(\Omega_{T})}\leq\Lambda$$\mathrm{g}$,,where A is independent of . Thus integrating (3.15) with respect to time variable,
we obtain$l$$||\hat{\theta}||_{L_{T}^{\infty}L^{l}}\leq||\theta_{0}^{\Lambda}||_{L^{l}}+\Lambda||\nabla \mathrm{u}_{t}||_{L_{T}^{2}L}\infty||\theta||_{L_{T}^{2}H^{2}}+\Lambda||\nabla \mathrm{u}_{t}||_{L_{T}^{2}L^{\infty}}^{2}$$\leq\Lambda+||\hat{\theta}_{0}||_{L^{\ell}}$

Since we have-norm of Vu, we have$\hat{\theta}_{0}\leq\theta_{0}(1+lM\Lambda)^{1/l}$, the desired result can be obtained. For the$W_{p}^{2,1}$$||\nabla \mathrm{u}||_{W^{\frac{.)}{p},1}(\Omega_{T})}.\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U_{7}(p)}+\Lambda||\theta||_{L^{\infty}(\Omega_{T})}^{r}||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}\{\Omega_{T})}$$+\Lambda||\overline{H}_{\nabla \mathrm{u}}(\nabla \mathrm{u})||_{L^{\infty}(\Omega_{T})}\leq$Afor $p\in(1, \infty)$ , by virtue of the maximal regularity (2.2). This completes the proof.
$\square$

The same procedure as in [8, Section 6] yields thatfor some
H\"oIder exponent $0<\alpha<1$ depending on ,and. Essentiallythe proof relies on the classical parabolic De Giorgi method. For more precise
information of this method we refer to [6, Chapter , \S 7] and [7, Chapter , Q12].
Here we note that Vu is Holder continuous because of Lemma 2.2$\theta\in \mathrm{C}^{\alpha,\alpha/2}(\overline{\Omega_{T}})$$T$$\sup_{\Omega}\theta_{0}$$\mathrm{I}\mathrm{I}$$||\theta||_{L}\infty(\Omega_{T})$$\mathrm{V}\mathrm{I}$135

Lemma 3.6 ([8, Lemma 6.1]). Assume that$k= \sup_{\Omega}\theta_{0}<\infty$A$||\nabla \mathrm{u}||_{W_{s}^{2,1}(\Omega_{T})}+||\theta||_{W_{2}^{2,1}(\Omega_{T})}+||\theta||_{L\infty(\Omega_{T})}\leq$holds for any $s\in(1, \infty)$ . Then 0 6
depending on A and .

. Suppose that
(3.16)

with Holder exponent$\mathrm{C}^{\alpha,\alpha/2}(\overline{\Omega_{T}})$$\alpha\in(\mathrm{O}, 1)$$k$Lemma 3.7. Assume that (3.16) holds. Then for any$5<p$ , $q<$ oo we have1%$(\mathrm{u}, \theta)||_{V_{T}(p,q)}=||\mathrm{u}||_{W_{p}^{4_{1}2}(\Omega_{T})}$

where A depends onand$||(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})||_{U(p,q)}$$T$q) and$(\mathrm{u}_{0}, \mathrm{u}_{1}, \theta_{0})\in U(p,$$||\theta||_{W_{q}^{2.1}(\Omega_{T})}\leq\Lambda$,.

Froof. By using Lemma 3.6 we have 0 is H\"older continuous. For brevity of nota-tion we denote $1-\theta G’(\theta)H$ (Vu) by (Vu, ) , and. Then the equation (1.2) can be rewritten as$\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})+\nu|\nabla \mathrm{u}_{t}|^{2}$$\theta$$c_{0}$by$R(\nabla \mathrm{u}, \theta)$$\mathrm{c}\mathrm{o}$(Vuo,$\theta_{0}$)$\theta_{t}-$Aft$=(c_{0}(\nabla \mathrm{u}_{0}, \theta_{0})-c_{0}$(Vu, )$\theta$$)\theta_{\mathrm{f}}+R(\nabla \mathrm{u}, \theta)$.

It follows from the assumptions that$||R(\nabla \mathrm{u}, \theta)||_{L^{q}(\Omega_{T})}\leq C||\theta||_{L^{\varpi}(\Omega_{T})}^{r}||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}(\Omega_{T})}||\nabla \mathrm{u}_{t}||_{L^{\mathrm{q}}(\Omega_{T})}+C||\nabla \mathrm{u}_{t}||_{L^{2q}(\Omega_{T})}^{2}$$\leq$A.

