数理解析研究所講究録
第 1905 巻 2014 年 61-72

61

On two phase problem:compressible-compressible model problem$*$

筑波大学数理物質系 久保 隆徹 (Takayuki Kubo)
Division of Mathematics, University bf Tsukuba
Abstract

We consider the model problem for the two phase problem in cases of compressible-compressible fluid flows without surface tension. In order to prove the local
in time existence theorem for our problem, the generation of analytic semigroup for
linearized problem and its maximal $L_{p}-L_{q}$ regularity are needed in our method.
The key step of our method is to prove the existence of -bounded solution operator
to the generalized resolvent problem corresponding to the linearized problem:$\mathcal{R}$1Introduction

Two phase problem appears in various situations. For example, in order to analyze
a motion of raindrops and air bubbles under water, we have to consider the two phaseproblem. Mathematical analysis for two phase problem has been studied by some mathematicians. We shall introduce the results corresponding to two phase problem.
In two phase problem of compressible and incompressible viscous fluid, Denisova [1]studied a local in time existence theorem for her problem under the technical condition.
Recently in Kubo, Shibata and Soga [2], the existence of -bounded solution operator
to generalized resolvent problem corresponding to two phase problem is shown under
the natural condition derived from physics. By Weis’ operator valued Fourier multiplier
theorem with -boundedness of solution operator, we can show the maximal regularity
for the linearized problem for two phase problem. A local in time existence theorem isobtained by applying the maximal regularity to proving the convergence of the successive
approximations.
On the other hand, in two phase problem of compressible and compressible viscous
fluid, Tani [4],[5] studied a local in time existence theorem under the natural condition
in H\"older space framework. In this article, we shall consider the two phase problem ofcompressible and compressible fluid in $L^{p}-L^{q}$ framework and prove the local in time
existence theorem of our problem in a similar way as [2].
For this purpose, we shall consider the model problem for the two phase problem incases of compressible-compressible fluid flows without surface tension. The key step of$\mathcal{R}$$\mathcal{R}$$-J$*This article is based on the ajoint work with Prof. Yoshihiro Shibata (Waseda University) and Prof.
Kohei Soga (CNRS-ENS Lyon).

62-bounded solution operator to the generalized
resolvent problem corresponding to the linearized problem:

our method is to prove the existence ofHere,$\rho\pm,$$\mathcal{R}$$\lambda\rho\pm+\gamma_{1}^{\pm}div\vec{u}\pm=f_{\pm}$in$\mathbb{R}_{\pm}^{N}$,$\lambda\vec{u}\pm-DivS_{\pm}(\vec{u}\pm, \rho_{\pm})=\vec{g}\pm$in$\mathbb{R}_{\pm}^{N}$

,

(1.1)
(1.2)

$\vec{u}_{+}|_{x_{N}=0+}-\vec{u}_{-}|_{x_{N}=0-}=\vec{k}$on$\mathbb{R}_{0}^{N}$

,

(1.3)

$S_{+}(\vec{u}_{+}, \rho_{+})\vec{n}|_{x_{N}=0+}-S_{-}(\vec{u}_{-}, \rho_{-})\vec{n}|_{x_{N}=0-}=-\vec{h} on\mathbb{R}_{0}^{N}$

.

(1.4)

$\vec{u}\pm=(u_{\pm,1}, \ldots, u_{\pm,N})(N\geq 2)$

are unknown mass density and unknown velocity

is stress tensor,
ismatrix called the Cauchy deformation tensor and denotes the $N\cross N$
identity matrix. Moreover for $N\cross N$ matrix function $M=(M_{ij})$ , the th component ofandis the unit outer normal to
DivM is defined by
are all constants satisfying

fields.

$D(\vec{u})=(\nabla\vec{u}+^{T}$$S_{\pm}(\vec{u}\pm, \rho_{\pm})=2\mu_{1}^{\pm}D(\vec{u}_{\pm})+(\mu_{2}^{\pm}div\vec{u}\pm-\gamma_{2}^{\pm}\rho_{\pm})I$$N\cross N$$\nabla\vec{u})/2$

$I$

$i$

$\sum_{j=1}^{N}\partial_{j}M_{ij}.\vec{n}=(0, \ldots, 0, -1)$$\mu_{i}^{\pm},$$\mathbb{R}^{\underline{n}}$$\gamma_{i}^{\pm}(i=1,2)$$\mu_{1}^{\pm}>0, \mu_{1}^{\pm}+\mu_{2}^{\pm}>0, \gamma_{1}^{\pm}, \gamma_{2}^{\pm}\geq 0$

.

(1.5)

are constants
andare 1st and 2nd viscosity constants, respectively, andproblem.
The resolvent parameter
appearing in the linearization of the original nonlinear
varies in
where
Here

$\mu_{1}^{\pm}$$\gamma_{1}^{\pm},$$\mu_{2}^{\pm}$$\lambda$$\gamma_{2}^{\pm}$$\Lambda_{\epsilon,\lambda_{0}}=\Sigma_{\epsilon,\lambda_{0}}\cap K_{\epsilon},$$\Sigma_{\epsilon,\lambda_{0}}=\{\lambda\in \mathbb{C}||\arg\lambda|\leq\pi-\epsilon, |\lambda|\geq\lambda_{0}\},$(1.6)$K_{\epsilon}=\{\lambda\in \mathbb{C}|({\rm Re}\lambda+\gamma_{m}+\epsilon)^{2}+^{\backslash }({\rm Im}\lambda)^{2}\geq(\gamma_{m}+\epsilon)^{2}\}$.
with
Before stating our main results, we shall introduce several symbols and functional, we use the following symbols:
spaces. For the diﬀerentations of $N$ -vector$\gamma_{m}=\max(\frac{\gamma^{+}\gamma^{+}}{\mu_{1}^{+}+\mu_{2}^{+}}, \hat{\mu_{1}^{-}}\gamma^{-}+\gamma^{-}\mu_{2}^{-}\Rightarrow)$$\vec{g}=(g_{1}, \ldots, g_{N})$$\nabla\vec{g}=(\partial_{i}f_{j}|i,j=1, \ldots, N) , \nabla^{2}\vec{g}=(\partial_{i}\partial_{j}g_{k}|i,j, k=1, \ldots, N)$.

For any domain
denote the usual Lebesgue space and Sobolev
andspace, while
denote their norms, respectively. For any two
andBanach spaces $X$ and $Y,$
denotes the set of all bounded linear operators from
$X$ to Y. $Ho1(U, X)$ denotes the set of all $X$ -valued holomorphic functions defined on $U.$anddenote the set of all natural and complex numbers, respectively, and we set$\Omega,$$\Vert$$L_{q}(\Omega)$$\Vert_{L_{q}(\Omega)}$$W_{q}^{m}(\Omega)$$\Vert$$\Vert_{W_{q}^{m}(\Omega)}$$\mathcal{L}(X, Y)$$\mathbb{N}$$\mathbb{C}$$\mathbb{N}_{0}=\mathbb{N}\cup\{0\}.$

Next we introduce the definition of$\mathcal{R}$-boundedness which is the key word in our

method.

