Symmetry and Monotonicity of Solutions to Mixed Local and Nonlocal Weighted Elliptic Equations

Abstract

This paper focuses on the symmetry and monotonicity of non-negative solutions to a mixed local and nonlocal weighted elliptic problem. This problem generalizes the ground-state representation of elliptic equations with the Hardy potential. The novelty of this research lies in developing the moving planes method for mixed local and nonlocal equations with a weighted function, thus clarifying the influence of the weighted function on the solution properties.

1. Introduction

The mixed local and nonlocal operators $\mathcal{L}=-\Delta +{(-\Delta )}^s$ emerge from the superposition of two stochastic processes with different scales; that is, the random walk and jump process [1,2], which is studied by many mathematicians. More precisely, Biagi et al. [3] established some qualitative property of solutions to the following mixed local and nonlocal elliptic problem:

$$\left\{\begin{array}{cc}-\Delta u\left(x\right)+{(-\Delta )}^{s}u\left(x\right)=f\left(x\right),\hfill & x\in \Omega ,\hfill \\ u\left(x\right)⩾0,\hfill & x\in \Omega ,\hfill \\ u\left(x\right)=0,\hfill & x\in {\mathbb{R}}^N\setminus \Omega .\hfill \end{array}\right.$$

Biagi et al. [4] ([Theorem 1.1]) obtained the symmetry properties of solutions to the following mixed operators by the moving planes method:

$$\left\{\begin{array}{cc}-\Delta u\left(x\right)+{(-\Delta )}^{s}u\left(x\right)=f\left(u\right),\hfill & x\in \Omega ,\hfill \\ u\left(x\right)⩾0,\hfill & x\in \Omega ,\hfill \\ u\left(x\right)=0,\hfill & x\in {\mathbb{R}}^N\setminus \Omega .\hfill \end{array}\right.$$

(1)

where f is a locally Lipschitz continuous function, and $\Omega \subset {\mathbb{R}}^N$ satisfies certain symmetry and convexity assumptions. Furthermore, [4] ([Theorem 1.2]) established the symmetry of the global solutions inspired by a Gibbons conjecture. Daiji et al. [5] ([Theorem 1.1]) extended the results of [4] ([Theorem 1.1]) to singular solutions to problem (1) with a singular set. Recently, Dipierro et al. [6,7,8] considered the existence of solutions to the superposition of a continuum of operators. An interesting feature of the superposition operators considered in [6,7,8] is that the type of operators may not only possibly involve uncountably many fractional operators but also that some of these operators may have the wrong sign. Some other results of mixed local and nonlocal operators can be found in [9,10,11,12,13,14,15,16,17] and the references therein.

It is well known that the Hardy potential has an important influence on the properties of solutions to elliptic equations. Compared to elliptic equations without a Hardy potential, the main difference is that the solutions to elliptic equations involving the Hardy potential must be unbounded, even if the right-hand side data belong to $L^{\infty }(\Omega )$. Furthermore, this critical value ${\Lambda }_{N,s}$, which is the sharp constant of the Hardy–Sobolev inequality, has an important effect on solvability. It is worth pointing out that the operators ${(-\Delta )}_{\alpha }^{s}u$ defined by (4) below appear in a natural way when dealing with the following problems:

$$\left\{\begin{array}{cc}{(-\Delta )}^{s}u\left(x\right)-\lambda {\displaystyle \frac{u\left(x\right)}{{\left|x\right|}^{2s}}}=f(x,u),\hfill & x\in \Omega ,\hfill \\ u\left(x\right)>0,\hfill & x\in \Omega ,\hfill \\ u\left(x\right)=0,\hfill & x\in {\mathbb{R}}^N\setminus \Omega ,\hfill \end{array}\right.$$

(2)

where $0<\lambda <{\Lambda }_{N,s}$. More precisely, by the ground state representation [18] ([Proposition 4.1]), it is well known that, for $\gamma \in (0,{\displaystyle \frac{N-2s}{2}})$, let $u\in C_0^{\infty }\left({\mathbb{R}}^N\right)$ and $v\left(x\right)={\left|x\right|}^{\gamma }u\left(x\right)$, then

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{{\mathbb{R}}^N}{\left|\xi \right|}^{2s}|\widehat{u}{|}^{2}d\xi -({\Lambda }_{N,s}+{\Phi }_{N,s}\left(\gamma \right)){\mathit{\int }}_{{\mathbb{R}}^N}{\left|x\right|}^{-2s}{\left|u\left(x\right)\right|}^{2}dx\hfill \\ \hfill =& a_{N,s}{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^N\times {\mathbb{R}}^N}{\displaystyle \frac{{\left|v\right(x)-v(y\left)\right|}^2}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dx}{{\left|x\right|}^{\gamma }}}{\displaystyle \frac{dy}{{\left|y\right|}^{\gamma }}},\hfill \end{array}$$

where

$${\Phi }_{N,s}\left(\gamma \right)=2^{2s}\left({\displaystyle \frac{\Gamma \left(\frac{\gamma +2s}{2}\right)\Gamma \left(\frac{N-\gamma }{2}\right)}{\Gamma \left(\frac{N-\gamma -2s}{2}\right)\Gamma \left(\frac{\gamma }{2}\right)}}-{\displaystyle \frac{{\Gamma }^2\left(\frac{N+2s}{4}\right)}{{\Gamma }^2\left(\frac{N-2s}{4}\right)}}\right).$$

$\Gamma$ is the gamma function. Therefore, if u is the solution to problem (2), then $v\left(x\right)={\left|x\right|}^{\gamma }u\left(x\right)$ satisfies

$$\left\{\begin{array}{cc}{L}_{\gamma }v\left(x\right)={\left|x\right|}^{-\gamma }{f(x,|x|}^{-\gamma }v\left(x\right)),\hfill & x\in \Omega ,\hfill \\ v\left(x\right)>0,\hfill & x\in \Omega ,\hfill \\ v\left(x\right)=0,\hfill & x\in {\mathbb{R}}^N\setminus \Omega ,\hfill \end{array}\right.$$

where

$$L_{\gamma }v\left(x\right)=a_{N,s}\mathrm{P}.\mathrm{V}.{\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{v\left(x\right)-v\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dy}{{\left|x\right|}^{\gamma }{\left|y\right|}^{\gamma }}}.$$

Similarly, the operators $-\mathrm{div}\left(\right|x{|}^{-2\alpha }\nabla u)$ appear when dealing with Laplace problems with a Hardy potential. Thus, it is of interest to deeply study the symmetry of solutions to mixed local and nonlocal weighted elliptic problems.

In this paper, we establish the symmetry of solutions to the following mixed local and nonlocal weighted semilinear elliptic problem:

$$\left\{\begin{array}{cc}\hfill -{\mathrm{div}\left(\right|x|}^{-2\alpha }{\nabla u)+(-\Delta )}_{\alpha }^{s}u={\displaystyle \frac{f\left(u\right)}{{\left|x\right|}^{2\alpha }}},& x\in \Omega ,\hfill \\ \hfill u\left(x\right)⩾0,& x\in \Omega ,\hfill \\ \hfill u\left(x\right)=0,& x\in {\mathbb{R}}^N\setminus \Omega ,\hfill \end{array}\right.$$

(3)

where $s\in (0,1),\alpha \in [0,{\displaystyle \frac{N-2s}{2}})$, $0\in \Omega \subset {\mathbb{R}}^N(N⩾3)$ is an open bounded set with a $C^1$ boundary, which is symmetric and convex with respect to the hyperplane $\{x_1=0\}$, and the weighted fractional Laplacian ${(-\Delta )}_{\alpha }^s$ is defined by

$$\begin{array}{c}\hfill {(-\Delta )}_{\alpha }^{s}u=C_{N,s}\mathrm{P}.\mathrm{V}.{\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{u\left(x\right)-u\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dy}{{\left|y\right|}^{\alpha }{\left|x\right|}^{\alpha }}},\end{array}$$

(4)

where $C_{N,s}$ is given by

$$\begin{array}{c}\hfill C_{N,s}={\left({\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{1-cos{\xi }_1}{{\left|\xi \right|}^{N+2s}}}d\xi \right)}^{-1}.\end{array}$$

Our main result is as follows.

Theorem 1. 

Assume that $N⩾3$, $s\in (0,1),\alpha \in [0,{\displaystyle \frac{N-2s}{2}})$. Let $f:\mathbb{R}\to \mathbb{R}$ be a locally Lipschitz continuous function, $\Omega \subset {\mathbb{R}}^N$ be an open bounded set with $C^1$ boundary, which is symmetric and convex with respect to the hyperplane $\{x_1=0\}$. Then, the non-zero weak solution $u\in C\left({\mathbb{R}}^N\right)$ to problem (3) is symmetric about the hyperplane $\{x_1=0\}$. Moreover, u is strictly increasing in the direction within the set $\Omega \cap \{x_1<0\}$.

