An integer degree for asymptotically conical self-expanders

Abstract

We establish the existence of an integer degree for the natural projection map from the space of parameterizations of asymptotically conical self-expanders to the space of parameterizations of the asymptotic cones when this map is proper. As an application we show that there is an open set in the space of cones in ${\mathbb {R}}^3$ for which each cone in the set has a strictly unstable self-expanding annuli asymptotic to it.

Appendix A. Poincaré inequality

Proposition A.1

Let $\Sigma \in \mathcal {ACH}_{n}^{k,\alpha }$ be a self-expander. For $\beta \ge \frac{1}{4}$ and $f\in W^{1}_{\beta }(\Sigma )$,

$$\begin{aligned} \int _{\Sigma } \left( 4 n +|{\textbf{x}}|^2\right) f^2 e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n \le 16 \int _{\Sigma }|\nabla _\Sigma f|^2 e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n. \end{aligned}$$

(A.1)

Likewise

$$\begin{aligned} (4\beta -1) \int _{\Sigma }|{\textbf{x}}^\top |^2 f^2 e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n \le 4 \int _{\Sigma }|\nabla _\Sigma f|^2 e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n. \end{aligned}$$

(A.2)

Proof

First observe that as $\Sigma$ is a self-expander one has that

$$\begin{aligned} {\mathscr {L}}_{\Sigma }^0 \left( |{\textbf{x}}|^2+2n\right) =|{\textbf{x}}|^2+2n. \end{aligned}$$

Now suppose that f has compact support. Integrating by parts gives

$$\begin{aligned}&\int _\Sigma \left( 2 n +|{\textbf{x}}|^2\right) f^2 e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n = \int _\Sigma \left( {\mathscr {L}}_\Sigma ^0 \left( 2 n +|{\textbf{x}}|^2\right) \right) f^2 e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n \\&= - \int _\Sigma \nabla _\Sigma \left( 2 n +|{\textbf{x}}|^2\right) \cdot \nabla _\Sigma \left( e^{\left( \beta -\frac{1}{4}\right) |{\textbf{x}}|^2} f^2 \right) e^{\frac{|{\textbf{x}}|^2}{4}}\, d\mathcal {H}^n \\&=-\int _{\Sigma } \left( (4\beta -1) |{\textbf{x}}^\top |^2 f^2 + 4 \phi {\textbf{x}}^\top \cdot \nabla _\Sigma f \right) e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n \\&= - (4\beta -1) \int _{\Sigma } |{\textbf{x}}^\top |^2 f^2 e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n -4 \int _{\Sigma } f {\textbf{x}}^\top \cdot \nabla _\Sigma f \, e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n. \end{aligned}$$

The absorbing inequality and the hypothesis that $\beta \ge \frac{1}{4}$ give

$$\begin{aligned} \int _\Sigma \left( 2 n +|{\textbf{x}}|^2\right) f^2 e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n \le \int _{\Sigma }\left( \frac{1}{2} |{\textbf{x}}^\top |^2 f^2 + 8| \nabla _\Sigma f|^2 \right) e^{\beta |{\textbf{x}}|^2}\, d\mathcal {H}^n. \end{aligned}$$

Rearranging this gives the first inequality. The second inequality follows from the same computation but using a different version of the absorbing inequality.

To see that these estimates hold for all $f\in W^{1}_{\beta }(\Sigma )$ one uses the compactly supported cutoff $\psi _R\in C^\infty ({\mathbb {R}}^{n+1})$ with the property that $0\le \psi _R\le 1$, $\psi _R=1$ in $B_R$, ${\textrm{spt}}\,\psi _R\subset B_{2R}$ and $|\nabla \psi _R|\le 2$. The estimates hold for $f_R=\psi _R f$ and hence, by the dominated convergence theorem, hold also for f. $\square$

