1. 弱调和映照
假设$(M,g)$, $(N,h)$是两个黎曼流形, 且$N \hookrightarrow \mathbb{R}^K $. 定义
\[
H^1(M,N):=\left\{ u\in L_{\mathrm{loc}}^1(M,\mathbb{R}^{K+1}):\int_{M}|\nabla u|^2<+\infty \& u(x)\in N \text{a.e.} \right\}.
\]
对$u\in H^1(M,N)$, 定义能量泛函
\[
E(u)=\int_{M}L(x,u,\nabla u)=\int_M\frac{1}{2}g^{\alpha\beta}\left\langle\frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta}\right\rangle.
\]
Definition 1. 我们称$u\in H^1(M,N)$是若调和映照, 如果$u$是$E(u)$在如下变分的临界点:
\[
\lim_{t\to0}\frac{E(P(u+tv))-E(u)}{t}=0,\quad\forall v\in H_0^1(M,\mathbb{R}^{K})\bigcap L^\infty(M,\mathbb{R}^{K}),
\]
其中$H_0^1$是$C_0^\infty$在$H^1$中的完备化, 而$P$是$N$的最近点投射.
下面, 我们来计算其欧拉—拉格朗日方程.
Lemma 2. $u\in H^1(M,N)$是若调和映照当且仅当如下方程在弱意义下成立
\[
\Delta_g u+g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right)=0,
\]
即对任何$v\in H_0^1(M,\mathbb{R}^{K})\bigcap L^\infty(M,\mathbb{R}^K)$, 有
\[
\int_{M}-g^{\alpha\beta}(x) \left\langle \frac{\partial u}{\partial x^\alpha},\frac{\partial v}{\partial x^\beta} \right\rangle +g^{\alpha\beta}(x)\left\langle A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right),v \right\rangle=0,
\]
其中$A$是$N \hookrightarrow \mathbb{R}^K$的第二基本型.
\[
\Delta_g u+g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right)=0,
\]
即对任何$v\in H_0^1(M,\mathbb{R}^{K})\bigcap L^\infty(M,\mathbb{R}^K)$, 有
\[
\int_{M}-g^{\alpha\beta}(x) \left\langle \frac{\partial u}{\partial x^\alpha},\frac{\partial v}{\partial x^\beta} \right\rangle +g^{\alpha\beta}(x)\left\langle A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right),v \right\rangle=0,
\]
其中$A$是$N \hookrightarrow \mathbb{R}^K$的第二基本型.
Proof . 我们直接计算得到
\begin{align*}
\left.\frac{d}{dt}\right\rvert_{t=0}\left( \frac{1}{2}|\nabla \Pi(u+tv)|^2 \right)
&=\left\langle\nabla u,\left.\frac{d}{dt}\right\rvert_{t=0}(\nabla\Pi(u+tv))\right\rangle\\
&=\left\langle \nabla u,\nabla [D\Pi(u)](v) \right\rangle\\
&=\left\langle \nabla u,\nabla v^\top \right\rangle,
\end{align*}
因此由$v\in H_0^1(M, \mathbb{R}^K)$并利用分部积分知
\[
\left.\frac{d}{dt}\right\rvert_{t=0}\int_{M}\frac{1}{2}|\nabla \Pi(u+tv)|^2
=\int_M\left\langle \nabla u,\nabla v^\top \right\rangle
=-\int_{M}\left\langle \Delta u,v^\top \right\rangle
=-\int_{M}\left\langle [\Delta u]^\top,v \right\rangle.
\]
现在假设$\set{\nu_i(u)}$是$u$处$T_uN^\perp$的幺正基底, 则注意到$\left\langle \nabla u,\nu_i(u) \right\rangle=0$, 我们知道
\begin{align*}
\left\langle \Delta u,\nu_i(u) \right\rangle &= \mathrm{div}\left( \left\langle \nabla u,\nu_i(u) \right\rangle \right)-\left\langle \nabla u,D(\nu_i(u)) \right\rangle
=-\left\langle \nabla u, [D\nu_i]( \nabla u) \right\rangle\\
&=-\left\langle \nabla u, D_{\nabla u}\nu_i\right\rangle
=-A^i(u)(\nabla u,\nabla u).
