On volume-preserving crystalline mean curvature flow

Abstract

In this work we consider the global existence of volume-preserving crystalline curvature flow in a non-convex setting. We show that a natural geometric property, associated with reflection symmetries of the Wulff shape, is preserved with the flow. Using this geometric property, we address global existence and regularity of the flow for smooth anisotropies. For the non-smooth case we establish global existence results for the types of anisotropies known to be globally well-posed.

Appendices

Appendix A: Geometric properties

Here we show several geometric properties used in the paper. First we show that ${{\mathcal {P}}}_\phi$ given below is a root system:

$$\begin{aligned} {{\mathcal {P}}}_\phi := \{p \in {{\mathbb {S}}}^{N-1} : \phi = \phi \circ \Psi _p \hbox { and } \psi = \psi \circ \Psi _p \}. \end{aligned}$$

By definition of ${{\mathcal {P}}}_\phi$ and the fact that $\Psi _p = \Psi _{-p}$, it can be shown that

$$\begin{aligned} p \in {{\mathcal {P}}}_\phi , \quad \hbox { if and only if } -p \in {{\mathcal {P}}}_\phi \end{aligned}$$

(A.1)

Recall the reflection with respect to a hyperplane containing the origin

$$\begin{aligned} \Psi _{p} = \Psi _{\Pi _p(0)} = I - 2 p \otimes p \end{aligned}$$

is a symmetric unitary operator and an involution. Furthermore, compositions of three (or any odd number of) reflections are also reflections. From this observation, we show that if p and q are directions of reflection symmetry, then $\pm \Psi _{q}(p)$ is also a direction of reflection symmetry.

Lemma A.1

If $p, q \in {{\mathcal {P}}}_\phi$, then $\pm \Psi _{q}(p) \in {{\mathcal {P}}}_\phi$. In particular, ${{\mathcal {P}}}_\phi$ is a root system.

Proof

As $\Psi _{q}$ is an involution and symmetric, we have

$$\begin{aligned} \Psi _{q} \Psi _{p} \Psi _{q} x&= \Psi _{q} \left( \Psi _{q} x - 2 ( \Psi _{q} x \cdot p) p \right) = x - 2 (x \cdot \Psi _{q} p) \Psi _{q} p = \Psi _{\Psi _{q}p} x. \end{aligned}$$

From $| \Psi _{q}(p) | = | p | = 1$ and (A.1), we conclude that $\pm \Psi _{p}(q) \in {{\mathcal {P}}}_\phi$. $\square$

Lemma A.2

The perimeter of a set E satisfying (1.4) and ${{\mathcal {B}}}_r \subset E \subset B_R(0)$ for some $r>0$ and $R>0$ is bounded by $C = C({{\mathcal {P}}}, r, R)>0$.

Proof

Set $F:= B_R(0) \setminus {{\mathcal {B}}}_r$. There exists a finite number of points $x_i, 1 \le i \le m$ in F such that

$$\begin{aligned} F \subset \bigcup _{1 \le i \le m} B_r(x_i). \end{aligned}$$

As E is a Lipschitz domain from Theorem 2.2, it suffices to show that ${{\mathcal {H}}}^{N-1}(\partial E \cap B_r(x_i))$ is uniformly bounded for $1 \le i \le m$. Here, ${{\mathcal {H}}}^{N-1}$ is the $(N-1)$-dimensional Hausdorff measure. Either $\partial E \cap B_r(x_i)$ is empty or it can be represented by a Lipschitz graph. In particular, from the cone condition in Theorem 2.2, the Lipschitz constant only depends on r and ${{\mathcal {P}}}$ and thus we conclude. $\square$

Next, let us recall the uniform density from [27, Definition 4], which is, roughly speaking, a quantitative definition for Ahlfors regular sets. Let $c \in (0,1)$ and $s_0 > 0$. We say that $\Omega \subset {{\mathbb {R}}^N}$ has $(s_0, c)$-uniform lower density if the estimate

$$\begin{aligned} 0 < c \le \frac{|B_s(x) \cap \Omega |}{|B_s(x)|} \end{aligned}$$

holds for all $s \in (0, s_0)$ and $x \in \partial \Omega$. Similarly, $\Omega$ is said to have $(s_0, c)$-uniform upper density if

$$\begin{aligned} \frac{|B_s(x) \cap \Omega |}{|B_s(x)|} \le 1 - c <1. \end{aligned}$$

When both conditions are satisfied together, $\Omega$ has $(s_0, c)$-uniform density.

