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:

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:


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

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 bounded.

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:

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:


 * $M = 0$