Homogenization of the Allen–Cahn equation with periodic mobility

Abstract

We analyze the sharp interface limit for the Allen–Cahn equation with an anisotropic, spatially periodic mobility coefficient and prove that the large-scale behavior of interfaces is determined by mean curvature flow with an effective mobility. Formally, the result follows from the asymptotics developed by Barles and Souganidis for bistable reaction–diffusion equations with periodic coefficients. However, we show that the corresponding cell problem is actually ill-posed when the normal direction is rational. To circumvent this issue, a number of new ideas are needed, both in the construction of mesoscopic sub- and supersolutions controlling the large-scale behavior of interfaces and in the proof that the interfaces obtained in the limit are actually described by the effective equation.

Appendices

Appendix A: Initialization

We are interested in proving an “initialization” type result for the phase field equation

$$\begin{aligned} m(x,{\widehat{Du}}) u_{t} - \Delta u + W'(u) = 0 \quad \text {in} \, \, {\mathbb {R}}^{d} \times (0,\infty ). \end{aligned}$$

Defining the modified forcing ${\bar{f}} : [-1,1] \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} {\bar{f}}(u) = \left\{ \begin{array}{ll} \Theta ^{-1} W'(u), &{}\quad \text {if} \, \, u \in [0,1], \\ \theta ^{-1} W'(u), &{}\quad \text {if} \, \, u \in [-1,0], \end{array} \right. \end{aligned}$$

we will construct a “universal subsolution” ${\tilde{\chi }}^{\epsilon }$ of the ODE

$$\begin{aligned} {\tilde{\chi }}^{\epsilon }_{t} \le -{\bar{f}}({\tilde{\chi }}^{\epsilon }) \quad \text {in} \, \, [-1,1] \end{aligned}$$

with the same properties as the specific subsolution used in [9,  Lemma 4.1]. As shown in Proposition 7 above, the choice of ${\bar{f}}$ enables us to build certain subsolutions of (1) that are used to analyze the development of sharp interfaces as $\epsilon \rightarrow 0^{+}$.

In this section, we build ${\tilde{\chi }}^{\epsilon }$ by proceeding by analogy with Chen’s paper [13,  Section 3]. The main result is Lemma 12 below.

1.1 Preliminaries

The assumptions on W imply that we can fix a $\mu \in (0,\frac{1}{8})$ such that

$$\begin{aligned} \frac{W''(-1)}{2} \le&W''(u) \le \frac{3 W''(-1)}{2} \quad \text {if} \, \, u \in [-1 - 2 \mu , -1 + 2 \mu ], \nonumber \\ \frac{W''(1)}{2} \le&W''(u) \le \frac{3 W''(1)}{2} \quad \text {if} \, \, u \in [1 - 2 \mu , 1], \nonumber \\ \frac{3 W''(0)}{2} \le&W''(u) \le \frac{W''(0)}{2} \quad \text {if} \, \, u \in [-\mu , \mu ]. \end{aligned}$$

(50)

From this, we deduce that

$$\begin{aligned} \theta ^{-1}W'(u) \le&W'(u) \le \Theta ^{-1} W'(u) \quad \text {if} \, \, u \in [0,1] \cup [-1 - 2 \mu ,-1], \\ \Theta ^{-1} W'(u) \le&W'(u) \le \theta ^{-1} W'(u) \quad \text {if} \, \, u \in [-1,0] \cup [1, 1 + 2\mu ]. \end{aligned}$$

1.2 Regularization

Fix $\epsilon \in (0,1)$. To start with, let ${\bar{f}} : [-3, 3] \rightarrow {\mathbb {R}}$ be the function defined by

$$\begin{aligned} {\bar{f}}(u) = \left\{ \begin{array}{ll} \Theta ^{-1} W'(u), &{} \quad \text {if} \, \, u \in [0,3] \cup [-3,-1], \\ \theta ^{-1} W'(u), &{}\quad \text {if} \, \, u \in [-1,0]. \end{array} \right. \end{aligned}$$

(51)

Notice that ${\bar{f}}$ is Lipschitz continuous.