Lemma A.3

[27, Theorem 4]. Let $\Omega \subset {{\mathbb {R}}^N}$ have $(s_0, c)$-uniform density. Then

$$\begin{aligned} | \{ x \in {{\mathbb {R}}^N}: 0< d(x, \Omega ) < s \} | \le C \left( 1 + \tfrac{1}{c} \right) ^{\frac{N-1}{N}} {\text {Per}}(\Omega ) s \hbox { for all } s \in (0, s_0). \end{aligned}$$

Here, C is a dimensional constant.

As a consequence of Theorem 2.2 and the above lemma, we conclude.

Proposition A.4

Suppose that E satisfies (1.4) and ${{\mathcal {B}}}_r \subset E \subset B_R(0)$ for some $r>0$ and $R>0$. Then, there exists $c = c({{\mathcal {P}}}, r, R)>0$ such that

$$\begin{aligned} | \{ x \in {{\mathbb {R}}^N}: 0< d(x, E) < s \} | \le c s \end{aligned}$$

(A.2)

for all $s \in (0,r)$. Here, ${{\mathcal {B}}}_r$ is given in (2.7)

Proof

We claim that E has $(r, \sigma _3^N)$-uniform density for $\sigma _3$ given in (2.6). For all $s \in (0,r)$, E has an s-interior cone and an s-exterior cone from Theorem 2.2. As s-interior and exterior cones are contained in a ball of radius s and contains a ball of radius $\sigma _3 s$ for $\sigma _3$ given in (2.6), we conclude that

$$\begin{aligned} \sigma _3^N \le \frac{|B_s(x) \cap E|}{|B_s(x)|} \le 1 - \sigma _3^N. \end{aligned}$$

Then, we apply Lemma A.3 and Lemma A.2 to conclude (A.2). $\square$

Lemma A.5

Suppose that a bounded set E satisfies (1.4) and ${{\mathcal {B}}}_r \subset E$ for some $r>0$. Then, there exist $C = C({{\mathcal {P}}})>0$ and $\varepsilon _0 = \varepsilon _0({{\mathcal {P}}}, r)>0$ such that

$$\begin{aligned} d(x, {{\mathcal {I}}}^- ) \le C \varepsilon \hbox { and } d(x, ({{\mathcal {I}}}^+)^{\textsf {c}}) \le C \varepsilon \quad \hbox { for all } x \in \partial E \hbox { and for all } \varepsilon \in (0, \varepsilon _0) \end{aligned}$$

(A.3)

where ${{\mathcal {I}}}^\pm := \{ x \in {{\mathbb {R}}^N}: {\text {sd}}(x, E) < \pm \varepsilon \}$.

Proof

Set $\varepsilon _0:= \sigma _3 r/2$ and fix $x \in \partial E$. Theorem 2.2 yields that E contains $x-{{\,\textrm{Cone}\,}}_r(A)$ given by some basis $A \subset {{\mathcal {P}}}$. This cone contains a ball of radius $s\sigma _3$ centered at $x_s := x - \frac{s}{2N} \sum _{p\in A} p$ for any $0 < s \le r$. For $\varepsilon \in (0, \varepsilon _0)$, choose $s := 2\varepsilon /\sigma _3$. As $B_{2\varepsilon }(x_s)$ is contained in E, we have $d(x_s, \partial E) > \varepsilon$ and thus $x_s \in {{\mathcal {I}}}^-$. This yields that

$$\begin{aligned} d(x, {{\mathcal {I}}}^-) \le |x - x_s| \le \frac{s}{2} = \frac{\varepsilon }{\sigma _3}. \end{aligned}$$

(A.4)

The second inequality can be handled similarly. $\square$

A function $f : {{\mathbb {R}}^N}\rightarrow {{\mathbb {R}}}$ is called positively one-homogeneous if

$$\begin{aligned} f(s\xi ) = s f(\xi ) \qquad \text {for all } \xi \in {{\mathbb {R}}}^N \text { and } s \ge 0. \end{aligned}$$

(A.5)

Recall the definition of the Wulff shape $W_f$ in (1.1).

Lemma A.6

For positively one-homogeneous functions $f, g : {{\mathbb {R}}^N}\rightarrow {{\mathbb {R}}}$ with $f \le g$ we have

$$\begin{aligned} {{\overline{B}}}_{m_f}(0) \subset W_f \subset W_g \subset {{\overline{B}}}_{M_g}(0), \end{aligned}$$

Here, $m_f$, $M_g$ are given in (5.6).

