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This is a classic result in differential geometry and is worth mentioning in these posts on minimal surfaces. Before we can talk about the deformation we need a definition.

Definition. A minimal surface described by the Weierstrass-Enneper data $(f,g)$ or $F(\tau)$ has an associated family of minimal surfaces given by, respectively, $(e^{it}f,g)$ or $e^{it}F(\tau).$

The catenoid has Weierstrass-Enneper representation $(f,g)=(-\frac{e^{-z}}{2},-e^z)$. Thus, the associated family of surfaces of the catenoid has Weierstrass-Enneper representation $(f,g)=(-\frac{e^{-z}}{2}e^{it},-e^z)$, which corresponds to the following standard parametrization.

$\textbf{x}(u,v)=(x^1(u,v),x^2(u,v),x^3(u,v))$, for any fixed $t$, where

$x^1(u, v) = \cos(t)\cos(v)\cosh(u) + \sin(t)\sin(v)\sinh(u)$

$x^2(u, v) = \cos(t)\cosh(u)\sin(v) - \cos(v)\sin(t)\sinh(u)$

$x^3(u, v) = u\cos(t) + v\sin(t)$

A very beautiful result in minimal surface theory. The catenoid can be continuously deformed into the helicoid by the transformation given above, where $t=0$ represents the catenoid and $t=\frac{\pi}{2}$ represents the helicoid. It should be pointed out that the parametrization above represents a minimal surface for any value of $t.$ That is, any surface in the associated family of a minimal surface is also minimal.

The surfaces below, plotted for different values of $t$, represent the associated family of minimal surfaces of the catenoid.

I’ve spent a few posts talking about the theory behind minimal surfaces. So what? Lets actually look at some.

Prior to the 18th century the only known minimal surface was the plane. This changed when Jean Baptiste Meusnier discovered the first non-planar minimal surfaces, the catenoid and the helicoid.

The catenoid may be parametrized as $\textbf{x}(u,v)=(a\cosh(v)\cos(u),a\cosh(v)\sin(u),av)$. This is a surface of revolution generated by rotating the catenary $y=a\cosh(\frac{z}{a})$ about the $z$-axis. It is easily checked that the mean curvature of $\textbf{x}(u,v)$ is zero. Thus, the catenoid is a minimal surface. It can be characterized as the only surface of revolution which is minimal. That is, if a surface of revolution $M$ is a minimal surface then $M$ is contained in either a plane or a catenoid.

the catenoid

The helicoid may be parametrized as $\textbf{x}(u,v)=(a\sinh(v)\cos(u),a\sinh(v)\sin(u),au)$. It is easily checked that the mean curvature of $\textbf{x}(u,v)$ is zero. Thus, the helicoid is a minimal surface. It can be characterized as the only minimal surface, other than the plane, which is also a ruled surface. That is, any ruled minimal surface in $\mathbb{R}^3$ is part of a plane or a helicoid.

the helicoid

Placing geometric restrictions on surfaces is a common theme in classification in the study of minimal surfaces. For example, assuming a surface has a parametrization of the form $\textbf{x}(u,v)=g(u)+h(v)$, one can explicitly solve the resulting differential equation to find the minimal surface solution $f(x,y)=\frac{1}{a}\ln (\frac{\cos ax}{\cos ay})$, which locally parametrizes Scherk’s minimal surface (discovered by Heinrich Ferdinand Scherk in 1835). Note that although this surface can be realized locally as a graph, its domain of definition is not the entire plane as it must be represented on patches of the form $-\frac{\pi}{2}<\frac{\pi}{2}$ and –$\frac{\pi}{2}<\frac{\pi}{2}$.

Scherk's surface

Let $\textbf{x}(u,v)$ be a regular parametrized surface and let $z=u+iv$ denote the corresponding complex coordinate. Since $u=\frac{z+\bar{z}}{2}$ and $v=\frac{-i(z-\bar{z})}{2}$, we may write $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right).$

We define the complex function $\phi$ as follows,

$\phi(z):=\left(\frac{1}{2}\left(\frac{\partial x^1}{\partial u}-i\frac{\partial x^1}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^2}{\partial u}-i\frac{\partial x^2}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^3}{\partial u}-i\frac{\partial x^3}{\partial v}\right)\right)$

Theorem. Let $M$ be a surface with patch $\textbf{x}(u,v)$ and let $\phi=\frac{\partial\textbf{x}}{\partial z}$. Then $\textbf{x}(u,v)$ is isothermal if, and only if, $(\phi^1)^2+(\phi^2)^2+(\phi^3)^2=0$.

