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Subsonic flows with a contact discontinuity in a finitely long axisymmetric cylinder

Shangkun Weng School of mathematics and statistics, Wuhan University, Wuhan, Hubei Province, 430072, People’s Republic of China. Email: [email protected] Zihao Zhang School of mathematics and statistics, Wuhan University, Wuhan, Hubei Province, 430072, People’s Republic of China. Email: [email protected]

Abstract

This paper concerns the structural stability of subsonic flows with a contact discontinuity in a finitely long axisymmetric cylinder. We establish the existence and uniqueness of axisymmetric subsonic flows with a contact discontinuity by prescribing the horizontal mass flux distribution, the swirl velocity, the entropy and the Bernoulli’s quantity at the entrance and the radial velocity at the exit. It can be formulated as a free boundary problem with the contact discontinuity to be determined simultaneously with the flows. Compared with the two-dimensional case, a new difficulty arises due to the singularity near the axis. One of the key points in the analysis is the introduction of an invertible modified Lagrangian transformation which can overcome this difficulty and straighten the contact discontinuity. Another one is to utilize the deformation-curl decomposition for the steady Euler system introduced in [17] to effectively decouple the hyperbolic and elliptic modes. Finally, the contact discontinuity will be located by using the implicit function theorem.

Mathematics Subject Classifications 2010: 35J15, 35L65, 76J25, 76N15.

Key words: contact discontinuity, structural stability, the modified Lagrangian transformation, the deformation-curl decomposition.

1 Introduction

In this paper, we are concerned with the structural stability of subsonic flows with a contact discontinuity governed by the three-dimensional steady full Euler system in a finitely long axisymmetric cylinder. The three-dimensional steady full Euler system for compressible inviscid gas is of the following form:

\displaystyle\begin{cases}{\partial}_{x_{1}}(\rho u_{1})+{\partial}_{x_{2}}(\rho u_{2})+{\partial}_{x_{3}}(\rho u_{3})=0,\\ {\partial}_{x_{1}}(\rho u_{1}^{2})+{\partial}_{x_{2}}(\rho u_{1}u_{2})+{\partial}_{x_{3}}(\rho u_{1}u_{3})+{\partial}_{x_{1}}P=0,\\ {\partial}_{x_{1}}(\rho u_{1}u_{2})+{\partial}_{x_{2}}(\rho u_{2}^{2})+{\partial}_{x_{3}}(\rho u_{2}u_{3})+{\partial}_{x_{2}}P=0,\\ {\partial}_{x_{1}}(\rho u_{1}u_{3})+{\partial}_{x_{2}}(\rho u_{2}u_{3})+{\partial}_{x_{3}}(\rho u_{3}^{2})+{\partial}_{x_{3}}P=0,\\ {\partial}_{x_{1}}(\rho u_{1}E+u_{1}P)+{\partial}_{x_{2}}(\rho u_{2}E+u_{2}P)+{\partial}_{x_{3}}(\rho u_{3}E+u_{3}P)=0,\\ \end{cases}

(1.1)

where $\bm{u}=(u_{1},u_{2},u_{3})$ is the velocity, $\rho$ is the density, $P$ is the pressure, $E$ is the energy, respectively. For polytropic gas, the equation of state and the energy are of the form

P=A(S)\rho^{\gamma},\quad{\rm{and}}\quad E=\frac{1}{2}|{\bm{u}}^{2}|+\frac{P}{(\gamma-1)\rho},

where $A(S)=Re^{S}$ and $\gamma\in(1,+\infty)$ , $R$ are positive constants. Denote the Bernoulli’s function and the local sonic speed by $B=\frac{1}{2}|{\textbf{u}}|^{2}+\frac{\gamma P}{(\gamma-1)\rho}$ and $c(\rho,A)=\sqrt{A\gamma}\rho^{\frac{\gamma-1}{2}}$ , respectively. Then the system (1.1) is hyperbolic for supersonic flows ( $|\textbf{u}|>c(\rho,A)$ ), and hyperbolic-elliptic coupled for subsonic flows ( $|\textbf{u}|<c(\rho,A)$ ).

To understand the contact discontinuity surface, we first give the definition of steady flows with a contact discontinuity. Let $\mathcal{D}\in\mathbb{R}^{3}$ be an open and connected domain. Suppose that a non-self-intersecting $C^{1}$ -curve $\Gamma$ divides $\mathcal{D}$ into two disjoint open subsets $\mathcal{D}^{\pm}$ such that $\mathcal{D}=\mathcal{D}^{-}\cup\Gamma\cup\mathcal{D}^{+}$ . Assume that $\bm{U}=(\rho,u_{1},u_{2},u_{3},P)$ satisfies the following properties:

(1)

$\bm{U}\in[L^{\infty}(\mathcal{D})\cap C^{1}_{loc}(\mathcal{D}^{\pm})\cap C^{0}_{loc}(\mathcal{D}^{\pm}\cup\Gamma)]^{5}$ ;

(2)

For any $\eta\in C_{0}^{\infty}(\mathcal{D})$ ,

\begin{cases}\int_{\mathcal{D}}(\rho u_{1}{\partial}_{x_{1}}\eta+\rho u_{2}{\partial}_{x_{2}}\eta+\rho u_{3}{\partial}_{x_{3}}\eta){\mathrm{d}}\mathbf{x}=0,\\ \int_{\mathcal{D}}((\rho u_{1}^{2}+P){\partial}_{x_{1}}\eta+\rho u_{1}u_{2}{\partial}_{x_{2}}\eta+\rho u_{1}u_{3}{\partial}_{x_{3}}\eta){\mathrm{d}}\mathbf{x}=0,\\ \int_{\mathcal{D}}(\rho u_{1}u_{2}{\partial}_{x_{1}}\eta+(\rho u_{2}^{2}+P){\partial}_{x_{2}}\eta+\rho u_{2}u_{3}{\partial}_{x_{3}}\eta){\mathrm{d}}\mathbf{x}=0,\\ \int_{\mathcal{D}}(\rho u_{1}u_{3}{\partial}_{x_{1}}\eta+\rho u_{2}u_{3}{\partial}_{x_{2}}\eta+(\rho u_{3}^{2}+P){\partial}_{x_{3}}\eta){\mathrm{d}}\mathbf{x}=0,\\ \int_{\mathcal{D}}(\rho u_{1}(E+\frac{P}{\rho}){\partial}_{x_{1}}\eta+\rho u_{2}(E+\frac{P}{\rho}){\partial}_{x_{2}}\eta+\rho u_{3}(E+\frac{P}{\rho}){\partial}_{x_{3}}\eta){\mathrm{d}}\mathbf{x}=0.\\ \end{cases}

(1.2)

By integration by parts, we get the Rankine-Hugoniot conditions:

\begin{cases}n_{1}[\rho u_{1}]+n_{2}[\rho u_{2}]+n_{2}[\rho u_{3}]=0,\\ n_{1}[\rho u_{1}^{2}]+n_{1}[P]+n_{2}[\rho u_{1}u_{2}]+n_{3}[\rho u_{1}u_{3}]=0,\\ n_{1}[\rho u_{1}u_{2}]+n_{2}[\rho u_{2}^{2}]+n_{2}[P]+n_{3}[\rho u_{2}u_{3}]=0,\\ n_{1}[\rho u_{1}u_{3}]+n_{2}[\rho u_{2}u_{3}]+n_{3}[\rho u_{3}^{2}]+n_{3}[P]=0,\\ n_{1}[\rho u_{1}(E+\frac{P}{\rho})]+n_{2}[\rho u_{2}(E+\frac{P}{\rho})]+n_{3}[\rho u_{3}(E+\frac{P}{\rho})]=0,\\ \end{cases}

(1.3)

where $\mathbf{n}=(n_{1},n_{2},n_{3})$ is the unit normal vector to $\Gamma$ , and $[F](\mathbf{x})=F_{+}(\mathbf{x})-F_{-}(\mathbf{x})$ denotes the jump across the surface $\Gamma$ for a piecewise smooth function $F$ .

Let $\bm{\tau}_{1}=(\tau_{11},\tau_{21},\tau_{31})$ and $\bm{\tau}_{2}=(\tau_{12},\tau_{22},\tau_{32})$ as the unit tangential vectors to $\Gamma$ , which means that $\mathbf{n}\cdot{\bm{\tau}_{k}}=0$ . Taking the dot product of $(\eqref{1-3}_{2},\eqref{1-3}_{3},\eqref{1-3}_{4})$ with $\mathbf{n}$ and $\bm{\tau}_{k}$ respectively, one has

[\rho(\mathbf{u}\cdot\mathbf{n})^{2}+P]_{\Gamma}=0,\quad\rho(\mathbf{u}\cdot\mathbf{n})[\mathbf{u}\cdot\bm{\tau}_{k}]_{\Gamma}=0,\quad{\rm{for}}\quad k=1,2.

(1.4)

Assume that $\rho>0$ in $\bar{\mathcal{D}}$ , (1.4) implies either $\mathbf{u}\cdot\mathbf{n}=0$ on $\Gamma$ or $[\mathbf{u}\cdot\bm{\tau}_{k}]_{\Gamma}=0$ . If $\mathbf{u}\cdot\mathbf{n}\neq 0$ and $[\mathbf{u}\cdot\bm{\tau}_{k}]_{\Gamma}=0$ hold on $\Gamma$ , the surface $\Gamma$ is called a shock; if the flow moves along both sides of $\Gamma$ such that $\mathbf{u}\cdot\mathbf{n}=0$ on $\Gamma$ , the surface $\Gamma$ is called a contact discontinuity. In the latter case, $\mathbf{u}\cdot\mathbf{n}=0$ and the first equation in (1.4) give $[P]=0$ . Then we get the R-H conditions corresponding to a contact discontinuity as follows:

\mathbf{u}\cdot\mathbf{n}=0\quad{\rm{and}}\quad[P]=0\quad{\rm{on}}\ \Gamma.

(1.5)

Definition 1.1.

We define $\bm{U}=(\rho,u_{1},u_{2},u_{3},P)$ to be a weak solution of the full Euler system (1.1) in $\mathcal{D}$ with a contact discontinuity $\Gamma$ if the the following properties hold:

(i)

$\Gamma$ is a non-self-intersecting $C^{1}$ -curve dividing $\mathcal{D}$ into two disjoint open subsets $\mathcal{D}^{\pm}$ such that $\mathcal{D}=\mathcal{D}^{-}\cup\Gamma\cup\mathcal{D}^{+}$ ;
(ii)

$\bm{U}$ satisfies $\rm{(1)}$ and $\rm{(2)}$ ;
(iii)

$\rho>0$ in $\bar{\mathcal{D}}$ ;
(iv)

$(\mathbf{u}|_{\bar{\mathcal{D}}_{-}\cap\Gamma}-\mathbf{u}|_{\bar{\mathcal{D}}_{+}\cap\Gamma})\neq\mathbf{0}$ holds for all $\mathbf{x}\in\Gamma$ ;
(v)

$\mathbf{u}\cdot\mathbf{n}=0$ and $[P]=0$ on $\Gamma$ .

This is a continuous work on the study of subsonic Euler flows with a contact discontinuity in a finitely long nozzle. In the previous work [20], we established the existence and uniqueness of subsonic flows with a contact discontinuity in a two-dimensional finitely long slightly curved nozzle. As an attempt to extend the results from [20] to the three dimensional case, this paper investigates the structural stability of subsonic flows with a contact discontinuity in a finitely long axisymmetric cylinder under the suitable axisymmetric perturbations of boundary conditions.

The study on the steady compressible flows with a contact discontinuity is not only of fundamental importance in developing the mathematical theory of partial differential equations arising from fluid dynamics, but also has important applications to engineering designs, such as rocket launching, sharp charged jet and so on. Up to now, there have been many works in the literature on the steady flows with a contact discontinuity. For the subsonic flow, the stability of flat contact discontinuity in infinite nozzles was established in [2] and [3, 4]. The authors in [3, 4] decompose the Rankine-Hugoniot conditions on the contact discontinuity via Helmholtz decomposition so that the compactness of approximated solutions can be achieved. The uniqueness and existence of the contact discontinuity in infinitely long nozzles, which is not a perturbation of the flat contact discontinuity, was obtained in [8]. The existence, uniqueness and stability of subsonic flows past an airfoil with a vortex line were obtained in [5]. The key idea in [5] is to use the implicit function theorem as the framework to solve the problem of subsonic flows past an airfoil with a vortex line. Inspired by [5], the stability of contact discontinuity in a finite nozzle was established in [20] by using the implicit function theorem. For the supersonic flow, the stability of flat contact discontinuity in finitely long nozzles was studied in [11]. The stability of three-dimensional supersonic contact discontinuity was investigated in [15, 16]. Recently, the stability of supersonic contact discontinuity for the two-dimensional steady rotating Euler system in a finitely curved nozzle has been established in [21]. For the transonic flow, the stability of flat contact discontinuity in finitely long nozzles was established in [12]. The stability of two-dimensional transonic contact discontinuity over a solid wedge and three-dimensional transonic contact discontinuity were established in [6, 7] and [14].

We make some comments on the new ingredients in our analysis for the contact discontinuity problem. Note that the contact discontinuity is part of the solution and is unknown, thus this is a free boundary that separates both subsonic flows in the inner and outer layers of the cylinder. The strategy to overcome this difficulty in the two dimensional case [20] is to introduce a Lagrangian transformation to straighten the contact discontinuity. The idea also applies to three dimensional steady axisymmetric Euler system. However, in the three dimensional axisymmetric setting, there is a singular term $r$ in the density equation. Inspired by [19], the singular term $r$ in the density equation is of order $O(r)$ near the axis $r=0$ , hence we can find a simple modified Lagrangian transformation such that it is invertible near the axis and also straightens the contact discontinuity. Another key issue is to decompose the hyperbolic and elliptic modes in the steady axisymmetric Euler system. It is well-known that the steady axisymmetric Euler system is hyperbolic-elliptic mixed in subsonic regions, whose effective decomposition of elliptic and hyperbolic modes is crucial for developing a well-defined iteration. Here we will use the deformation-curl decomposition introduced in [17, 18] to effectively decouple the hyperbolic and elliptic modes in subsonic regions.

The other key ingredient in our analysis is to employ the implicit function theorem to locate the contact discontinuity. The idea is inspired by the discussion of the airfoil problem in [5]. We choose a suitable Hölder space and design a proper map to verify the conditions in the implicit function theorem. However, it seems quite difficult to verify that the isomorphism of the differential of the map for general background flows with a straight contact discontinuity. Here we choose the background outer-layer flow is stagnant and restrict the perturbation only on the entrance of the inner-outer flow. In this case, the outer-layer flow is fixed and one can prove the isomorphism, the contact discontinuity can be located by the implicit function theorem.

This paper will be arranged as follows. In Section 2, we formulate the problem of subsonic flows with a contact discontinuity in a finitely long axisymmetric cylinder and state the main result. In Section 3, the modified Lagrange transformation is employed to straighten the contact discontinuity and reformulate the free boundary value problem 2.2. Then we use the deformation-curl decomposition in [17, 18] to derive an equivalent system. Finally, we state the main steps to solve the free boundary problem 3.1. In Section 4, we first linearize the nonlinear system and solve the linear system in a suitable weighted H $\ddot{\rm{o}}$ lder space. Then the framework of the contraction mapping theorem can be used to find the solution of the nonlinear system. In Section 5, we choose a suitable H $\ddot{\rm{o}}$ lder space and design a proper map to verify the conditions in the implicit function theorem. Then by using the implicit function theorem, we locate the contact discontinuity. In Section 6, we finish the proof of the main theorem.

2 Mathematical formulation of the problem

In this section, we first construct a special class of subsonic Euler flows with a straight contact discontinuity in a finitely long axisymmetric cylinder. Then we give a detailed formulation of the stability problem for these background flows with a contact discontinuity and state the main result.

2.1 The background solutions

The axisymmetric cylinder (Fig 1) of the length $L$ is given by

\mathcal{N}:=\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:0<x_{1}<L,\ 0\leq x_{2}^{2}+x_{3}^{2}<1\}.

Refer to caption — Fig 1: Subsonic flows with a contact discontinuity in an axisymmetric cylinder

Consider two layers of steady smooth Euler flows separated by the cylindrical surface $r=\sqrt{x_{2}^{2}+x_{3}^{2}}=\frac{1}{2}$ satisfying the following properties:

(i)

The velocity and density of the outer and inner layers are given by $(0,0,0),\rho_{b}^{+}$ and $(u_{b}^{-},0,0),\rho_{b}^{-}$ , where $u_{b}^{-}>0$ and $\rho_{b}^{\pm}>0$ ;
(ii)

the pressure of both the outer and inner layers is given by the same positive constant $P_{b}$ ;
(iii)

the flows in the outer and inner layers are subsonic, i.e.,

$(u_{b}^{-})^{2}<\frac{\gamma P_{b}}{\rho_{b}^{-}}.$

Then

\bm{U}_{b}=\begin{cases}\bm{U}_{b}^{+}:=(\rho_{b}^{+},0,0,0,P_{b}),\quad&{\rm{for}}\quad\frac{1}{2}<r<1,\\ \bm{U}_{b}^{-}:=(\rho_{b}^{-},u_{b}^{-},0,0,P_{b}),\quad&{\rm{for}}\quad 0\leq r<\frac{1}{2},\\ \end{cases}

(2.1)

with a contact discontinuity on the surface $r=\frac{1}{2}$ satisfy the steady Euler system (1.1) in the sense of Definition 1.1, which will be called the background solutions in this paper. This paper is going to establish the structural stability of these background solutions under the suitable axisymmetric perturbations of boundary conditions.

2.2 The stability problem and the main result

Let $(x,r,\theta)$ be the cylindrical coordinates of $(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$ , that is

x=x_{1},\ r=\sqrt{x_{2}^{2}+x_{3}^{2}},\ \theta=\arctan\frac{{{x_{3}}}}{{{x_{2}}}}.

Any function $f({\bf x})$ can be represented as $f({\bf x})=f(x,r,\theta)$ , and a vector-valued function ${\bf h}({\bf x})$ can be represented as ${\bf h}({\bf x})=h_{x}(x,r,\theta)\,{\bf e}_{x}+h_{r}(x,r,\theta)\,{\bf e}_{r}+h_{\theta}(x,r,\theta)\,{\bf e}_{\theta}$ , where

{\bf e}_{x}=(1,0,0),\quad{\bf e}_{r}=(0,\cos\theta,\sin\theta),\quad{\bf e}_{\theta}=(0,-\sin\theta,\cos\theta).

