# Oblique Injection of incompressible ideal fluid from a slot into a free
stream∗

Jianfeng Cheng1,2, Lili Du1,†

###### Abstract.

This paper deals with a two-phase fluid free boundary problem in a slot-film
cooling. We give two well-posedness results on the existence and uniqueness of
the incompressible inviscid two-phase fluid with a jump relation on free
interface. The problem formulates the oblique injection of an incompressible
ideal fluid from a slot into a free stream. From the mathematical point of
view, this work is motivated by the pioneer work [13] by A. Friedman, in which
some well-posedness results are obtained in some special case. Furthermore, A.
Friedman proposed an open problem in [14] on the existence and uniqueness of
the injection flow problem for more general case. The main results in this
paper solve the open problem and establish the well-posedness results on the
physical problem.

∗Cheng is supported in part by NSFC grant 12001387 and the Fundamental
Research Funds for the Central Universities No. YJ202046, Du is supported by
NSFC grant 11971331.

E-Mail<EMAIL_ADDRESS>(J. Cheng), E-mail<EMAIL_ADDRESS>(L. Du).

†Corresponding author.

1 Department of Mathematics, Sichuan University,

Chengdu 610064, P. R. China.

2 The Institute of Mathematical Sciences,

The Chinese University of Hong Kong,

Shatin, N.T., Hong Kong.

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

Key words: Existence and uniqueness, free boundary, two-phase fluid, contact
discontinuity.

##  1\. Introduction and main results

###  1.1. Introduction

This paper is concerned with a two-phase free boundary problem produced when a
secondary fluid (or injected or coolant) is injected obliquely at an angle
from a slot into a cross flow fluid (see Figure 1). One important physical
situation in which this problem arises in fuel injectors, smokestacks, the
cooling of gas-turbine blades, and dilution holes in gas turbine combustors.
Please see the review of this physical problem [17]. Many numerical
simulations on this problem were investigated in [11, 20, 21] and the
references therein.

Figure 1. Injection of flow from a slot into a cross flow Figure 2. Horizontal
blade surfaces

Mathematically, the motivation to investigate this free boundary problem
follows from the work [13] by A. Friedman. He first considered the two-
dimensional model and the simple situation of horizontal blade surface and the
secondary fluid injected perpendicularly into a free stream in two dimensions
(as in Figure 2). Here, for simplicity, we neglect the separation at the
trailing edge $\displaystyle B$ of the slot, such separation can be minimized
in practice by slightly around the trailing edge. Also, we have assumed that
the interface between the main stream flow and the secondary flow separates at
the leading edge $\displaystyle A$, since the viscosity effects are ignored.
Some existence and uniqueness of the solution to the two-phase fluid were
established for simple special case in [13]. And furthermore, A. Friedman
proposed an open problem in Page 69 in his survey [14], that

$\displaystyle\begin{array}[]{rl}\text{{\it Problem (1)}.}&\text{{\it Extend
the results of Theorem 9.1, 9.2 to more general geometries,}}\\\ &\text{{\it
such as in Figure 9.3}}.\end{array}$

Figure 3. Figure 9.3 in [14]

Please see Figure 3 for the Figure 9.3 in [14].

The main purpose in this paper is to establish the existence and uniqueness of
the free boundary problem on an incompressible inviscid fluid obliquely into a
free stream (as in Figure 4) and solve the open problem proposed by A.
Friedman.

Figure 4. Oblique injection of flow into a free stream

In general, there is a discontinuity in the magnitude of velocity across the
interface due to the Bernoulli’s law. Therefore, the standard method of
conformal mapping from the complex potential plane to the conjugate velocity
plane will not be fruitful because the interface is mapping into unknown curve
in the conjugate velocity plane. For the special case of
$\displaystyle\theta=\frac{\pi}{2}$, the free boundary problem was reduced to
a nonlinear singular integral differential equation in [22]. Along the
variational arguments introduced in [3, 4, 5, 6], A. Friedman established the
well-posedness results for the some special case ($\displaystyle b=0$ and
$\displaystyle\theta=\frac{\pi}{2}$) in Figure 2.

###  1.2. Notations and the free boundary problem

Before we state the main results in this paper, we will give the following
notations of the geometry of the blade surface.

Denote

$\displaystyle N_{1}=\\{(x,0)\mid x\leq 0\\},\ \ N_{2}=\\{(x,b)\mid x\geq
a\\}.$

Here, $\displaystyle a>0$, and we consider the general case that the blade
surfaces are not horizontal, namely, $\displaystyle b>0$. Let

$\displaystyle S_{1}=\\{(x,y)\mid x=y\cot\theta,y\leq 0\\}\ \ \text{and}\ \
S_{2}=\\{(x,y)\mid x=(y-b)\cot\theta+a,y\leq b\\},$

where $\displaystyle\theta\in\left(0,\frac{\pi}{2}\right]$. Furthermore, we
assume $\displaystyle a\sin\theta-b\cos\theta>0$, which excludes the
possibility of the intersection of $\displaystyle N_{1}$ and $\displaystyle
S_{2}$, $\displaystyle\theta$ is the inclination and the critical case
$\displaystyle\theta=\frac{\pi}{2}$ means the normal injection. The leading
edge $\displaystyle A=(0,0)$ and the trailing edge $\displaystyle B=(a,b)$.

Both of the mainstream flow and the secondary flow are assumed to be steady,
incompressible, inviscid and irrotational. Denote by
$\displaystyle(u_{+},v_{+}),p_{+},\rho_{+}$ the velocity, pressure and the
constant density of the mainstream flow in $\displaystyle\Omega^{+}$, and
$\displaystyle(u_{-},v_{-}),p_{-},\rho_{-}$ as the velocity, pressure and the
constant density of the secondary flow in $\displaystyle\Omega^{-}$. They are
separated by a streamline, denoted as $\displaystyle\Gamma$. The pressure
across the interface $\displaystyle\Gamma$ has to be continuous, i.e.,
$\displaystyle p_{+}=p_{-}$ on $\displaystyle\Gamma$. We assume that the
mainstream flow is horizontal and possesses a uniform speed $\displaystyle
U_{0}$ in upstream, without loss of generality, $\displaystyle
U_{0}=\frac{1}{\sqrt{\rho_{+}}}$. The secondary flow with mass flux
$\displaystyle Q_{0}$ emerges from a slot, where the magnitude of
$\displaystyle Q_{0}$ is unrestricted for the moment.

Denote by $\displaystyle\Omega$ the fluid field of the two-phase fluid,
composed of the following domains

$\displaystyle T_{1}=\\{x\leq 0,y\geq 0\\}$, $\displaystyle
T_{2}=\\{y\cot\theta\leq x\leq 0,y\leq 0\\}$,

and

$\displaystyle T_{3}=\\{0\leq x\leq\min\\{(y-b)\cot\theta+a,a\\}\\}\ \
\text{and}\ \ T_{4}=\\{x\geq a,y\geq b\\},$

namely, $\displaystyle\Omega=\text{int}(T_{1}\cup T_{2}\cup T_{3}\cup T_{4})$.

Define a stream function $\displaystyle\psi$ of the two-phase fluid as

$\displaystyle\frac{\partial\psi}{\partial x}=-\sqrt{\rho_{+}}v_{+}\ \
\text{and}\ \ \frac{\partial\psi}{\partial y}=\sqrt{\rho_{+}}u_{+}\ \ \text{in
the main fluid field $\displaystyle\Omega^{+}$,}$

and

$\displaystyle\frac{\partial\psi}{\partial x}=-\sqrt{\rho_{-}}v_{-}\ \
\text{and}\ \ \frac{\partial\psi}{\partial y}=\sqrt{\rho_{-}}u_{-}\ \ \text{in
the secondary fluid field $\displaystyle\Omega^{-}$}.$

On the solid boundaries, we impose that

$\text{$\displaystyle\psi=0$ on $\displaystyle N_{1}\cup S_{1}$, and
$\displaystyle\psi=-\frac{Q_{0}}{\sqrt{\rho_{-}}}$ on $\displaystyle N_{2}\cup
S_{2}$}.$ (1.1)

On the interface $\displaystyle\Gamma=\Omega\cap\\{\psi=0\\}$, the Bernoulli’s
equation gives that

$\rho_{-}(u_{-}^{2}+v_{-}^{2})-\rho_{+}(u_{+}^{2}+v_{+}^{2})=\text{constant,}\
\ \ \text{on}\ \ \Gamma,$ (1.2)

the jump constant is denoted as $\displaystyle\lambda$. It is easy to see that
$\displaystyle\lambda\in(-1,+\infty)$. The two-phase fluid we seek in this
paper is the vortex sheet solution and the jump condition (1.2) is in fact the
Rankine-Hugoniot jump condition to the vortex sheet. From the mathematical
point of view, to attack the well-posedness of the problem on the injection of
ideal fluid from a slot into a free stream in 1983, A. Friedman in [13] (see
also the Chapter 9 in [14]) introduced the injection flow problems in two
different situations.

The injection flow problem 1. For any given $\displaystyle Q_{0}>0$, does
there exist a unique injection flow $\displaystyle(\psi,\Gamma)$, such that
the mainstream flow possesses uniform speed in upstream, and the interface
$\displaystyle\Gamma$ connects at $\displaystyle A$ and extends to infinity?

The injection flow problem 2. For any given
$\displaystyle\lambda\in(-1,+\infty)$, does there exist a unique injection
flow $\displaystyle(\psi,\Gamma)$, such that the mainstream flow possesses
uniform speed in upstream, and the interface $\displaystyle\Gamma$ connects at
$\displaystyle A$ and extends to infinity?

Here, it is worth to mention that once the stream function $\displaystyle\psi$
is solved,

$\displaystyle(u_{+},v_{+})=\frac{1}{\sqrt{\rho_{+}}}\left(\frac{\partial\psi}{\partial
y},-\frac{\partial\psi}{\partial x}\right)\ \ \text{in the main fluid field
$\displaystyle\Omega^{+}$,}$

and

$\displaystyle(u_{-},v_{-})=\frac{1}{\sqrt{\rho_{-}}}\left(\frac{\partial\psi}{\partial
y},-\frac{\partial\psi}{\partial x}\right)\ \ \text{in the secondary fluid
field $\displaystyle\Omega^{-}$}$

will be solved by the stream function.

Meanwhile, the existence and uniqueness of the injection flow problem 1 and 2
in some geometric special situation were established in Theorem 3.2 and
Theorem 1.1 in [13], respectively. He assumed that the blade surface is
horizontal and the injection is vertical, namely, $\displaystyle b=0$ and
$\displaystyle\theta=\frac{\pi}{2}$ (see Figure 2). Moreover, he proposed an
open problem in [14] to extend the results in [13] to more general case as in
Figure 3. This is the main motivation to investigate the oblique injection
flow problem in this paper.

Next, we will define the solution to the injection flow problem 1 and problem
2, respectively.

######  Definition 1.1.

(A solution to the injection flow problem 1).
For any given $\displaystyle Q_{0}>0$, a vector $\displaystyle(\psi,\Gamma)$
is called a solution to the injection flow problem 1, provided that
(1) $\displaystyle\Delta\psi=0$ in $\displaystyle\Omega\setminus\Gamma$,
$\displaystyle\psi\in C^{0}(\bar{\Omega})$ and $\displaystyle\nabla\psi\in
L^{\infty}(\Omega\setminus B_{\varepsilon}(B))$ for any
$\displaystyle\varepsilon>0$.
(2) $\displaystyle\psi$ satisfies the Dirichlet boundary conditions (1.1).
(3) The free boundary $\displaystyle\Gamma:y=k(x)$ is $\displaystyle
C^{1}$-smooth strictly increasing function in $\displaystyle(0,+\infty)$, and
$\displaystyle k(x)>b$ for any $\displaystyle x\geq a$. Furthermore,

$k(0)=0,$ (1.3)

and there exists a $\displaystyle h\in(b,+\infty)$, such that

$\displaystyle\lim_{x\rightarrow+\infty}k(x)=h\ \ \text{and}\ \
\lim_{x\rightarrow+\infty}k^{\prime}(x)=0.$

(4) $\displaystyle\psi$ satisfies the Rankine-Hugoniot jump condition on
$\displaystyle\Gamma$, namely,

$\left(\frac{\partial\psi^{-}}{\partial\nu}\right)^{2}-\left(\frac{\partial\psi^{+}}{\partial\nu}\right)^{2}=\lambda~{}~{}\text{on}\
\Gamma,$ (1.4)

where
$\displaystyle\lambda=\frac{1}{\rho_{-}}\left(\frac{Q_{0}}{h-b}\right)^{2}-1$,
$\displaystyle\psi^{+}=\max\\{\psi,0\\}$,
$\displaystyle\psi^{-}=-\min\\{\psi,0\\}$ and $\displaystyle\nu$ is the normal
vector to $\displaystyle\Gamma$.
(5) $\displaystyle\Gamma$ is continuously differentiable at $\displaystyle A$
and

$k^{\prime}(0+0)=\left\\{\begin{array}[]{ll}\tan\theta,&\text{if
$\displaystyle\lambda>0$, (see Figure \ref{f4})}\\\ 0,&\text{if
$\displaystyle\lambda<0$, (see Figure \ref{f5})}\\\
\tan\frac{\theta}{2},&\text{if\ $\displaystyle\lambda=0$, (see Figure
\ref{f6})}.\end{array}\right.$ (1.5)

Figure 5. The case $\displaystyle\lambda<0$. Figure 6. The case
$\displaystyle\lambda=0$

(6) $\displaystyle\psi$ possesses the following asymptotic behaviors in far
field

$\displaystyle\psi(x,y)\rightarrow\left\\{\begin{array}[]{ll}\frac{Q_{0}(y-h)}{\sqrt{\rho_{-}}(h-b)},&\text{if
$\displaystyle b<y<h$, as $\displaystyle x\rightarrow+\infty$},\\\
y-h,&\text{if $\displaystyle h<y<C$, as $\displaystyle x\rightarrow+\infty$,
for any $\displaystyle C>0$},\end{array}\right.$
$\displaystyle\psi(x,y)\rightarrow y,\ \text{if $\displaystyle 0<y<C$, as
$\displaystyle x\rightarrow-\infty$, for any $\displaystyle C>0$},$

and

$\displaystyle\left|\psi(x,y)-\frac{Q_{0}(y\cos\theta-x\sin\theta)}{\sqrt{\rho_{-}}(a\sin\theta-b\cos\theta)}\right|\rightarrow
0\ \ \text{uniformly in any compact subset of $\displaystyle S$,}$

as $\displaystyle y\rightarrow-\infty$, where $\displaystyle S=\\{(x,y)\mid
y\cot\theta<x<(y-b)\cot\theta+a,-\infty<y<+\infty\\}$. Furthermore,

$\displaystyle\nabla\psi(x,y)\rightarrow(0,1),\ \text{if $\displaystyle
x^{2}+y^{2}\rightarrow+\infty$,
dist($\displaystyle(x,y),\Gamma$)$\displaystyle\rightarrow+\infty$ and with
$\displaystyle\psi(x,y)>0$},$

and

$\displaystyle\left|\nabla\psi(x,y)-\frac{Q_{0}}{\sqrt{\rho_{-}}(a\sin\theta-b\cos\theta)}\left(-\sin\theta,\cos\theta\right)\right|\rightarrow
0\ \ \text{uniformly in any compact subset of $\displaystyle S$,}$

as $\displaystyle y\rightarrow-\infty$.
(7) The following estimates hold,

$\displaystyle-h\leq\psi^{+}(x,y)-y\leq 0\ $ in $\displaystyle\
\Omega\cap\\{y>0\\}$. (1.6)

######  Definition 1.2.

(A solution to the injection flow problem 2). For some given appropriate
$\displaystyle\lambda\in(-1,+\infty)$, $\displaystyle(\psi,\Gamma)$ is called
a solution to the injection flow problem 2, provided that the conditions (1) -
(7) in Definition 1.1 hold.

######  Remark 1.1.

$\displaystyle k(0)=0$ is nothing but the continuous fit condition of the
interface $\displaystyle\Gamma$, which gives that the interface
$\displaystyle\Gamma$ initiates at the leading edge $\displaystyle A$. Since
the viscous effects are ignored here, and the boundary layer is not
considered, the continuous fit condition seems to be reasonable. Moreover, the
condition (1.5) is so-called smooth fit condition for
$\displaystyle\lambda\neq 0$ (please see Figure 4 and Figure 6).

######  Remark 1.2.

The conditions $\displaystyle\lim_{x\rightarrow+\infty}k(x)=h<+\infty$ and
$\displaystyle\lim_{x\rightarrow+\infty}k^{\prime}(x)=0$ in (1.3) imply that
the interface $\displaystyle\Gamma$ is flat and does not oscillate in
downstream.

######  Remark 1.3.

To attack the injection flow problem 1, we first regard the constant
$\displaystyle\lambda$ as an undetermined parameter, and then the parameter
$\displaystyle\lambda$ will be determined uniquely by the continuous fit
condition. It means that there exists a unique $\displaystyle\lambda$ such
that the interface connects at the leading edge point $\displaystyle A$. On
another hand, the asymptotic behavior in downstream gives the relation
$\displaystyle h=b+\frac{Q_{0}}{\sqrt{\rho_{-}(\lambda+1)}}$. Once the
constant $\displaystyle\lambda$ is fixed by the continuous fit condition, the
asymptotic width $\displaystyle h$ can be determined by the formula.

###  1.3. Main results

For the special case $\displaystyle b=0$ and
$\displaystyle\theta=\frac{\pi}{2}$, the existence and uniqueness were
established in [13], and we will give the existence and uniqueness results on
the injection flow problem in two situations in general case as follows.

######  Theorem 1.1.

For any $\displaystyle Q_{0}>0$, there exist a unique
$\displaystyle\lambda>-1$ and a unique solution $\displaystyle(\psi,\Gamma)$
to the injection flow problem 1. Furthermore, the interface
$\displaystyle\Gamma$ is analytic, $\displaystyle u_{\pm}>0$ in
$\displaystyle\Omega^{\pm}\cup\Gamma$, and $\displaystyle v_{\pm}>0$ in
$\displaystyle\Omega^{\pm}\cup\Gamma$.

######  Remark 1.4.

In [5], some well-posedness results on two fluids of steady incompressible
inviscid flows issuing from two nozzles were established (see Figure 7).
However, it is assumed that the two nozzles are symmetric with respect to
$\displaystyle x$-axis and the upper boundary of the nozzle I coincides with
the lower boundary of the nozzle II. Along the proof of Theorem 1.1 in this
paper, we can extend the existence and uniqueness of the two fluids in [5] to
the general case as Figure 8 (the nozzle are asymmetric and the nozzle walls
do not coincide) without any additional difficulties.

Figure 7. Two-phase fluid Figure 8. Two-phase fluid

######  Remark 1.5.

In the previous work [13], A. Friedman showed that the free boundary
$\displaystyle\Gamma$ is only $\displaystyle C^{1}$-smooth for $\displaystyle
b=0$ and $\displaystyle\theta=\frac{\pi}{2}$, and then the Rankine-Hugoniot
(1.4) holds in weak sense. However, we would like to emphasize that here we
showed that the free boundary is analytic and then the Rankine-Hugoniot (1.4)
holds in classical sense.

On another hand, to obtain the well-posedness results on the injection flow
problem 2, we will investigate the relationship of the constant
$\displaystyle\lambda$ and the flux of injection flow $\displaystyle Q_{0}>0$.
In fact, we show that $\displaystyle\lambda$ is strictly monotone increasing
and continuous with respect to $\displaystyle Q_{0}>0$, denoted as
$\displaystyle\lambda=\lambda(Q_{0})$.

######  Theorem 1.2.

For any $\displaystyle Q_{0}>0$, the solution
$\displaystyle(\psi,\Gamma,\lambda(Q_{0}))$ established in Theorem 1.1
satisfies that
(1) $\displaystyle\lambda(Q_{0})$ is strictly monotone increasing and
continuous with respect to $\displaystyle Q_{0}>0$.
(2) There exists a $\displaystyle\underline{\lambda}\in(-1,0)$, such that
$\displaystyle\lambda(Q_{0})\rightarrow\underline{\lambda}$ as $\displaystyle
Q_{0}\rightarrow 0$.
(3) There exists a $\displaystyle\kappa\in(0,+\infty)$, such that
$\displaystyle\frac{\lambda(Q_{0})}{Q_{0}^{2}}\rightarrow\kappa$ as
$\displaystyle Q\rightarrow+\infty$.

The second statement in Theorem 1.2 implies that the lower bound is
$\displaystyle\underline{\lambda}$, so we can establish the well-posedness
result to the injection flow problem 2.

######  Theorem 1.3.

There exists a $\displaystyle\underline{\lambda}>-1$
($\displaystyle\underline{\lambda}$ is given in Theorem 1.2), such that for
any $\displaystyle\lambda>\underline{\lambda}$, there exist a unique
$\displaystyle Q_{0}>0$ and a unique solution $\displaystyle(\psi,\Gamma)$ to
the injection flow problem 2. Furthermore, $\displaystyle u_{\pm}>0$ in
$\displaystyle\Omega^{\pm}\cup\Gamma$, and $\displaystyle v_{\pm}>0$ in
$\displaystyle\Omega^{\pm}\cup\Gamma$.

######  Remark 1.6.

Similar to the Theorem 1.1, when the constant $\displaystyle\lambda$ is
imposed, the flux of the injection flux $\displaystyle Q_{0}$ can be regarded
as a parameter to solve the injection flow problem 2. And the unique
solvability of the flux $\displaystyle Q_{0}$ can be determined by the
continuous fit condition. In particular, for $\displaystyle\lambda=0$, the
stream function is harmonic in the whole fluid field $\displaystyle\Omega$,
and the flux $\displaystyle Q_{0}>0$ is uniquely determined by the following
formula

$\displaystyle(a\sin\theta-b\cos\theta)^{\frac{\pi}{\theta}}=\left(\frac{\pi}{\theta}-1\right)\left(\frac{Q_{0}}{\sqrt{\rho_{-}}}\right)^{\frac{\pi}{\theta}}+\frac{b\pi}{\theta}\left(\frac{Q_{0}}{\sqrt{\rho_{-}}}\right)^{\frac{\pi}{\theta}-1},$

due to the conformal mapping in [18].