Definition 1.1. Let $X$ and be Banach spaces. A family of operatoris, if there exist constants $C>0$ and $p\in[1, \infty$ ) such thatcalled -bounded on
for anyand sequencesof independent,
symmetric, $\{-1, 1\}$ -valued random variables on , 1 there holds the inequality:$Y$$\mathcal{R}$$\mathcal{T}\subset \mathcal{L}(X, Y)$$\mathcal{L}(X, Y)$$n\in \mathbb{N},$$\{T_{j}\}_{j=1}^{n}\subset \mathcal{T},$$\{x_{j}\}_{j=1}^{n}\subset X$$\{r_{j}(u)\}_{j=1}^{n}$

$[0$

$]$

$\{\int_{0}^{1}\Vert\sum_{j=1}^{n}r_{j}(u)T_{j}x_{j}\Vert_{Y}^{p}du\}^{1/p}\leqC\{\int_{0}^{1}\Vert\sum_{j=1}^{n}r_{j}(u)x_{j}\Vert_{X}^{p}du\}^{1/p}$

The smallest such $C$ is called$\mathcal{R}$-bound of$\mathcal{T}$, which is denoted by$\mathcal{R}_{\mathcal{L}(X,Y)}(\mathcal{T})$.

63
Then we can obtain the following main result.
Theorem 1.2. Let $1<q<\infty,$ $0<\epsilon<\pi/2$ and. Set
defined in (1.6) and set$\lambda_{0}>$O. Let$\Sigma_{\epsilon,\lambda_{0}}$and$K_{\epsilon}$be the sets$\Lambda_{\epsilon,\lambda_{0}}=\Sigma_{\epsilon,\lambda_{0}}\cap K_{\epsilon}$$Y_{q}=\{(f_{+}, f_{-,\vec{9}+,\vec{9}-},\vec{h},\vec{k})|$$f_{\pm}\in W_{q}^{1}(\mathbb{R}_{\pm}^{N}) , \vec{g}\pm\in L_{q}(\mathbb{R}_{\pm}^{N})^{N}, \vec{h}\in W_{q}^{1}(\mathbb{R}^{N})^{N}, \vec{k}\in W_{q}^{2}(\mathbb{R}^{N})^{N}\},$$\mathcal{Y}_{q}=\{(F_{0+}, F_{0-}, F_{1+}, F_{1-}, F_{2}, F_{3}, F_{4}, F_{5}, F_{6})|F_{0\pm}\in W_{q}^{1}(\mathbb{R}_{\pm}^{N})$,$F_{1\pm}\in L_{q}(\mathbb{R}_{\pm}^{N})^{N}, F_{2}, F_{5’}\in L_{q}(\mathbb{R}^{N})^{N^{2}} F_{3}, F_{6}\in L_{q}(\mathbb{R}^{N})^{N}, F_{4}\in L_{q}(\mathbb{R}^{N})^{N^{3}}\}.$

Then, there exist operator families$\mathcal{P}_{\pm}(\lambda)\in Ho1(\Lambda_{\epsilon,\lambda_{0}}, \mathcal{L}(\mathcal{Y}_{q}, W_{q}^{1}(\mathbb{R}_{\pm}^{N}))) , \mathcal{U}_{\pm}(\lambda)\in Ho1(\Lambda_{\epsilon,\lambda_{0}}, \mathcal{L}(\mathcal{Y}_{q}, W_{q}^{2}(\mathbb{R}_{\pm}^{N})^{N}))$such that for any$(f_{+}, f_{-},\vec{g}_{+},\vec{g}-,\vec{h},\vec{k})\in Y_{q}$and$\lambda\in\Lambda_{\epsilon,\lambda_{0}},$$\rho_{\pm}=\mathcal{P}_{+}(\lambda)(f_{+}, f_{-},\vec{g}_{+},\vec{g}_{-}, \nabla\vec{h}, \lambda^{1/2}\vec{h}, \nabla^{2}\vec{k}, \lambda^{1/2}\nabla\vec{k}, \lambda\vec{k})$,$\vec{u}_{\pm}=\mathcal{U}_{\pm}(\lambda)(f_{+}, f_{-},\vec{g}_{+},\vec{g}-, \nabla\vec{h}, \lambda^{1/2}\vec{h}, \nabla^{2}\vec{k}, \lambda^{1/2}\nabla\vec{k}, \lambda\vec{k})$solve problem $(1.1)-(1.4)$ uniquely. Moreover, there exists a constantand $N$ such that$\lambda_{0},$

where

2

$C$

depending on$\epsilon,$$q$$\mathcal{R}_{\mathcal{L}(\mathcal{Y}_{q},W_{q}^{1}(\mathbb{R}_{+}^{N})^{2})}(\{(\tau\partial_{\tau})^{\ell}\{(\lambda, \gamma)\mathcal{P}_{\pm}(\lambda)\}|\lambda\in\Gamma_{\epsilon,\lambda_{0}}\})\leq C (\ell=0,1)$,$\mathcal{R}_{\mathcal{L}(\mathcal{Y}_{q},L_{q}(\mathbb{R}_{\pm}^{N})^{N^{3}+N^{2}+2N})}(\{(\tau\partial_{\tau})^{\ell}(G_{\lambda}\mathcal{U}_{\pm}(\lambda))|\lambda\in\Gamma_{\epsilon,\lambda_{0}}\})\leq C (\ell=0,1)$,$G_{\lambda}u=(\lambda u, \gamma u, \lambda^{1/2}\nabla u, \nabla^{2}u)$

and

(1.7)

$\lambda=\gamma+i\tau.$

Outline of the Proof of Theorem 1.2

In this section, we shall show the outline of the proof of Theorem 1.2. First step ofour method is to obtain the solution formula for $(1.1)-(1.4)$ by Fourier transform withrespect to $x’=(x_{1}, \ldots, x_{N-1})$ . Second step is to show the -boundedness for solution
operator by using solution formula with technical lemmas.
$\mathcal{R}$2.1

Solution formula

In this section, we shall show the solution formula for $(1.1)-(1.4)$ . For simplicity, we
consider the case where $f_{\pm}=0$ and. Substitute (1.1) into (1.2) and (1.4), we can$(1.1)-(1.4)$
reduceto the following equations:$\vec{g}\pm=\vec{0}$$\lambda v\pm-Div[2\mu_{1}^{\pm}D(v_{\pm})+(\mu_{2}^{\pm}+\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{\lambda})(div\vec{v}_{\pm})I]=0 in\mathbb{R}_{\pm}^{N}$(2.1)$v_{+,J}-v_{-,J}=k_{J} on\mathbb{R}_{0}^{N}$

(2.2)