Next, we make some adjustments. To start with, let $\rho : [-1,1] \rightarrow [0,\infty )$ be a smooth function with $\rho (s) = \rho (-s)$, $\rho (1) = \rho (-1) = 0$, and $\int _{-1}^{1} \rho (s) \, ds = 1$, and, given $\epsilon \in (0,1)$, define $\rho ^{\epsilon }(s) = \epsilon ^{-1} \rho (\epsilon ^{-1} s)$. Define ${\bar{f}}_{\epsilon }$ by ${\bar{f}}_{\epsilon } = \rho ^{\epsilon } * {\bar{f}} + 2 \text {Lip}({\bar{f}}; [-3,3]) \epsilon$. Notice that, by construction, for each $u \in [-2,2]$, we have

$$\begin{aligned} {\bar{f}}_{\epsilon }(u) - {\bar{f}}(u)&\ge \text {Lip}({\bar{f}};[-3,3]) \epsilon . \end{aligned}$$

In particular, ${\bar{f}}_{\epsilon } \ge {\bar{f}}$ in $[-2,2]$.

Finally, fix a cut-off function $\eta \in C^{\infty }({\mathbb {R}}; [0,1])$ such that

$$\begin{aligned} \eta (u) = 1 \, \, \text {if} \, \, u \in \left[ -\frac{1}{4},\infty \right) ,&\quad \eta (u) = 0 \, \, \text {if} \, \, u \in \left( -\infty ,-\frac{3}{4}\right] , \\ |\eta '(u)|&\le 4, \end{aligned}$$

and define $f_{\epsilon } : {\mathbb {R}} \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} f_{\epsilon }(u) = \eta (u) {\bar{f}}(u) + (1 - \eta (u)) {\bar{f}}_{\epsilon }(u). \end{aligned}$$

Some properties of $f_{\epsilon }$ are summarized next:

Lemma 10

There is a constant $M_{1} > 0$ and an $\epsilon _{0} > 0$ with $\epsilon _{0} < \frac{1}{2}$ such that if $\epsilon \in (0,\epsilon _{0})$, then the following statements hold:

• (i) $\frac{W''(-1)}{2 \Theta } \le f_{\epsilon }'(u) \le \frac{3W''(-1)}{2 \theta }$ if $u \in [-1 - \mu , -1 + \mu ]$.

• (ii) There is a $z_{\epsilon } \in [-1 - \mu ,-1]$ such that

  $$\begin{aligned} \{u \in [-1 - \mu ,1] \, \mid \, f_{\epsilon }(u) = 0\} = \{z_{\epsilon },0,1\}. \end{aligned}$$

  (52)

• (iii) $\Vert f'_{\epsilon }\Vert _{L^{\infty }([-(1 + \mu ),1])} \le M_{1}$.

The lemma follows directly from the properties of ${\bar{f}}$ and the definition of ${\bar{f}}_{\epsilon }$. Therefore, the proof is omitted.

1.3 Modification

In what follows, let $M = \theta ^{-1} \Vert W''\Vert _{L^{\infty }([-3,3])}$. Following [13,  Section 3], we now fix a family of cut-off functions $(\zeta _{\epsilon })_{\epsilon \in (0,1)} \subseteq C^{\infty }_{c}({\mathbb {R}}; [0,1])$ such that, for each $\epsilon \in (0,1)$,

• (a) $\zeta _{\epsilon }(u) = 1$ if $u \in [0,2 \epsilon |\log (\epsilon )|]$,

• (b) $\zeta _{\epsilon }(u) = 0$ if $u \in (-\infty ,-\frac{\epsilon }{M}] \cup [3 \epsilon |\log (\epsilon )|, \infty )$,

• (c) $\zeta _{\epsilon }'$ satisfies the bounds

  $$\begin{aligned} 0 \le \zeta _{\epsilon }'(u) \le \frac{2 M}{\epsilon } \qquad \text {if} \, \, u \in [-\frac{\epsilon }{M}, 0], \quad -\frac{2}{\epsilon |\log (\epsilon )|} \le \zeta _{\epsilon }'(s) \le 0 \qquad \text {if} \, \, u \in [0,3\epsilon |\log (\epsilon )|]. \end{aligned}$$

Now we define ${\tilde{f}}_{\epsilon } : {\mathbb {R}} \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u) = (1 - \zeta _{\epsilon }(u)) f_{\epsilon }(u) + \zeta _{\epsilon }(u) \left( \frac{\epsilon |\log (\epsilon )| - u}{|\log (\epsilon )|} \right) . \end{aligned}$$

To start with, we record some properties of the family $({\tilde{f}}_{\epsilon })_{\epsilon \in (0,1)}$:

Lemma 11

There are positive constants $c, M_{2}, \epsilon _{1} > 0$ such that if $\epsilon \in (0,\epsilon _{0} \wedge \epsilon _{1})$, then the following statements hold:

• (a) ${\tilde{f}}_{\epsilon } \ge f_{\epsilon } \ge {\bar{f}}$ in $[-1 - \mu , 1]$.