As we mentioned before, A. Friedman established the well-posedness results for
the simple case of horizontal blade surfaces ($\displaystyle b=0$), and
proposed an open problem on the general case as shown in Figure 3. However,
from the mathematical point of view, the extension to the present problem is
not straightforward, and involves some additional difficulties. For the
special case $\displaystyle b=0$ (see Figure 2), consider the critical case
$\displaystyle Q_{0}=0$, the injection flow vanishes and the mainstream flow
is nothing but a trivial uniform flow. The free boundary is the segment
connecting the leading edge $\displaystyle A$ and the trailing edge
$\displaystyle B$. However, for the general case ($\displaystyle b\neq 0$),
there does not exist a trivial flow for the critical case $\displaystyle
Q_{0}=0$. This is the one of main differences and the difficulties here. This
fact prevents us to establish the lower bound of $\displaystyle\lambda$ while
$\displaystyle Q_{0}\rightarrow 0$. To overcome this difficulty, we will
investigate the limiting flow ($\displaystyle Q_{0}\rightarrow 0$), and show
that the free boundary initiates smoothly at $\displaystyle A$ and terminated
at the wall $\displaystyle S_{2}$. Moreover, we will show that the
intersection of the free boundary $\displaystyle\Gamma$ and $\displaystyle
S_{2}$ must below the trailing edge $\displaystyle B$. Another difference is
that the domain is a star-sharped one with respect to $\displaystyle B$ for
the special case $\displaystyle b=0$, we can take a rescaling transform to
obtain the uniqueness. Furthermore, the property can not hold for the general
case, and we have to develop a new method to obtain the uniqueness.

The basic idea in this paper is to seek a two-phase fluid with a smooth
interface connecting at the leading edge $\displaystyle A$. A truncated
injection flow problem is presented in Section 2, and furthermore, we give a
result on existence and uniqueness in truncated fluid field. Section 3 studies
some useful properties of the minimizer and free boundary in the truncated
domain. In particular, we will establish the relationship between the jump
constant $\displaystyle\lambda$ and the injected flux $\displaystyle Q_{0}$,
which builds a bridge between the injection flow problem 1 and 2. Section 4 is
devoted to the solution of the injection flow problem using some uniform
estimates of the solution in truncated domain. The analysis reveals the
existence and uniqueness of the two-phase fluid with $\displaystyle
C^{1}$-smooth interface, the fact firstly proved in [13] for special case. Our
results solve the open problem on the well-posedness of an ideal fluid
injected obliquely from a slot into a stream.

##  2\. The truncated injection flow problem

To solve the injection flow problem, we first study the truncated injection
flow problem with finite height in this section. To simplify notation, denote

$\displaystyle Q=\frac{Q_{0}}{\sqrt{\rho_{-}}}.$

For any $\displaystyle L>b$, denote

$\displaystyle N_{L}=\\{(x,L)\mid-\infty<x<+\infty\\}\ \ \text{and}\ \
\Omega_{L}=\Omega\cap\\{y<L\\}.\ \ \text{(See Figure \ref{f9})}$

Figure 9. Truncated flow field

The definition of the truncated injection flow problem will be given in the
following.

The truncated injection flow problem 1 corresponding to the injection flow
problem 1 is as follows: For any given $\displaystyle Q>0$, does there exist a
unique $\displaystyle\lambda_{L}$ and a unique injection flow
$\displaystyle(\psi_{L},\Gamma_{L})$ in the truncated domain
$\displaystyle\Omega_{L}$, such that the mainstream flow possesses uniform
speed in upstream, and the interface $\displaystyle\Gamma_{L}$ connects at
$\displaystyle A$ and extends to infinity?

Next, we will give the definition of the solution to the truncated injection
flow problem 1.

######  Definition 2.1.

(A solution to the truncated injection flow problem 1).
For any $\displaystyle L>b$, a vector $\displaystyle(\psi_{L},\Gamma_{L})$ is
called a solution to the truncated injection flow problem 1, provided that
(1) $\displaystyle\Delta\psi_{L}=0$ in
$\displaystyle\Omega_{L}\setminus\Gamma_{L}$, $\displaystyle\psi_{L}\in
C^{0}(\bar{\Omega}_{L})$ and $\displaystyle\nabla\psi_{L}\in
L^{\infty}(\Omega_{L}\setminus B_{\varepsilon}(B))$ for any
$\displaystyle\varepsilon>0$.
(2) $\displaystyle\psi_{L}=0$ satisfies the Dirichlet boundary conditions
(1.1) and $\displaystyle\psi_{L}=L$ on $\displaystyle N_{L}$.
(3) The free boundary $\displaystyle\Gamma_{L}:y=k_{L}(x)$, and $\displaystyle
k_{L}(x)$ is a $\displaystyle C^{1}$-smooth strictly increasing function in
$\displaystyle(0,+\infty)$, and $\displaystyle k_{L}(x)>b$ for any
$\displaystyle x\geq a$. Furthermore,

$k_{L}(0)=0,$ (2.1)

and there exists a $\displaystyle h_{L}\in(b,L)$, such that

$\displaystyle\lim_{x\rightarrow+\infty}k_{L}(x)=h_{L}\ \ \text{and}\ \
\lim_{x\rightarrow+\infty}k_{L}^{\prime}(x)=0.$

(4) $\displaystyle\psi_{L}$ satisfies the Rankine-Hugoniot jump condition on
$\displaystyle\Gamma_{L}$, namely,

$\left(\frac{\partial\psi_{L}^{-}}{\partial\nu}\right)^{2}-\left(\frac{\partial\psi_{L}^{+}}{\partial\nu}\right)^{2}=\lambda_{L}\
\ \text{on}\ \ \Gamma_{L},$ (2.2)

where
$\displaystyle\lambda_{L}=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{L^{2}}{(L-h_{L})^{2}}$.
(5) $\displaystyle\Gamma_{L}$ is continuously differentiable at $\displaystyle
A$ and

$k_{L}^{\prime}(0+0)=\left\\{\begin{array}[]{ll}\tan\theta,&\text{if
$\displaystyle\lambda_{L}>0$},\\\ 0,&\text{if
$\displaystyle\lambda_{L}<0$},\\\ \tan\frac{\theta}{2},&\text{if}\
\lambda_{L}=0.\end{array}\right.$ (2.3)

(6) $\displaystyle\psi_{L}$ has the following asymptotic behaviors

$\displaystyle\psi_{L}(x,y)\rightarrow\left\\{\begin{array}[]{ll}\frac{Q(y-h_{L})}{h_{L}-b},&\text{if
$\displaystyle b<y<h_{L}$, as $\displaystyle x\rightarrow+\infty$},\\\
\frac{L(y-h_{L})}{L-h_{L}},&\text{if $\displaystyle h_{L}<y<L$, as
$\displaystyle x\rightarrow+\infty$},\end{array}\right.$
$\displaystyle\psi_{L}(x,y)\rightarrow y,\quad\text{if $\displaystyle 0<y<L$,
as $\displaystyle x\rightarrow-\infty$},$

and

$\displaystyle\left|\psi_{L}(x,y)-\frac{Q(y\cos\theta-x\sin\theta)}{a\sin\theta-b\cos\theta}\right|\rightarrow
0\ \ \text{uniformly in any compact subset of $\displaystyle S$,}$

as $\displaystyle y\rightarrow-\infty$, where $\displaystyle S=\\{(x,y)\mid
y\cot\theta<x<(y-b)\cot\theta+a,-\infty<y<+\infty\\}$.
(7) $\displaystyle\frac{L(y-h_{L})}{L-h_{L}}\leq\psi_{L}^{+}(x,y)\leq y$ in
$\displaystyle\Omega_{L}\cap\\{y>0\\}$.

######  Remark 2.1.

It should be noted that $\displaystyle
f(t)=\frac{Q^{2}}{(t-b)^{2}}-\frac{L^{2}}{(L-t)^{2}}$ is a strictly monotone
decreasing function for $\displaystyle t\in(b,L)$. Therefore, the asymptotic
height $\displaystyle h_{L}\in(b,L)$ of the free boundary can be determined
uniquely by
$\displaystyle\lambda=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{L^{2}}{(L-h_{L})^{2}}$.

###  2.1. Variational approach

To solve the truncated injection flow problem 1, as the first step, we
introduce a truncated variational problem for any given parameter
$\displaystyle\lambda\in(-\infty,+\infty)$ and $\displaystyle Q>0$. Secondly,
we will verify that there exists a unique parameter
$\displaystyle\lambda=\lambda_{L}$, such that the interface
$\displaystyle\Gamma_{L}$ connects at the leading edge $\displaystyle A$.
Finally, taking $\displaystyle L\rightarrow+\infty$ yields the existence of
solution to the injection flow problem 1.

For any $\displaystyle\mu>1$, denote

$\displaystyle\Omega_{L,\mu}=\Omega_{L}\cap\\{(x,y)\mid x>-\mu,y>-\mu\\}\ \
\text{and}\ \ \sigma_{L,\mu}=\\{(-\mu,y)\mid 0\leq y\leq L\\},$

and

$\displaystyle D_{1,L,\mu}=\Omega_{L,\mu}\cap\\{(x,y)\mid x<0,y<0\\}\ \
\text{and}\ \ D_{2,L,\mu}=\Omega_{L,\mu}\cap\\{(x,y)\mid x<0,y>0\\},$

and

$\displaystyle N_{1,\mu}=N_{1}\cap\\{x\geq-\mu\\},\ \
S_{1,\mu}=S_{1}\cap\\{y\geq-\mu\\}\ \ \text{and}\ \
N_{L,\mu}=N_{L}\cap\\{x\geq-\mu\\},$

and

$\displaystyle S_{\mu}=\\{-\mu\tan\theta<x<a-(\mu+b)\tan\theta,y=-\mu\\}\ \
\text{and}\ \ S_{2,\mu}=S_{2}\cap\\{y\geq-\mu\\}.$

Please see Figure 10.

Figure 10. The nozzle walls and the ground

As mentioned in Remark 2.1, for any given $\displaystyle\lambda$ and
$\displaystyle L>b$, we can obtain a unique asymptotic height $\displaystyle
h_{L}\in(b,L)$ of the interface $\displaystyle\Gamma_{\lambda,L}$. Then, we
can define $\displaystyle\lambda_{1,L}$, $\displaystyle\lambda_{2,L}$ and
$\displaystyle\lambda_{0,L}$ as follows

$\displaystyle\lambda_{1,L}=\frac{Q}{h_{L}-b},\ \
\lambda_{2,L}=\frac{L}{L-h_{L}}\ \ \text{and}\ \
\lambda_{0,L}=\min\\{\lambda_{1,L},\lambda_{2,L}\\}.$

Obviously, $\displaystyle\lambda=\lambda_{1,L}^{2}-\lambda_{2,L}^{2}$.
Moreover, we give the following functional

$\displaystyle
J_{\lambda,L,\mu}(\psi)=\int_{\Omega_{L,\mu}}\left|\nabla\psi-(\lambda_{1,L}I_{\\{\psi<0\\}}+\lambda_{2,L}I_{\\{\psi>0\\}}+\lambda_{0,L}I_{\\{\psi=0\\}})I_{\\{x>0\\}}e\right|^{2}dxdy$

where $\displaystyle e=(0,1)$ and $\displaystyle I_{D}$ is the characteristic
function of the set $\displaystyle D$. And the admissible set is defined as
follows

$\displaystyle\begin{array}[]{rl}K_{L,\mu}=\\{\psi\mid&\psi\in
H^{1}_{loc}(\Omega_{L,\mu}),\ -Q\leq\psi\leq L\ \text{a.e. in
$\displaystyle\Omega_{L,\mu}$},\ \ 0\leq\psi\leq L\ \text{a.e. in
$\displaystyle D_{2,L,\mu}$},\\\ &-Q\leq\psi\leq 0\ \text{a.e. in
$\displaystyle D_{1,L,\mu}$},\ \psi=-Q\ \text{on $\displaystyle S_{\mu}\cup
S_{2,\mu}\cup N_{2}$},\\\ &\ \psi=L\ \text{on $\displaystyle
N_{L,\mu}\cup\sigma_{L,\mu}$},\ \psi=\max\\{L-(x+\mu)L,0\\}\ \text{on
$\displaystyle N_{1,\mu}$},\\\ &\ \psi=\min\\{-Q+(y+\mu)Q,0\\}\ \text{on
$\displaystyle S_{1,\mu}$}\\}.\end{array}$

The truncated variational problem $\displaystyle(P_{\lambda,L,\mu})$: For any
$\displaystyle L>b$, $\displaystyle\mu>1$ and
$\displaystyle\lambda\in(-\infty,+\infty)$, find a
$\displaystyle\psi_{\lambda,L,\mu}\in K_{L,\mu}$ such that

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

Define the free boundary in the truncated domain as

$\displaystyle\Gamma_{\lambda,L,\mu}=\Omega_{L,\mu}\cap\\{x>0\\}\cap\\{\psi_{\lambda,L,\mu}=0\\}.$

For any $\displaystyle L>b$, $\displaystyle\mu>1$ and
$\displaystyle\lambda\in\mathbb{R}$, the existence and uniqueness of the
minimizer to the truncated variational problem
$\displaystyle(P_{\lambda,L,\mu})$ can be established along the proof of
Theorem 2.1 and Lemma 2.2 in [13]. We state the results in the following.

######  Proposition 2.1.

(Theorem 2.1 and Lemma 2.2 in [13]) For any $\displaystyle L>b$,
$\displaystyle\mu>1$ and $\displaystyle\lambda\in(-\infty,+\infty)$, there
exists a unique minimizer $\displaystyle\psi_{\lambda,L,\mu}$ to the truncated
variational problem $\displaystyle(P_{\lambda,L,\mu})$. Moreover,

$\displaystyle\begin{array}[]{rl}\Gamma_{\lambda,L,\mu}=&\Omega_{L,\mu}\cap\\{x>0\\}\cap\\{\psi_{\lambda,L,\mu}=0\\}\\\
=&\Omega_{L,\mu}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda,L,\mu}<0\\}\\\
=&\Omega_{L,\mu}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda,L,\mu}>0\\},\end{array}$

and $\displaystyle\psi_{\lambda,L,\mu}(x,y)$ is monotone increasing with
respect to $\displaystyle y$ and there exists a continuous function
$\displaystyle k_{\lambda,L,\mu}(x)$ for $\displaystyle x>0$, such that

$\displaystyle\Gamma_{\lambda,L,\mu}=\\{(x,y)\in\Omega_{L}\mid
x>0,y=k_{\lambda,L,\mu}(x)\\}.$

$\displaystyle\psi_{\lambda,L,\mu}$ satisfies the free boundary condition in
the weak sense, namely,

$\begin{array}[]{rl}\lim_{\varepsilon\rightarrow 0^{+},\delta\rightarrow
0^{+}}&\left(\int_{\Omega_{L,\mu}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda,L,\mu}>\varepsilon\\}}(|\nabla\psi_{\lambda,L,\mu}|^{2}-\lambda_{2,L}^{2})\eta\cdot\nu
dS\right.\\\
&\left.+\int_{\Omega_{L,\mu}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda,L,\mu}<-\delta\\}}(|\nabla\psi_{\lambda,L,\mu}|^{2}-\lambda_{1,L}^{2})\eta\cdot\nu
dS\right)=0.\end{array}$ (2.4)

Furthermore, if $\displaystyle\lambda<0$ and $\displaystyle|\lambda|$ is
sufficiently large, then we have
(1) $\displaystyle\psi_{\lambda,L,\mu}(x,y)$ is monotone decreasing with
respect to $\displaystyle x$.
(2) $\displaystyle k_{\lambda,L,\mu}(x)$ is monotone increasing with respect
to $\displaystyle x>0$.
(3) $\displaystyle k_{\lambda,L,\mu}(0)=\lim_{x\rightarrow
0^{+}}k_{\lambda,L,\mu}(x)$ exists and $\displaystyle 0\leq
k_{\lambda,L,\mu}(x)\leq L$.

###  2.2. The regularity of the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$

In Theorem 8.12 in [4], Alt, Caffarelli and Friedman proved that the free
boundary $\displaystyle\Gamma_{\lambda,L,\mu}$ of the minimizer
$\displaystyle\psi_{\lambda,L,\mu}$ is $\displaystyle C^{1}$-smooth. Based on
the significant work [7] by Caffarelli, we will obtain the higher regularity
of the free boundary of the minimizer in this subsection. First, we give the
definition of the weak solution of a free boundary problem as in [7].

######  Definition 2.2.

Assume that $\displaystyle G(t)$ is a continuous strictly monotone increasing
function with respect to $\displaystyle t\in\mathbb{R}$, which satisfies that
$\displaystyle G(t)\geq t$ and $\displaystyle t^{-C}G(t)$ is decreasing with
respect to $\displaystyle t>0$, for some large $\displaystyle C>0$. Let
$\displaystyle E$ be a bounded open set in
$\displaystyle\Omega_{L,\mu}\cap\\{x>0\\}$. A continuous function
$\displaystyle\psi$ in $\displaystyle E$ is called a weak solution of the free
boundary problem, provided that $\displaystyle\psi$ satisfies

(1) $\displaystyle\Delta\psi=0$ in $\displaystyle
E^{+}(\psi)=E\cap\\{\psi>0\\}$,

(2) $\displaystyle\Delta\psi=0$ in $\displaystyle
E^{-}(\psi)=\text{int}(E\cap\\{\psi\leq 0\\})$,

(3) (The weak free boundary condition) $\displaystyle\psi$ satisfies the free
boundary condition

$\displaystyle\psi^{+}_{\nu}=G(\psi_{\nu}^{-})\ \ \text{along}\ \
\mathcal{F}(\psi)=E\cap\partial\\{\psi>0\\},$

in the following sense.

For any $\displaystyle X_{0}\in\mathcal{F}(\psi)$, if
$\displaystyle\mathcal{F}(\psi)$ has an one-side tangent ball at
$\displaystyle X_{0}$ (i.e., there exists a ball $\displaystyle B_{r}(Y)$,
such that $\displaystyle X_{0}\in\partial B_{r}(Y)$ and $\displaystyle
B_{r}(X)$ is contained either in $\displaystyle E^{+}(\psi)$ or in
$\displaystyle E^{-}(\psi)$), then

$\displaystyle\psi(X)=\alpha<X-X_{0},\nu>^{+}-\beta<X-X_{0},\nu>^{-}+o(|X-X_{0}|),\
\ \text{ $\displaystyle\beta\geq 0$ and $\displaystyle\alpha=G(\beta)$},$

where $\displaystyle\nu$ is the unit radial direction of
$\displaystyle\partial B_{r}(Y)$ at $\displaystyle X_{0}$ pointing into
$\displaystyle E^{+}(\psi)$, $\displaystyle<p,q>^{+}=\max\\{p\cdot q,0\\}$ and
$\displaystyle<p,q>^{-}=\max\\{-p\cdot q,0\\}$.

Next, we will obtain the analyticity of the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$ in the following, which implies that the
Rankine-Hugoniot condition (2.2) on the free boundary holds in the classical
sense. The main idea borrows from the works [1, 7, 19].

######  Theorem 2.2.

The free boundary $\displaystyle\Gamma_{\lambda,L,\mu}$ is analytic.

###### Proof.

Step 1. In this step, we will show that the minimizer
$\displaystyle\psi_{\lambda,L,\mu}$ to the truncated variational problem
$\displaystyle(P_{\lambda,L,\mu})$ is a weak solution in Definition 2.2.

Similar to Theorem 2.2 in [4], it is easy to verify that the minimizer
$\displaystyle\psi_{\lambda,L,\mu}$ is harmonic in $\displaystyle
E\setminus\\{\psi_{\lambda,L,\mu}=0\\}$, where $\displaystyle E$ is a bounded
open set in $\displaystyle\Omega_{L,\mu}\cap\\{x>0\\}$, which implies that
$\displaystyle\psi_{\lambda,L,\mu}$ satisfies the conditions (1) and (2) in
Definition 2.2. Next, it suffices to verify the condition (3) in Definition
2.2. Without loss of generality, we assume that $\displaystyle\lambda\leq 0$.

For any $\displaystyle X_{0}\in\mathcal{F}(\psi_{\lambda,L,\mu})$, by means of
Theorem 7.4 in [4], we have

$\displaystyle\psi_{\lambda,L,\mu}(X)=\alpha<X-X_{0},\nu>^{+}-\beta<X-X_{0},\nu>^{-}+o(|X-X_{0}|),$

where $\displaystyle\alpha>0$, $\displaystyle\beta>0$ and
$\displaystyle\lambda=\beta^{2}-\alpha^{2}$.

Thus, $\displaystyle\alpha=G(\beta)=\sqrt{\beta^{2}-\lambda}$. It is easy to
see that $\displaystyle G(\beta)$ is strictly monotone increasing with respect
to $\displaystyle\beta$ and $\displaystyle\beta^{-1}G(\beta)$ is decreasing
with respect to $\displaystyle\beta$.

Hence, we conclude that the minimizer $\displaystyle\psi_{\lambda,L,\mu}$ is a
weak solution in Definition 2.2.

Step 2. Next, we will obtain the analyticity of the free boundary.

Since $\displaystyle\psi_{\lambda,L,\mu}$ is the weak solution in Definition
2.2, by using Theorem 1 in [7], we can conclude that the free boundary
$\displaystyle\Omega_{L,\mu}\cap\partial\\{\psi_{\lambda,L,\mu}>0\\}$ is
$\displaystyle C^{1,\alpha}$ for some $\displaystyle\alpha\in(0,1)$.

