Symmetry and Monotonicity of a Nonlinear Schrödinger Equation Involving the Fractional Laplacian

Abstract

In this paper, we consider a nonlinear Schrödinger equation involving the fractional Laplacian with Dirichlet condition:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{\alpha }{2}}u+A(x)u=f(x,u,\nabla u) \ \text{ in } \ \Omega ,\\ u>0, \ \text{ in }\ \Omega ; \ u\equiv 0, \ \text{ in }\ \mathbb R^n\backslash \Omega , \end{array}\right. } \end{aligned}$$

where $\Omega$ is a domain (bounded or unbounded) in $\mathbb R^n$ which is convex in $x_1$-direction. By using some ideas of maximum principle and the direct moving plane method, we prove that the solutions are strictly increasing in $x_1$-direction in the left half domain of $\Omega$. Symmetry of some solutions are also proved. Meanwhile, we obtain a Liouville type theorem on the half space $\mathbb R^n_+$.

1 Introduction

There are some literatures on the nonlinear Schrödinger equations involving the fractional Laplacian

$$\begin{aligned} (-\Delta )^{\alpha }u+A(x)u=f(x,u), \ \ x \in \mathbb {R}^n, \end{aligned}$$

(1.1)

where $n\ge 2$ and $\alpha \in (0,1)$. We recall that the fractional Laplacian is defined as

$$\begin{aligned} (-\Delta )^{\alpha }u(x)= C_{n,\alpha }\;\text {P.V.}\int _{\mathbb {R}^n} \frac{u(x)-u(y)}{|x-y|^{n+2\alpha }} dy, \end{aligned}$$

(1.2)

where $\text {P.V.}$ stands for the Principle Value and $C_{n,\alpha }>0$. For more backgrounds on $(-\Delta )^{\alpha }$, we refer the readers to [1][7][11][26], and etc.

For instance, Benhassine [4] took

$$\begin{aligned} \psi (x,t)=e^{i\omega t}u(x), \end{aligned}$$

where u is a real-valued function and f is assumed to satisfy $f(x,e^{i\omega t}u)=e^{-i\omega t}f(x,u)$. He proved the existence of solutions to (1.1) is equivalent to the existence of standing wave solutions for fractional Schrödinger equations of the form:

$$\begin{aligned} i \frac{\partial \psi }{\partial t}=(-\Delta )^\alpha \psi +(A(x)+\omega )\psi -f(x,\psi ), \end{aligned}$$

(1.3)

where $\omega$ is constant and $\alpha \in (0,1)$. That is to say, $\psi (x,t)$ solves (1.3) if and only if u(x) solves (1.1).

When $\alpha =1$, we have a nonlinear Schrödinger equation with the fraction Laplacian:

$$\begin{aligned} (-\Delta )u+A(x)u=f(x,u),x\in \mathbb {R}^n. \end{aligned}$$

Here, we mention that some earlier work were done by Floer and Weinstein [16], Rabinowitz [28], Wang [32], del Pino and Felmer [27], Azzollini[3], Secchi[29], and etc. In 2012, Felmer-Quaas-Tan [15] considered the case when $A(x)=1$ and studied the existence, regularity, decay and symmetry properties of positive solutions for the nonlinear Schrödinger equation involving the fractional Laplacian under the conditions $u>0, \ \text{ in }\ \mathbb {R}^n$ and $\lim _{|x|\rightarrow \infty }u(x)=0$. In 2016, Secchi [30] proved the existence of radially symmetric solutions for (1.1), where f(x, u) is replaced by g(u). In 2018, Servadei studied the nonlinear fractional Schrödinger equation under the condition $u\in H^{\alpha }(\mathbb {R}^n, \mathbb {R})$. There are also articles about Schrödinger equation under other conditions, such as [2, 28, 20, 24, 18, 19] and [22]. In this paper, we consider $u \in C^{1,1}_{loc} \cap L_{\alpha }(\mathbb R^n)$, where

$$\begin{aligned} \frac{}{}L_{\alpha }(\mathbb R^n) = \{ u \in L^1_{loc} \mid \int _{\mathbb {R}^n} \frac{|u(x)|}{1+|x|^{n+\alpha }} d x < \infty \}. \end{aligned}$$

On the other hand, Cheng-Huang-Li [14] considered the zero-Dirichlet problem with more generalized nonlinear term,

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{\alpha }{2}}u=f(x,u,\nabla u) \ \text{ in } \ \Omega \\ u>0 \ \text{ in }\ \Omega ; \ u\equiv 0 \ \text{ in }\ \mathbb R^n\backslash \Omega . \end{array}\right. } \end{aligned}$$

(1.4)

Note that in (1.4), $\Omega$ has been considered as a convex domain (bounded or unbounded) in $x_1$-direction in $\mathbb R^n$ (we say a domain $\Omega$ is convex in $x_1$-direction if and only if $(x_1,x^{\prime } ), (x_2,x^{\prime })\in \Omega$ imply that $((1-t)x_1+tx_2,x^{\prime })\in \Omega$ for any $t\in (0,1)$). Moreover, the nonlinear term f belongs to the function space $\digamma$, where $\digamma$ is defined as the collections of all functions $f(x,u,p):\mathbb {R}^n\times \mathbb {R}\times \mathbb {R}^n\rightarrow \mathbb {R}$ such that for any $M>0, u_1,u_2\in [-M,M]$ and any $x, p\in \mathbb {R}^n$,

$$\begin{aligned} |f(x,u_1,\mathbf{p} )-f(x,u_2,\mathbf{p} )|\le C_M|u_1-u_2|\ \text{ for } \text{ some } \ C_M>0. \end{aligned}$$

Under some other adaptable conditions on u and f, Cheng-Huang-Li[14] showed that the solution u for (1.4) is monotone and symmetric when $\Omega$ is bounded or unbounded respectively. We generalize the method of moving plane for the fractional Laplacian used by Li [23], Caffarelli and Silvestre [5], Cheng [13], Chen-Li-Li[10], Wang[31], Ma-Chen[25], Garofalo[17], Cao[6], Li [21], Chen [9], and etc.

