# Existence and uniqueness of axially symmetric compressible subsonic jet
impinging on an infinite wall

Jianfeng Cheng
Department of Mathematics, Sichuan University,
Chengdu 610065, P. R. China
<EMAIL_ADDRESS>

Lili Du
Department of Mathematics, Sichuan University,
Chengdu 610065, P. R. China
<EMAIL_ADDRESS>corresponding author.

Qin Zhang
Department of Mathematics, Chongqing Jiaotong University,
Chongqing 400074, P. R. China.
<EMAIL_ADDRESS>
###### Abstract

This paper is concerned with the well-posedness theory of the impact of a
subsonic axially symmetric jet emerging from a semi-infinitely long nozzle,
onto a rigid wall. The fluid motion is described by the steady isentropic
Euler system. We showed that there exists a critical value $M_{cr}>0$, if the
given mass flux is less than $M_{cr}$, there exists a unique smooth subsonic
axially symmetric jet issuing from the given semi-infinitely long nozzle and
hitting a given uneven wall. The surface of the axially symmetric impinging
jet is a free boundary, which detaches from the edge of the nozzle smoothly.
It is showed that a unique suitable choice of the pressure difference between
the chamber and the atmosphere guarantees the continuous fit condition of the
free boundary. Moreover, the asymptotic behaviors and the decay properties of
the impinging jet and the free surface in downstream were also obtained. The
main results in this paper solved the open problem on the well-posedness of
the compressible axially symmetric impinging jet, which has proposed by A.
Friedman in Chapter 16 in [FA2]. The key ingredient of our proof is based on
the variational method to the quasilinear elliptic equation with the
Bernoulli’s type free boundaries.

_2010 Mathematics Subject Classification:_ Primary 76N10, 76G25; Secondary
35Q31, 35J25.

_Keywords:_ Existence and uniqueness; free streamline; compressible Euler
system; subsonic impinging flow.

##  1 Introduction

The problem of a compressible jet falling from a channel and impacting on a
wall is a fascinating one, with very practical applications. The canonical
problem is of interest in a number of areas, such a flow is produced by the
downwards-directed jet from a vertical take-off aircraft spreading out over
the ground, or by a jet of water form a tap falling into a full sink. The
monographs of Birkhoff and Zarantonello in [BZ], Jacob in [Ja], Gurevich in
[Gu] and Milne-Thomson in [MT] gave good surveys of these flows.

The two-dimensional impinging jets have been considered by Helmholtz and
Kirchhoff in 1968. They constructed a solution to the steady irrotational
flows of ideal incompressible weightless fluid, bounded by the walls and free
streamlines. A first systematic well-posedness result on the incompressible
impinging jet was mentioned in Page 364 and Page 416 in [FA1] that A. Friedman
and L. Caffarelli established the existence of the incompressible irrotational
jet issuing from a two-dimensional semi-infinite channel and impinging on an
infinite plate (see also in Chapter 16.3 in [FA2]), see [ACF1, ACF2, ACF3,
CDW1, CDW3] in different settings. Furthermore, A. Friedman investigated the
compressible subsonic free surface flow theory on the sharped charge jets in
[FA2] and proposed several open problems on the impinging jets in two
dimensions. The one is

”Problem (4). Do the same for the compressible case.”

In the recent work [CDW2], the authors established the existence result on the
subsonic jets issuing from a convergent nozzle and impact on a flat plate for
some special atmospheric pressure (Please see Figure 1).

Figure 1: Subsonic impinging jet in two dimensions

As A. Friedman pointed out in [FA2], ”$\cdots$ the compressible axially
symmetric case is quite open $\cdots$”. In this paper, we will focus on
another open problem pointed out by A. Friedman in [FA2]:

”Problem (5). Extend the results to the axially symmetric flows.”

We will establish the existence and uniqueness of the compressible impinging
jet in axially symmetric case, and solve the open problem (5) pointed out by
A. Friedman. Many numerical simulations of the impact of a compressible flows
from a cylinder on a rigid wall are referred to [GE, GY, KO, SW, Ve1].

Figure 2: Subsonic axially symmetric impinging jet

The present paper treats the compressible impinging jet problem created by the
impingement of a subsonic axially symmetric jet emerging from a semi-
infinitely long nozzle on a solid curved wall (see Figure 2). The geometry
considered here is a semi-infinite nozzle in the form of a circular cylinder,
in an unbounded space. The infinite uneven wall is solid and undeformable. The
fluid is assumed to be steady, inviscid and irrotational throughout, and the
jet emerges from the orifice of the nozzle of circular cross-section bounded
by a stream surface, the nozzle wall and the curved wall.

###  1.1 Formulation of the physical problem

The steady isentropic compressible flow is governed by the following three-
dimensional Euler system

$\left\\{\begin{array}[]{rl}&\nabla\cdot(\rho U)=0,\\\ &(\rho
U\cdot\nabla)U+\nabla p=0,\end{array}\right.$ (1.1)

with the irrotational condition

$\nabla\times U=0.$ (1.2)

Here, $U=(u_{1},u_{2},u_{3})$ is the velocity, $\rho=\rho(x_{1},x_{2},x_{3})$
is the density and $p=p(\rho)$ denotes the pressure,
$(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$ is the space variable. Without loss of
generality, we assume that the flow is perfect polytropic gas satisfying the
$\gamma$-law

$p=\mathcal{A}\rho^{\gamma},$ (1.3)

with $\mathcal{A}>0$ and the adiabatic exponent $\gamma>1$. The sound speed of
the flow is defined as
$c(\rho)=\sqrt{p^{\prime}(\rho)}=\sqrt{\mathcal{A}\gamma\rho^{\gamma-1}}$, and
the flow is subsonic if and only if $|U|<c(\rho)$.

Here, we consider the axially symmetric flow in this paper, and take $y=x_{3}$
to be the axis of symmetry and $x=\sqrt{x_{1}^{2}+x^{2}_{2}}$. Let the fluid
density and velocity be $\rho(x,y)$ and $u(x,y),v(x,y),w(x,y)$ in cylindrical
coordinates, where $u,v,w$ are radial velocity, axially velocity and swirl
velocity respectively. Furthermore, we look for such an axisymmetric flow
without swirl in this paper, one has

$u_{1}(x_{1},x_{2},x_{3})=u(x,y)\frac{x_{1}}{x},\
u_{2}(x_{1},x_{2},x_{3})=u(x,y)\frac{x_{2}}{x},\
u_{3}(x_{1},x_{2},x_{3})=v(x,y).$

Then the governing system (1.1) and (1.2) are written in the cylindrical
coordinates as

$\left\\{\begin{array}[]{rl}&(x\rho u)_{x}+(x\rho v)_{y}=0,\\\ &(x\rho
u^{2})_{x}+(x\rho uv)_{y}+xp_{x}=0,\\\ &(x\rho uv)_{x}+(x\rho
v^{2})_{y}+xp_{y}=0,\end{array}\right.$ (1.4)

with the irrotational condition

$u_{y}-v_{x}=0.$ (1.5)

Figure 3: Axially symmetric impinging jet

In order to clarify the physical problem, we start with the notation and the
assumptions on the geometry of the nozzle and the impermeable wall as follows.
As shown in Figure 3, we denote the semi-infinite nozzle as

$N:y=g(x)\in
C^{2,\alpha}((a,b]),~{}~{}g(b)=1~{}~{}\text{and}~{}~{}\lim_{x\rightarrow
a^{+}}g(x)=+\infty,$ (1.6)

with $A=(b,1)$ being the endpoint of the nozzle, and $g(x)$ is decreasing in
$(a,a+\varepsilon_{0})$ for any small $\varepsilon_{0}>0$. Denote the uneven
wall as $N_{0}:~{}y=g_{0}(x)$ for $x\geq 0$ satisfying

$g_{0}(x)\in C^{2,\alpha}([0,+\infty)),~{}~{}g_{0}(0)=0,~{}~{}g(x)>g_{0}(x)\ \
\text{ for any $x\in(a,b]$ },$ (1.7)

and there exist a $\theta\in\left[0,\frac{\pi}{2}\right)$, and a $R_{0}>b$,
such that

$~{}~{}g_{0}^{\prime}(x)\rightarrow\tan\theta\ \ \text{ as
$x\rightarrow+\infty$, and}\ \ g_{0}^{\prime\prime}(x)\geq 0\ \ \text{for any
$x>R_{0}$}.$ (1.8)

The boundary conditions require that the nozzle wall $N$ and the uneven wall
$N_{0}$ are assumed to be impermeable, thus

$(u,v)\cdot\vec{n}=0~{}~{}~{}\text{on}~{}~{}N\cup N_{0},$ (1.9)

where $\vec{n}$ is the unit outward normal to $N\cup N_{0}$. We denote the
incoming mass flux as $M_{0}$ in the axially symmetric nozzle, namely

$-\int_{0}^{x_{0}}2\pi x\rho(x,y_{0})v(x,y_{0})dx=M_{0}>0,$ (1.10)

for any $y_{0}\in(1,+\infty)$, where $x_{0}=\inf\\{x\mid g(x)=y_{0}\\}$.

The well-known Bernoulli’s law gives that

$\frac{q^{2}}{2}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho^{\gamma-1}=\mathcal{B}\
\text{ in the fluid field,}$ (1.11)

where $q=\sqrt{u^{2}+v^{2}}$ is the flow speed and $\mathcal{B}$ is the
Bernoulli’s constant.

The free surface ${\Gamma}$ is defined as an interface between the fluid
issuing from the nozzle wall $N$ and the fluid outside. And then the fluid
still satisfies the slip boundary condition on the free surface ${\Gamma}$.
Moreover, the pressure on ${\Gamma}$ balances to the atmospheric pressure
$p_{atm}$, and thus we assume that

$p=p_{atm}~{}~{}\text{on}~{}~{}{\Gamma}.$ (1.12)

Hence, we can formulate the compressible subsonic impinging jet problem into
the following free boundary problem (FBP).

######  Definition 1.1.

The free boundary problem (FBP).
Given a semi-infinitely long nozzle wall $N$, an uneven wall $N_{0}$, for some
appropriate incoming mass flux $M_{0}>0$, whether there exists a unique
axially symmetric subsonic impinging jet flow, such that the free surface
${\Gamma}$ detaches smoothly from the endpoint of the nozzle wall $N$, and
goes to infinity in $x$-direction, and the pressure balances to the
atmospheric pressure $p_{atm}$ on the free surface?

Next, we give the definition of the subsonic solution to the FBP.

######  Definition 1.2.

(A subsonic solution to the FBP).
A vector $(u,v,\rho,{\Gamma})$ is called a subsonic solution to the FBP,
provided that
(1) the free surface ${\Gamma}$ is given by a $C^{1}$-smooth function $y=k(x)$
for $x\in(b,+\infty)$ with

$k(b+0)=g(b-0)=1,~{}~{}~{}k^{{}^{\prime}}(b+0)=g^{\prime}(b-0),$ (1.13)

and

$k^{\prime}(x)\rightarrow\tan\theta,~{}~{}~{}k(x)-g_{0}(x)\rightarrow
0~{}~{}\text{as}~{}~{}x\rightarrow+\infty.$

(2) $(u,v,\rho)\in~{}C^{1,\alpha}(\Omega_{0})\cap C(\bar{\Omega}_{0})$ solves
the compressible Euler system (1.4) in $\Omega_{0}$, where $\Omega_{0}$ is the
flow field bounded by $N$, $N_{0}$, $I$ and ${\Gamma}$;
(3) $\sup_{(x,y)\in\bar{\Omega}_{0}}\frac{\sqrt{u^{2}+v^{2}}}{c(\rho)}<1$ and
$p=p_{atm}$ on ${\Gamma}$.

######  Remark.

The conditions (1.13) are so-called continuous fit condition and smooth fit
condition to the impinging jet, which imply that the free surface ${\Gamma}$
initiates smoothly from the endpoint $A$ of the nozzle wall $N$.

###  1.2 Main results

Before we state the main results in this paper, we would like to emphasize
that the atmospheric pressure $p_{atm}$ is an arbitrary constant here. Once it
is fixed, we found that there exists an interval $(p_{1},p_{2})$ for the
constant pressure $p_{in}$ in the inlet, and then our results reveal that we
can impose a unique $p_{in}\in(p_{1},p_{2})$ to guarantee the unique existence
of the axially symmetric impinging jet. And the critical values $p_{1}$,
$p_{2}$ depend on the atmospheric pressure $p_{atm}$ and the mass flux
$M_{0}$, which can be determined uniquely by the following formulas,

$\frac{M^{2}_{0}}{2\pi^{2}a^{4}\left(\frac{p_{1}}{\mathcal{A}}\right)^{\frac{2}{\gamma}}}+\frac{\mathcal{A}\gamma}{\gamma-1}\left(\frac{p_{1}}{\mathcal{A}}\right)^{\frac{\gamma-1}{\gamma}}=\frac{\mathcal{A}\gamma}{\gamma-1}\left(\frac{p_{atm}}{\mathcal{A}}\right)^{\frac{\gamma-1}{\gamma}}$

and

$\frac{M^{2}_{0}}{2\pi^{2}a^{4}\left(\frac{p_{2}}{\mathcal{A}}\right)^{\frac{2}{\gamma}}}+\frac{\mathcal{A}\gamma}{\gamma-1}\left(\frac{p_{2}}{\mathcal{A}}\right)^{\frac{\gamma-1}{\gamma}}=\frac{\mathcal{A}\gamma(\gamma+1)}{2(\gamma-1)}\left(\frac{p_{atm}}{\mathcal{A}}\right)^{\frac{\gamma-1}{\gamma}}.$

Obviously, $p_{2}>p_{1}$ and the interval $(p_{1},p_{2})$ is well-defined.

The main results in this paper are stated as follows.

######  Theorem 1.3.

Assume that the semi-infinitely long nozzle wall $N$ and the uneven wall
$N_{0}$ satisfy the conditions (1.6)-(1.8), for any given atmospheric pressure
$p_{atm}>0$, then there exists a critical mass flux $M_{cr}$, such that for
any $M_{0}\in(0,M_{cr})$, there exist a unique incoming pressure
$p_{in}\in(p_{1},p_{2})$ and a unique subsonic solution $(u,v,\rho,{\Gamma})$
to the FBP. Moreover,
(1) the subsonic impinging jet flow satisfies the asymptotic behavior in
upstream as follows,

$(u,v,\rho)(x,y)\rightarrow(0,v_{in},\rho_{in})~{}~{}~{}\text{and}~{}~{}\nabla(u,v,\rho)\rightarrow
0,$

uniformly in any compact subset of $(0,a)$ as $y\rightarrow+\infty$, where
$v_{in}=-\frac{M_{0}}{\pi a^{2}\rho_{in}}$ and
$\rho_{in}=\left(\frac{p_{in}}{\mathcal{A}}\right)^{\frac{1}{\gamma}}$.

Similarly, the subsonic impinging jet flow satisfies the asymptotic behavior
in downstream as follows,

$(u,v,\rho)(x,y)\rightarrow\left(q_{0}\cos\theta,q_{0}\sin\theta,\rho_{0}\right),\
\ (x,y)\in\Omega_{0},$

as $x^{2}+y^{2}\rightarrow+\infty$, where
$\rho_{0}=\left(\frac{p_{atm}}{\mathcal{A}}\right)^{\frac{1}{\gamma}}$ and
$q_{0}=\sqrt{v^{2}_{in}+\frac{2\mathcal{A}\gamma}{\gamma-1}\left(\rho^{\gamma-1}_{in}-\rho_{0}^{\gamma-1}\right)}$;
(2) the free boundary ${\Gamma}$ converges to $N_{0}$ in the far field, and

$x(k(x)-g_{0}(x))\rightarrow\frac{M_{0}}{2\rho_{0}q_{0}\pi\cos\theta}~{}~{}~{}\text{as}~{}~{}x\rightarrow+\infty.$
(1.14)

Furthermore, the free boundary $\Gamma$ is analytic;
(3) the radial velocity $u>0$ in $\bar{\Omega}_{0}\setminus I$;
(4) $M_{cr}$ is the upper critical value of mass flux for the existence of
subsonic jet flow in the following sense: either

$\sup_{\bar{\Omega}_{0}}(u^{2}+v^{2}-c^{2}(\rho))\rightarrow 0\ \ \text{as}\ \
M_{0}\rightarrow M^{-}_{cr},$

or there is no $\sigma>0$, such that for any $M_{0}\in(M_{cr},M_{cr}+\sigma)$,
there exists a subsonic solution to the compressible jet flow problem and

$\sup_{M_{0}\in(M_{cr},M_{cr}+\sigma)}\left(\sup_{\bar{\Omega}_{0}}(u^{2}+v^{2}-c^{2}(\rho))\right)<0.$

######  Remark.

In view of the statement (4) in Theorem 1.3, we have that either

$\sup_{\bar{\Omega}_{0}}(u^{2}+v^{2}-c^{2}(\rho))\rightarrow 0\ \ \text{as}\ \
M_{0}\rightarrow M^{-}_{cr},$

or for any $M_{0}>M_{cr}$, there exists an incoming mass flux
$\tilde{M}_{0}\in[M_{cr},M_{0}]$, such that there are no an incoming pressure
$p_{in}\in(p_{1},p_{2})$ and a subsonic solution to the compressible jet flow
problem satisfying

$\sup_{\bar{\Omega}_{0}}(u^{2}+v^{2}-c^{2}(\rho))<0.$

######  Remark.

Our result indicates that the chamber pressure $p_{in}$ in the inlet is
determined uniquely by the continuous fit condition, provided that the
atmospheric pressure $p_{atm}$ is imposed. On another hand, the result implies
that there exists a unique pressure difference between the chamber and the
outside, such that there exists a unique axially symmetric impinging jet with
continuous fit condition.

######  Remark.

The result (1.14) in fact gives the convergence rate of the distance between
the free surface $\Gamma$ and the curved wall $N_{0}$ in the downstream, which
is quite different from the two-dimensional case.

######  Remark.

Here, we restrict the magnitude of the incoming mass flux $M_{0}$ to guarantee
the global subsonicity of the impinging jet, this idea is motivated by the
recent works [CDX1, CDW4, DX, DXX, DXY, XX1, XX2, XX3] on the global subsonic
flow in an infinitely long nozzle. This is also quite different from the
recent results on two-dimensional subsonic impinging jets in [CDW2].

Based on the significant work [ACF6] by Alt, Caffarelli and Friedman, we can
obtain the higher regularity of the free boundary near the end point $A$.

######  Theorem 1.4.

If $N$ is $C^{3,\alpha}$ near $A$, then the solution $(u,v,\rho,\Gamma)$
established in Theorem 1.3 satisfies that either
(1). $N\cup\Gamma$ is $C^{2}$ at $A$ or,
(2). the optimal regularity of $N\cup\Gamma$ at $A$ is only
$C^{1,\frac{1}{2}}$ and the curvature of $\Gamma$ goes to $\pm\infty$ as
$x\rightarrow b^{+}$.

######  Remark.

The results of Theorem 1.4 imply that either

the curvature along $\Gamma$ tends to the curvature of $N$ at $A$,

or

$\text{the curvature of $\Gamma$ tends to $\pm\infty$ in absolute value as one
approaches $A$ along $\Gamma$}.$

The second case is so-called abrupt separation. The proof of Theorem 1.4
follows from Theorem 1.1 in [ACF6] directly.

To investigate the well-posedness of the compressible subsonic impinging jet
in axially symmetric case, from the mathematical point of view, there are at
least three difficulties and key points here. The one is how to discover a
mechanism to guarantee the smoothness and the global subsonicity of the
impinging jet in the whole fluid field. In the first well-posedness result
[ACF5] on compressible subsonic jet, the authors suggested to constrain the
atmospheric pressure with subsonic condition and the convex geometry condition
of the nozzle wall. With the aid of geometry property, they can conclude that
the compressible jet achieves its maximal speed on the free boundary, and then
the subsonic condition on the free boundary implies the global subsonicity of
the jet. A similar idea has been adapted in the recent work on subsonic
impinging jet in two dimensions in [CDW2]. However, in the present work, our
idea is quite different from the one in [ACF5]. We do not restrict the
condition on the atmospheric pressure and the geometry condition on the nozzle
wall, and we find an upper critical value of the incoming mass flux and show
the regularity and global subsonicity of the impinging jet provided that the
incoming mass flux is less than the upper critical value.

The second difficulty is how to fulfill the continuous fit condition between
the nozzle wall and the free boundary. In the pioneer work [ACF5], the
continuous fit condition was fulfilled for special choice of the incoming mass
flux. Namely, they showed that there exists a unique incoming mass flux, such
that the free boundary connects smoothly at the endpoint of the nozzle wall.
Here, we choose the pressure in the inlet as a parameter and show that there
exists a unique pressure in the inlet lying in an appropriate interval
$(p_{1},p_{2})$, such that the continuous fit condition holds. As mentioned in
Remark Remark, the result implies that there exists a pressure difference
between the inlet and the outlet, such that the continuous fit condition is
fulfilled. This makes the result more reasonable from the physical point of
view. The third key point here is that for the 3D axially symmetric impinging
jet there is no uniform positive distance between the free surface $\Gamma$
and the uneven wall $N_{0}$. Our proof firstly focuses on the decay estimates
of the solution in far field. Moreover, with the optimal decay rate in hand,
we get the convergence rate of the distance between the free surface and the
ground. And then rescaling the impinging jet in downstream obtains many
important facts, such as the asymptotic behavior of the impinging jet in
downstream.

The rest of the paper is organized as follows. In Section 2, we introduce a
variational problem to solve the free boundary problem, and moreover,
establish some properties of the minimizer, such as the bounded gradient lemma
and the non-degeneracy lemma. The Section 3 is about the free boundary of the
minimizer, we prove the continuous dependence of the minimizer and the free
boundary with respect to the parameter $\lambda$, and obtain the continuous
fit condition of the free boundary. In Section 4, we establish the existence
and uniqueness of the subsonic solution to the axially symmetric impinging
flow problem, provided that the incoming mass flux is small enough. The
existence of critical mass flux is obtained in Section 5. In the final
section, we give a summary of the proof to the main results in this paper.