Proof. Suppose that $\textbf{x}(u,v)$ is isothermal. Note that $(\phi^i)^2=\frac{1}{4}[(x^i_u)^2-(x^i_v)^2-2ix^i_ux^i_v]$. Therefore,

$= (\phi)^2$

$= \frac{1}{4}\left[\sum_{i=1}^{3}(x^i_u)^2-\sum_{i=1}^{3}(x^i_v)^2-2i\sum_{i=1}^{3} x^i_ux^i_v\right]$

$= \frac{1}{4}\left(|\textbf{x}_u|^2-|\textbf{x}_v|^2-2i\langle\textbf{x}_u,\textbf{x}_v\rangle\right)$

$= \frac{1}{4}\left(E-G-2iF\right)$

$= 0$

Conversely, suppose that $(\phi)^2=0$. Then $\frac{1}{4}\left(E-G-2iF\right)=0$ and by properties of complex numbers this equations only holds if $E-G=0$ and $F=0$ which shows that $\textbf{x}(u,v)$ is isothermal. $\Box$

Theorem. Suppose that $M$ is a surface with patch $\textbf{x}(u,v)$. Let $\phi=\frac{\partial\textbf{x}}{\partial z}$ and suppose that $(\phi)^2=0$ (i.e. $\textbf{x}$ is isothermal). Then $M$ is minimal if, and only if, each $\phi^i$ is analytic.

Proof. Suppose that $M$ is minimal, then $\textbf{x}(u,v)$ is harmonic; that is, $\Delta\textbf{x}=0$. Thus, $\frac{\partial\phi}{\partial\bar{z}}=\frac{\partial}{\partial\bar{z}}\left(\frac{\partial\textbf{x}}{\partial z}\right)=\frac{1}{4}\Delta\textbf{x}=0$. Therefore, each $\phi^i=\frac{\partial\textbf{x}}{\partial z}$ is analytic. Conversely, the same calculation shows that if each $\phi^i$ is analytic, then each $x^i$ is harmonic, therefore, $M$ is minimal. $\Box$

Corollary. $x^i(z,\bar{z})=c_i+2\text{Re}\int\phi^i\;dz$.

Proof. Since $z=u+iv$, we may write $dz=du+idv$. Then

$\phi^idz=\frac{1}{2}\left[(x^i_u-ix^i_v)(du+idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv+i(x^i_udv-x^i_vdu)\right]$

$\bar{\phi}^idz=\frac{1}{2}\left[(x^i_u+ix^i_v)(du-idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv-i(x^i_udv-x^i_vdu)\right].$

We then have that $dx^i=\frac{\partial x^i}{\partial z}dz+\frac{\partial x^i}{\partial\bar{z}}d\bar{z}=\phi^idz+\bar{\phi}^id\bar{z}=2\text{Re}\phi dz$ and we can now integrate to get $x^i$. $\Box$

The problem of constructing minimal surfaces reduces to finding a tripple of analytic functions $\phi=(\phi^1,\phi^2,\phi^3)$ with $(\phi)^2=0$. A nice we of constructing such a $\phi$ is to take an analytic function $f$ and a meromorphic function $g$ with $fg^2$ analytic. Now, if we let $f=\phi^1-i\phi^2$ and $g=\frac{\phi^3}{(\phi^1-i\phi^2)}$ then we have,

$\phi^1=\frac{1}{2}f(1-g^2) \quad \phi^2=\frac{i}{2}f(1+g^2) \quad \phi^3=fg.$

Note that $f$ is analytic, $g$ is meromorphic, and $fg^2$ is analytic since $fg^2=-(\phi^1+i\phi^2)$. Furthermore, it is easily verified that $(\phi)^2=0$. Therefore, $\phi$ determines a minimal surface.

Theorem. (Weierstrass-Enneper Representation I) If $f$ is analytic on a domain $D,$ $g$ is meromorphic on $D$, and $fg^2$ is analytic on $D$ then a minimal surface is defined by $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right)$, where

$\displaystyle x^1(z,\bar{z})=\text{Re}\int f(1-g^2)\;dz$

$\displaystyle x^2(z,\bar{z})=\text{Re}\int if(1+g^2)\;dz$

$\displaystyle x^3(z,\bar{z})=\text{Re}2\int fg\;dz.$

Suppose that $g$ is analytic and has an inverse $g^{-1}$ in a domain $D$ which is analytic as well. Then we can consider $g$ as a new complex variable $\tau=g$ with $d\tau=g'dz$. Define $F(\tau)=\frac{f}{g'}$ and obtain $F(\tau)d\tau=fdz$. Therefore, if we replace $g$ by $\tau$ and $fdz$ by $F(\tau)d\tau$ we get the following.

