分数阶Hartree方程负解的对称性

Symmetry of Negative Solutions of Fractional Hartree Equations

摘要: 本文研究了分数阶Hartree型方程负解的对称性。我们采用的方法是移动平面法，首先我们对方程的解作Kelvin变换。然后，我们证明了无穷远处衰减性定理，移动平面可以由此起步。最后，通过狭窄区域定理，我们证明了方程的解具有径向对称性。

Abstract: In this paper, we study the symmetry of negative solutions of fractional Hartree type equations. The method we adopted was the moving plane method. First, we performed Kelvin transformation on the solution of the equation. Then, we prove the decay theorem at infinity, from which the moving plane can start. Finally, by the narrow region theorem, we prove that the solution of the equation has radial symmetry.

1. 引言

在本文中，我们主要考虑下列分数阶静态Hartree方程

$\{\begin{array}{l}{\left(-\Delta \right)}^{s}u\left(x\right)=\left(\frac{1}{{\left|x\right|}^q}\ast {\left|u\left(x\right)\right|}^p\right)u^{\sigma }\left(x\right),x\in {ℝ}^n,\\ u\in C_{loc}^{1,1}\cap L_s\left({ℝ}^n\right),x\in {ℝ}^n,\end{array}$ (1)

其中 $u\left(x\right)<0$， $0<s<1$， $0\le q\le 4s$， $n>\max \left\{2s,q\right\}$， $p=\frac{2n-q}{n-2s}$， $1\le \sigma \le \frac{n+2s-q}{n-2s}$， $\sigma$ 为奇数并且

$L_s\left({ℝ}^n\right):=\left\{u:{ℝ}^n\to ℝ|{\displaystyle {\int }_{{ℝ}^n}\frac{\left|u\left(x\right)\right|}{1+{\left|x\right|}^{n+2s}}\text{d}x}<\infty \right\}.$

分数阶拉普拉斯算子定义为(见 [1] [2])

${\left(-\Delta \right)}^{s}u\left(x\right)=C_{n,s}P.V.{\displaystyle {\int }_{{ℝ}^n}\frac{u\left(x\right)-u\left(y\right)}{{\left|x-y\right|}^{n+2s}}\text{d}y}.$ (2)

Hartree型方程在非相对论玻色子原子和分子大系统的量子理论中有许多有趣的应用，事实上，Hartree型方程是在研究这类系统的平均场极限时产生的，即在玻色子数量非常大，但它们之间的相互作用很弱的情况下，因此，通过数学方法研究Hartree型方程有重要意义，参见文献 [3] - [8] 及其引用的参考文献。近几年来，许多学者都对Hartree型方程产生兴趣并得到一系列结果。例如，当 $s=1$， $p=2$， $q=4$， $\sigma =1$ 时，刘书茂 [9] 通过方程(1)的等价积分形式对正解进行了分类。当 $0<s<1$， $p=2$， $q=4s$， $\sigma =1$ 时，戴蔚等人 [10] 使用积分形式的移动平面法对于方程(1)的正 $H^s\left({ℝ}^n\right)$ 弱解进行了分类。后来，戴蔚等人 [11] 通过直接移动平面法对于方程(1)的正解进行了分类，并且给出了解的具体形式。

本文我们扩展了戴蔚等人的系数范围 $0<s<1$， $p=2$， $q=4s$， $\sigma =1$，通过直接移动平面法来研究当 $u\left(x\right)<0$， $0<s<1$， $0\le q\le 4s$， $n>\max \left\{2s,q\right\}$， $p=\frac{2n-q}{n-2s}$， $1\le \sigma \le \frac{n+2s-q}{n-2s}$， $\sigma$ 为奇数时方程(1)

的负解的对称性。首先，我们通过证明无穷远处的衰减性定理来确保移动平面可以起步。然后，我们将平面移到极限位置，通过狭窄区域定理，我们可以得到方程(1)的负解是关于某点径向对称的。

2. 主要结果

本文关于静态Hartree方程的主要结论如下：

定理2.1设 $0<s<1$， $0\le q\le 4s$， $n>\max \left\{2s,q\right\}$， $p=\frac{2n-q}{n-2s}$， $1\le \sigma \le \frac{n+2s-q}{n-2s}$， $\sigma$ 为奇数。若u是方程(1)的负解，且存在某点 $x^0\in {ℝ}^n$ 使得 ${\lim }_{\left|x-x^0\right|\to \infty }{\left|x-x^0\right|}^{n-2s}u\left(x\right)=\infty$，则u关于点 $x^0$ 径向对称。