In this paper our goal is to study the nonlinear Schrödinger equation involving the fractional Laplacian:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{\alpha }{2}}u+A(x)u=f(x,u,\nabla u) \ \ \text{ in } \ \Omega \\ u>0 \ \ \text{ in }\ \Omega \\ u\equiv 0 \ \ \text{ in }\ \Omega ^c, \end{array}\right. } \end{aligned}$$

(1.5)

where $\alpha \in (0,2)$. We shall show that as long as A(x) and $f(x,u,\nabla u)$ in (1.5) satisfy certain conditions (these conditions will be exhibited in detail in the next theorem), the positive solutions in $\Omega$ will have symmetry and monotonicity. While in ${\mathbb R}^n_+$ similar results could not exist. Our main results are stated in the following three theorems.

Theorem 1.1

Let $\Omega$ be a bounded domain in $\mathbb {R}^n$ which is convex in $x_1$-direction and symmetric about $x_1=0$. We suppose that $u\in C(\mathbb {R}^n)\bigcap \mathcal C_{loc}^{1,1}(\Omega )$ solves (1.5) for $\alpha \in (0,2)$, $f(x,v,\mathbf{p} )\in \digamma$ satisfies

$$\begin{aligned} f(x_1,x',v,p_1,p_2,\cdots ,p_n)\le f(\overline{x}_1,x',v,-p_1,p_2,\cdots ,p_n), \ \forall -x_1, p_1\ge 0, \end{aligned}$$

and A(x) is bounded below and symmetric in $x_1$-direction and satisfies

$$\begin{aligned} A(x_1,x')<A(\overline{x}_1,x'), \; \forall x_1>\overline{x}_1,\ (x_1,x'), (\overline{x}_1,x')\in \Omega . \end{aligned}$$

Then $u(x_1,x')$ is strictly increasing in the left half of $\Omega$ in $x_1$-direction and

$$\begin{aligned} u(x_1,x')\le u(-x_1,x'), \; \forall x_1<0,\ (x_1,x')\in \Omega . \end{aligned}$$

Moreover, if $f(x_1,x',v,p_1,p_2,\cdot \cdot \cdot ,p_n)= f(-x_1,x',v,-p_1,p_2,\cdot \cdot \cdot ,p_n)$, then

$$\begin{aligned} u(x_1,x')= u(-x_1,x'). \end{aligned}$$

To state our second theorem in unbounded domain $\Omega$, we need to impose a growth condition on $f(x,v,\mathbf{p} )$:

$$\begin{aligned} \frac{|f(x,u_1,\mathbf{p} )-f(x,u_2,\mathbf{p} )|}{|u_1-u_2|}\le C(|u_1\mid ^s+|u_2|^s)\ as\ u_1,u_2\rightarrow 0,\ for\ some\ s>0. \end{aligned}$$

(1.6)

Theorem 1.2

Let $\Omega$ be an unbounded domain in $\mathbb {R}^n$, which is convex in $x_1$-direction and symmetric about $x_1=0$. Suppose that $u\in C(\mathbb {R}^n)\bigcap \mathcal C_{loc}^{1,1}(\Omega )\cap L_\alpha$ solves (1.5) for $\alpha \in (0,2)$ and $f(x,v,\mathbf{p} )\in \digamma$. If $f(x,v,\mathbf{p} )$ satisfies (??) (1.6) and u(x) satisfies

$$\begin{aligned} u(x)=o({|x|^{-\alpha / s}})\ \text{ as } \ |x|\rightarrow +\infty ,\; x_1<0 \end{aligned}$$

(1.7)

and A(x) is bounded below and symmetric in $x_1$-direction and satisfies

$$\begin{aligned} A(x_1,x')<A(\overline{x}_1,x'), \forall x_1>\overline{x}_1,\ (x_1,x'), (\overline{x}_1,x')\in \Omega . \end{aligned}$$

then there exists $u_0\le 0$ such that $u(x_1,x')$ is strictly increasing in $\Omega \cap \{x_1<\mu _0\}$ in $x_1$-direction and

$$\begin{aligned} u(x_1,x')\le u(2\mu _0-x_1,x'), (x_1,x')\in \Omega \cap \{x_1<\mu _0\}, \end{aligned}$$

(1.8)

where

$$\begin{aligned} \mu _0=\sup _{\mu \le 0}\{\mu \mid u(x_1,x')\le u(2\lambda -x_1,x'),\forall (x_1,x')\in \Omega \cap \{x_1<\lambda \},\forall \lambda \le \mu \}. \end{aligned}$$

Furthermore, if $\mu _0<0$, then $u(x_1,x')= u(2\mu -x_1,x'),(x_1,x')\in \Omega \cap \{x_1<\mu _0\}$.

We next turn our attention to the follow equation

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{\alpha }{2}}u+A(x)u=f(x,u,\nabla u), x\in \mathbb {R}^n_+,\\ u\equiv 0, x\in \mathbb {R}^n\backslash \mathbb {R}^n_+. \end{array}\right. } \end{aligned}$$

(1.9)

We have the next theorem.

Theorem 1.3

Suppose that $u\in C(\mathbb {R}^n_+)\bigcap \mathcal C_{loc}^{1,1}(\mathbb {R}^n_+)$ is a nonnegative solution of (1.9), $f(x,v,\mathbf{p} )\in \digamma$ satisfies

$$\begin{aligned} f(x',x_n,v,p_1,\cdots ,p_{n-1},p_n)\le f(x',\overline{x}_n,v,p_1,\cdots ,p_{n-1},-p_n), \end{aligned}$$

for any $x_n,\; p_n\ge 0$, $x_n\le \overline{x}_n$, and A(x) is bounded below and symmetric in $x_1$-direction and satisfies

$$\begin{aligned} A(x_1,x')<A(\overline{x}_1,x'), \forall x_1>\overline{x}_1,\ (x_1,x'), (\overline{x}_1,x')\in \Omega . \end{aligned}$$

Suppose that

$$\begin{aligned} \lim _{|x|\rightarrow \infty }u(x)=0 \end{aligned}$$

(1.10)

and

$$\begin{aligned} f(x,v,\mathbf {p})=0,\ if\ v=0. \end{aligned}$$

(1.11)

Then $u\equiv 0$.