Theorem. (Weierstrass-Enneper Representation II) For any analytic function $F(\tau)$, a minimal surface is defined by $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right)$, where

$\displaystyle x^1(z,\bar{z})=\text{Re}\int(1-\tau^2)F(\tau)\;d\tau$

$\displaystyle x^2(z,\bar{z})=\text{Re}\int i(1+g^2)F(\tau)\;d\tau$

$\displaystyle x^3(z,\bar{z})=\text{Re}2\int\tau F(\tau)\;d\tau.$

Note the corresponding $\phi=\left(\frac{1}{2}(1-\tau^2)F(\tau),\;\frac{i}{2}(1+\tau^2)F(\tau),\;\tau F(\tau)\right).$

This representaion tells us that any analytic function $F(\tau)$ defines a minimal surface.

We can now use the Weierstrass-Enneper representation to produce minimal surfaces. For example, if $(f, g) = (1, z)$ then we obtain a parametrization for Enneper’s surface. In fact, if $(f,g) = (1,z^n)$ then an nth order Enneper’s surface is obtained.

First order Enneper surface

The Weierstrass-Enneper representation leads to infinite families of minimal surfaces and has proved fundamental in relating the study of minimal surfaces to the theory of complex analysis.

I have decided to post all of my notes on minimal surfaces. These notes are essentially a summary of how I spent my summer. Each post will build on the previous and there will be four in total. 1 area functional, 2 harmonic function and isothermal coordinates, 3 Weierstrass-Enneper representation, 4 examples.

Definition. Let $\phi(x,y)$ be a real valued function of two real variables $x$ and $y$ defined on a domain $D$. The partial differential equation

$\displaystyle \Delta\phi:=\phi_{xx}(x,y)+\phi_{yy}(x,y)=0$

is known as Laplace’s equation. If $\phi$, $\phi_x$, $\phi_y$, $\phi_{xx}$, $\phi_{x,y}$, $\phi_{y,x}$, and $\phi_{yy}$ are all continuous and if $\phi(x,y)$ satisfies Laplace’s equation then $\phi(x,y)$ is harmonic on $D$.

An interesting relationship between minimal surfaces and harmonic functions comes about when the surface is parametrized by isothermal coordinates

Definition. A parameterization $\textbf{x}(u,v)$ is called isothermal if $E=\langle\textbf{x}_u,\textbf{x}_u\rangle=\langle\textbf{x}_v,\textbf{x}_v\rangle=G$ and $F=\langle\textbf{x}_u,\textbf{x}_v\rangle=0$

Theorem. Isothermal coordinates exist on any surface $M\subseteq\mathbb{R}^3$.

Proof. A Survey of Minimal Surfaces [Osserman]. $\Box$

When isothermal parameters are used, there is a close relationship between the Laplace operator $\Delta\textbf{x}=\textbf{x}_{uu}+\textbf{x}_{vv}$ and mean curvature. We have the following formulas for an orthogonal coordinate system

$\textbf{x}_{uu}=\frac{E_u}{2E}\textbf{x}_u-\frac{E_v}{2G}\textbf{x}_v+lU$

$\textbf{x}_{uv}=\frac{E_v}{2E}\textbf{x}_u-\frac{G_u}{2G}\textbf{x}_v+mU$

$\textbf{x}_{vv}=-\frac{G_u}{2E}\textbf{x}_u+\frac{G_v}{2G}\textbf{x}_v+nU.$

Theorem. If the patch $\textbf{x}(u,v)$ is isothermal then $\Delta\textbf{x}(u,v)=\textbf{x}_{uu}+\textbf{x}_{vv}=(2EH)U$.

Proof. Since $E=G$ and $F=0$, we have

$= \textbf{x}_{uu}+\textbf{x}_{vv}$

$= \left(\frac{E_u}{2E}\textbf{x}_u-\frac{E_v}{2G}\textbf{x}_v+lU\right)+\left(-\frac{G_u}{2E}\textbf{x}_u+\frac{G_v}{2G}\textbf{x}_v+nU\right)$

$= \frac{E_u}{2E}\textbf{x}_u-\frac{E_v}{2G}\textbf{x}_v+lU-\frac{E_u}{2E}\textbf{x}_u+\frac{E_v}{2E}\textbf{x}_v+nU$

$= (l+n)U$

$= 2E\left(\frac{l+n}{2E}\right)U.$

By examining the formula for mean curvature when $E=G$ and $F=0$, we see that

$\displaystyle H=\frac{lG-2mF+nE}{2EG-F^2}=\frac{lE+nE}{2E^2}=\frac{E(l+n)}{2E^2}=\frac{l+n}{2E}.$

Therefore, $\textbf{x}_{uu}+\textbf{x}_{vv}=(2EH)U$. $\Box$

Corollary. A surface $M:\textbf{x}(u,v)=\left(x^1(u,v),x^2(u,v),x^3(u,v)\right)$ with isothermal coordinates is minimal if, and only if, $x^1$, $x^2$, and $x^3$ are harmonic functions.