##  2 Mathematical setting on the FBP

###  2.1 Stream function setting

Based on the continuity equation, the stream function $\psi$ can be introduced
such that

$u=\frac{1}{x\rho}\psi_{y}\ \ \text{and}\ \ v=-\frac{1}{x\rho}\psi_{x}.$ (2.1)

Without loss of generality, we can impose the boundary condition as

$\psi=m_{0}\ \ \text{on $N\cup\Gamma$ and}\ \psi=0\ \ \text{on $N_{0}\cup
I$},$

where $m_{0}=\frac{M_{0}}{2\pi}$. Denote $\Omega$ as the possible fluid field
and $E=\Omega\cap\\{x>b\\}$. Define the free boundary of the stream function
as follows

$\Gamma=E\cap\partial\\{\psi<m_{0}\\}.$

Since the pressure is equal to the constant atmospheric pressure $p_{atm}>0$
on the free boundary $\Gamma$, it follows from Bernoulli’s law (1.11) that the
density $\rho$ and the momentum $\rho q$ are also constants on $\Gamma$,
denote

$\rho_{0}=\left(\frac{p_{atm}}{\mathcal{A}}\right)^{\frac{1}{\gamma}},\ \
q_{0}=\left(2\mathcal{B}-\frac{2\mathcal{A}\gamma}{\gamma-1}\rho_{0}^{\gamma-1}\right)^{\frac{1}{2}}\
\ \text{and}\ \ \lambda=\rho_{0}q_{0}\ \ \text{on}\ \ \Gamma.$ (2.2)

It is easy to see that

$\left|\frac{\nabla\psi}{x}\right|=\frac{1}{x}\frac{\partial\psi}{\partial\nu}=\lambda=\rho
q=\rho_{0}q_{0}=\rho_{0}\left(2\mathcal{B}-\frac{2\mathcal{A}\gamma}{\gamma-1}\rho_{0}^{\gamma-1}\right)^{\frac{1}{2}}~{}~{}\text{on}~{}~{}{\Gamma},$
(2.3)

where $\nu$ is the outer normal vector of $\Gamma$. Moreover, one has

$\frac{2m^{2}_{0}}{a^{4}\rho^{2}_{in}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho_{in}^{\gamma-1}=\frac{\lambda^{2}}{2\rho_{0}^{2}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho_{0}^{\gamma-1}=\mathcal{B}.$
(2.4)

By virtue of (2.4), $\lambda$ is uniquely determined by the density
$\rho_{in}$ in upstream, once $m_{0}$ and $\rho_{0}$ are fixed. Therefore, we
can take $\lambda$ as a parameter to solve the free boundary problem firstly,
and denote

$\mathcal{B}(\lambda^{2})=\frac{\lambda^{2}}{2\rho_{0}^{2}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho_{0}^{\gamma-1}.$
(2.5)

As we know, there exist some critical quantities,

$q_{\lambda,cr}=\left(2\mathcal{B}(\lambda^{2})\frac{\gamma-1}{\gamma+1}\right)^{\frac{1}{2}},\
\rho_{\lambda,cr}=\left(\frac{2\mathcal{B}(\lambda^{2})}{\mathcal{A}\gamma}\frac{\gamma-1}{\gamma+1}\right)^{\frac{1}{\gamma-1}},\
\rho_{\lambda,max}=\left(\frac{\mathcal{B}(\lambda^{2})}{\mathcal{A}\gamma}(\gamma-1)\right)^{\frac{1}{\gamma-1}},$

such that the flow is subsonic if and only if $q<q_{\lambda,cr}$ or
$\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$ (see also in [Bers2] and
[CFO]).

Let $t=\left|\frac{\nabla\psi}{x}\right|^{2}$ be the square norm of the
momentum and the Bernoulli’s law (1.11) gives that

$\frac{t}{2\rho^{2}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho^{\gamma-1}=\frac{\lambda^{2}}{2\rho_{0}^{2}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho_{0}^{\gamma-1}=\mathcal{B}(\lambda^{2}).$

Moreover, set

$\Pi_{\lambda}=\rho_{\lambda,cr}q_{\lambda,cr}$

and

$\mathcal{H}(t,\rho;\lambda^{2})=\frac{t}{2\rho^{2}}+\frac{\mathcal{A}\gamma}{\gamma-1}\rho^{\gamma-1}-\mathcal{B}(\lambda^{2})=0,$
(2.6)

for $t>0$, $\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$ and
$\lambda\in(0,\Pi_{\lambda})$.

A simple manipulation leads to that

$\frac{\partial\mathcal{H}}{\partial(\lambda^{2})}(t,\rho;\lambda^{2})=-\frac{1}{2\rho_{0}^{2}}\
\ \ \text{and}\ \ \ \frac{\partial\mathcal{H}}{\partial
t}(t,\rho;\lambda^{2})=\frac{1}{2\rho^{2}},$ (2.7)

and

$\begin{array}[]{rl}\frac{\partial\mathcal{H}}{\partial\rho}(t,\rho;\lambda^{2})&=\frac{1}{\rho}\left(\mathcal{A}\gamma\rho^{\gamma-1}-\frac{t}{\rho^{2}}\right)\\\
&=\frac{1}{\rho}\left(\frac{\gamma+1}{\gamma-1}\mathcal{A}\gamma\rho^{\gamma-1}-2\mathcal{B}(\lambda^{2})\right)\\\
&=\frac{1}{\rho}\frac{\gamma+1}{\gamma-1}\mathcal{A}\gamma\left(\rho^{\gamma-1}-\rho^{\gamma-1}_{\lambda,cr}\right)\\\
&>0,\end{array}$ (2.8)

for any $\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$ and
$\lambda\in(0,\Pi_{\lambda})$, where we used the equality (2.6). Thus,
noticing (2.6), the density $\rho$ can be described as a function of $t$ with
a parameter $\lambda\in(0,\Pi_{\lambda})$ saying $\rho(t;\lambda^{2})$,
provided that $\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$. Furthermore, it
follows from (2.7) and (2.8) that

$\frac{\partial\rho(t;\lambda^{2})}{\partial
t}=-\frac{\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial
t}}{\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial\rho}}=-\frac{1}{2\rho^{2}\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial\rho}}<0,$
(2.9)

and

$\frac{\partial\rho(t;\lambda^{2})}{\partial(\lambda^{2})}=-\frac{\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial(\lambda^{2})}}{\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial\rho}}=\frac{1}{2\rho_{0}^{2}\frac{\partial\mathcal{H}(t,\rho;\lambda^{2})}{\partial\rho}}>0,$
(2.10)

for any $\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$ and
$\lambda\in(0,\Pi_{\lambda})$. In view of (2.9), it is easy to check that for
any $\lambda\in(0,\Pi_{\lambda})$,

$\text{$\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}$ if and only if
$t\in(0,\Pi^{2}_{\lambda})$}.$ (2.11)

Thus we can conclude that $\rho(t;\lambda^{2})$ is a decreasing smooth
function with respect to $t\in[0,\Pi^{2}_{\lambda})$ for
$\lambda\in(0,\Pi_{\lambda})$. Moreover, the density $\rho$ for subsonic flow
has the following uniform estimates

$\left(\frac{2}{\gamma+1}\right)^{\frac{1}{\gamma-1}}\rho_{0}\leq\rho_{\lambda,cr}<\rho\leq\rho_{\lambda,max}\leq\left(\frac{\gamma+1}{2}\right)^{\frac{1}{\gamma-1}}\rho_{0}.$
(2.12)

After a direct computation, there exists a
$\lambda_{cr}=\left(\mathcal{A}\gamma\rho_{0}^{\gamma+1}\right)^{\frac{1}{2}}=(\mathcal{A}\gamma)^{\frac{1}{2}}\left(\frac{p_{atm}}{\mathcal{A}}\right)^{\frac{\gamma+1}{2\gamma}}$,
such that

$\lambda<\Pi_{\lambda}\ \ \text{for any $\lambda<\lambda_{cr}$, and}\ \
\lambda_{cr}=\Pi_{\lambda_{cr}}.$

In this paper, we assume that the parameter $\lambda<\lambda_{cr}$. In the
following, denote

$\rho_{1}(t;\lambda^{2})=\frac{\partial\rho(t;\lambda^{2})}{\partial t}\ \
\text{and}\ \
\rho_{2}(t;\lambda^{2})=\frac{\partial\rho(t;\lambda^{2})}{\partial(\lambda^{2})}$

the derivatives of the function $\rho(t;\lambda^{2})$ for any
$t\in(0,\Pi^{2}_{\lambda})$ and $\lambda\in(0,\Pi_{\lambda})$.

Furthermore, the irrotational condition (1.5) deduces to the governing
equation for the stream function in the flow field that

$Q_{\lambda}\psi=\nabla\cdot\left(\frac{\nabla\psi}{x\rho(|\frac{\nabla\psi}{x}|^{2};\lambda^{2})}\right)=0~{}~{}~{}~{}\text{in}~{}~{}\Omega_{0},$
(2.13)

where $\nabla=(\partial_{x},\partial_{y})$ and
$\Omega_{0}=\Omega\cap\\{\psi<m_{0}\\}$ is the fluid field.

It is not difficult to see that the equation in (2.13) becomes degenerate as
$\frac{|\nabla\psi|}{x}\rightarrow\Pi_{\lambda}$, in order to guarantee the
uniform ellipticity, at first we consider the following modified problem. The
essence of this idea has already been illustrated in the compressible subsonic
problem, seeing [ACF5, Bers1, Bers2, CDX, DD, DWX, DX, DXX, DXY, XX1, XX2,
XX3].

Let $\tilde{\rho}(t;\lambda^{2})$ be a smooth decreasing function satisfying

$\tilde{\rho}(t;\lambda^{2})=\left\\{\begin{array}[]{rl}\rho(t;\lambda^{2}),{}{}{}&\text{for}~{}~{}t\leq(\Pi_{\lambda}-2\tilde{\varepsilon})^{2},\\\
\rho((\Pi_{\lambda}-\tilde{\varepsilon})^{2};\lambda^{2}),{}{}{}&\text{for}~{}~{}t\geq(\Pi_{\lambda}-\tilde{\varepsilon})^{2},\end{array}\right.$
(2.14)

for any small $\tilde{\varepsilon}>0$, and

$-\frac{\tilde{\rho}_{1}(t;\lambda^{2})}{\tilde{\rho}^{2}(t;\lambda^{2})}\leq\frac{C_{\tilde{\varepsilon}}}{1+t}$
for $C_{\tilde{\varepsilon}}>0$. (2.15)

Here, $\tilde{\rho}_{1}(t,\lambda^{2})$ denotes the derivative function of
$\tilde{\rho}(t,\lambda^{2})$ with respect to $t$.

Hence, we first consider the following modified free boundary problem

$\left\\{\begin{array}[]{rl}\tilde{Q}_{\lambda}\psi=\nabla\cdot\left(\frac{\nabla\psi}{x\tilde{\rho}(|\frac{\nabla\psi}{x}|^{2};\lambda^{2})}\right)=0{}{}&\text{in}~{}~{}\Omega\cap\\{\psi<m_{0}\\},\\\
\frac{1}{x}\frac{\partial\psi}{\partial\nu}=\lambda{}{}&\text{on}~{}~{}{\Gamma},\\\
\psi=0~{}~{}~{}~{}\text{on}~{}~{}N_{0}\cup I,\ \
\psi=m_{0}{}{}{}{}&\text{on}~{}~{}N\cup{\Gamma}.\end{array}\right.$ (2.16)

In the end of this section, we will verify that
$\frac{|\nabla\psi|}{x}\leq\Pi_{\lambda}-2\tilde{\varepsilon}$ in
$\bar{\Omega}_{0}$, thus the subsonic cut-off can be taken away and
$\tilde{\rho}\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda^{2}\right)=\rho\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda^{2}\right)$.

###  2.2 Variational approach

To solve the free boundary value problem (2.16) with any parameter
$\lambda<\Pi_{\lambda}$, we will introduce the variational method, which has
been adapted to solve the compressible jet problem in [ACF5]. Next, we give
the corresponding variational problem as follows. Firstly, define an
admissible set (see Figure 3) as

$\begin{array}[]{rl}K=\\{\psi\in~{}H^{1}_{loc}(\mathbb{R}^{2})\mid&\psi\leq
m_{0}\ \text{a.e. in $\mathbb{R}^{2}$},\ \psi=m_{0}\ \text{lies above}\ N,\\\
&\psi=0\ \text{lies below $N_{0}$ and lies left $I$}\\}.\end{array}$

Denote

$F(t;\lambda)=\int_{0}^{t}\frac{1}{\tilde{\rho}(\tau;\lambda^{2})}d\tau,\ \
F_{1}(t;\lambda)=\frac{\partial F(t;\lambda)}{\partial t}\ \
\text{and}~{}~{}F_{11}(t;\lambda)=\frac{\partial^{2}F(t;\lambda)}{\partial
t^{2}},$

which together with (2.15) yield that

$F(0;\lambda)=0,\ \ F(\lambda^{2};\lambda)=\frac{1}{\rho_{0}},~{}~{}0\leq
F_{11}(t;\lambda)\leq\frac{C_{\tilde{\varepsilon}}}{1+t}.$ (2.17)

Set

$\Lambda=\Lambda(\lambda^{2})=\sqrt{2F_{1}(\lambda^{2};\lambda)\lambda^{2}-F(\lambda^{2};\lambda)}.$

In view of (2.12), it is easy to check that

$\begin{array}[]{rl}\frac{\Lambda^{2}(\lambda^{2})}{\lambda^{2}}=\frac{2}{\rho_{0}}-\frac{1}{\lambda^{2}}\int_{0}^{\lambda^{2}}\frac{1}{\tilde{\rho}(\tau;\lambda^{2})}d\tau\leq\frac{1}{\rho_{0}}\left(2-\left(\frac{2}{\gamma+1}\right)^{\frac{1}{\gamma-1}}\right),\end{array}$
(2.18)

and

$\begin{array}[]{rl}\frac{\Lambda^{2}(\lambda^{2})}{\lambda^{2}}=\frac{2}{\rho_{0}}-\frac{1}{\lambda^{2}}\int_{0}^{\lambda^{2}}\frac{1}{\tilde{\rho}(\tau;\lambda^{2})}d\tau\geq\frac{1}{\rho_{0}}\left(2-\left(\frac{\gamma+1}{2}\right)^{\frac{1}{\gamma-1}}\right)>0,\end{array}$
(2.19)

for any $\lambda\in[0,\Pi_{\lambda})$. Furthermore, it follows from (2.10)
that

$\frac{d\Lambda^{2}(\lambda^{2})}{d(\lambda^{2})}=\frac{1}{\rho_{0}}+\int_{0}^{\lambda^{2}}\frac{\rho_{2}(\tau;\lambda^{2})}{{\rho}^{2}(\tau;\lambda^{2})}d\tau>\frac{1}{\rho_{0}},$
(2.20)

for any $\lambda\in[0,\Pi_{\lambda}-2\tilde{\varepsilon})$, which implies that
$\Lambda(\lambda^{2})$ is uniquely determined by
$\lambda\in(0,\Pi_{\lambda}-2\tilde{\varepsilon})$.

Set

$f(\mathbf{\eta};\lambda)=F(|\mathbf{\eta}|^{2};\lambda)$ with
$\mathbf{\eta}=(\eta_{1},\eta_{2})\in\mathbb{R}^{2}$, (2.21)

it follows from (2.17) that $f(\mathbf{\eta};\lambda)$ is convex with respect
to $\eta$, and there exists a constant $\vartheta$ depending on $\lambda_{cr}$
and $\tilde{\varepsilon}$, such that

$\vartheta|\mathbf{\xi}|^{2}\leq\sum_{i,j=1}^{2}\frac{\partial^{2}f(\mathbf{\eta};\lambda)}{\partial\eta_{i}\partial\eta_{j}}\xi_{i}\xi_{j}\leq\vartheta^{-1}|\mathbf{\xi}|^{2}\
\ \text{for any $\mathbf{\xi}=(\xi_{1},\xi_{2})\in\mathbb{R}^{2}$},$

and

$\vartheta|\eta|^{2}\leq f_{\eta}(\eta;\lambda)\cdot\eta\ \ \text{and}\ \
\vartheta|\eta|^{2}\leq f(\eta;\lambda)\leq\vartheta^{-1}|\eta|^{2}.$

Define a function with any parameter $\lambda\in(0,\Pi_{\lambda})$ as follows,

$G(\nabla\psi,\psi,x;\lambda)=xF\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda\right)+\left(x\Lambda^{2}-2\lambda
F_{1}(\lambda^{2};\lambda)\nabla\psi\cdot e\right)\chi_{\\{\psi<m_{0}\\}\cap
E},$ (2.22)

where $\chi_{D}$ is the indicator function of the set $D$ and
$e=(-\sin\theta,\cos\theta)$. By virtue of the convexity of $F(t;\lambda)$
with respect to $t$, one has

$\begin{array}[]{rl}{}\hfil&G(\nabla\psi,\psi,x;\lambda)\\\
\geq&x\left(F\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda\right)-F(\lambda^{2};\lambda)-F_{1}(\lambda^{2};\lambda)\left(\left|\frac{\nabla\psi}{x}\right|^{2}-\lambda^{2}\right)\right)\chi_{\\{\psi<m_{0}\\}\cap
E}\\\ &+xF_{1}(\lambda^{2};\lambda)\left|\frac{\nabla\psi}{x}-\lambda
e\right|^{2}\chi_{\\{\psi<m_{0}\\}\cap E}\\\
\geq&xF_{1}(\lambda^{2};\lambda)\left|\frac{\nabla\psi}{x}-\lambda
e\right|^{2}\chi_{\\{\psi<m_{0}\\}\cap E},\end{array}$ (2.23)

and

$\begin{array}[]{rl}{}G(\nabla\psi,\psi,x;\lambda)\leq
Cx\left|\frac{\nabla\psi}{x}-\lambda e\right|^{2}\chi_{\\{\psi<m_{0}\\}\cap
E}+xF_{1}\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda\right)\chi_{\\{\psi<m_{0}\\}\setminus
E}.\end{array}$ (2.24)

Hence, we define a functional

$J_{\lambda}(\psi)=\int_{\Omega}G(\nabla\psi,\psi,x;\lambda)dxdy.$

Figure 4: Truncated domain

It follows from (2.23) that the functional $J_{\lambda}(\psi)$ is non-negative
for any $\psi\in K$. Obviously, $J_{\lambda}(\psi)$ is unbounded for any
$\psi\in K$. Thus we will truncate the domain as $\Omega_{\mu}$ for any
$\mu>1$ (see Figure 4), which is bounded by $N_{\mu},I_{\mu},N_{0},L_{\mu}$
and $T=\\{(b,y)\mid y>1\\}$, where

$\text{$N_{\mu}=N\cap\\{x\geq x_{\mu}\\}$, $I_{\mu}=I\cap\\{y\leq\mu\\}$, and
$L_{\mu}=\\{(x,y)\mid 0\leq x\leq x_{\mu},y=\mu\\}$},$

with $x_{\mu}=\min\\{x\mid g(x)=\mu\\}$. Define the following functional in
the truncated domain $\Omega_{\mu}$,

$J_{\lambda,\mu}(\psi)=\int_{\Omega_{\mu}}G(\nabla\psi,\psi,x;\lambda)dxdy.$

To overcome the singularity of the functional $J_{\lambda,\mu}$ near $y$-axis,
we first consider the following variational problem.

The truncated variational problem $(P^{\delta}_{\lambda,\mu})$: For any
$\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$, $\mu>1$ and small $\delta>0$,
find a $\psi_{\lambda,\mu}^{\delta}\in K_{\mu}^{\delta}$ such that

$J^{\delta}_{\lambda,\mu}(\psi^{\delta}_{\lambda,\mu})=\min_{\psi\in
K^{\delta}_{\mu}}J^{\delta}_{\lambda,\mu}(\psi),$

where

$J^{\delta}_{\lambda,\mu}(\psi)=\int_{\Omega_{\mu}}G(\nabla\psi,\psi,x+\delta;\lambda)dxdy$

and

$K_{\mu}^{\delta}=\left\\{\psi\in
K\mid\psi=\min\left\\{\frac{m_{0}}{x^{2}_{\mu}}((x+\delta)^{2}-\delta^{2}),m_{0}\right\\}~{}~{}~{}~{}\text{on}~{}~{}L_{\mu}\right\\}.$

######  Lemma 2.1.

The variational problem $(P^{\delta}_{\lambda,\mu})$ has a minimizer
$\psi^{\delta}_{\lambda,\mu}$ and $\psi_{\lambda,\mu}^{\delta}\in
C^{0,1}(\Omega_{\mu})$. Furthermore, the minimizer
$\psi^{\delta}_{\lambda,\mu}$ satisfies that

$\int_{\Omega_{\mu}}f_{\eta}\left(\frac{\nabla\psi^{\delta}_{\lambda,\mu}}{x+\delta};\lambda\right)\cdot\nabla\xi
dxdy\geq 0\ \ \text{for any $\xi\in C^{\infty}_{0}(\Omega_{\mu})$ and $\xi\geq
0$},$

and

$\int_{\Omega_{\mu}}f_{\eta}\left(\frac{\nabla\psi^{\delta}_{\lambda,\mu}}{x+\delta};\lambda\right)\cdot\nabla\zeta
dxdy=0\ \ \text{for any $\zeta\in
C^{\infty}_{0}(\Omega_{\mu}\cap\\{\psi^{\delta}_{\lambda,\mu}<m_{0}\\})$},$
(2.25)

where $f(\eta;\lambda)$ is defined in (2.21). Furthermore,

$0\leq\psi_{\lambda,\mu}^{\delta}(x,y)\leq\min\left\\{\frac{m_{0}}{x^{2}_{\mu}}((x+\delta)^{2}-\delta^{2}),m_{0}\right\\}\
$ in $\ \Omega_{\mu}$. (2.26)

###### Proof.

Define

$\psi_{0}(x,y)=\min\\{\lambda\max\\{(x+\delta)\left((y-g_{0}(R_{0}))\cos\theta-(x+\delta-
R_{0})\sin\theta\right),0\\},m_{0}\\},$

it follows from (1.8) that $\psi_{0}=0$ on $N_{0}\cap\\{x\geq R_{0}\\}$. Then
we can extend $\psi_{0}$ into the domain $\Omega_{\mu}\setminus\\{x\leq
R_{0}\\}$ so that it belongs to the admissible set $K^{\delta}_{\mu}$ and
$\\{\psi_{0}<m_{0}\\}\cap(\Omega_{\mu}\setminus\\{x\leq R_{0}\\})$ is bounded.
Hence, it suffices to verify that

$\int_{\Omega_{\mu}\cap\\{x>R_{0}\\}}G(\nabla\psi_{0},\psi_{0},x+\delta;\lambda)dxdy<+\infty.$

In fact, it follows from (2.24) that

$\begin{array}[]{rl}&\int_{\Omega_{\mu}\cap\\{x>R_{2}\\}}G(\nabla\psi_{0},\psi_{0},x+\delta;\lambda)dxdy\\\
&\leq C\int_{R_{2}}^{+\infty}\int_{g_{0}(R_{0})+(x+\delta-
R_{0})\tan\theta}^{g_{0}(R_{0})+(x+\delta-
R_{0})\tan\theta+\frac{m_{0}}{\lambda(x+\delta)\cos\theta}}(x+\delta)\left|\frac{\nabla\psi_{0}}{x+\delta}-\lambda
e\right|^{2}dydx\\\ &\leq
C\int_{R_{0}+\delta}^{+\infty}\int_{g_{0}(R_{0})+(x-R_{0})\tan\theta}^{g_{0}(R_{0})+(x-R_{0})\tan\theta+\frac{m_{0}}{\lambda
x\cos\theta}}\frac{\lambda^{2}\left((y-g_{0}(R_{0}))\cos\theta-(x-R_{0})\sin\theta\right)^{2}}{x}dydx\\\
&\leq\frac{C}{(R_{0}+\delta)^{3}}.\end{array}$

The existence of the minimizer to the variational problem
$(P_{\lambda,\mu}^{\delta})$ can be obtained via the similar arguments in
Lemma 1.1 in [ACF3] and Theorem 1.1 in [ACF4], and denote
$\psi^{\delta}=\psi^{\delta}_{\lambda,\mu}$ as the minimizer to the
variational problem $(P^{\delta}_{\lambda,\mu})$ for simplicity.

