Two-dimensional solutions to the c-Plateau problem in $\mathbb {R}^{3}$

Abstract

We give several regularity results for two-dimensional solutions to the c-Plateau problem in $\mathbb {R}^{3}:$ given $c>0$ and an integer one-rectifiable current $\varGamma$ without boundary in $\mathbb {R}^{3},$ we study integer two-rectifiable currents which minimize c-isoperimetric mass $\mathbf {M}^{c}(T):=\mathbf {M}(T)+c \mathbf {M}(\partial T)^{2},$ where $\mathbf {M}$ is the usual mass on currents, amongst all integer two-rectifiable currents T with boundary of the form $\partial T = \varGamma + \varSigma$ where $\varGamma ,\varSigma$ have disjoint supports. We show the following three results for $\mathbf {T}_{c}$ a solution to the c-Plateau problem with $\partial \mathbf {T}_{c} = \varGamma +\varSigma _{c}$: if $\varSigma _{c}$ is a smooth closed embedded curve, then $\varSigma _{c}$ parameterized by arc-length must have at some point large third derivative; $\mathbf {T}_{c}$ cannot have a tangent cone at a singular point of $\varSigma _{c}$ supported in a plane but with constant orientation; $\varSigma _{c}$ is regular wherever we can write the support of $\mathbf {T}_{c}$ as a finite union of $C^{1}$ surfaces-with-boundary.

Appendix

In this section we prove the ODE lemma used in the proof of Theorem 8.

Lemma 3

Suppose $\beta \in (0,1),$ and that with

$$\begin{aligned} \tilde{U} = \{ y \in \mathbb {R}^{2}: y_{1} > \delta ^{\beta } |y|^{1+\beta }, |y| < 1 \}, \end{aligned}$$

for some $\delta >0,$ we have a function

$$\begin{aligned} \tilde{u} \in C^{\infty }(\tilde{U}) \cap C^{1,\beta }({{\mathrm{clos}}}\tilde{U}) \end{aligned}$$

satisfying $\tilde{u}(0)=0,$ $D \tilde{u}(0)=0,$ and

$$\begin{aligned} \sup _{y \in \tilde{U}} \frac{|D^{2} \tilde{u}|}{|y|^{\beta -1}} < \infty . \end{aligned}$$

In addition, suppose we have two curves

$$\begin{aligned} \sigma _{1},\sigma _{2} \in C^{1}([0,\tilde{\epsilon });\{0\} \cup {{\mathrm{graph}}}_{\tilde{U}} \tilde{u}) \cap C^{\infty }((0,\tilde{\epsilon });{{\mathrm{graph}}}_{\tilde{U}} \tilde{u}) \end{aligned}$$

parameterized by arc length with $\sigma _{1}(0) = \sigma _{2}(0) = 0,$ $\sigma _{1}'(0) = \sigma _{2}'(0) = e_{1},$ and suppose there is $K \in (0,\infty )$ so that $|\sigma _{1}''(t)| = |\sigma _{2}''(t)| = K$ for each $t \in (0,\tilde{\epsilon }).$ If

$$\begin{aligned} \sigma _{1}''(t) \in T_{\sigma _{1}(t)} {{\mathrm{graph}}}_{\tilde{U}} \tilde{u}, \ \sigma _{2}''(t) \in T_{\sigma _{2}(t)} {{\mathrm{graph}}}_{\tilde{U}} \tilde{u} \end{aligned}$$

for each $t \in (0,\tilde{\epsilon }),$ and

$$\begin{aligned} \lim _{t \searrow 0} \sigma _{1}''(t) = \lim _{t \searrow 0} \sigma _{2}''(t) = -Ke_{2}, \end{aligned}$$

then $\sigma _{1}(t) = \sigma _{2}(t)$ for $t \in (0,\tilde{\epsilon }).$

Proof

We can take $C \in (0,\infty )$ so that

$$\begin{aligned} |D^{2} \tilde{u}(\mathbf {p}(x))| \le C |(\mathbf {p}(x))|^{\beta -1} \end{aligned}$$

whenever $x \in \sigma _{1}((0,\tilde{\epsilon })) \cup \sigma _{2}((0,\tilde{\epsilon })),$ where $\mathbf {p}:\mathbb {R}^{3} \rightarrow \mathbb {R}^{2}$ is the projection map. For $\nu$ the upward pointing unit normal of ${{\mathrm{graph}}}_{\tilde{U}} \tilde{u},$ we may also assume $C \in (0,\infty )$ is such that for $x,\tilde{x} \in \sigma _{1}((0,\tilde{\epsilon })) \cup \sigma _{2}((0,\tilde{\epsilon }))$

$$\begin{aligned} |\nu (x)-\nu (\tilde{x})| \le C |x-\tilde{x}| \cdot \left| |\mathbf {p}(x)| - |\mathbf {p}(x)-\mathbf {p}(\tilde{x})| \right| ^{\beta -1}. \end{aligned}$$

