一类分数阶调和映射迭代算法的收敛性

Convergence of Iterative Algorithm for a Class of Fractional Harmonic Maps

摘要: 基于能量泛函的极小值问题，本文给出了分数阶调和映射方程的一类迭代算法，我们证明了该算法的收敛性。

Abstract: This paper proposes an iterative algorithm for fractional harmonic maps and its convergence is proved.

1. 引言

设 $\Omega$ 是 ${ℝ}^N$ 中的光滑有界区域， $N=1,2,3$， ${\mathbb{S}}^2$ 是 ${ℝ}^{\text{3}}$ 中的标准单位球， $\text{0}<s<1$。定义映射 $u:\Omega \to {\mathbb{S}}^2$ 的非局部能量为

$E_s\left(u\right)=\frac{1}{2}{\displaystyle {\iint }_{\Omega \times \Omega }\frac{{\left|u\left(x\right)-u\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y},$ (1)

我们称(1)的临界点为s-阶调和映射。令 $\partial \Omega$ 表示 $\Omega$ 的边界， ${E^{\prime }}_s$ 表示 $E_s$ 在函数空间

$H^s\left(\Omega ,{ℝ}^3\right):=\left\{u:\Omega \to {ℝ}^3|E_s\left(u\right)+{\displaystyle {\int }_{\Omega }{\left|u\right|}^2\text{d}x}<\infty \right\}$

中的Frechet导数，即：对任意的 $u\in H^s\left(\Omega ,{\mathbb{S}}^2\right)$ 和 $v\in H_0^s\left(\Omega ,{ℝ}^3\right)$，

${E^{\prime }}_s\left(u\right)\left[v\right]={\displaystyle {\iint }_{\Omega \times \Omega }\frac{\left(u\left(x\right)-u\left(y\right)\right)\cdot \left(v\left(x\right)-v\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}.$

这里，

且

$H_0^s\left(\Omega ,{ℝ}^3\right)=\left\{u|u\in H^s\left(\Omega ,{ℝ}^3\right),{u|}_{\partial \Omega }=\stackrel{\to }{0}\right\}.$

于是具有固定边值条件的s-阶调和映射方程为：

(2)

这里 $n_0\left(x\right)$ 是一个固定的函数且对几乎处处的 $x\in \partial \Omega$， $\left|n_0\right|=1$。

设 $u\in {\stackrel{\dot{}}{H}}^s\left(\Omega ,\mathcal{N}\right)$，其中， $\mathcal{N}\subset {ℝ}^m$ 是无边的k-维光滑子流形。如果函数u是泛函 $E_s\left(u\right)$ 在扰动 $t\mapsto {\Pi }_{\mathcal{N}}\left(u+t\phi \right)$ 下的临界点，则称u为s-阶调和映射。其中， $\phi \in C_c^{\infty }\left(\Omega \right)$， ${\Pi }_{\mathcal{N}}:{ℝ}^m\to \mathcal{N}$ 是 ${ℝ}^m$ 到子流形 $\mathcal{N}$ 上的投影映射。当s = 1时，这是经典调和映射的概念，文献 [1] [2] 和 [3] 对调和映射的当前研究情况进行了系统的总结。近些年以来，人们对s为非整数的情形非常感兴趣，比如当靶流形 $\mathcal{N}$ 为标准黎曼球面时，文献 [4] 利用三交换子估计方法，证明了1/2-阶调和映射的光滑性；文献 [5] 得到了n/2-阶调和映射的光滑性；文献 [6] 分析了1/2-阶调和映射的紧致性和爆破性质；文献 [7] 证明了n/2-阶调和映射的Hölder连续性。当靶流形 $\mathcal{N}$ 为一般的紧致黎曼流形时，文献 [8] 得到了1/2-阶调和映射的Hölder连续性。文献 [9] 证明了靶流形分别为球和黎曼齐次流形时积分微分调和映射热流整体弱解的存在性。Alouges在文献 [10] 给出了一类经典调和映射能量极小问题的迭代算法，文献 [11] 讨论调和映射逼近格式的稳定性和收敛性，介绍了一种基于Alouges的迭代算法的有限元离散化方法，并证明了它仅对锐角型三角剖分是稳定和收敛的，并提出了一个后验准则，使我们能够得到一般三角剖分上调和映射弱收敛的充分条件和自适应网格细化的充分条件，数值模拟实验进一步表明，自适应策略可以自动细化典型奇异点附近的三角剖分，从而提高其运算效率。

在本文中，基于能量泛函(1)极小化问题，我们对s-阶调和映射(2)提出一类迭代算法，将经典的Alouges算法 [10] 推广到分数阶方程的情形(2)。具体地，我们有：

