本文中的公式均采用爱因斯坦求和约定

在推导极小曲面的方程之前,我们先考虑一个更一般的极值问题: 设$C$是空间以光滑闭曲线,$S$是过$C$的一曲面,它的参数表示为 $$ x^{\alpha}=x^{\alpha}(u^1,u^2),\ \alpha=1,2,3. $$ 给定关于$x^{\alpha}$和$\frac{\partial x^{\alpha}}{\partial u^i}(\alpha=1,2,3,i=1,2)$的函数$L(x^{\alpha},\frac{\partial x^{\alpha}}{\partial u^i})$,考虑积分 $$ I=\iint_{U}L(x^{\alpha},\frac{\partial x^{\alpha}}{\partial u^i})du^1du^2, $$ 其中$U$为$S$中曲线$C$所包围的区域.

现在问过曲线$C$的曲面$S$满足什么条件使得积分$I$取局部最小值?

考虑过曲线$C$的与$S$邻近的曲面 $$ \overline{S}:\overline{x}^\alpha=x^{\alpha}(u^1,u^2)+\varepsilon w^{\alpha}(u^1,u^2), $$ 其中$x^{\alpha}$为$C^{\infty}$类函数且在$C$上的取值为0.

设$\overline{S}$上由曲线$C$包围的区域为$\overline{U}$,则有积分 $$ \overline{I}=\iint_{U}L(x^{\alpha}+\varepsilon w^{\alpha},\frac{\partial x^{\alpha}}{\partial u^i}+\varepsilon \frac{\partial w^{\alpha}}{\partial u^i})du^1du^2, $$ 由于$I=\min \overline{I}$,所以 $$ \frac{d\overline{I}}{d\varepsilon }|_{\varepsilon=0}=0, $$ 即 $$ \iint_{\overline{U}}(\frac{\partial L}{\partial x^{\alpha}}w^{\alpha}+\frac{\partial L}{\partial x_{i}^{\alpha}}\frac{\partial w^{\alpha}}{\partial u^i})du^1du^2=0, $$ 其中,$x_{i}^{\alpha}=\frac{\partial x^{\alpha}}{\partial u^{i}}.$

上式又能改写成

$$ \iint w^{\alpha}(\frac{\partial L}{\partial x^{\alpha}}-\frac{\partial}{\partial u^{i}}(\frac{\partial L}{\partial x_{i}^{\alpha}}))du^1du^2+\iint \frac{\partial}{\partial u^{i}}(w^{\alpha}\frac{\partial L}{\partial x_{i}^{\alpha}})du^1du^2=0. $$ 由格林公式 $$ \iint_{U}(\frac{\partial Q}{\partial u^1}-\frac{\partial P}{\partial u^2})du^1du^2=\int_{\partial U}P(u^1,u^2)du^1+Q(u^1,u^2)du^2, $$ 有 $$ \iint [\frac{\partial}{\partial u^{1}}(w^{\alpha}\frac{\partial L}{\partial x_{1}^{\alpha}})+\frac{\partial}{\partial u^{2}}(w^{\alpha}\frac{\partial L}{\partial x_{2}^{\alpha}})]du^1du^2=\int_{C}-w^{\alpha}\frac{\partial L}{\partial x_{2}^{\alpha}}du^1+w^{\alpha}\frac{\partial L}{\partial x_{1}^{\alpha}}du^2=0. $$ 于是得到该极值问题的欧拉方程 $$ \frac{\partial}{\partial u^{i}}(\frac{\partial L}{\partial x_{i}^{\alpha}})-\frac{\partial L}{\partial x^{\alpha}}=0. $$ 现在我们希望找到使$C$所包围的区域$U$面积最小的曲面$S$,即$L=\sqrt{g}$,其中$g=\det(g_{ij})$.根据定义有 $$ g_{ij}=\vec{r}_i\cdot\vec{r}_j=\sum_{\alpha}\frac{\partial x^{\alpha}}{\partial u^{i}}\frac{\partial x^{\alpha}}{\partial u^{j}}, $$ 所以 $$ \frac{\partial \sqrt{g}}{\partial x_{i}^{\alpha}}=\frac{1}{2\sqrt{g}}\frac{\partial g}{\partial g_{jk}}\frac{\partial g_{jk}}{\partial x_{i}^{\alpha}}=\sqrt{g}g^{ij}x_{i}^{\alpha}, $$ 其中,$g^{ij}=(g^{-1})_{ij}$,

所以有 $$ \frac{\partial}{\partial u_{i}}(\frac{\partial \sqrt{g}}{\partial x_{i}^{\alpha}})=\sqrt{g}g^{ij}(\frac{\partial^2 x^{\alpha}}{\partial u^i \partial u^i}-\Gamma_{ji}^{k}x_{k}^{\alpha})=\sqrt{g}g^{ij}L_{ij}n^{\alpha}. $$ 其中,$\vec{n}=(n^1,n^2,n^3)$是曲面$S$的单位法向量,后一步来自高斯方程 $$\vec{r}_{ij}=\Gamma_{ji}^{k}\vec{r}_k+L_{ij}\vec{n}.$$ 又 $$ \frac{\partial \sqrt{g}}{\partial x^{\alpha}}=0, $$ 而平均曲率 $$H=\frac{1}{2}g^{ij}L_{ij}.$$ 因此欧拉方程简化为 $$ H=0. $$ 于是得到

定理:对于过空间光滑闭曲线$C$的曲面$S$来说,若果$C$所包围的曲面面积最小,则曲面$S$的平均曲率恒为0.即极小曲面的平均曲率为0.