For any nonnegative function $\xi\in C_{0}^{\infty}(\Omega_{\mu})$ and
$\varepsilon>0$, it is easy to check that
$\min\\{\psi^{\delta}+\varepsilon\xi,m_{0}\\}\in K_{\mu}$ and
$\\{\min\\{\psi^{\delta}+\varepsilon\xi,m_{0}\\}<m_{0}\\}\subset\\{\psi^{\delta}<m_{0}\\}$.
Thus, we have

$\begin{array}[]{rl}0\leq&J^{\delta}_{\lambda,\mu}\left(\min\\{\psi^{\delta}+\varepsilon\xi,m_{0}\\}\right)-J^{\delta}_{\lambda,\mu}(\psi^{\delta})\\\
=&\int_{\Omega_{\mu}}(x+\delta)F\left(\frac{|\nabla\min\\{\psi^{\delta}+\varepsilon\xi,m_{0}\\}|^{2}}{(x+\delta)^{2}};\lambda\right)-(x+\delta)F\left(\frac{|\nabla\psi^{\delta}|^{2}}{(x+\delta)^{2}};\lambda\right)dxdy\\\
\leq&\int_{\Omega_{\mu}\cap\\{\psi^{\delta}+\varepsilon\xi\leq
m_{0}\\}}(x+\delta)F\left(\frac{|\nabla(\psi^{\delta}+\varepsilon\xi)|^{2}}{(x+\delta)^{2}};\lambda\right)-(x+\delta)F\left(\frac{|\nabla\psi^{\delta}|^{2}}{(x+\delta)^{2}};\lambda\right)dxdy\\\
\leq&2\varepsilon\int_{\Omega_{\mu}\cap\\{\psi^{\delta}+\varepsilon\xi\leq
m_{0}\\}}F_{1}\left(\frac{|\nabla(\psi^{\delta}+\varepsilon\xi)|^{2}}{(x+\delta)^{2}};\lambda\right)\frac{\nabla\psi^{\delta}\cdot\nabla\xi}{x+\delta}dxdy+o(\varepsilon),\end{array}$

which implies that

$0\leq\int_{\Omega_{\mu}\cap\\{\psi^{\delta}+\varepsilon\xi\leq
m_{0}\\}}F_{1}\left(\frac{|\nabla(\psi^{\delta}+\varepsilon\xi)|^{2}}{(x+\delta)^{2}};\lambda\right)\frac{\nabla\psi^{\delta}\cdot\nabla\xi}{x+\delta}dxdy.$

Taking $\varepsilon\rightarrow 0$ in above inequality, we have

$0\leq\int_{\Omega_{\mu}}F_{1}\left(\frac{|\nabla\psi^{\delta}|^{2}}{(x+\delta)^{2}};\lambda\right)\frac{\nabla\psi^{\delta}\cdot\nabla\xi}{x+\delta}dxdy.$

Similarly, we can verify that (2.25) holds.

Next, we will show that

$\psi^{\delta}(x,y)\geq 0\ \ \ \text{in}\ \ \Omega_{\mu}.$ (2.27)

Denote
$\psi^{\delta}_{\varepsilon}=\psi^{\delta}-\varepsilon\min\\{\psi^{\delta},0\\}$
for $\varepsilon\in(0,1)$. It is easy to check that
$\psi_{\varepsilon}^{\delta}\in K_{\mu}^{\delta}$,

$\psi_{\varepsilon}^{\delta}>0\ \ \text{if and only if}\ \ \psi^{\delta}>0\ \
\text{and}\ \ \psi_{\varepsilon}^{\delta}\geq\psi^{\delta}\ \ \text{in}\ \
\Omega_{\mu}.$

Since $\psi^{\delta}$ is the minimizer to the truncated variational problem
$(P_{\lambda,\mu}^{\delta})$, one has

$0\leq
J^{\delta}_{\lambda,\mu}(\psi_{\varepsilon}^{\delta})-J^{\delta}_{\lambda,\mu}(\psi^{\delta}).$
(2.28)

For any sufficiently large $R>0$, denote
$\Omega_{\mu,R}=\Omega_{\mu}\cap\\{y<R\\}$ and
$E_{R}=\Omega_{\mu,R}\cap\\{x>b\\}$, we have

$\begin{array}[]{rl}&\int_{\Omega_{\mu,R}}G(\nabla\psi_{\varepsilon}^{\delta},\psi_{\varepsilon}^{\delta},x+\delta;\lambda)dxdy-\int_{\Omega_{\mu,R}}G(\nabla\psi^{\delta},\psi^{\delta},x+\delta;\lambda)dxdy\\\
=&\int_{\Omega_{\mu,R}}(x+\delta)F\left(\left|\frac{\nabla\psi_{\varepsilon}^{\delta}}{x+\delta}\right|^{2};\lambda\right)-(x+\delta)F\left(\left|\frac{\nabla\psi^{\delta}}{x+\delta}\right|^{2};\lambda\right)dxdy\\\
&-2\lambda F_{1}(\lambda^{2};\lambda)\int_{\Omega_{\mu,R}\cap
E}\nabla\psi_{\varepsilon}^{\delta}\cdot
e\chi_{\\{\psi^{\delta}_{\varepsilon}<m_{0}\\}}-\nabla\psi^{\delta}\cdot
e\chi_{\\{\psi^{\delta}<m_{0}\\}}dxdy\\\
\leq&\int_{\Omega_{\mu,R}}F_{1}\left(\left|\frac{\nabla\psi_{\varepsilon}^{\delta}}{x+\delta}\right|^{2};\lambda\right)\frac{|\nabla\psi_{\varepsilon}^{\delta}|^{2}-|\nabla\psi^{\delta}|^{2}}{x+\delta}dxdy\\\
&-2\lambda F_{1}(\lambda^{2};\lambda)\int_{\partial
E_{R}}(\psi_{\varepsilon}^{\delta}-\psi^{\delta})e\cdot\nu dS\\\
\leq&\int_{\Omega_{\mu,R}}F_{1}\left(\left|\frac{\nabla\psi_{\varepsilon}^{\delta}}{x+\delta}\right|^{2};\lambda\right)\frac{((1-\varepsilon)^{2}-1)|\nabla\min\\{\psi^{\delta},0\\}|^{2}}{x+\delta}dxdy.\end{array}$
(2.29)

Here, we have used the fact

$\int_{\partial
E_{R}}(\psi_{\varepsilon}^{\delta}-\psi_{\varepsilon})e\cdot\nu
dS=\sin\theta\int_{\partial
E_{R}\cap\\{x=b\\}}(\psi_{\varepsilon}^{\delta}-\psi_{\varepsilon})dS+\cos\theta\int_{\partial
E_{R}\cap\\{y=R\\}}(\psi_{\varepsilon}^{\delta}-\psi_{\varepsilon})dS\geq 0.$

Taking $R\rightarrow+\infty$ in (2.29), it follows from (2.28) that

$0\leq((1-\varepsilon)^{2}-1)\int_{\Omega_{\mu}}F_{1}\left(\left|\frac{\nabla\psi_{\varepsilon}^{\delta}}{x+\delta}\right|^{2};\lambda\right)\frac{|\nabla\min\\{\psi^{\delta},0\\}|^{2}}{x+\delta}dxdy,$

which implies that (2.27) holds.

Since $0\leq\psi^{\delta}_{\lambda,\mu}\leq m_{0}$ in $\Omega_{\mu}$, it
suffices to show that

$\psi^{\delta}_{\lambda,\mu}\leq\frac{m_{0}}{x_{\mu}^{2}}((x+\delta)^{2}-\delta^{2})\
~{}~{}\text{in}~{}~{}\
\Omega_{\mu}\cap\left\\{x\leq\sqrt{x^{2}_{\mu}+\delta^{2}}-\delta\right\\}.$
(2.30)

In view of (2.25), the maximum principle gives that the inequality (2.26)
holds.

∎

With the aid of Lemma 2.1, we consider the following truncated variational
problem.

The truncated variational Problem $(P_{\lambda,\mu})$: For any
$\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$ and $\mu>1$, find a $\psi\in
K_{\mu}$ such that

$J_{\lambda,\mu}(\psi_{\lambda,\mu})=\min_{\psi\in
K_{\mu}}J_{\lambda,\mu}(\psi),$

where

$K_{\mu}=\left\\{\psi\in
K\mid\psi=\frac{m_{0}}{x^{2}_{\mu}}x^{2}~{}~{}~{}~{}\text{on}~{}~{}L_{\mu}\ \
\text{and}\ \
\psi\leq\min\left\\{\frac{m_{0}}{x_{\mu}^{2}}x^{2},m_{0}\right\\}\ \
\text{a.e. in}\ \ \Omega_{\mu}\right\\}.$

###  2.3 Existence and fundamental properties of minimizer

######  Lemma 2.2.

There exists a minimizer $\psi_{\lambda,\mu}$ to the variational problem
$(P_{\lambda,\mu})$ and $\psi_{\lambda,\mu}\in C^{0,1}(\Omega_{\mu})$.
Moreover,
(1) the minimizer $\psi_{\lambda,\mu}$ satisfies that

$\int_{\Omega_{\mu}}f_{\eta}\left(\frac{\nabla\psi_{\lambda,\mu}}{x};\lambda\right)\cdot\nabla\xi
dxdy\geq 0\ \ \text{for any $\xi\in C^{\infty}_{0}(\Omega_{\mu})$ and $\xi\geq
0$},$

and

$\tilde{Q}_{\lambda}\psi_{\lambda,\mu}=0\ \ \text{in
$\Omega_{\mu}\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$ and $\psi_{\lambda,\mu}\in
C^{2,\alpha}(\Omega_{\mu}\cap\\{\psi_{\lambda,\mu}<m_{0}\\})$}.$

Furthermore,

$\psi_{\lambda,\mu}(x,y)\geq 0$ in $\Omega_{\mu}$. (2.31)

(2) the free boundary
$\Gamma_{\lambda,\mu}=E\cap\partial\\{\psi_{\lambda,\mu}<m_{0}\\}$ is
analytic, and

$\frac{1}{x}|\nabla\psi_{\lambda,\mu}|=\lambda\ \ \text{on}\ \
\Gamma_{\lambda,\mu},$

and

$\frac{1}{x}|\nabla\psi_{\lambda,\mu}|\geq\lambda\ \ \text{on}\ \ l,$

where $l$ is a segment with $l\subset
T\cap\partial\\{\psi_{\lambda,\mu}<m_{0}\\}$.

###### Proof.

(1) Along the similar arguments in the proof of Lemma 1.1 in [ACF3] and
Theorem 1.1 in [ACF4], one has that there exists a sequence $\\{\delta_{n}\\}$
with $\delta_{n}\rightarrow 0$ as $n\rightarrow+\infty$, such that

$\psi_{\lambda,\mu}^{\delta_{n}}\rightharpoonup\psi_{0}\ \ \text{in
$H^{1}_{loc}(\Omega_{\mu})$ and
$\psi_{\lambda,\mu}^{\delta_{n}}\rightarrow\psi_{0}$ uniformly in any compact
subset of $\Omega_{\mu}$}.$

It follows from (2.26) that

$0\leq\psi_{0}(x,y)\leq\min\left\\{\frac{x^{2}}{x_{\mu}^{2}}m_{0},m_{0}\right\\}\
\ \text{in}\ \ \Omega_{\mu},$

which together with (2.25) gives that

$\nabla\cdot f_{\eta}\left(\frac{\nabla\psi_{0}}{x};\lambda\right)=0\ \
\text{in}\ \ \Omega_{\mu}\cap\left\\{x<\frac{x_{\mu}}{2}\right\\}.$

Next, we will check that $J_{\lambda,\mu}(\psi_{0})<+\infty$. By virtue of the
proof of Lemma 2.1, it suffices to show that

$\frac{|\nabla\psi_{0}(X)|}{x}\leq C\ \ \text{near $I$}.$

For any $X_{0}=(x_{0},y_{0})\in\Omega_{\mu}$ with $x_{0}<\frac{x_{\mu}}{4}$,
denote $\phi(X)=\frac{\psi_{0}(X_{0}+r_{0}X)}{r^{2}_{0}}$ with
$r_{0}=\frac{x_{0}}{2}$. It is easy to check that

$\nabla\cdot f_{\eta}\left(\frac{\nabla\phi}{2+x};\lambda\right)=0\ \
\text{and}\ \ 0\leq\phi\leq(2+x)^{2}m_{0}\ \ \text{in}\ \ B_{1}(0).$

Thanks to the gradient estimate in Chapter 12 in [GT], one has

$|\nabla\phi(0)|\leq C,$

where the constant $C$ is independent of $x_{0}$. This gives that

$|\nabla\psi_{0}(X_{0})|=r_{0}|\nabla\phi(0)|\leq Cx_{0}.$

Since $J_{\lambda,\mu}(\psi_{0})<+\infty$ and $\psi_{0}\in K_{\mu}$, the
minimal functional $J_{\lambda,\mu}(\psi)$ is finite. By using the proof of
Lemma 1.1 in [ACF3], we can conclude that there exists a minimizer to the
variational problem $(P_{\lambda,\mu})$.

Denote $\psi_{\lambda,\mu}$ be the minimizer to the variational problem
$(P_{\lambda,\mu})$ and
$\Gamma_{\lambda,\mu}=E\cap\partial\\{\psi_{\lambda,\mu}<m_{0}\\}$ as the free
boundary of $\psi_{\lambda,\mu}$. Thanks to Lemma 2.1, we can show that
$\psi_{\lambda,\mu}\in C^{0,1}(\Omega_{\mu})$ satisfies the assertion (1) of
this lemma.

(2) Since $F(t;\lambda)$ is $C^{2}$-smooth with respect to $t\in[0,+\infty)$,
it follows from Theorem 6.3 in [ACF4] that the free boundary
$\Gamma_{\lambda,\mu}$ is $C^{1,\alpha}$, and thus $\psi_{\lambda,\mu}$ is
$C^{1,\alpha}$ up to the free boundary. Since
$\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$, the subsonic cut-off can be
removed near the free boundary. Then $F(t;\lambda)$ is analytic near
$\Gamma_{\lambda,\mu}$, the Remark 6.4 in [ACF4] gives that the free boundary
$\Gamma_{\lambda,\mu}$ is analytic. By using the similar arguments in the
proof of Lemma 9.1 in [CDW2], we can conclude that

$\frac{1}{x}|\nabla\psi_{\lambda,\mu}|=\lambda\ \ \text{on}\ \
\Gamma_{\lambda,\mu},\ \ \text{ and}\ \
\frac{1}{x}|\nabla\psi_{\lambda,\mu}|\geq\lambda\ \ \text{on}\ \ l,$

where $l$ is a segment with $l\subset
T\cap\partial\\{\psi_{\lambda,\mu}<m_{0}\\}$. ∎

Next, we will give the bounded gradient lemma in the following.

######  Lemma 2.3.

Let $X_{0}=(x_{0},y_{0})$ be a free boundary point and let
$B_{r}(X_{0})\subset B_{R}(X_{0})\subset E$ with $r<R$. Then

$|\nabla\psi_{\lambda,\mu}(X)|\leq C\Lambda x\ \ \text{in $B_{r}(X_{0})$,}$

where the constant $C$ depends only on $\vartheta,N,N_{0}$ and
$\left(1-\frac{r}{R}\right)^{-1}$, but not on $m_{0}$.

Figure 5: $B_{d(X_{1})}(X_{1})$ and $B_{R}(X_{0})$

###### Proof.

Step 1. In this step, we will show that

$m_{0}-\psi_{\lambda,\mu}(X)\leq C\Lambda xd(X)\ \ \text{for any $X\in
B_{r}(X_{0})$},$ (2.32)

where $d(X)=dist(X,\Gamma_{\lambda,\mu})$ and the constant $C$ depends only on
$\vartheta,b$ and $\left(1-\frac{r}{R}\right)^{-1}$.

Denote $d_{0}=R-r$. Suppose that $X_{1}=(x_{1},y_{1})\in
B_{r}(X_{0})\cap\\{\psi<m_{0}\\}$ and $d(X_{1})<d_{0}$. Thus
$B_{d(X_{1})}(X_{1})\subset B_{R}(X_{0})\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$
(please see Figure 5). Next, we assume that

$m_{0}-\psi_{\lambda,\mu}(X_{1})>Md(X_{1})x_{1},$ (2.33)

and we will derive an upper bound of $M$ in the following. Denote

$\phi(X)=\frac{m_{0}-\psi_{\lambda,\mu}(X_{1}+dX)}{dx_{1}}$ with $d=d(X_{1})$,
(2.34)

and one has

$\nabla\cdot f_{\eta}\left(\frac{x_{1}\nabla\phi}{x_{1}+dx};\lambda\right)=0\
\text{in}~{}~{}B_{1}(0).$

It follows from (2.33) that

$\phi(0)>M.$

It follows from Harnack’s inequality (see Theorem 8.20 in [GT]) that

$\phi(X)\geq cM\ \ \text{in $B_{\frac{3}{4}}(0)$ for some $c>0$},$ (2.35)

where the constant $c$ is independent of $d$ and $x_{1}$. On another hand,
there exists a $\tilde{X}=(\tilde{x},\tilde{y})\in\partial
B_{1}(0)\cap\Gamma_{\lambda,\mu}$. Define a function $\Psi$, which satisfies
that

$\left\\{\begin{array}[]{ll}&\nabla\cdot
f_{\eta}\left(\frac{x_{1}\nabla\Psi}{x_{1}+dx};\lambda\right)=0\
\text{in}~{}~{}B_{1}(\tilde{X}),\\\ &\Psi=\phi\ \ \ \text{outside}\ \
B_{1}(\tilde{X}).\end{array}\right.$

Since $\nabla\cdot
f_{\eta}\left(\frac{x_{1}\nabla\phi}{x_{1}+dx};\lambda\right)\geq 0$ in
$B_{1}(\tilde{X})$, the maximum principle gives that

$\text{$\phi\leq\Psi$ in $B_{1}(\tilde{X})$}.$ (2.36)

Then we have

$\begin{array}[]{rl}0\leq&\int_{B_{1}(\tilde{X})}(x_{1}+dx)\left\\{F\left(\frac{|x_{1}\nabla\Psi|^{2}}{(x_{1}+dx)^{2}};\lambda\right)-F\left(\frac{|x_{1}\nabla\phi|^{2}}{(x_{1}+dx)^{2}};\lambda\right)\right\\}dxdy\\\
&-2\lambda
F_{1}(\lambda^{2};\lambda)\int_{B_{1}(\tilde{X})}\nabla(\Psi-\phi)\cdot
edxdy+\Lambda^{2}\int_{B_{1}(\tilde{X})}(x_{1}+dx)(\chi_{\\{\Psi>0\\}}-\chi_{\\{\phi>0\\}})dxdy\\\
\leq&\int_{B_{1}(\tilde{X})}-\vartheta\frac{x_{1}^{2}|\nabla(\Psi-\phi)|^{2}}{x_{1}+dx}+f_{\eta}\left(\frac{x_{1}\nabla\Psi}{x_{1}+dx};\lambda\right)\cdot\nabla(\Psi-\phi)+\Lambda^{2}(x_{1}+dx)\chi_{\\{\phi=0\\}}dxdy\\\
=&-\vartheta\int_{B_{1}(\tilde{X})}\frac{x_{1}^{2}|\nabla(\Psi-\phi)|^{2}}{x_{1}+dx}dxdy+\Lambda^{2}\int_{B_{1}(\tilde{X})}(x_{1}+dx)\chi_{\\{\phi=0\\}}dxdy,\end{array}$

which implies that

$\begin{array}[]{rl}\int_{B_{1}(\tilde{X})}|\nabla(\Psi-\phi)|^{2}dxdy\leq&C\int_{B_{1}(\tilde{X})}\frac{x_{1}|\nabla(\Psi-\phi)|^{2}}{x_{1}+dx}dxdy\\\
\leq&\frac{C\Lambda^{2}}{x_{1}}\int_{B_{1}(\tilde{X})}(x_{1}+dx)\chi_{\\{\phi=0\\}}dxdy\\\
\leq&C\Lambda^{2}\int_{B_{1}(\tilde{X})}\chi_{\\{\phi=0\\}}dxdy,\end{array}$
(2.37)

where $C$ is a constant depending only on $\vartheta$ and
$\left(1-\frac{r}{R}\right)^{-1}$.

It follows from (2.35) and (2.36) that

$\Psi(X)\geq\phi(X)\geq cM\ \ \text{in}\ \ B_{\frac{3}{4}}(0)\cap
B_{1}(\tilde{X}).$

Applying Harnack’s inequality for $\Psi$ in $B_{1}(\tilde{X})$, one has

$\Psi(X)\geq C_{0}\ \ \text{in}\ \ B_{\frac{1}{2}}(\tilde{X}),\ \ \ C_{0}=cM.$
(2.38)

Define $\varphi(X)=C_{0}\left(e^{-\nu|X-\tilde{X}|^{2}-e^{-\nu}}\right)$,
after a direct computation, we have

$\nabla\cdot
f_{\eta}\left(\frac{x_{1}\nabla\varphi}{x_{1}+dx};\lambda\right)=\frac{2\nu
C_{0}x_{1}e^{-\nu|\xi|^{2}}}{x_{1}+dx}\left(f_{\eta_{i}\eta_{j}}(2\nu\xi_{i}\xi_{j}-\delta_{ij})+\frac{d(f_{\eta_{1}\eta_{1}}\xi_{1}+f_{\eta_{1}\eta_{2}}\xi_{2})}{x_{1}+dx}\right)>0$

in $B_{1}(\tilde{X})\setminus B_{\frac{1}{2}}(\tilde{X})$, provided that $\nu$
is large enough, where
$f_{\eta_{i}\eta_{j}}=f_{\eta_{i}\eta_{j}}\left(\frac{x_{1}\nabla\varphi}{x_{1}+dx};\lambda\right)$
and $\xi=(\xi_{1},\xi_{2})=X-\tilde{X}$ for $i,j=1,2$.

It is easy to check that

$\Psi\geq\varphi\ \ \text{on}\ \ \partial(B_{1}(\tilde{X})\setminus
B_{\frac{1}{2}}(\tilde{X})).$

The maximum principle gives that

$\Psi(X)\geq\varphi(X)=C_{0}\left(e^{-\nu|X-\tilde{X}|^{2}-e^{-\nu}}\right)\geq
cC_{0}(1-|X-\tilde{X}|)\ \ \text{in}\ \ B_{1}(\tilde{X})\setminus
B_{\frac{1}{2}}(\tilde{X}),$

which together with (2.38) gives that

$\Psi(X)\geq cM(1-|X-\tilde{X}|)\ \ \text{in}\ \ B_{1}(\tilde{X})\setminus
B_{\frac{1}{2}}(\tilde{X}).$ (2.39)

With the aid of (2.37) and (2.39), along the similar arguments in the proof of
Lemma 3.2 in [AC1] and Lemma 2.2 in [ACF4], one has

$M^{2}\leq C\Lambda^{2},$

where the constant $C$ depends only on $\vartheta,N,N_{0}$ and
$\left(1-\frac{r}{R}\right)^{-1}$. This implies that

$m_{0}-\psi_{\lambda,\mu}(X_{1})\leq C\Lambda d(X_{1})x_{1}.$ (2.40)

Take any point $X_{2}=(x_{2},y_{2})\in B_{r}(X_{0})$ such that
$d(X_{2})>d_{0}$ and there exists a point $X_{1}\in
B_{\frac{d_{0}}{2}}(X_{2})$ with $d(X_{1})<d_{0}$. By using Harnack’s
inequality for $m_{0}-\psi_{\lambda,\mu}$ in $B_{d_{0}}(X_{2})$ and (2.40),
one has

$m_{0}-\psi_{\lambda,\mu}(X_{2})\leq C(m_{0}-\psi_{\lambda,\mu}(X_{1}))\leq
C\Lambda d(X_{1})x_{1}\leq C\Lambda d(X_{2})x_{2}.$

For any $X\in B_{r}(X_{0})$, we can repeat this argument step by step, and
after a finite steps $N$ (depending only on $\vartheta,N,N_{0}$ and
$\left(1-\frac{r}{R}\right)^{-1}$), such that

$m_{0}-\psi_{\lambda,\mu}(X)\leq C\Lambda d(X)x.$

Hence, we complete the proof of (2.32).