Proof

$W_f \subset W_g$ is clear from the definition. For the ordering with ${{\overline{B}}}_{m_f}(0)$ and ${{\overline{B}}}_{M_g}(0)$, we note that ${{\overline{B}}}_r(0) = W_h$ for $h(p) := r|p|$ for any $r \ge 0$. $\square$

Appendix B: Technical lemmas

Lemma B.1

Suppose that $u_k: {{\mathbb {R}}^N}\times {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ is a locally-bounded sequence of upper semi-continuous functions and let $u := \mathop {\lim \,\sup {}^*}\limits \nolimits _{k\rightarrow \infty } u_k$. Let $r_k \in C({{\mathbb {R}}})$ be a sequence of non-negative continuous functions such that $r_k \rightarrow r$ locally uniformly and let $K \subset {{\mathbb {R}}^N}$ be a nonempty compact set. Then

$$\begin{aligned} \sup _{x + r(t) K} u(\cdot , t) = \mathop {\lim \,\sup }\limits _{\begin{array}{c} k\rightarrow \infty \\ (x_k,t_k) \rightarrow (x,t) \end{array}} \sup _{x_k + r_k(t_k)K} u_k(\cdot , t_k). \end{aligned}$$

(B.1)

Proof

Fix (x, t) and suppose $(x_k, t_k) \rightarrow (x,t)$.

Choose $y \in x + r(t) K$ such that $u(y, t) = \sup _{x + r(t) K} u(\cdot , t)$ and $y_k \in x_k + r_k(t_k) K$ such that $u_k(y_k, t_k) = \sup _{x_k + r_k(t_k) K} u(\cdot , t_k)$.

Since $\{y_k\}$ is clearly bounded, we can consider a subsequence indexed by $k_m$ so that $y_{k_m} \rightarrow z$ for some z and $\lim _m u_{k_m}(y_{k_m}, t_{k_m}) = \mathop {\lim \,\sup }\limits _k u_k(y_k, t_k)$. Suppose that $r(t) > 0$. We have $r_{k_m}(t_{k_m}) > 0$ for large m and therefore

$$\begin{aligned} \frac{r(t)}{r_{k_m}(t_{k_m})}(y_{k_m} - x_{k_m}) \in r(t) K. \end{aligned}$$

Since the left-hand side converges to $z - x$, we have $z - x\in r(t) K$. On the other hand, if $r(t) = 0$ then $r_{k_m}(t_{k_m}) \rightarrow 0$ and hence $z - x = \lim _m (y_{k_m}- x_{k_m}) = 0 \in r(t) K$. This implies that $u(z, t) \le u(y, t)$. Since $u = \mathop {\lim \,\sup {}^*}\limits u_k$, we conclude that

$$\begin{aligned} \mathop {\lim \,\sup }\limits _k u_k(y_k, t_k) = \lim _m u_{k_m}(y_{k_m}, t_{k_m}) \le u(z, t) \le u(y, t). \end{aligned}$$

As the sequence $\{(x_k, t_k)\}$ was arbitrary, we conclude that the inequality $\ge$ holds in (B.1).

To show the equality, we consider a maximizing sequence, i.e., we choose $(y_k, t_k) \rightarrow (y, t)$ such that $u(y,t) = \mathop {\lim \,\sup }\limits _k u_k(y_k, t_k)$. If $r(t) > 0$, we set $x_k = y_k - \frac{r_k(t_k)}{r(t)} (y - x)$. Clearly $y_k \in x_k + r_k(t_k) K$ and $x_k \rightarrow x$. If $r(t) = 0$, it is sufficient to take any $x_k \in y_k - r_k(t_k) K$ and we still have $x_k \rightarrow x = y$. We then have

$$\begin{aligned} \mathop {\lim \,\sup }\limits _k \sup _{x_k +r_k(t_k)K} u_k(\cdot , t_k) \ge \mathop {\lim \,\sup }\limits u_k(y_k, t_k) = u(y, t) = \sup _{x + r(t) K} u(\cdot , t). \end{aligned}$$

This finishes the proof. $\square$

Lemma B.2

Suppose that $\psi , \phi : {{\mathbb {R}}^N}\rightarrow [0, \infty )$ are positively one-homogeneous convex functions, with zero only at $p=0$ and suppose that $\phi \in C^2({{\mathbb {R}}^N}{\setminus } \{0\})$. Suppose that u is a viscosity subsolution of $u_t = \psi (-Du) (-{\text {div}}D\phi (-Du) + \lambda )$ for some $\lambda \in C({{\mathbb {R}}})$. Then for any positive $R \in C^1({{\mathbb {R}}})$, ${{\widehat{u}}}(\cdot ; R)$ from (3.3) is a viscosity subsolution of $u_t = \psi (-Du) (-{\text {div}}D\phi (-Du) + \lambda + R')$.

Proof

Without loss of generality we may assume that u is upper semi-continuous. To simplify the notation we write ${{\widehat{u}}}(x, t)$ instead of ${{\widehat{u}}}(x, t; R)$. Let $\varphi$ be a smooth test function such that ${{\widehat{u}}}- \varphi$ has a maximum 0 at $({\hat{x}}, {\hat{t}})$. Recall that we need to show $\varphi _t \le F^*({\hat{t}}, D\varphi , D^2\varphi )$ at $({\hat{x}}, {\hat{t}})$, where $F(t, p, X)\,{:=}\,\psi (-p)\left( \!{{\,\textrm{trace}\,}}[D_p^2\phi (-p) X]\,{+}\,\lambda \,{+}\,R'\!\right)$, $p \ne 0$.