3. 证明

由方程(1)可以得到 ${\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(x\right)\right|}^p}{{\left|x-x^0\right|}^q}\text{d}x}<\infty$，我们先对u关于点 $x^0$ 作Kelvin变换

$v\left(x\right)=\frac{1}{{\left|x-x^0\right|}^{n-2s}}u\left(\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0\right),x\in {ℝ}^n\backslash \left\{x^0\right\}.$

故 ${\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(x\right)\right|}^p}{{\left|x-x^0\right|}^q}\text{d}x}<\infty$ 等价于 ${\displaystyle {\int }_{{ℝ}^n}{\left|v\left(x\right)\right|}^p\text{d}x}<\infty$，从方程(1)中可得 $v\in C_{loc}^{1,1}\left({ℝ}^n\backslash \left\{x^0\right\}\right)\cap L_s\left({ℝ}^n\right)$。我们可

以由(1)和(2)推断出

$\begin{array}{c}{\left(-\Delta \right)}^{s}v\left(x\right)=\frac{1}{{\left|x-x^0\right|}^{n-2s}}{\left(-\Delta \right)}^{s}u\left(\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0\right)\\ =\frac{1}{{\left|x-x^0\right|}^{n-2s}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(y\right)\right|}^p}{{\left|\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0-y\right|}^q}\text{d}y}{u}^{\sigma }\left(\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0\right)\\ =\frac{1}{{\left|x-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}{v}^{\sigma }\left(x\right).\end{array}$

由于方程(1)在旋转下是不变的，所以我们任意选一个方向作为 $x_1$ 方向，并沿着 $x_1$ 轴进行移动平面的过程。方便起见，我们进行如下定义：

$x^{\lambda }:=\left(2\lambda -x_1,x^{\prime }\right)=\left(2\lambda -x_1,x_2,\cdots ,x_n\right),$

$T_{\lambda }:=\left\{x\in {ℝ}^n|x_1=\lambda \right\},$

${\Sigma }_{\lambda }:=\left\{x\in {ℝ}^n|x_1<\lambda \right\},$

$w_{\lambda }\left(x\right):=v\left(x\right)-v\left(x^{\lambda }\right)=v\left(x\right)-v^{\lambda }\left(x\right),$

${\Sigma }_{\lambda }^{-}:=\left\{x\in {\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}|w_{\lambda }\left(x\right)<0\right\}.$

Step 1. 从 $x_1=-\infty$ 附近开始移动平面，我们通过反证法证明

${\Sigma }_{\lambda }^{-}=\varnothing .$ (3)

假设(3)式不成立，那么 ${\Sigma }_{\lambda }^{-}\ne \varnothing$ 且 $w_{\lambda }$ 在 ${\Sigma }_{\lambda }$ 中存在负值，对于任意 $x\in {\Sigma }_{\lambda }^{-}$，我们有

$\begin{array}{c}{\left(-\Delta \right)}^s{w}_{\lambda }\left(x\right)=\frac{1}{{\left|x-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}{v}^{\sigma }\left(x\right)\\ -\frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}{v}^{\sigma }\left(x^{\lambda }\right)\\ \ge \frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}\left({\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}{v}^{\sigma }\left(x\right)-{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}{v}^{\sigma }\left(x^{\lambda }\right)\right)\end{array}$

$\begin{array}{l}=\frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}[v^{\sigma }\left(x\right){\displaystyle {\int }_{{\Sigma }_{\lambda }}\left(\frac{1}{{\left|x-y\right|}^q}-\frac{1}{{\left|x^{\lambda }-y\right|}^q}\right)\left({\left|v\left(y\right)\right|}^p-{\left|v\left(y^{\lambda }\right)\right|}^p\right)\text{d}y}\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\left(v^{\sigma }\left(x\right)-v^{\sigma }\left(x^{\lambda }\right)\right)]\\ \ge \frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}[v^{\sigma }\left(x\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|x-y\right|}^q}\left({\left|v\left(y\right)\right|}^p-{\left|v\left(y^{\lambda }\right)\right|}^p\right)\text{d}y}\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\sigma v^{\sigma -1}\left(x\right)w_{\lambda }\left(x\right)].\end{array}$ (4)

引理3.1 [11] 设u是方程(1)的正解，则u满足积分方程

$u\left(x\right)={\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|x-y\right|}^{n-2s}}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z{u}^{\sigma }\left(y\right)\text{d}y},$

其中 $R_{n,s}:=\frac{\Gamma \left(\frac{n}{2}-s\right)}{{\pi }^{\frac{n}{2}}{2}^{2s}\Gamma \left(s\right)}$ (见 [12])。

