Lecture 14 - Minimal surface condition

#math

Let $D$ be a bounded domain in the plane $R^{2}$ with piecewise boundary $\partial D$ . Let $X = (x, y, f (x, y))$ be our chart then:

\begin{aligned} X_{x} & = (1, 0, f_{x}) \\ X_{y} & = (0, 1, f_{y}) \\ X_{x} \times X_{y} & = (- f_{x}, - f_{y}, 1) \\ n & = \frac{X_{x} \times X_{y}}{| | X_{x} \times X_{y} | |} \\ E & =< \partial_{x} X, \partial_{x} X > \\ = 1 + f_{x}^{2} \\ F & = f_{x} f_{y} \\ G & = 1 + f_{y}^{2} \end{aligned}

Let us now calculate the terms of the second fundamental form:

\begin{aligned} L & = n \cdot X_{x x} \\ = \frac{f_{x x}}{\sqrt{1 + f_{x}^{2} + f_{y}^{2}}} \\ M & = \frac{f_{x y}}{\sqrt{1 + f_{x}^{2} + f_{y}^{2}}} \\ N & = \frac{f_{y y}}{\sqrt{1 + f_{x}^{2} + f_{y}^{2}}} \end{aligned}

Giving us:

\begin{aligned} H & = \frac{G L - 2 F M + E N}{2 (E G - F^{2})} \\ = \frac{1}{2} \frac{(1 + f_{x}^{2}) f_{y y} - 2 f_{x} f_{y} f_{x y} + (1 + f_{y}^{2}) f_{x x}}{(1 + f_{x}^{2} + f_{y}^{2})^{\frac{3}{2}}} \end{aligned}

Hence if $H = 0$ then we have that:

(1 + f_{x}^{2}) f_{y y} - 2 f_{x} f_{y} f_{x y} + (1 + f_{y}^{2}) f_{x x} = 0

Recall the area can be computed by:

A = \int_{D} \sqrt{E G - F^{2}} d u d v

Thus the area for our surface is:

A (f) = \int_{D} \sqrt{f_{x^{2}} + f_{y}^{2} + 1} d x d y = \int_{D} \sqrt{1 + | \nabla f |^{2}} d A

We are interested in finding the function $f$ for which the area is smallest among all nearby graphs over $D$ having the boundary values.
Consider the deformation of $f$ which are fixed on the boundary $\partial D$ :

s \in R ⟶ Area (f + s h) : h \in C^{2}

We are interested in varying s such that we obtain:

{\frac{d}{d s} |}_{s = 0} Area (f + s h) = 0

Thus:

\begin{aligned} {\frac{d}{d s} |}_{s = 0} Area (f + s h) & = \int_{D} {\frac{d}{d s} |}_{s = 0} \sqrt{1 + (f_{x} + s h_{x})^{2} + (f_{y} + s h_{y})^{2}} d x d y \\ = \int_{D} \frac{f_{x} h_{x} + f_{y} h_{y}}{\sqrt{1 + (\nabla f)^{2}}} d x d y \\ = \int_{D} (\frac{f_{x}}{\sqrt{1 + (\nabla f)^{2}}} h_{x} + \frac{f_{y}}{\sqrt{1 + (\nabla f)^{2}}} h_{y}) d x d y \\ = \int_{D} (\frac{\partial}{\partial x} (\frac{f_{x}}{\sqrt{1 + (\nabla f)^{2}}}) + \frac{\partial}{\partial y} (\frac{f_{y}}{\sqrt{1 + (\nabla f)^{2}}})) h d x d x \end{aligned}

Since we want this integral to be zero we obtain an equation which serves as the condition for which minimizes the area of a surface:

\begin{aligned} \frac{\partial}{\partial x} (\frac{f_{x}}{\sqrt{1 + (\nabla f)^{2}}}) + \frac{\partial}{\partial y} (\frac{f_{y}}{\sqrt{1 + (\nabla f)^{2}}}) & = 0 \\ ⟹ div (\frac{\nabla f}{\sqrt{1 + (\nabla f)^{2}}}) = 0 \\ ⟹ (1 + f_{x}^{2}) f_{y y} - 2 f_{x} f_{y} f_{x y} + (1 + f_{y}^{2}) f_{x x} & = 0 \end{aligned}