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

$\displaystyle\Gamma_{\lambda,L,\mu}=\Omega_{L,\mu}\cap\\{x>0\\}\cap\partial\\{\psi>0\\}=\Omega_{L,\mu}\cap\\{x>0\\}\cap\\{\psi=0\\},$

is $\displaystyle C^{1,\alpha}$-smooth, which implies that the
$\displaystyle\mathcal{L}^{2}$-measure of the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$ is zero. Therefore, we can use a
$\displaystyle C^{1,\alpha}$ transformation to flatten the free boundary. Then
reflect $\displaystyle\psi^{+}$ to the full neighborhood of the free boundary,
applying the Schauder estimates for elliptic equation in divergence form in
Section 9 in [1], we can obtain the $\displaystyle C^{1,\alpha}$ regularity of
$\displaystyle\psi^{+}$ up to the free boundary. Similarly, we can obtain the
$\displaystyle C^{1,\alpha}$ regularity of $\displaystyle\psi^{-}$ up to the
free boundary. Moreover, it follows from (2.4) that

$|\nabla\psi^{-}|^{2}-|\nabla\psi^{+}|^{2}=\lambda\ \ \text{on the free
boundary}.$ (2.5)

If $\displaystyle\lambda=0$, it follows from Theorem 2.2 in [4] that
$\displaystyle\psi$ is harmonic in $\displaystyle\Omega_{L,\mu}\cap\\{x>0\\}$.
By means of the monotonicity of $\displaystyle\psi(x,y)$ with respect to
$\displaystyle y$, the strong maximum principle gives that
$\displaystyle\partial_{y}\psi>0$ in
$\displaystyle\Omega_{L,\mu}\cap\\{x>0\\}$. Hence, the implicit function
theorem gives that the level set
$\displaystyle\Omega_{L,\mu}\cap\\{x>0\\}\cap\\{\psi=0\\}$ is analytic.

If $\displaystyle\lambda\neq 0$, without loss of generality, we assume that
$\displaystyle 0$ is a free boundary point of $\displaystyle\psi$,
$\displaystyle|\nabla\psi^{+}(0)|\neq 0$ and the inner normal to
$\displaystyle\Gamma_{\lambda,L,\mu}$ at $\displaystyle 0$ is in the direction
of the positive $\displaystyle y$-axis. Extend $\displaystyle\tilde{\psi}$ as
a $\displaystyle C^{1,\alpha}$ function into a full neighborhood of
$\displaystyle 0\in\Gamma_{\lambda,L,\mu}$, such that
$\displaystyle\tilde{\psi}=\psi$ in $\displaystyle\\{\psi>0\\}$. In view of
$\displaystyle|\nabla\psi^{+}(0)|\neq 0$, one has

$\tilde{\psi}_{y}(0)>0.$ (2.6)

Define a mapping as follows,

$\displaystyle S=TX=(s,t)\triangleq(x,\tilde{\psi}(x,y)),\ \ X=(x,y).$

By virtue of (2.6), it is easy to check that

$\displaystyle\text{det}\left(\frac{\partial S}{\partial
X}\right)=\tilde{\psi}_{y}(x,y)>0\ \ \text{in a small neighborhood of
$\displaystyle 0$}.$

And thus the mapping $\displaystyle T$ is a local diffeomorphism near
$\displaystyle 0$.

Denote the inverse transform as

$\left\\{\begin{array}[]{rl}&x=s\\\ &y=\phi(s,t).\end{array}\right.$ (2.7)

Therefore, the free boundary $\displaystyle\Gamma_{\lambda,L,\mu}$ is
transformed into $\displaystyle t=0$, and we have

$\displaystyle\left(\frac{\partial X}{\partial S}\right)=\left(\frac{\partial
S}{\partial X}\right)^{-1}=\left(\begin{matrix}1&0\\\
-\frac{\tilde{\psi}_{x}}{\tilde{\psi}_{y}}&\frac{1}{\tilde{\psi}_{y}}\end{matrix}\right).$

Consequently, one has

$\displaystyle\phi_{s}=\frac{\partial y}{\partial
s}=-\frac{\tilde{\psi}_{x}}{\tilde{\psi}_{y}},\ \ \phi_{t}=\frac{\partial
y}{\partial t}=\frac{1}{\tilde{\psi}_{y}},$

and

$\frac{\partial t}{\partial x}=\tilde{\psi}_{x}=-\frac{\phi_{s}}{\phi_{t}},\ \
\frac{\partial t}{\partial y}=\tilde{\psi}_{y}=\frac{1}{\phi_{t}}.$ (2.8)

It follows from (2.8) that

$\displaystyle\mathcal{Q}\phi=\partial_{s}\left(\frac{\phi_{s}}{\phi_{t}}\right)+\partial_{t}\left(-\frac{1+\phi^{2}_{s}}{2\phi^{2}_{t}}\right)=0\
\ \text{in the neighborhood of $\displaystyle 0$}.$

Denote $\displaystyle A_{1}(\phi_{s},\phi_{t})=\frac{\phi_{s}}{\phi_{t}}$ and
$\displaystyle
A_{2}(\phi_{s},\phi_{t})=-\frac{1+\phi^{2}_{s}}{2\phi^{2}_{t}}$. It is easy to
check that

$\displaystyle\mathcal{A}=\frac{\partial(A_{1},A_{2})}{\partial(\phi_{s},\phi_{t})}=\frac{1}{\phi_{t}^{3}}\left(\begin{matrix}\phi_{t}^{2}&-\phi_{s}\phi_{t}\\\
-\phi_{s}\phi_{t}&1+\phi_{s}^{2}\end{matrix}\right).$

Direct straightforward computations give that the matrix
$\displaystyle\mathcal{A}$ has two eigenvalues

$\displaystyle\lambda_{1}=\frac{1+\phi_{s}^{2}+\phi_{t}^{2}+\sqrt{((1+\phi_{t})^{2}+\phi_{s}^{2})((1-\phi_{t})^{2}+\phi_{s}^{2})}}{2\phi_{t}^{3}},$

and

$\displaystyle\lambda_{2}=\frac{2}{\left(1+\phi_{s}^{2}+\phi_{t}^{2}+\sqrt{((1+\phi_{t})^{2}+\phi_{s}^{2})((1-\phi_{t})^{2}+\phi_{s}^{2})}\right)\phi_{t}}.$

In view of (2.6), one has

$\displaystyle\lambda_{1}>0\ \ \text{and}\ \ \lambda_{2}>0\ \ \text{in a small
neighborhood of $\displaystyle 0$}.$

Thus, $\displaystyle\mathcal{Q}\phi=0$ is a quasilinear elliptic equation in a
neighborhood $\displaystyle E$ of $\displaystyle 0$. Furthermore,
$\displaystyle\phi$ satisfies the Neumann type boundary condition as follows,

$\left\\{\begin{array}[]{ll}\mathcal{Q}\phi=0{}{}{}{}&\text{in}\ \ \ D,\\\
\frac{\phi_{t}}{\sqrt{1+\phi_{s}^{2}}}=g(s)&\text{on}\ \
\bar{D}\cap\\{t=0\\},\end{array}\right.$ (2.9)

where $\displaystyle D=E\cap\\{t>0\\}$ and $\displaystyle
g(s,0)=\frac{1}{\sqrt{|\nabla\psi^{-}(s,0)|^{2}-\lambda}}$.

Noting that $\displaystyle\tilde{\psi}$ is in $\displaystyle C^{1,\alpha}$
near $\displaystyle 0$, we have that the coefficients of
$\displaystyle\mathcal{Q}$ and $\displaystyle g(s)$ are $\displaystyle
C^{0,\alpha}$. By using the elliptic regularity in Section 9 in [1], we obtain
that $\displaystyle\phi(S)$ is $\displaystyle C^{2,\alpha}$ near
$\displaystyle 0$. Furthermore, the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$ can be described by $\displaystyle
y=\phi(s,0)=\phi(x,0)$, and thus the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$ is $\displaystyle C^{2,\alpha}$ near
$\displaystyle 0$.

Applying the Schauder estimates for elliptic equations in [1], we can obtain
the $\displaystyle C^{2,\alpha}$ regularity of $\displaystyle\psi^{+}$ and
$\displaystyle\psi^{-}$ up to the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$. By using the above arguments, we can
conclude that the free boundary $\displaystyle\Gamma_{\lambda,L,\mu}$ is
$\displaystyle C^{3,\alpha}$.

Along the bootstrap arguments, the $\displaystyle C^{\infty}$ regularity of
the free boundary $\displaystyle\Gamma_{\lambda,L,\mu}$ can be established.
Finally, with the aid of the results in Section 6.7 in [19], we can obtain
that $\displaystyle\phi(s,t)$ is analytic in $\displaystyle t\geq 0$. Hence,
we obtain the analyticity of the free boundary
$\displaystyle\Gamma_{\lambda,L,\mu}$.

∎

It follows from Lemma 2.4 and Lemma 2.5 in [13] that we can obtain the
existence of $\displaystyle\lambda_{L,\mu}$, such that the continuous fit
condition of $\displaystyle\Gamma_{\lambda_{L,\mu},L,\mu}$ at $\displaystyle
A$ holds for any $\displaystyle L>b$ and $\displaystyle\mu>1$. Similar to the
arguments on the compressible subsonic flows in infinitely long nozzle in [8,
9, 10, 23, 24, 25], we can obtain the asymptotic behavior of the flows in
downstream and in upstream. We omit the details here.

######  Proposition 2.3.

For any $\displaystyle Q>0$, $\displaystyle L>b$ and $\displaystyle\mu>1$,
there exists a $\displaystyle\lambda_{L,\mu}\in\mathbb{R}$ with
$\displaystyle|\lambda_{L,\mu}|\leq C$ (the constant $\displaystyle C$ is
independent of $\displaystyle\mu$), such that $\displaystyle
k_{\lambda_{L,\mu},L,\mu}(0)=0$ and
$\displaystyle\psi_{\lambda_{L,\mu},L,\mu}$ is Lipschitz continuous in
$\displaystyle\bar{\Omega}_{L,\mu}\setminus B_{\varepsilon}(B)$ for any small
$\displaystyle\varepsilon>0$. Furthermore,

$\displaystyle k_{\lambda_{L,\mu},L,\mu}(x)\rightarrow
h_{\lambda_{L,\mu}}\in(b,L)$ and
$\displaystyle\psi_{\lambda_{L,\mu},L,\mu}(x,y)\rightarrow\psi_{0}(y)$

in any compact subset of $\displaystyle(b,L)$ as $\displaystyle
x\rightarrow+\infty$, where $\displaystyle h_{\lambda_{L},\mu}$ is determined
uniquely by

$\displaystyle\lambda_{L,\mu}=\frac{Q^{2}}{(h_{\lambda_{L,\mu}}-b)^{2}}-\frac{L^{2}}{(L-h_{\lambda_{L,\mu}})^{2}}\
\ \text{and}\ \
\psi_{0}(y)=\left\\{\begin{array}[]{ll}\frac{Q(y-h_{\lambda_{L,\mu}})}{h_{\lambda_{L,\mu}}-b},&\text{if
$\displaystyle b<y<h_{\lambda_{L,\mu}}$},\\\
\frac{L(y-h_{\lambda_{L,\mu}})}{L-h_{\lambda_{L,\mu}}},&\text{if
$\displaystyle h_{\lambda_{L,\mu}}<y<L$},\end{array}\right.$

###  2.3. The existence of the truncated injection flow problem 1

In this subsection, we will investigate the existence of the truncated
injection flow problem 1. Moreover, the positivity of horizontal velocity and
vertical velocity will be obtained.

######  Theorem 2.4.

For any $\displaystyle Q>0$ and $\displaystyle L>b$, there exist a
$\displaystyle\lambda_{L}$ and a solution
$\displaystyle(\psi_{\lambda_{L}},\Gamma_{\lambda_{L}})$ to the truncated
injection flow problem 1.

###### Proof.

Since $\displaystyle\psi_{\lambda_{L,\mu},L,\mu}$ is Lipschitz continuous in
$\displaystyle\Omega_{L,\mu}$ and $\displaystyle|\lambda_{L,\mu}|\leq C$( the
constant $\displaystyle C>0$ is independent of $\displaystyle\mu$), we can
take a sequence $\displaystyle\\{\mu_{n}\\}$ with
$\displaystyle\mu_{n}\rightarrow+\infty$, then there exist a
$\displaystyle\lambda_{L}$ and a $\displaystyle\psi_{\lambda_{L},L}\in
H_{loc}^{1}(\Omega_{L})$, such that

$\displaystyle\lambda_{L,\mu_{n}}\rightarrow\lambda_{L},$

and

$\displaystyle\psi_{\lambda_{L,\mu_{n}},L,\mu_{n}}\rightarrow\psi_{\lambda_{L},L}\
\text{in $\displaystyle H^{1}_{loc}(\Omega_{L})$ and uniformly in any compact
subset of $\displaystyle\Omega_{L}$},$

as $\displaystyle\mu_{n}\rightarrow+\infty$.

Next, we divide six steps to verify that
$\displaystyle(\psi_{\lambda_{L}},\Gamma_{\lambda_{L}})$ satisfies the
conditions in Definition 2.1.

Step 1. For any $\displaystyle Q>0$ and $\displaystyle L>b$, denote

$\displaystyle D_{1,L}=\Omega_{L}\cap\\{(x,y)\mid x<0,y<0\\}\ \ \text{and}\ \
D_{2,L}=\Omega_{L}\cap\\{(x,y)\mid x<0,y>0\\}.$

By virtue of Lemma 6.2 in [4], we can show that
$\displaystyle\psi_{\lambda_{L},L}$ is a local minimizer to the variational
problem $\displaystyle(P_{\lambda_{L},L})$, namely,

$\displaystyle P_{\lambda_{L},L}:\ \ J_{D}(\psi_{\lambda_{L},L})=\min
J_{D}(\psi)\ \ \text{for any $\displaystyle\psi\in K_{L}$ and
$\displaystyle\psi=\psi_{\lambda_{L},L}$ on $\displaystyle\partial D$},$

where

$\displaystyle
J_{D}(\psi)=\int_{D}\left|\nabla\psi-(\lambda_{1,L}I_{\\{\psi<0\\}}+\lambda_{2,L}I_{\\{\psi>0\\}}+\lambda_{0,L}I_{\\{\psi=0\\}})I_{\\{x>0\\}}e\right|^{2}dxdy$

and

$\displaystyle\begin{array}[]{rl}K_{L}=\\{\psi\mid&\psi\in
H^{1}_{loc}(\Omega_{L}),\ -Q\leq\psi\leq L\ \text{a.e. in
$\displaystyle\Omega_{L}$},\ \ 0\leq\psi\leq L\ \text{a.e. in $\displaystyle
D_{2,L}$},\\\ &-Q\leq\psi\leq 0\ \text{a.e. in $\displaystyle D_{1,L}$},\
\psi=-Q\ \text{on $\displaystyle S_{2}\cup N_{2}$},\\\ &\ \psi=L\ \text{on
$\displaystyle N_{L}$},\ \psi=0\ \text{on $\displaystyle N_{1}\cup
S_{1}$}\\}.\end{array}$

for any bounded domain $\displaystyle D\subset\Omega_{L}$. Therefore, the
conditions (1) and (2) in Definition 2.1 have be verified.

Step 2. We can conclude that $\displaystyle\psi_{\lambda_{L},L}(x,y)$ is
monotone increasing with respect to $\displaystyle y$ and decreasing with
respect to $\displaystyle x$, which follows from the monotonicity of
$\displaystyle\psi_{\lambda_{L,\mu},L,\mu}(x,y)$. Furthermore, the free
boundary

$\displaystyle\Gamma_{\lambda_{L},L}=\Omega_{L}\cap\\{x>0\\}\cap\\{\psi_{\lambda_{L},L}=0\\}$

of the minimizer $\displaystyle\psi_{\lambda_{L},L}$ is given by a continuous
function $\displaystyle y=k_{\lambda_{L},L}(x)$ for any $\displaystyle x>0$.
In particular, $\displaystyle k_{\lambda_{L},L}(0)=0$. In view of Proposition
2.3, one has

$\displaystyle k_{\lambda_{L},L}(x)\rightarrow h_{L}$ and
$\displaystyle\psi_{\lambda_{L},L}(x,y)\rightarrow\psi_{0}(y)$ (2.10)

in any compact subset of $\displaystyle(b,L)$ as $\displaystyle
x\rightarrow+\infty$, where $\displaystyle h_{L}$ is determined uniquely by

$\displaystyle\lambda_{L}=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{L^{2}}{(L-h_{L})^{2}}\
\ \text{and}\ \
\psi_{0}(x,y)=\left\\{\begin{array}[]{ll}\frac{Q(y-h_{L})}{h_{L}-b},&\text{if
$\displaystyle b<y<h_{L}$},\\\ \frac{L(y-h_{L})}{L-h_{L}},&\text{if
$\displaystyle h_{L}<y<L$}.\end{array}\right.$

Furthermore, by virtue of the analyticity of the free boundary
$\displaystyle\Gamma_{\lambda_{L},L}$ and (2.10), one has

$\displaystyle k^{\prime}_{\lambda_{L},L}(x)\rightarrow 0\ \ \text{as}\ \
x\rightarrow+\infty.$

Next, we will show that $\displaystyle k_{\lambda_{L},L}(x)$ is strictly
monotone increasing with respect to $\displaystyle x>0$. If not, there exist
$\displaystyle x_{1},x_{2}\in(0,+\infty)$ with $\displaystyle x_{1}<x_{2}$,
such that $\displaystyle k_{\lambda_{L},L}(x_{1})=k_{\lambda_{L},L}(x_{2})$.
The monotonicity of $\displaystyle\psi_{\lambda_{L},L}(x,y)$ with respect to
$\displaystyle x$ and $\displaystyle y$ gives that there exists a small
$\displaystyle r>0$, such that

$\displaystyle\psi_{\lambda_{L},L}>0$ in $\displaystyle
B_{r}(X_{0})\cap\\{y>y_{0}\\}$ and $\displaystyle\psi_{\lambda_{L},L}<0$ in
$\displaystyle B_{r}(X_{0})\cap\\{y<y_{0}\\}$,

where $\displaystyle X_{0}=(x_{0},y_{0})$ with $\displaystyle
x_{0}=\frac{x_{1}+x_{2}}{2}$ and $\displaystyle
y_{0}=k_{\lambda_{L},L}(x_{1})$. Denote $\displaystyle
I_{0}=\\{(x,y_{0})\mid-r<x-x_{0}<r\\}$, $\displaystyle
B^{+}=B_{2r}(X_{0})\cap\\{y>y_{0}\\}$ and $\displaystyle
B^{-}=B_{2r}(X_{0})\cap\\{y<y_{0}\\}$. Set
$\displaystyle\psi_{1}=\psi_{\lambda_{L},L}$ in $\displaystyle B^{-}$ and
$\displaystyle\psi_{2}=\psi_{\lambda_{L},L}$ in $\displaystyle B^{+}$. Since
$\displaystyle\psi_{\lambda_{L},L}(x,y)$ is decreasing with respect to
$\displaystyle x$, the strong maximum principle gives that

$\displaystyle\partial_{x}\psi_{1}<0\ \ \text{in $\displaystyle B^{-}$ and }\
\ \partial_{x}\psi_{2}<0\ \ \text{in $\displaystyle B^{+}$}.$

In view of that $\displaystyle\partial_{x}\psi_{1}=\partial_{x}\psi_{2}=0$ on
$\displaystyle I_{0}$, it follows from Hopf’s lemma that

$\frac{\partial}{\partial y}\partial_{x}\psi_{1}>0\ \ \text{and }\ \
\frac{\partial}{\partial y}\partial_{x}\psi_{2}<0\ \ \text{on $\displaystyle
I_{0}$}.$ (2.11)

Noting that $\displaystyle\psi_{1}=\psi_{2}=0$ on $\displaystyle I_{0}$,
thanks hopf’s lemma, one has

$\displaystyle\partial_{y}\psi_{1}>0\ \ \text{and }\ \ \partial_{y}\psi_{2}>0\
\ \text{on $\displaystyle I_{0}$},$

which together with (2.11) give that

$\displaystyle\frac{\partial(\partial_{y}\psi_{1})^{2}}{\partial x}>0\ \
\text{and }\ \ \frac{\partial(\partial_{y}\psi_{2})^{2}}{\partial x}<0\ \
\text{on $\displaystyle I_{0}$}.$

Thus one has

$\displaystyle\frac{\partial}{\partial
x}\left(|\nabla\psi^{-}_{\lambda_{L},L}|^{2}-|\nabla\psi^{+}_{\lambda_{L},L}|^{2}\right)=\frac{\partial}{\partial
x}\left((\partial_{y}\psi_{1})^{2}-(\partial_{y}\psi_{2})^{2}\right)>0\ \
\text{on $\displaystyle I_{0}$},$

which contradicts to the fact
$\displaystyle|\nabla\psi^{-}_{\lambda_{L},L}|^{2}-|\nabla\psi^{+}_{\lambda_{L},L}|^{2}=\lambda_{L}$
on $\displaystyle I_{0}$.

Step 3. It follows from (2.4) that

$\displaystyle\begin{array}[]{rl}\lim_{\varepsilon\rightarrow
0^{+},\delta\rightarrow
0^{+}}&\left(\int_{\Omega_{L}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda_{L},L}>\varepsilon\\}}(|\nabla\psi_{\lambda_{L},L}|^{2}-\lambda_{2,L}^{2})\eta\cdot\nu
dS\right.\\\
&\left.+\int_{\Omega_{L}\cap\\{x>0\\}\cap\partial\\{\psi_{\lambda_{L},L}<-\delta\\}}(|\nabla\psi_{\lambda_{L},L}|^{2}-\lambda_{1,L}^{2})\eta\cdot\nu
dS\right)=0.\end{array}$

Since the free boundary $\displaystyle\Gamma_{\lambda_{L},L}$ is analytic,
$\displaystyle\psi_{\lambda_{L},L}$ is $\displaystyle C^{2}$ up to the free
boundary $\displaystyle\Gamma_{\lambda_{L},L}$. Thus, the condition (4) in
Definition 2.1 holds.

Step 4. In this step, we will show that the free boundary
$\displaystyle\Gamma_{\lambda_{L},L}$ is continuous differentiable at
$\displaystyle A$, and (2.3) holds.

For any $\displaystyle r>0$, define a blow-up sequence
$\displaystyle\\{\psi_{n}\\}$, such that
$\displaystyle\psi_{n}(\tilde{X})=\frac{\psi_{\lambda_{L},L}(r_{n}\tilde{X})}{r_{n}}$
in $\displaystyle B_{r}(0)$ with
$\displaystyle\tilde{X}=(\tilde{x},\tilde{y})$ and $\displaystyle
r_{n}\rightarrow 0$. We next consider the following three cases.