Proposition A.2

Let $\Sigma \in \mathcal {ACH}_{n}^{k,\alpha }$ be a self-expander. Suppose $R_0\ge 1$ is such that on $\bar{E}_{R_0}$ the estimates (2.5) hold with constant $C_0$. There is an $R_1=R_1(C)\ge R_0$ so that: For $\beta \ge \frac{1}{4}$, $f \in W^{1}_{\beta }(\Sigma )$ and $R>R_1$ one has the estimate

$$\begin{aligned} R e^{\beta R^2} \int _{S_R} f^2\, d\mathcal {H}^{n-1} \le 2 \int _{\bar{E}_R} |\nabla _\Sigma f|^2 e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n. \end{aligned}$$

Proof

Suppose first that f has compact support. Let

$$\begin{aligned} {\textbf{Z}}= \frac{|{\textbf{x}}| {\textbf{x}}^\top }{|{\textbf{x}}^\top |}=\frac{\frac{1}{2}\nabla _\Sigma |{\textbf{x}}|^2}{|\nabla _\Sigma |{\textbf{x}}||}. \end{aligned}$$

By (2.5), $|\nabla _\Sigma |{\textbf{x}}||\ge \frac{1}{2}$ so this vector field is well-defined. Using (2.5), one sees that, by taking $R_1$ sufficiently large, on $\bar{E}_{R_1}$, one has

$$\begin{aligned} |{\textbf{x}}^\top | \ge |{\textbf{x}}|-\frac{C}{|{\textbf{x}}|^3}\ge |{\textbf{x}}| -\frac{1}{|{\textbf{x}}|} \text{ and } \textrm{div}_\Sigma {\textbf{Z}} \ge n-\frac{1}{2} \ge \frac{1}{2}. \end{aligned}$$

Hence, on $\bar{E}_R$,

$$\begin{aligned} \textrm{div}_\Sigma \left( f^2 e^{\beta |{\textbf{x}}|^2} {\textbf{Z}} \right)&= \left( f^2 \textrm{div}_\Sigma {\textbf{Z}}+ 2 f \nabla _\Sigma f \cdot {\textbf{Z}} + 2\beta f^2{\textbf{x}}^\top \cdot {\textbf{Z}}\right) e^{\beta |{\textbf{x}}|^2} \\&\ge \left( \frac{1}{2} f^2+ 2 f \nabla _\Sigma f \cdot {\textbf{Z}} + \frac{1}{2} f^2 |{\textbf{Z}}|\left( |{\textbf{x}}|-|{\textbf{x}}|^{-1}\right) \right) e^{\beta |{\textbf{x}}|^2} \\&\ge \left( \frac{1}{2} f^2-2 |\nabla _\Sigma f |^2 - \frac{1}{2} f^2 |{\textbf{Z}}|^2 + \frac{1}{2} f^2 |{\textbf{Z}}| \left( |{\textbf{x}}|-|{\textbf{x}}|^{-1}\right) \right) e^{\beta |{\textbf{x}}|^2} \end{aligned}$$

where the last estimate uses the absorbing inequality. As $|{\textbf{Z}}|=|{\textbf{x}}|$ one has

$$\begin{aligned} \textrm{div}_\Sigma \left( f^2 e^{\beta |{\textbf{x}}|^2} {\textbf{Z}} \right) \ge -2 | \nabla _\Sigma f |^2 e^{\beta |{\textbf{x}}|^2}. \end{aligned}$$

Integrating over $\bar{E}_R$ and appealing to the divergence theorem give

$$\begin{aligned} R e^{\beta R^2} \int _{S_R} f^2 \, d\mathcal {H}^{n-1}&= -\int _{S_R} f^2 {\textbf{Z}}\cdot \frac{-{\textbf{x}}^\top }{|{\textbf{x}}^\top |} e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^{n-1}\\&=\int _{\bar{E}_R} -\textrm{div}_\Sigma \left( f^2 e^{\beta |{\textbf{x}}|^2} {\textbf{Z}} \right) \, d\mathcal {H}^n \le 2 \int _{\bar{E}_R} |\nabla _\Sigma f |^2 e^{\beta |{\textbf{x}}|^2} \, d\mathcal {H}^n, \end{aligned}$$

which proves the claim for f of compact support. For general $f\in W^{1}_\beta (\Sigma )$ one uses cutoffs and the dominated convergence theorem. $\square$