A convergence rate of periodic homogenization for forced mean curvature flow of graphs in the laminar setting

Abstract

In this paper, we obtain the rate $O(\varepsilon ^{1/2})$ of convergence in periodic homogenization of forced graphical mean curvature flows in the laminated setting. We also discuss with an example that a faster rate cannot be obtained by utilizing Lipscthiz estimates.

Proof of Theorem 2.1

We prove Theorem 2.1 in this appendix. We will separate the steps into Propositions 1, 2, 3, whose statements are about the estimates of gradients, Hessians and time derivatives.

We state the short-time existence of classical solutions to (2.1). We skip the proof as known in the literature. The uniqueness follows from the standard comparison principle, for which we refer to [8]. See [1] for more general results in this direction. For the existence with gradient and Hessian estimates, we refer to [10] (in the absence of a forcing) and to [9, Appendix A] (when with a $C^2$ forcing term).

Proposition 1

Let $w_0$ be a globally Lipschitz function on $\mathbb {R}^n$ with $\Vert Dw_0\Vert _{L^{\infty }(\mathbb {R}^n)}$ $\leqslant N_0<+\infty$. Then, there exists $T^*=T^*(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0)>0$ such that (2.1) with $T=T^*$ has a unique classical solution $w=w(x,t)$. Moreover, for each $T\in (0,T^*)$, there exist $N=N(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,T)>0$ and $C=C(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,T)$ $>0$ such that

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [0,T])}\leqslant N, \end{aligned}$$

and

$$\begin{aligned} \Vert D^2w(\cdot ,t)\Vert _{L^{\infty }(\mathbb {R}^n)}\leqslant \frac{C}{\sqrt{t}} \end{aligned}$$

for $t\in (0,T]$.

Next, we state and prove a priori time derivative estimates based on the maximum principle. See [15, Lemma 3.1].

Proposition 2

Let $w_0$ be a globally Lipschitz function on $\mathbb {R}^n$ with $\Vert Dw_0\Vert _{L^{\infty }(\mathbb {R}^n)}$ $\leqslant N_0<+\infty$. Let $T^*=T^*(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0)>0$ be chosen such that (2.1) with $T=T^*$ has the unique classical solution $w=w(x,t)$. Then, for any $\tau \in (0,T^*)$, it holds that

$$\begin{aligned} \Vert w_t\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau ,T^*))} \leqslant \Vert w_t(\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)}<+\infty . \end{aligned}$$

Proof of Proposition 2

Let $T\in (\tau ,T^*)$. Then, by Proposition 1, there exist $N=N(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,T)>0$ and $C=C(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,T)>0$ such that

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [0,T])}\leqslant N, \end{aligned}$$

and

$$\begin{aligned} \Vert D^2w(\cdot ,t)\Vert _{L^{\infty }(\mathbb {R}^n)}\leqslant \frac{C}{\sqrt{t}} \end{aligned}$$

for $t\in (\tau ,T]$. Therefore, by the equation (2.1), we see that $\Vert w_t\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau ,T])}\leqslant K=K(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\tau ,T)$.

We aim to prove

$$\begin{aligned} \sup _{\mathbb {R}^n\times [\tau ,T]}w_t \leqslant \sup _{\mathbb {R}^n}w_t(\cdot ,\tau ). \end{aligned}$$

Suppose for the contrary that there exists $(x_0,t_0)\in \mathbb {R}^n\times (\tau ,T]$ such that

$$\begin{aligned} w_t(x_0,t_0)>\sup _{\mathbb {R}^n}w_t(\cdot ,\tau ). \end{aligned}$$

Then there would exist a number $\lambda \in (0,1)$ such that

$$\begin{aligned} w_t(x_0,t_0)-\lambda t_0>\sup _{\mathbb {R}^n}w_t (\cdot ,\tau )-\lambda \tau . \end{aligned}$$

We run the Bernstein method now with $\Phi (x,t):=w_t(x,t)-\lambda t$. Let $\Phi ^*(t):=\sup _{\mathbb {R}^n}\Phi (\cdot ,t)$ for each $t\in [\tau ,T]$. Then $\Phi ^*(t_0)>\Phi ^*(\tau )$. Fix a sequence $\{\varepsilon _j\}_j$ of positive numbers that converges to 0 as $j\rightarrow \infty$. For each $t\in [\tau ,T]$, let $x_j(t)$ be a maximizer of $\Phi _j(x,t)=\Phi (x,t)-\varepsilon _j|x|^2$. Then, $\Phi (x_j(t),t)\rightarrow \Phi ^*(t),\ D\Phi (x_j(t),t)\rightarrow 0$ as $j\rightarrow \infty$, and $\limsup _{j\rightarrow \infty }D^2\Phi (x_j(t),t)\leqslant 0$ in the sense that $\limsup _{j\rightarrow \infty }(D^2\Phi (x_j(t),t)v)\cdot v\leqslant 0$ for any $v\in \mathbb {R}^n$.