64
$\mu_{1}^{+}(D_{N}v_{+,j}+D_{j}v_{+,N})-\mu_{1}^{-}(D_{N}v_{-)j}+D_{j}v_{-)N})=-h_{j}$on$\mathbb{R}_{0}^{N}$

(2.3)

on

$\mathbb{R}_{0}^{N}$(2.4)$2 \mu_{1}^{+}D_{N^{V+)}N}+(\mu_{2}^{+}+\frac{\gamma_{1}^{+}\gamma_{2}^{+}}{\lambda})div\vec{v}_{+}$$-[2 \mu_{1}^{-}D_{N}v_{-,N}+(\mu_{2}^{-}+\frac{\gamma_{1}^{-}\gamma_{2}^{-}}{\lambda})div\vec{v}_{-}]=-h_{N}$Here and hereafter, and run from 1 through $N-1$ and $N$ and we set
for simplicity of notation. In order to obtain the solution formula of (2.1), we prepare
the following formula obtained by applying the divergence to (2.1):$j$$\delta_{\lambda}^{\pm}=\gamma_{1}^{\pm}\gamma_{2}^{\pm}/\lambda$$J$$[\lambda-(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})\triangle]divv\pm=0.$

By using the formula above and (2.1), we see that$[\lambda-(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})\triangle](\lambda-\mu_{1}^{\pm}\Delta)v_{\pm}=0$

.

(2.5)

In order to obtain the solution formula of $(2.1)-(2.4)$ , we use the partial Fouriertransform with respect to $x’=(x_{1}, \ldots, x_{N-1})$ and the partial inverse Fourier transformdefined by$\mathcal{F}_{x’}[v](\xi’, x_{N})=\hat{v}=\int_{\mathbb{R}^{N-1}}e^{-ix’\cdot\xi’}v(x’, x_{N})dx’,$$\mathcal{F}_{x’}^{-1}[w(\xi’, x_{N})](x’)=(\frac{1}{2\pi})^{N-1}\int_{\mathbb{R}^{N-1}}e^{ix’\cdot\xi’}w(\xi’, x_{N})d\xi’,$

respectively. Taking$2\mathcal{F}_{x’}[DivD(v_{j})](\xi’, x_{N})=-|\xi’|^{2}\hat{v_{j}}+D_{N}^{2}\hat{v_{j}}+i\xi_{j}(i\xi’\cdot\hat{v’}+D_{N}\hat{v_{N}})$,$2\mathcal{F}_{x’}[DivD(v_{N})](\xi’, x_{N})=-|\xi’|^{2}\hat{v_{N}}+D_{N}^{2}\hat{v_{N}}+D_{N}(i\xi’\cdot\hat{v’}+D_{N}\hat{v_{N}})$

into account, we obtain the following equations by applying the partial Fourier transform
to $(2.1)-(2.4)$ and (2.5):(2.6)$\{\begin{array}{l}\lambda\hat{v_{+,j}}-\mu_{1}^{+}[(D_{N}^{2}-|\xi’|^{2})\overline{v_{+,j}}+i\xi_{j}\underline{\overline{div\vec{v}_{+}}]}-(\mu_{2}^{+}+\delta_{\lambda}^{+})i\xi_{j}\overline{div\vec{v}_{+}}=0,\lambda\hat{v_{+,N}}-\mu_{1}^{+}[(D_{N}^{2}-|\xi’|^{2})\hat{v_{+,N}}+D_{N}div\vec{v}_{+}]-(\mu_{2}^{+}+\delta_{\lambda}^{+})D_{N}\overline{div\vec{v}+}=0,\lambda\hat{v_{-,j}}-\mu_{1}^{-}[(D_{N}^{2}-|\xi’|^{2})\overline{v_{-,j}}+i\xi_{j}\underline{\overline{div\vec{v}_{-}}]}-(\mu_{2}^{-}+\delta_{\lambda}^{-})i\xi_{j}\overline{div\vec{v}_{-}}=0,\lambda\hat{v_{-,N}}-\mu_{1}^{-}[(D_{N}^{2}-|\xi’|^{2})\hat{v_{-,N}}+D_{N}div\vec{v}_{-}]-(\mu_{2}^{-}+\delta_{\lambda}^{-})D_{N}\overline{div\vec{v}_{-}}=0\end{array}$and$[\lambda+(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm}))(|\xi’|^{2}-D_{N}^{2})][\lambda+\mu_{1}^{\pm}(|\xi’|^{2}-D_{N}^{2})]\hat{v_{\pm,J}}=0$.

By (2.7), we see that the characteristic roots of (2.6) are$A_{\pm}=\sqrt{(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})^{-1}\lambda+A^{2}},$$B_{\pm}=\sqrt{(\mu_{1}^{\pm})^{-1}\lambda+A^{2}},$$A=|\xi’|.$

(2.7)

65
By using $B\pm andA$ , we rewrite (2.6) as follows:$\overline{arrow}$$\mu_{1}^{+}(B_{+}^{2}-D_{N}^{2})\hat{v_{+,j}}-(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})i\xi_{j}di\underline{vv_{+}}=0,$$\mu_{1}^{+}(B_{+}^{2}-D_{N}^{2})\hat{v_{+,N}}-(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})D_{\underline{N}}div\vec{v}_{+}=0,$

(2.8)

$\{$

$\mu_{1}^{-}(B_{-}^{2}-D_{N}^{2})\hat{v_{-,J}\prime}-(\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})i\xi_{j}div\vec{v}_{-}=0,$$\mu_{1}^{-}(B_{-}^{2}-D_{N}^{2})\hat{v_{-,N}}-(\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})D_{N}\overline{div\vec{v}_{-}}=0.$

From now, we shall find the solution$\hat{v_{\pm,J}}$to (2.6) of the forms:$\hat{v_{+,J}}=\alpha_{J}^{+}(e^{-B_{\dagger}x_{N}}-e^{-A+x_{N}})+\beta_{J}^{+}e^{-B_{+}x_{N}},\hat{v_{-,J}}=\alpha_{J}^{-}(e^{B_{-}x_{N}}-e^{A_{-}x_{N}})+\beta_{J}^{-}e^{B_{-}x_{N}}.$(2.9)