This paper is organized as follows. In Sect. 2, we give the proof of Theorem 1.1. Section 3 is devoted to the proof of Theorem 1.2. While in Sect. 4, we will refer to the paper [8][12] and give the proof of Theorem 1.3.

2 Proof of Theorem1.1

Firstly, we define some frequently used notations in this paper. For $x'=(x_2,x_3,\cdots ,x_n)$ and $\lambda \in \mathbb {R}$, write

$$\begin{aligned} x=(x_{1},x')\in \mathbb {R}^n,\; x^{\lambda }=(2\lambda -x_{1},x{'})\in \mathbb {R}^n. \end{aligned}$$

Let $T_\lambda$ be a hyperplane in $\mathbb {R}^n$, which is defined as

$$\begin{aligned} T_\lambda =\{x=(x_1,x')\in \mathbb {R}^n|x_1=\lambda \} \end{aligned}$$

and

$$\begin{aligned} \Sigma _{\lambda }=\{(x_{1},x')\in \mathbb {R}^n\mid {x_{1}<\lambda }\}, \;\Sigma ^{c}_{\lambda }=\{(x_{1},x{'})\in \mathbb {R}^n\mid {x_{1}\ge \lambda }\}. \end{aligned}$$

We set

$$\begin{aligned} u_\lambda (x)=u(x^\lambda ),\ \omega _\lambda (x)=u_\lambda (x)-u(x),\ \forall x\in \mathbb {R}^n. \end{aligned}$$

Sometimes, we may write $\omega _\lambda (x)$ as $\omega (x)$ if there is no ambiguity. Now we introduce a lemma, which was established in [14].

Lemma 2.1

Let $\omega \in L_{\alpha }$ be a so called $\lambda$-antisymmetric function, i.e. $\omega (x_1,x')=-\omega (2\lambda -x_1,x')$. Suppose that there exists $x\in \Sigma _{\lambda }$ such that

$$\begin{aligned} \omega (x)=\underset{\Sigma _{\lambda }}{\inf } \omega (y)<0. \end{aligned}$$

If $\omega \in C^{1,1}$at x, then there exists some positive constant $C_{n,\alpha }$ such that

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}\omega (x)\le C_{n,\alpha }\left( \delta ^{-\alpha }\omega (x)-\delta \int _{\Sigma _{\lambda }} \frac{({\omega }(y)-\omega (x))(\lambda -y_{1})}{|x-y^{\lambda }|^{n+\alpha +2}} dy\right) , \end{aligned}$$

for all $\alpha \in (0,2)$, where $\delta =d(x,T_{\lambda })=|x_1-\lambda |.$

Proof of Theorem 1.1 We divide the proof into two steps. Firstly, we shall show that for $\lambda >-1$ which is sufficiently close to $-1$, we have

$$\begin{aligned} \omega _\lambda (x)\ge 0,\ \forall x\in \Sigma _\lambda . \end{aligned}$$

(2.1)

Next, we shall move the plane $T_\lambda$ along the $x_1$-axis to the right in the left half of $\Omega$ as long as inequality (2.1) holds. The plane will eventually stop at the limiting position $\lambda =0$.

Step 1. Since $\Omega$ is bounded and convex in $x_1$-direction, without loss of generality, we may assume that

$$\begin{aligned} \Omega \subset \{|x_1|\le 1\} \; \, \text{ and } \;\, \partial \Omega \cap \{x_1=-1\}\ne \emptyset . \end{aligned}$$

We claim that there exists $\delta >0$ small enough such that for all $\lambda \in [-1,-1+\delta )$, then

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _\lambda . \end{aligned}$$

(2.2)

If not, we set

$$\begin{aligned} \ell =\inf _{x\in \overline{\Sigma }_{ \begin{array}{c} \lambda \\ \lambda \end{array}\in [-1,-1+\delta )}}\omega _\lambda (x)<0. \end{aligned}$$

Since $0\le u(x)\in C(\mathbb R^n)$ and $u(x)\equiv 0, x\in \Omega ^c$, we know that for any $\delta \in (0,1)$, $\ell$ can obtained the infimum for some

$$\begin{aligned} (\lambda _0,x_0)\in \{(\lambda , x)|(\lambda , x)\in [-1, -1+\delta ]\times \overline{\Sigma _{\lambda }\cap \Omega }\}. \end{aligned}$$

Thus, $\omega _{\lambda _0}\ge 0$ on $\partial (\Omega \cap \Sigma _{\lambda _0})$, which yields that $x_0\in \Sigma _{\lambda _0}\cap \Omega$. Since $\lambda =-1$ implies $\Sigma _{-1}\cap \Omega =\partial \Sigma _{-1}\cap \Omega$, one gets $\omega _1(x)\ge 0$ in $\Sigma _{-1}\cap \Omega$. Therefore, $\lambda _0>-1$.