Note that $\{t\in [\tau ,T]:\Phi ^*(t)=\sup _{[\tau ,T]}\Phi ^*(\cdot )\}$ is a closed subinterval of $[\tau ,T]$ not containing $\tau$. Consequently, there exists $t^*\in (\tau ,T]$ such that $\Phi ^*(t^*)=\sup _{[\tau ,T]}\Phi ^*(\cdot ),$ $\Phi ^*(t)<\Phi ^*(t^*)$ for all $t\in [\tau ,t^*)$, and thus that $\liminf _{j\rightarrow \infty }\Phi _t(x_j(t^*),t^*)\geqslant 0$.

Differentiating the first line of (2.1) in t, we obtain

$$\begin{aligned} (w_t)_t-\text {tr}\{a(Dw)D^2w_t\}=\text {tr}\{a(Dw)_tD^2w\} +c\frac{Dw\cdot Dw_t}{\sqrt{1+|Dw|^2}}. \end{aligned}$$

Also, at $(x,t)\in \mathbb {R}^n\times [\tau ,T]$,

$$\begin{aligned} \text {tr}\{a(Dw)_tD^2w\}&=\text {tr}\{(D_pa(Dw)\odot Dw_t)D^2w\}\\&\leqslant \frac{4n^3}{\sqrt{1+|Dw|^2}}|Dw_t|\Vert D^2w\Vert \leqslant C(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\tau )|Dw_t| \end{aligned}$$

for some constant $C=C(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\tau )>0$ depending only on its argument. Here, we have used the fact that $\left| \frac{\partial }{\partial p_k}a^{ij}(p)\right| \leqslant \frac{4}{\sqrt{1+|p|^2}}$ for all $p\in \mathbb {R}^n$. Also,

$$\begin{aligned} c\frac{Dw\cdot Dw_t}{\sqrt{1+|Dw|^2}}\leqslant \Vert c\Vert _{L^{\infty } (\mathbb {R}^n)}|Dw_t|. \end{aligned}$$

Therefore, evaluated at $(x_j(t^*),t^*)$ in the following limit,

$$\begin{aligned} 0&\leqslant \liminf _{j\rightarrow \infty }\left( \Phi _t-\text {tr}\left\{ a(Dw) D^2\Phi \right\} \right) \\&\leqslant \liminf _{j\rightarrow \infty }\left( -\lambda +(w_t)_t-\text {tr} \{a(Dw)D^2w_t\}\right) \\&\leqslant -\lambda +\liminf _{j\rightarrow \infty }C(n,\Vert c\Vert _{C^2(\mathbb {R}^n)}, N_0,\tau )|Dw_t|\\&=-\lambda , \end{aligned}$$

a contradiction.

The statement $\inf _{\mathbb {R}^n\times [\tau ,T]}w_t \geqslant \inf _{\mathbb {R}^n}w_t(\cdot ,\tau )$ can be verified similarly, and $T\in (\tau ,T^*)$ can be chosen arbitrarily. Therefore, we complete the proof.

$\square$

We state and prove a priori gradient estimates. The point of the following proposition is to remove the dependency on $T\in (0,T^*)$ in the estimate of Proposition 1. We refer to [15, 17] regarding gradient estimates from the coercivity condition (A3).

Proposition 3

Let $w_0$ be a globally Lipschitz function on $\mathbb {R}^n$ with $\Vert Dw_0\Vert _{L^{\infty }(\mathbb {R}^n)}$ $\leqslant N_0<+\infty$. Let $T^*=T^*(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0)>0$ be chosen such that (2.1) with $T=T^*$ has the unique classical solution $w=w(x,t)$. Then, for any $\tau \in (0,T^*)$, there exists $M=M(n,\Vert c\Vert _{L^{\infty }(\mathbb {R}^n)},\Vert w_t(\cdot ,\tau ) \Vert _{L^{\infty }(\mathbb {R}^n)},\delta )>0$ such that

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau ,T^*))} \leqslant \max \{\Vert Dw(\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)},M\}. \end{aligned}$$

Here, $\delta >0$ is the number appearing in the condition (A3).

