3-D Axisymmetric Transonic Shock Solutions of the Full Euler System in Divergent Nozzles

Abstract

We establish the stability of 3-D axisymmetric transonic shock solutions of the steady full Euler system in divergent nozzles under small perturbations of an incoming radial supersonic flow and a constant pressure at the exit of the nozzles. To study 3-D axisymmetric transonic shock solutions of the full Euler system, we use a stream function formulation of the full Euler system for a 3-D axisymmetric flow. We resolve the singularity issue arising in stream function formulations of the full Euler system for a 3-D axisymmetric flow. We develop a new scheme to determine a shock location of a transonic shock solution of the steady full Euler system based on the stream function formulation.

1 Introduction

In [14, Chapter 147], the authors, using an approximate model, described a transonic shock phenomenon for a compressible invicid flow of an ideal polytropic gas in a convergent-divergent type nozzle called de Laval nozzle: if a subsonic flow accelerating as it passes through the convergent part of the nozzle reaches the sonic speed at the throat of the nozzle, then it becomes a supersonic flow right after the throat of the nozzle. It further accelerates as it passes through the divergent part of the nozzle. If an appropriately large exit pressure $p_e$ is imposed at the exit of the nozzle, then at a certain place of the divergent part of the nozzle, a shock front intervenes, the flow is compressed and slowed down to subsonic speed. The position and strength of the shock front are automatically adjusted so that the end pressure at the exit becomes $p_e$. This phenomenon was rigorously studied using radial solutions of the full Euler system in [31] (it was shown that in a divergent nozzle, for a given constant supersonic data on the entrance of the nozzle and appropriately large constant pressure on the exit of the nozzle, there exists a unique radial transonic shock solution satisfying these conditions). Motivated by this phenomenon, there were many studies on the stability of transonic shock solutions in divergent nozzles (structural stability of radial transonic shock solutions in divergent nozzles under multi-dimensional perturbations of an entrance supersonic data and exit pressure) and related problems.

The stability of one-dimensional transonic shock solutions in flat nozzles was studied first. This subject was studied using the potential flow model in [7,8,9, 26, 27] and further studied using the full Euler system in [5, 6, 11,12,13, 25, 29, 30]. These results showed that one-dimensional transonic shock solutions in flat nozzles are not stable under a perturbation of a physical boundary condition (supersonic data on the entrance or density, pressure or normal velocity on the exit) and, even if one-dimensional transonic shock solutions in flat nozzles are stable, their shock locations are not uniquely determined unless there exists the assumption that a shock location passes through some point on the wall of the nozzle, as it can be expected from the behavior of one-dimensional transonic shock solutions in flat nozzles (as a shock location changes, the value of the subsonic part of an one-dimensional transonic shock solution in a flat nozzle does not change). After that, the stability of radial transonic shock solutions in divergent nozzles was studied. This subject was first studied using the full Euler system in [19, 28]. In these results, the authors, by considering a perturbation of radial transonic shock solutions in divergent nozzles, could show that a shock location is uniquely determined for a given exit pressure without the assumption that a shock location passes through some point of the nozzle but they only had the result under the assumption that the tip angle of the nozzle is sufficiently small. The stability of radial transonic shock solutions in divergent nozzle without restriction on the tip angle of the nozzle was studied after that. In [3], the authors studied this subject using the non-isentropic potential model introduced in [3], and they obtained the stability result for radial transonic shock solutions in divergent nozzles. This subject was also studied using the full Euler system. For the 2-D case this was done in [18, 21, 24]. In these papers, the authors had the stability result for radial transonic shock solutions in divergent nozzles. In particular, the authors in [21] had the result for flows having $C^{1,\alpha }$ interior and $C^\alpha$ up to boundary regularity, so that they could consider a general perturbation of a nozzle. This for the 3-D case for axisymmetric flows with zero angular momentum components was done in [20]. The authors in this paper also had the same result. This, for the general 3-D case, was done in [10, 23]. The authors in [10, 23] also had the same result but under S-condition introduced in [10, 23]. Recently, this subject for the general 3-D case for flows having some friction term was studied in [32].

In this paper, we study the stability of radial transonic shock solutions in divergent nozzles under small perturbations of an incoming radial supersonic solution and a constant exit pressure using the full Euler system for the 3-D case for axisymmetric flows. We consider axisymmetric flows with non-zero angular momentum components. (This is a difference from [20].) We consider a divergent nozzle having no restriction on the tip angle of the nozzle and do not have any assumption on an incoming supersonic solution.

The main new feature in this paper is to develop a new iteration scheme to determine a shock location for a transonic shock solution of the steady full Euler system in a divergent nozzle and resolve the singularity issue arising in the stream function formulations of the full Euler system using an elliptic system approach.

To deal with the stability of 3-D axisymmetric transonic shock solutions of the full Euler system, we use a stream function formulation for the full Euler system for an axisymmetric flow. This formulation shows the fact that an initial shock position and a shape of a shock location (see the definitions below the proof of Theorem 2) are determined in different mechanisms clearly. Based on this formulation and using the fact that the entropy of the downstream subsonic solution of a radial transonic shock solution on a shock location monotonically increases as a shock location moves toward the exit of the nozzle (see Lemma 2), we develop a new scheme to determine a shock location of a transonic shock solution of the full Euler system in a divergent nozzle: 1. Pseudo Free Boundary Problem 2. Determination of a shape of a shock location (see below the proof of Theorem 2).

In technical part, we resolve the singularity issue arising in stream function formulations of the full Euler system. A stream function formulation when it is formulated by using the Stokes’ stream function (see (2.23)) has a singularity issue at the axis of symmetry. We resolve this singularity issue by formulating a stream function formulation using the vector potential form of the stream function (see § 2.4) and solving a singular elliptic equation appearing in this stream function formulation as an elliptic system (see § 3.2). The stream function formulation formulated by using the vector potential form of the stream function still contains a singular elliptic equation. We transform this singular elliptic equation into a form of an elliptic system, and solve the elliptic system form as an elliptic system to solve the singular elliptic equation. (We also use this approach to prove the orthogonal completeness of eigenfunction of an associated Legendre problem of type $m=1$ with a general domain (see Lemma 18).) Using the stream function formulation formulated by using the vector potential form of the stream function, we obtain the stability result for flows having $C^{1,\alpha }$ interior and $C^\alpha$ up to boundary regularity.

This paper is organized as follows: in Section 2, we present definitions and a basic lemma used throughout this paper and introduce our problem and result. In this section, we introduce the stream function formulation used in this paper. In Section 3, we solve the Pseudo Free Boundary Problem. In this section, we study a linear boundary value problem for a singular elliptic equation and an initial value problem of a transport equation appearing in the Pseudo Free Boundary Problem, and prove the unique existence of solutions of the Pseudo Free Boundary Problem. In Section 4, we show the existence and uniqueness of transonic shock solutions. In Section 5, we present some computations done by using the tensor notation given in § 3.2, and make up some fact needed for the well-definedness of the stream function defined in (2.23).

2 Problem and Theorem

2.1 Preliminaries

In this paper, we consider a 3-D steady compressible invicid flow of an ideal polytropic gas. The motion of this flow is governed by the full Euler system

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathrm{div}(\rho {\mathbf {u}})=0,\\ \mathrm{div}(\rho {\mathbf {u}}\otimes {\mathbf {u}}+p{{\mathbb {I}}})=0,\\ \mathrm{div}(\rho {\mathbf {u}}B)=0, \end{array}\right. } \end{aligned}$$

(2.1)

where $\rho$, ${\mathbf {u}}$ and p are the density, velocity and pressure of the flow, ${{\mathbb {I}}}$ is the $3\times 3$ identity matrix and B is the Bernoulli invariant of the flow given by

$$\begin{aligned} B=\frac{|{\mathbf {u}}|^2}{2} + \frac{ \gamma p}{(\gamma -1)\rho } \end{aligned}$$

(2.2)

for a constant $\gamma >1$. Types of the flow are classified by the quantity $M:=\frac{|{\mathbf {u}}|}{c}$ called Mach number where c is the sound speed of the flow given by

$$\begin{aligned} c:=\sqrt{\frac{\gamma p}{\rho }} \end{aligned}$$

for an ideal polytropic gas; if $M>1$, then a flow is called supersonic, if $M=1$, then it is called sonic and if $M<1$, then it is called subsonic. It is generally known that types of the system vary depending on the value of M. If $M>1$, then the system is a hyperbolic system and if $M<1$, then the system is an elliptic-hyperbolic coupled system.

When the flow passes through a domain having a certain geometric structure or satisfies a certain boundary condition, it may have a discontinuity across a surface in the domain in the direction of the flow. In this case, the surface is called a shock.

A shock solution of (2.1) is a solution of (2.1) having a shock in a domain, and is defined as follows:

Definition 1

(Shock solution) Let $\varOmega$ be an open connected set in $\mathbb {R}^3$. Assume that a $C^1$ surface $\varGamma$ in $\varOmega$ divides $\varOmega$ into two nonempty disjoint subset $\varOmega ^\pm$ such that $\varOmega =\varOmega ^+\cup \varGamma \cup \varOmega ^-$. Then a solution $(\rho ,{\mathbf {u}},p)$ of (2.1) is called a shock solution of (2.1) with a shock $\varGamma$ if

• (i) $(\rho ,{\mathbf {u}},p)$ is in $(C^0(\overline{\varOmega ^\pm })\cap C^1(\varOmega ^\pm ))^3$,

• (ii) $(\rho ,{\mathbf {u}},p)|_{\overline{\varOmega ^-}\cap \varGamma }\ne (\rho ,{\mathbf {u}},p)|_{\overline{\varOmega ^+}\cap \varGamma }$,

• (iii) $\rho {\mathbf {u}}|_{\overline{\varOmega ^-}} \cdot {\varvec{\nu }}\ne 0$ on $\varGamma$,

• (iv) and $(\rho ,{\mathbf {u}},p)$ satisfies (2.1) pointwisely in $\varOmega ^\pm$ and the Rankine-Hugoniot conditions

  $$\begin{aligned} {[\rho {\mathbf {u}}\cdot {\varvec{\nu }}]}_{\varGamma }=[{\mathbf {u}}\cdot {\varvec{\tau }_1}]_{\varGamma }=[{\mathbf {u}}\cdot {\varvec{\tau }_2}]_{\varGamma }=[\rho ({\mathbf {u}}\cdot {\varvec{\nu }})^2+p]_{\varGamma }=[B]_{\varGamma }=0, \end{aligned}$$

  (2.3)

  where ${\varvec{\nu }}$ is the unit normal vector field on $\varGamma$ pointing toward $\varOmega ^+$ and ${\varvec{\tau }_i}$ for $i=1,2$ are unit tangent vector fields on $\varGamma$ perpendicular to each other at each point on $\varGamma$ and

  $$\begin{aligned}{}[F]_\varGamma :=F|_{\overline{\varOmega ^-}}(x)-F|_{\overline{\varOmega ^+}}(x)\quad \text{ for }\quad x\in \varGamma . \end{aligned}$$

It is said that a shock solution is physically admissible if it satisfies the following entropy condition:

Definition 2

(Entropy condition) Let $(\rho ,{\mathbf {u}},p)$ be a shock solution of (2.1) defined in Definition 1. Without loss of generality, assume that ${\mathbf {u}}|_{\overline{\varOmega ^-}}\cdot {\varvec{\nu }}>0$ on $\varGamma$. Then $S|_{\overline{\varOmega ^+}\cap \varGamma }>S|_{\overline{\varOmega ^-}\cap \varGamma }$ where $S:=\frac{p}{\rho ^\gamma }$ is called the entropy condition.

Definition 3

In this paper, we call S the entropy of $(\rho ,{\mathbf {u}},p)$.

Using the definition of a shock solution, a transonic shock solution of (2.1) is defined as follows:

Definition 4

A shock solution of (2.1) in Definition 1 is called a transonic shock solution if it satisfies $M>1$ in $\overline{\varOmega ^-}$ and $M<1$ in $\overline{\varOmega ^+}$ or the otherway around.

Remark 1

If a transonic shock solution of (2.1) satisfies ${\mathbf {u}}|_{\overline{\varOmega ^-}}\cdot {\varvec{\nu }}>0$ on $\varGamma$, $M>1$ in $\overline{\varOmega ^-}$ and $M<1$ in $\overline{\varOmega ^+}$, then it satisfies the entropy condition.

In this paper, we deal with a 3-D axisymmetric transonic shock solution of (2.1). For precise statement, we define an axisymmetric domain and axisymmetric functions used in this paper. For later use, we present a lemma that can be used to deal with regularities of axisymmetric functions.

In this paper, we use the spherical coordinate system $(r,\theta ,\varphi )$ given by the relation

$$\begin{aligned} (x,y,z)=(r\sin \theta \cos \varphi ,r\sin \theta \sin \varphi ,r\cos \theta ), \end{aligned}$$

(2.4)

where (x, y, z) is the Cartesian coordinate system in $\mathbb {R}^3$. The unit vectors in this coordinate system are given by

$$\begin{aligned}&\mathbf{e}_r=\sin \theta \cos \varphi \mathbf{e}_1+\sin \theta \sin \varphi \mathbf{e}_2+\cos \theta \mathbf{e}_3,\\&\mathbf{e}_\theta =\cos \theta \cos \varphi \mathbf{e}_1+\cos \theta \sin \varphi \mathbf{e}_2-\sin \theta \mathbf{e}_3,\\&\mathbf{e}_{\varphi }=-\sin \varphi \mathbf{e}_1+\cos \varphi \mathbf{e}_2, \end{aligned}$$

where $\mathbf{e}_i$ for $i=1,2,3$ are the unit vector in the x, y and z direction, respectively.

Using this spherical coordinate system, an axisymmetric domain and axisymmetric functions are defined as follows:

Definition 5

Let $\varOmega \subset \mathbb {R}^3$. $\varOmega$ is called axisymmetric if $(x,y,z)\in \varOmega$, then

$$\begin{aligned} (\sqrt{x^2+y^2}\cos \varphi ,\sqrt{x^2+y^2}\sin \varphi ,z)\in \varOmega \end{aligned}$$

for $\varphi \in [0,2\pi )$. A function $f:\varOmega \rightarrow \mathbb {R}$ is called axisymmetric if f is independent of $\varphi$ as a function of the spherical coordinate system. A vector valued function ${\mathbf {u}}:\varOmega \rightarrow \mathbb {R}^3$ is called axisymmetric if $u_r={\mathbf {u}}\cdot \mathbf{e}_r$, $u_\theta ={\mathbf {u}}\cdot \mathbf{e}_\theta$ and $u_\varphi ={\mathbf {u}}\cdot \mathbf{e}_\varphi$ are axisymmetric.

Definition 6

In this paper, when a velocity field ${\mathbf {u}}$ is represented as ${\mathbf {u}}=u_r\mathbf{e}_r+u_\theta \mathbf{e}_\theta +u_\varphi \mathbf{e}_\varphi$, $u_\varphi \mathbf{e}_{\varphi }$ is called the angular momentum component of ${\mathbf {u}}$.

For later use, we present the following lemma that shows when an axisymmetric function in $C^k$ as a function of the spherical coordinate system is in $C^k$ as a function of the Cartesian coordinate system (this lemma is obtained from [22, Corollary 1]):

Lemma 1

Let $\varOmega$ be an axisymmetric connected open set in $\mathbb {R}^3$ that does not contain the origin. Suppose that a function f: $\varOmega \rightarrow \mathbb {R}$ is axisymmetric. Then

• (i) f and $f\mathbf{e}_r$ are in $C^k(\varOmega )$ for $k\in 0,1,2,\ldots$ if and only if f is in $C^k$ as a function of spherical coordinate system in $\varOmega$ and $\partial _{\theta }^{2m+1}f=0$ for all $0\le m \le \lfloor \frac{k-1}{2}\rfloor$.

• (ii) $f\mathbf{e}_\theta$ and $f\mathbf{e}_\varphi$ are in $C^k(\varOmega )$ for $k\in 0,1,2,\ldots$ if and only if f is in $C^k$ as a function of spherical coordinate system in $\varOmega$ and $\partial _{\theta }^{2m}f=0$ for all $0\le m \le \lfloor \frac{k}{2}\rfloor$.

In this paper, we use the same function notation when we represent an axisymmetric function as a function on the Cartesian coordinate system or spherical coordinate system.

2.2 Radial transonic shock solution

Let $r_0$, $r_1$ and $\theta _1$ be constants such that $0<r_0<r_1$ and $0\le \theta _1<\pi$. Define a divergent nozzle by

$$\begin{aligned} \mathcal {N}:=\{(x,y,z)\in \mathbb {R}^3\;|\, r_0<r<r_1,\;0\le \theta <\theta _1 \}. \end{aligned}$$

To introduce our problem and for our later analysis, we study a radial transonic shock solution of (2.1) in $\mathcal {N}$.

Fix positive constants $(\rho _{in},u_{in},p_{in})$ satisfying $M_{in}(:=u_{in}/\sqrt{\frac{\gamma p_{in}}{\rho _{in}}})>1$. Let $({\bar{\rho }},{\bar{u}}\mathbf{e}_r,{\bar{p}})$ be a radial shock solution of (2.1) in $\mathcal {N}$ with a shock $\varGamma _t:=\{r=t\}\cap \mathcal {N}$ for some $t\in [r_0,r_1]$ satisfying

$$\begin{aligned} ({\bar{\rho }},{\bar{u}}\mathbf{e}_r, {\bar{p}})=(\rho _{in},u_{in}\mathbf{e}_r,p_{in})\quad \text{ on }\quad \varGamma _{en}:=\partial \mathcal {N}\cap \{r=r_0,\;0\le \theta <\theta _1\}. \end{aligned}$$

(2.5)

Then, by (2.1) and (2.3), $({\bar{\rho }},{\bar{u}}, {\bar{p}})$ is a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} (r^2{\bar{\rho }}{\bar{u}})^\prime =0,\\ {\bar{\rho }} {\bar{u}} {\bar{u}}^\prime +{\bar{p}}^\prime =0,\\ {\bar{\rho }} {\bar{u}} {\bar{B}}^\prime =0 \end{array}\right. } \end{aligned}$$

(2.6)

with

$$\begin{aligned} ({\bar{\rho }},{\bar{u}},{\bar{p}})(r_0)=(\rho _{in},u_{in},p_{in}) \end{aligned}$$

(2.7)

in $\overline{D_t^-}$, where $D_t^-:=\{r_0<r<t\}$, ${\bar{B}}:=\frac{{\bar{u}}^2}{2}+\frac{\gamma {\bar{p}}}{(\gamma -1){\bar{\rho }}}$ and $^\prime$ is the derivative with respect to r, and is a solution of (2.6) with

$$\begin{aligned} {\left\{ \begin{array}{ll} {\bar{\rho }} {\bar{u}}(t)={\bar{\rho }}{\bar{u}}|_{\overline{D_t^-}}(t),\\ ({\bar{\rho }} {\bar{u}}^2+{\bar{p}})(t)=({\bar{\rho }} {\bar{u}}^2+{\bar{p}})|_{\overline{D_t^-}}(t),\\ {\bar{B}}(t)= {\bar{B}}|_{\overline{D_t^-}}(t) \end{array}\right. } \end{aligned}$$

(2.8)

in $\overline{D_t^+}$ where $D_t^+:=\{t<r<r_1\}$.

From (2.6), one can obtain

$$\begin{aligned} {\bar{S}}^\prime =0, \end{aligned}$$

(2.9)

where ${\bar{S}}:=\frac{{\bar{p}}}{{\bar{\rho }}^\gamma }$. Using this equation, one can see that a solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ of (2.6) with (2.7) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} r^2{\bar{\rho }} {\bar{u}}=m_0,\\ {\bar{S}}=S_{in},\\ {\bar{B}}=B_0, \end{array}\right. } \end{aligned}$$

(2.10)

where $m_0:=r_0^2\rho _{in} u_{in}$, $S_{in}:=\frac{p_{in}}{\rho _{in}^\gamma }$ and $B_0:=\frac{u_{in}^2}{2}+\frac{\gamma p_{in}}{(\gamma -1)\rho _{in}}$ on the domain where (2.6) with (2.7) has a unique solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$. Using this fact, the local unique existence theorem for ODE and the condition that $M_{in}>1$, it can be shown that (2.6) with (2.7) has a unique solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ satisfying ${\bar{M}}>1$ and ${\bar{M}}^\prime >0$ in $[r_0,r_1]$ where ${\bar{M}}:={\bar{u}}/\sqrt{\frac{\gamma {\bar{p}}}{{\bar{\rho }}}}$. From this, we have that (2.8) is well-defined for any $t\in [r_0,r_1]$.

From (2.8), we can obtain

$$\begin{aligned} {\bar{u}}(t)=\left. \frac{{\bar{K}}}{{\bar{u}}}\right| _{\overline{D_t^-}}(t), \end{aligned}$$

(2.11)

where ${\bar{K}}:=\frac{2(\gamma -1)}{\gamma +1} {\bar{B}}$, and

$$\begin{aligned} {\bar{S}}(t)=(g({{\bar{M}}}^2) {\bar{S}})|_{\overline{D_t^-}}(t), \end{aligned}$$

(2.12)

where

$$\begin{aligned} g(x):=\frac{1}{\gamma +1}(2\gamma x-(\gamma -1))\left( \frac{\gamma -1}{\gamma +1}+\frac{2}{\gamma +1} \frac{1}{x}\right) ^\gamma . \end{aligned}$$

(2.13)

Using (2.11), the third equations of (2.8) and (2.10) and ${\bar{M}}|_{\overline{D_t^-}}(t)>1$, it can be checked that $({\bar{\rho }},{\bar{u}},{\bar{p}})$ satisfying (2.8) satisfies ${\bar{M}}(t) <1$. By (2.9), (2.10) and (2.12), a solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ of (2.6) with (2.8) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} r^2{\bar{\rho }} {\bar{u}}=m_0,\\ {\bar{S}}=g({{\bar{M}}}|_{\overline{D_t^-}}^2(t))S_{in}\\ {\bar{B}}=B_0 \end{array}\right. } \end{aligned}$$

(2.14)

on the domain where (2.6) with (2.8) has a unique solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$. Using these two facts and the local unique existence theorem for ODE, it can be shown that (2.6) with (2.8) has a solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ satisfying ${\bar{M}}<1$ and ${\bar{M}}^\prime <0$ in $\overline{D_t^+}$. Combining this fact with the fact that (2.6) with (2.7) has a unique solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ satisfying ${\bar{M}}>1$ in $[r_0,r_1]$, we have that a radial shock solution $({\bar{\rho }},{\bar{u}}\mathbf{e}_r,{\bar{p}})$ uniquely exists in $\mathcal {N}$ for each $t\in [r_0,r_1]$ and is a radial transonic shock solution.

Remark 2

By (2.6) and the fact that a solution $({\bar{\rho }},{\bar{u}}, {\bar{p}})$ of (2.6) with (2.8) uniquely exists in $\overline{D_{t}^+}$ satisfying (2.14), we have that a solution $({\bar{\rho }},{\bar{u}},{\bar{p}})$ of (2.6) with (2.8) satisfies

$$\begin{aligned} {\bar{\rho }}^\prime =\frac{2{\bar{\rho }}}{r}\frac{{\bar{M}}^2}{1-{\bar{M}}^2} \quad \text{ in }\quad D_t^+. \end{aligned}$$

From this fact and the fact that ${\bar{M}}|_{\overline{D_{t}^+}}<1$ in $\overline{D_{t}^+}$, we obtain ${\bar{\rho }}|_{\overline{D_{t}^+}}^\prime >0$ in $\overline{D_{t}^+}$.

One can see that the values of $({\bar{\rho }},{\bar{u}},{\bar{p}})|_{\overline{D_t^+}}$ at a fixed location r in $\overline{D_{t}^+}$ are determined by the three conserved quantities on the right hand sides of the equations in (2.14). This combined with the fact that the conserved quantity for ${\bar{S}}$ in $\overline{D_t^+}$ given in (2.14) varies depending on t (obtained from (2.14) by using $({{\bar{M}}}|_{\overline{D_t^-}})^\prime >0$ in $\overline{D_t^-}$ for any $t\in [r_0,r_1]$) implies that the values of $({\bar{\rho }},{\bar{u}},{\bar{p}})|_{\overline{D_t^+}}$ at a fixed location r in $\overline{D_t^+}$ vary depending on t. To represent this dependence, we write $({\bar{\rho }},{\bar{u}},{\bar{p}})|_{\overline{D_t^+}}(r)$ and ${\bar{S}}|_{\overline{D_t^+}}(r)$ as $({\bar{\rho }},{\bar{u}},{\bar{p}})|_{\overline{D_t^+}}(r;t)$ and ${\bar{S}}|_{\overline{D_t^+}}(r;t)$, respectively.

By direct computation, one can see that the conserved quantity for ${\bar{S}}$ in $\overline{D_t^+}$ satisfies some monotonicity with respect to t. This is given in the following lemma:

Lemma 2

Let $r_0$, $r_1$, t be positive constants such that $r_0\le t\le r_1$. Suppose that $({\bar{\rho }},{\bar{u}},{\bar{p}})$ is as above. Then it holds that

$$\begin{aligned} \frac{d{\bar{S}}|_{\overline{D_t^+}}(t;t)}{\hbox {d}t}>0\quad \text{ for } \text{ any } t\in [r_0,r_1]. \end{aligned}$$

Proof

Differentiate ${\bar{S}}|_{\overline{D_t^+}}(t;t)$ with respect to t. Then we have

$$\begin{aligned} \frac{d{\bar{S}}|_{\overline{D_t^+}}(t;t)}{\hbox {d}t}=\left. \frac{d g(x)}{dx}\right| _{x=({\bar{M}}|_{\overline{D_t^-}})^2(t)}\frac{d({\bar{M}}|_{\overline{D_t^-}})^2(t)}{\hbox {d}t} S_{in}. \end{aligned}$$

(2.15)

One can easily check that $g(1)=1$ and $g^\prime (x)>0$ for all $x>1$. By this fact, ${\bar{M}}|_{\overline{D_t^-}}(t)>1$ and $({\bar{M}}|_{\overline{D_t^-}})^\prime (t)>0$ for any $t\in [r_0,r_1]$, we obtain from (2.15) the desired result. $\square$

From Lemma 2, we can obtain

Proposition 1

Suppose that $r_0$, $r_1$, t and $({\bar{\rho }},{\bar{u}},{\bar{p}})$ are as in Lemma 2. Then, for any $t\in [r_0,r_1]$,

$$\begin{aligned} \frac{d{\bar{p}}|_{\overline{D_t^+}}}{\hbox {d}t}(r_1;t)<0. \end{aligned}$$

Proof

By the definitions of ${\bar{B}}$ and ${\bar{S}}$ and the first and third equation of (2.14),

$$\begin{aligned} B_0=\left. \left( \frac{1}{2}(\frac{m_0}{r_1^2})^2 (\frac{{\bar{S}}}{{\bar{p}}})^{\frac{2}{\gamma }} +\frac{\gamma }{\gamma -1}{{\bar{p}}}^{1-\frac{1}{\gamma }} {{\bar{S}}}^{\frac{1}{\gamma }}\right) \right| _{\overline{D_t^+}}(r_1;t). \end{aligned}$$

Differentiate this with respect to t. Then we get

$$\begin{aligned} \frac{d{\bar{p}}|_{\overline{D_t^+}}(r_1;t)}{\hbox {d}t} =-\left. \left( \frac{(\frac{m_0}{r_1^2})^2 (\frac{{\bar{S}}}{{\bar{p}}})^{\frac{2}{\gamma }-1} +\frac{\gamma }{\gamma -1}{{\bar{p}}}^{2 -\frac{1}{\gamma }}{{\bar{S}}}^{\frac{1}{\gamma }-1}}{-(\frac{m_0}{r_1^2})^2(\frac{{\bar{S}}}{{\bar{p}}})^{\frac{2}{\gamma }} +\gamma {{\bar{p}}}^{1-\frac{1}{\gamma }}{{\bar{S}}}^{\frac{1}{\gamma }}}\right) \right| _{\overline{D_t^+}}(r_1;t)\frac{d{\bar{S}}|_{\overline{D_t^+}}(r_1;t)}{\hbox {d}t}. \end{aligned}$$

(2.16)

By ${\bar{M}}|_{\overline{D_t^+}}<1$ in $\overline{D_t^+}$, Lemma 2 and the second equation of (2.14), we obtain from (2.16) the desired result. $\square$

The above proposition implies that for any given $p_c\in [p_1,p_2]$ where $p_1:=p_0|_{\overline{D_{r_1}^+}}(r_1;r_1)$ and $p_2:=p_0|_{\overline{D_{r_0}^+}}(r_1;r_0)$, there is a unique shock location $\varGamma _t$ in $\overline{\mathcal {N}}$ such that $(\overline{\rho },\overline{u}\mathbf{e}_r,\overline{p})$ satisfies $\overline{p}|_{\overline{D_{t}^+}}(r_1;t)=p_c$. Hereafter, we fix a constant $p_c\in (p_1,p_2)$ and denote $t\in (r_0,r_1)$ such that a radial transonic shock solution of (2.1) satisfying (2.5) and having a shock location $\varGamma _t$ satisfies $p(r_1)=p_c$ by $r_s$. Also, we denote a solution $(\rho ,u,p)$ of (2.6) with (2.7) and a solution $(\rho ,u,p)$ of (2.6) with (2.8) for $t=r_s$ by $(\rho _0^-,u_0^-,p_0^-)$ and $(\rho _0^+,u_0^+,p_0^+)$, respectively, and denote $\frac{p_0^+}{{\rho _0^+}^\gamma }$ by $S_0^+$. By the local unique existence theorem for ODE, there exists a positive constant $\delta _1$ such that $(\rho _0^+,u_0^+,p_0^+)$ uniquely exists in $[r_s-\delta _1,r_s]$ satisfying $M_0^+(=u_0^+/\sqrt{\frac{\gamma p_0^+}{\rho _0^+}})<1$. Let $\delta _1$ be one such constant.

2.3 Problem

Using the radial transonic shock solution given in the previous subsection, we present our problem.

In this paper, we use the following weighted Hölder norm. For a bounded connected open set $\varOmega \subset \mathbb {R}^n$, let $\varGamma$ be a closed portion of $\partial \varOmega$. For $\mathrm{x},\mathrm{y}\in \varOmega$, set

$$\begin{aligned} \delta _{\mathrm{x}}:=\text{ dist }(\mathrm{x},\varGamma )\quad \text{ and }\quad \delta _{\mathrm{x},\mathrm{y}}:=\text{ min }(\delta _{\mathrm{x}},\delta _{\mathrm{y}}). \end{aligned}$$

For $k\in \mathbb {R}$, $\alpha \in (0,1)$ and $m\in \mathbb {Z}^+$, we define

$$\begin{aligned}&||u||^{(k,\varGamma )}_{m,0,\varOmega }:=\sum _{0\le |\beta |\le m}\underset{\mathrm{x}\in \varOmega }{\sup }\delta _{\mathrm{x}}^{\max (|\beta |+k,0)}|D^\beta u(\mathrm{x})|\\&[u]^{(k,\varGamma )}_{m,\alpha ,\varOmega }:=\sum _{|\beta |=m}\underset{\mathrm{x},\mathrm{y}\in \varOmega ,\mathrm{x}\ne \mathrm{y}}{\sup }\delta _{\mathrm{x},\mathrm{y}}^{\max (m+\alpha +k,0)}\frac{|D^\beta u(\mathrm{x})-D^\beta u(\mathrm{y})|}{|\mathrm{x}-\mathrm{y}|^\alpha }\\&||u||^{(k,\varGamma )}_{m,\alpha ,\varOmega }:=||u||^{(k,\varGamma )}_{m,0,\varOmega }+[u]^{(k,\varGamma )}_{m,\alpha ,\varOmega }, \end{aligned}$$

where $D^\beta :=\partial _{x_1}^{\beta _1}\ldots \partial _{x_n}^{\beta _n}$ for a multi-index $\beta =(\beta _1,\ldots ,\beta _n)$ with $\beta _i\in {{\mathbb {Z}}}^+$ for $i=1,\ldots ,n$ and $|\beta |=\sum _{i=1}^n\beta _i$. We denote the completion of a set of smooth functions under $||\cdot ||_{m,\alpha ,\varOmega }^{(k,\varGamma )}$ norm by $C^{m,\alpha }_{(k,\varGamma )}(\varOmega )$.

Our problem is given as follows:

Problem 1

(Transonic shock problem) Given an axisymmetric supersonic solution $(\rho _-,{\mathbf {u}}_-,p_-)$ of (2.1) in $\mathcal {N}$ satisfying the slip boundary condition

$$\begin{aligned} {\mathbf {u}}_-\cdot {\varvec{n}}_w=0\quad \text{ on }\quad \varGamma _w:=\partial \mathcal {N}\cap \{r_0<r<r_1,\;\theta =\theta _1\} \end{aligned}$$

(2.17)

where ${\varvec{n}}_w$ is the unit normal vector on $\varGamma _w$,

$$\begin{aligned} B=B_0\quad \text{ on }\quad \varGamma _{en} \end{aligned}$$

(2.18)

and

$$\begin{aligned} ||(\rho _-,{\mathbf {u}}_-,p_-)-(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-)||_{2,\alpha ,\mathcal {N}}\le \sigma \end{aligned}$$

(2.19)

and an axisymmetric exit pressure $p_{ex}$ on $\varGamma _{ex}:=\partial \mathcal {N}\cap \{r=r_1,0\le \theta <\theta _1\}$ satisfying

$$\begin{aligned} ||p_{ex}-p_c||_{1,\alpha ,\varGamma _{ex}}^{(-\alpha ,\partial \varGamma _{ex})}\le \sigma \end{aligned}$$

(2.20)

for a positive constant $\sigma$, find a shock location $\varGamma _f:=\mathcal {N}\cap \{r=f(\theta )\}$ and a corresponding subsonic solution $(\rho _+,{\mathbf {u}}_+,p_+)\in (C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+))^3$ of (2.1) satisfying

• (i) the system (2.1) in $\mathcal {N}_f^+:=\mathcal {N}\cap \{r>f(\theta )\}$,

• (ii) R-H conditions (2.3) on $\varGamma _f$,

• (iii) the slip boundary condition

  $$\begin{aligned} {\mathbf {u}}_+\cdot {\varvec{n}}_w=0\quad \text{ on }\quad \varGamma _w^+:=\varGamma _w\cap \{r>f(\theta _1)\}, \end{aligned}$$

  (2.21)

• (iv) and the exit pressure condition

  $$\begin{aligned} p_+=p_{ex}\quad \text{ on }\quad \varGamma _{ex}. \end{aligned}$$

  (2.22)

Remark 3

It is generally known that a supersonic solution of (2.1) is governed by a hyperbolic system. We assume that $(\rho _-,{\mathbf {u}}_-,p_-)$ in Problem 1 exists.

Remark 4

To simplify our argument, we assumed in Problem 1 that $(\rho _-,{\mathbf {u}}_-,$ $p_-)$ satisfies (2.18). This assumption will be used to reduce (2.1) and (2.3) (see § 2.5). The result for Problem 1 (Theorem 2) does not change if we consider a general perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-)$ in Problem 1.

We study Problem 1 using a stream function formulation of the full Euler system for an axisymmetric flow. We introduce a stream function formulation used in this paper in the next subsection.

2.4 Stream function formulation

Let $\varOmega$ be an open simply connected axisymmetric set in $\mathbb {R}^3$. Let $(\rho ,{\mathbf {u}})$ be axisymmetric $C^1$ functions in $\varOmega$ satisfying the first equation of (2.1) and $|\rho {\mathbf {u}}|>0$. For such $(\rho ,{\mathbf {u}})$, the Stokes’ stream function for an axisymmetric flow of the full Euler system is defined by

$$\begin{aligned} V(\mathrm x)= \int _{S_{\mathrm{x}}}\rho {\mathbf {u}}\cdot {\varvec{\nu }} d S \quad \hbox { for}\ \mathrm{x}\in \varOmega , \end{aligned}$$

(2.23)

where $S_{\mathrm{x}}$ is a simply connected $C^1$ surface in $\varOmega$ whose boundary is a circle centered at z-axis, parallel to xy-plane and passing through $\mathrm{x}$, and ${\varvec{\nu }}$ is the unit normal vector on $S_{\mathrm{x}}$ pointing outward direction with respect to the cone-like domain made by connecting $\partial S_{\mathrm{x}}$ and the origin by straight lines. By the first equation of (2.1), the value of this function at $\mathrm{x}$ is independent of the choice of $S_{\mathrm{x}}$. Since $\partial S_{\mathrm{x}}$ is axisymmetric, V is axisymmetric in $\varOmega$.

By the first equation of (2.1), this function is a constant on each stream surface of $\rho {\mathbf {u}}$ in $\varOmega$. Here, the stream surfaces of a vector field $\rho {\mathbf {u}}$ in $\varOmega$ is a set of surfaces made by collecting all the streamlines of $\rho {\mathbf {u}}$ initiating from a point on a circle in $\varOmega$ centered at z-axis and parallel to xy-plane. By $|\rho {\mathbf {u}}|>0$ in $\varOmega$, V on each different stream surface of vector field $\rho {\mathbf {u}}$ in $\varOmega$ is different from each other.

Using these facts, a vector field in $\varOmega$ tangential to the stream surfaces in $\varOmega$ and having no $\mathbf{e}_{\varphi }$ component can be obtained from V. Let $\nabla ^\perp =\frac{1}{2\pi r \sin \theta } \left( \mathbf{e}_r\frac{\partial _{\theta } }{r}-\right.$ $\left. \mathbf{e}_\theta \partial _r \right)$. Then $\nabla ^\perp$ satisfies $\nabla ^\perp h\cdot \nabla h=0$ and $\nabla ^\perp h \cdot \mathbf{e}_\varphi =0$ for a scalar function h. Using these properties of $\nabla ^\perp$, we have that if we apply $\nabla ^\perp$ to V, then we have a vector field in $\varOmega$ tangent to the stream surfaces in $\varOmega$ and having no $\mathbf{e}_{\varphi }$ component. Apply $\nabla ^\perp$ to V. Then we have

$$\begin{aligned} \nabla ^\perp V=\rho u_r \mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta , \end{aligned}$$

(2.24)

where $u_r={\mathbf {u}}\cdot \mathbf{e}_r$ and $u_\theta ={\mathbf {u}}\cdot \mathbf{e}_\theta$.

Using (2.24), we can reformulate the full Euler system for an axisymmetric flow. But if we do this, then there is a singularity issue that can be seen in the relation $||\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ||_{\alpha ,\mathcal {N}}=||\nabla ^\perp V||_{\alpha ,\mathcal {N}}\not \le C||V||_{1,\alpha ,\mathcal {N}}$ for any constant C. To avoid this issue, we use the following form of the stream function.

Let $\varPhi \mathbf{e}_\varphi$ be an axisymmetric vector field in $\varOmega$ satisfying

$$\begin{aligned} \oint _{\partial S_{\mathrm{x}}} \varPhi \mathbf{e}_\varphi \cdot d{\varvec{r}}=\int _{S_{\mathrm{x}}}\rho {\mathbf {u}}\cdot {\varvec{\nu }} dS, \end{aligned}$$

(2.25)

where ${\varvec{r}}$ is a parametrization of $\partial S_{\mathrm{x}}$ in a counter clockwise direction. Then by the definitions of $\varPhi \mathbf{e}_\varphi$ and V,

$$\begin{aligned} \varPhi =\frac{V}{2\pi r\sin \theta }. \end{aligned}$$

(2.26)

It is easily checked that

$$\begin{aligned} \nabla \times (\varPhi \mathbf{e}_\varphi )=\nabla ^\perp V. \end{aligned}$$

(2.27)

From this relation and (2.24) or (2.25) directly, we have

$$\begin{aligned} \nabla \times (\varPhi \mathbf{e}_\varphi )=\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta . \end{aligned}$$

(2.28)

We call $\varPhi \mathbf{e}_{\varphi }$ the vector potential form of the stream function.

We reformulate the full Euler system for an axisymmetric flow using (2.28). For our later analysis, when we reformulate the full Euler system using (2.28), we use the following form of the full Euler system representing the relation between $\nabla S$ and $\nabla \times {\mathbf {u}}$ clearly

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathrm{div}(\rho {\mathbf {u}})=0,\\ (\nabla \times {\mathbf {u}})\times {\mathbf {u}}=\frac{\rho ^{\gamma -1}}{\gamma -1}\nabla S-\nabla B,\\ \rho {\mathbf {u}}\cdot \nabla B=0 \end{array}\right. } \end{aligned}$$

(2.29)

which is obtained under the assumption that $(\rho ,{\mathbf {u}},p)\in C^1$ and $\rho >0$. We assume that $(\rho ,{\mathbf {u}},p)$ in (2.29) is axisymmetric and reformulate (2.29) using (2.28).

Rewrite (2.28) as ${\mathbf {u}}=\frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )+u_\varphi \mathbf{e}_\varphi .$ Substitute this into the second equation of (2.29). Then we obtain

$$\begin{aligned}&\left( \nabla \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )+u_\varphi \mathbf{e}_\varphi \right) \right) \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )+u_\varphi \mathbf{e}_\varphi \right) \nonumber \\&\quad =\frac{\rho ^{\gamma -1}}{\gamma -1}\nabla S-\nabla B. \end{aligned}$$

(2.30)

From this equation, three equations can be obtained. From $\mathbf{e}_\varphi$-components of (2.30), we get

$$\begin{aligned} (\nabla \times (u_\varphi \mathbf{e}_\varphi ))\times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\right) =0. \end{aligned}$$

(2.31)

Define $L:=2\pi r \sin \theta u_\varphi$ so that $\nabla \times (u_\varphi \mathbf{e}_\varphi )=\nabla ^\perp L$ (see (2.26) and (2.27)). Then, using this relation, (2.31) can be written as

$$\begin{aligned} \nabla ^\perp L\times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\right) =0. \end{aligned}$$

From this equation, we obtain

$$\begin{aligned} \nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \nabla L=0. \end{aligned}$$

(2.32)

From $\mathbf{e}_{\theta }$-components of (2.30), we get

$$\begin{aligned}&\left( \nabla \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\right) \right) \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \mathbf{e}_r+(\nabla \times (u_\varphi \mathbf{e}_\varphi ) \cdot \mathbf{e}_r)\mathbf{e}_r\times u_\varphi \mathbf{e}_\varphi \\&\quad =\left( \frac{\rho ^{\gamma -1}}{\gamma -1}\frac{\partial _\theta S}{r}-\frac{\partial _\theta B}{r}\right) \mathbf{e}_\theta . \end{aligned}$$

Using the definition of L, rewrite the above equation. And then move the rewritten term into the right hand side of the equation and multiply $\mathbf{e}_r\times$ to the resultant equation. Then we obtain

$$\begin{aligned}&\left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \nabla \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\right) \nonumber \\&\quad =\left( \frac{L}{2\pi r\sin \theta }\nabla \times \left( \frac{L}{2\pi r\sin \theta }\mathbf{e}_\varphi \right) \cdot \mathbf{e}_r+\frac{\rho ^{\gamma -1}}{\gamma -1} \frac{\partial _\theta S}{r}-\frac{\partial _\theta B}{r}\right) \mathbf{e}_\varphi .\qquad \end{aligned}$$

(2.33)

Using the third equation of (2.29), we obtain from ${\mathbf {u}}$-components of (2.30)

$$\begin{aligned} \rho {\mathbf {u}}\cdot \nabla S=0. \end{aligned}$$

(2.34)

Using the assumption that $(\rho ,{\mathbf {u}},p)$ are axisymmetric and (2.28), this equation can be written as

$$\begin{aligned} \nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \nabla S=0. \end{aligned}$$

(2.35)

Hence, we obtain from (2.30) three equations: (2.32), (2.33) and (2.35). Finally, in the same way that we obtained (2.35) from (2.34), we obtain from the third equation of (2.29)

$$\begin{aligned} \nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \nabla B=0. \end{aligned}$$

(2.36)

Combining (2.32), (2.33), (2.35) and (2.36), we have the following stream function formulation of the full Euler system for an axisymmetric flow

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \nabla \times \left( \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\right) \\ \quad \quad \quad \quad \quad \quad \quad =\left( \frac{L}{2\pi r\sin \theta }\nabla \times \left( \frac{L}{2\pi r\sin \theta }\mathbf{e}_\varphi \right) \cdot \mathbf{e}_r+\frac{\rho ^{\gamma -1}}{\gamma -1}\frac{\partial _\theta S}{r}-\frac{\partial _\theta B}{r}\right) \mathbf{e}_\varphi ,\\ \nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \nabla L=0,\\ \nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \nabla S=0,\\ \nabla \times (\varPhi \mathbf{e}_\varphi ) \cdot \nabla B=0. \end{array}\right. } \end{aligned}$$

(2.37)

Note that the first equation of (2.1) is omitted in (2.37) because for a 3-D axisymmetric flow, the first equation of (2.1) is reduced to $\mathrm{div}(\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta )=0$ and this equation is directly satisfied by $\mathrm{div}(\nabla \times (\varPhi \mathbf{e}_{\varphi }))=0$ if $\varPhi \in C^2(\varOmega )$. Also, note that when $(\rho ,{\mathbf {u}},p)$ in (2.1) is in a sufficiently small perturbation of $(\rho _0^+,u_0^+\mathbf{e}_r,p_0^+)$ or $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-)$, then unknowns of (2.37) can be $(\varPhi \mathbf{e}_\varphi ,L,S,B)$. This fact (for the first case) is checked via the following lemma:

Lemma 3

Let $\varOmega$ be an axisymmetric connected open subset of $\mathcal {N}_{r_s-\delta _1}^+$. There exist positive constants $\delta _{2,\varOmega }$ and $\delta _{3,\varOmega }$ and a function $\varrho : B_{\delta _{2,\varOmega },\varOmega }^{(1)}\rightarrow B_{\delta _{3,\varOmega },\varOmega }^{(2)}$, where

$$\begin{aligned} B_{\delta ,\varOmega }^{(1)}:= & {} \{(\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,u_\varphi \mathbf{e}_\varphi ,S,B)|\in (C^0(\overline{\varOmega }))^4\;|\\&\sup _{\varOmega }\{ |\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta -\rho _0^+u_0^+\mathbf{e}_r|+|u_\varphi \mathbf{e}_{\varphi }|+|S-S_0^+|+|B-B_0|\}\le \delta \}, \end{aligned}$$

and

$$\begin{aligned} B_{\delta ,\varOmega }^{(2)}:=\{\rho \in C^0(\overline{\varOmega })\;|\;\sup _{\varOmega } |\rho -\rho _0^+|\le \delta \}, \end{aligned}$$

such that

$$\begin{aligned} \rho ^2B-\frac{|\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta |^2}{2}-\frac{\rho ^2}{2}|u_\varphi \mathbf{e}_{\varphi }|^2-\frac{\gamma S \rho ^{\gamma +1}}{\gamma -1}=0 \end{aligned}$$

(2.38)

if and only if

$$\begin{aligned} \rho =\varrho (\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,u_\varphi \mathbf{e}_{\varphi },S,B), \end{aligned}$$

(2.39)

for all $(\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,u_\varphi \mathbf{e}_{\varphi },S,B)\in B_{\delta _{2,\varOmega },\varOmega }^{(1)}$ and $\rho \in B_{\delta _{3,\varOmega },\varOmega }^{(2)}$.

Proof

Using the definition of B given in (2.2), we define

$$\begin{aligned}&b(\rho ,\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,u_\varphi \mathbf{e}_\varphi ,S,B)\nonumber \\&\quad =\rho ^2B-\frac{|\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta |^2}{2}-\frac{\rho ^2}{2}|u_\varphi \mathbf{e}_\varphi |^2-\frac{\gamma S \rho ^{\gamma +1}}{\gamma -1}. \end{aligned}$$

(2.40)

Then $b\in C^\infty$ with respect to $(\rho ,\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,u_\varphi \mathbf{e}_{\varphi },S,B)$,

$$\begin{aligned} \partial _{\rho } b(\rho _0^+,\rho _0^+u_0^+\mathbf{e}_r,0,S_0^+,B_0)=\rho _0^+({u_0^+}^2-\gamma S_0^+{\rho _0^+}^{\gamma -1})<0, \end{aligned}$$

and by the third equation of (2.14),

$$\begin{aligned} b(\rho _0^+,\rho _0^+u_0^+\mathbf{e}_r,0,S_0^+,B_0)=0. \end{aligned}$$

With these facts, we apply the implicit function theorem to $b(\rho ,\rho u_r\mathbf{e}_r+\rho u_\theta \mathbf{e}_\theta ,$ $u_\varphi \mathbf{e}_{\varphi },S,B)$. Then we obtain the desired result. $\square$

Remark 5

Hereafter, $\delta _2$ and $\delta _3$ denote some constants $\delta _{2,\varOmega }$ and $\delta _{3,\varOmega }$ in Lemma 3 for $\varOmega =\mathcal {N}_{r_s-\delta _1}^+$. Hereafter, $\varrho$ denotes $\varrho$ in Lemma 3 for $\varOmega =\mathcal {N}_{r_s-\delta _1}^+$. One can see that if $\varOmega \subset \mathcal {N}_{r_s-\delta _1}^+$, then for all $(\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S,B)\in B_{\delta _{2},\varOmega }^{(1)}$ and $\rho \in B_{\delta _{3},\varOmega }^{(2)}$, (2.38) holds if and only if (2.39) holds.