We see that$(B_{\pm}^{2}-D_{N}^{2})\overline{v_{\pm,J}}=(A_{\pm}^{2}-B_{\pm}^{2})\alpha_{J}^{\pm}e^{\mp A\pm x_{N}}$and$\overline{arrow}$$divv_{+}=(i\xi’\cdot a_{+}’+i\xi’\cdot\beta_{+}’-B_{+}(\alpha_{N}^{+}+\beta_{N}^{+}))e^{-B_{+}x_{N}}+(A_{+}\alpha_{N}^{+}-i\xi’\cdot\alpha_{+}’)e^{-A_{+}x_{N}},$$\overline{div\vec{v}_{-}}=(i\xi’\cdot\alpha_{-}’+i\xi’\cdot\beta_{-}’+B_{-}(\alpha_{N}^{-}+\beta_{N}^{-}))e^{B_{-}x_{N}}-(A_{-}\alpha_{N}^{-}+i\xi’\cdot\alpha_{-}’)e^{A_{-}x_{N}}$

where
and
.
Substituting (2.10) into (2.8) and equating the coeﬃcients of$\alpha_{\pm}^{f}=(\alpha_{1}^{\pm}, \ldots, \alpha_{N-1}^{\pm})$,(2.10)$\beta_{\pm}’=(\beta_{1}^{\pm}, \ldots, \beta_{N-1}^{\pm})$$e^{\mp B\pm x_{N}},$ $e^{\mp A\pm x_{N}}$, we have$\{\begin{array}{l}i\xi’\cdot\alpha_{+}’+i\xi’\cdot\beta_{+}’-B_{+}(\alpha_{N}^{+}+\beta_{N}^{+})=0,i\xi’\cdot\alpha_{-}’+i\xi’\cdot\beta_{-}’+B_{-}(\alpha_{N}^{-}+\beta_{N}^{-})=0,\mu_{1}^{+}(A_{+}^{2}-B_{+}^{2})\alpha_{j}^{+}-(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})i\xi_{j}(A_{+}\alpha_{N}^{+}-i\xi’\cdot\alpha_{+}’)=0,\mu_{1}^{+}(A_{+}^{2}-B_{+}^{2})\alpha_{N}^{+}+(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})A_{+}(A_{+}\alpha_{N}^{+}-i\xi’\cdot\alpha_{+}’)=0,\mu_{1}^{-}(A_{-}^{2}-B \alpha_{j}^{-}+(\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})i\xi_{j}(A_{-}\alpha_{N}^{-}+i\xi’\cdot\alpha =0,\mu_{1}^{-}(A_{-}^{2}-B_{-}^{2})\alpha_{N}^{-}+(\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})A_{-}(A_{-}\alpha_{N}^{+}+i\xi’\cdot\alpha =0.\end{array}$Since(2.11), the fourth equation in. By the first equation in (2.11), we have$\mu_{1}^{+}(A_{+}^{2}-B_{+}^{2})+(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})A_{+}^{2}=(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})A^{2}$(2.11) implies that$\alpha_{N}^{+}=A^{-2}A_{+}i\xi’\cdot\alpha_{+}’$$i \xi’\cdot\alpha_{+}’=\frac{A^{2}}{B_{+}A_{+}-A^{2}}(i\xi’\cdot\beta_{+}’-B_{+}\beta_{N}^{+})$,$\alpha_{N}^{+}=\frac{A_{+}}{B_{+}A_{+}-A^{2}}(i\xi’\cdot\beta_{+}’-B_{+}\beta_{N}^{+})$. (2.12)Similarly, by the sixth equation and the second equation in (2.11), we obtain$i \xi’\cdot\alpha_{-}’=\frac{A^{2}}{B_{-}A_{-}-A^{2}}(i\xi’\cdot\beta_{-}’+B_{-}\beta_{N}^{-})$,$\alpha_{\overline{N}}=\frac{-A_{-}}{B_{+}A_{+}-A^{2}}(i\xi’\cdot\beta_{-}’+B_{-}\beta_{N}^{-})$. (2.13)

Next we consider the boundary condition $(2.2)-(2.4)$ . By applying the partial Fouriertransform to $(2.2)-(2.4)$ , we obtain$\beta_{J}^{+}-\beta_{J}^{-}=\hat{k_{J}}$,(2.14)$\mu_{1}^{+}((A_{+}-B_{+})\alpha_{j}^{+}-B_{+}\beta_{j}^{+}+i\xi_{j}\beta_{N}^{+})-\mu_{1}^{-}((B_{-}-A_{-})\alpha_{j}^{-}+B_{-}\beta_{j}^{-}\cdot+i\xi_{j}\beta_{N}^{-})=-\hat{h_{j}},$(2.15)$(2\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{\pm})(A_{+}-B_{+})\alpha_{N}^{+}-2\mu^{+}B_{+}\beta_{N}^{+}+(\mu_{2}^{+}+\delta_{\lambda}^{+})(i\xi’\cdot\beta_{+}’-B_{+}\beta_{N}^{+})$$-(2\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})(B_{-}-A_{-})\alpha_{N}^{-}-2\mu_{1}^{-}B_{-}\beta_{N}^{-}-(\mu_{2}^{-}+\delta_{\lambda}^{-})(i\xi’\cdot\beta_{-}’+B_{-}\beta_{N}^{-})=-\hat{h_{N}}.$(2.16)

66
Since by (2.12) and (2.13), we have$-i \xi’\cdot\hat{h}’=\mu_{1}^{+}(\frac{A_{+}(A^{2}-B_{+}^{2})}{B_{+}A_{+}-A^{2}}i\xi’\cdot\beta_{+}’-\frac{A^{2}(2A_{+}B_{+}-B_{+}^{2}-A^{2})}{B_{+}A_{+}-A^{2}}\beta_{N}^{+})$$- \mu_{1}^{-}(_{-}\frac{A_{-}(B_{-}^{2}-A^{2})}{B_{-}A_{-}-A^{2}}i\xi’\cdot\beta_{-}’+\frac{A^{2}(A^{2}+B_{-}^{2}-2A_{-}B_{-})}{B_{-}A_{-}-A^{2}}\beta_{N}^{-})$

and$- \hat{h_{N}}=\frac{1}{B_{+}A_{+}-A^{2}}[2\mu_{1}^{+}(A_{+}^{2}-A_{+}B_{+})+(\mu_{2}^{+}+\delta_{\lambda}^{+})(A_{+}^{2}-A^{2})]i\xi’\cdot\beta_{+}’$$-(2 \mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})\frac{A_{+}^{2}-A^{2}}{B_{+}A_{+}-A^{2}}B_{+}\beta_{N}^{+}-(2\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})\frac{A_{-}^{2}-A^{2}}{B_{-}A_{-}-A^{2}}B_{-}\beta_{N}^{-}$$- \frac{1}{B_{-}A_{-}-A^{2}}[2\mu_{1}^{-}(A_{-}^{2}-A_{-}B_{-})+(\mu_{2}^{-}+\delta_{\lambda}^{-})(A_{-}^{2}-A^{2})]i\xi’\cdot\beta$