Proof. If $M$ is minimal then $H=0$ and, by the previous theorem, $\textbf{x}_{uu}+\textbf{x}_{vv}=0.$ On the other hand, suppose that $x^1$, $x^2$, and $x^3$ are harmonic functions. Then $\textbf{x}(u,v)$ is harmonic so $\textbf{x}_{uu}+\textbf{x}_{vv}=0$ and, by the previous theorem, $(2EH)U=0$. Therefore, since $U$ is the unit normal and $E=\langle\textbf{x}_u,\textbf{x}_u\rangle\neq 0$, then $H=0$ and $M$ is minimal. $\Box$

Earlier this month I gave a lecture at the differential geometry seminar on the relation between minimal surfaces and the area functional. This is a summary of the lecture.

Definition. A regular parametrized surface is called minimal if its mean curvature is zero everywhere.

We shall try to understand why the word “minimal” is used for such surfaces. One of the original motivations for the development of the theory was the study of soap films formed when dipping closed wires into soapy water. These films tend to form surfaces of least area. In 1760 Joseph Lagrange recognized the connection between surfaces of least area and minimal surfaces and proposed the problem of showing the existence of minimal surfaces with a given boundary. This is now known as Plateau’s problem, named after the Belgian physicist Joseph Plateau for his experiments with soap films.

Formally, Plateau’s problem can be stated as follows. Given a curve $C$, find a minimal surface $M$ having $C$ as its boundary. As we shall see, least-area surfaces are minimal. Thus, another version of Plateau’s problem is to find a least-area surface having $C$ as its boundary.

What is a necessary condition that $M$ have least-area among all surfaces with boundary $C$? The answer may be found in a simplified version of the calculus of variations as follows.

Suppose that $M:z=f(x,y)$ is a surface of least-area with boundary $C$. Consider the nearby surfaces which look like slightly deformed versions of $M$, $M^t:z^t(x,y)=f(x,y)+tg(x,y).$ Here, $g$ is a function on the domain of $f$ which has the effect, when multiplied by a small $t$ and added to $f$, of moving points of $M$ a small bit and leaving $C$ fixed. That is, $g$ on $\partial C$ where $\partial C$ is the boundary of the domain of $f$ and $f(\partial C)=C$. Let us parametrize $M^t$ by $\textbf{x}^t(u,v)=(u,v,f(u,v)+tg(u,v))$ where $u=x$, $v=y$. Recall that the surface area of $z=f(u,v)+tg(u,v)$ is given by

$\displaystyle A(t)=\int_v\int_u\sqrt{1+f_u^2+f_v^2+2t(f_ug_u+f_vg_v)+t^2(g_u^2+g_v^2)}\,dudv.$

Now, taking the derivative with respect to $t$, which passes inside the integral, we obtain

$\displaystyle A'(t) = \int_v\int_u\frac{f_ug_u+f_vg_v+t(g_u^2+g_v^2)}{\sqrt{1+f_u^2+f_v^2+2t(f_ug_u+f_vg_v)+t^2(g_u^2+g_v^2)}}\,dudv.$

We assumed that $z=z_0$ was a minimum so $A'(0)=0$. Therefore, setting $t=0$ in the equation above, we get

$\displaystyle 0 = \int_v\int_u\frac{f_ug_u+f_vg_v}{\sqrt{1+f_u^2+f_v^2}}\,dudv.$

Now, let $P=\frac{f_ug}{\sqrt{1+f_u^2+f_v^2}}$ and $Q=\frac{f_vg}{\sqrt{1+f_u^2+f_v^2}}.$

Computing $\frac{\partial P}{\partial u}$, $\frac{\partial Q}{\partial v}$, and applying Green’s theorem we then get

$\displaystyle = \int_v\int_u \frac{f_ug_u+f_vg_v}{\sqrt{1+f_u^2+f_v^2}}\,dudv$

$\displaystyle + \int_v\int_u \frac{g[(1+f_u^2)f_{vv}-2f_uf_vf_{uv}+(1+f_v^2)f_{uu}]}{(1+f_u^2+f_v^2)^{\frac{3}{2}}}\,dudv$