In order to study 3-D axisymmetric transonic shock solution of the full Euler system using the stream function formulation of the full Euler system for an axisymmetric flow above, we reformulate (2.3) with respect to the variables in (2.37).

Assume that $\varGamma$ in (2.3) is an axisymmetric $C^1$ surface, and let ${\varvec{\tau }}_2$ and ${\varvec{\tau }}_1$ in (2.3) be $\mathbf{e}_\varphi$ and the unit tangent vector field on $\varGamma$ perpendicular to $\mathbf{e}_\varphi$ and satisfying ${\varvec{\nu }}\cdot ({\varvec{\tau }}_1\times \mathbf{e}_\varphi )>0$, respectively, where ${\varvec{\nu }}$ is the unit normal vector field on $\varGamma$ pointing toward $\varOmega ^+$. We reformulate this (2.3) with respect to the variables in (2.37).

By the definition of V given in (2.23), the first equation of (2.3) can be written as $[V]_{\varGamma }=0$. Rewrite this equation using (2.26). Then we have

$$\begin{aligned} {[\varPhi \mathbf{e}_\varphi ]}_\varGamma =0. \end{aligned}$$

Using (2.28), rewrite the second equation of (2.3). Then we get

$$\begin{aligned} \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot {\varvec{\tau }}_1={\mathbf {u}}_-\cdot {\varvec{\tau }}_1\quad \text{ on }\quad \varGamma . \end{aligned}$$

By the definition of L, the third equation of (2.3) can be written as

$$\begin{aligned} {[L]}_{\varGamma }=0. \end{aligned}$$

From the second, third and fifth equation of (2.3), we can obtain

$$\begin{aligned} {[B_s]}_\varGamma =0 \end{aligned}$$

(2.41)

where $B_s:=\frac{({\mathbf {u}}\cdot {\varvec{\nu }})^2}{2}+\frac{\gamma p}{(\gamma -1)\rho }$. In the same way that (2.12) is obtained from (2.8), we obtain from the first and fourth equation of (2.3) and (2.41)

$$\begin{aligned} S_+=g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}}{c_-}\right) ^2\right) S_-\quad \text{ on }\quad \varGamma , \end{aligned}$$

where g(x) is a function defined in (2.13) and variables with lower indices ± denote variables in $\varOmega ^\pm$, respectively. Combining these reformulated equations of (2.3) and the fifth equation of (2.3), we have the following stream function formulation of (2.3):

$$\begin{aligned} {\left\{ \begin{array}{ll} {[\varPhi \mathbf{e}_\varphi ]}_\varGamma =0,\\ \frac{1}{\rho }\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot {\varvec{\tau }}_1={\mathbf {u}}_-\cdot {\varvec{\tau }}_1\quad \text{ on }\quad \varGamma ,\\ {[L]}_{\varGamma }=0,\\ S_+=g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}}{c_-}\right) ^2\right) S_-\quad \text{ on }\quad \varGamma ,\\ {[B]}_\varGamma =0. \end{array}\right. } \end{aligned}$$

(2.42)

2.5 Restatement of Problem 1 using the stream function formulation and main result

Using the stream function formulation in the previous subsection, we restate Problem 1.

We first reduce (2.1) and (2.3) in Problem 1. In Problem 1, we assumed that a supersonic solution $(\rho _-,{\mathbf {u}}_-,p_-)$ of (2.1) satisfies (2.18). If this condition is satisfied, then by the third equation of (2.1) in $\mathcal {N}_f^-:=\mathcal {N}\cap \{r<f(\theta )\}$ and $\mathcal {N}_f^+$, the fifth equation of (2.3), (2.17) and (2.21), $(\rho _+,{\mathbf {u}}_+,p_+)$ we find in Problem 1 must satisfy $B=B_0$ in $\overline{\mathcal {N}_f^+}$. If the fact that the third equation of (2.1) in our case has a unique sufficiently regular solution is guaranteed, then it is not needed to solve the third equation of (2.1) to obtain $B=B_0$ in $\overline{\mathcal {N}_f^+}$. (The fact that the third equation of (2.1) in our case has a unique sufficiently regular solution is shown by Lemma 15.) To simplify our argument, we assume that $(\rho _+,{\mathbf {u}}_+,p_+)$ in Problem 1 satisfies $B=B_0$ in $\overline{\mathcal {N}_f^+}$. Under this assumption, (2.1) and (2.3) that $(\rho _+,{\mathbf {u}}_+,p_+)$ in Problem 1 satisfies are reduced to the first and second equation of (2.1) and the first, second, third and fourth equation of (2.3).

Then we present the stream function formulations of (2.1) and (2.3) satisfied by $(\rho _+,{\mathbf {u}}_+,p_+)$ in Problem 1. By (2.37) and (2.42), the stream function formulations of (2.1) and (2.3) satisfied by $(\rho _+,{\mathbf {u}}_+,p_+)$ in Problem 1 and reduced by using the assumption that $(\rho _+,{\mathbf {u}}_+,p_+)$ satisfies $B=B_0$ in $\overline{\mathcal {N}_f^+}$ are given as

$$\begin{aligned}&\left( \frac{1}{\rho _+}\nabla \times (\varPhi _+\mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \nabla \times \left( \frac{1}{\rho _+}\nabla \times (\varPhi _+\mathbf{e}_\varphi )\right) \nonumber \\&\quad \quad \quad =\left( \frac{L_+}{2\pi r\sin \theta }\nabla \times \left( \frac{L_+}{2\pi r\sin \theta }\mathbf{e}_\varphi \right) \cdot \mathbf{e}_r+\frac{\rho _+^{\gamma -1}}{\gamma -1}\frac{\partial _\theta S_+}{r}\right) \mathbf{e}_\varphi , \end{aligned}$$

(2.43)

$$\begin{aligned}&\nabla \times (\varPhi _+\mathbf{e}_\varphi )\cdot \nabla L_+=0,\end{aligned}$$

(2.44)

$$\begin{aligned}&\nabla \times (\varPhi _+\mathbf{e}_\varphi )\cdot \nabla S_+=0 \end{aligned}$$

(2.45)

in $\mathcal {N}_f^+$ and

$$\begin{aligned}&\varPhi _+\mathbf{e}_\varphi =\varPhi _-\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _f,\end{aligned}$$

(2.46)

$$\begin{aligned}&\frac{1}{\rho _+}\nabla \times (\varPhi _+\mathbf{e}_\varphi )\cdot {\varvec{\tau }}_f={\mathbf {u}}_-\cdot {\varvec{\tau }}_f\quad \text{ on }\quad \varGamma _f,\end{aligned}$$

(2.47)

$$\begin{aligned}&L_+=L_-\quad \text{ on }\quad \varGamma _f,\end{aligned}$$

(2.48)

$$\begin{aligned}&S_+=g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_f}{c_-}\right) ^2\right) S_-\quad \text{ on }\quad \varGamma _f, \end{aligned}$$

(2.49)

where ${\varvec{\nu }_f}$ is the unit normal vector on $\varGamma _f$ pointing toward $\mathcal {N}_f^+$ and ${\varvec{\tau }}_f$ is the unit tangential vector on $\varGamma _f$ perpendicular to $\mathbf{e}_\varphi$ and satisfying ${\varvec{\nu }}_f\cdot ({\varvec{\tau }}_f\times \mathbf{e}_\varphi )>0$. Here, $(\varPhi _-\mathbf{e}_\varphi ,L_-,S_-)$ and $(\varPhi _+\mathbf{e}_\varphi ,L_+,S_+)$ are $(\varPhi \mathbf{e}_\varphi ,L,S)$ given by the definitions of $\varPhi \mathbf{e}_{\varphi }$, L and S for $(\rho ,{\mathbf {u}},p)=(\rho _-,{\mathbf {u}}_-,p_-)$ and $(\rho _+,{\mathbf {u}}_+,p_+)$, respectively.

To use (2.43)–(2.49) to solve Problem 1, we determine $\rho _+$ in (2.43) and (2.47). In Problem 1, we will consider the case where $(\rho _-,{\mathbf {u}}_-,p_-)$ and $p_{ex}$ are in sufficiently small perturbations of $(\rho _0^-,$ $u_0^-\mathbf{e}_r, p_0^-)$ and $p_c$ so that $(\rho _+,{\mathbf {u}}_+,p_+)$ we will find in Problem 1 satisfies that $\rho _+$ is uniquely determined by $(\rho _+ u_{+,r}\mathbf{e}_r+\rho _+ u_{+,\theta }\mathbf{e}_{\theta },$ $u_{+,\varphi }\mathbf{e}_{\varphi },S_+,B_0)$. (If $(\rho _-,{\mathbf {u}}_-,p_-)$ and $p_{ex}$ are in sufficiently small perturbations of $(\rho _0^-,u_0^-\mathbf{e}_r, p_0^-)$ and $p_c$, then it is expected that the solution $(\rho _+,{\mathbf {u}}_+,p_+)$ in Problem 1 is in a small perturbation of $(\rho _0^+,u_0^+\mathbf{e}_r,p_0^+)$. And if $(\rho _+,{\mathbf {u}}_+,p_+)$ is in a small perturbation of $(\rho _0^+,$ $u_0^+\mathbf{e}_r,p_0^+)$, then $\rho _+$ can be uniquely determined by $(\rho _+ u_{+,r}\mathbf{e}_r+\rho _+ u_{+,\theta }\mathbf{e}_{\theta },$ $u_{+,\varphi }\mathbf{e}_{\varphi },S_+,B_0)$ where $u_{+,r}:={\mathbf {u}}_+\cdot \mathbf{e}_r$, $u_{+,\theta }:={\mathbf {u}}_+\cdot \mathbf{e}_{\theta }$ and $u_{+,\varphi }={\mathbf {u}}_+\cdot \mathbf{e}_{\varphi }$ (see Lemma 3). ) To find such $(\rho _+,{\mathbf {u}}_+,p_+)$ using (2.43)–(2.49), we set $\rho _+$ in (2.43) and (2.47) to be

$$\begin{aligned} \rho _+=\varrho (\nabla \times (\varPhi _+\mathbf{e}_\varphi ),\frac{L_+}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S_+,B_0), \end{aligned}$$

(2.50)

where $\varrho$ is a function given in Lemma 3. Hereafter, to simplify notation, we write $\varrho (\nabla \times (\varPhi \mathbf{e}_\varphi ),\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S,B_0)$ as $\varrho (\nabla \times (\varPhi \mathbf{e}_\varphi ),\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S)$.

We will find a subsonic solution of (2.1) using (2.43)–(2.45). To do this, we define a subsonic solution of (2.43)–(2.45). Using the definition of subsonic solution of (2.1), we define a subsonic solution of (2.43)–(2.45) by a solution $(\varPhi _+\mathbf{e}_{\varphi },L_+,S_+)$ of (2.43)–(2.45) satisfying

$$\begin{aligned}&\left| \frac{1}{\varrho (\nabla \times (\varPhi _+\mathbf{e}_\varphi ), \frac{L_+}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S_+)}\nabla \times (\varPhi _+\mathbf{e}_\varphi )+\frac{L_+}{2\pi r \sin \theta }\mathbf{e}_{\varphi }\right| ^2 \nonumber \\&\quad <\gamma S_+(\varrho (\nabla \times (\varPhi _+\mathbf{e}_\varphi ),\frac{L_+}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S_+))^{\gamma -1}. \end{aligned}$$

(2.51)

Next, we rewrite the boundary conditions in Problem 1 with respect to the variables in (2.37). Denote V given by the definition of V for $(\rho ,{\mathbf {u}},p)=(\rho _-,{\mathbf {u}}_-,p_-)$ and $(\rho _+,{\mathbf {u}}_+,p_+)$ by $V_-$ and $V_+$, respectively. By (2.24), (2.17) and (2.21) can be written as $V_-=V_-(r_0,\theta _1)$ on $\varGamma _w$ and $V_+=V_+(f(\theta _1),\theta _1)$ on $\varGamma _w^+.$ Rewrite this in the vector potential form. Then we have $\varPhi _-\mathbf{e}_\varphi =\frac{r_0\varPhi _-(r_0,\theta _1)}{r}\mathbf{e}_\varphi$ on $\varGamma _w$ and $\varPhi _+\mathbf{e}_\varphi =\frac{f(\theta _1)\varPhi _+(f(\theta _1),\theta _1)}{r}\mathbf{e}_\varphi$ on $\varGamma _w^+.$ Combine these relations with (2.46) and the continuity condition for $\varPhi \mathbf{e}_{\varphi }$ at $\overline{\varGamma _f}\cap \overline{\varGamma _w^+}$. Then we obtain

$$\begin{aligned} \varPhi _+\mathbf{e}_\varphi =\frac{r_0\varPhi _-(r_0,\theta _1)}{r}\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _w^+. \end{aligned}$$

(2.52)

Using the definition of S and (2.50), rewrite (2.22). Then we have

$$\begin{aligned} S_+(\varrho (\nabla \times (\varPhi _+ \mathbf{e}_\varphi ),\frac{L_+}{2\pi r\sin \theta }\mathbf{e}_\varphi ,S_+))^\gamma =p_{ex}\quad \text{ on }\quad \varGamma _{ex}. \end{aligned}$$

(2.53)

Finally, we add some continuity condition for $\varPhi _+\mathbf{e}_{\varphi }$ in the restatement of Problem 1 using the stream function formulation. By (2.1) and (2.21), $(\rho _+,{\mathbf {u}}_+)$ we find in Prolem 1 must satisfy that the total outgoing flux for $\rho _+{\mathbf {u}}_+$ on $\varGamma _{ex}$ is equal to the total incoming flux for $\rho _+{\mathbf {u}}_+$ on any cross section of $\mathcal {N}_f^+$ whose all boundary points intersect with $\varGamma _w^+$. Using (2.23), this statement can be expressed as $\lim _{\theta \rightarrow \theta _1}V_+(r_1,\theta )=\lim _{r\rightarrow r_1}V_+(r,\theta _1).$ In the vector potential form, this can be rewritten as

$$\begin{aligned} \lim _{\theta \rightarrow \theta _1}(\varPhi _+\mathbf{e}_\varphi )(r_1,\theta )=\lim _{r\rightarrow r_1}(\varPhi _+\mathbf{e}_\varphi )(r,\theta _1). \end{aligned}$$

(2.54)

Since this condition cannot be achieved by finding a solution satisfying (2.53) (this will be seen in § 3.1), we include this condition in the restatement of Problem 1 using the stream function formulation.

Using the equations and boundary conditions obtained above, Problem 1 is restated as follows:

Problem 2

Given an axisymmetric supersonic solution $(\rho _-,{\mathbf {u}}_-,p_-)$ of (2.1) and an axisymmetric exit pressure $p_{ex}$ as in Problem 1, find a shock location $\varGamma _f=\mathcal {N}\cap \{r=f(\theta )\}$ and a corresponding subsonic solution $(\varPhi _+\mathbf{e}_\varphi ,L_+,S_+)$ of (2.43)–(2.45) satisfying

• (i) the system (2.43)–(2.45) in $\mathcal {N}_f^+$,

• (ii) the R-H conditions (2.46)–(2.49),

• (iii) the slip boundary condition (2.52),

• (iv) the exit pressure condition (2.53),

• (v) the compatibility condition (2.54).

Let $S^{2,\theta _1}:=\{(x,y,z)\in \mathbb {R}^3\;|\; r=1,\; 0\le \theta <\theta _1 \}$. Then a function f representing a shock location $\varGamma _f$ can be considered as a function on $S^{2,\theta _1}$. Using this fact and the stereographic projection from $(0,0,-1)$ onto the plane $z=1$ passing through $S^{2,\theta _1}$, it is seen that f can be regarded as a function on $\varLambda$ where $\varLambda :=\{(x,y)\in \mathbb {R}^2\;|\; \sqrt{x^2+y^2}< 2\tan \frac{\theta _1}{2}\}$. Thus, f can be regarded as a function on $\varLambda$ or $(0,\theta _1)$. In this paper, we regard f in both ways. To simplify our notation, we use the same function notation when we represent f as a function on $\varLambda$ or $(0,\theta _1)$.

Hereafter, we denote $\varPhi \mathbf{e}_{\varphi }$, L and V given by the definitions of $\varPhi \mathbf{e}_{\varphi }$, L and V for $(\rho ,{\mathbf {u}},p)=(\rho _0^\pm ,u_0^\pm \mathbf{e}_r,p_0^\pm )$ by $\varPhi _0^\pm \mathbf{e}_{\varphi }$, $L_0^\pm$ and $V_0^\pm$, respectively. To simplify our notation, hereafter, we denote $(\varPhi _+\mathbf{e}_\varphi ,L_+,S_+)$ by $(\varPhi \mathbf{e}_\varphi ,L,S)$.

Our result of Problem 2, the main result in this paper, is given as follows.

Theorem 1

Let $\alpha \in (\frac{2}{3},1)$. There exists a positive constant $\sigma _1$ depending on $(\rho _{in},u_{in},p_{in})$, $p_c$, $\gamma$, $r_0$, $r_1$, $\theta _1$ and $\alpha$ such that if $\sigma \in (0,\sigma _1]$, then Problem 2 has a solution $(f,\varPhi \mathbf{e}_\varphi ,L,S)$ satisfying the estimate

$$\begin{aligned}&||f-r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )} +||\nabla \times ((\varPhi -\varPhi _0^+)\mathbf{e}_\varphi )||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha , \varGamma _w^+)} \nonumber \\&\quad +||\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||S-S_0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\le C \sigma \end{aligned}$$

(2.55)

where C is a positive constant depending on $(\rho _{in},u_{in},p_{in})$, $p_c$, $\gamma$, $r_0$, $r_1$, $\theta _1$ and $\alpha$. Furthermore, this solution is unique in the class of functions satisfying (2.55).

Remark 6

The condition $\alpha \in (\frac{2}{3}, 1)$ is a sufficient condition for obtaining $C^{1,\alpha }$ up to boundary estimate of solutions of the elliptic part of the linearized equations of (2.43)–(2.49), (2.52)–(2.54) (see the proof of Lemma 11). We do not know that this is an optimal condition.

Hereafter, we say that a constant depends on the data if a constant depends on $(\rho _{in},u_{in},p_{in})$, $p_c$, $\gamma$, $r_0$, $r_1$, $\theta _1$ and $\alpha$.

The following result of Problem 1 is obtained from Theorem 1.

Theorem 2

Let $\alpha \in (\frac{2}{3},1)$. There exists a positive constant $\sigma _2$ depending on the data such that if $\sigma \in (0,\sigma _2]$, then Problem 1 has a solution $(f,\rho _+,{\mathbf {u}}_+,p_+)$ satisfying the estimate

$$\begin{aligned}&||f-r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )} \nonumber \\&\quad +||\rho _+-\rho _0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||{\mathbf {u}}_+ -u_0^+\mathbf{e}_r||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||p_+\nonumber \\&\quad -p_0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\le C\sigma \end{aligned}$$

(2.56)

where C is a positive constant depending on the data. Furthermore, this solution is unique in the class of functions satisfying (2.56).

Proof

1. Let $(f,\varPhi \mathbf{e}_{\varphi },L,S)$ be a solution of Problem 2 given in Theorem 1 for $\sigma \in (0,\overline{\sigma }_2]$ where $\overline{\sigma }_2$ is a positive constant less than or equal to $\sigma _1$ and to be determined later. Then since $(f,\varPhi \mathbf{e}_{\varphi },L,S)$ is a solution of Problem 2, $\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S)$ is well-defined in $\overline{\mathcal {N}_f^+}$. Define $\rho _+:=\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S)$, ${\mathbf {u}}_+:=\frac{\nabla \times (\varPhi \mathbf{e}_{\varphi })}{\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S)}$ and $p_+:=S(\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),$ $\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S))^\gamma .$ Then since $(\varPhi \mathbf{e}_{\varphi },L,S)$ satisfies (2.43)–(2.45) in $\mathcal {N}_f^+$, (2.46)–(2.49), (2.52), (2.53), and

$$\begin{aligned} b(\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S),\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S,B_0)=0\quad \text{ in }\quad \overline{\mathcal {N}_f^+}, \end{aligned}$$

where b is a funtion defined in (2.40), $(\rho _+,{\mathbf {u}}_+,p_+)$ satisfies the second and third equation of (2.1) in $\mathcal {N}_f^+$, (2.3) on $\varGamma _f$, (2.21) and (2.22). Furthermore, since $\nabla \cdot (\nabla \times (\varPhi \mathbf{e}_{\varphi }))=0$ and $\nabla \cdot (\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S)\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi })=0$ in $\mathcal {N}_f^+$, $(\rho _+,{\mathbf {u}}_+,p_+)$ satisfies the first equation of (2.1) in $\mathcal {N}_f^+$. Since $(\varPhi \mathbf{e}_\varphi ,L,S)$ is a subsonic solution of (2.43)–(2.45), $(\rho _+,{\mathbf {u}}_+,p_+)$ is a subsonic solution of (2.1). Since $(\nabla \times (\varPhi \mathbf{e}_\varphi ),\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S)\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, $(\rho _+,{\mathbf {u}}_+,p_+)\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. From these facts, we have that $(f,\rho _+,{\mathbf {u}}_+,p_+)$ is a solution of Problem 1.

Obtain (2.56). By Lemma 3, $\rho _0^+$ can be written as $\rho _0^+=\varrho (\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi }),$ $\frac{L_0^+}{2\pi r \sin \theta }\mathbf{e}_{\varphi }(=0),S_0^+).$ Using this expression, we write $\rho _+-\rho _0^+$ as

$$\begin{aligned}&\int _0^1\nabla \varrho (t(\nabla \times (\varPhi \mathbf{e}_\varphi ), \frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S)+(1-t) (\nabla \times (\varPhi _0^+\mathbf{e}_\varphi ),0,S_0^+))\hbox {d}t\nonumber \\&\quad (\nabla \times ((\varPhi -\varPhi _0^+)\mathbf{e}_\varphi ),\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ,S-S_0^+). \end{aligned}$$

(2.57)

Since b is a $C^\infty$ function of $(\rho ,\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S,B)$, $\varrho$ is a $C^\infty$ function of $(\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S,B)$. With this fact, the fact that $(\rho _0^+,u_0^+\mathbf{e}_r,p_0^+)\in (C^\infty (\overline{\mathcal {N}_f^+}))^3$ and (2.55) satisfied by $(f,\varPhi \mathbf{e}_{\varphi },L,S)$ for $\sigma \in (0,\overline{\sigma }_2]$ for $\overline{\sigma }_2\le \sigma _1$, we estimate (2.57) in $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. Then we obtain

$$\begin{aligned} ||\rho _+-\rho _0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)} \le C\sigma \end{aligned}$$

(2.58)

where C is a positive constant depending on the data. By this estimate, there exists a positive constant $\overline{\sigma }_2^{(1)}$ depending on the data such that if $\sigma \in (0, \overline{\sigma }_2^{(1)}]$, then

$$\begin{aligned} \sup _{\mathcal {N}_f^+}|\frac{1}{\rho _+}|\le C, \end{aligned}$$

(2.59)

where C is a positive constant depending on the data. Take $\overline{\sigma }_2=\min (\sigma _1,\overline{\sigma }_2^{(1)})$ so that (2.59) holds. With (2.55) satisfied by $(f,\varPhi \mathbf{e}_{\varphi },L,S)$, (2.58) and (2.59), we estimate ${\mathbf {u}}_+ -u_0^+\mathbf{e}_r$ and $p_+ -p_0^+$ in $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. Then we obtain

$$\begin{aligned} ||{\mathbf {u}}_+ -u_0^+\mathbf{e}_r||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||p_+ -p_0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\le C\sigma , \end{aligned}$$

where C is a positive constant depending on the data. Combining this estimate, (2.55) and (2.58), we obtain (2.56).

2. Assume that for $\sigma \in (0,\underline{\sigma }_2]$ where $\underline{\sigma }_2$ is a positive constant to be determined later, there exist two solutions $(f^i,\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ for $i=1,2$ of Problem 1 satisfying the estimate (2.56). There exists a positive constant $\underline{\sigma }_2^{(1)}$ depending on the data such that if $\sigma \in (0,\underline{\sigma }_2^{(1)}]$, then 1)

$$\begin{aligned} \sup _{\mathcal {N}_{f_i}^+}|\frac{1}{\rho _+^i}|\le C \end{aligned}$$

(2.60)

for $i=1,2$ where C is a positive constant depending on the data and 2) $\rho _+^i$ for $i=1,2$ are uniquely determined by $(\rho _+^i(u_{+,r}^i\mathbf{e}_r+u_{+,\theta }^i\mathbf{e}_\theta ), u_{+,\varphi }^i\mathbf{e}_{\varphi }, S_+^i,B_0)$ where $u_{+,r}^i:={\mathbf {u}}_+^i\cdot \mathbf{e}_r$, $u_{+,\theta }^i:={\mathbf {u}}_+^i\cdot \mathbf{e}_\theta$, $u_{+,\varphi }^i:={\mathbf {u}}_+^i\cdot \mathbf{e}_{\varphi }$ and $S_+^i:=\frac{p_+^i}{(\rho _+^i)^\gamma }$ for $i=1,2$ (here we used Lemma 3 and the fact that if $(\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ are $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solutions of Problem 1, then $(\rho ,{\mathbf {u}},p)=(\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ satisfy $B=B_0$ in $\overline{\mathcal {N}_{f^i}^+}$).

Take $\underline{\sigma }_2= \underline{\sigma }_2^{(1)}$ so that 1) and 2) hold. Let $(\varPhi ^i\mathbf{e}_{\varphi },L^i,S^i)$ for $i=1,2$ be $(\varPhi \mathbf{e}_{\varphi },L,S)$ given by the definitions of $\varPhi$, L and S for $(\rho ,{\mathbf {u}},p)=(\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ for $i=1,2$. Then since $(f^i,\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ for $i=1,2$ are solutions of Problem 1, $(f^i,\varPhi ^i\mathbf{e}_{\varphi },L^i,S^i)$ for $i=1,2$ are solutions of Problem 2. Furthermore, since $(f^i,\rho _+^i,{\mathbf {u}}_+^i,p_+^i)$ for $i=1,2$ satisfy (2.56), $(f^i,\varPhi ^i\mathbf{e}_{\varphi },L^i,S^i)$ for $i=1,2$ satisfy (2.55) with C replaced by $C_1$ for some positive constant $C_1$ depending on the data (here we used (2.60)). Take $\underline{\sigma }_2=\min (\underline{\sigma }_2^{(1)},\frac{C\sigma _1}{C_1},\sigma _1)$ where C is C in (2.55) so that $(\rho _-,{\mathbf {u}}_-,p_-)$ and $p_{ex}$ satisfy (2.19) and (2.20), respectively, for $\sigma \in (0,\sigma _1]$ and $(f^i,\varPhi ^i\mathbf{e}_{\varphi },L^i,S^i)$ for $i=1,2$ satisfy (2.55) for $\sigma \in (0,\sigma _1]$. Then by Theorem 1, $(f^1,\varPhi ^1\mathbf{e}_{\varphi },L^1,S^1)=(f^2,\varPhi ^2\mathbf{e}_{\varphi },L^2,S^2)$. This implies

$$\begin{aligned} \rho _+^1(u_{+,r}^1\mathbf{e}_r+u_{+,\theta }^1\mathbf{e}_\theta )=\rho _+^2(u_{+,r}^2\mathbf{e}_r+u_{+,\theta }^2\mathbf{e}_\theta )\quad \text{ and }\quad u_{+,\varphi }^1\mathbf{e}_{\varphi }=u_{+,\varphi }^2\mathbf{e}_{\varphi }. \end{aligned}$$

By these relations, $S^1=S^2$ and 2), we have that $\rho _+^1=\rho _+^2$. With this relation, we can conclude that $(\rho _+^1,{\mathbf {u}}_+^1,p_+^1)=(\rho _+^2,{\mathbf {u}}_+^2,p_+^2)$. Let $\sigma _2=\min (\overline{\sigma }_2,\underline{\sigma }_2)$. This finishes the proof. $\square$

The rest of this paper is devoted to proving Theorem 1. For convenience, we describe our main process of proving Theorem 1 below.

To describe our process of proving Theorem 1, we define some terminologies. Let $f:[0,\theta _1]\rightarrow \mathbb {R}$ be a function representing an axisymmetric shock location $\varGamma _f$. Decompose f into $f(\theta )=f(0)+f_s(\theta )$. Then by $f_s(0)=0$, $f_s$ is uniquely determined by $f_s^\prime$. We call f(0) and $f_s^\prime$ the initial shock position and the shape of a shock location, respectively.

Using these terminologies, our process of proving Theorem 1 is described as follows:

• 1. For given an incoming supersonic solution, an exit pressure and a shape of a shock location $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )$ in a small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,$ $p_0^-,p_c,0)$, show that there exists a pair of an initial shock position f(0) and a subsonic solution $(\varPhi \mathbf{e}_{\varphi },L,S)$ of (2.43)–(2.45) satisfying all the conditions in Problem 2 except (2.47), and that this solution is unique in the class of functions in a small perturbation of $(r_s,\varPhi _0^+\mathbf{e}_{\varphi },L_0^+,S_0^+)$.

• 2. For given an incoming supersonic solution and an exit pressure as in Step 1 or in a much small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$ if necessary, show that there exists $f_s^\prime$ in a small perturbation of 0 as in Step 1 such that $(f(0),\varPhi \mathbf{e}_{\varphi },L,S)$ determined by $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )$ in Step 1 satisfies (2.47), and that for given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex})$ in a small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$, a solution $(f,\varPhi \mathbf{e}_{\varphi },L,S)$ of Problem 2 is unique in the class of functions in a small perturbation of $(r_s,\varPhi _0^+\mathbf{e}_{\varphi },L_0^+,S_0^+)$.

Once Step 1 and Step 2 are done, then $f=f(0)+f_s$ and $(\varPhi \mathbf{e}_{\varphi },L,S)$ obtained through Step 1 and Step 2 satisfies all the conditions in Problem 2. Thus, Theorem 1 is proved if Step 1 and Step 2 are done. We will deal with Step 1 and Step 2 in Sections 3 and 4, respectively.

Note that the fact that for transonic shock solutions of the full Euler system, an initial shock position and a shape of a shock location are determined in different mechanisms (an initial shock position is determined by the solvability condition related to the mass conservation law and a shape of a shock location is determined by the R-H conditions) was pointed out in [24] and the authors in [10, 18, 20, 21, 23, 24] proved the stability of transonic shock solutions of the full Euler system using iteration schemes based on this fact. In this paper, we also prove the stability of transonic shock solutions of the full Euler system using a scheme based on this fact, but we do this using a different scheme. In our scheme, a non-local elliptic equation appearing in [10, 18, 20, 21, 23, 24] does not appear.

3 Pseudo Free Boundary Problem

As a first step to prove Theorem 1, we will solve the Pseudo Free Boundary Problem below. This problem naturally arises from the requirement that a subsonic solution in Problem 2 must satisfy (2.54). From the linearized equation of (2.53), it is seen that an iteration scheme for a fixed boundary problem does not give a subsonic solution satisfying (2.54) in general. From this, we have that an iteration scheme for a fixed boundary problem is not a proper scheme to find a subsonic solution in Problem 2. To find a subsonic solution satisfying (2.54), an iteration scheme that gives a proper $(\rho ,{\mathbf {u}},p)$ on $\varGamma _{ex}$ is necessary. To find a subsonic solution satisfying (2.54), we let f(0) be an unknown to be determined simultaneously with a subsonic solution. Using this variable, we adjust the value of a subsonic solution so that this solution can satisfy (2.54). The main ingredient for this argument to hold is the monotonicity of the entropy of the downstream subsonic solution of a radial transonic shock solution on a shock location with respect to the shock location. This will be seen in the proof of Proposition 2.

Problem 3

(Pseudo Free Boundary Problem) Given an axisymmetric supersonic solution $(\rho _-,{\mathbf {u}}_-,p_-)$ of (2.1) and exit pressure $p_{ex}$ as in Problem 1 and a shape of a shock location $f_s^\prime \in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ satisfying $f_s^\prime (\theta )=0$ at $\theta =0, \theta _1$ and

$$\begin{aligned} ||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le \sigma \end{aligned}$$

(3.1)

for a sufficiently small $\sigma >0$, find an initial shock position f(0) and a corresponding subsonic solution $(\varPhi \mathbf{e}_\varphi ,L,S)$ of (2.43)–(2.45) satisfying

where $\rho =\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r\sin \theta }\mathbf{e}_{\varphi },S)$, and

where $f_s$ denotes $\int _0^\theta f_s^\prime$.

Hereafter, we denote a set of functions in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ having 0 value at $\theta =0,\theta _1$ by $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$. Hereafter, $f_s$ denotes $\int _0^\theta f_s^\prime$.

Our result of Problem 3 is given as follows:

Proposition 2

Let $\alpha \in (\frac{2}{3},1)$. There exists a positive constant $\sigma _3$ depending on the data such that if $\sigma \in (0, \sigma _3]$, then Problem 3 has a solution $(f(0),\varPhi \mathbf{e}_\varphi ,L,$ S) satisfying

$$\begin{aligned}&|f(0)-r_s|+||\nabla \times ((\varPhi -\varPhi _0^+)\mathbf{e}_\varphi )||_{1,\alpha , \mathcal {N}_{f(0)+f_s}^+}^{(-\alpha ,\varGamma _w^+)}\nonumber \\&\quad +||\frac{L}{2\pi r\sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_{f(0)+f_s}^+}^{(-\alpha ,\varGamma _w^+)} +||S-S_0^+||_{1,\alpha ,\mathcal {N}_{f(0)+f_s}^+}^{(-\alpha ,\varGamma _w^+)} \le C\sigma \end{aligned}$$

(3.2)

where C is a positive constant depending on the data. Furthermore, this solution is unique in the class of functions satisfying (3.2).

We will prove Proposition 2 using a fixed point argument. To do this, we linearize (A) with respect to (A) satisfied by $(\varPhi \mathbf{e}_\varphi ,L,S)=(\varPhi _0^+\mathbf{e}_{\varphi },L_0^+,S_0^+)$ and reformulate (B) in terms of $(\varPsi \mathbf{e}_{\varphi },A,T)$ where $(\varPsi ,A,T):=(\varPhi -\varPhi _0^+,L,S-S_0^+)$.

3.1 Linearization and reformulation of (A) and (B)

We linearize (A) with respect to (A) satisfied by $(\varPhi _0^+\mathbf{e}_{\varphi },L_0^+,S_0^+)$. Since $\rho$ in the first and fourth equation of (A) is given using an implicit relation, to obtain the linearized equations of (A), we first linearize $\rho$ with respect to $\rho _0^+$.

Lemma 4

Suppose that $f:\varLambda \rightarrow \mathbb {R}$ is an axisymmetric function in $C^{2,\alpha }_{(-1-\alpha ,\partial \varLambda )}$ $(\varLambda )$ satisfying

$$\begin{aligned} ||f-r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )}\le \delta _1. \end{aligned}$$

(3.3)

Also, suppose that $\varPsi \mathbf{e}_\varphi :\mathcal {N}_f^+\rightarrow \mathbb {R}^3$, $A: \mathcal {N}_f^+\rightarrow \mathbb {R}$ and $T:\mathcal {N}_f^+\rightarrow \mathbb {R}$ are axisymmetric functions in $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ and $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, respectively, and satisfy

$$\begin{aligned} ||\nabla \times (\varPsi \mathbf{e}_\varphi )||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+}+||\frac{A}{2\pi r\sin \theta }\mathbf{e}_\varphi ||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+}+||T||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+}\le \delta _2. \end{aligned}$$

(3.4)

Let $\rho =\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S)$. There holds

$$\begin{aligned}&\rho -\rho _0^+\nonumber \\&\quad =\frac{\nabla \times (\varPhi _0^+\mathbf{e}_\varphi )}{\rho _0^+({u_0^+}^2-{c_0^+}^2)}\cdot \nabla \times (\varPsi \mathbf{e}_\varphi )+\frac{\gamma {\rho _0^+}^{\gamma }}{(\gamma -1)({u_0^+}^2-{c_0^+}^2)}T \nonumber \\&\quad +g_1(\varPsi \mathbf{e}_\varphi ,A,T), \end{aligned}$$

(3.5)

where

$$\begin{aligned}&g_1(\varPsi \mathbf{e}_\varphi ,A,T)\nonumber \\&\quad =\frac{1}{\rho _0^+({u_0^+}^2-{c_0^+}^2)}\left( \int _0^1 \left( (2\rho _0^+B_0-\frac{\gamma (\gamma +1)}{\gamma -1}S_0^+{\rho _0^+}^\gamma ) \right. \right. \nonumber \\&\qquad \left. -(2(t\rho +(1-t)\rho _0^+)B_0 -\frac{\gamma (\gamma +1)}{\gamma -1}(S_0^++T)(t\rho +(1-t)\rho _0^+)^\gamma )\right) \hbox {d}t(\rho -\rho _0^+)\nonumber \\&\qquad +\int _0^1(-\nabla \times (\varPhi _0^+\mathbf{e}_\varphi )+(t\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )+(1-t)\nabla \times ( \varPhi _0^+\mathbf{e}_\varphi )))\hbox {d}t\nonumber \\&\qquad \cdot \nabla \times (\varPsi \mathbf{e}_\varphi ) \left. +\int _0^1 \frac{{\rho _0^+}^2 tA}{2\pi r\sin \theta }\mathbf{e}_\varphi \hbox {d}t\cdot \frac{A}{2\pi r \sin \theta }\mathbf{e}_\varphi \right) . \end{aligned}$$

(3.6)

$g_1(\varPsi \mathbf{e}_\varphi ,A,T)$ satisfies

$$\begin{aligned}&||g_1(\varPsi \mathbf{e}_\varphi ,A,T)||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)} \nonumber \\&\quad \le C(||\nabla \times (\varPsi \mathbf{e}_\varphi )||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha , \varGamma _w^+)} +||\frac{A}{2\pi r\sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\nonumber \\&\qquad +||T||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)})^2, \end{aligned}$$

(3.7)

where C is a positive constant depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$, $\alpha$, $\delta _1$ and $\delta _2$.

Proof

1. By the definition of $\varrho$, $\rho =\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S)$ satisfies

$$\begin{aligned} b\left( \rho ,\nabla \times (\varPhi \mathbf{e}_\varphi ), \frac{L}{2\pi r\sin \theta }\mathbf{e}_\varphi ,S,B_0\right) =0, \end{aligned}$$

where b is a function defined in (2.40). Subtract this equation from $b(\rho _0^+,\nabla \times (\varPhi _0^+\mathbf{e}_\varphi ),0,S_0^+,B_0)=0$ obtained from the third equation of (2.14). And then linearize the resultant equation. Then we obtain (3.5).

2. With the fact that $(\rho _0^+,u_0^+\mathbf{e}_r,p_0^+)\in (C^\infty (\overline{\mathcal {N}_f^+}))^3$ and (3.4), estimate $\rho -\rho _0^+$ in $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ in the way that we estimated $\rho _+-\rho _0^+$ in the proof of Theorem 2. Then we have

$$\begin{aligned}&||\rho -\rho _0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\nonumber \\&\quad \le C (||\nabla \times (\varPsi \mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||\frac{A}{2\pi r \sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)} \nonumber \\&\qquad +||T||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}), \end{aligned}$$

(3.8)

where C is a positive constant depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$, $\alpha$, $\delta _1$ and $\delta _2$. With this estimate, the fact that $(\rho _0^+,u_0^+\mathbf{e}_r,p_0^+)\in (C^\infty (\overline{\mathcal {N}_f^+}))^3$ and (3.4), estimate (3.6) in $C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. Then we obtain the desired result. $\square$

Then we linearize (A).

Subtract (2.43) satisfied by $(\varPhi \mathbf{e}_\varphi ,L,S)=(\varPhi _0^+\mathbf{e}_{\varphi },L_0^+,S_0^+)$ in $\mathcal {N}_f^+$ from (2.43) in $\mathcal {N}_f^+$ and then linearize the resultant equation (in this process, we express $\rho -\rho _0^+$ using (3.5)). Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (\varPsi \mathbf{e}_\varphi )\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}\left( 1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2}\right) \frac{\partial _\theta T}{r}\mathbf{e}_\varphi +{\varvec{F}}_1(\varPsi \mathbf{e}_\varphi ,A,T)\quad \text{ in }\quad \mathcal {N}_f^+ \nonumber \\&\qquad (=:{\varvec{F}}_2(\varPsi \mathbf{e}_\varphi ,A,T)), \end{aligned}$$

(3.9)

where

$$\begin{aligned}&{\varvec{F}}_1(\varPsi \mathbf{e}_\varphi ,A,T)\nonumber \\&\quad =\nabla \times \left( \frac{1}{\rho _0^+}g_1\nabla \times (\varPhi _0^+\mathbf{e}_\varphi )-\frac{1}{\rho {\rho _0^+}^2}g_2^2\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )+\frac{1}{{\rho _0^+}^2}g_2\nabla \times (\varPsi \mathbf{e}_\varphi )\right) \nonumber \\&\qquad -\frac{1}{u_0^+ }\left( -\frac{1}{\rho \rho _0^+}g_2\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )\cdot \mathbf{e}_r+\frac{1}{\rho _0^+}\nabla \times (\varPsi \mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \nonumber \\&\quad \quad \quad \nabla \times \left( -\frac{1}{\rho \rho _0^+}g_2\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )+\frac{1}{\rho _0^+}\nabla \times (\varPsi \mathbf{e}_\varphi )\right) \nonumber \\&\qquad +\frac{1}{u_0^+ }\left( \int _0^1(t\rho +(1-t)\rho _0^+)^{\gamma -2}\hbox {d}tg_2 \frac{\partial _\theta T}{r}\right) \mathbf{e}_\varphi \nonumber \\&\qquad +\frac{A}{u_0^+ 2\pi r \sin \theta }\left( \nabla \times (\frac{A}{2\pi r \sin \theta }\mathbf{e}_\varphi )\cdot \mathbf{e}_r\right) \mathbf{e}_\varphi , \end{aligned}$$

(3.10)

where $g_1$ and $g_2$ are $g_1(\varPsi \mathbf{e}_\varphi ,A,T)$ given in (3.6) and the right hand side of (3.5), respectively.

By the definition of $V_0^\pm$ and the first equations of (2.10) and (2.14), $V_0^+=V_0^-$ on $\varGamma _f.$ In the vector potential form, this is written as $\varPhi _0^+\mathbf{e}_{\varphi }=\varPhi _0^-\mathbf{e}_{\varphi }$ on $\varGamma _f$. Subtract this equation from (2.46). Then we obtain

$$\begin{aligned} \varPsi \mathbf{e}_\varphi =(\varPhi _--\varPhi _0^-)\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _{f}. \end{aligned}$$

(3.11)

By the definition of $V_0^\pm$ and the first equations of (2.10) and (2.14), $V_0^+=V_0^-(r_0,\theta _1)$ on $\varGamma _w^+.$ In the vector potential form, this is written as $\varPhi _0^+\mathbf{e}_\varphi =\frac{r_0\varPhi _0^-(r_0,\theta _1)\mathbf{e}_{\varphi }}{r}$ on $\varGamma _w^+$. By subtracting this equation from (2.52), we obtain

$$\begin{aligned} \varPsi \mathbf{e}_\varphi =\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r}\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _w^+. \end{aligned}$$

(3.12)

Rewrite (2.53) as $\rho =(\frac{p_{ex}}{S})^{\frac{1}{\gamma }}$ on $\varGamma _{ex}$. Subtract this equation from $\rho _0^+=(\frac{p_c}{S_0^+})^{\frac{1}{\gamma }}$ on $\varGamma _{ex}$. And then express $\rho -\rho _0^+$ using (3.5) and linearize $(\frac{p_{ex}}{S})^{\frac{1}{\gamma }}-(\frac{p_c}{S_0^+})^{\frac{1}{\gamma }}$. Multiply $\frac{ {u_0^+}^2-{c_0^+}^2}{u_0^+}r\sin \theta$ and then integrate the resultant equation from 0 to $\theta$. After that, divide $\sin \theta$ and multiply $\mathbf{e}_\varphi$ on both-hand sides of the equation. Then we obtain

$$\begin{aligned} \varPsi \mathbf{e}_\varphi= & {} \bigg (\frac{1}{r\sin \theta }\int _0^\theta \bigg (\mathfrak f_0(T,p_{ex})-\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}T\nonumber \\&\quad +\mathfrak f_1(\varPsi \mathbf{e}_\varphi ,A,T) \bigg )r^2\sin \xi \hbox {d}\xi \bigg )\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _{ex}, \end{aligned}$$

(3.13)

where

$$\begin{aligned} \mathfrak f_0(T,p_{ex})=\frac{{u_0^+}^2-{c_0^+}^2}{u_0^+} \left( \frac{1}{S_0^++T}\right) ^{\frac{1}{\gamma }}(p_{ex}^{\frac{1}{\gamma }} -p_c^{\frac{1}{\gamma }}) \end{aligned}$$

(3.14)

and

$$\begin{aligned}&\mathfrak f_1(\varPsi \mathbf{e}_\varphi ,A,T) \nonumber \\&\quad =-\frac{{p_c}^{\frac{1}{\gamma }}({u_0^+}^2-{c_0^+}^2)}{\gamma u_0^+}\int _0^1 \left( \left( \frac{1}{tS+(1-t)S_0^+}\right) ^{\frac{1}{\gamma }+1} -\left( \frac{1}{S_0^+}\right) ^{\frac{1}{\gamma }+1}\right) \hbox {d}t T \nonumber \\&\qquad -\frac{{u_0^+}^2-{c_0^+}^2}{u_0^+}g_1. \end{aligned}$$

(3.15)

By the definition of $V_0^\pm$ and the first equation of (2.14), $\lim _{\theta \rightarrow \theta _1}V_0^+(r_1,\theta )=\lim _{r\rightarrow r_1} V_0^+(r,\theta _1)$. In the vector potential form, this is written as $\lim _{\theta \rightarrow \theta _1}$ $(\varPhi _0^+\mathbf{e}_{\varphi })(r(r_1,\theta )=\lim _{r\rightarrow r_1}(\varPhi _0^+\mathbf{e}_{\varphi }),\theta _1)$. Subtract this equation from (2.54). Then we obtain

$$\begin{aligned} \lim _{\theta \rightarrow \theta _1}(\varPsi \mathbf{e}_\varphi )(r_1,\theta )=\lim _{r\rightarrow r_1} (\varPsi \mathbf{e}_\varphi )(r,\theta _1). \end{aligned}$$

Express this relation using (3.12) and (3.13). Then we have

$$\begin{aligned}&\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1} \bigg (\mathfrak f_0(T,p_{ex})-\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1) u_0^+S_0^+}T\nonumber \\&\quad \quad +\mathfrak f_1(\varPsi \mathbf{e}_\varphi ,A,T) \bigg )\bigg |_{r=r_1}r_1^2\sin \xi \hbox {d}\xi =\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1}. \end{aligned}$$

(3.16)

Combining (3.9), (3.11), (3.12), (3.13) and (3.16), we have the following linearized equations of (A):

For later use, we present the following estimates of ${\varvec{F}}_1(\varPsi \mathbf{e}_\varphi ,A,T)$ and $\mathfrak f_1(\varPsi \mathbf{e}_\varphi ,A,T)$.

Lemma 5

Let $\delta _4$ be a positive constant $\le \delta _2$ such that for f as in Lemma 4, if $(\varPsi \mathbf{e}_{\varphi },A,T)$ satisfies

$$\begin{aligned} ||\nabla \times (\varPsi \mathbf{e}_\varphi )||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+} +||\frac{A}{2\pi r\sin \theta }\mathbf{e}_\varphi ||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+} +||T||^{(-\alpha ,\varGamma _w^+)}_{1,\alpha ,\mathcal {N}_f^+}\le \delta _4, \end{aligned}$$

(3.17)

then

$$\begin{aligned} \sup _{\mathcal {N}_f^+}|\frac{1}{\varrho (\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S)}| \le C\quad \text{ and }\quad \sup _{\mathcal {N}_f^+}|\frac{1}{S}|\le C, \end{aligned}$$

where Cs are positive constants depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$ and $\delta _4$. Suppose that f is as in Lemma 4. Also, suppose that $(\varPsi \mathbf{e}_{\varphi },A,T)$ are as in Lemma 4 and satisfy (3.17). Then there hold

$$\begin{aligned}&||{\varvec{F}}_1(\varPsi \mathbf{e}_\varphi ,A,T)||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}\nonumber \\&\quad \le C (||\varPsi \mathbf{e}_\varphi ||_{2,\alpha ,\mathcal {N}_f^+}^{(-1-\alpha ,\varGamma _w^+)} +||\frac{A}{2\pi r \sin \theta }\mathbf{e}_\varphi ||_{1,\alpha , \mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}+||T||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)})^2 \end{aligned}$$

(3.18)

and

$$\begin{aligned}&||\mathfrak f_1(\varPsi \mathbf{e}_\varphi ,A,T)||_{1,\alpha ,\varGamma _{ex}}^{(-\alpha ,\partial \varGamma _{ex})} \nonumber \\&\quad \le C (||\varPsi \mathbf{e}_\varphi ||_{2,\alpha ,\varGamma _{ex}}^{(-1-\alpha ,\partial \varGamma _{ex})}+||\frac{A}{2\pi r \sin \theta }\mathbf{e}_\varphi |||_{1,\alpha ,\varGamma _{ex}}^{(-\alpha ,\partial \varGamma _{ex})} \nonumber \\&\qquad +||T||_{1,\alpha ,\varGamma _{ex}}^{(-\alpha ,\partial \varGamma _{ex})})^2 \end{aligned}$$

(3.19)

where Cs are positive constants depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$, $\alpha$, $\delta _1$, $\delta _2$ and $\delta _4$.

