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.