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.