We say that a function $f({\bf x})$ is axisymmetric if its value is independent of $\theta$ and that a vector-valued function ${\bf h}=(h_{x},h_{r},h_{\theta})$ is axisymmetric if each of functions $h_{x}({\bf x}),h_{r}({\bf x})$ and $h_{\theta}({\bf x})$ is axisymmetric.

Assume that

	$\displaystyle\rho({\textbf{x}})$	$\displaystyle=\rho(x,r),\quad P({\textbf{x}})=P(x,r),\quad A({\textbf{x}})=A(x,r),$
	$\displaystyle{\bf u}({\textbf{x}})$	$\displaystyle=u_{x}(x,r){\bf e}_{x}+u_{r}(x,r){\bf e}_{r}+u_{\theta}(x,r){\bf e}_{\theta},$

then (1.1) can be rewritten as

\begin{cases}\begin{aligned} &{\partial}_{x}(\rho u_{x})+{\partial}_{r}(\rho u_{r})+\frac{\rho u_{r}}{r}=0,\\ &\rho(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})u_{x}+{\partial}_{x}P=0,\\ &\rho(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})u_{r}-\frac{\rho u_{\theta}^{2}}{r}+{\partial}_{r}P=0,\\ &\rho(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})(ru_{\theta})=0,\\ &\rho(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})A=0.\\ \end{aligned}\end{cases}

(2.2)

The axis and boundary of the cylinder are denoted by $\Gamma_{a}$ and $\Gamma_{w}$ , i.e;

\Gamma_{a}:=\{(x,r)\in\mathbb{R}^{2}:0<x<L,r=0\},\quad\Gamma_{w}:=\{(x,r)\in\mathbb{R}^{2}:0<x<L,r=1\}.

(2.3)

The exit of the cylinder is denoted by

\Gamma_{L}:=\{(x,r):x=L,\ 0\leq r<1\}.

(2.4)

The entrance of the cylinder is separated into two parts:

\displaystyle\Gamma_{0}^{+}:=\{(x,r):x=0,\ \frac{1}{2}<r<1\},\quad\Gamma_{0}^{-}:=\{(x,r):x=0,\ 0\leq r<\frac{1}{2}\}.

(2.5)

At the entrance, we prescribe the boundary data for the horizontal mass distribution $J=\rho u_{x}$ , the swirl velocity $u_{\theta}$ , the entropy $A$ and the Bernoulli’s quantity $B$ :

(J,u_{\theta},A,B)(0,r)=\begin{cases}(0,0,A_{b}^{+},B_{b}^{+}),\quad&{\rm{on}}\quad\Gamma_{0}^{+},\\ (J_{0},\nu_{0},A_{0},B_{0})(r),\quad&{\rm{on}}\quad\Gamma_{0}^{-},\\ \end{cases}

(2.6)

where

A_{b}^{+}=\frac{P_{b}}{(\rho_{b}^{+})^{\gamma}},\quad B_{b}^{+}=\frac{\gamma P_{b}}{(\gamma-1)\rho_{b}^{+}},

and functions $(J_{0},\nu_{0},A_{0},B_{0})(r)\in\left(C^{1,\alpha}([0,\frac{1}{2}])\right)^{4}$ are close to the background solutions in some sense that will be clarified later. Moreover, the compatibility conditions hold:

\nu_{0}(0)={\partial}_{r}(\nu_{0},A_{0},{B}_{0})(0)=0,

(2.7)

since $(\nu_{0},A_{0},B_{0})(r)$ are $C^{1,\alpha}$ in $\Gamma_{0}^{-}$ . At the exit, the following boundary condition is satisfied:

{\bf u}\cdot{\bf e}_{r}=0,\quad{\rm{on}}\quad\Gamma_{L}.

(2.8)

We expect the flow in the cylinder will be separated by a contact discontinuity $\Gamma:=\{r=g_{cd}(x),0<x<L\}$ with $g_{cd}(0)=\frac{1}{2}$ , which divides the domain $\mathcal{N}$ into the subsonic and subsonic regions:

\mathcal{N}^{+}:=\mathcal{N}\cap\{g_{cd}(x)<r<1\},\quad\mathcal{N}^{-}:=\mathcal{N}\cap\{0\leq r<g_{cd}(x)\}.

(2.9)

Let

{\bm{U}}(x,r)=\begin{cases}{\bm{U}}^{+}(x,r):=(\rho^{+},u_{x}^{+},u_{r}^{+},u_{\theta}^{+},P^{+})(x,r)\quad{\rm{in}}\quad\mathcal{N}^{+},\\ {\bm{U}}^{-}(x,r):=(\rho^{-},u_{x}^{-},u_{r}^{-},u_{\theta}^{-},P^{-})(x,r)\quad{\rm{in}}\quad\mathcal{N}^{-}.\\ \end{cases}

Along the contact discontinuity $r=g_{cd}(x)$ , the following Rankine-Hugoniot conditions hold:

{\bf u}\cdot{\bf n}_{cd}=0,\quad P^{+}=P^{-},\quad{\rm{on}}\quad\Gamma,

(2.10)

where

{\bf n}_{cd}=\frac{-g_{cd}^{\prime}(x){\bf e}_{x}+{\bf e}_{r}}{\sqrt{1+|g_{cd}^{\prime}(x)|^{2}}}.

On the nozzle wall $\Gamma_{w}$ , the flow satisfies the slip condition $\textbf{u}^{+}\cdot\textbf{n}^{+}=0$ , where $\textbf{n}^{+}$ is the outer normal vector of the nozzle wall. Using cylindrical coordinates, the slip boundary condition can be rewritten as

u_{r}^{+}(x,1)=0,\quad{\rm{on}}\quad\Gamma_{w}.

(2.11)

Moreover, since the flow is smooth near the axis $\Gamma_{a}$ , thus we have the following compatibility conditions:

u_{r}^{-}(x,0)=u_{\theta}^{-}(x,0)=0,\quad\forall x\in[0,L].

(2.12)

In summary, we will investigate the following problem:

Problem 2.1.

Given functions $(J_{0},\nu_{0},A_{0},B_{0})(r)$ at the entrance satisfying (2.7), find a unique piecewise smooth axisymmetric subsonic solution $(\bm{U}^{+},\bm{U}^{-})$ defined on $\mathcal{N}^{+}$ and $\mathcal{N}^{-}$ respectively, with the contact discontinuity $\Gamma$ satisfying the axisymmetric Euler system (2.2) in the sense of Definition 1.1 and the Rankine-Hugoniot conditions in (2.10) and the slip boundary condition in (2.11) and the compatibility conditions (2.12).

It is easy to see that $\bm{U}_{b}^{+}$ satisfies the following properties:

•

$\rho_{b}^{+}>0$ and $0<\sqrt{\frac{\gamma P_{b}}{\rho_{b}^{+}}}$ ;
•

$A_{b}^{+}=\frac{P_{b}}{(\rho_{b}^{+})^{\gamma}}$ and $B_{b}^{+}=\frac{\gamma P_{b}}{(\gamma-1)\rho_{b}^{+}}$ ;
•

$\bm{0}\cdot\bm{v}=0$ for any vector $\bm{v}\in\mathbb{R}^{2}$ .

From this observation, we fix ${\bm{U}}^{+}(x_{1},x_{2})=\bm{U}_{b}^{+}$ in $\mathcal{N}^{+}$ and solve the following free boundary value problem:

Problem 2.2.

Under the assumptions of Problem 2.1, find a smooth axisymmetric subsonic solution $\bm{U}^{-}$ defined on $\mathcal{N}^{-}$ with the contact discontinuity $\Gamma:r=g_{cd}(x)$ such that the following hold.

(a)

$g_{cd}(0)=\frac{1}{2}$ .
(b)

The flow has positive density in the inner cylinder, i.e, $\rho^{-}>0$ .
(c)

Along the contact discontinuity $r=g_{cd}(x)$ , the following Rankine-Hugoniot conditions hold:

${\bf u}^{-}\cdot{\bf n}_{cd}=0,\quad P^{-}=P_{b},\quad{\rm{on}}\quad\Gamma.$ (2.13)
(d)

On the axis $\Gamma_{a}$ , the following compatibility conditions hold:

$u_{r}^{-}(x,0)=u_{\theta}^{-}(x,0)=0,\quad\forall x\in[0,L].$ (2.14)
(e)

On the exit, the following boundary condition holds:

$u_{r}^{-}(L,r)=0.$ (2.15)

Before we state our main result, some weighted Hölder norms are first introduced: For a bounded connected open set $\mathcal{D}\in\mathbb{R}^{3}$ , let $\mathcal{P}$ be a closed portion of ${\partial}\mathcal{D}$ . For $\mathbf{x},\tilde{\mathbf{x}}\in\mathcal{D}$ , define

\displaystyle\delta_{\mathbf{x}}:=\rm{dist}(\mathbf{x},\mathcal{P}),\quad\delta_{\textbf{x},\tilde{\textbf{x}}}:=\min\{\delta_{\mathbf{x}},\delta_{\tilde{\mathbf{x}}}\},

Given positive integer $m$ , $\alpha\in(0,1)$ and $k\in\mathbb{R}$ , we define

	$\displaystyle{\\|u\\|}_{m,0;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\sum_{0\leq\|\beta\|\leq m}\sup_{\bm{x}\in\mathcal{D}}\delta_{\mathbf{x}}^{\max(\|\beta\|+k,0)}\|D^{\beta}u(\bm{x})\|;$
	$\displaystyle[u]_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\sum_{\|\beta\|=m}\sup_{\bm{x},\tilde{\bm{x}}\in\mathcal{D},\bm{x}\neq\tilde{\bm{x}}}\delta_{\mathbf{x},\tilde{\mathbf{x}}}^{\max(m+\alpha+k,0)}\frac{\|D^{\beta}u({\bm{x}})-D^{\beta}u(\tilde{\bm{x}})\|}{\|\bm{x}-\tilde{\bm{x}}\|^{\alpha}};$
	$\displaystyle\\|u\\|_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\\|u\\|_{m,0;\mathcal{D}}^{k,\mathcal{P}}+[u]_{m,\alpha;\mathcal{D}}^{k,\mathcal{P}}$

with the corresponding function space defined as

C_{m,\alpha}^{(k,\mathcal{P})}(\mathcal{D})=\{u:\|u\|_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}<\infty\}.

For a vector function $\bm{u}=(u_{1},u_{2},\cdots,u_{n})$ , define

\|\bm{u}\|_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}:=\sum_{i=1}^{n}\|u_{i}\|_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}.

The main theorem of this paper can be stated as follows.

Theorem 2.3.

Given functions $(J_{0},\nu_{0},A_{0},B_{0})(r)$ at the entrance satisfying (2.7) and $\alpha\in(\frac{1}{2},1)$ , we define

\displaystyle\sigma(J_{0},\nu_{0},A_{0},B_{0}):

\displaystyle=\|(J_{0},\nu_{0},A_{0},B_{0})-(J_{b}^{-},0,A_{b}^{-},B_{b}^{-}\|_{1,\alpha;[0,\frac{1}{2}]},

(2.16)

where

J_{b}^{-}=\rho_{b}^{-}u_{b}^{-},\quad A_{b}^{-}=\frac{P_{b}}{(\rho_{b}^{-})^{\gamma}},\quad B_{b}^{-}=\frac{1}{2}(u_{b}^{-})^{2}+\frac{\gamma P_{b}}{(\gamma-1)\rho_{b}^{-}}.

There exist positive constants $\sigma_{1}$ and $\mathcal{C}^{\ast}$ depending only on $(\bm{U}_{b}^{-},L,\alpha)$ such that if

\displaystyle\sigma(J_{0},\nu_{0},A_{0},B_{0})\leq\sigma_{1},

(2.17)

$\mathbf{Problem\ 2.2}$ has a unique smooth axisymmetric subsonic solution $\bm{U}^{-}$ with the contact discontinuity $\Gamma:r=g_{cd}(x)$ satisfying the following properties:

(i)

The axisymmetric subsonic solution $\bm{U}^{-}\in C_{1,\alpha}^{(-\alpha,\Gamma)}(\mathcal{N}^{-})$ satisfies the following estimate:

\|\bm{U}^{-}-\bm{U}_{b}^{-}\|_{1,\alpha;\mathcal{N}^{-}}^{(-\alpha,\Gamma)}\leq\mathcal{C}^{\ast}\sigma(J_{0},\nu_{0},A_{0},B_{0}).

(2.18)

(ii)

The contact discontinuity surface $g_{cd}(x)\in C^{1,\alpha}([0,L])$ satisfies $g_{cd}(0)=\frac{1}{2}$ . Furthermore, it holds that

$\|g_{cd}-\frac{1}{2}\|_{1,\alpha;[0,L]}\leq\mathcal{C}^{\ast}\sigma(J_{0},\nu_{0},A_{0},B_{0}).$ (2.19)

Remark 2.4.

There are several differences between our result and the previous work [4]. The first one is that the boundary conditions imposed on the entrance and exit of the cylinder. We prescribe the boundary data for the horizontal mass distribution, the swirl velocity, the entropy and the Bernoulli’s quantity at the entrance, while [4] prescribes the entropy, the swirl velocity and the radial velocity. The second one is that the decomposition of the axisymmetric Euler system. In [4], the Helmholtz decomposition of the velocity field plays a crucial role. Instead, we utilize the deformation-curl decomposition developed in [17] for steady Euler system to effectively decouple the hyperbolic and elliptic modes. The last one is that the approach to locate contact discontinuity. The contact discontinuity in [4] is determined by an ordinary differential equation arising from the Rankine-Hugoniot conditions. In this paper, we employ the implicit function theorem to locate the contact discontinuity.

Remark 2.5.

In $\mathbf{Problem\ 2.2}$ , by fixing the outer-layer flow in $\mathcal{N}^{+}$ as the background flow $(\rho_{b}^{+},0,0,0,P_{b})$ , we seek a smooth axisymmetric subsonic solution $\bm{U}^{-}$ defined on $\mathcal{N}^{-}$ with the contact discontinuity $\Gamma:r=g_{cd}(x)$ . One can also find a smooth axisymmetric subsonic solution $\bm{U}^{+}$ defined on $\mathcal{N}^{+}$ with the contact discontinuity $\Gamma:r=g_{cd}(x)$ by fixing the inner-layer flow in $\mathcal{N}^{-}$ as the background flow $(\rho_{b}^{-},0,0,0,P_{b})$ . In fact, this case is even simpler than $\mathbf{Problem\ 2.2}$ since the singularity near the axis is not needed to be considered. Thus we can introduce the usual Lagrangian transformation and reduce the axisymmetric Euler system to a second order elliptic equation for the stream function as in [20] to obtain the solution $\bm{U}^{+}$ .

3 The reformulation of Problem 2.2

In this section, we first introduce the modified Lagrange transformation to straighten the contact discontinuity and reformulate the free boundary value problem 2.2. Then the deformation-curl decomposition in [17, 18] is employed to derive an equivalent system. Finally, we state the main steps to solve the free boundary problem 3.1.

3.1 Reformulation by the modified Lagrangian transformation

For steady Euler flows, the main advantage of the Euler-Lagrange coordinate transformation is to straighten the stream lines. However, in the three-dimensional axisymmetric setting, there is a singular term $r$ in the density equation. We introduce the modified Lagrange transformation to overcome this difficulty and apply this modified Lagrange transformation to straighten the contact discontinuity.

Let $(\bm{U}^{-}(x,r),g_{cd}(x))$ be a solution to $\mathbf{Problem\ 2.2}$ . Define

m^{2}=\int_{0}^{\frac{1}{2}}sJ_{0}(s){\mathrm{d}}s.

(3.1)

For any $x\in(0,L)$ , it follows from the first equation in (2.2) that

\int_{0}^{g_{cd}(x)}s\rho^{-}u_{x}^{-}(x,s){\mathrm{d}}s=m^{2}.

(3.2)

Define the modified Lagrangian transformation as

y_{1}=x,\quad y_{2}(x,r)=\left(\int_{0}^{r}s\rho^{-}u_{x}^{-}(x,s){\mathrm{d}}.s\right)^{\frac{1}{2}}.

(3.3)

Note that if $(\rho^{-},u_{x}^{-},u_{r}^{-},u_{\theta}^{-})$ is close to the background solutions $(\rho_{b}^{-},u_{b}^{-},0,0)$ , there exist two positive constants $C_{1}$ and $C_{2}$ , depending on the background solution, such that

C_{1}r^{2}\leq\int_{0}^{r}s\rho^{-}u_{x}^{-}(x,s){\mathrm{d}}s\leq C_{2}r^{2}.

Hence the Jacobian of the modified Lagrange transformation satisfies

\frac{{\partial}(y_{1},y_{2})}{{\partial}(x,r)}=\left|\begin{matrix}1&0\\ -\frac{r\rho^{-}u_{r}^{-}}{2y_{2}}&\frac{r\rho^{-}u_{x}^{-}}{2y_{2}}\end{matrix}\right|=\frac{r\rho^{-}u_{x}^{-}}{2y_{2}}\geq C_{3}>0.

(3.4)

That is invertible.

Under this transformation, the domain $\mathcal{N}^{-}$ becomes

\Omega:=\{(y_{1},y_{2}):0<y_{1}<L,\ 0<y_{2}<m\}.

The entrance and exit of $\Omega$ are defined as

\displaystyle\Sigma_{0}:=\{(y_{1},y_{2}):y_{1}=0,\ 0<y_{2}<m\},\quad\Sigma_{L}:=\{(y_{1},y_{2}):y_{1}=L,\ 0<y_{2}<m\}.

The axis $\Gamma_{a}$ is transformed into

\displaystyle\Sigma_{a}:=\{(y_{1},y_{2}):0<y_{1}<L,\ y_{2}=0\}.

Moreover, on $\Gamma$ , one has

y_{2}(x,g_{cd}(x))=\left(\int_{0}^{g_{cd}(x)}s\rho^{-}u_{x}^{-}(x,s){\mathrm{d}}s\right)^{\frac{1}{2}}=m.