Step 2. In this step, we will complete the proof of this lemma. For any
$X_{1}\in B_{r}(X_{0})$, denote $d_{0}=R-r$ and
$d(X)=dist(X,\Gamma_{\lambda,\mu})$, and we consider the following two cases.

Case 1. $d(X_{1})<d_{0}$. Then it follows from (2.32) that

$\nabla\cdot f_{\eta}\left(\frac{x_{1}\nabla\phi}{x_{1}+dx};\lambda\right)=0\
\ \text{and}\ \ 0\leq\phi\leq\frac{C\Lambda(x_{1}+dx)d(X_{1}+dX)}{dx_{1}}\leq
C\Lambda\ \text{in}~{}~{}B_{1}(0),$

where $\phi$ and $d$ are defined in (2.34), the constant $C$ depends only on
$\vartheta,N,N_{0}$ and $\left(1-\frac{r}{R}\right)^{-1}$. Applying the
elliptic estimate for the quasilinear equation in [GT], one has

$|\nabla\phi(0)|\leq C,$

which gives that

$|\nabla\psi_{\lambda,\mu}(X_{1})|=x_{1}|\nabla\phi(0)|\leq Cx_{1}.$

Case 2. $d(X_{1})\geq d_{0}$. Obviously, $B_{d_{0}}(X_{1})\subset
B_{R}(X_{0})\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$. Denote
$\phi_{0}(X_{1})=\frac{m_{0}-\psi_{\lambda,\mu}(X_{1}+d_{0}X)}{d_{0}x_{1}}$,
it follows from (2.34) that

$\nabla\cdot
f_{\eta}\left(\frac{x_{1}\nabla\phi_{0}}{x_{1}+d_{0}x};\lambda\right)=0\ \
\text{and}\ \
0\leq\phi_{0}\leq\frac{C\Lambda(x_{1}+d_{0}x)d(X_{1}+d_{0}X)}{d_{0}x_{1}}\leq
C\Lambda\ \text{in}~{}~{}B_{1}(0).$

By using the elliptic estimate for $\phi_{0}$ in $B_{1}(0)$, one has

$|\nabla\phi_{0}(0)|\leq C\ \ \text{and}\ \
|\nabla\psi_{\lambda,\mu}(X_{1})|=x_{1}|\nabla\phi_{0}(0)|\leq Cx_{1}.$

∎

With the aid of Lemma 2.3, applying the similar arguments in the proof of
Lemma 2.4 in [ACF4], we can obtain the following lemma.

######  Lemma 2.4.

There exists a positive constant $C^{*}$, such that for any disc
$B_{r}(X_{0})\subset\subset\Omega_{\mu}$ with $X_{0}=(x_{0},y_{0})$,
$r\leq\frac{x_{0}}{2}$, then

$\frac{1}{r}\fint_{\partial B_{r}(X_{0})}(m_{0}-\psi_{\lambda,\mu})dS\geq
C^{*}\Lambda x_{0},$

implies that

$\psi_{\lambda,\mu}<m_{0}~{}~{}~{}~{}\text{in}~{}~{}B_{r}(X_{0}).$

We next establish a non-degeneracy lemma.

######  Lemma 2.5.

There is a universal constant $c^{*}>0$ such that for any disc $B_{r}(X_{0})$
with center $X_{0}=(x_{0},y_{0})\in E$ and $r\leq\frac{x_{0}}{2}$, then

$\frac{1}{r}\left(\fint_{B_{r}(X_{0})}(m_{0}-\psi_{\lambda,\mu})^{2}dxdy\right)^{\frac{1}{2}}\leq
c^{*}\Lambda x_{0},$ (2.41)

implies that

$\psi_{\lambda,\mu}=m_{0}~{}~{}~{}~{}\text{in}~{}~{}B_{\frac{r}{8}}(X_{0})\cap
E.$

###### Proof.

It is easy to check that the set

$B_{\frac{r}{8}}(X_{0})\cap E$

can be covered by discs of the form

$B_{r_{1}}(X_{1})\subset B_{\frac{r}{4}}(X_{0})\cap E\ \ \text{with}\ \
r_{1}=\frac{r}{16}.$

Thus, it suffices to show that $\psi_{\lambda,\mu}=m_{0}$ in any discs
$B_{r_{1}}(X_{1})\subset B_{\frac{r}{4}}(X_{0})\cap E$, provided that the
assumption (2.41) holds. Let $\psi_{0}$ solves the following boundary value
problem

$\left\\{\begin{array}[]{ll}&\nabla\cdot
f_{\eta}\left(\frac{\nabla\psi_{0}}{x};\lambda\right)=0\ \ \text{in}\
B_{2r_{1}}(X_{1})\setminus B_{r_{1}}(X_{1}),\\\ &\psi_{0}=m_{0}\ \text{in}\
\overline{B_{r_{1}}(X_{1})},\ \psi_{0}=\psi_{\lambda,\mu}\ \text{outside of}\
B_{2r_{1}}(X_{1}).\end{array}\right.$ (2.42)

Obviously, $\max\\{\psi_{0},\psi_{\lambda,\mu}\\}\in K_{\mu}$, and thus

$\begin{array}[]{rl}0\leq&J_{\lambda,\mu}(\max\\{\psi_{\lambda,\mu},\psi_{0}\\})-J_{\lambda,\mu}(\psi_{\lambda,\mu})\\\
=&\int_{B_{2r_{1}}(X_{1})}xF\left(\left|\frac{\nabla\max\\{\psi_{\lambda,\mu},\psi_{0}\\}}{x}\right|^{2};\lambda\right)-xF\left(\left|\frac{\nabla\psi_{\lambda,\mu}}{x}\right|^{2};\lambda\right)dxdy\\\
&-2F_{1}(\lambda^{2};\lambda)\lambda\int_{B_{2r_{1}}(X_{1})}\nabla\max\\{\psi_{0}-\psi_{\lambda,\mu},0\\}\cdot
edxdy\\\
&+\Lambda^{2}\int_{B_{2r_{1}}(X_{1})}x\chi_{\\{\max\\{\psi_{\lambda,\mu},\psi_{0}\\}<m_{0}\\}}-x\chi_{\\{\psi_{\lambda,\mu}<m_{0}\\}}dxdy\\\
=&I_{1}+I_{2}+I_{3}.\end{array}$ (2.43)

For the first term on the right hand side of (2.43), one has

$\begin{array}[]{rl}I_{1}\leq&\int_{B_{2r_{1}}(X_{1})\setminus
B_{r_{1}}(X_{1})}\nabla\max\\{\psi_{0}-\psi_{\lambda,\mu},0\\}\cdot
f_{\eta}\left(\frac{\nabla\max\\{\psi_{\lambda,\mu},\psi_{0}\\}}{x};\lambda\right)dxdy\\\
&-\int_{B_{r_{1}}(X_{1})}xF\left(\left|\frac{\nabla\psi_{\lambda,\mu}}{x}\right|^{2};\lambda\right)dxdy\\\
\leq&-\int_{B_{r_{1}}(X_{1})}xF\left(\left|\frac{\nabla\psi_{\lambda,\mu}}{x}\right|^{2};\lambda\right)dxdy\\\
&+2\int_{\partial
B_{r_{1}}(X_{1})}(m_{0}-\psi_{\lambda,\mu})F_{1}\left(\left|\frac{\nabla\psi_{0}}{x}\right|^{2};\lambda\right)\frac{\nabla\psi_{0}\cdot\nu}{x}dS.\end{array}$
(2.44)

It is easy to check that $I_{2}=0$ and

$I_{3}=-\Lambda^{2}\int_{B_{2r_{1}}(X_{1})}x\chi_{\\{\psi_{\lambda,\mu}<\psi_{0}=m_{0}\\}}dxdy\leq-\Lambda^{2}\int_{B_{r_{1}}(X_{1})}x\chi_{\\{\psi_{\lambda,\mu}<m_{0}\\}}dxdy,$

which together with (2.43) and (2.44) gives that

$\begin{array}[]{rl}&\int_{B_{r_{1}}(X_{1})}xF\left(\left|\frac{\nabla\psi_{\lambda,\mu}}{x}\right|^{2};\lambda\right)dxdy+\Lambda^{2}\int_{B_{r_{1}}(X_{1})}x\chi_{\\{\psi_{\lambda,\mu}<m_{0}\\}}dxdy\\\
\leq&2\int_{\partial
B_{r_{1}}(X_{1})}(m_{0}-\psi_{\lambda,\mu})F_{1}\left(\left|\frac{\nabla\psi_{0}}{x}\right|^{2};\lambda\right)\frac{\nabla\psi_{0}\cdot\nu}{x}dS.\end{array}$
(2.45)

Set
$\tilde{\psi}(X)=\frac{m_{0}-\psi_{\lambda,\mu}(X_{1}+r_{1}X)}{r_{1}x_{0}}$
and $\tilde{\psi}_{0}(X)=\frac{m_{0}-\psi_{0}(X_{1}+r_{1}X)}{r_{1}x_{0}}$, one
has

$\nabla\cdot
f_{\eta}\left(\frac{x_{0}\nabla\tilde{\psi}}{x_{1}+r_{1}x};\lambda\right)\geq
0\ \ \text{in}\ B_{16}\left(\frac{X_{0}-X_{1}}{r_{1}}\right).$

Moreover, it follows from the assumption (2.41) that

$\left(\fint_{B_{16}\left(\frac{X_{0}-X_{1}}{r_{1}}\right)}\tilde{\psi}^{2}dxdy\right)^{\frac{1}{2}}\leq\delta\Lambda,$

where $\delta$ is to be chosen later on. By using the $L^{\infty}$ estimate in
Theorem 8.17 in [GT], one has

$\sup_{X\in B_{8}\left(\frac{X_{0}-X_{1}}{r_{1}}\right)}\tilde{\psi}(X)\leq
C\left(\fint_{B_{16}\left(\frac{X_{0}-X_{1}}{r_{1}}\right)}\tilde{\psi}^{2}dxdy\right)^{\frac{1}{2}}\leq
C\delta\Lambda,$ (2.46)

where $C$ is a constant depending only on $\vartheta$ and $a$. Since
$B_{2}(0)\subset B_{8}\left(\frac{X_{0}-X_{1}}{r_{1}}\right)$, it follows from
(2.46) that

$0\leq\tilde{\psi}_{0}=\tilde{\psi}\leq C\Lambda\delta\ \ \text{on}\ \
\partial B_{2}(0)\ \ \text{and}\ \ \tilde{\psi}_{0}=0\ \ \text{on}\ \ \partial
B_{1}(0).$ (2.47)

It is easy to check that

$\nabla\cdot
f_{\eta}\left(\frac{x_{0}\nabla\tilde{\psi}_{0}}{x_{1}+r_{1}x};\lambda\right)=0\
\ \text{in}\ B_{2}(0)\setminus B_{1}(0).$

Applying the boundary elliptic estimate in Lemma 6.10 in [GT], one has

$|\nabla\tilde{\psi}_{0}\cdot\nu|\leq C\delta\Lambda\ \ \text{on}\ \ \partial
B_{1}.$

In view of (2.45), (2.47) and the trace theorem, one has

$\begin{array}[]{rl}&\int_{B_{1}(0)}(x_{1}+r_{1}x)F\left(\frac{x_{0}^{2}|\nabla\tilde{\psi}|^{2}}{(x_{1}+r_{1}x)^{2}};\lambda\right)+\Lambda^{2}(x_{1}+r_{1}x)\chi_{\\{\tilde{\psi}>0\\}}dxdy\\\
\leq&2x_{0}\int_{\partial
B_{1}(0)}\tilde{\psi}F_{1}\left(\frac{x^{2}_{0}|\nabla\tilde{\psi}_{0}|^{2}}{|x_{1}+r_{1}x|^{2}};\lambda\right)\frac{x_{0}|\nabla\psi_{0}\cdot\nu|}{x_{1}+r_{1}x}dS\\\
\leq&Cx_{0}\delta\Lambda\int_{\partial B_{1}(0)}\tilde{\psi}dS\\\
\leq&Cx_{0}\delta\Lambda\left(\int_{B_{1}(0)}\tilde{\psi}dxdy+\int_{B_{1}(0)}|\nabla\tilde{\psi}|dxdy\right)\\\
\leq&Cx_{0}\delta\Lambda\left\\{\left(C\delta\Lambda+\frac{1}{4\varepsilon}\right)\int_{B_{1}(0)}\chi_{\\{\tilde{\psi}>0\\}}dxdy+\varepsilon\int_{B_{1}(0)}|\nabla\tilde{\psi}|^{2}dxdy\right\\},\end{array}$
(2.48)

where we have used the fact

$|\nabla\tilde{\psi}|\leq\varepsilon|\nabla\tilde{\psi}|^{2}+\frac{1}{4\varepsilon}\
\ \text{a.e. in $B_{1}\cap\\{\tilde{\psi}>0\\}$}.$

On the other hand, we have

$\begin{array}[]{rl}&x_{0}\int_{B_{1}(0)}|\nabla\tilde{\psi}|^{2}dxdy+x_{0}\int_{B_{1}(0)}\Lambda^{2}\chi_{\\{\tilde{\psi}>0\\}}dxdy\\\
\leq&C\int_{B_{1}(0)}(x_{1}+r_{1}x)F\left(\frac{x_{0}^{2}|\nabla\tilde{\psi}|^{2}}{(x_{1}+r_{1}x)^{2}};\lambda\right)+\Lambda^{2}(x_{1}+r_{1}x)\chi_{\\{\tilde{\psi}>0\\}}dxdy\\\
\leq&Cx_{0}\delta\Lambda\left\\{\left(C\delta\Lambda-\frac{1}{4\varepsilon}\right)\int_{B_{1}(0)}\chi_{\\{\tilde{\psi}>0\\}}dxdy+\varepsilon\int_{B_{1}(0)}|\nabla\tilde{\psi}|^{2}dxdy\right\\}.\end{array}$
(2.49)

Taking $\varepsilon=\frac{1}{C\delta\Lambda}$, it follows from (2.48) and
(2.49) that

$\begin{array}[]{rl}\left(1-C\delta^{2}\right)x_{0}\Lambda^{2}\int_{B_{1}(0)}\chi_{\\{\tilde{\psi}>0\\}}dxdy\leq
0,\end{array}$

which implies that

$\tilde{\psi}=0\ \ \text{in}\ \ \ B_{1}(0),$

provided that $\delta<\sqrt{\frac{1}{C}}$, where the constant $C$ depends on
$\vartheta$ and $b$. The proof is completed. ∎

######  Theorem 2.6.

The minimizer $\psi_{\lambda,\mu}$ is Lipschitz continuous in any compact
subset of $\bar{\Omega}_{\mu}$ that does not contain $A$ or the points where
$\partial\Omega_{\mu}$ is not $C^{1,\alpha}$.

###### Proof.

Denote $\psi=\psi_{\lambda,\mu}$ and $\Gamma=\Gamma_{\lambda,\mu}$ for
simplicity. The Lipschitz continuity of $\psi$ in any compact subset of
$\Omega_{\mu}$ follows from the proof of Lemma 2.3. On another hand, the
Lipschitz continuity of $\psi$ near $L_{\mu}\cup
N_{\mu}\cup(N_{0,\mu}\cap\left\\{x\leq\frac{b}{2}\right\\})$ can be obtained
by using the elliptic estimate. Along the similar arguments in the proof of
Lemma 2.2, we can obtain the Lipschitz continuity of $\psi$ near the symmetric
axis $I$.

We next consider the Lipschitz continuity of $\psi$ near $T$ or near the wall
$N_{0}\cap\left\\{x\geq\frac{b}{2}\right\\}$.

For $X=(x,y)\in\Omega_{\mu}$ with $y-1>\delta$, denote $X_{0}=(b,y)$,
$d(X)=dist(X,\Gamma_{\lambda,\mu})$ and $d_{1}(X)=dist(X,T)$. If $d(X)\leq
d_{1}(X)$, by using the similar arguments in the proof Lemma 2.3, we have
$|\nabla\psi(X)|\leq C$, where the constant $C$ depends on $\Lambda$ and
$\vartheta$.

For the case $d(X)>d_{1}(X)=x-b$, set
$r_{0}=\min\left\\{\frac{b}{2},y-1\right\\}$ and $B_{r_{0}}=B_{r_{0}}(X_{0})$.

Consider a function $\phi$, which solves the following boundary value problem

$\left\\{\begin{array}[]{ll}&\nabla\cdot
f_{\eta}\left(\frac{\nabla\phi}{x};\lambda\right)=0\
\text{in}~{}~{}B_{r_{0}}\cap\\{x>b\\},\\\ &\phi=0\ \ \ \text{on}\ \
B_{r_{0}}\cap\\{x=b\\},\ \ \phi=m_{0}-\psi\ \ \ \text{on}\ \ \partial
B_{r_{0}}\cap\\{x>b\\}.\end{array}\right.$

The maximum principle gives that

$m_{0}-\psi\leq\phi$ in $B_{r_{0}}\cap\\{x>b\\}$. (2.50)

Set $\tilde{\phi}(\tilde{X})=\frac{\phi(X_{0}+r_{0}\tilde{X})}{r_{0}}$ with
$\tilde{X}=(\tilde{x},\tilde{y})$. Noting $0\leq\phi\leq m_{0}$, one has

$\left\\{\begin{array}[]{ll}\nabla\cdot
f_{\eta}\left(\frac{\nabla\tilde{\phi}}{b+r_{0}\tilde{x}};\lambda\right)=0&\text{in}~{}~{}B_{1}(0)\cap\\{\tilde{x}>0\\},\\\
0\leq\tilde{\phi}\leq\frac{m_{0}}{r_{0}}&\text{on}\ \
\partial(B_{1}(0)\cap\\{\tilde{x}>0\\}).\end{array}\right.$

Applying the elliptic estimates for $\tilde{\phi}$ in
$B_{1}(0)\cap\\{\tilde{x}>0\\}$, one has

$\tilde{\phi}(\tilde{X})\leq C\frac{\tilde{x}}{r_{0}}\ \ \text{in
$B_{\frac{1}{2}}(0)\cap\\{\tilde{x}>0\\}$},$

which gives that

$\phi(X)=\tilde{\phi}\left(\frac{X-X_{0}}{r_{0}}\right)\leq
C\frac{x-b}{r_{0}}\ \ \text{in $B_{\frac{r_{0}}{2}}(X_{0})\cap\\{x>b\\}$}.$
(2.51)

If $r=d_{1}(x)=x-b<\frac{r_{0}}{4}$, we have $B_{r}(X)\subset
B_{\frac{r_{0}}{2}}(X_{0})\cap\\{x>b\\}$. Set
$\tilde{\psi}(\tilde{X})=\frac{m_{0}-\psi(X+r\tilde{X})}{r}$ with
$\tilde{X}=(\tilde{x},\tilde{y})$, it follows from (2.50) and (2.51) that

$\tilde{\psi}(\tilde{X})\leq\frac{\phi(X+r\tilde{X})}{r}\leq\frac{C(x+r\tilde{x}-b)}{rr_{0}}\leq\frac{C}{r_{0}}\
\ \text{in}\ \ B_{1}(0).$

By using the elliptic estimate, one has

$|\nabla\psi(X)|=|\nabla\tilde{\psi}(0)|\leq\frac{C}{r_{0}}.$

If $r=d_{1}(x)=x-b\geq\frac{r_{0}}{4}$, the elliptic estimate gives the
desired uniform bound for $\nabla\psi(X)$.

Finally, we consider the Lipschitz continuity of $\psi$ near
$N_{0}\cap\\{x\geq b\\}$. Since $N_{0}$ is $C^{2,\alpha}$ and
$\psi_{\lambda,\mu}=0$ on $N_{0}$, the Harnack’s inequality is still valid up
to the boundary $N_{0}\cap\left\\{x\geq\frac{b}{2}\right\\}$. It follows from
the similar arguments in the proof of Lemma 2.3 that

$\frac{|\nabla\psi(X)|}{x}\leq C\Lambda\ \ \text{near
$N_{0}\cap\left\\{x\geq\frac{b}{2}\right\\}$}.$

∎

##  3 The free boundary of the minimizer $\psi_{\lambda,\mu}$

In this section, we will show some important properties of the free boundary,
such as the continuity of the graph and the continuous fit condition.

###  3.1 Uniqueness and monotonicity of the minimizer

To obtain the continuous fit condition of the free boundary, we construct the
uniqueness and the monotonicity of the minimizer to the truncated variational
problem $(P_{\lambda,\mu})$.

######  Lemma 3.1.

For any $\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$ and $\mu>1$, the
minimizer $\psi_{\lambda,\mu}$ to the truncated variational problem
$(P_{\lambda,\mu})$ is unique, and
$\psi_{\lambda,\mu}(x,y_{1})\geq\psi_{\lambda,\mu}(x,y_{2})$ for any
$y_{1}>y_{2}$.

###### Proof.

Suppose that $\psi_{1}$ and $\psi_{2}$ are two minimizers to the truncated
variational problem $(P_{\lambda,\mu})$. Set

$\psi_{1}^{\varepsilon}(x,y)=\psi_{1}(x,y-\varepsilon)\ \ \text{for any}\ \
\varepsilon>0.$

Notice that $\psi_{1}^{\varepsilon}(x,y)$ is a minimizer of the functional
$J^{\varepsilon}_{\lambda,\mu}$ in $\Omega_{\mu}^{\varepsilon}$ with the
corresponding admissible set $K_{\mu}^{\varepsilon}$ as follows

$\Omega_{\mu}^{\varepsilon}=\\{(x,y)|~{}(x,y-\varepsilon)\in\Omega_{\mu}\\}~{}~{}\text{and}~{}~{}K_{\mu}^{\varepsilon}=\\{\psi^{\varepsilon}(x,y-\varepsilon)\in
K_{\mu}|~{}(x,y)\in\Omega_{\mu}^{\varepsilon}\\}.$

Extend $\psi_{2}(x,y)=\frac{m_{0}}{x_{\mu}^{2}}x^{2}$ in $\\{(x,y)\mid 0<x\leq
x_{\mu},\mu<y\leq\mu+\varepsilon\\}$ and denote

$\varphi_{1}=\min\\{\psi_{1}^{\varepsilon},\psi_{2}\\}\ \ \ \text{and}\ \ \
\varphi_{2}=\max\\{\psi_{1}^{\varepsilon},\psi_{2}\\}.$

Obviously, $\varphi_{1}\in K^{\varepsilon}_{\mu}$ and $\varphi_{2}\in
K_{\mu}$. For any sufficiently large $R>R_{0}$, denote
$\Omega_{\mu,R}=\Omega_{\mu}\cap\\{y<R\\}$ and
$\Omega_{\mu,R}^{\varepsilon}=\Omega_{\mu}^{\varepsilon}\cap\\{y<R\\}$. Since
$\varphi_{1}=\psi_{1}^{\varepsilon}$ in
$\Omega_{\mu,R}\setminus\Omega_{\mu,R}^{\varepsilon}$ and
$\varphi_{1}=\psi_{1}^{\varepsilon}$ in
$\Omega_{\mu,R}^{\varepsilon}\setminus\Omega_{\mu,R}$, it is easy to check
that

$\begin{array}[]{rl}&\int_{\Omega_{\mu,R}^{\varepsilon}}xF\left(\frac{|\nabla\varphi_{1}|^{2}}{x^{2}};\lambda\right)dxdy+\int_{\Omega_{\mu,R}}xF\left(\frac{|\nabla\varphi_{2}|^{2}}{x^{2}};\lambda\right)dxdy\\\
=&\int_{\Omega_{\mu,R}^{\varepsilon}}xF\left(\frac{|\nabla\psi_{1}^{\varepsilon}|^{2}}{x^{2}};\lambda\right)dxdy+\int_{\Omega_{\mu,R}}xF\left(\frac{|\nabla\psi_{2}|^{2}}{x^{2}};\lambda\right)dxdy,\end{array}$
(3.1)

and

$\begin{array}[]{rl}&\int_{E_{R}^{\varepsilon}}x\chi_{\\{\varphi_{1}<m_{0}\\}}dxdy+\int_{E_{R}}x\chi_{\\{\varphi_{2}<m_{0}\\}}dxdy\\\
=&\int_{E_{R}^{\varepsilon}}x\chi_{\\{\psi_{1}^{\varepsilon}<m_{0}\\}}dxdy+\int_{E_{R}}x\chi_{\\{\psi_{2}<m_{0}\\}}dxdy,\end{array}$
(3.2)

where $E_{R}=\Omega_{\mu,R}\cap\\{x>b\\}$ and
$E_{R}^{\varepsilon}=\Omega_{\mu,R}^{\varepsilon}\cap\\{x>b\\}$.