算法1 设 $u_0\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$，对 $n=0,1,\cdots$，直至收敛，第一步，定义 $K_u$ = { $w\in H_0^s\left(\Omega ,{ℝ}^3\right)$ 使得 $w\left(x\right)\cdot u\left(x\right)=0$ 几乎处处成立}，求解极小值问题

$\underset{w\in K_u}{\min }{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w\left(x\right)-u_n\left(y\right)+w\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$ (3)

设 $w_n=w\left(u_n\right)$ 为此问题的解。第二步，令 $u_{n+1}:=\frac{u_n-w_n}{\left|u_n-w_n\right|}$。

在算法1中，。于是我们有以下定理：

定理1 算法1中的序列 ${\left\{u_n\right\}}_{n=0}^{\infty }$ 在 $H^s\left(\Omega ,{ℝ}^3\right)$ 中弱收敛到s-阶调和映射 $u_{\infty }\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$。进一步， ${\left\{u_n\right\}}_{n=0}^{\infty }$ 在 $H_0^s\left(\Omega ,{ℝ}^3\right)$ 中强收敛到 $\stackrel{\to }{0}$。

注记1 在(2)两边同时取内积可知， $u\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$ 在弱解意义下是s-阶调和映射当且仅当

$0={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u\left(x\right)-u\left(y\right)\right)\cdot \left(\psi \left(x\right)-\psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}},\forall \psi \in K_u.$

2. 定理1的证明

我们将证明分成以下几步：

引理2.1 我们有

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-u_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}+{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.\end{array}$

证明：因为 $w_n$ 是问题(3)的解，对任意 $\psi \in K_{u_n}$ 我们有

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w\left(x\right)-u_n\left(y\right)+w\left(y\right)\right)\left(\psi \left(x\right)-\psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}=0.$

取 $\psi =w_n$，我们得到

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-u_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}+{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ +2{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w\left(x\right)-u_n\left(y\right)+w\left(y\right)\right)\left(\psi \left(x\right)-\psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}+{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.\end{array}$

引理2.2 我们有

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_{n+1}\left(x\right)-u_{n+1}\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

证明：因为 $u_n\left(x\right)\cdot w_n\left(x\right)=0$，我们有 ${\left|u_n\left(x\right)-w_n\left(x\right)\right|}^2=1+{\left|w_n\right|}^2\ge 1$，所以

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_{n+1}\left(x\right)-u_{n+1}\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|\frac{u_n-w_n}{\left|u_n-w_n\right|}\left(x\right)-\frac{u_n-w_n}{\left|u_n-w_n\right|}\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{2-2\frac{\left(u_n\left(x\right)-w_n\left(x\right)\right)\cdot \left(u_n\left(y\right)-w_n\left(y\right)\right)}{\left|u_n\left(x\right)-w_n\left(x\right)\right|\left|u_n\left(y\right)-w_n\left(y\right)\right|}}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ \le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{2\left|u_n\left(x\right)-w_n\left(x\right)\right|\left|u_n\left(y\right)-w_n\left(y\right)\right|-2\left(u_n\left(x\right)-w_n\left(x\right)\right)\cdot \left(u_n\left(y\right)-w_n\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ \le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.\end{array}$

引理2.3 $w_n$ 在空间 $H^s\left(\Omega ,{ℝ}^3\right)$ 中强收敛至 $\stackrel{\to }{0}$。

证明：根据引理2.1和引理2.2，我们有

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_{n+1}\left(x\right)-u_{n+1}\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-u_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}-{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.\end{array}$

从而，我们有

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_n\left(x\right)-u_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}-{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_{n+1}\left(x\right)-u_{n+1}\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

对这个不等式求和，设N为任意正整数，于是我们有

${\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}}\le {\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|u_0\left(x\right)-u_0\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

所以级数 ${\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_n\left(x\right)-w_n\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}}$ 收敛。

再根据Sobolev嵌入定理， $w_n$ 在空间 $H^s\left(\Omega ,{ℝ}^3\right)$ 中强收敛至 $\stackrel{\to }{0}$。

引理2.4 $u_n$ 在空间 $H^s\left(\Omega ,{ℝ}^3\right)$ 中弱收敛到 $u_{\infty }\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$ 且 $u_{\infty }$ 是s-阶调和映射。

证明：根据引理2.1可知， $u_n$ 在空间 $H^s\left(\Omega ,{ℝ}^3\right)$ 中有界，从而存在使得 $u_n$ 在空间 $H^s\left(\Omega ,{ℝ}^3\right)$ 中弱收敛到 $u_{\infty }$。因为 $u_n\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$ 且 $H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$ 在弱拓扑意义下是 $H_{n_0}^s\left(\Omega ,{ℝ}^3\right)$ 的闭子集，从而 $u_{\infty }\in H_{n_0}^s\left(\Omega ,{\mathbb{S}}^2\right)$。