Proof

With (3.7), (3.8), (3.17) and the fact that if an axisymmetric vector field ${\varvec{a}}$ on an axisymmetric connected open set $\varOmega$ is in $C^k(\varOmega )$, then $\nabla \times {\varvec{a}}\in C^{k-1}(\varOmega )$ and ${\varvec{a}}\cdot \mathbf{e}_r\in C^k(\varOmega )$ (the second one is obtained from Lemma 1), we estimate (3.10) and (3.15) in $C^{\alpha }_{(1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, and $C^{1,\alpha }_{(-\alpha ,\partial \varGamma _{ex})}(\varGamma _{ex})$, respectively. Then we obtain the desired result. $\square$

Next, we reformulate (B) in terms of $(\varPsi \mathbf{e}_\varphi ,A,T)$. Using the facts that $L_0^+=L_0^-=0$ in $\mathcal {N}_f^+$ and $S_0^+=(g({M_0^-}^2))(r_s)S_{in}$ in $\mathcal {N}_f^+$ (see (2.14)), we reformulate (B) in terms of $(\varPsi \mathbf{e}_\varphi ,A,T)$. Then we obtain

where

$$\begin{aligned} A_{en,f(0)+f_s}:=L_-\quad \text{ on }\quad \varGamma _{f(0)+f_s} \end{aligned}$$

(3.20)

and

$$\begin{aligned} T_{en,f(0)+f_s}:= g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f(0)+f_s}}{c_-}\right) ^2\right) S_- -\left( g({M_0^-}^2)\right) (r_s)S_{in}\quad \text{ on }\quad \varGamma _{f(0)+f_s}. \end{aligned}$$

(3.21)

For later use, we present the following estimate of $T_{en,f(0)+f_s}$:

Lemma 6

Let f(0) and $f_s^\prime$ be a constant and a function in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}$ $((0,\theta _1))$, respectively. Let $\delta _5$ be a positive constant such that if $(\rho _-,{\mathbf {u}}_-,p_-)$ satisfies (2.19) for $\sigma \in (0,\delta _5]$, then

$$\begin{aligned} \sup _{\mathcal {N}}|\frac{1}{c_-}|\le C, \end{aligned}$$

where C is a positive constant depending on $(\rho _0^-,u_0^-,p_0^-)$, $\gamma$, $r_0$, $r_1$ and $\delta _5$. Suppose that $f=f(0)+f_s$ satisfies (3.3). Also, suppose that $(\rho _-,{\mathbf {u}}_-,p_-)$ is an axisymmetric supersonic solution of (2.1) in $\mathcal {N}$ satisfying (2.19) for $\sigma \in (0,\delta _5]$. Then it holds that

$$\begin{aligned} ||T_{en,f}||_{1,\alpha ,\varGamma _f}^{(-\alpha ,\partial \varGamma _f)}\le C(|f(0)-r_s|+||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha , \{\theta =\theta _1\})}) +C\sigma , \end{aligned}$$

(3.22)

where Cs are positive constants depending on $(\rho _0^-,u_0^-,p_0^-)$, $\gamma$, $r_0$, $r_s$, $r_1$, $\theta _1$, $\alpha$, $\delta _1$ and $\delta _5$.

Proof

In this proof, Cs denote positive constants depending on the whole or a part of $(\rho _0^-,u_0^-,p_0^-)$, $\gamma$, $r_0$, $r_s$, $r_1$, $\theta _1$, $\alpha$, $\delta _1$ and $\delta _5$. Each C in different situations differs from each other.

By the fact that $(\rho _-,{\mathbf {u}}_-,p_-)$ and $f=f(0)+f_s$ are axisymmetric, $T_{en,f}$ defined in (3.21) can be regarded as a function of $\theta$. As a function of $\theta$, $T_{en,f}$ can be written as

$$\begin{aligned} T_{en,f}&=\int _0^1(g({M_0^-}^2))^\prime (t f(\theta )+(1-t)r_s)S_{in}\hbox {d}t (f(\theta )-r_s)\\&\quad +\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_f(\theta )}{c_-}\right) ^2\right) S_-\right) (f(\theta ),\theta ) -\left( g({M_0^-}^2)\right) (f(\theta ),\theta )S_{in}\\&=:(a)+(b). \end{aligned}$$

To estimate $||T_{en,f}||_{1,\alpha ,\varGamma _f}^{(-\alpha ,\partial \varGamma _f)}$, we estimate (a) and (b) in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$, respectively.

Since an estimate of (a) in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ is obtained directly:

$$\begin{aligned} ||(a)||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C (|f(0)-r_s|+||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}), \end{aligned}$$

(3.23)

where we used $||f_s||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}$, we only estimate (b) in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$.

Estimate of (b) in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$:

Decompose (b) into two parts:

$$\begin{aligned} \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot \mathbf{e}_r}{c_-}\right) ^2\right) S_-\right) (f(\theta ),\theta ) -\left( g({M_0^-}^2)\right) (f(\theta ),\theta )S_{in}=:(b)_1 \end{aligned}$$

and

$$\begin{aligned} \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_f}{c_-}\right) ^2\right) S_-\right) (f(\theta ),\theta ) -\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot \mathbf{e}_r}{c_-}\right) ^2\right) S_-\right) (f(\theta ),\theta )=:(b)_2. \end{aligned}$$

With (2.19) for $\sigma \in (0,\delta _5]$ and (3.3), we estimate $(b)_1$ in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$. Then we obtain

$$\begin{aligned} ||(b)_1||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C\sigma . \end{aligned}$$

(3.24)

Write $(b)_2$ as

$$\begin{aligned}&\int _0^1 g^\prime \left( \left( \frac{{\mathbf {u}}_-}{c_-}\cdot (t{\varvec{\nu }}_f+(1-t)\mathbf{e}_r)\right) ^2\right) 2\left( \frac{{\mathbf {u}}_-}{c_-}\cdot (t{\varvec{\nu }}_f+(1-t)\mathbf{e}_r)\right) S_-\hbox {d}t \nonumber \\&\qquad \frac{{\mathbf {u}}_-}{c_-}\cdot ({\varvec{\nu }}_f-\mathbf{e}_r). \end{aligned}$$

(3.25)

By ${\varvec{\nu }_f}=\frac{\mathbf{e}_r-\frac{f^\prime }{f}\mathbf{e}_\theta }{\sqrt{1+(\frac{f^\prime }{f})^2}}$, ${\varvec{\nu }}_f-\mathbf{e}_r$ can be written as

$$\begin{aligned} {\varvec{\nu }}_f-\mathbf{e}_r =\left( \int _0^1-\frac{1}{2} \left( 1+t\left( \frac{f^\prime }{f}\right) ^2\right) ^{-\frac{3}{2}}\hbox {d}t\left( \frac{f^\prime }{f}\right) ^2\right) \mathbf{e}_r-\frac{\frac{f^\prime }{f}}{\sqrt{1+\left( \frac{f^\prime }{f}\right) ^2}}\mathbf{e}_{\theta }. \end{aligned}$$

Substitute this expression of ${\varvec{\nu }}_f-\mathbf{e}_r$ into ${\varvec{\nu }}_f-\mathbf{e}_r$ in (3.25) and then estimate (3.25) in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ with (2.19) for $\sigma \in (0,\delta _5]$ and (3.3). Then we obtain

$$\begin{aligned} ||(b)_2||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C (||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})})^2+C\delta _5||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}. \end{aligned}$$

Using the fact that $||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,(0,\theta _1))}\le C||f-r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )}\le C\delta _1$, we get from this estimate

$$\begin{aligned} ||(b)_2||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}. \end{aligned}$$

(3.26)

Combining (3.24) and (3.26), we obtain

$$\begin{aligned} ||(b)||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C ||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,(0,\theta _1))}+C\sigma . \end{aligned}$$

(3.27)

From the facts that $\partial _\theta T_{en,f}(0)=0$ and $(a), (b)\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ (obtained from (3.23) and (3.27)), we see that $T_{en,f}\in C^{1,\alpha }_{(-\alpha ,\partial \varGamma _f)}$ $(\varGamma _f)$ (see Lemma 1). So, $||T_{en,f}||_{1,\alpha ,\varGamma _f}^{(-\alpha ,\partial \varGamma _f)}$ can be estimated by estimating $||T_{en,f}||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}$. Using this fact, we obtain from (3.23) and (3.27) (3.22). This finishes the proof. $\square$

From (A$^\prime$), (B$^\prime$), the Pseudo Free Boundary Problem is naturally derived. We explain this below.

For a given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex})$, find $(\varPsi \mathbf{e}_{\varphi },A,T)$ satisfying (A$^\prime$), (B$^\prime$) using an iteration scheme for a fixed boundary problem (for example, in a fixed domain $\mathcal {N}_{f(0)+f_s}^+$, for a given $\varPsi \mathbf{e}_{\varphi }$, solve (B$^\prime$), substitute the resultant A and T and the previously given $\varPsi \mathbf{e}_\varphi$ into the right hand sides of (A$^\prime$), obtain a new $\varPsi \mathbf{e}_{\varphi }$ by solving the resultant (A$^\prime$) and show that a new $\varPsi \mathbf{e}_{\varphi }$ is equal to the given $\varPsi \mathbf{e}_{\varphi }$ using a fixed point argument). Then $(\varPsi \mathbf{e}_{\varphi },A,T)$ we find in this way does not satisfy the fifth equation of (A$^\prime$) in general. Thus, this kind of iteration scheme does not give a subsonic solution of (2.43)–(2.45) satisfying (2.54) in general. To find a subsonic solution satisfying (2.54), an iteration scheme that gives a proper $(\varPsi \mathbf{e}_{\varphi },A,T)$ on $\varGamma _{ex}$ is necessary. From the fifth equation of (A$^\prime$) and the facts that the entropy at a point on a shock location in the subsonic side is conserved along the streamline passing through that point and the entropy of the downstream subsonic solution of a radial transonic shock solution in a divergent nozzle on a shock location monotonically increases as a shock location moves toward the exit (see Lemma 2), it is seen that we can find $(\varPsi \mathbf{e}_{\varphi },A,T)$ satisfying the fifth equation of (A$^\prime$) by adjusting f(0) to vary S on $\varGamma _{ex}$. From this fact, Problem 3 is derived.

(A$^\prime$) and (B$^\prime$) are of the form of one linear boundary value problem for a singular elliptic equation (this will be seen in the next subsection) and two initial value problems of a transport equation whose coefficient is an axisymmetric and divergence-free vector field, respectively. We will study these problems, seperately, in § 3.2 and § 3.3.

3.2 Linear boundary value problem for a singular elliptic equation

Fix the right hand sides of the first and fourth equation in (A$^\prime$) with the fifth equation in (A$^\prime$) satisfied. Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ( \varPsi \mathbf{e}_{\varphi })\right) ={\varvec{F}}\quad \text{ in }\quad \mathcal {N}_f^+,\end{aligned}$$

(3.28)

$$\begin{aligned}&\varPsi \mathbf{e}_{\varphi }={\left\{ \begin{array}{ll} h_1\mathbf{e}_{\varphi }\quad \text{ on } \quad \varGamma _f,\\ \frac{f(\theta _1)h_1(f(\theta _1),\theta _1)}{r}\mathbf{e}_{\varphi } \quad \text{ on }\quad \varGamma _w^+,\\ h_2\mathbf{e}_{\varphi }\quad \text{ on } \quad \varGamma _{ex} \end{array}\right. } \end{aligned}$$

(3.29)

where f, ${\varvec{F}}$ and $h_i\mathbf{e}_{\varphi }$ for $i=1,2$ are functions given in Lemma 7. Since (3.28) is expressed as

$$\begin{aligned}&\bigg (-\frac{1}{\rho _0^+}\left( \varDelta \varPsi -\frac{\varPsi }{r^2\sin ^2\theta }\right) +\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\partial _r(r\varPsi ) \nonumber \\&\quad -\frac{{u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)r^2} \left( \frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }\varPsi ) -\frac{\varPsi }{\sin ^2\theta }\right) \bigg )\mathbf{e}_{\varphi }={\varvec{F}}, \end{aligned}$$

(3.30)

(3.28), (3.29) is a linear boundary value problem for a singular equation as a problem for $\varPsi$. Thus, the standard elliptic theorems cannot be applied to this problem as a problem for $\varPsi$. We resolve this problem by dealing with (3.28), (3.29) as a boundary value problem for an elliptic system.

The following is the main result in this subsection.

Lemma 7

Let $\alpha \in (\frac{2}{3},1)$. Suppose that f is as in Lemma 4 and satisfy $f^\prime (\theta _1)=0$. Also, suppose that ${\varvec{F}}:\mathcal {N}_f^+\rightarrow \mathbb {R}^3$ is a function in $C^{\alpha }_{(1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ having the form

$$\begin{aligned} {\varvec{F}}=\sum _i A^i \partial _r B^i \mathbf{e}_{\varphi }+\sum _i C^i\partial _{\theta }D^i \mathbf{e}_{\varphi }+E\frac{\partial _\theta (F\sin \theta )}{\sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

(3.31)

where $A^i$, $B^i$, $C^i$, $D^i$, E and F are axisymmetric functions satisfying

$$\begin{aligned}&A^i\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+),\; B^i\mathbf{e}_\theta \in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+), \; C^i\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+), \nonumber \\&D^i\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+),\; E\mathbf{e}_{\varphi }\in C^{1,\alpha } (\overline{\mathcal {N}_f^+})\;\text{ and }\; F\mathbf{e}_{\varphi }\in C^{1,\alpha } (\overline{\mathcal {N}_f^+}). \end{aligned}$$

(3.32)

Finally, suppose that $h_1\mathbf{e}_{\varphi }:\varGamma _f\rightarrow \mathbb {R}^3$ and $h_2\mathbf{e}_{\varphi }:\varGamma _{ex}\rightarrow \mathbb {R}^3$ are axisymmetric functions in $C^{2,\alpha }_{(-1-\alpha ,\partial \varGamma _f)}(\varGamma _f)$ and $C^{2,\alpha }_{(-1-\alpha ,\partial \varGamma _{ex})}(\varGamma _{ex})$, respectively, and satisfy $\frac{f(\theta _1)h_1(f(\theta _1),\theta _1)}{r_1}=h_2(r_1,\theta _1)$. Then the boundary value problem (3.28), (3.29) has a unique axisymmetric $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution $\varPsi \mathbf{e}_{\varphi }$. Furthermore, the solution $\varPsi \mathbf{e}_{\varphi }$ satisfies

$$\begin{aligned}&||\varPsi \mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_f^+}^{(-1-\alpha ,\varGamma _w^+)} \nonumber \\&\quad \le C\left( ||{\varvec{F}}||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}+\sum _{i=1,2,3}F_i +||h_1\mathbf{e}_{\varphi }||_{2,\alpha ,\varGamma _f}^{(-1-\alpha ,\partial \varGamma _f)}\right. \nonumber \\&\qquad \left. +||h_2\mathbf{e}_{\varphi }||_{2,\alpha ,\varGamma _{ex}}^{(-1-\alpha ,\partial \varGamma _{ex})}\right) ,\nonumber \\ \end{aligned}$$

(3.33)

where C is a positive constant depending only on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$, $\theta _1$ and $\alpha$, and

$$\begin{aligned}&F_1=\sum _i ||A^i||_{W^{1,3}(\mathcal {N}_f^+)\cap L^\infty (\mathcal {N}_f^+)}||B^i\mathbf{e}_{\theta }||_{\alpha ,\mathcal {N}_f^+}\\&F_2=\sum _i||C^i||_{W^{1,3}(\mathcal {N}_f^+)\cap L^\infty (\mathcal {N}_f^+)} ||D^i||_{\alpha ,\mathcal {N}_f^+},\\&F_3=||E\frac{\partial _\theta (F\sin \theta )}{\sin \theta }\mathbf{e}_{\varphi }||_{L^q(\mathcal {N}_f^+)} \end{aligned}$$

for $q=\frac{3}{1-\alpha }$ with $||\cdot ||_{W^{1,3}(\mathcal {N}_f^+)\cap L^\infty (\mathcal {N}_f^+)}:= ||\cdot ||_{W^{1,3}(\mathcal {N}_f^+)}+||\cdot ||_{L^\infty (\mathcal {N}_f^+)}$.

Remark 7

The form of ${\varvec{F}}$ given in (3.31) is obtained from ${\varvec{F}}_2(\varPsi \mathbf{e}_\varphi ,A,T)$ in (3.9). This form will be used in the proof of Lemma 11.

To avoid the singularity issue in (3.28), (3.29), we deal with (3.28), (3.29) as a boundary value problem for a vector equation. From $\nabla \times (\nabla \times (\varPsi \mathbf{e}_{\varphi }))=-\varDelta (\varPsi \mathbf{e}_{\varphi })$, we expected that (3.28) can be transformed into a form of an elliptic system. We, motivated by the work in [4], thought that if (3.28) can be transformed into a solvable elliptic system form, then the unique existence and regularity of solutions of (3.28), (3.29) can be obtained by obtaining those of solutions of the elliptic system form of (3.28), (3.29) as a boundary value problem for an elliptic system.

For this argument to hold, it is needed to find a solvable elliptic system form of (3.28). For computational convenience to find such a form and for our later argument (reflection argument in the proof of Lemma 10 and Lemma 11), we use the following tensor notation:

Tensor notation

Let ${\varvec{a}}\otimes {\varvec{b}}={\varvec{a}}{\varvec{b}}^T$ for ${\varvec{a}}$, ${\varvec{b}}\in \mathbb {R}^3$. Then ${\varvec{a}}\otimes {\varvec{b}}$ is a linear map from $\mathbb {R}^3$ to $\mathbb {R}^3$ and any linear map from $\mathbb {R}^3$ to $\mathbb {R}^3$ can be represented using this operator. This notation can be extended to represent any linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$. For any ${\varvec{a}}$, ${\varvec{b}}$, ${\varvec{c}}$, ${\varvec{d}}\in \mathbb {R}^3$, let ${\varvec{a}}\otimes {\varvec{b}}\otimes {\varvec{c}}\otimes {\varvec{d}}$ be an operator satisfying

$$\begin{aligned} ({\varvec{a}}\otimes {\varvec{b}}\otimes {\varvec{c}}\otimes {\varvec{d}})({\varvec{e}}\otimes {\varvec{f}})=({\varvec{d}}\cdot {\varvec{e}})({\varvec{c}}\cdot {\varvec{f}}){\varvec{a}}\otimes {\varvec{b}}, \end{aligned}$$

(3.34)

where ${\varvec{e}}$, ${\varvec{f}}\in \mathbb {R}^3$. Then ${\varvec{a}}\otimes {\varvec{b}}\otimes {\varvec{c}}\otimes {\varvec{d}}$ is a linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ and any linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ can be represented using this operator.

By direct computation done by using the above tensor notation, we found an elliptic system form of (3.28). We found the following form of (3.28)

$$\begin{aligned}&\mathrm{div}\left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\left( {\varvec{I}}-\frac{ {u_0^+}^2}{{c_0^+}^2}(\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I} )\right) D(\varPsi \mathbf{e}_{\varphi })\right) \nonumber \\&\quad -\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\varPsi \mathbf{e}_{\varphi }=-{\varvec{F}}\quad \text{ in }\quad \mathcal {N}_f^+, \end{aligned}$$

(3.35)

where ${\varvec{I}}$ is the identity map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ and $\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I}$ is a linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ satisfing

$$\begin{aligned} (\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r\otimes \mathcal {I})({\varvec{a}}\otimes {\varvec{b}})=({\varvec{b}}\cdot \mathbf{e}_r){\varvec{a}}\otimes \mathbf{e}_r \end{aligned}$$

for any ${\varvec{a}}$, ${\varvec{b}}\in R^3$ (see the definition of $\mathcal {I}\otimes {\varvec{a}}\otimes {\varvec{b}} \otimes \mathcal {I}$ for any ${\varvec{a}}$, ${\varvec{b}}\in \mathbb {R}^3$ in (5.4)). By $M_0^+<1$ in $\mathcal {N}_f^+$ and the boundedness of $(\rho _0^+,u_0^+,p_0^+)$ in $\mathcal {N}_f^+$ for $\mathcal {N}_f^+\subset \mathcal {N}_{r_s-\delta _1}^+$, there exist positive constants $\mu$ and $\mathcal {M}$ such that

$$\begin{aligned} \mu |{\varvec{\xi }}|^2\le & {} \sum _{\alpha ,\beta ,i,j=1}^3 \left[ \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\left( {\varvec{I}}-\frac{ {u_0^+}^2}{{c_0^+}^2}(\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I} )\right) \right] _{ij}^{\alpha \beta }\xi ^i_\alpha \xi ^j_\beta \nonumber \\\le & {} \mathcal {M}|{\varvec{\xi }}|^2 \quad \text{ in }\quad \mathcal {N}_f^+ \end{aligned}$$

(3.36)

for any ${\varvec{\xi }}\in \mathbb {R}^{3\times 3}$ where $|{\varvec{\xi }}|=\sqrt{\sum _{i,j=1}^3|\xi _j^i|^2}$ with ${\varvec{\xi }}=[\xi _j^i]$. And by $\partial _r \rho _0^+>0$ in $\mathcal {N}_f^+$ for $\mathcal {N}_f^+\subset \mathcal {N}_{r_s-\delta _1}^+$ (see Remark 2),

$$\begin{aligned} \frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}> 0\quad \text{ in }\quad \mathcal {N}_f^+. \end{aligned}$$

Hence, (3.35) is a form of a solvable elliptic system for a Dirichlet boundary condition.

We obtain the unique existence and regularity of solutions of (3.28), (3.29) by obtaining those of solutions of (3.35), (3.29) as a boundary value problem for an elliptic system. The result of the unique existence and regularity of solutions of (3.35), (3.29) as a boundary value problem for an elliptic system is given in the following lemma:

Lemma 8

Under the assumptions as in Proposition 7, the boundary value problem

$$\begin{aligned}&\mathrm{div}\left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\left( {\varvec{I}}-\frac{ {u_0^+}^2}{{c_0^+}^2}(\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I} )\right) D{\varvec{U}}\right) -\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}{\varvec{U}}=-{\varvec{F}}\nonumber \\&\text{ in }\quad \mathcal {N}_f^+,\end{aligned}$$

(3.37)

$$\begin{aligned}&{{\varvec{U}}}={\left\{ \begin{array}{ll} h_1\mathbf{e}_{\varphi }\quad \text{ on } \quad \varGamma _f,\\ \frac{f(\theta _1)h_1(f(\theta _1),\theta _1)}{r}\mathbf{e}_{\varphi } \quad \text{ on }\quad \varGamma _w^+,\\ h_2\mathbf{e}_{\varphi }\quad \text{ on } \quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(3.38)

has a unique $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution ${\varvec{U}}$, and this solution ${\varvec{U}}$ satisfies

$$\begin{aligned}&||{\varvec{U}}||_{2,\alpha ,\mathcal {N}_f^+}^{(-1-\alpha ,\varGamma _w^+)} \nonumber \\&\quad \le \underbrace{C\left( ||{\varvec{F}}||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha , \varGamma _w^+)}+\sum _{i=1,2,3}F_i +||h_1\mathbf{e}_{\varphi }||_{2,\alpha ,\varGamma _f}^{(-1-\alpha ,\partial \varGamma _f)}+||h_2\mathbf{e}_{\varphi }||_{2,\alpha ,\varGamma _{ex}}^{(-1-\alpha ,\partial \varGamma _{ex})}\right) }_{=:C^*},\nonumber \\ \end{aligned}$$

(3.39)

where C is a positive constant depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $r_s$, $r_1$, $\theta _1$ and $\alpha$ and $F_i$ for $i=1,2,3$ are constants given in Lemma 11. This solution ${\varvec{U}}$ is of the form $\varPsi (r,\theta )\mathbf{e}_{\varphi }$.

One can see that that Lemma 7 is obtained from Lemma 8. To prove Lemma 7, in the remainder of this subsection, we prove Lemma 8.

To prove Lemma 8, we transform (3.37), (3.38) into the ${\varvec{0}}$ boundary problem

$$\begin{aligned}&\mathrm{div}({\varvec{A}}D {\varvec{U}}^\sharp )-d {\varvec{U}}^\sharp =-{\varvec{F}}-\mathrm{div}({\varvec{A}}D {\varvec{h}})+ d{\varvec{h}} (=:{\varvec{F}}^\sharp ) \quad \text{ in }\quad \mathcal {N}_{f}^+,\end{aligned}$$

(3.40)

$$\begin{aligned}&{\varvec{U}}^\sharp =0\quad \text{ on }\quad \partial \mathcal {N}_{f}^+, \end{aligned}$$

(3.41)

where ${\varvec{U}}^\sharp :={\varvec{U}}-{\varvec{h}}$, ${\varvec{A}}=\frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\left( {\varvec{I}}-\frac{ {u_0^+}^2}{{c_0^+}^2}(\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I} )\right) ,$ $d:=\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}$ and ${\varvec{h}}:=\frac{(r-f(\theta ))\frac{r_1}{r} h_2\mathbf{e}_{\varphi }+(r_1-r)\frac{f(\theta )}{r} h_1\mathbf{e}_{\varphi }}{r_1-f(\theta )}$. One can see that Lemma 8 can be proved by showing (3.40), (3.41) has a unique weak solution, the weak solution of (3.40), (3.41) is in $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ and the $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41) is of the form $\varPsi (r,\theta )\mathbf{e}_\varphi$. Hereafter, we prove these statements.

We first prove the unique existence of weak solution of (3.40), (3.41).

Lemma 9

Under the assumptions as in Proposition 7, the boundary value problem (3.40), (3.41) has a unique weak solution ${\varvec{U}}^\sharp \in H_0^1(\mathcal {N}_f^+)$. Furthermore, ${\varvec{U}}^\sharp$ satisfies

$$\begin{aligned} ||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_{f}^+)}\le C||{\varvec{F}}^\sharp ||_{L^2(\mathcal {N}_{f}^+)}, \end{aligned}$$

where C is a positive constant depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$ and $\mathcal {N}_{f}^+$.

Proof

Write (3.40), (3.41) in the form

$$\begin{aligned} (B[{\varvec{U}}^\sharp ,{\varvec{\xi }}]:=)\int _{\mathcal {N}_{f}^+}{\varvec{A}}D{\varvec{U}}^\sharp D{\varvec{\xi }}+d{\varvec{U}}^\sharp {\varvec{\xi }} =\int _{\mathcal {N}_{f}^+}{\varvec{F}}^\sharp {\varvec{\xi }}(=:<{\varvec{F}}^\sharp ,{\varvec{\xi }}>) \end{aligned}$$

(3.42)

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{f}^+)$. Then B is a bilinear map satisfying $B[{\varvec{U}}^\sharp ,{\varvec{\xi }}]\le C$ $||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_{f}^+)} ||{\varvec{\xi }}||_{H^1(\mathcal {N}_{f}^+)}$ for a constant $C>0$ and

$$\begin{aligned} {\underline{\mu }}||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_{f}^+)}^2\le B[{\varvec{U}}^\sharp ,{\varvec{U}}^\sharp ], \end{aligned}$$

where ${\underline{\mu }}=\min ( \mu ,\min _{\mathcal {N}_f^+}d)>0$ with $\mu$ given in (3.36). And since $h_1 \mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\partial \varGamma _f)}$ $(\varGamma _f)$, $h_2 \mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\partial \varGamma _{ex})}(\varGamma _{ex})$ and ${\varvec{F}}\in C^{\alpha }_{(1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ for $\alpha \in (\frac{2}{3},1)$, ${\varvec{F}}^\sharp \in L^2(\mathcal {N}_{f}^+)$ and thus $<{\varvec{F}}^\sharp ,{\varvec{\xi }}>$ is a bounded linear functional on $H_0^1(\mathcal {N}_f^+)$. With these facts, we apply the Lax-Milgram Theorem to (3.42). Then we obtain that there exists a unique ${\varvec{U}}^\sharp \in H_0^1(\mathcal {N}_f^+)$ such that (3.42) holds for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{f}^+)$. This finishes the proof. $\square$

We next prove that this weak solution is in $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. For this, we prove that the weak solution of (3.40), (3.41) is in $C^\beta (\overline{\mathcal {N}_f^+})$ for any $\beta \in (0,1)$ and $C^{1,\alpha }(\overline{\mathcal {N}_f^+})$.

Lemma 10

Under the assumptions as in Proposition 7, let ${\varvec{U}}^\sharp$ be the weak solution of the boundary value problem (3.40), (3.41). Then for any $\beta \in (0,1)$,

$$\begin{aligned} ||{\varvec{U}}^\sharp ||_{\beta ,\mathcal {N}_{f}^+}\le C\left( ||{\varvec{F}}^\sharp ||_{L^p(\mathcal {N}_{f}^+)}+||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}\right) \end{aligned}$$

for $p=\frac{3}{2-\beta }$ where C is a positive constant depending on $\mu$, $\mathcal {M}$, $\tau$, $||d||_{L^3(\mathcal {N}_f^+)}$ and $\mathcal {N}_f^+$, and $\tau$ is the modulus of continuity of ${\varvec{A}}$ in $\mathcal {N}_f^+$ given as

$$\begin{aligned} \tau (t)=\mathop {\sup _{\mathrm{x},\mathrm{y}\in \mathcal {N}_f^+,}}_{|\mathrm{x}-\mathrm{y}|\le t}(\sum |A^{\alpha \beta }_{ij}(\mathrm{x})-A^{\alpha \beta }_{ij}(\mathrm{y})|^2)^{\frac{1}{2}}. \end{aligned}$$

(3.43)

Lemma 11

Under the assumptions as in Proposition 7, let ${\varvec{U}}^\sharp$ be the weak solution of (3.40), (3.41). Then

$$\begin{aligned} ||{\varvec{U}}^\sharp ||_{1,\alpha ,\mathcal {N}_f^+}\le C\left( \sum _{i=1,2,3}F_i +||{\varvec{h}}||_{1,\alpha ,\mathcal {N}_f^+ }+||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}\right) , \end{aligned}$$

where C is a positive constant depending on $\mu$, $\mathcal {M}$, $||{\varvec{A}}||_{\alpha ,\mathcal {N}_f^+}$, $||d||_{L^q(\mathcal {N}_f^+)}$ with $q=\frac{3}{1-\alpha }$ and $\mathcal {N}_f^+$ and $F_i$ for $i=1,2,3$ are constants given in Lemma 11.

We will prove Lemmas 10 and 11 using the method of freezing the coefficients (Korn’s device of freezing the coefficients) (see [15, Chapter 3]). Since $\mathcal {N}_f^+$ is a Lipshitz domain, ${\varvec{U}}^\sharp \in C^\beta (\overline{\mathcal {N}_f^+})$ and ${\varvec{U}}^\sharp \in C^{1,\alpha }(\overline{\mathcal {N}_f^+})$ can be proved by showing that (i) there are positive constants C and R such that

$$\begin{aligned} \int _{D_{t}(\mathrm{x}_0)}|D{\varvec{U}}^\sharp |^2\le C {t}^{3-2+2\beta }\quad \text{ for } \text{ any }\quad 0<t<R \end{aligned}$$

(3.44)

for all $\mathrm{x}_0\in \overline{\mathcal {N}_{f}^+}$, and (ii) there are positive constants C and R such that

$$\begin{aligned} \int _{D_{t}(\mathrm{x}_0)} |D{\varvec{U}}^\sharp -(D{\varvec{U}}^\sharp )_{\mathrm{x}_0,{t}}|^2\le C{t}^{3+2\alpha }\quad \text{ for } \text{ any }\quad 0<{t}< R \end{aligned}$$

(3.45)

for all $\mathrm{x}_0\in \overline{\mathcal {N}_{f}^+}$ where $D_{t}(\mathrm{x}_0):=B_{t}(\mathrm{x}_0)\cap \mathcal {N}_{f}^+$ with $B_t(\mathrm{x}_0):=\{\mathrm{x}\in \mathbb {R}^3:|\mathrm{x}-\mathrm{x}_0|<t\}$ and $(D{\varvec{U}}^\sharp )_{\mathrm{x}_0,{t}}:=\frac{1}{|D_{t}(\mathrm{x}_0)|}\int _{D_{t}(\mathrm{x}_0)}D{\varvec{U}}^\sharp$. We prove (i) and (ii) by obtaining (3.44) and (3.45) at each point $\mathrm{x}_0$ in $\overline{\mathcal {N}_f^+}$ for C and R independent of $\mathrm{x}_0$ using the method of freezing the coefficients. When we do this, there exists some difficulty. For the case of $\mathrm{x}_0\in \mathcal {N}_f^+$ or $\varGamma _f\cup \varGamma _w^+\cup \varGamma _{ex}$, we can obtain the integral estimates for the fixed coefficients equation using the Cacciopolli inequality and the quotient difference method, and obtain (3.44) and (3.45) at $x_0\in \mathcal {N}_f^+$ or $\varGamma _f\cup \varGamma _w^+\cup \varGamma _{ex}$ using these estimates and the method of freezing the coefficients (see [1, Chapter 6]). But for the case of $\mathrm{x}_0\in \overline{\varGamma _f}\cap \overline{\varGamma _w^+}$ or $\overline{\varGamma _w^+}\cap \overline{\varGamma _{ex}}$, we cannot obtain the integral estimates for the fixed coefficients equation using the Cacciopolli inequality and the quotient difference method. Thus, we cannot obtain (3.44) and (3.45) at $\mathrm{x}_0\in \overline{\varGamma _f}\cap \overline{\varGamma _w^+}$ or $\overline{\varGamma _w^+}\cap \overline{\varGamma _{ex}}$ using the standard method of freezing the coefficients. We resolve this difficulty by developing some reflection argument that holds for a linear boundary value problem for an elliptic system in the case that the domain is a Lipshitz domain whose all non-differentiable parts are perpendicular corners, and the domain part of the principal coefficients of the elliptic system is diagonal with respect to the coordinate systems representing the walls near the corners of the domain. This will be seen in the proof of Lemmas 10 and 11.

Hereafter, we use the following notation:

$$\begin{aligned}&\mathrm{div}_{(r,\theta ,\varphi )}:=(\mathbf{e}_1\partial _r +\mathbf{e}_2\partial _\theta +\mathbf{e}_3\partial _\varphi )\cdot ,\quad D_{(r,\theta ,\varphi )}:=\mathbf{e}_1\partial _r + \mathbf{e}_2\partial _\theta + \mathbf{e}_3\partial _\varphi \\&{\mathbb {T}}:\;\text{ a } \text{ one } \text{ dimensional } \text{ torus } \text{ with } \text{ period }\;2\pi ,\\&\mathcal {N}_{a}^{+,*}:=\{(r,\theta ,\varphi )\in \mathbb {R}^3\;|\;a<r<r_1,0<\theta<\theta _1, \varphi \in {\mathbb {T}}\},\\&\varGamma _{a}^{*}:=\{(r,\theta ,\varphi )\in \mathbb {R}^3\;|\;r=a,0<\theta<\theta _1, \varphi \in {\mathbb {T}}\},\\&\varGamma _{w,a}^{+,*}:=\{(r,\theta ,\varphi )\in \mathbb {R}^3\;|\;a<r<r_1,\theta =\theta _1, \varphi \in {\mathbb {T}}\}\quad \text{ for }\quad 0<a<r_1\\&\varGamma _w^{+,*}:=\{(r,\theta ,\varphi )\in \mathbb {R}^3\;|\;f(\theta _1)<r<r_1,\theta =\theta _1, \varphi \in {\mathbb {T}}\},\\&\varGamma _{ex}^*:=\{(r,\theta ,\varphi )\in \mathbb {R}^3\;|\;r=r_1,0<\theta<\theta _1, \varphi \in {\mathbb {T}}\} \\&{\mathfrak r}, {\mathfrak t}:\; \text{ a } \text{ radius } \text{ of } \text{ a } \text{ ball } \text{ in } \text{ the } \text{ spherical } \text{ coordinate } \text{ system },\\&B_{\mathfrak r}^*(\mathrm{x}^*):=\{(r,\theta ,\varphi )\in \mathbb {R}^+\times [0,\pi ]\times \ {\mathbb {T}}:\\&\qquad \qquad \quad |r-r^*|^2+|\theta -\theta ^*|^2+|\varphi -\varphi ^*|^2<{\mathfrak r}^2\},\\&D_{\mathfrak r}^*(\mathrm{x}^*):=B_{\mathfrak r}^*(\mathrm{x}^*)\cap \mathcal {N}_{f(\theta _1)}^{+,*} \quad \text{ for }\quad \mathrm{x}^*=(r^*,\theta ^*,\varphi ^*)\in \mathbb {R}^+\times (0,\pi )\times \ {\mathbb {T}}. \end{aligned}$$

To prove Lemma 10 and Lemma 11, we prove the following:

Lemma 12

Let $\mathrm{x}_0^*= (f(\theta _1),\theta _1,\varphi _0)$ for some $\varphi _0\in {\mathbb {T}}$. Let $0<\mathfrak r< \min (\theta _1,\pi ,$ $r_1-f(\theta _1))$. Suppose that ${\varvec{W}}\in H^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$ is a weak solution of

$$\begin{aligned}&\mathrm{div}_{(r,\theta ,\varphi )}\left( \left. \bigg (\frac{r^2\sin \theta }{\rho _0^+}\right| _{r=f(\theta _1),\theta =\theta _1}\mathcal {I} \otimes \mathbf{e}_1\otimes \mathbf{e}_1\otimes \mathcal {I}\right. \nonumber \\&\qquad +\left. \frac{\sin \theta }{\rho _0^+} (\frac{{c_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\right| _{r=f(\theta _1),\theta =\theta _1} \mathcal {I}\otimes \mathbf{e}_2\otimes \mathbf{e}_2\otimes \mathcal {I}\nonumber \\&\qquad \left. +\left. \frac{1}{\sin \theta \rho _0^+}(\frac{{c_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\right| _{r=f(\theta _1),\theta =\theta _1} \mathcal {I}\otimes \mathbf{e}_3\otimes \mathbf{e}_3 \otimes \mathcal {I} \bigg ) D_{(r,\theta ,\varphi )} {\varvec{W}}\right) =0\nonumber \\&\qquad \text{ in }\quad D_{\mathfrak r}^*(\mathrm{x}_0^*),\end{aligned}$$

(3.46)

$$\begin{aligned}&{\varvec{W}}=\mathbf{0} \quad \text{ on }\quad \partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}}). \end{aligned}$$

(3.47)

Then for any $\mathfrak t$ such that $0<\mathfrak t\le \mathfrak r$, there hold

$$\begin{aligned} \int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} {\varvec{W}}|^2\le C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^3\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} {\varvec{W}}|^2, \end{aligned}$$

(3.48)

and

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}{\varvec{W}}-(D_{(r,\theta ,\varphi )}{\varvec{W}})^*_{\mathrm{x}_0^*,\mathfrak t}|^2 \nonumber \\&\quad \le C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^{5}\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}{\varvec{W}}-(D_{(r,\theta ,\varphi )}{\varvec{W}})^*_{\mathrm{x}_0^*,\mathfrak r}|^2, \end{aligned}$$

(3.49)

where C is a positive constant depending on $\mu$, $\mathcal {M}$, $f(\theta _1)$ and $\theta _1$ and $(D_{(r,\theta ,\varphi )}$ ${\varvec{W}})^*_{\mathrm{x}_0^*,\mathfrak r}:=\frac{1}{|D_{\mathfrak r}^*(\mathrm{x}_0^*)|}\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}D_{(r,\theta ,\varphi )}{\varvec{W}}$.

Proof

The result is obtained by using the reflection argument.

Extend (3.46) in $B_{\mathfrak r}^*(\mathrm{x}_0^*)$:

$$\begin{aligned}&\mathrm{div}_{(r,\theta ,\varphi )}\left( \left. \bigg (\frac{r^2\sin \theta }{\rho _0^+}\right| _{r=f(\theta _1),\theta =\theta _1}\right. \mathcal {I}\otimes \mathbf{e}_1\otimes \mathbf{e}_1\otimes \mathcal {I} \nonumber \\&\qquad +\left. \frac{\sin \theta }{\rho _0^+}\left( \frac{{c_0^+}^2}{{c_0^+}^2-{u_0^+}^2}\right) \right| _{r=f(\theta _1),\theta =\theta _1}\mathcal {I}\otimes \mathbf{e}_2\otimes \mathbf{e}_2 \otimes \mathcal {I}\nonumber \\&\qquad \left. +\left. \frac{1}{\sin \theta \rho _0^+}\left( \frac{{c_0^+}^2}{{c_0^+}^2-{u_0^+}^2}\right) \right| _{r=f(\theta _1),\theta =\theta _1}\mathcal {I}\otimes \mathbf{e}_3\otimes \mathbf{e}_3 \otimes \mathcal {I}\bigg ) D_{(r,\theta ,\varphi )} {\varvec{W}}\right) =0\nonumber \\&\qquad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*), \end{aligned}$$

(3.50)

and extend a weak solution ${\varvec{W}}$ of (3.46), (3.47) in $B_{\mathfrak r}^*(\mathrm{x}_0^*)$:

$$\begin{aligned} {\varvec{W}}_{ext}={\left\{ \begin{array}{ll}{\varvec{W}}(r, \theta ,\varphi )\quad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*)\cap \{r\ge f(\theta _1),\theta \le \theta _1\} \\ -{\varvec{W}}(2f(\theta _1)-r,\theta ,\varphi )\quad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*)\cap \{r< f(\theta _1),\theta \le \theta _1\}\\ -{\varvec{W}}(r,2\theta _1-\theta ,\varphi )\quad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*)\cap \{r\ge f(\theta _1),\theta>\theta _1\}\\ {\varvec{W}}(2f(\theta _1)-r,2\theta _1-\theta ,\varphi )\quad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*)\cap \{r< f(\theta _1),\theta > \theta _1\}. \end{array}\right. } \end{aligned}$$

Then ${{\varvec{W}}}_{ext}\in H^1(B_{\mathfrak r}^*(\mathrm{x}_0^*))$. One can see that there exists a unique weak solution of (3.50),

$$\begin{aligned} {\varvec{W}}={\varvec{W}}_{ext}\quad \text{ on }\quad \partial B_{\mathfrak r}^*(\mathrm{x}_0^*). \end{aligned}$$

(3.51)

We denote the weak solution of (3.50), (3.51) by $\overline{{\varvec{W}}}$.

One can check that $-\overline{\varvec{W}}(2f(\theta _1)-r,\theta ,\varphi )$ and $-\overline{\varvec{W}}(r,2\theta _1-\theta ,\varphi )$ are also weak solutions of (3.50), (3.51). By this fact and the uniqueness of weak solutions of (3.50), (3.51),

$$\begin{aligned} \overline{\varvec{W}}(r, \theta ,\varphi )= -\overline{\varvec{W}}(2f(\theta _1)-r,\theta ,\varphi )=-\overline{\varvec{W}}(r,2\theta _1-\theta ,\varphi ) \quad \text{ in }\quad B_{\mathfrak r}^*(\mathrm{x}_0^*). \end{aligned}$$

(3.52)

From (3.52), we have $\overline{\varvec{W}}=0$ on $\partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}})$. By this fact and the fact that $\overline{{\varvec{W}}}$ is a unique weak solution of (3.50), (3.51), $\overline{\varvec{W}}$ is a weak solution of (3.46),

$$\begin{aligned} \overline{\varvec{W}}={\left\{ \begin{array}{ll} {\varvec{W}}\quad \text{ on }\quad \partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\setminus (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}}),\\ 0\quad \text{ on }\quad \partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}}). \end{array}\right. } \end{aligned}$$

By this fact and the uniqueness of weak solutions of (3.46) satisfying ${\varvec{W}}={\varvec{W}}_{ext}$ on $\partial D_{\mathfrak r}^*(\mathrm{x}_0^*)$ $\setminus (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}})$ and ${\varvec{W}}=0$ on $\partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}})$, we have

$$\begin{aligned} \overline{{\varvec{W}}}={\varvec{W}}\quad \text{ in }\quad D_{\mathfrak r}^*(\mathrm{x}_0^*). \end{aligned}$$

(3.53)

By [15, Theorem 2.1, Chapter 3], $\overline{{\varvec{W}}}$ satisfies for any $0<\mathfrak t\le \mathfrak r$,

$$\begin{aligned} \int _{B_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} \overline{\varvec{W}} |^2\le C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^3\int _{B_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} \overline{\varvec{W}} |^2 \end{aligned}$$

(3.54)

and

$$\begin{aligned}&\int _{B_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\overline{\varvec{W}} -(D_{(r,\theta ,\varphi )}\overline{\varvec{W}} )^{**}_{\tilde{\mathrm{x}}_0^{*},\mathfrak t}|^2\nonumber \\&\qquad \le C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^{5}\int _{B_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\overline{\varvec{W}} -(D_{(r,\theta ,\varphi )}\overline{\varvec{W}} )^{**}_{\tilde{\mathrm{x}}_0^{*},\mathfrak r}|^2, \end{aligned}$$

(3.55)

where C is a positive constant depending on $\mu$, $\mathcal {M}$, $f(\theta _1)$ and $\theta _1$ and $(D_{(r,\theta ,\varphi )}$ $\overline{\varvec{W}} )^{**}_{\mathrm{x}_0^*,\mathfrak r}:=\frac{1}{|B_{\mathfrak r}^*(\mathrm{x}_0^*)|}\int _{B_{\mathfrak r}^*(\mathrm{x}_0^*)}D_{(r,\theta ,\varphi )}\overline{\varvec{W}}$. By (3.52) and (3.53), we obtain from (3.54) and (3.55) for any $0<\mathfrak t\le \mathfrak r$,

$$\begin{aligned} 4\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} {\varvec{W}}|^2\le 4C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^3\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )} {\varvec{W}}|^2 \end{aligned}$$

and

$$\begin{aligned}&4\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}{\varvec{W}}-(D_{(r,\theta ,\varphi )}{\varvec{W}})^*_{\mathrm{x}_0^*,\mathfrak t}|^2\\&\qquad \qquad \le 4C\left( \frac{\mathfrak t}{\mathfrak r}\right) ^{5}\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}{\varvec{W}}-(D_{(r,\theta ,\varphi )}{\varvec{W}})^*_{\mathrm{x}_0^*,\mathfrak r}|^2, \end{aligned}$$

where we used

$$\begin{aligned} (D_{(r,\theta ,\varphi )}\overline{\varvec{W}})^{**}_{\mathrm{x}_0^*,\mathfrak r}=\frac{1}{|B_{\mathfrak r}^*(\mathrm{x}_0^*)|}\int _{B_{\mathfrak r}^*(\mathrm{x}_0^*)}D_{(r,\theta ,\varphi )}\overline{\varvec{W}}=\frac{1}{|D_{\mathfrak r}^*(\mathrm{x}_0^*)|}\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}D_{(r,\theta ,\varphi )}{\varvec{W}}. \end{aligned}$$

This finishes the proof. $\square$

The next corollary is obtained from Lemma 12 in the same way that Corollary 3.11 is obtained from Lemma 3.10 in [17]. We omit the proof.

Corollary 1

Suppose that ${\varvec{W}}$ is as in Lemma 12. Let $\tilde{{\varvec{U}}}^*$ be any function in $H^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$ for $0<\mathfrak r< \min (\theta _1,\pi ,r_1-f(\theta _1))$. Then for any $\mathfrak t$ and $\mathfrak r$ such that $0<\mathfrak t\le \mathfrak r< \min (\theta _1,\pi ,r_1-f(\theta _1))$, it holds that

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^*|^2\\&\quad \le C\left( \left( \frac{\mathfrak t}{\mathfrak r}\right) ^3\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^*|^2+\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}(\tilde{{\varvec{U}}}^*-\mathbf{W})|^2\right) \end{aligned}$$

and

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^*-(D_{(r,\theta ,\varphi )} \tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak t}|^2\\&\quad \le C\bigg (\left( \frac{\mathfrak t}{\mathfrak r}\right) ^5 \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^* -(D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak r}|^2\\&\qquad \quad +\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}(\tilde{{\varvec{U}}}^*-\mathbf{W})|^2\bigg ), \end{aligned}$$

where C is a positive constant depending on $\mu$, $\mathcal {M}$, $f(\theta _1)$ and $\theta _1$.