From Holder continuity it follows that$||c_{0}(\nabla \mathrm{u}_{0}, \theta_{0})-c_{0}$(Vu, )$\theta$,$||_{L\infty(\Omega_{T_{1}})}\leq KT_{1}^{\frac{\alpha}{2}}$where $K$ is H\"older constant independent of . Here $T_{1}<<T$ will be determinedlater.
is Holder continuous with respect to the
Next we show that$[0,T]$
. We remark thatspace variable for fixed in$T_{1}$$1/c_{0}(\nabla \mathrm{u}, \theta)(x, T_{2})$$T_{2}$$\mathcal{G}(y):=yG’(y)\leq M$

and ($;\in \mathrm{C}^{1}$is Lipschitz continuous. Then we have$| \frac{1}{c_{0}}(x, T_{2})-\frac{1}{c_{0}}(x’, T_{2})|$$=| \frac{\mathcal{G}(\theta(x’,T_{2}))H(\nabla \mathrm{u}(x’,T_{2}))-\mathcal{G}(\theta(x,T_{2}))H(\nabla \mathrm{u}(x,T_{2})}{\{1-\mathcal{G}(\theta(x,T_{2}))H(\nabla \mathrm{u}(x,T_{2}))\}\{1-\mathcal{G}(\theta(x’,T_{2}))H(\nabla \mathrm{u}(x’},\frac{)}{T_{2}))\}}|$$\leq|${ ( ($\mathcal{G}$$\theta$$x’$,$T_{2}$)) $H$ (Vu $(x’,$$T_{2}))-\mathcal{G}(\theta(x$,$T_{2}))H(\nabla \mathrm{u}(x’$,$T_{2}))$}$+\{\mathcal{G}(\theta(x, T_{2}))H(\nabla \mathrm{u}(x’, T_{2}))-\mathcal{G}(\theta(x, T_{2}))H(\nabla \mathrm{u}(x, T_{2}))\}|$$\leq|H(\nabla \mathrm{u}(x’, T_{2}))||\mathcal{G}(\theta(x’, T_{2}))-\mathcal{G}(\theta(x, T_{2}))|$$+|\mathcal{G}(\theta(x, T_{2}))||H(\nabla \mathrm{u}(x’, T_{2}))-H(\nabla \mathrm{u}(x, T_{2}))|$$\leq\Lambda K|x-x’|^{\alpha}+CAI|x$ $-x’|^{\alpha}$$\leq\Lambda|x-x’|^{\alpha}$,136(Vu, )] $(x, T_{2})$ is H\"older continuous
where A is independent of . Therefore [$T_{2}\in[0,T]$(Vu, ] $\geq 1/(1+M\Lambda)$ . These[. Moreover, we havefor any
A has the maximal regularity property according to (2.3).
assure that
$1/c_{0}$$T_{2}$$\sup_{\Omega_{T}}$$\theta$$1/c_{0}$$\theta)$$\frac{1}{c\mathrm{o}\langle\nabla \mathrm{u}(T_{2}),\theta(T_{2}\})}$

Hence, taking$T_{1}=( \frac{1}{2\Lambda(K,M,T)K})\frac{1}{\alpha}$, we have(Vu, )$\theta$$||\theta||_{W_{q}^{2,1}(\Omega_{T_{1}})}\leq\Lambda(K, M, T)||c_{0}(\nabla \mathrm{u}_{0}, \theta_{0})-c_{0}$$||_{L(\Omega_{T_{1}}\rangle}\infty||\theta_{t}||_{L^{q}(\Omega_{T_{1}})}$$+\Lambda(K, M, T)||R(\nabla \mathrm{u}, \ )$ $||_{L^{q}(\Omega_{T_{1}})}+C||\theta_{0}||_{B_{q}^{\frac{9}{\mathrm{q}}}(\Omega)}$,$\leq\frac{1}{2}||\theta_{t}||_{L^{q}\{\Omega_{T_{1}})}+\Lambda+\Lambda||\theta_{0}||_{B_{q,q}^{2-2/q}(\Omega)}$,

which yields$||\theta||_{W_{\mathrm{q}}^{2,1}(\Omega_{T_{1}})}\leq\Lambda+\Lambda||\theta_{0}||_{B_{q,q}^{2-2/q}(\Omega)}$.

Here we remark that$||\theta(T_{1})||_{B_{q,q}^{2-2/\mathrm{q}}}\leq C(T_{1})||\theta||_{W_{q}^{2,1}\{\Omega_{T_{1}})}\leq C(T_{1})(\Lambda+\Lambda||\mathrm{u}_{0}||_{B_{q,q}^{2-2/\mathrm{q}}})$

thanks to the embedding
for the interval
we have$W_{q}^{2_{\}}1}(\Omega_{T_{1}})\mathrm{c}arrow BUC([0, T_{1}], B_{q,q}^{2-\frac{2}{q}})$(see [2]). Then similarly$[T_{1},2T_{1}]$$||\theta||_{W_{q}^{2,1}\acute{(}\Omega_{[T_{1},2T_{1}]})}\leq \mathrm{A}+\Lambda||\mathrm{u}(T_{1})||_{B_{q,q}^{2-2/q}}\leq\Lambda+\Lambda||\mathrm{u}_{0}||_{B_{q,q}^{2-2/q}}\leq$A.

