题目
问题 4. 曲面的面积为
S=∬D1+ux2+uy2dxdyS = \iint_{D} \sqrt{1 + u_x^2 + u_y^2} dxdyS=∬D1+ux2+uy2 dxdy (5)
其中 z=u(x,y) z = u(x, y) z=u(x,y), (x,y)∈D(x, y) \in D (x,y)∈D 是曲面的方程。
(a) 写出最小面积曲面的欧拉-拉格朗日偏微分方程（边界条件为 u(x,y)=ϕ(x,y) u(x, y) = \phi(x, y) u(x,y)=ϕ(x,y) 当 (x,y)∈Γ(x, y) \in \Gamma (x,y)∈Γ，其中 Γ\GammaΓ 是 D D D 的边界）。
(b) 如果势能为
E=kS−∬DfudxdyE = kS - \iint_{D} f u dxdyE=kS−∬Dfudxdy (6)
其中 S S S 由 (5) 定义，f f f 是外力的面密度。写出最小能量曲面的欧拉-拉格朗日偏微分方程。
解题过程
部分 (a)
我们需要最小化曲面面积泛函：
I[u]=∬D1+ux2+uy2dxdy I[u] = \iint_{D} \sqrt{1 + u_x^2 + u_y^2} dxdy I[u]=∬D1+ux2+uy2 dxdy
定义被积函数 F=1+ux2+uy2 F = \sqrt{1 + u_x^2 + u_y^2} F=1+ux2+uy2 。由于 F F F 不显式依赖于 u u u，即 ∂F∂u=0 \frac{\partial F}{\partial u} = 0 ∂u∂F=0，欧拉-拉格朗日方程为：
∂F∂u−∂∂x(∂F∂ux)−∂∂y(∂F∂uy)=0 \frac{\partial F}{\partial u} - \frac{\partial}{\partial x} \left( \frac{\partial F}{\partial u_x} \right) - \frac{\partial}{\partial y} \left( \frac{\partial F}{\partial u_y} \right) = 0 ∂u∂F−∂x∂(∂ux∂F)−∂y∂(∂uy∂F)=0
计算偏导数：
∂F∂ux=ux1+ux2+uy2,∂F∂uy=uy1+ux2+uy2 \frac{\partial F}{\partial u_x} = \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}}, \quad \frac{\partial F}{\partial u_y} = \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} ∂ux∂F=1+ux2+uy2 ux,∂uy∂F=1+ux2+uy2 uy
因此，欧拉-拉格朗日方程变为：
−∂∂x(ux1+ux2+uy2)−∂∂y(uy1+ux2+uy2)=0 -\frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) - \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = 0 −∂x∂⎝⎛1+ux2+uy2 ux⎠⎞−∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=0
或等价地：
∂∂x(ux1+ux2+uy2)+∂∂y(uy1+ux2+uy2)=0 \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) + \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = 0 ∂x∂⎝⎛1+ux2+uy2 ux⎠⎞+∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=0
这是最小面积曲面的欧拉-拉格朗日 PDE。边界条件为 u(x,y)=ϕ(x,y) u(x, y) = \phi(x, y) u(x,y)=ϕ(x,y) on Γ \Gamma Γ.
部分 (b)
势能泛函为：
E=kS−∬Dfudxdy=∬D[k1+ux2+uy2−fu]dxdy E = kS - \iint_{D} f u dxdy = \iint_{D} \left[ k \sqrt{1 + u_x^2 + u_y^2} - f u \right] dxdy E=kS−∬Dfudxdy=∬D[k1+ux2+uy2 −fu]dxdy
定义被积函数 F=k1+ux2+uy2−fu F = k \sqrt{1 + u_x^2 + u_y^2} - f u F=k1+ux2+uy2 −fu。欧拉-拉格朗日方程为：
∂F∂u−∂∂x(∂F∂ux)−∂∂y(∂F∂uy)=0 \frac{\partial F}{\partial u} - \frac{\partial}{\partial x} \left( \frac{\partial F}{\partial u_x} \right) - \frac{\partial}{\partial y} \left( \frac{\partial F}{\partial u_y} \right) = 0 ∂u∂F−∂x∂(∂ux∂F)−∂y∂(∂uy∂F)=0
计算偏导数：
∂F∂u=−f \frac{\partial F}{\partial u} = -f ∂u∂F=−f
∂F∂ux=k⋅ux1+ux2+uy2,∂F∂uy=k⋅uy1+ux2+uy2 \frac{\partial F}{\partial u_x} = k \cdot \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}}, \quad \frac{\partial F}{\partial u_y} = k \cdot \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} ∂ux∂F=k⋅1+ux2+uy2 ux,∂uy∂F=k⋅1+ux2+uy2 uy
因此，
∂∂x(∂F∂ux)=k∂∂x(ux1+ux2+uy2),∂∂y(∂F∂uy)=k∂∂y(uy1+ux2+uy2) \frac{\partial}{\partial x} \left( \frac{\partial F}{\partial u_x} \right) = k \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right), \quad \frac{\partial}{\partial y} \left( \frac{\partial F}{\partial u_y} \right) = k \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) ∂x∂(∂ux∂F)=k∂x∂⎝⎛1+ux2+uy2 ux⎠⎞,∂y∂(∂uy∂F)=k∂y∂⎝⎛1+ux2+uy2 uy⎠⎞
代入欧拉-拉格朗日方程：
−f−k∂∂x(ux1+ux2+uy2)−k∂∂y(uy1+ux2+uy2)=0 -f - k \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) - k \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = 0 −f−k∂x∂⎝⎛1+ux2+uy2 ux⎠⎞−k∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=0
整理得：
k[∂∂x(ux1+ux2+uy2)+∂∂y(uy1+ux2+uy2)]=−f k \left[ \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) + \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) \right] = -f k⎣⎡∂x∂⎝⎛1+ux2+uy2 ux⎠⎞+∂y∂⎝⎛1+ux2+uy2 uy⎠⎞⎦⎤=−f
或
∂∂x(ux1+ux2+uy2)+∂∂y(uy1+ux2+uy2)=−fk \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) + \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = -\frac{f}{k} ∂x∂⎝⎛1+ux2+uy2 ux⎠⎞+∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=−kf
这是最小能量曲面的欧拉-拉格朗日 PDE。边界条件同样为 u(x,y)=ϕ(x,y) u(x, y) = \phi(x, y) u(x,y)=ϕ(x,y) on Γ \Gamma Γ.
答案
(a) 最小面积曲面的欧拉-拉格朗日 PDE 为：
∂∂x(ux1+ux2+uy2)+∂∂y(uy1+ux2+uy2)=0 \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) + \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = 0 ∂x∂⎝⎛1+ux2+uy2 ux⎠⎞+∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=0
(b) 最小能量曲面的欧拉-拉格朗日 PDE 为：
∂∂x(ux1+ux2+uy2)+∂∂y(uy1+ux2+uy2)=−fk \frac{\partial}{\partial x} \left( \frac{u_x}{\sqrt{1 + u_x^2 + u_y^2}} \right) + \frac{\partial}{\partial y} \left( \frac{u_y}{\sqrt{1 + u_x^2 + u_y^2}} \right) = -\frac{f}{k} ∂x∂⎝⎛1+ux2+uy2 ux⎠⎞+∂y∂⎝⎛1+ux2+uy2 uy⎠⎞=−kf