对于本文中的负解u，我们令 $u^{*}=-u$，再应用引理3.1可以得到u满足同样的结论。

由引理3.1可得

$\begin{array}{c}v\left(x\right)=\frac{1}{{\left|x-x^0\right|}^{n-2s}}u\left(\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0\right)\\ =\frac{1}{{\left|x-x^0\right|}^{n-2s}}{\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0-y\right|}^{n-2s}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{u}^{\sigma }\left(y\right)\text{d}y}\\ ={\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|x-y\right|}^{n-2s}}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y\right){\left|y-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\text{d}y,\end{array}$ (5)

由引理3.1，当 $\left|x\right|$ 充分大时，有

$\begin{array}{c}u\left(x\right)={\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|x-y\right|}^{n-2s}}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|u\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{u}^{\sigma }\left(y\right)\text{d}y\\ \le {\displaystyle {\int }_{B_3\left(x^0\right)\backslash B_2\left(x^0\right)}\frac{R_{n,s}}{{\left|x-y\right|}^{n-2s}}}{\displaystyle {\int }_{B_1\left(x^0\right)}\frac{{\left|u\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{u}^{\sigma }\left(y\right)\text{d}y\\ \le \frac{-C}{{\left|x-x^0\right|}^{n-2s}}.\end{array}$

因此，对于充分小的 $\epsilon$，当 $x\in B_{\epsilon }\left(x^0\right)\backslash \left\{x^0\right\}$ 时，有 $v\left(x\right)=\frac{1}{{\left|x-x^0\right|}^{n-2s}}u\left(\frac{x-x^0}{{\left|x-x^0\right|}^2}+x^0\right)\le -C$。又因为当 $\lambda$ 充分负且 $x\in B_{\epsilon }\left({\left(x^0\right)}^{\lambda }\right)\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 时， $v\left(x\right)>-\frac{C}{2}$，故此时 $w_{\lambda }\left(x\right)=v\left(x\right)-v\left(x^{\lambda }\right)\ge \frac{C}{2}>0$。

现在我们需要下列定理：

定理3.2 (无穷远处的衰减性定理)设 $\lambda <x_1^0$， $w_{\lambda }\in C_{loc}^{1,1}\left({\Sigma }_{\lambda }^{-}\right)\cap L_s\left({ℝ}^n\right)$ 且 $w_{\lambda }$ 负最小值在 ${\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 内部达到，则存在 $R_0>0$ (依赖u且与 $\lambda$ 无关)，使得如果 $\hat{x}\in {\Sigma }_{\lambda }^{-}$ 满足 $w_{\lambda }\left(\hat{x}\right)={\min }_{{\Sigma }_{\lambda }^{-}}{w}_{\lambda }\left(x\right)<0$ 那么 $\left|\hat{x}-x^0\right|\le R_0$。

证明：反设对任意 $R_0>0$，存在 $\hat{x}\in {\Sigma }_{\lambda }^{-}$ 且 $\left|\hat{x}-x^0\right|>R_0$，使得 $w_{\lambda }\left(\hat{x}\right)={\min }_{{\Sigma }_{\lambda }^{-}}{w}_{\lambda }\left(x\right)<0$。

一方面，由(2)我们有

$\begin{array}{c}{\left(-\Delta \right)}^s{w}_{\lambda }\left(\hat{x}\right)=C_{n,s}P.V.{\displaystyle {\int }_{{ℝ}^n}\frac{w_{\lambda }\left(\hat{x}\right)-w_{\lambda }\left(y\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}\\ =C_{n,s}\left(P.V.{\displaystyle {\int }_{{\Sigma }_{\lambda }}\frac{w_{\lambda }\left(\hat{x}\right)-w_{\lambda }\left(y\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}+{\displaystyle {\int }_{{\Sigma }_{\lambda }^c}\frac{w_{\lambda }\left(\hat{x}\right)-w_{\lambda }\left(y\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}\right)\\ \le C_{n,s}\left({\displaystyle {\int }_{{\Sigma }_{\lambda }}\frac{w_{\lambda }\left(\hat{x}\right)-w_{\lambda }\left(y\right)}{{\left|\hat{x}-y^{\lambda }\right|}^{n+2s}}\text{d}y}+{\displaystyle {\int }_{{\Sigma }_{\lambda }^c}\frac{w_{\lambda }\left(\hat{x}\right)-w_{\lambda }\left(y\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}\right)\\ =2C_{n,s}{\displaystyle {\int }_{{\Sigma }_{\lambda }^c}\frac{w_{\lambda }\left(\hat{x}\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}\\ \le C{\displaystyle {\int }_{B_{\left|\hat{x}-x^0\right|}\left({\hat{x}}_1+2\left|\hat{x}-x^0\right|,{\left(\hat{x}\right)}^{\prime }\right)}\frac{w_{\lambda }\left(\hat{x}\right)}{{\left|\hat{x}-y\right|}^{n+2s}}\text{d}y}\\ \le \frac{C}{{\left|\hat{x}-x^0\right|}^{2s}}{w}_{\lambda }\left(\hat{x}\right),\end{array}$ (6)

