# Steady collision of two jets issuing from two axially symmetric channels†

Lili Dua,1, Yongfu Wangb,2

###### Abstract.

In the classical survey (Chapter 16.2, Mathematics in industrial problem, Vol.
24, Springer-Verlag, New York, 1989), A. Friedman proposed an open problem on
the collision of two incompressible jets emerging from two axially symmetric
nozzles. In this paper, we concerned with the mathematical theory on this
collision problem, and establish the well-posedness theory on hydrodynamic
impinging outgoing jets issuing from two coaxial axially symmetric nozzles.
More precisely, we showed that for any given mass fluxes $M_{1}>0$ and
$M_{2}<0$ in two nozzles respectively, that there exists an incompressible,
inviscid impinging outgoing jet with contact discontinuity, which issues from
two given semi-infinitely long axially symmetric nozzles and extends to
infinity. Moreover, the constant pressure free stream surfaces of the
impinging jet initiate smoothly from the mouths of the two nozzles and shrink
to some asymptotic conical surface. There exists a smooth surface separating
the two incompressible fluids and the contact discontinuity occurs on the
surface. Furthermore, we showed that there is no stagnation point in the flow
field and its closure, except one point on the symmetric axis. Some asymptotic
behavior of the impinging jet in upstream and downstream, geometric properties
of the free stream surfaces are also obtained. The main results in this paper
solved the open problem on the collision of two incompressible axially
symmetric jets in [24].

† Du is supported by NSFC grant 11971331 and Sichuan Youth Science and
Technology Foundation (No. 21CXTD0076). Wang is supported by NSFC grant
11801460 and the Applied Fundamental Research Plan of Sichuan Province (No.
2021YJ0520).

1 E-Mail<EMAIL_ADDRESS>2 E-mail<EMAIL_ADDRESS>Corresponding author

a Department of Mathematics, Sichuan University,

Chengdu 610064, P. R. China.

b School of Economic Mathematics,

Southwestern University of Finance and Economics,

Chengdu 611130, P. R. China.

2010 Mathematics Subject Classification: 76B10; 76B03; 35Q31; 35J25.

Key words: Impinging outgoing jets, incompressible flows, free boundary,
existence and uniqueness, contact discontinuity.

##  1\. Introduction

The three-dimensional incompressible, stationary and inviscid flow is governed
by the Euler equations

$\left\\{\begin{array}[]{l}\sum_{i=1}^{3}\frac{\partial u_{i}}{\partial
x_{i}}=0,\\\ \sum_{i=1}^{3}u_{i}\frac{\partial u_{j}}{\partial
x_{i}}+\frac{1}{\rho}\frac{\partial P}{\partial x_{j}}=0,\quad\text{for}\quad
j=1,2,3,\end{array}\right.$ (1.1)

with the irrotational condition

$\displaystyle\nabla\times(u_{1},u_{2},u_{3})=0.$

Here, $\displaystyle(u_{1},u_{2},u_{3})$ is the velocity, $\displaystyle P$
denotes the pressure of the flow, and positive constant $\displaystyle\rho$
denotes density.

In this paper, we shall be concerned with steady, irrotational incompressible
impinging jets issuing from two semi-infinitely long axially symmetric nozzles
with variable cross-section. We will investigate the well-posedness theory of
the collision problem of two jets issuing from two general axially symmetric
nozzles, and solve the open problem (1) proposed by A. Friedman in 1989.

Consider the axially symmetric flows in this paper and let $\displaystyle
U(x,r)$, $\displaystyle V(x,r)$ and $\displaystyle W(x,r)$ be the axial
velocity, the radial velocity and the swirl velocity respectively,
$\displaystyle x=x_{1}$ and $\displaystyle r=\sqrt{x_{2}^{2}+x_{3}^{2}}$.
Furthermore, we seek such an axially symmetric flow without swirl, one has

$u_{1}=U(x,r),\quad u_{2}=V(x,r)\frac{x_{2}}{r},\quad
u_{3}=V(x,r)\frac{x_{3}}{r}.$ (1.2)

Then, instead of (1.1), we have

$\left\\{\begin{array}[]{l}\left(rU\right)_{x}+\left(rV\right)_{r}=0,\\\
\left(r\rho U^{2}\right)_{x}+\left(r\rho UV\right)_{r}+rP_{x}=0,\\\
\left(r\rho UV\right)_{x}+\left(r\rho V^{2}\right)_{r}+rP_{r}=0.\\\
\end{array}\right.$ (1.3)

Consider the flow issuing from the two semi-infinitely long nozzles as (see
Figure 1)

$\displaystyle\mathcal{N}_{1}=\left\\{(x,r)\in{\mathbb{R}}_{+}^{2}\left|f_{1}(r)<x<-1,\
r<R_{1}\right.\right\\},$

and

$\displaystyle\mathcal{N}_{2}=\left\\{(x,r)\in{\mathbb{R}}_{+}^{2}\left|1<x<f_{2}(r),\
r<R_{2}\right.\right\\},$

where $\displaystyle{\mathbb{R}}_{+}^{2}={\mathbb{R}}^{1}\times[0,+\infty)$,
$\displaystyle R_{1}$, $\displaystyle R_{2}>0$, $\displaystyle f_{1}(r)$ and
$\displaystyle f_{2}(r)$ are smooth functions and satisfy that

$f_{i}(r)=(-1)^{i}\infty,\quad\quad r\leq r_{i},\quad(r_{i}>0)$ (1.4)

and

$f_{i}(R_{i})=(-1)^{i},$ (1.5)

for $\displaystyle R_{i}>r_{i}$ and $\displaystyle i=1,2$. Without loss of
generality, we assume that $\displaystyle R_{1}=R_{2}=R$.

Figure 1. Two axially symmetric semi-infinitely long nozzles

For convenience, we denote the symmetric axis

$\displaystyle N_{0}=\\{(x,r)|r=0,-\infty<x<+\infty\\},$

the nozzle walls

$\displaystyle N_{1}=\\{(x,r)|x=f_{1}(r),\ r_{1}<r<R\\},\quad
N_{2}=\\{(x,r)|x=f_{2}(r),\ r_{2}<r<R\\},$

and the edge points of the nozzle walls $\displaystyle A_{1}=(-1,R)$ and
$\displaystyle A_{2}=(1,R)$.

In this paper, we consider two ideal, nonmiscible, irrotational fluids
$\displaystyle(U_{1},V_{1},P_{1},\rho_{1})$ and
$\displaystyle(U_{2},V_{2},P_{2},\rho_{2})$ issuing from two semi-infinite
axisymmetric nozzles. We designate by
$\displaystyle(U_{1},V_{1},P_{1},\rho_{1})$ and
$\displaystyle(U_{2},V_{2},P_{2},\rho_{2})$ be the axial velocity, the radial
velocity and the pressure of the fluid I and the fluid II, respectively.
Denote $\displaystyle\Omega_{i}$ as the fluid field of the $\displaystyle
i$-th fluid for $\displaystyle i=1,2$, and

$\displaystyle(U,V,P,\rho)=\left\\{\begin{array}[]{l}(U_{1},V_{1},P_{1},\rho_{1})\
\ \ \text{in}\ \ \Omega_{1},\\\ (U_{2},V_{2},P_{2},\rho_{2})\ \ \ \text{in}\ \
\Omega_{2},\end{array}\right.\text{as the two-phase fluids}.$

In this paper, we seek a contact discontinuity $\displaystyle(U,V,P,\rho)$
with a smooth interface $\displaystyle\Gamma$: $\displaystyle\\{x=g(r)\\}$
between the two fluids. And $\displaystyle(U,V,P,\rho)$ is a week solution of
(1.3) in the distributional sense and
$\displaystyle(U_{i},V_{i},P_{i},\rho_{i})$ solves the incompressible Euler
system (1.3) classically in $\displaystyle\Omega_{i}$ for $\displaystyle
i=1,2$. (Please see Figure 3).

Figure 2. Collision of two jets

Figure 3. Impinging outgoing jet

Then the Rankine-Hugoniot jump conditions on $\displaystyle\Gamma$ become

$-\left[\begin{array}[]{l}\rho U\\\ \rho U^{2}+P\\\ \rho UV\\\
\end{array}\right]+g^{\prime}(r)\left[\begin{array}[]{l}\rho V\\\ \rho UV\\\
\rho V^{2}+P\\\ \end{array}\right]=0,$ (1.6)

where $\displaystyle[\cdot]$ denotes the jump of a related function crossing
the interface $\displaystyle\Gamma$.

Set
$\displaystyle\mathfrak{m}_{i}=\rho_{i}\left(g^{\prime}(r)V_{i}-U_{i}\right)$
($\displaystyle i=1,2$) be the mass flux across the interface, if
$\displaystyle\mathfrak{m}_{1}=\mathfrak{m}_{2}=0$ on the interface
$\displaystyle\Gamma$, then $\displaystyle(U,V,P,\rho)$ is a contact
discontinuity. The Rankine-Hugoniot conditions (1.6) read as

$-U_{1}+g^{\prime}(r)V_{1}=0,\ \ -U_{2}+g^{\prime}(r)V_{2}=0\ \text{and}\
P_{1}=P_{2}.$ (1.7)

The condition (1.7) implies that the normal velocities on both sides of the
interface $\displaystyle\Gamma$ vanish, while the tangential velocity on both
side of $\displaystyle\Gamma$ may have nontrivial jump.

Furthermore, the well-known Bernoulli’s law can be written as

$\displaystyle(U_{i},V_{i})\cdot\nabla\left(\frac{1}{2}(U_{i}^{2}+V_{i}^{2})+\frac{P_{i}}{\rho_{i}}\right)=0,\
\ \text{for}\ \ i=1,2,$

namely,

$\displaystyle\frac{U_{1}^{2}+V_{1}^{2}}{2}+\frac{P_{1}}{\rho_{1}}=\mathfrak{B}_{1}\
\ \text{and}\ \
\frac{U_{2}^{2}+V_{2}^{2}}{2}+\frac{P_{2}}{\rho_{2}}=\mathfrak{B}_{2},$

where $\displaystyle\mathfrak{B}_{1}$ and $\displaystyle\mathfrak{B}_{2}$
denote the Bernoulli’s constants of the two fluids, respectively, in view of
(1.7), then

$\rho_{1}\left(U_{1}^{2}+V_{1}^{2}\right)-\rho_{2}\left(U_{2}^{2}+V_{2}^{2}\right)=2\left(\rho_{1}\mathfrak{B}_{1}-\rho_{2}\mathfrak{B}_{2}\right)\triangleq\Lambda\
\ \ \text{on}\ \ \Gamma,$ (1.8)

without loss of generality, we assume $\displaystyle\Lambda\geq 0$.

On another hand, on the free surfaces $\displaystyle\Gamma_{1}$ and
$\displaystyle\Gamma_{2}$, the pressure is assumed to be the constant
atmosphere pressure $\displaystyle P_{at}$ (in absence of gravity and surface
tension), namely,

$P=P_{at}\ \ \text{on $\displaystyle\Gamma_{1}\cup\Gamma_{2}$.}$ (1.9)

Here is our problem of fluid mechanics: determine an impinging outgoing jet
$\displaystyle(U,V,P,\rho)$ issuing from two nozzles
$\displaystyle\mathcal{N}_{1}$ and $\displaystyle\mathcal{N}_{2}$ with two
mass fluxes $\displaystyle M_{1}$ and $\displaystyle M_{2}$, bounded by two
free surfaces $\displaystyle\Gamma_{1}$ and $\displaystyle\Gamma_{2}$ on which
the pressure is a constant $\displaystyle P_{at}$. Furthermore, on the
interface, the Rankine-Hugoniot conditions (1.7) and (1.8) hold.

On the solid walls $\displaystyle N_{1}$ and $\displaystyle N_{2}$, the flow
satisfies the slip-boundary condition,

$(U_{i},V_{i})\cdot\vec{n}_{i}=0,\ \ \ \ \ \text{on}\ \ N_{i},$ (1.10)

where $\displaystyle\vec{n}_{i}$ is the unit outer normal of the wall
$\displaystyle N_{i}$, for $\displaystyle i=1,2$. Moreover, on the symmetry
axis $\displaystyle N_{0}$,

$V_{i}=0.$ (1.11)

Denote $\displaystyle M_{1}$ and $\displaystyle M_{2}$ as the mass fluxes in
nozzles $\displaystyle\mathcal{N}_{1}$ and $\displaystyle\mathcal{N}_{2}$,
respectively, then

$\int_{\Sigma_{i}}(rU,rV,0)\cdot\vec{l_{i}}dS=\frac{M_{i}}{2\pi},$ (1.12)

where $\displaystyle\Sigma_{i}$ is any curve transversal to the $\displaystyle
x$-axis direction and $\displaystyle\vec{l_{i}}$ is the normal of
$\displaystyle\Sigma_{i}$ in the positive $\displaystyle x$-axis direction for
$\displaystyle i=1,2$.

###  1.1. Impinging outgoing jet problem and main results

We define the axially symmetric impinging outgoing jet problem as follows.

Axially symmetric impinging outgoing jet problem. For given any mass fluxes
$\displaystyle M_{1}>0$ and $\displaystyle M_{2}<0$ in the two semi-infinitely
long axially symmetric nozzles $\displaystyle\mathcal{N}_{1}$ and
$\displaystyle\mathcal{N}_{2}$, respectively, there exists an axially
symmetric impinging outgoing jet extending to the infinity, the free stream
surfaces initiate at the edges of the nozzles smoothly and shrink to some
conical surface at the far field, and a smooth interface separates the two
jets, furthermore, the pressure remains a constant on free stream surfaces
(see Figure 3).

Next, we give the definition of the solution to the impinging outgoing jet
problem.

A solution to the axially symmetric impinging outgoing jet problem. A
quintuple $\displaystyle(U,V,P,\Gamma_{1},\Gamma_{2})$ is called a solution to
the axially symmetric impinging outgoing jet problem, provided that

(1). The smooth surfaces $\displaystyle\Gamma_{1}$ and
$\displaystyle\Gamma_{2}$ are given by two functions $\displaystyle
x=g_{1}(r)\in C^{1}{((R,+\infty))}$ and $\displaystyle x=g_{2}(r)\in
C^{1}{((R,+\infty))}$ with $\displaystyle g_{1}(r)<g_{2}(r)$, and

$g_{1}(R+0)=f_{1}(R-0),\quad g_{2}(R+0)=f_{2}(R-0)\quad\text{({\it continuous
fit conditions}),}$ (1.13)

and

$g_{1}^{\prime}(R+0)=f_{1}^{\prime}(R-0),\quad
g_{2}^{\prime}(R+0)=f_{2}^{\prime}(R-0)\quad\text{({\it smooth fit
conditions})}.$ (1.14)

Moreover, there exists an asymptotic direction
$\displaystyle\nu=(\cos\theta,\sin\theta)$ with
$\displaystyle\theta\in(0,\pi)$, such that $\displaystyle g_{1}$ and
$\displaystyle g_{2}$ are close to the asymptotic direction $\displaystyle\nu$
at far field (See Figure 5), ie.,

$\lim_{r\rightarrow\infty}\left(g_{2}(r)-g_{1}(r)\right)=0\quad\text{and}\quad\lim_{r\rightarrow\infty}g^{\prime}_{1}(r)=\lim_{r\rightarrow\infty}g^{\prime}_{2}(r)=\cot\theta,$
(1.15)

the angle $\displaystyle\theta$ is called the asymptotic deflection angle of
the impinging outgoing jet.

(2). Denote the flow field $\displaystyle G$ bounded by the symmetric axis
$\displaystyle N_{0}$, the nozzle walls $\displaystyle N_{1},N_{2}$ and the
free boundaries $\displaystyle\Gamma_{1},\Gamma_{2}$.
$\displaystyle(U,V,P)\in\left(C^{1,\alpha}(G)\cap
C^{0}(\overline{G})\right)^{3}$ solves the steady incompressible Euler system
(1.3), the boundary condition (1.10), the Rankine-Hugoniot conditions (1.7)
and the mass flux conditions (1.12);

(3). The radial velocity $\displaystyle V$ is positive in flow field and its
closure, except the symmetric axis and interface $\displaystyle\Gamma$,
namely, $\displaystyle V>0$ in
$\displaystyle\bar{G}\setminus\left(N_{0}\cup\Gamma\right)$;

(4). $\displaystyle P=P_{at}$ on $\displaystyle\Gamma_{1}\cup\Gamma_{2}$;

(5). The interface $\displaystyle\Gamma$ satisfies the condition (1.8).

The first result in this paper is the existence of the impinging outgoing jet
as follows.

######  Theorem 1.1.

For any given atmosphere pressure $\displaystyle P_{at}$, mass fluxes
$\displaystyle M_{1}>0$, $\displaystyle M_{2}<0$ and $\displaystyle\Lambda\geq
0$ issuing from the two axially symmetric nozzles
$\displaystyle\mathcal{N}_{1}$ and $\displaystyle\mathcal{N}_{2}$,
respectively, there exists a solution
$\displaystyle(U,V,P,\Gamma_{1},\Gamma_{2})$ to the axially symmetric
impinging outgoing jet problem. Furthermore, there exists a $\displaystyle
C^{1}$-smooth surface $\displaystyle x=g(r)$ satisfying $\displaystyle
g_{1}(r)<g(r)<g_{2}(r)$ for $\displaystyle R<r<\infty$, which separates the
two fluids and initiates at the branching point on the symmetric axis (Figure
5). Furthermore, there exists a positive constant $\displaystyle\lambda$, such
that

$r(g(r)-g_{1}(r))\rightarrow\frac{M_{1}}{2\pi\sqrt{\rho_{1}(\Lambda+\lambda})\sin\theta}\
\ \text{and}\ \
r(g(r)-g_{2}(r))\rightarrow\frac{M_{2}}{2\pi\sqrt{\rho_{2}\lambda}\sin\theta}\quad\quad\text{as}\quad
r\rightarrow+\infty.$ (1.16)

We would like to give the following comments on the existence theorem.

######  Remark 1.1.

One of key points in this work is the appearance of the interface between the
two fluids, which is also a free boundary and is determined by the solution
itself. In this paper, the impinging outgoing jets possess a smooth surface
separating the two immiscible fluids, which intersects the symmetric axis at a
unique point, called the _branching point_. However, the appearance of the
interface takes many essential difficulties to solve the free boundary problem
in mathematics. The first one is the non-trivial jump of the velocity field on
the interface (see (1.8)), which leads that we have to seek a non-smooth
solution in the whole fluid field. The second one is that the interface is
common boundaries of two fluids, which is totally free. And we shall define
the interface as the level set of the stream function and show that it is
indeed a smooth curve. The third one is the regularity of the two-phase fluids
near the branching point.

Figure 4. Axisymmetric impinging outgoing jet flow in cylindrical coordinates

Figure 5. Impinging outgoing jet and the interface $\displaystyle\Gamma$

######  Remark 1.2.

There are many numerical results on the impinging free jets in absence of
rigid nozzle walls, such as unsymmetrically impinging jets in [29], impinging
free jets in [28], compressible impinging jets in [10]. However, here we have
to consider the geometry of both solid boundaries and free boundaries, the one
of main difficulties is to verify the continuous fit and smooth fit
conditions. In present work, an essential point is that we can choose a
suitable pair of parameters $\displaystyle(\lambda,\theta)$, such that the
continuous fit conditions are fulfilled. In other word, the parameters
$\displaystyle\lambda$ and $\displaystyle\theta$ can be determined by the
continuous fit conditions, which is the main difference from the analysis of
impinging free jets without rigid boundaries. Therefore, we first solve the
free boundary problem for any $\displaystyle\lambda$ and
$\displaystyle\theta$, and then show the existence of a suitable pair of
parameters $\displaystyle(\lambda,\theta)$ to guarantee the continuous fit
conditions of impinging outgoing jet. Furthermore, we can show that the
continuous fit conditions imply the smooth fit conditions.

Theorem 1.1 gives that there exists a pair of parameters
$\displaystyle(\lambda,\theta)$ to guarantee the existence of the axially
symmetric impinging outgoing jet. However, to the best of our knowledge, the
uniqueness of the jet with two free boundaries is totally open. Next, for
$\displaystyle\Lambda=0$, we give the uniqueness results on the axially
symmetric impinging outgoing jet, the idea borrows from the recent work [11]
on the uniqueness of the asymmetric incompressible jet.

######  Theorem 1.2.

(Uniqueness of the axially impinging outgoing jet)(1) Given any parameters
$\displaystyle(\lambda,\theta)$, such that the continuous fit conditions
(1.13) hold, then the axially symmetric impinging outgoing jet
$\displaystyle(U,V,P,\Gamma_{1},\Gamma_{2})$ established in Theorem 1.1 is
unique.
(2) Suppose that there exist two pairs of the parameters
$\displaystyle(\lambda,\theta)$ and $\displaystyle(\lambda,\tilde{\theta})$,
such that the continuous fit conditions (1.13) to the axially symmetric
impinging outgoing jet hold, then $\displaystyle\theta=\tilde{\theta}$.

Next, we give the asymptotic behaviors and the decay rate of the impinging
outgoing jet in the far field.