Repeating the same operation, we obtain$||\theta||_{W_{q}^{2,1}(\Omega_{[kT_{1\prime}(h+1)T_{1}]})}.\leq$A.

Summing the inequalities from $k=0$ to $k=m$ satisfying $(m+1)T_{1}>T$ and, we conclude thatA.

$mT_{1}\leq T$$||\theta||_{W_{q}^{2,1}(\Omega_{T})}\leq$

Next we estimate the norm$||\mathrm{u}||_{W_{p}^{4,2}(\Omega_{T})}$. From Lemma 2.2 it follows that$||\nabla\theta||_{L(\Omega_{T}\}}\infty+||\triangle \mathrm{u}||_{L^{\infty}\langle\Omega_{T})}\leq\Lambda$

for $q>5$ . Therefore, by virtue of the maximal regularity (2.1) we have$||\mathrm{u}||_{W_{p}^{4,2}(\Omega_{T}\}}\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U(p,q)}+C||\nabla\cdot(G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u}))||_{L^{p}(\Omega_{T})}$$+C\downarrow|\nabla$.

$\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{p}(\Omega_{T}\}}$$\leq C||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U(p,q)}+\Lambda||\nabla\theta||_{L^{\infty}(\Omega_{T}\rangle}||G’(\theta)||_{L(\Omega_{T})}\infty||H_{\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L\infty(\Omega_{2’})}$$+\Lambda||\theta||_{L^{\infty}(\Omega_{T})}^{r}||\triangle \mathrm{u}||_{L^{\infty}(\Omega_{T})}||H_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L\infty(\Omega_{T})}$

$+$

A A
$||$

$\mathrm{u}||_{L^{\infty}(\Omega_{T})}||\overline{H}_{\nabla \mathrm{u}\nabla \mathrm{u}},(\nabla \mathrm{u})||_{L^{\infty}(\Omega_{T})}$$\leq\Lambda(||(\mathrm{u}_{0}, \mathrm{u}_{1},0)||_{U(p,q\rangle})$,which completes the proof.
$\square$137
Here we note that we assume thatin all the lemmas of this section.
sufficiently
for
the
such
non-negativity
smooth solution
Theof 0 is assured. Hence, we can not proceed the above arguments,
asdirectly. One of the solvents for this problem is the following. We first consider the:truncated problem$\theta\geq 0$$(\mathrm{u}_{\mathit{3}}\theta)$$(\mathrm{u},$$\ )$$\in W_{p}^{4,2}(\Omega_{T})\}<L_{T}^{\infty}L^{2}$$(TE)_{3}^{L}$$\mathrm{u}_{tt}+\triangle^{2}\mathrm{u}-\nu\triangle \mathrm{u}_{t}=\Gamma_{L}(\nabla\cdot[G(\theta)H_{\nabla \mathrm{u}},(\nabla \mathrm{u})+\overline{H}_{\nabla \mathrm{u}},(\nabla \mathrm{u})])$$\theta_{t}-\triangle\theta=\theta G’(\theta)\theta_{t}H(\nabla \mathrm{u})$$+\theta G’(\theta)\partial_{t}H(\nabla \mathrm{u})$(3.17),

in$+\nu|\nabla \mathrm{u}_{t}|^{2}$on$\mathrm{u}=\triangle \mathrm{u}=\nabla\theta\cdot \mathrm{n}=0$$\mathrm{u}(0_{j}\cdot)=\mathrm{u}_{0}$,$\mathrm{u}_{t}(0, \cdot)=\mathrm{u}_{1}$,in$\theta(0_{2}\cdot)=\theta_{0}\geq 0$

where$\Gamma_{L}(x)=\{$$xL \frac{x}{|x|}$$\Omega_{T}$$\mathit{3}_{T}$$\Omega$,(3.18),,

if $|x|\leq L$ ,if $|x|\geq L$ .

for $L>0$ . Then the solution satisfies also the
We construct the solution
sufficiently
large truncation size because a priorifor(1.1)-(1.4)
systemoriginal
estimates obtained in this section are independent of . More precisely, we refer to[12].
$(\mathrm{u}_{L}, \theta_{L})$

$L$

$L$

Acknowledgment, The first author is partly supported by the Research Fellowships of the Japan Society of Promotion of Science (JSPS) for Young Scientists.

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