(33)

Since $\sigma _{\ell }''(t) \in T_{\sigma _{\ell }(t)} {{\mathrm{graph}}}_{\tilde{U}} \tilde{u}$ for each $t \in (0,\tilde{\epsilon })$ and $\ell =1,2,$ $\sigma _{1}'(0) = \sigma _{2}'(0)= e_{1}$, $\lim _{t \searrow 0} \sigma _{1}''(t) = \lim _{t \searrow 0} \sigma _{2}''(t) = -Ke_{2},$ and $\nu (0)=e_{3},$ then

$$\begin{aligned} \sigma _{1}''(t) = K \sigma _{1}'(t) \times \nu (\sigma _{1}(t))\quad \text { and }\; \sigma _{2}''(t) = K \sigma _{2}'(t) \times \nu (\sigma _{1}(t)) \end{aligned}$$

for $t \in (0,\tilde{\epsilon }),$ so that

$$\begin{aligned} |\sigma _{1}''(t)-\sigma _{2}''(t)| \le K \left[ |\sigma _{1}'(t)-\sigma _{2}'(t)| + | \nu (\sigma _{1}(t))- \nu (\sigma _{2}(t))| \right] . \end{aligned}$$

Letting $d(t) = \sigma _{1}(t)-\sigma _{2}(t),$ then (33) implies for $t \in (0,\epsilon )$

$$\begin{aligned} |d''(t)| \le K \left[ |d'(t)| + C |d(t)| \cdot \left| |\mathbf {p}(\sigma _{1}(t))| - |\mathbf {p}(\sigma _{1}(t))- \mathbf {p}(\sigma _{2}(t)) | \right| ^{\beta -1} \right] . \end{aligned}$$

(34)

Since $\lim _{\searrow 0} d''(t) =0,$ then we can choose $\tau \in (0,\min \{ K,\frac{1}{2K}, \frac{1}{(2C)^{1/\beta }},\tilde{\epsilon } \})$ so that $|d''(t)| < \frac{1}{8K}$ for $t \in (0,\tau ).$ Therefore,

$$\begin{aligned} |d'(t)| \le \frac{t}{8K}\quad \text { and }\; |d(t)| \le \frac{t^{2}}{16K} \end{aligned}$$

for $t \in (0,\tau ).$ Since $\sigma _{1}'(0) = e_{1}$ and $\sigma _{1}$ is parameterized by arc-length, we can also assume $\sigma _{1}(t) \cdot e_{1} > \frac{t}{2}$ for each $t \in (0,\tau ).$

We conclude that for $t \in (0,\tau )$

$$\begin{aligned} \frac{1}{2} \sigma _{1}(t) \cdot e_{1} > K^{1/2} |d(t)|^{1/2} \ge |d(t)|. \end{aligned}$$

This implies for $t \in (0,\tau )$

$$\begin{aligned} \left| |\mathbf {p}(\sigma _{1}(t))|-|\mathbf {p}(\sigma _{1}(t))-\mathbf {p}(\sigma _{2}(t))| \right| ^{\beta -1} \le \left( \frac{1}{2} \sigma _{1}(t) \cdot e_{1} \right) ^{\beta -1}. \end{aligned}$$

(35)

From (34), (35) we compute for $t \in (0,\tau )$

$$\begin{aligned} |d''(t)|\le & {} K \left[ |d'(t)| + C |d(t)| \cdot (K^{1/2} |d(t)|^{1/2})^{\beta -1} \right] \\\le & {} K \left[ |d'(t)| + C K^{\frac{\beta -1}{2}} |d(t)|^{\frac{\beta +1}{2}} \right] \\\le & {} \frac{t}{8} + \frac{C t^{\beta +1}}{4^{\beta +1}} \\\le & {} \frac{(1+C \tau ^{\beta }) t}{4} < t. \end{aligned}$$