Now we set $u_{x_i}(x)=\frac{\partial u}{\partial {x_i}}(x)$ for $i=1,2,\ldots ,n$. Since $(\lambda _0,x_0)$ is point of minimum, we know that for $\lambda _0\in (-1,-1+\delta )$,

$$\begin{aligned} \left. \frac{\partial \omega _\lambda (x)}{\partial \lambda }\right| _{(\lambda _0,x_0)}=0, \;\text {and}\; \left. \frac{\partial \omega _\lambda (x)}{\partial \lambda }\right| _{(-1+\delta ,x_0)}\le 0. \end{aligned}$$

Moreover, we also have (a) $u_{x_1}(x_0^{\lambda _0})\le 0$. This is because that

$$\begin{aligned}&0\ge \left. \frac{\partial \omega _\lambda (x)}{\partial \lambda }\right| _{(\lambda _0,x_0)}\\&\quad =\left. \frac{\partial }{\partial \lambda }(u(x^\lambda )-u(x))\right| _{(\lambda _0,x_0)}\\&\quad =2u_{x_1}(x^\lambda )|_{(\lambda _0,x_0)}; \end{aligned}$$

(b) $\nabla _x\omega (x_0^{\lambda _0}) =0$, where $\nabla _x$ denotes the gradient operator with respective to x. From (b), we know that

$$\begin{aligned} \nabla _x u(x_0^{\lambda _0})=\nabla _x u(x_0) , \end{aligned}$$

which yields

$$\begin{aligned} u_{x_i}(x_0^{\lambda _0})=u_{x_i}(x_0) \ \text{ for }\ i=2, \cdots ,n \end{aligned}$$

(2.3)

and

$$\begin{aligned} u_{x_1}(x_0^{\lambda _0})=-u_{x_1}(x_0). \end{aligned}$$

(2.4)

Moreover, (a) implies that

$$\begin{aligned} u_{x_n}(x_0)\ge 0. \end{aligned}$$

(2.5)

By using Property (1.6) of $f(x,u,\mathbf{p} )$, Conditions (2.3)-(2.5) and the fact that A(x) satisfies $A(x_1,x')<A(\overline{x}_1,x')$ for all $x_1>\overline{x}_1, \ (x_1,x'), (\overline{x}_1,x')\in \Omega ,$ we know from (1.5) that

$$\begin{aligned} \nonumber (-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})&=f(x_0^{\lambda _0},u_{\lambda _0}(x_0),\nabla _xu(x_0^{\lambda _0}))-f(x_0,u(x_0),\nabla _xu(x_0))\\ \nonumber&\quad +A(x_{0})u(x_{0})-A(x_0^{\lambda _{0}})u(x_0^{\lambda _{0}})\\ \nonumber&\ge f(x_0,u_{\lambda _0}(x_0),\nabla _xu(x_0))-f(x_0,u(x_0),\nabla _xu(x_0))\\ \nonumber&\quad +A(x_{0})u(x_{0})-A(x_0)u(x_0^{\lambda _{0}})\\&=c(x_0)\omega _{\lambda _0}(x_0)-A(x_0)\omega _{\lambda _{0}}(x_{0}), \end{aligned}$$

(2.6)

where

$$\begin{aligned} c(x_0)=\frac{f(x_0,u_{\lambda _0}(x_0),\nabla _xu(x_0))-f(x_0,u(x_0),\nabla _xu(x_0))}{u_{\lambda _0}(x_0)-u(x_0)}. \end{aligned}$$

Since $u(x)\in C(\mathbb {R}^n)$ with compact support and $f(x,u,\mathbf{p} )\in \digamma$, we get that c(x) is uniformly bounded.

Moreover, since A(x) is bounded from below, by using Lemma 1 we have

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0) \nonumber \\\le (\frac{C}{\delta ^\alpha }+A(x_{0})-c(x_0))\omega _{\lambda _0}(x_0)<0. \end{aligned}$$

(2.7)

Take $\delta$ small enough, we arrive at a contradiction with (2.6). Hence, Claim (2.2) holds.

Step 2. Set

$$\begin{aligned} \lambda _0=\sup _{-1\le \lambda \le 0}\{\lambda \mid \omega _\mu (x)\ge 0,\forall x\in \Sigma _\mu ,\forall \mu \le \lambda \}. \end{aligned}$$

Then we have $\lambda _0\le 0$. Indeed, we must have $\lambda _0=0$. Otherwise, suppose that $\lambda _0<0$, since $\omega _{\lambda _0}\not \equiv 0$, we will have

$$\begin{aligned} \omega _{\lambda _0}>0,\; x\in \Sigma _{\lambda _0}\cap \Omega . \end{aligned}$$

(2.8)

If not, there would exist $x_0\in \Omega \cap \Sigma _{\lambda _0}$ such that $\omega _{\lambda _0}(x_0)=0$. By using (2.6) we have

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0) \ge 0. \end{aligned}$$

On the other hand, we have

$$\begin{aligned}&(-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})\nonumber \\&=C_{n,\alpha } \text {P.V.} \int _{\mathbb {R}^n} \frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y|^{n+\alpha }} dy\nonumber \\&=C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y|^{n+\alpha }} dy+C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }^c}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y|^{n+\alpha }} dy\nonumber \\&=C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y|^{n+\alpha }} dy+C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y^{\lambda })}{|x-y^{\lambda }|^{n+\alpha }} dy\nonumber \\&=C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y|^{n+\alpha }} dy+C_{n,\alpha }P.V.\int _{ \Sigma _{\lambda }}\frac{\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)}{|x-y^{\lambda }|^{n+\alpha }} dy\nonumber \\&=C_{n,\alpha }P.V.\int _{\Sigma _{\lambda }}\Big (\frac{1}{|x-y|^{n+\alpha }}-\frac{1}{|x-y^{\lambda }|^{n+\alpha }}\Big ) \Big (\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)\Big )dy\nonumber \\&\quad +C_{n,\alpha }\int _{\Sigma _{\lambda }}\frac{2\omega _{\lambda _{0}}(x_{0})}{|x-y^{\lambda }|^{n+\alpha }}dy\nonumber \\&:=I_1+I_2. \end{aligned}$$

(2.9)

The fact $\omega _{\lambda _0}(x_0)=0$ gives $I_2=0$. Then

$$\begin{aligned}&(-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0)\\&\le C_{n,\alpha }P.V.\int _{\Sigma _{\lambda }}\Big (\frac{1}{|x-y|^{n+\alpha }}-\frac{1}{|x-y^{\lambda }|^{n+\alpha }}\Big ) \Big (\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)\Big )dy <0, \end{aligned}$$

where

$$\begin{aligned} \frac{1}{|x-y|^{n+\alpha }}>\frac{1}{|x-y^{\lambda }|^{n+\alpha }},\;\forall x,y\in \Sigma _{\lambda }. \end{aligned}$$

Since $\omega _{\lambda _0}(y)\ge 0$ and $\omega _{\lambda _0}(y)\not \equiv 0$, which yields a contradiction, thus (2.8) holds.

