Local solutions to a free boundary problem for the Willmore functional

Abstract

We consider a free boundary problem for the Willmore functional $\mathcal{W}(f) = \frac{1}{4} \int _\Sigma H^2\,d\mu _f$. Given a smooth bounded domain $\Omega \subset {\mathbb R}^3$, we construct Willmore disks which are critical in the class of surfaces meeting $\partial \Omega$ at a right angle along their boundary and having small prescribed area. Using rescaling and the implicit function theorem, we first obtain constrained solutions with prescribed barycenter on $\partial \Omega$. We then study the variation of that barycenter.

Appendix: Construction of the barycenter

The concept of Riemannian barycenter is due to Karcher [11]. For our purposes we only need a local version, which does not involve e.g. Riemannian comparison theory. Let $U = D_\delta (0) \subset {\mathbb R}^2$, $V = B_{\frac{3}{2}}(0) \subset {\mathbb R}^3$. For $x \in U$, $v \in V$ we put

$$\begin{aligned} c_{x,v}:[0,1] \rightarrow Z_2,\,c_{x,v}(t) = x + t v. \end{aligned}$$

Further let $X = \{\phi \in C^2([0,1],{\mathbb R}^3): \phi (0) = \phi '(0) = 0\}$ and

$$\begin{aligned} X_\varepsilon = \{\phi \in X: \Vert \phi \Vert _{C^0([0,1])} < \varepsilon \}. \end{aligned}$$

We finally put $G_\varepsilon = \{\tilde{g} \in C^l(\bar{Z_2},{\mathbb R}^{3 \times 3}): \Vert \tilde{g} - \delta \Vert _{C^l(Z_2)} < \varepsilon \}$ for $l \ge 1$, and consider

$$\begin{aligned} F:U \times V \times X_\varepsilon \times G_\varepsilon \rightarrow C^0([0,1],{\mathbb R}^3),\, F[x,v,\phi ,\tilde{g}] = c'' + \tilde{\Gamma } \circ c (c',c')|_{c = c_{x,v} +\phi }. \end{aligned}$$

We claim that F is of class $C^{l-1}$. Write $F = F_2 \circ F_1$ where $F_1$ is the affine map

$$\begin{aligned} F_1:U \times V \times X_\varepsilon \rightarrow C^2([0,1],{\mathbb R}^3),\,F_1[x,v,\phi ] = c_{x,v} + \phi . \end{aligned}$$

$F_1$ is continuous and hence smooth. The nonlinear map $F_2$ is given by

$$\begin{aligned} F_2:C^2([0,1],Z_2) \times G_\varepsilon \rightarrow C^0([0,1],{\mathbb R}^3),\, F_2[c,\tilde{g}] = c'' + \tilde{\Gamma } \circ c (c',c'). \end{aligned}$$

The composition $C^2 \times C^{l-1} \rightarrow C^0$, $(c,\tilde{\Gamma }) \mapsto \tilde{\Gamma } \circ c$, is of class $C^{l-1}$. Namely differentiating $l-1$ times with respect to c leaves exactly a $C^0$ function. Since we can build $F_2$ from $\tilde{\Gamma }\circ c$ by linear or bilinear operations, it is also of class $C^{l-1}$. Assuming from now on $l \ge 2$, we have

$$\begin{aligned} D_c F_2[c,\tilde{g}] \phi= & {} \phi '' + 2\, \tilde{\Gamma } \circ c(\phi ',c') + (D\tilde{\Gamma })\circ c (\phi ,c',c')\\ D_{\tilde{g}} F_2[c,\tilde{g}] h= & {} \frac{1}{2} \left( \tilde{g}^{kp}(\partial _i h_{jp} + \partial _j h_{ip} - \partial _p h_{ij}) \right) \circ c\, (c^i)' (c^j)' e_k\\&- \frac{1}{2} \left( \tilde{g}^{km} h_{mn} \tilde{g}^{np} \left( \partial _i \tilde{g}_{jp} + \partial _j \tilde{g}_{ip} - \partial _p \tilde{g}_{ij}\right) \right) \circ c\, (c^i)' (c^j)' e_k. \end{aligned}$$

In particular

$$\begin{aligned} F[x,v,0,\delta ] = 0 \quad \text{ and } \quad D_\phi F[x,v,0,\delta ] \psi = \psi ''. \end{aligned}$$