Integration by parts, one has

$\begin{array}[]{rl}&\int_{E_{R}^{\varepsilon}}\nabla\varphi_{1}\cdot
e\chi_{\\{\varphi_{1}<m_{0}\\}}dxdy+\int_{E_{R}}\nabla\varphi_{2}\cdot
e\chi_{\\{\varphi_{2}<m_{0}\\}}dxdy\\\
&-\int_{E_{R}^{\varepsilon}}\nabla\psi_{1}^{\varepsilon}\cdot
e\chi_{\\{\varphi_{1}<m_{0}\\}}dxdy+\int_{E_{R}}\nabla\psi_{2}\cdot
e\chi_{\\{\psi_{2}<m_{0}\\}}dxdy\\\ =&\int_{\partial
E_{R}^{\varepsilon}}(\varphi_{1}-\psi_{1}^{\varepsilon})e\cdot\nu
dS+\int_{E_{R}}(\varphi_{2}-\psi_{2})e\cdot\nu dS\\\ =&\int_{\partial
E_{R}^{\varepsilon}\cap\partial
E_{R}}(\varphi_{1}+\varphi_{2}-\psi_{1}^{\varepsilon}-\psi_{2})e\cdot\nu dS\\\
=&0,\end{array}$ (3.3)

where we have used the facts $\varphi_{2}=\psi_{2}=0$ in $E_{R}\setminus
E_{R}^{\varepsilon}$.

In view of (3.1) - (3.3), one has

$\begin{array}[]{rl}&\int_{\Omega_{\mu,R}^{\varepsilon}}G(\nabla\varphi_{1},\varphi_{1},x;\lambda)dxdy+\int_{\Omega_{\mu,R}}G(\nabla\varphi_{1},\varphi_{1},x;\lambda)dxdy\\\
=&\int_{\Omega_{\mu,R}^{\varepsilon}}G(\nabla\psi^{\varepsilon}_{1},\psi^{\varepsilon}_{1},x;\lambda)dxdy+\int_{\Omega_{\mu,R}}G(\nabla\psi_{2},\psi_{2},x;\lambda)dxdy.\end{array}$
(3.4)

Taking $R\rightarrow+\infty$ in (3.4) yields that

$\begin{array}[]{rl}J_{\lambda,\mu}^{\varepsilon}(\psi_{1}^{\varepsilon})+J_{\lambda,\mu}(\varphi_{2})=J_{\lambda,\mu}^{\varepsilon}(\varphi_{1})+J_{\lambda,\mu}(\psi_{2}).\end{array}$
(3.5)

Since $\psi_{1}^{\varepsilon}$ and $\psi_{2}$ are minimizers, it follows from
(3.5) that

$J_{\lambda,\mu}^{\varepsilon}(\psi_{1}^{\varepsilon})=J_{\lambda,\mu}^{\varepsilon}(\varphi_{1})\
\ \text{and}\ \ J_{\lambda,\mu}(\psi_{2})=J_{\lambda,\mu}(\varphi_{2}).$ (3.6)

Next, we claim that

$\psi_{1}^{\varepsilon}(x,y)<\psi_{2}(x,y)\ \ \text{in $D$},$ (3.7)

where $D$ is the maximal connected component of
$\Omega_{\mu}\cap\\{\psi_{2}<m_{0}\\}$, which contains an
$\Omega_{\mu}$-neighborhood of $N_{\mu}$.

Suppose not, note that $\psi_{1}^{\varepsilon}<m_{0}=\psi_{2}$ on $N_{\mu}$,
then there exists a dist $B_{1}$, such that

$\psi_{1}^{\varepsilon}<\psi_{2}\ \ \text{in}\ \ B_{1},\ \
\bar{B}_{1}\subset\Omega_{\mu}\cap\\{\psi_{2}<m_{0}\\},$

and

$\psi_{1}^{\varepsilon}=\psi_{2}\ \ \text{at some points $X_{0}\in\partial
B_{1}\cap(\Omega_{\mu}\cap\\{\psi_{2}<m_{0}\\})$}.$

Thanks to Hopf’s lemma, one has

$\frac{\partial(\psi_{1}^{\varepsilon}-\psi_{2})}{\partial\nu}>0\ \ \text{at
$X_{0}$},$

where $\nu$ is the outer normal vector of $\partial B_{1}$ at $X_{0}$. This
implies that the level set
$\\{\psi_{1}^{\varepsilon}=\psi_{2}=\psi_{2}(X_{0})\\}$ is smooth curve in a
neighborhood of $X_{0}$. Then there exists a smooth curve
$\Gamma_{0}=\\{X\mid\psi_{1}^{\varepsilon}(X)=\psi_{2}(X)=\psi_{2}(X_{0})\\}$
passing through $X_{0}$ and a disc $B_{2}$, such that

$\psi_{1}^{\varepsilon}>\psi_{2}\ \ \text{in $B_{2}$ and
$X_{0}\in\Gamma_{0}\cap\partial B_{2}\cap\partial B_{1}$}.$

Hence, one has

$\frac{\partial(\varphi_{1}-\psi_{2})(X)}{\partial\nu}=\frac{\partial(\psi_{1}^{\varepsilon}-\psi_{2})(X)}{\partial\nu}\rightarrow\frac{\partial(\psi_{1}^{\varepsilon}-\psi_{2})(X_{0})}{\partial\nu}>0\
\ \text{as $X\rightarrow X_{0}$, $X\in B_{1}$},$

and

$\frac{\partial(\varphi_{1}-\psi_{2})(X)}{\partial\nu}=\frac{\partial(\psi_{2}-\psi_{2})(X)}{\partial\nu}=0,\
\ \text{$X\in B_{2}$},$

which implies that $\varphi_{1}$ is not $C^{1}$-smooth in a neighborhood of
$X_{0}$, due to that $\psi_{2}$ is smooth at $X_{0}$. On the other hand, it
follows from (3.6) that $\varphi_{1}$ is a minimizer, and
$\varphi_{1}(X_{0})<m_{0}$. By virtue of the elliptic regularity, we can
conclude that $\varphi_{1}$ is smooth in a neighborhood of $X_{0}$. This leads
a contradiction. Hence, we complete the proof of the claim (3.7).

We next show that

$\psi_{1}(x,y)$ is monotone increasing with respect to $y$ in $\Omega_{\mu}$.
(3.8)

Choosing $\psi_{1}=\psi_{2}$ in (3.7), it implies that

$\frac{\partial\psi_{1}}{\partial y}\geq 0$ in $D$. (3.9)

To obtain (3.8), it suffices to show that

$\text{$D=\Omega_{\mu}\cap\\{\psi_{1}<m_{0}\\}$}.$

Suppose not, it follows from (3.7) that $D\cap\\{x>b\\}=\\{x>b,y<\phi(x)\\}$.
As a part of the free boundary of the graph $\phi$, we can conclude that
$\phi(x)$ is continuous (see the proof of Lemma 3.3 later). Define
$\psi_{0}=\psi_{1}$ in $D$ and $\psi_{0}=m_{0}$ in
$\Omega_{\mu}\cap\\{x>b,y\geq\phi(x)\\}$, it is easy to check that $\psi\in
K_{\mu}$. Therefore, we have

$J_{\lambda,\mu}(\psi_{0})-J_{\lambda,\mu}(\psi_{1})=-\int_{\Omega_{\mu}\setminus
D}G(\nabla\psi_{1},\psi_{1},x;\lambda)dxdy<0,$

which leads a contradiction.

Similarly, we can obtain that

$\psi_{2}(x,y)$ is monotone increasing with respect to $y$ in $\Omega_{\mu}$,

which together with (3.8) gives that

$\text{$\Omega_{\mu}\cap\\{\psi_{1}<m_{0}\\}$ and
$\Omega_{\mu}\cap\\{\psi_{2}<m_{0}\\}$ are connected}.$

In view of (3.7), one has

$\psi_{1}^{\varepsilon}(x,y)\leq\psi_{2}(x,y)~{}~{}\text{in}~{}~{}~{}\Omega_{\mu}.$

Taking $\varepsilon\rightarrow 0$ in above inequality, we have

$\psi_{1}\leq\psi_{2}\ \ \text{in}\ \ \Omega_{\mu}.$

Similarly, we can obtain that

$\psi_{1}\geq\psi_{2}\ \ \text{in}\ \ \Omega_{\mu}.$

Hence, $\psi_{1}=\psi_{2}$ and the minimizer to the variational problem
$(P_{\lambda,\mu})$ is unique.

∎

###  3.2 Fundamental properties of the free boundary

In this subsection, we show some significant properties of the free boundary
${\Gamma}_{\lambda,\mu}$ to the truncated variational problem
$(P_{\lambda,\mu})$. Thanks to the monotonicity of the minimizer
$\psi_{\lambda,\mu}(x,y)$ with respect to $y$, there exists a mapping
$y=k_{\lambda,\mu}(x)$, such that

$E\cap\\{\psi_{\lambda,\mu}<m_{0}\\}=\\{(x,y)\mid
b<x<\infty,g_{0}(x)<y<k_{\lambda,\mu}(x)\\}.$

To obtain the continuity of the function $k_{\lambda,\mu}(x)$, we need the
following non-oscillation lemma and the proof can be found in Lemma 4.4 in
[ACF5].

######  Lemma 3.2.

Let G be a domain in $E\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$, bounded by two
disjointed arcs $\gamma_{1}$, $\gamma_{2}$ of the free boundary,
${y=\beta_{1}}$, ${y=\beta_{2}}$. Suppose that the arcs $\gamma_{i}$ ($i=1,2$)
lie in $\\{\beta_{1}<y<\beta_{2}\\}$ with the endpoints
$(\alpha_{i},\beta_{1})$ and $(\zeta_{i},\beta_{2})$. Suppose the distant
$d=dist(G,B)>0$, then

$|\beta_{2}-\beta_{1}|\leq
C\max\\{|\alpha_{1}-\alpha_{2}|,~{}~{}|\zeta_{1}-\zeta_{2}|\\},$

where $C$ is a constant depending only on $\Lambda,\vartheta$, $d,N,N_{0}$ and
$m_{0}$.

######  Remark.

The nonoscillation Lemma 3.2 remains true provided that one of the arcs
$\gamma_{2}$ is a line segment on $T=\\{(b,y)\mid y\geq 1\\}$, and

$\frac{1}{x}\frac{\partial\psi_{\lambda,\mu}}{\partial\nu}\geq\lambda~{}~{}\text{on}~{}~{}\gamma_{2}.$

######  Lemma 3.3.

The function $y=k_{\lambda,\mu}(x)$ is continuous for $x\in(b,+\infty)$.
Moreover, $k_{\lambda,\mu}(b)=\lim_{x\rightarrow b^{+}}k_{\lambda,\mu}(x)$
exists and is finite.

Figure 6: The domain $D_{n}$

###### Proof.

We first consider the existence of the limit $\lim_{x\rightarrow
b^{+}}k_{\lambda,\mu}(x)$.

Suppose not, one has that $\liminf_{x\rightarrow
b^{+}}k_{\lambda,\mu}(x)<\limsup_{x\rightarrow b^{+}}k_{\lambda,\mu}(x)$, then
we consider the following two cases for $\underline{y}=\lim_{x\rightarrow
b^{+}}k_{\lambda,\mu}(x)$.

Case 1. $\underline{y}<1$. Denote $\delta=\frac{1-\underline{y}}{4}$ and
$I_{\delta}=\left\\{(1,y)\mid 1-\frac{5}{2}\delta\leq y\leq
1-\frac{3}{2}\delta\right\\}$. Then there exist two sequences $\\{x_{n}\\}$
and $\\{\tilde{x}_{n}\\}$ with $x_{n}\downarrow b$ and
$\tilde{x}_{n}\downarrow b$, such that

$k_{\lambda,\mu}(x_{n})\rightarrow 1-\delta\ \ \text{and}\ \
k_{\lambda,\mu}(x_{n})\rightarrow 1-3\delta,\ \ \delta>0,$ (3.10)

and

$\psi_{\lambda,\mu}(x_{n},y)=m_{0},\ \
\psi_{\lambda,\mu}(\tilde{x}_{n},y)<m_{0}\ \ \text{for}\ \
|y-1+2\delta|\leq\frac{\delta}{2},\ \ x_{n+1}<\tilde{x}_{n}<x_{n}.$ (3.11)

By virtue of Lemma 2.3 and the monotonicity of $\psi_{\lambda,\mu}$, we have
that $\psi_{\lambda,\mu}$ is Lipschitz continuous in an $\bar{E}$-neighborhood
of $I_{\delta}$ and $\psi_{\lambda,\mu}=m_{0}$ on $I_{\delta}$.

It follows from (3.11) that there exists a domain $D_{n}\subset
E\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$ (please see Figure 6), which is bounded
by the arcs $y_{1}=1-\frac{3}{2}\delta$, $y_{2}=1-\frac{5}{2}\delta$,
$\gamma_{n}^{1}$ and $\gamma_{n}^{2}$. Here, $\gamma_{n}^{1}$ and
$\gamma_{n}^{2}$ are parts of free boundary $\Gamma_{\lambda,\mu}\cap\\{x\leq
x_{n-1}\\}$, and the curve $\gamma_{n}^{1}$ lies the right of the curve
$\gamma_{n}^{2}$. Denote $h_{n}=\text{dist}(\gamma_{n}^{1},\gamma_{n}^{2})$.
By virtue of that $x_{n}\rightarrow b^{+}$, one has

$h_{n}\rightarrow 0\ \ \text{as}\ \ n\rightarrow+\infty.$ (3.12)

Thanks to the non-oscillation Lemma 3.2, we have

$0<\delta\leq Ch_{n},$

which contradicts to (3.12), provided that $n$ is sufficiently large.

Case 2. $\underline{y}\geq 1$. Take a constant $\delta>0$, such that
$\delta\leq\frac{\limsup_{x\rightarrow
b^{+}}k_{\lambda,\mu}(x)-\underline{y}}{4}$. Denote
$\gamma_{\delta}=\left\\{(b,y)\mid\underline{y}+\frac{3}{2}\delta\leq
y\leq\underline{y}+\frac{5}{2}\delta\right\\}$. In an $E$-neighborhood of
$\gamma_{\delta}$, we can obtain a contradiction by using the non-oscillation
Lemma 3.2.

Similarly, we can show that the limits $\lim_{x\rightarrow
x_{0}^{+}}k_{\lambda,\mu}(x)$ and $\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)$ exist for any $x_{0}\in(b,+\infty)$.

Step 2. $\lim_{x\rightarrow x_{0}^{+}}k_{\lambda,\mu}(x)=\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)$ for any $x_{0}\in(b,+\infty)$.

Suppose that there exists a $x_{0}\in(b,+\infty)$, such that
$\lim_{x\rightarrow x_{0}^{+}}k_{\lambda,\mu}(x)\neq\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)$. Without loss of generality, we assume that
$\lim_{x\rightarrow x_{0}^{+}}k_{\lambda,\mu}(x)>\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)$. Denote $\gamma=\\{(x_{0},y)\mid y_{1}\leq y\leq
y_{2}\\}$ with $\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)<y_{1}<y_{2}<\lim_{x\rightarrow
x_{0}^{-}}k_{\lambda,\mu}(x)$. The monotonicity and Lipschitz continuity of
$\psi_{\lambda,\mu}$ give that $\gamma$ is the free boundary of
$\psi_{\lambda,\mu}$ and $\psi_{\lambda,\mu}<m_{0}$ in
$E_{\varepsilon}=\\{(x,y)\mid x_{0}<x<x_{0}+\varepsilon,y_{1}<y<y_{2}\\}$ for
small $\varepsilon>0$. Then we have

$\tilde{Q}_{\lambda}\psi_{\lambda,\mu}=0\ \ \text{in}\ \ E_{\varepsilon},\ \
\psi_{\lambda,\mu}=m_{0}\ \ \text{and}\ \
\frac{1}{x}\frac{\psi_{\lambda,\mu}}{\partial x}=-\lambda\ \ \text{on}\ \
\gamma.$

Since $\psi_{\lambda,\mu}$ is analytic in $E_{\varepsilon}$ for small
$\varepsilon>0$, thanks to Cauchy-Kovalevskaya theorem, one has

$\psi_{\lambda,\mu}(x,y)=-\lambda(x^{2}-x^{2}_{0})+m_{0}\ \ \text{in}\ \
\Omega_{\mu}\cap\\{x_{0}<x<x_{0}+\varepsilon\\},$

which contradicts to $\psi_{\lambda,\mu}=0$ on $N_{0}$.

Step 3. $k_{\lambda,\mu}(x)<+\infty$ for any $x\in[b,+\infty)$. We first show
that

$\text{the free boundary $\Gamma_{\lambda,\mu}$ is non-empty in $E$}.$ (3.13)

Suppose that $\Gamma_{\lambda,\mu}$ is empty, it implies that

$\psi_{\lambda,\mu}<m_{0}\ \ \text{in}\ \ E.$ (3.14)

For any $R>0$, there exists a disc $B_{R}(X_{0})\subset E$ with
$X_{0}=(x_{0},y_{0})$, such that

$\frac{1}{R}\left(\fint_{B_{R}(X_{0})}(m_{0}-\psi_{\lambda,\mu})^{2}dxdy\right)^{\frac{1}{2}}\leq\frac{m_{0}}{R}\leq
c^{*}\Lambda x_{0},$

for sufficiently large $R$. It follows from non-degeneracy Lemma 2.5 that
$\psi_{\lambda,\mu}=m_{0}$ in $B_{\frac{R}{8}}(X_{0})$. This contradicts to
(3.14). With the aid of (3.13), we can take a maximal interval
$(\gamma_{1},\gamma_{2})\subset(b,+\infty)$, such that

$k_{\lambda,\mu}(x)\ \ \text{is finite in $(\gamma_{1},\gamma_{2})$ and
$k_{\lambda,\mu}(\gamma_{1}+0)=k_{\lambda,\mu}(\gamma_{2}-0)=+\infty$}.$

We first show that

$\gamma_{2}=+\infty.$ (3.15)

If not, then $\gamma_{2}<+\infty$. By using the proof of (3.13), we can
conclude that $\Gamma_{\lambda,\mu}$ is non-empty in
$E\cap\\{x>\gamma_{2}\\}$, and there exists a
$\gamma_{3}\in[\gamma_{2},+\infty)$, such that

$\Gamma_{\lambda,\mu}\cap\\{\gamma_{2}<x<\gamma_{3}\\}=\varnothing\ \
\text{and $k_{\lambda,\mu}(x)$ is finite in
$(\gamma_{3},\gamma_{3}+\varepsilon)$ for small $\varepsilon>0$}.$

Denote
$D_{R}=\left\\{(x,y)\mid\frac{\gamma_{1}+\gamma_{2}}{2}<x<\gamma_{3}+\frac{\varepsilon}{2},R<y<2R\right\\}$
for large $R$, applying the non-oscillation Lemma 3.2 for $\psi_{\lambda,\mu}$
in $D_{R}$, one has

$R\leq
C\left|\frac{\gamma_{1}+\gamma_{2}}{2}-\gamma_{3}-\frac{\varepsilon}{2}\right|,$

where the constant $C$ is independent of $R$. This leads a contradiction for
sufficiently large $R>0$.

Next, we will show that

$\gamma_{1}=b.$

If not, then $\gamma_{1}>b$ and we consider the following two cases.

Case 1. $k_{\lambda,\mu}(x)=\infty$ for any $x\in(b,\gamma_{1})$. It follows
from Lemma 2.2 that
$\frac{1}{x}\frac{\partial\psi_{\lambda,\mu}}{\partial\nu}\geq\lambda$ on
$T_{R}=\\{(b,y)\mid R<y<2R\\}$ for large $R>0$. Denote
$D_{R}=\left\\{(x,y)\mid b<x<\frac{\gamma_{1}+\gamma_{2}}{2},R<y<2R\right\\}$,
by using the non-oscillation Lemma 3.2 and Remark Remark for
$\psi_{\lambda,\mu}$ in $D_{R}$, one has

$R\leq C\frac{\gamma_{1}+\gamma_{2}-2b}{2},$

which leads a contradiction for sufficiently large $R>0$.

Case 2. There exists a $\gamma_{0}\in(b,\gamma_{1}]$, such that

$\Gamma_{\lambda,\mu}\cap\\{\gamma_{0}<x<\gamma_{1}\\}=\varnothing\ \
\text{and $k_{\lambda,\mu}(x)$ is finite in
$(\gamma_{0}-\varepsilon,\gamma_{0})$ for small $\varepsilon>0$}.$

Using the non-oscillation Lemma 3.2 for $\psi_{\lambda,\mu}$ in $D_{R}$ leads
a contradiction for sufficiently large $R>0$, where
$D_{R}=\left\\{(x,y)\mid\gamma_{0}-\frac{\varepsilon}{2}<x<\frac{\gamma_{1}+\gamma_{2}}{2},R<y<2R\right\\}$.

Finally, we can show that $k_{\lambda,\mu}(b)<+\infty$ by using the non-
oscillation Lemma 3.2 and Remark Remark.

∎

In the following, we will show some important properties, such as, the optimal
decay rate of the free boundary, the convergence rate and the asymptotic
behavior of the subsonic impinging jet in downstream.