Hence the free boundary $\Gamma$ becomes the following fixed straight line

\Sigma:=\{(y_{1},y_{2}):0<y_{1}<L,\ y_{2}=m\}.

(3.5)

In the following, the superscript “-” in $\rho^{-},u_{x}^{-},u_{r}^{-},u_{\theta}^{-},P^{-}$ will be ignored to simplify the notations. Let

{\bm{U}}^{-}(y_{1},y_{2}):=(\rho,u_{x},u_{r},u_{\theta},P)(y_{1},r(y_{1},y_{2}))\quad{\rm{in}}\quad\Omega.

(3.6)

Then the axisymmetric Euler system (2.2) in the new coordinates can be rewritten as

\begin{cases}\begin{aligned} &{\partial}_{y_{1}}\left(\frac{2y_{2}}{r\rho u_{x}}\right)-{\partial}_{y_{2}}\left(\frac{u_{r}}{u_{x}}\right)=0,\\ &{\partial}_{y_{1}}\left(u_{x}+\frac{P}{\rho u_{x}}\right)-\frac{r}{2y_{2}}{\partial}_{y_{2}}\left(\frac{Pu_{r}}{u_{x}}\right)-\frac{Pu_{r}}{r\rho u_{x}^{2}}=0,\\ &{\partial}_{y_{1}}u_{r}+\frac{r}{2y_{2}}{\partial}_{y_{2}}P-\frac{u_{\theta}^{2}}{ru_{x}}=0,\\ &{\partial}_{y_{1}}(ru_{\theta})=0,\\ &{\partial}_{y_{1}}A=0.\\ \end{aligned}\end{cases}

(3.7)

The background solutions in the Lagrange coordinates are

{\bm{U}}_{b}^{-}:=(\rho_{b}^{-},u_{b}^{-},0,0,P_{b}),\quad{\rm{in}}\quad\Omega_{b},

(3.8)

where $\Omega_{b}:=\{(y_{1},y_{2}):0<y_{1}<L,\ 0<y_{2}<m_{b}^{-}\}$ and $m_{b}^{-}=\sqrt{\frac{\rho_{b}^{-}u_{b}^{-}}{8}}$ . Without loss of generality, we assume that

\rho_{b}^{-}u_{b}^{-}=2\quad{\rm{and}}\quad m=m_{b}^{-}=\frac{1}{2}.

Furthermore, under the modified Lagrangian transformation, $r$ as a function of $(y_{1},y_{2})$ becomes nonlinear and nonlocal in the new coordinates. In fact, it follows from the inverse transformation that

\displaystyle\frac{{\partial}r}{{\partial}y_{1}}=\frac{u_{r}}{u_{x}},\quad\frac{{\partial}r}{{\partial}y_{2}}=\frac{2y_{2}}{r\rho u_{x}},\quad r(y_{1},0)=0.

Thus one derives

r(y_{1},y_{2})=\left(2\int_{0}^{y_{2}}\frac{2s}{\rho u_{x}(y_{1},s)}{\mathrm{d}}s\right)^{\frac{1}{2}},\ {\rm{in}}\quad\Omega.\\

(3.9)

In particular, for the background solutions $(\rho_{b}^{-},u_{b}^{-},0,0)$ , one has

r_{b}(y_{2})=\sqrt{{\frac{2}{\rho_{b}^{-}u_{b}^{-}}}}y_{2}=y_{2},\ {\rm{in}}\quad\Omega.

(3.10)

In the new coordinates, the boundary data (2.6) at the entrance is given by

(J,u_{\theta},A,B)(0,y_{2})=(\tilde{J}_{0},\tilde{\nu}_{0},\tilde{A}_{0},\tilde{B}_{0})(y_{2}),\quad{\rm{on}}\quad\Sigma_{0},

(3.11)

where

(\tilde{J}_{0},\tilde{\nu}_{0},\tilde{A}_{0},\tilde{B}_{0})(y_{2})=(J_{0},\nu_{0},A_{0},B_{0})\left(\left(\int_{0}^{y_{2}}\frac{2s}{J_{0}(s)}{\mathrm{d}}s\right)^{\frac{1}{2}}\right).

The Rankine-Hugoniot conditions in (2.13) become

\frac{u_{r}}{u_{x}}(y_{1},\frac{1}{2})=g_{cd}^{\prime}(y_{1}),

(3.12)

and

P(y_{1},\frac{1}{2})=P_{b}.

(3.13)

3.2 The deformation-curl decomposition for axisymmetric Euler system

It is well-known that the steady Euler system is elliptic-hyperbolic coupled in subsonic region, to construct a well-defined iteration scheme, one should decompose the hyperbolic and elliptic modes effectively. Different from the pervious decomposition in [20], we will employ the deformation-curl decomposition developed in [17, 18] to deal with the elliptic-hyperbolic coupled structure in the axisymmetric Euler system.

First, using the Bernoulli’s function, the density $\rho$ can be represented as

\rho=H(B,A,|\textbf{u}|^{2})=\left(\frac{\gamma-1}{A\gamma}(B-\frac{1}{2}|\textbf{u}|^{2})\right)^{\frac{1}{\gamma-1}}.

(3.14)

Define the vorticity $\omega={\rm curl\,}\textbf{u}=\omega_{x}\mathbf{e}_{x}+\omega_{r}\mathbf{e}_{r}+\omega_{\theta}\mathbf{e}_{\theta}$ , where

\omega_{x}=\frac{1}{r}{\partial}_{r}(ru_{\theta}),\ \omega_{r}=-{\partial}_{x}u_{\theta},\ \omega_{\theta}={\partial}_{x}u_{r}-{\partial}_{r}u_{x}.

From the third equation in (2.2) and the Bernoulli’s law, one derives that

\omega_{\theta}=\frac{u_{\theta}{\partial}_{r}u_{\theta}+\frac{u_{\theta}^{2}}{r}+\frac{(B-\frac{1}{2}|{\bf u}|^{2})}{A\gamma}{\partial}_{r}A-{\partial}_{r}B}{u_{x}}.

(3.15)

Substituting (3.14) into the density equation in (2.2), the axisymmetric Euler system (2.2) is equivalent to the following system:

\begin{cases}\begin{aligned} &(c^{2}(H,A)-u_{x}^{2}){\partial}_{x}u_{x}+(c^{2}(H,A)-u_{r}^{2}){\partial}_{r}u_{x}-u_{x}u_{r}({\partial}_{x}u_{r}+{\partial}_{r}u_{x})+u_{r}\frac{c^{2}(H,A)+u_{\theta}^{2}}{r}=0,\\ &u_{x}({\partial}_{x}u_{r}-{\partial}_{r}u_{x})=u_{\theta}{\partial}_{r}u_{\theta}+\frac{u_{\theta}^{2}}{r}+\frac{(B-\frac{1}{2}|{\bf u}|^{2})}{A\gamma}{\partial}_{r}A-{\partial}_{r}B,\\ &(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})(ru_{\theta})=0,\\ &(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})A=0,\\ &(u_{x}{\partial}_{x}+u_{r}{\partial}_{r})B=0.\\ \end{aligned}\end{cases}

(3.16)

Under the transformation (3.3), $A$ and $B$ satisfy the following transport equations:

\begin{cases}\begin{aligned} &{\partial}_{y_{1}}A=0,\\ &{\partial}_{y_{1}}B=0.\\ \end{aligned}\end{cases}

(3.17)

Thus one has

A(y_{1},y_{2})=\tilde{A}_{0}(y_{2}),\quad B(y_{1},y_{2})=\tilde{B}_{0}(y_{2}).

(3.18)

Next, it follows from (3.3) and (3.18) that $u_{x}$ , $u_{r}$ and $u_{\theta}$ satisfy the following system:

\begin{cases}\begin{aligned} &(c^{2}(\rho,\tilde{A}_{0})-u_{x}^{2})\left({\partial}_{y_{1}}u_{x}-\frac{r\rho u_{r}}{2y_{2}}{\partial}_{y_{2}}u_{x}\right)+(c^{2}(\rho,\tilde{A}_{0})-u_{r}^{2})\left(\frac{r\rho u_{x}}{2y_{2}}{\partial}_{y_{2}}u_{r}\right)\\ &+\frac{c^{2}(\rho,\tilde{A}_{0})}{r}u_{r}+\frac{u_{\theta}^{2}}{r}u_{r}=u_{x}u_{r}\left({\partial}_{y_{1}}u_{r}-\frac{r\rho u_{r}}{2y_{2}}{\partial}_{y_{2}}u_{r}+\frac{r\rho u_{x}}{2y_{2}}{\partial}_{y_{2}}u_{x}\right),\\ &u_{x}\left({\partial}_{y_{1}}u_{r}-\frac{r\rho u_{r}}{2y_{2}}{\partial}_{y_{2}}u_{r}-\frac{r\rho u_{x}}{2y_{2}}{\partial}_{y_{2}}u_{x}\right)\\ &=\frac{r\rho u_{x}}{2y_{2}}u_{\theta}{\partial}_{y_{2}}u_{\theta}+\frac{u_{\theta}^{2}}{r}+\frac{r\rho u_{x}}{2y_{2}}\frac{\rho^{\gamma-1}}{\gamma-1}{\partial}_{y_{2}}\tilde{A}_{0}-\frac{r\rho u_{x}}{2y_{2}}{\partial}_{y_{2}}\tilde{B}_{0},\\ &{\partial}_{y_{1}}(ru_{\theta})=0,\\ \end{aligned}\end{cases}

(3.19)

with the following boundary conditions:

\begin{cases}\rho u_{x}(0,y_{2})=\tilde{J}_{0}(y_{2}),\ u_{\theta}(0,y_{2})=\tilde{\nu}_{0}(y_{2}),\quad&{\rm{on}}\quad\Sigma_{0},\\ u_{r}(L,y_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L},\\ u_{r}(y_{1},\frac{1}{2})=u_{x}(y_{1},\frac{1}{2})g_{cd}^{\prime}(y_{1}),\quad&{\rm{on}}\quad\Sigma,\\ u_{r}(y_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}.\\ \end{cases}

(3.20)

Furthermore, by (3.13), one obtains

\tilde{A}_{0}\left(\frac{1}{2}\right)\left(\rho(u_{x},u_{r},u_{\theta},\tilde{A}_{0},\tilde{B}_{0})\right)^{\gamma}(y_{1},\frac{1}{2})=P_{b}.

(3.21)

Therefore $\mathbf{Problem\ 2.2}$ is reformulated as follows.

Problem 3.1.

Given functions $(J_{0},\nu_{0},A_{0},B_{0})$ at the entrance satisfying (2.7), find a unique smooth subsonic solution $(u_{x},u_{r},u_{\theta};g_{cd})$ satisfying (3.19) and (3.20) and the Rankine-Hugoniot condition (3.21).

3.3 Solving the free boundary problem 3.1

Note that $\mathbf{Problem\ 3.1}$ is a free boundary problem since the function $g_{cd}$ is unknown, this free boundary problem will be solved by using the implicit function theorem. We follow the steps below to solve $\mathbf{Problem\ 3.1}$ :

(a)

Given any function $g_{cd}(y_{1})=\int_{0}^{y_{1}}w(s){\mathrm{d}}s+\frac{1}{2}$ belonging to some suitable function classes, we will solve the nonlinear system (3.19) with mixed boundary condition (3.20) in $\Omega$ . This will be achieved by decomposing the system (3.19) into two boundary value problems with different inhomogeneous terms and employing the standard elliptic theory. The detailed analysis will be given in Section 4.
(b)

We use the implicit function theorem to locate the contact discontinuity. More precisely, define the map $\mathcal{Q}(\bm{\varphi}_{0},w):=\tilde{A}_{0}\left(\frac{1}{2}\right)\left(\rho(u_{x},u_{r},u_{\theta},\tilde{A}_{0},\tilde{B}_{0})\right)^{\gamma}(y_{1},\frac{1}{2})-P_{b}$ , we need to compute the Fréchet derivative $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ of the functional $\mathcal{Q}(\bm{\varphi}_{0},w)$ with respect to $w$ and show that $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)$ is an isomorphism. This step will be achieved in Section 5.

4 The solution to a fixed boundary value problem in $\Omega$

In this section, given any function $g_{cd}(y_{1})=\int_{0}^{y_{1}}w(s){\mathrm{d}}s+\frac{1}{2}\in C^{1,\alpha}([0,L])$ satisfying $g_{cd}^{\prime}(L)=0$ , we will solve the nonlinear system (3.19) with mixed boundary condition (3.20) in $\Omega$ .

4.1 Linearization

To solve nonlinear system (3.19) in the domain $\Omega$ , we first linearize (3.19) and then solve the linear system in the domain $\Omega$ . Define

W_{1}=u_{x}-u_{b}^{-},\quad W_{2}=u_{r},\quad W_{3}=u_{\theta},\quad(\hat{J},\hat{A},\hat{B})=(\tilde{J}_{0},\tilde{A}_{0},\tilde{B}_{0})-(J_{b}^{-},A_{b}^{-},B_{b}^{-}).

Denoting the solution space by $\mathcal{J}(\delta_{1})$ , which is defined as

\mathcal{J}(\delta_{1})=\{{\bf W}=(W_{1},W_{2},W_{3}):\sum_{j=1}^{3}\|W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\delta_{1},\ W_{2}(y_{1},0)=W_{2}(L,y_{2})=W_{3}(y_{1},0)=0\}.

(4.1)

Here $\delta_{1}$ is a positive constant to be determined later.

Given $\hat{{\bf W}}\in\mathcal{J}(\delta_{1})$ , it follows from the third equation in (3.18) that $W_{3}$ can be solved as follows

W_{3}(y_{1},y_{2})=\frac{\hat{\Lambda}(y_{2})}{\hat{r}(y_{1},y_{2})},

(4.2)

where

		$\displaystyle\hat{\Lambda}(y_{2})=r(0,y_{2})\tilde{\nu}_{0}(y_{2})=\left(2\int_{0}^{y_{2}}\frac{2s}{\tilde{J}_{0}(s)}{\mathrm{d}}s\right)^{\frac{1}{2}}\tilde{\nu}_{0}(y_{2}),$
		$\displaystyle\hat{r}(y_{1},y_{2})=\left(2\int_{0}^{y_{2}}\frac{2s}{\hat{\rho}(\hat{W}_{1}+u_{b}^{-},\hat{W}_{2},\hat{W}_{3},\tilde{A}_{0},\tilde{B}_{0})(\hat{W}_{1}+u_{b}^{-})(y_{1},s)}{\mathrm{d}}s\right)^{\frac{1}{2}}.$

Then one derives that

\|W_{3}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq C\sigma_{cd},

(4.3)

where $\sigma_{cd}=\sigma(J_{0},\nu_{0},A_{0},B_{0})$ is defined in (2.16) and $C>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

In the following, we turn to concern $W_{1}$ and $W_{2}$ . It follows from the first and second equations in (3.19) that $W_{1}$ and $W_{2}$ satisfy the following first order elliptic system:

\begin{cases}\begin{aligned} &(c^{2}(\rho_{b}^{-},A_{b}^{-})-(u_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}+c^{2}(\rho_{b}^{-},A_{b}^{-}){\partial}_{y_{2}}W_{2}\\ &+\frac{c^{2}(\rho_{b}^{-},A_{b}^{-})}{r_{b}}W_{2}=F_{1}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}W_{2}-\frac{r_{b}\rho_{b}^{-}u_{b}^{-}}{2y_{2}}{\partial}_{y_{2}}W_{1}=F_{2}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B}),\\ \end{aligned}\end{cases}

(4.4)

where

		$\displaystyle F_{1}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B})$
		$\displaystyle=(c^{2}(\hat{\rho},\tilde{A}_{0})-(u_{b}^{-}+\hat{W}_{1})^{2})\frac{\hat{r}\hat{\rho}\hat{W}_{2}}{2y_{2}}{\partial}_{y_{2}}\hat{W}_{1}+\hat{W}_{2}^{2}\frac{\hat{r}\hat{\rho}(u_{b}^{-}+\hat{W}_{1})}{2y_{2}}{\partial}_{y_{2}}\hat{W}_{2}-\left(\frac{c^{2}(\hat{\rho},\tilde{A}_{0})}{\hat{r}}-\frac{c^{2}(\rho_{b}^{-},A_{b}^{-})}{r_{b}}\right)\hat{W}_{2}$
		$\displaystyle\quad-\left((\gamma-1)\hat{B}-\frac{\gamma+1}{2}\hat{W}_{1}^{2}-\frac{\gamma-1}{2}\hat{W}_{2}^{2}-\frac{\gamma-1}{2}\hat{W}_{3}^{2}-(\gamma+1)u_{b}^{-}\hat{W}_{1}\right){\partial}_{y_{1}}\hat{W}_{1}$
		$\displaystyle\quad-\left((\gamma-1)\hat{B}-\frac{\gamma-1}{2}\hat{W}_{1}^{2}-\frac{\gamma-1}{2}\hat{W}_{2}^{2}-\frac{\gamma-1}{2}\hat{W}_{3}^{2}-(\gamma-1)u_{b}^{-}\hat{W}_{1}\right){\partial}_{y_{2}}\hat{W}_{2}$
		$\displaystyle\quad-\frac{\hat{W}_{3}^{2}}{\hat{r}}\hat{W}_{2}+(\hat{W}_{1}+u_{b}^{-})\hat{W}_{2}\left({\partial}_{y_{1}}\hat{W}_{2}-\frac{\hat{r}\hat{\rho}\hat{W}_{2}}{2y_{2}}{\partial}_{y_{2}}\hat{W}_{2}+\frac{\hat{r}\hat{\rho}(\hat{W}_{1}+u_{b}^{-})}{2y_{2}}{\partial}_{y_{2}}\hat{W}_{1}\right),$
		$\displaystyle F_{2}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B})$
		$\displaystyle=\frac{1}{u_{b}^{-}+\hat{W}_{1}}\left(\frac{\hat{r}\hat{\rho}(\hat{W}_{1}+u_{b}^{-})}{2y_{2}}\hat{W}_{3}{\partial}_{y_{2}}\hat{W}_{3}+\frac{\hat{W}_{3}^{2}}{\hat{r}}+\frac{\hat{r}\hat{\rho}(\hat{W}_{1}+u_{b}^{-})}{2y_{2}}\frac{\hat{\rho}^{\gamma-1}}{\gamma-1}{\partial}_{y_{2}}\hat{A}-\frac{\hat{r}\rho(\hat{W}_{1}+u_{b}^{-})}{2y_{2}}{\partial}_{y_{2}}\hat{B}\right)$
		$\displaystyle\quad+\frac{1}{u_{b}^{-}+\hat{W}_{1}}\left(\frac{\hat{r}\hat{\rho}\hat{W}_{2}}{2y_{2}}{\partial}_{y_{2}}\hat{W}_{2}\right).$

Recalling that ${\partial}_{r}(A_{0},{B}_{0})(0)=0$ , then one obtains

{\partial}_{y_{2}}(\hat{A},\hat{B})(0)={\partial}_{y_{2}}(\tilde{A}_{0},\tilde{B}_{0})(0)=0.