• (b) The following inequalities hold away from 0:

  $$\begin{aligned} {\tilde{f}}_{\epsilon }(u) \le - c \epsilon \quad \text {if} \quad u \in [2 \epsilon |\log (\epsilon )|, 3 \epsilon |\log (\epsilon )|], \quad {\tilde{f}}_{\epsilon }(u)&\ge c \epsilon \quad \text {if} \quad u \in \left[ - \frac{\epsilon }{M},0\right] . \end{aligned}$$

• (c) $\Vert {\tilde{f}}_{\epsilon }'\Vert _{L^{\infty }([-1 - \mu ,1])} \le M_{2}$.

Proof

To see that (a) holds, observe that the identity $f_{\epsilon }(0) = {\bar{f}}(0) = 0$ implies we can write

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u) = f_{\epsilon }(u) + \zeta _{\epsilon }(u) \left( \epsilon - \left( \frac{1}{|\log (\epsilon )|} + \frac{f_{\epsilon }(u) - f_{\epsilon }(0)}{u} \right) u \right) . \end{aligned}$$

(53)

Recall that the $\zeta _{\epsilon }$ term only has to be dealt with when $u \in [-\epsilon /M,0] \cup [0,3\epsilon |\log (\epsilon )|]$.

Fix $\epsilon _{1}' > 0$ such that $\epsilon /M \le 1/4$ if $\epsilon \in (0,\epsilon _{1}')$. If $\epsilon \in (0,\epsilon _{1}')$ and $u \in [-\epsilon /M,0]$, then the definition of M gives

$$\begin{aligned} f_{\epsilon }(u) = |f_{\epsilon }(u)|&= |{\bar{f}}(u)| \le \theta ^{-1} \Vert W''\Vert _{L^{\infty }([-3,3])} |u| = M |u|. \end{aligned}$$

Hence $f_{\epsilon }(u) \le M |u| \le \epsilon$, which gives

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u) = f_{\epsilon }(u) + \zeta _{\epsilon }(u) \left( \epsilon - f_{\epsilon }(u) - \frac{u}{|\log (\epsilon )|} \right) \ge f_{\epsilon }(u). \end{aligned}$$

Making $\epsilon _{1}'$ smaller if necessary, we can assume that $3 \epsilon |\log (\epsilon )| \le \mu$ if $\epsilon \in (0,\epsilon _{1}')$. Now note that if $u \in [2\epsilon |\log (\epsilon )|,3 \epsilon |\log (\epsilon )|]$, then we can write

$$\begin{aligned} -\left( \frac{1}{|\log (\epsilon )|} + \frac{f_{\epsilon }(u) - f_{\epsilon }(0)}{u} \right) u&\ge -3 \epsilon - f_{\epsilon }(u) \\&\ge -3 \epsilon - \frac{W''(0)}{2 \Theta } u \\&\ge -3 \epsilon + \frac{|W''(0)| \epsilon |\log (\epsilon )| }{\Theta } . \end{aligned}$$

Finally, we let $\epsilon _{1}'' = \exp (-\frac{3 \Theta }{|W''(0)|})$ and $\epsilon _{1} = \epsilon _{1}' \wedge \epsilon _{1}'' \wedge \frac{1}{2}$. The previous string of inequalities implies that if $\epsilon \in (0,\epsilon _{1} \wedge \epsilon _{0})$, then

$$\begin{aligned} -\left( \frac{1}{|\log (\epsilon )|} + \frac{f_{\epsilon }(u) - f_{\epsilon }(0)}{u} \right) u \ge 0 \quad \text {if} \quad u \in [2\epsilon |\log (\epsilon )|,3 \epsilon |\log (\epsilon )|]. \end{aligned}$$

From this and (53), it follows that ${\tilde{f}}_{\epsilon }(u) \ge f_{\epsilon }(u)$ for all $u \in [2 \epsilon |\log (\epsilon )|, 3 \epsilon |\log (\epsilon )|]$.