We first prove Lemma 10.

Proof of Lemma 10

We prove Lemma 10 by proving that (3.44) holds for all $\mathrm{x}_0\in \overline{\mathcal {N}_{f}^+}$ for some positive constants C and R.

1. Transform (3.40), (3.41) into a problem in $\mathcal {N}_{f(\theta _1)}^+$.

Define a map

$$\begin{aligned} \varPi _{ab}^*(r,\theta ,\varphi ):= \left( \frac{r_1-b}{r_1-a}(r-a)+b,\theta ,\varphi \right) \end{aligned}$$

(3.56)

for $0<a,b<r_1$. Then $\varPi _{ab}^*(r,\theta ,\varphi )$ maps $(a,r_1)\times (0,\theta _1)\times {\mathbb {T}}$ to $(b,r_1)\times (0,\theta _1)\times {\mathbb {T}}$. $\varPi ^*_{ab}$ naturally induces a map from $\mathcal {N}_a^+$ to $\mathcal {N}_b^+$ as a map between two Cartesian coordinate systems. Denote this map by $\varPi _{ab}$. Using $\varPi :=\varPi _{f(\theta )f(\theta _1)}$, we transform the boundary value problem (3.40), (3.41) into the boundary value problem

$$\begin{aligned}&\mathrm{div}_{\mathrm{y}} (\tilde{\varvec{A}}D_{\mathrm{y}} {\tilde{{\varvec{U}}}}^\sharp )-\tilde{d} {\tilde{{\varvec{U}}}}^\sharp =\tilde{\varvec{F}}^\sharp \quad \text{ in }\quad \mathcal {N}_{f(\theta _1)}^+,\end{aligned}$$

(3.57)

$$\begin{aligned}&\tilde{{\varvec{U}}}^\sharp =\mathbf{0}\quad \text{ on }\quad \partial \mathcal {N}_{f(\theta _1)}^+, \end{aligned}$$

(3.58)

where $\tilde{\varvec{A}}:=\frac{1}{\det (\frac{\partial \varPi }{\partial \mathrm{x}})}(\frac{\partial \mathrm{\varPi }}{\partial \mathrm{x}})^T( {\varvec{A}}\circ \varPi ^{-1})(\frac{\partial \mathrm{\varPi }}{\partial \mathrm{x}})$, $\tilde{d}:=\frac{d\circ \varPi ^{-1}}{\det (\frac{\partial \mathrm{\varPi }}{\partial \mathrm{x}})}$, ${\tilde{{\varvec{U}}}}^\sharp :={\varvec{U}}^\sharp \circ \varPi ^{-1}$, $\tilde{\varvec{F}}^\sharp :=\frac{ {\varvec{F}}^\sharp \circ \varPi ^{-1}}{\det (\frac{\partial \varPi }{\partial \mathrm{x}})}$ and $\mathrm{x}$ and $\mathrm{y}$ are the Cartesian coordinate systems for $\mathcal {N}_f^+$ and $\mathcal {N}_{f(\theta _1)}^+$, respectively.

2. Transform the weak formulation of (3.57), (3.58) near $\mathrm{x}_0\in \overline{\varGamma _{f(\theta _1)}}\cap \overline{\varGamma _w^+}$ into the weak formulation of the spherical coordinate representation of (3.57), (3.58).

Write (3.57), (3.58) in the form

$$\begin{aligned} \int _{\mathcal {N}_{f(\theta _1)}^+}\tilde{\varvec{A}}D_{\mathrm{y}}{\tilde{{\varvec{U}}}}^\sharp D_{\mathrm{y}}{\varvec{\xi }}+\tilde{d} {\tilde{{\varvec{U}}}}^\sharp {\varvec{\xi }}=-\int _{\mathcal {N}_{f(\theta _1)}^+}\tilde{\varvec{F}}^\sharp {\varvec{\xi }} \end{aligned}$$

(3.59)

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{f(\theta _1)}^+)$. Let $(\tilde{r},\tilde{\theta },{\tilde{\varphi }})$ be the spherical coordinate system for $\mathrm{y}$ and $\varXi$ be the map from $\mathrm{y}$ to $(\tilde{r},\tilde{\theta },{\tilde{\varphi }})$. Choose any $\mathrm{x}_0^*:= (f(\theta _1),\theta _1,\varphi _0)$ for some $\varphi _0\in {\mathbb {T}}$ and set ${\varvec{\xi }}=0$ outside of $\varXi ^{-1}(D_{{\mathfrak r}}^*(\mathrm{x}_0^*))$ for $0<\mathfrak r<\min (\theta _1,\pi ,r_1-f(\theta _1))$ in (3.59). Then we obtain

$$\begin{aligned} \int _{\varXi ^{-1}(D_{\mathfrak r}^*(\mathrm{x}_0^*))}\tilde{\varvec{A}}D_{\mathrm{y}}{\tilde{{\varvec{U}}}}^\sharp D_{\mathrm{y}}{\varvec{\xi }}+\tilde{d} {\tilde{{\varvec{U}}}}^\sharp {\varvec{\xi }}=-\int _{\varXi ^{-1}(D_{\mathfrak r}^*(\mathrm{x}_0^*))}\tilde{\varvec{F}}^\sharp {\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(\varXi ^{-1}(D_{\mathfrak r}^*(\mathrm{x}_0^*)))$. Using $\varXi$, transform this equation. Then we have

$$\begin{aligned} \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }}+\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{\xi }}=-\int _{D_{r}^*(\mathrm{x}_0^*)}\tilde{\varvec{F}}^*{\varvec{\xi }} \end{aligned}$$

(3.60)

for all ${\varvec{\xi }}\in H_0^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$ where ${\tilde{{\varvec{U}}}}^*={\tilde{{\varvec{U}}}}^\sharp \circ \varXi ^{-1}$, $\tilde{d}^*=\frac{\tilde{d}\circ \varXi ^{-1}}{\det {\tilde{M}}}$, $\tilde{\varvec{F}}^*=\frac{\tilde{\varvec{F}}^\sharp \circ \varXi ^{-1}}{\det {\tilde{M}}}$ and ${\tilde{M}}=\frac{\partial \varXi }{\partial y}$.

3. Obtain (3.44) at $\mathrm{x}_0\in \overline{\varGamma _f}\cap \overline{\varGamma _w^+}$.

Rewrite (3.60) as

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}} \circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})} \tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{\xi }}\\&\quad =\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)} (\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}})D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{\xi }}\\&\qquad -\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{\xi }}-\tilde{\varvec{F}}^*{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$. Fix the principal coefficients of the left hand side of the resultant equation at $\mathrm{x}_0^*$. Then we obtain

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*)D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})} \tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{\xi }} \nonumber \\&\quad =\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}\left( \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*) -\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) \nonumber \\&\qquad D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})} \tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{\xi }}\nonumber \\&\qquad + \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}} \circ \varXi ^{-1}{\tilde{M}}-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}} \circ \varXi ^{-1}{\tilde{M}}\right) D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{\xi }}\nonumber \\&\qquad -\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{\xi }}-\tilde{\varvec{F}}^*{\varvec{\xi }} \end{aligned}$$

(3.61)

for all ${\varvec{\xi }}\in H_0^1(D_{\mathfrak r}^*(\tilde{\mathrm{x}}^*_0))$. Using the argument in Appendix, it can be checked that $(\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}) (\mathrm{x}_0^*)$ is equal to the principal coefficients of the equation in (3.46).

Let ${\varvec{W}}$ be the weak solution of

$$\begin{aligned}&\mathrm{div}_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}((\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}})(\mathrm{x}_0^*) D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{W}})=0\quad \text{ in }\quad D_{\mathfrak r}^{*}(\mathrm{x}_0^*),\end{aligned}$$

(3.62)

$$\begin{aligned}&{\varvec{W}}={\left\{ \begin{array}{ll} \tilde{{\varvec{U}}}^*\quad \text{ on }\quad \partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap \mathcal {N}_{f(\theta _1)}^{+,*}, \\ \mathbf{0} \quad \text{ on }\quad \partial D_{\mathfrak r}^*(\mathrm{x}_0^*)\cap (\overline{\varGamma _{f(\theta _1)}^{*}}\cup \overline{\varGamma _{w,f(\theta _1)}^{+,*}}). \end{array}\right. } \end{aligned}$$

(3.63)

Subtract the weak formulation of (3.62), (3.63) from (3.61) and then take ${\varvec{\xi }}={\varvec{V}}$ to the resultant equation. Then we obtain

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}(\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}) (\mathrm{x}_0^*)D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})} {\varvec{V}}D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{V}} \\&\quad =\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}((\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}) (\mathrm{x}_0^*)-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}})D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})} \tilde{{\varvec{U}}}^*\\&D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{V}} + (\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}})\\&\qquad D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{V}}-\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{V}}-\tilde{\varvec{F}}^*{\varvec{V}}, \end{aligned}$$

where ${\varvec{V}}:={\tilde{{\varvec{U}}}}^*-{\varvec{W}}\in H_0^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$. Using the Sobolev and Hölder inequality, we obtain from this equation

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\varvec{V}}|^2\le C\left( (\tau _1^2(\mathfrak r)+\tau _2^2(\mathfrak r) )\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\tilde{{\varvec{U}}}}^*|^2\right. \nonumber \\&\qquad \left. +(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{d}^*|^3)^{\frac{2}{3}}\int _{D_r^*(\mathrm{x}_0^*)}|{\tilde{{\varvec{U}}}}^*|^2+(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{\varvec{F}}^*|^{\frac{6}{5}})^{\frac{5}{3}}\right) , \end{aligned}$$

(3.64)

where

$$\begin{aligned} \tau _1(\mathfrak r)= & {} \mathop {\sup _{\mathrm{x}^*,\mathrm{y}^*\in \mathcal {N}_f^{+,*}}}_{|\mathrm{x}^*-\mathrm{y}^*|\le {\mathfrak r}}\left| \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}^*)\right. \\&\left. \qquad -\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{y}^*)\right| \end{aligned}$$

and

$$\begin{aligned} \tau _2(\mathfrak r):=\sup _{|\theta -\theta _1|\le \mathfrak r} \left\{ |f(\theta )-f(\theta _1)|+|f^\prime (\theta )|\right\} \end{aligned}$$

(note that $f^\prime (\theta _1)=0$). By Corollary 1 and (3.64), we have for any $0<\mathfrak t\le \mathfrak r$,

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\le C \Bigg (\left( \left( \frac{\mathfrak t}{\mathfrak r}\right) ^3+\tau _1^2(\mathfrak r)+\tau _2^2(\mathfrak r)\right) \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\\&\qquad +(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{d}^*|^3)^{\frac{2}{3}}\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|{\tilde{{\varvec{U}}}}^*|^2+(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{\varvec{F}}^*|^{\frac{6}{5}})^{\frac{5}{3}}\Bigg ). \end{aligned}$$

Using the Hölder and Poincaré inequality, we get from this inequality

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\le C \left( \left( \left( \frac{\mathfrak t}{\mathfrak r}\right) ^3+\tau _1^2(\mathfrak r)+\tau _2^2(\mathfrak r)\right) \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\right. \nonumber \\&\qquad \left. +(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{d}^*|^3)^{\frac{2}{3}}{\mathfrak r}^2\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}{\tilde{{\varvec{U}}}}^*|^2+(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{\varvec{F}}^*|^p)^{\frac{2}{p}}{\mathfrak r}^{3-2+2\beta }\right) ,\nonumber \\ \end{aligned}$$

(3.65)

where $\beta =2-\frac{3}{p}\in (0,1)$ if $p\in (\frac{3}{2},3)$.

Depending on the value of $\beta$, we consider two cases.

Case 1: $3-2+2\beta \le 2$.

Apply Lemma 2.1 in [15, Chapter 3] to (3.65). Then we obtain that there exists $R_1\in (0,\min (\theta _1,\pi ,r_1-f(\theta _1)))$ such that for any $0<\mathfrak r\le R_1$,

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\nonumber \\&\quad \le C{\mathfrak r}^{3-2+2\beta }\left( \int _{D_{R_1}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2+(\int _{D_{R_1}^*(\mathrm{x}_0^*)}|\tilde{\varvec{F}}^*|^p)^\frac{2}{p}\right) . \end{aligned}$$

(3.66)

Case 2: $3-2+2\beta >2$.

As in the Case 1, we apply Lemma 2.1 in [15, Chapter 3] to (3.65). Then we obtain that there exists $R_2\in (0,\min (\theta _1,\pi ,r_1-f(\theta _1)))$ such that for any $0<\mathfrak r\le R_2$,

$$\begin{aligned} \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\le C{\mathfrak r}^2\left( \int _{D_{R_2}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta }, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2+(\int _{D_{R_2}^*(\mathrm{x}_0^*)}|\tilde{\varvec{F}}^*|^p)^\frac{2}{p}\right) . \end{aligned}$$

Substitute this into (3.65). After that, apply Lemma 2.1 in [15, Chapter 3] again. Then we obtain that there exists $R_3\le R_2$ such that for any $0<\mathfrak r\le R_3$, (3.66) holds with $R_1$ replaced by $R_2$.

From this result, we obtain that for any $\beta \in (0,1)$, there exists $R>0$ such that

$$\begin{aligned}&\int _{D_{ t}(\mathrm{x}_0) }|D {\varvec{U}}^\sharp |^2\le C{t}^{3-2+2\beta } \left( ||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^p(\mathcal {N}_f^+)}^2\right) \\&\quad \quad \quad \text{ for } \text{ any }\ 0<t<R \end{aligned}$$

for $\mathrm{x}_0\in \overline{\varGamma _f}\cap \overline{\varGamma _w^+}$. Since ${\varvec{F}}^\sharp \in C^{\alpha }_{(1-\alpha ,\varGamma _{w}^+)}(\mathcal {N}_{f}^{+})$ for $\alpha \in (\frac{2}{3},1)$, ${\varvec{F}}^\sharp \in L^3(\mathcal {N}_{f}^+)$. This implies that ${\varvec{F}}^\sharp \in L^p(\mathcal {N}_{f}^+)$ for any $\beta \in (0,1)$ where $p=\frac{3}{2-\beta }$. Hence, we obtain (3.44) with C replaced by $C_1=C(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^p(\mathcal {N}_f^+)}^2)$ for $\mathrm{x}_0\in \overline{\varGamma _f}\cap \overline{\varGamma _w^+}$.

4. When $\mathrm{x}_0$ is in $\overline{\varGamma _w^+}\cap \overline{\varGamma _{ex}}$, we obtain (3.44) with C replaced by $C_1$ using similar arguments without the process of transforming $\mathcal {N}_f^+$ into $\mathcal {N}_{f(\theta _1)}^+$. When $\mathrm{x}_0$ is in $\mathcal {N}_f^+$ or $\varGamma _f\cup \varGamma _w^+\cup \varGamma _{ex}$ and far away from the corners $\overline{\varGamma _f}\cap \overline{\varGamma _w^+}$ and $\overline{\varGamma _w^+}\cap \overline{\varGamma _{ex}}$, we obtain (3.44) with C replaced by $C_1$ using the standard method of freezing the coefficients. When $\mathrm{x}_0$ is in $\mathcal {N}_f^+$ or $\varGamma _f\cup \varGamma _w^+\cup \varGamma _{ex}$ and near $\overline{\varGamma _f}\cap \overline{\varGamma _w^+}$ or $\overline{\varGamma _w^+}\cap \overline{\varGamma _{ex}}$, we obtain (3.44) with C replaced by $C_1$ using the arguments in the proof of [16, Theorem 5.21] and arguments similar to the ones in Step 1-Step 3 above. Combining these results, we obtain that there exist a positive constant R such that ${\varvec{U}}^\sharp$ satisfies (3.44) for all $\mathrm{x}_0\in \overline{\mathcal {N}_f^+}$ for $C=C_1$. This finishes the proof. $\square$

Next, we prove Lemma 11. We prove Lemma 11 using the method of freezing the coefficients and the reflection arguments as we did in the proof of Lemma 10. When we do this, there exists some problem: since ${\varvec{F}}^\sharp$ is not in $L^p(\mathcal {N}_f^+)$ for $q=\frac{3}{1-\alpha }$ nor has the form $\mathrm{div}{\varvec{G}}$ with ${\varvec{G}}\in C^{\alpha }(\overline{\mathcal {N}_f^+})$, we cannot get the required power of t for the $C^{1,\alpha }$ estimate (see (3.45)) from the integral estimate of ${\varvec{F}}^\sharp$ directly. We obtain this power by delivering $\theta$-derivatives imposed on some functions in ${\varvec{F}}^\sharp$ to the functions multiplied to those functions in the integral form of ${\varvec{F}}^\sharp$ using integration by parts and estimating the resultant integral form of ${\varvec{F}}^\sharp$. To make our argument clear, we present the detailed proof.

Proof of Lemma 11

Using (3.31), write (3.40), (3.41) in the form

$$\begin{aligned}&\int _{\mathcal {N}_f^+}{\varvec{A}}D {\varvec{U}}^\sharp D{\varvec{\xi }} +d{\varvec{U}}^\sharp {\varvec{\xi }}\\&\quad =\int _{\mathcal {N}_f^+}\sum _i A^i\partial _r (B^i-B^i(\mathrm{x}_0)) \xi _\varphi +\sum _i C^i \partial _\theta (D^i-D^i(\mathrm{x}_0))\xi _\varphi \\&\qquad +E\frac{\partial _\theta (F\sin \theta )}{\sin \theta }\xi _\varphi +\mathrm{div}({\varvec{A}}D{\varvec{h}} ){\varvec{\xi }}-d{\varvec{h}}{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_f^+)$ where $\xi _\varphi ={\varvec{\xi }}\cdot \mathbf{e}_\varphi$. Using integration by parts, we change this equation into

$$\begin{aligned}&\int _{\mathcal {N}_f^+}{\varvec{A}}D {\varvec{U}}^\sharp D{\varvec{\xi }} +d{\varvec{U}}^\sharp {\varvec{\xi }}\\&\quad =\int _{\mathcal {N}_f^+}-\sum _i\left( \partial _r A^i (B^i-B^i(\mathrm{x}_0)) \xi _\varphi +A^i(B^i-B^i(\mathrm{x}_0))\frac{1}{r^2}\partial _r (r^2\xi _\varphi )\right) \\&\qquad -\sum _i\left( \partial _\theta C^i (D^i-D^i(\mathrm{x}_0))\xi _\varphi +C^i(D^i-D^i(\mathrm{x}_0))\frac{1}{\sin \theta }\partial _\theta (\xi _\varphi \sin \theta )\right) \\&\qquad +E\frac{\partial _\theta (F\sin \theta )}{\sin \theta }\xi _\varphi -{\varvec{A}}D{\varvec{h}}D{\varvec{\xi }} -d{\varvec{h}}{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_f^+)$. Using $\varPi$ defined in Step 1 in the proof of Lemma 10, transform this equation. Then we obtain

$$\begin{aligned} \int _{\mathcal {N}_{f(\theta _1)}^+}\tilde{\varvec{A}}D {\tilde{{\varvec{U}}}}^\sharp D{\varvec{\xi }} +\tilde{d}{\tilde{{\varvec{U}}}}^\sharp {\varvec{\xi }}= \int _{\mathcal {N}_{f(\theta _1)}^+}(a) \frac{1}{\det (\frac{d\mathrm{y}}{ d\mathrm{x}})} -\tilde{\varvec{A}}D\tilde{\varvec{h}}D{\varvec{\xi }} -\tilde{d}\tilde{\varvec{h}}{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{f(\theta _1)}^+)$ where

$$\begin{aligned} (a)=&-\sum _i \left( \frac{\partial \tilde{r}}{\partial r}\partial _{\tilde{r}}{\tilde{A}}^i({\tilde{B}}^i-B^i(\mathrm{x}_0)) \xi _\varphi +{\tilde{A}}^i({\tilde{B}}^i-B^i(\mathrm{x}_0))\frac{1}{r^2}\frac{\partial \tilde{r}}{\partial r}\partial _{\tilde{r}}(r^2\xi _\varphi ) \right) \\&-\sum _i\bigg ( (\frac{\partial \tilde{r}}{\partial \theta }\partial _{\tilde{r}}+\partial _{\tilde{\theta }}){\tilde{C}}^i ({\tilde{D}}^i-D^i(\mathrm{x}_0))\xi _\varphi +{\tilde{C}}^i({\tilde{D}}^i-D^i(\mathrm{x}_0))\\&\frac{1}{\sin \tilde{\theta }}(\frac{\partial \tilde{r}}{\partial \theta }\partial _{\tilde{r}}+\partial _{\tilde{\theta }})(\xi _\varphi \sin \tilde{\theta })\bigg ) +{\tilde{E}}\frac{1}{\sin {\tilde{\theta }}}(\frac{\partial \tilde{r}}{\partial \theta }\partial _{\tilde{r}}+\partial _{\tilde{\theta }})({\tilde{F}}\sin \tilde{\theta })\xi _\varphi , \end{aligned}$$

$\tilde{\varvec{A}}$, $\tilde{d}$ and ${\tilde{{\varvec{U}}}}^\sharp$ are functions given below (3.58),

$$\begin{aligned}&{\tilde{A}}^{i}:=A^i\circ \varPi ^{-1},\; {\tilde{B}}^i:=B^i \circ \varPi ^{-1},\; {\tilde{C}}^{i}:=C^i\circ \varPi ^{-1},\\&{\tilde{D}}^i=D^i\circ \varPi ^{-1},\;{\tilde{E}}:=E\circ \varPi ^{-1}, {\tilde{F}}:=F\circ \varPi ^{-1},\; \tilde{\varvec{h}}={\varvec{h}}\circ \varPi ^{-1}, \end{aligned}$$

$\mathrm{x}$ is the Cartesian coordinate representing $\mathcal {N}_f^+$, $\mathrm{y}=\varPi (\mathrm{x})$ and $(r,\theta ,\varphi )$ and $(\tilde{r},\tilde{\theta },{\tilde{\varphi }})$ are the spherical coordinate systems for $\mathrm{x}$ and $\mathrm{y}$, respectively. As we did in Step 2 in the proof of Lemma 10, set ${\varvec{\xi }}=0$ outside of $\varXi ^{-1}(D_{{\mathfrak r}}^*(\mathrm{x}_0^*))$ for $0<\mathfrak r<\min (\theta _1,\pi ,r_1-f(\theta _1))$ where $\mathrm{x}_0^*= (f(\theta _1),\theta _1,\varphi _0)$ for some $\varphi _0\in {\mathbb {T}}$ and then transform this equation using $\varXi$. Then we obtain

$$\begin{aligned} \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}&\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }}+\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{\xi }}\\&=\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}(a) \frac{\tilde{r}^2\sin \tilde{\theta }}{\det (\frac{d\mathrm{y}}{ d\mathrm{x}})}-\bigg (\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*\\&\quad -(\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*)(\mathrm{x}_0^*)\bigg )D_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }} -\tilde{d}^*\tilde{\varvec{h}}^*{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$, where ${\tilde{{\varvec{U}}}}^*$, $\tilde{d}^*$, and ${\tilde{M}}$ are functions defined below (3.60) and we used $\mathrm{div}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \left( (\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*)(\mathrm{x}_0^*)\right) =0$. Fix the principal coefficients of the left hand side of the above equation. Then we get

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)} \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*)D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }}\\&\quad =\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)} \left( \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*) -\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*\\&\qquad D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }} +\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}} -\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*\\&\qquad D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }} -\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{\xi }}+(a) \frac{\tilde{r}^2\sin \tilde{\theta }}{\det \left( \frac{d\mathrm{y}}{ d\mathrm{x}}\right) } -\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*\right. \\&\qquad \left. -\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*\right) (\mathrm{x}_0^*)\right) D_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})}{\varvec{\xi }} -\tilde{d}^*\tilde{\varvec{h}}^*{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1(D_{\mathfrak r}^*(\mathrm{x}_0^*))$.

Let ${\varvec{W}}$ be the weak solution of (3.62), (3.63). As we did in Step 3 in the proof of Lemma 10, subtracting the weak formulation of (3.62), (3.63) from the above equation. And then take ${\varvec{\xi }}={\varvec{V}}$ where ${\varvec{V}}=\tilde{\varvec{U}}^*-{\varvec{W}}$ to the resultant equation. Then we have

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*)D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}D_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}\\&\quad =\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)} \left( \left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) (\mathrm{x}_0^*)-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})} \tilde{{\varvec{U}}}^*\\&\qquad D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}+\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}-\frac{1}{\det {\tilde{M}}}{\tilde{M}}^T\tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{M}}\right) D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*\\&\qquad D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}-\tilde{d}^*\tilde{{\varvec{U}}}^*{\varvec{V}}+(a^\prime ) \frac{\tilde{r}^2\sin \tilde{\theta }}{\det (\frac{d\mathrm{y}}{ d\mathrm{x}})} -\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*\right. \\&\qquad \left. -\left( \frac{1}{\det {\tilde{M}}}{\tilde{M}}^T \tilde{\varvec{A}}\circ \varXi ^{-1}{\tilde{MD}}_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})} \tilde{\varvec{h}}^*\right) (\mathrm{x}_0^*)\right) D_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}-\tilde{d}^*\tilde{\varvec{h}}^*{\varvec{V}}, \end{aligned}$$

where $V_\varphi ={\varvec{V}}\cdot \mathbf{e}_{\varphi }$ and $(a^\prime )$ is (a) with $\xi _\varphi$ replaced by $V_\varphi$. Using the Sobolev and Hölder inequality and the facts that ${\varvec{A}}\in C^{\alpha }(\overline{\mathcal {N}_f^+})$, $f\in C^{1,\alpha }(\overline{\varLambda })$, $\varvec{h}\in C^{1,\alpha }(\overline{\mathcal {N}_f^+})$ and

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\frac{1}{r^2}\partial _{\tilde{r}}(r^2V_\varphi )|^2,\; \int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\frac{1}{\sin {\tilde{\theta }}}\partial _{{\tilde{\theta }}}(V_\varphi \sin {\tilde{\theta }})|^2\\ {}&\le C\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D(V_\varphi \mathbf{e}_{\varphi })|^2 \quad (\text{ See }\ (5.7))\\&\quad \le C\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r}, \tilde{\theta },{\tilde{\varphi }})}{\varvec{V}}|^2 \end{aligned}$$

we obtain from the above equation

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}{\varvec{V}}|^2\le C \left( {\mathfrak r}^{2\alpha }\int _{D_r^*(\mathrm{x}_0^*)}|D_{(r,\theta ,\varphi )}\tilde{{\varvec{U}}}^*|^2\right. \nonumber \\&\quad \left. +(\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{d}^*|^q)^{\frac{2}{q}}{\mathfrak r}^{2\alpha }\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|\tilde{{\varvec{U}}}^*|^2 +{\mathfrak r}^{3+2\alpha }F^\flat \right) \end{aligned}$$

(3.67)

for $q=\frac{3}{1-\alpha }$, where

$$\begin{aligned} F^\flat =&\sum _i ||B^i\mathbf{e}_{\theta }||_{\alpha ,\mathcal {N}_f^+}^2||A^i||^2_{W^{1,3}(\mathcal {N}_f^+)\cap L^\infty (\mathcal {N}_f^+)}\\&+\sum _i ||D^{i}||^2_{\alpha ,\mathcal {N}_f^+}||C^{i}||^2_{W^{1,3}(\mathcal {N}_f^+)\cap L^\infty (\mathcal {N}_f^+)}\\&+||E\frac{\partial _\theta (F\sin \theta )}{\sin \theta } \mathbf{e}_{\varphi }||_{L^q(\mathcal {N}_f^+)}^2+|| {\varvec{h}}||^2_{1,\alpha ,\mathcal {N}_f^+} \end{aligned}$$

with $||\cdot ||_{W^{1,3}(\varOmega )\cap L^\infty (\varOmega )}:=||\cdot ||_{W^{1,3}(\varOmega )}+||\cdot ||_{L^\infty (\varOmega )}$. By Corollory 1 and (3.67), we have, for any $0<{\mathfrak t}\le {\mathfrak r}$,

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^* -(D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak t}|^2\nonumber \\&\quad \le C\left( \left( \frac{\mathfrak t}{\mathfrak r}\right) ^5\int _{D_{\mathfrak r}^*(\tilde{\mathrm{x}}_0)}|D_{(\tilde{r},{\tilde{\theta }}, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*-(D_{(\tilde{r},{\tilde{\theta }}, {\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak r}|^2\right. \nonumber \\&\quad \left. +{\mathfrak r}^{2\alpha }\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2+{\mathfrak r}^{2\alpha }\int _{D_{\mathfrak r}^*( \mathrm{x}_0^*)}|\tilde{{\varvec{U}}}^*|^2 +{\mathfrak r}^{3+2\alpha }F^\flat \right) . \end{aligned}$$

(3.68)

In the proof of Lemma 10, we showed that for any $\varepsilon \in (0,1)$, there exists $R_4>0$ such that

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*|^2\le C{\mathfrak r}^{3-2\varepsilon }(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2)\\&\qquad \text{ for } \text{ any } \ 0<\mathfrak r\le R_4. \end{aligned}$$

Using this inequality and ${\varvec{U}}^\sharp \in C^0(\overline{\mathcal {N}_f^+})$ obtained from Lemma 10, we apply Lemma 2.1 in [15, Chapter 3] to (3.68). Then we obtain

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*-(D_{(\tilde{r},{\tilde{\theta }},{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,{\mathfrak r}}|^2 \nonumber \\&\quad \le C(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2+{F}^\flat ){\mathfrak r}^{3+2\alpha -2\varepsilon } \nonumber \\&\qquad \quad \text{ for } \text{ any }\ 0<\mathfrak r<R_5 \end{aligned}$$

(3.69)

for a constant $R_5>0$.

When $\mathrm{x}_0^*$ is in $\overline{\varGamma _w^{+.*}}\cap \overline{\varGamma _{ex}^*}$, we obtain (3.69) using similar argument without the process of transforming $\mathcal {N}_f^+$ to $\mathcal {N}_{f(\theta _1)}^+$. When $\mathrm{x}_0^*$ is in $\mathcal {N}_f^{+,*}\cap \{\theta \ge \frac{ \theta _1}{3}\}$ or $(\varGamma _f^*\cup \varGamma _w^{+,*}\cup \varGamma _{ex}^*)\cap \{\theta \ge \frac{ \theta _1}{3}\}$ and far away from the corners $\overline{\varGamma _f^*}\cap \overline{\varGamma _w^{+,*}}$ and $\overline{\varGamma _w^{+,*}}\cap \overline{\varGamma _{ex}^*}$, we obtain (3.69) using the standard method of freezing coefficients to the spherical coordinate representation of (3.40), (3.41) with integration by parts argument above. When $\mathrm{x}_0^*$ is in $\mathcal {N}_f^{+,*}\cap \{\theta \ge \frac{ \theta _1}{3}\}$ or $\varGamma _f^*\cup \varGamma _w^{+,*}\cup \varGamma _{ex}^*\cap \{\theta \ge \frac{ \theta _1}{3}\}$ and near $\overline{\varGamma _f^*}\cap \overline{\varGamma _w^{+_*}}$ or $\overline{\varGamma _w^{+,*}}\cap \overline{\varGamma _{ex}^*}$, we obtain (3.69) using the arguments in [16, Theorem 5.21] and arguments similar to the ones above. Then we obtain $D{\varvec{U}}^\sharp \in C^{\alpha -\varepsilon }([f(\theta ),r_1]\times [\frac{\theta _1}{3},\theta _1]\times {\mathbb {T}})$. When $\mathrm{x}_0$ is in $\mathcal {N}_f^+\cap \{\theta \le \frac{ 2\theta _1}{3}\}$ or $(\varGamma _f\cup \varGamma _w^+\cup \varGamma _{ex})\cap \{\theta \le \frac{2 \theta _1}{3}\}$, we obtain (3.45) with $\alpha$ and C replaced by $\alpha -\varepsilon$ for any $\varepsilon \in (0,1)$ and $C(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2+{F}^\flat )$ using the standard method of freezing coefficients to (3.40), (3.41) with integration by parts argument above. Note that if we estimate the integral form of ${\varvec{F}}^\sharp$ using integration by parts argument above, then there is no singularity issue. From this result, we obtain $D{\varvec{U}}^\sharp \in C^{\alpha -\varepsilon }(\overline{\mathcal {N}_f^+}\cap \{\theta \le \frac{2\theta _1}{3}\})$. Combining these two regularity results for $D{\varvec{U}}^\sharp$, we obtain $D{\varvec{U}}^\sharp \in C^{\alpha -\varepsilon }(\overline{\mathcal {N}_f^+})$.

Using the regularity result for $D{\varvec{U}}^\sharp$ and ${\varvec{U}}^\sharp \in C^0(\overline{\mathcal {N}_f^+})$, we obtain from (3.68)

$$\begin{aligned}&\int _{D_{\mathfrak t}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*-(D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak t}|^2\nonumber \\&\quad \le C\bigg (\left( \frac{\mathfrak t}{\mathfrak r}\right) ^5\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*-(D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,\mathfrak r}|^2 \nonumber \\&\quad +{\mathfrak r}^{3+2\alpha }\left( ||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2 +{F}^\flat \right) \bigg ) \end{aligned}$$

(3.70)

(here $\mathrm{x}_0^*\in \overline{\varGamma _f^*}\cap \overline{\varGamma _w^{+,*}}$). Apply Lemma 2.1 in [15, Chapter 3] to (3.70). Then we have

$$\begin{aligned}&\int _{D_{\mathfrak r}^*(\mathrm{x}_0^*)}|D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*-(D_{(\tilde{r},\tilde{\theta },{\tilde{\varphi }})}\tilde{{\varvec{U}}}^*)^*_{\mathrm{x}_0^*,{\mathfrak r}}|^2 \nonumber \\&\quad \le C(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2+{F}^\flat ){\mathfrak r}^{3+2\alpha } \nonumber \\&\qquad \quad \text{ for } \text{ any }\ 0<\mathfrak r<R_6 \end{aligned}$$

(3.71)

for a constant $R_6>0$. Using the arguments used when we obtained $D{\varvec{U}}\in C^{\alpha -\varepsilon }(\overline{\mathcal {N}_f^+})$, we obtain (3.71) at $\mathrm{x}_0^*\in \overline{\mathcal {N}_f^{+,*}}\cap \{\theta \ge \frac{\theta _1}{3}\}$ and (3.45) with C replaced by $C(||{\varvec{U}}^\sharp ||_{H^1(\mathcal {N}_f^+)}^2+||{\varvec{F}}^\sharp ||_{L^3(\mathcal {N}_f^+)}^2+{F}^\flat )$ at $\mathrm{x}_0\in \overline{\mathcal {N}_f^+}\cap \{\theta \le \frac{2\theta _1}{3}\}$. From this result, we obtain the desired result. This finishes the proof. $\square$

Using the scailing argument given in the proof of Proposition 3.1 in [2] with the results in Theorem 5.21 in [16] and Lemma 11, we can obtain that the weak solution of (3.40), (3.41) is in $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ and satisfies the estimate in the following lemma (we omit the proof):

Lemma 13

Under the assumption as in Lemma 7, let ${\varvec{U}}^\sharp$ be the weak solution of (3.40), (3.41). Then ${\varvec{U}}^\sharp \in C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. Furthermore, ${\varvec{U}}^\sharp$ satisfies

$$\begin{aligned} ||{\varvec{U}}^\sharp ||_{2,\alpha ,\mathcal {N}_f^+}^{(-1-\alpha ,\varGamma _w^+)}\le C(||{\varvec{F}}^\sharp ||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}+||{\varvec{U}}^\sharp ||_{1,\alpha ,\mathcal {N}_f^+}), \end{aligned}$$

where C is a positive constant depending on $(\rho _0^+,u_0^+,p_0^+)$, $\gamma$, $\mathcal {N}_f^+$ and $\alpha$.

Finally, we prove that the $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41) is of the form $\varPsi (r,\theta )\mathbf{e}_{\varphi }$. This statement is proved using the argument as in the proof of Proposition 3.3 in [4] (Method II). Although the arguments to prove this statement are almost the same with those in the proof of Proposition 3.3 in [4] (Method II), since (3.40), (3.41) is different from the problem in Proposition 3.3 in [4] and similar arguments will be used later in the proof of Lemma 18, we present the detailed proof.

Lemma 14

Under the assumption as in Lemma 7, the $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41) is of the form $\varPsi (r,\theta )\mathbf{e}_{\varphi }$.

Proof

Let ${\varvec{U}}=U_r\mathbf{e}_r+U_\theta \mathbf{e}_\theta +U_\varphi \mathbf{e}_\varphi$ be the $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41). Then, $(U_r,U_\theta ,U_\varphi )$ is in $(C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}(\mathcal {N}_{f}^{+,*}))^3$ satisfying

$$\begin{aligned}&||U_r||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})},\; ||U_\theta ||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})},\; ||U_\varphi ||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})}\nonumber \\&\quad \le C||{\varvec{U}}||_{2,\alpha ,\mathcal {N}_f^+}^{(-1-\alpha ,\varGamma _w^+)} \le CC^*, \end{aligned}$$

(3.72)

where C and $C^*$ are positive constants depending on $\mathcal {N}_f^+$ and $\alpha$ and given in (3.39), respectively, and satisfies the following spherical coordinate representation of (3.40), (3.41)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\varDelta U_r-\frac{2U_r}{r^2}-\frac{2}{r^2\sin \theta }\partial _\theta (U_\theta \sin \theta )-\frac{2}{r^2\sin \theta }\partial _\varphi U_\varphi )\right. \\ \left. \quad \quad \quad \quad -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\frac{1}{r^2}\partial _r (r^2\partial _r U_r)-\frac{\partial _r \rho _0^+}{{\rho _0^+}^2}\frac{\partial _r (r U_r)}{r}\right) =0,\\ \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\varDelta U_\theta -\frac{U_\theta }{r^2\sin ^2\theta }+\frac{2}{r^2}\partial _\theta U_r-\frac{2\cos \theta }{r^2\sin ^2\theta }\partial _\varphi U_\varphi )\right. \\ \left. \quad \quad \quad \quad \quad \quad \quad \quad -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\frac{1}{r^2}\partial _r (r^2 \partial _r U_\theta )-\frac{\partial _r\rho _0^+}{{\rho _0^+}^2}\frac{\partial _r (r U_\theta )}{r}\right) =0,\\ \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\varDelta U_\varphi -\frac{U_\varphi }{r^2\sin ^2\theta }+\frac{2}{r^2\sin \theta }\partial _\varphi U_r+\frac{2\cos \theta }{r^2\sin ^2\theta }\partial _\varphi U_\theta )\right. \\ \left. \quad \quad \quad -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)} \frac{1}{r^2}\partial _r (r^2\partial _r U_\varphi ) -\frac{\partial _r \rho _0^+}{{\rho _0^+}^2}\partial _r (r U_\varphi )\right) =- {\mathcal {F}}(r,\theta ) \end{array}\right. }\nonumber \\&\text{ in } \mathcal {N}_f^{+,*},\end{aligned}$$

(3.73)

$$\begin{aligned}&(U_r,U_\theta ,U_\varphi )= (0,0,0) \quad \text{ on }\quad \varGamma _f^*,\varGamma _w^{+,*},\varGamma _{ex}^*, \end{aligned}$$

(3.74)

where ${\mathcal {F}}={\varvec{F}}^\sharp \cdot \mathbf{e}_{\varphi }$ and $\varDelta U_k=\frac{1}{r^2}\partial _r(r^2\partial _r U_k)+\frac{1}{r^2\sin \theta }\partial _\theta (\sin \theta \partial _\theta U_k)+\frac{1}{r^2\sin ^2\theta }$ $\partial _\varphi ^2 U_k$ for $k=r,\theta ,\varphi$.

Define

$$\begin{aligned} U_k^n:=\frac{1}{2^n}\sum _{k=0}^{2^n-1}U_k\left( r,\theta ,\varphi +\frac{2\pi k}{2^n}\right) \end{aligned}$$

for $k=r,\theta ,\varphi$. Then by the definition of $U_k^n$ for $k=r,\theta ,\varphi$ and (3.72),

$$\begin{aligned} ||U_r^n||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})},\; ||U_\theta ^n||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})},\; ||U_\varphi ^n||_{2,\alpha ,\mathcal {N}_{f}^{+,*}}^{(-1-\alpha ,\{\theta =\theta _1\})}\le CC^*. \end{aligned}$$

(3.75)

By (3.75) and $C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}(\mathcal {N}_f^{+,*})\Subset C^{2,\alpha }_{(-1-\frac{\alpha }{2},\{\theta =\theta _1\})}(\mathcal {N}_f^{+,*})$, there exists a subsequence $(U_r^{n_k},$ $U_\theta ^{n_k},U_\varphi ^{n_k})$ of $(U_r^{n},U_\theta ^{n},U_\varphi ^{n})$ such that $(U_r^{n_k},U_\theta ^{n_k},U_\varphi ^{n_k})$ converges in $C^{2,\alpha }_{(-1-\frac{\alpha }{2},\{\theta =\theta _1\})}(\mathcal {N}_f^{+,*})$ as $n_k\rightarrow \infty$. Denote its limit by $(U_r^*,U_\theta ^*,U_\varphi ^*)$. Then $(U_r^*,U_\theta ^*,U_\varphi ^*)$ is independent of $\varphi$ and $(U_r^*,U_\theta ^*,U_\varphi ^*)\in (C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}$ $(\mathcal {N}_f^{+,*}))^3$.

Since the coefficients of (3.73), ${\mathcal {F}}$ and the boundary conditions in (3.74) are independent of $\varphi$, $(U_r^n,U_\theta ^n,U_\varphi ^n)$ satisfies (3.73), (3.74) for all $n\in {\mathbb {N}}\cup \{0\}$. By this fact and the definition of $(U_r^*,U_\theta ^*,U_\varphi ^*)$, $(U_r^*,U_\theta ^*,U_\varphi ^*)$ satisfies (3.73), (3.74). Since $(U_r^*,U_\theta ^*,U_\varphi ^*)$ is independent of $\varphi$, (3.73) satisfied by $(U_r^*,U_\theta ^*,U_\varphi ^*)$ is given as

$$\begin{aligned}&{\left\{ \begin{array}{ll} \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\frac{1}{r^2}\partial _r(r^2\partial _r U_r^*)+\frac{1}{r^2\sin \theta }\partial _\theta (\sin \theta \partial _\theta U_r^*)-\frac{2U_r^*}{r^2}\right. \\ \quad -\frac{2}{r^2\sin \theta }\partial _\theta (U_\theta ^*\sin \theta )) \left. -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\frac{1}{r^2}\partial _r (r^2\partial _r U_r^*)-\frac{\partial _r \rho _0^+}{{\rho _0^+}^2}\frac{\partial _r (r U_r^*)}{r}\right) =0,\\ \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\frac{1}{r^2}\partial _r(r^2\partial _r U_\theta ^*)+\frac{1}{r^2\sin \theta }\partial _\theta (\sin \theta \partial _\theta U_\theta ^*)-\frac{U_\theta ^*}{r^2\sin ^2\theta }+\frac{2}{r^2}\partial _\theta U_r^*)\right. \\ \left. \quad \quad \quad \quad \quad \quad \quad \quad -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\frac{1}{r^2}\partial _r (r^2 \partial _r U_\theta ^*)-\frac{\partial _r\rho _0^+}{{\rho _0^+}^2}\frac{\partial _r (r U_\theta ^*)}{r}\right) =0,\\ \left( \frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}(\frac{1}{r^2}\partial _r(r^2\partial _r U_\varphi ^*)+\frac{1}{r^2\sin \theta }\partial _\theta (\sin \theta \partial _\theta U_\varphi ^*)-\frac{U_\varphi ^*}{r^2\sin ^2\theta })\right. \\ \left. \quad \quad \quad \quad \quad \quad -\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)} \frac{1}{r^2}\partial _r (r^2\partial _r U_\varphi ^*) -\frac{\partial _r \rho _0^+}{{\rho _0^+}^2}\partial _r (r U_\varphi ^*)\right) =- {\mathcal {F}}(r,\theta ) \end{array}\right. }\nonumber \\&\text{ in } \ \mathcal {N}_f^{+,*}. \end{aligned}$$

(3.76)

Note that the first and second equation of (3.76) is not coupled with the third equation of (3.76).

Let $\varPsi =U_\varphi ^*$. Using the third equation of (3.76) and the facts that $U_\varphi ^*\in C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}(\mathcal {N}_f^{+,*})$ and ${\varvec{F}}^\sharp$ is an axisymmetric function in $C^{\alpha }_{(1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, it can be checked that $\varPsi =\partial _{\theta \theta }\varPsi =0$ on $\theta =0$. By this fact and $U_\varphi ^*\in C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}(\mathcal {N}_f^{+,*})$, we have that $\varPsi \mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$. From (3.76) and (3.74), one can see that $(0,0,U_\varphi ^*)$ is a solution of (3.73), (3.74). Combining this fact with the fact that $\varPsi \mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$, we have that $\varPsi \mathbf{e}_\varphi$ is a $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41). By Lemma 9, a $C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+)$ solution of (3.40), (3.41) is unique. Therefore, ${\varvec{U}}=\varPsi \mathbf{e}_{\varphi }$. This finishes the proof. $\square$

3.3 Initial value problem of a transport equation with an axisymmetric divergence-free coefficient

The initial value problems of a transport equation in (B$^\prime$) are of the form

$$\begin{aligned}&\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )\cdot \nabla Q=0\quad \text{ in }\quad \mathcal {N}_f^+,\end{aligned}$$

(3.77)

$$\begin{aligned}&Q=Q_{en}\quad \text{ on }\quad \varGamma _f, \end{aligned}$$

(3.78)

where $\varPsi \mathbf{e}_\varphi :\mathcal {N}_f^+\rightarrow \mathbb {R}^3$ and $Q_{en}:\varGamma _f\rightarrow \mathbb {R}$ are axisymmetric functions. Here, (3.77) is a transport equation whose coefficient is an axisymmetric and divergence-free vector field. Thus, the stream function of the coefficient vector field of (3.77) can be defined as in (2.23). We will find a solution of (3.77), (3.78) and obtain the regularity and uniqueness of solutions of (3.77), (3.78) using the stream function of the coefficient vector field of (3.77) and the solution expression given by using the stream function. This will be done in the proof of the following lemma:

Lemma 15

Suppose that f is as in Lemma 4. Let $\delta _6$ be a positive constant such that for such f, if $||\nabla \times (\varPsi \mathbf{e}_\varphi )||_{0,0,\mathcal {N}_f^+}\le \delta _6,$ then

$$\begin{aligned}&\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )\cdot \mathbf{e}_r>c^*\quad \text{ in }\quad \overline{\mathcal {N}_f^+}\nonumber \\&\qquad \text{ and }\quad \nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_\varphi )\cdot {\varvec{\nu }}_f >c^*\quad \text{ on }\quad \overline{\varGamma _f}, \end{aligned}$$

(3.79)

where $c^*$ is a positive constant depending on $\rho _0^+$, $u_0^+$, $r_s$, $r_1$, $\delta _1$ and $\delta _6$, and ${\varvec{\nu }}_f$ is the unit normal vector on $\varGamma _f$ pointing toward $\mathcal {N}_f^+$. Suppose that $\varPsi \mathbf{e}_{\varphi }:\mathcal {N}_f^+\rightarrow \mathbb {R}^3$ is an axisymmetric function in $C^{1,\alpha }(\overline{\mathcal {N}_f^+})$ satisfying

$$\begin{aligned} \varPsi =\frac{f(\theta _1)\varPsi (f(\theta _1),\theta _1)}{r}\quad \text{ on }\quad \overline{\varGamma _w^+} \end{aligned}$$

(3.80)

and

$$\begin{aligned} ||\nabla \times (\varPsi \mathbf{e}_\varphi )||_{0,\alpha ,\mathcal {N}_f^+}\le \delta _6. \end{aligned}$$

(3.81)

Suppose finally that $Q_{en}:\varGamma _f\rightarrow \mathbb {R}$ is an axisymmetric function in $C^{1,\alpha }_{(-\alpha ,\partial \varGamma _f)}$ $(\varGamma _f)$. Then the problem (3.77), (3.78) has a unique axisymmetric $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution

$$\begin{aligned} Q=Q_{en}({\mathcal {L}}) \quad \text{ in }\quad \mathcal {N}_f^+, \end{aligned}$$

(3.82)

where ${\mathcal {L}}=k^{-1}\circ V$ with $V=2\pi r \sin \theta (\varPhi _0^++\varPsi )$ and $k(\theta )=V(f(\theta ),\theta )$. Furthermore, this solution Q satisfies

$$\begin{aligned} ||Q||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\le C ||Q_{en}||_{1,\alpha ,\varGamma _f}^{(-\alpha ,\partial \varGamma _f)}, \end{aligned}$$

where C is a positive constant depending on $\rho _0^+$, $u_0^+$, $r_s$, $r_1$, $\theta _1$, $\alpha$, $\delta _1$ and $\delta _6$.

