证明 由于$F$可微且$\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z}\ne 0$，由隐函数定理知方程$F(x,y,z)=F(x_0,y_0,z_0)$唯一确定了三个可微函数$x=x(y,z),y=y(x,z),z=z(x,y)$，且满足$$\begin{array}{}\text{(4)}& \begin{array}{rl}{\left(\frac{\partial x}{\partial y}\right)}_z& =-{\left(\frac{\partial F}{\partial x}\right)}^{-1}\left(\frac{\partial F}{\partial y}\right)\\ {\left(\frac{\partial y}{\partial z}\right)}_x& =-{\left(\frac{\partial F}{\partial y}\right)}^{-1}\left(\frac{\partial F}{\partial z}\right)\\ {\left(\frac{\partial z}{\partial x}\right)}_y& =-{\left(\frac{\partial F}{\partial z}\right)}^{-1}\left(\frac{\partial F}{\partial x}\right)\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(5)}& {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x{\left(\frac{\partial z}{\partial x}\right)}_y=-1\end{array}$$ $◻$

证明 (1) 可直接通过全微分算子的协变变换证明：$$\begin{array}{}\text{(7)}& \frac{\partial }{\partial x_k}\frac{\partial }{\partial x_k}=\frac{\partial }{\partial y_i}{A}_{ik}\frac{\partial }{\partial y_j}{A}_{jk}=\frac{\partial }{\partial y_i}(A{A}^{\mathrm{T}}{)}_{ij}\frac{\partial }{\partial y_j}=\frac{\partial }{\partial y_i}{\delta }_{ij}\frac{\partial }{\partial y_j}=\frac{\partial }{\partial y_i}\frac{\partial }{\partial y_i}\end{array}$$

(2) 可借助Hessian矩阵⁴证明⁵：$$\begin{array}{}\text{(8)}& H_F(\mathit{x})={\partial }_{\mathit{x}}({\mathrm{\nabla }}_{\mathit{x}}F)(\mathit{x})=A^{-1}{\partial }_{\mathit{y}}({\mathrm{\nabla }}_{\mathit{y}}f)(\mathit{y})A=A^{-1}{H}_f(\mathit{y})A\end{array}$$ 故$$\begin{array}{}\text{(9)}& \sum _{k=1}^m\frac{{\partial }^{2}F}{\partial x_k^2}=\mathrm{tr}{H}_F(\mathit{x})=\mathrm{tr}{H}_f(\mathit{y})=\sum _{k=1}^m\frac{{\partial }^{2}f}{\partial y_k^2}\end{array}$$ $◻$

证明 可微函数$z(u,v)=x(u,v)+\mathrm{i}y(u,v)$在单连通域$\mathbb{C}$上满足Cauchy-Riemann方程，从而解析。

(1) $w$在单连通域${\mathbb{R}}^2$上调和，故存在解析函数$f$使得$w(x,y)=\mathrm{\Re }f(x+\mathrm{i}y)$，因此$f\circ z$亦为解析函数，从而$\tilde{w}(u,v)=\mathrm{\Re }f(z(u+\mathrm{i}v))$调和。

(2) 由(1)知$\tilde{w}=xy$调和。 $◻$

证明 令$\xi (x.y)=\frac{x}{x^2+y^2}$、$\eta (x,y)=-\frac{y}{x^2+y^2}$，其满足CR方程，由上例可知$\tilde{u}=f(\xi ,\eta )$调和。由于$(\xi ,\eta )\mapsto (\xi ,-\eta )$是正交变换，故$u$调和。 $◻$

证明 设$u_1,u_2$均满足该方程和边界条件，令$u=u_1-u_2$，则$u$满足$$\begin{array}{}\text{(12)}& \{\begin{array}{ll}\mathrm{\Delta }u=cu,& \mathit{r}\in D\\ u=0,& \mathit{r}\in \partial D\end{array}\end{array}$$

当$c\ne 0$时，令$v=\frac{u}{c}$，则$v$满足$$\begin{array}{}\text{(13)}& \{\begin{array}{ll}\mathrm{\Delta }v=v,& \mathit{r}\in D\\ v=0,& \mathit{r}\in \partial D\end{array}\end{array}$$ $v$在有界闭集$\bar{D}$上存在最小值点${\mathit{x}}_0$，显然$v({\mathit{x}}_0)<0$。若$v({\mathit{x}}_0)<0$，则有${\mathit{x}}_0\notin \partial D$，故${\mathit{x}}_0$为极小值点，从而$\mathrm{\Delta }v({\mathit{x}}_0)=\mathrm{tr}{H}_v({\mathit{x}}_0)\ge 0>v({\mathit{x}}_0)$，矛盾！故$v(\mathit{x})\ge 0$对任意$\mathit{x}\in \bar{D}$成立。同理可得$v(\mathit{x})\le 0$，故$v\equiv 0$，即$u_1\equiv u_2$。

当$c=0$时，注意到$$\begin{array}{}\text{(14)}& {\int }_D\parallel \mathrm{\nabla }u{\parallel }^2\mathrm{d}V={\int }_D[\mathrm{\nabla }\cdot (u\mathrm{\nabla }u)-{\mathrm{\nabla }}^{2}u]\mathrm{d}V={\oint }_{\partial D}u\mathrm{\nabla }u\cdot \mathit{n}\mathrm{d}S-{\int }_D{\mathrm{\nabla }}^{2}u\mathrm{d}V=0\end{array}$$ 由$u\in \mathcal{C}(\bar{D})$可得$u\equiv 0$，即$u_1\equiv u_2$。

综上所述，原边值问题具有唯一解。 $◻$