# On minimal surfaces

Let $D \subseteq \mathbb{R}^2$ be a domain. $f: D \rightarrow \mathbb{R}$, $f \in C^2(D)$.

Recall: from last talk, Zhenghe described the Lagrange’s Equation, in this case the equation is written: (we denote $\frac{\partial f}{\partial x}$ as $f_x$)

$(1+ f_y^2)f_{xx}-2f-xf-yf_{xy}+(1+f_x^2)f_{yy}=0$

Theorem: The graph of $f$ is area-minimizing then $f$ satisfies Lagrange’s equation.

Proof: Since $f$ is area-minimizing,

$A(f) = \int_D(1+f_x^2+f_y^2)^{\frac{1}{2}}dx dy$

is minimized by $f$ for given boundary values. Hence the variation $\delta A$ of $A$ due to an infinitesimal $\delta f$ of $f$ where $\delta f |_{\partial D} = 0$. i.e.

$\delta A = \int_D \frac{1}{2}(1+f_x^2+f_y^2)^{-\frac{1}{2}}(2 f_x \delta f_x + 2 f_y \delta f_y) dxdy$

$= \int_D \{ [ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_x ] \delta f_x$

$+ [ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_y ] \delta f_y \}dxdy =0$

Let $G(f) = - \frac{\partial}{\partial x}[ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_x ] - \frac{\partial}{\partial y}[ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_y ]$

$= -(1 + f_x^2 + f_y^2)^{-\frac{3}{2}} [(1+f_y^2)f_{xx}$

$- 2f_x f_y f_{xy}+(1+f_x^2)f_{yy}]$

Apply integration by parts, since $\delta f$ vanishes on $\partial D$, the constant term vanishes, we have:

$\int_D G(f) \delta f dxdy= 0$ for all $\delta f$ with $\delta f |_{\partial D} = \bar{0}$, hence $G(f) = \bar{0}$. i.e.

$(1+ f_y^2)f_{xx}-2f-xf-yf_{xy}+(1+f_x^2)f_{yy}=0$

which is the Lagrange’s equation.

We should note that the converse of the theorem is, in general, not true.

Example: two rectangles, star-shaped 4-gon.

Theorem: For $D$ convex, any $f$ satisfying Lagrange’s equation has area-minimizing graph.

Let $\varphi$ be 2-form in $\mathbb{R}^3$ s.t. $d \varphi = 0$ and $\sup(\varphi) = 1$. i.e. $\varphi$ acts on the unit Grassmannian space of oriented planes in $\mathbb{R}^3$.

Definition: An immersed surface $S$ is calibrated by $\varphi$ if $\varphi(P) = 1$ for all $P$ in the unit tangent bundle of $S$.

*All calibrated surfaces are automatically area-minimizing.

Let $\varphi: D \times \mathbb{R} \rightarrow (\Lambda^2 \mathbb{R}^3)^\ast$ be the two-form

$\varphi(x,y,z) =\frac{ -f_x dydz - f_y dzdx + dxdy}{f_x^2+f_y^2+1}$

By construction, for all $(x,y,z) \in D \times \mathbb{R}$, $v, w \in S^2$, we have $\varphi(x,y,z)(v,w) \leq 1$, $\varphi(x,y,z)(v,w) = 1$ when the plane spanned by $v, w$ is tangent to the graph of $f$ at $(x,y,f(x,y))$.

$d \varphi = - \frac{\partial}{\partial x}f_x(f_x^2+f_y^2+1)^{-\frac{1}{2}}- \frac{\partial}{\partial y}f_y(f_x^2+f_y^2+1)^{-\frac{1}{2}}$

$= -(1 + f_x^2 + f_y^2)^{-\frac{3}{2}} [(1+f_y^2)f_{xx}$

$- 2f_x f_y f_{xy}+(1+f_x^2)f_{yy}]$

which is $0$ by Lagrange’s equation. Hence $\varphi$ is closed.

Let $S$ be the graph of $f$, since $\varphi(p)(v,w) = 1$ whenever the plane spanned by $v, w$ is tangent to $S$ at $p$, we have

$A(S) = \int_S \varphi$

Suppose $S$ is not area-minimizing, there exists 2-chain $T$ with $\partial T = \partial S$ with smaller area than that of $S$.

Since $D$ is convex, any $T$ not contained in $D \times \mathbb{R}$ cannot be area-minimizing (by projecting to the cylinder). Hence we may assume $T \subseteq D \times \mathbb{R}$ (So that $\varphi$ is well-defined on $T$)

Since $S-T$ bounds a 3-chain, $\varphi$ is closed, hence $\int_S \varphi = \int_T \varphi$

Beacuse $\varphi(p)(v,w) \leq 1$ hence $\int_T \varphi \leq A(T)$.

Therefore we have $A(S) \leq A(T)$. i.e. $S$ is area-minimizing.

Definition: A minimal surface in $\mathbb{R}^3$ is a smoothly immersed surface which is locally the graph of a solution to the Lagrange’s equation.

Note that small pieces of minimal surfaces are area-minimizing but lager pieces may not be.

Example: Enneper’s surface

Theorem: Let $C$ be a rectifiable Jordan curve in $\mathbb{R}^3$, there is a area-minimizing 2-chain $S \subseteq \mathbb{R}^3$ with $\partial S = C$

Sketch of proof:

There exists rectifiable 2-chain with boundary being $C$. -Take a point in $\mathbb{R}^3$ and take the cone of the curve.

Define flat norm on the space of 2-chains in $\mathbb{R}^3$ by $F(T) = \inf\{A(C_2)+V(C_3) \ | \ T = C_2+\partial C_3 \}$ i.e. if two chains are close together, they would almost bound a 3-chain with small volume, hence the difference has small norm.

Fact: $\mathcal{T} = \{ T \subseteq B^3(\bar{0}, R) \ | \ A(T) < K$ and $l(\partial T)0$, for any chain $T$, we may find a chain $T'$ inside the grid of mesh $\delta$ where $F(T-T') < \varepsilon$ (hence the area of $T'$ is also bounded). Since there are only finitely many such chains, we have:

$\mathcal{T}$ is totally bounded under the flat norm $F$.

Hence $\mathcal{T}$ is compact.

Now we choose sequence $(T_n)$ of rectifiable chains with boundary $C$ and area decreasing to $\inf \{ A(T) \ | \ \partial T = C \}$

Choose $R$ large enough s.t. $C \subseteq B^3(\bar{0}, R)$. Project $\mathbb{R}^3 \backslash B^3(\bar{0}, R)$ radially onto $S^2(\bar{0}, R)$ the projection does not increase area.

Hence $A(\pi(T_i)) \leq A(T_i)$ for all $i$. i.e. $(A(\pi(T_i))) \rightarrow \inf \{ A(T) \ | \ \partial T = C \}$ and $\partial(\pi(T_i)) = \partial (T_i) = C$.

Since $\mathcal{T} = \{ T \subseteq B^3(\bar{0}, R) \ | \ A(T) < K$ and $l(\partial T)< K \}$ is compact, there exists subsequence $(T_{n_i})$ converging to a rectifiable chain $S \in \mathcal{T}$.

We can prove that: $\partial S = C$ (continuity of $\partial$ under the flat norm).

$A(S) = \inf \{ A(T) \ | \ \partial T = C \}$ (lower-semicontinuity of area under the flat norm).

Therefore $S$ is an area-minimizing surface with $\partial S = C$.

## One thought on “On minimal surfaces”

1. Mengyi Shen says:

You get up so early…amazing. When will you leave?

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