We claim that there exists $\epsilon >0$ small enough such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _\lambda \cap \Omega ,\forall \lambda \in [\lambda _0,\lambda _0+\epsilon ). \end{aligned}$$

(2.10)

Now we prove (2.10) is true by contradiction. Indeed, combining (2.8) and the fact that $\omega _\lambda (x)$ is lower semi-continuous in $\Omega$, for any $\delta >0$ we have

$$\begin{aligned} \omega _{\lambda _0}(x)\ge C_\delta >0,\forall x\in \Sigma _{\lambda _0-\delta }\cap \Omega . \end{aligned}$$

By the continuity of $\omega _\lambda$ with respect to $\lambda$, there exist $\epsilon >0$ such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _{\lambda _0-\delta }\cap \Omega ,\forall \lambda \in [\lambda _0,\lambda _0+\epsilon ). \end{aligned}$$

(2.11)

Suppose (2.10) is not true, then for all $\epsilon >0$ we have

$$\begin{aligned} \ell _\epsilon =\inf _{x\in \overline{\begin{array}{c} {\Sigma _{\lambda }}\\ \lambda \end{array}}\in [\lambda _0,\lambda _0+\epsilon ]}\omega _\lambda (x)<0. \end{aligned}$$

Since $\Omega$ is a bounded domain, then by (2.11), $\ell _\epsilon$ can obtain the infimum for some point $(\mu ,x_0)\in \{(\lambda ,x)|(\lambda ,x)\in [\lambda _0,\lambda _0+\epsilon ]\times \overline{(\Sigma _\lambda \backslash \Sigma _{{\lambda _0}-\delta })\cap \Omega }\}$, i.e.,

$$\begin{aligned} \omega _\mu (x_0)=\ell _\epsilon <0. \end{aligned}$$

Obviously we have $\omega _{\mu }\ge 0$ on $\partial (\Sigma _\mu \backslash \Sigma _{{\lambda _0}-\delta }\cap \Omega )$, thus $x_0\in \Sigma _\mu \backslash \Sigma _{{\lambda _0}-\delta }\cap \Omega$. By similar arguments as in Step 1 (2.6)-(2.7), we have

$$\begin{aligned} \nonumber 0&\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0)\\&\nonumber \le (\frac{C}{(\delta +\epsilon )^\alpha }+A(x_{0})-c(x_0))\omega _\mu (x_0)<0, \end{aligned}$$

if $\delta$, $\epsilon$ are chosen small enough. This yields a contradiction, so (2.10) holds. Therefore, we must have $\lambda _0=0$. It follows that

$$\begin{aligned} \omega _0(x)\ge 0,\; x\in \Sigma _0, \end{aligned}$$

which implies

$$\begin{aligned} u(x_1,x')\le u(-x_1,x'), \forall x_1\le 0,\ (x_1,x')\in \Omega . \end{aligned}$$

The proof of Theorem 1.1 is completed. $\square$

3 Proof of Theorem 1.2

We divide the proof into two steps. To begin with, we shall show that for $\lambda <0$,

$$\begin{aligned} \omega _\lambda (x)\ge 0,\ \forall x\in \Sigma _\lambda . \end{aligned}$$

(3.1)

Then we move the plane $T_\lambda$ along the $x_1$-axis to the right in the left half of $\Omega$ as long as inequality (3.1) holds. The plane will eventually stop at some limit position $\lambda =\lambda _0<0$ or $\lambda =0$.

Step 1. Start moving the plane $T_\lambda$ along the $x_1$-axis from near $-\infty$ to the right.

Claim: There exists $R_0>0$ large enough such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _\lambda ,\forall \lambda \le -R_0. \end{aligned}$$

(3.2)

Suppose the claim is not true, then there exist $\lambda _k\rightarrow -\infty$ such that

$$\begin{aligned} \ell =\inf _{x\in \overline{\begin{array}{c} {\Sigma }_{\mu _k}\\ \lambda \end{array}}\in (-\infty ,\lambda _k]}\omega _\lambda (x)<0. \end{aligned}$$

Since u(x) vanishes at infinity, then we have $\omega _\lambda (x)=u_\lambda (x)-u(x)\ge -u(x)\ge \frac{\ell }{2}$ for any $|x|\ge R_k$ and $R_k$ large enough. Hence, we can get

$$\begin{aligned} \omega _{\mu _k}(x^k)=\ell \ \text {for some}\ \mu _k\in [-|x^k|,\lambda _k]\;\text {and}\;|x^k|\le R_k. \end{aligned}$$

On the one hand, similarly as (2.9), we may write $(-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _k}(x^{k}):=I_1+I_2$. We also have $I_1\le 0$, then by similar arguments as in (2.9)-(2.11), we get

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _k}(x^{k}) \le I_2 \le 2c_0C_{n,\alpha }\int _{\Sigma _{\mu _k}}\frac{\omega _{\mu _k}(x^k)}{|x^k-y^{\mu _k}|^{n+\alpha }}dy \le \frac{C\omega _{\mu _k}(x^k)}{|x^k|^\alpha }. \end{aligned}$$

(3.3)