Substituting$\beta_{J}^{+}=\beta_{J}^{-}+\hat{k_{J}}$by (2.14) into the formula of$-i\xi’\cdot\hat{h’}$and$-\hat{h_{N}}$, we obtain$- \mu_{1}^{+}\frac{A_{+}(A^{2}-B_{+}^{2})}{B_{+}A_{+}-A^{2}}i\xi’\cdot\hat{k’}+\mu_{1}^{-}\frac{A^{2}(2A_{+}B_{+}-B_{+}^{2}-A^{2})}{B_{+}A_{+}-A^{2}}\hat{k_{N}}-i\xi’\cdot\hat{h’}$$=-[ \mu_{1}^{+}\frac{A_{+}(B_{+}^{2}-A^{2})}{B_{+}A_{+}-A^{2}}+\mu_{1}^{-}\frac{A_{-}(B_{-}^{2}-A^{2})}{B_{-}A_{-}-A^{2}}]i\xi’\cdot\beta_{-}’$$-A^{2}[ \mu_{1}^{+}\frac{2A_{+}B_{+}-B_{+}^{2}-A^{2}}{B_{+}A_{+}-A^{2}}-\mu_{1}^{-}\frac{2A_{-}B_{-}-A^{2}-B_{-}^{2}}{B_{-}A_{-}-A^{2}}]\beta_{N}^{-}$

and$\frac{-i\xi’\cdot\hat{k’}}{B_{+}A_{+}-A^{2}}[2\mu_{1}^{+}(A_{+}^{2}-A_{+}B_{+})+(\mu_{2}^{+}+\delta_{\lambda}^{+})(A_{+}^{2}-A^{2})]$$+(2 \mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})\frac{(A_{+}^{2}-A^{2})B_{+}\hat{k_{N}}}{B_{+}A_{+}-A^{2}}-\hat{h_{N}}$$= \frac{i\xi’\cdot\beta_{-}’}{B_{+}A_{+}-A^{2}}[2\mu_{1}^{+}(A_{+}^{2}-A_{+}B_{+})+(\mu_{2}^{+}+\delta_{\lambda}^{+})(A_{+}^{2}-A^{2})]$$- \frac{i\xi’\cdot\beta_{-}’}{B_{-}A_{-}-A^{2}}[2\mu_{1}^{-}(A_{-}^{2}-A_{-}B_{-})+(\mu_{2}^{-}+\delta_{\lambda}^{-})(A_{-}^{2}-A^{2})]$$-(2 \mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})\frac{(A_{+}^{2}-A^{2})B_{+}\beta_{N}^{-}}{B_{+}A_{+}-A^{2}}-(2\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})\frac{(A_{-}^{2}-A^{2})B_{-}\beta_{N}^{-}}{B_{-}A_{-}-A^{2}}.$

Here setting$L_{11}^{+}=- \mu_{1}^{+}\frac{A_{+}(B_{+}^{2}-A^{2})}{B_{+}A_{+}-A^{2}},$$-A_{-}(B^{\underline{2}}-A^{2})-A^{2}$$L_{11}^{-}=-\mu_{1}\overline{B_{-}A_{-}}$ ’$L_{12}^{+}=- \mu_{1}^{+}A^{2}\frac{2A_{+}B_{+}-B_{+}^{2}-A^{2}}{B_{+}A_{+}-A^{2}},$$L_{12}^{-}= \mu_{1}^{-}A^{2}\frac{2A_{-}B_{-}-A^{2}-B_{-}^{2}}{B_{-}A_{-}-A^{2}},$

67$L_{21}^{+}= \frac{1}{B_{+}A_{+}-A^{2}}[2\mu_{1}^{+}(A_{+}^{2}-A_{+}B_{+})+(\mu_{2}^{+}+\delta_{\lambda}^{+})(A_{+}^{2}-A^{2})],$$L_{21}^{-}=- \frac{1}{B_{-}A_{-}-A^{2}}[2\mu_{1}^{-}(A_{-}^{2}-A_{-}B_{-})+(\mu_{2}^{-}+\delta_{\lambda}^{-})(A_{-}^{2}-A^{2})],$$L_{22}^{+}=-(2 \mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})\frac{A_{+}^{2}-A^{2}}{B_{+}A_{+}-A^{2}}B_{+},$

and$L_{ij}=L_{ij}^{+}+L_{ij}^{-},$$L=(\begin{array}{ll}L_{11} L_{12}L_{21} L_{22}\end{array})$$L_{22}^{-}=-(2 \mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})\frac{A_{-}^{2}-A^{2}}{B_{-}A_{-}-A^{2}}B_{-}$, we obtain$L(\begin{array}{ll}i\xi’ \beta_{-}’\beta_{N}^{-} \end{array})=(^{-i\xi’\cdot\hat{h’}-L_{11}^{+}i\xi’\cdot\hat{k’}-L_{12}^{+}\hat{k_{N}}}-\hat{h_{N}}-L_{21}^{+}i\xi’\cdot\hat{k’}-L_{22}^{+}\hat{k_{N}})$If $\det L\neq 0$ , we have the inverse of$L$.

(2.17)and obtain$(\begin{array}{ll}i\xi’ \beta_{-}’\beta_{N}^{-} \end{array})=\frac{1}{\det L}(\begin{array}{ll}L_{22} -L_{12}-L_{21} L_{11}\end{array})(^{-i\xi’\cdot\hat{h’}-L_{11}^{+}i\xi’\cdot\hat{k’}-L_{12}^{+}\hat{k_{N}}}- \hat{h_{N}}-L_{21}^{+}i\xi’\cdot\hat{k’}-L_{22}^{+}\hat{k_{N}})$

Then we get the formula ofandby (2.12), (2.13) and (2.14). Since wehave the formula ofby (2.11), we can obtain the solution formula of $(1.1)-(1.4)$ if$\det L\neq$ O.
In next section, we shall consider the Lopantinski determinant $\det L$ when$i\xi’\cdot\alpha_{\pm}’,$$\alpha_{N}^{\pm}$$\beta_{J}^{+}$$\alpha_{j}^{\pm}$$\lambda\in\Lambda_{\epsilon,\lambda_{0}}=\Sigma_{\epsilon,\lambda_{0}}\cap K_{\epsilon}.$

2.2

Analysis of Lopatinski determinant

In order to analyze Lopatinski determinant, we shall prove the following lemma, which
is one of the essential steps in this article.

Lemma 2.1. Let$L$

be the matrix defined in section 2.1.