(4.5)

Thus for $\hat{{\bf W}}\in\mathcal{J}(\delta_{1})$ , it is easy to check that $F_{2}(y_{1},0)=0$ . Furthermore, there exist positive constants $\kappa_{1}$ and $\kappa_{2}$ depending only on the background solutions $\bm{U}_{b}^{-}$ such that

\kappa_{1}y_{2}\leq\hat{r}\leq\kappa_{2}y_{2},

which implies that

\kappa_{1}\leq\frac{\hat{r}}{y_{2}}\leq\kappa_{2}.

(4.6)

Hence one can derive

\displaystyle\sum_{j=1}^{2}\|F_{j}\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}\leq C\left(\bigg{(}\sum_{j=1}^{3}\|\hat{W_{j}}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\bigg{)}^{2}+\sigma_{cd}\right)\leq C\left(\delta_{1}^{2}+\sigma_{cd}\right),

(4.7)

where $C>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

Next, we derive the boundary conditions for ${\bf W}$ . It follows from (3.19) that

\begin{cases}W_{1}(0,y_{2})=F_{3}(\hat{{\bf W}},\hat{J},\hat{A},\hat{B})(0,y_{2}),\quad&{\rm{on}}\quad\Sigma_{0},\\ W_{2}(L,y_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L},\\ W_{2}(y_{1},\frac{1}{2})=F_{4}(\hat{{\bf W}},g_{cd})(y_{1},\frac{1}{2}),\quad&{\rm{on}}\quad\Sigma,\\ W_{2}(y_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a},\\ \end{cases}

(4.8)

where

		$\displaystyle F_{3}(\hat{{\bf W}},\hat{J},\hat{A},\hat{B})$
		$\displaystyle=\frac{\hat{J}}{\rho_{b}^{-}(1-(M_{b}^{-})^{2})}-\frac{\hat{W}_{1}}{\rho_{b}^{-}(1-(M_{b}^{-})^{2})}\bigg{(}H(\hat{B}+B_{b}^{-},\hat{A}+A_{b}^{-},(u_{b}^{-}+\hat{W}_{1})^{2}+\hat{W}_{2}^{2}+\hat{W}_{3}^{2})$
		$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-H(B_{b}^{-},A_{b}^{-},(u_{b}^{-})^{2})\bigg{)}$
		$\displaystyle\quad-\frac{u_{b}^{-}}{\rho_{b}^{-}(1-(M_{b}^{-})^{2})}\left(H(\hat{B}+B_{b}^{-},\hat{A}+A_{b}^{-},(u_{b}^{-}+\hat{W}_{1})^{2}+\hat{W}_{2}^{2}+\hat{W}_{3}^{2})\right.$
		$\displaystyle\qquad\qquad\qquad\qquad\qquad\left.-H(B_{b}^{-},A_{b}^{-},(u_{b}^{-})^{2})+\frac{J_{b}^{-}}{c^{2}(\rho_{b}^{-},A_{b}^{-})}\hat{W}_{1}\right),$
		$\displaystyle(M_{b}^{-})^{2}=\frac{(u_{b}^{-})^{2}}{c^{2}(\rho_{b}^{-},A_{b}^{-})},\quad F_{4}(\hat{{\bf W}},g_{cd})=(u_{b}^{-}+\hat{W}_{1})g_{cd}^{\prime}.$

Then a direct computation yields

		$\displaystyle\\|F_{3}\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha,\{\frac{1}{2}\})}+\\|F_{4}\\|_{0,\alpha;[0,L]}$		(4.9)
		$\displaystyle\leq C\left(\bigg{(}\sum_{j=1}^{3}\\|\hat{W_{j}}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\bigg{)}^{2}+\\|\hat{W_{1}}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\\|g_{cd}^{\prime}\\|_{0,\alpha;[0,L]}+\\|g_{cd}^{\prime}\\|_{0,\alpha;[0,L]}+\sigma_{cd}\right)$
		$\displaystyle\leq C\left(\delta_{1}^{2}+\delta_{1}\\|w\\|_{0,\alpha;[0,L]}+\\|w\\|_{0,\alpha;[0,L]}+\sigma_{cd}\right),$

where $C>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

4.2 Solving the linear boundary value problem

In this subsection, we consider the following linear boundary value problem:

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}+{\partial}_{y_{2}}W_{2}+\frac{1}{y_{2}}W_{2}=\mathcal{F}_{1}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}W_{2}-{\partial}_{y_{2}}W_{1}=\mathcal{F}_{2}(\hat{{\bf W}},\nabla{\hat{\bf W}},\hat{A},\hat{B}),\\ &W_{1}(0,y_{2})=\mathcal{F}_{3}(\hat{{\bf W}},\hat{J},\hat{\Lambda},\hat{A},\hat{B})(0,y_{2}),\quad&{\rm{on}}\quad\Sigma_{0},\\ &W_{2}(L,y_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L},\\ &W_{2}(y_{1},\frac{1}{2})=\mathcal{F}_{4}(\hat{{\bf W}},g_{cd})(y_{1},\frac{1}{2}),\quad&{\rm{on}}\quad\Sigma,\\ &W_{2}(y_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a},\\ \end{aligned}\end{cases}

(4.10)

where

\mathcal{F}_{1}=\frac{1}{c^{2}(\rho_{b}^{-},A_{b}^{-})}F_{1},\quad\mathcal{F}_{i}=F_{i},i=2,3,4.

For the problem (4.10), we have the following conclusion:

Lemma 4.1.

Let $\alpha\in(\frac{1}{2},1)$ . For given $\mathcal{F}_{j}\in C_{0,\alpha}^{(1-\alpha,\Sigma)}(\Omega)$ , $j=1,2$ , $\mathcal{F}_{2}(y_{1},0)=0$ , $\mathcal{F}_{3}\in C_{1,\alpha}^{(-\alpha,\{\frac{1}{2}\})}([0,\frac{1}{2}))$ , $\mathcal{F}_{4}\in C^{0,\alpha}([0,L])$ , the boundary value problem (4.10) has a unique solution $(W_{1},W_{2})\in\left(C_{1,\alpha}^{(-\alpha,\Sigma)}(\Omega)\right)^{2}$ satisfying

\sum_{j=1}^{2}\|W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq C\left(\sum_{j=1}^{2}\|\mathcal{F}_{j}\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}+\|\mathcal{F}_{3}\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha,\{\frac{1}{2}\})}+\|\mathcal{F}_{4}\|_{0,\alpha;[0,L]}\right),

(4.11)

where $C>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

Proof.

We divide the proof into four steps.

Step 1: In this step, in order to solve (4.10), we first introduce a transformation to reduce it to a typical form.

Let

\begin{array}[]{ll}\left\{\begin{aligned} &z_{1}=\sqrt{\frac{1}{1-(M_{b}^{-})^{2}}}y_{1},\\ &z_{2}=y_{2},\\ \end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &V_{1}=\sqrt{{1-(M_{b}^{-})^{2}}}W_{1},\\ &V_{2}=W_{2}.\\ \end{aligned}\right.\end{array}

(4.12)

The domain $\Omega$ becomes

\Omega^{\ast}=\{(z_{1},z_{2}):0<z_{1}<L^{\ast},0<y_{2}<\frac{1}{2}\},

where $L^{\ast}=\sqrt{\frac{1}{1-(M_{b}^{-})^{2}}}L$ , and its boundaries are transformed into

		$\displaystyle\Sigma_{0}^{\ast}:=\{(z_{1},z_{2}):z_{1}=0,\ 0\leq z_{2}<\frac{1}{2}\},\quad\Sigma_{L}^{\ast}:=\{(z_{1},z_{2}):z_{1}=L^{\ast},\ 0\leq z_{2}<\frac{1}{2}\},$
		$\displaystyle\Sigma_{a}^{\ast}:=\{(z_{1},z_{2}):0<z_{1}<L^{\ast},\ z_{2}=0\},\quad\Sigma^{\ast}:=\{(z_{1},z_{2}):0<z_{1}<L^{\ast},\ z_{2}=\frac{1}{2}\}.$

Then the system (4.10) is reformulated as

\begin{cases}\begin{aligned} &{\partial}_{z_{1}}(z_{2}V_{1})+{\partial}_{z_{2}}(z_{2}V_{2})=z_{2}\mathcal{F}_{1},\\ &{\partial}_{z_{1}}V_{2}-{\partial}_{z_{2}}V_{1}=\sqrt{{1-(M_{b}^{-})^{2}}}\mathcal{F}_{2}:=\tilde{\mathcal{F}}_{2},\\ &V_{1}(0,z_{2})=\sqrt{{1-(M_{b}^{-})^{2}}}\mathcal{F}_{3}:=\tilde{\mathcal{F}}_{3},\quad&{\rm{on}}\quad\Sigma_{0}^{\ast},\\ &V_{2}(L^{\ast},z_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L}^{\ast},\\ &V_{2}(z_{1},\frac{1}{2})=\mathcal{F}_{4},\quad&{\rm{on}}\quad\Sigma^{\ast},\\ &V_{2}(z_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}^{\ast}.\\ \end{aligned}\end{cases}

(4.13)

Next, we decompose the problem (4.13) into two boundary value problems with different inhomogeneous terms as follows. Let $(V_{1},V_{2})^{T}=(H_{1},H_{2})^{T}+(K_{1},K_{2})^{T}$ , where $(H_{1},H_{2})^{T}$ is the solution to the problem

\begin{cases}\begin{aligned} &{\partial}_{z_{1}}(z_{2}H_{1})+{\partial}_{z_{2}}(z_{2}H_{2})=0,\\ &{\partial}_{z_{1}}H_{2}-{\partial}_{z_{2}}H_{1}=\tilde{\mathcal{F}}_{2},\\ &H_{1}(0,z_{2})=0,\quad&{\rm{on}}\quad\Sigma_{0}^{\ast},\\ &H_{2}(L^{\ast},z_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L}^{\ast},\\ &H_{2}(z_{1},\frac{1}{2})=0,\quad&{\rm{on}}\quad\Sigma^{\ast},\\ &H_{2}(z_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}^{\ast},\\ \end{aligned}\end{cases}

(4.14)

and $(K_{1},K_{2})^{T}$ satisfies the following problem

\begin{cases}\begin{aligned} &{\partial}_{z_{1}}(z_{2}K_{1})+{\partial}_{z_{2}}(z_{2}K_{2})=z_{2}\mathcal{F}_{1},\\ &{\partial}_{z_{1}}K_{2}-{\partial}_{z_{2}}K_{1}=0,\\ &K_{1}(0,z_{2})=\tilde{\mathcal{F}}_{3},\quad&{\rm{on}}\quad\Sigma_{0}^{\ast},\\ &K_{2}(L^{\ast},z_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L}^{\ast},\\ &K_{2}(z_{1},\frac{1}{2})=\mathcal{F}_{4},\quad&{\rm{on}}\quad\Sigma^{\ast},\\ &K_{2}(z_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}^{\ast}.\\ \end{aligned}\end{cases}

(4.15)

Step 2: In this step, we are going to solve (4.14). The first equation in (4.14) implies that there exists a potential function $\phi_{1}$ such that

({\partial}_{z_{1}}\phi_{1},{\partial}_{z_{2}}\phi_{1})=(z_{2}H_{2},-z_{2}H_{1}).

(4.16)

Let $\Phi_{1}=\frac{\phi_{1}}{z_{2}}$ . Then (4.16) yields that

(H_{1},H_{2})=(-({\partial}_{z_{2}}\Phi_{1}+\frac{\Phi_{1}}{z_{2}}),{\partial}_{z_{1}}\Phi_{1}).

(4.17)

Without loss of generality, we assume that $\Phi_{1}(0,0)=0$ . Thus (4.14) can be rewritten as the following equation for $\Phi_{1}$ :

\begin{cases}\begin{aligned} &{\partial}_{z_{1}}^{2}\Phi_{1}+{\partial}_{z_{2}}^{2}\Phi_{1}+\frac{{\partial}_{z_{2}}\Phi_{1}}{z_{2}}-\frac{\Phi_{1}}{z_{2}^{2}}=\tilde{\mathcal{F}}_{2},\\ &{\partial}_{z_{1}}\Phi_{1}=0,\quad{\rm{on}}\quad\Sigma_{L}^{\ast},\\ &\Phi_{1}=0,\qquad\ {\rm{on}}\quad\Sigma_{a}^{\ast}\cup\Sigma^{\ast}\cup\Sigma_{0}^{\ast}.\\ \end{aligned}\end{cases}

(4.18)

Obviously, the coefficients of equation (4.18) tends to infinity as $z_{2}$ goes to $0$ . By applying the idea of Proposition 3.3 in [1], we rewrite (4.18) as a boundary value problem in $\mathbb{R}^{5}$ so that the singular term in (4.18) can be removed from the equation for $\Phi_{1}$ . Set

\Phi^{\ast}=\frac{\Phi_{1}}{z_{2}},\quad\mathcal{F}_{2}^{\ast}=\frac{\tilde{\mathcal{F}}_{2}}{z_{2}}.

(4.19)

We regard $\Phi^{\ast}$ and $\mathcal{F}_{2}^{\ast}$ as functions defined in

\mathcal{D}^{\ast}:=\{(z_{1},\mathbf{z}_{2}):z_{1}\in(0,L^{\ast}),\mathbf{z}_{2}\in\mathbb{R}^{4}:|\mathbf{z}_{2}|\leq\frac{1}{2}\}\subset\mathbb{R}^{5},

(4.20)

where $\mathbf{z}_{2}=(z_{21},z_{22},z_{23},z_{24})$ and $\sum_{j=1}^{4}z_{2j}^{2}=|\mathbf{z}_{2}|^{2}$ . Define

\bm{\mathcal{G}}^{\ast}(z_{1},\mathbf{z}_{2})=(0,\mathcal{G}^{\ast}(z_{1},\mathbf{z}_{2})z_{21},\mathcal{G}^{\ast}(z_{1},\mathbf{z}_{2})z_{22},\mathcal{G}^{\ast}(z_{1},\mathbf{z}_{2})z_{23},\mathcal{G}^{\ast}(z_{1},\mathbf{z}_{2})z_{24}),\ \forall\mathbf{z}=(z_{1},\mathbf{z}_{2})\in\mathcal{D}^{\ast},

with

\mathcal{G}^{\ast}(z_{1},\mathbf{z}_{2})=\frac{1}{|\mathbf{z}_{2}|^{4}}\int_{0}^{\mathbf{z}_{2}}s^{3}\mathcal{F}_{2}^{\ast}(z_{1},s){\mathrm{d}}s.

Then it follows from (4.18) that

\begin{cases}\begin{aligned} &\triangle_{\mathbf{z}}\Phi^{\ast}=\text{div}_{\mathbf{z}}\bm{\mathcal{G}}^{\ast},\qquad{\rm{in}}\quad\mathcal{D}^{\ast},\\ &{\partial}_{z_{1}}\Phi^{\ast}(L^{\ast},\mathbf{z}_{2})=0,\quad{\rm{on}}\quad B_{L^{\ast}}:=\{L^{\ast}\}\times\{\mathbf{z}_{2}\in\mathbb{R}^{4}:|\mathbf{z}_{2}|\leq\frac{1}{2}\},\\ &\Phi^{\ast}(0,\mathbf{z}_{2})=0,\qquad\ \ {\rm{on}}\quad B_{0}:=\{0\}\times\{\mathbf{z}_{2}\in\mathbb{R}^{4}:|\mathbf{z}_{2}|\leq\frac{1}{2}\},\\ &\Phi^{\ast}(z_{1},\mathbf{z}_{2})=0,\qquad\ \ {\rm{on}}\quad B_{w}:=[0,L^{\ast}]\times\{\mathbf{z}_{2}\in\mathbb{R}^{4}:|\mathbf{z}_{2}|=\frac{1}{2}\}.\\ \end{aligned}\end{cases}

(4.21)

The standard elliptic theory in [9] yields that (4.21) has a unique weak solution $\Phi^{\ast}\in H^{1}(\mathcal{D}^{\ast})$ satisfying

\mathcal{L}[\Phi^{\ast},\xi]=(\bm{\mathcal{G}}^{\ast},\xi)\quad{\rm{for}}\ {\rm{all}}\ \xi\in\{\xi\in H^{1}(\mathcal{D}^{\ast}):\xi=0\quad{\rm{on}}\quad B_{0}\cup B_{w}\},

(4.22)

where

\displaystyle\mathcal{L}[\Phi^{\ast},\xi]=\int_{\mathcal{D}^{\ast}}\nabla\Phi^{\ast}\nabla\xi{\mathrm{d}}\mathbf{z},\quad(\bm{\mathcal{G}}^{\ast},\xi)=\int_{\mathcal{D}^{\ast}}\bm{\mathcal{G}}^{\ast}\nabla\xi{\mathrm{d}}\mathbf{z}.

Furthermore, $\Phi^{\ast}$ satisfies

\|\Phi^{\ast}\|_{H^{1}(\mathcal{D}^{\ast})}\leq C\|\bm{\mathcal{G}}^{\ast}\|_{L^{2}(\mathcal{D}^{\ast})}.

(4.23)

Next, we prove

\|\bm{\mathcal{G}}^{\ast}\|_{L^{2}(\mathcal{D}^{\ast})}\leq C\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{(1-\alpha,{\partial}\mathcal{D}^{\ast})}.