Next, we note that if $u \in [0,2 \epsilon |\log (\epsilon )|]$ and $\epsilon \in (0,\epsilon _{1} \wedge \epsilon _{0})$, then a direct computation shows that

$$\begin{aligned} f_{\epsilon }(u) \le - \frac{|W''(0)| u}{2 \Theta } \le \epsilon - \frac{u}{|\log (\epsilon )|} = {\tilde{f}}_{\epsilon }(u). \end{aligned}$$

This completes the proof that ${\tilde{f}}_{\epsilon } \ge f_{\epsilon }$ in $[-2,2]$ and then the inequality $f_{\epsilon } \ge {\bar{f}}$ in the same interval follows from the construction of $f_{\epsilon }$.

Next, we prove (b). Recall that if $u \in [0,\mu ]$, then

$$\begin{aligned} f_{\epsilon }(u) \le - \frac{|W''(0)|}{2 \Theta }u \end{aligned}$$

and, thus, for all $u \in [2 \epsilon |\log (\epsilon )|,3 \epsilon |\log (\epsilon )|]$ and $\epsilon \in (0,\epsilon _{1} \wedge \epsilon _{0})$,

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u)&\le -(1 - \zeta _{\epsilon }(u)) \Theta ^{-1} |W''(0)| \epsilon |\log (\epsilon )| - \zeta _{\epsilon }(u) \epsilon \le - \epsilon . \end{aligned}$$

Let $\epsilon _{1}''' = M\mu$. If $u \in [-\frac{\epsilon }{M},0]$ and $\epsilon \in (0,\epsilon _{1}''')$, then

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u)&\ge (1 - \zeta _{\epsilon }(u)) \frac{|W''(0)|}{2 \Theta } |u| + \zeta _{\epsilon }(u)\epsilon . \end{aligned}$$

When $u \in [-\frac{\epsilon }{4M},0]$, this gives (by property (c) of $\zeta _{\epsilon }$ above),

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u) \ge \left( 1 - \frac{2M}{\epsilon } \cdot \frac{\epsilon }{4M} \right) \epsilon = \frac{\epsilon }{2} \end{aligned}$$

while the case $u \in [-\frac{\epsilon }{M},-\frac{\epsilon }{4M}]$ yields

$$\begin{aligned} {\tilde{f}}_{\epsilon }(u) \ge (1 - \zeta _{\epsilon }(u)) \frac{|W''(0)| \epsilon }{8M \Theta }+ \zeta _{\epsilon }(u)\epsilon \ge \frac{|W''(0)| \epsilon }{8M \Theta }. \end{aligned}$$

Therefore, if we replace $\epsilon _{1}$ above by $\epsilon _{1} \wedge \epsilon _{1}'''$, we conclude that there is a $c > 0$ such that (b) holds.

(c) follows directly from the choice of $\zeta _{\epsilon }$, conclusion (b) of Lemma 10, and (50). $\square$

Henceforth, we let $\chi ^{\epsilon } : [-1 - \mu ,1] \times [0,\infty ) \rightarrow [-1 - \mu ,1]$ denote the solution map of the ODE associated with $-{\tilde{f}}_{\epsilon }$, that is,

$$\begin{aligned} \left\{ \begin{array}{ll} \chi ^{\epsilon }_{s}(\xi ,s) + {\tilde{f}}_{\epsilon }(\chi ^{\epsilon }(\xi ,s)) = 0 &{}\quad \text {if} \, \, (\xi ,s) \in [-1 - \mu ,1] \times (0,\infty ), \\ \chi ^{\epsilon }(\xi ,0) = \xi &{}\quad \text {if} \, \, \xi \in [-1 - \mu ,1]. \end{array} \right. \end{aligned}$$

Lemma 12

• (i) For each $\beta > 0$, there is an $\epsilon (\beta ), \tau (\beta ) > 0$ such that if $\epsilon \in (0,\epsilon (\beta ))$, then

  $$\begin{aligned} \chi ^{\epsilon }(\xi ,s) \ge 1 - \beta \epsilon \quad \text {if} \quad \xi \ge 3 \epsilon |\log (\epsilon )|, \, \, s \ge \tau (\beta )|\log (\epsilon )|. \end{aligned}$$

  (54)

• (ii) $\chi ^{\epsilon }_{\xi } > 0$ in $[-1 - \mu ,1] \times [0,\infty )$, independently of $\epsilon > 0$.