(I) there exists a positive constant$\omega$depending on$\mu_{1}^{\pm},$$\mu_{2}^{\pm},$$\lambda_{0}$$\epsilon,$

and$\delta_{0}$such that$|A\det L|\geq\omega(|\lambda|^{1/2}+A)^{3}$for any(2.18)$(\lambda, \xi’)\in\tilde{\Gamma}_{\epsilon,\lambda_{0}}.$(II) For any multi-index$\kappa’\in \mathbb{N}_{0}^{N-1}$

and$(\lambda, \xi’)\in\tilde{\Gamma}_{\epsilon,\lambda_{0}}$, the following inequalities hold:$|\partial_{\xi}^{\kappa’},\{(\tau\partial_{\tau})^{\ell}(A\det L)^{-1}\}|\leq C_{\kappa’}(|\lambda|^{1/2}+A)^{-3}A^{-|\kappa’|}, (\ell=0,1)$

Proof. Since we can prove (2.19) by using Leibniz rule and the Bell formula:$\partial_{\xi}^{\kappa’},f(g(\xi’))=\sum_{\ell=1}^{|\kappa’|}f^{(\ell)}(g(\xi’))\sum_{\kappa_{1}’+\cdots+\kappa_{\ell}’=\kappa’}, |\kappa_{i}’|\geq 1^{\Gamma_{\kappa_{1\rangle\rangle}\kappa_{\ell}’}^{\kappa’}(\partial_{\xi}^{\kappa_{1}’},g(\xi’))\ldots(\partial_{\xi}^{\kappa_{\ell}},g(\xi’))}$(2.19)

68
with $f(t)=1/t$ and $g(\xi’)=A\det L$ with (2.18), it is suﬃcient to prove (2.18). In order, (iii), (ii)
to prove (2.18), we consider the three cases: (i)andfor large. We see thatwith large
First we consider the case:and $\beta>0$ by elemental calculation. By
for anya very small positive constant suchthere
existsusing this inequality, we notice thatforthat
$B\pm=A(1+O(\delta_{3}))$$A_{\pm}=A(1+O(\delta_{3}))$.
we canTherefore
,
as
small
Therefore we haveobtain$R_{2}A\leq|\lambda|^{1/2}$$R_{1}|\lambda|^{1/2}\leq A$$R_{2}^{-1}|\lambda|^{1/2}\leq A\leq R_{1}|\lambda|^{1/2}$$R_{2}.$$R_{1}$$R_{1}\geq 1$$R_{1}|\lambda|^{1/2}\leq A$$(\sin\epsilon/2)(\alpha|\lambda|+\beta)$$\lambda\in\Sigma_{\epsilon},$$\xi\in \mathbb{R}^{N}$$|\alpha\lambda+\beta|\geq$$\alpha,$$\delta_{3}$$|(s_{1}\mu_{1}^{\pm}+s_{2}\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})^{-1}\lambda A^{-2}|\leq(\sin(\epsilon/2))^{-1}(s_{1}\mu_{1}^{\pm}+s_{2}\mu_{2}^{\pm})^{-1}R_{1}^{-2}\leq\delta_{3}$$s_{1},$$s_{2}\in \mathbb{R}.$$\delta_{3}$$L_{11}^{\pm}=- \frac{\mu_{1}^{\pm}(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})}{3\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm}}A(2+O(\delta_{3}))$$L_{21}^{\pm}= \mp\frac{2(\mu_{1}^{\pm})^{2}}{3\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm}}(1+O(\delta_{3}))$,,$L_{12}^{\pm}= \mp\frac{2(\mu_{1}^{\pm})^{2}}{3\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm}}A^{2}(1+O(\delta_{3}))$,$L_{22}^{\pm}=- \frac{2(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})(\mu_{1}^{\pm})}{3\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm}}A(1+O(\delta_{3}))$,(2.20)

which imply that$\det L=(\mu_{1}^{+}+\frac{\mu_{1}^{-}(\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-})}{3\mu_{1}^{-}+\mu_{2}^{-}+\delta_{\lambda}^{-}})(\frac{\mu_{1}^{+}(\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+})}{3\mu_{1}^{+}+\mu_{2}^{+}+\delta_{\lambda}^{+}}+\mu_{1}^{-})A^{2}(4+O(\delta_{3}))$

Taking the fact:$\mu_{1}^{\pm}>0,$ $\mu_{1}^{\pm}+\mu_{2}^{\pm}>0$.

into account, we see$| \mu_{1}^{\pm}+\frac{\mu_{1}^{\mp}(\mu_{1}^{\mp}+\mu_{2}^{\mp}+\delta_{\lambda}^{\mp})}{3\mu_{1}^{\mp}+\mu_{2}^{\mp}+\delta_{\lambda}^{\mp}}|=|\frac{\mu_{1}^{\pm}(3\mu_{1}^{\mp}+\mu_{2}^{\mp})+\mu_{1}^{\mp}(\mu_{1}^{\mp}+\mu_{2}^{\mp})+\delta_{\lambda}^{\mp}(\mu_{1}^{\pm}+\mu_{1}^{\mp})}{3\mu_{1}^{\mp}+\mu_{2}^{\mp}+\delta_{\lambda}^{\mp}}|>0.$

Summing up, we can show that there exists a positive constant such that $|\det L|\geq\omega A^{2}.$is shown in a similar way to the casefor large
Since the case
we omit the caseandFinally, we consider the caseSet$\omega$$R_{2}A\leq|\lambda|^{1/2}$$A\geq R_{1}|\lambda|^{1/2},$$R_{2}$$R_{2}A\leq|\lambda|^{1/2}.$$\tilde{\lambda}=\lambda/(|\lambda|^{1/2}+A)^{2}$$R_{2}^{-1}|\lambda|^{1/2}\leq A\leq R_{1}|\lambda|^{1/2}.$$\overline{A\pm}=\sqrt{(2\mu_{1}^{\pm}+\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})^{-1}\tilde{\lambda}+\tilde{A}^{2}},$$\overline{B\pm}=\sqrt{(\mu_{1}^{\pm})^{-1}\tilde{\lambda}+\tilde{A}^{2}}$$\tilde{A}=\frac{A}{|\lambda|^{1/2}+A},$

and$D(R_{1}, R_{2})$$=\{(\tilde{\lambda},\tilde{A})|(1+R_{1})^{-2}\leq|\tilde{\lambda}|\leq R_{2}^{2}(1+R_{2})^{2}, (1+R_{2})^{-1}\leq\tilde{A}\leq R_{1}(1+R_{1})^{-1}\}.$

satisfies the conditionbyandby replacing
and , respectively. And we set$\det
L=(|\lambda|^{1/2}+A)^{2}\det\tilde{L}.$and then we haveprovided that
andby
We shall prove that$\det\tilde{L}=0$
$\det
L=0$
, namely. In this case, incontradiction. To this end, we assume thatview of (2.17) that $\vec{w}_{\pm}(x_{N})=(w_{\pm,1}(x_{N}), \ldots, w_{\pm,N}(x_{N}))\neq(0, \ldots, 0)$ satisfy $(2.1)-(2.4)$
We remarkWe also define$(\tilde{\lambda},\tilde{A})\in D(R_{1}, R_{2})$$\tilde{L}_{ij}$if$R_{2}^{-1}|\lambda|^{1/2}\leq A\leq R_{1}|\lambda|^{1/2}.$$(\lambda, \xi’)$$A\pm,$ $A$$B_{\pm}$$\overline{A\pm},$$\tilde{A}$$\tilde{B}$$\det\tilde{L}=\tilde{L}_{11}\tilde{L}_{22}-\tilde{L}_{12}\tilde{L}_{21}$$\det\tilde{L}\neq 0$$(\tilde{\lambda},\tilde{A})\in D(R_{1}, R_{2})$$\tilde{\lambda}\in\Sigma_{\epsilon}$