Remark 1. 

On the one hand, we generalize the corresponding symmetry results of mixed local and nonlocal equations without the weighted function established by Biagi et al. [4] ([Theorem 1.1]). On the other hand, we develop the moving planes method for mixed local and nonlocal equations with a weighted function.

This paper is organized as follows: Section 2 gathers some definitions and lemmas. The proof of Theorem 1 is given in Section 3. Section 4 presents a comprehensive summary of the principal findings and outcomes achieved in this paper.

2. Preliminaries

In this section, we introduce some preliminaries to prove Theorem 1.

2.1. Some Definitions

Without loss of generality, we may assume that

$$\begin{array}{c}\hfill \underset{x\in \Omega }{inf}{x}_1=-a(a>0).\end{array}$$

(5)

Definition 1. 

Let $s\in (0,1)$ define the mixing local and nonlocal Sobolev space

$$\begin{array}{c}\hfill {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)=\{u\in L^2({\mathbb{R}}^N{,\left|x\right|}^{2\alpha })\mathit{is}\mathit{measurable}:{\left[u\right]}_{{\mathcal{H}}_{\alpha }^s}<+\infty \},\end{array}$$

where

$$\begin{array}{c}\hfill {\left[u\right]}_{{\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)}={\left({\mathit{\int }}_{{\mathbb{R}}^N}{|\nabla u\left(x\right)|}^2{\displaystyle \frac{dx}{{\left|x\right|}^{2\alpha }}}\right)}^{{\displaystyle \frac{1}{2}}}+{\left({\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\right)}^{{\displaystyle \frac{1}{2}}}.\end{array}$$

(6)

Define

$$\begin{array}{c}\hfill {\mathcal{H}}_{\alpha }^s(\Omega )=\{u\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right):u\equiv 0,x\in {\mathbb{R}}^N\setminus \Omega \}.\end{array}$$

(7)

Now we give the definition of weak solutions to problem (3).

Definition 2. 

We say that $u\in {\mathcal{H}}_{\alpha }^s(\Omega )$ is a weak solution to problem (3) if

1. 

$u>0a.e.x\in \Omega$.

2. 

For any $\phi \in {\mathcal{H}}_{\alpha }^s(\Omega )$,

$$\begin{array}{c}\hfill \mathcal{B}(u,\phi )={\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{f\left(u\right)}{{\left|x\right|}^{2\alpha }}}\phi dx,\end{array}$$

(8)

where

$$\begin{array}{c}\hfill \mathcal{B}(u,\phi )={\mathit{\int }}_{\Omega }{\left|x\right|}^{-2\alpha }\nabla u·\nabla \phi dx+{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{\left(u\right(x)-u(y\left)\right)\left(\phi \right(x)-\phi (y\left)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}.\end{array}$$

(9)

Let $H\subset {\mathbb{R}}^N$ be an open and affine half space. We denote by $Q:{\mathbb{R}}^N\to {\mathbb{R}}^N$ the reflection with respect to $\partial H$ and set

$$\begin{array}{c}\hfill \overline{x}=Q\left(x\right),x\in {\mathbb{R}}^N.\end{array}$$

(10)

Definition 3. 

A function $v:{\mathbb{R}}^N\to \mathbb{R}$ is antisymmetric with respect to Q if

$$\begin{array}{c}\hfill v\left(\overline{x}\right)=-v\left(x\right),x\in {\mathbb{R}}^N.\end{array}$$

(11)

2.2. Some Useful Lemmas

The main tool employed in our analysis is the moving planes method introduced by Sciunzi [19,20]. By carefully applying this method to the mixed local and nonlocal equations with a weighted function, we are able to prove the symmetry and monotonicity of the solutions. The moving planes method is a powerful and influential technique in the field of partial differential equations, primarily utilized to explore the symmetry and monotonicity properties of solutions.

The following is the process of applying the moving plane method to the solution to problem (3). Let $u\in C(\Omega )$ be a weak solution to problem (3). For every $\lambda \in (-a,a)$ (where a is given by (5)), define the $x_{\lambda }$ as the symmetry of x with respect to hyperplane $T_{\lambda }=\{x_1=\lambda \}$, so that $x_{\lambda }=(2\lambda -x_1,x_2,\cdots x_N)$ and

$$\begin{array}{c}\hfill u_{\lambda }\left(x\right)=u\left(x_{\lambda }\right).\end{array}$$

(12)

According to problem (3), we find that $u_{\lambda }$ satisfies

$$\begin{array}{c}\hfill -\mathrm{div}\left(\right|x_{\lambda }{|}^{-2\alpha }\nabla u_{\lambda })+{\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{u_{\lambda }\left(x\right)-u_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dy}{|y_{\lambda }{|}^{\alpha }{\left|x_{\lambda }\right|}^{\alpha }}}={\displaystyle \frac{f\left(u_{\lambda }\right)}{|x_{\lambda }{|}^{2\alpha }}},x\in \Omega .\end{array}$$

(13)

Combining (3) and (13), we obtain the following equation

$$\begin{array}{cc}& -{\mathrm{div}\left(\right|x|}^{-2\alpha }{\nabla u\left(x\right))+(-\Delta )}_{\alpha }^{s}u\left(x\right)-[-\mathrm{div}(|x_{\lambda }{|}^{-2\alpha }\nabla u_{\lambda }\left(x\right))+{(-\Delta )}_{\alpha }^s{u}_{\lambda }\left(x\right)]\hfill \\ \hfill ⩾& c_0\left(x\right)(u_{\lambda }\left(x\right)-u\left(x\right)),\hfill \end{array}$$

(14)

where

$$\begin{array}{c}\hfill c_0\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{{\left|x\right|}^{-2\alpha }f\left(u_{\lambda }\right)-{\left|x_{\lambda }\right|}^{-2\alpha }f\left(u\right)}{{\left|x\right|}^{-2\alpha }{\left|x_{\lambda }\right|}^{-2\alpha }(u_{\lambda }-u)}},\hfill & \lambda <0,\hfill \\ 0,\hfill & \lambda ⩾0.\hfill \end{array}\right.\end{array}$$

(15)

We remember

$$\begin{array}{c}\hfill {\Sigma }_{\lambda }=\left\{\begin{array}{cc}\{x\in {\mathbb{R}}^N:x_1<\lambda \},\hfill & \lambda <0,\hfill \\ \{x\in {\mathbb{R}}^N:x_1>\lambda \},\hfill & \lambda ⩾0,\hfill \end{array}\right.\end{array}$$

(16)

and

$$\begin{array}{c}\hfill {\Omega }_{\lambda }=\Omega \cap {\Sigma }_{\lambda }.\end{array}$$

(17)

Let $u\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)$ be any non-identically vanishing weak solution to problem (3). For any $\lambda \in (-a,0)$, where a is given by (5), define $w_{\lambda }:{\mathbb{R}}^N\to \mathbb{R}$ as

$$\begin{array}{c}\hfill w_{\lambda }\left(x\right)=\left\{\begin{array}{cc}{(u-u_{\lambda })}^{+}\left(x\right),\hfill & x\in {\Sigma }_{\lambda },\hfill \\ {(u-u_{\lambda })}^{-}\left(x\right),\hfill & x\in {\mathbb{R}}^N\setminus {\Sigma }_{\lambda },\hfill \end{array}\right.\end{array}$$

(18)

where

$$\begin{array}{c}\hfill {(u-u_{\lambda })}^{+}=\max \{u-u_{\lambda },0\},{(u-u_{\lambda })}^{-}=\min \{u-u_{\lambda },0\}.\end{array}$$

It is well known that $u\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)$. Thus, $u_{\lambda }\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)$ and

$$\begin{array}{c}\hfill {(u-u_{\lambda })}_{{\chi }_{\left({\Sigma }_{\lambda }\right)}}^{+}\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right),{(u-u_{\lambda })}_{{\chi }_{({\mathbb{R}}^N\setminus {\Sigma }_{\lambda })}}^{-}\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right).\end{array}$$

where

$$\begin{array}{c}\hfill u_{{\chi }_{(\Sigma )}}=\left\{\begin{array}{cc}u,\hfill & x\in \Sigma ,\hfill \\ 0,\hfill & x\in {\mathbb{R}}^N\setminus \Sigma .\hfill \end{array}\right.\end{array}$$