另一方面，由(4)我们可以得出

$\begin{array}{c}{\left(-\Delta \right)}^s{w}_{\lambda }\left(\hat{x}\right)\ge \frac{1}{{\left|{\left(\hat{x}\right)}^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}[v^{\sigma }\left(\hat{x}\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|\hat{x}-y\right|}^q}\left({\left|v\left(y\right)\right|}^p-{\left|v\left(y^{\lambda }\right)\right|}^p\right)\text{d}y}\begin{array}{c}\begin{array}{c}\\ \end{array}\\ \end{array}\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|{\left(\hat{x}\right)}^{\lambda }-y\right|}^q}\text{d}y}\sigma v^{\sigma -1}\left(\hat{x}\right)w_{\lambda }\left(\hat{x}\right)]\\ \ge c\left(x\right)w_{\lambda }\left(\hat{x}\right)\ge c\left(\hat{x}\right)w_{\lambda }\left(\hat{x}\right),\end{array}$ (7)

其中 $c\left(x\right)=\frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}\left[\sigma v^{\sigma -1}\left(x\right){\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}-p{v}^{\sigma }\left(x\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\right]$。

下面我们需要估计 $c\left(x\right)$ 的范围，当 $\left|x\right|$ 充分大时，我们有如下估计：

$\begin{array}{c}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\le {\displaystyle {\int }_{\begin{array}{l}\\ \left|x^{\lambda }-y\right|\ge \frac{\left|x^{\lambda }-x^0\right|}{2}\end{array}}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x^{\lambda }-y\right|<\frac{\left|x^{\lambda }-x^0\right|}{2}\end{array}}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\\ \le \frac{C}{{\left|x^{\lambda }-x^0\right|}^q}+\frac{C}{{\left|x-x^0\right|}^n}\le \frac{C}{{\left|x-x^0\right|}^q},\end{array}$ (8)

以及

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^n}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\\ \le {\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|x-x^0\right|}{2}\\ \left|y-x^0\right|<\frac{\left|x-x^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|x-x^0\right|}{2}\\ \left|y-x^0\right|\ge \frac{\left|x-x^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|<\frac{\left|x-x^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\\ \le {\left({\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|x-x^0\right|}{2}\\ \left|y-x^0\right|<\frac{\left|x-x^0\right|}{2}\end{array}}{\left({\left|v\left(y\right)\right|}^{p-1}\right)}^{\frac{p}{p-1}}\text{d}y}\right)}^{\frac{p-1}{p}}{\left({\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|x-x^0\right|}{2}\\ \left|y-x^0\right|<\frac{\left|x-x^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^{pq}}\text{d}y}\right)}^{\frac{1}{p}}\\ +{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|x-x^0\right|}{2}\\ \left|y-x^0\right|\ge \frac{\left|x-x^0\right|}{2}\end{array}}{\left(\frac{3}{\left|y-x^0\right|}\right)}^q{\left(\frac{C}{{\left|y-x^0\right|}^{n-2s}}\right)}^{p-1}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|<\frac{\left|x-x^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left(\frac{C}{{\left|y-x^0\right|}^{n-2s}}\right)}^{p-1}\text{d}y}\\ \le \frac{C}{{\left|x-x^0\right|}^{q-\frac{n}{p}}}+\frac{C}{{\left|x-x^0\right|}^{2s}},\end{array}$ (9)

因此，由(8)和(9)我们可以得出

$0<c\left(x\right)\le \frac{C}{{\left|x-x^0\right|}^{n\left(1-\frac{1}{p}\right)+2s}}+\frac{C}{{\left|x-x^0\right|}^{4s}},$ (10)

联合(6) (7) (10)三式，我们有

$\left(\frac{C}{{\left|\hat{x}-x^0\right|}^{n\left(1-\frac{1}{p}\right)+2s}}+\frac{C}{{\left|\hat{x}-x^0\right|}^{4s}}\right)w_{\lambda }\left(\hat{x}\right)\le {\left(-\Delta \right)}^s{w}_{\lambda }\left(\hat{x}\right)\le \frac{C}{{\left|\hat{x}-x^0\right|}^{2s}}{w}_{\lambda }\left(\hat{x}\right),$