• (iii) For each $a > 0$, there is an $\epsilon (a) > 0$ and $B(a) > 0$ such that if $\epsilon \in (0,\epsilon (a))$, then

  $$\begin{aligned} \left| \frac{\chi _{\xi \xi }^{\epsilon }(\xi ,s)}{\chi _{\xi }^{\epsilon }(\xi ,s)} \right| \le \frac{B(a)}{\epsilon } \quad \text {if} \quad (\xi , s) \in [-1 - \mu ,1] \times [0,a |\log (\epsilon )|]. \end{aligned}$$

  (55)

Proof

The proof proceeds exactly as in [13,  Lemma 3.1]. The main difference is ${\tilde{f}}''_{\epsilon }$ can grow like $C\epsilon ^{-1}$ near $-1$, which, upon inspection of the proof in [13], only has the effect of increasing the constant B(a) in (55). $\square$

Appendix B: Comparison principle

After a multiplication by $m^{-1}$, (1) is a special case of the following class of equations:

$$\begin{aligned} \left\{ \begin{array}{ll} u_{t} - G(x,Du) \text {tr}(D^{2}u) + B(x,u,Du) = 0 &{}\quad \text {in} \, \, {\mathbb {R}}^{d} \times (0,T), \\ u = u_{0} &{}\quad \text {on} \, \, {\mathbb {R}}^{d} \times \{0\}. \end{array} \right. \end{aligned}$$

(56)

In what follows, we assume that $G : {\mathbb {R}}^{d} \times {\mathbb {R}}^{d} \rightarrow (0,\infty )$ and $B : {\mathbb {R}}^{d} \times {\mathbb {R}}^{d} \rightarrow {\mathbb {R}}$ are bounded, continuous functions for which there are constants $C_{1}, K, m, M > 0$ such that, for each $(y,u,v) \in {\mathbb {R}}^{d} \times {\mathbb {R}} \times {\mathbb {R}}^{d}$, $y' \in {\mathbb {R}}^{d}$, and $u' \in {\mathbb {R}}$, we have

$$\begin{aligned}&|G(y,v) - G(y',v)| + |B(y,v,u) - B(y',v,u)| \le C_{1} \Vert y - y'\Vert , \end{aligned}$$

(57)

$$\begin{aligned}&m \le G(y,v) \le M, \end{aligned}$$

(58)

$$\begin{aligned}&|B(y,u,v) - B(y,u',v)| \le K |u - u'|. \end{aligned}$$

(59)

As is standard in the theory of viscosity solutions, to obtain well-posedness of (56), we start with a comparison principle:

Theorem 6

Fix $T > 0$. If u is a bounded, upper semi-continuous function satisfying $u_{t} - G(x,Du) \text {tr}(D^{2}u) + B(x,u,Du) \le 0$ in ${\mathbb {R}}^{d} \times (0,T)$ and v is a bounded, lower semi-continuous function satisfying $v_{t} - G(x,Dv) \text {tr}(D^{2}v) + B(x,v,Dv) \ge 0$ in ${\mathbb {R}}^{d} \times (0,T)$ and if M is defined by

$$\begin{aligned} M= \lim _{\delta \rightarrow 0^{+}} \sup \left\{ u^{*}(x,0) - v_{*}(y,0) \, \mid \, x,y \in {\mathbb {R}}^{d}, \, \, \Vert x-y\Vert \le \delta \right\} , \end{aligned}$$

then

$$\begin{aligned} u(x,t) - v(x,t) \le M e^{Kt} \vee 0 \quad \text {for all} \, \, (x,t) \in {\mathbb {R}}^{d} \times (0,T). \end{aligned}$$

Sketch of the proof

The Lipschitz assumption on G implies that a comparison argument can be carried out in the spirit of [16]. To get the exponential bound, we note that, given $\delta > 0$ and K as in (59), if we define the functions ${\tilde{u}}^{\delta }$ and ${\tilde{v}}^{\delta }$ by ${\tilde{u}}^{\delta }(x,t) = e^{-(K + \delta ) t} u(x,t)$ and ${\tilde{v}}(x,t) = e^{-(K + \delta )t} v(x,t)$ and write ${\tilde{w}}^{\delta } = {\tilde{u}}^{\delta } - {\tilde{v}}^{\delta }$, then a standard argument shows that ${\tilde{w}}^{\delta }$ satisfies ${\tilde{w}}^{\delta } \le M \vee 0$ in ${\mathbb {R}}^{d} \times (0,T)$. The result is recovered after sending $\delta \rightarrow 0^{+}$. $\square$

Existence now follows using Perron’s Method and regularization:

Corollary 1

Given $u_{0} \in BUC({\mathbb {R}}^{d})$, there is a unique, bounded viscosity solution of (56).