(4.24)

Note that $\tilde{\mathcal{F}}_{2}(z_{1},0)=0$ . Then it holds that

\mathcal{F}_{2}^{\ast}(z_{1},z_{2})=\frac{\tilde{\mathcal{F}}_{2}(z_{1},z_{2})-\tilde{\mathcal{F}}_{2}(z_{1},0)}{z_{2}}=\frac{\tilde{\mathcal{F}}_{2}(z_{1},z_{2})-\tilde{\mathcal{F}}_{2}(z_{1},0)}{z_{2}^{\alpha}}z_{2}^{\alpha-1}.

(4.25)

Since $\tilde{\mathcal{F}}_{2}\in C_{0,\alpha}^{(1-\alpha,\Sigma^{\ast})}(\Omega^{\ast})$ , it is easy to check that

\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{(1-\alpha,{\partial}\mathcal{D}^{\ast})}\leq C\|\tilde{\mathcal{F}}_{2}\|_{0,\alpha;\Omega^{\ast}}^{(1-\alpha,\Sigma^{\ast})}.

(4.26)

By the weighted H $\ddot{\rm{o}}$ lder norm, one derives

|\bm{\mathcal{G}}^{\ast}(\mathbf{z})|\leq\delta_{\mathbf{z}}^{\alpha-1}\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\tilde{\mathcal{D}}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})},

(4.27)

with $\delta_{\mathbf{z}}={\rm{dist}}(\mathbf{z},{\partial}\mathcal{D}^{\ast})$ . Thus for $\alpha\in(\frac{1}{2},1)$ , one has

\int_{\mathcal{D}^{\ast}}|\bm{\mathcal{G}}^{\ast}|^{2}{\mathrm{d}}\mathbf{z}\leq C\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}.

Next, we improve the regularity of $\Phi^{\ast}$ . For $\mathbf{z}_{0}\in{\mathcal{D}^{\ast}}$ and $\eta\in\mathbb{R}$ with $0<\eta<\frac{1}{10}$ , set

		$\displaystyle B_{\eta}(\mathbf{z}_{0}):=\{\mathbf{z}\in\mathbb{R}^{5}:\|\mathbf{z}_{0}-\mathbf{z}\|<\eta\},\quad D_{\eta}(\mathbf{z}_{0}):=B_{\eta}(\mathbf{z}_{0})\cap\mathcal{D}^{\ast},$
		$\displaystyle\Phi^{\ast}_{\mathbf{z}_{0},\eta}:=\frac{1}{\|D_{\eta}(\mathbf{z}_{0})\|}\int_{D_{\eta}(\mathbf{z}_{0})}\Phi^{\ast}{\mathrm{d}}\mathbf{z}.$

Note that there exists a constant $\lambda_{0}\in(0,1/10)$ such that

\lambda_{0}\leq\frac{|D_{\eta}(\mathbf{z}_{0})|}{|B_{\eta}(\mathbf{z}_{0})|}\leq\frac{1}{\lambda_{0}}.

Hence we follow the proof in Theorem 3.8 of [10] to get

\int_{D_{\eta}(\mathbf{z})}|\Phi^{\ast}-\Phi^{\ast}_{\mathbf{z},\eta}|^{2}{\mathrm{d}}\mathbf{z}\leq C\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\eta^{5+2\alpha}

(4.28)

for any $\mathbf{z}\in\overline{\mathcal{D}^{\ast}}$ . Once (4.28) is obtained, it follows from Theorem 3.1 in [10] that

\|\Phi^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}\leq C\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}.

(4.29)

We proof (4.28) only for the case $\mathbf{z}\in B_{0}\cap B_{w}$ , since the other cases can be treated similarly. Fix $\mathbf{z}_{0}\in B_{0}\cap B_{w}$ and $\chi\in\mathbb{R}$ with $0<\chi<\frac{1}{10}$ . Let $\Phi^{\ast}_{h}$ be a weak solution of the following problem:

\begin{cases}\begin{aligned} &\triangle_{\mathbf{z}}\Phi_{h}^{\ast}=0,\qquad{\rm{in}}\quad D_{\chi}(\mathbf{z}_{0}),\\ &\Phi_{h}^{\ast}=\Phi^{\ast},\qquad\ \ {\rm{in}}\quad{\partial}D_{\chi}(\mathbf{z}_{0})\cap\mathcal{D}^{\ast}.\\ \end{aligned}\end{cases}

(4.30)

Then $\tilde{\Phi}_{h}^{\ast}=\Phi^{\ast}-\Phi_{h}^{\ast}$ satisfies

\int_{D_{\chi}(\mathbf{z}_{0})}\nabla\tilde{\Phi}_{h}^{\ast}\nabla\xi{\mathrm{d}}\mathbf{z}=\int_{D_{\chi}(\mathbf{z}_{0})}\bm{\mathcal{G}}^{\ast}\nabla\xi{\mathrm{d}}\mathbf{z},

for any $\xi\in\{\xi\in H^{1}(D_{\chi}(\mathbf{z}_{0}):\xi=0\ {\rm{on}}\ {\partial}D_{\chi}(\mathbf{z}_{0})\cap(B_{0}\cup B_{w})\}$ . Taking the test function $\xi=\tilde{\Phi}_{h}^{\ast}$ and using the Hölder inequality to yield that

\int_{D_{\chi}(\mathbf{z}_{0})}|\nabla\tilde{\Phi}_{h}^{\ast}|^{2}{\mathrm{d}}\mathbf{z}\leq\int_{D_{\chi}(\mathbf{z}_{0})}|\bm{\mathcal{G}}^{\ast}|^{2}{\mathrm{d}}\mathbf{z}.

(4.31)

Due to $\bm{\mathcal{G}}^{\ast}\in C_{0,\alpha}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}(\mathcal{D}^{\ast})$ , one gets

\int_{D_{\chi}(\mathbf{z}_{0})}|\nabla\tilde{\Phi}_{h}^{\ast}|^{2}{\mathrm{d}}\mathbf{z}\leq\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\int_{D_{\chi}(\mathbf{z}_{0})}\delta_{\mathbf{z}}^{2(\alpha-1)}{\mathrm{d}}\mathbf{z}\leq\chi^{3+2\alpha}\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}.

(4.32)

By Corollary 3.11 in [10], for $0<\eta<\chi$ , one has

\displaystyle\int_{D_{\eta}(\mathbf{z}_{0})}|\nabla\Phi^{\ast}|^{2}{\mathrm{d}}\mathbf{z}

\displaystyle\leq C\left(\frac{\eta}{\chi}\right)^{5}\int_{D_{\chi}(\mathbf{z}_{0})}|\nabla\Phi^{\ast}|^{2}{\mathrm{d}}\mathbf{z}+C\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\chi^{3+2\alpha}.

(4.33)

Then it follows from Lemma 3.4 in [10] that one obtains

\displaystyle\int_{D_{\eta}(\mathbf{z}_{0})}|\nabla\Phi^{\ast}|^{2}{\mathrm{d}}\mathbf{z}

\displaystyle\leq C\left(\frac{1}{\chi^{3+2\alpha}}\int_{\mathcal{D}^{\ast}}|\nabla\Phi^{\ast}|^{2}{\mathrm{d}}\mathbf{z}+\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\right)\eta^{3+2\alpha}.

(4.34)

By applying Poincaré inequality, one can derive

\int_{D_{\eta}(\mathbf{z}_{0})}|\Phi^{\ast}-\Phi^{\ast}_{\mathbf{z},\eta}|^{2}{\mathrm{d}}\mathbf{z}\leq C\left(\frac{1}{\chi^{3+2\alpha}}\int_{\mathcal{D}^{\ast}}|\nabla\Phi^{\ast}|^{2}{\mathrm{d}}\mathbf{z}+\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\right)\eta^{5+2\alpha}.

(4.35)

Then it follows from (4.23) and (4.24) that

\int_{D_{\eta}(\mathbf{z}_{0})}|\Phi^{\ast}-\Phi^{\ast}_{\mathbf{z},\eta}|^{2}{\mathrm{d}}\mathbf{z}\leq C\left(\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{({1-\alpha;{\partial}\mathcal{D}^{\ast}})}\right)^{2}\eta^{5+2\alpha}.

Hence the proof of (4.29) is completed.

Therefore, by the scaling argument and the Schauder estimate in [9], one has

\|\Phi^{\ast}\|_{1,\alpha;\mathcal{D}^{\ast}}^{(-\alpha,{\partial}\mathcal{D}^{\ast})}\leq C\|\bm{\mathcal{G}}^{\ast}\|_{0,\alpha;\mathcal{D}^{\ast}}^{(1-\alpha,{\partial}\mathcal{D}^{\ast})}\leq C\|\tilde{\mathcal{F}}_{2}\|_{0,\alpha;\Omega^{\ast}}^{(1-\alpha,\Sigma^{\ast})}.

(4.36)

Furthermore, the rotational invariance of the boundary value problem (4.21) and the uniqueness of the solution $\Phi^{\ast}$ and the estimate (4.36) imply that

\|\Phi^{\ast}\|_{1,\alpha;\Omega^{\ast}}^{(-\alpha,\Sigma^{\ast})}\leq C\|\tilde{\mathcal{F}}_{2}\|_{0,\alpha;\Omega^{\ast}}^{(1-\alpha,\Sigma^{\ast})}.

Using the first equation in (4.18), one can verify

{\partial}_{z_{1}}^{2}\Phi_{1}+{\partial}_{z_{2}}^{2}\Phi_{1}=\tilde{\mathcal{F}}_{2}-\frac{{\partial}_{z_{2}}\Phi_{1}}{z_{2}}+\frac{\Phi_{1}}{z_{2}^{2}}=\tilde{\mathcal{F}}_{2}-{\partial}_{z_{2}}\Phi^{\ast}\in H_{0,\alpha}^{(1-\alpha,\Sigma^{\ast})}(\Omega^{\ast}).

(4.37)

By the Schauder estimate in Theorem 4.6 of [13], we obtain

\|\Phi_{1}\|_{2,\alpha;\Omega^{\ast}}^{(-1-\alpha,\Sigma^{\ast})}\leq C\|\tilde{\mathcal{F}}_{2}\|_{0,\alpha;\Omega^{\ast}}^{(1-\alpha,\Sigma^{\ast})}.

(4.38)

Finally, by the definition of $\Phi_{1}$ , one has

\sum_{j=1}^{2}\|H_{j}\|_{1,\alpha;\Omega^{\ast}}^{(-\alpha,\Sigma^{\ast})}\leq C\|\tilde{\mathcal{F}}_{2}\|_{0,\alpha;\Omega^{\ast}}^{(1-\alpha,\Sigma^{\ast})}.

(4.39)

Step 3: In this step, we are going to solve (4.15). The second equation in (4.15) implies that there exists a potential function $\phi_{2}$ such that

({\partial}_{z_{1}}\phi_{2},{\partial}_{z_{2}}\phi_{2})=(K_{1},K_{2}),\quad\phi_{2}(L^{\ast},0)=0.

(4.40)

Then (4.15) can be rewritten as the following equation for $\phi_{2}$ :

\begin{cases}\begin{aligned} &{\partial}_{z_{1}}(z_{2}{\partial}_{z_{1}}\phi_{2})+{\partial}_{z_{2}}(z_{2}{\partial}_{z_{2}}\phi_{2})=z_{2}\mathcal{F}_{1},\\ &{\partial}_{z_{1}}\phi_{2}(0,z_{2})=\tilde{\mathcal{F}}_{3},\quad&{\rm{on}}\quad\Sigma_{0}^{\ast},\\ &\phi_{2}(L^{\ast},z_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L}^{\ast},\\ &{\partial}_{z_{2}}\phi_{2}(z_{1},\frac{1}{2})=\mathcal{F}_{4},\quad&{\rm{on}}\quad\Sigma^{\ast},\\ &{\partial}_{z_{2}}\phi_{2}(z_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}^{\ast}.\\ \end{aligned}\end{cases}

(4.41)

In order to deal with the singularity near $z_{2}=0$ , we rewrite the problem (4.41) in the three dimensional setting. Define

\zeta_{1}=z_{1},\ \zeta_{2}=z_{2}\cos\tau,\ \zeta_{3}=z_{2}\sin\tau,\ \tau\in[0,2\pi],

and

		$\displaystyle E_{1}=\{(\zeta_{1},\zeta_{2},\zeta_{3}):0<\zeta_{1}<L^{\ast},\zeta_{2}^{2}+\zeta_{3}^{2}\leq\frac{1}{2}\},\quad E_{2}=\{(\zeta_{2},\zeta_{3}):\zeta_{2}^{2}+\zeta_{3}^{2}\leq\frac{1}{2}\},$
		$\displaystyle\Gamma_{w,\zeta}=[0,L^{\ast}]\times\{(\zeta_{2},\zeta_{3}):\zeta_{2}^{2}+\zeta_{3}^{2}=\frac{1}{2}\},$
		$\displaystyle\Gamma_{0,\zeta}=\{0\}\times\{(\zeta_{2},\zeta_{3}):\zeta_{2}^{2}+\zeta_{3}^{2}\leq\frac{1}{2}\},\quad\Gamma_{L^{\ast},\zeta}=\{L^{\ast}\}\times\{(\zeta_{2},\zeta_{3}):\zeta_{2}^{2}+\zeta_{3}^{2}\leq\frac{1}{2}\},$
		$\displaystyle\Psi(\bm{\zeta})=\phi_{2}(\zeta_{1},\sqrt{\zeta_{2}^{2}+\zeta_{3}^{2}})=\Psi(\zeta_{1},\|\zeta^{\prime}\|).$

Then $\Psi$ solves the following problem

\begin{cases}\begin{aligned} &\Delta\Psi=\mathcal{F}_{1}(\zeta_{1},|\zeta^{\prime}|),\\ &{\partial}_{\zeta_{1}}\Psi(0,|\zeta^{\prime}|)=\tilde{\mathcal{F}}_{3}(|\zeta^{\prime}|),\quad&{\rm{on}}\quad\Gamma_{0,\zeta},\\ &\Psi(L^{\ast},|\zeta^{\prime}|)=0,\quad&{\rm{on}}\quad\Gamma_{L^{\ast},\zeta},\\ &(\zeta_{2}{\partial}_{\zeta_{2}}+\zeta_{3}{\partial}_{\zeta_{3}})\Psi(\zeta_{1},|\zeta^{\prime}|)=\frac{1}{2}\mathcal{F}_{4}(\zeta_{1}),\quad&{\rm{on}}\quad\Gamma_{w,\zeta}.\\ \end{aligned}\end{cases}

(4.42)

First, the weak solution to (4.42) can be obtained as follows. $\Psi\in H^{1}(E_{1})$ is said to be a weak solution to (4.42) if the following holds

\mathcal{B}(\Psi,\psi)=\mathcal{P}(\psi),\quad{\rm{for}}\ {\rm{all}}\ \psi\in\{\psi\in H^{1}(E_{1}):\psi=0\quad{\rm{on}}\quad\Gamma_{L^{\ast},\zeta}\},

(4.43)

where

		$\displaystyle\mathcal{B}(\Psi,\psi)=\int_{E_{1}}\nabla\Psi\nabla\psi{\mathrm{d}}\bm{\zeta},$
		$\displaystyle\mathcal{P}(\psi)=-\int_{E_{1}}\mathcal{F}_{1}\psi{\mathrm{d}}\bm{\zeta}+\int_{E_{2}}\tilde{\mathcal{F}}_{3}\psi{\mathrm{d}}\zeta_{2}{\mathrm{d}}\zeta_{3}+\int_{0}^{L^{\ast}}\frac{1}{2}\mathcal{F}_{4}(s)\psi(s,\frac{1}{2}){\mathrm{d}}s.$

By the Lax-Milgram theorem, there exists a unique weak solution $\Psi\in H^{1}(E)$ . Then multiplying $\Psi$ on the sides of the equation (4.42) and integrating over $E_{1}$ yield that

\|\nabla\Psi\|_{L^{2}(E_{1})}^{2}\leq C\|\mathcal{F}_{1}\|_{0,\alpha;E_{1}}^{({1-\alpha;\Gamma_{w,\zeta}})}\|\Psi\|_{L^{2}(E_{1})}+C\|\tilde{\mathcal{F}}_{3}\|_{L^{2}(E_{2})}\|\Psi\|_{L^{2}(E_{2})}+C\|\mathcal{F}_{4}\|_{L^{2}[0,L^{\ast}]}\|\Psi\|_{{L^{2}[0,L^{\ast}]}}.

(4.44)

Then it follows from Poincaré inequality and the trace theorem that one obtains

\|\Psi\|_{H^{1}(E_{1})}\leq C\left(\|\mathcal{F}_{1}\|_{0,\alpha;E_{1}}^{({1-\alpha;\Gamma_{w,\zeta}})}+\|\tilde{\mathcal{F}}_{3}\|_{1,\alpha;E_{2}}^{(-\alpha,\Gamma_{w,\zeta})}+\|\mathcal{F}_{4}\|_{0,\alpha;[0,L^{\ast}]}\right).

(4.45)

Next, we can follow the analogous argument as in Step 2 to obtain $C^{0,\alpha}$ estimate for $\Psi$ . Then the Schauder estimate in Theorem 4.6 of [13] implies that

\|\Psi\|_{2,\alpha;E_{1}}^{(-1-\alpha,\Gamma_{w,\zeta})}\leq C\left(\|\mathcal{F}_{1}\|_{0,\alpha;E_{1}}^{({1-\alpha;\Gamma_{w,\zeta}})}+\|\tilde{\mathcal{F}}_{3}\|_{1,\alpha;E_{2}}^{(-\alpha,\Gamma_{w,\zeta})}+\|\mathcal{F}_{4}\|_{0,\alpha;[0,L^{\ast}]}\right).

(4.46)

Finally, it follows from the definition of $\phi_{2}$ that

\sum_{j=1}^{2}\|K_{j}\|_{1,\alpha;\Omega^{\ast}}^{(-\alpha,\Sigma^{\ast})}\leq C\left(\|\mathcal{F}_{1}\|_{0,\alpha;\Omega^{\ast}}^{({1-\alpha;\Sigma^{\ast}})}+\|\tilde{\mathcal{F}}_{3}\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}+\|\mathcal{F}_{4}\|_{0,\alpha;[0,L^{\ast}]}\right).

