Minimal Smooth Surface Spanned by Contour

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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.

Then the smooth surface of least area spanned by the contour $\Gamma$ has to satisfy the following Euler's equation:

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

where:

\(\ds p\) \(=\) \(\ds z_x\)
\(\ds q\) \(=\) \(\ds z_y\)
\(\ds r\) \(=\) \(\ds z_{xx}\)
\(\ds s\) \(=\) \(\ds z_{xy}\)
\(\ds t\) \(=\) \(\ds z_{yy}\)

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:

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


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It follows that:

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

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:

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

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:

$M = 0$

$\blacksquare$


Sources

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