######  Theorem 1.3.

The impinging outgoing jet flow $\displaystyle(U,V,P,\Gamma_{1},\Gamma_{2})$
established in Theorem 1.1 satisfies the following asymptotic behavior in far
fields,

$(U(x,r),V(x,r),P(x,r))\rightarrow\left(\frac{M_{1}}{\pi\rho_{1}r_{1}^{2}},0,\frac{\lambda+\Lambda}{2\rho_{1}}+P_{at}-\frac{M_{1}^{2}}{2\rho_{1}\pi^{2}r_{1}^{4}}\right),$
(1.17)

and

$\nabla U\rightarrow 0,\ \ \nabla V\rightarrow 0,\ \ \nabla P\rightarrow 0,$
(1.18)

as $\displaystyle x\rightarrow-\infty$, in any compact subset of
$\displaystyle(0,r_{1})$ and

$(U(x,r),V(x,r),P(x,r))\rightarrow\left(\frac{M_{2}}{\pi\rho_{2}r_{1}^{2}},0,\frac{\lambda}{2\rho_{2}}+P_{at}-\frac{M_{2}^{2}}{2\rho_{2}\pi^{2}r_{2}^{4}}\right),$
(1.19)

and

$\nabla U\rightarrow 0,\ \ \nabla V\rightarrow 0,\ \ \nabla P\rightarrow 0,$
(1.20)

as $\displaystyle x\rightarrow+\infty$, in any compact subset of
$\displaystyle(0,r_{2})$, and in the downstream,

$(U(x,r),V(x,r),P(x,r))\rightarrow\left(\sqrt{\frac{\Lambda+\lambda}{\rho_{1}}}\cos\theta,\sqrt{\frac{\Lambda+\lambda}{\rho_{1}}}\sin\theta,P_{at}\right),$
(1.21)

uniformly in any compact subset of $\displaystyle\Omega_{1}$ as $\displaystyle
r\rightarrow+\infty$, and

$(U(x,r),V(x,r),P(x,r))\rightarrow\left(\sqrt{\frac{\lambda}{\rho_{2}}}\cos\theta,\sqrt{\frac{\lambda}{\rho_{2}}}\sin\theta,P_{at}\right),$
(1.22)

uniformly in any compact subset of $\displaystyle\Omega_{2}$ as $\displaystyle
r\rightarrow+\infty$, and

$\nabla U\rightarrow 0,\ \ \nabla V\rightarrow 0,\ \ \nabla P\rightarrow 0,$
(1.23)

uniformly in any compact subset of $\displaystyle\Omega_{1}\cup\Omega_{2}$ as
$\displaystyle r\rightarrow+\infty$.

Furthermore, for any $\displaystyle\alpha\in(0,2)$, one has

$r^{\alpha}\left(\left|U_{1}(x,r)-\sqrt{\frac{\Lambda+\lambda}{\rho_{1}}}\cos\theta\right|+\left|V_{1}(x,r)-\sqrt{\frac{\Lambda+\lambda}{\rho_{1}}}\sin\theta\right|\right)\rightarrow
0,$ (1.24)
$r^{\alpha}\left(\left|U_{2}(x,r)-\sqrt{\frac{\lambda}{\rho_{2}}}\cos\theta\right|+\left|V_{2}(x,r)-\sqrt{\frac{\lambda}{\rho_{2}}}\sin\theta\right|\right)\rightarrow
0,$ (1.25)

as $\displaystyle r\rightarrow+\infty$.

######  Remark 1.3.

The one of main differences between the impinging outgoing jet in two-
dimensional and axially symmetric case is that the two-dimensional outgoing
jet possesses a uniform positive width in far field, and however, the distance
of free streamlines goes to zero in downstream in axially symmetric case.
Here, we have to establish the decay estimates of outgoing jets and the free
boundaries in far field. Indeed, the facts (1.16) and (1.24) give the decay
rates of the velocity field and the distance of the two free stream surfaces
in downstream. In particular, (1.16) implies that the optimal decay rate of
the distance of two free stream surfaces is $\displaystyle\frac{1}{r}$ in
downstream.

###  1.2. Motivation and history of the problem

The motivation to investigate the impinging outgoing jets from two nozzles
comes from Chapter V. $\displaystyle\S$ 5 in the classical book [9] by G.
Birkhoff and E. H. Zarantonello, in which the impinging outgoing jets from two
plane symmetric cylinders were considered. Except for some simple channel
geometries, the impinging problem of two jets can not be solved analytically,
as was shown in the monographs [9] and [27]. Here, we consider the general
case that the impinging jet issuing from two axially symmetric nozzles with
variable cross-section.

Figure 6. Collision of two jets (Figure 16.7 in [24])

Another motivation to investigate the impinging outgoing jets issuing from two
nozzles comes from the Chapter 16 in famous survey [24]. The physical problem
is also related to the shaped charge question in [13]. As mentioned in Page
152 [24], ”… _we can formulate this problem as a collision of two jets, say a
garden hose and a fire hose; see Figure 16.7._ ” (see Figure 6) and A.
Friedman proposed an open problem that

“ _Problem (1). Analyze the axially symmetric free boundary problem associated
with the flow in Figure 16.7 in incompressible case._ ”

On another side, there are many numerical results on this impinging outgoing
jet problem, such as the incompressible problem for an arbitrary polygonal
nozzle in [14], and the incompressible jet with gravity in [15] and so on.
Moreover, Hurean and Weber in [28] considered the impinging of two
incompressible ideal free jets (in the absence of rigid nozzle walls)
numerically, and some existence results on two compressible free jets were
also investigated in [10]. However, we also would like to mention the
numerical result on asymmetric impinging free jets in [29].

The study of liquid jets issuing from containers is centuries old. As far back
as 1868, Helmholtz and Kirchhoff introduced the classical theory of free
streamlines in two-dimensional jets. The steady irrotational flows of ideal
incompressible fluid, bounded by nozzle walls and free streamlines were
investigated. The following decades saw extensions of a great many different
kinds of two-dimensional flows, on the basis of the complex analysis methods
by Planck, Joukowsky, Réthy, Levi-Civita, Greenhill and others.

Some substantial post-war monographs are those of Birkhoff-Zarantonello [9],
Gurevich [27], Milne-Thomson [30]. For two-dimensional irrotational case, a
generalized Schwarz-Christoffel transformation, combined with a Fourier
technique to formulate a free boundary problem into a nonlinear integral-
differential equation, some existence results on jets in special nozzles have
been constructed. However, two-dimensional jets have been given most of the
attention in the existence theory, and the limited amount of work on
axisymmetric jets has been confined. The reason is that the complex analysis
method which has been adapted to two-dimensional jets has noneffective in the
axially symmetric case. A first breakthrough work on the axially symmetric
free streamline was due to Garabedian, Lewy and Schiffer in [25] in 1952, in
which some existence results on the axially symmetric cavity were established
by variational approach. Furthermore, Alt, Caffarelli and Friedman developed
the variational method to deal the free streamlines problem in their elegant
works [1, 8]. Based on their framework, some well-posedness results on axially
symmetric jet in [4], asymmetric jet in [2], jet with gravity in [3], and jets
with two fluids in [6, 7] have been established. In this paper, some
fundamental ideas on the existence theory are still borrowed from the
variational method in [1]. Recently, if we assume that the fluid is smooth
across the interface $\displaystyle\Gamma$ (ie. $\displaystyle\Lambda=0$)
apriorily, some existence results for incompressible plane symmetric impinging
outgoing jets has been obtained in [17]. In fact, the interface
$\displaystyle\Gamma$ is a contact discontinuity and the jump
$\displaystyle\Lambda$ is always non-trivial and non-zero generally. As we
mentioned before, we have to investigate the non-zero jump
$\displaystyle\Lambda\neq 0$, which is one of main differences between this
paper to the previous paper [17]. As a first step to attack the original
problem on impinging outgoing jets with nontrivial jump, Wang and Xiang in
[31] considered a toy model on the incompressible fluids issuing from two
infinity long co-axis and symmetric nozzles without jet free boundary and
established some properties on the contact discontinuity between the two
fluids. Therefore, the objective of the present paper is to establish the
well-posedness theory on the impinging outgoing jet problem and solve the open
problem proposed by A. Frideman in 1989.

###  1.3. Methodology

From the physical problem here, the main difference and difficulty here stems
from the shape of nozzle walls, we have to find a mechanism, such that the
free boundaries of the jets connecting smoothly at the edge of the nozzle
walls (so-called _continuous fit and smooth fit conditions_). Another main
difficulty is how to analyze the interface with contact discontinuity between
the two incompressible ideal fluids.

We would like to comment the main ideas of the proof as follows. The
variational method developed by H. Alt, L. Caffarelli and A. Friedman in 1980s
has been shown to be powerful and effective to solve the free streamline
theory for more general models. In two-dimensional case, the stream function
is harmonic in the fluid domain, while in axially symmetric case, it solves a
linear elliptic equation with some lower-order term, and we have to deal with
the regularity near the symmetric axis. This is the first difficulty in this
paper. The second one is that the distance of the two free boundaries
converges to a positive constant in two-dimensional case (see [17]), and
however, the distance goes to zero in axially symmetric jet here. Therefore,
we can not borrow some uniform special flow to show some fundamental
properties of the free boundaries as in 2D case, such as the vanishing and
non-vanishing of free boundary, and the asymptotic behaviors of the jet in far
fields, and so on. Here, our strategy is to establish firstly the decay rate
of distance of the two free boundaries and the impinging outgoing jet in far
field, and then to obtain the desired properties via some rescaling arguments.
The third principal difficulty in this paper centers about how to choose the
suitable parameters $\displaystyle(\lambda,\theta)$ to assure the continuous
fit conditions in impinging outgoing jets. Some continuous dependent
relationships and monotonic properties to the impinging outgoing jets with
respect to the parameters $\displaystyle(\lambda,\theta)$ are established and
guarantee the fact to be fulfilled. The fourth difficulty here is the
occurrence of three free boundaries $\displaystyle\Gamma_{1}$,
$\displaystyle\Gamma_{2}$ and $\displaystyle\Gamma$, which is main difference
to the previous results, such as [17, 31]. In particular, the solution is not
smooth across the interface $\displaystyle\Gamma$ and it is a contact
discontinuity between the two fluids. To our knowledge, this is the first
well-posedness work on the jet flows problem with three free boundaries.

The remain of this paper is organized as follows. First, we establish the free
boundary value problem to the physical problem in Section 2. The solvability
of the free boundary value problem follows from the standard variational
approach, which has been developed by Alt, Caffarelli and Friedman in the
celebrated works [1, 2, 4]. Moreover, some properties of the free boundaries
will be obtained and we can verify the continuous fit and smooth fit
conditions for suitable parameters $\displaystyle\lambda$ and
$\displaystyle\theta$. Additionally, we will investigate the existence and
properties of the interface between the two fluids. Hence, we can obtain the
existence of impinging outgoing jet in Section 3. In Section 4, we will give
the uniqueness of the impinging outgoing jet and the parameters. In Section 5,
the asymptotic behavior of the impinging outgoing jet is obtained along the
blow-up argument, which has been used to deal with the subsonic compressible
flows in infinitely long nozzles in [16, 18, 19, 20, 21, 22, 32, 33, 34]. Some
results on the variational problem are given in the Appendix.

##  2\. Mathematical settings of the free boundary problem

Due to the continuity equation (1.3), one can introduce stream functions
$\displaystyle\Psi_{i}(x,r)$ ($\displaystyle i=1,2$) such that

$\partial_{x}\Psi_{i}=-r\rho_{i}V_{i},\ \ \ \
\partial_{r}\Psi_{i}=r\rho_{i}U_{i}.$ (2.1)

In order to convenient to the analysis, we introduce the scaled stream
function $\displaystyle\psi_{i}=\frac{\Psi_{i}}{\sqrt{\rho_{i}}}$
($\displaystyle i=1,2$), and denote $\displaystyle\psi$ as

$\displaystyle\psi=\left\\{\begin{array}[]{l}\psi_{1}\ \ \ \text{in}\ \
\Omega_{1},\\\ \psi_{2}\ \ \ \text{in}\ \ \Omega_{2}.\end{array}\right.$

This together with the irrotational condition gives

$\Delta\psi-\frac{1}{r}\frac{\partial\psi}{\partial r}=0\ \ \text{in
$\displaystyle\Omega_{1}\cup\Omega_{2}$}.$ (2.2)

Here and after, $\displaystyle\Omega_{i}$ denotes the flow field, bounded by
the nozzle walls $\displaystyle N_{i}$, the symmetric axis $\displaystyle
N_{0}$, the interface $\displaystyle\Gamma$ and the free boundaries
$\displaystyle\Gamma_{i}$ ($\displaystyle i=1,2$).

In this paper, we expect to seek an axially symmetric impinging outgoing jet
flow with positive vertical velocity, and thus denote $\displaystyle\Omega$
bounded by $\displaystyle N_{i}$, $\displaystyle N_{0}$ and $\displaystyle
L_{i}$ ($\displaystyle i=1,2$) as the possible flow field of impinging
outgoing jet (see Figure 7), where

$\displaystyle L_{1}=\\{(x,r)\mid r=R,x<-1\\}\ \ \text{and}\ \
L_{2}=\\{(x,r)\mid r=R,x>1\\}.$

Figure 7. The possible flow field $\displaystyle\Omega$

Moreover, the nozzles $\displaystyle N_{1}\cup N_{2}$ and the free boundaries
$\displaystyle\Gamma_{1}\cup\Gamma_{2}$ are streamlines, thus
$\displaystyle\psi$ remains some constant on those boundaries, without loss of
generality, we can impose $\displaystyle\psi=m_{1}$ on $\displaystyle
N_{1}\cup\Gamma_{1}$ and $\displaystyle\psi=m_{2}$ on $\displaystyle
N_{2}\cup\Gamma_{2}$. Thus, the free boundaries $\displaystyle\Gamma_{1}$ and
$\displaystyle\Gamma_{2}$ can be defined as

$\displaystyle\Gamma_{1}=\Omega\cap\partial\left\\{\psi<m_{1}\right\\},\ \
\Gamma_{2}=\Omega\cap\partial\left\\{\psi>m_{2}\right\\},$

respectively, where $\displaystyle\Omega$ is the possible flow field defined
before and $\displaystyle m_{i}=\frac{M_{i}}{2\pi\sqrt{\rho_{i}}}$
($\displaystyle i=1,2$). And the constant pressure boundary condition on free
boundaries can be rewritten as

$\left|\frac{1}{r}\frac{\partial\psi}{\partial\nu}\right|=\sqrt{\Lambda+\lambda}\
\ \text{on}~{}~{}\Gamma_{1},\ \ \ \ \
\left|\frac{1}{r}\frac{\partial\psi}{\partial\nu}\right|=\sqrt{\lambda}\ \
\text{on}~{}~{}\Gamma_{2},$ (2.3)

where $\displaystyle\nu$ is unit outward normal to the free stream surfaces
$\displaystyle\Gamma_{1}$ and $\displaystyle\Gamma_{2}$.

Hence, we formulate the boundary value problem to the stream function
$\displaystyle\psi$,

$\left\\{\begin{array}[]{ll}\Delta\psi-\frac{1}{r}\frac{\partial\psi}{\partial
r}=0,&\text{in}\ \ \ \Omega_{1}\cup\Omega_{2},\\\
\left|\frac{1}{r}\frac{\partial\psi}{\partial\nu}\right|=\sqrt{\Lambda+\lambda},&\text{on}\
\ \ \Gamma_{1},\ \ \ \ \
\left|\frac{1}{r}\frac{\partial\psi}{\partial\nu}\right|=\sqrt{\lambda},\ \ \
\ \text{on}\ \ \ \Gamma_{2},\\\
\left|\frac{\nabla\psi^{+}}{r}\right|^{2}-\left|\frac{\nabla\psi^{-}}{r}\right|^{2}=\Lambda,&\text{on}\
\ \ \Gamma,\\\ \psi=m_{1},&\text{on}\ \ \ N_{1}\cup\Gamma_{1},\ \ \ \ \
\psi=m_{2},\ \ \ \text{on}\ \ N_{2}\cup\Gamma_{2},\\\ \psi=0,&\text{on}\ \
N_{0}\cup\Gamma,\end{array}\right.$ (2.4)

where $\displaystyle\psi^{\pm}(X_{0})$ ($\displaystyle X_{0}\in\Gamma$)
denotes the limit of $\displaystyle\psi(X)$ with $\displaystyle
X\in\\{\pm\psi>0\\}$, as $\displaystyle X\rightarrow X_{0}$.

We would like to emphasize that the undetermined constants
$\displaystyle\lambda$ and $\displaystyle\theta$ are regarded as two
parameters to solve the free boundary problem. We will solve the free boundary
problem for any $\displaystyle\lambda$ and $\displaystyle\theta$, and then to
show the existence of suitable parameters to guarantee the continuous fit
conditions.

##  3\. Existence of the impinging outgoing jets

###  3.1. Truncated variational problem

In order to solve the free boundary value problem (2.4), we first introduce
some notations and two auxiliary functions as follows. Define domain
$\displaystyle D$ as

$\displaystyle D=\Omega\cap\left\\{r<R\right\\}.$

Next, we define two bounded functions $\displaystyle\Phi_{1}$ and
$\displaystyle\Phi_{2}$ as follows,

$\displaystyle\text{$\displaystyle\Delta\Phi_{1}-\frac{1}{r}\frac{\partial\Phi_{1}}{\partial
r}=0$ in $\displaystyle D$, and $\displaystyle m_{2}<\Phi_{1}<m_{1}$ in
$\displaystyle D$},$
$\displaystyle\Phi_{1}(x,r)=\left\\{\begin{array}[]{ll}0,&\text{if}\ \
(x,r)\in N_{0},\\\ m_{2},&\text{if}\ \ r_{2}<r\leq R,~{}(x,r)~{}\text{lies
right}~{}N_{2},\\\ m_{1},&\text{if}\ \ r_{1}<r\leq R,~{}(x,r)~{}\text{lies
left}~{}N_{1},\\\ m_{1},&\text{if}\ \ (x,r)\in\Omega\cap\left\\{r\geq
R\right\\},\end{array}\right.$ (3.1)

and

$\displaystyle\text{$\displaystyle\Delta\Phi_{2}-\frac{1}{r}\frac{\partial\Phi_{2}}{\partial
r}=0$ in $\displaystyle D$, $\displaystyle m_{2}<\Phi_{2}<m_{1}$ in
$\displaystyle D$},$
$\displaystyle\Phi_{2}(x,r)=\left\\{\begin{array}[]{ll}0,&\text{if}\ \
(x,r)\in N_{0},\\\ m_{2},&\text{if}\ \ r_{2}<r\leq R,~{}(x,r)~{}\text{lies
right}~{}N_{2},\\\ m_{1},&\text{if}\ \ r_{1}<r\leq R,~{}(x,r)~{}\text{lies
left}~{}N_{1},\\\ m_{2},&\text{if}\ \ (x,r)\in\Omega\cap\left\\{r\geq
R\right\\}.\end{array}\right.$ (3.2)

We introduce the admissible set as

$\displaystyle K=\left\\{\psi\in
H_{loc}^{1}(\Omega)|~{}\Phi_{2}\leq\psi\leq\Phi_{1}\right\\},$

set $\displaystyle e=\left(-\sin\theta,\cos\theta\right)$ with
$\displaystyle\theta\in[0,\pi]$, and a functional

$J_{\lambda,\theta}(\psi)=\int_{\Omega}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dxdr.$ (3.3)

Since the functional $\displaystyle J_{\lambda,\theta}$ is unbounded for any
$\displaystyle\psi\in K$, we have to truncate the possible flow field
$\displaystyle\Omega$ and formulate truncated problems as follows.

For any $\displaystyle\mu>1$ and $\displaystyle i=1,2$, we define

$\displaystyle r_{i,\mu}=\min\left\\{r\right|(-1)^{i}\mu=f_{i}(r)\\},\quad\
H_{i,\mu}=\left\\{\left((-1)^{i}\mu,r\right)|~{}0\leq r\leq
r_{i,\mu}\right\\},$ (3.4) $\displaystyle
N_{0,\mu}=N_{0}\cap\\{-\mu<x<\mu\\},\ \text{and}\
N_{i,\mu}=\left\\{(x,r)|~{}x=f_{i}(r),r_{i,\mu}<r\leq R\right\\}.$

Moreover, we introduce a cutoff domain $\displaystyle\Omega_{\mu}$ (see Figure
8) as

$\text{$\displaystyle\Omega_{\mu}$ is bounded by $\displaystyle N_{i,\mu}$,
$\displaystyle L_{i}$, $\displaystyle N_{0,\mu}$ and $\displaystyle
H_{i,\mu}$},$ (3.5)

and denote

$\displaystyle D_{\mu}=\Omega_{\mu}\cap\left\\{r<R\right\\}.$

Figure 8. The truncated domain $\displaystyle\Omega_{\mu}$

We also introduce an admissible set

$\displaystyle K_{\mu}=\left\\{\psi\in
K\left|\psi=\frac{m_{1}}{r_{1,\mu}^{2}}r^{2}~{}\text{on}~{}H_{1,\mu},\right.~{}~{}\psi=\frac{m_{2}}{r_{2,\mu}^{2}}r^{2}~{}\text{on}~{}H_{2,\mu}\right\\},$

and an auxiliary functional

$\displaystyle
J_{\lambda,\theta,\mu}(\psi)=\int_{\Omega_{\mu}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX,\ \ \psi\in K_{\mu},$

where $\displaystyle\chi_{E}$ is the indicator function of the set
$\displaystyle E$. Here and after, we denote $\displaystyle dX=dxdr$ and
$\displaystyle X=(x,r)$ for simplicity.