Remark 8

The existence of $\delta _6$ is obtained from the facts that $\nabla \times (\varPhi _0^+\mathbf{e}_\varphi )\cdot \mathbf{e}_r =\rho _0^+u_0^+>0$ in $\overline{\mathcal {N}_{r_s-\delta _1}^+}$ and ${\varvec{\nu }}_f\cdot \mathbf{e}_r>0$ on $\overline{\varGamma _f}$ for f given in Lemma 4.

Remark 9

If $\varPsi \mathbf{e}_\varphi$ satisfies (3.79), then $V=2\pi r \sin \theta (\varPhi _0^++\varPsi )$ satisfies

$$\begin{aligned} \partial _\theta V>0\quad \text{ in }\quad \mathcal {N}_f^+\setminus \{\theta =0\}\quad \text{ and }\quad \partial _\theta (V(f(\theta ),\theta ))>0\quad \text{ for }\quad \theta \in (0,\theta _1). \end{aligned}$$

This condition will be used to construct the stream surfaces of $\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_{\varphi })$ in $\mathcal {N}_f^+$ in the proof of Lemma 15.

Proof

1. Construct the stream surfaces of $\nabla \times ((\varPhi _0^++\varPsi )\mathbf{e}_{\varphi })$ in $\overline{\mathcal {N}_f^+}$.

Let us define $V:=2\pi r \sin \theta (\varPhi _0^++\varPsi )$. By the fact that $\varPsi \mathbf{e}_{\varphi }$ and $\varPhi _0^+\mathbf{e}_{\varphi }$ are axisymmetric functions in $C^{1,\alpha }(\overline{\mathcal {N}_f^+})$, V is an axisymmetric function in $C^{1,\alpha }(\overline{\mathcal {N}_f^+})$ (see Lemma 1). By (3.79), $\partial _\theta V>0$ in $\mathcal {N}_f^+\setminus \{x=y=0\}$. Using these facts, we apply the implicit function theorem to V. Then we obtain that for any $(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )\in \mathcal {N}_f^{+,*}$, there exists a unique $C^1$ surface $\theta =h_{(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )}(r,\varphi )$ defined near $(r^\sharp ,\varphi ^\sharp )$ such that $\theta ^\sharp =h_{(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )}(r^\sharp ,\varphi ^\sharp )$ and $V(r,h_{(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )}(r,\varphi ),\varphi )=V(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )$. Since V is axisymmetric, this surface is axisymmetric. We denote $\theta =h_{(r^\sharp ,\theta ^\sharp ,\varphi ^\sharp )}(r,\varphi )$ by $\theta =h_{(r^\sharp ,\theta ^\sharp )}(r)$.

By $V_0^+=V_0^+(f(\theta _1),\theta _1)$ on $\overline{\varGamma _w^{+}}$ and (3.80), $V=V(f(\theta _1),\theta _1)$ on $\overline{\varGamma _w^{+}}$. By this fact, $V=0$ on $\overline{\mathcal {N}_f^{+}}\cap \{x=y=0\}$ and $\partial _\theta V>0$ in $\mathcal {N}_f^{+}\setminus \{x=y=0\}$, $\theta =h_{(r^\sharp ,\theta ^\sharp )}(r)$ is defined until it reaches $\varGamma _f$ or $\varGamma _{ex}$ not touching $\overline{\mathcal {N}_f^{+}}\cap \{x=y=0\}$ or $\overline{\varGamma _w^{+}}$. Note that by the facts that $\partial _\theta (V(f(\theta ),\theta ))>0$ for $\theta \in (0,\theta _1)$ and $\partial _\theta V(r_1,\theta )>0$ for $\theta \in (0,\theta _1)$ obtained from (3.79), the surface $\theta =h_{(r^\sharp ,\theta ^\sharp )}(r)$ intersects with $\varGamma _f$ and $\varGamma _{ex}$ once, respectively. Collect $\theta =h_{(r^\sharp ,\theta ^\sharp )}(r)$ for all $(r^\sharp ,\theta ^\sharp )\in (f(\theta ^\sharp ),r_1)\times (0,\theta _1)$, $\overline{\mathcal {N}_f^{+}}\cap \{x=y=0\}$ and $\overline{\varGamma _w^{+}}$. Then we have the entire level surfaces of V in $\overline{\mathcal {N}_f^+}$. By $\partial _\theta V>0$ in $\mathcal {N}_f^{+}\setminus \{x=y=0\}$, the values of V on distinct level surfaces of V in $\overline{\mathcal {N}_f^+}$ are different from each other.

2. Find a solution of (3.77), (3.78).

By (2.27), (3.77) can be written as

$$\begin{aligned} \nabla ^\perp V\cdot \nabla Q=0. \end{aligned}$$

(3.83)

Using this form of (3.77), it can be checked that 1) if Q is in $C^1(\mathcal {N}_f^+)$ and $Q=\text{ constant }$ on any curve on any level surface of V in $\mathcal {N}_f^+$ whose $\varphi$ argument is fixed (in the case when a level surface of V is $\mathcal {N}_f^+\cap \{x=y=0\}$, $Q=\text{ constant }$ on $\mathcal {N}_f^+\cap \{x=y=0\}$), then Q is a solution of (3.77) and that 2) if Q is a $C^1$ solution of (3.77), then $Q=\text{ constant }$ on any curve on any level surface of V in $\mathcal {N}_f^+$ whose $\varphi$ argument is fixed. Denote the $\theta$-argument of the intersection points of $\varGamma _f$ and the level surface of V in $\overline{\mathcal {N}_f^+}$ passing through $\mathrm{x}\in \overline{\mathcal {N}_f^+}$ by ${\mathcal {L}}(\mathrm{x})$. Since each level surface of V in $\overline{\mathcal {N}_f^+}$ intersects with $\overline{\varGamma _f}$ where the value of V on $\overline{\varGamma _f}$ is equal to the value of V on the level surface, ${\mathcal {L}}(\mathrm{x})$ is given by

$$\begin{aligned} {\mathcal {L}}(\mathrm{x}):=k^{-1}\circ V(\mathrm{x}), \end{aligned}$$

where $k(\theta )=V(f(\theta ),\theta )$.

By 2), a $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution of (3.77), (3.78) has the form

$$\begin{aligned} Q(r,\theta ,\varphi )={\left\{ \begin{array}{ll} Q_{en}({\mathcal {L}}(r,\theta ),\varphi )\quad \text{ if } (r,\theta )\in [f(\theta ),r_1]\times (0,\theta _1]\\ Q_{en}({\mathcal {L}}(r,0),0)\quad \text{ if } (r,\theta )\in [f(0),r_1]\times \{0\}. \end{array}\right. } \end{aligned}$$

Since $Q_{en}$ is axisymmetric, this can be written as

$$\begin{aligned} Q=Q_{en}({\mathcal {L}}). \end{aligned}$$

Let $Q=Q_{en}({\mathcal {L}})$. One can see that Q is a constant on any level surface of V in $\overline{\mathcal {N}_f^+}$ and satisfies (3.78). Thus, by 1), if $Q\in C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$, then Q is a $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution of (3.77), (3.78).

3. Estimate $||Q||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}$.

Since ${\mathcal {L}}\in C^0(\overline{\mathcal {N}_f^+})$ and $Q_{en}\in C^0(\overline{\varGamma _f})$, Q is in $C^0(\overline{\mathcal {N}_f^+})$. It directly follows from the definition of Q

$$\begin{aligned} ||Q||_{0,0,\mathcal {N}_f^+}=||Q_{en}||_{0,0,\varGamma _f}. \end{aligned}$$

Thus, to estimate $||Q||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}$, it is enough to estimate $||DQ||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}$. We estimate $||DQ||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}$.

By direct computation,

$$\begin{aligned} DQ&=Q_{en}^\prime ({\mathcal {L}})D{\mathcal {L}}\\&=Q_{en}^\prime ({\mathcal {L}})\frac{DV}{(\partial _r V(f(\vartheta ),\vartheta )f^\prime (\vartheta )+ \partial _\vartheta V(f(\vartheta ),\vartheta ))|_{\vartheta ={\mathcal {L}}}}. \end{aligned}$$

To estimate $||DQ||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}$, we estimate $||D{\mathcal {L}}||_{0,\alpha ,\mathcal {N}_f^+}$.

Write $D{\mathcal {L}}$ as

$$\begin{aligned} \frac{\frac{DV}{2\pi \sin \theta } }{\left. (\nabla ^\perp V)(f(\vartheta ),\vartheta )\cdot {\varvec{\nu }}_f(\vartheta )f^2(\vartheta )\sqrt{1+\left( \frac{f^\prime (\vartheta )}{f(\vartheta )}\right) ^2}\right| _{\vartheta ={\mathcal {L}}}}\cdot \frac{\sin \theta }{\sin {\mathcal {L}}} \end{aligned}$$

Here, we used the definition of $\nabla ^\perp V$ and the spherical coordinate expression of ${\varvec{\nu }}_f$. Using (3.3) and (3.79), it is easily seen that

$$\begin{aligned} \sup _{\mathcal {N}_f^+}\left| \left. (\nabla ^\perp V)(f(\vartheta ),\vartheta )\cdot {\varvec{\nu }}_f(\vartheta )f^2(\vartheta )\sqrt{1+\left( \frac{f^\prime (\vartheta )}{f(\vartheta )}\right) ^2}\right| _{\vartheta ={\mathcal {L}}}\right| >c^{**}, \end{aligned}$$

(3.84)

where $c^{**}$ is a positive constant depending on $c^*$, $r_s$ and $\delta _1$. Using (3.3) and (3.81), it is also easily seen that

$$\begin{aligned} ||\frac{DV}{2\pi \sin \theta }||_{0,\alpha ,\mathcal {N}_f^+}\le C \end{aligned}$$

(3.85)

and

$$\begin{aligned}&||\left. (\nabla ^\perp V)(f(\vartheta ),\vartheta )\cdot {\varvec{\nu }}_f(\vartheta )f^2(\vartheta )\sqrt{1+\left( \frac{f^\prime (\vartheta )}{f(\vartheta )}\right) ^2}\right| _{\vartheta ={\mathcal {L}}}||_{0,\alpha ,\mathcal {N}_f^+}\nonumber \\&\quad \le C||D{\mathcal {L}}||_{0,0,\mathcal {N}_f^+}, \end{aligned}$$

(3.86)

where Cs are positive constants depending on $\rho _0^+$, $u_0^+$, $r_s$, $\alpha$, $\delta _1$ and $\delta _6$. By these three estimates, one can see that to estimate $||D{\mathcal {L}}||_{0,\alpha ,\mathcal {N}_f^+}$, it is enough to estimate $||\frac{\theta }{{\mathcal {L}}}||_{0,\alpha ,\mathcal {N}_f^+}$. To estimate $||D{\mathcal {L}}||_{0,\alpha ,\mathcal {N}_f^+}$, we prove the following claim:

Claim. There exists a positive constant C depending on $\rho _0^+$, $u_0^+$, $r_s$, $r_1$, $\theta _1$, $\alpha$, $\delta _1$ and $\delta _6$ such that

$$\begin{aligned} ||\frac{\theta }{{\mathcal {L}}}||_{0,\alpha ,\mathcal {N}_f^+}\le C. \end{aligned}$$

(3.87)

Proof of Claim

To simplify our argument, we assume that $f=r_s$.

First, we estimate $||\frac{\theta }{{\mathcal {L}}}||_{0,0,\mathcal {N}_{r_s}^+}$. By the definition of ${\mathcal {L}}$,

$$\begin{aligned} V(r_s, {\mathcal {L}}(r,\theta ))=V(r,\theta ) \end{aligned}$$

for all $(r,\theta )\in [r_s,r_1]\times [0,\theta _1]$. Write this as

$$\begin{aligned}&\int _0^\theta (\rho _0^+u_0^++\nabla \times (\varPsi \mathbf{e}_\varphi )\cdot \mathbf{e}_r)(r,\xi ) r^2\sin \xi \hbox {d}\xi \nonumber \\&\quad =\int _0^{{\mathcal {L}}(r,\theta )}(\rho _0^+u_0^++\nabla \times (\varPsi \mathbf{e}_\varphi )\cdot \mathbf{e}_r)(r_s,\xi )r_s^2\sin \xi \hbox {d}\xi . \end{aligned}$$

By (3.79), (3.81) and the first equation of (2.14), we have from this equation

$$\begin{aligned} \int _0^\theta c^*r^2\sin \xi \hbox {d}\xi \le \int _0^{{\mathcal {L}}(r,\theta )}(m_0+\delta _6r_s^2)\sin \xi \hbox {d}\xi . \end{aligned}$$

Using $\frac{\sin \theta _1}{\theta _1}\xi \le \sin \xi \le \xi$ for $\xi \in [0,\theta _1]$, change $\sin \xi$ in the integrands in the left and right hand side of the above inequalities to $\frac{\sin \theta _1}{\theta _1}\xi$ and $\xi$, respectively, and then integrate the resultant terms. Then we obtain

$$\begin{aligned} \frac{\sin \theta _1}{\theta _1}c^*r^2\frac{\theta ^2}{2}\le (m_0+\delta _6r_s^2)\frac{{\mathcal {L}}^2(r,\theta )}{2}. \end{aligned}$$

From this inequality, we have

$$\begin{aligned} \sqrt{\frac{\sin \theta _1}{\theta _1}\frac{c^*r_s^2}{m_0+\delta _6r_s^2}}\theta \le {\mathcal {L}}(r,\theta ). \end{aligned}$$

This holds for all $(r,\theta )\in [r_s,r_1]\times [0,\theta _1]$. Hence,

$$\begin{aligned} ||\frac{\theta }{{\mathcal {L}}}||_{0,0,\mathcal {N}_{r_s}^+}\le C, \end{aligned}$$

(3.88)

where C is a positive constant depending on $\rho _0^+$, $u_0^+$, $r_s$, $\theta _1$, $\delta _6$ and $c^*$.

Next, we estimate $[\frac{\theta }{{\mathcal {L}}}]_{0,\alpha ,\mathcal {N}_{r_s}^+}$. By (3.88), we can obtain an estimate of $[\frac{\theta }{{\mathcal {L}}}]_{0,\alpha ,\mathcal {N}_{r_s}^+}$ by obtaining

$$\begin{aligned} \left| \frac{\theta }{{\mathcal {L}}(r,\theta )}-\frac{\theta ^\prime }{{\mathcal {L}}(r^\prime ,\theta ^\prime )}\right| \le C\varepsilon ^\alpha \end{aligned}$$

(3.89)

for all $(r,\theta )$, $(r^\prime ,\theta ^\prime )\in [r_s,r_1]\times [0,\theta _1]$ satisfying $\varepsilon \le \varepsilon _0$ for a positive constant $\varepsilon _0$ and a positive constant C where $\varepsilon :=\sqrt{|r^\prime -r|^2+|\theta ^\prime -\theta |^2}$. We obtain this estimate.

Hereafter, to simplify our notation, we denote ${\mathcal {L}}(r^\prime ,\theta ^\prime )$, ${\mathcal {L}}(r,\theta )$ and $\varPhi _0^++\varPsi$ by ${\mathcal {L}}^\prime$, ${\mathcal {L}}$ and $\varPhi$, respectively.

By the definition of ${\mathcal {L}}$,

$$\begin{aligned} V(r_s,{\mathcal {L}}^\prime )-V(r_s,{\mathcal {L}})=V(r^\prime ,\theta ^\prime )-V(r,\theta ). \end{aligned}$$

Write this as

$$\begin{aligned}&\int _{{\mathcal {L}}}^{{\mathcal {L}}^\prime }(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r )(r_s,\xi )r_s^2 \sin \xi \hbox {d}\xi \nonumber \\&\quad =\int _\theta ^{\theta ^\prime }(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r)(r,\xi ) r^2\sin \xi d \xi \nonumber \\&\qquad +\int _r^{r^\prime }(\nabla \times (\varPhi \mathbf{e}_\varphi ))(t,\theta ^\prime )\cdot \mathbf{e}_\theta (\theta ^\prime ) t sin\theta ^\prime d t. \end{aligned}$$

(3.90)

Using (3.90), we find an upper bound of ${\mathcal {L}}^\prime$. Using (3.81), the fact that $\varPhi _0^+\mathbf{e}_\varphi \in C^\infty (\overline{\mathcal {N}_{r_s}^+})$ and $\frac{\sin \theta _1}{\theta _1}\xi \le \sin \xi \le \xi$ for $\xi \in [0,\theta _1]$, we have, for ${\mathcal {L}}^\prime \ge {\mathcal {L}}$,

$$\begin{aligned}&\int _{{\mathcal {L}}}^{{\mathcal {L}}^\prime } \left( (\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r )(r_s,{\mathcal {L}})-C (\xi -{\mathcal {L}})^\alpha \right) r_s^2 \frac{\sin \theta _1}{\theta _1 } \xi \hbox {d}\xi \\&\quad \le (\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r)(r,\theta ) r^2\frac{{\theta ^\prime }^2-\theta ^2}{2}\\&\qquad +C r^2\left( \frac{1}{\alpha +1}|\theta ^\prime -\theta |^{\alpha +1}\theta ^\prime -\frac{1}{(\alpha +1)(\alpha +2)}|\theta ^\prime -\theta |^{\alpha +2}\right) \\&\qquad +(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_\theta )(r,\theta ^\prime )\sin \theta ^\prime \frac{ {r^\prime }^2-r^2}{2}\\&\qquad +C \sin \theta ^\prime \left( \frac{1}{\alpha +1} |{r^\prime }-r|^{\alpha +1}r^\prime -\frac{1}{(\alpha +1)(\alpha +2)}|r^\prime -r|^{\alpha +2}\right) (=:(u)), \end{aligned}$$

where the right hand side is an upper bound of the right hand side of (3.90) and C is a positive constant depending on $\rho _0^+$, $u_0^+$, $\mathcal {N}_{r_s}^+$, $\alpha$ and $\delta _6$. From this inequality, we get

$$\begin{aligned}&(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r )(r_s,{\mathcal {L}})\frac{\sin \theta _1}{\theta _1} r_s^2 \frac{{{\mathcal {L}}^\prime }^2-{\mathcal {L}}^2}{2}\\&\quad -C r_s^2\frac{\sin \theta _1}{\theta _1} \left( \frac{1}{\alpha +1}({\mathcal {L}}^\prime -{\mathcal {L}})^{\alpha +1}{\mathcal {L}}^\prime -\frac{1}{(\alpha +1)(\alpha +2)}({\mathcal {L}}^\prime -{\mathcal {L}})^{\alpha +2}\right) \le (u). \end{aligned}$$

Using this inequality, we have that for each $(r,\theta )\in [r_s,r_1]\times [0,\theta _1]$, there exists positive constants $\varepsilon _{(r,\theta )}< 1$ and $C_{(r,\theta )}$ such that for any $(r^\prime ,\theta ^\prime )$ satisfying $\varepsilon \le \varepsilon _{(r,\theta )}$,

$$\begin{aligned} {\mathcal {L}}^\prime \le \theta ^\prime \left( \frac{{\mathcal {L}}}{\theta }+C_{(r,\theta )}\varepsilon ^\alpha \right) . \end{aligned}$$

Note that when $(r,\theta )\in [r_s,r_1]\times \{0\}$, there exists a positive constant $\varepsilon _{(r,\theta )}<1$ such that for any $(r^\prime ,\theta ^\prime )$ satisfying $\varepsilon \le \varepsilon _{(r,\theta )}$,

$$\begin{aligned} {\mathcal {L}}^\prime \le \theta ^\prime \left( \sqrt{\frac{r^2(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r )(r,0)}{r_s^2(\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot \mathbf{e}_r )(r_s,0)}}+C_{(r,\theta )}\varepsilon ^\alpha \right) . \end{aligned}$$

Let $\overline{\varepsilon }_0=\inf _{(r,\theta )\in [r_s,r_1]\times [0,\theta _1]}\varepsilon _{(r,\theta )}$ and $C=\sup _{(r,\theta )\in [r_s,r_1]\times [0,\theta _1]}C_{(r,\theta )}$. Then by the above statement,

$$\begin{aligned} \frac{{\mathcal {L}}^\prime }{\theta ^\prime }\le \frac{{\mathcal {L}}}{\theta }+C\varepsilon ^\alpha \end{aligned}$$

for all $(r,\theta )$, $(r^\prime ,\theta ^\prime )$ satisfying $\varepsilon \le \overline{\varepsilon }_0$. This gives

$$\begin{aligned} \frac{\theta }{{\mathcal {L}}}-\frac{\theta ^\prime }{{\mathcal {L}}^\prime }\le C\varepsilon ^\alpha \end{aligned}$$

for all $(r,\theta )$, $(r^\prime ,\theta ^\prime )$ satisfying $\varepsilon \le \overline{\varepsilon }_0$ for a positive constant C.

Similarly, we can obtain

$$\begin{aligned} - C\varepsilon ^\alpha \le \frac{\theta }{{\mathcal {L}}}-\frac{\theta ^\prime }{{\mathcal {L}}^\prime } \end{aligned}$$

for all $(r,\theta )$, $(r^\prime ,\theta ^\prime )$ satisfying $\varepsilon \le \underline{\varepsilon }_0$ for a positive constant C and a positive constant $\underline{\varepsilon }_0$. Let $\varepsilon _0=\min (\overline{\varepsilon }_0,\underline{\varepsilon }_0)$. Choose $\varepsilon _{(r,\theta )}$ so that it can depend on $\rho _0^+$, $u_0^+$, $\mathcal {N}_{r_s}^+$, $\alpha$ and $\delta _6$. Then $\overline{\varepsilon }_0$ depends on $\rho _0^+$, $u_0^+$, $\mathcal {N}_{r_s}^+$, $\alpha$ and $\delta _6$. In the same way, we have $\underline{\varepsilon }_0$ depends on $\rho _0^+$, $u_0^+$, $\mathcal {N}_{r_s}^+$, $\alpha$ and $\delta _6$. Thus, $\varepsilon _0$ depends on $\rho _0^+$, $u_0^+$, $\mathcal {N}_{r_s}^+$, $\alpha$ and $\delta _6$. This finishes the proof of Claim. $\square$

By (3.84)–(3.87), we have

$$\begin{aligned} ||D{\mathcal {L}}||_{0,\alpha ,\mathcal {N}_f^+}\le C \end{aligned}$$

for a positive constant C. It can be easily shown that there exists a positive constant C such that

$$\begin{aligned} |\theta -\theta _1|\le C|{\mathcal {L}}(r,\theta )-\theta _1| \end{aligned}$$

for all $(r,\theta )\in [f(\theta ),r_1]\times [0,\theta _1]$. Using these two estimates, estimate ||DQ|| $_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}$. Then we obtain

$$\begin{aligned} ||DQ||_{\alpha ,\mathcal {N}_f^+}^{(1-\alpha ,\varGamma _w^+)}\le C, \end{aligned}$$

where C is a positive constant depending on $\rho _0^+$, $u_0^+$, $r_s$, $r_1$, $\alpha$, $\delta _1$ and $\delta _6$.

4. By the result in Step 3, Q is a $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution of (3.77), (3.78). By 2), a $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution of (3.77), (3.78) is unique. This finishes the proof. $\square$

3.4 Proof of Proposition 2.

Using the results in § 3.2 and § 3.3, we prove Proposition 2.

Proof of Proposition 2 (Existence) Suppose that $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )$ is as in Problem 3 for $\alpha \in (\frac{2}{3},1)$ and $\sigma \in (0,\overline{\sigma }_3]$ where $\overline{\sigma }_3$ is a positive constant to be determined later.

For the same $\alpha$ and $\sigma$ and $M_1>0$ to be determined later, define

$$\begin{aligned} {\mathcal {P}}(M_1)= & {} \{(f(0),\varPsi \mathbf{e}_{\varphi })\in \mathbb {R}\times C^{2,\alpha }_{(-1-\alpha ,\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)\;|\;\\&\varPsi \mathbf{e}_{\varphi }=\frac{f(\theta _1)\varPsi (f(\theta _1),\theta _1)}{r}\mathbf{e}_{\varphi }\;\text{ on }\;\varGamma _{w,r_s+f_s}^+,\\&|f(0)-r_s|+||\varPsi \mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_{r_s+f_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s+f_s}^+)}\le M_1\sigma \}. \end{aligned}$$

By the definition of ${\mathcal {P}}(M_1)$, ${\mathcal {P}}(M_1)$ is a compact convex subset of $\mathbb {R}\times C$ $^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)$. We will prove the existence part of Proposition 2 by constructing a continuous map of ${\mathcal {P}}(M_1)$ into itself as a map from $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)$ to $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)$ and applying the Schauder fixed point theorem.

In this proof, Cs and $C_i$ for $i=1,2,\ldots$ denote positive constants depending on the whole or a part of the data, $\delta _1$, $\delta _2$, $\delta _4$, $\delta _5$ and $\delta _6$ unless otherwise specified. Each C in different situations differs from each other.

1. For a fixed $(f(0),\varPsi \mathbf{e}_\varphi )$, solve (B$^\prime$).

Take $(f(0)^*,{\tilde{\varPsi }}^*\mathbf{e}_{\varphi })\in {\mathcal {P}}(M_1)$. By the definition of $f_s$ and the assumption that $f_s^\prime$ satisfies (3.1),

$$\begin{aligned} ||f_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )}\le C_1\sigma . \end{aligned}$$

Choose $\overline{\sigma }_3=\min (\frac{\delta _1}{2M_1},\frac{\delta _1}{2C_1}) (=:\overline{\sigma }_3^{(1)})$ so that

$$\begin{aligned} |f(0)^*-r_s|\le \frac{\delta _1}{2}\quad \text{ and }\quad ||f_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )} \le \frac{\delta _1}{2}. \end{aligned}$$

(3.91)

Then, using $\varPi _{f(0)^*+f_sr_s+f_s}$, extend ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }$ to a function in $\mathcal {N}_{f(0)^*+f_s}^+$:

$$\begin{aligned} \varPsi ^*\mathbf{e}_{\varphi }:=\frac{W^*}{2\pi r \sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

where $W^*:={\tilde{W}}^*(\varPi _{f(0)^*+f_sr_s+f_s})$ with ${\tilde{W}}^*:=2\pi r \sin \theta {\tilde{\varPsi }}^*$ and $\varPi _{ab}$ for $0<a,b<r_1$ is a map from $\mathcal {N}_a^+$ to $\mathcal {N}_b^+$ defined in Step 1 in the proof of Lemma 10. By (3.91) and the fact that $||{\tilde{\varPsi }}^*\mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_{r_s+f_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s+f_s}^+)}\le M_1\sigma$,

$$\begin{aligned} ||\nabla \times (\varPsi ^*\mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_{f(0)^*+f_s}^+}^{(-\alpha ,\varGamma _{w,f(0)^*+f_s}^+)}\le C_2 M_1\sigma . \end{aligned}$$

(3.92)

Choose $\overline{\sigma }_3= \min (\frac{\delta _6}{C_2M_1}, \overline{\sigma }_3^{(1)})(=:\overline{\sigma }_3^{(2)})$ where $\delta _6$ is a positive constant given in Lemma 15 so that $\varPsi \mathbf{e}_{\varphi }=\varPsi ^*\mathbf{e}_{\varphi }$ satisfies (3.81) for $f=f(0)^*+f_s$.

For $(f(0),\varPsi \mathbf{e}_{\varphi })=(f(0)^*,\varPsi ^*\mathbf{e}_{\varphi })$, solve (B$^\prime$) : solve

$$\begin{aligned}&{\left\{ \begin{array}{ll} \nabla \times ((\varPhi _0^++\varPsi ^*)\mathbf{e}_{\varphi })\cdot \nabla A=0\quad \text{ in }\quad \mathcal {N}_{f(0)^*+f_s}^+,\\ A=A_{en,f(0)^*+f_s}\quad \text{ on }\quad \varGamma _{f(0)^*+f_s}, \end{array}\right. } \end{aligned}$$

(3.93)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \nabla \times ((\varPhi _0^++\varPsi ^*)\mathbf{e}_{\varphi })\cdot \nabla T=0\quad \text{ in }\quad \mathcal {N}_{f(0)^*+f_s}^+,\\ T=T_{en,f(0)^*+f_s}\quad \text{ on }\quad \varGamma _{f(0)^*+f_s}, \end{array}\right. } \end{aligned}$$

(3.94)

where $A_{en,f(0)^*+f_s}$ and $T_{en,f(0)^*+f_s}$ are $A_{en,f(0)+f_s}$ and $T_{en,f(0)+f_s}$ given in (3.20) and (3.21), respectively, for $f(0)=f(0)^*$.

By $(\rho _-,{\mathbf {u}}_-,p_-,f_s^\prime )\in (C^{2,\alpha }(\overline{\mathcal {N}}))^3$ $\times C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$,

$$\begin{aligned}&A_{en,f(0)^*+f_s}\in C^{2,\alpha }_{(-1-\alpha ,\partial \varGamma _{f(0)^*+f_s})}(\varGamma _{f(0)^*+f_s})\\&\quad \text{ and }\quad T_{en,f(0)^*+f_s}\in C^{1,\alpha }_{(-\alpha ,\partial \varGamma _{f(0)^*+f_s})}(\varGamma _{f(0)^*+f_s}). \end{aligned}$$

Take $\overline{\sigma }_3 =\min (\overline{\sigma }_3^{(2)},\delta _5)(=:\overline{\sigma }_3^{(3)})$ where $\delta _5$ is a positive constant given in Lemma 6 so that $(\rho _-,{\mathbf {u}}_-,p_-)$ satisfies (2.19) for $\sigma \le \delta _5$. Then by Lemma 6, (3.1) and the fact that $f(0)^*$ satisfies $|f(0)^*-r_s|\le M_1\sigma$,

$$\begin{aligned} ||T_{en,f^*}||_{1,\alpha ,\varGamma _{f^*}}^{(-\alpha ,\partial \varGamma _{f^*})}\le CM_1\sigma +C\sigma , \end{aligned}$$

(3.95)

where $f^*:=f^*(0)+f_s$. Apply Lemma 15 to (3.93), (3.94). Then we obtain that each (3.93) and (3.94) has a unique axisymmetric $C^0(\overline{\mathcal {N}_f^+})\cap C^1(\mathcal {N}_f^+)$ solution

$$\begin{aligned} A^*=A_{en,f(0)^*+f_s}({\mathcal {L}}^*)\quad \text{ and }\quad T^*=T_{en,f(0)^*+f_s}({\mathcal {L}}^*), \end{aligned}$$

respectively, where ${\mathcal {L}}^*$ is ${\mathcal {L}}$ defined in Lemma 15 for $V=2\pi r \sin \theta (\varPhi _0^++\varPsi ^*)$ and $f=f(0)^*+f_s$. Furthermore, we have that $T^*$ satisfies

$$\begin{aligned} ||T^*||_{1,\alpha ,\mathcal {N}_{f^*}^+}^{(-\alpha ,\varGamma _{w,f^*}^+)}\le CM_1\sigma +C\sigma , \end{aligned}$$

(3.96)

where we used (3.95). Using the solution expression of $A^*$, estimate $\frac{A^*}{2\pi r \sin \theta }\mathbf{e}_{\varphi }$ in $C^{1,\alpha }(\overline{\mathcal {N}_{f^*}^+})$. Using (3.20), write $\frac{A^*}{2\pi r \sin \theta }\mathbf{e}_{\varphi }$ as

$$\begin{aligned} \frac{2\pi f^*({\mathcal {L}}^*)\sin ({\mathcal {L}}^*)u_{-,\varphi }({\mathcal {L}}^*)}{2\pi r \sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

(3.97)

where $u_{-,\varphi }={\mathbf {u}}_-\cdot \mathbf{e}_\varphi$. Using arguments similar to the ones in the proof of Claim in Lemma 15, we can obtain

$$\begin{aligned} ||\frac{{\mathcal {L}}^*}{\theta }||_{0,\alpha ,(f^*(\theta ),r_1)\times (0,\theta _1)}\le C. \end{aligned}$$

With this estimate, $||{\mathcal {L}}^*||_{1,\alpha ,(f^*(\theta ),r_1)\times (0,\theta _1)}\le C$, (2.19) and (3.91), we estimate (3.97) in $C^{1,\alpha }(\overline{\mathcal {N}_{f^*}^+})$. Then we have

$$\begin{aligned} ||\frac{A^*}{2\pi r \sin \theta }\mathbf{e}_{\varphi }||_{1,\alpha ,\mathcal {N}_{f^*}^+} \le C\sigma . \end{aligned}$$

(3.98)

2. By substituting an extension of $(\varPsi ^*\mathbf{e}_{\varphi },A^*,T^*)$ into nonlinear parts of (A$^\prime$), obtain a linear problem having unknowns $(f(0),\varPsi \mathbf{e}_{\varphi })$.

Let $f(0)^\sharp$ be a point in $[r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$. By the choice of $\overline{\sigma }_3$,

$$\begin{aligned} ||f^\sharp -r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )} \le \delta _1, \end{aligned}$$

where $f^\sharp :=f(0)^\sharp +f_s$. Extend $\varPsi ^*\mathbf{e}_{\varphi }$ $A^*$ and $T^*$ to functions in $\mathcal {N}_{f^\sharp }^+$:

$$\begin{aligned} \varPsi ^{\sharp ,*}\mathbf{e}_{\varphi }:=\frac{W^{\sharp ,*}}{2\pi r \sin \theta }\mathbf{e}_{\varphi },\quad A^{\sharp ,*}:=A^*(\varPi _{f^\sharp f^*})\quad \text{ and }\quad T^{\sharp ,*}:=T^*(\varPi _{f^\sharp f^*}), \end{aligned}$$

where $W^{\sharp ,*}=W^*(\varPi _{f^\sharp f^*})$. By (3.92), (3.96) and (3.98),

$$\begin{aligned}&||\nabla \times (\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_{f^\sharp }^+}^{(-\alpha ,\varGamma _{w,f^\sharp }^+)}+||\frac{A^{\sharp ,*}}{2\pi r \sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_{f^\sharp }^+}+||T^{\sharp ,*}||_{1,\alpha ,\mathcal {N}_{f^\sharp }^+}^{(-\alpha ,\varGamma _{w,f^\sharp }^+)}\nonumber \\&\qquad \le C_3M_1\sigma +C_4\sigma \end{aligned}$$

(3.99)

for all $f^\sharp (0)\in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$. Take $\overline{\sigma }_3= \min (\overline{\sigma }_3^{(3)}, \frac{\delta _4}{C_3M_1+C_4})(=:\overline{\sigma }_3^{(4)})$ where $\delta _4$ is a positive constant given in Lemma 5 so that $\varrho (\nabla \times ((\varPhi _0^++\varPsi ^{\sharp ,*})\mathbf{e}_{\varphi }),$ $\frac{A^{\sharp ,*}}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S_0^++T^{\sharp ,*})$ is well-defined in $\mathcal {N}_{f^\sharp }^+$ and $\varrho (\nabla \times ((\varPhi _0^++\varPsi ^{\sharp ,*})\mathbf{e}_{\varphi }),\frac{A^{\sharp ,*}}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S_0^++T^{\sharp ,*})$ and $S_0^++T^{\sharp ,*}$ are strictly positive in $\mathcal {N}_{f^\sharp }^+$ for all $f^\sharp (0)\in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$. By substituting $f(0)^\sharp$ and $(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*})$ into the place of f(0) in (A$^\prime$) and $(\varPsi \mathbf{e}_{\varphi },A,T)$ in ${\varvec{F}}_1(\varPsi \mathbf{e}_{\varphi },A,T)$, $\mathfrak f_0(T,p_{ex})$ and $\mathfrak f_1(\varPsi \mathbf{e}_{\varphi },A,T)$ in (A$^\prime$), we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (\varPsi \mathbf{e}_{\varphi })\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta T}{r}\mathbf{e}_{\varphi }\nonumber \\&\qquad +{\varvec{F}}_1(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*})\quad \text{ in }\quad \mathcal {N}_{f^\sharp }^+, \end{aligned}$$

(3.100)

$$\begin{aligned}&\varPsi \mathbf{e}_{\varphi }={\left\{ \begin{array}{ll} (\varPhi _--\varPhi _0^-)\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{f^\sharp },\\ \frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r}\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{w,f^\sharp }^+:=\varGamma _w\cap \{r>f^\sharp \},\\ \bigg (\frac{1}{r_1\sin \theta }\int _0^\theta \bigg (\mathfrak f_0(T^{\sharp ,*},p_{ex}) -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}T\\ +\mathfrak f_1(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*}) \bigg )r_1^2\sin \xi \hbox {d}\xi \bigg )\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(3.101)

$$\begin{aligned}&\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1} \bigg (\mathfrak f_0(T^{\sharp ,*},p_{ex}) -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}T\nonumber \\&\quad +\mathfrak f_1(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*}) \bigg )\bigg |_{r=r_1}r_1^2\sin \xi \hbox {d}\xi =\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1}.\nonumber \\ \end{aligned}$$

(3.102)

We let $f(0)^\sharp$ be an unknown in this problem and let T in (3.100)–(3.102) be a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \nabla \times ( (\varPhi _0^++\varPsi ^{\sharp ,*})\mathbf{e}_{\varphi })\cdot \nabla T=0\quad \text{ in }\quad \mathcal {N}_{f^\sharp }^+,\\ T=T_{en,f^\sharp }\quad \text{ on }\quad \varGamma _{f^\sharp }. \end{array}\right. } \end{aligned}$$

(3.103)

Then since a solution T of (3.103) is uniquely determined by $f(0)^\sharp$ (see Step 3), unknowns of (3.100)–(3.102) become $(f^\sharp (0),\varPsi \mathbf{e}_\varphi )$. We denote an unknown $\varPsi \mathbf{e}_{\varphi }$ of (3.100)–(3.102) by $\varPsi ^\sharp \mathbf{e}_{\varphi }$.

3. Find $f(0)^\sharp$ using (3.102).

By the definition of $\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi }$ and (2.27), the transport equation in (3.103) can be written as

$$\begin{aligned} \nabla ^\perp (V_0^++W^{\sharp ,*})\cdot \nabla T=0\quad \text{ in }\quad \mathcal {N}_{f^\sharp }^+. \end{aligned}$$

From this form of the transport equation in (3.103), we see that the stream surface of the vector field $\nabla \times ( (\varPhi _0^++\varPsi ^{\sharp ,*})\mathbf{e}_{\varphi })$ in $\mathcal {N}_{f^\sharp }^+$ is obtained by stretching or contracting the stream surface of the vector field $\nabla \times ((\varPhi _0^++\varPsi ^*)\mathbf{e}_{\varphi })$ in $\mathcal {N}_{f^*}^+$ in r-direction. Using this fact, we obtain that the solution of (3.103) is given by

$$\begin{aligned} T=T_{en,f^\sharp }({\mathcal {L}}^\sharp ), \end{aligned}$$

where ${\mathcal {L}}^\sharp ={\mathcal {L}}^*(\varPi _{f^\sharp f^*})$. We denote this solution by $T^\sharp$.

By (3.21), $T^\sharp$ is expressed as

$$\begin{aligned} T^{\sharp }=\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }_{f^\sharp }({\mathcal {L}}^\sharp )}}{c_-}\right) ^2\right) S_-\right) (f(0)^\sharp +f_s({\mathcal {L}}^\sharp ),{\mathcal {L}}^\sharp )-(g({M_0^-}^2))(r_s)S_{in}. \end{aligned}$$

Substituting this expression of $T^\sharp$ into the place of T in (3.102) using the fact that $\varPi _{f^\sharp f^*}(r_1,\theta )=(r_1,\theta )$, we obtain

$$\begin{aligned} (L)&:=\frac{1}{r_1\sin \theta _1} \int _0^{\theta _1}\left. \frac{\rho _0^+((\gamma -1){u_0^+}^2 +{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}\right| _{r=r_1}(a)_{1} r_1^2\sin \xi \hbox {d}\xi \nonumber \\&=-\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1}+\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1} \left( \mathfrak f_0(T^*,p_{ex})\right. \nonumber \\&\quad -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}\left( (a)_{2}+ \left. \left. (a)_{3}\right) +\mathfrak f_1(\varPsi ^{*}\mathbf{e}_{\varphi },A^{*}, T^{*}) \right) \right| _{r=r_1}\nonumber \\&\qquad r_1^2\sin \xi \hbox {d}\xi \nonumber \\&=:(R), \end{aligned}$$

(3.104)

where

$$\begin{aligned} (a)_{1}&= (g({M_0^-}^2))(f(0)^\sharp +f_s({\mathcal {L}}^{*}(r_1,\theta )))S_{in}\\&\quad -(g({M_0^-}^2))(r_s+f_s({\mathcal {L}}^{*}(r_1,\theta )))S_{in},\\ (a)_{2}&=\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }_{f^\sharp }({\mathcal {L}}^{*}(r_1,\theta ))}}{c_-}\right) ^2\right) S_-\right) (f(0)^\sharp +f_s({\mathcal {L}}^{*}(r_1,\theta )),{\mathcal {L}}^{*}(r_1,\theta ))\\&\quad -(g({M_0^-}^2))(f(0)^\sharp +f_s({\mathcal {L}}^{*}(r_1,\theta )))S_{in},\\ (a)_{3}&=(g({M_0^-}^2))(r_s+f_s({\mathcal {L}}^{*}(r_1,\theta )))S_{in}-(g({M_0^-}^2))(r_s)S_{in}. \end{aligned}$$

We find $f(0)^\sharp$ satisfying (3.104). For this, we estimate |(R)|.

Estimate |(R)|:

With (2.19) and (3.1), we estimate $(a)_2$, $(a)_3$ and $\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1}$. Then we have

$$\begin{aligned} \sup _{\theta \in (0,\theta _1)}|(a)_{2}|&\le C\sigma |f(0)^\sharp -r_s|+C\sigma \nonumber \\&\le C\frac{\delta _1}{2}\sigma +C\sigma \end{aligned}$$

(3.105)

for all $f^\sharp (0)\in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$,

$$\begin{aligned} \sup _{\theta \in (0,\theta _1)}|(a)_{3}|\le C\sigma \end{aligned}$$

(3.106)

and

$$\begin{aligned} \left| \frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1}\right| \le C\sigma . \end{aligned}$$

(3.107)

With (2.20), we estimate $\mathfrak f_0(T^*,p_{ex})(r_1,\theta )$. Then we have

$$\begin{aligned} \sup _{\theta \in (0,\theta _1)}|\mathfrak f_0(T^*,p_{ex})(r_1,\theta )|\le C \sigma . \end{aligned}$$

(3.108)

By Lemma 5 and using (3.92), (3.96) and (3.98), we estimate $\mathfrak f_1(\varPsi ^{*}\mathbf{e}_{\varphi },A^{*},T^*)$ $(r_1,\theta )$. Then we have

$$\begin{aligned} \sup _{\theta \in (0,\theta _1)}|\mathfrak f_1(\varPsi ^{*}\mathbf{e}_{\varphi },A^{*},T^*)(r_1,\theta )| \le C(M_1+1)^2\sigma ^2. \end{aligned}$$

(3.109)

With (3.105)–(3.109), we estimate |(R)|. Then we obtain

$$\begin{aligned} |(R)|\le C_5(M_1+1)^2\sigma ^2 +C_6\sigma \end{aligned}$$

(3.110)

for all $f^\sharp (0)\in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$.

Then we find $f(0)^\sharp$. By Lemma 2, there exists a positve constant $\lambda$ such that

$$\begin{aligned} (L)^\prime (f(0)^\sharp )\ge \lambda \end{aligned}$$

for all $f^\sharp (0)\in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$. By this fact and $(L)(r_s)=0$,

$$\begin{aligned} (L)\left( \frac{\delta _1}{2}\right) \ge \frac{\delta _1\lambda }{2}\quad \text{ and }\quad (L)\left( -\frac{\delta _1}{2}\right) \le -\frac{\delta _1\lambda }{2}. \end{aligned}$$

(3.111)

Choose $\overline{\sigma }_3= \min (\overline{\sigma }_3^{(4)}, \frac{C_6}{C_5(M_1+1)^2},\frac{\delta _1\lambda }{8C_6}) (=:\overline{\sigma }_3^{(5)})$. Then since

$$\begin{aligned} |(R)|\le \frac{\delta _1\lambda }{4} \end{aligned}$$

for all $f(0)^\sharp \in [r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$ and (3.111) holds, by the intermediate value theorem, there exists $f(0)^\sharp$ satisfying (3.104) in $[r_s-\frac{\delta _1}{2},r_s+\frac{\delta _1}{2}]$. Such $f(0)^\sharp$ is unique because (L) is a monotone function of $f(0)^\sharp$. Since at $f(0)^\sharp$ where (3.104) holds it holds that

$$\begin{aligned} \lambda |f(0)^\sharp -r_s|\le |(R)(f(0)^\sharp )|, \end{aligned}$$

we have that

$$\begin{aligned} |f(0)^\sharp -r_s|\le C(M_1+1)^2\sigma ^2+C\sigma . \end{aligned}$$

(3.112)

4. Find $\varPsi ^\sharp \mathbf{e}_{\varphi }$.

Fix $f(0)^\sharp$ obtained in Step 3. By (3.112), Lemma 6 and Lemma 15, $T^{\sharp }$ determined by $f(0)^\sharp$ satisfies

$$\begin{aligned} ||T^{\sharp }||_{1,\alpha ,\mathcal {N}_{f^\sharp }^+}^{(-\alpha ,\varGamma _{w,f^\sharp }^+)}\le C(M_1+1)^2\sigma ^2+C\sigma . \end{aligned}$$

(3.113)

Substitute this $T^{\sharp }$ into the place of T in (3.100) and (3.101). Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (\varPsi \mathbf{e}_{\varphi })\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta T^\sharp }{r}\mathbf{e}_{\varphi }+{\varvec{F}}_1(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*}) \quad \text{ in }\quad \mathcal {N}_{f^\sharp }^+, \end{aligned}$$

(3.114)

$$\begin{aligned}&\varPsi \mathbf{e}_{\varphi }={\left\{ \begin{array}{ll} (\varPhi _--\varPhi _0^-)\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{f^\sharp },\\ \frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r}\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{w,f^\sharp }^+,\\ \bigg (\frac{1}{r_1\sin \theta }\int _0^\theta \bigg (\mathfrak f_0(T^{\sharp ,*},p_{ex})-\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}T^\sharp \\ \quad \quad +\mathfrak f_1(\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi },A^{\sharp ,*},T^{\sharp ,*}) \bigg )r_1^2\sin \xi \hbox {d}\xi \bigg )\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{ex}. \end{array}\right. } \end{aligned}$$

(3.115)

Since $f(0)^\sharp$ is chosen for $T=T^\sharp$ to satisfy (3.102), (3.115) is a continuous boundary condition. We apply Lemma 7 to (3.114), (3.115) with (2.19), (2.20), (3.1), (3.99) and (3.113). Then we obtain that (3.114), (3.115) has a unique $C^{2,\alpha }_{(-1-\alpha ,\varGamma _{f^\sharp })}(\mathcal {N}_{f^\sharp }^+)$ solution $\varPsi ^\sharp \mathbf{e}_{\varphi }$ and this solution satisfies the estimate

$$\begin{aligned} ||\varPsi ^\sharp \mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_{f^\sharp }^+}^{(-1-\alpha ,\varGamma _{w,f^\sharp }^+)}\le C(M_1+1)^2\sigma ^2+C\sigma . \end{aligned}$$

(3.116)

5. Using $\varPi _{r_s+f_sf^\sharp }$, transform $\varPsi ^\sharp \mathbf{e}_{\varphi }$ into a function in $\mathcal {N}_{r_s+f_s}^+$:

$$\begin{aligned} {\tilde{\varPsi }}^\sharp \mathbf{e}_{\varphi }=\frac{{\tilde{W}}^\sharp }{2\pi r \sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

where ${\tilde{W}}^\sharp =W^\sharp (\varPi _{r_s+f_sf^\sharp })$ with $W^\sharp =2\pi r \sin \theta \varPsi ^\sharp$. By (3.112) and (3.116),

$$\begin{aligned} ||{\tilde{\varPsi }}^\sharp \mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_{r_s+f_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s+f_s}^+)}\le C(M_1+1)^2\sigma ^2+C\sigma . \end{aligned}$$

Combining this with (3.112), we have

$$\begin{aligned} |f(0)^\sharp -r_s|+||{\tilde{\varPsi }}^\sharp \mathbf{e}_{\varphi }||_{2,\alpha ,\mathcal {N}_{r_s+f_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s+f_s}^+)}\le C_7(M_1+1)^2\sigma ^2+C_8\sigma . \end{aligned}$$

Take $M_1=2C_8$ and $\overline{\sigma }_3= \min (\overline{\sigma }_3^{(5)},\frac{C_8}{C_7(M_1+1)^2})(=:\overline{\sigma }_3^{(6)})$. And then define a map ${\mathcal {J}}$ from $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)$ to $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s+f_s}^+)}(\mathcal {N}_{r_s+f_s}^+)$ by

$$\begin{aligned} {\mathcal {J}}(f(0)^*,{\tilde{\varPsi }}^*\mathbf{e}_{\varphi })=(f(0)^\sharp , {\tilde{\varPsi }}^\sharp \mathbf{e}_{\varphi }). \end{aligned}$$

Then by the choice of $M_1$ and $\overline{\sigma }_3$, ${\mathcal {J}}$ is a map of ${\mathcal {P}}(M_1)$ into itself. Using the standard argument, one can easily check that ${\mathcal {J}}$ is continuous. Thus, the Schauder fixed point theorem can be applied to ${\mathcal {J}}$. We apply the Schauder fixed point theorem to ${\mathcal {J}}$. Then we obtain that there exists a fixed point $(f^\flat (0),{\tilde{\varPsi }}^\flat \mathbf{e}_{\varphi })\in {\mathcal {P}}(M_1)$ of ${\mathcal {J}}$.