Consequently,

$$\begin{array}{c}\hfill w_{\lambda }={(u-u_{\lambda })}_{{\chi }_{\left({\Sigma }_{\lambda }\right)}}^{+}+{(u-u_{\lambda })}_{{\chi }_{({\mathbb{R}}^N\setminus {\Sigma }_{\lambda })}}^{-}\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right).\end{array}$$

This fact implies that

$$\begin{array}{c}\hfill {\left|x\right|}^{2\alpha }{w}_{\lambda }\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right),{\left|x_{\lambda }\right|}^{2\alpha }{w}_{\lambda }\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right).\end{array}$$

(19)

Note that $w_{\lambda }\left(x\right)={(0-u_{\lambda }\left(x\right))}^{+}=0$ if $x\in {\Sigma }_{\lambda }\setminus {\Omega }_{\lambda }$ and $w_{\lambda }\left(x_{\lambda }\right)={(u\left(x\right)-u\left(x_{\lambda }\right))}^{-}=0$ if $x_{\lambda }\in {\Sigma }_{\lambda }\setminus {\Omega }_{\lambda }$. Thus,

$$\begin{array}{c}\hfill w_{\lambda }\equiv 0,x\in {\mathbb{R}}^N\setminus ({\Omega }_{\lambda }\cup Q_{\lambda }\left({\Omega }_{\lambda }\right))=({\Sigma }_{\lambda }\setminus {\Omega }_{\lambda })\cup Q_{\lambda }({\Sigma }_{\lambda }\setminus {\Omega }_{\lambda }).\end{array}$$

(20)

Now we give the following lemma.

Lemma 1. 

Under the conditions of Theorem 1, we have

$$\begin{array}{cc}\hfill & {\langle (-\Delta )}_{\alpha }^s{u,\left|x\right|}^{2\alpha }{w}_{\lambda }{\rangle -\langle (-\Delta )}_{\alpha }^s{u}_{\lambda },|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\rangle \hfill \\ \hfill ⩾& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)){\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}.\hfill \end{array}$$

(21)

Proof. 

According to (19) and (20), we have

$$\begin{array}{c}\hfill {\langle (-\Delta )}_{\alpha }^s{u,\left|x\right|}^{2\alpha }{w}_{\lambda }\rangle ={\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\lambda }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\lambda }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}},\end{array}$$

(22)

and

$$\begin{array}{cc}\hfill & \langle {(-\Delta )}_{\alpha }^s{u}_{\lambda },|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\rangle \hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u_{\lambda }\left(x\right)-u_{\lambda }\left(y\right))\left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)-|y_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\lambda }{|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}.\hfill \end{array}$$

(23)

Thus, (22) and (23) lead to

$$\begin{array}{cc}\hfill & {\langle (-\Delta )}_{\alpha }^s{u,\left|x\right|}^{2\alpha }{w}_{\lambda }{\rangle -\langle (-\Delta )}_{\alpha }^s{u}_{\lambda },|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\rangle \hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{\left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))\right]{\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{u_{\lambda }\left(x\right)-u_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}\left[\left({\displaystyle \frac{{\left|x\right|}^{\alpha }}{{\left|y\right|}^{\alpha }}}-{\displaystyle \frac{|x_{\lambda }{|}^{\alpha }}{|y_{\lambda }{|}^{\alpha }}}\right)w_{\lambda }\left(x\right)-\left({\displaystyle \frac{|y_{\lambda }{|}^{\alpha }}{|x_{\lambda }{|}^{\alpha }}}-{\displaystyle \frac{{\left|y\right|}^{\alpha }}{{\left|x\right|}^{\alpha }}}\right)w_{\lambda }\left(y\right)\right]dxdy\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{\left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))\right]{\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)){\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dx}{{\left|x\right|}^{\alpha }}}{\displaystyle \frac{dy}{{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}},\hfill \end{array}$$

(24)

where

$$\begin{array}{c}\hfill g(x,y)=\left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))-(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))\right]{\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)),\end{array}$$

we have used the fact that

$$\begin{array}{c}\hfill |x_{\lambda }-y_{\lambda }{|}^{N+2s}={|x-y|}^{N+2s}\mathrm{and}{\displaystyle \frac{{\left|x\right|}^{\alpha }}{{\left|y\right|}^{\alpha }}}-{\displaystyle \frac{|x_{\lambda }{|}^{\alpha }}{|y_{\lambda }{|}^{\alpha }}}=0.\end{array}$$

Now we show that

$$\begin{array}{c}\hfill {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}⩾0.\end{array}$$

(25)

This fact, together with (24) shows that (21) holds.

In order to show that (25) holds, use the decomposition

$$\begin{array}{c}\hfill {\mathbb{R}}^N\times {\mathbb{R}}^N=(S_{\lambda }\cup S_{\lambda }^c\cup D_{\lambda }\cup D_{\lambda }^c)\times (S_{\lambda }\cup S_{\lambda }^c\cup D_{\lambda }\cup D_{\lambda }^c),\end{array}$$

where

$$\begin{array}{cc}\hfill S_{\lambda }& =\mathrm{supp}{w}_{\lambda }\left(x\right)\cap {\Sigma }_{\lambda },S_{\lambda }^c={\Sigma }_{\lambda }\setminus S_{\lambda },\hfill \\ \hfill D_{\lambda }& =\mathrm{supp}{w}_{\lambda }\left(x\right)\cap ({\mathbb{R}}^N\setminus {\Sigma }_{\lambda }),D_{\lambda }^c=({\mathbb{R}}^N\setminus {\Sigma }_{\lambda })\setminus D_{\lambda }.\hfill \end{array}$$

By the definition of $w_{\lambda }\left(x\right)$ (see (18) for more details), we find that $w_{\lambda }\left(x\right)=0$ for $x\in S_{\lambda }^c\cup D_{\lambda }^c$ and $w_{\lambda }\left(y\right)=u\left(y\right)-u_{\lambda }\left(y\right)$ for $y\in S_{\lambda }\cup D_{\lambda }$. Thus, for any $(x,y)\in (S_{\lambda }^c\cup D_{\lambda }^c)\times (S_{\lambda }\cup D_{\lambda })$,

$$\begin{array}{cc}\hfill & g(x,y)\hfill \\ \hfill =& \left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))-(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))\right]{\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))\hfill \\ \hfill =& \left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))+w_{\lambda }\left(y\right)\right]{(-|y|}^{2\alpha }{w}_{\lambda }\left(y\right))\hfill \\ \hfill =& -\left[u\left(x\right)-u_{\lambda }\left(x\right)\right]{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right).\hfill \end{array}$$

(26)

It is easily seen that $u\left(x\right)-u_{\lambda }\left(x\right)⩽0$ if $x\in S_{\lambda }^c$ and ${\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)⩾0$ if $y\in S_{\lambda }$, then $g(x,y)⩾0,(x,y)\in S_{\lambda }^c\times S_{\lambda }$.

Therefore, using the fact that $D_{\lambda }$ is the reflection of $S_{\lambda }$ and $w_{\lambda }\left(y_{\lambda }\right)=-w_{\lambda }\left(y\right)$, we have

$$\begin{array}{cc}\hfill & {\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}+{\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y_{\lambda }\right|}^{2\alpha }{w}_{\lambda }\left(y_{\lambda }\right)}{|x-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y_{\lambda }\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{|x-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& -{\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}\left[u\left(x\right)-u_{\lambda }\left(x\right)\right]w_{\lambda }\left(y\right)\left[{\displaystyle \frac{{\left|y\right|}^{2\alpha }}{{|x-y|}^{N+2s}{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}-{\displaystyle \frac{|y_{\lambda }{|}^{2\alpha }}{|x-y_{\lambda }{|}^{N+2s}{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\right]dxdy\hfill \\ \hfill =& -{\mathit{\int }\mathit{\int }}_{S_{\lambda }^c\times S_{\lambda }}\left[u\left(x\right)-u_{\lambda }\left(x\right)\right]w_{\lambda }\left(y\right)\left[{\displaystyle \frac{{\left|y\right|}^{\alpha }}{{|x-y|}^{N+2s}{\left|x\right|}^{\alpha }}}-{\displaystyle \frac{|y_{\lambda }{|}^{\alpha }}{|x-y_{\lambda }{|}^{N+2s}{\left|x\right|}^{\alpha }}}\right]dxdy\hfill \\ \hfill ⩾& 0,\hfill \end{array}$$