$\displaystyle = \int_C\frac{f_ug}{\sqrt{1+f_u^2+f_v^2}}\,dv-\frac{f_vg}{\sqrt{1+f_u^2+f_v^2}}\,du$

$\displaystyle = 0$

since $g=0$ on $\partial C$. Of course the first integral is zero as well, so we end up with

$\displaystyle 0=\int_v\int_u \frac{g[(1+f_u^2)f_{vv}-2f_uf_vf_{uv}+(1+f_v^2)f_{uu}]}{(1+f_u^2+f_v^2)^{\frac{3}{2}}}\,dudv.$

Since this is true for all such $g$, by the fundamental lemma of the calculus of variations, we must have

$\displaystyle 0=(1+f_u^2)f_{vv}-2f_uf_vf_{uv}+(1+f_v^2)f_{uu}.$

But this shows that that mean curvature is identically zero! Therefore, we have shown the following necessary condition for a surface to be area minimizing: if $M$ is area minimizing then $M$ is minimal.

I particularly like the previous result as it has a very nice application of Green’s theorem. Unfortunately, the previous result is somewhat limiting since it only considers if our surface is parametrized as a graph. By dropping this restriction we can arrive at a slightly more general result.

We define the normal variation of a surface $M$ in $\mathbb{R}^3$ to be a family of surfaces $t\rightarrow M(t)$ representing how $M$ changes when pulled in a normal direction. Let $A(t)$ denote the area of $M(t)$. We show that the mean curvature of $M$ vanishes if, and only if, the first derivative of $t\rightarrow A(t)$ vanishes at $M$.

Let $\textbf{x}:U\subset\mathbb{R}^2\rightarrow\mathbb{R}^3$ be a regular parametrized surface and choose a bounded domain $D\subset U$. Suppose that $h:\bar{D}\rightarrow\mathbb{R}$ is differentiable and $\epsilon>0$, where $\bar{D}$ is the union of the domain $D$ with its boundary $\partial D$. Let $N$ denote a unit vector field such that $N(u,v)$ is perpendicular to $\textbf{x}$ for all $(u,v)\in U$; that is, $\langle N,\textbf{x}_u\rangle=\langle N,\textbf{x}_v\rangle=0$.

Definition. The normal variation of $\textbf{x}(\bar{D})$, determined by $h$ is the map $\varphi:\bar{D}\times(-\epsilon,\epsilon)\rightarrow\mathbb{R}^3$ given by $\varphi(u,v,t)=\textbf{x}(u,v)+th(u,v)N(u,v)$ for $(u,v)\in\bar{D}$ and $t\in(-\epsilon,\epsilon)$. For each fixed $t\in(-\epsilon,\epsilon)$, the map $\textbf{x}^t:D\rightarrow\mathbb{R}^3$ given by $\textbf{x}^t(u,v)=\varphi(u,v,t)$ is a parametrized surface.

We denote by $E^t$, $F^t$, and $G^t$ the coefficients of the first fundamental form of $\textbf{x}^t.$ Then the area $A(t)$ of $\textbf{x}^t(\bar{D})$ is $A(t)=\iint_{\bar{D}}\sqrt{E^tG^t-(F^t)^2}\;dudv.$

Lemma. We have $A'(0)=-2\iint_{\bar{D}}hH\sqrt{EG-F^2}\;dudv$ where $H$ denotes the mean curvature of $M$.

Theorem. Let $\textbf{x}:U\rightarrow\mathbb{R}^3$ be a regular parametrized surface and let $D\subset U$ be a bounded domain in $U$. Then $\textbf{x}$ is minimal on $D$ if, and only if, $A'(0)=0$ for all such $D$ and all normal variations of $\textbf{x}(\bar{D})$.

Proof. If $\textbf{x}$ is minimal then $H$ is identically zero and so $A'(0)=0$ for any $h$. Conversely, assume that $A'(0)=0$ for any $h$, but that there exists some $q\in D$ for which $H(q)\neq0$. We choose $h:\bar{D}\rightarrow\mathbb{R}$ such that $h(q)=H(q)$ and $h$ is identically zero outside a small neighborhood of $q$. But then, $A'(0)<0$ for the variation determined by this $h$. This contradiction shows that $H(q)=0$. Since $q$ is arbitrary, $\textbf{x}$ is minimal. $\Box$

It should be pointed out that we have said nothing about the second derivative of $A$ at $0$, so that a minimal surface, although a critical point of $A$, may not actually be a minimum. This is a question concerning the stability of minimal surfaces and might possibly be discussed in a later post.