Case 1. $\displaystyle\lambda_{L}<0$. Taking a sequence
$\displaystyle\\{X_{n}\\}$ with $\displaystyle X_{n}\in\Gamma_{\lambda_{L},L}$
and $\displaystyle X_{n}\rightarrow 0$, set $\displaystyle r_{n}=|X_{n}|$. By
virtue of the non-degeneracy Theorem 3.1 in [4], one has

$\frac{1}{r_{n}}\fint_{\partial
B_{\frac{r_{n}}{2}}(X_{n})}\psi_{\lambda_{L},L}^{+}(X)dS_{X}\geq
c|\lambda_{L}|^{\frac{1}{2}},$ (2.12)

where $\displaystyle c>0$ is a constant independent of $\displaystyle n$.
Denote $\displaystyle\tilde{Y}_{n}=\frac{X_{n}}{r_{n}}$, it follows from
(2.12) that

$\fint_{\partial
B_{\frac{1}{2}}(\tilde{Y}_{n})}\psi_{n}^{+}(\tilde{X})dS_{\tilde{X}}\geq
c|\lambda_{L}|^{\frac{1}{2}}\ \ \text{and}\ \ |\tilde{Y}_{n}|=1.$ (2.13)

It follows from the similar arguments in Pages 444-445 in [4] that there exist
a subsequence $\displaystyle\\{\psi_{n}\\}$ and a blow-up limit
$\displaystyle\psi_{0}\in H_{loc}^{1}(\mathbb{R}^{2})$, such that

$\psi_{n}(\tilde{X})\rightarrow\psi_{0}(\tilde{X})\ \text{uniformly in bounded
sets, and $\displaystyle\nabla\psi_{n}\rightarrow\nabla\psi_{0}$ a.e. in
$\displaystyle\mathbb{R}^{2}$}.$ (2.14)

Furthermore,

$\partial\\{\psi_{n}>0\\}\rightarrow\partial\\{\psi_{0}>0\\}\ \text{locally in
the Hausdorff metric}.$ (2.15)

In particular, $\displaystyle 0$ is the free boundary point of
$\displaystyle\psi_{0}$.

In view of (2.13) and (2.14), there exist two subsequences
$\displaystyle\\{\tilde{Y}_{n}\\}$ and $\displaystyle\\{\psi_{n}\\}$, such
that $\displaystyle\tilde{Y}_{n}\rightarrow\tilde{Y}_{0}$ and

$\fint_{\partial
B_{\frac{1}{2}}(\tilde{Y}_{0})}\psi_{0}^{+}(\tilde{X})dS_{\tilde{X}}\geq
c|\lambda_{L}|^{\frac{1}{2}}\ \ \text{and}\ \ |\tilde{Y}_{0}|=1.$ (2.16)

By using the monotonicity formula lemma 5.1 in [4], one has

$\displaystyle\frac{1}{r^{4}}\int_{B_{r}(\tilde{Y}_{0})}|\nabla\psi_{0}^{+}(\tilde{X})|^{2}d\tilde{X}\cdot\int_{B_{r}(\tilde{Y}_{0})}|\nabla\psi_{0}^{-}(\tilde{X})|^{2}d\tilde{X}=\gamma\geq
0,$

for any $\displaystyle r>0$, which together with Lemma 6.6 in [4] gives that

$\displaystyle\psi_{0}$ is either a 2-plane solution, or a 1-plane solution,
or identically zero. (2.17)

Since $\displaystyle\theta>0$, one has

$\psi_{\lambda_{L},L}\equiv 0\ \ \text{in $\displaystyle S$ and}\
\frac{\mathcal{L}^{2}(B_{r}(A)\cap
S)}{\mathcal{L}^{2}(B_{r}(A))}=\frac{\theta}{2\pi}>0,$ (2.18)

for any $\displaystyle r>0$, where $\displaystyle S=\\{(x,y)\mid x\leq
y\cot\theta,y\leq 0\\}$. By virtue of (2.17) and (2.18), one has

$\displaystyle\psi_{0}$ is either a 1-plane solution, or identically zero,

which together with (2.16) gives that

$\psi_{0}\geq 0\ \ \text{and}\ \ \ \psi_{0}\not\equiv 0.$ (2.19)

Since $\displaystyle\psi_{\lambda_{L},L}(X)\leq 0$ in
$\displaystyle\\{(x,y)\mid y\leq 0\\}$ and
$\displaystyle\psi_{\lambda_{L},L}(X)>0$ in $\displaystyle\\{(x,y)\mid
x<0,y>0\\}$, by virtue of (2.14) and (2.19), we have

$\psi_{0}(\tilde{x},\tilde{y})\equiv 0\ \ \text{in
$\displaystyle\\{(\tilde{x},\tilde{y})\mid\tilde{y}\leq 0\\}$},$ (2.20)

and

$\psi_{0}(\tilde{x},\tilde{y})>0\ \ \text{in
$\displaystyle\\{(\tilde{x},\tilde{y})\mid\tilde{x}<0,\tilde{y}>0\\}$}.$
(2.21)

Thus, it follows from (2.21) that

$\psi_{0}(\tilde{x},\tilde{y})=\max\\{\alpha\tilde{y},0\\}\ \ \text{and}\ \
\alpha>0.$ (2.22)

By virtue of the proof of Lemma 6.2 in [4], we can conclude that
$\displaystyle\psi_{0}$ is a local minimizer for the variational problem

$\displaystyle J_{R}(\psi_{0})=\min J_{R}(\psi)\ \ \text{for any
$\displaystyle\psi-\psi_{0}\in H_{0}^{1}(B_{R})$ and $\displaystyle R>0$},$

where $\displaystyle B_{R}=B_{R}(0)$ and the functional

$\displaystyle
J_{R}(\psi)=\int_{B_{R}}\left|\nabla\psi-(\lambda_{1,L}I_{\\{\psi\leq
0\\}}+\lambda_{2,L}I_{\\{\psi>0\\}})I_{\\{\tilde{x}>0\\}}e\right|^{2}d\tilde{x}d\tilde{y},$

with $\displaystyle\lambda_{1,L}=\frac{Q}{h_{L}-b}$ and
$\displaystyle\lambda_{2,L}=\frac{L}{L-h_{L}}$. Next, we claim that
$\displaystyle\psi_{0}$ is a local minimizer for the variational problem,
namely,

$\displaystyle J_{R}^{0}(\psi_{0})=\min J_{R}^{0}(\psi)\ \ \text{for any
$\displaystyle\psi-\psi_{0}\in H_{0}^{1}(B_{R})$ and $\displaystyle R>0$},$

where the functional

$\displaystyle
J^{0}_{R}(\psi)=\int_{B_{R}}|\nabla\psi|^{2}+(\lambda_{2,L}^{2}-\lambda_{1,L}^{2})I_{\\{\psi>0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}.$

In fact, for any $\displaystyle\psi-\psi_{0}\in H_{0}^{1}(B_{R})$, one has

$\displaystyle\begin{array}[]{rl}&\int_{B_{R}}|\nabla\psi_{0}|^{2}-\lambda_{L}^{2}I_{\\{\psi_{0}>0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}-\int_{B_{R}}|\nabla\psi|^{2}-\lambda_{L}^{2}I_{\\{\psi>0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}\\\
=&\int_{B_{R}}|\nabla\psi_{0}|^{2}+(\lambda^{2}_{1,L}I_{\\{\psi_{0}\leq
0\\}}+\lambda^{2}_{2,L}I_{\\{\psi_{0}>0\\}})I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}\\\
&-\int_{B_{R}}|\nabla\psi|^{2}+(\lambda^{2}_{1,L}I_{\\{\psi\leq
0\\}}+\lambda^{2}_{2,L}I_{\\{\psi>0\\}})I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}\\\
=&\int_{B_{R}}\left|\nabla\psi_{0}-(\lambda_{1,L}I_{\\{\psi_{0}\leq
0\\}}+\lambda_{2,L}I_{\\{\psi_{0}>0\\}})I_{\\{\tilde{x}>0\\}}e\right|^{2}d\tilde{x}d\tilde{y}\\\
&-\int_{B_{R}}\left|\nabla\psi-(\lambda_{1,L}I_{\\{\psi\leq
0\\}}+\lambda_{2,L}I_{\\{\psi>0\\}})I_{\\{\tilde{x}>0\\}}e\right|^{2}d\tilde{x}d\tilde{y}\\\
\leq&0,\end{array}$

where we have used the fact

$\displaystyle\int_{B_{R}}\nabla\psi_{0}\cdot eI_{\\{\psi_{0}\leq
0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}=\int_{B_{R}}\nabla\psi\cdot
eI_{\\{\psi\leq 0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y},$

and

$\displaystyle\int_{B_{R}}\nabla\psi_{0}\cdot
eI_{\\{\psi_{0}>0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}=\int_{B_{R}}\nabla\psi\cdot
eI_{\\{\psi>0\\}}I_{\\{\tilde{x}>0\\}}d\tilde{x}d\tilde{y}.$

For the minimal functional $\displaystyle J^{0}_{R}(\psi_{0})$, it follows
from Theorem 2.5 in [2] that

$\displaystyle|\nabla\psi_{0}|^{2}=\lambda_{2,L}^{2}-\lambda_{1,L}^{2}=-\lambda_{L}\
\ \text{on the free boundary of $\displaystyle\psi_{0}$},$

which implies that

$\displaystyle\alpha=\sqrt{-\lambda_{L}}.$

Hence, one has

$\displaystyle\psi_{n}(\tilde{x},\tilde{y})\rightarrow\sqrt{-\lambda_{L}}\tilde{y}^{+}\
\ \text{as $\displaystyle r_{n}\rightarrow 0$.}$

By using the similar arguments in Lemma 11.2 in Chapter 3 in [12], we can
conclude that $\displaystyle k^{\prime}_{\lambda_{L},L}(0+0)=0$.

Case 2. $\displaystyle\lambda_{L}>0$. Similar to Case 1, we can conclude that
$\displaystyle k^{\prime}_{\lambda_{L},L}(0+0)=\tan\theta$.

Case 3. $\displaystyle\lambda_{L}=0$. It is easy to check that
$\displaystyle\psi_{\lambda_{L},L}$ is a harmonic function across the free
boundary in $\displaystyle\Omega_{L}$. By using a conformal mapping
$\displaystyle\tilde{\psi}(z)=\psi_{\lambda_{L},L}\left(z^{\frac{\pi}{2\pi-\theta}}\right)$
with $\displaystyle z=x+iy$, such that $\displaystyle\tilde{\psi}(z)$ becomes
a harmonic function in $\displaystyle B_{r}(0)\cap\\{\text{Im}z>0\\}$,
$\displaystyle\tilde{\psi}=0$ on $\displaystyle
B_{r}(0)\cap\partial\\{\text{Im}z>0\\}$, where $\displaystyle z=x+iy$.
Furthermore, $\displaystyle N_{1}\cup S_{1}$ is mapped into the real axis and
the free boundary $\displaystyle\Gamma_{\lambda_{L},L}$ is mapped into a
continuous arc $\displaystyle\gamma$ initiating at $\displaystyle 0$. Then the
harmonic function $\displaystyle\tilde{\psi}(z)$ has harmonic continuation
across Im$\displaystyle z=0$ in $\displaystyle B_{r}(0)$. It follows that the
level set $\displaystyle\\{\tilde{\psi}=0\\}$ consists arcs forming equal
angles at $\displaystyle A$. Since $\displaystyle\tilde{\psi}$ vanishes only
on $\displaystyle\gamma$, which implies that the continuous arc
$\displaystyle\gamma$ must intersect $\displaystyle\partial\\{\text{Im}z>0\\}$
orthogonally at $\displaystyle A$, namely, $\displaystyle
k_{\lambda_{L},L}^{\prime}(0+0)=\tan\frac{\theta}{2}$.

Hence, the condition (5) in Definition 2.1 is obtained.

Step 5. In this step, we will verify that $\displaystyle\psi_{\lambda_{L},L}$
satisfies the condition (6) in Definition 2.1. The asymptotic behavior of
$\displaystyle\psi_{\lambda_{L},L}$ in downstream has been obtained in Step
2\. Next, we consider the asymptotic behavior of
$\displaystyle\psi_{\lambda_{L},L}$ in upstream. Define a blow-up sequence
$\displaystyle\psi_{n}(x,y)=\psi_{\lambda_{L},L}(x-n,y)$ for $\displaystyle
x<\frac{n}{2}$. By using the elliptic regularity in [16], there exists a
subsequence $\displaystyle\\{\psi_{n}\\}$, such that

$\displaystyle\psi_{n}(x,y)\rightarrow\bar{\psi}_{0}(x,y)\ \ \text{in}\ \
C^{2,\alpha}(D),$

for any compact subset $\displaystyle D$ of $\displaystyle
E=\\{-\infty<x<+\infty\\}\times\\{0<y<L\\}$, and

$\displaystyle\left\\{\begin{array}[]{ll}&\Delta\bar{\psi}_{0}=0\
\text{in}~{}~{}E,\\\ &\bar{\psi}_{0}(x,0)=0\ \text{and}\
\bar{\psi}_{0}(x,L)=L\ \ \text{for $\displaystyle-\infty<x<+\infty$},\\\
&0\leq\bar{\psi}_{0}(x,y)\leq L\ \ \text{in}~{}~{}E.\end{array}\right.$

Then the above boundary value problem has a unique solution

$\displaystyle\bar{\psi}_{0}(x,y)=y\ \ \text{in}\ \ E,$

which implies that

$\psi_{\lambda_{L},L}(x,y)\rightarrow y\ \ \text{for}\ \ 0<y<L,\ \ \text{as}\
\ x\rightarrow-\infty.$ (2.23)

Denote $\displaystyle\tilde{x}=x\sin\theta-y\cos\theta$ and
$\displaystyle\tilde{y}=y\sin\theta+x\cos\theta$, let

$\displaystyle\tilde{\psi}(\tilde{x},\tilde{y})=\psi_{\lambda_{L},L}(\tilde{x}\sin\theta+\tilde{y}\cos\theta,\tilde{y}\sin\theta-\tilde{x}\cos\theta)$

and
$\displaystyle\tilde{\psi}_{n}(\tilde{x},\tilde{y})=\tilde{\psi}(\tilde{x},\tilde{y}-n)$.
By virtue of the elliptic regularity, there exists a subsequence
$\displaystyle\\{\tilde{\psi}_{n}\\}$, such that

$\displaystyle\tilde{\psi}_{n}(\tilde{x},\tilde{y})\rightarrow\tilde{\psi}_{0}(\tilde{x},\tilde{y})\
\ \text{in}\ \ C^{2,\alpha}(D),$

for any compact subset $\displaystyle D$ of
$\displaystyle\tilde{E}=\\{0<\tilde{x}<a\sin\theta-b\cos\theta\\}\times\\{-\infty<\tilde{y}<+\infty\\}$,
and $\displaystyle\tilde{\psi}_{0}$ satisfies that

$\displaystyle\left\\{\begin{array}[]{ll}&\Delta\tilde{\psi}_{0}=0\
\text{in}~{}~{}\tilde{E},\\\ &\tilde{\psi}_{0}(0,\tilde{y})=0\ \text{and}\
\tilde{\psi}_{0}(a\sin\theta-b\cos\theta,\tilde{y})=-Q\ \ \text{for
$\displaystyle-\infty<\tilde{y}<+\infty$},\\\
&-Q\leq\tilde{\psi}_{0}(\tilde{x},\tilde{y})\leq 0\ \
\text{in}~{}~{}\tilde{E}.\end{array}\right.$

Then one has

$\displaystyle\tilde{\psi}_{0}(\tilde{x},\tilde{y})=-\frac{Q}{a\sin\theta-b\cos\theta}\tilde{x}\
\ \text{in}\ \ \tilde{E},$

which implies that

$\left|\psi_{\lambda_{L},L}(x,y)-\frac{Q(y\cos\theta-x\sin\theta)}{a\sin\theta-b\cos\theta}\right|\rightarrow
0\ \ \text{uniformly in any compact subset of $\displaystyle S$,}$ (2.24)

as $\displaystyle y\rightarrow-\infty$, where $\displaystyle S=\\{(x,y)\mid
y\cot\theta<x<(y-b)\cot\theta+a,-\infty<y<+\infty\\}$.

Step 6. Finally, we will verify the condition (7) in Definition 2.1 and
complete the proof. For any $\displaystyle\varepsilon>0$, by virtue of (2.10)
and (2.23), there exists a large $\displaystyle\mu_{0}>0$, such that

$\displaystyle\psi_{\lambda_{L},L}-y\leq\varepsilon\ \ \text{in
$\displaystyle\Omega_{L}^{+}\cap\\{x\leq-\mu_{0}\\}$ and
$\displaystyle\psi_{\lambda_{L},L}\leq y\ $ in $\displaystyle\
\Omega_{L}^{+}\cap\\{x\geq\mu_{0}\\}$},$

where
$\displaystyle\Omega_{L}^{+}=\Omega_{L}\cap\\{\psi_{\lambda_{L},L}>0\\}$. This
together with the maximum principle gives that

$\displaystyle\psi_{\lambda_{L},L}-y\leq\varepsilon\ \ \text{in
$\displaystyle\Omega_{L}^{+}\cap\\{-\mu_{0}\leq x\leq\mu_{0}\\}$}.$

Therefore, we have

$\psi_{\lambda_{L},L}-y\leq\varepsilon\ \ \text{in $\displaystyle\
\Omega_{L}^{+}$}.$ (2.25)

Taking $\displaystyle\varepsilon\rightarrow 0$ in (2.25), one has

$\displaystyle\psi_{\lambda_{L},L}^{+}(x,y)\leq y$ in
$\displaystyle\Omega_{L}\cap\\{y>0\\}$.

Similarly, we can show that

$\displaystyle\psi^{+}_{\lambda_{L},L}(x,y)\geq\frac{L(y-h_{L})}{L-h_{L}}\ $
in $\displaystyle\ \Omega_{L}\cap\\{y>0\\}$.

∎

Finally, we will obtain the positivity of horizontal velocity and vertical
velocity in the following.

######  Lemma 2.5.

The horizontal velocity and vertical velocity are positive in
$\displaystyle\Omega_{L}$, namely,

$\displaystyle u>0\ \text{and}\ v>0\ \ \text{in}\ \ \
\Omega_{L}^{-}\cup\Gamma_{\lambda_{L},L},\ \ \ u>0\ \ \text{and}\ v>0\
\text{in}\ \ \ \Omega_{L}^{+}\cup\Gamma_{\lambda_{L},L},$

where $\displaystyle\Omega_{L}^{-}=\Omega_{L}\cap\\{\psi_{\lambda_{L},L}<0\\}$
and $\displaystyle\Omega_{L}^{+}=\Omega_{L}\cap\\{\psi_{\lambda_{L},L}>0\\}$.

###### Proof.

Denote $\displaystyle\omega_{1}(x,y)=\partial_{y}\psi_{\lambda_{L},L}(x,y)$ in
$\displaystyle\Omega_{L}^{-}$ and
$\displaystyle\omega_{2}(x,y)=\partial_{y}\psi_{\lambda_{L},L}(x,y)$ in
$\displaystyle\Omega_{L}^{+}$, it is easy to check that

$\displaystyle\Delta\omega_{1}=0\ \ \text{in}\ \ \Omega_{L}^{-},\ \
\text{and}\ \ \Delta\omega_{2}=0\ \ \text{in}\ \ \Omega_{L}^{+}.$

Since $\displaystyle\psi_{\lambda_{L},L}(x,y)$ is monotone increasing with
respect to $\displaystyle y$, which together with the strong maximum principle
gives that

$\displaystyle\text{$\displaystyle\omega_{1}(x,y)>0$ in
$\displaystyle\Omega_{L}^{-}$, and $\displaystyle\omega_{2}(x,y)>0$ in
$\displaystyle\Omega_{L}^{+}$}.$

Next, we claim that

$\displaystyle\omega_{1}(x,y)>0\ \ \text{and}\ \ \omega_{2}(x,y)>0\ \
\text{on}\ \ \Gamma_{\lambda_{L},L}.$

Suppose not, without loss of generality, we assume that there exists an
$\displaystyle x_{0}\in(0,+\infty)$, such that
$\displaystyle\omega_{1}(X_{0})=0$ with $\displaystyle
X_{0}=(x_{0},k_{\lambda_{L},L}(x_{0}))$. We consider the following two cases.

Case 1. $\displaystyle\lambda_{L}=0$. Then we have that
$\displaystyle\psi_{\lambda_{L},L}$ is harmonic in $\displaystyle\Omega_{L}$,
the strong maximum principle gives that $\displaystyle\omega_{1}(X_{0})>0$,
which contradicts to our assumption.

Case 2. $\displaystyle\lambda_{L}\neq 0$. Since the free boundary
$\displaystyle\Gamma_{\lambda_{L},L}$ is analytic at $\displaystyle X_{0}$,
$\displaystyle\omega_{1}(X_{0})=0$ implies that the normal vector of
$\displaystyle\Gamma_{\lambda_{L},L}$ is parallel to $\displaystyle(1,0)$,
which implies that $\displaystyle\omega_{2}(X_{0})$ is also zero. Thus, it
follows from (2.5) that
$\displaystyle|\partial_{x}\psi_{\lambda_{L},L}^{-}(X_{0})|^{2}-|\partial_{x}\psi_{\lambda_{L},L}^{+}(X_{0})|^{2}=\lambda_{L}$.