Due to the assumption we have

$$\begin{aligned} \varphi (x,t)\ge {{\widehat{u}}}(x,t)= \max _{x - R(t)W_\psi } u(\cdot , t) \end{aligned}$$

with equality at $({\hat{x}}, {\hat{t}})$. We now fix ${\hat{y}} \in {\hat{x}} - R({\hat{t}}) W_\psi$ such that $u({\hat{y}}, {\hat{t}}) = {{\widehat{u}}}({\hat{x}}, {\hat{t}})$. Note that from the definition $R(t) \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})} \in R(t) W_\psi$ and so $x - R(t) \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})} \in x - R(t) W_\psi$, which yields

$$\begin{aligned} \varphi (x,t) \ge {{\widehat{u}}}(x,t) \ge u(x - R(t) \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})}, t) \end{aligned}$$

for all x, t with equality at $({\hat{x}}, {\hat{t}})$. Thus we deduce

$$\begin{aligned} {\hat{\varphi }}(x, t) := \varphi (x + R(t) \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})},t) \ge u(x, t) \end{aligned}$$

for all x, t with equality at $({\hat{y}}, {\hat{t}})$. In particular, $u - {\hat{\varphi }}$ has a local maximum at $({\hat{y}}, {\hat{t}})$.

Now a direct computation yields $D{\hat{\varphi }}({\hat{y}}, {\hat{t}}) = D\varphi ({\hat{x}}, {\hat{t}})$, $D^2 {\hat{\varphi }}({\hat{y}}, {\hat{t}}) = D^2 \varphi ({\hat{x}}, {\hat{t}})$ and

$$\begin{aligned} {\hat{\varphi }}_t ({\hat{y}}, {\hat{t}})= \varphi _t ({\hat{x}}, {\hat{t}})+ D\varphi ({\hat{x}}, {\hat{t}})\cdot \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})} R'({\hat{t}}), \end{aligned}$$

(B.2)

If $D\varphi ({\hat{x}}, {\hat{t}} ) = 0$ this simply yields $\hat{\varphi }_t({\hat{y}}, {\hat{t}})= \varphi _t({\hat{x}}, {\hat{t}})$ and we conclude that the correct viscosity solution condition is satisfied for $\varphi$ since u is a viscosity solution with right-hand side $F(t, p, X) - \psi (-p) R'(t)$ by assumption.

Now suppose that $D\varphi ({\hat{x}}, {\hat{t}} ) \ne 0$. As

$$\begin{aligned} \varphi (x, {\hat{t}}) \ge {{\widehat{u}}}(x, {\hat{t}}) \ge u({\hat{y}}, {\hat{t}}) = \varphi ({\hat{x}}, {\hat{t}}) \qquad \text {for }x \in {\hat{y}} + R({\hat{t}}) W_\psi , \end{aligned}$$

we deduce that ${\hat{y}} + R({\hat{t}})W_{\psi } \subset \{\varphi (\cdot , {\hat{t}}) \ge \varphi ({\hat{x}}, {\hat{t}})\}$. In particular, $-D\varphi ({\hat{x}}, {\hat{t}})$ is an outer normal to $W_\psi$ at $({\hat{x}} - {\hat{y}})/ R({\hat{t}})$. Therefore, by definition of $W_\psi = \{x : x\cdot p \le \psi (p) \ \forall p\}$ and the fact that $\psi$ is positively one-homogeneous and convex,

$$\begin{aligned} -D\varphi ({\hat{x}}, {\hat{t}}) \cdot \tfrac{{\hat{x}} - {\hat{y}}}{R({\hat{t}})} = \psi (-D\varphi ({\hat{x}}, {\hat{t}})). \end{aligned}$$

which yields together with (B.2)

$$\begin{aligned} {\hat{\varphi }}_t ({\hat{y}}, {\hat{t}})= \varphi _t ({\hat{x}}, {\hat{t}})- \psi (-D\varphi ({\hat{x}}, {\hat{t}})) R'({\hat{t}}), \end{aligned}$$

again yielding the correct viscosity condition for $\varphi$ at $({\hat{x}}, {\hat{t}})$ from the viscosity solution condition that $\hat{\varphi }$ satisfies at $({\hat{y}}, {\hat{t}})$.

We conclude that ${{\widehat{u}}}$ is a viscosity subsolution of $u_t = \psi (-Du) (-{\text {div}}D\phi (-Du) + \lambda + R')$. $\square$