Proof of Proposition 3

Let $T\in (\tau ,T^*)$. By Proposition 1, there exists $N=N(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,T)>0$ such that

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau ,T])}\leqslant N. \end{aligned}$$

(A.1)

The goal of this proof is to make this estimate independent of $T\in (\tau ,T^*)$.

We now run the Bernstein method with $\Phi (x,t):=z(x,t)$. Let $\Phi ^*(t):=\sup _{\mathbb {R}^n}w(\cdot ,t)$ for $t\in [\tau ,T]$. Let $\{\varepsilon _j\}_j$ be a sequence of positive numbers that converges to 0 as $j\rightarrow \infty$. For each $t\in [\tau ,T]$, a maximizer $\{x_j(t)\}_j$ of $\Phi _j(x,t):=\Phi (x,t)-\varepsilon _j|x|^2$ satisfies that $\Phi (x_j(t),t)\rightarrow \Phi ^*(t),\ D\Phi (x_j(t),t)\rightarrow 0$ as $j\rightarrow \infty$, and that $\limsup _{j\rightarrow \infty }D^2\Phi (x_j(t),t)\leqslant 0$. Here, we are using the estimate (A.1).

If $\{t\in [\tau ,T]:\Phi ^*(t)=\sup _{[\tau ,T]}\Phi ^*(\cdot )\}$ contains $\tau$, we obtain the conclusion. We assume the other case so that there exists $t_1\in (\tau ,T]$ such that $\Phi ^*(t)<\Phi ^*(t_1)=\sup _{[\tau ,T]}\Phi ^*(\cdot )$ for all $t\in [\tau ,t_1)$. Then, it holds that $\liminf _{j\rightarrow \infty }\Phi _t(x_j(t_1),t_1)\geqslant 0$.

We differentiate the first line of (2.1) in $x_k$ and multiply by $w_{x_k}$ and then sum over $k=1,\cdots ,n.$ We get, as a result,

$$\begin{aligned} zz_t-z\text {tr}\{a(Dw)D^2z\}&=z\text {tr}\{(D_pa(Dw)\odot Dz)D^2w\}-\text {tr}\{(a(Dw)D^2w)^2\}\nonumber \\&\quad +zDc\cdot Dw+cDw\cdot Dz. \end{aligned}$$

(A.2)

We estimate the term $\text {tr}\{(a(Dw)D^2w)^2\}$. Using the fact that $Dz=z^{-1}D^2wDw$, we see that

$$\begin{aligned}&\text {tr}\{(a(Dw)D^2w)^2\}\nonumber \\&\quad =\text {tr}\{a(Dw)D^2wI_nD^2w\}-\text {tr}\{a(Dw) D^2w\frac{Dw}{z}\otimes \frac{Dw}{z}D^2w\}\nonumber \\&\quad =\text {tr}\{a(Dw)(D^2w)^2\}-\text {tr}\{a(Dw)Dz\otimes Dz\}. \end{aligned}$$

(A.3)

Recall the Cauchy-Schwarz’s inequality; for two square matrices $\alpha ,\beta$ of the same size, we have

$$\begin{aligned} \text {tr}\{\alpha \beta ^t\}^2\leqslant \Vert \alpha \Vert ^2\Vert \beta \Vert ^2 \end{aligned}$$

Assume $n\geqslant 2$. We put $\alpha =(a(Dw))^{1/2}D^2w,\ \beta =(a(Dw))^{1/2}$ to obtain

$$\begin{aligned}&\text {tr}\{a(Dw)(D^2w)^2\}\nonumber \\&\quad \geqslant \frac{1}{n-1+\frac{1}{z^2}} \text {tr}\{a(Dw)D^2w\}^2\nonumber \\&\quad \geqslant \frac{1}{n-1}(w_t-cz)^2 -\frac{1}{(n-1)^2z^2}(w_t-cz)^2\nonumber \\&\quad \geqslant \frac{c^2}{n-1}z^2-\frac{2\Vert w_t (\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)} \Vert c\Vert _{L^{\infty }(\mathbb {R}^n)}}{n-1}z-C \end{aligned}$$

(A.4)

for some constant $C=C(n,\Vert c\Vert _{L^{\infty }(\mathbb {R}^n)}, \Vert w_t(\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)})>0$. Here, we have used Proposition 2.