(27)

where we have used that fact that

$$\begin{array}{c}\hfill {\displaystyle \frac{{\left|y\right|}^{\alpha }}{{|x-y|}^{N+2s}}}-{\displaystyle \frac{|y_{\lambda }{|}^{\alpha }}{|x-y_{\lambda }{|}^{N+2s}}}⩾0,\end{array}$$

since $\left|y\right|⩾|y_{\lambda }|$ and $|x-y|⩽|x-y_{\lambda }|$ if $(x,y)\in S_{\lambda }^c\times S_{\lambda }$. Now we consider $(x,y)\in (D_{\lambda }^c\times D_{\lambda })\cup (D_{\lambda }^c\times S_{\lambda })$. It is obvious that $u\left(x\right)-u_{\lambda }\left(x\right)⩾0$ if $x\in D_{\lambda }^c$ and ${\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)\le 0$ if $y\in D_{\lambda }$. Furthermore, for $\alpha \in [0,{\displaystyle \frac{N-2s}{2}})$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{{\left|y\right|}^{\alpha }}{{|x-y|}^{N+2s}}}-{\displaystyle \frac{|y_{\lambda }{|}^{\alpha }}{|x-y_{\lambda }{|}^{N+2s}}}⩾0,\end{array}$$

since ${\displaystyle \frac{|y_{\lambda }|}{\left|y\right|}}={\displaystyle \frac{|x-y_{\lambda }|}{|x-y|}}$ and $|y_{\lambda }|⩾|y|$ for $(x,y)\in D_{\lambda }^c\times D_{\lambda }$.

Therefore, using $D_{\lambda }$ as the reflection of $S_{\lambda }$ and $w_{\lambda }\left(y_{\lambda }\right)=-w_{\lambda }\left(y\right)$ again, we have

$$\begin{array}{cc}\hfill & {\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}+{\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times S_{\lambda }}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y_{\lambda }\right|}^{2\alpha }{w}_{\lambda }\left(y_{\lambda }\right)}{|x-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{-(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}{\displaystyle \frac{(u\left(x\right)-u_{\lambda }\left(x\right)){\left|y_{\lambda }\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{|x-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& -{\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}\left[(u\left(x\right)-u_{\lambda }\left(x\right))w_{\lambda }\left(y\right)\right]\left[{\displaystyle \frac{{\left|y\right|}^{2\alpha }}{{|x-y|}^{N+2s}{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}-{\displaystyle \frac{|y_{\lambda }{|}^{2\alpha }}{|x-y_{\lambda }{|}^{N+2s}{\left|x\right|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\right]dxdy\hfill \\ \hfill =& -{\mathit{\int }\mathit{\int }}_{D_{\lambda }^c\times D_{\lambda }}\left[(u\left(x\right)-u_{\lambda }\left(x\right))w_{\lambda }\left(y\right)\right]\left[{\displaystyle \frac{{\left|y\right|}^{\alpha }}{{|x-y|}^{N+2s}{\left|x\right|}^{\alpha }}}-{\displaystyle \frac{|y_{\lambda }{|}^{\alpha }}{|x-y_{\lambda }{|}^{N+2s}{\left|x\right|}^{\alpha }}}\right]dxdy\hfill \\ \hfill ⩾& 0.\hfill \end{array}$$

(28)

Similarly, for $(x,y)\in (S_{\lambda }\cup D_{\lambda })\times (S_{\lambda }^c\cup D_{\lambda }^c)$,

$$\begin{array}{cc}\hfill & g(x,y)\hfill \\ \hfill =& \left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))-(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))\right]{\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))\hfill \\ \hfill =& \left[(u\left(x\right)-u_{\lambda }\left(x\right))-(u\left(y\right)-u_{\lambda }\left(y\right))-w_{\lambda }\left(x\right)\right]{\left|x\right|}^{2\alpha }{w}_{\lambda }\left(x\right)\hfill \\ \hfill =& -\left[u\left(y\right)-u_{\lambda }\left(y\right)\right]{\left|x\right|}^{2\alpha }{w}_{\lambda }\left(x\right).\hfill \end{array}$$

Analysis similar to that in the proof of (27) and (28) shows

$$\begin{array}{c}\hfill {\mathit{\int }\mathit{\int }}_{S_{\lambda }\times S_{\lambda }^c}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}+{\mathit{\int }\mathit{\int }}_{D_{\lambda }\times S_{\lambda }^c}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}⩾0,\end{array}$$

(29)

and

$$\begin{array}{c}\hfill {\mathit{\int }\mathit{\int }}_{S_{\lambda }\times D_{\lambda }^c}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}+{\mathit{\int }\mathit{\int }}_{D_{\lambda }\times D_{\lambda }^c}{\displaystyle \frac{g(x,y)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}⩾0.\end{array}$$

(30)

Furthermore,

$$\begin{array}{cc}\hfill g(x,y)& =0,\mathrm{elsewhere}.\hfill \end{array}$$

(31)

Combining (27)–(31) yields (25) holds. □

Lemma 2. 

Under the conditions of Lemma 1, we have

$$\begin{array}{c}\hfill {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)){\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}⩾0.\end{array}$$

(32)

Proof. 

Obviously,

$$\begin{array}{cc}\hfill & {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)){\left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}h(x,y){\displaystyle \frac{{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))}^2}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}},\hfill \end{array}$$

where

$$\begin{array}{c}\hfill h(x,y)={\displaystyle \frac{{\left|x\right|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|y\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)}}.\end{array}$$

An easy computation shows that

$$\begin{array}{cc}\hfill h(x,y)& =0,(x,y)\in (S_{\lambda }^c\cup D_{\lambda }^c)\times (S_{\lambda }^c\cup D_{\lambda }^c),\hfill \\ \hfill h(x,y)& ={\left|x\right|}^{2\alpha }⩾0,(x,y)\in (S_{\lambda }\cup D_{\lambda })\times (S_{\lambda }^c\cup D_{\lambda }^c),\hfill \\ \hfill h(x,y)& ={\left|y\right|}^{2\alpha }⩾0,(x,y)\in (S_{\lambda }^c\cup D_{\lambda }^c)\times (S_{\lambda }\cup D_{\lambda }).\hfill \end{array}$$

(33)

For $(x,y)\in (S_{\lambda }\cup D_{\lambda })\times (S_{\lambda }\cup D_{\lambda })$ with $\left|x\right|⩾\left|y\right|$, we have

$$\begin{array}{c}\hfill h(x,y)⩾{\displaystyle \frac{{\left|x\right|}^{2\alpha }{w}_{\lambda }\left(x\right)-{\left|x\right|}^{2\alpha }{w}_{\lambda }\left(y\right)}{w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right)}}={\left|x\right|}^{2\alpha }⩾0.\end{array}$$

(34)

The above inequality holds also for $(x,y)\in (S_{\lambda }\cup D_{\lambda })\times (S_{\lambda }\cup D_{\lambda })$ with $\left|x\right|⩽\left|y\right|$ by symmetry. Combining (33) and (34) gives (32) holds. □

Lemma 3. 

Suppose that $H\subset {\mathbb{R}}^N$ is an open and affine half space, $U^{\prime }\subset {\mathbb{R}}^N$ is an open set satisfying $Q\left(U^{\prime }\right)=U^{\prime }$, $U\subset H$ is open and bounded set, and $\overline{{\Omega }_{\lambda }}\subset H\cap U^{\prime }$. Let $u_{\lambda }-u\in {\mathcal{H}}_{\alpha }^s\left(U^{\prime }\right)$ be an antisymmetric function such that

$$\begin{array}{c}\hfill u_{\lambda }-u⩾0,x\in H\setminus U.\end{array}$$

(35)

Then

$$\begin{array}{cc}\hfill I:=& {\mathit{\int }}_{{\Omega }_{\lambda }}{\left|x\right|}^{-2\alpha }{|\nabla w_{\lambda }|}^{2}dx+{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))}^2}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & +{\mathit{\int }}_{{\Omega }_{\lambda }}{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\lambda }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\lambda }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & -{\mathit{\int }}_{{\Omega }_{\lambda }}|x_{\lambda }{|}^{-2\alpha }\nabla u_{\lambda }·\nabla \left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill & -{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u_{\lambda }\left(x\right)-u_{\lambda }\left(y\right))\left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)-|y_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(y\right))}{|x_{\lambda }-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\lambda }{|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill ⩾& 0.\hfill \end{array}$$

(36)

Proof. 