69
with

$\hat{h}_{j}(0)=0,$ $\hat{h}_{N}(0)=0$

and$\hat{k}_{J}(0)=0$, that is, they satisfy the following homogeneous

equations:$\lambda w_{\pm,j}-\mu_{1}^{\pm}\sum_{\ell=1}^{N-1}i\xi_{l}^{(}i\xi_{j}w_{\pm,l}+i\xi_{\ell}w_{\pm,j})$$-\mu_{1}^{\pm}D_{N}(i\xi_{j}w_{\pm,N}+D_{N}w_{\pm,j})-(\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})i\xi_{j}(i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N})=0$,(2.21)$\lambda w_{\pm,N}-\mu_{1}^{\pm}\sum_{\ell=1}^{N-1}i\xi_{\ell}(D_{N}w\pm,\ell+i\xi_{\ell}w_{\pm,N})-2\mu_{1}^{\pm}D_{N}^{2}w_{\pm,N}$,$\mu_{1}^{+}(D_{N}w_{+,j}+i\xi_{j}w_{+,N})|_{x_{N}=0}-\mu_{1}^{-}(D_{N}w_{-,j}+i\xi_{j}w_{-,N})|_{x_{N}=0}=0$ ,(2.22)(2.23)$-(\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})D_{N}(i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N})=0$$2\mu_{1}^{-}D_{N}w_{+,N}+(\mu_{2}^{+}+\delta_{\lambda}^{+})(i\xi’\cdot w_{+}’+D_{N}w_{+,N})|_{x_{N}=0}$$-(2\mu_{1}^{-}D_{N}w_{-,N}+(\mu_{2}^{-}+\delta_{\lambda}^{-})(i\xi’\cdot w_{-}’+D_{N}w_{-,N})|_{x_{N}=0}=0$.

(2.24)

Here we set$(a, b)_{\pm}= \pm\int_{0}^{\pm\infty}a(x_{N})\overline{b(x_{N})}dx_{N}, \Vert a\Vert_{\pm}=\sqrt{(a,a)_{\pm}}.$Multipling (2.21) by$\overline{w_{\pm,j}}$and (2.22) by$\overline{w_{\pm,N}}$and by integration by parts, we obtain$\lambda\Vert w_{\pm,j}\Vert_{\pm}^{2}+\mu_{1}^{\pm}\sum_{\ell=1}^{N-1}((i\xi_{\ell}, w_{\pm,\ell}, i\xi_{j\pm,j}w)_{\pm}+\Vert i\xi_{\ell}w_{\pm,j}\Vert_{\pm}^{2})+\mu_{1}^{\pm}(i\xi_{j}w_{\pm,N}, D_{N}w_{\pm,j})_{\pm}$$+\mu_{1}^{\pm}\Vert D_{N}w_{\pm,j}\Vert_{\pm}^{2}+(\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})((i\xi’\cdot w_{\pm}’, i\xi_{j}w_{\pm,j})_{\pm}+(D_{N}w_{\pm,N}, i\xi_{j}w_{\pm,j})_{\pm})=0$

and$\lambda\Vert w_{\pm,N}\Vert_{\pm}^{2}+\mu_{1}^{\pm}\sum_{\ell=1}^{N-1}((D_{N}w_{\ell,\ell}, i\xi_{\ell\pm,N}w)_{\pm}+\Vert i\xi_{\ell}w_{\pm,N}\Vert_{\pm}^{2})+2\mu_{1}^{\pm}\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2}$$+(\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})((i\xi’\cdot w_{\pm}’, D_{N}w_{\pm,N})_{\pm}+\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2})=0.$

Summing up, we see$0= \lambda\sum_{j=1}^{N}\Vert w_{\pm,j}\Vert_{\pm}^{2}$$+ \mu_{1}^{\pm}(\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+\sum_{\ell,j=1}^{N-1} IIi\xi_{\ell\pm,j}w\Vert_{\pm}^{2}+\sum_{j=1}^{N-1}(i\xi_{\ell}w_{\pm,N},D_{N}w_{\pm,j})_{\pm}$$+ \sum_{j=1}^{N-1}\Vert D_{N}w_{\pm,j}\Vert_{\pm}^{2}+\sum_{l=1}^{N-1}((D_{N}w_{\pm,\ell},i\xi_{\ell\pm,N}w)_{\pm}+\Vert i\xi_{\ell}w_{\pm,N}\Vert_{\pm}^{2})+2\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2})$$+(\mu_{2}^{\pm}+\delta_{\lambda}^{\pm})(\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+(i\xi’, w_{\pm}’,D_{N}w_{\pm,N})_{\pm}+(D_{N}w_{\pm,N},i\xi’\cdot w_{\pm}’)_{\pm}+\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2})$(2.25)

70
Taking the fact$\delta_{\lambda}^{\pm}=\perp\gamma^{\pm}\gamma^{\pm}|\lambda|\#({\rm Re}\lambda-{\rm Im}\lambda)$

and$\Vert i\xi_{j}w_{\pm,j}+D_{N}w_{\pm,j}\Vert_{\pm}^{2}$$=(i\xi_{\ell}w_{\pm,N}, D_{N}w_{\pm,j})_{\pm}+\Vert D_{N}w_{\pm,j}\Vert_{\pm}^{2}+(D_{N}w_{\pm,j}, i\xi_{j}w_{\pm,N})_{\pm}+\Vert i\xi_{j}w_{\pm,N}\Vert_{\pm}^{2},$$\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}^{2}$$=\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+(i\xi’\cdot w_{\pm}’, D_{N}w_{\pm,N})_{\pm}+(D_{N}w_{\pm,N}, i\xi’\cdot w_{\pm}’)_{\pm}+\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2},$into account and taking the real part and the imaginary part in (2.25), we have
(2.26)$({\rm Im} \lambda)(\sum_{j=1}^{N}\Vert w_{\pm,j}\Vert_{\pm}^{2}-\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{|\lambda|^{2}}\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}^{2})=0$

and${\rm Re} \lambda\sum_{j=1}^{N}\Vert w_{\pm,j}\Vert_{\pm}^{2}$$+ \mu_{1}^{\pm}(\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+\sum_{\ell,j=1}^{N-1}\Vert i\xi_{l}w_{\pm,j}\Vert_{\pm}^{2}+\sum_{j=1}^{N-1}\Vert i\xi_{j}w_{\pm,j}+D_{N}w_{\pm,j}\Vert_{\pm}^{2}+2\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2})$$+( \mu_{2}^{\pm}+\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{|\lambda|^{2}}{\rm Re}\lambda)\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}^{2}=0$

When${\rm Im}\lambda=0$$w\pm\neq 0$. Whenand${\rm Re}\lambda\geq 0$${\rm Im}\lambda\neq 0$, we see$\Vert w_{\pm,j}\Vert_{\pm}=0$.