(4.47)

Step 4: By recalling the transformation (4.12) and combining the estimates (4.39) and (4.47), we conclude that the boundary value problem (4.10) has a unique solution $(W_{1},W_{2})\in\left(C_{1,\alpha}^{(-\alpha,\Sigma)}(\Omega)\right)^{2}$ satisfying

\sum_{j=1}^{2}\|W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq C\left(\sum_{j=1}^{2}\|\mathcal{F}_{j}\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}+\|\mathcal{F}_{3}\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}+\|\mathcal{F}_{4}\|_{0,\alpha;[0,L]}\right).

Thus, the proof of Lemma 4.1 is completed. ∎

4.3 Solving the nonlinear boundary value problem

For a given $\hat{{\bf W}}\in\mathcal{J}(\delta_{1})$ , it follows from Lemma 4.1 that the problem (4.10) has a unique solution $(W_{1},W_{2})\in\left(C_{1,\alpha}^{(-\alpha,\Sigma)}(\Omega)\right)^{2}$ satisfying the estimate (4.11). Define a map $\mathcal{T}$ as follows

\mathcal{T}(\hat{{\bf W}})=({{\bf W}}),\quad{\rm{for}}\ \hat{{\bf W}}\in\mathcal{J}(\delta_{1}).

(4.48)

The estimate (4.11), together with (4.3), (4.7) and (4.9), yields

\displaystyle\sum_{j=1}^{3}\|W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\mathcal{C}_{1}\left(\delta_{1}^{2}+\delta_{1}\|w\|_{0,\alpha;[0,L]}+\|w\|_{0,\alpha;[0,L]}+\sigma_{cd}\right),

(4.49)

where $\mathcal{C}_{1}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

We assume that

\|w\|_{0,\alpha;[0,L]}\leq\delta_{2},

(4.50)

where $\mathcal{C}_{1}\delta_{2}\leq\frac{\delta_{1}}{4}$ with $\mathcal{C}_{1}$ given in (4.49). Let $\sigma_{2}=\frac{1}{4(\mathcal{C}_{1}^{2}+\mathcal{C}_{1})}$ and choose $\delta_{1}=4\mathcal{C}_{1}\sigma_{cd}$ . Then if $\sigma_{cd}\leq\sigma_{2}$ , one has

\sum_{j=1}^{3}\|W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\delta_{1}.

(4.51)

Hence $\mathcal{T}$ maps $\mathcal{J}(\delta_{1})$ into itself.

In the following, we will show that $\mathcal{T}$ is a contraction in $\mathcal{J}(\delta_{1})$ . Let $\hat{{\bf W}}^{k}\in\mathcal{J}(\delta_{1}),k=1,2$ , one has ${\bf W}^{k}=\mathcal{T}(\hat{{\bf W}}^{k})$ for $k=1,2$ . Define

{\bf Y}={\bf W}^{1}-{\bf W}^{2},\quad\quad\hat{{\bf Y}}=\hat{{\bf W}}^{1}-\hat{{\bf W}}^{2}.

Then it follows from (4.2) that

Y_{3}(y_{1},y_{2})={\hat{\Lambda}(y_{2})}\left(\frac{1}{\hat{r}^{1}(y_{1},y_{2})}-\frac{1}{\hat{r}^{2}(y_{1},y_{2})}\right),

(4.52)

where

\hat{r}^{i}(y_{1},y_{2})=\left(2\int_{0}^{y_{2}}\frac{2s}{\hat{\rho}(\hat{W}_{1}^{i}+u_{b}^{-},\hat{W}_{2}^{i},\hat{W}_{3}^{i},\tilde{A}_{0},\tilde{B}_{0})(\hat{W}_{1}^{i}+u_{b}^{-})(y_{1},s)}{\mathrm{d}}s\right)^{\frac{1}{2}}.

Next, we obtain that $(Y_{1},Y_{2})$ satisfies

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}Y_{1}+{\partial}_{y_{2}}Y_{2}+\frac{1}{y_{2}}Y_{2}=\mathcal{F}_{1}(\hat{{\bf W}}^{1},\nabla{\hat{\bf W}}^{1},\hat{A},\hat{B})-\mathcal{F}_{1}(\hat{{\bf W}}^{2},\nabla{\hat{\bf W}}^{2},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}Y_{2}-{\partial}_{y_{2}}Y_{1}=\mathcal{F}_{2}(\hat{{\bf W}}^{1},\nabla{\hat{\bf W}}^{1},\hat{A},\hat{B})-\mathcal{F}_{2}(\hat{{\bf W}}^{2},\nabla{\hat{\bf W}}^{2},\hat{A},\hat{B}),\\ &Y_{1}(0,y_{2})=(\mathcal{F}_{3}(\hat{{\bf W}}^{1},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}(\hat{{\bf W}}^{2},\hat{J},\hat{A},\hat{B}))(0,y_{2}),\\ &Y_{2}(L,y_{2})=0,\\ &Y_{2}(y_{1},\frac{1}{2})=(\mathcal{F}_{4}(\hat{{\bf W}}^{1},g_{cd})-\mathcal{F}_{4}(\hat{{\bf W}}^{2},g_{cd}))(y_{1},\frac{1}{2}),\\ &Y_{2}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(4.53)

where $\mathcal{F}_{j}(\hat{{\bf W}}^{k}),j=1,2,3,4,k=1,2,3$ are functions defined in (4.10) by replacing $\hat{{\bf W}}$ with $\hat{{\bf W}}^{k}$ respectively. Then it follows from Lemma 4.1 that one can derive

$\displaystyle\sum_{j=1}^{3}\\|Y_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	$\displaystyle\leq\mathcal{C}_{2}\left(\sum_{j=1}^{2}\\|\mathcal{F}_{j}(\hat{{\bf W}}^{1},\nabla{\hat{\bf W}}^{1},\hat{A},\hat{B})-\mathcal{F}_{j}(\hat{{\bf W}}^{2},\nabla{\hat{\bf W}}^{2},\hat{A},\hat{B})\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}\right.$	(4.54)
	$\displaystyle\qquad\quad\left.+\\|\mathcal{F}_{3}(\hat{{\bf W}}^{1},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}(\hat{{\bf W}}^{2},\hat{J},\hat{A},\hat{B})\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}\right.$
	$\displaystyle\qquad\quad\left.+\\|\mathcal{F}_{4}(\hat{{\bf W}}^{1},g_{cd})-\mathcal{F}_{4}(\hat{{\bf W}}^{2},g_{cd})\\|_{0,\alpha;[0,L]}+\left\\|{\hat{\Lambda}}\left(\frac{1}{\hat{r}^{1}}-\frac{1}{\hat{r}^{2}}\right)\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\right)$
	$\displaystyle\leq\mathcal{C}_{2}(\delta_{1}+\delta_{2}+\sigma_{cd})\sum_{j=1}^{3}\\|\hat{Y}_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)},$

where $\mathcal{C}_{2}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ . Setting

\sigma_{3}=\min\left\{\sigma_{2},\frac{1}{4\mathcal{C}_{2}(4\mathcal{C}_{1}+2)}\right\}.

(4.55)

Then for $\sigma_{cd}\leq\sigma_{3}$ , one has $\mathcal{C}_{2}(\delta_{1}+\delta_{2}+\sigma_{cd})\leq\mathcal{C}_{2}(4\mathcal{C}_{1}+2)\sigma_{cd}\leq\frac{1}{4}$ . Hence the mapping $\mathcal{T}$ is a contraction mapping so that $\mathcal{T}$ has a unique fixed point in $\mathcal{J}(\delta_{1})$ .

5 The construction of the contact discontinuity surface

Up to now, for a given function $g_{cd}(y_{1})=\int_{0}^{y_{1}}w(s){\mathrm{d}}s+\frac{1}{2}$ satisfying $g_{cd}^{\prime}(L)=0$ , we have obtained the solution $(u_{x},u_{r},u_{\theta})$ for the nonlinear boundary value problem (3.19)-(3.20). To complete the proof of Theorem 2.3, we will use the implicit function theorem to find the contact discontinuity $g_{cd}(y_{1})$ such that (3.21) is satisfied.

First, define a Banach space

\mathcal{V}=\{w:\|w\|_{0,\alpha;[0,L]}<\infty\}.

Set

\mathcal{V}_{1}=\{w:w(L)=0,\|w\|_{0,\alpha;[0,L]}<\infty\}

(5.1)

and

\mathcal{V}_{1}(\delta_{2})=\{w\in\mathcal{V}_{1}:\|w\|_{0,\alpha;[0,L]}\leq\delta_{2}\},

(5.2)

where $\delta_{2}$ is defined in (4.50). Then for any $w\in\mathcal{V}_{1}(\delta_{2})$ , the nonlinear boundary value problem (3.19)-(3.20) has a unique solution $(u_{x},u_{r},u_{\theta})$ satisfying

\|u_{x}-u_{b}^{-}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|u_{r}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|u_{\theta}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq 4\mathcal{C}_{1}\sigma_{cd}.

(5.3)

Let

\displaystyle\mathcal{V}_{0}=C^{1,\alpha}([0,1/2])\times C^{1,\alpha}([0,1/2])\times C^{1,\alpha}([0,1/2])\times C^{1,\alpha}([0,1/2]).

Then we set

\mathcal{V}_{2}(\delta_{3})=\{\bm{\varphi}_{0}\in\mathcal{V}_{0}:\|\bm{\varphi}_{0}-\bm{\varphi}_{b}\|_{\mathcal{V}_{0}}\leq\delta_{3}\},

(5.4)

where

\bm{\varphi}_{0}=(J_{0},\nu_{0},A_{0},B_{0})\quad{\rm{and}}\quad\bm{\varphi}_{b}=(J_{b}^{-},0,A_{b}^{-},B_{b}^{-}).

Define a map $\mathcal{Q}:\mathcal{V}_{2}(\delta_{3})\times\mathcal{V}_{1}(\delta_{2})\rightarrow\mathcal{V}$ by

\mathcal{Q}(\bm{\varphi}_{0},w):=N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})-P_{b},

(5.5)

where

N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})=\tilde{A}_{0}\left(\frac{1}{2}\right)\left(\rho(W_{1}+u_{b}^{-},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})\right)^{\gamma}(y_{1},\frac{1}{2}).

Hence (3.21) can be written as the equation

\mathcal{Q}(\bm{\varphi}_{0},w)=0,

(5.6)

which will be solved (5.6) by employing the implicit function theorem. For the precise statement of the implicit function theorem, one can see Theorem 3.3 in [20]. We will verify the conditions $\rm(i)$ , $\rm(ii)$ and $\rm(iii)$ in Theorem 3.3 of [20].

Obviously,

\mathcal{Q}(\bm{\varphi}_{b},0)=0.

Next, the proof is divided into two steps.

Step 1. Differentiability of $\mathcal{Q}$ .

Given any $w\in\mathcal{V}_{1}(\delta_{2}),w_{1}\in\mathcal{V}_{1}$ , and $\tau>0$ , let $(u_{x},u_{r},u_{\theta})=(W_{1}+u_{b}^{-},W_{2},W_{3})$ be the solution of (3.19) with the following boundary conditions:

\begin{cases}\rho u_{x}(0,y_{2})=\tilde{J}_{0}(y_{2}),\ u_{\theta}(0,y_{2})=\tilde{\nu}_{0}(y_{2}),\quad&{\rm{on}}\quad\Sigma_{0},\\ u_{r}(L,y_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L},\\ u_{r}(y_{1},\frac{1}{2})=\tilde{u}_{x}(y_{1},\frac{1}{2})w(y_{1}),\quad&{\rm{on}}\quad\Sigma,\\ u_{r}(y_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a},\\ \end{cases}

and $(\tilde{u}_{x},\tilde{u}_{r},\tilde{u}_{\theta})=(\tilde{W}_{1}+u_{b}^{-},\tilde{W}_{2},\tilde{W}_{3})$ be the solution of (3.19) with the following boundary conditions:

\begin{cases}\tilde{\rho}\tilde{u}_{x}(0,y_{2})=\tilde{J}_{0}(y_{2}),\ \tilde{u}_{\theta}(0,y_{2})=\tilde{\nu}_{0}(y_{2}),\quad&{\rm{on}}\quad\Sigma_{0},\\ \tilde{u}_{r}(L,y_{2})=0,\quad&{\rm{on}}\quad\Sigma_{L},\\ \tilde{u}_{r}(y_{1},\frac{1}{2})=\tilde{u}_{x}(y_{1},\frac{1}{2})(w+\tau w_{1})(y_{1}),\quad&{\rm{on}}\quad\Sigma,\\ \tilde{u}_{r}(y_{1},0)=0,\quad&{\rm{on}}\quad\Sigma_{a}.\\ \end{cases}

Then it follows from Section 4 that

\tilde{W}_{3}=\mathcal{F}_{0}(\tilde{\bf W},\hat{\Lambda},\hat{A},\hat{B}),\quad{\rm{and}}\quad\ W_{3}=\mathcal{F}_{0}({\bf W},\hat{\Lambda},\hat{A},\hat{B}),

(5.7)

where

	$\displaystyle\mathcal{F}_{0}(\tilde{\bf W},\hat{\Lambda},\hat{A},\hat{B}):=\frac{\hat{\Lambda}}{\tilde{r}}=\frac{\hat{\Lambda}}{\left(2\int_{0}^{y_{2}}\frac{2s}{\rho(\tilde{W}_{1}+u_{b}^{-},\tilde{W}_{2},\tilde{W}_{3},\tilde{A}_{0},\tilde{B}_{0})(\tilde{W}_{1}+u_{b}^{-})(y_{1},s)}{\mathrm{d}}s\right)^{\frac{1}{2}}},$
	$\displaystyle\mathcal{F}_{0}({\bf W},\hat{\Lambda},\hat{A},\hat{B}):=\frac{\hat{\Lambda}}{r}=\frac{\hat{\Lambda}}{\left(2\int_{0}^{y_{2}}\frac{2s}{\rho(W_{1}+u_{b}^{-},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(W_{1}+u_{b}^{-})(y_{1},s)}{\mathrm{d}}s\right)^{\frac{1}{2}}}.$

Furthermore, $(W_{1},W_{2})$ and $(\tilde{W}_{1},\tilde{W}_{2})$ satisfy

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}+{\partial}_{y_{2}}W_{2}+\frac{1}{y_{2}}W_{2}=\mathcal{F}_{1}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}W_{2}-{\partial}_{y_{2}}W_{1}=\mathcal{F}_{2}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B}),\\ &W_{1}(0,y_{2})=\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})(0,y_{2}),\\ &W_{2}(L,y_{2})=0,\\ &W_{2}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+W_{1}(y_{1},\frac{1}{2})\right)w(y_{1}),\\ &W_{2}(y_{1},0)=0,\\ \end{aligned}\end{cases}

(5.8)

and

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}\tilde{W}_{1}+{\partial}_{y_{2}}\tilde{W}_{2}+\frac{1}{y_{2}}\tilde{W}_{2}=\mathcal{F}_{1}(\tilde{\bf W},\nabla{\tilde{\bf W}},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}\tilde{W}_{2}-{\partial}_{y_{2}}\tilde{W}_{1}=\mathcal{F}_{2}(\tilde{{\bf W}},\nabla{\tilde{\bf W}},\hat{A},\hat{B}),\\ &\tilde{W}_{1}(0,y_{2})=\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})(0,y_{2}),\\ &\tilde{W}_{2}(L,y_{2})=0,\\ &\tilde{W}_{2}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+\tilde{W}_{1}(y_{1},\frac{1}{2})\right)(w+\tau w_{1})(y_{1}),\\ &\tilde{W}_{2}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.9)

Choosing $\tau_{1}>0$ such that $\tau_{1}\|w_{1}\|_{0,\alpha;[0,L]}\leq\delta_{2}$ . Then for $\tau\in(0,\tau_{1})$ , it follows from Section 4 that one gets

\sum_{j=1}^{3}\|\tilde{W}_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq 4\mathcal{C}_{1}\sigma_{cd}.

(5.10)

Denote

{\bf W}^{\tau}=\frac{\tilde{{\bf W}}-{{\bf W}}}{\tau}.

It follows from (5.7)-(5.8) that one obtains

\begin{cases}\begin{aligned} &W_{3}^{\tau}=\frac{\tilde{\mathcal{F}}_{0}(\tilde{\bf W},\hat{\Lambda},\hat{A},\hat{B})-\tilde{\mathcal{F}}_{0}({\bf W},\hat{\Lambda},\hat{A},\hat{B})}{\tau},\\ &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{\tau}+{\partial}_{y_{2}}W_{2}^{\tau}+\frac{1}{y_{2}}W_{2}^{\tau}=\frac{\mathcal{F}_{1}(\tilde{{\bf W}},\nabla{\tilde{\bf W}},\hat{A},\hat{B})-\mathcal{F}_{1}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau},\\ &{\partial}_{y_{1}}W_{2}^{\tau}-{\partial}_{y_{2}}W_{1}^{\tau}=\frac{\mathcal{F}_{2}(\tilde{{\bf W}},\nabla{\tilde{\bf W}},\hat{A},\hat{B})-\mathcal{F}_{2}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau},\\ &W_{1}^{\tau}(0,y_{2})=\frac{\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})}{\tau}(0,y_{2}),\\ &W_{2}^{\tau}(L,y_{2})=0,\\ &W_{2}^{\tau}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+\tilde{W}_{1}(y_{1},\frac{1}{2})\right)w_{1}(y_{1})+W_{1}^{\tau}(y_{1},\frac{1}{2})w(y_{1}),\\ &W_{2}^{\tau}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.11)

Then for $\tau\in(0,\tau_{1})$ , one can apply (5.10) and (4.11) to obtain the following estimate:

$\displaystyle\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	$\displaystyle\leq\mathcal{C}_{3}\left\\|\frac{\mathcal{F}_{0}(\tilde{{\bf W}},\hat{\Lambda},\hat{A},\hat{B})-\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B})}{\tau}\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	(5.12)
	$\displaystyle\quad+\mathcal{C}_{3}\left(\sum_{j=1}^{2}\left\\|\frac{\mathcal{F}_{j}(\tilde{{\bf W}},\nabla{\tilde{\bf W}},\hat{A},\hat{B})-\mathcal{F}_{j}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau}\right\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}\right)$
	$\displaystyle\quad+\mathcal{C}_{3}\left\\|\frac{\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})}{\tau}\right\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|(u_{b}^{-}+\tilde{W}_{1})w_{1}+W_{1}^{\tau}w\\|_{0,\alpha;[0,L]}$
	$\displaystyle\leq\mathcal{C}_{3}\left(\delta_{2}+4\mathcal{C}_{1}\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+4\mathcal{C}_{3}\mathcal{C}_{1}\sigma_{cd}\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\leq\mathcal{C}_{3}((4\mathcal{C}_{1}+1)\sigma_{cd})\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{3}((4\mathcal{C}_{1}+1)\sigma_{cd})\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|w_{1}\\|_{0,\alpha;[0,L]},$

where $\mathcal{C}_{3}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ . Setting

\sigma_{4}=\min\left\{\sigma_{3},\frac{1}{4\mathcal{C}_{3}(1+4\mathcal{C}_{1})}\right\},

(5.13)

where $\sigma_{3}$ is defined in (4.55). Then for $\tau\in(0,\tau_{1})$ and $\sigma_{cd}\leq\sigma_{4}$ , one has

\sum_{j=1}^{3}\|W_{j}^{\tau}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq(2\mathcal{C}_{3}+2)\|w_{1}\|_{0,\alpha;[0,L]}.