One can see that if $(f(0)^*,{\tilde{\varPsi }}^*\mathbf{e}_{\varphi })=(f(0)^\sharp , {\tilde{\varPsi }}^\sharp \mathbf{e}_{\varphi })$, then $\varPsi ^*\mathbf{e}_{\varphi } =\varPsi ^{\sharp ,*}\mathbf{e}_{\varphi }=\varPsi ^\sharp \mathbf{e}_{\varphi }$, $A^{\sharp ,*}=A^*$ and $T^*=T^{\sharp ,*}=T^\sharp$. From this fact, we see that $(f(0),\varPhi \mathbf{e}_{\varphi },L,$ $S)=(f(0)^\flat ,(\varPhi _0^++\varPsi ^\flat )\mathbf{e}_{\varphi },A^\flat , S_0^++T^\flat )$, where

$$\begin{aligned} \varPsi ^\flat \mathbf{e}_{\varphi }:=\frac{(2\pi r \sin \theta {\tilde{\varPsi }}^\flat )(\varPi _{f(0)^\flat +f_s r_s+f_s})}{2\pi r \sin \theta }\mathbf{e}_{\varphi } \end{aligned}$$

and $A^\flat$ and $T^\flat$ are solutions of (3.93) and (3.94) for given $(f(0)^*,\varPsi ^*)=(f(0)^\flat ,\varPsi ^\flat )$, respectively, is a solution of (A), (B). By $(f(0)^\flat , {\tilde{\varPsi }}^\flat \mathbf{e}_{\varphi })\in {\mathcal {P}}(M_1)$, (3.96) and (3.98),

$$\begin{aligned}&|f(0)^\flat -r_s|\nonumber \\&\quad +||\nabla \times (\varPsi ^\flat \mathbf{e}_{\varphi })||_{1,\alpha , \mathcal {N}_{f^\flat }^+}^{(-\alpha ,\varGamma _{w,f^\flat }^+)}+||\frac{A^\flat }{2\pi r \sin \theta }||_{1,\alpha ,\mathcal {N}_{f^\flat }^+} \nonumber \\&\qquad +||T^\flat ||_{1,\alpha ,\mathcal {N}_{f^\flat }^+}^{(-\alpha ,\varGamma _{w,f^\flat }^+)}\le C\sigma , \end{aligned}$$

(3.117)

where $f^\flat :=f(0)^\flat +f_s$. One can easily see that there exists a positive constant $\delta _7$ such that for any $\mathcal {N}_f^+\subset \mathcal {N}_{r_s-\delta _1}^+$, if $(\nabla \times (\varPhi \mathbf{e}_{\varphi }),\frac{L}{2\pi r \sin \theta }\mathbf{e}_{\varphi },S,B_0)\in B^{(1)}_{\delta _7,\mathcal {N}_f^+}$ where $B^{(1)}_{\delta ,\varOmega }$ for $\delta >0$ and $\varOmega \subset \mathbb {R}^3$ is a neighborhood of $(\nabla \times (\varPhi _0^+\mathbf{e}_\varphi ),0,S_0^+,B_0)$ defined in Lemma 3, then (2.51) holds in $\mathcal {N}_f^+$. Let $\delta _7$ be one such constant. Take $\overline{\sigma }_3=\min (\overline{\sigma }_3^{(6)},\frac{\delta _7}{C})$ where C is C in (3.117). Then $(f(0)^\flat ,(\varPhi _0^++\varPsi ^\flat )\mathbf{e}_{\varphi },A^\flat , S_0^++T^\flat )$ is a subsonic solution of (2.43)–(2.45) in $\mathcal {N}_{f^\flat }^+$. Choose $\delta _1$, $\delta _2$, $\delta _4$, $\delta _5$, $\delta _6$ and $\delta _7$ so that they can depend on the data. Then $M_1$ and $\overline{\sigma }_3$ depend on the data. This finishes the proof. $\square$

Proof of Proposition 2(Uniqueness) Let $\alpha \in (\frac{2}{3},1)$. Let $\sigma _3$ be a positive constant $\le \overline{\sigma }_3$ and to be determined later. Suppose that there exist two solutions $(f_i(0), \varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ for $i=1,2$ of Problem 3 for $\sigma \le \sigma _3$ satisfying the estimate (3.2).

Let $(\varPsi _i,A_i,T_i):=(\varPhi _i-\varPhi _0^+,L_i,S_i-S_0^+)$ for $i=1,2$. We will prove that there exists a positive constant $\underline{\sigma }_3$ such that if $\sigma _3=\underline{\sigma }_3$, then

$$\begin{aligned} (f_1(0), \varPsi _1\mathbf{e}_{\varphi },A_1,T_1)=(f_2(0),\varPsi _2\mathbf{e}_{\varphi }, A_2,T_2), \end{aligned}$$

by constructing a contraction map in a low regularity space.

In this proof, Cs denote positive constants depending on the data unless otherwise specified. Each C in different situations differs from each other.

1. By subtracting (A$^\prime$) satisfied by $(f_1(0), \varPsi _1\mathbf{e}_{\varphi },A_1,T_1)$ from (A$^\prime$) satisfied by $(f_2(0), \varPsi _2\mathbf{e}_{\varphi },$ $A_2,T_2)$, obtain the equations that will give a contraction map.

Let $f_i:=f_i(0)+f_s$ for $i=1,2$. From

$$\begin{aligned} (\nabla ^\perp a)(\varPi _{f_2f_1})=N\nabla ^\perp (a(\varPi _{f_2f_1}))\quad \text{ and }\quad \nabla \times \left( \frac{a\mathbf{e}_{\varphi }}{2\pi r \sin \theta }\right) =\nabla ^\perp a \end{aligned}$$

for an axisymmetric scalar function a, we can obtain

$$\begin{aligned}&(\nabla \times (\varPsi _1\mathbf{e}_{\varphi }))(\varPi _{f_2f_1})=N\nabla \times ( {\tilde{\varPsi }}_1\mathbf{e}_{\varphi })\nonumber \\&\quad \text{ and }\quad \left( \nabla \times \left( \frac{A_1}{2\pi r \sin \theta }\mathbf{e}_{\varphi }\right) \right) (\varPi _{f_2f_1})= N\nabla \times \left( \frac{{\tilde{A}}_1}{2\pi r \sin \theta }\mathbf{e}_{\varphi }\right) , \end{aligned}$$

(3.118)

where ${\tilde{\varPsi }}_1:=\frac{{\tilde{W}}_1}{2\pi r \sin \theta }$ with ${\tilde{W}}_1:=W_1(\varPi _{f_2f_1})$ and $W_1:=2\pi r \sin \theta \varPsi _1$, ${\tilde{A}}_1:=A_1(\varPi _{f_2f_1})$ and

$$\begin{aligned} N= & {} \frac{r^2}{(\varPi _{f_2f_1}^r)^2}\mathbf{e}_r\otimes \mathbf{e}_r-\frac{\left( \partial _{\tilde{\theta }}\varPi _{f_1f_2}^{r}\right) (\varPi _{f_2f_1}) }{(\varPi _{f_2f_1}^r)^2}r\mathbf{e}_r\otimes \mathbf{e}_\theta \\&+\frac{\left( \partial _{\tilde{r}}\varPi _{f_1f_2}^{r}\right) (\varPi _{f_2f_1}) }{\varPi _{f_2f_1}^r}r\mathbf{e}_\theta \otimes \mathbf{e}_\theta , \end{aligned}$$

with $\varPi _{f_2f_1}^r$ and $\varPi _{f_1f_2}^{r}$, r-components of $\varPi _{f_2f_1}^*$ and $\varPi _{f_1f_2}^*$, respectively (see the definition of $\varPi _{ab}^*$ in (3.56)), $(\tilde{r},\tilde{\theta })$, $(r,\theta )$ coordinates for the Cartesian coordinate for $\mathcal {N}_{f_1(0)+f_s}^+$, and $(r,\theta )=\varPi _{f_1f_2}^*(\tilde{r},\tilde{\theta })$. Using $\varPi _{f_2f_1}$ and (3.118), transform (A$^\prime$) satisfied by $(f_1(0), \varPsi _1\mathbf{e}_{\varphi },A_1,T_1)$. And then subtract the resultant equations from (A$^\prime$) satisfied by $(f_2(0), \varPsi _2\mathbf{e}_{\varphi },A_2,T_2)$. Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ((\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi })\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta (T_2-{\tilde{T}}_1)}{r}\mathbf{e}_{\varphi }+{\varvec{F}}_3\quad \text{ in }\quad \mathcal {N}_{f_2(0)+f_s}^+ \end{aligned}$$

(3.119)

$$\begin{aligned}&(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }\nonumber \\&\quad = {\left\{ \begin{array}{ll} (\varPhi _--\varPhi _0^-)\mathbf{e}_{\varphi }-\frac{\varPi _{f_2f_1}^r(\varPhi _--\varPhi _0^-)(\varPi _{f_2f_1})}{r}\mathbf{e}_{\varphi }(=:\overline{{\varvec{h}}}_1)\quad \text{ on }\quad \varGamma _{f_2(0)+f_s},\\ {\varvec{0}}\quad \text{ on }\quad \varGamma _{w,f_2(0)+f_s}^+:=\varGamma _w\cap \{r>f_2(0)+f_s\},\\ \bigg (\frac{1}{r_1\sin \theta } \int _0^\theta \bigg (\mathfrak f_0(T_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex}) -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}(T_2-{\tilde{T}}_1)\\ +\mathfrak f_1(\varPsi _2\mathbf{e}_{\varphi },A_2,T_2) -\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1)\bigg )r_1^2\sin \xi \hbox {d}\xi \bigg )\mathbf{e}_{\varphi }(=:\overline{{\varvec{h}}}_2)\\ \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \nonumber \\\end{aligned}$$

(3.120)

$$\begin{aligned}&0=\frac{1}{r_1\sin \theta _1} \int _0^{\theta _1} \bigg (\mathfrak f_0(T_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex}) -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}\nonumber \\&\times (T_2-{\tilde{T}}_1)\nonumber \\&\qquad \left. +\mathfrak f_1(\varPsi _2\mathbf{e}_{\varphi },A_2,T_2) -\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1)\bigg )\right| _{r=r_1}r_1^2\sin \xi \hbox {d}\xi , \end{aligned}$$

(3.121)

where

$$\begin{aligned} {\varvec{F}}_3&= -\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ({\tilde{\varPsi }}_1\mathbf{e}_{\varphi })\right) \\&\quad +M\nabla \times \left( \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\right) (\varPi _{f_2f_1})N\nabla \times ( {\tilde{\varPsi }}_1\mathbf{e}_{\varphi })\right) \\&\quad +\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{1}{r}\partial _\theta {\tilde{T}}_1 \mathbf{e}_{\varphi }\\&\quad -\left( \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{1}{r}\right) (\varPi _{f_2f_1})\left( \frac{\partial \varPi _{f_1f_2}^{r}}{\partial {\tilde{\theta }}}(\varPi _{f_2f_1})\partial _r + \partial _\theta \right) {\tilde{T}}_1\mathbf{e}_{\varphi }\\&\quad +{\varvec{F}}_1(\varPsi _2\mathbf{e}_{\varphi },A_2,T_2)-\tilde{\varvec{F}}_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1), \end{aligned}$$

${\tilde{T}}_1=T_1(\varPi _{f_2f_1})$,

$$\begin{aligned} M= & {} \left( \partial _{\tilde{r}}\varPi _{f_1f_2}^{r}\right) (\varPi _{f_2f_1})\mathbf{e}_r\otimes \mathbf{e}_r +\frac{\left( \partial _{\tilde{\theta }}\varPi _{f_1f_2}^{r}\right) (\varPi _{f_2f_1})}{\varPi _{f_2f_1}^r} \mathbf{e}_\theta \otimes \mathbf{e}_r\\&+ \frac{r}{\varPi _{f_2f_1}^r}\mathbf{e}_\theta \otimes \mathbf{e}_\theta + \frac{r}{\varPi _{f_2f_1}^r}\mathbf{e}_{\varphi }\otimes \mathbf{e}_{\varphi }, \end{aligned}$$

and $\tilde{\varvec{F}}_1$ is ${\varvec{F}}_1$ changed by using the transformation $\varPi _{f_2f_1}$. We will construct a contraction map using (3.119)–(3.121). For this, we estimate $||T_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}$ for $\beta =1-\frac{3}{q}\in (0,\alpha )$ where $q\in (3,\frac{1}{1-\alpha })$ and $||(\frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{f_2}^+}$.

2. Estimate $||T_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}$.

Since $T_i$ for $i=1,2$ are solutions of

$$\begin{aligned} \nabla \times ((\varPhi _0^++\varPsi _i)\mathbf{e}_{\varphi })\cdot \nabla T=0\quad \text{ in }\quad \mathcal {N}_{f_i(0)+f_s}^+,\quad T=T_{en,f_i(0)+f_s}\quad \text{ on }\quad \varGamma _{f_i(0)+f_s} \end{aligned}$$

for $i=1,2$, respectively, where $T_{en,f_i(0)+f_s}$ is $T_{en,f(0)+f_s}$ given in (3.21) for $f(0)=f_i(0)$, by Lemma 15,

$$\begin{aligned} T_i=T_{en,f_i(0)+f_s}({\mathcal {L}}_i) \end{aligned}$$

(3.122)

for $i=1,2$ where ${\mathcal {L}}_i$ are ${\mathcal {L}}$ given in Lemma 15 for $V=V_0^++ W_i(=:V_i)$ and $f=f_i(0)+f_s$ with $W_i:=2\pi r \sin \theta \varPsi _i$.

By (3.21), (3.122) and the definition of ${\tilde{T}}_1$, $T_2-{\tilde{T}}_1$ can be written as

$$\begin{aligned}&\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_2}({\mathcal {L}}_2)}{c_-}\right) ^2\right) S_- \right) (f_2({\mathcal {L}}_2),{{\mathcal {L}}}_2)\\&\qquad -\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1) \end{aligned}$$

where ${\varvec{\nu }}_{f_i}$ for $i=1,2$ are the unit normal vectors on $\varGamma _{f_i}$ pointing toward $\mathcal {N}_{f_i }^+$, respectively, and $\tilde{{\mathcal {L}}}_1:={\mathcal {L}}_1(\varPi _{f_2f_1})$. This can be decomposed into $(a)+(b)$ where

$$\begin{aligned} (a):= & {} \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_2}({\mathcal {L}}_2)}{c_-}\right) ^2\right) S_- \right) (f_2({\mathcal {L}}_2),{{\mathcal {L}}}_2)\\&-\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2({\mathcal {L}}_2),{{\mathcal {L}}}_2) \end{aligned}$$

and

$$\begin{aligned} (b):= & {} \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2({\mathcal {L}}_2),{{\mathcal {L}}}_2)\\&-\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1). \end{aligned}$$

To estimate $T_2-{\tilde{T}}_1$ in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, we estimate (a) and (b) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, respectively. To obtain an estimate of (a) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ and later to obtain an estimate of $||(\frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{f_2}^+}$, we prove the following claim (before we prove the Claim, we take $\sigma _3=\min (\overline{\sigma }_3,\frac{\delta _6}{C})(=:\sigma _3^{(1)})$ where C is C in (3.2) so that $\varPsi \mathbf{e}_\varphi =\varPsi _i\mathbf{e}_\varphi$ for $i=1,2$ satisfy (3.79) for $f=f_i(0)+f_s$):

Claim. Let $h:[0,\theta _1]\rightarrow \mathbb {R}$ be a function in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$. It holds that

$$\begin{aligned}&||\int _0^1 h^\prime (t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1)||_{0,\beta ,\mathcal {N}_{f_2}^+}\\&\quad \le C||h||_{1,\beta ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

Proof of Claim

By the definitions of ${\mathcal {L}}_2$ and $\tilde{{\mathcal {L}}}_1$,

$$\begin{aligned} {\mathcal {L}}_2=k_2^{-1}\circ V_2\quad \text{ and }\quad \tilde{{\mathcal {L}}}_1=(k_1^{-1}\circ V_1)(\varPi _{f_2f_1}), \end{aligned}$$

where $k_i(\theta )=V_i(f_i(\theta ),\theta )$ for $i=1,2$. By $V_0^+(\varPi _{f_2f_1})=V_0^+$ and ${\tilde{W}}_1(f_2(\theta ),\theta )=W_1(f_1(\theta ),\theta )$ where ${\tilde{W}}_1$ is defined below (3.118), $\tilde{{\mathcal {L}}}_1$ can be written as

$$\begin{aligned} \tilde{{\mathcal {L}}}_1={\tilde{k}}_1^{-1}\circ {\tilde{V}}_1, \end{aligned}$$

where ${\tilde{V}}_1=V_0^++{\tilde{W}}_1$ and ${\tilde{k}}_1={\tilde{V}}_1(f_2(\theta ),\theta )$. Since ${\tilde{V}}_1(r,\theta )\in [0,V_-(r_0,\theta _1)]$ for $(r,\theta )\in [f_2(\theta ),r_1]\times [0,\theta _1]$ and $V_2(f_2(\theta ),\theta )\in [0,V_-(r_0,\theta _1)]$ for $\theta \in [0,\theta _1]$ where $V_-=2\pi r \sin \theta \varPhi _-$, $k_2^{-1}\circ {\tilde{V}}_1$ is well-defined in $\overline{\mathcal {N}_{f_2}^+}$. With this fact, we write ${\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1$ as

$$\begin{aligned} {\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1&=(k_2^{-1}\circ V_2-k_2^{-1}\circ {\tilde{V}}_1)+(k_2^{-1}\circ {\tilde{V}}_1-{\tilde{k}}_1^{-1}\circ {\tilde{V}}_1)\\&=\int _0^1\frac{1}{k_2^\prime (k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1))}\hbox {d}t (V_2-{\tilde{V}}_1)\\&\qquad +\int _0^1 \frac{1}{{\tilde{k}}_1^\prime ({\tilde{k}}_1^{-1}\circ (t {\tilde{k}}_1(\vartheta )+(1-t)k_2(\vartheta )))} \hbox {d}t ({\tilde{k}}_1(\vartheta )-k_2(\vartheta )), \end{aligned}$$

where $\vartheta =k_2^{-1}\circ {\tilde{V}}_1$ and we used the fact that $k_2^{-1}\circ {\tilde{V}}_1-{\tilde{k}}_1^{-1}\circ {\tilde{V}}_1={\tilde{k}}_1^{-1}\circ {\tilde{k}}_1(\vartheta )-{\tilde{k}}_1^{-1}\circ k_2(\vartheta )$.

Substitute the above expression of ${\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1$ into the place of ${\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1$ in $\int _0^1 h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1)$. Then we have

$$\begin{aligned}&\int _0^1 h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1)\\&\quad =\int _0^1 h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t\int _0^1\frac{1}{k_2^\prime (k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1))}\hbox {d}t (V_2-{\tilde{V}}_1)\\&\qquad +\int _0^1 h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t \int _0^1\frac{1}{{\tilde{k}}_1^\prime ({\tilde{k}}_1^{-1}\circ (t {\tilde{k}}_1(\vartheta )+(1-t)k_2(\vartheta )))}\hbox {d}t \\&\qquad \quad ({\tilde{k}}_1(\vartheta )-k_2(\vartheta )) =:(c)+(d). \end{aligned}$$

To estimate $||\int _0^1 h^\prime (t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1)||_{0,\beta ,\mathcal {N}_{f_2}^+}$, we estimate (c) and (d) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, respectively.

Estimate (c) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$:

With the definitions of $k_2$, $\varPsi _2$ and ${\tilde{\varPsi }}_1$, write (c) as

$$\begin{aligned} (c)= & {} \int _0^1 (\theta -\theta _1) h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t\nonumber \\&\int _0^1\frac{r \sin \theta }{J\sin (k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1))} \hbox {d}t\int _0^1 \partial _{\theta }\nonumber \\&\quad (\varPsi _2-{\tilde{\varPsi }}_1)(r,t\theta +(1-t)\theta _1) \hbox {d}t, \end{aligned}$$

(3.123)

where

$$\begin{aligned} J:= & {} \left( f_2^2\sqrt{1+\left( \frac{f_2^\prime }{f_2}\right) ^2}\right) (\cdot )\\&(\nabla \times ((\varPhi _0^++\varPsi _2)\mathbf{e}_{\varphi }))(f(\cdot ),\cdot )\cdot {\varvec{\nu }}_{f_2}(\cdot )\bigg |_{\cdot =k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1)}. \end{aligned}$$

By the choice of $\sigma _3$, $\varPsi \mathbf{e}_\varphi =\varPsi _i\mathbf{e}_\varphi$ for $i=1,2$ satisfy (3.79) for $f=f_i(0)+f_s$. By this fact and $||f_2-r_s||_{2,\alpha ,\varLambda }^{(-1-\alpha ,\partial \varLambda )}\le \delta _1$ and the fact that $k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1)$ maps $\overline{\mathcal {N}_{f_2}^+}$ to $[0,\theta _1]$,

$$\begin{aligned} J>{\tilde{c}}^*\quad \text{ in }\quad \mathcal {N}_{f_2}^+ \end{aligned}$$

(3.124)

for all $t\in [0,1]$ for some positive constant ${\tilde{c}}^*$. Using arguments similar to the ones in the proof of Claim in Lemam 15, we can obtain

$$\begin{aligned} ||\frac{\theta -\theta _1}{(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)-\theta _1}||_{0,\beta ,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C \end{aligned}$$

(3.125)

and

$$\begin{aligned} ||\frac{\theta }{k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1)}||_{0,\beta ,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C \end{aligned}$$

(3.126)

for any $t\in [0,1]$. With (3.124)–(3.126), $||ab||_{0,\beta ,\varOmega }\le ||a||_{0,\beta ,\varOmega }||b||_{0,0,\varOmega }+ ||a||_{0,0,\varOmega }$ $||b||_{0,\beta ,\varOmega }$, (3.2) satisfied by $(f_i(0), \varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ for $i=1,2$,

$h^{\prime }\in C^\alpha _{(1-\alpha ,\{\theta =\theta _1\})}$ $((0,\theta _1))$,

$$\begin{aligned} ||{\mathcal {L}}_2||_{1,0,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C,\; ||\tilde{{\mathcal {L}}}_1||_{1,0,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C \end{aligned}$$

(3.127)

and $||k_2^{-1}\circ (tV_2+(1-t){\tilde{V}}_1)||_{1,0,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C$, we estimate the right hand side of (3.123) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Then we obtain

$$\begin{aligned} ||(c)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C||h||_{1,\beta ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.128)

(d) can be estimated in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ in a similar way. As we do this, ${\tilde{k}}_1$ and $k_2$ play the role of $V_2$ and ${\tilde{V}}_1$ in the estimate of (e) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ and $\vartheta$ is regarded as the argument of ${\tilde{k}}_1$ and $k_2$.

We have

$$\begin{aligned} ||(d)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C||h||_{1,\beta ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.129)

Combining (3.128) and (3.129), we obtain the desired result. $\square$

Note that if we change ${\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1$ in the way that we changed ${\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1$ in the estimate of $\int _0^1 h^{\prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1)$ in the proof of Claim and estimate the resultant terms in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ without changing $V_2-{\tilde{V}}_1$ and ${\tilde{k}}_1(\vartheta )-k_2(\vartheta )$ into $2\pi r \sin \theta (\theta -\theta _1)\int _0^1\partial _{\theta }(\varPsi _2-{\tilde{\varPsi }}_1)(r,t\theta +(1-t)\theta _1)\hbox {d}t$ and $2\pi f_2(\vartheta )\sin \vartheta (\vartheta -\theta _1)\int _0^1\partial _{\theta }(({\tilde{\varPsi }}_1-\varPsi _2)(f_2(\theta ),\theta ))|_{\theta =t\vartheta +(1-t)\theta _1}\hbox {d}t$, then we obtain

$$\begin{aligned} ||{\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{0,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.130)

With the above Claim, we estimate (a) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Write (a) as

$$\begin{aligned}&\int _0^1 \left( g^\prime \left( \left( \frac{{\mathbf {u}}_-}{c_-}\cdot (t{\varvec{\nu }}_{f_2}({\mathcal {L}}_2) +(1-t){\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1))\right) ^2\right) \right. \nonumber \\&\quad \left. 2\left( \frac{{\mathbf {u}}_-}{c_-}\cdot (t{\varvec{\nu }}_{f_2}({\mathcal {L}}_2)+(1-t){\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1))\right) \frac{{\mathbf {u}}_-}{c_-}S_-\right) (f_2({\mathcal {L}}_2),{\mathcal {L}}_2)\hbox {d}t\nonumber \\&\qquad \cdot ({\varvec{\nu }}_{f_2}({\mathcal {L}}_2)-{\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)). \end{aligned}$$

(3.131)

By ${\varvec{\nu }}_{f_i}=\frac{\mathbf{e}_r-\frac{f_s^\prime }{f_i(0)+f_s}\mathbf{e}_\theta }{\sqrt{1+(\frac{f_s^\prime }{f_i(0)+f_s})^2}}$ for $i=1,2$, ${\varvec{\nu }}_{f_2}({\mathcal {L}}_2)-{\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)$ can written as

$$\begin{aligned} t_1\mathbf{e}_r+t_2\mathbf{e}_\theta , \end{aligned}$$

(3.132)

where

$$\begin{aligned} t_1= & {} \int _0^1 \left. \nabla _{(f(0),f_s,f_s^\prime )}\left( \frac{1}{\sqrt{1+(\frac{f_s^\prime }{f(0)+f_s})^2}}\right) \right| _{(f(0),f_s,f_s^\prime )=\chi }\hbox {d}t\\&\cdot (f_2(0)-f_1(0),f_s({\mathcal {L}}_2)-f_s(\tilde{{\mathcal {L}}}_1),f_s^\prime ({\mathcal {L}}_2)-f_s^\prime (\tilde{{\mathcal {L}}}_1)) \end{aligned}$$

and

$$\begin{aligned} t_2= & {} \int _0^1 \left. \nabla _{(f(0),f_s,f_s^\prime )} \left( \frac{-\frac{f_s^\prime }{f(0)+f_s}}{\sqrt{1+(\frac{f_s^\prime }{f(0)+f_s})^2}}\right) \right| _{(f(0),f_s,f_s^\prime )=\chi }\hbox {d}t\\&\cdot (f_2(0)-f_1(0),f_s({\mathcal {L}}_2)-f_s(\tilde{{\mathcal {L}}}_1),f_s^\prime ({\mathcal {L}}_2)-f_s^\prime (\tilde{{\mathcal {L}}}_1)) \end{aligned}$$

with $\chi =(tf_2(0)+(1-t)f_1(0),t f_s({\mathcal {L}}_2)+(1-t)f_s(\tilde{{\mathcal {L}}}_1),tf_s^\prime ({\mathcal {L}}_2)+(1-t)f_s^\prime (\tilde{{\mathcal {L}}}_1))$. To estimate (3.131) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, we estimate $f_s({\mathcal {L}}_2)-f_s(\tilde{{\mathcal {L}}}_1)$ and $f_s^\prime ({\mathcal {L}}_2)-f_s^\prime (\tilde{{\mathcal {L}}}_1)$ in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, respectively.

Express $f_s({\mathcal {L}}_2)-f_s(\tilde{{\mathcal {L}}}_1)$ and $f_s^\prime ({\mathcal {L}}_2)-f_s^\prime (\tilde{{\mathcal {L}}}_1)$ as

$$\begin{aligned} \int _0^1 f_s^\prime (t {\mathcal {L}}_2+(1-t){\mathcal {L}}_1)\hbox {d}t ({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1) \end{aligned}$$

(3.133)

and

$$\begin{aligned} \int _0^1 f_s^{\prime \prime }(t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1)\hbox {d}t({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1), \end{aligned}$$

(3.134)

respectively. Since $f_s^\prime \in C^\alpha ([0,\theta _1])$, (3.133) can be estimated in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ directly. With (3.1), (3.127) and (3.130), we estimate (3.133) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ directly. Then we obtain

$$\begin{aligned} ||f_s({\mathcal {L}}_2)-f_s(\tilde{{\mathcal {L}}}_1)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C \sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{0,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.135)

Since $f_s^{\prime \prime }\in C^{\alpha }_{(1-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$, we cannot estimate (3.134) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ directly. With Claim and (3.1), we estimate (3.134) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Then we obtain

$$\begin{aligned} ||f_s^\prime ({\mathcal {L}}_2)-f_s^\prime (\tilde{{\mathcal {L}}}_1)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.136)

With (2.19), (3.1), (3.127), the expression of ${\varvec{\nu }}_{f_2}({\mathcal {L}}_2)-{\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)$ given in (3.132), (3.135) and (3.136), we estimate (3.131) in $C^{\beta }(\overline{\mathcal {N}_{f_2}^+})$. Then we have

$$\begin{aligned} ||(a)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C\sigma |f_2(0)-f_1(0)|+C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.137)

Next, we estimate (b) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Divide (b) into two parts:

$$\begin{aligned}&\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2({\mathcal {L}}_2),{{\mathcal {L}}}_2)\\&\qquad -\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1)=:(b)_1 \end{aligned}$$

and

$$\begin{aligned}&\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1)\nonumber \\&\qquad -\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1) =:(b)_2. \end{aligned}$$

Write $(b)_1$ and $(b)_2$ as

$$\begin{aligned} \int _0^1\partial _\theta \left. \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (f_2(\theta ),\theta )\right| _{\theta =t{\mathcal {L}}_2+(1-t)\tilde{{\mathcal {L}}}_1}\hbox {d}t ({\mathcal {L}}_2-\tilde{{\mathcal {L}}}_1) \end{aligned}$$

and

$$\begin{aligned}&\int _0^1\partial _r \left. \left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1)}{c_-}\right) ^2\right) S_- \right) (r,\tilde{{\mathcal {L}}}_1)\right| _{r=tf_2(\tilde{{\mathcal {L}}}_1)+(1-t)f_1(\tilde{{\mathcal {L}}}_1)}\hbox {d}t\\&\quad (f_2 (0)-f_1(0)). \end{aligned}$$

Integrands in both the expressions are in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Thus, $(b)_1$ and $(b)_2$ can be estimated in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$ directly. With (2.19), (3.1), (3.127) and (3.130), we estiamte $(b)_1$ and $(b)_2$ in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$, respectively. Then we obtain

$$\begin{aligned} ||(b)_1||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{0,\beta ,\mathcal {N}_{f_2}^+} \end{aligned}$$

(3.138)

and

$$\begin{aligned} ||(b)_2||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C|f_2(0)-f_1(0)|. \end{aligned}$$

(3.139)

Combining (3.138) and (3.139), we have

$$\begin{aligned} ||(b)||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C|f_2(0)-f_1(0)|+C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{0,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.140)

Finally, combine (3.137) and (3.140). Then we have

$$\begin{aligned} ||T_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C|f_2(0)-f_1(0)|+C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.141)

3. Estimate $||(\frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{f_2}^+}$.

Since $A_i$ for $i=1,2$ are solutions of

$$\begin{aligned} \nabla \times ((\varPhi _0^++\varPsi _i)\mathbf{e}_{\varphi })\cdot \nabla A=0\;\text{ in }\; \mathcal {N}_{f_i(0)+f_s}^+,\; A=A_{en,f_i(0)+f_s}\;\text{ on }\; \varGamma _{f_i(0)+f_s} \end{aligned}$$

for $i=1,2$, respectively, where $A_{en,f_i}$ is $A_{en,f(0)+f_s}$ given in (3.20) for $f(0)=f_i(0)$, by Lemma 15,

$$\begin{aligned} A_i=A_{en,f_i}({\mathcal {L}}_i) \end{aligned}$$

for $i=1,2$. With these solution expressions, express $(\frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }$ as

$$\begin{aligned} \frac{f_2({\mathcal {L}}_2)\sin ({\mathcal {L}}_2)u_{-,\varphi }(f_2({\mathcal {L}}_2),{\mathcal {L}}_2)-f_1(\tilde{{\mathcal {L}}}_1)\sin (\tilde{{\mathcal {L}}}_1)u_{-,\varphi }(f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1)}{r\sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

(3.142)

where $u_{-,\varphi }={\mathbf {u}}_-\cdot \mathbf{e}_{\varphi }$. With (2.19), (3.1), $||\frac{{\mathcal {L}}_2}{\theta }||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C$ and $||\frac{\tilde{{\mathcal {L}}}_1}{\theta }||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C$ obtained using arguments similar to the ones in the proof of Claim in Lemma 15, $||{\mathcal {L}}_2||_{1,\beta ,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C$, $||\tilde{ {\mathcal {L}}}_1||_{1,\beta ,(f_2(\theta ),r_1)\times (0,\theta _1)}\le C$ and Claim, we estimate (3.142) in $C^\beta (\overline{\mathcal {N}_{f_2}^+})$. Then we obtain

$$\begin{aligned}&||(\frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{f_2}^+}\nonumber \\&\quad \le C\sigma |f_2(0)-f_1(0)|+C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.143)

4. Estimate $|f_2(0)-f_1(0)|$.

Substitute $T_2-{\tilde{T}}_1=(a)+(b)_1+(b)_2$ into (3.121). Then we have

$$\begin{aligned}&\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1}\left. \frac{\rho _0^+((\gamma -1) {u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}(b)_2 \right| _{r=r_1}r_1^2\sin \xi \hbox {d}\xi \nonumber \\&\quad =\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1}\left( \mathfrak f_0(T_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex})\right. -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}\nonumber \\&\quad \left( (a)+(b)_1\right) \left. \left. +\mathfrak f_1(\varPsi _2\mathbf{e}_{\varphi },A_2,T_2) -\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1)\right) \right| _{r=r_1}r_1^2\sin \xi \hbox {d}\xi . \end{aligned}$$

(3.144)

Using (3.2) satisfied by $(f_i(0), \varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ for $i=1,2$, we estimate $\mathfrak f_1(\varPsi _2\mathbf{e}_{\varphi },A_2,$ $T_2) -\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1)$ in $C^\beta (\overline{\varGamma _{ex}})$. Then we obtain

$$\begin{aligned}&||\mathfrak f_1(\varPsi _2\mathbf{e}_{\varphi },A_2,T_2) -\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_{\varphi },{\tilde{A}}_1,{\tilde{T}}_1)||_{0,\beta ,\varGamma _{ex}}\le C\sigma (||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\varGamma _{ex}}\\&\quad +||\left( \frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta }\right) \mathbf{e}_{\varphi }||_{0,\beta ,\varGamma _{ex}}+||T_2-{\tilde{T}}_1||_{0,\beta ,\varGamma _{ex}}). \end{aligned}$$

With this estimate, (2.19), (2.20), (3.1) and (3.2) satisfied by $(f_i(0), \varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ for $i=1,2$, (3.137), (3.138), (3.141), (3.143) and the fact that $(g({M_0^-}^2))^\prime S_{in}$ is strictly positive in $[r_0,r_1]$ (see Lemma 2), we estimate $f_2(0)-f_1(0)$ in (3.144). Then we get

$$\begin{aligned} |f_2(0)-f_1(0)|\le C_1\sigma |f_2(0)-f_1(0)|+ C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}, \end{aligned}$$

where $C_1$ is a positive constant depending on the data. Take $\sigma _3=\min (\frac{1}{2C_1},$ $\sigma _3^{(1)})(=:\sigma _3^{(2)})$. Then we have

$$\begin{aligned} |f_2(0)-f_1(0)|\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

(3.145)

5. With the fact that $\varPsi _2\mathbf{e}_{\varphi }$ and ${\tilde{\varPsi }}_1\mathbf{e}_{\varphi }$ are in $C^{2,\alpha }_{(-1-\alpha ,\varGamma _{w,f_2}^+)}(\mathcal {N}_{f_2}^+)$, transform (3.119) into an elliptic system form (see (3.35)). Then transform the resultant equation with (3.120) into the following ${\varvec{0}}$ boundary value problem

$$\begin{aligned}&\mathrm{div}\left( {\varvec{A}}D((\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }-\overline{\varvec{h}})\right) -d((\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }-\overline{\varvec{h}})\nonumber \\&\quad \quad =\underbrace{-\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta (T_2-{\tilde{T}}_1)}{r}\mathbf{e}_{\varphi }-{\varvec{F}}_3}_{\overline{\varvec{F}}}\nonumber \\&\quad \quad \quad -\mathrm{div}\left( {\varvec{A}}D\overline{\varvec{h}}\right) +d\overline{\varvec{h}} \quad \text{ in }\quad \mathcal {N}_{f_2(0)+f_s}^+, \end{aligned}$$

(3.146)

$$\begin{aligned}&(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }-\overline{\varvec{h}}= {\varvec{0}}\quad \text{ on } \quad \partial \mathcal {N}_{f_2(0)+f_s}^+, \end{aligned}$$

(3.147)

where ${\varvec{A}}$ and d are a (2, 2) tensor and scalar function defined below (3.41) and

$$\begin{aligned} \overline{\varvec{h}}:=\frac{(r-f(\theta ))\frac{r_1}{r} \overline{{\varvec{h}}}_2+(r_1-r)\frac{f(\theta )}{r} \overline{{\varvec{h}}}_1}{r_1-f(\theta )}. \end{aligned}$$

Here, $\overline{\varvec{F}}$ is of the form

$$\begin{aligned}&\sum _i (A^i_2-A^i_1)\partial _r B_2^i\mathbf{e}_{\varphi }+\sum _i A_1^i\partial _r (B_2^i-B_1^i)\mathbf{e}_{\varphi }+\sum _i C_1^i \partial _{\theta }(D_2^i-D_1^i)\mathbf{e}_{\varphi }\\&\quad +(E_2-E_1)\frac{\partial _{\theta }(F_2\sin \theta )}{\sin \theta }\mathbf{e}_{\varphi }+E_1\frac{\partial _{\theta }((F_2-F_1)\sin \theta )}{\sin \theta }\mathbf{e}_{\varphi }, \end{aligned}$$

where

$$\begin{aligned}&A^i_j\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+),\; B^i_j\mathbf{e}_\theta \in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+), \; C^i_j\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+),\\&\quad D^i_j\in C^{1,\alpha }_{(-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+),\; E_j\mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)}(\mathcal {N}_f^+) \;\text{ and }\; F_j\mathbf{e}_{\varphi }\in C^{2,\alpha }_{(-1-\alpha ,\varGamma _w^+)} (\mathcal {N}_f^+) \end{aligned}$$

for $j=1,2$. With arguments similar to the ones in the proof of Lemma 11 using (2.19), (2.20), (3.1), (3.2) satisfied by $(f_i(0), \varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ for $i=1,2$, (3.145) and

$$\begin{aligned} ||T_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi ||_{1,\beta ,\mathcal {N}_{f_2}^+} \end{aligned}$$

(3.148)

and

$$\begin{aligned} \left| \left| \left( \frac{A_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta }\right) \mathbf{e}_\varphi \right| \right| _{1,0,\mathcal {N}_{f_2}^+}\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi ||_{1,\beta ,\mathcal {N}_{f_2^+}} \end{aligned}$$

(3.149)

obtained from (3.141) and (3.143) using (3.145), we estimate $(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi$ in (3.146), (3.147) in $C^{1,\beta }(\overline{\mathcal {N}_{f_2}^+})$ (in this argument, we only change $\int _{\mathcal {N}_{f_2}^+}\sum _i A_1^i\partial _r (B_2^i-B_1^i)\mathbf{e}_{\varphi }{\varvec{\xi }}$, $\int _{\mathcal {N}_{f_2}^+} \sum _iC_1^i\partial _{\theta }(D_2^i-D_1^i)\mathbf{e}_\varphi {\varvec{\xi }}$ and $\int _{\mathcal {N}_{f_2}^+}\mathrm{div}({\varvec{A}}D{\overline{\varvec{h}}}){\varvec{\xi }}$ using integration by parts). Then we obtain

$$\begin{aligned} ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi ||_{1,\beta ,\mathcal {N}_{f_2}^+}\le C\sigma ||(\varPsi _2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi ||_{1,\beta ,\mathcal {N}_{f_2}^+}. \end{aligned}$$

Take $\sigma _3=\min (\frac{1}{2C},\sigma _3^{(2)})(=:\underline{\sigma }_3)$. Then we have $\varPsi _2\mathbf{e}_\varphi ={\tilde{\varPsi }}_1\mathbf{e}_\varphi$. Using this fact, we obtain from (3.145), (3.148) and (3.149) $f_2(0)=f_1(0)$, $T_2={\tilde{T}}_1$ and $A_2={\tilde{A}}_1$. One can see that $\underline{\sigma }_3$ depends on the data. This finishes the proof. $\square$

4 Determination of a Shape of a Shock Location

In the previous section, for given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )$ in a small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-$ $,p_c,0)$, we found $(f(0),\varPhi \mathbf{e}_\varphi ,L,S)$ satisfying all the conditions in Problem 2 except (2.47). In this section, to finish the proof of Theorem 1, for given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex})$ in a small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$ as in the previous section or in a much small perturbation of $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$ if necessary, we find $f_s^\prime$ in a small perturbation of 0 as in the previous section such that a solution of Problem 3 for given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,f_s^\prime )$ satisfies (2.47).

4.1 Proof of Theorem 1 (Existence)

For a constant $\sigma >0$, we define

$$\begin{aligned} {\mathcal {B}}^{(1)}_{\sigma }&:=\{(\rho _-,{\mathbf {u}}_-,p_-)\in (C^{2,\alpha } (\overline{\mathcal {N}}))^3\;|\;\\&\quad ||\rho _- -\rho _0^-||_{2,\alpha ,\mathcal {N}} +||{\mathbf {u}}_- -u_0^-\mathbf{e}_r||_{2,\alpha ,\mathcal {N}}+||p_- -p_0^-||_{2,\alpha ,\mathcal {N}}\le \sigma \} \\ {\mathcal {B}}^{(2)}_{\sigma }&:=\{p_{ex}\in C^{1,\alpha }_{(-\alpha ,\partial \varGamma _{ex})}(\varGamma _{ex})\;|\; ||p_{ex}-p_c||_{1,\alpha ,\varGamma _{ex}}^{(-\alpha ,\partial \varGamma _{ex})}\le \sigma \},\\ {\mathcal {B}}^{(3)}_{\sigma }&:=\{h\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))\;|\; ||h||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le \sigma \},\\ {\mathcal {B}}^{(4)}_{\sigma }&:={\mathcal {B}}^{(1)}_{\sigma }\times {\mathcal {B}}^{(2)}_{\sigma }\times {\mathcal {B}}^{(3)}_{\sigma }. \end{aligned}$$

For given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )\in {\mathcal {B}}_{\sigma }^{(4)}$ for $\sigma \le \sigma _3$, let $(f(0),\varPhi \mathbf{e}_{\varphi },L,S)$ be a solution of Problem 3 satisfying (3.2) given in Proposition 2. We define

$$\begin{aligned}&\mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )\nonumber \\&\quad :=\left. \left( \frac{1}{\varrho (\nabla \times (\varPhi \mathbf{e}_\varphi ),\frac{L}{2\pi r\sin \theta }\mathbf{e}_\varphi ,S)}\nabla \times (\varPhi \mathbf{e}_\varphi )\cdot {\varvec{\tau }}_{f(0)+f_s}- {\mathbf {u}}_-\cdot {\varvec{\tau }}_{f(0)+f_s}\right) \right| _{\varGamma _{f(0)+f_s}}\nonumber \\&\qquad \quad \circ \varPi _{r_s f(0)+f_s}, \end{aligned}$$

(4.1)

where ${\varvec{\tau }}_{f(0)+f_s}$ is the unit tangent vector on $\varGamma _{f(0)+f_s}$ perpendicular to $\mathbf{e}_\varphi$ and satisfying $({\varvec{\tau }}_{f(0)+f_s}\times \mathbf{e}_\varphi )\cdot {\varvec{\nu }}_{f(0)+f_s}>0$ and ${\varvec{\nu }}_{f(0)+f_s}$ is the unit normal vector field on $\varGamma _{f(0)+f_s}$. Then $\mathcal {A}$ satisfies $\mathcal {A}(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,0)=0$ and $\mathcal {A}$ is a map from ${\mathcal {B}}^{(4)}_{\sigma }$ to ${\mathcal {B}}^{(3)}_{C\sigma }$ where C is a positive constant depending on the data. If for given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex})\in {\mathcal {B}}^{(1)}_\sigma \times {\mathcal {B}}^{(2)}_\sigma$ for $\sigma \le \sigma _3$, we find $f_s^\prime \in {\mathcal {B}}^{(3)}_{C\sigma }$ for $C\sigma \le \sigma _3$ such that $\mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=0$, then $(f(0)+f_s,\varPhi \mathbf{e}_{\varphi },L,S)$ satisfies all the conditions in Problem 2 and thus the existence part of Theorem 1 is proved. We will find such $f_s^\prime$ using the weak implicit function theorem introduced in [3]. To apply the weak implicit function theorem, we need to prove that $\mathcal {A}$ is continuous, $\mathcal {A}$ is Fréchet differentiable at $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,0)(=:\zeta _0)$ and the partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$ is invertible. We will prove these in the following lemmas.

We first prove that $\mathcal {A}$ is continuous.

Lemma 16

$\mathcal {A}$ is continuous in ${\mathcal {B}}^{(4)}_{\sigma _3^*}$ for a positive constant $\sigma _3^*\le \sigma _3$ in the sense that if $\zeta ^{(k)}:=(\rho _-^{(k)},{\mathbf {u}}_-^{(k)},p_-^{(k)},p_{ex}^{(k)},(f_s^\prime )^{(k)})\in {\mathcal {B}}^{(4)}_{\sigma _3^*}$ converges to $\zeta ^{(\infty )}:=(\rho _-^{(\infty )},{\mathbf {u}}_-^{(\infty )},p_-^{(\infty )},p_{ex}^{(\infty )},(f_s^\prime )^{(\infty )})\in {\mathcal {B}}^{(4)}_{\sigma _3^*}$ in $(C^{2,\frac{\alpha }{2}}(\overline{\mathcal {N}}))^3\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\partial \varGamma _{ex})}(\varGamma _{ex})\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$, then $\mathcal {A}(\zeta ^{(k)})$ converges to $\mathcal {A}(\zeta ^{(\infty )})$ in $C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}$ $((0,\theta _1))$.

Proof

The result is obtained by using the standard argument.

Let $\sigma _3^*$ be a positive constant $\le \sigma _3$ and to be determined later. Let $\zeta ^{(k)}$ for $k=1,2,\ldots$ be a sequence in ${\mathcal {B}}^{(4)}_{\sigma _3^*}$ that converges to $\zeta ^{(\infty )}\in {\mathcal {B}}^{(4)}_{\sigma _3^*}$ in $(C^{2,\frac{\alpha }{2}}(\overline{\mathcal {N}}))^3\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\partial \varGamma _{ex})}(\varGamma _{ex})\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$. By Proposition 2, for each given $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=\zeta ^{(k)}$ for $k=1,2,\dots$ and $\infty$, there exists a unique ${\mathcal {U}}_k:=(f(0)^{(k)}, \varPhi ^{(k)}\mathbf{e}_{\varphi },L^{(k)},S^{(k)})$ satisfying (A), (B) and the estimate (3.2) with $\sigma$ replaced by $\sigma _3^*$. Let $({\tilde{\varPhi }}^{(k)},{\tilde{L}}^{(k)},{\tilde{S}}^{(k)}):=(\varPhi ^{(k)},L^{(k)},S^{(k)})$ $(\varPi _{r_sf(0)^{(k)}+f_s^{(k)}})$ for $k=1,2,\ldots$ and $\infty$ and $\tilde{{\mathcal {U}}}_k:=(f(0)^{(k)},{\tilde{\varPhi }}^{(k)}\mathbf{e}_{\varphi },{\tilde{L}}^{(k)},{\tilde{S}}^{(k)})$ for $k=1,2,\ldots$ and $\infty$. To prove that $\mathcal {A}(\zeta ^{(k)})$ converges to $\mathcal {A}(\zeta ^{(\infty )})$ in C $^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$, we show that $\tilde{{\mathcal {U}}}_k$ converges to $\tilde{{\mathcal {U}}}_{\infty }$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$.