The truncated variational problem ($\displaystyle P_{\lambda,\theta,\mu}$):
Find a $\displaystyle\psi_{\lambda,\theta,\mu}\in K_{\mu}$, such that

$J_{\lambda,\theta,\mu}(\psi_{\lambda,\theta,\mu})=\mathop{\min}\limits_{\psi\in
K_{\mu}}J_{\lambda,\theta,\mu}(\psi).$ (3.6)

Furthermore, the free boundaries of the truncated variational problem
($\displaystyle P_{\lambda,\theta,\mu}$) are defined as follows.

######  Definition 3.1.

The set

$\displaystyle\Gamma_{1,\mu}=\Omega_{\mu}\cap\partial\left\\{\psi_{\lambda,\theta,\mu}<m_{1}\right\\},$

is called the left free boundary, and

$\displaystyle\Gamma_{2,\mu}=\Omega_{\mu}\cap\partial\left\\{\psi_{\lambda,\theta,\mu}>m_{2}\right\\},$

is called the right free boundary.

Furthermore, define

$\displaystyle\Gamma_{\mu}=\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}=0\\},$

be the interface separating the two fluids.

###  3.2. Existence of minimizer to the truncated variational problem

First, we give the existence of the minimizer to the truncated variational
problem.

######  Proposition 3.1.

For any $\displaystyle\lambda>0$, $\displaystyle\theta\in[0,\pi]$ and
$\displaystyle\mu>1$, there exists a minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}\in K_{\mu}$ to the truncated
variational problem ($\displaystyle P_{\lambda,\theta,\mu}$).

###### Proof.

Due to Theorem 1.3 in [1], it suffices to construct a function
$\displaystyle\psi_{0}\in K_{\mu}$ such that $\displaystyle
J_{\lambda,\theta,\mu}(\psi_{0})<+\infty$.

Case 1: For $\displaystyle\theta\in(0,\pi)$ in $\displaystyle\Omega_{\mu}$.

Indeed, for some sufficiently large

$\displaystyle
R_{0}>\max\left\\{1,\frac{m_{1}}{\sqrt{\Lambda+\lambda}(R+1)\sin\theta}-(R+1)\cot\theta,-\frac{m_{2}}{\sqrt{\lambda}(R+1)\sin\theta}+(R+1)\cot\theta\right\\},$

and taking $\displaystyle\overline{\psi}$ be a smooth function such that
$\displaystyle\psi_{0}\in K_{\mu}$. Define $\displaystyle\psi_{0}$ in
$\displaystyle\Omega_{\mu}$ as follows

$\displaystyle\psi_{0}(X)=\left\\{\begin{array}[]{ll}\min\left\\{\max\left\\{\sqrt{\Lambda+\lambda}r\left(r\cos\theta-x\sin\theta\right),0\right\\},m_{1}\right\\},&\text{if}\
\ r\geq R+1,\ \ r\cos\theta-x\sin\theta\geq 0,\\\
\max\left\\{\min\left\\{\sqrt{\lambda}r\left(r\cos\theta-x\sin\theta\right),0\right\\},m_{2}\right\\},&\text{if}\
\ r\geq R+1,\ \ r\cos\theta-x\sin\theta\leq 0,\\\ m_{1},&\text{if}\ \ x\leq-
R_{0},\ \ R\leq r\leq R+1,\\\ m_{2},&\text{if}\ \ x\geq R_{0},\ \ R\leq r\leq
R+1,\\\ \overline{\psi}(X),&\text{if
$\displaystyle(x,r)\in\Omega_{\mu,R_{0}}$},\\\
\eta(x)\frac{m_{1}}{r_{1,\mu}^{2}}r^{2}+(1-\eta(x))\frac{m_{2}}{r_{2,\mu}^{2}}r^{2},&\text{if
$\displaystyle(x,r)\in\Omega^{\prime}_{\mu}$}.\end{array}\right.$

Here,
$\displaystyle\Omega^{\prime}_{\mu}=\Omega_{\mu}\cap\\{r\leq\min\\{r_{1,\mu},r_{2,\mu}\\}\\}$,
$\displaystyle\eta(x)$ be a cut-off function satisfying

$\eta(x)=1\ \ \text{for}\ \ x\leq-\mu,\ \ \eta(x)=\frac{\mu-x}{2\mu}\ \
\text{for}\ \ -\mu\leq x\leq\mu,\ \ \eta(x)=0\ \ \text{for}\ \ x\geq\mu,$
(3.7)

and

$\displaystyle\Omega_{\mu,R_{0}}=\Omega_{\mu}\cap\left\\{\min\\{r_{1,\mu},r_{2,\mu}\\}\leq
r\leq R\right\\}\cup\left\\{-R_{0}\leq x\leq R_{0},R\leq r\leq R+1\right\\}.$

It is easy to check that $\displaystyle
J_{\lambda,\theta,\mu}(\psi_{0})<+\infty$.

Case 2: For $\displaystyle\theta=0$ or $\displaystyle\pi$.

Without loss of generality, assume $\displaystyle\theta=0$. It suffice to
define a function $\displaystyle\psi_{0}(X)$ as follows. Set

$\displaystyle\tilde{\Omega}_{\mu,R_{0}}=\Omega_{\mu}\cap\left\\{\min\\{r_{1,\mu},r_{2,\mu}\\}\leq
r\leq R\right\\}\cup\left\\{-2\leq x\leq 2,R\leq r\leq R_{0}\right\\},$

for some sufficiently large $\displaystyle
R_{0}>\sqrt{R^{2}+\frac{2m_{1}}{\sqrt{\Lambda+\lambda}}-\frac{2m_{2}}{\sqrt{\lambda}}}$,
and define a function $\displaystyle\psi_{0}$ as

$\displaystyle\psi_{0}(X)=\left\\{\begin{array}[]{ll}\frac{\sqrt{\lambda}(r^{2}-R^{2})}{2}+m_{2},&\text{if}\
\ x\geq 2,\ \ R\leq r\leq\sqrt{R^{2}-\frac{2m_{2}}{\sqrt{\lambda}}},\\\
\min\left\\{\frac{\sqrt{\Lambda+\lambda}}{2}\left(r^{2}+\frac{2m_{2}}{\sqrt{\lambda}}-R^{2}\right),m_{1}\right\\},&\text{if}\
\ x\geq 2,\ \ r\geq\sqrt{R^{2}-\frac{2m_{2}}{\sqrt{\lambda}}},\\\
m_{1},&\text{if}\ \ x\leq 2,\ \ r\geq R_{0},\\\ m_{1},&\text{if}\ \ x\leq-2,\
\ R\leq r\leq R_{0},\\\ \overline{\psi}(X),&\text{if}\ \
(x,r)\in\tilde{\Omega}_{\mu,R_{0}},\\\
\eta(x)\frac{m_{1}}{r_{1,\mu}^{2}}r^{2}+(1-\eta(x))\frac{m_{2}}{r_{2,\mu}^{2}}r^{2},&\text{if
$\displaystyle(x,r)\in\Omega_{\mu}\cap\left\\{r\leq\\{r_{1,\mu},r_{2,\mu}\\}\right\\}$},\end{array}\right.$

where $\displaystyle\eta(x)$ is defined as (3.7), and
$\displaystyle\overline{\psi}$ be a smooth function such that
$\displaystyle\psi_{0}\in K_{\mu}$.

Therefore, we finish the proof of Proposition 3.1. ∎

Next, we will obtain the regularity of the minimizer.

######  Proposition 3.2.

Let $\displaystyle\psi_{\lambda,\theta,\mu}$ be a minimizer to the truncated
variational problem ($\displaystyle P_{\lambda,\theta,\mu}$), and for any open
subset
$\displaystyle\Omega_{0}\subset\subset\Omega_{\mu}\cap\\{m_{2}<\psi_{\lambda,\theta,\mu}<m_{1}\\}\cap\\{\psi_{\lambda,\theta,\mu}\neq
0\\}$, then $\displaystyle\psi_{\lambda,\theta,\mu}\in C^{0,1}(\Omega_{\mu})$,
$\displaystyle\psi_{\lambda,\theta,\mu}\in C^{2,\sigma}(\Omega_{0})$ and
$\displaystyle\psi_{\lambda,\theta,\mu}\in C^{1,\sigma}(\Omega_{0}\cup
N_{1,\mu}\cup N_{2,\mu})$ for some $\displaystyle 0<\sigma<1$.

###### Proof.

Firstly, $\displaystyle\psi_{\lambda,\theta,\mu}\in C^{0,1}(\Omega_{\mu})$
follows in the same manner as Corollary 4.4 in [6].

Next, the standard interior estimates to linear elliptic equation in Chapter 8
in [26] gives $\displaystyle\psi_{\lambda,\theta,\mu}\in
C^{2,\sigma}(\Omega_{0})$ and $\displaystyle\psi_{\lambda,\theta,\mu}\in
C^{1,\sigma}(N_{1,\mu}\cup N_{2,\mu})$.

The regularity of $\displaystyle\psi_{\lambda,\theta,\mu}$ near the axis
$\displaystyle N_{0,\mu}$ can be obtained by the standard arguments as in [16]
and [33].

Therefore, we finish the proof of Lemma 3.2. ∎

###  3.3. Uniqueness and monotonicity of the minimizer

Firstly, we will give a lower bound and an upper bound to the minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}$.

######  Lemma 3.3.

For any minimizer $\displaystyle\psi_{\lambda,\theta,\mu}$ to the variational
problem ($\displaystyle P_{\lambda,\theta,\mu}$), one has

$\max\left\\{\frac{m_{2}}{r_{2,\mu}^{2}}r^{2},m_{2}\right\\}\leq\psi_{\lambda,\theta,\mu}(x,r)\leq\min\left\\{\frac{m_{1}}{r_{1,\mu}^{2}}r^{2},m_{1}\right\\}\
\ \ \ \text{in}\ \ \ \Omega_{\mu},$ (3.8)

where $\displaystyle r_{1,\mu}$ and $\displaystyle r_{2,\mu}$ are defined in
(3.4).

###### Proof.

Firstly, consider the lower bound of $\displaystyle\psi_{\lambda,\theta,\mu}$.

Set
$\displaystyle\phi_{1}=\max\left\\{\frac{m_{2}}{r_{2,\mu}^{2}}r^{2},m_{2}\right\\}$,
$\displaystyle\phi_{2}=\min\left\\{\frac{m_{1}}{r_{1,\mu}^{2}}r^{2},m_{1}\right\\}$
and $\displaystyle\psi=\psi_{\lambda,\theta,\mu}$ for simplicity.

Firstly, since $\displaystyle\psi\in K_{\mu}$, one has

$m_{2}\leq\psi\leq m_{1}\ \ \text{in}\ \ \Omega_{\mu}.$ (3.9)

Next, we shall prove that

$\displaystyle\phi_{1}\leq\psi\ \ \text{in $\displaystyle\Omega_{\mu}$}.$

Due to the fact that $\displaystyle\max\left\\{\psi,\phi_{1}\right\\}\in
K_{\mu}$, we have

$\displaystyle J_{\lambda,\theta,\mu}(\psi)\leq
J_{\lambda,\theta,\mu}(\max\left\\{\psi,\phi_{1}\right\\}).$

Furthermore, the fact

$\phi_{1}=m_{2}\ \ \text{for $\displaystyle r\geq r_{2,\mu}$},$ (3.10)

gives that

$\displaystyle\psi\geq\phi_{1}\ \ \text{in}\ \ \Omega_{\mu}\cap\\{r\geq
r_{2,\mu}\\}.$

Therefore, we obtain

$\displaystyle 0\geq$
$\displaystyle\int_{\Omega_{\mu,r_{2,\mu}}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)\cdot e\right|^{2}dX$ (3.11)
$\displaystyle-\int_{\Omega_{\mu},r_{2,\mu}}r\left|\frac{\nabla\max\left\\{\psi,\phi_{1}\right\\}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\max\left\\{\psi,\phi_{1}\right\\}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\max\left\\{\psi,\phi_{1}\right\\}\leq
0\\}}\right)\cdot e\right|^{2}dX,$

here, $\displaystyle\Omega_{\mu,r_{2,\mu}}=\Omega_{\mu}\cap\\{r<r_{2,\mu}\\}$.
This together with the similar arguments as Lemma 3.4 in [31] yields to

$\displaystyle\int_{\Omega_{\mu,r_{2,\mu}}}\left|\nabla\min\left(\psi-\phi_{1},0\right)\right|^{2}dX\leq
0,$

which implies that

$\displaystyle\psi-\phi_{1}=\text{constant}\ \ \ \
\text{in}~{}\Omega_{\mu,r_{2,\mu}}.$

Since $\displaystyle\psi\geq\phi_{1}$ on
$\displaystyle\partial\Omega_{\mu,r_{2,\mu}}$, we conclude that

$\displaystyle\psi\geq\phi_{1}\ \ \ \ \text{in}~{}\Omega_{\mu,r_{2,\mu}}.$

Therefore, we obtain the lower bound of
$\displaystyle\psi_{\lambda,\theta,\mu}$ in (3.8).

Next, we can now proceed as before to show the upper bound of
$\displaystyle\psi_{\lambda,\theta,\mu}$.

Similarly, it suffices to prove that the upper bound holds in
$\displaystyle\Omega_{\mu,r_{1,\mu}}=\Omega_{\mu}\cap\left\\{r<r_{1,\mu}\right\\}$.
Due to (6.2), one has

$\displaystyle\Delta\psi_{\lambda,\theta,\mu}-\frac{1}{r}\frac{\partial\psi_{\lambda,\theta,\mu}}{\partial
r}\geq 0\ \ \text{in $\displaystyle\Omega_{\mu,r_{1,\mu}}$ in a weak sense,}$

and
$\displaystyle\psi_{\lambda,\theta,\mu}\leq\frac{m_{1}}{r_{1,\mu}^{2}}r^{2}$
on $\displaystyle\partial\Omega_{\mu,r_{1,\mu}}$, then the maximum principle
implies
$\displaystyle\psi_{\lambda,\theta,\mu}(x,r)\leq\frac{m_{1}}{r_{1,\mu}^{2}}r^{2}$
in $\displaystyle\Omega_{\mu,r_{1,\mu}}$. We complete the proof of Lemma 3.3.
∎

Next, in view of Lemma 3.3, using the similar arguments Proposition 3.5 in
[31], we will establish the uniqueness and some monotonicity of the minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}$ to the variational problem
($\displaystyle P_{\lambda,\theta,\mu}$), and we omit the proof here.

######  Proposition 3.4.

For any $\displaystyle\lambda\in(0,+\infty)$ and
$\displaystyle\theta\in[0,\pi]$, the minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}$ to the truncated variational problem
($\displaystyle P_{\lambda,\theta,\mu}$) is unique. Furthermore, the solution
$\displaystyle\psi_{\lambda,\theta,\mu}$ is monotonic with respect to
$\displaystyle x$, namely

$\psi_{\lambda,\theta,\mu}(x,r)\leq\psi_{\lambda,\theta,\mu}(\tilde{x},r)\ \ \
\ \text{for any}~{}~{}x\geq\tilde{x}.$ (3.12)

###  3.4. Some properties of the free boundaries

####  3.4.1. Preliminaries

Before we investigate the properties of the free boundaries, we give some
important auxiliary lemmas, and we refer the proofs in [1, 2, 23]. So we only
state the result and omit the proof as follows.

######  Lemma 3.5.

There exists a universal constant $\displaystyle c>0$, such that for
$\displaystyle
X_{0}=(x_{0},r_{0})\in\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}<0\\}$ and
$\displaystyle
B_{r}(X_{0})\subset\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}<0\\}$ with

$\displaystyle\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(\psi_{\lambda,\theta,\mu}-m_{2}\right)dS\geq\sqrt{\lambda}cr_{0},$

then we have $\displaystyle\psi_{\lambda,\theta,\mu}>m_{2}$ in $\displaystyle
B_{r}(X_{0})$; Similarly, $\displaystyle
X_{0}=(x_{0},r_{0})\in\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}>0\\}$ and
$\displaystyle
B_{r}(X_{0})\subset\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}>0\\}$, if

$\displaystyle\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(m_{1}-\psi_{\lambda,\theta,\mu}\right)dS\geq\sqrt{\lambda+\Lambda}cr_{0},$

then we have $\displaystyle\psi_{\lambda,\theta,\mu}<m_{1}$ in $\displaystyle
B_{r}(X_{0})$. Here and after, $\displaystyle B_{r}(X)$ denotes some ball with
radius $\displaystyle r>0$ and center $\displaystyle X\in\Omega_{\mu}$.

Next, we will establish a non-degeneracy lemma to
$\displaystyle\psi_{\lambda,\theta,\mu}-m_{2}$ and $\displaystyle
m_{1}-\psi_{\lambda,\theta,\mu}$ as follows.

######  Lemma 3.6.

(Non-degeneracy lemma) For any $\displaystyle 0<\kappa_{1}<1$, there exists a
positive constant $\displaystyle c$ (depending on $\displaystyle\kappa_{1}$),
if $\displaystyle
B_{r}(X_{0})\subset\Omega_{\mu}\cap\\{\psi_{\lambda,\theta,\mu}<0\\}$
($\displaystyle X_{0}=(x_{0},r_{0})$) and

$\displaystyle\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(\psi_{\lambda,\theta,\mu}-m_{2}\right)dS\leq\sqrt{\lambda}cr_{0},~{}~{}and~{}~{}\psi_{\lambda,\theta,\mu}<0~{}~{}in~{}~{}B_{r}(X_{0}),$

then $\displaystyle\psi_{\lambda,\theta,\mu}=m_{2}$ in $\displaystyle
B_{\kappa_{1}r}(X_{0})$; Similarly, for any $\displaystyle 0<\kappa_{2}<1$,
there exists a positive constant $\displaystyle c$ (depending on
$\displaystyle\kappa_{2}$) and

$\displaystyle\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(m_{1}-\psi_{\lambda,\theta,\mu}\right)dS\leq\sqrt{\lambda+\Lambda}cr_{0},~{}~{}and~{}~{}\psi_{\lambda,\theta,\mu}>0~{}~{}in~{}~{}B_{r}(X_{0}),$

then $\displaystyle\psi_{\lambda,\theta,\mu}=m_{1}$ in $\displaystyle
B_{\kappa_{2}r}(X_{0})$.

A direct application of Lemma 3.6 gives the following lemma.

######  Lemma 3.7.

Suppose that $\displaystyle
X_{0}=(x_{0},r_{0})\in\overline{\\{\psi_{\lambda,\theta,\mu}>m_{2}\\}\cap(\Omega_{\mu}\backslash
D_{\mu})}$ and $\displaystyle\psi_{\lambda,\theta,\mu}<0$ in $\displaystyle
B_{r}(X_{0})$ for some $\displaystyle r>0$, then

$\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(\psi_{\lambda,\theta,\mu}-m_{2}\right)dS\geq\sqrt{\lambda}cr_{0}.$
(3.13)

In particular,

$\sup_{\partial
B_{r}(X_{0})}\left(\psi_{\lambda,\theta,\mu}-m_{2}\right)\geq\sqrt{\lambda}cr_{0}r.$
(3.14)

We shall establish a non-oscillation lemma, which implies that the free
boundary $\displaystyle\Gamma_{i,\mu}$ for $\displaystyle i=1,2$ cannot
oscillate near the solid boundaries. Without loss of generality, consider the
right free boundary $\displaystyle\Gamma_{2,\mu}$, and introduce a domain
$\displaystyle G\subset\Omega_{\mu}\backslash D_{\mu}$ bounded by

$\displaystyle x=x_{1},\quad x=x_{1}+h\ (h>0),$

and

$\displaystyle\gamma_{1}:X=X^{1}(t)=(x^{1}(t),r^{1}(t)),\quad\gamma_{2}:X=X^{2}(t)=(x^{2}(t),r^{2}(t)),$

where $\displaystyle 0\leq t\leq T$ with

$\displaystyle x_{1}<x^{i}(t)<x_{1}+h\ \ \text{for}\ \ 0<t<T,$

and

$\displaystyle x^{i}(0)=x_{1},\ \ x^{i}(T)=x_{1}+h,\ \ r_{1}\leq r^{i}(t)\leq
r_{1}+\delta,\ \ i=1,2.$

Furthermore, the arc $\displaystyle\gamma_{2}$ lies above the arc
$\displaystyle\gamma_{1}$, this implies that $\displaystyle
r^{1}(0)<r^{2}(0)$, $\displaystyle\gamma_{1}$ and $\displaystyle\gamma_{2}$ do
not intersect, $\displaystyle\gamma_{2}$ is contained in
$\displaystyle\Gamma_{2,\mu}$, either

$\displaystyle\text{Case}\ 1.\ \gamma_{1}\ \ \text{is contained in}\
\Gamma_{2,\mu},\quad\text{(see Figure \ref{fi7})}$

or

$\displaystyle\text{Case}\ 2.\ \gamma_{1}\ \ \text{lies on}\ \\{r=R,x>1\\},\
\text{and then}\ \ r_{1}=R,\ x_{1}\geq 1.\quad\text{(see Figure \ref{fi8})}$

Figure 9. Case 1

Figure 10. Case 2

Let the domain $\displaystyle G\subset\\{\psi_{\lambda,\theta,\mu}>m_{2}\\}$
be a neighborhood of $\displaystyle\gamma_{1}$ and $\displaystyle\gamma_{2}$,
and $\displaystyle\psi_{\lambda,\theta,\mu}<0$ in $\displaystyle G$ and for
some $\displaystyle c^{*}>0$, we have

$\displaystyle\text{dist}(G,\overline{A_{1}A_{2}})>c^{*}.$

######  Lemma 3.8.

(Non-oscillation lemma) Under the foregoing assumptions, there exists a
positive constant $\displaystyle C$ depending only on $\displaystyle\lambda$,
$\displaystyle m_{2}$ and $\displaystyle c^{*}$ such that

$h\leq C\delta.$ (3.15)

The proof is similar to Lemma 4.1 in [2] and Lemma 5.6 [4], we omit it here.