\end{align*}
最后一个等号中, 我们用了第二基本型$A$的定义. 这样, 我们得到
\[
\Delta u-[\Delta u]^\top=[\Delta u]^\perp=\left\langle \Delta u,\nu_i \right\rangle\nu_i=-A^i(u)(\nabla u,\nabla u)\nu_i=-A(u)(\nabla u,\nabla u).
\]
这表明
\[
\left.\frac{d}{dt}\right\rvert_{t=0}\int_{M}\frac{1}{2}|\nabla \Pi(u+tv)|^2=-\int_{M}\left\langle [\Delta u]^\top,v \right\rangle
=-\int_M \left\langle \Delta u+A(u)(\nabla u,\nabla u), v \right\rangle.
\]
由此即得其欧拉—拉格朗日方程为
\[
\Delta u+A(u)(\nabla u,\nabla u)=0.
\]
最后注意到$A$的双线性性, 以及$\nabla u=g^{\alpha\delta}\frac{\partial u}{\partial x^\alpha}\frac{\partial }{\partial x^\delta}$
\[
A(u)(\nabla u,\nabla u)=g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right).
\]
事实上, 一个仔细的计算表明
\begin{align*}
\left\langle \Delta u, \nu_i(u) \right\rangle_N \nu_i(u)
&=\left\langle \mathrm{div}_g(\nabla_g u),\nu_i(u) \right\rangle_N \nu_i(u)\\
&=\mathrm{div}_g\left\langle \nabla_gu,\nu_i(u) \right\rangle_N\nu_i(u)
-g^{\alpha\beta}\left\langle \nabla_{\frac{\partial}{\partial x^\alpha}}u,D_{\frac{\partial}{\partial x^\beta}}\nu_i(u) \right\rangle_{ N}\nu_i(u)\\
&=-g^{\alpha\beta}\left\langle \frac{\partial u}{\partial x^\alpha},(D\nu_i)(\nabla_{\frac{\partial}{\partial x^\beta}}u) \right\rangle_{N}\nu_i(u)\\
&=-g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right).
\end{align*}
\begin{align*}
\left.\frac{d}{dt}\right\rvert_{t=0}\left( \frac{1}{2}|\nabla \Pi(u+tv)|^2 \right)
&=\left\langle\nabla u,\left.\frac{d}{dt}\right\rvert_{t=0}(\nabla\Pi(u+tv))\right\rangle\\
&=\left\langle \nabla u,\nabla [D\Pi(u)](v) \right\rangle\\
&=\left\langle \nabla u,\nabla v^\top \right\rangle,
\end{align*}
因此由$v\in H_0^1(M, \mathbb{R}^K)$并利用分部积分知
\[
\left.\frac{d}{dt}\right\rvert_{t=0}\int_{M}\frac{1}{2}|\nabla \Pi(u+tv)|^2
=\int_M\left\langle \nabla u,\nabla v^\top \right\rangle
=-\int_{M}\left\langle \Delta u,v^\top \right\rangle
=-\int_{M}\left\langle [\Delta u]^\top,v \right\rangle.
\]
现在假设$\set{\nu_i(u)}$是$u$处$T_uN^\perp$的幺正基底, 则注意到$\left\langle \nabla u,\nu_i(u) \right\rangle=0$, 我们知道
\begin{align*}
\left\langle \Delta u,\nu_i(u) \right\rangle &= \mathrm{div}\left( \left\langle \nabla u,\nu_i(u) \right\rangle \right)-\left\langle \nabla u,D(\nu_i(u)) \right\rangle
=-\left\langle \nabla u, [D\nu_i]( \nabla u) \right\rangle\\
&=-\left\langle \nabla u, D_{\nabla u}\nu_i\right\rangle
=-A^i(u)(\nabla u,\nabla u).
\end{align*}
最后一个等号中, 我们用了第二基本型$A$的定义. 这样, 我们得到
\[
\Delta u-[\Delta u]^\top=[\Delta u]^\perp=\left\langle \Delta u,\nu_i \right\rangle\nu_i=-A^i(u)(\nabla u,\nabla u)\nu_i=-A(u)(\nabla u,\nabla u).