To do this, we show that if a subsequence of $\tilde{{\mathcal {U}}}_k$ converges in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$, then it converges to $\tilde{{\mathcal {U}}}_{\infty }$. Assume that a subsequence $\tilde{{\mathcal {U}}}_{k_j}$ of $\tilde{{\mathcal {U}}}_k$ converges to a function $\tilde{{\mathcal {U}}}^*=(f(0)^*,{\tilde{\varPhi }}^*\mathbf{e}_{\varphi },{\tilde{L}}^*,{\tilde{S}}^*)$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$. Let (A)$^{(k)}$ and (B)$^{(k)}$ be (A) and (B) satisfied by ${\mathcal {U}}_k$, respectively, and let $\tilde{\text{(A) }}^{(k)}$ and $\tilde{\text{(B) }}^{(k)}$ be (A)$^{(k)}$ and (B)$^{(k)}$ transformed by using $\varPi _{r_sf(0)^{(k)}+f_s^{(k)}}$, respectively. Since ${{\mathcal {U}}}_{k_j}$ satisfies (A)$^{(k_j)}$, (B)$^{(k_j)}$, $\tilde{{\mathcal {U}}}_{k_j}$ satisfies $\tilde{\text{(A) }}^{(k_j)}$, $\tilde{\text{(B) }}^{(k_j)}$. Take $j\rightarrow \infty$ to $\tilde{\text{(A) }}^{(k_j)}$, $\tilde{\text{(B) }}^{(k_j)}$. Then since $\zeta ^{(k)}\rightarrow \zeta ^{(\infty )}$ in $(C^{2,\frac{\alpha }{2}}(\overline{\mathcal {N}}))^3\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\partial \varGamma _{ex})}(\varGamma _{ex})\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$ and $\tilde{{\mathcal {U}}}_{k_j}\rightarrow \tilde{{\mathcal {U}}}^*$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$, we obtain 1) $\tilde{\text{(A) }}^{(\infty )}$, $\tilde{\text{(B) }}^{(\infty )}$ with $\tilde{{\mathcal {U}}}_{\infty }$ replaced by $\tilde{{\mathcal {U}}}^*$. By the facts that ${{\mathcal {U}}}_{k}$ for $k=1,2,\ldots$ satisfy (3.2) with $\sigma$ replaced by $\sigma _3^*$ and $(f_s^\prime )^{(k)}$ for $k=1,2,\ldots$ are in ${\mathcal {B}}^{(3)}_{\sigma _3^*}$,

$$\begin{aligned}&|f(0)^{(k)}-r_s|+||\nabla \times (({\tilde{\varPhi }}^{(k)}-{\tilde{\varPhi }}_0^{+,(k)})\mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}\\&\quad +||\frac{{\tilde{L}}^{(k)}}{2\pi \varPi _{r_sf(0)^{(k)}+f_s^{(k)}}^r \sin \theta }\mathbf{e}_{\varphi }||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}+||{\tilde{S}}^{(k)}-{\tilde{S}}_0^{+,(k)}||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}\le C\sigma _3^*, \end{aligned}$$

where $({\tilde{\varPhi }}_0^{+,(k)},{\tilde{S}}_0^{+,(k)}):=(\varPhi _0^+,S_0^+)(\varPi _{r_s f(0)^{(k)}+f_s^{(k)}})$ for $k=1,2,\ldots$ and C is a positive constant depending on the data and independent of k. Using the facts that $(f_s^\prime )^{(k)}\rightarrow (f_s^\prime )^{(\infty )}$ in $C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$ and $\tilde{{\mathcal {U}}}_{k_j}\rightarrow \tilde{{\mathcal {U}}}^*$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$, we obtain from this inequality

$$\begin{aligned}&|f(0)^{*}-r_s|+||\nabla \times (({\tilde{\varPhi }}^{*}-{\tilde{\varPhi }}_0^{+,(\infty )})\mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}\nonumber \\&\quad +||\frac{{\tilde{L}}^{*}}{2\pi \varPi _{r_sf(0)^{(\infty )}+f_s^{(\infty )}}^r \sin \theta }\mathbf{e}_{\varphi }||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}+||{\tilde{S}}^{*}-{\tilde{S}}_0^{+,(\infty )}||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}\le C\sigma _3^*, \end{aligned}$$

where $({\tilde{\varPhi }}_0^{+,(\infty )},{\tilde{S}}_0^{+,(\infty )}):=(\varPhi _0^+,S_0^+)(\varPi _{r_s f(0)^{(\infty )}+f_s^{(\infty )}})$. From this inequality, we have

$$\begin{aligned}&|f(0)^*-r_s|+||\nabla \times ( (\varPhi ^*-\varPhi _0^+)\mathbf{e}_{\varphi })||_{1,\alpha ,\mathcal {N}_{f(0)^*+f_s^{(\infty )}}^+}^{(-\alpha ,\varGamma _{w,f(0)^*+f_s^{(\infty )}}^+)}\\&\quad +||\frac{L^*}{2\pi r \sin \theta }\mathbf{e}_{\varphi }||_{1,\alpha ,\mathcal {N}_{f(0)^*+f_s^{(\infty )}}^+}^{(-\alpha ,\varGamma _{w,f(0)^*+f_s^{(\infty )}}^+)}+|| S^*-S_0^+||_{1,\alpha ,\mathcal {N}_{f(0)^*+f_s^{(\infty )}}^+}^{(-\alpha ,\varGamma _{w,f(0)^*+f_s^{(\infty )}}^+)}\le C_1\sigma _3^*, \end{aligned}$$

where $(\varPhi ^*,L^*,S^*):=({\tilde{\varPhi }}^*,{\tilde{L}}^*,{\tilde{S}}^*)(\varPi _{f(0)^*+f_s^{(\infty )} r_s})$ and $C_1$ is a positive constant depending on the data. Take $\sigma _3^*=\min (\frac{C\sigma _3}{C_1},\sigma _3)$ where C is C in (3.2). Then we obtain that 2) $(f(0)^*,\varPhi ^*\mathbf{e}_\varphi ,L^*,S^*)$ satisfies (3.2) with $\sigma$ replaced by $\sigma _3$. By 1) and 2), $(f(0)^*,\varPhi ^*\mathbf{e}_\varphi ,L^*,S^*)$ satisfies (A), (B) for $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=\zeta ^{(\infty )}$ and (3.2) with $\sigma$ replaced by $\sigma _3$. Then since by Proposition 2$(f(0),\varPhi \mathbf{e}_{\varphi },L,S)$ satisfying (A), (B) for $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=\zeta ^{(\infty )}$ and (3.2) with $\sigma$ replaced by $\sigma _3$ is unique, $(f(0)^*,\varPhi ^*\mathbf{e}_\varphi ,L^*,S^*)=(f(0)^{(\infty )},\varPhi ^{(\infty )}\mathbf{e}_\varphi ,L^{(\infty )},S^{(\infty )})$. Therefore, $\tilde{{\mathcal {U}}}^*=\tilde{{\mathcal {U}}}_\infty$.

Using the fact that we showed above, we prove that $\tilde{{\mathcal {U}}}_k$ converges to $\tilde{{\mathcal {U}}}_{\infty }$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$. By $\mathbb {R}\times C^{2,\alpha }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\alpha }_{(-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2\Subset \mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$, every subsequence of $\tilde{{\mathcal {U}}}_k$ has a convergent subsequence in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$. By the fact that we showed above, this convergent subsequence must converge to $\tilde{{\mathcal {U}}}_\infty$. From this fact, we have $\tilde{{\mathcal {U}}}_k\rightarrow \tilde{{\mathcal {U}}}_\infty$ in $\mathbb {R}\times C^{2,\frac{\alpha }{2}}_{(-1-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)\times (C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+))^2$.

Using this fact, we can conclude that $\mathcal {A}(\zeta ^{(k)})\rightarrow \mathcal {A}(\zeta ^{(\infty )})$ in $C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}$ $((0,\theta _1))$ as $\zeta ^{(k)}\in {\mathcal {B}}^{(4)}_{\sigma _3^*}$ converges to $\zeta ^{(\infty )}\in {\mathcal {B}}^{(4)}_{\sigma _3^*}$ in $(C^{2,\frac{\alpha }{2}}(\overline{\mathcal {N}}))^3\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\partial \varGamma _{ex})}$ $(\varGamma _{ex})$ $\times C^{1,\frac{\alpha }{2}}_{(-\frac{\alpha }{2},\{\theta =\theta _1\}),0}((0,\theta _1))$. This finishes the proof. $\square$

Next, we prove that $\mathcal {A}$ is Fréchet differentiable.

Lemma 17

• (i) The mapping $\mathcal {A}$ defined in (4.1) is Fréchet differentiable at $\zeta _0:=(\rho _0^-,{\mathbf {u}}_0^-,p_0^-,p_c,0)$ as a map from $(C^{1,\alpha }_{(-\alpha ,\varGamma _w)}(\mathcal {N}))^3\times C^{1,\alpha }_{(-\alpha ,\partial \varGamma _{ex})}(\varGamma _{ex})\times C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ to $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$.

• (ii) The partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$ is given by

  $$\begin{aligned} D_{f_s^\prime }\mathcal {A}(\zeta _0) {\tilde{f_s^\prime }} = \left. \frac{1}{\rho _0^+}\nabla \times ({\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi )\cdot \mathbf{e}_\theta \right| _{r=r_s} -(u_0^- -u_0^+)(r_s)\frac{{\tilde{f_s^\prime }}}{r_s} \end{aligned}$$

  (4.2)

  for ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ where ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi$ is a solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }}$.

Proof

In this proof, we prove that ${\mathcal {A}}$ is Fréchet differentiable as a function of $f_s^\prime$ at 0 with the other variables fixed at $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$. The Fréchet differentiability of ${\mathcal {A}}$ as a function of $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )$ at $\zeta _0$ can be proved in a similar way.

In this proof, Cs denote positive constants depending on the data. Each C in different inequalities differs from each other.

1. Let $(f(0),\varPhi \mathbf{e}_\varphi ,L,S)$ be a solution of Problem 3 for $\sigma \in (0,\sigma _3]$ satisfying (3.2).

Let $(\varPsi ,A,T)=(\varPhi -\varPhi _0^+,L,S-S_0^+)$ and $({\tilde{\varPsi }},{\tilde{A}},{\tilde{T}}):=(\varPsi ,A,T)(\varPi _{r_s f(0)+f_s}$ ). Find the partial Fréchet derivatives of f(0), ${\tilde{\varPsi }}\mathbf{e}_{\varphi }$, ${\tilde{A}}$ and ${\tilde{T}}$ with respect to $f_s^\prime$ at $\zeta _0$.

Let ${\tilde{f}}_s^\prime$ be a function in $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ satisfying $||{\tilde{f}}_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}=1$. By Proposition 2, for each $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,\varepsilon {\tilde{f}}_s^\prime )$ for $\varepsilon \in [0,\sigma _3]$, there exists a unique $(f_\varepsilon (0),\varPhi _\varepsilon \mathbf{e}_{\varphi },L_\varepsilon ,S_\varepsilon )$ satisfying (A), (B) and the estimate (3.2). Define $(\varPsi _\varepsilon ,A_\varepsilon ,T_\varepsilon ):=(\varPhi _\varepsilon -\varPhi _0^+,L_\varepsilon ,S_\varepsilon -S_0^+)$. Since each $(f_\varepsilon (0),\varPhi _\varepsilon \mathbf{e}_{\varphi },L_\varepsilon ,S_\varepsilon )$ satisfies (A), (B) for $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,$ $p_c,\varepsilon {\tilde{f}}_s^\prime )$, each $(f_\varepsilon (0),\varPsi _\varepsilon \mathbf{e}_{\varphi },A_\varepsilon ,T_\varepsilon )$ satisfies (A$^\prime$), (B$^\prime$) for $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=(\rho _0^-,$ $u_0^-\mathbf{e}_r,p_0^-,p_c,\varepsilon {\tilde{f}}_s^\prime )$. Let

$$\begin{aligned}&{\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}:=\lim _{\varepsilon \rightarrow 0^+}\frac{f_\varepsilon (0)-f_0(0)}{\varepsilon },\quad {\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }=\lim _{\varepsilon \rightarrow 0^+}\frac{{\tilde{\varPsi }}_\varepsilon \mathbf{e}_\varphi -\varPsi _0\mathbf{e}_{\varphi }}{\varepsilon },\nonumber \\&{\tilde{A}}^{({\tilde{f}}_s^\prime )}:=\lim _{\varepsilon \rightarrow 0^+}\frac{{\tilde{A}}_\varepsilon -A_0}{\varepsilon }\quad \text{ and }\quad {\tilde{T}}^{({\tilde{f}}_s^\prime )}:=\lim _{\varepsilon \rightarrow 0^+}\frac{{\tilde{T}}_\varepsilon -T_0}{\varepsilon }, \end{aligned}$$

(4.3)

where ${\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}$, ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$, ${\tilde{A}}^{({\tilde{f}}_s^\prime )}$ and ${\tilde{T}}^{({\tilde{f}}_s^\prime )}$ represent the Gâteaux derivatives of f(0), ${\tilde{\varPsi }}\mathbf{e}_{\varphi }$, ${\tilde{A}}$ and ${\tilde{T}}$ in the direction of ${\tilde{f_s^\prime }}$ at $\zeta _0$. To find the Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$, we find ${\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}$, ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$, ${\tilde{A}}^{({\tilde{f}}_s^\prime )}$ and ${\tilde{T}}^{({\tilde{f}}_s^\prime )}$.

Transform (A$^\prime$), (B$^\prime$) satisfied by $(f_\varepsilon (0),\varPsi _\varepsilon \mathbf{e}_{\varphi },A_\varepsilon , T_\varepsilon )$ into the equations in $\mathcal {N}_{r_s}^+$ or on a part of $\partial \mathcal {N}_{r_s}^+$ by using $\varPi _{r_sf_\varepsilon }$ where $f_\varepsilon :=f_\varepsilon (0)+\varepsilon {\tilde{f}}_s$ with ${\tilde{f}}_s=\int _0^\theta {\tilde{f_s^\prime }}$. And then subtract the resultant equations from the same equations satisfied by $(f_0 (0),\varPsi _0 \mathbf{e}_{\varphi },A_0, T_0)$. Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (({\tilde{\varPsi }}_\varepsilon -\varPsi _0)\mathbf{e}_{\varphi })\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta ({\tilde{T}}_\varepsilon -T_0)}{r}\mathbf{e}_\varphi +F_0\nonumber \\&\quad \quad \quad +\tilde{\varvec{F}}_1({\tilde{\varPsi }}_\varepsilon \mathbf{e}_{\varphi },{\tilde{A}}_\varepsilon , {\tilde{T}}_\varepsilon )\quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.4)

$$\begin{aligned}&({\tilde{\varPsi }}_\varepsilon -\varPsi _0)\mathbf{e}_{\varphi }= {\left\{ \begin{array}{ll} {\varvec{0}}\quad \text{ on }\quad \varGamma _{r_s},\varGamma _{w,r_s}^+,\\ \frac{1}{r_1\sin \theta }\int _0^\theta \bigg (- \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_\varepsilon -T_0)\\ \quad \quad \quad \quad +\mathfrak f_1({\tilde{\varPsi }}_\varepsilon \mathbf{e}_{\varphi },{\tilde{A}}_\varepsilon , {\tilde{T}}_\varepsilon )\bigg )r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(4.5)

$$\begin{aligned}&\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1} \bigg (- \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_\varepsilon -T_0)\nonumber \\&\quad +\mathfrak f_1({\tilde{\varPsi }}_\varepsilon \mathbf{e}_{\varphi },{\tilde{A}}_\varepsilon , {\tilde{T}}_\varepsilon )\bigg )\bigg |_{r=r_1}r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }={\varvec{0}}, \end{aligned}$$

(4.6)

$$\begin{aligned}&{\left\{ \begin{array}{ll} (M_\varepsilon ^TM_\varepsilon \nabla \times (({\tilde{\varPhi }}_0^++{\tilde{\varPsi }}_\varepsilon )\mathbf{e}_{\varphi })-\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi }))\cdot \nabla {\tilde{A}}_\varepsilon \\ \quad \quad \quad \quad +\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi })\cdot \nabla ({\tilde{A}}_\varepsilon -A_0)=0\quad \text{ in }\quad \mathcal {N}_{r_s}^+,\\ {\tilde{A}}_\varepsilon -A_0=0\quad \text{ on }\quad \varGamma _{r_s} \end{array}\right. }\\&{\left\{ \begin{array}{ll} (M_\varepsilon ^TM_\varepsilon \nabla \times (({\tilde{\varPhi }}_0^++{\tilde{\varPsi }}_\varepsilon )\mathbf{e}_{\varphi })-\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi }))\cdot \nabla {\tilde{T}}_\varepsilon \\ \quad \quad \quad \quad +\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi })\cdot \nabla ({\tilde{T}}_\varepsilon -T_0)=0\quad \text{ in }\quad \mathcal {N}_{r_s}^+,\\ {\tilde{T}}_\varepsilon -T_0\\ \quad \quad =\left. g\left( (M_0^-\mathbf{e}_r\cdot {\varvec{\nu }_{f_\varepsilon }})^2\right) \right| _{\varGamma _{f_\varepsilon }}S_{in}\circ \varPi _{r_sf_\varepsilon }-(g({M_0^-}^2))(r_s)S_{in}\quad \text{ on }\quad \varGamma _{r_s}, \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} F_0=&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ({\tilde{\varPsi }}_\varepsilon \mathbf{e}_{\varphi })\right) \\&-M_\varepsilon \nabla \times \left( \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\right) (\varPi _{r_sf_\varepsilon })M_\varepsilon \nabla \times ({\tilde{\varPsi }}_\varepsilon \mathbf{e}_{\varphi })\right) \\&-\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta {\tilde{T}}_\varepsilon }{r}\mathbf{e}_\varphi \\&+\left( \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{1}{r}\right) (\varPi _{r_sf_\varepsilon })\\&\quad \left( \partial _{{\tilde{\theta }}}\varPi _{f_\varepsilon r_s}^r(\varPi _{r_sf_\varepsilon })\partial _r{\tilde{T}}_\varepsilon +\partial _\theta {\tilde{T}}_\varepsilon \right) \mathbf{e}_\varphi , \end{aligned}$$

$({\tilde{\varPsi }}_\varepsilon ,{\tilde{A}}_\varepsilon , {\tilde{T}}_\varepsilon ):=(\varPsi _\varepsilon ,A_\varepsilon ,T_\varepsilon )\circ \varPi _{r_sf_\varepsilon }$, ${\tilde{\varPhi }}_0^+=\varPhi _0^+\circ \varPi _{r_sf_\varepsilon }$, $M_\varepsilon =\left( \frac{\partial \varPi _{f_\varepsilon r_s}}{\partial \mathrm{y}}\right) (\varPi _{r_sf_\varepsilon })$, $\mathrm{y}$ is the Cartesian coordinate system representing $\mathcal {N}_{f_\varepsilon }^+$, $\tilde{\theta }$ is $\theta$ coordinate for $\mathrm{y}$, $(r,\theta )$ is $(r,\theta )$ coordinates for $\mathcal {N}_{r_s}^+$, and $\varPi _{f_\varepsilon r_s}^r$ is the r-component of $\varPi _{f_\varepsilon r_s}^*$, ${\varvec{\nu }}_{f_\varepsilon }$ is the unit normal vector field on $\varGamma _{f_\varepsilon }$ pointing toward $\mathcal {N}_{f_\varepsilon }^+$ and $\tilde{\varvec{F}}_1$ is ${\varvec{F}}_1$ changed by using the transformation $\varPi _{f_\varepsilon r_s}$. Divide the above equations by $\varepsilon$ and formally take $\varepsilon \rightarrow 0^+$ using

$$\begin{aligned}&(f_\varepsilon (0),{\tilde{\varPsi }}_\varepsilon ,{\tilde{A}}_\varepsilon ,{\tilde{T}}_\varepsilon )\rightarrow (r_s,\varPsi _0,A_0,T_0)=(r_s,0,0,0)\quad \text{ as }\quad \varepsilon \rightarrow 0^+,\nonumber \\&\varepsilon {\tilde{f}}_s^\prime \rightarrow 0\quad \text{ as }\quad \varepsilon \rightarrow 0^+ \end{aligned}$$

(4.7)

and (4.3). Then we have

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ({\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi })\right) \nonumber \\&\quad \quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta {\tilde{T}}^{({\tilde{f}}_s^\prime )}}{r}\mathbf{e}_\varphi \quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.8)

$$\begin{aligned}&{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }= {\left\{ \begin{array}{ll} {\varvec{0}}\quad \text{ on }\quad \varGamma _{r_s},\;\varGamma _{w,r_s}^+,\\ -\frac{1}{r_1\sin \theta }\int _0^\theta \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}{\tilde{T}}^{({\tilde{f}}_s^\prime )}r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(4.9)

$$\begin{aligned}&-\frac{1}{r_1\sin \theta _1}\int _0^{\theta _1}\left. \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}{\tilde{T}}^{({\tilde{f}}_s^\prime )}\right| _{r=r_1}r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }={\varvec{0}}, \end{aligned}$$

(4.10)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _r {\tilde{A}}^{({\tilde{f}}_s^\prime )}=0\quad \text{ in }\quad \mathcal {N}_{r_s}^+,\\ {\tilde{A}}^{({\tilde{f}}_s^\prime )}=0\quad \text{ on }\quad \varGamma _{r_s}, \end{array}\right. } \end{aligned}$$

(4.11)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \partial _r {\tilde{T}}^{({\tilde{f}}_s^\prime )}=0\quad \text{ in }\quad \mathcal {N}_{r_s}^+,\\ {\tilde{T}}^{({\tilde{f}}_s^\prime )}=(g({M_0^-}^2))^\prime (r_s)S_{in}({\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}+{\tilde{f}}_s)\quad \text{ on }\quad \varGamma _{r_s}. \end{array}\right. } \end{aligned}$$

(4.12)

We solve this system for given ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$.

By (4.11) and (4.12),

$$\begin{aligned} {\tilde{A}}^{({\tilde{f}}_s^\prime )}=0\quad \text{ in }\quad \mathcal {N}_{r_s}^+ \end{aligned}$$

(4.13)

and

$$\begin{aligned} {\tilde{T}}^{({\tilde{f}}_s^\prime )}=(g({M_0^-}^2))^\prime (r_s)S_{in}({\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}+{\tilde{f}}_s)\quad \text{ in }\quad \mathcal {N}_{r_s}^+. \end{aligned}$$

(4.14)

Substitute (4.14) into (4.10). And then find ${\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}$ in the resultant equation. Then we obtain

$$\begin{aligned} {\tilde{f}}(0)^{({\tilde{f}}_s^\prime )}=\frac{-\int _0^{\theta _1}{\tilde{f}}_s\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }. \end{aligned}$$

(4.15)

Substitute this into (4.14) again. Then we have

$$\begin{aligned} {\tilde{T}}^{({\tilde{f}}_s^\prime )}=(g({M_0^-}^2))^\prime (r_s) S_{in}(-\frac{\int _0^{\theta _1}{\tilde{f}}_s\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }+{\tilde{f}}_s)\quad \text{ in }\quad \mathcal {N}_{r_s}^+. \end{aligned}$$

(4.16)

Substituting ${\tilde{T}}^{({\tilde{f}}_s^\prime )}$ given in (4.16) into (4.8) and (4.9), we get

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ({\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi })\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})(g({M_0^-}^2))^\prime (r_s) S_{in}\frac{{\tilde{f}}_s^\prime }{r}\mathbf{e}_\varphi \quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.17)

$$\begin{aligned}&{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }= {\left\{ \begin{array}{ll} {\varvec{0}}\quad \text{ on }\quad \varGamma _{r_s},\;\varGamma _{w,r_s}^+,\\ -\frac{1}{r_1\sin \theta }\int _0^\theta \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}(g({M_0^-}^2))^\prime (r_s) S_{in}\\ \quad \quad \quad \quad \quad \quad (-\frac{\int _0^{\theta _1}{\tilde{f}}_s\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }+{\tilde{f}}_s)r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{ex}. \end{array}\right. } \end{aligned}$$

(4.18)

Note that since (4.18) is a $C^0$ boundary condition, Lemma 7 can be applied to (4.17), (4.18). Using Lemma 7 and the arguments used to prove Lemma 13 (here we can obtain more higher regularity of solutions of (4.17), (4.18) than that of solutions of (3.28), (3.29) in Lemma 7 because ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$), we obtain that there exists a unique $C^{3,\alpha }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)$ solution ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$ of (4.17), (4.18) and this solution satisfies

$$\begin{aligned} ||{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }||_{3,\alpha ,\mathcal {N}_{r_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s}^+)}\le C \end{aligned}$$

(4.19)

for a positive constant C independent of ${\tilde{f}}_s^\prime$.

Using (3.2) satisfied by $(f_\varepsilon (0),\varPhi _\varepsilon \mathbf{e}_{\varphi },L_\varepsilon ,S_\varepsilon )$, (4.4)–(4.6), (4.8)–(4.10), (4.13), (4.15), (4.16), the solution expressions of ${\tilde{A}}_\varepsilon$ and ${\tilde{T}}_\varepsilon$ obtained by solving (B$^\prime$) for $(f(0),\varPsi )=(f_\varepsilon (0),\varPsi _\varepsilon )$ and $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )=(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,\varepsilon {\tilde{f}}_s^\prime )$ using Lemma 15, and Lemma 7, we can obtain

$$\begin{aligned}&|f_\varepsilon (0)-f_0(0)-\varepsilon {\tilde{f}}(0)^{({\tilde{f_s^\prime }})}|\le C\varepsilon ^2,\nonumber \\&||{\tilde{\varPsi }}_\varepsilon -\varPsi _0-\varepsilon {\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}||_{2,\alpha ,\mathcal {N}_{r_s}^+}^{(-1-\alpha ,\varGamma _{w,r_s}^+)}\le C\varepsilon ^2,\nonumber \\&||{\tilde{A}}_\varepsilon -A_0-\varepsilon {\tilde{A}}^{({\tilde{f_s^\prime }})}||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}=0,\nonumber \\&||{\tilde{T}}_\varepsilon -T_0-\varepsilon {\tilde{T}}^{({\tilde{f_s^\prime }})}||_{1,\alpha ,\mathcal {N}_{r_s}^+}^{(-\alpha ,\varGamma _{w,r_s}^+)}\le C\varepsilon ^2 \end{aligned}$$

(4.20)

for all ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ satisfying $||{\tilde{f}}_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}=1$ and $\varepsilon \in [0,\sigma _3]$. By (4.13) and (4.15)–(4.18), ${\tilde{f}}(0)^{({\tilde{f_s^\prime }})}$, ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$, ${\tilde{A}}^{({\tilde{f}}_s^\prime )}$ and ${\tilde{T}}^{({\tilde{f}}_s^\prime )}$ are bounded linear maps from $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ to $\mathbb {R}$, $C^{2,\alpha }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)$, $C^{1,\alpha }_{(-\alpha ,\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)$ and $C^{1,\alpha }_{(-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)$, respectively. From the above inequalities and this fact, we have that f(0), ${\tilde{\varPsi }}\mathbf{e}_{\varphi }$, ${\tilde{A}}$ and ${\tilde{T}}$ are Fréchet differentiable as a function of $f_s^\prime$ at 0 with the other variables fixed at $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$ and ${\tilde{f}}(0)^{({\tilde{f_s^\prime }})}$, ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$, ${\tilde{A}}^{({\tilde{f}}_s^\prime )}$ and ${\tilde{T}}^{({\tilde{f}}_s^\prime )}$ are the partial Fréchet derivatives of f(0), ${\tilde{\varPsi }}\mathbf{e}_{\varphi }$, ${\tilde{A}}$ and ${\tilde{T}}$ with respect to $f_s^\prime$ at $\zeta _0$, respectively.

2. Show that ${\mathcal {A}}$ is Fréchet differentiable as a function of $f_s^\prime$ at 0 with the other variables fixed at $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$.

By definition, the Gâteaux derivative of $\mathcal {A}$ in the direction $(0,0,0,0,{\tilde{f}}_s^\prime )$ at $\zeta _0$ is given by

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0^+}&\frac{\mathcal {A}(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,\varepsilon {\tilde{f}}_s^\prime )-\mathcal {A}(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,0)}{\varepsilon }\\&= \lim _{\varepsilon \rightarrow 0^+}\frac{1}{\varepsilon }\left( \frac{1}{\varrho (\nabla \times ((\varPhi _0^++\varPsi _\varepsilon )\mathbf{e}_{\varphi }),\frac{A_\varepsilon }{2\pi r\sin \theta }\mathbf{e}_\varphi ,S_0^++T_\varepsilon )}\right. \\&\quad \nabla \times ( (\varPhi _0^++\varPsi _\varepsilon )\mathbf{e}_{\varphi })\cdot {\varvec{\tau }_{f_\varepsilon }}\bigg |_{\varGamma _{f_\varepsilon }}\circ \varPi _{r_sf_\varepsilon } -\left. u_0^-\mathbf{e}_r\cdot {\varvec{\tau }_{f_\varepsilon }}\right| _{\varGamma _{f_\varepsilon }}\circ \varPi _{r_sf_\varepsilon }\\&-\frac{1}{\varrho (\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi }),0,S_0^+)}\nabla \times (\varPhi _0^+\mathbf{e}_{\varphi })\cdot \mathbf{e}_\theta \bigg |_{\varGamma _{r_s}} +\left. u_0^-\mathbf{e}_r\cdot \mathbf{e}_\theta \right| _{\varGamma _{r_s}}\bigg ) \end{aligned}$$

where ${\varvec{\tau }}_{f_\varepsilon }$ is the unit tangent vector on $\varGamma _{f_\varepsilon }$ perpendicular to $\mathbf{e}_\varphi$ and satisfying $({\varvec{\tau }}_{f_\varepsilon }\times \mathbf{e}_\varphi )\cdot {\varvec{\nu }}_{f_\varepsilon }>0$ and ${\varvec{\nu }}_{f_\varepsilon }$ is the unit normal vector field on $\varGamma _{f_\varepsilon }$. As we did in Step 1, formally take $\varepsilon \rightarrow 0^+$ using (4.3) and (4.7) to the right hand side of the above equation. Then we obtain

$$\begin{aligned} \left. \frac{1}{\rho _0^+}\nabla \times ({\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi })\cdot \mathbf{e}_\theta \right| _{r=r_s}-(u_0^- -u_0^+)(r_s)\frac{{\tilde{f}}_s^\prime }{r_s}. \end{aligned}$$

Define a map L by

$$\begin{aligned} L{\tilde{f}}_s^\prime :=\left. \frac{1}{\rho _0^+}\nabla \times ({\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }) \cdot \mathbf{e}_\theta \right| _{r=r_s}-(u_0^- -u_0^+)(r_s)\frac{{\tilde{f}}_s^\prime }{r_s} \end{aligned}$$

for ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ satisfying $||{\tilde{f}}_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}=1$ where ${\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}\mathbf{e}_{\varphi }$ is a solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }} \in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$. Then L is a bounded linear map from $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ to $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$. Using (3.2) satisfied by $(f(0),\varPhi _\varepsilon \mathbf{e}_\varphi ,L_\varepsilon ,S_\varepsilon )$ and (4.20), one can check that

$$\begin{aligned} ||\mathcal {A}(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,\varepsilon {\tilde{f}}_s^\prime )-\mathcal {A}(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c,0)-\varepsilon L{\tilde{f}}_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C\varepsilon ^2 \end{aligned}$$

(4.21)

for any ${\tilde{f}}_s^\prime \in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ satisfying $||{\tilde{f}}_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}=1$ and $\varepsilon \in [0,\sigma _3]$. Therefore, ${\mathcal {A}}$ is Fréchet differentiable as a function of $f_s^\prime$ at 0 with the other variables fixed at $(\rho _0^-,u_0^-\mathbf{e}_r,p_0^-,p_c)$ and L is the partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$. This finishes the proof. $\square$

Finally, we prove that the partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$ is invertible. When we prove the invertibility of the partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$, we use eigenfunction expansions of ${\tilde{f_s^\prime }}$ and ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}$. The eigenfunctions used to express ${\tilde{f_s^\prime }}$ and ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}$ are eigenfunctions of the eigenvalue problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }q)-\frac{q}{\sin ^2\theta }=-\lambda q\quad \text{ in }\quad \theta \in (0,\theta _1),\\ q=0\quad \text{ on }\quad \theta =0,\theta _1 \end{array}\right. } \end{aligned}$$

(4.22)

which arises from $\theta$-part of the spherical coordinate representation of (4.17), (4.18) (note that this is the associated Legendre equation of type $m=1$ with a general domain that is a singular Sturm-Liouville problem). To express ${\tilde{f_s^\prime }}$ and ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}$ using eigenfunctions of (4.22), we need to prove the orthogonal completeness of the set of eigenfunctions of (4.22). We prove this in the following lemma:

Lemma 18

The eigenvalue problem (4.22) has infinitely countable eigenvalues $\lambda _j$ for $j=1,2,\ldots$ satisfying $\lambda _j\rightarrow \infty$ as $j\rightarrow \infty$ and $\lambda _j>0$. A set of eigenfunctions of (4.22) forms a complete orthorgonal set in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$.

Proof

1. For given $f\in L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$, we consider

$$\begin{aligned} {\left\{ \begin{array}{ll}\frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }q)-\frac{q}{\sin ^2\theta }=-f\quad \text{ in }\quad \theta \in (0,\theta _1),\\ q=0\quad \text{ on }\quad \theta =0,\theta _1. \end{array}\right. } \end{aligned}$$

(4.23)

Write this equation in the form

$$\begin{aligned} \int _0^{\theta _1}(\partial _\theta q\partial _\theta \xi +\frac{q\xi }{\sin ^2\theta })\sin \theta d\theta =\int _0^{\theta _1}f\xi \sin \theta \hbox {d}\theta \quad \end{aligned}$$

(4.24)

for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$. Assume for a moment that there exists a unique $q\in H^1_0((0,\theta _1),\sin \theta \hbox {d}\theta )$ satisfying (4.24) for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$ and this q satisfies

$$\begin{aligned} ||q||_{H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )}\le C||f||_{L^2((0,\theta _1),\sin \theta \hbox {d}\theta )} \end{aligned}$$

(4.25)

for some positive constant C. Using this q, we define a map ${\mathcal {S}}: L^2((0,\theta _1),$ $\sin \theta \hbox {d}\theta )\rightarrow L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$ by

$$\begin{aligned} {\mathcal {S}}f=q. \end{aligned}$$

Then ${\mathcal {S}}$ is a self-adjoint and compact linear operator. Hence, by the spectral theorem for compact self-adjoint operators, ${\mathcal {S}}$ has contable infinite eigenvalues $\mu _j$ satisfying $\mu _j\rightarrow 0$ as $j\rightarrow \infty$ and the set of eigenfunctions $q_j$ of ${\mathcal {S}}$ corresponding to $\mu _j$ forms a complete orthorgonal set in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$. From this fact, we obtain that (4.22) has infinitely countable eigenvalues $\lambda _j\rightarrow \infty$ as $j\rightarrow \infty$ and the set of eigenfunctions of (4.22) forms a complete orthorgonal set in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$.

2. Show that there exists a unique $q\in H^1_0((0,\theta _1),\sin \theta \hbox {d}\theta )$ satisfying (4.24) for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$.

For given $f\in L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$, we consider

$$\begin{aligned} \int _{{\mathcal {D}}} \delta {\varvec{U}}\delta {\varvec{\xi }} =\int _{{\mathcal {D}}} f(\theta )\mathbf{e}_{\varphi }{\varvec{\xi }} \end{aligned}$$

(4.26)

for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ where ${\varvec{U}}:{\mathcal {D}}\rightarrow \mathbb {R}^3$, ${\varvec{\xi }}:{\mathcal {D}}\rightarrow \mathbb {R}^3$, ${\mathcal {D}}:=\{(x,y,z)\in S^2:\frac{z}{\sqrt{x^2+y^2+z^2}}\ge \cos \theta _1\}$ and $\delta$ is the tangential gradient operator on $S^2$. By the Lax-Milgram theorem, there exists a unique ${\varvec{U}}\in H_0^1({\mathcal {D}})$ satisfying (4.26) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ and this ${\varvec{U}}$ satisfies

$$\begin{aligned} ||{\varvec{U}}||_{H_0^1({\mathcal {D}})}\le C||f\mathbf{e}_\varphi ||_{L^2({\mathcal {D}})} \end{aligned}$$

(4.27)

for some positive constant C depending on ${\mathcal {D}}$. Let ${\varvec{U}}$ be a function in $H_0^1({\mathcal {D}})$ satisfying (4.26) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$. Using the standard argument, it can be shown that ${\varvec{U}}$ satisfies

$$\begin{aligned} ||{\varvec{U}}||_{H^2({\mathcal {D}})}\le C||f\mathbf{e}_\varphi ||_{L^2({\mathcal {D}})}. \end{aligned}$$

From this fact, we see that ${\varvec{U}}$ satisfies

$$\begin{aligned} \int _{{\mathcal {D}}}\varDelta _{S^2}{\varvec{U}}{\varvec{\xi }}=-\int _{{\mathcal {D}}}f(\theta )\mathbf{e}_{\varphi }{\varvec{\xi }} \end{aligned}$$

for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$. Using this fact and the fact that the coefficients of $\varDelta _{S^2}{\varvec{U}}$ in the spherical coordinate system are independent of $\varphi$, we apply arguments similar to the ones in the proof of Lemma 14 to ${\varvec{U}}$ (here, we use the facts that a bounded sequence in $H^2({\mathcal {D}})$ contains a weakly convergent subsequence and a $H^1_0({\mathcal {D}})$ function satisfying (4.26) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ is unique). Then we have that ${\varvec{U}}$ only has the form $u_\varphi (\theta )\mathbf{e}_{\varphi }$.

One can see that if ${\varvec{U}}=u_\varphi (\theta )\mathbf{e}_{\varphi }\in H^1_0({\mathcal {D}})$ satisfies (4.26) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$, then $u_\varphi \in H^1_0((0,\theta _1),\sin \theta \hbox {d}\theta )$ satisfies (4.24) for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$ and that if $u_\varphi \in H^1_0((0,\theta _1),\sin \theta \hbox {d}\theta )$ satisfies (4.24) for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$, then ${\varvec{U}}=u_\varphi (\theta )\mathbf{e}_{\varphi }\in H^1_0({\mathcal {D}})$ satisfies (4.26) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ having the form $\xi (\theta ) \mathbf{e}_{\varphi }$. Using this fact, we can deduce that there exists a unique $q\in H^1_0((0,\theta _1),\sin \theta \hbox {d}\theta )$ satisfying (4.24) for all $\xi \in H_0^1((0,\theta _1),\sin \theta \hbox {d}\theta )$. By (4.27), this solution satisfies (4.25).

3. Show that eigenvalues $\lambda _j$ of (4.22) are positive.

If q is an eigenfunction of ${\mathcal {S}}$ corresponding to an eigenvalue $\mu =\frac{1}{\lambda }$, then it holds that

$$\begin{aligned} \int _{{\mathcal {D}}}\delta (q\mathbf{e}_\varphi )\delta {\varvec{\xi }}=\int _{{\mathcal {D}}}\lambda q \mathbf{e}_\varphi {\varvec{\xi }} \end{aligned}$$

(4.28)

for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ having the form $\xi (\theta ) \mathbf{e}_{\varphi }$. Using the weak maximum principle, we can have that for $q\mathbf{e}_\varphi \in H_0^1({\mathcal {D}})$ to be a nonzero function satisfying (4.28) for all ${\varvec{\xi }}\in H_0^1({\mathcal {D}})$ having the form $\xi (\theta ) \mathbf{e}_{\varphi }$, $\lambda$ must be positive. Hence, eigenvalues of (4.22) are positive. This finishes the proof. $\square$

Then we prove the invertibility of the partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$.

Lemma 19

The partial Fréchet derivative of $\mathcal {A}$ with respect to $f_s^\prime$ at $\zeta _0$ given by (4.2) is an invertible map from $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$ to $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1)$ ).

Proof

By (4.19) and $C^{2,\alpha }_{(-1-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))\Subset C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1))$, $D_{f_s^\prime } {\mathcal {A}}(\zeta _0)$ given by (4.2) is of the form $c I-K$ where c is a constant and K is a compact linear map from $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$ to $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$. By the Fredholm alternative, this implies that if $\ker D_{f_s^\prime }\mathcal {A}(\zeta _0)=\{0\}$, then $D_{f_s^\prime }\mathcal {A}(\zeta _0)$ is invertible. In this proof, we show that $\ker D_{f_s^\prime }\mathcal {A}(\zeta _0)=\{0\}$. Since it is obvious that $D_{f_s^\prime }\mathcal {A}(\zeta _0)0=0$, we show that $D_{f_s^\prime }\mathcal {A}(\zeta _0){\tilde{f_s^\prime }}=0$ only if ${\tilde{f_s^\prime }}=0$.

1. Assume that for a nonzero ${\tilde{f_s^\prime }}\in C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\}),0}((0,\theta _1))$, $D_{f_s^\prime }\mathcal {A}(\zeta _0){\tilde{f_s^\prime }}=0$. Then it holds that

$$\begin{aligned} \left. \frac{1}{\rho _0^+} \nabla \times (\varPsi ^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi )\cdot \mathbf{e}_\theta \right| _{r=r_s} -(u_0^- -u_0^+)(r_s)\frac{{\tilde{f_s^\prime }}}{r_s}=0, \end{aligned}$$

(4.29)

where ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi }$ is the $C^{3,\alpha }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)$ solution of

$$\begin{aligned}&-\frac{1}{\rho _0^+ r^2}\partial _r (r^2\partial _r {\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )})+\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\partial _r(r{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )})\nonumber \\&\qquad -\frac{1}{\rho _0^+ r^2}\left( \frac{ {c_0^+}^2}{ {c_0^+}^2-{u_0^+}^2}\right) \left( \frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )})-\frac{{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}}{\sin ^2\theta }\right) \nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}\left( 1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2}\right) \frac{\left( g( {M_0^-}^2)\right) ^\prime (r_s) S_{in}}{r}{\tilde{f}}_s^\prime \quad \text{ in }\quad \mathcal {N}_{r_s}^{+,*}, \end{aligned}$$

(4.30)

$$\begin{aligned}&{\tilde{\varPsi }}^{({\tilde{f}}_s^\prime )}={\left\{ \begin{array}{ll} 0\quad \text{ on }\quad \varGamma _{r_s}^*,\;\varGamma _{w,r_s}^{+,*},\\ -\frac{1}{r_1\sin \theta }\int _0^\theta \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}\left( g({M_0^-}^2)\right) ^\prime (r_s) S_{in}\\ \quad \quad \quad \quad \quad \left( -\frac{\int _0^{\theta _1}{\tilde{f}}_s\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }+{\tilde{f}}_s\right) r_1^2\sin \xi \hbox {d}\xi \quad \text{ on }\quad \varGamma _{ex}^*, \end{array}\right. } \end{aligned}$$

(4.31)

which is the spherical coordinate representation of (4.17), (4.18), for given ${\tilde{f_s^\prime }}$.

Using $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta _1)) \subset L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$ and Lemma 18, we express ${\tilde{f_s^\prime }}$ as

$$\begin{aligned} {\tilde{f_s^\prime }}=\sum _{j=1}^\infty c_j q_j \end{aligned}$$

(4.32)

where $c_j$ are constants and $q_j$ are eigenfunctions of (4.22) corresponding to eigenvalues $\lambda _j$ of (4.22). Define

$$\begin{aligned} {\tilde{f}}^\prime _{s,m}:=\sum _{j=1}^m c_j q_j, \end{aligned}$$

(4.33)

and ${\tilde{f}}_{s,m}:=\int _0^\theta {\tilde{f}}_{s,m}^\prime$. We consider (4.30), (4.31) for given ${\tilde{f_s^\prime }}={\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_s={\tilde{f}}_{s,m}$.

Using the fact that $q_j\in C^\infty ([0,\theta _1])$ and $\frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }q)-\frac{q}{\sin ^2\theta }=\partial _{\theta }$ $\left( \frac{\partial _{\theta }(q\sin \theta )}{\sin \theta }\right)$, we can have

$$\begin{aligned} {\tilde{f}}_{s,m}=\sum _{j=1}^m -\frac{c_j}{\lambda _j}\left( \frac{\partial _{\theta }(q_j\sin \theta )}{\sin \theta }-\left. \frac{\partial _{\theta }(q_j\sin \theta )}{\sin \theta }\right| _{\theta =0}\right) . \end{aligned}$$

(4.34)

Here, $\left. \frac{\partial _{\theta }(q_j\sin \theta )}{\sin \theta }\right| _{\theta =0}$ is bounded because $q_j=0$ on $\theta =0$ and $q_j\in C^\infty ([0,\theta _1])$.

Substitute ${\tilde{\varPsi }}_m=\sum _{j =1}^m p_j(r)q_j(\theta )$, ${\tilde{f}}_{s,m}^\prime$ given in (4.33) and ${\tilde{f}}_{s,m}$ given in (4.34) into the places of ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}$, ${\tilde{f_s^\prime }}$ and ${\tilde{f}}_s$ in (4.30), (4.31), respectively. Then we obtain

$$\begin{aligned}&(L_1:=)-\frac{1}{r^2}\left( \frac{1}{\rho _0^+}r^2p_j^\prime \right) ^\prime +\left( \frac{1}{\rho _0^+r^2}\left( 1+\frac{ {u_0^+}^2}{ {c_0^+}^2-{u_0^+}^2}\right) \lambda _j+\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\right) p_j\nonumber \\&\quad =\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}\left( 1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2}\right) \frac{\left( g( {M_0^-})^2\right) ^\prime (r_s) S_{in}}{r}c_j\quad \text{ in }\quad (r_s,r_1), \end{aligned}$$

(4.35)

$$\begin{aligned}&p_j={\left\{ \begin{array}{ll} 0\quad \text{ on }\quad r=r_s,\\ \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)(g( {M_0^-}^2))^\prime (r_s)S_{in}r_1}{\gamma (\gamma -1)u_0^+S_0^+\lambda _j}c_j\quad \text{ on }\quad r=r_1 \end{array}\right. } \end{aligned}$$

(4.36)

for $j=1,\ldots ,m$. One can see that $L_1$ is of the form $L_0-c_0I$ where $L_0(=\frac{1}{r^2}(\frac{1}{\rho _0^+}r^2p_j^\prime )^\prime )$ is an invertible operator and $c_0$ is a positive constant. Using the Fredholm alternative and the maximum principle, we can obtain that (4.35), (4.36) for each j has a unique $C^\infty ([r_s,r_1])$ solution. Let $p_j$ be the solution of (4.35), (4.36). Then by the fact that $q_j\mathbf{e}_\varphi \in C^\infty (\overline{{\mathcal {D}}})$ where ${\mathcal {D}}$ is a domain defined in the proof of Lemma 18, we have that ${\tilde{\varPsi }}_m \mathbf{e}_\varphi =\sum _{j=1}^m p_jq_j\mathbf{e}_\varphi$ is a $C^\infty (\overline{\mathcal {N}_{r_s}^+})$ solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }}={\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_s={\tilde{f}}_{s,m}$.

2. Show that there exists a subsequence ${\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi }$ of ${\tilde{\varPsi }}_m\mathbf{e}_{\varphi }$ such that ${\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi }$ and $D({\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi })$ weakly converge to ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi$ and $D({\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi })$ in $L^2(\mathcal {N}_{r_s}^+)$, respectively.