Finally, we give the uniform bound of the gradient to the minimizer, which is
independent of $\displaystyle m_{1}$ and $\displaystyle m_{2}$. Please see
Lemma 8.1 in [4] and Lemma 5.2 in [2] for the proof.

######  Lemma 3.9.

Let $\displaystyle X_{0}=(x_{0},r_{0})$ be a free boundary point in
$\displaystyle\Omega_{\mu}\setminus\overline{D}_{\mu}$ and $\displaystyle G$
be a bounded domain with $\displaystyle X_{0}\in G$,
$\displaystyle\overline{G}\subset{\Omega_{\mu}\setminus\overline{D}_{\mu}}$.
There exists a constant $\displaystyle C>0$ depending only on
$\displaystyle\lambda$, $\displaystyle G$ and $\displaystyle\Lambda$, such
that

$\frac{\left|\nabla\psi_{\lambda,\theta,\mu}\right|}{r}\leq
C~{}~{}\text{in}~{}~{}G.$ (3.16)

####  3.4.2. Some properties of the free boundaries

It follows from the monotonicity of $\displaystyle\psi_{\lambda,\theta,\mu}$
with respect to $\displaystyle x$ that the free boundaries are $\displaystyle
r$-graph, namely, the free boundaries $\displaystyle\Gamma_{i,\mu}$
($\displaystyle i=1,2$) intersect $\displaystyle r=r_{0}$ either one single
point or a segment for any $\displaystyle r_{0}\in(R,+\infty)$. Thus, there
exist four mappings $\displaystyle g_{1,\lambda,\theta,\mu}(r)$ with
$\displaystyle r>R$, $\displaystyle g_{2,\lambda,\theta,\mu}(r)$ with
$\displaystyle r>R$, $\displaystyle g_{\lambda,\theta,\mu}(r)$ with
$\displaystyle r>0$ and $\displaystyle\tilde{g}_{\lambda,\theta,\mu}(r)$ with
$\displaystyle r>0$ such that

$\left\\{0<\psi_{\lambda,\theta,\mu}<m_{1}\right\\}\cap\Omega_{\mu}=\left\\{\tilde{g}_{1,\lambda,\theta,\mu}(r)<x<g_{\lambda,\theta,\mu}(r)\right\\}\cap\Omega_{\mu},$
(3.17)

and

$\left\\{m_{2}<\psi_{\lambda,\theta,\mu}<0\right\\}\cap\Omega_{\mu}=\left\\{\tilde{g}_{\lambda,\theta,\mu}(r)<x<\tilde{g}_{2,\lambda,\theta,\mu}(r)\right\\}\cap\Omega_{\mu},$
(3.18)

where

$\displaystyle\tilde{g}_{1,\lambda,\theta,\mu}(r)=\left\\{\begin{array}[]{ll}f_{1}(r)&\text{for}~{}0<r\leq
R,\\\ g_{1,\lambda,\theta,\mu}(r)&\text{for
}~{}R<r<+\infty,\end{array}\right.$

and

$\displaystyle\tilde{g}_{2,\lambda,\theta,\mu}(r)=\left\\{\begin{array}[]{ll}f_{2}(r)&\text{for}~{}0<r\leq
R,\\\ g_{2,\lambda,\theta,\mu}(r)&\text{for
}~{}R<r<+\infty.\end{array}\right.$

Indeed, along similar arguments as in [2], we obtain that $\displaystyle
g_{i,\lambda,\theta,\mu}(r)$ is indeed a general continuous function in
$\displaystyle[R,+\infty)$, and $\displaystyle g_{i,\lambda,\theta,\mu}(R)$ is
defined as $\displaystyle\lim_{r\rightarrow R^{+}}g_{i,\lambda,\theta,\mu}(r)$
for $\displaystyle i=1,2$. Furthermore, due to Lemma 3.3 in [7] and
Proposition 4.1 in [31], the interface $\displaystyle
g_{\lambda,\theta,\mu}(r)\equiv\tilde{g}_{\lambda,\theta,\mu}(y)$ is indeed a
continuous function in $\displaystyle[0,+\infty)$, and we omit the proof here.

######  Lemma 3.10.

The free boundary $\displaystyle\Gamma_{i,\mu}:x=g_{i,\lambda,\theta,\mu}(r)$
is a generalized continuous function in $\displaystyle R\leq r<+\infty$ with
values in $\displaystyle[-\infty,+\infty]$ ($\displaystyle i=1,2$),
respectively. Furthermore, the interface $\displaystyle\Gamma_{\mu}$:
$\displaystyle x=g_{\lambda,\theta,\mu}(r)$ is bounded continuous functions in
$\displaystyle 0<r<+\infty$, $\displaystyle
g_{\lambda,\theta,\mu}(0+0)\triangleq\lim_{r\rightarrow
0^{+}}g_{\lambda,\theta,\mu}(r)$ exists and is finite.

In order to study the limit behavior of the solution as $\displaystyle
r\rightarrow+\infty$, we first establish the decay estimate of the minimizer
in far field as follows. This is one of the crucial parts in this paper.

######  Lemma 3.11.

For any $\displaystyle\theta\in(0,\pi)$ and $\displaystyle r_{0}>2R$, there
exists a constant $\displaystyle C$ (independent of $\displaystyle r_{0}$)
such that

$\int_{\Omega_{\mu}\cap\\{r>r_{0}\\}}r\left|\frac{\nabla\psi_{\lambda,\theta,\mu}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{\lambda,\mu,\theta}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\theta,\mu}\leq
0\\}}\right)e\right|^{2}dX\leq\frac{C}{r_{0}^{3}}.$ (3.19)

###### Proof.

Denote $\displaystyle\psi(x,r)=\psi_{\lambda,\theta,\mu}(x,r)$, $\displaystyle
g(r)=g_{\lambda,\theta,\mu}(r)$ and $\displaystyle
g_{i}(r)=g_{i,\lambda,\theta,\mu}(r)$ ($\displaystyle i=1,2$) for simplicity.

For any $\displaystyle r_{0}>2R$, define

$\displaystyle
S(r_{0})=\int_{\Omega_{\mu}\cap\\{\frac{r_{0}}{2}<r<r_{0}\\}}r\left|\frac{\nabla\psi_{\lambda,\theta,\mu}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\mu,\theta}<0\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\theta,\mu}\leq
0\\}}\right)e\right|^{2}dX.$

Taking advantage of the mean value theorem, there exists some
$\displaystyle\tilde{r}\in\left(\frac{r_{0}}{2},r_{0}\right)$ such that

$\displaystyle S(r_{0})=$
$\displaystyle\frac{r_{0}}{2\tilde{r}}\left\\{\int_{g_{1}(\tilde{r})}^{g(\tilde{r})}\left|\nabla\psi(x,\tilde{r})-\sqrt{\Lambda+\lambda}\tilde{r}e\right|^{2}dx+\int_{g(\tilde{r})}^{g_{2}(\tilde{r})}\left|\nabla\psi(x,\tilde{r})-\sqrt{\lambda}\tilde{r}e\right|^{2}dx\right\\}$
(3.20) $\displaystyle\geq$
$\displaystyle\frac{1}{2}\left\\{\int_{g_{1}(\tilde{r})}^{g(\tilde{r})}\left|\nabla\psi(x,\tilde{r})-\sqrt{\Lambda+\lambda}\tilde{r}e\right|^{2}dx+\int_{g(\tilde{r})}^{g_{2}(\tilde{r})}\left|\nabla\psi(x,\tilde{r})-\sqrt{\lambda}\tilde{r}e\right|^{2}dx\right\\}.$

We choose a function $\displaystyle w(x,r)$ as follows

$w(x,r)=\left\\{\begin{array}[]{ll}\psi(x,r),&\text{in}\ \
\Omega_{\mu}\cap\\{r\leq\tilde{r}\\},\\\
\eta(r)\bar{\psi}(x,r)+(1-\eta(r))\phi(x,r),&\text{in}\ \
\Omega_{\mu}\cap\\{r\geq\tilde{r}\\},\end{array}\right.$ (3.21)

where
$\displaystyle\eta(r)=\max\left\\{0,\frac{\bar{r}-r}{\bar{r}-\tilde{r}}\right\\}$
with $\displaystyle\bar{r}=\tilde{r}+\frac{1}{\tilde{r}}$,

$\displaystyle\bar{\psi}(x,r)=\psi(x-(r-\tilde{r})\cot\theta,\tilde{r}),$

and

$\displaystyle\phi(x,r)=$
$\displaystyle\min\left\\{\max\left\\{\sqrt{\Lambda+\lambda}r\left((r-\tilde{r})\cos\theta-(x-g(\tilde{r}))\sin\theta\right),0\right\\},m_{1}\right\\}$
$\displaystyle+\max\left\\{\min\left\\{\sqrt{\lambda}r\left((r-\tilde{r})\cos\theta-(x-g(\tilde{r}))\sin\theta\right),0\right\\},m_{2}\right\\}.$

It’s easy to check that $\displaystyle w(x,r)\in K_{\mu}$, then $\displaystyle
J_{\lambda,\theta,\mu}(\psi)\leq J_{\lambda,\theta,\mu}(w)$, which implies

$\displaystyle\int_{\Omega_{\mu}\cap\\{r>\tilde{r}\\}}r\left|\frac{\nabla\psi_{\lambda,\theta,\mu}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\mu,\theta}<0\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\theta,\mu}\leq
0\\}}\right)e\right|^{2}dX$ (3.22) $\displaystyle\leq$
$\displaystyle\int_{\Omega_{\mu}\cap\\{r>\tilde{r}\\}}r\left|\frac{\nabla
w}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{m_{2}<w<0\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<w\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle=$
$\displaystyle\int_{\Omega_{\mu}\cap\\{\tilde{r}<r\leq\bar{r}\\}}r\left|\frac{\nabla
w}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{m_{2}<w<0\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<w\leq
0\\}}\right)e\right|^{2}dX$
$\displaystyle+\int_{\Omega_{\mu}\cap\\{r\geq\bar{r}\\}}r\left|\frac{\nabla
w}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{m_{2}<w<0\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<w\leq
0\\}}\right)e\right|^{2}dX.$

First, similar arguments as Lemma 3.10 in [10] and Lemma 4.1 in [11]. We
obtain

$\int_{\Omega_{\mu}\cap\\{r>\tilde{r}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX\leq\frac{C}{r_{0}^{3}}+\frac{S(r_{0})}{16},$ (3.23)

where $\displaystyle C$ is independent of $\displaystyle r_{0}$.

Next, $\displaystyle S(2r_{0})$ can be calculated as follows,

$\displaystyle S(2r_{0})=$
$\displaystyle\int_{\Omega_{\mu}\cap\\{r_{0}<r<2r_{0}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ (3.24) $\displaystyle\leq$
$\displaystyle\int_{\Omega_{\mu}\cap\\{r>\tilde{r}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle\leq$
$\displaystyle\frac{C}{r_{0}^{3}}+\frac{S(r_{0})}{16},$

where we have used
$\displaystyle\tilde{r}\in\left(\frac{r_{0}}{2},r_{0}\right)$ and (3.23).

Using mathematical induction for any $\displaystyle n\in\mathbb{N}$ and
(3.24), one has

$S(2^{n+1}R)\leq\frac{2C}{\left(2^{n}R\right)^{3}},\ \ n=0,1....$ (3.25)

Indeed, (3.25) holds for $\displaystyle n=0$ when choose $\displaystyle C$
large enough. If (3.25) holds for $\displaystyle n$, one has

$\displaystyle S(2^{n+2}R)=S(2\cdot
2^{n+1}R)\leq\frac{C}{(2^{n+1}R)^{3}}+\frac{S(2^{n+1}R)}{16}\leq\frac{C}{(2^{n+1}R)^{3}}+\frac{1}{16}\frac{2C}{(2^{n}R)^{3}}=\frac{2C}{\left(2^{n+1}R\right)^{3}},$

which implies (3.25) holds for $\displaystyle n+1$.

Therefore, for any $\displaystyle r_{0}>2R$, there exists a $\displaystyle
n_{0}$ such that $\displaystyle 2^{n_{0}}R\leq r_{0}\leq 2^{n_{0}+1}R$, this
together with (3.25) yields to

$\displaystyle\int_{\Omega_{\mu}\cap\\{r>r_{0}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle\leq$
$\displaystyle\int_{\Omega_{\mu}\cap\\{r>2^{n_{0}}R\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle\leq$
$\displaystyle\sum_{j=n_{0}}^{+\infty}S(2^{j+1}R)\leq\frac{\tilde{C}}{R^{3}r_{0}^{3}},$

where have used the following fact

$\displaystyle\\\ \sum_{j=n_{0}}^{+\infty}S(2^{j+1}R)\leq
2C\sum_{j=n_{0}}^{+\infty}\frac{1}{\left(2^{j}R\right)^{3}}\leq\frac{32C}{\left(2^{n_{0}+1}R\right)^{3}}\leq\frac{\tilde{C}}{r_{0}^{3}}.$

This completes the proof of Lemma 3.11.

∎

Firstly, we will show that some convergence of the minimizer in far field and
the free boundaries approach to the asymptotic direction
$\displaystyle\theta\in(0,\pi)$ as $\displaystyle r\rightarrow+\infty$ in the
far field.

######  Lemma 3.12.

Let $\displaystyle\theta\in(0,\pi)$,
$\displaystyle\psi_{n}(\tilde{x},\tilde{r})=\psi_{\lambda,\theta,\mu}\left(x_{n}+\frac{\tilde{x}}{r_{n}},r_{n}+\frac{\tilde{r}}{r_{n}}\right)$
with $\displaystyle X_{n}=(x_{n},r_{n})\in\Gamma_{1,\mu}$ and $\displaystyle
r_{n}\rightarrow+\infty$,
$\displaystyle\tilde{X}=(\tilde{x},\tilde{r})\in\mathbb{R}^{2}$, then for a
subsequence

$\displaystyle\psi_{n}(\tilde{x},\tilde{r})\rightarrow\Theta(\tilde{x},\tilde{r})\triangleq\left\\{\begin{array}[]{ll}m_{1},&\text{if}\quad\tilde{r}\cos\theta-\tilde{x}\sin\theta\geq
0,\\\
m_{1}+\sqrt{\Lambda+\lambda}(\tilde{r}\cos\theta-\tilde{x}\sin\theta),&\text{if
}\quad-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}\leq\tilde{r}\cos\theta-\tilde{x}\sin\theta\leq
0,\\\
\frac{m_{1}\sqrt{\lambda}}{\sqrt{\Lambda+\lambda}}+\sqrt{\lambda}(\tilde{r}\cos\theta-\tilde{x}\sin\theta),&\text{if
}\quad\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}\leq\tilde{r}\cos\theta-\tilde{x}\sin\theta\leq-\frac{m_{1}}{\sqrt{\Lambda+\lambda}},\\\
m_{2},&\text{if}\quad\tilde{r}\cos\theta-\tilde{x}\sin\theta\leq\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}},\end{array}\right.$
(3.26)

uniformly in any compact subset of $\displaystyle\mathbb{R}^{2}$. Furthermore,

$\displaystyle g^{\prime}_{1,\lambda,\theta,\mu}(r)\rightarrow\cot\theta\ \
\text{as}\ \ r\rightarrow+\infty.$

The similar conclusion holds for $\displaystyle X_{n}\in\Gamma_{2,\mu}$.

###### Proof.

Set $\displaystyle\psi=\psi_{\lambda,\theta,\mu}$, $\displaystyle
g(r)=g_{\lambda,\theta,\mu}(r)$ and $\displaystyle
g_{i}(r)=g_{i,\lambda,\theta,\mu}(r)$ ($\displaystyle i=1,2$) for simplicity.
Define $\displaystyle x=x_{n}+\frac{\tilde{x}}{r_{n}}$ and $\displaystyle
r=r_{n}+\frac{\tilde{r}}{r_{n}}$.

For any $\displaystyle R_{0}>0$, one has

$\displaystyle\int_{\\{|r-r_{n}|<R_{0}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ (3.27) $\displaystyle=$
$\displaystyle\int_{\\{|\tilde{r}|<R_{0}r_{n}\\}}\left(\frac{1}{r_{n}}+\frac{\tilde{r}}{r_{n}^{3}}\right)\left|\frac{\tilde{\nabla}\psi_{n}}{1+\frac{\tilde{r}}{r_{n}^{2}}}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{n}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{n}\leq
0\\}}\right)e\right|^{2}d\tilde{X},$

where
$\displaystyle\tilde{\nabla}=(\partial_{\tilde{x}},\partial_{\tilde{r}})$.

For large $\displaystyle r_{n}>R_{0}+2R$, in view of (3.19), we obtain

$\displaystyle\int_{\\{|r-r_{n}|<R_{0}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX\leq\frac{C}{\left(r_{n}-R_{0}\right)^{3}},$

which together with (3.27) implies that

$\int_{\Omega_{\mu}\cap\\{|\tilde{r}|<R_{0}r_{n}\\}}\left(1+\frac{\tilde{r}}{r_{n}^{2}}\right)\left|\frac{\tilde{\nabla}\psi_{n}}{1+\frac{\tilde{r}}{r_{n}^{2}}}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{n}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{n}\leq
0\\}}\right)e\right|^{2}d\tilde{X}\leq\frac{Cr_{n}}{\left(r_{n}-R_{0}\right)^{3}}.$
(3.28)

Recalling Proposition 3.2 and $\displaystyle\Theta(\tilde{x},\tilde{r})\in
H_{loc}^{1}(\mathbb{R}^{2})$, then there exist a subsequence
$\displaystyle\psi_{n_{k}}$ and two functions
$\displaystyle\gamma_{1},\gamma_{2}\in[0,1]$ such that

$\displaystyle\psi_{n_{k}}(\tilde{x},\tilde{r})\rightarrow\Theta(\tilde{x},\tilde{r})\
\ \text{weakly in $\displaystyle H_{loc}^{1}(\mathbb{R}^{2})$},$
$\displaystyle\psi_{n_{k}}(\tilde{x},\tilde{r})\rightarrow\Theta(\tilde{x},\tilde{r})\
\ \text{a.e. in $\displaystyle\mathbb{R}^{2}$},$
$\displaystyle\chi_{\\{0<\psi_{n_{k}}<m_{1}\\}}\rightarrow\gamma_{1}\ \
\text{weak-star in $\displaystyle L_{loc}^{\infty}(\mathbb{R}^{2})$,
$\displaystyle\gamma_{1}=1$ a.e. on
$\displaystyle\\{0<\Theta(\tilde{x},\tilde{r})<m_{1}\\}$},$

and

$\displaystyle\chi_{\\{m_{2}<\psi_{n_{k}}\leq 0\\}}\rightarrow\gamma_{2}\ \
\text{weak-star in $\displaystyle L_{loc}^{\infty}(\mathbb{R}^{2})$,
$\displaystyle\gamma_{2}=1$ a.e. on
$\displaystyle\\{m_{2}<\Theta(\tilde{x},\tilde{r})\leq 0\\}$},$

as $\displaystyle k\rightarrow+\infty$. This together with (3.28) gives that

$\tilde{\nabla}\Theta=\sqrt{\Lambda+\lambda}e\chi_{\\{0<\psi_{0}<m_{1}\\}}+\sqrt{\lambda}e\chi_{\\{m_{2}<\psi_{0}\leq
0\\}}\ \ \text{a.e.},$ (3.29)

in any compact subset $\displaystyle\Omega^{\prime}$ of
$\displaystyle\mathbb{R}^{2}$.

Lemma 3.9 implies that for sufficiently large $\displaystyle n$

$\displaystyle|\tilde{\nabla}\psi_{n}(\tilde{x},\tilde{r})|=\left|\frac{1}{r_{n}}\nabla\psi\left(x_{n}+\frac{\tilde{x}}{\tilde{r}_{n}},r_{n}+\frac{\tilde{r}}{r_{n}}\right)\right|\leq
c_{0},$

where the constant $\displaystyle c_{0}$ is independent of $\displaystyle
R_{0}$. Hence, we conclude that there exists a subsequence
$\displaystyle\psi_{n_{k}}\rightarrow\Theta(\tilde{x},\tilde{r})$ uniformly in
any compact subset of $\displaystyle\mathbb{R}^{2}$ and

$\displaystyle
m_{1}-\psi_{n}(\tilde{x},\tilde{r})=\psi(x_{n},r_{n})-\psi\left(x_{n}+\frac{\tilde{x}}{\tilde{r}_{n}},r_{n}+\frac{\tilde{r}}{r_{n}}\right)\leq|\tilde{\nabla}\psi_{n}||\tilde{X}|,\
\ \text{for}\ \ |\tilde{X}|<\frac{m_{1}}{c_{0}},$

which implies $\displaystyle\psi_{n}(\tilde{x},\tilde{r})>0$.