The map $D_\phi F[x,v,0,\delta ]:X \rightarrow C^0([0,1],{\mathbb R}^3)$ is an isomorphism, in fact the equation $\psi '' = f$ has the unique solution $\psi \in X$ given by

$$\begin{aligned} \psi (u) = \int _0^u \int _0^s f(t)dt\,ds. \end{aligned}$$

By the implicit function theorem, the set of solutions of $F[x,v,\phi ,\tilde{g}] = 0$ near $[0,v_0,0,\delta ]$ is given as a $C^{l-1}$ graph

$$\begin{aligned} \mathbf{\phi } = \mathbf{\phi }[x,v,\tilde{g}], \end{aligned}$$

i.e. the corresponding curves $\mathbf{c}[x,v,\tilde{g}] = c_{x,v} + {\phi }[x,v,\tilde{g}]$ are geodesics with respect to $\tilde{g}$ having initial data $c(0) = x$, $c'(0) = v$. The exponential mapping is now given by

$$\begin{aligned} \exp :U \times V \times G_\varepsilon \rightarrow Z_2,\,\exp _x^{\tilde{g}}(v) = \mathbf{c}[x,v,\tilde{g}](1). \end{aligned}$$

Now $D\exp _x^{\delta } = \mathrm{Id}_{{\mathbb R}^3}$, Thus for $l \ge 2$ and $\varepsilon > 0$ small we get

$$\begin{aligned} \Vert D\exp ^{\tilde{g}}_x - \mathrm{Id}_{{\mathbb R}^3}\Vert _{C^0(V)} \le \varepsilon _0 \quad \text{ for } \quad \tilde{g} \in G_\varepsilon ,\,x \in U = D_\delta (0). \end{aligned}$$

This gives for $v,w \in V$

$$\begin{aligned} \left| \exp ^{\tilde{g}}_x (v) - \exp ^{\tilde{g}}_x (w)\right|= & {} \left| \int _0^1 D\exp ^{\tilde{g}}_x((1-t)w + tv)\cdot (v-w)\,dt \right| \\\ge & {} |v-w| - \left| \int _0^1 \left( D\exp ^{\tilde{g}}_x((1-t)w + tv) - \mathrm{Id}_{{\mathbb R}^3} \right) \cdot (v-w)\,dt \right| \\\ge & {} (1-\varepsilon _0) |v-w|. \end{aligned}$$

This shows that $\exp _x^{\tilde{g}}$ is injective on $V = B_{\frac{3}{2}}(0)$. We further estimate

$$\begin{aligned} \left| \exp ^{\tilde{g}}_x(v) - v\right| = \left| \int _0^1 \left( D\exp ^{\tilde{g}}_x(tv) - \mathrm{Id}_{{\mathbb R}^3} \right) \,dt \cdot v \right| \le \varepsilon _0\,|v|. \end{aligned}$$

We now show that $\exp _x^{\tilde{g}}(V) \cap B_{\frac{5}{4}}(0)$ is a closed subset of $B_{\frac{5}{4}}(0)$. Assume that $\exp _x^{\tilde{g}}(v_k) \rightarrow p \in B_{\frac{5}{4}}(0)$. From the above we then have

$$\begin{aligned} |v_k| - \frac{5}{4} < |v_k| - |\exp _x^{\tilde{g}}(v_k)| \le |v_k - \exp _x^{\tilde{g}}(v_k)| \le \varepsilon _0 |v_k|, \end{aligned}$$

which implies $|v_k| \le (1-\varepsilon _0)^{-1}\, \frac{5}{4} < \frac{3}{2}$ for appropriate $\varepsilon _ 0 > 0$. Up to a subsequence, we thus have $v_k \rightarrow v \in V$ and $\exp _x^{\tilde{g}}(v) = p$. Now $\exp _x^{\tilde{g}}(V) \cap B_{\frac{5}{4}}(0)$ is also open by the inverse function theorem, hence we have $B_{\frac{5}{4}}(0) \subset \exp _x^{\tilde{g}}(V)$, and we obtain the inverse

$$\begin{aligned} (\exp _x^{\tilde{g}})^{-1}: B_{\frac{5}{4}}(0) \rightarrow V. \end{aligned}$$

Of course we are not claiming that $\exp _x^{\tilde{g}}$ maps all of V into $B_{\frac{5}{4}}(0)$. The inverse is of class $C^{l-1}$ in all variables $x \in U$, $p \in B_{\frac{5}{4}}(0)$ and $\tilde{g} \in C^l(Z_2)$. Namely let $\exp _{x_0}^{\tilde{g}_0}(v_0) = p_0 \in B_{\frac{5}{4}}(0)$, where $v_0 \in V$. Consider the equation

$$\begin{aligned} \exp _x^{\tilde{g}}(v) - p = 0. \end{aligned}$$

By the implicit function theorem, the set of solutions has a local representation $v = v[x,p,\tilde{g}]$ which is of class $C^{l-1}$. But the local inverse equals the global inverse, and hence also the global inverse is of class $C^{l-1}$ as claimed.