(2.27), namely $w\pm=0$ , which contradict to, by (2.26), (2.27) and$\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+\sum_{\ell,j=1}^{N-1}\Vert i\xi_{\ell}w_{\pm,j}\Vert_{\pm}^{2}+2\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2}\geq 2(\Vert i\xi’\cdot w_{\pm}’\Vert_{\pm}^{2}+\Vert D_{N}w_{\pm,N}\Vert_{\pm}^{2})$$\geq\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}^{2},$we obtain$\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}^{2}(2{\rm Re}\lambda\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{|\lambda|^{2}}+\mu_{1}^{\pm}+\mu_{2}^{\pm})+\mu_{1}^{\pm}\sum_{j=1}^{N-1}\Vert i\xi_{j}w_{\pm,j}+D_{N}w_{\pm,j}\Vert_{\pm}^{2}\leq 0.$

since$\mu_{1}^{\pm}>0$

and$2{\rm Re} \lambda\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{|\lambda|^{2}}+\mu_{1}^{\pm}+\mu_{2}^{\pm}=\frac{\mu_{1}^{\pm}+\mu_{2}^{\pm}}{|\lambda|^{2}}(({\rm Re}\lambda+\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{\mu_{1}^{\pm}+\mu_{2}^{\pm}})^{2}+({\rm Im}\lambda)^{2}-(\frac{\gamma_{1}^{\pm}\gamma_{2}^{\pm}}{\mu_{1}^{\pm}+\mu_{2}^{\pm}}))$,

namely $w\pm=0$ by (2.26) which
impliesthe condition$w\pm\neq0$. Therefore we see that there exists a positive constant such that
contradict to. Therefore we obtain $|\det L|\geq c(|\lambda|^{1/2}+A)^{2}$ , which implies Lemma 2.1.
$\lambda\in K_{\epsilon,\lambda_{0}}$$\Vert i\xi’\cdot w_{\pm}’+D_{N}w_{\pm,N}\Vert_{\pm}=0$$c$$|\det\tilde{L}|\geq c$$\square$71
2.3

Technical Lemma

In this section, we shall introduce one of technical lemmas which we need to prove Theorem 1.2. In order to prove the -boundedness of solution operator, we use the following
lemmas which is proven by Kubo, Shibata and Soga [2].
$\mathcal{R}$Lemma 2.2. Let
be a domain inand set$(i=1,2)$ be multipliers defined onsuch that$\Lambda$$\mathbb{C}$$\tilde{\Lambda}=\Lambda\cross(\mathbb{R}^{N-1}\backslash \{0\})$. Let$n_{i}(\lambda, \xi’)$$\tilde{\Lambda}$$|\partial_{\xi}^{\kappa’},\{(\tau\partial_{\tau})^{\ell}n_{1}(\lambda, \xi \leqC_{\kappa’}(|\lambda|^{1/2}+A)^{-2}A^{-|\kappa’|},$$|\partial_{\xi}^{\kappa’},\{(\tau\partial_{\tau})^{p}n_{2}(\lambda, \xi \leq C_{\kappa’}(|\lambda|^{1/2}+A)^{-1-|\kappa’|}$$(\ell=0,1)$

for any$\kappa’\in \mathbb{N}_{0}^{N-1}$

and$(\lambda, \xi’)\in\tilde{\Lambda}$. Let $K_{i}^{\pm}(i=1,2,3,4)$ be operators defined$by$$K_{1}^{\pm}( \lambda)_{9}=\pm\int_{0}^{\pm\infty}\mathcal{F}_{\xi’}^{-1}[n_{1}(\lambda, \xi’)AA_{\pm}M_{\pm}(x_{N}+y_{N})\hat{g}(\xi’, y_{N})](x’)dy_{N},$$K_{2}^{\pm}( \lambda)g=\pm\int_{0}^{\pm\infty}\mathcal{F}_{\xi’}^{-1}[n_{1}(\lambda, \xi’)Ae^{\mp B\pm(x_{N}+y_{N})}\hat{9}(\xi’, y_{N})](x’)dy_{N},$$K_{3}^{\pm}( \lambda)_{9}=\pm\int_{0}^{\pm\infty}\mathcal{F}_{\xi’}^{-1}[n_{2}(\lambda, \xi’)A_{\pm}M_{\pm}(x_{N}+y_{N})\hat{g}(\xi’, y_{N})](x’)dy_{N},$$K_{4}^{\pm}( \lambda)g=\pm\int_{0}^{\pm\infty}\mathcal{F}_{\xi’}^{-1}[n_{2}(\lambda, \xi’)e^{\mp B\pm(x_{N}+y_{N})}\hat{g}(\xi’, y_{N})](x’)dy_{N},$where$M_{\pm}(x_{N})= \frac{e^{\mp B\pm x_{N}}-e^{\mp A\pm x_{N}}}{B_{\pm}-A_{\pm}}.$Then, there exists a constant$C$such that$\mathcal{R}_{\mathcal{L}(L_{q}(\pi_{\pm}^{N}),L_{q}(\pi_{\pm}^{N})^{1+N+N^{2}})}(\{(\tau\partial_{\tau})^{p}G_{\lambda}K_{i}^{\pm}(\lambda)|\lambda\in\Lambda\})\leq C$

where$G_{\lambda}$

is an operator defined by$(\ell=0,1, i=1,2,3,4)$ ,$G_{\lambda}u=(\lambda u, \gamma u, \lambda^{1/2}\nabla u, \nabla^{2}u)$.

By the Volevich trick;$a(x_{N})b(0)=- \int_{0}^{\infty}\{a’(x_{N}+y_{N})b(\dot{y}_{N})+a(x_{N}+y_{N})b’(y_{N})\}dy_{N}$$= \int_{-\infty}^{0}\{a’(x_{N}+y_{N})b(y_{N})+a(x_{N}+y_{N})\dot{b}’(y_{N})\}dy_{N},$

we can reduce the solution formula obtained in section 2.1 into the form which we canapply Lemma 2.2. We can check that the multipliers in the solution formula satisfy thecondition of Lemma 2.2. Therefore we can prove the -boundedness of solution operator
to problem $(1.1)-(1.4)$ , namely we can show the main theorem.
$\mathcal{R}$72

References[1] I. V. Denisova, “Evolution of compressible and imcompressible fluids separated by aclosed interface Interface Free Bound. 2, No.3, (2000), 283-312.

[2] T. Kubo, Y. Shibata and K. Soga, “On the -boundedness for the Two phaseprolem:compressible incompressible model prolem (submitted).
$\mathcal{R}$[3] T. Kubo, Y. Shibata and K. Soga, “On two phase problem:compressible compressiblemodel prolem (preprint)[4] A. Tani, “On the free boundary value problem for compressible viscous fluid motion
J. Math. Kyoto Univ., 21,(1981), 839-859.

[5] A. Tani, “Two-phase free boundary problem for compressible viscous fluid motion
J. Math. Kyoto Univ., 24, (1984), 243-267.