The non-degeneracy lemma 3.6 implies that

$\displaystyle\frac{1}{r}\fint_{\partial
B_{r}(0)}(m_{1}-\psi_{n}(\tilde{x},\tilde{r}))d\tilde{S}=\frac{1}{r_{n}}\frac{1}{\frac{r}{r_{n}}}\fint_{\partial
B_{\frac{r}{r_{n}}}(X_{n})}(m_{1}-\psi(x,r))dS\geq c\sqrt{\Lambda+\lambda},\ \
\text{for}\ \ r<\frac{m_{1}}{c_{0}},$

taking $\displaystyle n\rightarrow+\infty$, which implies that
$\displaystyle\Theta\not\equiv m_{1}$ in $\displaystyle B_{r}(0)$ and
$\displaystyle\Theta(0)=m_{1}$.

Define

$t=\tilde{x}\cos\theta+\tilde{r}\sin\theta\ \ \text{and}\ \
s=\tilde{r}\cos\theta-\tilde{x}\sin\theta,$ (3.30)

and $\displaystyle w(t,s)=\Theta(\tilde{x},\tilde{r})$, then (3.29) implies
that

$\displaystyle\frac{\partial w}{\partial t}=0,\ \ \frac{\partial w}{\partial
s}=\sqrt{\Lambda+\lambda}e\chi_{\\{0<w<m_{1}\\}}+\sqrt{\lambda}e\chi_{\\{m_{2}<w\leq
0\\}}\ \ \text{a.e. in $\displaystyle\Omega^{\prime}_{\mu}$}\ \ \text{and}\ \
w(0)=m_{1}.$

A direction computation gives that

$\displaystyle w(t,s)=\left\\{\begin{array}[]{ll}m_{1}&\text{if}~{}~{}s\geq
0,\\\ m_{1}+\sqrt{\Lambda+\lambda}s,&\text{if
}~{}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}\leq s\leq 0,\\\
\frac{m_{1}\sqrt{\lambda}}{\sqrt{\Lambda+\lambda}}+\sqrt{\lambda}s,&\text{if
}\quad\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}\leq
s\leq-\frac{m_{1}}{\sqrt{\Lambda+\lambda}},\\\
m_{2},&\text{if}~{}~{}s\leq\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}},\end{array}\right.$

which yields (3.26).

Next, let $\displaystyle X_{0}=(t,s)$ with $\displaystyle s>0$ or
$\displaystyle
s<\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}$, for
small $\displaystyle r>0$, then

$\displaystyle\lim_{n\rightarrow+\infty}\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}(m_{1}-\psi_{n})dS=0,\ \ \text{or}\ \
\lim_{n\rightarrow+\infty}\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}(\psi_{n}-m_{2})dS=0,$

respectively. Then, the non-degeneracy lemma implies that $\displaystyle
X_{0}$ is not a free boundary point for sufficiently large $\displaystyle n$.

Similarly, for the case
$\displaystyle\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}<s<0$,
one gets

$\displaystyle\lim_{n\rightarrow+\infty}\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}(m_{1}-\psi_{n})dS\rightarrow+\infty,\ \ \text{or}\ \
\lim_{n\rightarrow+\infty}\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}(\psi_{n}-m_{2})dS\rightarrow+\infty,\ \ \text{as}\ \
r\rightarrow 0,$

and then $\displaystyle X_{0}$ is not a free boundary point for sufficiently
large $\displaystyle n$.

Then, one has

$\partial\\{\psi_{n}>0\\}\rightarrow\left\\{s=\frac{m_{2}}{\sqrt{\lambda}}-\frac{m_{1}}{\sqrt{\Lambda+\lambda}}\right\\}\
\ \text{and}\ \ \partial\\{\psi_{n}<m_{1}\\}\rightarrow\left\\{s=0\right\\},$
(3.31)

locally in Hausdorff distance (see Definition 3.1 in [23]).

Noticing the flatness conditions in Section 7 in [1] for the free boundaries,
there exists a
$\displaystyle\xi_{1,n}\in\left(\min\left\\{r_{n},r_{n}+\frac{\tilde{r}}{r_{n}}\right\\},\max\left\\{r_{n},r_{n}+\frac{\tilde{r}}{r_{n}}\right\\}\right)$
such that

$\displaystyle\tilde{x}=r_{n}\left(g_{1}\left(r_{n}+\frac{\tilde{r}}{r_{n}}\right)-x_{n}\right)=r_{n}\left(g_{1}\left(r_{n}+\frac{\tilde{r}}{r_{n}}\right)-g_{1}(r_{n})\right)=g_{1}^{\prime}(\xi_{1,n})\tilde{r},$

thus, we obtain

$\displaystyle
g_{1}^{\prime}\left(r_{n}+\frac{\tilde{r}}{r_{n}}\right)\rightarrow\cot\theta,\
\ \text{as}\ \ n\rightarrow+\infty.$

Therefore, we complete the proof of Lemma 3.12. ∎

Next, for the critical cases $\displaystyle\theta=0$ or
$\displaystyle\theta=\pi$, we have the following facts.

######  Proposition 3.13.

Assume that there exist some free boundary points $\displaystyle
X_{n}=(x_{n},r_{n})\in\Gamma_{2,\mu}$, such that $\displaystyle
r_{n}\rightarrow\xi>R$ and $\displaystyle x_{n}\rightarrow+\infty$, where
$\displaystyle\xi$ is a finite positive number, then $\displaystyle\theta=0$.
Moreover, let
$\displaystyle\psi_{n}(\tilde{X})=\psi_{\lambda,\theta,\mu}(x_{n}+\tilde{x},r_{n}+\tilde{r})$,
then

$\displaystyle\psi_{n}(\tilde{X})\rightarrow\min\left\\{\max\left\\{\sqrt{\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}+\frac{\tilde{r}^{2}}{2}+\xi\tilde{r}\right),m_{2}\right\\},0\right\\}+\max\left\\{\min\left\\{\sqrt{\Lambda+\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}+\frac{\tilde{r}^{2}}{2}+\xi\tilde{r}\right),m_{1}\right\\},0\right\\},$

uniformly in any compact subset of
$\displaystyle\\{(\tilde{x},\tilde{r})\mid\tilde{r}>R-\xi\\}$. If
$\displaystyle
r_{n}\rightarrow\xi>\sqrt{R^{2}+\frac{2m_{1}}{\sqrt{\Lambda+\lambda}}-\frac{2m_{2}}{\sqrt{\lambda}}}$
and $\displaystyle x_{n}\rightarrow-\infty$, $\displaystyle\xi$ is a finite
positive number, then $\displaystyle\theta=\pi$ and

$\displaystyle\psi_{n}(\tilde{X})\rightarrow\min\left\\{\max\left\\{\sqrt{\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}-\frac{\tilde{r}^{2}}{2}-\xi\tilde{r}\right),m_{2}\right\\},0\right\\}+\max\left\\{\min\left\\{\sqrt{\Lambda+\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}-\frac{\tilde{r}^{2}}{2}-\xi\tilde{r}\right),m_{1}\right\\},0\right\\},$

uniformly in any compact subset of
$\displaystyle\\{(\tilde{x},\tilde{r})\mid\tilde{r}>R-\xi\\}$. The similar
assertion holds for $\displaystyle X_{n}=(x_{n},r_{n})\in\Gamma_{1,\mu}$.

###### Proof.

If $\displaystyle X_{n}=(x_{n},r_{n})\in\Gamma_{2,\mu}$ with $\displaystyle
r_{n}\rightarrow\xi$ ($\displaystyle\xi$ is a finite positive number) and
$\displaystyle x_{n}\rightarrow+\infty$. Set $\displaystyle x=x_{n}+\tilde{x}$
and $\displaystyle r=r_{n}+\tilde{r}$. For any large $\displaystyle R_{0}>0$,
the boundedness of $\displaystyle
J_{\lambda,\theta,\mu}(\psi_{\lambda,\theta,\mu})$ gives that

$\displaystyle\int_{\Omega_{\mu}\cap\\{|x-x_{n}|<R_{0}\\}\cap\\{R-\xi<r-r_{n}<R_{0}\\}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle=$
$\displaystyle\int_{\tilde{\Omega}_{\mu}\cap\\{|\tilde{x}|<R_{0}\\}\cap\\{R-\xi<\tilde{r}<R_{0}\\}}\left(r_{n}+\tilde{r}\right)\left|\frac{\tilde{\nabla}\psi_{n}}{r_{n}+\tilde{r}}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{n}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{n}\leq
0\\}}\right)e\right|^{2}d\tilde{X}\rightarrow 0,$

as $\displaystyle n\rightarrow+\infty$.

Along the similar arguments in Lemma 3.12, then there exists a subsequence
$\displaystyle\psi_{n_{k}}$ such that

$\displaystyle\psi_{n_{k}}\rightarrow\psi_{0}\ \ \text{weakly in
$\displaystyle H_{loc}(\Omega^{\prime}_{\mu})$},$

and

$\displaystyle\psi_{n_{k}}\rightarrow\psi_{0}\ \ \text{a.e. in
$\displaystyle\Omega^{\prime}_{\mu}$},$

as $\displaystyle k\rightarrow+\infty$, and

$\displaystyle\nabla\psi_{0}=(\tilde{r}+\xi)\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{0}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{0}\leq
0\\}}\right)(-\sin\theta,\cos\theta)\ \ \text{a.e.},$

in any compact subset $\displaystyle\Omega^{\prime}$ of
$\displaystyle\\{(\tilde{x},\tilde{r})\mid\tilde{r}>R-\xi\\}$. Furthermore,
it’s easy to see that $\displaystyle\psi_{0}(0,0)=m_{2}$,
$\displaystyle\psi_{0}\not\equiv m_{2}$ in any neighborhood of
$\displaystyle(0,0)$ and

$\psi(x_{n}+\tilde{x},R-\xi+r_{n})\rightarrow\psi_{0}(\tilde{x},R-\xi)=m_{2}\
\ \text{if}\ \ x_{n}\rightarrow+\infty.$ (3.32)

Next, we claim $\displaystyle\theta=0$. Suppose not, if
$\displaystyle\theta=\pi$, indeed, similar arguments as Lemma 3.12, we obtain

$\displaystyle\psi_{0}(\tilde{x},\tilde{r})=\min\left\\{\max\left\\{\sqrt{\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}-\frac{\tilde{r}^{2}}{2}-\xi\tilde{r}\right),m_{2}\right\\},0\right\\}+\max\left\\{\min\left\\{\sqrt{\Lambda+\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}-\frac{\tilde{r}^{2}}{2}-\xi\tilde{r}\right),m_{1}\right\\},0\right\\},$

which contradicts with (3.32).

If $\displaystyle\theta\in(0,\pi)$, since $\displaystyle\psi_{0}$ is smooth in
any compact subset of $\displaystyle
G\subset\\{m_{2}<\psi_{0}<0\\}\cap\\{\tilde{r}\geq R-\xi\\}$, one has

$\displaystyle\frac{\partial^{2}\psi_{0}}{\partial\tilde{x}\partial\tilde{r}}=-\sqrt{\lambda}\sin\theta,\
\ \frac{\partial^{2}\psi_{0}}{\partial\tilde{r}\partial\tilde{x}}=0,$

which derives a contradiction with $\displaystyle\theta\in(0,\pi)$.

Therefore, we obtain $\displaystyle\theta=0$. Along the similar arguments in
Lemma 3.12, one has

$\displaystyle\psi_{n}(\tilde{X})\rightarrow\min\left\\{\max\left\\{\sqrt{\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}+\frac{\tilde{r}^{2}}{2}+\xi\tilde{r}\right),m_{2}\right\\},0\right\\}+\max\left\\{\min\left\\{\sqrt{\Lambda+\lambda}\left(\frac{m_{2}}{\sqrt{\lambda}}+\frac{\tilde{r}^{2}}{2}+\xi\tilde{r}\right),m_{1}\right\\},0\right\\},$

uniformly in any compact subset of
$\displaystyle\\{(\tilde{x},\tilde{r})\mid\tilde{r}>R-\xi\\}$. The similar
conclusion holds if $\displaystyle X_{n}=(x_{n},r_{n})\in\Gamma_{1,\mu}$ with
$\displaystyle r_{n}\rightarrow\xi>R$ and $\displaystyle
x_{n}\rightarrow-\infty$.

Similarly, we can obtain the conclusion for $\displaystyle\theta=\pi$. Thus,
we complete the proof of Proposition 3.13.

∎

Now, we can obtain the convergence rate of distance of the two free boundaries
and the minimizer as follows.

######  Lemma 3.14.

For any $\displaystyle\theta\in(0,\pi)$ and $\displaystyle\alpha\in(0,2)$, the
free boundaries $\displaystyle x=g_{1,\lambda,\theta,\mu}(r)$, $\displaystyle
x=g_{2,\lambda,\theta,\mu}(r)$, the interface $\displaystyle
x=g_{\lambda,\theta,\mu}(r)$ and the minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}$ satisfy

$r(g_{\lambda,\theta,\mu}(r)-g_{1,\lambda,\theta,\mu}(r))\rightarrow\frac{m_{1}}{\sqrt{\Lambda+\lambda}\sin\theta},$
(3.33)
$r(g_{\lambda,\theta,\mu}(r)-g_{2,\lambda,\theta,\mu}(r))\rightarrow\frac{m_{2}}{\sqrt{\lambda}\sin\theta},$
(3.34)

and

$r^{\alpha}\left(\frac{\nabla\psi_{\lambda,\theta,\mu}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{\lambda,\theta,\mu}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\theta,\mu}\leq
0\\}}\right)e\right)\rightarrow 0,$ (3.35)

as $\displaystyle r\rightarrow+\infty$.

###### Proof.

Define
$\displaystyle\psi_{n}(\tilde{x},\tilde{r})=\psi_{\lambda,\theta,\mu}\left(x_{n}+\frac{\tilde{x}}{r_{n}},r_{n}+\frac{\tilde{r}}{r_{n}}\right)$
with $\displaystyle(x_{n},r_{n})\in\Gamma_{1,\lambda,\theta,\mu}$,
$\displaystyle r_{n}\rightarrow+\infty$. Set $\displaystyle
x=x_{n}+\frac{\tilde{x}}{r_{n}}$ and $\displaystyle
r=r_{n}+\frac{\tilde{r}}{r_{n}}$. The free boundaries and interface of
$\displaystyle\psi_{n}(\tilde{x},\tilde{r})$ are given by

$\left\\{(\tilde{x},\tilde{r})\mid
x_{n}+\frac{\tilde{x}}{r_{n}}=g_{i,\lambda,\theta,\mu}\left(r_{n}+\frac{\tilde{r}}{r_{n}}\right),i=1,2\right\\}\
\ \text{and}\ \ \left\\{(\tilde{x},\tilde{r})\mid
x_{n}+\frac{\tilde{x}}{r_{n}}=g_{\lambda,\theta,\mu}\left(r_{n}+\frac{\tilde{r}}{r_{n}}\right)\right\\}.$
(3.36)

Thanks to Lemma 3.12, we have

$\displaystyle\partial\\{0<\psi_{n}<m_{1}\\}$ converges to
$\displaystyle\partial\\{0<\Theta<m_{1}\\}$ locally in Hausdorff distance.

This together with (3.36) and (3.26), taking $\displaystyle\tilde{r}=0$,
yields that

$\displaystyle\tilde{x}\sin\theta=r_{n}\left(g_{\lambda,\theta,\mu}(r_{n})-x_{n}\right)\sin\theta=r_{n}\left(g_{\lambda,\theta,\mu}(r_{n})-g_{1,\lambda,\theta,\mu}(r_{n})\right)\sin\theta\rightarrow\frac{m_{1}}{\sqrt{\Lambda+\lambda}}.$

Furthermore, set
$\displaystyle\psi_{n}(\tilde{x},\tilde{r})=\psi_{\lambda,\theta,\mu}\left(x_{n}+\tilde{x},r_{n}+\tilde{r}\right)$
with $\displaystyle(x_{n},r_{n})\in\Gamma_{1,\lambda,\theta,\mu}$,
$\displaystyle r_{n}\rightarrow+\infty$, for any $\displaystyle\tilde{R}>0$,
and large $\displaystyle r_{n}>\tilde{R}+2R$, similarly in Lemma 3.11, one has

$\displaystyle\int_{\Omega_{\mu}\cap\\{|r-r_{n}|<\tilde{R}\\}}r^{2\alpha}\left|\frac{\nabla\psi_{\lambda,\theta,\mu}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{\lambda,\theta,\mu}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{\lambda,\theta,\mu}\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle=$
$\displaystyle\int_{\\{|\tilde{r}|<\tilde{R}\\}}\left(\tilde{r}+r_{n}\right)^{2\alpha}\left|\frac{\tilde{\nabla}\psi_{n}}{\tilde{r}+r_{n}}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{n}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{n}\leq
0\\}}\right)e\right|^{2}d\tilde{X}$ $\displaystyle\leq$
$\displaystyle\frac{C(r_{n}+\tilde{R})^{2\alpha-1}}{\left(r_{n}-\tilde{R}\right)^{3}}.$

Hence for any $\displaystyle\alpha\in(0,2)$ and
$\displaystyle(\tilde{x},\tilde{r})\in\mathbb{R}^{2}$, one has

$\displaystyle\left(\tilde{r}+r_{n}\right)^{\alpha}\left|\frac{\tilde{\nabla}\psi_{n}}{\tilde{r}+r_{n}}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi_{n}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi_{n}\leq
0\\}}\right)e\right|\rightarrow 0\ \ \text{as $\displaystyle
n\rightarrow+\infty$},$

taking $\displaystyle\tilde{r}=0$, $\displaystyle x=x_{n}+\tilde{x}$ and
$\displaystyle r=r_{n}$ yields to the desired estimate (3.35).

Therefore, we complete the proof of Lemma 3.14. ∎

Next, we will prove that one of free boundaries will vanish, provided that the
asymptotic direction of the outgoing jet is horizontal. We call that
$\displaystyle\Gamma_{1,\mu}$ vanishes in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$, means
$\displaystyle\psi_{\lambda,\theta,\mu}<m_{1}$ in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$, and similarly, we call that the free
boundary $\displaystyle\Gamma_{2,\mu}$ vanishes in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$ means that
$\displaystyle\psi_{\lambda,\theta,\mu}>m_{2}$ in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$.

######  Proposition 3.15.

(1). If $\displaystyle\theta=\pi$, then the left free boundary
$\displaystyle\Gamma_{1,\mu}$ vanishes in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$;

(2). If $\displaystyle\theta=0$, then the right free boundary
$\displaystyle\Gamma_{2,\mu}$ vanishes in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$.

###### Proof.

Denote $\displaystyle\psi=\psi_{\lambda,\theta,\mu}$ for simplicity.

For $\displaystyle\theta=\pi$, then $\displaystyle e=(0,-1)$, for
$\displaystyle(x,r)\in\Omega_{\mu}$, set

$\displaystyle\psi_{0}(x,r)=\max\left\\{\min\left\\{m_{1}-\frac{\sqrt{\Lambda+\lambda}(r^{2}-R^{2})}{2},m_{1}\right\\},0\right\\}+\min\left\\{\max\left\\{\frac{\sqrt{\lambda}(R^{2}+\frac{2m_{1}}{\sqrt{\Lambda+\lambda}}-r^{2})}{2},m_{2}\right\\},0\right\\}.$

Next, we claim that

$\psi(x,r)\leq\psi_{0}(x,r)\ \ \text{in $\displaystyle\Omega_{\mu}$}.$ (3.37)

Suppose that the assertion (3.37) is not true, recalling that
$\displaystyle\min\\{\psi,\psi_{0}\\}\in K_{\mu}$ and the uniqueness of
minimizer, we obtain

$\displaystyle
J_{\lambda,\theta,\mu}(\psi)<J_{\lambda,\theta,\mu}(\min\\{\psi,\psi_{0}\\}).$

This implies that there exists some sufficiently large $\displaystyle
R_{0}>\max\left\\{\sqrt{\frac{2m_{1}}{\sqrt{\Lambda+\lambda}}-\frac{2m_{2}}{\sqrt{\lambda}}+R^{2}},1\right\\}$,
and

$\displaystyle 0>$
$\displaystyle\int_{\Omega_{\mu,R_{0}}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$ (3.38)
$\displaystyle-\int_{\Omega_{\mu,R_{0}}}r\left|\frac{\nabla\min\\{\psi,\psi_{0}\\}}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\min\\{\psi,\psi_{0}\\}<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\min\\{\psi,\psi_{0}\\}\leq
0\\}}\right)e\right|^{2}dX$ $\displaystyle=$
$\displaystyle\int_{\Omega_{\mu,R_{0}}}\frac{\nabla\max\\{\psi-\psi_{0},0\\}\cdot\nabla(\psi+\psi_{0})}{r}dX$
$\displaystyle-2\sqrt{\Lambda+\lambda}\int_{\Omega_{\mu,R_{0}}}\nabla\psi\cdot
e\chi_{\\{0<\psi<m_{1}\\}}-\nabla\min\left\\{\psi,\psi_{0}\right\\}\cdot
e\chi_{\\{0<\min\left\\{\psi,\psi_{0}\right\\}<m_{1}\\}}dX$
$\displaystyle-2\sqrt{\lambda}\int_{\Omega_{\mu,R_{0}}}\nabla\psi\cdot
e\chi_{\\{m_{2}<\psi\leq 0\\}}-\nabla\min\left\\{\psi,\psi_{0}\right\\}\cdot
e\chi_{\\{m_{2}<\min\left\\{\psi,\psi_{0}\right\\}\leq 0\\}}dX$
$\displaystyle+\int_{\Omega_{\mu,R_{0}}}(\Lambda+\lambda)r\left(\chi_{\\{0<\psi<m_{1}\\}}-\chi_{\\{0<\min\\{\psi,\psi_{0}\\}<m_{1}\\}}\right)+\lambda
r\left(\chi_{\\{m_{2}<\psi\leq
0\\}}-\chi_{\\{m_{2}<\min\\{\psi,\psi_{0}\\}\leq 0\\}}\right)dX$
$\displaystyle=$ $\displaystyle I_{1}+I_{2}+I_{3}+I_{4},$

where $\displaystyle\Omega_{\mu,R_{0}}$ is bounded by $\displaystyle
N_{i,\mu}$, $\displaystyle N_{0,\mu}$, $\displaystyle L_{i}$, $\displaystyle
H_{i,\mu}$, $\displaystyle\left\\{((-1)^{i},r)\mid R\leq r\leq R_{0}\right\\}$
and $\displaystyle\\{(x,R_{0})\mid-R_{0}\leq x\leq R_{0}\\}$ for
$\displaystyle i=1,2$.

