Lecture 15 - Minimal Surfaces

#math

Recall that $(1 + f_{y}^{2}) f_{x x} - 2 f_{x} f_{y} f_{x y} + (1 + f_{x}^{2}) f_{y y} = 0$ is the minimal surface equation. Assume that $f (x) = g (x) + h (y)$ then the minima surface equation transforms to:

g^{″} (x) [1 + h^{'} (y)]^{2} + h^{″} (y) (1 + g^{'} (x)^{2}) = 0

Thus we obtain:

\begin{aligned} - \frac{1 + g^{'} (x)^{2}}{g^{″} (y)} = \frac{1 + h (x)^{2}}{h^{″} (y)} \\ ⟹ & \frac{- 1 + g^{'} (x)^{2}}{g^{″} (x)} = κ \\ ⟹ & 1 + g^{'} (x)^{2} = - κ g^{″} (x) \\ Let ϕ (x) = g^{'} (x) \\ ⟹ & \int d x = - κ \int \frac{d ϕ}{1 + ϕ^{2}} \\ ⟹ & x = - κ a r c t a n ϕ + c \end{aligned}

For convenience $c = 0, κ = 1$

\begin{aligned} g (x) & = l n (c o s x) \\ h (y) & = - l n (c o s y) \\ ∴ f (x, y) & = l n (\frac{c o s (x)}{c o s (y)}) \end{aligned}

This is Scherks double periodic surface. For reference (application) check out Randall Kamien (condensed matter physicist).

The Weierstrass Enneper representation

Proposition

Let $ϕ : V \to R^{3}$ be an isothermal parameterization if $E = G = λ^{2} (u, v)$ and $F = 0$ then $ϕ$ satisfies:

ϕ_{u u} + ϕ_{v v} = 2 λ^{2} H \vec{n}

Proof

Since $ϕ$ is isothermal with $E = G = λ^{2}$ and $F = 0$ then:

< ϕ_{u}, ϕ_{u} >=< ϕ_{v}, ϕ_{v} >

Suppose we differentiate wrt $u$ then:

\begin{matrix} (1) & < ϕ_{u u}, ϕ_{u} >=< ϕ_{u v}, ϕ_{v} > \end{matrix}

since $F = 0 ⟹ < ϕ_{u}, ϕ_{v} >= 0$ then differentiating wrt $v$ gives us:

\begin{matrix} (2) & < ϕ_{u v}, ϕ_{v} > + < ϕ_{u}, ϕ_{v v} >= 0 \end{matrix}

$(1) + (2)$ gives us that:

< ϕ_{u u}, ϕ_{u} >= - < ϕ_{v v}, ϕ_{u} >

that is:

< ϕ_{u u} + ϕ_{v v}, ϕ_{u} >= 0

Similarly we can show that $< ϕ_{u u} + ϕ_{v v}, ϕ_{v} >= 0$ thus $ϕ_{u u} + ϕ_{v v}$ must be parallel to the normal.
We know that $H = \frac{L + N}{2 λ^{2}}$ thus:

\begin{aligned} 2 λ^{2} H & = L + N \\ = n \cdot (ϕ_{u u} + ϕ_{v v}) \end{aligned}

That is:

ϕ_{u u} + ϕ_{v v} = 2 λ^{2} H \vec{n}

Examples of Isothermal surfaces

Catenoid: $ϕ (u, v) = (a c o s h (v) c o s (u), a c o s h (v) s i n (u), a v)$
Helicoid: $ϕ (u, v) = (a s i n h (v) c o s (u), a s i n h (v) s i n (u), a u)$
Ruled Surface

Definition

Let $ϕ$ be a function of one complex variable having the form

ϕ (z) = x (u, v) + i y (u, v)

$ϕ$ is called harmonic if:

ϕ_{u u} + ϕ_{v v} = 0.