Lemma 12

(two-dimensional barycenter) Assume $w:{\mathbb S}^2_{+} \rightarrow {\mathbb R}$, $\tilde{g}:Z_2 \rightarrow {\mathbb R}^{3 \times 3}$ belong to the neighborhoods $W_\varepsilon$, $G_\varepsilon$ given by

$$\begin{aligned} \Vert w\Vert _{C^1({\mathbb S}^2_{+})} < \varepsilon \quad \text{ and } \quad \Vert \tilde{g}-\delta \Vert _{C^l(Z_2)} < \varepsilon \quad \text{ where } l\ge 2. \end{aligned}$$

For $\varepsilon > 0$ small we then have a welldefined function

$$\begin{aligned} X[w,\tilde{g}]:U \rightarrow {\mathbb R}^2,\,X[w,\tilde{g}](x) = -\pi _{{\mathbb R}^2} \left( \int _{{\mathbb S}^2_{+}} (\exp ^{\tilde{g}}_x)^{-1} (f(\omega ))\,d\mu _g(\omega ) \right) , \end{aligned}$$

and there is a unique point $x \in U$ with $X[w,\tilde{g}](x) = 0$. This point $x = C[w,\tilde{g}]$ is called the two-dimensional barycenter of (the radial graph of) w with respect to $\tilde{g}$. The map $C[w,\tilde{g}]$ is of class $C^{l-1}$.

Proof

Let $f(\omega ) = \omega + w(\omega )$. Fixing a coordinate system on ${\mathbb S}^2_{+}$, we consider the map

$$\begin{aligned} U \times W_\varepsilon \times G_\varepsilon \rightarrow C^0({\mathbb S}^2_{+},{\mathbb R}^3),\,[x,w,\tilde{g}] \mapsto (\exp ^{\tilde{g}}_x)^{-1} \circ f\, \sqrt{\det g}. \end{aligned}$$

(5.1)

By standard rules for product and composition, the right hand side belongs to $C^0({\mathbb S}^2_{+},{\mathbb R}^3)$; in particular $X[w,\tilde{g}]$ is well-defined. We claim that the map (5.1) is of class $C^{l-1}$ in all three variables. For this we recall that $\Psi [x,p,\tilde{g}] = (\exp _x^{\tilde{g}})^{-1}(p)$ is of class $C^{l-1}$. For $\omega \in {\mathbb S}^2_{+}$ fixed we have the $C^{l-1}$ composition

$$\begin{aligned} \begin{array}{ccccc} U \times W_\varepsilon \times G_\varepsilon &{} \mathop {\rightarrow }\limits ^{C^\infty } &{} U \times B_{\frac{5}{4}}(0) \times G_\varepsilon &{} \mathop {\rightarrow }\limits ^{\Psi } &{} V\\ (x,w,\tilde{g}) &{} \mapsto &{} (x,f(\omega ),\tilde{g}) &{}\mapsto &{} \Psi [x,f(\omega ),\tilde{g}]. \end{array} \end{aligned}$$

Now all derivatives with respect to $x,w,\tilde{g}$ up to order $l-1$ depend also continuously on $\omega$, which yields the claim. For $\tilde{g} = \delta$ we have $(\exp _x)^{-1}(p) = p-x$ which implies

in particular $X[0,\delta ](0) = 0$ and $D_x X[0,\delta ](x) = 2\pi \, \mathrm{Id}_{{\mathbb R}^2}$. Thus by the implicit function theorem there is a unique point $x \in U$ with $X[w,\tilde{g}](x) = 0$, and the resulting map $x = C[w,\tilde{g}]$ is of class $C^{l-1}$. $\square$

From the proof we note the explicit formula

(5.2)

We consider the two coordinates $C^i[f,\tilde{g}]$ of the barycenter as functionals depending on w resp. f, and we now compute the corresponding $L^2$ gradient. Consider a compactly supported variation of f in direction $\phi = \varphi \nu$. Then we have

$$\begin{aligned} \frac{\partial }{\partial \varepsilon } (g_{\varepsilon })_{ij}|_{\varepsilon = 0} = - 2\varphi h_{ij} \quad \text{ and } \quad \frac{\partial }{\partial \varepsilon } d\mu _{g_\varepsilon }|_{\varepsilon = 0} = - \varphi H \,d\mu _g. \end{aligned}$$

The first variation of $X[f,\tilde{g}]$ is then

$$\begin{aligned} \frac{\partial }{\partial \varepsilon } X[f_\varepsilon ,\tilde{g}](x)|_{\varepsilon =0}= & {} -\, \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D ((\exp ^{\tilde{g}}_x)^{-1})(f(\omega )) \cdot \phi (\omega )\,d\mu _g(\omega )\\&+\, \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} (\exp ^{\tilde{g}}_x)^{-1}(f(\omega )) H(\omega ) \varphi (\omega )\,d\mu _g(\omega ). \end{aligned}$$