Note that

$$\begin{aligned} z\text {tr}\{(D_pa(Dw)\odot Dz)D^2w\} \leqslant 4n^3|Dz|\Vert D^2w\Vert _{L^{\infty }(\mathbb {R}^n \times [\tau ,T])}\leqslant C|Dz| \end{aligned}$$

for some constant $C=C(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\tau ,T)>0$ from the Hessian estimate in Proposition 1. We also have used the fact that $\left| \frac{\partial }{\partial p^k}a^{ij}(p)\right| \leqslant \frac{4}{\sqrt{1+|p|^2}}$ for $p\in \mathbb {R}^n$.

From (A.2), (A.3), (A.4), we have

$$\begin{aligned}&zz_t-z\text {tr}\{a(Dw)D^2w\}\\&\quad \leqslant C(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\tau ,T)|Dz| -\left( \frac{c^2}{n-1}-|Dc|\right) z^2\\&\qquad +\frac{2\Vert w_t(\cdot ,\tau )\Vert _{L^{\infty } (\mathbb {R}^n)}\Vert c\Vert _{L^{\infty }(\mathbb {R}^n)}}{n-1}z +|Dz|^2+\Vert c\Vert _{L^{\infty }(\mathbb {R}^n)}|Dz|z+C \end{aligned}$$

for some constant $C=C(n,\Vert c\Vert _{L^{\infty } (\mathbb {R}^n)},\Vert w_t(\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)})>0$. Evaluate at $(x_j(t_1),t_1)$ and let $j\rightarrow \infty$ to obtain

$$\begin{aligned} 0\leqslant -\delta \Phi ^*(t_1)^2+C\Phi ^*(t_1)+C \end{aligned}$$

for some constant $C=C(n,\Vert c\Vert _{L^{\infty }(\mathbb {R}^n)}, \Vert w_t(\cdot ,\tau )\Vert _{L^{\infty }(\mathbb {R}^n)})>0$ (taking a larger one than the previous lines if necessary) depending only on its arguments. This completes the proof when $n\geqslant 2$. In the case of $n=1$, the estimate can be carried out similarly. $\square$

Combining Propositions 1, 2, 3, we obtain the long-time existence, proved in the following.

Proof of Theorem 2.1

The uniqueness is standard [1, 8]. We prove the existence of classical solutions for all time.

Let $T^*=T^*(\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0)>0$ be chosen as in Proposition 1. Fix $\tau _0=\frac{1}{2}T^*$. Tracking the dependency on parameters using Propositions 1, 2, 3, we see that there exists $M=M(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\delta )>0$ such that

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau _0,T^*))}\leqslant M. \end{aligned}$$

Let $\tau _1=T^*-\varepsilon \in (\tau _0,T^*)$. Starting from $w(\cdot ,\tau _1)$ at $t=\tau _1$, seen as an initial data, we can extend the solution on time interval $[\tau _1,\tau _1+\frac{1}{(C_1+1)\sqrt{1+M^2}})$ (see the proof of Proposition 1 for this explicit expression). As $\varepsilon >0$ can be arbitrarily small, the solution exists on time interval $[0,T_1^*)$ with $T_1^*=T^*+\frac{1}{(C_1+1)\sqrt{1+M^2}}$. Not changing the choice $\tau _0=\frac{1}{2}T^*$, we still have

$$\begin{aligned} \Vert Dw\Vert _{L^{\infty }(\mathbb {R}^n\times [\tau _0,T_1^*))}\leqslant M. \end{aligned}$$

with the same constant $M=M(n,\Vert c\Vert _{C^2(\mathbb {R}^n)},N_0,\delta )>0$ by applying the proofs of Propositions 2, 3. Then, we can extend the solution on time interval $[0,T_2^*)$ with $T_2^*=T^*+\frac{2}{(C_1+1)\sqrt{1+M^2}}$ as we just did from $[0,T^*)$ to $[0,T_1^*)$. We inductively proceed to conclude the solution exists for all time.

The estimate (2.2) is a simple consequence of Propositions 1, 2. $\square$