(5.14)

Hence there exists a subsequence $\{\tau_{k}\}_{k=1}^{\infty}$ such that $(W_{1}^{\tau_{k}},W_{2}^{\tau_{k}},W_{3}^{\tau_{k}})$ converges to $(W_{1}^{0},W_{2}^{0},W_{3}^{0})$ in $C_{1,\alpha^{\prime}}^{(-\alpha^{\prime},\Sigma)}(\Omega)$ as $\tau_{k}\rightarrow 0$ for some $0<\alpha^{\prime}<\alpha$ . The estimate (5.14) also implies that $(W_{1}^{0},W_{2}^{0},W_{3}^{0})\in\left(C_{1,\alpha}^{(-\alpha,\Sigma)}(\Omega)\right)^{3}$ and

\sum_{j=1}^{3}\|W_{j}^{0}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq(2\mathcal{C}_{3}+2)\|w_{1}\|_{0,\alpha;[0,L]}.

(5.15)

Define a map $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ by

\displaystyle D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)(w_{1})=\sum_{j=1}^{3}{\partial}_{W_{j}}N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})W_{j}^{0}(y_{1},\frac{1}{2}).

(5.16)

Then (5.15) implies that $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ is a linear mapping from $\mathcal{V}_{1}$ to $\mathcal{V}$ . Next, we need to show that $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ is the Fréchet derivative of the functional $\mathcal{Q}(\bm{\varphi}_{0},w)$ with respect to $w$ . To this end, It follows from (5.11) that ${\bf W}^{0}$ satisfies

\begin{cases}\begin{aligned} &W_{3}^{0}=\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B})W_{j}^{0},\\ &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{0}+{\partial}_{y_{2}}W_{2}^{0}+\frac{1}{y_{2}}W_{2}^{0}\\ &=\sum_{j=1}^{3}\left({\partial}_{W_{j}}\mathcal{F}_{1}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})W_{j}^{0}+{\partial}_{\nabla W_{j}}\mathcal{F}_{1}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})\nabla W_{j}^{0}\right),\\ &{\partial}_{y_{1}}W_{2}^{0}-{\partial}_{y_{2}}W_{1}^{0}\\ &=\sum_{j=1}^{3}\left({\partial}_{W_{j}}\mathcal{F}_{2}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})W_{j}^{0}+{\partial}_{\nabla W_{j}}\mathcal{F}_{2}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})\nabla W_{j}^{0}\right),\\ &W_{1}^{0}(0,y_{2})=\left(\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})W_{j}^{0}\right)(0,y_{2}),\\ &W_{2}^{0}(L,y_{2})=0,\ W_{2}^{0}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+W_{1}(y_{1},\frac{1}{2})\right)w_{1}(y_{1})+W_{1}^{0}(y_{1},\frac{1}{2})w(y_{1}),\ W_{2}^{0}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.17)

By taking difference of (5.11) and (5.17), the following estimate can be derived:

		$\displaystyle\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$		(5.18)
		$\displaystyle\leq\mathcal{C}_{4}\left\\|\frac{\mathcal{F}_{0}(\tilde{\bf W},\hat{\Lambda},\hat{A},\hat{B})-\mathcal{F}_{0}({\bf W},\hat{\Lambda},\hat{A},\hat{B})}{\tau}-\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B})W_{j}^{0}\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$
		$\displaystyle\quad+\mathcal{C}_{4}\sum_{i=1}^{2}\left\\|\frac{\mathcal{F}_{i}(\tilde{{\bf W}},\nabla{\tilde{{\bf W}}},\hat{A},\hat{B})-\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau}\right.$
		$\displaystyle\qquad\qquad\quad\left.-\sum_{j=1}^{3}\left({\partial}_{W_{j}}\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})W_{j}^{0}+{\partial}_{\nabla W_{j}}\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})\nabla W_{j}^{0}\right)\right\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}$
		$\displaystyle\quad+\mathcal{C}_{4}\left\\|\frac{\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})}{\tau}-\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})W_{j}^{0}\right\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}$
		$\displaystyle\quad+\mathcal{C}_{4}\\|\tau W_{1}^{\tau}w_{1}+(W_{1}^{\tau}-W_{1}^{0})w\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\mathcal{C}_{4}\left(\delta_{2}+4\mathcal{C}_{1}\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{4}(2\mathcal{C}_{3}+2)\tau(\\|w_{1}\\|_{0,\alpha;[0,L]})^{2}$
		$\displaystyle\leq\mathcal{C}_{4}\left((4\mathcal{C}_{1}+1)\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{4}(2\mathcal{C}_{3}+2)\tau(\\|w_{1}\\|_{0,\alpha;[0,L]})^{2},$

where $\mathcal{C}_{4}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ . Setting

\sigma_{5}=\min\left\{\sigma_{4},\frac{1}{8\mathcal{C}_{4}(1+4\mathcal{C}_{1})}\right\}.

(5.19)

Then for $\sigma_{cd}\leq\sigma_{5}$ , one has

\sum_{j=1}^{3}\|W_{j}^{\tau}-W_{j}^{0}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq 4\mathcal{C}_{4}(2\mathcal{C}_{3}+2)\tau\left(\|w_{1}\|_{0,\alpha;[0,L]}\right)^{2}.

(5.20)

By the definition of $\mathcal{Q}(\bm{\varphi}_{0},w)$ and $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ , one gets

		$\displaystyle\left\\|\mathcal{Q}(\bm{\varphi}_{0},w+\tau w_{1})-\mathcal{Q}(\bm{\varphi}_{0},w)-D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)(\tau w_{1})\right\\|_{0,\alpha;[0,L]}$		(5.21)
		$\displaystyle=\left\\|N(\tilde{W}_{1},\tilde{W}_{2},\tilde{W}_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})-N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})\right.$
		$\displaystyle\quad\quad-\left.\tau\left(\sum_{j=1}^{3}{\partial}_{W_{j}}N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})W_{j}^{0}\right)(y_{1},\frac{1}{2})\right\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\left\\|N(\tilde{W}_{1},\tilde{W}_{2},\tilde{W}_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})-N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(y_{1},\frac{1}{2})\right.$
		$\displaystyle\quad\quad-\left.\tau\left(\sum_{j=1}^{3}{\partial}_{W_{j}}N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})W_{j}^{\tau}\right)(y_{1},\frac{1}{2})\right\\|_{0,\alpha;[0,L]}$
		$\displaystyle\quad+\left\\|\tau\left(\sum_{j=1}^{3}{\partial}_{W_{j}}N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})(W_{j}^{\tau}-W_{j}^{0})\right)(y_{1},\frac{1}{2})\right\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\mathcal{C}\tau^{2}\left(\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\right)^{2}+\mathcal{C}\tau\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$
		$\displaystyle\leq\mathcal{C}\tau^{2}\left(\\|w_{1}\\|_{0,\alpha;[0,L]}\right)^{2},$

where $\mathcal{C}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ . Thus it holds that

\frac{\left\|\mathcal{Q}(\bm{\varphi}_{0},w+\tau w_{1})-\mathcal{Q}(\bm{\varphi}_{0},w)-D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)(\tau w_{1})\right\|_{0,\alpha;[0,L]}}{\tau\|w_{1}\|_{0,\alpha;[0,L]}}\rightarrow 0

as $\tau\rightarrow 0$ . Hence $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ is the Fréchet derivative of the functional $\mathcal{Q}(\bm{\varphi}_{0},w)$ with respect to $w$ .

It remains to prove the continuity of the map $\mathcal{Q}(\bm{\varphi}_{0},w)$ and $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ . For any fixed $(\bm{\varphi}_{0},w)\in\mathcal{V}_{0}\times\mathcal{V}_{1}(\delta_{2})$ , we assume that $(\bm{\varphi}_{0}^{k},w^{k})\rightarrow(\bm{\varphi}_{0},w)$ in $\mathcal{V}_{0}\times\mathcal{V}_{1}(\delta_{2})$ as $k\rightarrow\infty$ . Then we first show that as $k\rightarrow\infty$ ,

\mathcal{Q}(\bm{\varphi}_{0}^{k},w^{k})\rightarrow\mathcal{Q}(\bm{\varphi}_{0},w),\quad{\rm{in}}\quad\mathcal{V}.

(5.22)

By (5.7) and (5.8), ${\bf W}^{k}$ satisfies the following problem:

\begin{cases}\begin{aligned} &W_{3}^{k}=\mathcal{F}_{0}({{\bf W}}^{k},\hat{\Lambda}^{k},\hat{A}^{k},\hat{B}^{k}),\\ &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{k}+{\partial}_{y_{2}}W_{2}^{k}+\frac{1}{y_{2}}W_{2}^{k}=\mathcal{F}_{1}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k}),\\ &{\partial}_{y_{1}}W_{2}^{k}-{\partial}_{y_{2}}W_{1}^{k}=\mathcal{F}_{2}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k}),\\ &W_{1}^{k}(0,y_{2})=\mathcal{F}_{3}({{\bf W}}^{k},\hat{J}^{k},\hat{A}^{k},\hat{B}^{k})(0,y_{2}),\\ &W_{2}^{k}(L,y_{2})=0,\\ &W_{2}^{k}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+W_{1}^{k}(y_{1},\frac{1}{2})\right)w^{k}(y_{1}),\\ &W_{2}^{k}(y_{1},0)=0,\\ \end{aligned}\end{cases}

(5.23)

where

(\hat{J}^{k},\hat{\Lambda}^{k},\hat{A}^{k},\hat{B}^{k})=(\tilde{J}_{0}^{k},\tilde{\nu}_{0}^{k},\tilde{A}_{0}^{k},\tilde{B}_{0}^{k})-(J_{b}^{-},0,A_{b}^{-},B_{b}^{-}).

Taking the difference of (5.8) and (5.23), one obtains that

\begin{cases}\begin{aligned} &W_{3}^{k}-W_{3}^{0}=\mathcal{F}_{0}({{\bf W}}^{k},\hat{\Lambda}^{k},\hat{A}^{k},\hat{B}^{k})-\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B}),\\ &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}(W_{1}^{k}-W_{1})+{\partial}_{y_{2}}(W_{2}^{k}-W_{2})+\frac{1}{y_{2}}(W_{2}^{k}-W_{2})\\ &=\mathcal{F}_{1}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})-\mathcal{F}_{1}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B}),\\ &{\partial}_{y_{1}}(W_{2}^{k}-W_{2})-{\partial}_{y_{2}}(W_{1}^{k}-W_{1})\\ &=\mathcal{F}_{2}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})-\mathcal{F}_{2}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B}),\\ &(W_{1}^{k}-W_{1})(0,y_{2})=\left(\mathcal{F}_{3}({{\bf W}}^{k},\hat{J}^{k},\hat{A}^{k},\hat{B}^{k})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})\right)(0,y_{2}),\\ &(W_{2}^{k}-W_{2})(L,y_{2})=0,\\ &(W_{2}^{k}-W_{2})(y_{1},\frac{1}{2})=u_{b}^{-}(w^{k}-w)(y_{1})+W_{1}^{k}(y_{1},\frac{1}{2})w^{k}(y_{1})-W_{1}(y_{1},\frac{1}{2})w(y_{1}),\\ &(W_{2}^{k}-W_{2})(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.24)

By similar estimate in (5.12), one can infer that

\sum_{j=1}^{3}\|W_{j}^{k}-W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\mathcal{C}\left(\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\|_{\mathcal{V}_{0}}+\|w^{k}-{w}\|_{0,\alpha;[0,L]}\right).

(5.25)

Therefore,

		$\displaystyle\\|\mathcal{Q}(\bm{\varphi}_{0}^{k},w^{k})-\mathcal{Q}(\bm{\varphi}_{0},w)\\|_{0,\alpha;[0,L]}$		(5.26)
		$\displaystyle=\left\\|N(W_{1}^{k},W_{2}^{k},W_{3}^{k},\tilde{A}_{0}^{k},\tilde{B}_{0}^{k})-N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})\right\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\mathcal{C}\left(\sum_{j=1}^{3}\\|W_{j}^{k}-W_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\\|_{\mathcal{V}_{0}}\right)$
		$\displaystyle\leq\mathcal{C}\left(\\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\\|_{\mathcal{V}_{0}}+\\|w^{k}-{w}\\|_{0,\alpha;[0,L]}\right),$

which yields (5.22).

Next, we prove the continuity of the map $D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)$ , i.e. to show that as $k\rightarrow\infty$ ,

D_{w}\mathcal{Q}(\bm{\varphi}_{0}^{k},w^{k})\rightarrow D_{w}\mathcal{Q}(\bm{\varphi}_{0},w),\quad{\rm{in}}\quad\mathcal{V}.

(5.27)

It follows from (5.17) that ${\bf W}^{0,k}$ satisfies the following problem:

\begin{cases}\begin{aligned} &W_{3}^{0,k}=\sum_{j=1}^{3}{\partial}_{W_{j}^{k}}\mathcal{F}_{0}({{\bf W}}^{k},\hat{\Lambda}^{k},\hat{A}^{k},\hat{B}^{k})W_{j}^{0,k},\\ &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{0,k}+{\partial}_{y_{2}}W_{2}^{0,k}+\frac{1}{y_{2}}W_{2}^{0,k}\\ &=\sum_{j=1}^{3}\left({\partial}_{W_{j}^{k}}\mathcal{F}_{1}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})W_{j}^{0,k}-{\partial}_{\nabla W_{j}^{k}}\mathcal{F}_{1}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})\nabla W_{j}^{0,k}\right),\\ &{\partial}_{y_{1}}W_{2}^{0,k}-{\partial}_{y_{2}}W_{1}^{0,k}\\ &=\sum_{j=1}^{3}\left({\partial}_{W_{j}^{k}}\mathcal{F}_{2}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})W_{j}^{0,k}-{\partial}_{\nabla W_{j}^{k}}\mathcal{F}_{2}({{\bf W}}^{k},\nabla{{\bf W}}^{k},\hat{A}^{k},\hat{B}^{k})\nabla W_{j}^{0,k}\right),\\ &W_{1}^{0,k}(0,y_{2})=\left(\sum_{j=1}^{2}{\partial}_{W_{j}^{k}}\mathcal{F}_{3}({{\bf W}}^{k},\hat{J}^{k},\hat{A}^{k},\hat{B}^{k})W_{j}^{0,k}\right)(0,y_{2}),\\ &W_{2}^{0,k}(L,y_{2})=0,\\ &W_{2}^{0,k}(y_{1},\frac{1}{2})=\left(u_{b}^{-}+W_{1}^{k}(y_{1},\frac{1}{2})\right)w_{1}(y_{1})+W_{1}^{0,k}(y_{1},\frac{1}{2})w^{k}(y_{1}),\\ &W_{2}^{0,k}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.28)

Taking the difference of (5.17) and (5.28), one has

\displaystyle\sum_{j=1}^{3}\|W_{j}^{0,k}-W_{j}^{0}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\mathcal{C}\left(\sum_{j=1}^{3}\|W_{j}^{k}-W_{j}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\|_{\mathcal{V}_{0}}+\|w^{k}-{w}\|_{0,\alpha;[0,L]}\right).

Combining (5.25) yields that

\displaystyle\sum_{j=1}^{3}\|W_{j}^{0,k}-W_{j}^{0}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\mathcal{C}\left(\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\|_{\mathcal{V}_{0}}+\|w^{k}-{w}\|_{0,\alpha;[0,L]}\right).

(5.29)

Then it holds that

		$\displaystyle\\|D_{w}\mathcal{Q}(\bm{\varphi}_{0}^{k},w^{k})(w_{1})-D_{w}\mathcal{Q}(\bm{\varphi}_{0},w)(w_{1})\\|_{0,\alpha;[0,L]}$		(5.30)
		$\displaystyle=\left\\|\sum_{j=1}^{3}\left({\partial}_{W_{j}^{k}}N(W_{1}^{k},W_{2}^{k},W_{3}^{k},\tilde{A}_{0}^{k},\tilde{B}_{0}^{k})W_{j}^{0,k}-{\partial}_{W_{j}}N(W_{1},W_{2},W_{3},\tilde{A}_{0},\tilde{B}_{0})W_{j}^{0}\right)\right\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\mathcal{C}\left(\sum_{j=1}^{3}\\|W_{j}^{0,k}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\sum_{j=1}^{3}\\|W_{j}^{k}-W_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\\|_{\mathcal{V}_{0}}\right)$
		$\displaystyle\leq\mathcal{C}\left(\\|\bm{\varphi}_{0}^{k}-\bm{\varphi}_{0}\\|_{\mathcal{V}_{0}}+\\|w^{k}-{w}\\|_{0,\alpha;[0,L]}\right),$

which implies that (5.27) holds.