\]
这表明
\[
\left.\frac{d}{dt}\right\rvert_{t=0}\int_{M}\frac{1}{2}|\nabla \Pi(u+tv)|^2=-\int_{M}\left\langle [\Delta u]^\top,v \right\rangle
=-\int_M \left\langle \Delta u+A(u)(\nabla u,\nabla u), v \right\rangle.
\]
由此即得其欧拉—拉格朗日方程为
\[
\Delta u+A(u)(\nabla u,\nabla u)=0.
\]
最后注意到$A$的双线性性, 以及$\nabla u=g^{\alpha\delta}\frac{\partial u}{\partial x^\alpha}\frac{\partial }{\partial x^\delta}$
\[
A(u)(\nabla u,\nabla u)=g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right).
\]
事实上, 一个仔细的计算表明
\begin{align*}
\left\langle \Delta u, \nu_i(u) \right\rangle_N \nu_i(u)
&=\left\langle \mathrm{div}_g(\nabla_g u),\nu_i(u) \right\rangle_N \nu_i(u)\\
&=\mathrm{div}_g\left\langle \nabla_gu,\nu_i(u) \right\rangle_N\nu_i(u)
-g^{\alpha\beta}\left\langle \nabla_{\frac{\partial}{\partial x^\alpha}}u,D_{\frac{\partial}{\partial x^\beta}}\nu_i(u) \right\rangle_{ N}\nu_i(u)\\
&=-g^{\alpha\beta}\left\langle \frac{\partial u}{\partial x^\alpha},(D\nu_i)(\nabla_{\frac{\partial}{\partial x^\beta}}u) \right\rangle_{N}\nu_i(u)\\
&=-g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right).
\end{align*}
特别地, 当$N=S^n$时, 我们有
Corollary 3. 假设$u\in H^1(M,S^n)$是若调和映照, 则其欧拉—拉格朗日方程是
\[
\Delta_g u+|\nabla_gu|^2u=0.
\]
\[
\Delta_g u+|\nabla_gu|^2u=0.
\]
Proof . 若$N=S^n \hookrightarrow \mathbb{R}^{n+1}$, 则我们可以将$S^n$的外法向$\nu=u$径向延拓为$\mathbb{R}^{n+1}$中的单位向量场$\tilde u(y)=u(y/|y|)$. 这样, 计算其第二基本型有, 对$X,Y\in T_uS^n$,
\[
A(u)(X,Y)= \left\langle D_X\nu,Y \right\rangle \nu = \left\langle \bar D_X \tilde u,Y \right\rangle \nu
= \left\langle (d\tilde u)(X),Y \right\rangle u
=\left\langle X,Y \right\rangle u.
\]
其中, 我们首先利用了嵌入子流形诱导联络的基本关系$D_X\nu=[\bar D_X\nu]^\top$, 这里$\bar D$是欧氏空间$\mathbb{R}^{n+1}$中的Levi-Civita联络, 其实它就是方向导数(可以验证其度量相容性与无挠性). 即$D_X \tilde u=(d\tilde u)(X)$. 又注意到, $\tilde u(y)=y$其实就是该电处的位置向量. 而关于位置向量, 我们容易得到$d(\tilde u)(X)=X$.
\[
A(u)(X,Y)= \left\langle D_X\nu,Y \right\rangle \nu = \left\langle \bar D_X \tilde u,Y \right\rangle \nu
= \left\langle (d\tilde u)(X),Y \right\rangle u
=\left\langle X,Y \right\rangle u.
\]
其中, 我们首先利用了嵌入子流形诱导联络的基本关系$D_X\nu=[\bar D_X\nu]^\top$, 这里$\bar D$是欧氏空间$\mathbb{R}^{n+1}$中的Levi-Civita联络, 其实它就是方向导数(可以验证其度量相容性与无挠性). 即$D_X \tilde u=(d\tilde u)(X)$. 又注意到, $\tilde u(y)=y$其实就是该电处的位置向量. 而关于位置向量, 我们容易得到$d(\tilde u)(X)=X$.
由此, 我们知道
\[
g^{\alpha\beta}A(u)\left( \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right)=g^{\alpha\beta}\left\langle \frac{\partial u}{\partial x^\alpha},\frac{\partial u}{\partial x^\beta} \right\rangle u
=|\nabla_g u|^2u.
\]