Thanks to the definition of $w_{\lambda }$, it is evident that

$$\begin{array}{c}\hfill \nabla u·{\nabla ln\left|x\right|}^{2\alpha }-\nabla u_{\lambda }·\nabla ln{\left|x_{\lambda }\right|}^{2\alpha }=0,\end{array}$$

which implies that

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{{\Omega }_{\lambda }}{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda })dx-{\mathit{\int }}_{{\Omega }_{\lambda }}|x_{\lambda }{|}^{-2\alpha }\nabla u_{\lambda }·\nabla \left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }}\nabla u·\nabla w_{\lambda }+w_{\lambda }\nabla u·{\nabla ln\left|x\right|}^{2\alpha }dx-{\mathit{\int }}_{{\Omega }_{\lambda }}\nabla u_{\lambda }·\nabla w_{\lambda }+w_{\lambda }\nabla u_{\lambda }·\nabla ln{\left|x_{\lambda }\right|}^{2\alpha }dx\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }}\nabla u_{\lambda }·\nabla w_{\lambda }dx-{\mathit{\int }}_{{\Omega }_{\lambda }}\nabla u·\nabla w_{\lambda }dx\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }}\nabla (u-u_{\lambda })·\nabla w_{\lambda }dx,\hfill \end{array}$$

(37)

the Lemma 1 and 2 are applied to complete the proof. □

Inspired by [4], now we consider the first mixed eigenvalue of the operator $-{\mathrm{div}\left(\right|x|}^{-2\alpha }{\nabla u)+(-\Delta )}_{\alpha }^{s}u$, which is defined as

$$\begin{array}{c}\hfill {\Lambda }_1(\Omega )=\underset{u\in {\mathcal{H}}_{\alpha }^s(\Omega )}{inf}{\displaystyle \frac{\mathcal{B}(u,u)}{{\left|\right|u\left|\right|}_{L^2(\Omega )}^2}},\end{array}$$

(38)

where $\mathcal{B}$ is defined by (9), we see that

$$\begin{array}{c}\hfill {\Lambda }_1(\Omega )⩾{\Lambda }_{{(-\mathrm{div}(\left|x\right|}^{-2\alpha }\nabla u\left)\right)}(\Omega ),\end{array}$$

(39)

where ${\Lambda }_{{(-\mathrm{div}(\left|x\right|}^{-2\alpha }\nabla u\left)\right)}(\Omega )$ stands for the first eigenvalue of $-\mathrm{div}\left(\right|x{|}^{-2\alpha }\nabla u)$ in $\Omega$ with homogeneous Dirichlet boundary conditions. Define

$$\begin{array}{c}\hfill {\Lambda }_1\left(r\right)=\{{\Lambda }_1(\Omega ):\Omega \subset {\mathbb{R}}^N\mathrm{with}|\Omega |=r\},r>0.\end{array}$$

Recalling that

$$\begin{array}{c}\hfill {\Lambda }_{{(-\mathrm{div}(\left|x\right|}^{-2\alpha }\nabla u\left)\right)}(\Omega )\to +\infty \mathrm{as}|\Omega |\to 0,\end{array}$$

which, together with (39), gives

$$\begin{array}{c}\hfill {\Lambda }_1\left(r\right)\to +\infty \mathrm{as}r\to 0^{+}.\end{array}$$

(40)

Lemma 4. 

Let ${\Omega }_{\lambda }\subset {\mathbb{R}}^N$ be an open and bounded set with $\overline{{\Omega }_{\lambda }}\subset H$ and

$$\begin{array}{c}\hfill c\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(u_{\lambda }\right)-f\left(u\right)}{u_{\lambda }-u}},\hfill & \lambda <0,\hfill \\ 0,\hfill & \lambda ⩾0.\hfill \end{array}\right.\end{array}$$

Let λ be small enough such that

$$\begin{array}{c}\hfill \parallel c^{-}{\parallel }_{L^{\infty }}\left({\Omega }_{\lambda }\right)>{\Lambda }_1\left({\Omega }_{\lambda }\right),\end{array}$$

(41)

where $c^{-}\left(x\right)=max\{-c\left(x\right),0\}$ denotes the negative part of a function $c\left(x\right)$, and ${\Lambda }_1\left({\Omega }_{\lambda }\right)$ is defined as (38). Then, $u_{\lambda }\left(x\right)-u\left(x\right)⩾0$ for a.e. $x\in H$.

Proof. 

Firstly, we claim that

$$\begin{array}{c}\hfill w_{\lambda }\equiv 0.\end{array}$$

(42)

where $w_{\lambda }$ is defined as (18).

We prove this claim by contradiction. Suppose that $\parallel w_{\lambda }{\parallel }_{L^2\left({\Omega }_{\lambda }\right)}\ne 0$. From this, (36), (38) and (41), we conclude that

$$\begin{array}{cc}\hfill & {\Lambda }_1\left({\Omega }_{\lambda }\right){\parallel w_{\lambda }\parallel }_{L^2\left({\Omega }_{\lambda }\right)}^2\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }}{\left|x\right|}^{-2\alpha }{|\nabla w_{\lambda }|}^{2}dx+{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{{(w_{\lambda }\left(x\right)-w_{\lambda }\left(y\right))}^2}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill ⩾& {\mathit{\int }}_{{\Omega }_{\lambda }}|x_{\lambda }{|}^{-2\alpha }\nabla u_{\lambda }·\nabla (|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx-{\mathit{\int }}_{{\Omega }_{\lambda }}{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u_{\lambda }\left(x\right)-u_{\lambda }\left(y\right))\left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)-|y_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(y\right))}{|x_{\lambda }-y_{\lambda }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\lambda }{|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill & -{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\lambda }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\lambda }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }}{\displaystyle \frac{f\left(u_{\lambda }\right)-f\left(u\right)}{u_{\lambda }-u}}(u_{\lambda }-u)\left(x\right)w_{\lambda }\left(x\right)dx\hfill \\ \hfill =& -{\mathit{\int }}_{{\Omega }_{\lambda }}c\left(x\right)w_{\lambda }^2\left(x\right)dx\hfill \\ \hfill ⩾& \parallel c^{-}{\parallel }_{L^{\infty }\left({\Omega }_{\lambda }\right)}{\parallel w_{\lambda }\parallel }_{L^2\left({\Omega }_{\lambda }\right)}^2\hfill \\ \hfill >& {\Lambda }_1\left({\Omega }_{\lambda }\right){\parallel w_{\lambda }\parallel }_{L^2\left({\Omega }_{\lambda }\right)}^2,\hfill \end{array}$$

which is a contradiction. This shows that (42) holds. □

We now establish a strong maximum principle for antisymmetric supersolutions, which is the counterpart in the setting of mixed local-nonlocal operators of [4] ([Propossition 2.7]).

Lemma 5. 

Suppose that ${\Omega }_{\lambda }\subset H$ is an open and bounded set, $c\in L^{\infty }\left({\Omega }_{\lambda }\right)$, and

$$\begin{array}{c}\hfill u_{\lambda }\left(x\right)-u\left(x\right)⩾0a.e.x\in H.\end{array}$$

(43)

Then, either $u_{\lambda }\left(x\right)-u\left(x\right)\equiv 0,x\in {\mathbb{R}}^N$ or

$$\begin{array}{c}\hfill \mathrm{ess}{inf}_K(u_{\lambda }\left(x\right)-u\left(x\right))>0,\mathit{for}\mathit{any}\mathit{compact}\mathit{set}K\subset {\Omega }_{\lambda }.\end{array}$$

Proof. 

Now assume that

$$\begin{array}{c}\hfill u_{\lambda }\left(x\right)-u\left(x\right)\not\equiv 0,x\in {\mathbb{R}}^N.\end{array}$$

(44)

For a fixed $x_0\in {\Omega }_{\lambda }$, we show that

$$\begin{array}{c}\hfill \mathrm{ess}{inf}_{B_r\left(x_0\right)}(u_{\lambda }\left(x\right)-u\left(x\right))>0,\end{array}$$

(45)

for some radius $r>0$ is small enough.

First of all, since (43) and (44), together with the fact that $u_{\lambda }\left(x\right)-u\left(x\right)$ is antisymmetric, we know there exists a bounded set $M\subset H$ with $\left|M\right|>0$, which does not contain a small neighbourhood of $x_0$ such that

$$\begin{array}{c}\hfill \delta :={inf}_M(u_{\lambda }-u)\left(x\right)>0.\end{array}$$

(46)

Set

$$\begin{array}{c}\hfill r\in \left(0,{\displaystyle \frac{dist\left(x_0;\left({\mathbb{R}}^N\setminus H\right)\cup M\right)}{4}}\right),\end{array}$$

(47)

such that

$$\begin{array}{c}\hfill {\Lambda }_1\left(B_{2r}\left(x_0\right)\right)<{\parallel c\parallel }_{L^{\infty }\left(U\right)}.\end{array}$$

(48)

Let g be a decreasing function satisfying $g\in C_0^2\left({\mathbb{R}}^N\right)\to [0,1]$ and

$$\begin{array}{c}\hfill g\left(x\right)=\left\{\begin{array}{c}1,x\in B_r\left(x_0\right),\hfill \\ 0,x\in {\mathbb{R}}^N\setminus B_{2r}\left(x_0\right).\hfill \end{array}\right.\end{array}$$

Moreover, for a given $b>0$ to be selected at a later time, define

$$\begin{array}{c}\hfill h:{\mathbb{R}}^N\to \mathbb{R},h\left(x\right)=g\left(x\right)-g\left(\overline{x}\right)+b\left({\chi }_M\left(x\right)-{\chi }_M\left(\overline{x}\right)\right).\end{array}$$

(49)

Write $U_0=B_{2r}\left(x_0\right)$ and $U_0^{\prime }=B_{3r}\left(x_0\right)\cup Q\left(B_{3r}\left(x_0\right)\right)$. Obviously, $(M\cup Q\left(M\right))\cap U_0^{\prime }=\varnothing$.