Proof

First, assume that $u_{0}$ is a bounded function in ${\mathbb {R}}^{2}$ with bounded second derivative. It follows that ${\bar{u}}(x,t) = u_{0}(x) + Ct$ and ${\underline{u}}(x,t) = u_{0}(x) - Ct$ define super- and subsolutions of (56) provided $C > 0$ is large enough. Thus, an application of Perron’s Method gives a bounded, continuous solution u with $u(\cdot ,0) = u_{0}$ in ${\mathbb {R}}^{d}$.

If $u_{0}$ is not so regular, then nonetheless we can find a sequence $(u_{0,n})_{n \in {\mathbb {N}}}$ of functions as above such that $\Vert u_{0,n} - u_{0}\Vert _{L^{\infty }({\mathbb {R}}^{d})} \rightarrow 0$ as $n \rightarrow \infty$. The bound in Theorem 6 implies that the associated solutions $(u_{n})_{n \in {\mathbb {N}}}$ are uniformly Cauchy in ${\mathbb {R}}^{d} \times [0,T]$. Therefore, their limit $u = \lim _{n \rightarrow \infty } u_{n}$ exists and, by stability, is a solution of (56). $\square$

Appendix C: Construction of correctors

In this section, we discuss the standing waves and correctors used in the ansatz (19). Throughout we assume that W satisfies satisfies (14), (15), and (16) as in Theorem 1.

1.1 Standing waves

We begin by recalling some standard facts concerning the standing waves of the Allen–Cahn equation. Up to translations, this is the function $q : {\mathbb {R}} \rightarrow (-1,1)$ such that

$$\begin{aligned} -\ddot{q} + W'(q) = 0 \quad \text {in} \, \, {\mathbb {R}}, \quad \lim _{s \rightarrow \pm \infty } q(s) = \pm 1, \quad q(0) = 0. \end{aligned}$$

(60)

Proposition 13

There is a unique, strictly increasing $q : {\mathbb {R}} \rightarrow (-1,1)$ satisfying (60). Further, there is a constant $C > 0$ such that

$$\begin{aligned} |q(s) - 1| \le C \exp \left( -\frac{s}{C} \right) , \quad |q(s) + 1| \le C \exp \left( \frac{s}{C} \right) . \end{aligned}$$

Proof

(60) has a Hamiltonian structure. In particular, the expression $\frac{1}{2} {\dot{q}}(s)^{2} - W(q(s))$ is independent of $s \in {\mathbb {R}}$. Since this should clearly be zero at infinity, we deduce that $\frac{1}{2} {\dot{q}}(0)^{2} = W(0)$. Thus, $|{\dot{q}}(0)|$ is uniquely determined a priori. We solve the ODE $\ddot{q} = W'(q)$ in ${\mathbb {R}}$ with $q(0) = 0$ and ${\dot{q}}(0) = \sqrt{2 W(0))}$, and then an exercise shows that the solution satisfies $\lim _{s \rightarrow \pm \infty } q(s) = \pm 1$. Further, the identity ${\dot{q}}(s) = \sqrt{2 W(q(s))}$ implies q is strictly increasing.

The exponential convergence to $\pm 1$ can be proved using a stability analysis or the maximum principle. Here we are using (15). $\square$

The standing wave q generates the solutions of (8). More precisely, for each $e \in S^{d-1}$, the function $U_{e}(s,y) = q(s)$ is a solution of (8) with $a \equiv \text {Id}$. The penalized correctors constructed in Sect. C.3 will be approximate solutions of (9). It will therefore be helpful to know some properties of the principal eigenfunction $\partial _{s} U_{e}$, which in this case equals ${\dot{q}}$.