当 $\left|\hat{x}-x^0\right|>R_0$， $R_0$ 充分大时，此不等式不成立，因此导出矛盾假设错误，故定理得证。

通过定理3.2我们可以得到当 $\lambda$ 充分负且 $\lambda -x_1^0<-R_0$ 的时候，(3)式成立。

Step 2. 将平面 $T_{\lambda }$ 向 $x_1$ 轴正方向移动并保持 $w_{\lambda }\left(x\right)\ge 0,x\in {\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 成立至极限位置。令

${\lambda }_0=\sup \left\{\lambda <x_1^0|w_{\mu }\left(x\right)\ge 0,x\in {\Sigma }_{\mu }\backslash \left\{{\left(x^0\right)}^{\mu }\right\},\mu \le \lambda \right\},$ (11)

我们想要得到 ${\lambda }_0=x_1^0$，在这里我们使用反证法，假设 ${\lambda }_0<x_1^0$。

若 $w_{{\lambda }_0}\left(x\right)\equiv 0,x\in {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$，则 $x^0$ 不是v的奇点，因此 ${\lim }_{\left|x-x^0\right|\to \infty }{\left|x-x^0\right|}^{n-2s}u\left(x\right)<\infty$，这与定理

2.1条件矛盾。

若存在 $x^{*}\in {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$，使得 $w_{{\lambda }_0}\left(x^{*}\right)>0$。

由连续性，存在 $\delta >0$ 充分小及常数 $C>0$ 使得 $w_{{\lambda }_0}\left(x\right)\ge C>0$， $x\in B_{\delta }\left(x^{*}\right)\subset {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$。通过(5)和引理3.1我们可以计算得到对任意 $x\in {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$，有

$\begin{array}{l}{w}_{{\lambda }_0}\left(x\right)=v\left(x\right)-v\left(x^{{\lambda }_0}\right)\\ ={\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|x-y\right|}^{n-2s}}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y\right){\left|y-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\text{d}y\\ -{\displaystyle {\int }_{{ℝ}^n}\frac{R_{n,s}}{{\left|x^{{\lambda }_0}-y\right|}^{n-2s}}}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y\right){\left|y-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\text{d}y\\ =R_{n,s}{\displaystyle {\int }_{{\Sigma }_{{\lambda }_0}}\left(\frac{1}{{\left|x-y\right|}^{n-2s}}-\frac{1}{{\left|x^{{\lambda }_0}-y\right|}^{n-2s}}\right)}({\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y\right){\left|y-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\text{d}y\\ -{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y^{{\lambda }_0}-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y^{{\lambda }_0}\right){\left|y^{{\lambda }_0}-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\text{d}y)\end{array}$

$\begin{array}{l}\ge R_{n,s}{\displaystyle {\int }_{{\Sigma }_{{\lambda }_0}}\left(\frac{1}{{\left|x-y\right|}^{n-2s}}-\frac{1}{{\left|x^{{\lambda }_0}-y\right|}^{n-2s}}\right){\left|y^{{\lambda }_0}-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}}\\ \times \left({\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y\right)\text{d}y-{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y^{{\lambda }_0}-z\right|}^q}\text{d}z}{v}^{\sigma }\left(y^{{\lambda }_0}\right)\text{d}y\right)\\ =R_{n,s}{\displaystyle {\int }_{{\Sigma }_{{\lambda }_0}}\left(\frac{1}{{\left|x-y\right|}^{n-2s}}-\frac{1}{{\left|x^{{\lambda }_0}-y\right|}^{n-2s}}\right)}{\left|y^{{\lambda }_0}-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\\ \times [v^{\sigma }\left(y\right){\displaystyle {\int }_{{\Sigma }_{{\lambda }_0}}\left(\frac{1}{{\left|y-z\right|}^q}-\frac{1}{{\left|y^{{\lambda }_0}-z\right|}^q}\right)}\left({\left|v\left(z\right)\right|}^p-{\left|v\left(z^{{\lambda }_0}\right)\right|}^p\right)\text{d}z\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y^{{\lambda }_0}-z\right|}^q}\text{d}z}\left(v^{\sigma }\left(y\right)-v^{\sigma }\left(y^{{\lambda }_0}\right)\right)]\text{d}y\end{array}$