It is easily seen that $h\in L^2\left({\mathbb{R}}^N\right)\cap {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)$ is antisymmetric,

$$\begin{array}{c}\hfill h\equiv 0,x\in H\setminus (U_0\cup M)\mathrm{and}h\equiv b,x\in M.\end{array}$$

(50)

Now we show that there exists a constant $C>0$, depending on g, such that

$$\begin{array}{c}\hfill \mathcal{B}(g,\phi )⩽C{\mathit{\int }}_{U_0}{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{2\alpha }}}dx,\forall \phi \in {\mathcal{H}}_{\alpha }^s\left(U_0\right)\mathrm{with}\phi ⩾0.\end{array}$$

(51)

Indeed, for any $\phi \in {\mathcal{H}}_{\alpha }^s\left(U_0\right)$ with $\phi ⩾0$, We have

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{U_0}{\left|x\right|}^{-2\alpha }\nabla g·\nabla \phi dx\hfill \\ \hfill =& {\mathit{\int }}_{\partial U_0}{\left|x\right|}^{-2\alpha }\nabla g·\phi dx-{\mathit{\int }}_{U_0}\mathrm{div}\left(\right|x{|}^{-2\alpha }\nabla g)·\phi \left(x\right)dx\hfill \\ \hfill =& -{\mathit{\int }}_{U_0}\mathrm{div}\left(\right|x{|}^{-2\alpha }\nabla g)·\phi \left(x\right)dx\hfill \\ \hfill =& -{\mathit{\int }}_{U_0}\nabla g·\nabla \left(\right|x{|}^{-2\alpha })\phi \left(x\right)dx-{\mathit{\int }}_{U_0}\Delta g·{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{2\alpha }}}dx\hfill \\ \hfill ⩽& C{\mathit{\int }}_{U_0}{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{2\alpha }}}dx,\hfill \end{array}$$

(52)

and

$$\begin{array}{cc}\hfill & {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{\left(g\right(x)-g(y\left)\right)\left(\phi \right(x)-\phi (y\left)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dx}{{\left|x\right|}^{\alpha }}}{\displaystyle \frac{dy}{{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{U_0}{(-\Delta )}_{\alpha }^{s}g\left(x\right)\phi \left(x\right)dx\hfill \\ \hfill ⩽& C{\mathit{\int }}_{U_0}{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{\alpha }}}dx,\hfill \end{array}$$

this fact, together with (52), implies that (51) holds. Similarly,

$$\begin{array}{c}\hfill \mathcal{B}(g\left(\overline{x}\right),\phi )⩽C{\mathit{\int }}_{U_0}{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{2\alpha }}}dx\mathrm{for}\mathrm{every}\phi \in {\mathcal{H}}_{\alpha }^s\left(U_0\right)\mathrm{with}\phi ⩾0,\end{array}$$

(53)

combining (49), (51) and (53), we conclude that, for any $\phi \in {\mathcal{H}}_{\alpha }^s\left(U_0\right)$ with $\phi ⩾0$,

$$\begin{array}{cc}\hfill \mathcal{B}(h,\phi )=& \mathcal{B}(g\left(x\right),\phi )+\mathcal{B}(g\left(\overline{x}\right),\phi )\hfill \\ \hfill & +b{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(({\chi }_M\left(x\right)-{\chi }_M\left(\overline{x}\right))-({\chi }_M\left(y\right)-{\chi }_M\left(\overline{y}\right)))(\phi \left(x\right)-\phi \left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill ⩽& C{\mathit{\int }}_{U_0}{\displaystyle \frac{\phi \left(x\right)}{{\left|x\right|}^{2\alpha }}}dx.\hfill \end{array}$$

(54)

Note that C in (54) depended on b. Choose $b>0$ such that

$$\begin{array}{c}\hfill C<-{\parallel c\left(x\right)\parallel }_{L^{\infty }\left(U_0\right)},\end{array}$$

which, together with (54), yields

$$\begin{array}{c}\hfill \mathcal{B}(h,\phi )⩽-{\parallel c\left(x\right)\parallel }_{L^{\infty }\left(U_0\right)}{\mathit{\int }}_{U_0}\phi \left(x\right)dx⩽{\mathit{\int }}_{U_0}c\left(x\right)h\left(x\right)\phi \left(x\right)dx,\end{array}$$

(55)

since $h\left(x\right)=g\left(x\right)\in [0,1]$ for every $x\in U_0$. Clearly,

$$\begin{array}{c}\hfill (u_{\lambda }-u)\left(x\right)-{\displaystyle \frac{\delta }{b}}h\left(x\right)\in L^2\left({\mathbb{R}}^N\right)\cap {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right),\end{array}$$

(56)

and

$$\begin{array}{c}\hfill (u_{\lambda }-u)\left(x\right)-{\displaystyle \frac{\delta }{b}}h\left(x\right)⩾0,x\in H\setminus U_0.\end{array}$$

According to (14) and (55), we find, for any $\phi \in {\mathcal{H}}_{\alpha }^s\left(U_0\right)$ with $\phi ⩾0$,

$$\begin{array}{cc}\hfill & \mathcal{B}(u_{\lambda },\phi )-\mathcal{B}(u,\phi )-{\displaystyle \frac{\delta }{b}}\mathcal{B}(h,\phi )\hfill \\ \hfill ⩾& {\mathit{\int }}_{U_0}c\left(x\right)(u_{\lambda }-u)\left(x\right)\phi \left(x\right)dx-{\displaystyle \frac{\delta }{b}}{\mathit{\int }}_{U_0}c\left(x\right)h\left(x\right)\phi \left(x\right)dx\hfill \\ \hfill =& {\mathit{\int }}_{U_0}c\left(x\right)((u_{\lambda }-u)\left(x\right)-{\displaystyle \frac{\delta }{b}}h\left(x\right))\phi \left(x\right)dx.\hfill \end{array}$$

This fact, together with Lemma 4, leads to $(u_{\lambda }-u)\left(x\right)-{\displaystyle \frac{\delta }{b}}h\left(x\right)⩾0$ a.e. $x\in U_0$. Consequently,

$$\begin{array}{c}\hfill u_{\lambda }\left(x\right)-u\left(x\right)⩾{\displaystyle \frac{\delta }{b}}>0\mathrm{a}.\mathrm{e}.x\in B_r\left(x_0\right),\end{array}$$

which shows that (45) holds. □

We came to a conclusion.

Lemma 6. 

Let u be a weak solution to (3). If there exists some $\lambda \in (-a,0)$ such that $u-u_{\lambda }\equiv 0,x\in {\mathbb{R}}^N$, then

$$\begin{array}{c}\hfill u\equiv 0,x\in \Omega (\mathit{hence}u\equiv 0,x\in {\mathbb{R}}^N).\end{array}$$

(57)

Proof. 

For any fixed $\lambda \in (-a,0)$, obviously ${\Omega }_{\lambda }\ne \varnothing$. The convexity of $\Omega$ in the $x_1$ direction implies that ${\Omega }_{\lambda }\subseteq Q_{\lambda }(\Omega )$, where $Q_{\lambda }(\Omega )$ is the symmetric region of $\Omega$ with respect to the hyperplane $T_{\lambda }$.

Clearly, $\ell =a+\lambda \in (0,a)$ since $\lambda \in (-a,0)$. Therefore, $Q_{\lambda }\left({\Omega }_{\ell }\right)\cap \overline{\Omega }=\varnothing$. Thus, $u=0,x\in Q_{\lambda }\left({\Omega }_{\ell }\right)$ and $u_{\lambda }\equiv 0,x\in {\Omega }_{\ell }$, which leads to

$$\begin{array}{c}\hfill u\equiv u_{\lambda }\equiv 0,x\in {\Omega }_{\ell }.\end{array}$$

(58)

Set $\eta =a+\lambda /2$.