To obtain the last inequality, we notice that $0<u_{\mu _k}(x^k)<u(x^k)$. For $x^k\in \Sigma _{\mu _k}$ and $|x^k|$ sufficiently large, let $\Sigma _{\mu _k}^c=\mathbb {R}^n\backslash \Sigma _{\mu _k},$ choose a point in $\Sigma _{\mu _k}^c:x^m=(3|x^k|+x_1^k,(x^k)')$, then $B_{|x^k|}(x^m)\subset \Sigma _{\mu _k}^c$. There exists a $C>0$ such that

$$\begin{aligned} \int _{\Sigma _{\mu _k}}\frac{1}{|x^k-y^{\mu _k}|^{n+\alpha }}dy&\ge \int _{\Sigma _{\mu _k}^c}\frac{1}{|x^k-y|^{n+\alpha }}dy\\ \nonumber&\ge \int _{B_{|x^k|}(x^m)}\frac{1}{|x^k-y|^{n+\alpha }}dy\sim \frac{C}{|x^k|^\alpha }. \end{aligned}$$

(3.4)

On the other hand, from (2.6), we have

$$\begin{aligned} c(x_k)\omega _{\mu _k}(x_k)-A(x_k)\omega _{\mu _k}(x_{k})\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _k}(x^{k}), \end{aligned}$$

where

$$\begin{aligned} c(x^k)&=\frac{f(x^k,u_{\mu _k}(x^k),\nabla u(x^k))-f(x_0,u(x^k),\nabla u(x^k))}{u_{\mu _k}(x^k)-u(x^k)}\\&\le C(|u_{\mu _k}(x^k)|^s+|u(x^k)|^s)\le C|u(x^k)|^s, \end{aligned}$$

and A(x) is bounded below, then we have

$$\begin{aligned} c(x^k)-A(x^k)&\le C|u(x^k)|^s. \end{aligned}$$

(3.5)

Then from (1.7) and (3.3)-(3.4), we can get

$$\begin{aligned} C|u(x^k)|^s\omega _{\mu _k}(x^k)\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _k}(x^{k})\le \frac{C\omega _{\mu _k}(x^k)}{|x^k|^\alpha }, \end{aligned}$$

that is

$$\begin{aligned} o(1)=C|x^k|^\alpha u^s(x^k)\ge \widetilde{C}_{n,\alpha }, \end{aligned}$$

since $|x^k|\ge |\lambda _k|\rightarrow \infty$. This yields a contradiction and ends the proof of Step 1.

Step 2. Set

$$\begin{aligned} \lambda _0=\sup _{\lambda \le 0}\{\lambda \mid \omega _\mu (x)\ge 0,\forall x\in \Sigma _\mu ,\forall \mu \le \lambda \}. \end{aligned}$$

By the definition of $\lambda _0$ and the continuity of u(x), we have $\omega _{\lambda _0}(x)\ge 0$ for all $x\in \Sigma _{\lambda _0}.$ Then we must have

$$\begin{aligned} \lambda _0=0 \end{aligned}$$

or

$$\begin{aligned} \ \lambda _0<0\ \text {and} \ \omega _{\lambda _0}(x)\equiv 0. \end{aligned}$$

The second case happens only when $\sup _{x\in \Omega }|x_1|=+\infty$. If not, we suppose that $\lambda _0<0$ but $\omega _{\lambda _0}(x)\not \equiv 0$. We claim that there exists $\delta >0$ small enough such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _\lambda ,\forall \lambda \in [\lambda _0,\lambda _0+\delta ). \end{aligned}$$

(3.6)

Indeed, for $x\in \Sigma _{\lambda _0}\cap \Omega ^c$, it is easy to see that $\omega _{\lambda _0}(x)\ge 0$. Then by similar arguments as in the proof of (2.8), we have

$$\begin{aligned} \omega _{\lambda _0}>0,\; x\in \Sigma _{\lambda _0}\cap B_{R_0}(0)\cap \Omega . \end{aligned}$$

It follows that for any positive real number $\sigma$,

$$\begin{aligned} \omega _{\lambda _0}(x)\ge C_\delta >0,\forall x\in \Sigma _{\lambda _0-\sigma }\cap B_{R_0}(0)\cap \Omega , \end{aligned}$$

(3.7)

where $R_0$ is defined in Step 1. Since $\omega _\lambda$ is continuous with respect to $\lambda$, for all sufficiently small $\delta >0$, we have

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _{\lambda _0-\sigma }\cap B_{R_0}(0)\cap \Omega ,\ \forall \lambda \in [\lambda _0,\lambda _0+\delta ). \end{aligned}$$

(3.8)

Suppose that (3.6) is false, then we have for any $\delta >0$,

$$\begin{aligned} \ell =\inf _{x\in \overline{\begin{array}{c} {\Sigma _\lambda }\\ \lambda \end{array}}\in [\lambda _0,\lambda _0+\delta ]}\omega _\lambda (x)<0. \end{aligned}$$

By Step 1, we know that the point of minimum can not be obtained in $B_{R_0}^c$. By using (3.8), $\ell$ can obtain the minimum at some point

$$\begin{aligned} (\mu _k,x_0)\in \{(\lambda ,x)|(\lambda ,x)\in [\lambda _0,\lambda _0+\epsilon ]\times \overline{(\Sigma _\lambda \backslash \Sigma _{{\lambda _0}-\sigma })\cap B_{R_0}(0)}\}. \end{aligned}$$

Obviously we have $\omega _{\lambda _0}\ge \frac{\ell }{2}$ on $\partial ((\Sigma _\lambda \backslash \Sigma _{{\lambda _0}-\delta })\cap B_{R_0})$, thus

$$\begin{aligned} x_0\in (\Sigma _\lambda \backslash \Sigma _{{\lambda _0}-\delta })\cap B_{R_0}(0). \end{aligned}$$

By virtue of similar arguments as in Theorem 1 (2.6)-(2.7), one can see that

$$\begin{aligned} 0\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _0}(x^{0})\le (\frac{C}{{(\delta +\sigma )}^\alpha }+A(x^{0})-c(x^0))\omega _{\mu _0}(x^0)<0, \end{aligned}$$

if $\delta$, $\sigma$ are chosen small enough. This yields a contradiction, hence (3.6) holds, which contradicts with the definition of $\lambda _0$. The strict monotonicity follows from the fact that $\omega _\lambda (x)>0$ in $\Sigma _\lambda \cap \Omega$ for all $\lambda <\lambda _0$.

