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:

 $\displaystyle p$ $=$ $\displaystyle z_x$ $\displaystyle q$ $=$ $\displaystyle z_y$ $\displaystyle r$ $=$ $\displaystyle z_{xx}$ $\displaystyle s$ $=$ $\displaystyle z_{xy}$ $\displaystyle t$ $=$ $\displaystyle z_{yy}$

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:

 $\displaystyle \dfrac \d {\d x} \frac \partial {\partial z_x} \sqrt {1 + z_x^2 + z_y^2}$ $=$ $\displaystyle \dfrac \d {\d x} \frac {z_x} {\sqrt {1 + z_x^2 + z_y^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {z_{xx} + z_y^2 z_{xx} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} }$ $\displaystyle$ $=$ $\displaystyle \frac {r \paren {1 + q^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} }$
 $\displaystyle \dfrac \d {\d y} \frac \partial {\partial z_y} \sqrt {1 + z_x^2 + z_y^2}$ $=$ $\displaystyle \dfrac \d {\d y} \frac {z_y} {\sqrt {1 + z_x^2 + z_y^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {z_{yy} + z_x^2 z_{yy} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} }$ $\displaystyle$ $=$ $\displaystyle \frac {t \paren {1 + p^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} }$
$\dfrac {r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} } {\paren {1 + p^2 + q^2}^{\frac 3 2} } = 0$

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:

 $\displaystyle E$ $=$ $\displaystyle 1 + p^2$ $\displaystyle F$ $=$ $\displaystyle p q$ $\displaystyle G$ $=$ $\displaystyle 1 + q^2$ $\displaystyle e$ $=$ $\displaystyle \dfrac r {\sqrt {1 + p^2 + q ^2} }$ $\displaystyle f$ $=$ $\displaystyle \dfrac s {\sqrt {1 + p^2 + q^2} }$ $\displaystyle g$ $=$ $\displaystyle \dfrac t {\sqrt {1 + p^2 + q^2} }$

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:

$M = 0$

$\blacksquare$