Without loss of generality, we assume that the outer normal vector of
$\displaystyle\partial\\{\psi_{\lambda_{L},L}>0\\}$ at $\displaystyle X_{0}$
is $\displaystyle\nu=(1,0)$. Thanks to Hopf’s lemma, one has

$\partial_{x}\psi_{\lambda_{L},L}^{+}=\frac{\partial\psi_{\lambda_{L},L}^{+}}{\partial\nu}<0\
\ \text{and}\ \
\partial_{x}\psi_{\lambda_{L},L}^{-}=\frac{\partial\psi_{\lambda_{L},L}^{-}}{\partial\nu}>0\
\ \text{on}\ \ X_{0},$ (2.26)

and

$\partial^{2}_{xy}\psi_{\lambda_{L},L}^{+}=\partial_{x}\omega_{1}=\frac{\partial\omega_{1}}{\partial\nu}<0\
\ \text{and}\ \
\partial^{2}_{xy}\psi_{\lambda_{L},L}^{-}=\partial_{x}\omega_{2}=\frac{\partial\omega_{2}}{\partial\nu}<0\
\ \text{on}\ \ X_{0}.$ (2.27)

Since
$\displaystyle|\nabla\psi_{\lambda_{L},L}^{-}|^{2}-|\nabla\psi_{\lambda_{L},L}^{+}|^{2}=\lambda_{L}$
on the free boundary $\displaystyle\Gamma_{\lambda_{L},L}$, one has

$0=\frac{\partial(|\nabla\psi_{\lambda_{L},L}^{-}|^{2}-|\nabla\psi_{\lambda_{L},L}^{+}|^{2})}{\partial
s}=2(\partial_{xy}\psi_{\lambda_{L},L}^{-}\partial_{x}\psi_{\lambda_{L},L}^{-}-\partial_{xy}\psi_{\lambda_{L},L}^{+}\partial_{x}\psi_{\lambda_{L},L}^{+})\
\ \text{at}\ \ X_{0},$ (2.28)

where $\displaystyle s=(0,1)$ is the tangential direction of
$\displaystyle\Gamma_{\lambda_{L},L}$ at $\displaystyle X_{0}$. On the other
hand, it follows from (2.26) and (2.27) that

$\displaystyle\partial^{2}_{xy}\psi_{\lambda_{L},L}^{-}\partial_{x}\psi_{\lambda_{L},L}^{-}-\partial^{2}_{xy}\psi_{\lambda_{L},L}^{+}\partial_{x}\psi_{\lambda_{L},L}^{+}<0\
\ \text{at}\ \ X_{0},$

which contradicts to (2.28).

Similarly, we can show that

$\displaystyle v>0\ \ \text{in}\ \ \ \Omega_{L}^{-}\cup\Gamma_{\lambda_{L},L}\
\ \text{and}\ \ v>0\ \text{in}\ \ \ \Omega_{L}^{+}\cup\Gamma_{\lambda_{L},L}.$

∎

###  2.4. The uniqueness of the truncated injection flow problem 1

In this subsection, we will obtain the uniqueness of the truncated injection
flow problem 1 for any given $\displaystyle Q>0$ and $\displaystyle L>b$.

######  Lemma 2.6.

For any $\displaystyle Q>0$ and $\displaystyle L>b$, there exist a unique
$\displaystyle\lambda_{L}$ and a unique solution
$\displaystyle(\psi_{\lambda_{L},L},\Gamma_{\lambda_{L},L})$ to the truncated
injection flow problem 1.

###### Proof.

Suppose that there exist two different solutions
$\displaystyle\psi=\psi_{\lambda_{L},L}$ and
$\displaystyle\tilde{\psi}=\tilde{\psi}_{\tilde{\lambda}_{L},L}$. We divide
two steps to complete the proof.

Step 1. First, we show that

$\displaystyle\lambda_{L}=\tilde{\lambda}_{L}.$

Suppose not, without loss of generality, we assume that
$\displaystyle\lambda_{L}<\tilde{\lambda}_{L}$. Noting that

$\displaystyle\lambda_{L}=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{L^{2}}{(L-h_{L})^{2}}\
\ \text{and}\ \
\tilde{\lambda}_{L}=\frac{Q^{2}}{(\tilde{h}_{L}-b)^{2}}-\frac{L^{2}}{(L-\tilde{h}_{L})^{2}},$

this together with Remark 2.1 implies that

$\lim_{x\rightarrow+\infty}k_{\lambda_{L},L}(x)=h_{L}>\tilde{h}_{L}=\lim_{x\rightarrow+\infty}\tilde{k}_{\tilde{\lambda}_{L},L}(x).$
(2.29)

Then,

$k_{\lambda_{L},L}(x)>\tilde{k}_{\tilde{\lambda}_{L},L}(x)\ \ \text{for
sufficiently large $\displaystyle x>0$}.$ (2.30)

Define a function $\displaystyle\psi_{\varepsilon}(x,y)=\psi(x,y-\varepsilon)$
for any $\displaystyle\varepsilon\geq 0$, and
$\displaystyle\Gamma_{\lambda_{L},L}^{\varepsilon}:y=k_{\lambda_{L},L}(x)+\varepsilon$
as the free boundary of $\displaystyle\psi_{\varepsilon}$. Take
$\displaystyle\varepsilon_{0}\geq 0$ to be the smallest one such that

$\psi_{\varepsilon_{0}}(X)\leq\tilde{\psi}(X)\ \ \text{in
$\displaystyle\Omega_{L}$ and
$\displaystyle\psi_{\varepsilon_{0}}(X_{0})=\tilde{\psi}(X_{0})$ for some
$\displaystyle X_{0}\in\bar{\Omega}_{L}$.}$ (2.31)

We consider the following two cases for $\displaystyle\varepsilon_{0}$.

Case 1. $\displaystyle\varepsilon_{0}>0$. The strong maximum principle gives
that $\displaystyle
X_{0}\notin\Omega_{L}\cap(\\{\tilde{\psi}<0\\}\cup\\{\psi_{\varepsilon_{0}}>0\\})$.
Suppose not, without loss of generality, we assume that there exists
$\displaystyle X_{0}\in\Omega_{L}\cap\\{\tilde{\psi}<0\\}$, such that
$\displaystyle-Q<\psi_{\varepsilon_{0}}(X_{0})=\tilde{\psi}(X_{0})<0$. The
continuity of $\displaystyle\psi_{\varepsilon_{0}}$ and
$\displaystyle\tilde{\psi}$ implies that there exists a small $\displaystyle
r>0$, such that

$\displaystyle-Q<\psi_{\varepsilon_{0}}(X)<0\ \ \text{and}\ \
-Q<\tilde{\psi}(X)<0\ \ \text{in $\displaystyle B_{r}(X_{0})$}.$

Since $\displaystyle\psi_{\varepsilon_{0}}$ and $\displaystyle\tilde{\psi}$
are harmonic in $\displaystyle B_{r}(X_{0})$, the strong maximum principle
gives that $\displaystyle\psi_{\varepsilon_{0}}\equiv\tilde{\psi}$ in
$\displaystyle B_{r}(X_{0})$, due to
$\displaystyle\psi_{\varepsilon_{0}}(X_{0})=\tilde{\psi}(X_{0})$. Applying the
strong maximum principle again, we can obtain a contradiction to the boundary
value of $\displaystyle\tilde{\psi}$.

Since $\displaystyle\varepsilon_{0}>0$, it follows from (2.30) that
$\displaystyle|X_{0}|<+\infty$. Therefore, choose $\displaystyle X_{0}$ to be
a free boundary point of $\displaystyle\psi_{\varepsilon_{0}}$ and
$\displaystyle\tilde{\psi}$, and one has
$\displaystyle\psi_{\varepsilon_{0}}(X_{0})=\tilde{\psi}(X_{0})=0$. In view of
(2.31), the strong maximum principle gives that

$\displaystyle\psi_{\varepsilon_{0}}<\tilde{\psi}\ \ \text{in}\ \
\Omega_{L}\cap\\{\tilde{\psi}<0\\}\ \ \text{and}\ \
\psi_{\varepsilon_{0}}<\tilde{\psi}\ \ \text{in}\ \
\Omega_{L}\cap\\{\psi_{\varepsilon_{0}}>0\\}.$

Since the free boundaries
$\displaystyle\Gamma_{\lambda_{L},L}^{\varepsilon_{0}}$ and
$\displaystyle\tilde{\Gamma}_{\tilde{\lambda}_{L},L}$ are analytic at
$\displaystyle X_{0}$, thanks to Hopf’s lemma, one has

$\displaystyle|\nabla\psi_{\varepsilon_{0}}^{-}|=-\frac{\partial\psi^{-}_{\varepsilon_{0}}}{\partial\nu}>-\frac{\partial\tilde{\psi}^{-}}{\partial\nu}=|\nabla\tilde{\psi}^{-}|\
\ \text{and}\ \
|\nabla\psi_{\varepsilon_{0}}^{+}|=\frac{\partial\psi^{+}_{\varepsilon_{0}}}{\partial\nu}<\frac{\partial\tilde{\psi}^{+}}{\partial\nu}=|\nabla\tilde{\psi}^{+}|\
\ \text{at}\ \ X_{0},$

where $\displaystyle\nu$ is the inner normal vector to
$\displaystyle\partial\\{\tilde{\psi}>0\\}$ at $\displaystyle X_{0}$. Those
give that

$\displaystyle\lambda_{L}=|\nabla\psi_{\varepsilon_{0}}^{-}|^{2}-|\nabla\psi_{\varepsilon_{0}}^{+}|^{2}>|\nabla\tilde{\psi}^{-}|^{2}-|\nabla\tilde{\psi}^{+}|^{2}=\tilde{\lambda}_{L}\
\ \text{at}\ \ X_{0},$

which contradicts to our assumption.

Case 2. $\displaystyle\varepsilon_{0}=0$. We first claim that

$\lambda_{L}\cdot\tilde{\lambda}_{L}>0.$ (2.32)

Suppose not, if $\displaystyle\tilde{\lambda}_{L}>0\geq\lambda_{L}$, by virtue
of (2.3), one has

$\displaystyle\tilde{k}_{\lambda_{L},L}^{\prime}(0+0)=\tan\theta>\tan\frac{\theta}{2}\geq
k^{\prime}_{\tilde{\lambda}_{L},L}(0+0),$

which implies that $\displaystyle
k_{\lambda_{L},L}(x)<\tilde{k}_{\tilde{\lambda}_{L},L}(x)$ for small
$\displaystyle x>0$. This leads a contradiction to the fact
$\displaystyle\psi\geq\tilde{\psi}$ in $\displaystyle\Omega$. Similarly, we
can obtain a contradiction if
$\displaystyle\tilde{\lambda}_{L}=0>\lambda_{L}$.

In view of the claim (2.32), without loss of generality, we assume that
$\displaystyle 0>\tilde{\lambda}_{L}>\lambda_{L}$ and take $\displaystyle
X_{0}=A$. Define two blow-up sequences $\displaystyle\\{\psi_{n}\\}$ and
$\displaystyle\\{\tilde{\psi}_{n}\\}$ with
$\displaystyle\psi_{n}(\tilde{X})=\frac{\psi(r_{n}\tilde{X})}{r_{n}}$ and
$\displaystyle\tilde{\psi}_{n}(\tilde{X})=\frac{\tilde{\psi}(r_{n}\tilde{X})}{r_{n}}$.
Let $\displaystyle\psi_{0}$ and $\displaystyle\tilde{\psi}_{0}$ be the blow-up
limits of $\displaystyle\psi_{n}$ and $\displaystyle\tilde{\psi}_{n}$ as
$\displaystyle r_{n}\rightarrow 0$, respectively.

It follows from the similar arguments in Step 3 in the proof of Theorem 2.4
that $\displaystyle\psi_{0}$ and $\displaystyle\tilde{\psi}_{0}$ satisfy that

$\psi_{0}(\tilde{x},\tilde{y})=\max\\{\alpha\tilde{y},0\\},\ \ \ \alpha>0\ \
\text{and}\ \ \alpha^{2}=-\lambda_{L},$ (2.33)

and

$\tilde{\psi}_{0}(\tilde{x},\tilde{y})=\max\\{\tilde{\alpha}\tilde{y},0\\},\ \
\tilde{\alpha}>0\ \ \text{and}\ \ \tilde{\alpha}^{2}=-\tilde{\lambda}_{L}.$
(2.34)

The fact (2.31) implies that

$\displaystyle\tilde{\psi}_{0}\geq\psi_{0}\ \ \text{in $\displaystyle
B_{1}(0)\cap\\{\tilde{\psi}_{0}>0\\}$},$

which together with (2.33) and (2.34) implies that

$\displaystyle\sqrt{-\tilde{\lambda}_{L}}=\frac{\partial\tilde{\psi}_{0}}{\partial\nu}\geq\frac{\partial\psi_{0}}{\partial\nu}=\sqrt{-\lambda_{L}}\
\ \text{at}\ \ 0,$

where $\displaystyle\nu=(0,1)$ is inner normal vector. This contradicts to our
assumption $\displaystyle 0>\tilde{\lambda}_{L}>\lambda_{L}$.

Step 2. In this step, we will show that $\displaystyle\psi=\tilde{\psi}$. It
follows from the asymptotic behavior of $\displaystyle\psi$ and
$\displaystyle\tilde{\psi}$ that

$\displaystyle\lim_{x\rightarrow+\infty}k_{\lambda_{L},L}(x)=\lim_{x\rightarrow+\infty}\tilde{k}_{\lambda_{L},L}(x).$

Without loss of generality, we assume that there exists $\displaystyle
x_{0}>0$, such that

$k_{\lambda_{L},L}(x_{0})<\tilde{k}_{\lambda_{L},L}(x_{0}).$ (2.35)

Consider a function
$\displaystyle\psi_{\varepsilon}(x,y)=\psi(x,y-\varepsilon)$ for
$\displaystyle\varepsilon\geq 0$, and choosing the smallest
$\displaystyle\varepsilon_{0}\geq 0$ such that

$\displaystyle\psi_{\varepsilon_{0}}(X)\leq\tilde{\psi}(X)\ \text{in
$\displaystyle\Omega$, and}\
\psi_{\varepsilon_{0}}(X_{0})=\tilde{\psi}(X_{0})\ \text{for some
$\displaystyle X_{0}\in\bar{\Omega}$}.$

It follows from (2.35) that $\displaystyle\varepsilon_{0}>0$, which implies
that $\displaystyle|X_{0}|<+\infty$. By using the similar arguments in Step 1,
we can let $\displaystyle X_{0}$ be the free boundary point of
$\displaystyle\psi_{\varepsilon_{0}}$ and $\displaystyle\tilde{\psi}$.
Applying Hopf’s lemma at $\displaystyle X_{0}$, one has

$\displaystyle\lambda_{L}=|\nabla\psi_{\varepsilon_{0}}^{-}|^{2}-|\nabla\psi_{\varepsilon_{0}}^{+}|^{2}>|\nabla\tilde{\psi}^{-}|^{2}-|\nabla\tilde{\psi}^{+}|^{2}=\lambda_{L}\
\ \text{at}\ \ X_{0},$

which is impossible.

Hence, we obtain the uniqueness of the solution to the truncated injection
flow problem 1 for any $\displaystyle L>b$. ∎

###  2.5. The relation between $\displaystyle\lambda_{L}$ and $\displaystyle
Q$

In Lemma 2.6, the uniqueness of $\displaystyle\lambda_{L}$ is obtained for any
$\displaystyle Q>0$ and $\displaystyle L>b$. Then we can denote
$\displaystyle\lambda_{L}=\lambda_{L}(Q)$ for any $\displaystyle Q>0$ and
$\displaystyle L>b$. We investigate the relation between
$\displaystyle\lambda_{L}(Q)$ and $\displaystyle Q$ for any fixed
$\displaystyle L>b$, and show that $\displaystyle\lambda_{L}(Q)$ is strictly
monotone increasing and continuous with respect to $\displaystyle Q>0$ for any
$\displaystyle L>b$.

######  Lemma 2.7.

For any $\displaystyle L>b$, $\displaystyle\lambda_{L}(Q)$ is strictly
monotone increasing and continuous with respect to $\displaystyle Q$.

###### Proof.

For any $\displaystyle Q_{1}>Q_{2}>0$, there exist a unique
$\displaystyle\lambda_{L}(Q_{1})$ and a unique
$\displaystyle\lambda_{L}(Q_{2})$, such that
$\displaystyle(\psi_{\lambda_{L}(Q_{1}),L},\Gamma_{\lambda_{L}(Q_{1}),L})$ and
$\displaystyle(\psi_{\lambda_{L}(Q_{2}),L},\Gamma_{\lambda_{L}(Q_{1}),L})$ are
the solutions to the truncated injection flow problem 1. We next show that

$\displaystyle\lambda_{L}(Q_{1})>\lambda_{L}(Q_{2})\ \ \text{for any
$\displaystyle Q_{1}>Q_{2}>0$}.$

If not, suppose that there exist $\displaystyle Q_{1}>Q_{2}>0$, such that
$\displaystyle\lambda_{L}(Q_{1})\leq\lambda_{L}(Q_{2})$. Denote
$\displaystyle\psi_{1}=\psi_{\lambda_{L}(Q_{1}),L}$ and
$\displaystyle\psi_{2}=\psi_{\lambda_{L}(Q_{2}),L}$ for simplicity. We next
consider the following two cases.

Case 1. $\displaystyle\lambda_{L}(Q_{1})=\lambda_{L}(Q_{2})=0$. Then
$\displaystyle\psi_{1}$ and $\displaystyle\psi_{2}$ are harmonic functions in
$\displaystyle\Omega_{L}$. For the harmonic function $\displaystyle\psi_{1}$
in $\displaystyle\Omega_{L}$, along the conformal mapping, it follows from (7)
in pp.292 in [18] that

$\displaystyle\left(\frac{U_{1}}{U}\right)^{1-\frac{\pi}{\theta}}\frac{d_{1}}{d}-\left(\frac{U_{2}}{U}\right)^{1-\frac{\pi}{\theta}}\frac{d_{2}}{d}=1,$

where $\displaystyle d=L-b$, $\displaystyle d_{1}=L$, $\displaystyle
d_{2}=a\sin\theta-b\cos\theta$, $\displaystyle U=\frac{L+Q_{1}}{d}$,
$\displaystyle U_{1}=\frac{L}{d_{1}}$ and $\displaystyle
U_{2}=\frac{Q_{1}}{d_{2}}$. Then we have

$d_{2}^{\frac{\pi}{\theta}}Q_{1}^{1-\frac{\pi}{\theta}}=L-(L-b)^{\frac{\pi}{\theta}}(L+Q_{1})^{1-\frac{\pi}{\theta}}.$
(2.36)

Define $\displaystyle
f(Q)=d_{2}^{\frac{\pi}{\theta}}Q^{1-\frac{\pi}{\theta}}-L+(L-b)^{\frac{\pi}{\theta}}(L+Q)^{1-\frac{\pi}{\theta}}$
for $\displaystyle Q>0$, it is easy to check that $\displaystyle f(Q)$ is
strictly monotone decreasing with respect to $\displaystyle Q>0$, which
implies that $\displaystyle Q_{1}>0$ is uniquely determined by (2.36).
Similarly, for the harmonic function $\displaystyle\psi_{2}$ in
$\displaystyle\Omega_{L}$, one has

$\displaystyle
d_{2}^{\frac{\pi}{\theta}}Q_{2}^{1-\frac{\pi}{\theta}}=L-(L-b)^{\frac{\pi}{\theta}}(L+Q_{2})^{1-\frac{\pi}{\theta}},$

which implies that $\displaystyle Q_{1}=Q_{2}$. This leads a contradiction.

Case 2. $\displaystyle\lambda_{L}(Q_{1})\neq 0$ or
$\displaystyle\lambda_{L}(Q_{2})\neq 0$.

Recalling Remark 2.1, there exists a unique $\displaystyle h_{i,L}$
($\displaystyle i=1,2$), such that

$\displaystyle\lambda_{L}(Q_{1})=\frac{Q_{1}^{2}}{(h_{1,L}-b)^{2}}-\frac{L^{2}}{(L-h_{1,L})^{2}}\
\ \text{and}\ \
\lambda_{L}(Q_{2})=\frac{Q_{2}^{2}}{(h_{2,L}-b)^{2}}-\frac{L^{2}}{(L-h_{2,L})^{2}}.$

Moreover, $\displaystyle Q_{1}>Q_{2}$ implies that

$\displaystyle h_{1,L}>h_{2,L}$

and furthermore,

$\displaystyle k_{\lambda_{L}(Q_{1}),L}(x)>k_{\lambda_{L}(Q_{2}),L}(x)\ \
\text{for sufficiently large $\displaystyle x>0$}.$

Define a function
$\displaystyle\psi_{1,\varepsilon}(x,y)=\psi_{1}(x,y-\varepsilon)$ for
$\displaystyle\varepsilon\geq 0$, and let $\displaystyle\varepsilon_{0}\geq 0$
be the smallest one such that

$\displaystyle\psi_{1,\varepsilon_{0}}(X)\leq\psi_{2}(X)\ \ \text{in
$\displaystyle\Omega_{L}$ and
$\displaystyle\psi_{1,\varepsilon_{0}}(X_{0})=\psi_{2}(X_{0})$ for some
$\displaystyle X_{0}\in\bar{\Omega}_{L}$.}$

We consider the following two subcases.

Subcase 2.1. $\displaystyle\varepsilon_{0}>0$. Similar to Case 1 in the proof
of Lemma 2.6, we can conclude that $\displaystyle
X_{0}\notin\Omega_{L}\cap(\\{\psi_{2}<0\\}\cup\\{\psi_{1,\varepsilon_{0}}>0\\})$
and $\displaystyle|X_{0}|<+\infty$. Therefore, choose $\displaystyle X_{0}$ be
a free boundary point of $\displaystyle\psi_{1,\varepsilon_{0}}$ and
$\displaystyle\psi_{2}$. Thanks to Hopf’s lemma, one has

$\displaystyle\lambda_{L}(Q_{1})=|\nabla\psi_{1,\varepsilon_{0}}^{-}|^{2}-|\nabla\psi_{1,\varepsilon_{0}}^{+}|^{2}>|\nabla\psi_{2}^{-}|^{2}-|\nabla\psi_{2}^{+}|^{2}=\lambda_{L}(Q_{2})$
at $\displaystyle X_{0}$,

which contradicts to our assumption
$\displaystyle\lambda_{L}(Q_{1})\leq\lambda_{L}(Q_{2})$.