By ${\Omega }_{\eta }\cup Q_{\eta }\left({\Omega }_{\eta }\right)\subseteq {\Omega }_{\ell }$, we have

$$\begin{array}{c}\hfill u-u_{\eta }=0,x\in {\Omega }_{\eta }.\end{array}$$

Using Lemma 5 again with $H={\Sigma }_{\eta },U={\Omega }_{\eta }$, we deduce that

$$\begin{array}{c}\hfill u_{\eta }-u\equiv 0,x\in {\mathbb{R}}^N.\end{array}$$

Therefore, u has two different parallel symmetry hyperplanes $\partial {\Sigma }_{\lambda }=\{x_1=\lambda \}$ and $\partial {\Sigma }_{\eta }=\{x_1=\eta \}$. By $u\equiv 0,{\mathbb{R}}^N\setminus \Omega$ and $Q_{\lambda }(\Omega \setminus ({\Omega }_{\lambda }\cup Q_{\lambda }\left({\Omega }_{\lambda }\right)))\cap \Omega =\varnothing$, we find that

$$\begin{array}{c}\hfill u\equiv 0,x\in \mathcal{O}=\Omega \setminus ({\Omega }_{\lambda }\cup Q_{\lambda }\left({\Omega }_{\lambda }\right)).\end{array}$$

(59)

Note that $\partial {\Sigma }_{\eta }$ is a symmetry hyperplane for u and $Q_{\eta }({\Omega }_{\lambda }\cup Q_{\lambda }\left({\Omega }_{\lambda }\right)\subseteq \mathcal{O}$, we infer that

$$\begin{array}{c}\hfill u\equiv 0,x\in {\Omega }_{\lambda }\cup Q_{\lambda }\left({\Omega }_{\lambda }\right).\end{array}$$

(60)

Combining (59) with (60), (57) holds. □

The following lemma will be used in the proof of the main theorem.

Lemma 7 

([21]). [Lemma 2.10] Let $N⩾2$ and $\Omega \subseteq {\mathbb{R}}^N$ be an open and bounded set with a Lipschitz boundary. There exists a real constant $C=C_n$, independent of Ω, such that

$$\begin{array}{c}\hfill {\parallel u\parallel }_{L^2(\Omega )}⩽{C|\Omega |}^{1/N}{\parallel \nabla u\parallel }_{L^2(\Omega )}.\end{array}$$

3. Proof of Theorem 1

Now we give the proof of Theorem 1.

Proof of Theorem 1. 

In the following, we show that every solution $u\in {\mathcal{H}}_{\alpha }^s\left({\mathbb{R}}^N\right)$ to problem (3) is actually symmetric and monotone decreasing around the origin by the moving planes method. For any $\lambda \in \mathbb{R}$, the definitions of ${\Sigma }_{\lambda }$ and $T_{\lambda }$ are given in (16), respectively. Moreover, $u_{\lambda }\left(x\right)=u\left(x_{\lambda }\right)$ is defined by (12) for $x_{\lambda }=(2\lambda -x_1,x_2,\cdots x_N)\in {\mathbb{R}}^N$.

In the rest of the proof, we assume ${\displaystyle \frac{C_{N,s}}{2}}=1$ for simplicity of notation.

By (19) and (20), we know that ${\left|x\right|}^{2\alpha }{w}_{\lambda }\left(x\right)$ and $|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)$ are admissible test functions for problem (3) and (13), respectively. Thus,

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right))dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\lambda }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\lambda }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\mathbb{R}}^N}f\left(u\right)w_{\lambda }dx,\hfill \end{array}$$

(61)

and

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }|x_{\lambda }{|}^{-2\alpha }\nabla u\left(x_{\lambda }\right)·\nabla \left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{(u\left(x_{\lambda }\right)-u\left(y_{\lambda }\right))\left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)-|y_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\lambda }{|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\mathbb{R}}^N}f\left(u_{\lambda }\right)w_{\lambda }dx.\hfill \end{array}$$

(62)

Subtracting from (61) and (62), we obtain

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right))dx-{\mathit{\int }}_{\Omega }|x_{\lambda }{|}^{-2\alpha }\nabla u\left(x_{\lambda }\right)·\nabla \left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\lambda }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\lambda }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill & -{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x_{\lambda }\right)-u\left(y_{\lambda }\right))\left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(x\right)-|y_{\lambda }{|}^{2\alpha }{w}_{\lambda }\left(y\right))}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\lambda }{|}^{\alpha }{\left|y_{\lambda }\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\mathbb{R}}^N}f\left(u\right)w_{\lambda }dx-{\mathit{\int }}_{{\mathbb{R}}^N}f\left(u_{\lambda }\right)w_{\lambda }dx.\hfill \end{array}$$

(63)

It is evident that $\nabla u·{\nabla ln\left|x\right|}^{2\alpha }-\nabla u_{\lambda }·\nabla ln{\left|x_{\lambda }\right|}^{2\alpha }=0$ for $x\in \Omega$, which leads to

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }{\left|x\right|}^{-2\alpha }\nabla u·{\nabla \left(\right|x|}^{2\alpha }{w}_{\lambda }\left(x\right))dx-{\mathit{\int }}_{\Omega }|x_{\lambda }{|}^{-2\alpha }\nabla u\left(x_{\lambda }\right)·\nabla \left(\right|x_{\lambda }{|}^{2\alpha }{w}_{\lambda })dx\hfill \\ \hfill =& {\mathit{\int }}_{\Omega }\nabla u·\nabla w_{\lambda }+w_{\lambda }\nabla u·{\nabla ln\left|x\right|}^{2\alpha }dx-{\mathit{\int }}_{\Omega }\nabla u_{\lambda }·\nabla w_{\lambda }+w_{\lambda }\nabla u_{\lambda }·\nabla ln{\left|x_{\lambda }\right|}^{2\alpha }dx\hfill \\ \hfill =& {\mathit{\int }}_{\Omega }\nabla (u-u_{\lambda })·\nabla w_{\lambda }+\left[\nabla u·{\nabla ln\left|x\right|}^{2\alpha }-\nabla u_{\lambda }·\nabla ln{\left|x_{\lambda }\right|}^{2\alpha }\right]w_{\lambda }\left(x\right)dx\hfill \\ \hfill =& {\mathit{\int }}_{\Omega }\nabla (u-u_{\lambda })·\nabla w_{\lambda }dx\hfill \\ \hfill =& {\mathit{\int }}_{\Omega }{|\nabla w_{\lambda }|}^{2}dx.\hfill \end{array}$$

(64)

Taking into account Lemmas 1 and 2, (63) and (64), we yield

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{|\nabla w_{\lambda }|}^{2}dx\hfill \\ \hfill ⩽& {\mathit{\int }}_{{\mathbb{R}}^N}(f\left(u\right)-f\left(u_{\lambda }\right))w_{\lambda }dx\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}(f\left(u\right)-f\left(u_{\lambda }\right))w_{\lambda }dx\hfill \\ \hfill =& {\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{\displaystyle \frac{f\left(u\right)-f\left(u_{\lambda }\right)}{(u\left(x\right)-u_{\lambda }\left(x\right))}}{w}_{\lambda }^{2}dx\hfill \\ \hfill ⩽& C{\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{w}_{\lambda }^{2}dx,\hfill \end{array}$$

(65)

where we have used the fact that

$$\begin{array}{c}\hfill w_{\lambda }=0,x\in ({\Sigma }_{\lambda }\setminus \Omega )\cup (\Omega \setminus [Q_{\lambda }\left({\Omega }_{\lambda }\right)\cup ({\mathbb{R}}^N\setminus {\Sigma }_{\lambda })\setminus \Omega ).\end{array}$$

(66)

Using Lemma 7 and (65), we obtain that

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{|\nabla w_{\lambda }|}^{2}dx\hfill \\ \hfill ⩽& C|{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\}){|}^{1/N}{\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{|\nabla w_{\lambda }|}^{2}dx.\hfill \end{array}$$

(67)

Let $\lambda +a$ be small enough such that

$$\begin{array}{c}\hfill C|{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\}){|}^{1/N}<{\displaystyle \frac{1}{2}},\end{array}$$

which, combined with (67), gives that

$$\begin{array}{c}\hfill {\mathit{\int }}_{{\Omega }_{\lambda }\cup (Q_{\lambda }\left({\Omega }_{\lambda }\right)\setminus \left\{0\right\})}{|\nabla w_{\lambda }|}^{2}dx=0.\end{array}$$