Since ${\tilde{\varPsi }}_m \mathbf{e}_\varphi$ is a $C^\infty (\overline{\mathcal {N}_{r_s}^+})$ solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }}={\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_s={\tilde{f}}_{s,m}$, (4.17), (4.18) satisfied by ${\tilde{\varPsi }}_m \mathbf{e}_\varphi$ can be transformed into the boundary value problem for an elliptic system

$$\begin{aligned}&\mathrm{div}\left( {\varvec{A}}D({\tilde{\varPsi }}_m\mathbf{e}_{\varphi })\right) -d({\tilde{\varPsi }}_m\mathbf{e}_{\varphi })\nonumber \\&\quad =-\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2}) \nonumber \\&\qquad (g({M_0^-}^2))^\prime (r_s)S_{in}\frac{{\tilde{f}}_{s,m}^\prime }{r}\mathbf{e}_\varphi \quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.37)

where ${\varvec{A}}$ and d are a (2, 2)-tensor and scalar function defined below (3.41),

$$\begin{aligned} {\tilde{\varPsi }}_m\mathbf{e}_{\varphi }={\left\{ \begin{array}{ll} {\varvec{0}}\quad \text{ on }\quad \varGamma _{r_s},\;\varGamma _{w,r_s}^+,\\ -\frac{1}{r_1\sin \theta }\int _0^\theta \frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}(g({M_0^-}^2))^\prime (r_s)S_{in}\\ \quad \quad \quad (-\frac{\int _0^{\theta _1}{\tilde{f}}_{s,m}\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }+{\tilde{f}}_{s,m})r_1^2\sin \xi \hbox {d}\xi \mathbf{e}_{\varphi }(=:h_m\mathbf{e}_{\varphi })\quad \text{ on }\quad \varGamma _{ex}. \end{array}\right. } \end{aligned}$$

(4.38)

Set ${\varvec{h}}_m:=\frac{(r-r_s)r_1}{(r_1-r_s)r}h_m\mathbf{e}_\varphi .$ Transform (4.37), (4.38) into a ${\varvec{0}}$ boundary value problem by substituting ${\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi }+{\varvec{h}}_m$ into the place of ${\tilde{\varPsi }}_m\mathbf{e}_{\varphi }$ in (4.37), (4.38). Write the resultant problem in the following form:

$$\begin{aligned}&\int _{\mathcal {N}_{r_s}^+}{\varvec{A}}D({\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi }) D{\varvec{\xi }}+d{\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi } {\varvec{\xi }} \nonumber \\&\quad =\int _{\mathcal {N}_{r_s}^+}\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})(g({M_0^-}^2))^\prime (r_s)S_{in}\frac{{\tilde{f}}_{s,m}^\prime }{r}\mathbf{e}_\varphi {\varvec{\xi }} \nonumber \\&\qquad -{\varvec{A}}D{\varvec{h}}_m D{\varvec{\xi }}-d{\varvec{h}}_m{\varvec{\xi }}, \end{aligned}$$

(4.39)

for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{r_s}^+)$ where ${\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi }:={\tilde{\varPsi }}_m\mathbf{e}_{\varphi }-{\varvec{h}}_m$. Using Lemma 9, we obtain from (4.39)

$$\begin{aligned} ||{\tilde{\varPsi }}_m^* \mathbf{e}_\varphi ||_{H_0^1(\mathcal {N}_{r_s}^+)}\le C||{\tilde{f}}^\prime _{s,m}\mathbf{e}_\varphi ||_{L^2(\mathcal {N}_{r_s}^+)}. \end{aligned}$$

Since $||{\tilde{f}}^\prime _{s,m}\mathbf{e}_\varphi ||_{L^2(\mathcal {N}_{r_s}^+)}\le ||{\tilde{f_s^\prime }}\mathbf{e}_\varphi ||_{L^2(\mathcal {N}_{r_s}^+)}$, ${\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi }$ is a bounded sequence in $H_0^1$ $(\mathcal {N}_{r_s}^+)$. Hence, there exists a subsequence ${\tilde{\varPsi }}_{m_l}^*\mathbf{e}_{\varphi }$ of ${\tilde{\varPsi }}_m^*\mathbf{e}_{\varphi }$ and some function ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }\in H_0^1(\mathcal {N}_{r_s}^+)$ such that ${\tilde{\varPsi }}_{m_l}^*\mathbf{e}_{\varphi }$ and $D({\tilde{\varPsi }}_{m_l}^*\mathbf{e}_{\varphi })$ weakly converge to ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }$ and $D({\tilde{\varPsi }}^*\mathbf{e}_{\varphi })$ in $L^2(\mathcal {N}_{r_s}^+)$, respectively. Take $l\rightarrow \infty$ to (4.39) for $m=m_l$. Then by ${\tilde{\varPsi }}_{m_l}^*\mathbf{e}_{\varphi }\rightharpoonup {\tilde{\varPsi }}^*\mathbf{e}_{\varphi }$ in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$, $D({\tilde{\varPsi }}_{m_l}^*\mathbf{e}_{\varphi })\rightharpoonup D({\tilde{\varPsi }}^*\mathbf{e}_{\varphi })$ in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$ and ${\tilde{f}}_{s,m}^\prime \rightarrow {\tilde{f}}_s^\prime$ in $L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$, one has (4.39) with ${\tilde{f}}_{s,m}^\prime$, ${\tilde{f}}_{s,m}$ and ${\tilde{\varPsi }}_m^*$ replaced by ${\tilde{f}}_s^\prime$, ${\tilde{f}}_s$ and ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }$, respectively. Thus, ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }$ is a $H_0^1(\mathcal {N}_{r_s}^+)$ function satisfying (4.39) with ${\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_{s,m}$ replaced by ${\tilde{f}}_s^\prime$ and ${\tilde{f}}_s$, respectively, for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{r_s}^+)$. Let ${\varvec{h}}$ be ${\varvec{h}}_m$ with ${\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_{s,m}$ replaced by ${\tilde{f}}_s^\prime$ and ${\tilde{f}}_s$, respectively. One can see that ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi }-{\varvec{h}}$ is a $H_0^1(\mathcal {N}_{r_s}^+)$ function satisfying (4.39) with ${\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_{s,m}$ replaced by ${\tilde{f}}_s^\prime$ and ${\tilde{f}}_s$, respectively, for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{r_s}^+)$. By Lemma 9, a $H_0^1(\mathcal {N}_{r_s}^+)$ function satisfying (4.39) with ${\tilde{f}}_{s,m}^\prime$ and ${\tilde{f}}_{s,m}$ replaced by ${\tilde{f}}_s^\prime$ and ${\tilde{f}}_s$, respectively, for all ${\varvec{\xi }}\in H_0^1(\mathcal {N}_{r_s}^+)$ is unique. Hence, ${\tilde{\varPsi }}^*\mathbf{e}_{\varphi }={\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi }-{\varvec{h}}$. From this, we can conclude that ${\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi }$ and $D({\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi })$ weakly converge to ${\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi$ and $D({\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi })$ in $L^2(\mathcal {N}_{r_s}^+)$, respectively.

3. Show that $D_{f_s^\prime }\mathcal {A}(\zeta _0){\tilde{f_s^\prime }}=0$ only if ${\tilde{f_s^\prime }}=0$.

Since ${\tilde{f}}_s^\prime \ne 0$ by the assumption, there exists $k\in {{\mathbb {N}}}$ such that $c_k>0$ or $c_k<0$ in the expression of ${\tilde{f_s^\prime }}$ in (4.32). Without loss of generality, assume that $c_k>0$ for some $k\in {{\mathbb {N}}}$. Then since $\lambda _k>0$ by Lemma 18, $p_k$, that is, the solution of (4.35), (4.36) for $j=k$, satisfies $p_k(r_1)>0$. Using this fact, $p_k(r_s)=0$ and the form of (4.35), we can deduce that $p_k \ge 0$ in $[r_s,r_1]$. Thus, $p_k^\prime (r_s)\ge 0$.

Write (4.29) in the form

$$\begin{aligned} \int _0^{\theta _1}\left( \left. -\frac{1}{\rho _0^+}\frac{1}{r}\partial _r (r{\tilde{\varPsi }}^{({\tilde{f_s^\prime }})})\right| _{r=r_s}-(u_0^- -u_0^+)(r_s)\frac{{\tilde{f}}_s^\prime }{r_s}\right) \xi \sin \theta d \theta =0 \end{aligned}$$

for all $\xi \in L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$.

Rewrite this as

$$\begin{aligned}&\int _0^{\theta _1}\left( -\frac{1}{\rho _0^+}\sum _{j=1}^{m_l}\frac{\partial _r(rp_j)}{r}q_j-(u_0^- -u_0^+)(r_s)\sum _{j=1}^{m_l}\frac{c_jq_j}{r_s}\right. \nonumber \\&\quad \left. -\frac{1}{\rho _0^+}\frac{1}{r}\partial _r (r{\tilde{\varPsi }}^{({\tilde{f_s^\prime }})})\right| _{r=r_s} -(u_0^- -u_0^+)(r_s)\frac{{\tilde{f}}_s^\prime }{r_s}\nonumber \\&\quad \left. -\left( -\frac{1}{\rho _0^+}\sum _{j=1}^{m_l}\frac{\partial _r(rp_j)}{r}q_j-(u_0^- -u_0^+)(r_s)\sum _{j=1}^{m_l}\frac{c_jq_j}{r_s}\right) \right) \xi \sin \theta d\theta =0 \nonumber \\ \end{aligned}$$

(4.40)

for all $\xi \in L^2((0,\theta _1),\sin \theta \hbox {d}\theta )$. Since $p_k^\prime (r_s)\ge 0$, $p_k(r_s)=0$, $(u_0^--u_0^+)(r_s)>0$, ${\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi }\rightharpoonup {\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_\varphi$ in $L^2(\mathcal {N}_{r_s}^+)$ and $D({\tilde{\varPsi }}_{m_l}\mathbf{e}_{\varphi })\rightharpoonup D({\tilde{\varPsi }}^{({\tilde{f_s^\prime }})}\mathbf{e}_{\varphi })$ in $L^2(\mathcal {N}_{r_s}^+)$, for a sufficiently large l such that $m_l\ge k$, if we take $\xi =q_k$, then the left hand side of (4.40) becomes a negative number (here we used the trace theorem). This contradicts to the assumption that $D_{f_s^\prime }\mathcal {A}(\zeta _0){\tilde{f_s^\prime }}=0$. This finishes our proof. $\square$

Applying the weak implicit function theorem introduced in [3] with the results in Lemmas 16, 17 and 19, we obtain the result of the existence part of Theorem 1.

4.2 Proof of Theorem 1 (Uniqueness)

Finally, we prove the uniqueness part of Theorem 1.

Proof of Theorem 1(Uniqueness) Let $\overline{\sigma }_1$ be a positive constant $\le \sigma _3$ obtained in the previous subsection such that if $\sigma \le \overline{\sigma }_1$, then Problem 2 has a solution satisfying

$$\begin{aligned}&|f(0)-r_s|+||f_s^\prime ||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})} +||\nabla \times ((\varPhi -\varPhi _0^+)\mathbf{e}_\varphi )||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha , \varGamma _w^+)}\nonumber \\&\quad +||\frac{L}{2\pi r \sin \theta }\mathbf{e}_\varphi ||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)} +||S-S_0^+||_{1,\alpha ,\mathcal {N}_f^+}^{(-\alpha ,\varGamma _w^+)}\le C \sigma , \end{aligned}$$

(4.41)

where C is a positive constant depending on the data. Let $\sigma _1$ be a positive constant $\le \overline{\sigma }_1$ and to be determined later. Suppose that there exist two solutions $(f_i,\varPhi _i\mathbf{e}_\varphi ,L_i,S_i)$ for $i=1,2$ of Problem 2 for $\sigma \le \sigma _1$ satisfing (4.41).

We will prove that there exists a positive constant $\underline{\sigma _1}\le \overline{\sigma }_1$ such that if $\sigma _1=\underline{\sigma }_1$, then

$$\begin{aligned} (f_1,\varPhi _1\mathbf{e}_\varphi ,L_1,S_1)=(f_2,\varPhi _2\mathbf{e}_\varphi ,L_2,S_2). \end{aligned}$$

(4.42)

In this proof, Cs and $C_i$ for $i=1,2,\ldots$ denote positive constants depending on the data. Each C in different inequalities differs from each other. In this proof, when we estimate quantities, we will use all or a part of the conditions $(\rho _-,{\mathbf {u}}_-,p_-)\in {\mathcal {B}}_{\sigma }^{(1)}$, $p_{ex}\in {\mathcal {B}}^{(3)}_\sigma$, (4.41) satisfied by $(f_i,\varPhi _i\mathbf{e}_\varphi ,L_i,S_i)$ for $i=1,2$ for $\sigma \le \sigma _3$ without mentioning that we use these conditions.

Case 1. $f_{s,1}=f_{s,2}$ where $f_{s,i}:=f_i-f_i(0)$ for $i=1,2$.

By (4.41) satisfied by $(f_i,\varPhi _i\mathbf{e}_\varphi ,L_i,S_i)$,

$$\begin{aligned} ||f^\prime _{s,i}||_{1,\alpha ,(0,\theta _1)}^{(-\alpha ,\{\theta =\theta _1\})}\le C_1\sigma \end{aligned}$$

for $i=1,2$. Choose $\sigma _1=\min (\overline{\sigma }_1,\frac{C_3\sigma _3}{C_2},\frac{\sigma _3}{C_1})(=:\sigma _1^{(1)})$ where $C_2$ and $C_3$ are C in (4.41) and (3.2), respectively. Then $(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_s^\prime )\in {\mathcal {B}}^{(4)}_{\sigma }$ for $\sigma \in (0,\sigma _3]$ and $(f_i(0),\varPhi _i\mathbf{e}_{\varphi },L_i,S_i)$ satisfies (3.2) for $f_s=f_{s,i}$, so the hypothesis in Proposition 2 is satisfied. If $f_{s,1}=f_{s,2}$, then by Proposition 2, $(f_1(0),\varPhi _1\mathbf{e}_\varphi ,L_1,S_1)=(f_2(0),\varPhi _2\mathbf{e}_\varphi ,L_2,S_2)$. From this, we have (4.42).

Case 2. General case.

1. Let $(\varPsi _i,A_i,T_i):=(\varPhi _i-\varPhi _0^+,L_i,S_i-S_0^+)$ and $({\tilde{\varPsi }}_i,{\tilde{A}}_i, {\tilde{T}}_i):=(\frac{{\tilde{W}}_i}{2\pi r \sin \theta },$ $A_i(\varPi _{r_s f_i}),T_i(\varPi _{r_s f_i}))$ for $i=1,2$ with ${\tilde{W}}_i:=W_i(\varPi _{r_s f_i})$ and $W_i:=2\pi r \sin \theta \varPsi _i$. Show that

$$\begin{aligned}&|f_2(0)-f_1(0)|+||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi ||_{1,\beta ,\mathcal {N}_{r_s}^+}\nonumber \\&\qquad +||\frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta }||_{0,\beta ,\mathcal {N}_{r_s}^+}+||{\tilde{T}}_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{r_s}^+} \nonumber \\&\quad \le C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}, \end{aligned}$$

(4.43)

where $\beta$ is a positive constant given in the proof of the uniqueness part of Proposition 2.

By the assumption, $(f(0),f_s,\varPsi \mathbf{e}_{\varphi },L,T)= (f_i(0),f_{s,i},\varPsi _i\mathbf{e}_{\varphi },A_i,T_i)$ for $i=1,2$ satisfy (A$^\prime$). Transform (A$^\prime$) satisfied by $(f_i(0),f_{s,i},\varPsi _i\mathbf{e}_{\varphi },A_i,T_i)$ for $i=1,2$ into equations in $\mathcal {N}_{r_s}^+$ or on a part of $\partial \mathcal {N}_{r_s}^+$ by using $\varPi _{r_s f_i}$ in the way that we transformed (A$^\prime$) satisfied by $(f_1(0),\varPsi _1\mathbf{e}_{\varphi },A_1,T_1)$ in the proof of the uniqueness part of Proposition 2. Then we obtain

$$\begin{aligned}&M_i \nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})(\varPi _{r_s f_i})N_i\nabla \times ({\tilde{\varPsi }}_i\mathbf{e}_\varphi )\right) \nonumber \\&\quad =\left( \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\right) (\varPi _{r_s f_i})\frac{(\partial _{{\tilde{\theta }}_i}\varPi _{f_i r_s}^{*,r})(\varPi _{r_s f_i})\partial _r {\tilde{T}}_i+\partial _\theta {\tilde{T}}_i}{\varPi _{r_s f_i}^{*,r}}\mathbf{e}_\varphi \nonumber \\&\quad \quad \quad +\tilde{\varvec{F}}_{1i}({\tilde{\varPsi }}_i\mathbf{e}_\varphi ,{\tilde{A}}_i,{\tilde{T}}_i)\quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.44)

$$\begin{aligned}&{\tilde{\varPsi }}_i\mathbf{e}_\varphi = {\left\{ \begin{array}{ll}\frac{\varPi _{r_s f_i}^{*,r} (\varPhi _--\varPhi _0^-)(\varPi _{r_s f_i})}{r}\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{r_s},\\ \frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r }\mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{w,r_s}^+:=\varGamma _w\cap \{r>r_s\},\\ \bigg (\frac{1}{r\sin \theta }\int _0^\theta \bigg (\mathfrak f_0({\tilde{T}}_i,p_{ex})-\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}{\tilde{T}}_i\\ \quad \quad \quad +\mathfrak f_1({\tilde{\varPsi }}_i\mathbf{e}_\varphi ,{\tilde{A}}_i,{\tilde{T}}_i) \bigg )r^2\sin \xi \hbox {d}\xi \bigg )\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(4.45)

$$\begin{aligned}&\frac{1}{r\sin \theta }\int _0^\theta \bigg (\mathfrak f_0({\tilde{T}}_i,p_{ex})-\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}{\tilde{T}}_i\nonumber \\&\quad +\mathfrak f_1({\tilde{\varPsi }}_i\mathbf{e}_\varphi ,{\tilde{A}}_i,{\tilde{T}}_i) \bigg )r^2\sin \xi \hbox {d}\xi =\frac{r_0(\varPhi _--\varPhi _0^-)(r_0,\theta _1)}{r_1} \end{aligned}$$

(4.46)

for $i=1,2$ where $M_i=\left( \frac{\partial \varPi _{f_i r_s}}{\partial \mathrm{y}_i}\right) (\varPi _{r_s f_i})$ for $i=1,2$, $N_i=\frac{r^2}{(\varPi _{r_sf_i}^{*,r})^2}\mathbf{e}_r\otimes \mathbf{e}_r-\frac{\left( \partial _{\tilde{\theta }_i}\varPi _{f_ir_s}^{*,r}\right) (\varPi _{r_sf_i}) }{(\varPi _{r_sf_i}^{*,r})^2}r\mathbf{e}_r\otimes \mathbf{e}_\theta +\frac{\left( \partial _{\tilde{r}_i}\varPi _{f_i r_s}^{*,r}\right) (\varPi _{r_sf_i}) }{\varPi _{r_sf_i}^{*,r}}r\mathbf{e}_\theta \otimes \mathbf{e}_\theta$, $\mathrm{y}_i$ for $i=1,2$ are the Cartesian coordinate systems representing $\mathcal {N}_{f_i}^+$, respectively, $(\tilde{r}_i,\tilde{\theta }_i)$ are $(r,\theta )$ coordinates for $\mathrm{y}_i$, respectively, $(r,\theta )=\varPi _{f_i r_s}^*(\tilde{r}_i,\tilde{\theta }_i)$, $\varPi _{r_s f_i}^{*,r}$ and $\varPi _{f_i r_s}^{*,r}$ are the r-components of $\varPi _{r_s f_i}^{*}$ and $\varPi _{f_i r_s}^{*}$, respectively, and $\tilde{\varvec{F}}_{1i}$ for $i=1,2$ are ${\varvec{F}}_1$ changed by using the transformation $\varPi _{r_sf_i}$ for $i=1,2$ using the relations $(\nabla \times (\varPsi _i\mathbf{e}_{\varphi }))(\varPi _{r_s f_i})=N_i\nabla \times ({\tilde{\varPsi }}_i \mathbf{e}_{\varphi })$ and $\left( \nabla \times \left( \frac{A}{2\pi r \sin \theta }\mathbf{e}_{\varphi }\right) \right) (\varPi _{r_s f_i})=N_i\nabla \times \left( \frac{{\tilde{A}}_i}{2\pi r \sin \theta }\mathbf{e}_{\varphi }\right)$. Subtract (4.44)–(4.46) for $i=1$ from the same equations for $i=2$. Then we have

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi )\right) \nonumber \\&\quad = \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta ({\tilde{T}}_2-{\tilde{T}}_1)}{r}\mathbf{e}_\varphi +A_2-A_1+B_2-B_1\nonumber \\&\qquad +\tilde{\varvec{F}}_{12}({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)-\tilde{\varvec{F}}_{11}({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1)(=:\tilde{\varvec{F}})\quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.47)

$$\begin{aligned}&({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_\varphi \nonumber \\&\quad ={\left\{ \begin{array}{ll} \left( \frac{\varPi _{r_s f_2}^{*,r} (\varPhi _--\varPhi _0^-)(\varPi _{r_s f_2})-\varPi _{r_s f_1}^{*,r} (\varPhi _--\varPhi _0^-)(\varPi _{r_s f_1})}{r} \right) \mathbf{e}_{\varphi }(=:\tilde{\varvec{h}}_1)\quad \text{ on }\quad \varGamma _{r_s},\\ {\varvec{0}} \quad \text{ on }\quad \varGamma _{w,r_s}^+,\\ \left( \left. \frac{1}{r\sin \theta }\int _0^\theta \left( \mathfrak f_0({\tilde{T}}_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex})\right. \right. \right. -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_2-{\tilde{T}}_1)\\ \left. \left. +\mathfrak f_1({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)-\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1) \right) r^2\sin \xi \hbox {d}\xi \right) \mathbf{e}_\varphi (=:\tilde{\varvec{h}}_2)\quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(4.48)

$$\begin{aligned}&\left. \frac{1}{r\sin \theta }\int _0^\theta \left( \mathfrak f_0({\tilde{T}}_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex})\right. \right. -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_2-{\tilde{T}}_1)\nonumber \\&\qquad \left. +\mathfrak f_1({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)-\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1) \right) r^2\sin \xi \hbox {d}\xi =0, \end{aligned}$$

(4.49)

where

$$\begin{aligned} A_i=&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times ({\tilde{\varPsi }}_i\mathbf{e}_\varphi )\right) \\&\quad \quad \quad -M_i \nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})(\varPi _{r_sf_i})N_i\nabla \times ({\tilde{\varPsi }}_i\mathbf{e}_\varphi )\right) \end{aligned}$$

and

$$\begin{aligned} B_i&=-\frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta {\tilde{T}}_i}{r}\mathbf{e}_\varphi \\&\quad +\left( \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1+\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\right) (\varPi _{r_sf_i})\frac{(\partial _{{\tilde{\theta }}_i}\varPi _{f_i r_s}^{*,r})(\varPi _{r_sf_i})\partial _r {\tilde{T}}_i+\partial _\theta {\tilde{T}}_i}{\varPi _{r_sf_i}^r}\mathbf{e}_\varphi \end{aligned}$$

for $i=1,2$. We will estimate $||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}$ using (4.47)–(4.49). For this, we estimate $||\frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta }||_{1,0,\mathcal {N}_{r_s}^+}$ and $||{\tilde{T}}_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{r_s}^+}$.

Estimate $||\frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta }||_{0,\beta ,\mathcal {N}_{r_s}^+}$ and $||{\tilde{T}}_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{r_s}^+}$:

Since $A_i$ and $T_i$ for $i=1,2$ are solutions of (B$^\prime$) for $(\varPsi ,f(0),f_s)=(\varPsi _i,f_i(0),f_{s,i})$, by Lemma 15, $A_i$ and $T_i$ are represented as

$$\begin{aligned} A_i=2\pi f_i({\mathcal {L}}_i) \sin {\mathcal {L}}_i u_{-,\varphi }(f_i({\mathcal {L}}_i),{\mathcal {L}}_i) \end{aligned}$$

and

$$\begin{aligned} T_i=\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_i}({\mathcal {L}}_i) }{c_-}\right) ^2\right) S_-\right) (f_i({\mathcal {L}}_i),{\mathcal {L}}_i)-(g({M_0^-}^2))(r_s)S_{in} \end{aligned}$$

where $u_{-,\varphi }={\mathbf {u}}_-\cdot \mathbf{e}_{\varphi }$, ${\mathcal {L}}_i$ for $i=1,2$ are ${\mathcal {L}}$ given in Lemma 15 for $V=2\pi r \sin \theta (\varPhi _0^++\varPsi _i)$ and $f=f_i$, respectively, and ${\varvec{\nu }}_{f_i}$ for $i=1,2$ are the unit normal vectors on $\varGamma _{f_i}$ pointing toward $\mathcal {N}_{f_i}^+$, respectively. Using these solution expressions, express $(\frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta })\mathbf{e}_\varphi$ and ${\tilde{T}}_2-{\tilde{T}}_1$ as

$$\begin{aligned} \left( \frac{2\pi f_2(\tilde{{\mathcal {L}}}_2) \sin \tilde{{\mathcal {L}}}_2 u_{\varphi ,-}(f_2(\tilde{{\mathcal {L}}}_2),\tilde{{\mathcal {L}}}_2)}{2\pi r\sin \theta }-\frac{2\pi f_1(\tilde{{\mathcal {L}}}_1) \sin \tilde{{\mathcal {L}}}_1 u_{\varphi ,-}(f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1)}{2\pi r\sin \theta }\right) \mathbf{e}_\varphi \end{aligned}$$

(4.50)

and

$$\begin{aligned}&\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_2}(\tilde{{\mathcal {L}}}_2) }{c_-}\right) ^2\right) S_-\right) (f_2(\tilde{{\mathcal {L}}}_2),\tilde{{\mathcal {L}}}_2)\nonumber \\&\quad -\left( g\left( \left( \frac{{\mathbf {u}}_-\cdot {\varvec{\nu }}_{f_1}(\tilde{{\mathcal {L}}}_1) }{c_-}\right) ^2\right) S_-\right) (f_1(\tilde{{\mathcal {L}}}_1),\tilde{{\mathcal {L}}}_1), \end{aligned}$$

(4.51)

respectively, where $\tilde{{\mathcal {L}}}_i:={\mathcal {L}}_i(\varPi _{r_s f_i})$ for $i=1,2$. Using arguments similar to the ones used to estimate $||(\frac{A_2}{2\pi r \sin \theta }-\frac{{\tilde{A}}_1}{2\pi r \sin \theta })\mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{f_2}^+}$ and $||T_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{f_2}^+}$ in the proof of the uniqueness part of Proposition 2, we estimate (4.50) and (4.51) in $C^\beta (\overline{\mathcal {N}_{r_s}^+})$. Then we obtain

$$\begin{aligned}&||\left( \frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta }\right) \mathbf{e}_{\varphi }||_{1,0,\mathcal {N}_{r_s}^+}\nonumber \\&\quad \le C\sigma \left( ||f_2-f_1||_{1,\beta ,\varLambda }+||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}\right) \end{aligned}$$

(4.52)

and

$$\begin{aligned} ||{\tilde{T}}_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{r_s}^+} \le C||f_2-f_1||_{1,\beta ,\varLambda }+ C\sigma ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}. \end{aligned}$$

(4.53)

Using these estimates, we estimate $||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}.$ Substitute (4.51) into (4.49). And then using (4.52) and (4.53), estimate $f_2(0)-f_1(0)$ in the resultant equation (see Step 4 in the proof of the uniqueness part of Proposition 2). Then we obtain

$$\begin{aligned}&|f_2(0)-f_1(0)|\\&\quad \le C_4\sigma |f_2(0)-f_1(0)|+ C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)} +C\sigma ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}. \end{aligned}$$

Take $\sigma _1=\min (\sigma _1^{(1)}, \frac{1}{2C_4})(=:\sigma _1^{(2)})$. Then we have

$$\begin{aligned} |f_2(0)-f_1(0)|\le C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)} +C\sigma ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}. \end{aligned}$$

(4.54)

Using (4.54), we obtain from (4.52) and (4.53) that

$$\begin{aligned}&||\frac{{\tilde{A}}_2}{2\pi r\sin \theta }-\frac{{\tilde{A}}_1}{2\pi r\sin \theta }||_{0,\beta ,\mathcal {N}_{r_s}^+}\nonumber \\&\quad \le C\sigma \left( ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}+||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{0,\beta ,\mathcal {N}_{r_s}^+}\right) \end{aligned}$$

(4.55)

and

$$\begin{aligned} ||{\tilde{T}}_2-{\tilde{T}}_1||_{0,\beta ,\mathcal {N}_{r_s}^+} \le C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}+ C\sigma ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}. \end{aligned}$$

(4.56)

Using these two estimates, (4.54) and arguments similar to the ones in Step 5 in the proof of the uniqueness part of Proposition 2, we estimate $({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }$ in (4.47), (4.48) in $C^{1,\beta }(\overline{\mathcal {N}_{r_s}^+})$. Then we obtain

$$\begin{aligned} ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}\le C_5\sigma ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}+C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

Take $\sigma _1=\min (\sigma _1^{(2)}, \frac{1}{2C_5})(=:\sigma _1^{(3)})$. Then we have

$$\begin{aligned} ||({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1)\mathbf{e}_{\varphi }||_{1,\beta ,\mathcal {N}_{r_s}^+}\le C||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)} \end{aligned}$$

(4.57)

Combining (4.54), (4.55), (4.56) and (4.57), we obtain (4.43).

2. Using the arguments in Step 1 in the proof of Lemma 17, we can see that the system (4.8)–(4.12) has a unique solution for given ${\tilde{f}}_s^\prime =f_{s,2}^\prime -f_{s,1}^\prime \in C_0^\beta ([0,\theta _1]):=\{f\in C^\beta ([0,\theta _1])\;|\; f^\prime (0)=f^\prime (\theta _1)=0\}$:

$$\begin{aligned}&f(0)^{(f_{s,2}^\prime -f_{s,1}^\prime )}=\frac{-\int _0^{\theta _1} (f_{s,2}-f_{s,1})\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }, \end{aligned}$$

(4.58)

$$\begin{aligned}&{\tilde{A}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}=0, \end{aligned}$$

(4.59)

$$\begin{aligned}&{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}=(g({M_0^-}^2))^\prime (r_s)S_{in}\left( \frac{-\int _0^{\theta _1} (f_{s,2}-f_{s,1})\sin \zeta d\zeta }{\int _0^{\theta _1}\sin \zeta d\zeta }+f_{s,2}- f_{s,1}\right) \end{aligned}$$

(4.60)

and ${\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}\mathbf{e}_{\varphi }$, the unique $C^{2,\beta }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)$ solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }}=f_{s,2}^\prime -f_{s,1}^\prime \in C^\beta _0([0,\theta _1])$ (here we had ${\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}\mathbf{e}_{\varphi }\in C^{2,\beta }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}$ $(\mathcal {N}_{r_s}^+)$ because $f_{s,2}^\prime -f_{s,1}^\prime \in C_0^\beta ([0,\theta _1])$). Subtract (4.8)–(4.10) for given ${\tilde{f}}_s^\prime =f_{s,2}^\prime -f_{s,1}^\prime \in C_0^\beta ([0,\theta _1])$ from (4.47)–(4.49). Then we obtain

$$\begin{aligned}&\nabla \times \left( \frac{1}{\rho _0^+}(1+\frac{{u_0^+}\mathbf{e}_r\otimes u_0^+\mathbf{e}_r}{{c_0^+}^2-{u_0^+}^2})\nabla \times (({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1-{\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )})\mathbf{e}_\varphi )\right) \nonumber \\&\quad = \frac{{\rho _0^+}^{\gamma -1}}{(\gamma -1)u_0^+}(1 +\frac{\gamma {u_0^+}^2}{{c_0^+}^2-{u_0^+}^2})\frac{\partial _\theta ({\tilde{T}}_2-{\tilde{T}}_1-{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )})}{r}\mathbf{e}_\varphi \nonumber \\&\qquad +A_2-A_1+B_2-B_1+\tilde{\varvec{F}}_{12}({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)\nonumber \\&\qquad -\tilde{\varvec{F}}_{11}({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1)\quad \text{ in }\quad \mathcal {N}_{r_s}^+, \end{aligned}$$

(4.61)

$$\begin{aligned}&({\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1-{\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )})\mathbf{e}_\varphi \nonumber \\&\quad ={\left\{ \begin{array}{ll} \left( \frac{\varPi _{r_s f_2}^{*,r} (\varPhi _--\varPhi _0^-)(\varPi _{r_s f_2})-\varPi _{r_s f_1}^{*,r} (\varPhi _--\varPhi _0^-)(\varPi _{r_s f_1})}{r} \right) \mathbf{e}_{\varphi }\quad \text{ on }\quad \varGamma _{r_s},\\ {\varvec{0}} \quad \text{ on }\quad \varGamma _{w,r_s}^+,\\ \Bigg (\frac{1}{r\sin \theta }\int _0^\theta \bigg (\mathfrak f_0({\tilde{T}}_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex})\\ \quad \quad \quad \quad \quad -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_2-{\tilde{T}}_1-{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )})\\ +\mathfrak f_1({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)-\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1) \bigg )r^2\sin \xi \hbox {d}\xi \Bigg )\mathbf{e}_\varphi \quad \text{ on }\quad \varGamma _{ex}, \end{array}\right. } \end{aligned}$$

(4.62)

$$\begin{aligned}&\frac{1}{r\sin \theta }\int _0^\theta \bigg (\mathfrak f_0({\tilde{T}}_2,p_{ex})-\mathfrak f_0({\tilde{T}}_1,p_{ex}) \nonumber \\&\qquad -\frac{\rho _0^+((\gamma -1){u_0^+}^2+{c_0^+}^2)}{\gamma (\gamma -1)u_0^+S_0^+}({\tilde{T}}_2-{\tilde{T}}_1-{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}) \nonumber \\&\qquad +\mathfrak f_1({\tilde{\varPsi }}_2\mathbf{e}_\varphi ,{\tilde{A}}_2,{\tilde{T}}_2)-\mathfrak f_1({\tilde{\varPsi }}_1\mathbf{e}_\varphi ,{\tilde{A}}_1,{\tilde{T}}_1) \bigg )r^2\sin \xi \hbox {d}\xi =0. \end{aligned}$$

(4.63)

Estimate $||{\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1-{\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}||_{1,\beta ,\mathcal {N}_{r_s}^+}$ using (4.61)–(4.63).

Write ${\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ as

$$\begin{aligned} (g({M_0^-}^2))^\prime (r_s)S_{in}\left( f(0)^{(f_{s,2}^\prime -f_{s,1}^\prime )}+f_{s,2}- f_{s,1}\right) . \end{aligned}$$

(4.64)

Substitute this ${\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ and (4.51) into the places of ${\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ and ${\tilde{T}}_2-{\tilde{T}}_1$ in (4.63), respectively. And then using (4.43), estimate $f_2(0)-f_1(0)-f(0)^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ in the resultant equation. Then we obtain

$$\begin{aligned} |f_2(0)-f_1(0)-f(0)^{(f_{s,2}^\prime -f_{s,1}^\prime )}|\le C\sigma ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

Using this estimate and (4.43), we estimate ${\tilde{T}}_2-{\tilde{T}}_1-{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ given by (4.51) and (4.64) in $C^{\beta }(\overline{\mathcal {N}_{r_s}^+})$. Then we obtain

$$\begin{aligned} ||{\tilde{T}}_2-{\tilde{T}}_1-{\tilde{T}}^{(f_{s,2}^\prime -f_{s,1}^\prime )}||_{0,\beta ,\mathcal {N}_{r_s}^+}\le C\sigma ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

(4.65)

Using arguments similar to the ones in Step 5 in the proof of the uniqueness part of Proposition 2 with (4.43) and (4.65), we estimate ${\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1-{\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}$ in (4.61), (4.62) in $C^{1,\beta }(\overline{\mathcal {N}_{r_s}^+})$. Then we have

$$\begin{aligned} ||{\tilde{\varPsi }}_2-{\tilde{\varPsi }}_1-{\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}||_{1,\beta ,\mathcal {N}_{r_s}^+} \le C\sigma ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

(4.66)

3. By the assumption, $(f_i,\varPhi _i\mathbf{e}_\varphi ,L_i,S_i)$ for $i=1,2$ satisfy (2.47). This implies that $\mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_{s,i}^\prime )=0$ for $i=1,2$ where $\mathcal {A}$ is a map defined in (4.1). Subtract these two equations. Then we have

$$\begin{aligned} \mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_{s,2}^\prime )-\mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_{s,1}^\prime )=0. \end{aligned}$$

Write this as

$$\begin{aligned} 0= D_{f_s^\prime } \mathcal {A}(\zeta _0)(f_{s,2}^\prime -f_{s,1}^\prime )+R, \end{aligned}$$

(4.67)

where

$$\begin{aligned} R= & {} \mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_{s,2}^\prime ) -\mathcal {A}(\rho _-,{\mathbf {u}}_-,p_-,p_{ex},f_{s,1}^\prime )\\&-D_{f_s^\prime } \mathcal {A}(\zeta _0)(f_{s,2}^\prime -f_{s,1}^\prime ), \end{aligned}$$

and $D_{f_s^\prime } \mathcal {A}(\zeta _0)$ is a map given in (4.2) as a map from $C_0^\beta ([0,\theta _1])$ to $C_0^\beta ([0,\theta _1])$ (here $D_{f_s^\prime } \mathcal {A}(\zeta _0)$ takes $C^\beta _0([0,\theta _1])$ functions and ${\tilde{\varPsi }}^{(f_{s,2}^\prime -f_{s,1}^\prime )}\mathbf{e}_{\varphi }$ in $D_{f_s^\prime } \mathcal {A}(\zeta _0)(f_{s,2}^\prime$ $-f_{s,1}^\prime )$ is the $C^{2,\beta }_{(-1-\alpha ,\varGamma _{w,r_s}^+)}(\mathcal {N}_{r_s}^+)$ solution of (4.17), (4.18) for given ${\tilde{f_s^\prime }}=f_{s,2}^\prime -f_{s,1}^\prime \in C^\beta _0([0,\theta _1])$).

Using (4.43) and (4.66), we can obtain

$$\begin{aligned} ||R||_{0,\beta ,(0,\theta _1)}\le C\sigma ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

(4.68)

In the way that we proved the invertiblity of $D_{f_s^\prime } \mathcal {A}(\zeta _0)$ as a map from C $^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})} ((0,\theta ))$ to $C^{1,\alpha }_{(-\alpha ,\{\theta =\theta _1\})}((0,\theta ))$ in the the proof of Lemma 19, we can prove that $D_{f_s^\prime } \mathcal {A}(\zeta _0)$ is an invertible map as a map from $C_0^\beta ([0,\theta _1])$ to $C_0^\beta ([0,\theta _1])$. Using this fact and (4.68), we obtain, from (4.67),

$$\begin{aligned} ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}\le C_6\sigma ||f_{s,2}^\prime -f_{s,1}^\prime ||_{0,\beta ,(0,\theta _1)}. \end{aligned}$$

Take $\sigma _1=\min (\sigma _1^{(3)},\frac{1}{2C_6})(=:\underline{\sigma }_1)$. Then we have $f_{s,2}^\prime =f_{s,1}^\prime$. One can see that $\underline{\sigma }_1$ depends on the data. This finishes the proof. $\square$

Appendix

In this section, we present some computations done by using the tensor notation given in § 3.2, and make up some fact needed for the well-definedness of V defined in (2.23). We explain how we transformed an elliptic system on the Cartesian coordinate system into a system on the spherical coordinate system in the proof of Lemma 10 and Lemma 11, and show that (3.35) is equivalent to (3.28) and that V defined in (2.23) is independent of the choice of $S_{\mathrm{x}}$.

Let $(q_1,q_2,q_3)$ be an orthogonal coordinate system in $\mathbb {R}^3$. The unit vectors in this coordinate system in the direction of $q_i$ for $i=1,2,3$ are given as $\frac{1}{h_i}\frac{\partial {\varvec{x}}}{\partial q_i}(=:\mathbf{e}_{q_i})$ for $i=1,2,3$ where ${\varvec{x}}=x\mathbf{e}_1+y\mathbf{e}_2+z\mathbf{e}_3$ and $h_i:=|\frac{\partial {\varvec{x}}}{\partial q_i}|$. By $\nabla =\sum _{i=1}^3 \frac{\mathbf{e}_{q_i}}{h_i}\partial _{q_i}$, $\nabla {\varvec{U}}$ where ${\varvec{U}}: \mathbb {R}^3\rightarrow \mathbb {R}^3$ can be written as

$$\begin{aligned} \nabla {\varvec{U}}=\sum _{i=1}^3\frac{\partial _{q_i}{\varvec{U}}}{h_i}\otimes \mathbf{e}_{q_i}. \end{aligned}$$

(5.1)

Here, ${\varvec{a}}\otimes {\varvec{b}}$ for ${\varvec{a}}, {\varvec{b}}\in \mathbb {R}^3$ denotes ${\varvec{a}}{\varvec{b}}^T$. By (3.34), the multiplication of the tensor notation defined in § 3.2 is defined by

$$\begin{aligned} ({\varvec{a}}\otimes {\varvec{b}}\otimes {\varvec{c}}\otimes {\varvec{d}}) ({\varvec{e}}\otimes {\varvec{f}}\otimes {\varvec{g}}\otimes {\varvec{h}})=({\varvec{d}}\cdot {\varvec{e}})({\varvec{c}}\cdot {\varvec{f}}) ({\varvec{a}}\otimes {\varvec{b}}\otimes {\varvec{g}}\otimes {\varvec{h}}), \end{aligned}$$

(5.2)

where ${\varvec{a}}$, ${\varvec{b}}$, ${\varvec{c}}$, ${\varvec{d}}$, ${\varvec{e}}$, ${\varvec{f}}$, ${\varvec{g}}$, ${\varvec{h}}\in \mathbb {R}^3$. By (5.1) and (5.2), M in

$$\begin{aligned} \nabla {\varvec{U}}=M\nabla _{(q_1,q_2,q_3)}{\varvec{U}}, \end{aligned}$$

where $\nabla _{(q_1,q_2,q_3)}{\varvec{U}}=\sum _{i=1}^3\partial _{q_i}{\varvec{U}} \otimes \mathbf{e}_i$ and $M=\frac{\partial (q_1,q_2,q_3)}{\partial (x,y,z) }$, can be expressed as

$$\begin{aligned} M=\sum _{i=1}^3\frac{1}{h_i}\mathcal {I}\otimes \mathbf{e}_{q_i}\otimes \mathbf{e}_i\otimes \mathcal {I}. \end{aligned}$$

(5.3)

Here, $\mathcal {I}\otimes {\varvec{a}}\otimes {\varvec{b}}\otimes \mathcal {I}$ for ${\varvec{a}}$, ${\varvec{b}}\in R^3$ is a linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ defined by

$$\begin{aligned} \mathcal {I}\otimes {\varvec{a}}\otimes {\varvec{b}}\otimes \mathcal {I}:=\sum _{i=1}^3\mathbf{e}_i\otimes {\varvec{a}}\otimes {\varvec{b}}\otimes \mathbf{e}_i. \end{aligned}$$

(5.4)

(Note that $\mathcal {I}\otimes {\varvec{a}}\otimes {\varvec{b}}\otimes \mathcal {I}$ satisfies

$$\begin{aligned} (\mathcal {I}\otimes {\varvec{a}}\otimes {\varvec{b}}\otimes \mathcal {I})({\varvec{c}}\otimes {\varvec{d}})=({\varvec{b}}\cdot {\varvec{d}}){\varvec{c}}\otimes {\varvec{a}} \end{aligned}$$

(5.5)

for ${\varvec{c}}$, ${\varvec{d}}\in \mathbb {R}^3$.) Using these facts, we can transform an elliptic system on the Cartesian coordinate system into a system on another coordinate system. In the standard way, we transform $\mathrm{div}({\varvec{B}}\nabla {\varvec{U}})=0$ on the Cartesian coordinate system where ${\varvec{B}}$ is a linear map from $\mathbb {R}^{3\times 3}$ to $\mathbb {R}^{3\times 3}$ into an equation on $(q_1,q_2,q_3)$-coordinate system. Then we obtain

$$\begin{aligned} \mathrm{div}_{(q_1,q_2,q_3)}\left( \frac{1}{\det M}M^T {\varvec{B}}\circ \chi ^{-1}M\nabla _{(q_1,q_2,q_3)}({\varvec{U}}\circ \chi ^{-1})\right) =0, \end{aligned}$$

(5.6)

where $\mathrm{div}_{(q_1,q_2,q_3)}:=\nabla _{(q_1,q_2,q_3)}\cdot$, $\nabla _{(q_1,q_2,q_3)}$ and M are as above, $\chi$ is the map from (x, y, z) to $(q_1,q_2,q_3)$ and $\det M=\frac{1}{h_1h_2h_3}$. Using (5.2) and (5.3), we compute $M^T{\varvec{B}}\circ \chi ^{-1} M$. Then we obtain the explicit form of (5.6). In this way, we can obtain the explicit form of the spherical coordinate representation of (3.57).

Next, we show that (3.35) is equivalent to (3.28).

By (5.1),

$$\begin{aligned} \nabla (\varPsi \mathbf{e}_{\varphi })=\partial _r \varPsi \mathbf{e}_{\varphi }\otimes \mathbf{e}_r+\frac{\partial _{\theta }\varPsi }{r}\mathbf{e}_{\varphi }\otimes \mathbf{e}_{\theta }-\frac{\varPsi }{r}\mathbf{e}_r\otimes \mathbf{e}_{\varphi }-\frac{\cos \theta }{r\sin \theta }\varPsi \mathbf{e}_{\theta }\otimes \mathbf{e}_{\varphi }. \end{aligned}$$

(5.7)

Using (5.5) and (5.7), we have

$$\begin{aligned}&\frac{{c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\left( {\varvec{I}}-\frac{ {u_0^+}^2}{{c_0^+}^2}(\mathcal {I}\otimes \mathbf{e}_r\otimes \mathbf{e}_r \otimes \mathcal {I} )\right) \nabla (\varPsi \mathbf{e}_\varphi )\\&\quad =\frac{ {c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\nabla (\varPsi \mathbf{e}_{\varphi })-\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}\partial _r\varPsi \mathbf{e}_{\varphi }\otimes \mathbf{e}_r. \end{aligned}$$

Using this relation and the relation $\mathrm{div}({\varvec{a}}\otimes {\varvec{b}})=\nabla {\varvec{a}}{\varvec{b}}+{\varvec{a}}\mathrm{div}{\varvec{b}}$, we compute the left hand side of (3.35). Then we get

$$\begin{aligned}&a(r)\varDelta (\varPsi \mathbf{e}_{\varphi })+\nabla (\varPsi \mathbf{e}_{\varphi })\nabla a(r)-\nabla \mathbf{e}_{\varphi }b(r)\partial _r\varPsi \mathbf{e}_r\nonumber \\&\quad -\mathbf{e}_{\varphi }\mathrm{div}(b(r)\partial _r\varPsi \mathbf{e}_r)-\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\varPsi \mathbf{e}_\varphi , \end{aligned}$$

(5.8)

where $a(r):=\frac{ {c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}$ and $b(r):=\frac{ {u_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)}$. By direct computation, one can check that $\nabla \mathbf{e}_{\varphi }b(r)\partial _r\varPsi \mathbf{e}_r=0$. Using this fact and (5.7), we compute (5.8). Then we have

$$\begin{aligned}&\left( a(r)(\varDelta \varPsi -\frac{\varPsi }{r^2\sin ^2\theta })+a^\prime (r)\partial _r\varPsi -b(r)(\frac{1}{r^2}\partial _r(r^2\partial _r\varPsi ))-b^\prime (r)\partial _r\varPsi \right) \mathbf{e}_{\varphi }\nonumber \\&\qquad \quad -\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\varPsi \mathbf{e}_\varphi . \end{aligned}$$

(5.9)

By $a-b=\frac{1}{\rho _0^+}$, $(a-b)^\prime =-\frac{\partial _r\rho _0^+}{{\rho _0^+}^2}$. Substitute this relation into (5.9) and then express the resultant equation in the spherical coordinate system. Then we obtain

$$\begin{aligned}&\left( \frac{1}{\rho _0^+r^2}\partial _r(r^2\partial _r \varPsi )-\frac{\partial _r\rho _0^+}{{\rho _0^+}^2r}\partial _r(r\varPsi )\right. \\&\quad \left. +\frac{ {c_0^+}^2}{\rho _0^+({c_0^+}^2-{u_0^+}^2)r^2}\left( \frac{1}{\sin \theta }\partial _{\theta }(\sin \theta \partial _{\theta }\varPsi )-\frac{\varPsi }{\sin ^2\theta }\right) \right) \mathbf{e}_{\varphi }. \end{aligned}$$

This after multiplied by $(-1)$ is equal to the left hand side of (3.30). By this fact, (3.35) is equivalent to (3.28).

Finally, for the well-definedness of V defined in (2.23), we show that V defined in (2.23) is independent of the choice of $S_{\mathrm{x}}$.

Lemma 20

Let $\varOmega$ be an open simply connected axisymmetric set in $\mathbb {R}^3$. Let $S_{\mathrm{x}}^1$ and $S_{\mathrm{x}}^2$ be two different simply connected open bounded $C^1$ surfaces in $\varOmega$ having the same boundary. Then

$$\begin{aligned} \int _{S_{\mathrm{x}}^1}\rho {\mathbf {u}}\cdot {\varvec{\nu }}_1dS=\int _{S_{\mathrm{x}}^2}\rho {\mathbf {u}}\cdot {\varvec{\nu }}_2dS, \end{aligned}$$

(5.10)

where ${\varvec{\nu }}_i$ for $i=1,2$ are the unit normal vector on $S_{\mathrm{x}}^i$ pointing outward direction with respect to the cone-like domain made by connecting $\partial S_{\mathrm{x}}^i$ and the origin by straight lines, respectively.

Proof

The result can be obtained by using the divergence theorem.

Suppose that there exists a simply connected open bounded $C^1$ surface in $\varOmega$ whose boundary is equal to the boundary of $S_{\mathrm{x}}^1$ or $S_{\mathrm{x}}^2$, and not intersecting with $S_{\mathrm{x}}^1$ or $S_{\mathrm{x}}^2$ except the boundary of the surface. Let us denote this surface by $S^*$. Then by the first equation of (2.1) and the divergence theorem, we have that

$$\begin{aligned} \int _{S_{\mathrm{x}}^i}\rho {\mathbf {u}}\cdot {\varvec{\nu }}_idS=\int _{S^*}\rho {\mathbf {u}}\cdot {\varvec{\nu }}^*dS \end{aligned}$$

for $i=1,2$, where ${\varvec{\nu }}^*$ is the unit normal vector on $S^*$ pointing outward direction with respect to the cone-like domain made by connecting $\partial S^*$ and the origin by straight lines. From this, we obtain (5.10). The existence of $S^*$ can be obtained by using the fact that $\varOmega$ is open. This finishes the proof. $\square$