The proof of Theorem 1.2 is completed.

4 Proof of Theorem 1.3

We first claim that

$$\begin{aligned} u(x)>0,x\in \mathbb {R}^n_+ \ or\ u(x)\equiv 0,x\in \mathbb {R}^n_+. \end{aligned}$$

(4.1)

To prove (4.1), we assume that there exist $x_0\in \mathbb {R}^n_+$ such that $u(x_0)=0$, then by (1.11), we have

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}u(x_{0})+A(x_{0})u(x_{0})=f(x,u,\nabla u)=0. \end{aligned}$$

On the other hand, by the definition (1.2), we have

$$\begin{aligned} (-\Delta )^{\frac{\alpha }{2}}u(x_{0})&= C_{n,\alpha } P.V. \int _{\mathbb {R}^n} \frac{u(x_0)-u(y)}{|x_0-y|^{n+\alpha }} dy\\&=-C_{n,\alpha }P.V.\int _{\mathbb {R}^n_+}\frac{u(y)}{|x_0-y|^{n+\alpha }}dy <0, \end{aligned}$$

where the last inequality follows from the fact that $u(y)\ge 0$ and $u(y)\not \equiv 0$ with $u\in C(\mathbb {R}^n_{+})$. This yields a contradiction, so (4.1) holds. Now we assume that

$$\begin{aligned} u(x)>0,x\in \mathbb {R}^n_+. \end{aligned}$$

(4.2)

We carry out the method of moving planes on the solution u along $x_n$-direction. Let $T_\lambda$ be a hyperplane in $\mathbb {R}^n$ defined by

$$\begin{aligned} T_\lambda =\{x=(x',x_n)\in \mathbb {R}^n|x_n=\lambda ,\lambda >0\}, \end{aligned}$$

where $x'=(x_1,x_2,\cdots ,x_{n-1})$. Let $x^\lambda =(x_1,\cdots ,x_{n-1},2\lambda -x_n)$ be the reflection of x with respect to the hyperplane $T_\lambda$. Set

$$\begin{aligned} \Sigma _\lambda = \{x\in \mathbb {R}^n|0<x_n<\lambda \} \end{aligned}$$

and

$$\begin{aligned} u_\lambda (x)=u(x^\lambda ),\ \omega _\lambda (x)=u_\lambda (x)-u(x),\ \forall x\in \mathbb {R}^n. \end{aligned}$$

We divide the proof into two steps: Step 1. We claim that there exists $\delta > 0$ small enough such that

$$\begin{aligned} \omega _\lambda (x)\ge 0, \,\ \forall x\in \Sigma _\lambda ,\ \forall \lambda \in [0,\delta ]. \end{aligned}$$

(4.3)

Suppose that the claim is not true, then we have for any $\delta >0$ (without loss of generality, we can restrict $\delta \in (0,1)$),

$$\begin{aligned} \ell =\inf _{x\in \overline{\begin{array}{c} {\Sigma _\lambda }\\ \lambda \end{array}}\in [0,\delta ]}\omega _\lambda (x)<0. \end{aligned}$$

(4.4)

First we know that $\omega _\lambda (x)=u_\lambda (x)-u(x)\ge -u(x)$ uniformly with respect to $\lambda$, so by (1.10), we have

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\omega _\lambda (x)\ge \lim _{|x|\rightarrow \infty }-u(x)=0 \end{aligned}$$

uniformly with respect to $\lambda$. Then there must exist $R_0>0$ large enough such that $\omega _{\lambda }(x)\ge \frac{\ell }{2}, x\in B_{R_0}^c(0)$ holds uniformly with respect to $\lambda$. So $\ell$ in (4.4) can obtain the infimum at some point

$$\begin{aligned} (\lambda _0,x_0)\in \{(\lambda ,x)|(\lambda ,x)\in [0,\delta ]\times \overline{(\Sigma _\lambda \backslash \Sigma _0)\cap B_{R_0}(0)}\}. \end{aligned}$$

Notice that $u\equiv 0 \ \text{ in }\ \mathbb {R}^n_{-},\; u\ge 0 \ \text{ in } \ \mathbb {R}^n$ and $\omega _{\lambda _0}(x)\ge \frac{\ell }{2}, x\in B_{R_0}^c(0)$, we have $\omega _{\lambda _0}\ge \frac{\ell }{2}$ on $\partial ((\Sigma _{\lambda _0}\backslash \Sigma _0)\cap B_{R_0}(0))$, so $x_0$ must be in $B_{R_0}(0)\cap (\Sigma _{\lambda _0}\backslash \Sigma _0)$. On the other hand, it is easy to see that $\lambda _0\in (0, \delta ]$. Since $(\lambda _0,x_0)$ is a point of minimum, we can get (a) $u_{x_n}(x_0^{\lambda _0})\le 0$; (b) $\nabla _x\omega (x_0^{\lambda _0}) =0$.

Then by similar arguments as in the proof of Theorem 1 Step 1 (2.6)-(2.7), we get

$$\begin{aligned} 0&\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0)\nonumber \\&\le (\frac{C}{\delta ^\alpha }+A(x_{0})-c(x_0))\omega _{\lambda _0}(x_0)<0, \end{aligned}$$

(4.5)

if we choose $\delta$ small enough. This yields a contradiction, so (4.3) is true.