$\begin{array}{l}\ge R_{n,s}{\displaystyle {\int }_{B_{\delta }\left(x^{*}\right)}\left(\frac{1}{{\left|x-y\right|}^{n-2s}}-\frac{1}{{\left|x^{{\lambda }_0}-y\right|}^{n-2s}}\right)}{\left|y^{{\lambda }_0}-x^0\right|}^{\sigma \left(n-2s\right)+q-n-2s}\\ \times {\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(z\right)\right|}^p}{{\left|y^{{\lambda }_0}-z\right|}^q}\text{d}z}\left(v^{\sigma }\left(y\right)-v^{\sigma }\left(y^{{\lambda }_0}\right)\right)\text{d}y\\ >0.\end{array}$ (12)

我们可由(12)得到存在 $0<\eta <\left|{\lambda }_0-x_1^0\right|$ 充分小，使得

$w_{{\lambda }_0}\left(x\right)\ge C>0,x\in B_{\eta }\left({\left(x^0\right)}^{{\lambda }_0}\right)\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}.$ (13)

这等价于

$v\left(x^{{\lambda }_0}\right)-v\left(x\right)\ge C>0,x\in B_{\eta }\left(x^0\right)\backslash \left\{x^0\right\}.$

又因为v在 ${\bar{B}}_{\eta +\frac{\left|{\lambda }_0-x_1^0\right|}{2}}\left({\left(x^0\right)}^{{\lambda }_0}\right)$ 内是一致连续的，所以存在 $0<{\epsilon }_1<\frac{\left|{\lambda }_0-x_1^0\right|}{4}$ 充分小，使得对于任意的 $\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_1\right)$，有

$v\left(x^{\lambda }\right)-v\left(x\right)\ge \frac{C}{2}>0,x\in B_{\eta }\left(x^0\right)\backslash \left\{x^0\right\}.$

这等价于

$w_{\lambda }\left(x\right)=v\left(x\right)-v\left(x^{\lambda }\right)\ge \frac{C}{2}>0,x\in B_{\eta }\left({\left(x^0\right)}^{\lambda }\right)\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}.$

因此，对任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_1\right)$，有 ${\text{Σ}}_{\lambda }^{-}\cap B_{\eta }\left({\left(x^0\right)}^{\lambda }\right)=\varnothing$，故 $w_{\lambda }\in C_{loc}^{1,1}\left({\Sigma }_{\lambda }^{-}\right)\cap L_s\left({ℝ}^n\right)$ 并且

(F)若 ${\Sigma }_{\lambda }^{-}\ne \varnothing$，则 $w_{\lambda }$ 负最小值在 ${\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 内部达到。

由此结合定理3.2，我们可以推导出对于任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_1\right)$，有

(H) $w_{\lambda }$ 的负最小值不可能在 $\left({\Sigma }_{\lambda }\cap {\bar{B}}_{R_0}^c\left(x^0\right)\right)\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 内取到。

现在，我们需要下列定理

定理3.3 (狭窄区域定理)设 $\Omega$ 为 $\left\{x\in {ℝ}^n|\lambda -<x_1<\lambda <x_1^0\right\}\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 中的狭窄区域，l充分小， $w_{\lambda }\in C_{loc}^{1,1}\left(\Omega \right)\cap L_s\left({ℝ}^n\right)$ 且满足

(a) $\begin{array}{l}{\left(-\Delta \right)}^s{w}_{\lambda }\left(x\right)\ge \frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}[-p{v}^{\sigma }\left(x\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}{w}_{\lambda }\left(y\right)\text{d}y}\begin{array}{c}\\ \\ \end{array}\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\sigma v^{\sigma -1}\left(x\right)w_{\lambda }\left(x\right)],x\in \Omega \cap {\Sigma }_{\lambda }^{-}\end{array}$ ；

(b) 若 ${\Sigma }_{\lambda }^{-}\ne \varnothing$，则 $w_{\lambda }$ 负最小值在 ${\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 内部达到；

(c) $w_{\lambda }$ 负最小值不在 $\left({\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}\right)\backslash \Omega$ 内部达到；

则存在 $l_0$ 充分小(关于 $\lambda$ 连续)，使得对于任意 $0<l\le l_0$，任意 $x\in \Omega$，有 $w_{\lambda }\left(x\right)\ge 0$。

证明：反设对任意 $l_0$ 充分小，存在 $0<l\le l_0$，存在 $\bar{x}\in \Omega$，使得 $w_{\lambda }\left(\bar{x}\right)={\min }_{{\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}}{w}_{\lambda }\left(x\right)<0$。

一方面，类似 [1] 中式子22的计算步骤我们可以得到

${\left(-\Delta \right)}^s{w}_{\lambda }\left(\bar{x}\right)\le 2C_{n,s}{\displaystyle {\int }_{{\Sigma }_{\lambda }^c}\frac{w_{\lambda }\left(\bar{x}\right)}{{\left|\bar{x}-y\right|}^{n+2s}}\text{d}y}\le \frac{2C^{*}}{l^{2s}}{w}_{\lambda }\left(\bar{x}\right)$ (14)

