Minimal Smooth Surface Spanned by Contour

Theorem
Let $\map z {x, y}: \R^2 \to \R$ be a real-valued function.

Let $\Gamma$ be a closed contour in $3$-dimensional Euclidean space.

Suppose this surface is smooth for every $x$ and $y$.

Then it has to satisfy the following Euler's equation:


 * $r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$

where:


 * $p = z_x$


 * $q = z_y$


 * $r = z_{xx}$


 * $s = z_{xy}$


 * $t = z_{yy}$

with subscript denoting respective partial derivatives.

In other words, its mean curvature has to vanish.

Proof
The surface area for a smooth surface embedded in $3$-dimensional Euclidean space is given by:


 * $\displaystyle A \sqbrk z = \iint_\Gamma \sqrt {1 + z_x^2 + z_y^2} \rd x \rd y$

It follows that:

By Euler's equation:


 * $\dfrac {r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} } {\paren {1 + p^2 + q^2}^{\frac 3 2} } = 0$

Due to the smoothness of the surface, $1 + p^2 + q^2$ is finite.

Hence, the following equation is sufficient:


 * $r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$

Introduce the following change of variables:


 * $E = 1 + p^2$


 * $F = p q$


 * $G = 1 + q^2$


 * $e = \dfrac r {\sqrt {1 + p^2 + q ^2} }$


 * $f = \dfrac s {\sqrt {1 + p^2 + q^2} }$


 * $g = \dfrac t {\sqrt {1 + p^2 + q^2} }$

Then Euler's equation can be rewritten as:


 * $\dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} } = 0$

By definition, mean curvature is:


 * $M = \dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} }$

Hence:


 * $M = 0$