######  Lemma 3.4.

The minimizer $\psi_{\lambda,\mu}$ and the free boundary
$y=k_{\lambda,\mu}(x)$ satisfy that
(1) for any sufficiently large $x_{0}>b$, there exists a constant C
(independent of $x_{0}$) such that

$\int_{\Omega_{\mu}\cap\\{x\geq
x_{0}\\}}G(\nabla\psi,\psi,x;\lambda)dxdy\leq\frac{C}{x_{0}^{3}},$ (3.16)

where

$G(\nabla\psi,\psi,x;\lambda)=xF\left(\left|\frac{\nabla\psi}{x}\right|^{2};\lambda\right)+\left(x\Lambda^{2}-2\lambda
F_{1}(\lambda^{2};\lambda)\nabla\psi\cdot e\right)\chi_{\\{\psi<m_{0}\\}\cap
E}.$

(2) in the downstream,

$x(k_{\lambda,\mu}(x)-g_{0}(x))\rightarrow\frac{m_{0}}{\lambda\cos\theta}~{}~{}\text{as}~{}~{}x\rightarrow+\infty,$
(3.17)

and

$\frac{\nabla\psi_{\lambda,\mu}(x,y)}{x}\rightarrow(-\lambda\sin\theta,\lambda\cos\theta)~{}~{}\text{for
$(x,y)\in\Omega_{\mu}\cap\\{\psi_{\lambda,\mu}<m_{0}\\}$, as
$x\rightarrow+\infty$.}$ (3.18)

###### Proof.

(1) By using the inequality (2.24), we have

$\int_{\Omega_{\mu}\cap\\{x\geq x_{0}\\}}G(\nabla\psi,\psi,x;\lambda)dxdy\leq
C\int_{\Omega_{\mu}\cap\\{x\geq
x_{0}\\}}x\left|\frac{\nabla\psi_{\lambda,\mu}}{x}-\lambda
e\chi_{\\{\psi<m_{0}\\}}\right|^{2}dxdy,$ (3.19)

for $x_{0}>b$.

It follows from the proof of Proposition 4.4 in [CD] that

$\int_{\Omega_{\mu}\cap\\{x\geq
x_{0}\\}}x\left|\frac{\nabla\psi_{\lambda,\mu}}{x}-\lambda
e\chi_{\\{\psi<m_{0}\\}}\right|^{2}dxdy\leq\frac{C}{x_{0}^{3}},$

for sufficiently large $x_{0}>b$, where $C$ is a constant independent of
$x_{0}$. This together with (3.19) gives (3.16).

(2) For a sequence $\\{x_{n}\\}$ with $x_{n}\rightarrow+\infty$, set
$\psi_{n}(\tilde{X})=\psi_{\lambda,\mu}\left(X_{n}+\frac{\tilde{X}}{x_{n}}\right)$
with $X_{n}=(x_{n},g_{0}(x_{n}))$ and $\tilde{X}=(\tilde{x},\tilde{y})$.
Obviously,
$\nabla\psi_{n}(\tilde{X})=\frac{1}{x_{n}}\nabla\psi_{\lambda,\mu}\left(X_{n}+\frac{\tilde{X}}{x_{n}}\right)$.
For any $R>0$, thanks to (3.16), we have

$\begin{array}[]{rl}I_{n}=&\int_{\\{|\tilde{x}|<Rx_{n}\\}}\left(1+\frac{\tilde{x}}{x^{2}_{n}}\right)\left|\frac{\nabla\psi_{n}}{1+\frac{\tilde{x}}{x_{n}^{2}}}-\lambda
e\chi_{\\{\psi_{n}<m_{0}\\}}\right|^{2}d\tilde{x}d\tilde{y}\\\
=&x_{n}\int_{\Omega_{\mu}\cap\\{x_{n}-R|<x<x_{n}+R\\}}x\left|\frac{\nabla\psi_{\lambda,\mu}}{x}-\lambda
e\chi_{\\{\psi_{\lambda,\mu}<m_{0}\\}}\right|^{2}dxdy\\\
\leq&\frac{Cx_{n}}{(x_{n}-R)^{3}}\rightarrow 0\ \ \text{as
$x_{n}\rightarrow+\infty$}.\end{array}$

Without loss of generality, we may assume that

$\psi_{n}\rightarrow\psi_{0}$ weakly in $H_{loc}^{1}(\mathbb{R}^{2})$ and
$a.e.$ in $\mathbb{R}^{2}$, (3.20)

and

$\chi_{\\{\psi_{n}<m_{0}\\}}\rightarrow\gamma\ \ \text{weakly star in
$L_{loc}^{\infty}(\mathbb{R}^{2})$ and
$\chi_{\\{\psi_{0}<m_{0}\\}}\leq\gamma\leq 1$}.$

Furthermore, one has

$\begin{array}[]{rl}\int_{\mathbb{R}^{2}}\left|\nabla\psi_{0}-\lambda
e\chi_{\\{\psi_{0}<m_{0}\\}}\right|^{2}d\tilde{x}d\tilde{y}\leq\int_{\mathbb{R}^{2}}\left|\nabla\psi_{0}-\lambda
e\gamma\right|^{2}d\bar{x}d\bar{y}\leq\liminf_{n\rightarrow+\infty}I_{n}=0,\end{array}$

which gives that

$\nabla\psi_{0}=\lambda\chi_{\\{\psi_{0}<m_{0}\\}}e\ \ \ \text{a.e. in
$\mathbb{R}^{2}$ and}~{}~{}\psi_{0}(0)=0.$ (3.21)

Denote $\omega(s,t)=\psi_{0}(\tilde{x},\tilde{y})$ with
$s=\tilde{x}\cos\theta+\tilde{y}\sin\theta$ and
$t=\tilde{y}\cos\theta-\tilde{x}\sin\theta$, one has

$\frac{\partial\omega}{\partial
s}=\frac{\partial\psi_{0}}{\partial\tilde{x}}\cos\theta+\frac{\partial\psi_{0}}{\partial\tilde{y}}\sin\theta\
\ \text{and}\ \ \frac{\partial\omega}{\partial
t}=-\frac{\partial\psi_{0}}{\partial\tilde{x}}\sin\theta+\frac{\partial\psi_{0}}{\partial\tilde{y}}\cos\theta.$

By virtue of (3.21), one has

$\frac{\partial\omega(s,t)}{\partial s}=0\ \ \text{and}\ \
\frac{\partial\omega(s,t)}{\partial t}=\lambda\chi_{\\{\omega<m_{0}\\}}\ \
\text{a.e. in $\mathbb{R}^{2}$},$ (3.22)

which imply that $\omega(s,t)$ is only a function of $t$. Moreover,
$\omega(s,t)$ is monotone increasing with respect to $t$. In view of
$\omega(0,0)=0$ and $0\leq\omega\leq m_{0}$, it follows from (3.22) that

$\psi_{0}(\tilde{x},\tilde{y})=\omega(t)=\min\\{\max\\{\lambda
t,0\\},m_{0}\\}=\min\\{\max\\{\lambda(\tilde{y}\cos\theta-\tilde{x}\sin\theta),0\\},m_{0}\\}\
\text{ in $\mathbb{R}^{2}$}.$

To obtain the asymptotic behavior of the free boundary $\Gamma_{\lambda,\mu}$,
we first show that

$\partial\\{\psi_{n}<m_{0}\\}\ \ \text{converges to}\ \
\partial\\{\psi_{0}<m_{0}\\}\ \text{locally in Hausdorff metric,}\ \text{as}\
\ n\rightarrow+\infty.$ (3.23)

Definition of Hausdorff distance $d(D,F)$ between two sets $D$ and $F$ is as
follows

$d(D,F)=\inf\left\\{\varepsilon>0\mid D\subset\bigcup_{X\in
F}B_{\varepsilon}(X)\ \ \text{and}\ \ F\subset\bigcup_{X\in
D}B_{\varepsilon}(X)\right\\}.$

For any $X_{0}=(x_{0},y_{0})\notin\partial\\{\psi_{0}<m_{0}\\}$, the
continuity of $\psi_{0}$ gives that $\psi_{0}(X_{0})<m_{0}$ or
$\psi_{0}(X_{0})=m_{0}$. If $\psi_{0}<m_{0}$ in $B_{r}(X_{0})$, it follows
from (3.20) that

$\lim_{n\rightarrow+\infty}\fint_{\partial B_{r}(X_{0})}(m_{0}-\psi_{n})dS>0,$

which implies that

$\frac{1}{r}\fint_{\partial
B_{\frac{r}{x_{n}}}\left(X_{n}+\frac{X_{0}}{x_{n}}\right)}(m_{0}-\psi_{\lambda,\mu})dS=\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}(m_{0}-\psi_{n})dS>C\Lambda,$

for sufficiently large $n$ and small $r>0$. Thanks to Lemma 2.4, one has

$\psi_{\lambda,\mu}<m_{0}\ \ \text{in
$B_{\frac{r}{x_{n}}}\left(X_{n}+\frac{X_{0}}{x_{n}}\right)$, namely,
$\psi_{n}<m_{0}$ in $B_{r}(X_{0})$},$

for sufficiently large $n$ and small $r>0$.

If $\psi_{0}=m_{0}$ in $B_{r}(X_{0})$, it follows from (3.20) that for a.e
$r>0$,

$\lim_{n\rightarrow+\infty}\frac{2}{r}\left(\fint_{B_{\frac{r}{2}}(X_{0})}(m_{0}-\psi_{n})^{2}dxdy\right)^{\frac{1}{2}}=0,$

which together with Lemma 2.5 gives that $\psi_{n}=m_{0}\ \text{in}\
B_{\frac{r}{8}}(X_{0})$ for sufficiently large $n$.

Hence, we have the convergence of the free boundary in the Hausdorff distance.

For any $R>0$ and small $\varepsilon>0$, it follows from (3.23) that there
exists a large $N_{\varepsilon,R}$, such that the free boundary
$B_{R}(0)\cap\partial\\{\psi_{n}<m_{0}\\}$ and
$B_{R}(0)\cap\partial\\{\psi_{0}<m_{0}\\}$ lie each within an
$\varepsilon$-neighborhood of one another, provided that
$n>N_{\varepsilon,R}$. Thus we can check that the free boundary
$B_{R}(0)\cap\partial\\{\psi_{n}<m_{0}\\}$ satisfies the flatness condition
(see Section 5 in [ACF4]), it follows from Theorem 6.3 in [ACF4] that

$B_{R}(0)\cap\partial\\{\psi_{n}<m_{0}\\}\rightarrow
B_{R}(0)\cap\partial\\{\psi_{0}<m_{0}\\}\ \ \text{in}\ \ C^{1,\alpha},$

which implies that

$k_{\lambda,\mu}^{\prime}(x_{n}+\tilde{x})\rightarrow\tan\theta\ \ \text{as}\
\ x_{n}\rightarrow+\infty.$

Since
$\tilde{y}=x_{n}\left(k_{\lambda,\mu}\left(x_{n}+\frac{\tilde{x}}{x_{n}}\right)-g_{0}(x_{n})\right)$
is the free boundary of $\psi_{n}$, we have

$\frac{d\tilde{y}}{d\tilde{x}}=k^{\prime}_{\lambda,\mu}\left(x_{n}+\frac{\tilde{x}}{x_{n}}\right)\rightarrow\tan\theta\
\ \text{and}\ \
x_{n}\left(k_{\lambda,\mu}\left(x_{n}+\frac{\tilde{x}}{x_{n}}\right)-g_{0}(x_{n})\right)\rightarrow\tilde{x}\tan\theta+\frac{m_{0}}{\lambda\cos\theta},$

which imply that

$k^{\prime}_{\lambda,\mu}\left(x_{n}\right)\rightarrow\tan\theta\ \
\text{and}\ \
x_{n}\left(k_{\lambda,\mu}\left(x_{n}\right)-g_{0}(x_{n})\right)\rightarrow\frac{m_{0}}{\lambda\cos\theta}.$

The compactness of $\psi_{n}$ gives that

$\nabla\psi_{n}\rightarrow\nabla\psi_{0}=\lambda(-\cos\theta,\sin\theta)\ \
\text{uniformly in any compact subset of $S$},$

where $S=\left\\{(\tilde{x},\tilde{y})\mid
0<\tilde{y}\cos\theta-\tilde{x}\sin\theta<\frac{m_{0}}{\lambda}\right\\}$, and
thus (3.18) holds.

∎

Finally, we will give the gradient estimate of $\psi_{\lambda,\mu}$ near the
initial point of $\Gamma_{\lambda,\mu}$ in the following.

######  Lemma 3.5.

Let $P=(b,k_{\lambda,L}(b))$, then

$|\nabla\psi_{\lambda,\mu}(X)|\leq C\ \ \text{ in $B_{r}(P)$},$

for some $r>0$, where $C$ is constant depending only on $\Lambda,b$ and
$\vartheta$, but not on $m_{0}$.

###### Proof.

Without loss of generality, we assume that $P=A$. For any small $\gamma>0$, it
suffices to show that

$|\nabla\psi_{\lambda,\mu}(X)|\leq C\ \ \text{in}\ \
\Omega\cap\\{\gamma<r<2\gamma\\},$

where $r=|X-A|$. Denote

$\psi_{\gamma}(X)=\frac{m_{0}-\psi_{\lambda,\mu}(A+\gamma X)}{\gamma}\ \
\text{in}\ \ D_{1}=\left\\{X\mid\frac{1}{2}<|X|<\frac{5}{2},A+\gamma
X\in\Omega\right\\},$

and

$D_{2}=\left\\{X\mid 1<|X|<2,A+\gamma X\in\Omega\right\\}.$

Then we have

$\nabla\cdot f_{\eta}\left(\frac{\nabla\psi_{\gamma}}{b+\gamma
x};\lambda\right)=0\ \text{in}~{}~{}D_{1}\cap\\{\psi_{\gamma}>0\\}.$

It follows from Lemma 2.1 that

$\psi_{\gamma}=0,\ \ \frac{1}{b+\gamma
x}\frac{\partial\psi_{\gamma}}{\partial\nu}=\lambda\ \ \text{on the free
boundary of $\psi_{\gamma}$, and
$0\leq\psi_{\gamma}\leq\frac{m_{0}}{\gamma}$}.$

Since $\partial D_{1}\cap\\{X\mid A+\gamma X\in\partial\Omega_{\mu}\\}$ is
$C^{2,\alpha}$-smooth and $\psi_{\gamma}=0$ on $\partial D_{1}\cap\\{X\mid
A+\gamma X\in\partial\Omega_{\mu}\\}$, the Harnack’s inequality is still valid
up to this part of the boundary. By using the similar arguments in the proof
of Lemma 2.3, we have

$|\nabla\psi_{\lambda,\mu}(A+\gamma X)|=|\nabla\psi_{\gamma}(X)|\leq C\ \
\text{in}\ \ D_{2},$

where $C$ is a constant depending on $\Lambda,b$ and $\vartheta$, but not on
$\frac{m_{0}}{\gamma}$. Therefore, we obtain the assertion of this lemma.

∎

###  3.3  Continuous dependence of $\psi_{\lambda,\mu}$ and
$\Gamma_{\lambda,\mu}$ with respect to $\lambda$

To obtain the continuous fit condition, we will show that the minimizer
$\psi_{\lambda,\mu}$ and the free boundary $\Gamma_{\lambda,\mu}$ are
continuous dependence with respect to the parameter
$\lambda\leq\Pi_{\lambda}-4\tilde{\varepsilon}$.

######  Lemma 3.6.

Let $\psi_{\lambda_{n},\mu}$ be a minimizer to the variational problem
$(P_{\lambda_{n},\mu})$ with the admissible set $K_{\mu}$ and
$\lambda_{n}\leq\Pi_{\lambda_{n}}-4\tilde{\varepsilon}$, then we have

$\psi_{\lambda_{n},\mu}\rightharpoonup\psi_{\lambda,\mu}\ \ \text{in}\ \
H^{1}_{loc}(\Omega_{\mu})\ \text{and}\
\psi_{\lambda_{n},\mu}\rightarrow\psi_{\lambda,\mu}\ \ \text{a.e. in}\ \
\Omega_{\mu},$

as $\lambda_{n}\rightarrow\lambda\leq\Pi_{\lambda}-4\tilde{\varepsilon}$,
where $\psi_{\lambda,\mu}$ is the minimizer to the variational problem
$(P_{\lambda,\mu})$.

###### Proof.

Denote $\psi_{n}=\psi_{\lambda_{n},\mu}$ for simplicity. By virtue of
Proposition 2.1, we have $|\nabla\psi_{n}|\leq C$ in any compact subset of
$\Omega_{\mu}$, where the constant $C$ is independent of $n$. Then there
exists a subsequence $\\{\psi_{n}\\}$ such that

$\psi_{n}\rightharpoonup\omega\ \ \text{weakly in}\ \
H_{loc}^{1}(\Omega_{\mu})\ \text{and}\ \psi_{n}\rightarrow\omega\ \text{in}\
C_{loc}^{\alpha}(\Omega_{\mu})\ \ \text{for all}\ \ 0<\alpha<1,$ (3.24)

and

$\nabla\psi_{n}\rightarrow\nabla\omega\ \ \text{weakly star in}\ \
L_{loc}^{\infty}(\Omega_{\mu}).$

Step 1. $E\cap\partial\\{\psi_{n}<m_{0}\\}\rightarrow
E\cap\partial\\{\omega<m_{0}\\}$ locally in the Hausdorff distance in
$\Omega_{\mu}$.

For any $X_{0}=(x_{0},y_{0})\notin E\cap\partial\\{\omega<m_{0}\\}$, the
continuity of $\omega$ gives that there exists a small $r>0$, such that
$B_{r}(X_{0})\cap\partial\\{\omega<m_{0}\\}=\varnothing$. We next claim that

$B_{\frac{r}{16}}(X_{0})\cap\partial\\{\psi_{n}<m_{0}\\}=\varnothing$ for
sufficiently large $n$. (3.25)

If $\omega<m_{0}$ in $B_{r}(X_{0})$, (3.24) implies that the claim (3.25) is
valid. If $\omega=m_{0}$ in $B_{r}(X_{0})$, for any small $\varepsilon>0$, it
follows from (3.24) that there exists a $N_{\varepsilon}$ such that

$\text{$|m_{0}-\psi_{n}(X)|<\varepsilon$ in $B_{\frac{r}{2}}(X_{0})$ for
$n>N_{\varepsilon}$},$

which implies that

$\frac{2}{r}\left(\fint_{B_{\frac{r}{2}}(X_{0})}(m_{0}-\psi_{n})^{2}dxdy\right)^{\frac{1}{2}}<\frac{2\varepsilon}{r}\leq\lambda
c^{*}\Lambda x_{0}\ \text{for $n>N_{\varepsilon}$}.$

Thanks to Lemma 2.5 for $\psi_{n}$, one has $\psi_{n}=m_{0}\ \text{in}\
B_{\frac{r}{16}}(X_{0})$ for sufficiently large $n$, and the claim (3.25)
holds.

On the other hand, for any $X_{0}=(x_{0},y_{0})\notin
E\cap\partial\\{\psi_{n}<m_{0}\\}$. Then
$B_{r}(X_{0})\cap\partial\\{\psi_{n}<m_{0}\\}=\varnothing$ for small $r>0$,
and we claim that

$B_{\frac{r}{2}}(X_{0})\cap\partial\\{\omega<m_{0}\\}=\varnothing.$ (3.26)

If $\psi_{n}<m_{0}$ in $B_{r}(X_{0})$ for a subsequence $\\{\psi_{n}\\}$, one
has

$\tilde{Q}_{\lambda_{n}}\psi_{n}=0\ \ \text{in}\ \ B_{r}(X_{0}),$

which implies that

$\tilde{Q}_{\lambda}\omega=0\ \text{in}\ B_{r}(X_{0}),\ \omega\leq m_{0}\
\text{in}\ B_{r}(X_{0}).$

The strong maximum principle yields that $\omega=m_{0}$ or $\omega<m_{0}$ in
$B_{r}(X_{0})$. Thus, the claim (3.26) holds.

It is easy to check that the claim (3.26) is valid, if $\psi_{n}=m_{0}$ in
$B_{r}(X_{0})$ for a subsequence $\\{\psi_{n}\\}$.

Hence, we complete the proof of the convergence of the free boundary in the
Hausdorff distance.

Step 2. $\chi_{\\{\psi_{n}<m_{0}\\}\cap
E}\rightarrow\chi_{\\{\omega<m_{0}\\}\cap E}$ locally in
$L^{1}(\Omega_{\mu})$.

In view of (3.24), we can deduce that Lemma 2.4 and Lemma 2.5 still hold for
$\omega$ by taking the limit $n\rightarrow\infty$. By using Theorem 2.8 in
[ACF4], one has

$\mathcal{L}^{2}(E\cap\partial\\{\omega<m_{0}\\}\cap B_{R})=0,\ \text{for any
$R>0$},$

where $\mathcal{L}^{2}$ is the Lebesgue measure in $\mathbb{R}^{2}$.

Applying the results in Step 1, there exists a sequence
$\\{\varepsilon_{n}\\}$ with $\varepsilon_{n}\rightarrow 0$, such that

$E\cap\\{\psi_{n}<m_{0}\\}\subset O_{\varepsilon_{n}},$

where $O_{\varepsilon_{n}}$ be an $\varepsilon_{n}$-neighborhood of
$E\cap\partial\\{\omega<m_{0}\\}$. Then we have

$\int_{\Omega_{\mu}\cap B_{R}}\left|\chi_{\\{\psi_{n}<m_{0}\\}\cap
E}-\chi_{\\{\omega<m_{0}\\}\cap E}\right|dxdy\leq\int_{\Omega_{\mu}\cap
B_{R}\cap O_{\varepsilon_{n}}}dxdy\rightarrow 0,$

for any $R>0$.

Step 3. $\nabla\psi_{n}\rightarrow\nabla\omega$ a.e. locally in
$\Omega_{\mu}$.

Let $D$ be any compact subset of $\Omega_{\mu}\cap\\{\omega<m_{0}\\}$. Then
one has

$\tilde{Q}_{\lambda_{n}}\psi_{n}=0\ \ \text{in}\ \ D\ \ \text{for sufficiently
large}\ n.$

The elliptic estimates for $\psi_{n}$ gives that

$\nabla\psi_{n}\rightarrow\nabla\omega\ \ \text{in}\ \ D.$ (3.27)

Next, we claim that

$\nabla\psi_{n}\rightarrow\nabla\omega\ \ \text{a.e. in
$\Omega_{\mu}\cap\\{\omega=m_{0}\\}$}.$ (3.28)

Since $\\{\omega=m_{0}\\}$ is $\mathcal{L}^{2}$-measurable, it follows from
Corollary 3 in [EV] that

$\lim_{r\rightarrow
0}\frac{\mathcal{L}^{2}(B_{r}(X)\cap\\{\omega=m_{0}\\})}{\mathcal{L}^{2}(B_{r}(X))}=1\
\ \text{for $\mathcal{L}^{2}$ a.e. $X\in\\{\omega=m_{0}\\}$}.$

Denote

$\mathcal{S}=\left\\{X\in\\{\omega=m_{0}\\}\mid\lim_{r\rightarrow
0}\frac{\mathcal{L}^{2}(B_{r}(X)\cap\\{\omega=m_{0}\\})}{\mathcal{L}^{2}(B_{r}(X))}=1\right\\}.$

We next show that

$\text{$m_{0}-\omega(X_{0}+X)=o(|X|)$ for any $X_{0}\in\mathcal{S}$}.$ (3.29)

In fact, if $m_{0}-\omega(Y)>kr$ for some $Y\in B_{r}(X_{0})$ with
$r\rightarrow 0$ and $k>0$. The Lipschitz continuity of $\omega$ gives that

$m_{0}-\omega(X)>\frac{k}{2}r\ \ \text{in $B_{\varepsilon kr}(Y)$ for some
small $\varepsilon>0$},$

which implies that $\\{\omega<m_{0}\\}$ has positive density at $X_{0}$, and
then it contradicts to $X_{0}\in\mathcal{S}$.