In particular, at the background state,

\displaystyle D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)(w_{1})=-J_{b}^{-}W_{1}^{b}(y_{1},\frac{1}{2})=-2W_{1}^{b}(y_{1},\frac{1}{2}),

(5.31)

where ${\bf W}^{b}$ satisfies

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{b}+{\partial}_{y_{2}}W_{2}^{b}+\frac{1}{y_{2}}W_{2}^{b}=0,\\ &{\partial}_{y_{1}}W_{2}^{b}-{\partial}_{y_{2}}W_{1}^{b}=0,\\ &W_{3}^{b}=0,\\ &W_{1}^{b}(0,y_{2})=0,\\ &W_{2}^{b}(L,y_{2})=0,\\ &W_{2}^{b}(y_{1},\frac{1}{2})=u_{b}^{-}w_{1}(y_{1}),\\ &W_{2}^{b}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.32)

Define

\mathcal{V}_{b}=\{v\in\mathcal{V}:v(0)=0\}.

(5.33)

Then $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)$ is a continuous mapping from $\mathcal{V}_{1}$ to $\mathcal{V}_{b}\subset\mathcal{V}$ .

Step 2. The isomorphism of $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)$ .

To prove the isomorphism of $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)$ , we need to show that for any given function $P^{\ast}\in\mathcal{V}_{b}$ , there exists a unique $w^{\ast}\in\mathcal{V}_{1}$ such that $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)(w^{\ast})=P^{\ast}$ , i.e.,

\displaystyle P^{\ast}(y_{1})=-2W_{1}^{\ast}(y_{1},\frac{1}{2}).

(5.34)

It follows from (5.32) that the solution $(W_{1}^{\ast},W_{2}^{\ast},0)$ satisfies

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{y_{1}}W_{1}^{\ast}+{\partial}_{y_{2}}W_{2}^{\ast}+\frac{1}{y_{2}}W_{2}^{\ast}=0,\\ &{\partial}_{y_{1}}W_{2}^{\ast}-{\partial}_{y_{2}}W_{1}^{\ast}=0,\\ &W_{1}^{\ast}(0,y_{2})=0,\\ &W_{2}^{\ast}(L,y_{2})=0,\\ &W_{1}^{\ast}(y_{1},\frac{1}{2})=-\frac{P^{\ast}(y_{1})}{2},\\ &W_{2}^{\ast}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.35)

The second equation in (5.35) implies that there exists a potential function $\phi^{\ast}$ such that

({\partial}_{y_{1}}\phi^{\ast},{\partial}_{y_{2}}\phi^{\ast})=(W_{1}^{\ast},W_{2}^{\ast}),\quad\phi^{\ast}(L,0)=0.

(5.36)

Then (5.35) can be written as

\begin{cases}\begin{aligned} &{\partial}_{y_{1}}((1-(M_{b}^{-})^{2})y_{2}{\partial}_{y_{1}}\phi^{\ast})+{\partial}_{y_{2}}(y_{2}{\partial}_{y_{2}}\phi^{\ast})=0,\\ &{\partial}_{y_{1}}\phi^{\ast}(0,y_{2})=0,\\ &\phi^{\ast}(L,y_{2})=0,\\ &\phi^{\ast}(y_{1},\frac{1}{2})=\int_{L}^{y_{1}}-\frac{P^{\ast}(s)}{2}{\mathrm{d}}s,\\ &{\partial}_{y_{2}}\phi^{\ast}(y_{1},0)=0.\\ \end{aligned}\end{cases}

(5.37)

To deal with the singularity near $y_{2}=0$ , we rewrite the problem (5.37) by using the cylindrical coordinate transformation again. Define

\lambda_{1}=y_{1},\ \lambda_{2}=y_{2}\cos\tau,\ \lambda_{3}=y_{2}\sin\tau,\ \tau\in[0,2\pi],

and

		$\displaystyle D_{1}=\{(\lambda_{1},\lambda_{2},\lambda_{3}):0<\lambda_{1}<L,\lambda_{2}^{2}+\lambda_{3}^{2}\leq\frac{1}{2}\},\quad D_{2}=\{(\lambda_{2},\lambda_{3}):\lambda_{2}^{2}+\lambda_{3}^{2}\leq\frac{1}{2}\},$
		$\displaystyle\Gamma_{w,\lambda}=[0,L]\times\{(\lambda_{2},\lambda_{3}):\lambda_{2}^{2}+\lambda_{3}^{2}=\frac{1}{2}\},$
		$\displaystyle\Gamma_{0,\lambda}=\{0\}\times\{(\lambda_{2},\lambda_{3}):\lambda_{2}^{2}+\lambda_{3}^{2}\leq\frac{1}{2}\},\quad\Gamma_{L,\lambda}=\{L\}\times\{(\lambda_{2},\lambda_{3}):\lambda_{2}^{2}+\lambda_{3}^{2}\leq\frac{1}{2}\},$
		$\displaystyle\Psi^{\ast}(\bm{\lambda})=\phi^{\ast}(\lambda_{1},\sqrt{\lambda_{2}^{2}+\lambda_{3}^{2}})=\Psi^{\ast}(\lambda_{1},\|\lambda^{\prime}\|).$

Then $\Psi^{\ast}$ solves the following problem

\begin{cases}\begin{aligned} &(1-(M_{b}^{-})^{2}){\partial}_{\lambda_{1}}^{2}\Psi^{\ast}+{\partial}_{\lambda_{2}}^{2}\Psi^{\ast}+{\partial}_{\lambda_{3}}^{2}\Psi^{\ast}=0,\\ &{\partial}_{\lambda_{1}}\Psi^{\ast}(0,|\lambda^{\prime}|)=0,\quad&{\rm{on}}\quad\Gamma_{0,\lambda},\\ &\Psi^{\ast}(L,|\lambda^{\prime}|)=0,\quad&{\rm{on}}\quad\Gamma_{L,\lambda},\\ &\Psi^{\ast}(\lambda_{1},|\lambda^{\prime}|)=\int_{L}^{\lambda_{1}}-\frac{P^{\ast}(s)}{2}{\mathrm{d}}s,\quad&{\rm{on}}\quad\Gamma_{w,\lambda}.\\ \end{aligned}\end{cases}

(5.38)

By similar arguments as in the Step 3 of Lemma 4.1, (5.38) has a unique solution $\Psi^{\ast}\in C_{2,\alpha}^{(-1-\alpha,\Gamma_{w,\lambda})}(D_{1})$ satisfying

\|\Psi^{\ast}\|_{2,\alpha;D_{1}}^{(-1-\alpha,\Gamma_{w,\lambda})}\leq\mathcal{C}\|P^{\ast}\|_{0,\alpha;[0,L]}.

(5.39)

Thus

\sum_{i=1}^{2}\|W_{i}^{\ast}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\leq\mathcal{C}\|P^{\ast}\|_{0,\alpha;[0,L]}.

(5.40)

Set $w^{\ast}(y_{1})=u_{b}^{-}W_{2}^{\ast}(y_{1},\frac{1}{2})$ , then (5.40) shows that $w^{\ast}\in\mathcal{V}_{1}$ . Hence we have shown there exists a unique $w^{\ast}\in\mathcal{V}_{1}$ such that $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)(w^{\ast})=P^{\ast}$ . The proof of the isomorphism of $D_{w}\mathcal{Q}(\bm{\varphi}_{b},0)$ is completed.

6 Proof of Theorem 2.3

Now, by the implicit function theorem, there exist positive constants $\sigma_{6}$ and $\mathcal{C}$ depending only on $(\bm{U}_{b}^{-},L,\alpha)$ such that for $\delta_{3}\leq\sigma_{6}$ , the equation $\mathcal{Q}(\bm{\varphi}_{0},w)=0$ has a unique solution $w=w(\bm{\varphi}_{0})$ satisfying

\|w\|_{0,\alpha;[0,L]}\leq\mathcal{C}\|\bm{\varphi}_{0}-\bm{\varphi}_{b}\|_{\mathcal{V}_{0}}=\mathcal{C}\sigma_{cd}.

(6.1)

Here $\delta_{3}$ is defined in (5.4). Hence the contact discontinuity $g_{cd}(y_{1})=\int_{0}^{y_{1}}w(s){\mathrm{d}}s+\frac{1}{2}$ is determined and (6.1) implies that

\|g_{cd}-\frac{1}{2}\|_{1,\alpha;[0,L]}\leq\mathcal{C}_{5}\|\bm{\varphi}_{0}-\bm{\varphi}_{b}\|_{\mathcal{V}_{0}}=\mathcal{C}_{5}\sigma_{cd},

(6.2)

where $\mathcal{C}_{5}>0$ depends only on $(\bm{U}_{b}^{-},L,\alpha)$ .

We choose $\sigma_{1}$ and $\mathcal{C}^{\ast}$ as

\sigma_{1}=\min\{\sigma_{5},\sigma_{6}\}\quad{\rm{and}}\quad\mathcal{C}^{\ast}=4\mathcal{C}_{1}+\mathcal{C}_{5},

(6.3)

where $\sigma_{5}$ is defined in (5.19) and $\mathcal{C}_{1}$ is defined in (4.49). Then if $\sigma_{cd}\leq\sigma_{1}$ , $\mathbf{Problem\ 3.1}$ has a unique smooth subsonic solution $(u_{x},u_{r},u_{\theta};g_{cd})$ satisfying

\|u_{x}-u_{b}^{-}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|u_{r}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|u_{\theta}\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\|g_{cd}-\frac{1}{2}\|_{1,\alpha;[0,L]}\leq\mathcal{C}^{\ast}\sigma_{cd}.

(6.4)

Furthermore, one has

\|(A,B)-(A_{b}^{-},B_{b}^{-})\|_{1,\alpha;\overline{\Omega}}=\|(\tilde{A}_{0},\tilde{B}_{0})-(A_{b}^{-},B_{b}^{-})\|_{1,\alpha;\overline{\Omega}}\leq\mathcal{C}^{\ast}\sigma_{cd}.

(6.5)

Since the modified Lagrangian transformation (3.3) is invertible, thus the solution transformed back in $(x,r)$ -coordinates solves $\mathbf{Problem\ 2.2}$ and the estimates (6.4) and (6.5) imply that the estimates (2.18) and (2.19) in Theorem 2.3 hold. Thus the proof of Theorem 2.3 is completed.

Acknowledgement. Weng is partially supported by National Natural Science Foundation of China 11971307, 12071359, 12221001.

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	$\displaystyle{\\|u\\|}_{m,0;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\sum_{0\leq\|\beta\|\leq m}\sup_{\bm{x}\in\mathcal{D}}\delta_{\mathbf{x}}^{\max(\|\beta\|+k,0)}\|D^{\beta}u(\bm{x})\|;$
	$\displaystyle[u]_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\sum_{\|\beta\|=m}\sup_{\bm{x},\tilde{\bm{x}}\in\mathcal{D},\bm{x}\neq\tilde{\bm{x}}}\delta_{\mathbf{x},\tilde{\mathbf{x}}}^{\max(m+\alpha+k,0)}\frac{\|D^{\beta}u({\bm{x}})-D^{\beta}u(\tilde{\bm{x}})\|}{\|\bm{x}-\tilde{\bm{x}}\|^{\alpha}};$
	$\displaystyle\\|u\\|_{m,\alpha;\mathcal{D}}^{(k,\mathcal{P})}$	$\displaystyle:=\\|u\\|_{m,0;\mathcal{D}}^{k,\mathcal{P}}+[u]_{m,\alpha;\mathcal{D}}^{k,\mathcal{P}}$

		$\displaystyle\\|F_{3}\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha,\{\frac{1}{2}\})}+\\|F_{4}\\|_{0,\alpha;[0,L]}$		(4.9)
		$\displaystyle\leq C\left(\bigg{(}\sum_{j=1}^{3}\\|\hat{W_{j}}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\bigg{)}^{2}+\\|\hat{W_{1}}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\\|g_{cd}^{\prime}\\|_{0,\alpha;[0,L]}+\\|g_{cd}^{\prime}\\|_{0,\alpha;[0,L]}+\sigma_{cd}\right)$
		$\displaystyle\leq C\left(\delta_{1}^{2}+\delta_{1}\\|w\\|_{0,\alpha;[0,L]}+\\|w\\|_{0,\alpha;[0,L]}+\sigma_{cd}\right),$

$\displaystyle\sum_{j=1}^{3}\\|Y_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	$\displaystyle\leq\mathcal{C}_{2}\left(\sum_{j=1}^{2}\\|\mathcal{F}_{j}(\hat{{\bf W}}^{1},\nabla{\hat{\bf W}}^{1},\hat{A},\hat{B})-\mathcal{F}_{j}(\hat{{\bf W}}^{2},\nabla{\hat{\bf W}}^{2},\hat{A},\hat{B})\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}\right.$	(4.54)
	$\displaystyle\qquad\quad\left.+\\|\mathcal{F}_{3}(\hat{{\bf W}}^{1},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}(\hat{{\bf W}}^{2},\hat{J},\hat{A},\hat{B})\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}\right.$
	$\displaystyle\qquad\quad\left.+\\|\mathcal{F}_{4}(\hat{{\bf W}}^{1},g_{cd})-\mathcal{F}_{4}(\hat{{\bf W}}^{2},g_{cd})\\|_{0,\alpha;[0,L]}+\left\\|{\hat{\Lambda}}\left(\frac{1}{\hat{r}^{1}}-\frac{1}{\hat{r}^{2}}\right)\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}\right)$
	$\displaystyle\leq\mathcal{C}_{2}(\delta_{1}+\delta_{2}+\sigma_{cd})\sum_{j=1}^{3}\\|\hat{Y}_{j}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)},$

$\displaystyle\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	$\displaystyle\leq\mathcal{C}_{3}\left\\|\frac{\mathcal{F}_{0}(\tilde{{\bf W}},\hat{\Lambda},\hat{A},\hat{B})-\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B})}{\tau}\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$	(5.12)
	$\displaystyle\quad+\mathcal{C}_{3}\left(\sum_{j=1}^{2}\left\\|\frac{\mathcal{F}_{j}(\tilde{{\bf W}},\nabla{\tilde{\bf W}},\hat{A},\hat{B})-\mathcal{F}_{j}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau}\right\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}\right)$
	$\displaystyle\quad+\mathcal{C}_{3}\left\\|\frac{\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})}{\tau}\right\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|(u_{b}^{-}+\tilde{W}_{1})w_{1}+W_{1}^{\tau}w\\|_{0,\alpha;[0,L]}$
	$\displaystyle\leq\mathcal{C}_{3}\left(\delta_{2}+4\mathcal{C}_{1}\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+4\mathcal{C}_{3}\mathcal{C}_{1}\sigma_{cd}\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\leq\mathcal{C}_{3}((4\mathcal{C}_{1}+1)\sigma_{cd})\sum_{j=1}^{3}\\|W_{j}^{\tau}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{3}((4\mathcal{C}_{1}+1)\sigma_{cd})\\|w_{1}\\|_{0,\alpha;[0,L]}$
	$\displaystyle\quad+\mathcal{C}_{3}\\|w_{1}\\|_{0,\alpha;[0,L]},$

		$\displaystyle\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$		(5.18)
		$\displaystyle\leq\mathcal{C}_{4}\left\\|\frac{\mathcal{F}_{0}(\tilde{\bf W},\hat{\Lambda},\hat{A},\hat{B})-\mathcal{F}_{0}({\bf W},\hat{\Lambda},\hat{A},\hat{B})}{\tau}-\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{0}({{\bf W}},\hat{\Lambda},\hat{A},\hat{B})W_{j}^{0}\right\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}$
		$\displaystyle\quad+\mathcal{C}_{4}\sum_{i=1}^{2}\left\\|\frac{\mathcal{F}_{i}(\tilde{{\bf W}},\nabla{\tilde{{\bf W}}},\hat{A},\hat{B})-\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})}{\tau}\right.$
		$\displaystyle\qquad\qquad\quad\left.-\sum_{j=1}^{3}\left({\partial}_{W_{j}}\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})W_{j}^{0}+{\partial}_{\nabla W_{j}}\mathcal{F}_{i}({{\bf W}},\nabla{{\bf W}},\hat{A},\hat{B})\nabla W_{j}^{0}\right)\right\\|_{0,\alpha;\Omega}^{(1-\alpha,\Sigma)}$
		$\displaystyle\quad+\mathcal{C}_{4}\left\\|\frac{\mathcal{F}_{3}(\tilde{{\bf W}},\hat{J},\hat{A},\hat{B})-\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})}{\tau}-\sum_{j=1}^{3}{\partial}_{W_{j}}\mathcal{F}_{3}({{\bf W}},\hat{J},\hat{A},\hat{B})W_{j}^{0}\right\\|_{1,\alpha;[0,\frac{1}{2})}^{(-\alpha;\{\frac{1}{2}\})}$
		$\displaystyle\quad+\mathcal{C}_{4}\\|\tau W_{1}^{\tau}w_{1}+(W_{1}^{\tau}-W_{1}^{0})w\\|_{0,\alpha;[0,L]}$
		$\displaystyle\leq\mathcal{C}_{4}\left(\delta_{2}+4\mathcal{C}_{1}\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{4}(2\mathcal{C}_{3}+2)\tau(\\|w_{1}\\|_{0,\alpha;[0,L]})^{2}$
		$\displaystyle\leq\mathcal{C}_{4}\left((4\mathcal{C}_{1}+1)\sigma_{cd}\right)\sum_{j=1}^{3}\\|W_{j}^{\tau}-W_{j}^{0}\\|_{1,\alpha;\Omega}^{(-\alpha,\Sigma)}+\mathcal{C}_{4}(2\mathcal{C}_{3}+2)\tau(\\|w_{1}\\|_{0,\alpha;[0,L]})^{2},$

Subsonic flows with a contact discontinuity in a finitely long axisymmetric cylinder

Abstract

1 Introduction

Definition 1.1.

2 Mathematical formulation of the problem

2.1 The background solutions

2.2 The stability problem and the main result

Problem 2.1.

Problem 2.2.

Theorem 2.3.

Remark 2.4.

Remark 2.5.

3 The reformulation of Problem 2.2

3.1 Reformulation by the modified Lagrangian transformation

3.2 The deformation-curl decomposition for axisymmetric Euler system

Problem 3.1.

3.3 Solving the free boundary problem 3.1

4 The solution to a fixed boundary value problem in Ω\Omega

4.1 Linearization

4.2 Solving the linear boundary value problem

Lemma 4.1.

Proof.

4.3 Solving the nonlinear boundary value problem

5 The construction of the contact discontinuity surface

6 Proof of Theorem 2.3

References

4 The solution to a fixed boundary value problem in $\Omega$