By definition of the barycenter we have

$$\begin{aligned} 0= & {} \frac{\partial }{\partial \varepsilon } X \left[ f_\varepsilon ,\tilde{g}\right] (C[f_\varepsilon ,\tilde{g}])|_{\varepsilon =0}\\= & {} -\,\pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D \left( (\exp ^{\tilde{g}}_x)^{-1} \right) (f(\omega )) \phi (\omega )\,d\mu _g(\omega )|_{x = C[f,\tilde{g}]}\\&+\, \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} (\exp ^{\tilde{g}}_x)^{-1}(f(\omega )) H(\omega ) \varphi (\omega )\,d\mu _g(\omega )|_{x = C[f,\tilde{g}]}\\&-\, \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D_x (\exp ^{\tilde{g}}_x)^{-1} (f(\omega ))\,d\mu _g(\omega )|_{x = C[f,\tilde{g}]} \cdot \frac{\partial }{\partial \varepsilon } C[f_\varepsilon ,\tilde{g}]|_{\varepsilon =0}. \end{aligned}$$

This implies the formula

$$\begin{aligned} \frac{\partial }{\partial \varepsilon }C[f_\varepsilon ,\tilde{g}]|_{\varepsilon =0}= & {} \left( \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D_x (\exp ^{\tilde{g}}_x)^{-1}(f(\omega ))\,d\mu _g(\omega ) \right) ^{-1}|_{x = C[f,\tilde{g}]} \\&\cdot \left( -\pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D \left( (\exp ^{\tilde{g}}_x)^{-1} \right) (f(\omega )) \phi (\omega )\,d\mu _g(\omega )|_{x = C[f,\tilde{g}]}\right. \\&\left. +\, \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} (\exp ^{\tilde{g}}_x)^{-1}(f(\omega )) H(\omega ) \varphi (\omega )\,d\mu _g(\omega )|_{x = C[f,\tilde{g}]} \right) . \end{aligned}$$

Under reparametrizations of f the barycenter remains the same, hence the $L^2$ gradient of $C^i[f,\tilde{g}]$ is normal along f. Taking the $\tilde{g}$ inner product with $\nu$ yields a scalar function, which we denote by $\mathrm{grad}_{L^2}\,C^i[w,\tilde{g}]$ in slight abuse of notation. We now conclude

$$\begin{aligned} \sum _{i=1}^2 \mathrm{grad}_{L^2}\,C^i[f,\tilde{g}] e_i&= \left( \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D_x (\exp _x^{\tilde{g}})^{-1} (f(\omega ))\,d\mu _g(\omega ) \right) ^{-1}|_{x = C[f,\tilde{g}]}\nonumber \\&\quad \cdot \pi _{{\mathbb R}^2} \left( (\exp _x^{\tilde{g}})^{-1}(f) H - D \left( (\exp _x^{\tilde{g}})^{-1} \right) (f) \nu \right) |_{x = C[f,\tilde{g}]}. \end{aligned}$$

(5.3)

In the Euclidean case $\tilde{g} = \delta$ we have $\exp _x v = x+v$, which yields for $i = 1,2$

$$\begin{aligned} \pi _{{\mathbb R}^2} \int _{{\mathbb S}^2_{+}} D_x (\exp _x)^{-1}(f(\omega ))\,d\mu _g(\omega )= & {} - \mu _g({\mathbb S}^2_{+})\, \mathrm{Id}_{{\mathbb R}^2},\\ \mathrm{grad}_{L^2}\,C^i[f,\delta ]= & {} \frac{1}{\mu _g({\mathbb S}^2_{+})} \left\langle \nu - (f - C[f,\delta ]) H, e_i \right\rangle . \end{aligned}$$

Specializing further to $f_0(\omega ) = \omega$, we see

$$\begin{aligned} \mathrm{grad}_{L^2}\,C^i[f_0,\delta ](\omega ) = - \frac{3}{2\pi } \langle \omega ,e_i \rangle . \end{aligned}$$

(5.4)

For $w \in C^{k,\alpha }({\mathbb S}^2_{+})$ and $\tilde{g} \in C^l(\overline{Z}_2,{\mathbb R}^{3 \times 3})$ where $l \ge k+1$, one deduces $\mathrm{grad}_{L^2} C^i[w,\tilde{g}] \in C^{k-2,\alpha }({\mathbb S}^2_{+})$. Moreover as a functional into $C^{k-4,\alpha }({\mathbb S}^2_{+})$, it is of class $C^{l-k+1}$.