The first term $\displaystyle I_{1}$ can be estimated as follows,

$\displaystyle I_{1}=$
$\displaystyle\int_{\Omega_{\mu,R_{0}}}\frac{|\nabla\max\\{\psi-\psi_{0},0\\}|^{2}}{r}dX+2\int_{\Omega_{\mu,R_{0}}}\frac{\nabla\max\\{\psi-\psi_{0},0\\}\cdot\nabla\psi_{0}}{r}dX$
(3.39) $\displaystyle=$
$\displaystyle\int_{\Omega_{\mu,R_{0}}}\frac{|\nabla\max\\{\psi-\psi_{0},0\\}|^{2}}{r}dX-2\sqrt{\Lambda+\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}=0\\}}\max\\{\psi-\psi_{0},0\\}dx$
$\displaystyle+2\sqrt{\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}=0\\}}\max\\{\psi-\psi_{0},0\\}dx-2\sqrt{\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}<0\\}\cap\\{r=R_{0}\\}}\max\\{\psi-\psi_{0},0\\}dx.$

Furthermore, for the second term $\displaystyle I_{2}$, one has

$\displaystyle I_{2}=$
$\displaystyle-2\sqrt{\Lambda+\lambda}\left\\{\int_{\Omega_{\mu,R_{0}}\cap\\{0<\psi_{0}<m_{1}\\}\cap\\{0<\psi<m_{1}\\}}\nabla\max\left\\{\psi-\psi_{0},0\right\\}\cdot
edX\right.$ (3.40)
$\displaystyle+\left.\int_{\Omega_{\mu,R_{0}}\cap\\{\psi_{0}<0\\}\cap\\{0<\psi<m_{1}\\}}\nabla(\psi-\psi_{0})\cdot
edX-\int_{\Omega_{\mu,R_{0}}\cap\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}\nabla(m_{1}-\psi_{0})\cdot
edX\right\\}$ $\displaystyle=$ $\displaystyle
2\sqrt{\Lambda+\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}=0\\}}\max\\{\psi-\psi_{0},0\\}dx.$

Similarly, we obtain

$I_{3}=-2\sqrt{\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}=0\\}}\max\\{\psi-\psi_{0},0\\}dx+2\sqrt{\lambda}\int_{\bar{\Omega}_{\mu,R_{0}}\cap\\{\psi_{0}<0\\}\cap\\{r=R_{0}\\}}\max\\{\psi-\psi_{0},0\\}dx.$
(3.41)

Finally, we have

$\displaystyle I_{4}\geq$
$\displaystyle(\Lambda+\lambda)\int_{\Omega_{\mu,R_{0}}}r\left(\chi_{\\{0<\psi<m_{1}\\}\cap\\{\psi_{0}\leq
0\\}}-\chi_{\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}\right)dX-\lambda\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{m_{2}<\psi_{0}\leq
0\\}\cap\\{\psi>0\\}}dX$ (3.42) $\displaystyle\geq$
$\displaystyle-(\Lambda+\lambda)\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}dX-\lambda\int_{\Omega_{\mu,R_{0}}\cap\\{m_{2}<\psi_{0}\leq
0\\}}r\left(\chi_{\\{\psi>0\\}}-\chi_{\\{0<\psi<m_{1}\\}}\right)dX$
$\displaystyle\geq$
$\displaystyle-(\Lambda+\lambda)\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}dX-\lambda\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{m_{2}<\psi_{0}\leq
0\\}\cap\\{\psi=m_{1}\\}}dX.$

Inserting (3.39)-(3.42) into (3.38) yields

$\displaystyle 0>$
$\displaystyle\int_{\Omega_{\mu,R_{0}}}\frac{|\nabla\max\\{\psi-\psi_{0},0\\}|^{2}}{r}dX-(\Lambda+\lambda)\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}dX$
$\displaystyle-\lambda\int_{\Omega_{\mu,R_{0}}}r\chi_{\\{m_{2}<\psi_{0}\leq
0\\}\cap\\{\psi=m_{1}\\}}dX$ $\displaystyle=$
$\displaystyle\int_{\Omega_{\mu,R_{0}}\cap\\{0<\psi_{0}<m_{1}\\}\cap\\{\psi=m_{1}\\}}\left(\frac{|\nabla\psi_{0}|^{2}}{r}-(\Lambda+\lambda)r\right)dX+\int_{\Omega_{\mu,R_{0}}\cap\\{0<\psi_{0}<\psi<m_{1}\\}}\frac{|\nabla(\psi-\psi_{0})|^{2}}{r}dX$
$\displaystyle+\int_{\Omega_{\mu,R_{0}}\cap\\{m_{2}<\psi_{0}\leq
0\\}\cap\\{\psi=m_{1}\\}}\left(\frac{|\nabla\psi_{0}|^{2}}{r}-\lambda
r\right)dX+\int_{\Omega_{\mu,R_{0}}\cap\\{m_{2}<\psi_{0}\leq
0\\}\cap\\{\psi_{0}<\psi<m_{1}\\}}\frac{|\nabla(\psi-\psi_{0})|^{2}}{r}dX$
$\displaystyle=$
$\displaystyle\int_{\Omega_{\mu,R_{0}}\cap\\{m_{2}<\psi_{0}<\psi<m_{1}\\}}\frac{|\nabla(\psi-\psi_{0})|^{2}}{r}dX,$

which derives a contradiction. Hence, (3.37) holds, it implies that

$\displaystyle\psi(x,r)<m_{1}\ \ \text{in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$},$

and thus, this gives that the free boundaries $\displaystyle\Gamma_{1,\mu}$
vanishes.

For $\displaystyle\theta=0$, taking

$\displaystyle\psi_{0}=\min\left\\{\max\left\\{\frac{\sqrt{\lambda}(r^{2}-R^{2})}{2}+m_{2},m_{2}\right\\},0\right\\}+\max\left\\{\min\left\\{\frac{\sqrt{\Lambda+\lambda}(r^{2}-R^{2}+\frac{2m_{2}}{\sqrt{\lambda}})}{2},m_{1}\right\\},0\right\\}.$

Similar arguments as before, yield that

$\displaystyle\psi(x,r)\geq\psi_{0}(x,r)\ \ \text{in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$},$

which implies that the free boundary $\displaystyle\Gamma_{2,\mu}$ is empty.

Therefore, we complete the proof of Proposition 3.15. ∎

######  Remark 3.1.

Furthermore, we define $\displaystyle g_{1,\lambda,\theta,\mu}(R)=-\infty$ for
$\displaystyle\theta=\pi$, $\displaystyle g_{2,\lambda,\theta,\mu}(R)=+\infty$
for $\displaystyle\theta=0$, respectively.

Proposition 3.15 implies that the one of free boundaries vanishes for
horizontal asymptotic direction, and on another side, we will show that the
both of two free boundaries are non-empty, for non-horizontal asymptotic
direction.

######  Lemma 3.16.

If $\displaystyle\theta\in(0,\pi)$, then $\displaystyle\Gamma_{i,\mu}$ is non-
empty and a connected curve, $\displaystyle x=g_{i,\lambda,\theta,\mu}(r)$ is
continuous in $\displaystyle(R,+\infty)$. And $\displaystyle\lim_{r\rightarrow
R^{+}}g_{i,\lambda,\theta,\mu}(r)$ exists and denoted as $\displaystyle
g_{i,\lambda,\theta,\mu}(R+0)$ for $\displaystyle i=1,2$.

###### Proof.

Step 1. We will show that $\displaystyle\Gamma_{i,\mu}$ is non-empty for
$\displaystyle i=1,2$.

Firstly, we claim that there exists a constant $\displaystyle R_{0}>0$, such
that

$\displaystyle B_{R_{0}}(X_{0})\subset\Omega_{\mu}\cap\\{r>R_{0}\\}$

contains a free boundary point $\displaystyle
X_{0}=(x_{0},R+R_{0})\in\Omega_{\mu}$ for any $\displaystyle
0<\psi_{\lambda,\theta,\mu}(X_{0})<m_{1}$.

Indeed, suppose not, we have $\displaystyle
B_{R_{0}}(X_{0})\cap\Gamma_{1,\mu}=\varnothing$. Similar arguments as Lemma
3.7, one gets

$\displaystyle\sup_{\partial
B_{R_{0}}(X_{0})}\left(m_{1}-\psi_{\lambda,\theta,\mu}\right)\geq
c\sqrt{\Lambda+\lambda}R(R+R_{0}),$

which implies $\displaystyle R_{0}\leq\frac{m_{1}}{c\sqrt{\Lambda+\lambda}R}$.
This is impossible for sufficiently large $\displaystyle R_{0}$. Hence, the
claim holds.

Without loss of generality, we assume that $\displaystyle\Gamma_{1,\mu}$ is
empty, then we obtain $\displaystyle\psi_{\lambda,\theta,\mu}<m_{1}$ in
$\displaystyle\Omega_{\mu}\cap\\{r>R\\}$.

In view of the claim, there is a sequence $\displaystyle
X_{n}=(x_{n},r_{n})\in\Gamma_{2,\mu}$ such that $\displaystyle R<r_{n}\leq c$
and $\displaystyle x_{n}\rightarrow-\infty$. Hence, there exists a subsequence
$\displaystyle X_{n_{k}}=(x_{n_{k}},r_{n_{k}})\in\Gamma_{2,\mu}$ and
$\displaystyle r_{n_{k}}\rightarrow\xi$, $\displaystyle
x_{n_{k}}\rightarrow-\infty$ as $\displaystyle k\rightarrow+\infty$. Due to
Proposition 3.13, we can prove that
$\displaystyle\psi_{\lambda,\theta,\mu}(X+X_{n_{k}})\rightarrow\psi_{0}(X)$
uniformly in any compact subset of $\displaystyle\\{(x,r)|r>R-\xi\\}$ as
$\displaystyle k\rightarrow+\infty$, where $\displaystyle\psi_{0}$ is a
constant flow with deflection angle $\displaystyle\theta=\pi$. This
contradicts with $\displaystyle\theta\in(0,\pi)$. Thus, the free boundaries
$\displaystyle\Gamma_{1,\mu}$ and $\displaystyle\Gamma_{2,\mu}$ are non-empty.

Step 2. We will verify that $\displaystyle\Gamma_{i,\mu}$ is a connected curve
and $\displaystyle x=g_{i,\lambda,\theta,\mu}(r)$ is a continuous function in
$\displaystyle[R,+\infty)$, $\displaystyle i=1,2$.

Without loss of generality, we consider the left free boundary. Let
$\displaystyle(\alpha,\beta)$ be the maximal interval such that $\displaystyle
x=g_{1,\lambda,\theta,\mu}(r)$ is finite-valued for all
$\displaystyle[R,+\infty)$.

Similar arguments as Section 5 in [2], we obtain $\displaystyle\alpha=R$, and
the limit $\displaystyle\lim_{r\rightarrow R}g_{1,\lambda,\theta,\mu}(r)$
exists.

If $\displaystyle\beta<+\infty$, one has

$\displaystyle x=g_{1,\lambda,\theta,\mu}(r)\rightarrow+\infty\quad\text{
or}\quad x=g_{1,\lambda,\theta,\mu}(r)\rightarrow-\infty\ \ \text{as}\ \
r\rightarrow\beta,$

which together with Proposition 3.13 implies $\displaystyle\theta=0$ or
$\displaystyle\theta=\pi$. This leads a contradiction to the assumption
$\displaystyle\theta\in(0,\pi)$.

Therefore, we complete the proof of Lemma 3.16. ∎

###  3.5. Monotonicity with respect to the parameter $\displaystyle\theta$

Next, we will establish a fact that the minimizer
$\displaystyle\psi_{\lambda,\theta,\mu}$ and free boundary $\displaystyle
x=g_{i,\lambda,\theta,\mu}(r)$ ($\displaystyle i=1,2$) are monotonic with
respect to the asymptotic deflection angle $\displaystyle\theta$.

######  Proposition 3.17.

Suppose that $\displaystyle\theta_{1},\theta_{2}\in[0,\pi]$ with
$\displaystyle\theta_{1}<\theta_{2}$,
$\displaystyle\psi_{\lambda,\theta_{1},\mu}$ and
$\displaystyle\psi_{\lambda,\theta_{2},\mu}$ are minimizers to the truncated
variational problem ($\displaystyle P_{\lambda,\theta_{1},\mu}$) and
($\displaystyle P_{\lambda,\theta_{2},\mu}$), and $\displaystyle
x=g_{i,\lambda,\theta_{1},\mu}(r)$ and $\displaystyle
x=g_{i,\lambda,\theta_{2},\mu}(r)$ be the free boundary of
$\displaystyle\psi_{\lambda,\theta_{1},\mu}$ and
$\displaystyle\psi_{\lambda,\theta_{2},\mu}$, respectively, then

$\psi_{\lambda,\theta_{1},\mu}\geq\psi_{\lambda,\theta_{2},\mu}\ \ \text{for
$\displaystyle(x,r)\in\Omega_{\mu}$},$ (3.43)

and

$g_{i,\lambda,\theta_{1},\mu}(r)>g_{i,\lambda,\theta_{2},\mu}(r)\ \ \text{for
$\displaystyle r\geq R$, $\displaystyle i=1,2$}.$ (3.44)

###### Proof.

Denote $\displaystyle\psi_{1}=\psi_{\lambda,\theta_{1},\mu}$ and
$\displaystyle\psi_{2}=\psi_{\lambda,\theta_{2},\mu}$ for simplicity, and set
$\displaystyle v_{1}=\max\left\\{\psi_{1},\psi_{2}\right\\}$ and
$\displaystyle v_{2}=\min\left\\{\psi_{1},\psi_{2}\right\\}$.

For $\displaystyle\theta_{1}<\theta_{2}$, as is customary Lemma 8.1 in [2], we
obtain

$\displaystyle
J_{\lambda,\theta_{1},\mu}(\psi_{1})=J_{\lambda,\theta_{1},\mu}(v_{1})\ \
\text{and}\ \
J_{\lambda,\theta_{2},\mu}(\psi_{2})=J_{\lambda,\theta_{2},\mu}(v_{2}).$

Since $\displaystyle\psi_{1}$ and $\displaystyle\psi_{2}$ are the minimizers
to the functionals $\displaystyle J_{\lambda,\theta_{1},\mu}$ and
$\displaystyle J_{\lambda,\theta_{2},\mu}$, respectively, we can now proceed
as in Theorem 7.1 in [4] to obtain that

$\displaystyle\text{either}~{}~{}\psi_{1}\geq\psi_{2}~{}~{}\text{or}~{}~{}\psi_{1}\leq\psi_{2}~{}\text{in}~{}\Omega_{\mu}.$

However, noticing that $\displaystyle\psi_{1}\geq\psi_{2}$ in
$\displaystyle\Omega_{\mu}\cap\\{r>R_{0}\\}$ for some sufficiently large
$\displaystyle R_{0}>R$, we conclude that the case
$\displaystyle\psi_{1}\geq\psi_{2}$ in $\displaystyle\Omega_{\mu}$.

Next, without loss of generality, we prove that (3.44) holds for
$\displaystyle i=1$, namely

$g_{1,\lambda,\theta_{1},\mu}(r)>g_{1,\lambda,\theta_{2},\mu}(r)\ \ \text{for
$\displaystyle r\geq R$}.$ (3.45)

Indeed, in view of (3.43), one has

$g_{1,\lambda,\theta_{1},\mu}(r)\geq g_{1,\lambda,\theta_{2},\mu}(r)\ \
\text{for}\ \ r\geq R.$ (3.46)

For any $\displaystyle r>R$, suppose not, there exists a point $\displaystyle
X_{0}=\left(x_{0},r_{0}\right)$ with $\displaystyle r_{0}>R$ such that

$\displaystyle
x_{0}=g_{1,\lambda,\theta_{1},\mu}(r_{0})=g_{1,\lambda,\theta_{2},\mu}(r_{0}).$

Since the free boundary $\displaystyle x=g_{1,\lambda,\theta_{1},\mu}(r)$ is
analytic in $\displaystyle r>R$, and applying Hopf’s lemma yields that

$\displaystyle\frac{\partial}{\partial\nu}\left(\psi_{\lambda,\theta_{1},\mu}-\psi_{\lambda,\theta_{2},\mu}\right)<0\
\ \text{at}~{}~{}X_{0},$

where $\displaystyle\nu$ is the unit outward normal vector of $\displaystyle
x=g_{1,\lambda,\theta_{1},\mu}(r)$ at $\displaystyle X_{0}$. This contradicts
to the free boundary conditions
$\displaystyle\sqrt{\lambda+\Lambda}=\frac{1}{r}\frac{\partial\psi_{\lambda,\theta_{1},\mu}}{\partial\nu}<\frac{1}{r}\frac{\partial\psi_{\lambda,\theta_{2},\mu}}{\partial\nu}=\sqrt{\lambda+\Lambda}$
at $\displaystyle X_{0}$.

On another side, for $\displaystyle r=R$, suppose that $\displaystyle
g_{1,\lambda,\theta_{1},\mu}(R)=g_{1,\lambda,\theta_{2},\mu}(R)$ and
$\displaystyle X_{0}=(g_{1,\lambda,\theta_{1},\mu}(R),R)$. If $\displaystyle
g_{1,\lambda,\theta_{1},\mu}(R)\leq-1$, let $\displaystyle G_{\delta}$ be a
domain bounded by $\displaystyle N_{1}$, $\displaystyle L_{1}$,
$\displaystyle\Gamma_{1,\lambda,\theta_{1},\mu}$ and $\displaystyle\partial
B_{\delta}(X_{0})$, and $\displaystyle
G_{\delta}\subset\\{0<\psi_{\lambda,\theta_{1},\mu}<m_{1}\\}$. If
$\displaystyle g_{1,\lambda_{\mu},\theta_{1},\mu}(R)>-1$, set $\displaystyle
G_{\delta}$ be a domain bounded by $\displaystyle N_{1}$,
$\displaystyle\\{(x,R)\mid-1\leq x\leq
g_{1,\lambda_{\mu},\theta_{1},\mu}(R)\\}$,
$\displaystyle\Gamma_{1,\lambda,\theta_{1},\mu}$ and $\displaystyle\partial
B_{\delta}(X_{0})$, and $\displaystyle
G_{\delta}\subset\\{0<\psi_{\lambda,\theta_{1},\mu}<m_{1}\\}$.

Set

$\displaystyle\tilde{\psi}_{\varepsilon}=(1+\varepsilon)(m_{1}-\psi_{\lambda,\theta_{1},\mu})-(m_{1}-\psi_{\lambda,\theta_{2},\mu})\
\ \text{for some $\displaystyle\varepsilon>0$}.$

Recalling the fact $\displaystyle\psi_{1}\geq\psi_{2}$ in
$\displaystyle\Omega_{\mu}$, we can choose $\displaystyle\delta>0$
sufficiently small such that

$\displaystyle
G_{\delta}\subset\\{0<\psi_{\lambda,\theta_{1},\mu}<m_{1}\\}\cap\\{0<\psi_{\lambda,\theta_{2},\mu}<m_{1}\\}.$

It follows from the similar arguments as in Corollary 11.5 [23] that there
exists a small $\displaystyle\varepsilon>0$ such that

$\tilde{\psi}_{\varepsilon}<0\ \ \text{in $\displaystyle G_{\delta}$}.$ (3.47)

Hence, we obtain

$\displaystyle\frac{1}{R}\frac{\partial\tilde{\psi}_{\varepsilon}\left(g_{1,\lambda,\theta_{1},\mu}(R),R\right)}{\partial\nu}\geq
0,$

where $\displaystyle\nu$ is the unit normal vector of the left free boundary
$\displaystyle\Gamma_{1,\lambda,\theta_{1},\mu}$ at
$\displaystyle\left(g_{1,\lambda,\theta_{1},\mu}(R),R\right)$, then

$\displaystyle(1+\varepsilon)\sqrt{\lambda+\Lambda}\leq\sqrt{\lambda+\Lambda}.$

This leads a contradiction and then the inequality (3.45) holds for
$\displaystyle r=R$.