We now argue iteratively. Suppose we have shown for $n \ge 1$ that

$$\begin{aligned} |d''(t)| < K^{n-1} t^{n} \end{aligned}$$

whenever $t \in (0,\tau ).$ This implies

$$\begin{aligned} |d'(t)| \le \frac{K^{n-1}}{n+1} t^{n+1}\quad \text { and }\; |d(t)| \le \frac{K^{n-1}}{(n+1)(n+2)} t^{n+2}, \end{aligned}$$

which gives

$$\begin{aligned} \sigma _{1}(t) \cdot e_{1} \ge \frac{1}{2} \left( \frac{(n+1)(n+2)}{K^{n-1}} \right) ^{\frac{1}{n+2}} |d(t)|^{\frac{1}{n+2}}. \end{aligned}$$

Using (34), (35) again we conclude for $t \in (0,\tau )$

$$\begin{aligned} |d''(t)|\le & {} K \left[ |d'(t)| + C|d(t)| \cdot (\sigma _{1}(t) \cdot e_{1}/2)^{\beta -1} \right] \\\le & {} K \left[ |d'(t)| + 4^{1-\beta } C \left( \frac{(n+1)(n+2)}{K^{n-1}} \right) ^{\frac{\beta -1}{n+2}} |d(t)|^{1+\frac{\beta -1}{n+2}} \right] \\\le & {} K \left[ \frac{K^{n-1}}{n+1} t^{n+1} + 4^{1-\beta } C \left( \frac{K^{n-1}}{(n+1)(n+2)} \right) t^{n+1+\beta } \right] \\\le & {} \left[ \frac{1}{n+1} + \frac{4^{1-\beta } C \tau ^{\beta }}{(n+1)(n+2)} \right] K^{n} t^{n+1} \\< & {} K^{n} t^{n+1}, \end{aligned}$$

since $C \tau ^{\beta } < \frac{1}{2}$ and $n \ge 1.$

We have thus shown $|d''(t)| \le K^{n-1} t^{n}$ for each $n \ge 1$ whenever $t \in (0,\tau ).$ Since $\tau \le \frac{1}{2K},$ then $|d''(t)| \le \frac{1}{K} (1/2)^{n}$ for each $n \ge 1,$ which means $d''(t) = 0$ for $t \in (0,\tau ).$ The lemma follows by considering $\sigma _{1}(\tau )=\sigma _{2}(\tau ).$

The proof of Lemma 3 also gives the following slight modification.

Lemma 4

Suppose $M \subset B_{\rho }(0)$ is a $C^{2}$ surface-with-boundary, with $0 \in {{\mathrm{clos}}}M$ and $T_{0} M = \mathbb {R}^{2}$ (where possibly $0 \in \partial M$). Also, suppose we have two curves

$$\begin{aligned} \sigma _{1},\sigma _{2} \in C^{1}([0,\epsilon ); B_{\rho }(0) \cap {{\mathrm{clos}}}M) \cap C^{2}((0,\epsilon );B_{\rho }(0) \cap {{\mathrm{clos}}}M) \end{aligned}$$

parameterized by arc length with $\sigma _{1}(0) = \sigma _{2}(0) = 0,$ $\sigma _{1}'(0) = \sigma _{2}'(0) = e_{1},$ and suppose there is $K \in (0,\infty )$ so that $|\sigma _{1}''(t)| = |\sigma _{2}''(t)| = K$ for each $t \in (0,\epsilon ).$ If

$$\begin{aligned} \sigma _{1}''(t) \in T_{\sigma _{1}(t)} M, \ \sigma _{2}''(t) \in T_{\sigma _{2}(t)} M \end{aligned}$$

for each $t \in (0,\epsilon )$ (including when $\sigma _{1}(t) \in {{\mathrm{clos}}}M$ or $\sigma _{2}(t) \in {{\mathrm{clos}}}M$), and

$$\begin{aligned} \sigma _{1}''(0) = \sigma _{2}''(0) = -Ke_{2}, \end{aligned}$$

then $\sigma _{1}(t) = \sigma _{2}(t)$ for $t \in (0,\epsilon ).$

Proof

Follows by the proof of Lemma 3, essentially by noting we can take $\beta = 1.$ $\square$