Therefore, $w_{\lambda }=0$, $x\in ({\Omega }_{\lambda }\cup \left(Q_{\lambda }\left({\Omega }_{\lambda }\right)\right))$ if $\lambda$ is sufficiently close to $-a$. This fact, together with (67), implies that

$$\begin{array}{c}\hfill u⩽u_{\lambda },x\in {\Sigma }_{\lambda }.\end{array}$$

(68)

Define

$$\begin{array}{c}\hfill {\Lambda }_0=\{\lambda \in (-a,0):u⩽u_t,x\in {\Sigma }_t,\forall t\in (-a,\lambda ]\}.\end{array}$$

(69)

and

$$\begin{array}{c}\hfill \overline{\lambda }=\sup {\Lambda }_0.\end{array}$$

(70)

Now we show that

$$\begin{array}{c}\hfill \overline{\lambda }=0.\end{array}$$

(71)

We argue by contradiction and assume that

$$\begin{array}{c}\hfill \overline{\lambda }<0.\end{array}$$

By the definition of $w_{\lambda }$ in (18) and the local continuity of $w_{\lambda }$ with $\lambda$, we find that $w_{\overline{\lambda }}=0,x\in {\Sigma }_{\overline{\lambda }}$, which implies that

$$\begin{array}{c}\hfill u_{\overline{\lambda }}-u⩾0,x\in {\Sigma }_{\overline{\lambda }}.\end{array}$$

This fact, together with $u\not\equiv 0$ and Lemma 6, leads to $u_{\overline{\lambda }}-u\not\equiv 0,x\in {\mathbb{R}}^N$. Consequently, Lemma 5 (with $H={\Sigma }_{\overline{\lambda }},U={\Omega }_{\overline{\lambda }}$) yields

$$\begin{array}{c}\hfill u_{\overline{\lambda }}-u>0,x\in {\Omega }_{\overline{\lambda }}.\end{array}$$

Let $K\subseteq {\Omega }_{\overline{\lambda }}$ be a given compact set, which is to be chosen later. By the local continuity of $u_{\lambda }-u$ with $\lambda$, there exists a suitable $\overline{\tau }=\overline{\tau }\left(K\right)>0$ such that

$$\begin{array}{c}\hfill u_{\overline{\lambda }+\tau }\left(x\right)-u\left(x\right)>0,x\in K,\forall \tau \in (0,\overline{\tau }),\end{array}$$

(72)

that is

$$\begin{array}{c}\hfill w_{\overline{\lambda }+\tau }\left(x\right)=0,x\in K,\forall \tau \in (0,\overline{\tau }).\end{array}$$

For every fixed $\tau \in (0,\overline{\tau })$, take ${\left|x\right|}^{2\alpha }{w}_{\overline{\lambda }+\tau }$ as an admissible test function in (8), by which we have

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }\nabla u·\nabla w_{\overline{\lambda }+\tau }+w_{\overline{\lambda }+\tau }\nabla u·\nabla ln{\left|x\right|}^{2\alpha }dx\hfill \\ \hfill & +{\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^{2N}}{\displaystyle \frac{(u\left(x\right)-u\left(y\right))\left(\right|x{|}^{2\alpha }{w}_{\overline{\lambda }+\tau }\left(x\right)-\left|y{|}^{2\alpha }{w}_{\overline{\lambda }+\tau }\left(y\right)\right)}{{|x-y|}^{N+2s}}}{\displaystyle \frac{dxdy}{{\left|x\right|}^{\alpha }{\left|y\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\mathbb{R}}^N}f\left(u\left(x\right)\right)w_{\overline{\lambda }+\tau }\left(x\right)dx.\hfill \end{array}$$

Take $|x_{\overline{\lambda }+\tau }{|}^{2\alpha }{w}_{\overline{\lambda }+\tau }$ as a test function in (62). We find that

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{\Omega }\nabla u_{\overline{\lambda }+\tau }·\nabla w_{\overline{\lambda }+\tau }+w_{\overline{\lambda }+\tau }\nabla u_{\overline{\lambda }+\tau }·\nabla ln{\left|x_{\overline{\lambda }+\tau }\right|}^{2\alpha }dx\hfill \\ \hfill +& {\mathit{\int }\mathit{\int }}_{{\mathbb{R}}^N}{\displaystyle \frac{(u\left(x_{\overline{\lambda }+\tau }\right)-u\left(y_{\overline{\lambda }+\tau }\right))\left(\right|x_{\overline{\lambda }+\tau }{|}^{2\alpha }{w}_{\overline{\lambda }+\tau }\left(x\right)-|y_{\overline{\lambda }+\tau }{|}^{2\alpha }{w}_{\overline{\lambda }+\tau }\left(y\right))}{|x_{\overline{\lambda }+\tau }-y_{\overline{\lambda }+\tau }{|}^{N+2s}}}{\displaystyle \frac{dxdy}{|x_{\overline{\lambda }+\tau }{|}^{\alpha }{\left|y_{\overline{\lambda }+\tau }\right|}^{\alpha }}}\hfill \\ \hfill =& {\mathit{\int }}_{{\mathbb{R}}^N}f\left(u_{\overline{\lambda }+\tau }\right)w_{\overline{\lambda }+\tau }dx.\hfill \end{array}$$

Repeating the previous argument leads to

$$\begin{array}{cc}\hfill & {\mathit{\int }}_{({\Omega }_{\overline{\lambda }+\tau }\setminus K)\cup Q_{\overline{\lambda }+\tau }({\Omega }_{\overline{\lambda }+\tau }\setminus K)}{|\nabla w_{\overline{\lambda }+\tau }|}^{2}dx\hfill \\ \hfill ⩽& C|({\Omega }_{\overline{\lambda }+\tau }\setminus K)\cup Q_{\overline{\lambda }+\tau }({\Omega }_{\overline{\lambda }+\tau }\setminus K){|}^{1/N}{\mathit{\int }}_{({\Omega }_{\overline{\lambda }+\tau }\setminus K)\cup Q_{\overline{\lambda }+\tau }({\Omega }_{\overline{\lambda }+\tau }\setminus K)}{|\nabla w_{\overline{\lambda }+\tau }|}^{2}dx.\hfill \end{array}$$

(73)

Let the compact K be big enough and $\overline{\tau }$ be small enough such that

$$\begin{array}{c}\hfill C|({\Omega }_{\overline{\lambda }+\tau }\setminus K)\cup Q_{\overline{\lambda }+\tau }({\Omega }_{\overline{\lambda }+\tau }\setminus K){|}^{1/N}<1,\end{array}$$

which, combined with (73), implies that

$$\begin{array}{c}\hfill {\mathit{\int }}_{({\Omega }_{\overline{\lambda }+\tau }\setminus K)\cup Q_{\overline{\lambda }+\tau }({\Omega }_{\overline{\lambda }+\tau }\setminus K)}{|\nabla w_{\overline{\lambda }+\tau }|}^{2}dx=0.\end{array}$$

Thus, $w_{\overline{\lambda }+\tau }\equiv 0$, $x\in {\Omega }_{\overline{\lambda }+\tau }$; that is,

$$\begin{array}{c}\hfill u⩽u_{\overline{\lambda }+\tau },x\in {\Omega }_{\overline{\lambda }+\tau }.\end{array}$$

for every $\tau \in (0,\overline{\tau })$, provided that $\overline{\tau }>0$ is small enough, which contradicts the definition of $\overline{\lambda }$. Thus, (71) holds. This completes the proof. □

4. Conclusions

In this paper, by the moving planes method, we have successfully established the symmetry and monotonicity of non-negative solutions to the mixed local and nonlocal weighted elliptic problem. We have overcome the challenges brought about by the weighted function and derived significant results.

The main theorem (Theorem 1) indicates that under specific conditions, non-identically vanishing weak solutions u to the given equation are symmetric with respect to the hyperplane $\{x_1=0\}$ and strictly increase in the $x_1-$ direction in $\Omega \cap \{x_1<0\}$. This not only enriches our understanding of the qualitative behavior of solutions to the given equation but also extends the existing symmetry results of mixed local and nonlocal equations without weighted functions, as established by Biagi et al. Moreover, the development of the moving planes method for equations with a weighted function offers a powerful tool for dealing with similar weighted problems in the future.

However, this study also paves the way for future research directions. For instance, it would be interesting to explore the symmetry and monotonicity properties of solutions in more general domains or under more complex nonlinearities. Additionally, investigating the impact of different types of weighted functions on the solutions can further deepen our understanding of the mixed local and nonlocal elliptic equations. Overall, our research serves as a stepping stone for further in-depth studies in this area.