Minimal Surfaces

Exercise 3.1
Example 3.1
Exercise 3.2

Definition 3.2
Example3.3
Example 3.4
Example 3.5
Exercise 3.3

Area of regular parametrized surfaces

Recall from Exercise 2.3 that the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ is given by $\|\vec{u}\times\vec{v}\|$ . This observation serves as the motivation for the following formula for the area of regular parametrized surfaces.

To get an estimate of the area of a regular parametrized surface $\vec{x}(u, v), (u, v)\in I_1\times I_2$ , we may first subdivide the surface by coordinate curves into small pieces, such that each is approximately a parallelogram spanned by the tangents of coordinate curves, and then sum the area of each approximating parallelograms. As the subdivision gets finer, the area approximation will tend to the actual area. This limiting process, which is the same principle behind calculating the area under the graph of a smooth single-variable function by means of integration, can be expressed as a double integral

$\text{Area of }\vec{x}=\int_{I_2}\int_{I_1}\|\vec{x}_u\times\vec{x}_v\|dudv$ The above double integral can be evaluated first by computing the inner integral, and then the outer one.

Exercise 3.1 Prove that $\|\vec{x}_u\times\vec{x}_v\|=\sqrt{EG-F^2}$ , and hence

$\text{Area of }\vec{x}=\int_{I_2}\int_{I_1}\sqrt{EG-F^2}dudv$
Solution

Example 3.1 As in Example 2.7, the parametrization of the unit sphere minus the two poles is given by

$\vec{x}(u, v)=(\sin u\cos v, \sin u\sin v, \cos u), (u, v)\in (0, \pi)\times[0, 2\pi]$ Its surface area is $\begin{align*} \int_0^{2\pi}\int_0^{\pi}\|\vec{x}_u\times\vec{x}_v\|dudv&=\int_0^{2\pi}\int_0^{\pi}\sin ududv\\ &=\int_0^{2\pi}2dv\\ &=4\pi \end{align*}$ Since the two poles are points which account for no area, the surface area of the unit sphere is also $4\pi$ .

Exercise 3.2 Compute the surface area of the torus as in Example 2.8.
Solution

The mean curvature condition

Suppose that $\vec{x}(u, v)$ , $(u, v)\in I_1\times I_2$ is the surface with minimal area among those whose boundary coincides with that of $\vec{x}$ . Let $t(u, v)$ be any smooth function on $I_1\times I_2$ such that it vanishes on the boundary of $\vec{x}$ .

Consider the following family of regular parametrized surfaces

$\vec{x}^\varepsilon(u, v)=\vec{x}(u, v)+\varepsilon t(u, v)N, -\lambda<\varepsilon<\lambda$ for some small $\lambda$ . These surfaces can be thought of as being obtained by varying $\vec{x}$ along its normals, while fixing its boundary. Note that
$\begin{align*} E^\varepsilon&=\langle\vec{x}_u^\varepsilon, \vec{x}_u^\varepsilon\rangle\\ &=E+2\left\langle\vec{x}_u, \varepsilon\left(\frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}\right)\right\rangle+\varepsilon^2\left\|\frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}\right\|^2 \end{align*}$
$\begin{align*} F^\varepsilon&=\langle\vec{x}_u^\varepsilon, \vec{x}_v^\varepsilon\rangle\\ &=F+\left\langle \vec{x}_u, \varepsilon\left(\frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right)\right\rangle+\left\langle\vec{x}_v, \varepsilon\left(\frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}\right)\right\rangle+\\ &\varepsilon^2\left\langle \frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}, \frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right\rangle \end{align*}$
$\begin{align*} G^\varepsilon&=\langle\vec{x}_v^\varepsilon, \vec{x}_v^\varepsilon\rangle\\ &=G+2\left\langle\vec{x}_v, \varepsilon\left(\frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right)\right\rangle+\varepsilon^2\left\|\frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right\|^2\\ \end{align*}$
Let $A(\varepsilon)=\text{Area of }\vec{x}^\varepsilon$ . If $\vec{x}^0=\vec{x}$ achieves minimal area, then $\left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0}A(\varepsilon)=0$ . So
$\begin{align*} 0&=\left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0}\int_{I_2}\int_{I_1}\sqrt{E^\varepsilon G^\varepsilon-(F^\varepsilon)^2}dudv\\ &=\int_{I_2}\int_{I_1}\left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0}\sqrt{E^\varepsilon G^\varepsilon-(F^\varepsilon)^2}dudv \end{align*}$
$\begin{align*} &=\int_{I_2}\int_{I_1}\frac{1}{2\sqrt{EG-F^2}}\left(2E\left\langle\vec{x}_v, \frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right\rangle+2G\left\langle\vec{x}_u, \frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}\right\rangle-\right.\\ &\left.2F\left(\left\langle\vec{x}_u, \frac{\partial t}{\partial v}N+t\frac{\partial N}{\partial v}\right\rangle+\left\langle\vec{x}_v, \frac{\partial t}{\partial u}N+t\frac{\partial N}{\partial u}\right\rangle\right)\right)dudv \end{align*}$
$\begin{align*} &=\int_{I_2}\int_{I_1}\frac{t}{\sqrt{EG-F^2}}(-E\langle N, \vec{x}_{vv}\rangle-G\langle N, \vec{x}_{uu}\rangle+2F\langle N, \vec{x}_{uv}\rangle)dudv\\ &=-\int_{I_2}\int_{I_1}2tH\sqrt{EG-F^2}dudv \end{align*}$
If we take $t(u, v)=H(u, v)$ , then we have $-\int_{I_2}\int_{I_1}2H^2\sqrt{EG-F^2}dudv=0$ Note that $2H^2\sqrt{EG-F^2}\geq 0$ and $\sqrt{EG-F^2}>0$ (as $\sqrt{EG-F^2}=\|\vec{x}_u\times\vec{x}_v\|\neq 0$ , if $\vec{x}$ is regular). So $H(u, v)=0, \text{for all }(u, v)\in I_1\times I_2$ which is precisely the necessary condition we want.