Therefore, we finish the proof of the Proposition 3.17. ∎

###  3.6. Continuous dependence to the parameters $\displaystyle\lambda$ and
$\displaystyle\theta$

In this subsection, a convergence result to the parameters
$\displaystyle\lambda$ and $\displaystyle\theta$ will be stated as follows.

######  Proposition 3.18.

For any $\displaystyle\lambda>0$ and $\displaystyle\theta\in[0,\pi]$, and
sequences $\displaystyle\lambda_{n}\rightarrow\lambda$,
$\displaystyle\theta_{n}\rightarrow\theta$ with
$\displaystyle\theta_{n}\in[0,\pi]$, let
$\displaystyle\psi_{\lambda_{n},\theta_{n},\mu}$ be the minimizer to the
variational problem ($\displaystyle P_{\lambda_{n},\theta_{n},\mu}$),
$\displaystyle x=g_{i,\lambda_{n},\theta_{n},\mu}(r)$ and $\displaystyle
x=g_{\lambda_{n},\theta_{n},\mu}(r)$ be the free boundary of
$\displaystyle\psi_{\lambda_{n},\theta_{n},\mu}$ and interface, respectively.
Then there exist three subsequences still labeled as
$\displaystyle\psi_{\lambda_{n},\theta_{n},\mu}$, $\displaystyle
g_{i,\lambda_{n},\theta_{n},\mu}(r)$ and $\displaystyle
g_{\lambda_{n},\theta_{n},\mu}(r)$ such that

$\psi_{\lambda_{n},\theta_{n},\mu}\rightarrow\psi_{\lambda,\theta,\mu}\ \
\text{weakly in}~{}~{}H_{loc}^{1}(\Omega_{\mu})~{}~{}\text{and pointwise}\
\text{in}\ \Omega_{\mu},$ (3.48)
$g_{i,\lambda_{n},\theta_{n},\mu}(r)\rightarrow g_{i,\lambda,\theta,\mu}(r)\ \
\text{uniformly for $\displaystyle r\geq R$},$ (3.49)

and

$g_{\lambda_{n},\theta_{n},\mu}(r)\rightarrow g_{\lambda,\theta,\mu}(r)\ \
\text{uniformly for $\displaystyle r\geq 0$}.$ (3.50)

Here, $\displaystyle\psi_{\lambda,\theta,\mu}$ is the minimizer to the
variational problem ($\displaystyle P_{\lambda,\theta,\mu}$) and
$\displaystyle x=g_{i,\lambda,\theta,\mu}(r)$ and $\displaystyle
x=g_{\lambda,\theta,\mu}(r)$ are the free boundary and interface of
$\displaystyle\psi_{\lambda,\theta,\mu}$ for $\displaystyle i=1,2$,
respectively.

###### Proof.

Firstly, recalling the following facts

$\displaystyle\psi_{\lambda_{n},\theta_{n},\mu}\in H_{loc}^{1}(\Omega_{\mu}),\
\ \left|\nabla\psi_{\lambda_{n},\theta_{n},\mu}\right|\leq C,$

and using diagonal procedure gives that there exists a subsequence
$\displaystyle\left\\{\psi_{\lambda_{n},\theta_{n},\mu}\right\\}_{n=1}^{\infty}$
and a function $\displaystyle\omega\in H_{loc}^{1}(\Omega_{\mu})$ for some
$\displaystyle 0<\alpha<1$ such that

$\displaystyle\psi_{\lambda_{n},\theta_{n},\mu}\rightarrow\omega\ \
\text{weakly
in}~{}~{}H_{loc}^{1}(\Omega_{\mu}),~{}~{}C_{loc}^{\alpha}(\Omega_{\mu})~{}~{}\text{
and pointwise in}~{}~{}\Omega_{\mu}.$

Along the similar arguments as Lemma 9.2 in [2], we obtain that
$\displaystyle\omega$ is indeed a minimizer to the truncated variational
problem ($\displaystyle P_{\lambda,\theta,\mu}$). Due to the uniqueness of
minimizer to the truncated variational problem ($\displaystyle
P_{\lambda,\theta,\mu}$), we have
$\displaystyle\omega=\psi_{\lambda,\theta,\mu}$. Therefore, we obtain the
convergence of (3.48).

Secondly, we will show the statement (3.49) for
$\displaystyle\Gamma_{1,\lambda_{n},\theta_{n},\mu}$. Indeed, for any
$\displaystyle r_{n}>R$, let

$\displaystyle
X_{n}=\left(g_{1,\lambda_{n},\theta_{n},\mu}(r_{n}),r_{n}\right)\in\Gamma_{1,\lambda_{n},\theta_{n},\mu},\
\ \text{and}\ \ X_{n}\rightarrow
X_{0}=(x_{0},r_{0}),~{}~{}\text{as}~{}~{}n\rightarrow+\infty.$

Then, for any small $\displaystyle r>0$, the non-degeneracy lemma implies that
there exist two positive constants $\displaystyle C_{1}$ and $\displaystyle
C_{2}$, such that

$\displaystyle C_{1}\lambda_{n}r_{n}\leq\frac{1}{r}\fint_{\partial
B_{r}(X_{n})}\left(m_{1}-\psi_{\lambda_{n},\theta_{n},\mu}\right)dS\leq
C_{2}\lambda_{n}r_{n}.$

Letting $\displaystyle n\rightarrow+\infty$ gives

$\displaystyle C_{1}\lambda r_{0}\leq\frac{1}{r}\fint_{\partial
B_{r}(X_{0})}\left(m_{1}-\psi_{\lambda,\theta,\mu}\right)dS\leq C_{2}\lambda
r_{0}.$

Moreover, recalling the non-degeneracy Lemma 3.5 and Lemma 3.6 yields that
$\displaystyle X_{0}\in\Gamma_{1,\mu}$. Hence, we obtain the assertion (3.49)
for $\displaystyle r\in(R,+\infty)$.

Using Lemma 10.4 in [23], we can obtain the result for $\displaystyle r=R$,
namely,

$\displaystyle g_{1,\lambda_{n},\theta_{n},\mu}(R)\rightarrow
g_{1,\lambda,\theta,\mu}(R)\ \ \text{as $\displaystyle n\rightarrow+\infty$}.$

Similarly, (3.49) holds for the right free boundary
$\displaystyle\Gamma_{2,\lambda_{n},\theta_{n},\mu}$.

Finally, similar arguments as Theorem 7.1 in [6], we can obtain (3.50). ∎

###  3.7. Continuous and smooth fit conditions of the free boundaries

In this subsection, we will verify that there exist two parameters
$\displaystyle\lambda$ and $\displaystyle\theta$, such that the free
boundaries $\displaystyle\Gamma_{i,\mu}$ connect smoothly at the end points
$\displaystyle A_{i}$ of the nozzles $\displaystyle N_{i}$ ($\displaystyle
i=1,2$), respectively. Namely, for any $\displaystyle\mu>0$, there exists a
pair of parameters $\displaystyle(\lambda_{\mu},\theta_{\mu})$ with
$\displaystyle\lambda_{\mu}>0$, $\displaystyle\theta_{\mu}\in(0,\pi)$, such
that

$\displaystyle g_{1,\lambda_{\mu},\theta_{\mu},\mu}(R)=-1\ \ \text{and}\ \
g_{2,\lambda_{\mu},\theta_{\mu},\mu}(R)=1.$

As already mentioned before, this is the main difference to the impinging free
jet without rigid nozzle walls.

To see this, we first define a set $\displaystyle\Sigma_{\mu}$ as

$\Sigma_{\mu}=\\{\lambda\mid\lambda\geq 0,\text{there exists a
$\displaystyle\theta\in(0,\pi)$, such that $\displaystyle
g_{1,\lambda,\theta,\mu}(R)<-1$ and $\displaystyle
g_{2,\lambda,\theta,\mu}(R)>1$}\\}.$ (3.51)

The following lemma implies that $\displaystyle\Sigma_{\mu}$ is non-empty.

######  Lemma 3.19.

There exists $\displaystyle\theta_{0}\in(0,\pi)$ such that

$g_{1,\lambda,\theta_{0},\mu}(R)<-1\ \ \text{and}\ \
g_{2,\lambda,\theta_{0},\mu}(R)>1,$ (3.52)

for sufficiently small $\displaystyle\lambda>0$.

###### Proof.

For any
$\displaystyle\Omega_{0}\subset\subset\Omega_{\mu}\cap\\{r<R\\}\cap\\{m_{2}<\psi_{\lambda,\theta,\mu}<0\\}$,
firstly, it follows from Lemma 5.2 in [2] that there exists a positive
constant $\displaystyle C$ (depending only on $\displaystyle\Omega_{0}$), such
that

$|\nabla\psi_{\lambda,\theta,\mu}|\leq C\lambda\ \ \text{in}\ \ \Omega_{0},$
(3.53)

provided that $\displaystyle\Omega_{0}$ contains a free boundary point.

For $\displaystyle\theta\in(0,\pi)$, suppose not, without loss of generality,
suppose $\displaystyle g_{2,\lambda,\theta,\mu}(R)\leq 1$.

Indeed, it follows from the monotonicity of
$\displaystyle\psi_{\lambda,\theta,\mu}(x,r)$ with respect to $\displaystyle
x$, that there exists a point $\displaystyle X_{1}\in\Omega_{\mu}$, such that

$\displaystyle\psi_{\lambda,\theta,\mu}(X_{1})=\frac{m_{2}}{2},\ \
\text{with}\ \ X_{1}=(x_{1},R),\ \ \text{and}\ \
g_{\lambda,\theta,\mu}(R)<x_{1}<1.$

Denote $\displaystyle X_{2}=(x_{2},R)$ as the initial point of the right free
boundary $\displaystyle x=g_{2,\lambda,\theta,\mu}(r)$, due to the
monotonicity of $\displaystyle\psi_{\lambda,\theta,\mu}$ with respect to
$\displaystyle x$, one has $\displaystyle x_{2}>x_{1}$. Taking $\displaystyle
X_{3}=\left(\frac{3x_{2}-x_{1}}{2},R\right)$, an arc
$\displaystyle\gamma\in\Omega_{\mu}\cap\\{r>R\\}$ connecting $\displaystyle
X_{1}$ to $\displaystyle X_{3}$ and $\displaystyle|\gamma|\leq
C|X_{2}-X_{1}|=C|x_{2}-x_{1}|$, which intersects $\displaystyle\Gamma_{2,\mu}$
at $\displaystyle X_{4}$, $\displaystyle\gamma_{0}$ denotes the arc part
$\displaystyle\gamma$ from $\displaystyle X_{1}$ to $\displaystyle X_{4}$.

Let $\displaystyle\Omega_{0}$ be bounded by $\displaystyle\gamma_{0}$,
$\displaystyle r=R$ and $\displaystyle\Gamma_{2,\mu}$, it follows from (3.53)
that

$|\nabla\psi_{\lambda,\theta,\mu}|\leq C\lambda\ \text{in}\ \
\Omega_{0}\setminus B_{\delta}(A_{2}),\ \ \text{for sufficiently small
$\displaystyle\delta>0$}.$ (3.54)

Hence, it follows from (3.54) that

$\displaystyle-\frac{m_{2}}{2}=\psi_{\lambda,\theta,\mu}(X_{1})-\psi_{\lambda,\theta,\mu}(X_{4})\leq\int_{\gamma_{0}}|\nabla\psi_{\lambda,\theta,\mu}|dl\leq
C\lambda|X_{1}-X_{2}|\leq C\lambda\left(1-g_{\lambda,\theta,\mu}(R)\right),$

which is impossible with sufficiently small $\displaystyle\lambda$. Therefore,
for some sufficiently small $\displaystyle\lambda>0$, one has $\displaystyle
g_{2,\lambda,\theta,\mu}(R)>1$.

Indeed, due to Proposition 3.18 and Remark 3.1, we obtain that there exists an
$\displaystyle\theta_{0}$ ($\displaystyle\pi-\theta_{0}\ll 1$) such that
(3.52) holds.

Hence, we complete the proof of this lemma. ∎

Next, the following lemma implies that the set $\displaystyle\Sigma_{\mu}$ has
a uniform positive lower bound.

######  Lemma 3.20.

If $\displaystyle\theta\in(0,\pi)$, we have

$\min\left\\{-g_{1,\lambda,\theta,\mu}(R),g_{2,\lambda,\theta,\mu}(R)\right\\}<1,$
(3.55)

for sufficiently large $\displaystyle\lambda$.

###### Proof.

Indeed, it suffices to prove that the free boundaries
$\displaystyle\Gamma_{1,\mu}:x=g_{1,\lambda,\theta,\mu}(r)$ and
$\displaystyle\Gamma_{2,\mu}:x=g_{2,\lambda,\theta,\mu}(r)$ with
$\displaystyle R\leq r\leq 2R$ are contained in a neighborhood of
$\displaystyle\Gamma_{\mu}:x=g_{\lambda,\theta,\mu}(r)$ for sufficiently large
$\displaystyle\lambda$.

Firstly, we prove that the free boundary $\displaystyle\Gamma_{1,\mu}$ with
$\displaystyle R\leq r\leq 2R$ is contained in a neighborhood of
$\displaystyle\Gamma_{\mu}$ for sufficiently large $\displaystyle\lambda$.

Suppose not, then there exists a small and fixed $\displaystyle r_{0}>0$ and
$\displaystyle\tilde{X}=(\tilde{x},\tilde{r})\in\Gamma_{1,\mu}\cap\left\\{R\leq
r\leq 2R\right\\}$, and $\displaystyle
B_{r_{0}}(\tilde{X})\subset\Omega_{\mu}\cap\\{r>R\\}\cap\\{0<\psi_{\lambda,\theta,\mu}<m_{1}\\}$,
such that for any $\displaystyle\lambda>0$,

$\displaystyle B_{r_{0}}(\tilde{X})\cap\Gamma_{\mu}=\varnothing.$

Thus, the non-degeneracy Lemma 3.6 implies that

$\displaystyle\sqrt{\lambda+\Lambda}C\tilde{r}\leq\frac{1}{r_{0}}\fint_{\partial
B_{r_{0}}(\tilde{X})}\left(m_{1}-\psi_{\lambda,\theta,\mu}\right)dS\leq\frac{m_{1}}{r_{0}},$

which yields

$\displaystyle\sqrt{\lambda+\Lambda}\leq\frac{m_{1}}{Cr_{0}R}.$

This leads to a contradiction for sufficiently large $\displaystyle\lambda>0$.

Similarly, we can prove that the free boundaries $\displaystyle\Gamma_{2,\mu}$
with $\displaystyle R\leq r\leq 2R$ is contained in a neighborhood of
$\displaystyle\Gamma_{\mu}$ for sufficiently large $\displaystyle\lambda$.

Therefore, we finish the proof of Lemma 3.20. ∎

Define

$\lambda_{\mu}=\sup\left\\{\lambda\mid\lambda\in\Sigma_{\mu}\right\\},$ (3.56)

Lemma 3.20 implies that there exists a positive constant $\displaystyle C$
independent of $\displaystyle\mu$, such that

$\displaystyle\lambda_{\mu}\leq C.$

Finally, we will check that there exists a
$\displaystyle\theta_{\mu}\in(0,\pi)$ such that

$g_{1,\lambda_{\mu},\theta_{\mu},\mu}(R)=-1\ \ \text{and}\ \
g_{2,\lambda_{\mu},\theta_{\mu},\mu}(R)=1.$ (3.57)

######  Proposition 3.21.

There exists a $\displaystyle\theta_{\mu}\in(0,\pi)$ such that (3.57) holds.
Furthermore, $\displaystyle
N_{i}\cup\Gamma_{i,\lambda_{\mu},\theta_{\mu},\mu}$ is $\displaystyle
C^{1}$-smooth in a neighborhood of $\displaystyle A_{i}$, for $\displaystyle
i=1,2$.

###### Proof.

Taking a sequence $\displaystyle(\lambda_{n},\theta_{n})$ such that

$\displaystyle g_{1,\lambda_{n},\theta_{n},\mu}(R)<-1,\ \
g_{2,\lambda_{n},\theta_{n},\mu}(R)>1,$

and

$\displaystyle\lambda_{n}\rightarrow\lambda_{\mu}>0,\ \
\theta_{n}\rightarrow\theta_{\mu}\in[0,\pi].$

Noticing the fact that $\displaystyle x=g_{i,\lambda,\theta,\mu}(r)$
$\displaystyle(i=1,2)$ is continuous with respect to the parameters
$\displaystyle\lambda$ and $\displaystyle\theta$, then we have

$g_{1,\lambda_{\mu},\theta_{\mu},\mu}(R)\leq-1\ \ \text{and}\ \
g_{2,\lambda_{\mu},\theta_{\mu},\mu}(R)\geq 1.$ (3.58)

Firstly, we claim that

$0<\theta_{\mu}<\pi.$ (3.59)

Without loss of generality, we suppose $\displaystyle\theta_{\mu}=0$, then for
any $\displaystyle\tilde{\theta}>0$, the monotonicity in Proposition 3.17
gives that

$\psi_{\lambda_{\mu},\tilde{\theta},\mu}\leq\psi_{\lambda_{\mu},0,\mu},$
(3.60)

and

$g_{1,\lambda_{\mu},\tilde{\theta},\mu}(r)<g_{1,\lambda_{\mu},0,\mu}(r)\quad\text{for}\
r\geq R.$ (3.61)

It follows from $\displaystyle\theta_{\mu}=0$ that $\displaystyle
g_{2,\lambda_{\mu},0,\mu}(R)=+\infty$, choosing
$\displaystyle\tilde{\theta}>0$ be sufficiently small, due to the convergence
of the free boundaries (3.49), one gets

$\displaystyle g_{2,\lambda_{\mu},\tilde{\theta},\mu}(R)\geq 2.$

Furthermore, (3.61) implies that

$\displaystyle g_{1,\lambda_{\mu},\tilde{\theta},\mu}(R)<-1.$

Then, we can choose a $\displaystyle\lambda_{0}>\lambda_{\mu}$ and
$\displaystyle\lambda_{0}-\lambda_{\mu}$ suitably small such that

$\displaystyle g_{1,\lambda_{0},\tilde{\theta},\mu}(R)<-1\ \ \text{and}\ \
g_{2,\lambda_{0},\tilde{\theta},\mu}(R)>1,$

which implies $\displaystyle\lambda_{0}\in\Sigma_{\mu}$. This contradicts with
the definition of $\displaystyle\lambda_{\mu}$.

Consequently, we obtain $\displaystyle\theta_{\mu}>0$. Similarly, we can prove
$\displaystyle\theta_{\mu}<\pi$. Hence, the claim (3.59) holds.

Moreover, we will verify the continuous fit conditions (3.57). Indeed, suppose
not, without loss of generality, we assume that

$\displaystyle g_{1,\lambda_{\mu},\theta_{\mu},\mu}(R)<-1.$

Taking $\displaystyle\tilde{\theta}\in(0,\theta_{\mu})$ with
$\displaystyle\theta_{\mu}-\tilde{\theta}$ being suitably small, then the
continuity of $\displaystyle g_{1,\lambda_{\mu},\theta_{\mu},\mu}(R)$ with
respect to $\displaystyle\theta$ gives

$\displaystyle g_{1,\lambda_{\mu},\tilde{\theta},\mu}(R)<-1.$

Similar to (3.61), we have

$\displaystyle 1\leq
g_{2,\lambda_{\mu},\theta_{\mu},\mu}(R)<g_{2,\lambda_{\mu},\tilde{\theta},\mu}(R).$

Hence,

$\displaystyle g_{1,\lambda_{\mu},\tilde{\theta},\mu}(R)<-1\ \ \text{and}\ \
g_{2,\lambda_{\mu},\tilde{\theta},\mu}(R)>1.$

Therefore, similar to the above arguments for
$\displaystyle\theta=\tilde{\theta}$, we can choose a
$\displaystyle\lambda_{0}>\lambda_{\mu}$ and
$\displaystyle\lambda_{0}-\lambda_{\mu}$ being sufficiently small, and
$\displaystyle\lambda_{0}\in\Sigma_{\mu}$, it leads a contradiction to the
definition of $\displaystyle\lambda_{\mu}$.

Thus, we obtain the continuous fit conditions (3.57).