下面我们证明 $u_{\infty }$ 是s-阶调和映射。注意到 $w_n$ 满足的欧拉–拉格朗日方程为

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w\left(y\right)\right)\left(\Psi \left(x\right)-\Psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}=0,\Psi \in K_{u_n}.$

取 $\Psi =\varphi \times u_n$，这里 $\varphi \in C_0^{\infty }\left(\Omega ,{ℝ}^3\right)$，于是我们有

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right)\left(\varphi \left(x\right)\times u_n\left(x\right)-\varphi \left(y\right)\times u_n\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}=0.$

进一步，我们有

$\begin{array}{c}0={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right)\left(\varphi \left(x\right)\times u_n\left(x\right)-\varphi \left(y\right)\times u_n\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right)\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\times u_n\left(x\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ +{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-w_n\left(x\right)-u_n\left(y\right)+w_n\left(y\right)\right)\left[\varphi \left(y\right)\times \left(u_n\left(x\right)-u_n\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_n\left(x\right)-u_n\left(y\right)\right)\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\times u_n\left(x\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ -{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(w_n\left(x\right)-w_n\left(y\right)\right)\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\times u_n\left(x\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\end{array}$

$\begin{array}{l}-{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(w_n\left(x\right)-w_n\left(y\right)\right)\left[\varphi \left(y\right)\times \left(u_n\left(x\right)-u_n\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ ={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\cdot \left[u_n\left(x\right)\times \left(u_n\left(x\right)-u_n\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ -{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\cdot \left[u_n\left(x\right)\times \left(w_n\left(x\right)-w_n\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}\\ -{\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\varphi \left(y\right)\cdot \left[\left(u_n\left(x\right)-u_n\left(y\right)\right)\times \left(w_n\left(x\right)-w_n\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}},\end{array}$

取极限我们得到

$0={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(\varphi \left(x\right)-\varphi \left(y\right)\right)\cdot \left[u_{\infty }\left(x\right)\times \left(u_{\infty }\left(x\right)-u_{\infty }\left(y\right)\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

这等价于

$0={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_{\infty }\left(x\right)-u_{\infty }\left(y\right)\right)\cdot \left[u_{\infty }\left(x\right)\times \varphi \left(x\right)-u_{\infty }\left(y\right)\times \varphi \left(y\right)\right]}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

最后，令 $\varphi =u_{\infty }\times \psi$， $\psi \in C_0^{\infty }\left(\Omega ,{ℝ}^3\right)$ 且 $\psi \in K_{u_{\infty }}$，于是根据向量恒等式：对于任意的 $a,b,c\in {ℝ}^3$， $a\times \left(b\times c\right)=\langle a,c\rangle b-\langle a,b\rangle c$，我们有

$\begin{array}{c}{u}_{\infty }\left(x\right)\times \varphi \left(x\right)=u_{\infty }\left(x\right)\times \left(u_{\infty }\left(x\right)\times \psi \left(x\right)\right)\\ =\left(u_{\infty }\left(x\right)\cdot \psi \left(x\right)\right)u_{\infty }\left(x\right)-\left(u_{\infty }\left(x\right)\cdot u_{\infty }\left(x\right)\right)\psi \left(x\right)\\ =\psi \left(x\right),\end{array}$

所以

$0={\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u_{\infty }\left(x\right)-u_{\infty }\left(y\right)\right)\cdot \left(\psi \left(x\right)-\psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}.$

因此，根据注记1可知 $u_{\infty }$ 在弱解意义下是s-阶调和映射。

综上所述，定理1得证。

注记2 设 $w\left(u\right)$ 是极小值问题(3)的解，则它是唯一的。事实上，设 $w_1$ 和 $w_2$ 都是解，则

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u\left(x\right)-w_1\left(x\right)-u\left(y\right)+w_1\left(y\right)\right)\cdot \left(w_1\left(x\right)-w_2\left(x\right)-w_1\left(y\right)+w_2\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}$

和

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u\left(x\right)-w_2\left(x\right)-u\left(y\right)+w_2\left(y\right)\right)\cdot \left(w_1\left(x\right)-w_2\left(x\right)-w_1\left(y\right)+w_2\left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}$

成立，于是我们有 ${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{{\left|w_1\left(x\right)-w_2\left(x\right)-w_1\left(y\right)+w_2\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}=0$。这意味着 $w_1=w_2$。

注记3 设 $w\left(u\right)$ 是极小值问题(3)的解，则u是s-阶调和映射当且仅当 $w\left(u\right)=0$。事实上，若 $w\left(u\right)=0$，我们有

${\displaystyle {\int }_{\Omega }{\displaystyle {\int }_{\Omega }\frac{\left(u\left(x\right)-u\left(y\right)\right)\cdot \left(\Psi \left(x\right)-\Psi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}=0,\Psi \in K_u.$

从而u是s-阶调和映射。上述过程可逆，从而充分条件成立。