With the aid of (3.24) and (3.29), for any $\varepsilon>0$, we have

$\frac{m_{0}-\psi_{n}}{r}<\varepsilon\ \ \text{in $B_{r}(X_{0})$ for small
$r$},$

provided that $n$ is sufficiently large, that is $n>N(\varepsilon,r)$. It
follows from the non-degeneracy Lemma 2.5 that $\psi_{n}\equiv m_{0}$ in
$B_{\frac{r}{8}}(X_{0})$, which implies that $\omega\equiv m_{0}$ in
$B_{\frac{r}{10}}(X_{0})$. Then we have that the set $\mathcal{S}$ is open.
Furthermore, one has

$\psi_{n}\equiv\omega$ in any compact subset of $\mathcal{S}$, provided that
$n$ is sufficiently large.

This completes the proof of the claim (3.28).

Since $\mathcal{L}^{2}(E\cap\partial\\{\omega<m_{0}\\})=0$, it follows from
(3.27) and (3.28) that $\nabla\psi_{n}\rightarrow\nabla\omega$ a.e. in
$\mathbb{R}^{2}$.

Step 4. $\omega=\psi$, where $\psi=\psi_{\lambda,\mu}$ is the minimizer to the
truncated variational problem $(P_{\lambda,\mu})$.

First, we will show that

$J_{R}(\omega)\leq J_{R}(\phi)\ \ \text{for any $\phi\in K_{\mu}$ and
$\phi=\omega$ on $\partial\Omega_{R}$},$ (3.30)

where $J_{R}(\phi)=\int_{\Omega_{R}}G(\nabla\phi,\phi,x;\lambda)dxdy$ and
$\Omega_{R}=\Omega_{\mu}\cap B_{R}(0)$ for sufficiently large $R>0$.

For any $\phi\in K_{\mu}$ and $\phi=\omega$ on $\partial\Omega_{R}$, set

$\phi_{n}=\phi+(1-\xi_{\delta})(\psi_{n}-\omega),$

where
$\xi_{\delta}(X)=\min\left\\{\frac{\text{dist}(X,\mathbb{R}^{2}\setminus\Omega_{R})}{\delta},1\right\\}$.
Obviously, $\phi_{n}=\psi_{n}$ on $\partial\Omega_{R}$ and extend
$\phi_{n}=\psi_{n}$ outside $\Omega_{R}$, such that $\phi_{n}\in K_{\mu}$.
Then one has

$\int_{\Omega_{R}}G(\nabla\psi_{n},\psi_{n},x;\lambda_{n})dxdy\leq\int_{\Omega_{R}}G(\nabla\phi_{n},\phi_{n},x;\lambda_{n})dxdy.$

By using the convergence results in Step 2 and Step 3, taking
$n\rightarrow\infty$ in the above inequality, we have

$\begin{array}[]{rl}\int_{\Omega_{R}}G(\nabla\omega,\omega,x;\lambda)dxdy\leq&\int_{\Omega_{R}\cap\\{\xi_{\delta}=1\\}}G(\nabla\phi,\phi,x;\lambda)dxdy\\\
&+\int_{\Omega_{R}\cap\\{\xi_{\delta}<1\\}}G(\nabla\phi,\phi,x;\lambda)dxdy.\end{array}$
(3.31)

Taking $\delta\rightarrow 0$ in (3.31) yields that

$\begin{array}[]{rl}\int_{\Omega_{R}}G(\nabla\omega,\omega,x;\lambda)dxdy\leq\int_{\Omega_{R}}G(\nabla\phi,\phi,x;\lambda)dxdy.\end{array}$
(3.32)

Hence, (3.30) can be obtained by taking $R\rightarrow+\infty$ in (3.32).

It follows from (3.16) that

$\int_{\Omega_{\mu}\cap\\{x\geq
x_{0}\\}}G(\nabla\omega,\omega,x;\lambda)dxdy\leq\frac{C}{x_{0}^{3}},$

for sufficiently large $x_{0}>0$, which implies that the results (2) in Lemma
3.4 still be valid for $\omega$.

For any $\varepsilon>0$, Set $\psi_{\varepsilon}(x,y)=\psi(x,y-\varepsilon)$
and
$\Omega_{\mu}^{\varepsilon}=\\{(x,y)\mid(x,y-\varepsilon)\in\Omega_{\mu}\\}$.
Extend $\omega(x,y)=\frac{m_{0}}{x_{\mu}}x^{2}$ in $\\{(x,y)\mid 0\leq x\leq
x_{\mu},\mu\leq y\leq\mu+\varepsilon\\}$. It is easy to check that
$\min\\{\psi_{\varepsilon},\omega\\}\in K_{\mu}^{\varepsilon}$ and
$\max\\{\psi_{\varepsilon},\omega\\}\in K_{\mu}$. Therefore, one has

$J^{\varepsilon}_{\lambda,\mu}(\psi_{\varepsilon})\leq
J^{\varepsilon}_{\lambda,\mu}(\min\\{\psi_{\varepsilon},\omega\\}).$ (3.33)

Similar to the proof of Theorem 4.1 in [CD], we can check that

$J_{\lambda,\mu}(\omega)+J^{\varepsilon}_{\lambda,\mu}(\psi_{\varepsilon})=J_{\lambda,\mu}(\max\\{\psi_{\varepsilon},\omega\\})+J^{\varepsilon}_{\lambda,\mu}(\min\\{\psi_{\varepsilon},\omega\\}).$
(3.34)

With the aid of the asymptotic behaviors of $\psi$ and $\omega$ in Lemma 3.4,
we have

$\psi_{\varepsilon}(x,y)\leq\omega(x,y)\ \ \text{in}\ \ \Omega_{\mu}\setminus
B_{R}\ \ \text{for any sufficiently large $R>0$}.$

Thus $\max\\{\psi_{\varepsilon},\omega\\}=\omega$ on $\partial\Omega_{R}$, it
follows from (3.30) that

$J_{R}(\omega)\leq J_{R}(\max\\{\psi_{\varepsilon},\omega\\}).$

Taking $R\rightarrow+\infty$ in the above inequality yields that

$J_{\lambda,\mu}(\omega)\leq
J_{\lambda,\mu}(\max\\{\psi_{\varepsilon},\omega\\}),$

which together with (3.33) and (3.34) gives that

$J^{\varepsilon}_{\lambda,\mu}(\psi_{\varepsilon})=J^{\varepsilon}_{\lambda,\mu}(\min\\{\psi_{\varepsilon},\omega\\}).$

Since the minimizer $\psi_{\varepsilon}$ to the variational problem
$(P_{\lambda,\mu}^{\varepsilon})$ is unique, one has

$\psi(x,y-\varepsilon)=\psi_{\varepsilon}(x,y)=\min\\{\psi_{\varepsilon},\omega\\}\leq\omega(x,y)\
\ \text{in}\ \ \Omega_{\mu}.$ (3.35)

Similarly, we can show that

$\psi(x,y+\varepsilon)\geq\omega(x,y)\ \ \text{in}\ \ \Omega_{\mu}.$ (3.36)

Taking $\varepsilon\rightarrow 0$ in (3.35) and (3.36), we have that
$\psi=\omega$ in $\Omega_{\mu}$.

∎

Next, we will obtain the continuous dependence of the free boundary
$\Gamma_{\lambda,\mu}$ with respect to $\lambda$. We remark that even though
the ideas of the proof borrow from the one for incompressible jet as done in
Theorem 3.1 in [ACF3] and Theorem 6.1 in Chapter 3 in [FA1], due to the
difference of the governing equations and the functional, we have to overcome
several additional difficulties. Actually, the stream function satisfies the
linear elliptic equation for the incompressible flows, and here we have to
deal with a quasilinear elliptic equation.

######  Lemma 3.7.

The free boundary $y=k_{\lambda_{n},\mu}(x)$ of the minimizer
$\psi_{\lambda_{n},\mu}$ with
$\lambda_{n}\leq\Pi_{\lambda_{n}}-4\tilde{\varepsilon}$ satisfies that

$k_{\lambda_{n},\mu}(x)\rightarrow k_{\lambda,\mu}(x)\ \ \text{for any
$x\in(b,+\infty)$},$

as $\lambda_{n}\rightarrow\lambda$, where $y=k_{\lambda,\mu}(x)$ is the free
boundary of the minimizer $\psi_{\lambda,\mu}$.

###### Proof.

For any fixed $x\in(b,+\infty)$, the convergence of $k_{\lambda_{n},\mu}(x)$
can be obtained by using the Step 1 in the proof of Lemma 3.6.

Next, we will consider the initial point of the free boundary and show that
$k_{\lambda_{n},\mu}(b)\rightarrow k_{\lambda,\mu}(b)$ as
$\lambda_{n}\rightarrow\lambda$. Suppose not, then there exists a subsequence
$\\{k_{\lambda_{n},\mu}(b)\\}$, such that $k_{\lambda_{n},\mu}(b)\rightarrow
k_{\lambda,\mu}(b)+\beta$ and $\beta\neq 0$. We will show that it is
impossible case by case based on the sign of $\beta$.

Case 1. $\beta<0$. The monotonicity of $\psi_{\lambda,\mu}(x,y)$ with respect
to $y$ gives that $k_{\lambda,\mu}(b)+\beta\geq 1$. In fact, if
$k_{\lambda,\mu}(b)+\beta<1$, it follows from Lemma 3.6 that

$\psi_{\lambda,\mu}(b,y)=m_{0}\ \ \ \text{for}\ \ \
k_{\lambda,\mu}(b)+\beta\leq y\leq\min\\{k_{\lambda,\mu}(b),1\\},$

which contradicts to the fact $\psi_{\lambda,\mu}<m_{0}$ for
$y<\min\\{k_{\lambda,\mu}(b),1\\}$, due to the fact that
$\psi_{\lambda,\mu}(x,y)$ is monotone increasing with respect to $y$.

Denote

$T_{\beta}=\left\\{(b,y)\mid
k_{\lambda,\mu}(b)+\frac{3\beta}{4}<y<k_{\lambda,\mu}(b)+\frac{\beta}{4}\right\\}.$

Next, we claim that

$\frac{\partial\psi_{\lambda,\mu}(b+0,y)}{\partial x}=-\lambda\ \ \text{on}\ \
T_{\beta}.$ (3.37)

For small $\varepsilon>0$, set

$U_{\varepsilon}=\left\\{(x,y)\mid
b-\varepsilon<x<b+\varepsilon,k_{\lambda,\mu}(b)+\frac{3\beta}{4}<y<k_{\lambda,\mu}(b)+\frac{\beta}{4}\right\\}.$

It is easy to check that

$T_{\beta}=U_{\varepsilon}\cap\partial\\{\psi_{\lambda,\mu}<m_{0}\\}.$

Set $\phi_{n}=m_{0}-\psi_{\lambda_{n},\mu}$ and
$\phi=m_{0}-\psi_{\lambda,\mu}$, then one has

$\left\\{\begin{array}[]{ll}&\tilde{Q}_{\lambda_{n}}\phi_{n}=0\ \ \text{in}\
U_{\varepsilon}\cap\\{\phi_{n}>0\\},\\\
&\frac{1}{x}\frac{\partial\phi_{n}}{\partial\nu_{n}}=\lambda_{n}\ \text{on}\
U_{\varepsilon}\cap\partial\\{\phi_{n}>0\\},\end{array}\right.$

where $\nu_{n}$ is the inner normal vector to
$U_{\varepsilon}\cap\partial\\{\phi_{n}>0\\}$. Now, in order to show the claim
(3.37), it suffices to check that $\phi$ satisfies the following boundary
value problem,

$\left\\{\begin{array}[]{ll}&\tilde{Q}_{\lambda}\phi=0\ \ \text{in}\
U_{\varepsilon}\cap\\{\phi>0\\},\\\
&\frac{1}{x}\frac{\partial\phi}{\partial\nu}=\lambda\ \text{on}\
U_{\varepsilon}\cap\partial\\{\phi>0\\},\end{array}\right.$ (3.38)

where $\nu=(1,0)$ is the inner normal vector to $\partial\\{\phi>0\\}$ at
$T_{\beta}$.

We divide the proof into two steps to show that (3.38) holds.

Step 1. In this step, we will verify that

$\frac{1}{x}\frac{\partial\phi}{\partial\nu}\geq\lambda\ \ \text{on}\ \
U_{\varepsilon}\cap\partial\\{\phi>0\\}.$ (3.39)

It follows from (3.24) that for $\alpha\in(0,1)$,

$\phi_{n}\rightarrow\phi\ \ \text{uniformly in
$C^{\alpha}(U_{\varepsilon})$},$

and $U_{\varepsilon}\cap\\{\phi_{n}>0\\}\rightarrow
U_{\varepsilon}\cap\\{\phi>0\\}$ in the Hausdorff metric space, and

$\nabla\phi_{n}\rightarrow\nabla\phi\ \ \text{weakly in
$L^{2}(U_{\varepsilon})$},$

as $n\rightarrow+\infty$. Moreover, by virtue of the bounded gradient Lemma
2.3 and the Step 4 in the proof of Lemma 3.6, one has

$|\nabla\phi_{n}|\leq C\Lambda x\ \ \text{in $U_{\varepsilon}$ and}\ \
\nabla\phi_{n}\rightarrow\nabla\phi\ \ \text{a.e. in $U_{\varepsilon}$, as
$n\rightarrow\infty$},$ (3.40)

where the constant $C$ is independent of $n$.

Since $F_{1}(t;\lambda)=\frac{1}{\tilde{\rho}(t;\lambda^{2})}$ is
$C^{1,\alpha}$, it follows from (3.40) that

$\begin{array}[]{rl}&\lim_{n\rightarrow+\infty}F_{1}(\lambda_{n}^{2};\lambda_{n})\lambda_{n}\int_{U_{\varepsilon}\cap\partial\\{\phi_{n}>0\\}}\xi
dS\\\
=&-\lim_{n\rightarrow+\infty}\int_{U_{\varepsilon}\cap\\{\phi_{n}>0\\}}\frac{1}{x}F_{1}\left(\left|\frac{\nabla\phi_{n}}{x}\right|^{2};\lambda_{n}\right)\nabla\phi_{n}\cdot\nabla\xi
dxdy\\\
=&-\int_{U_{\varepsilon}\cap\\{\phi>0\\}}\frac{1}{x}F_{1}\left(\left|\frac{\nabla\phi}{x}\right|^{2};\lambda\right)\nabla\phi\cdot\nabla\xi
dxdy\\\
=&\int_{U_{\varepsilon}\cap\partial\\{\phi>0\\}}\frac{1}{x}F_{1}\left(\left|\frac{\nabla\phi}{x}\right|^{2};\lambda\right)\frac{\partial\phi}{\partial\nu}\xi
dS,\end{array}$ (3.41)

for any non-negative $\xi\in C_{0}^{\infty}(U_{\varepsilon})$.

On other hand, it follows from (3.6) in Chapter 3 in [FA1] that

$\begin{array}[]{rl}\int_{U_{\varepsilon}\cap\partial\\{\phi>0\\}}\xi
dS\leq\liminf_{n\rightarrow+\infty}\int_{U_{\varepsilon}\cap\partial\\{\phi_{n}>0\\}}\xi
dS.\end{array}$ (3.42)

In view of (3.41) and (3.42), one has

$\begin{array}[]{rl}F_{1}(\lambda^{2};\lambda)\lambda\int_{U_{\varepsilon}\cap\partial\\{\phi>0\\}}\xi
dS\leq\int_{U_{\varepsilon}\cap\partial\\{\phi>0\\}}\frac{1}{x}F_{1}\left(\left|\frac{\nabla\phi}{x}\right|^{2};\lambda\right)\frac{\partial\phi}{\partial\nu}\xi
dS,\end{array}$ (3.43)

for any non-negative $\xi\in C_{0}^{\infty}(U)$. Since $\phi$ is
$C^{1,\alpha}$ up to the boundary $\partial\\{\phi>0\\}$, it follows from
(3.43) that

$F_{1}\left(\left|\frac{\nabla\phi}{x}\right|^{2};\lambda\right)\left|\frac{\nabla\phi}{x}\right|=F_{1}\left(\left|\frac{\nabla\phi}{x}\right|^{2};\lambda\right)\frac{1}{x}\frac{\partial\phi}{\partial\nu}\geq
F_{1}\left(\lambda^{2};\lambda\right)\lambda\ \ \text{on}\ \
U_{\varepsilon}\cap\partial\\{\phi>0\\}.$ (3.44)

Since $F_{1}(t^{2};\lambda)t$ is increasing with respect to $t$, (3.44) gives
that (3.39) holds.

Step 2.  In this step, we will check that

$\frac{1}{x}\frac{\partial\phi}{\partial\nu}\leq\lambda\ \ \text{on}\ \
U_{\varepsilon}\cap\partial\\{\phi>0\\}.$ (3.45)

By virtue of the non-oscillation Lemma (3.2) and the flatness of the free
boundary in Section 5 in [ACF4], we have that

the free boundary $\Gamma_{\lambda_{n},\mu}$ is a $y$-graph in
$U_{\varepsilon}$ for sufficiently large $n$,

where we denote $\Gamma_{\lambda_{n},\mu}:x=f_{\lambda_{n},\mu}(y)$ in the
region $U_{\varepsilon}$. It follows from the result in Theorem 6.3 and Remark
6.4 in [ACF4] that

$\left|f_{\lambda_{n},\mu}^{(j)}(y)\right|\leq C\ \ \text{for}\ \
k_{\lambda,\mu}(b)+\frac{3\beta}{4}<y<k_{\lambda,\mu}(b)+\frac{\beta}{4},\ \
j=1,2.$

Then we have

$U_{\varepsilon}\cap\partial\\{\phi_{n}>0\\}\rightarrow
U_{\varepsilon}\cap\partial\\{\phi>0\\}$ in $C^{1,\alpha}$ for some
$\alpha\in(0,1)$.

For any fixed $X_{0}=(x_{0},y_{0})\in
U_{\varepsilon}\cap\partial\\{\phi>0\\}$, it follows from Lemma 3.6 that there
exists a sequence $X_{n}\in U_{\varepsilon}$ with $\phi_{n}(X_{n})=0$, such
that $X_{n}\rightarrow X_{0}$ as $n\rightarrow+\infty$. Take a small $r>0$ and
domain $E_{n}\subset B_{r}(X_{0})\subset U_{\varepsilon}$, such that

$B_{r}(X_{0})\cap\partial E_{n}\rightarrow
B_{r}(X_{0})\cap\partial\\{\phi>0\\}\ \text{in}\ C^{1,\gamma},\ \phi_{n}>0\
\text{in}\ E_{n}\ \text{and}\ X_{n}\notin E_{n},$ (3.46)

for $0<\gamma<\alpha$. We can take a sequence $\\{\varepsilon_{n}\\}$ with
$\varepsilon_{n}\downarrow 0$, such that

$E_{n}=B_{r}(X_{0})\cap\\{x>b+\varepsilon_{n}\\}.$

Define a function $f_{\delta,n}(y)$ as follows

$f_{\delta,n}(y)=b+\varepsilon_{n}-\delta\eta\left(\frac{2(y-y_{0})}{r}\right),\
\ \delta>0,$

where

$\eta(y)=\left\\{\begin{array}[]{ll}e^{-\frac{y^{2}}{1-y^{2}}}&\text{for}\ \
|y|<1,\\\ 0&\text{for}\ \ |y|\geq 1.\end{array}\right.$

Denote the domain $E_{\delta,n}=B_{r}(X_{0})\cap\\{x>f_{\delta,n}(y)\\}$. It
is easy to check that $E_{0,n}=E_{n}$. By virtue of the definitions of $E_{n}$
and $E_{\delta,n}$, there exists the largest number $\delta=\delta_{n}$, such
that $\phi_{n}>0$ in $E_{\delta_{n},n}$, and $B_{r}(X_{0})\cap\partial
E_{\delta_{n},n}$ contains a point of the free boundary of $\phi_{n}$, which
is denoted as $\tilde{X}_{n}=(\tilde{x}_{n},\tilde{y}_{n})$ with
$\tilde{x}_{n}=f_{\delta_{n},n}(\tilde{y}_{n})$. Furthermore,

$\delta_{n}\rightarrow 0\ \ \text{as}\ \ n\rightarrow+\infty.$

Let $\omega_{n}$ be the solution of the following Dirichlet problem

$\left\\{\begin{array}[]{ll}&\tilde{Q}_{\lambda_{n}}\omega_{n}=0\ \ \text{in}\
E_{\delta_{n},n},\\\ &\omega_{n}=0\ \text{on}\ \partial E_{\delta_{n},n}\cap
B_{\frac{r}{2}}(X_{0}),\ \omega_{n}=\zeta\phi_{n}\ \text{on}\ \partial
E_{\delta_{n},n}\cap(B_{r}(X_{0})\setminus B_{\frac{r}{2}}(X_{0})),\\\
&\omega_{n}=\phi_{n}\ \text{on}\ \partial E_{\delta_{n},n}\cap\partial
B_{r}(X_{0}),\end{array}\right.$

where
$\zeta(X)=\min\left\\{\max\left\\{\frac{2|X-X_{0}|-r}{r},0\right\\},1\right\\}$.
Since $\phi_{n}$ satisfies the quasilinear equation
$\tilde{Q}_{\lambda_{n}}\phi_{n}=0$ in $E_{\delta_{n},n}$ and
$\omega_{n}\leq\phi_{n}$ on $\partial E_{\delta_{n},n}$, the maximum principle
implies that $\omega_{n}\leq\phi_{n}$ in $E_{\delta_{n},n}$. Hence, one has

$\lambda_{n}=\frac{1}{\tilde{x}_{n}}\frac{\partial\phi_{n}(\tilde{X}_{n})}{\partial\nu_{n}}\geq\frac{1}{\tilde{x}_{n}}\frac{\partial\omega_{n}(\tilde{X}_{n})}{\partial\nu_{n}},$
(3.47)

where $\nu_{n}$ is the inner normal vector to $\partial E_{\delta_{n},n}$ at
$\tilde{X}_{n}$.

It follows from the fact (3.46) that

$f_{\delta_{n},n}(y)\rightarrow b\ \ \text{in}\ \ C^{1,\gamma}.$

Thanks to the standard estimates of the solutions of the elliptic equation of
second order, we conclude that $\omega_{n}$ in $E_{\delta_{n},n}\cap
B_{\frac{r}{2}}(X_{0})$ converges to $\phi$ in $\\{\phi>0\\}\cap
B_{\frac{r}{2}}(X_{0})$ in $C^{1,\gamma}$-sense. Suppose that

$\tilde{X}_{n}\rightarrow\tilde{X}=(\tilde{x},\tilde{y})\in\partial\\{\phi>0\\}\cap
B_{\frac{r}{2}}(X_{0}).$

This together with (3.47) gives that

$\frac{1}{\tilde{x}_{n}}\frac{\partial\omega_{n}(\tilde{X}_{n})}{\partial\nu}\rightarrow\frac{1}{\tilde{x}}\frac{\partial\phi(\tilde{X})}{\partial\nu}\
\ \text{and}\ \
\frac{1}{\tilde{x}}\frac{\partial\phi(\tilde{X})}{\partial\nu}\leq\lambda.$

Taking $r\rightarrow 0$, we have $\tilde{X}\rightarrow X_{0}$ and

$\lambda\geq\frac{1}{\tilde{x}}\frac{\partial\phi(\tilde{X})}{\partial\nu}\rightarrow\frac{1}{x_{0}}\frac{\partial\phi(X_{0})}{\partial\nu},\
\ X_{0}\in U_{\varepsilon}\cap\partial\\{\phi>0\\},$

which gives the inequality (3.45) holds.