另一方面，由(a)可得

$\begin{array}{c}{\left(-\Delta \right)}^s{w}_{\lambda }\left(\bar{x}\right)\ge \frac{1}{{\left|{\bar{x}}^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}[-p{v}^{\sigma }\left(\bar{x}\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|\bar{x}-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\begin{array}{c}\\ \\ \end{array}\\ +{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|{\bar{x}}^{\lambda }-y\right|}^q}\text{d}y}\sigma v^{\sigma -1}\left(\bar{x}\right)]w_{\lambda }\left(\bar{x}\right),\end{array}$ (15)

记 $h\left(x\right)=\frac{1}{{\left|x^{\lambda }-x^0\right|}^{n+2s-\sigma \left(n-2s\right)-q}}\left[-p{v}^{\sigma }\left(x\right){\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}+{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}\sigma v^{\sigma -1}\left(x\right)\right]$。

下面我们需要估计 $h\left(x\right)$ 的范围，类似(8) (9)的过程，我们可以得到对于任意 $x\in \text{Ω}$ 有如下估计：

$\begin{array}{c}{\displaystyle {\int }_{{ℝ}^n}\frac{{\left|v\left(y\right)\right|}^p}{{\left|x^{\lambda }-y\right|}^q}\text{d}y}\le {\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|<\frac{\left|\lambda -x_1^0\right|}{2}\end{array}}\frac{{\left|v\left(y^{\lambda }\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|\ge \frac{\left|\lambda -x_1^0\right|}{2}\end{array}}\frac{{\left|v\left(y^{\lambda }\right)\right|}^p}{{\left|x-y\right|}^q}\text{d}y}\\ \le \frac{C}{{\left|\lambda -x_1^0\right|}^n}{\max }_{\left|x-x^0\right|\le \frac{4}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^p+\frac{C}{{\left|\lambda -x_1^0\right|}^q},\end{array}$ (16)

以及

$\begin{array}{l}{\displaystyle {\int }_{{\Sigma }_{\lambda }^{-}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\\ \le {\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|<\frac{\left|\lambda -x_1^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|>\frac{\left|\lambda -x_1^0\right|}{2}\\ \left|y-x^0\right|<\frac{\left|\lambda -x_1^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}+{\displaystyle {\int }_{\begin{array}{l}\\ \left|x-y\right|>\frac{\left|\lambda -x_1^0\right|}{2}\\ \left|y-x^0\right|\ge \frac{\left|\lambda -x_1^0\right|}{2}\end{array}}\frac{1}{{\left|x-y\right|}^q}{\left|v\left(y\right)\right|}^{p-1}\text{d}y}\\ \le \frac{C}{{\left|\lambda -x_1^0\right|}^s}{\max }_{\left|x-x^0\right|\le \frac{4}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^{p-1}+\frac{C}{{\left|\lambda -x_1^0\right|}^{q-\frac{n}{p}}}+\frac{C}{{\left|\lambda -x_1^0\right|}^{2s}}{\max }_{\left|x-x^0\right|\le \frac{2}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^{p-1},\end{array}$ (17)

故当 $x\in \Omega$ 时，由(16)和(17)可得

$0<h\left(x\right)\le C_{\lambda },$ (18)

其中

$\begin{array}{c}{C}_{\lambda }=\frac{C}{{\left|\lambda -x_1^0\right|}^{2s+n\left(1-\frac{1}{p}\right)}}{\max }_{\left|x-x^0\right|\le \frac{4}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^{\sigma }+\frac{C}{{\left|\lambda -x_1^0\right|}^{4s+n-q}}{\max }_{\left|x-x^0\right|\le \frac{4}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^{\sigma +p-1}\\ +\frac{C}{{\left|\lambda -x_1^0\right|}^{4s}}{\max }_{\left|x-x^0\right|\le \frac{4}{\left|\lambda -x_1^0\right|}}{\left|u\left(x\right)\right|}^{\sigma -1}.\end{array}$

因此存在 $l_0\left(\lambda \right)$ 充分小使得 $C^{*}{l}_0^{-2s}=C_{\lambda }$，联合(14) (15) (18)，我们有

