Minimal surface having two plane curves as its boundary

Question

Let $\alpha$ and $\beta$ be closed, simple plane curves contained in $\mathbb{R}^2\times\{0\}$. For $d>0$, define $\beta_d(t):=\beta(t)+d(0,0,1)$, the translation of $\beta$ along the $z$-axis. Define

$$S:=\{d\in\mathbb{R}_+\mid\exists\text{ a minimal surface }\Sigma\subset\mathbb{R}^3\text{ such that }\partial\Sigma=\alpha\cup\beta_d\}$$

Prove that $\sup S<\infty$

[hint: compare $\Sigma$ with an appropriate catenoid and use the maximum principle]

I probably don't know what this hint means. I know that minimal surfaces with isothermal parametrization have harmonic coordinate functions. Is the hint suggesting to use the maximum principle for harmonic functions? How would I do that?

Answer 1

Here are the assumptions. This is just one of hundreds of other possible assumptions:

• $\alpha, \beta, \Sigma \in \mathbb{R}^3$,

Here assuming the curves are enough smooth for us, i.e. both curves are homeomorphic to the unit 2d ball circumference $S^2$, with the proper z coordinates:

• $\alpha:\{(x,y,z),(x,y) \cong S^2, z=z_1=1\}$,
• $\beta:\{(x,y,z),(x,y) \cong S^2, z=z_2=d,d>0\}$,

Here describing the surface we want. Basically a 2D shape containing our curves. Lets describe it as a non pathological surface, i.e. a set homeomorphic to the unit 3D ball surface $S^3$, the boundary of a simply connected set, without holes, orientable.

• $\Sigma:\{(x,y,z) \in \partial V, V \in \mathbb{R}^3, V \cong S^3 \alpha \ \beta\in \partial V\}$.

Hence we are interested in the area of our surface:

• $S=A(\Sigma)$

Please check if all these assumptions are properly adressing the problem.

The following proof is simply geometrical, exploiting the fact every defined shape is a stretching of some unit ball. Hence there is no pathological situation and hence the area is bounded for these conditions. Note that not every surface not every curve for the problem will render the proposition valid!. Only under some smoothness conditions.

First, lets note each curves are homeomorphic to the unit 2d ball circumference, hence $\alpha$ and $\beta$ are homeomorphic between them:

• $\alpha \cong S^2: \exists f_{\alpha}:\alpha \rightarrow S^2, f_{\alpha} \ \text{bi}, C^1$,
• $\beta \cong S^2: \exists f_{\beta}:\alpha \rightarrow S^2, f_{\beta} \ \text{bi}, C^1$,
• $\alpha \cong \beta: \exists f=f_{\alpha} f_{\beta}^{-1}:\alpha \rightarrow \beta, f \ \text{bi}, C^1$.

Now lets check our surface. We built it homeomorphic to the unit 3d ball surface:

• $\Sigma \cong S^3: \exists f_{\Sigma}:\Sigma \rightarrow S^3, f_{\Sigma} \ \text{bi}, C^1$,

We know the unit 3d ball surface is simply connected, hence every closed path over its surface can be continuously transformed into any other closed path. Those closed paths are, again, homeomorphic between them and with the unit 2d ball circumference:

• $\forall \gamma_i, i=1,2 \cong S^2, \gamma_i \in S^3, \exists f_{\gamma_i}: \gamma_i \rightarrow S^3, f_{\gamma_i} \ \text{bi}, C^1$,
• $\forall \gamma_1 \cong \gamma_2: \exists f_{\gamma_{ij}}=f_{\gamma_1}f_{\gamma_2}^{-1}: \gamma_1 \rightarrow \gamma_2, f_{\gamma_{ij}} \ \text{bi}, C^1$.

In particular, our curves are two of these paths.

Now we have to realize that our surface is bounded; again by construction, the infinity, i.e. an arbitrary large 3d surface ball, do not crosses of our shapes:

• $\forall n \in \mathbb{R} \exists X_n \in \Sigma, X_n \in S^3(n) \rightarrow \Sigma \ \neg \text{compact}$
• $\Sigma \cong S^3, \Sigma \ \text{compact}$
• $\rightarrow\leftarrow$

Also, lets realize each closed path inside our surface is bounded in length. That is, there are no infinite paths curves. So go and assume there is an infinite path curve, as we built, homeomorphic to the 2d ball curve. If there are one curve with an infinite length, there is a sequence inside a bounded 3d surface, with a given limit, and we always could get the path out of that limit. A contradiction again:

• $\exists \gamma, \exists f:\gamma \rightarrow S^2, f \text{bi}, C^1$,
• $\forall y \in S^2 \exists x in \gamma, |y-f(x)|<\epsilon$,
• $\forall n in \mathbb{N} \exists x_n \in \gamma, x_0 \in \gamma, |x_n-x_0|<\epsilon$,
• $\gamma \text{ infinity length}:\forall n \in \mathbb{N} \exists \epsilon' \in \mathbb{R}, \gamma' = \gamma \cap (B(\epsilon')^2 \cap \Sigma), x_0 \neg \in \gamma', x_n \in \gamma, |x_n-x_0|>\epsilon'$,
• $\gamma \text{compact}: \forall n \in N, x_n \in \gamma, \exists x_0 \in \gamma, |x_n-x_0|<\epsilon$,
• $\rightarrow\leftarrow$.

Hence the surface is bounded, every path is length bounded, and thus, every ball over every point in $\Sigma$ it will have a bounded area. Moreover, because our surface is simply connected, there exist a continuous mapping between $\alpha$ and $\beta$, passing through $\Sigma$.

• $\exists F:\sigma \times [0,1] \rightarrow \Sigma, F(\Sigma,0)=\alpha, F(\Sigma,1)=\beta$

Because the unit 2d ball is homeomorphic to the [0,1) interval under any trigonometric function properly suited, and properly setting up the $t$ parameters of our original curves, we can parametrize this further:

• $\exists F(t,s):[0,1) \times [0,1] \rightarrow \Sigma, F(t,0)=\alpha(t), F(t,1)=\beta(t)$

Moreover, the surface can be obtained as an integral, and the integral exists.

$$ S=\int_{\Sigma}d\Sigma=\int_0^1\int_0^1dsdt $$