Subcase 2.2. $\displaystyle\varepsilon_{0}=0$. Along the proof of the claim
(2.32), we have that
$\displaystyle\lambda_{L}(Q_{1})\cdot\lambda_{L}(Q_{2})>0$. Without loss of
generality, we assume that $\displaystyle
0>\lambda_{L}(Q_{2})\geq\lambda_{L}(Q_{1})$. Taking $\displaystyle X_{0}=A$,
the strong maximum principle gives that

$\displaystyle\psi_{1}<\psi_{2}\ \ \text{in}\ \ \Omega\cap\\{\psi_{1}>0\\}.$

We next show that

$k_{\lambda_{L}(Q_{1}),L}(x)>k_{\lambda_{L}(Q_{2}),L}(x)\ \ \text{for any
$\displaystyle x>0$}.$ (2.37)

If not, there exists an $\displaystyle x_{1}\in(0,+\infty)$, such that
$\displaystyle
k_{\lambda_{L}(Q_{1}),L}(x_{1})=\tilde{k}_{\lambda_{L}(Q_{2}),L}(x_{1})$.
Taking $\displaystyle X_{0}=(x_{1},k_{\lambda_{L}(Q_{1}),L})$ as the free
boundary point, we can obtain a contradiction by using the similar arguments
in Subcase 2.1,

Since $\displaystyle N_{1}$ is $\displaystyle C^{2,\alpha}$-smooth, by using
Hopf’s lemma, one has

$\frac{\partial\psi_{1}}{\partial\nu}<\frac{\partial\psi_{2}}{\partial\nu}\ \
\text{on $\displaystyle N_{1}\cap\\{x<0\\}$, $\displaystyle\nu$ is the inner
normal vector of $\displaystyle N_{1}$}.$ (2.38)

In view of (2.37) and (2.38), for small $\displaystyle r>0$, there exists a
small $\displaystyle\delta>0$, such that

$\displaystyle(1+\delta)\psi_{1}\leq\psi_{2}\ \ \text{on
$\displaystyle\partial(B_{r}(0)\cap\\{\psi_{1}>0\\})$}.$

The maximum principle gives that

$(1+\delta)\psi_{1}\leq\psi_{2}\ \ \text{in $\displaystyle
B_{r}(0)\cap\\{\psi_{1}>0\\}$}.$ (2.39)

Define two blow-up sequences $\displaystyle\\{\psi_{1,n}\\}$ and
$\displaystyle\\{\psi_{2,n}\\}$ with
$\displaystyle\psi_{1,n}(\tilde{X})=\frac{\psi_{1}(r_{n}\tilde{X})}{r_{n}}$
and
$\displaystyle\psi_{2,n}(\tilde{X})=\frac{\psi_{2}(r_{n}\tilde{X})}{r_{n}}$.
Since $\displaystyle\psi_{1,n}$ and $\displaystyle\tilde{\psi}_{1,n}$ are
Lipschitz continuous, we can denote $\displaystyle\psi_{1,0}$ and
$\displaystyle\psi_{2,0}$ as the blow-up limit of $\displaystyle\psi_{1,n}$
and $\displaystyle\psi_{2,n}$, respectively. Furthermore,

$\displaystyle\psi_{1,0}(\tilde{x},\tilde{y})=\max\\{\sqrt{-\lambda_{L}(Q_{1})}\tilde{y},0\\}\
\ \ \text{and}\ \
\psi_{2,0}(\tilde{x},\tilde{y})=\max\\{\sqrt{-\lambda_{L}(Q_{2})}\tilde{y},0\\},$

and

$\displaystyle\psi_{2,0}\geq(1+\delta)\psi_{1,0}\ \ \text{in
$\displaystyle\\{\tilde{y}>0\\}$}.$

This give that

$\displaystyle\sqrt{-\tilde{\lambda}_{L}(Q_{2})}=\frac{\partial\psi_{2,0}}{\partial\nu}\geq(1+\delta)\frac{\partial\psi_{1,0}}{\partial\nu}=(1+\delta)\sqrt{-\lambda_{L}(Q_{1})}\
\ \text{at}\ \ 0,$

where $\displaystyle\nu=(0,1)$ is the inner normal vector. This contradicts to
our assumption $\displaystyle 0>\lambda_{L}(Q_{2})\geq\lambda_{L}(Q_{1})$.

Next, we will show that $\displaystyle\lambda_{L}(Q)$ is continuous with
respect to $\displaystyle Q$. Since $\displaystyle\lambda_{L}(Q)$ is strictly
monotone increasing with respect to $\displaystyle Q$, it suffices to show
that

$\displaystyle\lambda_{L}(Q+0)=\lambda_{L}(Q-0)$ for any $\displaystyle Q>0$,

where $\displaystyle\lambda_{L}(Q+0)=\lim_{Q_{n}\rightarrow
Q^{+}}\lambda_{L}(Q_{n})$ and
$\displaystyle\lambda_{L}(Q-0)=\lim_{Q_{n}\rightarrow
Q^{-}}\lambda_{L}(Q_{n})$.

Suppose not, then there exists a $\displaystyle Q_{0}>0$, such that
$\displaystyle\lambda_{L}(Q_{0}+0)>\lambda_{L}(Q_{0}-0)$. For a sequence
$\displaystyle\\{Q_{n}\\}$ with $\displaystyle Q_{n}\downarrow Q_{0}$, there
exist a unique $\displaystyle\lambda_{L}(Q_{n})$ and a unique solution
$\displaystyle\psi_{\lambda_{L}(Q_{n}),L}$ to the truncated injection flow
problem 1. Then there exists a subsequence $\displaystyle\\{Q_{n}\\}$, such
that

$\displaystyle\lambda_{L}(Q_{n})\rightarrow\lambda_{L}(Q_{0}+0),$

and

$\displaystyle\psi_{\lambda_{L}(Q_{n}),L}\rightarrow\psi_{\lambda_{L}(Q_{0}+0),L}\
\text{in $\displaystyle H^{1}_{loc}(\Omega_{L})$ and uniformly in any compact
subset of $\displaystyle\Omega_{L}$}.$

It is easy to check that
$\displaystyle(\psi_{\lambda_{L}(Q_{0}+0),L},\Gamma_{\lambda_{L}(Q_{0}+0),L})$
is a solution to the truncated injection flow problem 1.

Similarly, there exists a solution
$\displaystyle(\psi_{\lambda_{L}(Q_{0}-0),L},\Gamma_{\lambda_{L}(Q_{0}-0),L})$
to the truncated injection flow problem 1.

For the given $\displaystyle Q_{0}>0$, the uniqueness of
$\displaystyle\lambda_{L}$ and $\displaystyle\psi_{\lambda_{L},L}$ gives that
$\displaystyle\lambda_{L}(Q_{0}+0)=\lambda_{L}(Q_{0}-0)$ and
$\displaystyle\psi_{\lambda_{L}(Q_{0}+0),L}=\psi_{\lambda_{L}(Q_{0}-0),L}$,
which leads a contradiction.

∎

Next, we will obtain the upper bound and the lower bound of
$\displaystyle\lambda_{L}(Q)$. It should be noted that the monotonicity of
$\displaystyle\lambda_{L}(Q)$ implies that the lower bound of
$\displaystyle\lambda_{L}(Q)$ follows from the limit
$\displaystyle\lim_{Q\rightarrow 0}\lambda_{L}(Q)$, which means the injection
flow vanishes. To see this, we have to investigate the one-phase flow above a
blade surface with unit velocity in upstream. In the special case
$\displaystyle b=0$ (see Figure 2), the problem is so simple and the one-phase
flow for $\displaystyle Q=0$ is nothing but the uniform flow
$\displaystyle(u,v)=\frac{1}{\sqrt{\rho_{+}}}(1,0)$ with free boundary
$\displaystyle\Gamma=\\{0<x<a,y=0\\}$. And it is clear that the limit
$\displaystyle\lambda_{L}(Q)\rightarrow-1$ as $\displaystyle Q\rightarrow 0$.
However, at the present situation ($\displaystyle b>0$), the one-phase fluid
problem is unclear and complicated, which is the one of main differences and
difficulties here. Therefore, there is an important observation that for the
limit case $\displaystyle Q=0$, the free boundary initiates at the leading
edge $\displaystyle A$ and touches the boundary $\displaystyle S_{2}$ below
the trailing edge $\displaystyle B$ (see Figure 11). Based on this important
observation, we will show that the limit $\displaystyle\lim_{Q\rightarrow
0}\lambda_{L}(Q)$ is not $\displaystyle-1$ but a constant
$\displaystyle\lambda_{L}(0)>-1$. This is also a difference from the special
case $\displaystyle b=0$.

Figure 11. The critical case $\displaystyle Q=0$

######  Lemma 2.8.

For any $\displaystyle L>b$, there exist a
$\displaystyle\lambda_{L}(0)\in(-1,0)$ and a
$\displaystyle\kappa_{L}\in(0,+\infty)$, such that

$\displaystyle\lambda_{L}(Q)\rightarrow\lambda_{L}(0)\ \ \text{as}\ \
Q\rightarrow 0,$

and

$\displaystyle\frac{\lambda_{L}(Q)}{Q^{2}}\rightarrow\kappa_{L}\ \ \text{as}\
\ Q\rightarrow+\infty.$

Furthermore, if $\displaystyle L$ is sufficiently large,
$\displaystyle\kappa_{L}$ is a uniform constant independent of $\displaystyle
L$.

###### Proof.

Step 1. The limit $\displaystyle Q\rightarrow 0$. For any sequence
$\displaystyle\\{Q_{n}\\}$ with $\displaystyle Q_{n}>0$ and $\displaystyle
Q_{n}\rightarrow 0$, there exists a subsequence $\displaystyle\\{Q_{n}\\}$,
such that

$\displaystyle\lambda_{L}(Q_{n})\rightarrow\lambda_{L}(0),$

and

$\displaystyle\psi_{\lambda_{L}(Q_{n}),L}\rightarrow\psi_{\lambda_{L}(0),L}\
\text{in $\displaystyle H^{1}_{loc}(\Omega_{L})$ and uniformly in any compact
subset of $\displaystyle\Omega_{L}$},$

as $\displaystyle Q_{n}\rightarrow 0$. Furthermore,
$\displaystyle\psi_{\lambda_{L}(0),L}(x,y)$ is monotone increasing with
respect to $\displaystyle y$ and decreasing with respect to $\displaystyle x$.
The monotonicity of $\displaystyle\psi_{\lambda_{L}(0),L}(x,y)$ implies that
there exists a monotone increasing function $\displaystyle
y=k_{\lambda_{L}(0),L}(x)$ for $\displaystyle x>0$, such that

$\displaystyle\Omega_{L}\cap\\{x>0\\}\cap\\{\psi_{\lambda_{L}(0),L}>0\\}=\Omega_{L}\cap\\{x>0\\}\cap\\{y>k_{\lambda_{L}(0),L}(x)\\},$

and

$\displaystyle k_{\lambda_{L}(0),L}(0)=0.$

We first claim that

$k_{\lambda_{L}(0),L}(x)\equiv b\ \ \text{for any $\displaystyle
x\in(a,+\infty)$}.$ (2.40)

Suppose not, there exists an $\displaystyle x_{0}\in(a,+\infty)$, such that
$\displaystyle k_{\lambda_{L}(0),L}(x_{0})=b$ and $\displaystyle
k_{\lambda_{L}(0),L}(x)>b$ for any $\displaystyle x\in(x_{0},+\infty)$. By
using the asymptotic behavior of $\displaystyle\psi_{\lambda_{L}(0),L}$ in the
downstream, one has

$\lim_{x\rightarrow+\infty}k_{\lambda_{L}(0),L}(x)=h_{0}\in(b,L)\ \
\text{and}\ \ -\lambda_{L}(0)=\frac{L^{2}}{(L-h_{0})^{2}}.$ (2.41)

It follows from the results in Section 9 in [3] and Section 11 in Chapter 3 in
[12] that the continuous fit condition implies the smooth fit condition,
namely, $\displaystyle N_{1}\cup\Gamma_{\lambda_{L}(0),L}$ is $\displaystyle
C^{1}$-smooth at $\displaystyle A$ and $\displaystyle
k^{\prime}_{\lambda_{L}(0),L}(0+0)=0$. Furthermore,
$\displaystyle\nabla\psi_{\lambda_{L}(0),L}$ is uniformly continuous in a
$\displaystyle\\{\psi_{\lambda_{L}(0),L}>0\\}$-neighborhood of $\displaystyle
A$.

Define $\displaystyle\omega(y)=\max\\{y,0\\}$ for $\displaystyle
y\in(-\infty,L)$, it is easy to check that
$\displaystyle\omega(y)\geq\psi_{\lambda_{L}(0),L}(x,y)$ in
$\displaystyle\Omega_{L}$. In view of
$\displaystyle\omega(0)=\psi_{\lambda_{L}(0),L}(0,0)=0$, one has

$\displaystyle
1=\frac{\partial\omega}{\partial\nu}\geq\frac{\psi_{\lambda_{L}(0),L}}{\partial\nu}=\sqrt{-\lambda_{L}(0)}\
\ \text{at}\ \ A,$

where $\displaystyle\nu=(0,1)$ is the inner normal vector. This contradicts to
the fact that $\displaystyle\lambda_{L}(0)=-\frac{L^{2}}{(L-h_{0})^{2}}<-1$ in
(2.41).

Next, we will show that

$\lambda_{L}(0)>-1.$ (2.42)

If not, we assume that $\displaystyle\lambda_{L}(0)\leq-1$. For any small
$\displaystyle r>0$, it follows from the proof of (2.39) that there exists a
small $\displaystyle\delta>0$, such that

$\displaystyle\omega\geq(1+\delta)\psi_{\lambda_{L}(0),L}\ \ \text{in}\ \
B_{r}(A)\cap\\{\psi_{\lambda_{L}(0),L}>0\\},$

which gives that

$\displaystyle
1=\frac{\partial\omega}{\partial\nu}\geq(1+\delta)\frac{\psi_{\lambda_{L}(0),L}}{\partial\nu}=(1+\delta)\sqrt{-\lambda_{L}(0)}\
\ \text{at}\ \ A.$

This contradicts to our assumption $\displaystyle\lambda_{L}(0)\leq-1$.

Moreover, we will show that

$\bar{\Gamma}_{\lambda_{L}(0),L}\cap N_{2}=\varnothing.$ (2.43)

If not, it follows from (2.40) that $\displaystyle k_{\lambda_{L}(0),L}(a)=b$.
Similarly, we have that $\displaystyle N_{2}\cup\Gamma_{\lambda_{L}(0),L}$ is
$\displaystyle C^{1}$-smooth at $\displaystyle B$ and
$\displaystyle\nabla\psi_{\lambda_{L}(0),L}$ is uniformly continuous in a
$\displaystyle\\{\psi_{\lambda_{L}(0),L}>0\\}$-neighborhood of $\displaystyle
B$. Define $\displaystyle\omega_{1}(y)=\frac{L}{L-b}\max\\{y-b,0\\}$, it is
easy to check that
$\displaystyle\omega_{1}(y)\leq\psi_{\lambda_{L}(0),L}(x,y)$ in
$\displaystyle\Omega_{L}$, and thus

$\displaystyle
1<\frac{L}{L-b}=\frac{\partial\omega_{1}}{\partial\nu}\leq\frac{\psi_{\lambda_{L}(0),L}}{\partial\nu}=\sqrt{-\lambda_{L}(0)}\
\ \text{at}\ \ B,$

where $\displaystyle\nu=(0,1)$ is the inner normal vector. This contradicts to
(2.42).

Since $\displaystyle\lambda_{L}(Q)$ is strictly decreasing with respect to
$\displaystyle Q$, we can obtain the uniqueness of
$\displaystyle\lambda_{L}(0)$.

Step 2. The limit $\displaystyle Q\rightarrow+\infty$. We will show that there
exists a positive constant $\displaystyle\kappa_{L}$, such that

$\displaystyle\frac{\lambda_{L}(Q)}{Q^{2}}\rightarrow\kappa_{L}\ \ \ \
\text{as}\ \ Q\rightarrow+\infty.$

For any fixed $\displaystyle L>b$, set
$\displaystyle\psi_{Q}=\frac{\psi_{\lambda_{L}(Q),L}}{Q}$ and
$\displaystyle\lambda_{Q}=\frac{\lambda_{L}(Q)}{Q^{2}}$. Then
$\displaystyle\psi_{Q}$ solves the following free boundary value problem

$\displaystyle\left\\{\begin{array}[]{ll}&\Delta\psi_{Q}=0\
\text{in}~{}~{}\Omega_{L}\cap\\{\psi_{Q}<0\\},\ \ \Delta\psi_{Q}=0\
\text{in}~{}~{}\Omega_{L}\cap\\{\psi_{Q}>0\\},\\\
&|\nabla\psi_{Q}^{-}|^{2}-|\nabla\psi_{Q}^{+}|^{2}=\lambda_{Q}\ \
\text{on}~{}~{}\Gamma_{\lambda_{L}(Q),L},\\\ &\psi_{Q}=0\ \text{on}\ N_{1}\cup
S_{1}\cup\Gamma_{\lambda_{L}(Q),L},\ \psi_{Q}=-1\ \text{on}\ N_{2}\cup S_{2},\
\ \psi_{Q}=\frac{L}{Q}\ \text{on}\ N_{L}.\end{array}\right.$

By virtue of non-degeneracy Theorem 3.1 in [4], we have that if
$\displaystyle\lambda_{L}(Q)>0$, then

$\frac{1}{r}\fint_{\partial B_{r}(X_{0})}\psi^{-}_{\lambda_{L}(Q),L}dS\leq
c\sqrt{\lambda_{L}(Q)}\ \ \text{implies
$\displaystyle\psi_{\lambda_{L}(Q),L}\equiv 0$ in $\displaystyle
B_{\frac{r}{2}}(X_{0})$},$ (2.44)

and if $\displaystyle\lambda_{L}(Q)<0$, then

$\frac{1}{r}\fint_{\partial B_{r}(X_{0})}\psi^{+}_{\lambda_{L}(Q),L}dS\leq
c\sqrt{-\lambda_{L}(Q)}\ \ \text{implies
$\displaystyle\psi_{\lambda_{L}(Q),L}\equiv 0$ in $\displaystyle
B_{\frac{r}{2}}(X_{0})$},$ (2.45)

for any disc $\displaystyle B_{r}(X_{0})\subset\Omega_{L}$ with $\displaystyle
B_{\frac{r}{2}}(X_{0})\subset\Omega_{L}\cap\\{(x,y)\mid x>0,y>0\\}$. Here,
$\displaystyle c>0$ is a constant independent of $\displaystyle\lambda_{L}(Q)$
and $\displaystyle L$. Therefore, there exists a constant $\displaystyle
r_{0}>0$ independent of $\displaystyle\lambda_{L}(Q)$ and $\displaystyle L$,
such that $\displaystyle B_{r_{0}}(X_{0})\subset\Omega_{L}$ with
$\displaystyle B_{\frac{r_{0}}{2}}(X_{0})\subset\Omega_{L}\cap\\{(x,y)\mid
x>0,y>0\\}$ and $\displaystyle
B_{\frac{r_{0}}{2}}(X_{0})\cap\Gamma_{\lambda_{L}(Q),L}\neq\varnothing$, and
it follows from (2.44) and (2.45) that

$\displaystyle\frac{Q}{r_{0}}\geq\frac{1}{r_{0}}\left|\fint_{\partial
B_{r_{0}}(X_{0})}\psi_{\lambda_{L}(Q),L}dS\right|\geq
c|\lambda_{L}(Q)|^{\frac{1}{2}},$

for any $\displaystyle Q>0$. This implies that

$|\lambda_{Q}|\leq C,\ \ \text{$\displaystyle C>0$ is a constant independent
of $\displaystyle Q$ and $\displaystyle L$}.$ (2.46)

For any sequence $\displaystyle\\{Q_{n}\\}$ with $\displaystyle
Q_{n}\rightarrow+\infty$, there exists a subsequence
$\displaystyle\\{Q_{n}\\}$, such that

$\displaystyle\lambda_{Q_{n}}\rightarrow\kappa_{L},$

and

$\displaystyle\psi_{Q_{n}}\rightarrow\bar{\psi}_{\kappa_{L}}\ \text{in
$\displaystyle H^{1}_{loc}(\Omega_{L})$ and uniformly in any compact subset of
$\displaystyle\Omega_{L}$},$

as $\displaystyle Q_{n}\rightarrow+\infty$. The monotonicity of
$\displaystyle\psi_{\lambda_{L}(Q),L}(x,y)$ with respect to $\displaystyle x$
and $\displaystyle y$ gives that $\displaystyle\bar{\psi}_{\kappa_{L}}(x,y)$
is monotone increasing with respect to $\displaystyle y$ and decreasing with
respect to $\displaystyle x$.

Since $\displaystyle 0\leq\psi_{Q}\leq\frac{L}{Q}$ in
$\displaystyle\Omega_{L}\cap\\{(x,y)\mid x\leq 0,y\geq 0\\}$, we have that
$\displaystyle\bar{\psi}_{\kappa_{L}}=0$ in
$\displaystyle\Omega_{L}\cap\\{(x,y)\mid x\leq 0,y\geq 0\\}$. Denote
$\displaystyle E_{L}=\Omega_{L}\setminus(\\{(x,y)\mid x\leq 0,y\geq 0\\})$,
then $\displaystyle\bar{\psi}_{\kappa_{L}}$ is a solution of the following
free boundary value problem

$\left\\{\begin{array}[]{ll}&\Delta\bar{\psi}_{\kappa_{L}}=0\
\text{in}~{}~{}E_{L}\cap\\{\bar{\psi}_{\kappa_{L}}<0\\},\\\
&\bar{\psi}_{\kappa_{L}}=0,\ \
\left|\frac{\partial\bar{\psi}_{\kappa_{L}}}{\partial\nu}\right|^{2}=\kappa_{L}\
\ \text{on $\displaystyle\Gamma_{\kappa_{L}}$},\\\ &\bar{\psi}_{\kappa_{L}}=0\
\text{on}\ N_{L}^{+}\cup I_{L}\cup S_{1},\ \bar{\psi}_{\kappa_{L}}=-1\
\text{on}\ N_{2}\cup S_{2},\end{array}\right.$ (2.47)

where $\displaystyle N_{L}^{+}=N_{L}\cap\\{x\geq 0\\}$, $\displaystyle
I_{L}=\\{(0,y)\mid 0\leq y\leq L\\}$ and
$\displaystyle\Gamma_{\kappa_{L}}=E_{L}\cap\partial\\{\bar{\psi}_{\kappa_{L}}<0\\}$
is the free boundary of $\displaystyle\bar{\psi}_{\kappa_{L}}$. Furthermore,
the free boundary $\displaystyle\Gamma_{\kappa_{L}}$ is $\displaystyle
C^{1}$-smooth at the initial point $\displaystyle A$, and which is given by

$\displaystyle\Gamma_{\kappa_{L}}=\\{(x,y)\mid
x=g_{\kappa_{L}}(y),0<y<h_{L}\\},\ \ \text{$\displaystyle g_{\kappa_{L}}(y)$
is increasing with respect to $\displaystyle y$,}$

where either $\displaystyle h_{L}<L$, $\displaystyle
g_{\kappa_{L}}(h_{L}-0)=+\infty$ or $\displaystyle h_{L}=L$, $\displaystyle
g_{\kappa_{L}}(h_{L}-0)\leq+\infty$.