$\mathbf{Step}\ 2.$ Set

$$\begin{aligned} \lambda _0=\sup \{\lambda >0|\omega _\mu (x)\ge 0,\forall x\in \Sigma _\mu ,\forall \mu \le \lambda \}. \end{aligned}$$

We claim that

$$\begin{aligned} \lambda _0=\infty . \end{aligned}$$

(4.6)

Otherwise, if $\lambda _0<\infty$, then by the definition of $\lambda _0$, we have

$$\begin{aligned} \omega _{\lambda _0}>0,x\in \Sigma _{\lambda _0}\ or\ \omega _{\lambda _0}\equiv 0,x\in \Sigma _{\lambda _0}. \end{aligned}$$

(4.7)

To prove (4.7), we assume there exists $x_0\in \Sigma _{\lambda _0}$, such that $\omega _{\lambda _0}(x_0)=0$. Then by using (2.6), we have

$$\begin{aligned}&(-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0)\ge 0. \end{aligned}$$

On the other hand, by using Lemma 2.1 and $\omega _{\lambda _0}\ge 0$, we can write $(-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})=I_1+I_2$. The fact $\omega _{\lambda _0}(x_0)=0$ gives $I_2=0$. We have

$$\begin{aligned}&(-\Delta )^{\frac{\alpha }{2}}\omega _{\lambda _{0}}(x_{0})+A(x_0)\omega _{\lambda _{0}}(x_{0})-c(x_0)\omega _{\lambda _0}(x_0)\\&\quad \le C_{n,\alpha }P.V.\int _{\Sigma _{\lambda }}\Big (\frac{1}{|x-y|^{n+\alpha }}-\frac{1}{|x-y^{\lambda }|^{n+\alpha }}\big ) \Big (\omega _{\lambda _{0}}(x_{0})-\omega _{\lambda _{0}}(y)\Big )dy <0. \end{aligned}$$

Since $\omega _{\lambda _0}(y)\not \equiv 0$, it yields a contradiction. So (4.7) holds.

Now we suppose that

$$\begin{aligned} \omega _{\lambda _0}(x)>0,\; x\in \Sigma _{\lambda _0}. \end{aligned}$$

(4.8)

We first claim that there exists $\epsilon >0$ small enough such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \Sigma _\lambda ,\forall \lambda \in [\lambda _0,\lambda _0+\epsilon ). \end{aligned}$$

(4.9)

Indeed, since $\omega _\lambda (x)$ is lower semi-continuous in $\Omega$, it follows by (4.8) that for any $\delta >0$ and $R>0$,

$$\begin{aligned} \omega _{\lambda _0}(x)\ge C_\delta >0,\forall x\in \overline{\Sigma _{\lambda _0-\delta }\cap B_R(0)}. \end{aligned}$$

By the continuity of $\omega _\lambda$ with respect to $\lambda$, there exist $\epsilon >0$ such that

$$\begin{aligned} \omega _\lambda (x)\ge 0,\forall x\in \overline{\Sigma _{\lambda _0-\delta }\cap B_R(0)},\forall \lambda \in [\lambda _0,\lambda _0+\epsilon ). \end{aligned}$$

(4.10)

Suppose (4.9) is not true, then for any $\epsilon >0$ we have

$$\begin{aligned} \ell =\inf _{x\in \overline{\begin{array}{c} {\Sigma _\lambda }\\ \lambda \end{array}}\in [\lambda _0,\lambda _0+\epsilon ]}\omega _\lambda (x)<0. \end{aligned}$$

Since $\omega _\lambda (x)=u_\lambda (x)-u(x)\ge -u(x)$, so by (1.10), we have

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\omega _\lambda (x)\ge \lim _{|x|\rightarrow \infty }-u(x)=0. \end{aligned}$$

So there must exist $R_0>0$ large enough such that

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\omega _{\lambda _0}(x)\ge 0, x\in B_{R_0}^c(0)\cap \Sigma _\lambda . \end{aligned}$$

By virtue of (4.10), $\ell$ can obtain the infimum at some point

$$\begin{aligned} (\mu _k,x_0)\in \{(\lambda ,x)|(\lambda ,x)\in [\lambda _0,\lambda _0+\epsilon ]\times \overline{(\Sigma _\lambda \backslash \Sigma _{{\lambda _0}-\delta })\cap B_{R_0}(0)}\}.\end{aligned}$$

By the same arguments as in Theorem 1 (2.6)-(2.7), it is easy to see that

$$\begin{aligned} 0\le (-\Delta )^{\frac{\alpha }{2}}\omega _{\mu _k}(x_0)-c(x_0)\omega _{\mu _k}(x_0)\le \Big (\frac{C}{(\delta +\epsilon )^\alpha }+A(x_{0})-c(x_0)\Big )\omega _{\mu _k}(x_0)<0, \end{aligned}$$

if $\delta$, $\epsilon$ are chosen small enough. This yields a contradiction and proves claim (4.9). However, this contradicts with the definition of $\lambda _0$. So (4.8) is not true. Therefore, we must have

$$\begin{aligned} \omega _{\lambda _0}\equiv 0,\;&x\in \Sigma _{\lambda _0}, \end{aligned}$$

or equivalently,

$$\begin{aligned} u_{\lambda _0}(x)=u(x),\;&x\in \Sigma _{\lambda _0}. \end{aligned}$$

If we choose the point $\bar{x}=(x_1,x_2,\cdots ,x_{n-1},0)$ in the hyperplane $\{x_n=0\}$, then $\bar{x}^{\lambda }\in \mathbb R^{n}_+$ and

$$\begin{aligned} u(\bar{x}^{\lambda })=u(x_1,x_2,\cdots ,x_{n-1},2\lambda _0)=u(x_1,x_2,\cdots ,x_{n-1},0)=0. \end{aligned}$$

This contradict with the assumption (4.2).

Therefore, we have proved Claim (4.6): $\lambda _0=\infty$. Consequently, the solution u(x) is monotonically increasing with respect to $x_n$. Recall that the condition (1.10) tells us $\lim _{|x|\rightarrow \infty }u(x)=0$. So Claim (4.6) is not true, thus $u(x)\equiv 0, x\in \mathbb {R}^n_+$.

The proof of Theorem 1.3 is completed. $\square$