Definition 3.2 A smooth surface with vanishing mean curvature is called a minimal surface.

Example 3.3 Let $\vec{x}(u, v)=(u, v, f(u, v))$ be the graph of $f(u, v)$ , a smooth function on $(u, v)\in I_1\times I_2$ . Then $\vec{x}$ is a minimal surface if

$\left(1+\left(\frac{\partial f}{\partial u}\right)^2\right) \frac{\partial^2 f}{\partial v^2}+\left(1+\left(\frac{\partial f}{\partial v}\right)^2\right)\frac{\partial^2 f}{\partial u^2}-2\frac{\partial f}{\partial u}\frac{\partial f}{\partial v}\frac{\partial^2 f}{\partial u\partial v}=0$ by Example 2.20. The above equation is called the minimal surface equation.

Example 3.4 The catenoid. $\vec{x}(u, v)= (\cosh u\cos v, \cosh u\sin v, u)$ .

$\begin{align*} \vec{x}_u&=(\sinh u\cos v, \sinh u\sin v, 1)\\ \vec{x}_v&=(-\cosh u\sin v, \cosh u\cos v, 0)\\ \vec{x}_u\times\vec{x}_v&=(-\cosh u\cos v, -\cosh u\sin v, \sinh u\cosh u)\\ \|\vec{x}_u\times\vec{x}_v\|=\cosh^2 u\\ N(u, v)&=\left(-\frac{\cos v}{\cosh u}, -\frac{\sin v}{\cosh u}, \tanh u\right) \end{align*}$ $E=G=\cosh^2 u, F=0$ $\begin{align*} e&=\langle N, \vec{x}_{uu}\rangle\\ &=-1\\ f&=\langle N, \vec{x}_{uv}\rangle\\ &=0\\ g&=\langle N, \vec{x}_{vv}\rangle\\ &=1 \end{align*}$ $H=\frac{Eg-2Ff+Ge}{2(EG-F^2)}=0$ Hence the catenoid is a minimal surface.

Example 3.5 Enneper surface. $\vec{x}(u, v)= \left(u-\frac{1}{3}u^3+uv^2, -v+\frac{1}{3}v^3-vu^2, u^2-v^2\right)$ .

$\begin{align*} \vec{x}_u&=(1-u^2+v^2, -2uv, 2u)\\ \vec{x}_v&=(2uv, -1+v^2-u^2, -2v)\\ \vec{x}_u\times\vec{x}_v&=(2u(u^2+v^2+1), 2v(u^2+v^2+1), (u^2+v^2-1)(u^2+v^2+1))\\ \|\vec{x}_u\times\vec{x}_v\|=(u^2+v^2+1)^2\\ N(u, v)&=\left(\frac{2u}{u^2+v^2+1}, \frac{2v}{u^2+v^2+1}, \frac{u^2+v^2-1}{u^2+v^2+1}\right) \end{align*}$ $E=G=(u^2+v^2+1)^2, F=0$ $e=-2, f=0, g=2$ Hence $H=0$ , and Enneper surface is a minimal surface.

Enneper surface

Exercise 3.3 Show that the helicoid as in Exercise 2.9 is a minimal surface.
Solution