We first show that

$\kappa_{L}>0.$ (2.48)

Suppose that $\displaystyle\kappa_{L}=0$. By virtue of (4.6) in [5], for any
free boundary point $\displaystyle X_{0}$ and $\displaystyle\varepsilon>0$, we
have

$\frac{1}{r}\left|\fint_{\partial B_{r}(X_{0})}\psi_{Q_{n}}dS\right|\leq
C|\lambda_{Q_{n}}|^{\frac{1}{2}},$ (2.49)

if $\displaystyle
B_{r}(X_{0})\subset\Omega_{L}\cap\\{\varepsilon<y<L-\varepsilon\\}$ and
$\displaystyle n$ is sufficiently large, where $\displaystyle C>0$ is a
constant depending only on $\displaystyle\varepsilon$. Taking $\displaystyle
Q_{n}\rightarrow+\infty$ in (2.49) yields that

$\displaystyle\frac{1}{r}\left|\fint_{\partial
B_{r}(X_{0})}\bar{\psi}_{\kappa_{L}}dS\right|\leq
C|\kappa_{L}|^{\frac{1}{2}}=0,$

which together with $\displaystyle\bar{\psi}_{\kappa_{L}}\equiv 0$ in
$\displaystyle\Omega_{L}\cap\\{x<0,y>0\\}$ implies that

$\displaystyle\bar{\psi}_{\kappa_{L}}\equiv 0\ \ \text{in}\ \ \\{(x,y)\mid
0<x<\varepsilon,\varepsilon<y<L-\varepsilon\\}.$

By using the unique continuation, we can conclude that
$\displaystyle\psi_{\kappa_{L}}\equiv 0$ in $\displaystyle E_{L}$, which
contradicts to the fact $\displaystyle\bar{\psi}_{\kappa_{L}}=-1$ on
$\displaystyle N_{2}$.

Finally, we will investigate the relation between $\displaystyle\kappa_{L}$
and $\displaystyle h_{\kappa_{L}}$, where $\displaystyle
h_{\kappa_{L}}\in(b,L]$ is the asymptotic height of the free boundary
$\displaystyle\Gamma_{\kappa_{L}}$. Consider the following two cases.

Case 1. $\displaystyle h_{\kappa_{L}}<L$ and $\displaystyle
g_{\kappa_{L}}(h_{\kappa_{L}}-0)=+\infty$. (See Figure 12)

Figure 12. Case 1

Similar to Step 2 in the proof of Theorem 2.4 , we can obtain that

$\kappa_{L}=\frac{1}{(h_{\kappa_{L}}-b)^{2}}.$ (2.50)

Case 2. $\displaystyle h_{\kappa_{L}}=L$ and $\displaystyle
g_{\kappa_{L}}(L-0)\in(0,+\infty)$. (See Figure 13).

Figure 13. Case 2

Denote $\displaystyle X_{0}=(g_{\kappa_{L}}(L-0),L)$. Similarly, we have that
$\displaystyle N_{L}\cup\Gamma_{\kappa_{L}}$ is $\displaystyle C^{1}$-smooth
at $\displaystyle X_{0}$ and its tangent is in the direction of positive
$\displaystyle x$-axis. Moreover, $\displaystyle\nabla\bar{\psi}_{\kappa_{L}}$
is uniformly continuous in a
$\displaystyle\\{\bar{\psi}_{\kappa_{L}}<0\\}$-neighborhood of $\displaystyle
X_{0}$. Define $\displaystyle\omega(y)=\frac{1}{L-b}\max\\{y-b,0\\}-1$, it is
easy to check that

$\displaystyle\bar{\psi}_{\kappa_{L}}(x,y)\geq\omega(y)\ \ \text{in}\ \
E_{L},$

which implies that

$\displaystyle\sqrt{\kappa_{L}}=\frac{\partial\bar{\psi}_{\kappa_{L}}}{\partial\nu}\leq\frac{\partial\omega}{\partial\nu}=\frac{1}{L-b}\
\ \text{at}\ \ X_{0},$

where $\displaystyle\nu=(0,1)$ is the outer normal vector. This implies that

$\kappa_{L}\leq\frac{1}{(L-b)^{2}}.$ (2.51)

By using the similar arguments in the proof of Lemma 2.6, we can obtain the
uniqueness of $\displaystyle\kappa_{L}$ and
$\displaystyle\bar{\psi}_{\kappa_{L}}$ to the free boundary problem (2.47).
Hence, one has

$\displaystyle\frac{\lambda_{L}(Q)}{Q^{2}}\rightarrow\kappa_{L},$

and

$\displaystyle\frac{\psi_{\lambda_{L}(Q),L}}{Q}\rightarrow\bar{\psi}_{\kappa_{L}}\
\text{uniformly in $\displaystyle E_{L}$},$

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

Step 3. Finally, we will show that $\displaystyle\kappa_{L}$ is a uniform
constant for any large $\displaystyle L$, namely, there exists a
$\displaystyle L_{0}$, such that $\displaystyle\kappa_{L_{1}}=\kappa_{L_{2}}$
for any $\displaystyle L_{2}>L_{1}>L_{0}$. It follows from (2.46) that there
exists a positive constant $\displaystyle C_{2}$ independent of $\displaystyle
L$, such that

$\kappa_{L}\leq C_{2}.$ (2.52)

By using the bounded gradient lemma 5.1 in Chapter 3 in [12], one has

$|\nabla\bar{\psi}_{\kappa_{L}}|\leq C\sqrt{\kappa_{L}}\ \ \ \text{in}\ \
D\subset\Omega_{L},$ (2.53)

where $\displaystyle D\cap\Gamma_{\kappa_{L}}\neq\varnothing$ and the constant
$\displaystyle C$ depends only on $\displaystyle D$. Denote $\displaystyle
D=\Omega_{L}\cap B_{2a}(0)$, it is easy to check that $\displaystyle
D\cap\Gamma_{\kappa_{L}}\neq\varnothing$. Then there exist two points
$\displaystyle X_{1}\in\bar{D}\cap S_{2}$ and $\displaystyle X_{2}\in
D\cap\Gamma_{\kappa_{L}}$, such that $\displaystyle X_{t}=tX_{1}+(1-t)X_{2}\in
D$ for any $\displaystyle t\in(0,1)$. It follows from (2.53) that

$\displaystyle
1=\bar{\psi}_{\kappa_{L}}(X_{2})-\bar{\psi}_{\kappa_{L}}(X_{1})\leq|\nabla\bar{\psi}_{\kappa_{L}}(X_{t_{0}})||X_{1}-X_{2}|\leq
C\sqrt{\kappa_{L}},$

for $\displaystyle t_{0}\in(0,1)$, where $\displaystyle C$ is a constant
independent of $\displaystyle L$. This implies that there exists a positive
constant $\displaystyle C_{1}$ independent of $\displaystyle L$, such that

$\kappa_{L}\geq C_{1}>0.$ (2.54)

It follows from (2.50)- (2.54) that

$h_{\kappa_{L}}\leq b+\frac{1}{\sqrt{\kappa_{L}}}\leq L_{0},$ (2.55)

where $\displaystyle L_{0}$ is a constant independent of $\displaystyle L$.

Suppose that there exist two solutions
$\displaystyle(\bar{\psi}_{\kappa_{L_{1}}},\kappa_{L_{1}})$ and
$\displaystyle(\bar{\psi}_{\kappa_{L_{2}}},\kappa_{L_{2}})$ to the free
boundary problem (2.47), with $\displaystyle L_{2}>L_{1}>L_{0}$.

By virtue of (2.55), we have

$\displaystyle\text{the free boundary of
$\displaystyle\bar{\psi}_{\kappa_{L_{1}}}$ lies below
$\displaystyle\\{y=L_{1}\\}$},$

and

$\displaystyle\text{the free boundary of
$\displaystyle\bar{\psi}_{\kappa_{L_{2}}}$ lies below
$\displaystyle\\{y=L_{1}\\}$}.$

Applying the similar arguments in the proof of Lemma 2.6, we can obtain that
$\displaystyle\bar{\psi}_{\kappa_{L_{1}}}=\bar{\psi}_{\kappa_{L_{2}}}$ and
$\displaystyle\kappa_{L_{1}}=\kappa_{L_{2}}$.

∎

######  Remark 2.2.

By virtue of Lemma 2.8, there exists a constant
$\displaystyle\kappa\in(0,+\infty)$, such that

$\displaystyle\frac{\lambda_{L}(Q)}{Q^{2}}\rightarrow\kappa\ \ \text{and}\ \
\frac{\psi_{\lambda_{L}(Q),L}}{Q}\rightarrow\bar{\psi}_{\kappa}\
\text{uniformly in $\displaystyle E_{L}$},$

as $\displaystyle Q\rightarrow+\infty$, for any $\displaystyle L>L_{0}$, where
$\displaystyle E_{L}$ is defined as in (2.47).

Next, we will give the uniform estimate of the asymptotic height
$\displaystyle h_{L}$ of the free boundary.

######  Lemma 2.9.

For any $\displaystyle Q>0$, there exists a positive constant $\displaystyle
C$ independent of $\displaystyle L$, such that

$\displaystyle h_{L}\leq C,$

where $\displaystyle h_{L}$ is the asymptotic height of the free boundary of
$\displaystyle\psi_{\lambda_{L}(Q),L}$.

###### Proof.

Suppose not, we assume that there exists a sequence $\displaystyle\\{L_{n}\\}$
with $\displaystyle L_{n}\rightarrow+\infty$, such that $\displaystyle
h_{L_{n}}\rightarrow+\infty$. Note that

$\lambda_{L_{n}}(Q)=\frac{Q^{2}}{(h_{L_{n}}-b)^{2}}-\frac{L_{n}^{2}}{(L_{n}-h_{L_{n}})^{2}}.$
(2.56)

Denote $\displaystyle(\psi_{\lambda_{L_{n}}(Q),L_{n}},\lambda_{L_{n}}(Q))$ as
the corresponding solution to the truncated injection flow problem 1 for any
$\displaystyle Q>0$. By virtue of (2.46), there exists a subsequence
$\displaystyle\\{L_{n}\\}$, such that

$\displaystyle\lambda_{L_{n}}(Q)\rightarrow\lambda,$

and

$\displaystyle\psi_{\lambda_{L_{n}}(Q),L_{n}}\rightarrow\psi_{\lambda}\
\text{in $\displaystyle H^{1}_{loc}(\Omega)$ and uniformly in any compact
subset of $\displaystyle\Omega$},$

as $\displaystyle L_{n}\rightarrow+\infty$. Moreover,
$\displaystyle\psi_{\lambda}(x,y)$ is monotone increasing with respect to
$\displaystyle y$ and decreasing with respect to $\displaystyle x$, which
implies that the free boundary of $\displaystyle\psi_{\lambda}$ can be denoted
as

$\displaystyle\Gamma_{\lambda}=\Omega\cap\\{x>0\\}\cap\\{\psi_{\lambda}=0\\}:y=k_{\lambda}(x)\
\ \text{for any $\displaystyle x>0$}.$

Here, $\displaystyle k_{\lambda}(x)$ is continuous and strictly monotone
increasing with respect to $\displaystyle x$, $\displaystyle k_{\lambda}(0)=0$
and $\displaystyle k_{\lambda}(x)\rightarrow+\infty$ as $\displaystyle
x\rightarrow+\infty$. It follows from (2.56) that

$\lambda\leq-1.$ (2.57)

Furthermore, the free boundary $\displaystyle\Gamma_{\lambda}$ is continuous
differentiable at $\displaystyle A$, namely, $\displaystyle
k_{\lambda}^{\prime}(0+0)=0$.

In view of the condition (7) in Definition 2.1, one has

$\displaystyle\psi_{\lambda_{L_{n}}(Q),L_{n}}(x,y)\leq\max\\{y,0\\}\ \
\text{in}\ \ \Omega_{L_{n}}.$

The strong maximum principle gives that

$\displaystyle\psi_{\lambda}(x,y)<y\ \text{in}\ \
\Omega\cap\\{\psi_{\lambda}>0\\}.$

Therefore, for small $\displaystyle r>0$, there exists a small
$\displaystyle\delta>0$, such that

$\displaystyle\max\\{y,0\\}\geq(1+\delta)\psi_{\lambda}\ \ \text{on}\ \
\partial(B_{r}(0)\cap\\{\psi_{\lambda}>0\\}).$

It follows from the maximum principle that

$\max\\{y,0\\}\geq(1+\delta)\psi_{\lambda}\ \ \text{in}\ \
B_{r}(0)\cap\\{\psi_{\lambda}>0\\}.$ (2.58)

Define a blow-up sequence
$\displaystyle\tilde{\psi}_{n}(\tilde{X})=\frac{\psi_{\lambda}(r_{n}\tilde{X})}{r_{n}}$
with $\displaystyle r_{n}\rightarrow 0$, it follows from (2.58) that

$\max\\{\tilde{y},0\\}\geq(1+\delta)\tilde{\psi}_{n}(\tilde{X})\ \ \text{in}\
\ B_{1}(0)\cap\\{\psi_{n}>0\\}.$ (2.59)

Denote $\displaystyle\tilde{\psi}_{0}$ as the blow-up limit of
$\displaystyle\tilde{\psi}_{n}$, it follows from (2.59) and the similar
arguments in the proof of Theorem 2.4 that

$\displaystyle\tilde{\psi}_{0}(\tilde{X})=\max\\{\sqrt{-\lambda}\tilde{y},0\\}\
\ \text{and}\ \
\max\\{\tilde{y},0\\}\geq(1+\delta)\tilde{\psi}_{0}(\tilde{X})\ \ \text{in}\ \
B_{1}(0)\cap\\{\psi_{0}>0\\}.$

This gives that

$\displaystyle
1\geq(1+\delta)\frac{\partial\tilde{\psi}_{0}}{\partial\nu}=(1+\delta)\sqrt{-\lambda}\
\ \text{at}\ \ 0,$

where $\displaystyle\nu=(0,1)$ is the inner normal vector. It leads a
contradiction with (2.57).

∎

##  3\. The proof of The main results

Based on the results in previous sections, we will complete the proof of
Theorem 1.1 \- Theorem 1.3 in this section.

######  Theorem 3.1.

For any $\displaystyle Q>0$, there exist a unique $\displaystyle\lambda>-1$
and a unique solution $\displaystyle(\psi_{\lambda},\Gamma_{\lambda})$ to the
injection flow problem 1.

###### Proof.

Step 1. It follows from (2.46) that there exists a positive constant
$\displaystyle C$ independent of $\displaystyle Q$ and $\displaystyle L$, such
that

$\displaystyle|\lambda_{L}|\leq CQ^{2}\ \ \text{for any $\displaystyle L>b$
and $\displaystyle Q>0$}.$

By virtue of Lemma 2.9, one has

$\displaystyle h_{L}\leq C\ \ \text{and}\ \ \
\lambda_{L}=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{L^{2}}{(L-h_{L})^{2}}=\frac{Q^{2}}{(h_{L}-b)^{2}}-\frac{1}{\left(1-\frac{h_{L}}{L}\right)^{2}}.$

Then there exist a sequence $\displaystyle\\{L_{n}\\}$, a constant
$\displaystyle\lambda$ and a $\displaystyle h>b$, such that

$\displaystyle\lambda_{L_{n}}\rightarrow\lambda,\ \ h_{L_{n}}\rightarrow h,$

and

$\displaystyle\psi_{\lambda_{L_{n}},L_{n}}\rightarrow\psi_{\lambda}\ \text{in
$\displaystyle H^{1}_{loc}(\Omega)$ and uniformly in any compact subset of
$\displaystyle\Omega$},$

as $\displaystyle L_{n}\rightarrow+\infty$. Obviously,
$\displaystyle\lambda=\frac{Q^{2}}{(h-b)^{2}}-1$. By using the similar
arguments in Lemma 6.2 in [4], we can show that $\displaystyle\psi_{\lambda}$
is a local minimizer to the variational problem $\displaystyle(P_{\lambda})$,
namely,

$\displaystyle P_{\lambda}:\ \ J_{D}(\psi_{\lambda})=\min J_{D}(\psi)\ \
\text{for any $\displaystyle\psi\in K$ and $\displaystyle\psi=\psi_{\lambda}$
on $\displaystyle\partial D$},$

where

$\displaystyle
J_{D}(\psi)=\int_{D}\left|\nabla\psi-(\lambda_{1}I_{\\{\psi<0\\}}+\lambda_{2}I_{\\{\psi>0\\}}+\lambda_{0}I_{\\{\psi=0\\}})I_{\\{x>0\\}}e\right|^{2}dxdy$

for any bounded domain $\displaystyle D\subset\Omega$, where
$\displaystyle\lambda_{1}=\frac{Q}{h-b}$ and $\displaystyle\lambda_{2}=1$.

Step 2. Since $\displaystyle\psi_{\lambda}$ is a local minimizer, we can
conclude that $\displaystyle\psi_{\lambda}$ is a harmonic in
$\displaystyle\Omega\setminus\Gamma$. Moreover, the free boundary
$\displaystyle\Gamma_{\lambda}:y=k_{\lambda}(x)$ of
$\displaystyle\psi_{\lambda}$ satisfies the continuous fit condition
$\displaystyle k_{\lambda}(0)=0$ and the smooth fit condition (1.5), where
$\displaystyle k_{\lambda}(x)$ is continuous and strictly monotone increasing
with respect to $\displaystyle x$, and $\displaystyle
k_{\lambda}(x)\rightarrow h$ as $\displaystyle x\rightarrow+\infty$. It
follows from the condition (7) in Definition 2.1 that

$y-h\leq\psi_{\lambda}^{+}(x,y)\leq y\ \ \text{in
$\displaystyle\Omega\cap\\{y>0\\}$}.$ (3.1)

Hence, the conditions (1)-(5) and (7) in Definition 1.1 hold.

Step 3. In this step, we will verify the condition (6) in Definition 1.1.
Denote $\displaystyle\phi(x,y)=\psi_{\lambda}(x,y)-y$ and
$\displaystyle\phi_{n}(x,y)=\phi(x-n,y)$, it follows from (3.1) that

$\displaystyle\Delta\phi_{n}=0\ \ \text{and}\ \ -h\leq\phi_{n}\leq 0\ \
\text{in}\ \ \\{x<n,y>0\\}.$

By using the elliptic estimate, there exists a subsequence
$\displaystyle\\{\phi_{n}\\}$, such that

$\displaystyle\phi_{n}\rightarrow\phi_{0}\ \ \text{in}\ \
\\{-\infty<x<+\infty,0<y<+\infty\\},$

and $\displaystyle\phi_{0}$ satisfies

$\displaystyle\Delta\phi_{0}=0\ \ \text{and}\ \ -h\leq\phi_{0}\leq 0\ \
\text{in}\ \ \\{-\infty<x<+\infty,0<y<+\infty\\},\ \ \text{and}\ \
\phi_{0}(x,0)=0.$

Then $\displaystyle\phi_{0}\equiv 0$ in
$\displaystyle\\{-\infty<x<+\infty,0<y<+\infty\\}$, and thus

$\psi_{\lambda}(x,y)\rightarrow y\ \ \text{uniformly in any compact subset of
$\displaystyle(0,+\infty)$, as $\displaystyle x\rightarrow-\infty$},$ (3.2)

Along the similar arguments in the proof of (2.24), one has

$\left|\psi_{\lambda}(x,y)-\frac{Q(y\cos\theta-x\sin\theta)}{a\sin\theta-b\cos\theta}\right|\rightarrow
0\ \ \text{uniformly in any compact subset of $\displaystyle S$,}$ (3.3)

as $\displaystyle y\rightarrow-\infty$, where $\displaystyle S=\\{(x,y)\mid
y\cot\theta<x<(y-b)\cot\theta+a,-\infty<y<+\infty\\}$.

Next, we consider the asymptotic behavior of $\displaystyle\psi_{\lambda}$ in
the downstream. For any blow-up sequence
$\displaystyle\psi_{n}(x,y)=\psi_{\lambda_{L},L}(x+n,y)$ for $\displaystyle
x>-\frac{n}{2}$, such that

$\displaystyle\psi_{n}(x,y)\rightarrow\psi_{0}(x,y)\ \ \text{uniformly in any
compact subset of $\displaystyle(0,+\infty)$, as $\displaystyle
x\rightarrow+\infty$},$

and $\displaystyle\psi_{0}$ satisfies

$\displaystyle\left\\{\begin{array}[]{ll}&\Delta\psi_{0}=0\
\text{in}~{}~{}\mathbb{R}_{b}^{2}\setminus\\{y=h\\},\\\ &\psi_{0}(x,b)=-Q\
\text{and}\ \psi_{0}(x,h)=0\ \ \text{for $\displaystyle-\infty<x<+\infty$},\\\
&0\leq\psi_{0}(x,y)\leq h\ \ \text{in}\ \
\\{-\infty<x<+\infty\\}\times\\{0<y<h\\},\end{array}\right.$

where $\displaystyle\mathbb{R}^{2}_{b}=\\{(x,y)\mid-\infty<x<+\infty,y>b\\}$.