Furthermore, the similar proof to the jet flow problem in [4] implies that the
free boundaries are $\displaystyle C^{1}$-smooth at the end points of the
nozzles $\displaystyle A_{i}$ $\displaystyle(i=1,2)$, we omit it here. ∎

###  3.8. Existence of the impinging outgoing jet

In order to obtain the existence of the impinging outgoing jet, we take a
sequence $\displaystyle\mu=\mu_{n}\rightarrow+\infty$, and the corresponding
($\displaystyle\lambda_{\mu_{n}},\theta_{\mu_{n}}$) with
$\displaystyle\lambda_{\mu_{n}}>0$ and
$\displaystyle\theta_{\mu_{n}}\in(0,\pi)$,

$\displaystyle g_{1,\lambda_{\mu_{n},\theta_{\mu_{n}},\mu_{n}}}(R)=-1,\ \
\text{and}\ \ g_{2,\lambda_{\mu_{n},\theta_{\mu_{n}},\mu_{n}}}(R)=1,$

then there exist a $\displaystyle\lambda\geq 0$ and
$\displaystyle\theta\in[0,\pi]$ and a subsequence $\displaystyle\mu_{n}$, such
that $\displaystyle\lambda_{\mu_{n}}\rightarrow\lambda$,
$\displaystyle\theta_{\mu_{n}}\rightarrow\theta$ and

$\displaystyle\psi_{\lambda_{\mu_{n}},\theta_{\mu_{n}},\mu_{n}}\rightarrow\psi_{\lambda,\theta}\
\ \text{weakly in}~{}~{}H_{loc}^{1}(\Omega)~{}~{}\text{and a.e
in}~{}~{}\Omega.$

The similar arguments as in Proposition 3.18 imply that
$\displaystyle\psi_{\lambda,\theta}$ is a local minimizer to the variational
problem $\displaystyle J_{\lambda,\theta}$, namely,

$\displaystyle J_{\Omega_{0}}(\psi_{\lambda,\theta})\leq J_{\Omega_{0}}(v)\ \
\text{for any $\displaystyle\Omega_{0}\subset\subset\Omega$ and $\displaystyle
v-\psi_{\lambda,\theta}\in H_{0}^{1}(\Omega_{0})$},$

where $\displaystyle
J_{\Omega_{0}}(v)=\int_{\Omega_{0}}r\left|\frac{\nabla\psi}{r}-\left(\sqrt{\Lambda+\lambda}\chi_{\\{0<\psi<m_{1}\\}}+\sqrt{\lambda}\chi_{\\{m_{2}<\psi\leq
0\\}}\right)e\right|^{2}dX$.

Furthermore, along the similar arguments in Proposition 6.1, we can check that
$\displaystyle\psi_{\lambda,\theta}$ is a weak solution to the boundary value
problem (2.4).

Since

$\displaystyle m_{2}\leq\psi_{\lambda,\theta}\leq m_{1}\ \
\text{in}~{}~{}\Omega,$

and

$\psi_{\lambda,\theta}(x,r)\geq\psi_{\lambda,\theta}(\tilde{x},r)\ \ \ \
\text{for any}~{}~{}x<\tilde{x},$ (3.62)

using the same arguments as before, there exist two $\displaystyle
C^{1}$-smooth functions $\displaystyle x=g_{1,\lambda,\theta}(r)$ and
$\displaystyle x=g_{2,\lambda,\theta}(r)$ such that

$g_{1,\lambda_{\mu_{n}},\theta_{\mu_{n}},{\mu_{n}}}(r)\rightarrow
g_{1,\lambda,\theta}(r)\ \ \text{for any $\displaystyle r\in[R,+\infty)$},$
(3.63)

and

$g_{2,\lambda_{\mu_{n}},\theta_{\mu_{n}},{\mu_{n}}}(r)\rightarrow
g_{2,\lambda,\theta}(r)\ \ \text{for any $\displaystyle r\in[R,+\infty)$}.$
(3.64) $g_{1,\lambda,\theta}(R)=-1,\quad\text{and}\quad
g_{2,\lambda,\theta}(R)=1.$ (3.65)

Furthermore, along the similar arguments as Lemma 3.19 and 3.20, we assert
that

$\lambda>0\ \ \text{and}\ \ 0<\theta<\pi,$ (3.66)

and the smooth fit condition of $\displaystyle x=g_{i,\lambda,\theta}(r)$ at
$\displaystyle A_{i}$ follows immediately from the arguments in Proposition
3.21 for $\displaystyle i=1,2$.

Using the standard elliptic estimates yields that
$\displaystyle\psi_{\lambda,\theta}\in
C^{2,\sigma}(\Omega_{1}\cup\Omega_{2})\cap
C^{0}(\overline{\Omega_{1}\cup\Omega_{2}})$ for some
$\displaystyle\sigma\in(0,1)$ and it solves the boundary value problem (2.4).

Hence, the existence of the impinging outgoing jets in Theorem 1.1 has been
established.

Next, we will show the positivity of radial velocity to the axially symmetric
impinging outgoing jets.

######  Proposition 3.22.

Let $\displaystyle\psi_{\lambda,\theta}$ be the solution to the boundary value
problem (2.4), then

$m_{2}<\psi_{\lambda,\theta}<m_{1}\ \ \text{in}~{}~{}G,$ (3.67)

and

$V=-\frac{1}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial x}>0\ \ \
\text{in}~{}~{}\overline{G}\setminus\left(N_{0}\cup\Gamma\right),$ (3.68)

where $\displaystyle G$ is bounded by $\displaystyle N_{i}$,
$\displaystyle\Gamma_{i}$ and $\displaystyle N_{0}$ for $\displaystyle i=1,2$.

###### Proof.

Noting that

$\displaystyle\Delta\psi_{\lambda,\theta}-\frac{1}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial
r}=0\ \text{in any bounded connected smooth open subdomain $\displaystyle
G_{0}\subset G\setminus\Gamma$},$

and $\displaystyle m_{2}\leq\psi_{\lambda,\theta}\leq m_{1}$ on
$\displaystyle\partial G_{0}$, then, the strong maximum principle implies that

$\displaystyle m_{2}<\psi_{\lambda,\theta}<m_{1}\ \ \text{in}~{}~{}G_{0}.$

The arbitrariness of domain $\displaystyle G_{0}\subset G$ yields to (3.67).

Next, since $\displaystyle N_{1}\cup\Gamma_{1}\in C^{1}$, there exists a
bounded smooth subdomain $\displaystyle
G_{0}\subset\\{0<\psi_{\lambda,\theta}<m_{1}\\}$ with
$\displaystyle\overline{G_{0}}\cap N_{1}=X_{0}$ (or $\displaystyle
G_{0}\subset\\{m_{2}<\psi_{\lambda,\theta}<0\\}$ with
$\displaystyle\overline{G_{0}}\cap N_{2}=X_{0}$). Then, $\displaystyle
w=-\frac{\partial\psi_{\lambda,\theta}}{\partial x}$ satisfies

$\displaystyle\Delta w-\frac{1}{r}\frac{\partial w}{\partial r}=0\ \ \
\text{in}~{}~{}G_{0}.$

Since $\displaystyle\psi_{\lambda,\theta}=m_{1}$ on $\displaystyle N_{1}$, the
slip boundary condition (1.10) implies that

$\displaystyle\partial_{x}\psi_{\lambda,\theta}(f_{1}(r),r)f_{1}^{\prime}(r)+\partial_{r}\psi_{\lambda,\theta}(f_{1}(r),r)=0.$

This implies that the unit outward normal derivative satisfies

$\displaystyle\frac{\partial\psi_{\lambda,\theta}}{\partial\nu}\left(f_{1}(r),r)\right)=-\partial_{x}\psi_{\lambda,\theta}\left(f_{1}(r),r)\right)\sqrt{1+f_{1}^{\prime}(r)^{2}}.$

On another hand, $\displaystyle\psi_{\lambda,\theta}$ attains its maximum on
$\displaystyle N_{1}$, and Hopf’s lemma gives that

$\displaystyle w=-\partial_{x}\psi_{\lambda,\theta}>0,\ \ \
\text{on}~{}~{}N_{1}.$

Similarly, one gets

$\displaystyle w=-\partial_{x}\psi_{\lambda,\theta}>0,\ \ \
\text{on}~{}~{}N_{2}.$

Then, in view of (3.62), one has

$\displaystyle w\geq 0\ \ \ \text{on}~{}~{}\partial G_{0},\ \ \text{and}\ \
w(X_{0})>0,$

and applying the maximum principle to $\displaystyle
w=-\frac{\partial\psi_{\lambda,\theta}}{\partial x}$ yields that

$\displaystyle w>0\ \ \ \text{in any subdomain}~{}~{}G_{0}\subset G.$

Finally, we claim that (3.68) holds on
$\displaystyle\Gamma_{1}\cup\Gamma_{2}$.

Recalling the fact

$\displaystyle
w=\frac{\sqrt{\lambda+\Lambda}r}{\sqrt{1+(g_{1,\lambda,\theta}^{\prime}(r))^{2}}}\geq
0\ \ \text{on}~{}~{}\Gamma_{1},\ \ \text{and}\ \
w=\frac{\sqrt{\lambda}r}{\sqrt{1+(g_{2,\lambda,\theta}^{\prime}(r))^{2}}}\geq
0\ \ \text{on}~{}~{}\Gamma_{2},$

Suppose that the claim can not hold, then, without loss of generality, there
exists a $\displaystyle r_{0}\geq R$ such that $\displaystyle
g_{1,\lambda,\theta}^{\prime}(r_{0})=+\infty$ or $\displaystyle-\infty$,
$\displaystyle\partial_{x}\psi_{\lambda,\theta}(g_{1,\lambda,\theta}(r_{0}),r_{0})=0$
and $\displaystyle w(g_{1,\lambda,\theta}(r_{0}),r_{0})=0$.

Thanks to the fact that
$\displaystyle\frac{|\nabla\psi_{\lambda,\theta}|}{r}=\sqrt{\lambda+\Lambda}$
on $\displaystyle\Gamma_{1}$ (or
$\displaystyle\frac{|\nabla\psi_{\lambda,\theta}|}{r}=\sqrt{\lambda}$ on
$\displaystyle\Gamma_{2}$), then

$\displaystyle\frac{\partial}{\partial
s}\left(\frac{|\nabla\psi_{\lambda,\theta}|^{2}}{r^{2}}\right)=0\ \ \text{on
$\displaystyle\Gamma_{1}\cup\Gamma_{2}$},$

where $\displaystyle s=(1,0)$ is the tangential vector of
$\displaystyle\Gamma_{1}\cup\Gamma_{2}$. This implies that

$\displaystyle\left(\frac{\partial}{\partial
x}\left(\frac{|\nabla\psi_{\lambda,\theta}|^{2}}{r^{2}}\right),\frac{\partial}{\partial
r}\left(\frac{|\nabla\psi_{\lambda,\theta}|^{2}}{r^{2}}\right)\right)\cdot\left(1,0\right)=0\
\ \text{on $\displaystyle\Gamma_{1}\cup\Gamma_{2}$}.$

Then, one has

$\partial_{xr}\psi_{\lambda,\theta}(g_{1,\lambda,\theta}(r_{0}),r_{0})=\partial_{r}w(g_{1,\lambda,\theta}(r_{0}),r_{0})=0.$
(3.69)

However, the Hopf’s Lemma gives that

$\displaystyle\left|\frac{\partial w}{\partial\nu}\right|=\left|\frac{\partial
w}{\partial r}\right|>0\ \ \text{at
$\displaystyle(g_{1,\lambda,\theta}(r_{0}),r_{0})$},$

which contradicts with (3.69). Then, we prove that (3.68) holds on
$\displaystyle\Gamma_{1}\cup\Gamma_{2}$.

Therefore, we obtain the positivity of the vertical velocity and complete the
proof of Proposition 3.22. ∎

###  3.9. The properties of the interface

In this subsection, we will show that there exists a $\displaystyle
C^{1}$-smooth curve
$\displaystyle\Gamma:\\{\psi_{\lambda,\theta}=0\\}\cap\\{r>0\\}$ separating
the two fluids, and the axially symmetric impinging outgoing jet established
here possesses a unique branching point on the symmetric axis $\displaystyle
N_{0}$. For $\displaystyle\Lambda=0$, the proof is similar to Section 4.10 in
[17], we omit it here.

Next, it suffices to prove that the results hold for $\displaystyle\Lambda>0$.

Indeed, taking subsequence $\displaystyle\mu_{n}\rightarrow+\infty$, one has

$\displaystyle g_{\lambda_{\mu_{n}},\theta_{\mu_{n}},{\mu_{n}}}(r)\rightarrow
g_{\lambda,\theta}(r)\ \ \text{for any $\displaystyle r>0$}.$

Then, the similar arguments as Lemma 3.10 implies that $\displaystyle
x=g_{\lambda,\theta}(r)\in[-\infty,+\infty]$ is generalized continuous
function in $\displaystyle[0,+\infty)$, we need to prove that $\displaystyle
x=g_{\lambda,\theta}(r)$ is finite valued for any $\displaystyle
r\in[0,+\infty)$.

$g_{\lambda_{\mu_{n}},\theta_{\mu_{n}},{\mu_{n}}}(r)\rightarrow
g_{\lambda,\theta}(r)\ \ \text{for any $\displaystyle r\in(0,+\infty)$}.$
(3.70)

Since $\displaystyle g_{\lambda,\theta}(r)<g_{2,\lambda,\theta}(r)$ for any
$\displaystyle r>R$ and $\displaystyle
g_{\lambda,\theta}(r)>g_{1,\lambda,\theta}(r)$ for any $\displaystyle r>R$,
then it suffices to prove that $\displaystyle g_{\lambda,\theta}(r)$ is finite
valued for $\displaystyle 0\leq r\leq M_{0}$, where
$\displaystyle\max\\{r_{1},r_{2}\\}\leq M_{0}\leq R$. Denote by
$\displaystyle(\alpha_{i},\beta_{i})\subset[0,M_{0}]$ ($\displaystyle
i=1,2,...$, $\displaystyle\alpha_{i}<\beta_{i}$ and
$\displaystyle\beta_{i}\leq\alpha_{i+1}$) the maximum intervals where
$\displaystyle x=g_{\lambda,\theta}(r)$ is finite valued.

Similar arguments as Proposition 5.1 in [31], we can prove that the number of
intervals $\displaystyle(\alpha_{i},\beta_{i})$ is one, denote
$\displaystyle(\alpha,+\infty)$ for simplicity.

Next, we will prove that $\displaystyle\alpha=0$.

Suppose $\displaystyle\alpha>0$ and
$\displaystyle\lim_{r\rightarrow\alpha}g_{\lambda,\theta}(r)=-\infty$. For
some sufficiently large $\displaystyle M>0$, Set

$\displaystyle R_{0}=\max\\{x\mid
x=g_{\lambda,\theta}(r),\alpha<r<r_{1}\\},r_{3}=\min\\{r\mid
g_{\lambda,\theta}(r)=R_{0}-M\\},$ $\displaystyle r_{4}=\max\\{r\mid
g_{\lambda,\theta}(r)=R_{0},\alpha<r<r_{1}\\},\text{such that $\displaystyle
R_{0}-M\leq g_{\lambda,\theta}(r)\leq R_{0}$\ \ with $\displaystyle r_{3}\leq
r\leq r_{4}$}.$

It follows from Lemma 6.1 in [6] that

$\displaystyle M\leq C(r_{4}-r_{3})\leq Cr_{1},$

and this is impossible when $\displaystyle M$ sufficiently large.

Furthermore, the case $\displaystyle\alpha>0$ and
$\displaystyle\lim_{r\rightarrow\alpha}g_{\lambda,\theta}(r)=+\infty$, there
exists a sufficiently large $\displaystyle R_{0}>0$ and for any $\displaystyle
M>0$, define

$\displaystyle H_{R_{0}}=\min\\{r\mid g_{\lambda,\theta}(r)=R_{0}\\},\ \
l_{R_{0}}=\left\\{(R_{0},r)\mid 0\leq r\leq H_{R_{0}}\right\\},$
$\displaystyle H_{R_{0}+M}=\min\\{r\mid g_{\lambda,\theta}(r)=R_{0}+M\\},\ \
\text{and}\ \ l_{R_{0}+M}=\left\\{(R_{0},r)\mid 0\leq r\leq
H_{R_{0}+M}\right\\}.$

We define a domain $\displaystyle G_{R_{0},M}$, which is bounded by
$\displaystyle l_{R_{0}}$, $\displaystyle l_{R_{0}+M}$, $\displaystyle\Gamma$
and $\displaystyle x$-axis.

Applying Green’s formula in $\displaystyle G_{R_{0},M}$ and
$\displaystyle|\psi_{\lambda,\theta}|\leq Cr^{2}$ in
$\displaystyle\Omega\cap\\{r<\min\\{r_{1},r_{2}\\}\\}$ (using the fact (3.8)),
we find that

$\displaystyle-\int_{\partial
G_{R_{0},M}}\frac{x-R_{0}}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial\nu}dS$
$\displaystyle=-\int_{\partial
G_{R_{0},M}}\frac{\partial(x-R_{0})}{\partial\nu}\frac{\psi_{\lambda,\theta}}{r}dS$
(3.71) $\displaystyle=\int_{\partial
G_{R_{0},M}\cap\\{x=R_{0}\\}}\frac{\psi_{\lambda,\theta}}{r}dS-\int_{\partial
G_{R_{0},M}\cap\\{x=R_{0}+M\\}}\frac{\psi_{\lambda,\theta}}{r}dS$
$\displaystyle\leq CH_{R_{0}}^{2},$

where $\displaystyle\nu$ is the unit normal vector.

Indeed, in view of
$\displaystyle\frac{|\nabla\psi_{\lambda,\theta}^{+}|}{r}\geq\sqrt{\Lambda}$
on $\displaystyle\Gamma\cap\\{R_{0}\leq x\leq R_{0}+M\\}$, and
$\displaystyle\frac{\partial\psi_{\lambda,\theta}}{\partial r}\geq 0$ on
$\displaystyle x$-axis with $\displaystyle x\geq R_{0}$, the left hand side of
(3.71) is estimated as

$\displaystyle-\int_{\partial
G_{R_{0},M}}\frac{x-R_{0}}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial\nu}dS$
$\displaystyle=-\int_{\partial
G_{R_{0},M}\cap(\Gamma\cup\\{r=0\\}\cup\\{x=R_{0}+M\\})}\frac{x-R_{0}}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial\nu}dS$
$\displaystyle\geq\sqrt{\Lambda}\int_{\partial
G_{R_{0},M}\cap\Gamma}(x-R_{0})dS-M\int_{\partial
G_{R_{0},M}\cap\\{x=R_{0}+M\\}}\frac{1}{r}\frac{\partial\psi_{\lambda,\theta}}{\partial\nu}dS$
$\displaystyle\geq c\sqrt{\Lambda}M^{2}-CMH_{R_{0}},$

due to (3.16). Here $\displaystyle\nu$ is parallel to
$\displaystyle\nabla\psi_{\lambda,\theta}$ on free streamlines, the positive
constants $\displaystyle c$, $\displaystyle C$ are independent of
$\displaystyle M$ and $\displaystyle H_{R_{0}}$.

This together with (3.71) gives that

$\displaystyle M\leq CH_{R_{0}},$

then we derives a contradiction for sufficiently large $\displaystyle M$.

Hence, we obtain $\displaystyle\alpha=0$.

Finally, it suffices to prove that
$\displaystyle|g_{\lambda,\theta}(0+0)|<+\infty$.

Suppose not, without loss of generality, we assume $\displaystyle
g_{\lambda,\theta}(0+0)=+\infty$. Similar to the above arguments, we only need
to construct domain $\displaystyle G$ bounded by $\displaystyle x=R_{0}$ with
some $\displaystyle R_{0}>0$ sufficiently large, $\displaystyle x=R_{0}$,
$\displaystyle x=R_{0}+M$, $\displaystyle r=0$ and interface
$\displaystyle\Gamma$, then

$\displaystyle M\leq C,$

which derives a contradiction for sufficiently large $\displaystyle M$.

Similarly, we can exclude that $\displaystyle g_{\lambda,\theta}(0)=-\infty$.
Therefore, we conclude that $\displaystyle x=g_{\lambda,\theta}(r)$ is finite
for any $\displaystyle r\in[0,+\infty)$.

Now, collecting all results obtained above, we complete the proof of Theorem
1.1.

##  4\. Uniqueness of the impinging outgoing jet

In this section, we will investigate the uniqueness of the impinging outgoing
jet and the parameters when $\displaystyle\Lambda=0$.

Proof of Theorem 1.2. Let $\displaystyle\psi_{\lambda,\theta}$ and
$\displaystyle\tilde{\psi}_{\lambda,\theta}$ be two solutions to the boundary
value problem (2.4), and
$\displaystyle\Gamma_{i,\lambda,\theta}:x=g_{i,\lambda,\theta}(r)$ and
$\displaystyle\tilde{\Gamma}_{i,\lambda,\theta}:x=\tilde{g}_{i,\lambda,\theta}(r)$
be the corresponding free boundaries for $\displaystyle i=1,2$. Due to the
continuous fit conditions, one has

$\displaystyle
f_{1}(R)=g_{1,\lambda,\theta}(R)=\tilde{g}_{1,\lambda,\theta}(R)\ \
\text{and}\ \
f_{2}(R)=g_{2,\lambda,\theta}(R)=\tilde{g}_{2,\lambda,\theta}(R).$

Without loss of generality, we assume

$\displaystyle\lim_{r\rightarrow+\infty}(g_{1,\lambda,\theta}(r)-\tilde{g}_{1,\lambda,\theta}(r))\geq
0.$

Set $\displaystyle\psi^{\varepsilon}=\psi_{\lambda,\theta}(x+\varepsilon,r)$
for some $\displaystyle\varepsilon\geq 0$ and choose a smallest
$\displaystyle\varepsilon_{0}\geq 0$ such that

$\displaystyle\psi^{\varepsilon_{0}}\leq\tilde{\psi}_{\lambda,\theta}\ \
\text{in $\displaystyle\Omega$ and
$\displaystyle\psi^{\varepsilon_{0}}(X_{0})=\tilde{\psi}_{\lambda,\theta}(X_{0})$},$

for some $\displaystyle