$C_{\lambda }{w}_{\lambda }\left(\bar{x}\right)\le h\left(x\right)w_{\lambda }\left(\bar{x}\right)\le {\left(-\Delta \right)}^s{w}_{\lambda }\left(\hat{x}\right)\le \frac{2C^{*}}{l^{2s}}{w}_{\lambda }\left(\bar{x}\right)\le \frac{2C^{*}}{l_0^{2s}}{w}_{\lambda }\left(\bar{x}\right)=2C_{\lambda }{w}_{\lambda }\left(\bar{x}\right)<0.$

此不等式矛盾，故假设错误，定理得证。

现在我们定义

$\overline{l_0}:={\min }_{\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_1\right]}{l}_0\left(\lambda \right)>0,$ (19)

给定 $0<{\delta }_0<\overline{l_0}$ 充分小，联合(12)(13)可以得到 $w_{{\lambda }_0}\left(x\right)\ge C>0,x\in \overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$，

这等价于

$v\left(x^{{\lambda }_0}\right)-v(x)\ge C>0,x\in {\left(\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\right)}^{{\lambda }_0}\backslash \left\{x^0\right\},$

其中 ${\left(\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\right)}^{{\lambda }_0}$ 表示 ${\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)$ 关于超平面 $T_{{\lambda }_0}$ 的反射。因为v在闭区域 $\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}$ 内一致连续，故存在 $0<{\epsilon }_2\left({\delta }_0\right)<\frac{{\delta }_0}{2}$ 充分小，对于任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_2\right)$，有

$v\left(x^{\lambda }\right)-v\left(x\right)\ge \frac{C}{2}>0,x\in {\left(\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\right)}^{{\lambda }_0}\backslash \left\{x^0\right\},$

这等价于

$w_{\lambda }\left(x\right)=v\left(x\right)-v\left(x^{\lambda }\right)\ge \frac{C}{2}>0,x\in {\left[{\left(\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\right)}^{{\lambda }_0}\right]}^{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\},$

容易验证 $\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}\subset {\left[{\left(\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\right)}^{{\lambda }_0}\right]}^{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$。

因此，对于任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+{\epsilon }_2\right)$，我们有

$w_{\lambda }\left(x\right)=v\left(x\right)-v\left(x^{\lambda }\right)\ge \frac{C}{2}>0,x\in \overline{{\Sigma }_{{\lambda }_0-{\delta }_0}\cap B_{2R_0}\left(x^0\right)}\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}.$ (20)

定义 $\epsilon :=\min \left\{{\epsilon }_1,{\epsilon }_2\right\}>0$，我们结合(H)以及(20)可以得到，对于任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+\epsilon \right)$，

(G) $w_{\lambda }$ 负最小值不能在 $\overline{{\Sigma }_{{\lambda }_0-{\delta }_0}}\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$ 内取到。

令 $\Omega ={\Sigma }_{\lambda }^{-}\backslash \overline{{\Sigma }_{{\lambda }_0-{\delta }_0}}$，当给定 $0<{\delta }_0<\overline{l_0}$ 时，联合(4) (F) (19) (G)以及定理3.3，我们可以证明对任意 $\lambda \in \left[{\lambda }_0,{\lambda }_0+\epsilon \right)$，有 $w_{\lambda }\left(x\right)\ge 0$， $x\in {\Sigma }_{\lambda }\backslash \left\{{\left(x^0\right)}^{\lambda }\right\}$，这与(11)矛盾，因此我们得到 ${\lambda }_0=x_1^0$， $w_{{\lambda }_0}\left(x\right)\ge 0,x\in {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$。

同样的方法，我们将平面从 $x_1$ 轴充分正的地方向 $x_1$ 轴负方向移动平面。因此我们可以得到 ${\lambda }_0=x_1^0$， $w_{{\lambda }_0}\left(x\right)\equiv 0,x\in {\Sigma }_{{\lambda }_0}\backslash \left\{{\left(x^0\right)}^{{\lambda }_0}\right\}$。因为方程(1)是旋转不变的，因此我们得到定理2.1。

4. 总结

本文通过使用 [12] 中的移动平面法，增加了新的条件，即存在某点 $x^0\in {ℝ}^n$ 使得式子

${\lim }_{\left|x-x^0\right|\to \infty }{\left|x-x^0\right|}^{n-2s}u\left(x\right)=\infty$ 成立，考虑了方程(1)的负解，同时将 [12] 中的系数范围 $0<s<1$， $p=2$， $q=4s$， $\sigma =1$ 推广到了 $0<s<1$， $0\le q\le 4s$， $n>\max \left\{2s,q\right\}$， $p=\frac{2n-q}{n-2s}$， $1\le \sigma \le \frac{n+2s-q}{n-2s}$，且 $\sigma$ 为

奇数。最后也得到了解的径向对称性的结论。