Then one has that $\displaystyle\\{y=h\\}$ is the free boundary of
$\displaystyle\psi_{0}$ and $\displaystyle\psi_{0}=\frac{Q(y-b)}{h-b}-Q$ in
$\displaystyle\\{-\infty<x<+\infty\\}\times\\{0<y<h\\}$. In view of the
condition (3) in Definition 1.1, we can conclude that
$\displaystyle\frac{\partial\psi_{0}(x,h+0)}{\partial y}=1$, and thus
$\displaystyle\psi_{0}(x,y)=y-h$ in
$\displaystyle\\{-\infty<x<+\infty\\}\times\\{y>h\\}$. Therefore, the boundary
value problem above possesses a unique solution

$\displaystyle\psi_{0}(x,y)=\left\\{\begin{array}[]{ll}\frac{Q(h-y)}{h-b},&\text{if
$\displaystyle b<y<h$},\\\ y-h,&\text{if $\displaystyle
h<y<+\infty$}.\end{array}\right.$

Finally, we will verify the convergence of $\displaystyle\nabla\psi_{\lambda}$
in the far field. For any sequence $\displaystyle
X_{n}=(x_{n},y_{n})\in\Omega\cap\\{\psi_{\lambda}>0\\}$ with
$\displaystyle\rho_{n}=|X_{n}|\rightarrow+\infty$, we next consider the
following two cases.

Case 1. $\displaystyle y_{n}>\varepsilon|x_{n}|$ for
$\displaystyle\varepsilon>0$, or $\displaystyle x_{n}<0$ and
$\displaystyle\frac{y_{n}}{x_{n}}\rightarrow 0$. Define
$\displaystyle\tilde{Y}_{n}=\frac{X_{n}}{\rho_{n}}$ and a blow-up sequence

$\displaystyle\tilde{\psi}_{\rho_{n}}(\tilde{X})=\frac{\psi_{\lambda}(\rho_{n}\tilde{X})}{\rho_{n}}.$

Then one has

$\displaystyle\tilde{Y}_{n}\rightarrow\tilde{Y}_{0}=(\tilde{x}_{0},\tilde{y}_{0})\
\text{and}\ \tilde{\psi}_{\rho_{n}}\rightarrow\tilde{\phi}\ \ \text{uniformly
in any compact subset of $\displaystyle\mathbb{R}^{2}\cap\\{\tilde{y}>0\\}$}.$

By virtue of (3.1), one has

$\displaystyle\tilde{y}-\frac{h}{\rho_{n}}\leq\tilde{\psi}_{\rho_{n}}(\tilde{X})\leq\tilde{y}\
\ \text{in}\ \ \mathbb{R}^{2}\cap\\{\psi_{\rho_{n}}>0\\},$

which implies that
$\displaystyle\tilde{\phi}(\tilde{X})=\max\\{\tilde{y},0\\}$.

If $\displaystyle y_{n}>\varepsilon|x_{n}|$ for $\displaystyle\varepsilon>0$,
it is easy to check that

$\displaystyle|\tilde{Y}_{0}|=1\ \ \text{and}\ \
\tilde{y}_{0}\geq\frac{\varepsilon}{\sqrt{1+\varepsilon^{2}}}.$

By virtue of elliptic regularity, one has

$\displaystyle\tilde{\psi}_{\rho_{n}}\rightarrow\tilde{\phi}\ \ \text{in
$\displaystyle C^{2,\alpha}(B_{r}(\tilde{Y}))$},\ \ \alpha\in(0,1),$

for $\displaystyle 0<r<\frac{\varepsilon}{4\sqrt{1+\varepsilon^{2}}}$. Thus
one has

$\displaystyle\nabla\tilde{\psi}_{\rho_{n}}\left(\frac{X_{n}}{\rho_{n}}\right)\rightarrow\nabla\tilde{\phi}(\tilde{Y}_{0})=(0,1).$

If $\displaystyle x_{n}<0$ and $\displaystyle\frac{y_{n}}{x_{n}}\rightarrow
0$, one has

$\displaystyle\tilde{Y}_{0}=(-1,0).$

Applying elliptic estimates, one has

$\displaystyle\tilde{\psi}_{\rho_{n}}\rightarrow\tilde{\phi}\ \ \text{in
$\displaystyle C^{2,\alpha}(D\cup T)$},\ \ \alpha\in(0,1),$

where $\displaystyle D=B_{2r}(\tilde{Y})\cap\\{\tilde{y}>0\\}$ and
$\displaystyle T=\\{(\tilde{x},0)\mid|\tilde{x}+1|<r\\}$ for $\displaystyle
r>0$. Consequently,

$\displaystyle\nabla\tilde{\psi}_{\rho_{n}}\left(\frac{X_{n}}{\rho_{n}}\right)\rightarrow\nabla\tilde{\phi}(\tilde{Y}_{0})=(0,1).$

Case 2. $\displaystyle x_{n}>0$ and
$\displaystyle\frac{y_{n}}{x_{n}}\rightarrow 0$ and $\displaystyle
y_{n}-k_{\lambda}(x_{n})\rightarrow+\infty$ as $\displaystyle
n\rightarrow+\infty$. Define a blow-up sequence
$\displaystyle\psi_{r_{n}}(\tilde{X})=\frac{\psi_{\lambda}(Z_{n}+r_{n}\tilde{X})}{r_{n}}$
and $\displaystyle\psi_{0}$ is the blow-up limit of
$\displaystyle\psi_{r_{n}}$, where $\displaystyle
r_{n}=y_{n}-k_{\lambda}(x_{n})$ and $\displaystyle
Z_{n}=(x_{n},k_{\lambda}(x_{n}))$. The inequality (3.1) gives that

$\displaystyle\tilde{y}+\frac{k_{\lambda}(x_{n})-h}{r_{n}}\leq\psi^{+}_{r_{n}}(\tilde{X})\leq\tilde{y}+\frac{k_{\lambda}(x_{n})}{r_{n}}\
\ \text{in}\ \ \\{\tilde{y}>0\\},$

which implies that

$\displaystyle\psi_{0}(\tilde{X})=\max\\{\tilde{y},0\\}\ \ \text{in}\ \
\\{\tilde{y}>0\\}.$

Since $\displaystyle-Q\leq\psi_{\lambda}<0$ in
$\displaystyle\Omega\cap\\{\psi_{\lambda}<0\\}$, which implies that
$\displaystyle\psi^{-}_{0}=0$. Therefore, $\displaystyle\psi_{0}$ is 1-plane
solution, and $\displaystyle\psi_{0}(\tilde{X})=\max\\{\tilde{y},0\\}$. The
elliptic regularity gives that

$\displaystyle\psi_{r_{n}}\rightarrow\psi_{0}\ \ \text{in $\displaystyle
C^{2,\alpha}(B_{\frac{1}{4}}(X_{1}))$},\ \ \alpha\in(0,1),\ \ X_{1}=(0,1).$

Thus one has

$\displaystyle\nabla\psi_{\lambda}(X_{n})=\nabla\psi_{r_{n}}(X_{1})\rightarrow(0,1)\
\ \text{as}\ \ n\rightarrow+\infty.$

This gives that
$\displaystyle\nabla\psi_{\lambda}(x,y)\rightarrow\nabla\psi(X_{1})=(0,1)$ as
$\displaystyle x^{2}+y^{2}\rightarrow+\infty$ with
dist($\displaystyle(x,y),\Gamma$)$\displaystyle\rightarrow+\infty$ and
$\displaystyle x>0$.

Step 2. In this step, we will obtain the uniqueness of the injection flow
problem 1. Suppose that there exists another different solution
$\displaystyle(\tilde{\psi}_{\tilde{\lambda}},\tilde{\lambda})$ to the
injection flow problem 1. In view of (1.6), one has

$\psi_{\lambda}(X)-y=o(|X|),\ \ \psi_{\lambda}(X)>0,\text{as
$\displaystyle|X|\rightarrow+\infty$,}$ (3.4)

and

$\tilde{\psi}_{\tilde{\lambda}}(X)-y=o(|X|),\ \
\tilde{\psi}_{\tilde{\lambda}}(X)>0,\text{as
$\displaystyle|X|\rightarrow+\infty$.}$ (3.5)

Without loss of generality, we assume that
$\displaystyle\lambda\leq\tilde{\lambda}$. It is easy to check that

$\lim_{x\rightarrow+\infty}k_{\lambda}(x)=h=\frac{Q}{\sqrt{1+\lambda}}+b\geq\frac{Q}{\sqrt{1+\tilde{\lambda}}}+b=\tilde{h}=\lim_{x\rightarrow+\infty}\tilde{k}_{\tilde{\lambda}}(x).$
(3.6)

Define
$\displaystyle\psi_{\lambda,\varepsilon}(x,y)=\psi_{\lambda}(x,y-\varepsilon)$
for any $\displaystyle\varepsilon\geq 0$. Since the asymptotic heights of the
free boundaries $\displaystyle\Gamma_{\lambda}$ and
$\displaystyle\tilde{\Gamma}_{\tilde{\lambda}}$ are finite, it follows from
(3.6) that we can take $\displaystyle\varepsilon_{0}\geq 0$ to be the smallest
one, such that

$\text{the free boundary of $\displaystyle\psi_{\lambda,\varepsilon_{0}}$ lies
above the free boundary of $\displaystyle\tilde{\psi}_{\tilde{\lambda}}$}.$
(3.7)

Denote
$\displaystyle\Omega^{+}=\Omega\cap\\{\psi_{\lambda,\varepsilon_{0}}>0\\}\cap\\{\tilde{\psi}_{\tilde{\lambda}}>0\\}$
and
$\displaystyle\omega(X)=\tilde{\psi}_{\tilde{\lambda}}-\psi_{\lambda,\varepsilon_{0}}$,
it follows from (3.4), (3.5) and (3.7) that

$\displaystyle\omega(X)\geq 0\ \ \text{on}\ \ \partial\Omega^{+}\ \
\text{and}\ \ \lim_{r\rightarrow+\infty}\frac{m(r)}{r}\rightarrow 0,$

where $\displaystyle r=|X|$ and $\displaystyle m(r)=\min_{|X|=r}\omega(X)$.
Applying the Phragm$\displaystyle\grave{\text{e}}$n-Lindelöf theorem in [15],
one has

$\displaystyle\omega(X)\geq 0\ \ \text{in}\ \ \Omega^{+},$

which implies that

$\tilde{\psi}_{\tilde{\lambda}}\geq\psi_{\lambda,\varepsilon_{0}}\ \ \text{in
$\displaystyle\Omega\cap\\{\psi_{\lambda,\varepsilon_{0}}>0\\}$}.$ (3.8)

By virtue of the asymptotic behavior of
$\displaystyle\psi_{\lambda,\varepsilon_{0}}$ and
$\displaystyle\tilde{\psi}_{\tilde{\lambda}}$, it follows from the similar
arguments in the step 4 in the proof of Theorem 2.4 that

$\tilde{\psi}_{\tilde{\lambda}}\geq\psi_{\lambda,\varepsilon_{0}}\ \ \text{in
$\displaystyle\Omega\cap\\{\tilde{\psi}_{\tilde{\lambda}}<0\\}$}.$ (3.9)

Next, we consider two cases in the following.

Case 1. $\displaystyle\varepsilon_{0}>0$. In view of (3.7), we can take a free
boundary point $\displaystyle X_{0}$ with $\displaystyle|X_{0}|<+\infty$.
Applying the strong maximum principle, one has

$\displaystyle\tilde{\psi}_{\tilde{\lambda}}>\psi_{\lambda,\varepsilon_{0}}\ \
\text{in $\displaystyle\Omega\cap\\{\psi_{\lambda,\varepsilon_{0}}>0\\}$ and}\
\ \tilde{\psi}_{\tilde{\lambda}}>\psi_{\lambda,\varepsilon_{0}}\ \ \text{in
$\displaystyle\Omega\cap\\{\tilde{\psi}_{\tilde{\lambda}}<0\\}$}.$

Since the free boundary $\displaystyle\tilde{\Gamma}_{\tilde{\lambda}}$ and
$\displaystyle\Gamma^{\varepsilon_{0}}_{\lambda}$ are analytic at
$\displaystyle X_{0}$, it follows from Hopf’s lemma that

$\displaystyle|\nabla\psi_{\lambda,\varepsilon_{0}}^{-}|=-\frac{\partial\psi^{-}_{\lambda,\varepsilon_{0}}}{\partial\nu}>-\frac{\partial\tilde{\psi}_{\tilde{\lambda}}^{-}}{\partial\nu}=|\nabla\tilde{\psi}_{\tilde{\lambda}}^{-}|\
\ \text{and}\ \
|\nabla\psi_{\lambda,\varepsilon_{0}}^{+}|=\frac{\partial\psi^{+}_{\lambda,\varepsilon_{0}}}{\partial\nu}<\frac{\partial\tilde{\psi}_{\tilde{\lambda}}^{+}}{\partial\nu}=|\nabla\tilde{\psi}_{\tilde{\lambda}}^{+}|\
\ \text{at}\ \ X_{0},$

where $\displaystyle\nu$ is the inner normal vector to
$\displaystyle\partial\\{\tilde{\psi}_{\tilde{\lambda}}>0\\}$ at
$\displaystyle X_{0}$. Those give that

$\displaystyle\lambda=|\nabla\psi_{\lambda,\varepsilon_{0}}^{-}|^{2}-|\nabla\psi_{\lambda,\varepsilon_{0}}^{+}|^{2}>|\nabla\tilde{\psi}_{\tilde{\lambda}}^{-}|^{2}-|\nabla\tilde{\psi}_{\tilde{\lambda}}^{+}|^{2}=\tilde{\lambda}\
\ \text{at}\ \ X_{0},$

which contradicts to our assumption $\displaystyle\lambda\leq\tilde{\lambda}$.

Case 2. $\displaystyle\varepsilon_{0}=0$. Similar to (2.32), we can show that
$\displaystyle\lambda\cdot\tilde{\lambda}>0$. Without loss of generality, one
assume that $\displaystyle 0>\tilde{\lambda}\geq\lambda$. Therefore, we can
obtain a contradiction by using the similar arguments in Subcase 2.1 in the
proof of Lemma 2.5.

∎

By virtue of Theorem 3.1, we complete the proof of Theorem 1.1.

Due to the uniqueness of $\displaystyle\lambda$ for any $\displaystyle Q>0$,
we can define a function $\displaystyle\lambda=\lambda(Q)$ for any
$\displaystyle Q>0$. We next consider the relation between
$\displaystyle\lambda(Q)$ and $\displaystyle Q>0$, and complete the proof of
Theorem 1.2.

###### Proof of Theorem 1.2.

(1). For any $\displaystyle Q_{1}>Q_{2}>0$, there exist two solutions
$\displaystyle(\psi_{\lambda(Q_{1})},\lambda(Q_{1}))$ and
$\displaystyle(\psi_{\lambda(Q_{2})},\lambda(Q_{2}))$ to the injection flow
problem 1. We next show that

$\displaystyle\lambda(Q_{1})>\lambda(Q_{2})\ \ \text{for any $\displaystyle
Q_{1}>Q_{2}>0$}.$

If not, then there exist $\displaystyle Q_{1}>Q_{2}>0$, such that
$\displaystyle\lambda(Q_{1})\leq\lambda(Q_{2})$, and we consider the following
two cases.

Case 1. $\displaystyle\lambda(Q_{1})=\lambda(Q_{2})=0$. Since
$\displaystyle\lambda(Q_{1})$ and $\displaystyle\psi_{\lambda(Q_{1})}$ are
unique for any given $\displaystyle Q_{1}>0$, there exists a sequence
$\displaystyle\lambda_{L_{n}}(Q_{1})$ with
$\displaystyle\lambda_{L_{n}}(Q_{1})=0$, such that

$\displaystyle\psi_{\lambda_{L_{n}}(Q_{1}),L_{n}}\rightarrow\psi_{\lambda(Q_{1})}\
\text{in $\displaystyle H^{1}_{loc}(\Omega)$ and uniformly in any compact
subset of $\displaystyle\Omega$},$

as $\displaystyle L_{n}\rightarrow+\infty$. By virtue of (2.36), one has

$\displaystyle(a\sin\theta-b\cos\theta)^{\frac{\pi}{\theta}}Q_{1}^{1-\frac{\pi}{\theta}}=L_{n}-(L_{n}-b)^{\frac{\pi}{\theta}}(L_{n}+Q_{1})^{1-\frac{\pi}{\theta}}.$

Set $\displaystyle t_{n}=L_{n}+Q_{1}$, one has

$\displaystyle(a\sin\theta-b\cos\theta)^{\frac{\pi}{\theta}}Q_{1}^{1-\frac{\pi}{\theta}}=t_{n}\left(1-\left(1-\frac{Q_{1}+b}{t_{n}}\right)^{\frac{\pi}{\theta}}\right)-Q_{1}\rightarrow\left(\frac{\pi}{\theta}-1\right)Q_{1}+\frac{b\pi}{\theta},$

as $\displaystyle t_{n}\rightarrow+\infty$. Then one has

$(a\sin\theta-b\cos\theta)^{\frac{\pi}{\theta}}=\left(\frac{\pi}{\theta}-1\right)Q_{1}^{\frac{\pi}{\theta}}+\frac{b\pi}{\theta}Q_{1}^{\frac{\pi}{\theta}-1}.$
(3.10)

Similarly, we have

$\displaystyle(a\sin\theta-b\cos\theta)^{\frac{\pi}{\theta}}=\left(\frac{\pi}{\theta}-1\right)Q_{2}^{\frac{\pi}{\theta}}+\frac{b\pi}{\theta}Q_{2}^{\frac{\pi}{\theta}-1},$

which together with (3.10) implies that $\displaystyle Q_{1}=Q_{2}$. This
leads a contradiction.

Case 2. $\displaystyle\lambda(Q_{1})\neq 0$ or
$\displaystyle\lambda(Q_{2})\neq 0$.

Since $\displaystyle Q_{1}>Q_{2}$, one has

$h_{1}=\frac{Q_{1}}{\sqrt{\lambda(Q_{1})+1}}+b>\frac{Q_{2}}{\sqrt{\lambda(Q_{2})+1}}+b=h_{2}\
\ \text{and}\ \ k_{\lambda(Q_{1})}(x)>k_{\lambda(Q_{2})}(x)$ (3.11)

for sufficiently large $\displaystyle x>0$.

Define a function
$\displaystyle\psi_{\lambda(Q_{1}),\varepsilon}(x,y)=\psi_{\lambda(Q_{1})}(x,y-\varepsilon)$
for $\displaystyle\varepsilon\geq 0$. In view of (3.11), let
$\displaystyle\varepsilon_{0}\geq 0$ be the smallest one, such that

$\displaystyle\text{the free boundary of
$\displaystyle\psi_{\lambda(Q_{1}),\varepsilon_{0}}$ lies above the free
boundary of $\displaystyle\psi_{\lambda(Q_{2})}$}.$

Similar to the proof of Lemma 3.1, by using the
Phragm$\displaystyle\grave{\text{e}}$n-Lindelöf theorem in [15] and the
asymptotic behavior of $\displaystyle\psi_{\lambda(Q_{1}),\varepsilon_{0}}$
and $\displaystyle\psi_{\lambda(Q_{2})}$, we have

$\displaystyle\psi_{\lambda(Q_{1}),\varepsilon_{0}}\leq\psi_{\lambda(Q_{2})}\
\text{in
$\displaystyle\Omega\cap\\{\psi_{\lambda(Q_{1}),\varepsilon_{0}}>0\\}$ and
}\psi_{\lambda(Q_{1}),\varepsilon_{0}}\leq\psi_{\lambda(Q_{2})}\ \ \text{in
$\displaystyle\Omega\cap\\{\psi_{\lambda(Q_{2})}<0\\}$}.$

Then we can obtain a contradiction by using the similar arguments in the proof
of Lemma 3.1.

By virtue of the uniqueness of the solution
$\displaystyle(\psi_{\lambda},\lambda)$, it follows from the similar arguments
in the proof of Lemma 2.7 that $\displaystyle\lambda(Q)$ is continuous for any
$\displaystyle Q>0$.

(2). Next, we will show that there exists a
$\displaystyle\underline{\lambda}\in(-1,0)$, such that
$\displaystyle\lambda(Q)\rightarrow\underline{\lambda}$ as $\displaystyle
Q\rightarrow 0$. The monotonicity of $\displaystyle\lambda(Q)$ gives that
there exists $\displaystyle\underline{\lambda}\geq-1$, such that
$\displaystyle\lambda(Q)\rightarrow\underline{\lambda}$ as $\displaystyle
Q\rightarrow 0$. It suffices to exclude the case $\displaystyle\lambda=-1$.
For any sequence $\displaystyle\\{Q_{n}\\}$ with $\displaystyle Q_{n}>0$ and
$\displaystyle Q_{n}\rightarrow 0$, such that

$\displaystyle\lambda(Q_{n})\rightarrow\underline{\lambda},\
\psi_{\lambda(Q_{n})}\rightarrow\psi_{\underline{\lambda}}\ \text{in
$\displaystyle H^{1}_{loc}(\Omega)$ and uniformly in any compact subset of
$\displaystyle\Omega$},$

as $\displaystyle Q_{n}\rightarrow 0$. Moreover,
$\displaystyle\partial_{x}\psi_{\underline{\lambda}}\leq 0$ and
$\displaystyle\partial_{y}\psi_{\underline{\lambda}}\geq 0$ in
$\displaystyle\Omega\cap\\{\psi_{\underline{\lambda}}>0\\}$, which implies
that

$\displaystyle\Omega\cap\\{x>0\\}\cap\\{\psi_{\underline{\lambda}}>0\\}=\Omega\cap\\{x>0\\}\cap\\{y>k_{\underline{\lambda}}(x)\\},$

where $\displaystyle k_{\underline{\lambda}}(x)$ is monotone increasing for
$\displaystyle x>0$, and $\displaystyle k_{\underline{\lambda}}(0)=0$.

Suppose that $\displaystyle\underline{\lambda}=-1$. Define
$\displaystyle\omega(y)=\max\\{y,0\\}$, it follows from the condition (7) in
Definition 1.1 that
$\displaystyle\omega(y)\geq\psi_{\underline{\lambda}}(x,y)$ in
$\displaystyle\Omega$. The strong maximum principle gives that

$\displaystyle\psi_{\underline{\lambda}}<y\ \ \text{in}\ \
\Omega\cap\\{\psi_{\underline{\lambda}}>0\\}.$

For any small $\displaystyle r>0$, it follows from the proof of (2.39) that
there exists a small $\displaystyle\delta>0$, such that