Definition

Suppose $ϕ : ω \subset C \to C$ then $ϕ$ is said to be meromorphic if there exists isolated points $z_{i}$ such that:

\tilde{ϕ} : Ω / {z_{i}} \overset{h o l o m o r p h i c}{\to} C_{i} .

In addition the points ${z_{i}}$ are called poles of the function. The order of the pole $z_{i}$ is the number of terms that vanish in that vanish Laurent series around $z_{i}$ .

Since we are interested in $ϕ (z) = x (u, v) + i y (u, v)$ then we have the Writenger derivative:

\begin{aligned} \frac{\partial ϕ}{\partial z} & = \frac{1}{2} (\frac{\partial x}{\partial u} + \frac{\partial y}{\partial v}) + \frac{i}{2} (\frac{\partial y}{\partial u} + \frac{\partial x}{\partial v}) \\ \frac{\partial ϕ}{\partial \bar{z}} & = \frac{1}{2} (\frac{\partial x}{\partial u} - \frac{\partial y}{\partial v}) + \frac{i}{2} (\frac{\partial y}{\partial u} + \frac{\partial x}{\partial v}) \end{aligned}

Using these we obtain:

4 (\frac{\partial}{\partial \bar{z}} (\frac{\partial ϕ}{\partial z})) = ϕ_{u u} + ϕ_{v v}

Thus if our function is harmonic then it is holomorphic.

Proposition

Let $ϕ (u, v) = (x (u, v), y (u, v), z (u, v))$
Suppose $ϕ_{k} = \frac{\partial x_{k}}{\partial z}$ then $ϕ$ is minimal iff

ϕ_{1}^{2} + ϕ_{2}^{2} + ϕ_{3}^{2} = 0

and that $ϕ_{k}$ are holomorphic functions.

Proof

Note that $ϕ_{1} = x_{u} - i x_{v}$ , $ϕ_{2} = y_{u} - i y_{v}$ , $ϕ_{3} = z_{u} - i z_{v}$ then:

\begin{aligned} ϕ_{1}^{2} + ϕ_{2}^{2} + ϕ_{3}^{2} & = (x_{u} - i x_{v}) \cdot (x_{u} - i x_{v}) + (y_{u} - i y_{v}) \cdot (y_{u} - i y_{v}) + (z_{u} - i z_{v}) \cdot (z_{u} - i z_{v}) \\ = (x_{u} \cdot x_{u} + y_{u} \cdot y_{u} + z_{u} \cdot z_{u}) - (x_{v} \cdot x_{v} + y_{v} \cdot y_{v} + z_{v} \cdot z_{v}) - 2 i (x_{u} \cdot x_{v} + y_{u} \cdot y_{v} + z_{u} \cdot z_{v}) \\ = E - G - 2 i F \\ = 0 (Since ϕ satisfies the isothermal surface condition) \end{aligned}

Lemma

Let $D$ be a domain in the complex plane, $g (z)$ an arbitrary meromorphic function in $D$ having the property that at each point where $g (z)$ has a pole of order $k$ , f(z) has a zero zero order of atleast $2 k$ then the function:

ϕ_{1} = \frac{1}{2} f (1 - g^{2}), ϕ_{2} = \frac{i}{2} f (1 + g^{2}), ϕ_{3} = f g

will be holomorphic in $D$ satisfying $ϕ_{1}^{2} + ϕ_{2}^{2} + ϕ_{3}^{2} = 0$

Proposition

Every regular minimal surface has locally an isothermal paramterization of the form equal to:

\begin{array}{r} (R e \int^{z} f (1 + g^{2}) d w, R e \int^{z} - i f (1 - g^{2}) d w, R e \int^{z} - 2 i f g d w) \end{array}

$f$ vanishes at the pole of $g$ and having a zero of order of at least $2 k$ wherever $g$ has a pole of order $k$ , and:

| ϕ |^{2} = 4 | f | (1 + g^{2}) \neq 0