Therefore, the claim (3.38) follows from (3.39) and (3.45).

With the aid of the claim (3.37), $\psi_{\lambda,\mu}$ satisfies

$\tilde{Q}_{\lambda}\psi_{\lambda,\mu}=0\ \ \text{in}\ E_{\varepsilon},\ \
\frac{1}{x}\frac{\partial\psi_{\lambda,\mu}}{\partial x}=-\lambda\ \
\text{and}\ \ \psi_{\lambda,\mu}=m_{0}\ \ \text{on}\ \partial
E_{\varepsilon}\cap\\{x=b\\},$

for small $\varepsilon>0$, where $E_{\varepsilon}=\left\\{(x,y)\mid
b<x<b+\varepsilon,k_{\lambda,\mu}(b)+\frac{3\beta}{4}<y<k_{\lambda,\mu}(b)+\frac{\beta}{4}\right\\}$.

It follows from the Cauchy-Kovalevskaya theorem that

$\psi_{\lambda,\mu}=-\lambda(x^{2}-b^{2})+m_{0}\ \text{in}\ \left\\{(x,y)\mid
b<x<b+\varepsilon,g_{0}(b)<y<+\infty\right\\}\cap\Omega_{\mu}.$

This contradicts to the fact $\psi_{\lambda,\mu}=0$ on $N_{0}$.

Case 2. $\beta>0$ and $k_{\lambda,\mu}(b)<1$. By using the similar arguments
in Case 1, we can conclude that

$\frac{1}{b}\frac{\partial\psi_{\lambda,\mu}(b-0,y)}{\partial x}=\lambda\ \
\text{if}\ \
k_{\lambda,\mu}(b)+\frac{\beta}{4}<y<\min\left\\{k_{\lambda,\mu}(b)+\frac{3\beta}{4},1\right\\}.$

We can obtain a contradiction by using Cauchy-Kovalevskaya theorem as in Case
1 again.

Case 3. $\beta>0$ and $k_{\lambda,\mu}(b)\geq 1$. It follows from the
arguments in Lemma 2.2 that

$\frac{1}{b}\frac{\partial\psi_{\lambda_{n},\mu}(b+0,y)}{\partial
x}\leq-\lambda_{n}\ \ \text{on}\ \
\left\\{x=b,k_{\lambda,\mu}(b)+\frac{\beta}{4}<y<k_{\lambda,\mu}(b)+\frac{3\beta}{4}\right\\},$

for sufficiently large $n$. Let $D_{n}$ be bounded by $x=b$,
$y=k_{\lambda_{n},\mu}(x)$, $y=k_{\lambda,\mu}(b)+\frac{\beta}{4}$ and
$y=k_{\lambda,\mu}(b)+\frac{3\beta}{4}$. Furthermore, we have

$x_{n}=\min\left\\{x\mid
k_{\lambda_{n},\mu}(x)=k_{\lambda,\mu}(b)+\frac{3\beta}{4}\right\\}\rightarrow
b\ \ \text{as}\ \ n\rightarrow+\infty.$

Thanks to the non-oscillation Lemma 4.4 in [ACF5] for $\psi_{\lambda_{n},\mu}$
in $D_{n}$, one has

$\frac{\beta}{2}\leq C(x_{n}-b),\ \ \text{the constant $C$ is independent of
$n$},$

which leads a contradiction for sufficiently large $n$.

∎

###  3.4 The continuous fit condition of the free boundary
$\Gamma_{\lambda,\mu}$

In the following, we will consider the continuous fit condition of the free
boundary at $A$.

######  Proposition 3.8.

For any $\mu>1$ and small $m_{0}>0$, there exists a
$\lambda_{\mu}\leq\Pi_{\lambda_{\mu}}-4\tilde{\varepsilon}$, such that the
free boundary $\Gamma_{\lambda_{\mu},\mu}$ satisfies the continuous fit
condition at $A$, namely, $k_{\lambda_{\mu},\mu}(b)=1$. Moreover,

$\lambda_{\mu}\leq C_{0}m_{0},\ \ \text{the constant $C_{0}$ is independent of
$\mu$ and $m_{0}$}.$

###### Proof.

Step 1. In this step, we will show that

for $m_{0}>0$, if $\lambda>0$ is sufficiently small, then
$k_{\lambda,\mu}(b)>1$.

Suppose that there exists a small $\lambda_{0}$, such that
$k_{\lambda_{0},\mu}(b)\leq 1$. Let $S$ be a ring centered at
$P=(b,k_{\lambda_{0},\mu}(b))$ with some suitable radius $R_{1}$ and $R_{2}$
which are independent of $m_{0}$ and $R_{1}<R_{2}$, such that
${\Gamma}_{\lambda_{0},\mu}\cap S\cap\\{x>b\\}$ and
$N_{0}\cap\bar{S}\cap\\{x>b\\}$ are nonempty.

It follows from Lemma 2.3 that there exists a constant $C_{0}$ depending on
$N,N_{0},\vartheta$ (independent of $m_{0}$), such that

$\left|D\psi_{\lambda_{0},\mu}\right|\leq
C_{0}\Lambda(\lambda_{0})~{}~{}\text{in}~{}~{}~{}S\cap\\{x>b\\},$

where
$\Lambda(\lambda_{0})=\sqrt{2F_{1}(\lambda_{0}^{2};\lambda_{0})\lambda_{0}^{2}-F(\lambda_{0}^{2};\lambda_{0})}$.

Choosing $X_{1}\in{\Gamma}_{\lambda_{0},\mu}\cap\bar{S}\cap\\{x>b\\}$,
$X_{2}\in N_{0}\cap\bar{S}\cap\\{x>b\\}$ with $|X_{1}-P|=|X_{2}-P|$, shows
that

$m_{0}=\int_{\gamma}\left|\frac{\partial\psi_{\lambda_{0}}}{\partial
s}\right|dS\leq|D\psi_{\lambda,\mu}|2\pi R_{2}\leq C\Lambda(\lambda^{2}_{0}),$
(3.48)

where $\gamma\subset S$ is disc curve which connects $X_{1}$ and $X_{2}$, $s$
is the unit tangent vector of $\gamma$. On another hand, it follows from
(2.18) that

$\Lambda(\lambda_{0})\leq\lambda_{0}\sqrt{\frac{1}{\rho_{0}}\left(2-\left(\frac{2}{\gamma+1}\right)^{\frac{1}{\gamma-1}}\right)},$

which together with (3.48) gives that

$0<m_{0}\leq C\lambda_{0}.$

This is impossible for sufficiently small $\lambda_{0}$.

Step 2. For $\lambda=\frac{\lambda_{cr}}{2}$, we will show that

if $m_{0}>0$ is sufficiently small, then $k_{\lambda,\mu}(b)<1$.

Suppose that $k_{\lambda,\mu}(b)\geq 1$ for some small $m_{0}>0$, then there
exists a disc $B_{R}(X_{0})\subset\Omega_{\mu}$ ($R$ is fixed), such that
$X_{0}\in E$ and
$B_{\frac{R}{8}}(X_{0})\cap\Gamma_{\lambda,\mu}\neq\varnothing$. According to
the non-degeneracy Lemma 2.5, we have

$\frac{m_{0}}{R}\geq\frac{1}{R}\left(\fint_{B_{R}(X_{0})}(m_{0}-\psi_{\lambda,\mu})^{2}dxdy\right)^{\frac{1}{2}}\geq
c\Lambda(\lambda^{2}),$

which together with (2.19) gives that

$m_{0}\geq cR\Lambda(\lambda^{2})\geq C\lambda=\frac{C\lambda_{cr}}{2}>0.$

This leads a contradiction for sufficiently small $m_{0}$.

Step 3. In this step, we will show that there exists a
$\lambda_{\mu}\leq\Pi_{\lambda_{\mu}}-4\tilde{\varepsilon}$, such that
$k_{\lambda_{\mu},\mu}(b)=1$.

For any small $m_{0}>0$, with the aid of the results in Step 1, we can define
a set

$\Sigma_{\mu}=\\{\lambda\mid k_{\lambda,\mu}(b)>1\\}.$

Define

$\lambda_{\mu}=\sup_{\lambda\in\Sigma_{\mu}}\lambda.$ (3.49)

The result in Step 2 gives that

$\lambda_{\mu}\leq C_{0}m_{0},$ (3.50)

where $C_{0}$ is a constant depending on $N,N_{0}$ and $\vartheta$,
independent of $m_{0}$ and $\mu$. It is easy to check that there exists a
$C_{0}>0$ (independent of $\mu$), such that

$\lambda_{\mu}\leq C_{0}m_{0}\leq\Pi_{\lambda_{\mu}}-4\tilde{\varepsilon},$

for sufficiently small $m_{0}$. The continuous dependence of
$k_{\lambda,\mu}(b)$ with respect to $\lambda$ gives that

$k_{\lambda_{\mu},\mu}(b)=1.$

If not, the definition of $\lambda_{\mu}$ implies that
$k_{\lambda_{\mu},\mu}(b)>1$. By using the continuous dependence of
$k_{\lambda,\mu}(b)$ with respect to $\lambda$ in Lemma 3.7, there exists a
$\lambda\in(\lambda_{\mu},\Pi_{\lambda}-3\tilde{\varepsilon})$, such that

$\text{$\lambda-\lambda_{\mu}$ is small and $k_{\lambda,\mu}(b)>1$}.$

Therefore, $\lambda\in\Sigma_{\mu}$, which contradicts to the definition of
$\lambda_{\mu}$ in (3.49).

∎

##  4 The existence of subsonic solution to the impinging jet flow problem

To establish the existence of subsonic solution to the impinging jet flow
problem, we will take limit $\mu\rightarrow\infty$ to the solution
$\psi_{\lambda_{\mu},\mu}$ of the truncated variational problem
$(P_{\lambda_{\mu},\mu})$ and show the limit $\psi_{\lambda}$ is indeed a
solution to the following variational problem $(P_{\lambda})$ stated as
follows in the whole domain $\Omega$.

The variational problem $(P_{\lambda})$: For any bounded domain
$D\subset\Omega$, find a $\psi_{\lambda}\in K$ such that

$J_{D}(\psi_{\lambda})\leq J_{D}(\psi),$

for any $\psi\in K_{0}$ with $\psi=\psi_{\lambda}$ on $\partial D$, where
$J_{D}(\psi)=\int_{D}G(\nabla\psi,\psi,x;\lambda)dxdy$ and

$K_{0}=\left\\{\psi\in
K\mid\psi\leq\min\left\\{\frac{m_{0}}{a^{2}}x^{2},m_{0}\right\\}\ \ \text{a.e.
in}\ \ \Omega_{\mu}\right\\}.$

######  Theorem 4.1.

If $m_{0}>0$ is sufficiently small, there exist a
$\lambda\leq\Pi_{\lambda}-4\tilde{\varepsilon}$ and a subsonic solution
$(\psi_{\lambda},\Gamma_{\lambda})$ to the compressible impinging jet flow.

###### Proof.

By virtue of the uniform gradient estimate
$|\nabla\psi_{\lambda_{\mu_{n}},\mu_{n}}|\leq C$ in any compact subset of
$\mathbb{R}^{2}$, it follows from the similar arguments in the proof of Lemma
3.6 that there exist a subsequence $\\{\psi_{\lambda_{\mu_{n}},\mu_{n}}\\}$
and $\\{\lambda_{\mu_{n}}\\}$ with
$\lambda_{\mu_{n}}\leq\Pi_{\lambda_{\mu_{n}}}-4\tilde{\varepsilon}$ and
$k_{\lambda_{\mu_{n}},\mu_{n}}(b)=1$, such that

$\lambda_{\mu_{n}}\rightarrow\lambda\leq\Pi_{\lambda}-4\tilde{\varepsilon},$

and

$\psi_{\lambda_{\mu_{n}},\mu_{n}}\rightarrow\psi_{\lambda}\ \text{weakly~{}
in}~{}~{}H_{loc}^{1}(\mathbb{R}^{2})~{}\text{and uniformly in any compact
subset of $\mathbb{R}^{2}$},$

as $n\rightarrow\infty$. Moreover, it follows from (3.50) that

$\lambda\leq C_{0}m_{0},\ \ \text{the constant $C_{0}$ depends on
$N,N_{0},\vartheta$, not on $m_{0}$}.$ (4.1)

Step 1. $\psi_{\lambda}$ is a subsonic solution of the free boundary problem
(2.16).

By using the similar arguments in Step 4 in the proof of Lemma 3.6, we can
check that $\psi_{\lambda}$ is a minimizer to the variational problem
$(P_{\lambda})$. Moreover, the monotonicity of
$\psi_{\lambda_{\mu_{n}},\mu_{n}}(x,y)$ in Lemma 3.1 gives that
$\psi_{\lambda}(x,y)$ is monotonic increasing with respect to $y$, which
implies that the free boundary of $\psi_{\lambda}$ is $x$-graph. Applying the
similar arguments in the proof of Lemma 3.3, there exists a continuous
function $k_{\lambda}(x)$, such that the free boundary $\Gamma_{\lambda}$ of
the minimizer $\psi_{\lambda}$ can be described as

$\Gamma_{\lambda}=E\cap\partial\\{\psi_{\lambda}<m_{0}\\}:y=k_{\lambda}(x)\ \
\text{for}\ \ x\in(b,+\infty).$

Furthermore, it follows from the similar arguments in Lemma 3.7 that

$k_{\lambda}(x)=\lim_{n\rightarrow\infty}k_{\lambda_{\mu_{n}},\mu_{n}}(x)\ \
\text{for any $x\in[b,+\infty)$}.$

In particular, one has

$k_{\lambda}(b)=1,$

which is the continuous fit condition to the axially symmetric compressible
subsonic impinging jet flow.

In Theorem 6.1 in [ACF4] and Section 3.11 in [FA1], Alt, Caffarelli and
Friedman obtained that the continuous fit condition implies the smooth fit
condition at the detachment point $A$, namely, $N\cup\Gamma_{\lambda}$ is
$C^{1}$ at $A$. Since $\psi_{\lambda}$ is a minimizer to the variational
problem $(P_{\lambda})$, it follows from Theorem 6.3 in [ACF4] that the free
boundary $\Gamma_{\lambda}$ is $C^{1,\alpha}$, and thus $\psi_{\lambda}$ is
$C^{1,\alpha}$-smooth up to the free boundary $\Gamma_{\lambda}$. In view of
$\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$, the subsonic cut-off can be
removed near $\Gamma_{\lambda}$. Then $F(t;\lambda)$ is analytic with respect
to $t$, near the free boundary $\Gamma_{\lambda}$. Recalling Remark 6.4 in
[ACF4], we can conclude that the free boundary $\Gamma_{\lambda}$ is analytic.
By using the similar arguments in the proof of Theorem 9.1 in [CDW2], we can
conclude that

$\frac{1}{x}\frac{\partial\psi_{\lambda}}{\partial\nu}=\left|\frac{\nabla\psi_{\lambda}}{x}\right|=\lambda$
on $\Gamma_{\lambda}$,

where $\nu$ is outer normal vector to $\Gamma_{\lambda}$. Utilizing Lemma 6.4
in [ACF5], one has

$\text{$\nabla\psi_{\lambda}$ is uniformly continuous in a
$\\{\psi_{\lambda}<m_{0}\\}$-neighborhood of $A$},$

and

$\left|\frac{\nabla\psi_{\lambda}}{x}\right|=\lambda$ at $A$.

Since $\psi_{\lambda}$ is a minimizer to the variational problem
$(P_{\lambda})$, applying the results (2) in Lemma 2.2, one has

$\tilde{Q}_{\lambda}\psi_{\lambda}=0\ \ \text{in}\ \
\Omega\cap\\{\psi_{\lambda}<m_{0}\\}.$

Hence, the minimizer $\psi_{\lambda}$ is a solution of the truncated free
boundary problem (2.16).

Step 2. The asymptotic behavior of $\psi_{\lambda}$ will be obtained. For the
asymptotic behavior of $\psi_{\lambda}$ in upstream, we can use the blow-up
arguments in the proof of Lemma 5 in [XX2], and obtain that

$\nabla\psi_{\lambda}(x,y)\rightarrow(-v_{in},0),~{}~{}\tilde{\rho}\rightarrow\rho_{in}\
\
\text{and}~{}~{}\nabla(\partial_{x}\psi_{\lambda},\partial_{y}\psi_{\lambda},\tilde{\rho})\rightarrow
0,$

uniformly in any compact subset of $(0,a)$, as $y\rightarrow+\infty$, where
$v_{in}=-\frac{2m_{0}}{\rho_{0}a^{2}}$ and
$\rho_{in}=\tilde{\rho}\left(\frac{4m_{0}^{2}}{a^{4}};\lambda^{2}\right)$.

By virtue of (3.16), there exists a constant $C>0$, such that

$\int_{\Omega\cap\\{x>x_{0}\\}}x\left|\frac{\nabla\psi}{x}-\lambda
e\chi_{\\{\psi<m_{0}\\}}\right|^{2}dxdy\leq\frac{C}{x_{0}^{3}},$ (4.2)

for sufficiently large $x_{0}>b$, where the constant $C$ is independent of
$x_{0}$.

With the aid of (4.2), by using the similar arguments in the proof of Lemma
3.4, we can show that

$x(k_{\lambda}(x)-g_{0}(x))\rightarrow\frac{m_{0}}{\lambda\cos\theta}~{}~{}\text{as}~{}~{}x\rightarrow+\infty,$
(4.3)

and

$\frac{\nabla\psi_{\lambda}(x,y)}{x}\rightarrow(-\lambda\sin\theta,\lambda\cos\theta)~{}~{}\text{for
$(x,y)\in\Omega_{\mu}\cap\\{\psi_{\lambda}<m_{0}\\}$, as
$x\rightarrow+\infty$.}$ (4.4)

Step 3. In this step, we will remove the subsonic cut-off for
$\tilde{\rho}_{\lambda}$ in (2.14).

For $\lambda\leq\Pi_{\lambda}-3\tilde{\varepsilon}$, it is easy to check that
$\tilde{\rho}\left(\left|\frac{\nabla\psi_{\lambda}}{x}\right|^{2};\lambda^{2}\right)$
is a monotonic decreasing function of
$\left|\frac{\nabla\psi_{\lambda}}{x}\right|\in(0,\Pi_{\lambda})$. Moreover,

$\frac{|\nabla\psi_{\lambda}|}{x}$ takes the maximum at $X_{0}$ if and only if
$q$ takes the maximum at $X_{0}$, (4.5)

where $q=\sqrt{u^{2}+v^{2}}$ is the speed of the fluid.

By using the similar arguments in Page 114 in [ACF5], one has

$\mathcal{Q}q^{2}=D_{i}\left(e^{\gamma
q^{2}}a_{ij}(X;\lambda)D_{j}q^{2}\right)\geq 0\ \ \text{in the fluid region
$\Omega_{0}$,}$

for some $\gamma>0$, where

$a_{ij}(X;\lambda)=\tilde{\rho}(|\nabla\varphi(X)|^{2};\lambda^{2})\delta_{ij}+2\rho_{1}(|\nabla\varphi(X)|^{2};\lambda^{2})D_{i}\varphi(X)D_{j}\varphi(X)$

and the potential function $\varphi$ satisfies that $\nabla\varphi=(u,v)$. In
view of the maximum principle for $q^{2}$ in $\Omega_{0}$, we have that
$q^{2}$ cannot take its maximum in $\Omega_{0}$. Since the flow is assumed to
be symmetric with respect to the symmetric axis $I$, thus $I$ can be regarded
as the interior of the fluid field by the even extension of $\varphi$. Thus we
conclude that the speed $q$ cannot take its maximum at the symmetric axis $I$.
By virtue of (4.5), $\frac{|\nabla\psi_{\lambda}(X)|}{x}$ takes its maximal
value at $N\cup\Gamma_{\lambda}\cup N_{0}$ or in the far field. Next, we
consider the following three cases for the maximal momentum
$\frac{|\nabla\psi_{\lambda}(X)|}{x}$.

Case 1. $\frac{|\nabla\psi_{\lambda}(X)|}{x}$ takes its maximum in the far
field or on the free boundary $\Gamma_{\lambda}$. By virtue of (4.1), it
follows from the asymptotic behavior and the free boundary condition that

$\sup_{X\in\bar{\Omega}_{0}}\frac{|\nabla\psi_{\lambda}(X)|}{x}=\max\left\\{\frac{m_{0}}{a},\lambda\right\\}\leq
C_{0}m_{0},$ (4.6)

where $C_{0}$ is a constant depending only on $N,N_{0}$ and $\vartheta$.

Case 2. $\frac{|\nabla\psi_{\lambda}(X)|}{x}$ takes its maximum on walls
$N_{0}\cap\left\\{x\leq\frac{a+b}{2}\right\\}$ or on
$N\cap\left\\{x\leq\frac{a+b}{2}\right\\}$. By using the similar arguments in
Section 3 in [XX2], we have

$\sup_{X\in\bar{\Omega}_{0}}\frac{|\nabla\psi_{\lambda}(X)|}{x}\leq
C_{0}m_{0},$ (4.7)

where $C_{0}$ is a constant depending only on $N,N_{0}$ and $\vartheta$.

Case 3. $\frac{|\nabla\psi_{\lambda}(X)|}{x}$ takes its maximum at the nozzle
wall $N_{0}\cap\left\\{x\geq\frac{a+b}{2}\right\\}$ or
$N\cap\left\\{x\geq\frac{a+b}{2}\right\\}$. Applying the similar arguments in
the proof of Theorem 2.3 and Lemma 3.5, we have

$\sup_{X\in\bar{\Omega}_{0}}\frac{|\nabla\psi_{\lambda}(X)|}{x}\leq
C\lambda\leq C_{0}m_{0},$ (4.8)

where $C_{0}$ is a constant depending only on $N,N_{0}$ and $\vartheta$.

It follows from (4.7)-(4.8) that

$\frac{|\nabla\psi_{\lambda}(X)|}{x}\leq C_{0}m_{0}\ \ \text{in
$\Omega_{0}$},$

which implies that $C_{0}m_{0}\leq\Pi_{\lambda}-4\tilde{\varepsilon}$ for
sufficiently small $m_{0}>0$. Then the subsonic cut-off can be taken away
$\rho(t;\lambda^{2})=\tilde{\rho}(t;\lambda^{2})$.

Step 4. In this step, the positivity of horizontal velocity will be obtained,
namely, then

$\frac{\partial\psi_{\lambda}}{\partial
y}>0~{}~{}\text{in}~{}~{}\bar{\Omega}_{0}\setminus I,$ (4.9)

where $\Omega_{0}=\Omega\cap\\{\psi_{\lambda}<m_{0}\\}$.

Set $\omega=\partial_{y}\psi_{\lambda}$, which solves

$\partial_{i}\left(\frac{\partial_{i}\omega
x^{2}\rho-2\rho_{1}\partial_{i}\psi_{\lambda}\partial_{j}\psi_{\lambda}\partial_{j}\omega}{x^{3}\rho^{2}}\right)=0\
\ \text{in}\ \ \Omega\cap\\{\psi_{\lambda}<m_{0}\\}.$

Since $\omega\geq 0$ in $\Omega_{0}$, the strong maximum principle gives that
$\omega>0$ in $\Omega_{0}$.

Owing to that $\psi_{\lambda}$ attains its maximal value $m_{0}$ on $N$, it
follows from Hopf’s lemma that