Proposition 14

${\dot{q}} \in C^{2,\alpha }({\mathbb {R}})$ solves the linearized Allen–Cahn equation:

$$\begin{aligned} - \ddot{q} + W''(q(s)) {\dot{q}} = 0 \quad \text {in} \, \, {\mathbb {R}}. \end{aligned}$$

Furthermore, there is a constant $C > 0$ such that $|{\dot{q}}(s)| \le C \exp (-C^{-1}|s|).$

Proof

The PDE is obtained directly by differentiating q. The exponential convergence can be proved using the maximum principle and (15) (cf. [32,  Proposition 31]).$\square$

Finally, we will need the following fact concerning the drift appearing in the renormalized Allen–Cahn operator:

Proposition 15

The function $s \mapsto \frac{\ddot{q}(s)}{{\dot{q}}(s)}$ is bounded and uniformly Lipschitz continuous in ${\mathbb {R}}$.

Proof

Applying Schauder estimates to ${\dot{q}}$, we find a constant $C > 0$ such that

$$\begin{aligned} |\ddot{q}(s)| + |\dddot{q}(s)| \le C ( \sup \left\{ {\dot{q}}(s') \, \mid \, s' \in (s - 1, s + 1) \right\} ). \end{aligned}$$

Further, by the Harnack inequality, there is no loss of generality in assuming that

$$\begin{aligned} \sup \left\{ {\dot{q}}(s') \, \mid \, s' \in (s - 1,s + 1) \right\} \le C {\dot{q}}(s). \end{aligned}$$

Thus, $\frac{|\ddot{q}(s)|}{{\dot{q}}(s)} + \frac{|\ddot{q}(s)|}{{\dot{q}}(s)} \le C^{2}$. This gives the boundedness of $\frac{\ddot{q}}{{\dot{q}}}$ directly and the uniform Lipschitz continuity after differentiation. $\square$

1.2 Linearized Allen–Cahn equation

In the construction of approximate correctors, we used the linearized Allen–Cahn operator $-\partial ^{2}_{s} + W''(q(s))$. In particular, the next solvability result was used:

Proposition 16

If $f \in C^{\alpha }({\mathbb {R}})$ for some $\alpha \in (0,1)$, then there is a unique $P^{f} \in C^{2,\alpha }({\mathbb {R}})$ and a unique ${\overline{f}} \in {\mathbb {R}}$ solving the PDE:

$$\begin{aligned} - {\ddot{P}}^{f} + W''(q(s)) P^{f} = (f(s) - {\overline{f}}){\dot{q}}(s) \quad \text {in} \, \, {\mathbb {R}}, \quad P^{f}(0) = 0. \end{aligned}$$

Furthermore, there is a $C > 0$ such that $|P^{f}(s)| \le \Vert P^{f}\Vert _{L^{\infty }({\mathbb {R}})} \exp (-C^{-1}|s|)$.

Proof

The operator ${\mathcal {L}} = -\partial _{s}^{2} + W''(q(s))$ with domain $H^{2}({\mathbb {R}})$ is self-adjoint in $L^{2}({\mathbb {R}})$ with closed range. Therefore, $\text {Ran}({\mathcal {L}}) = \text {Ker}({\mathcal {L}})^{\perp }$. Since ${\dot{q}}$ is a positive eigenfunction, it follows that, for each $g \in H^{2}({\mathbb {R}})$,

$$\begin{aligned} \langle {\mathcal {L}}g,g \rangle _{L^{2}({\mathbb {R}})} = \int _{-\infty }^{\infty } {\dot{h}}(s)^{2} {\dot{q}}(s)^{2} \, d s, \end{aligned}$$

where $h(s) = {\dot{q}}(s)^{-1} g(s)$. From this, it follows that $\text {Ker}({\mathcal {L}}) = \text {span}\{{\dot{q}}\}$.

Thus, if $f \in C^{\alpha }({\mathbb {R}})$, then $f {\dot{q}} \in L^{2}({\mathbb {R}})$ and there is a $P^{f} \in H^{2}({\mathbb {R}})$ with ${\mathcal {L}} P^{f} = (f - {\overline{f}}){\dot{q}}$ provided ${\overline{f}} \in {\mathbb {R}}$ is given by

$$\begin{aligned} {\overline{f}} = c_{W}^{-1} \int _{- \infty }^{\infty } f(s) {\dot{q}}(s)^{2} \, ds. \end{aligned}$$

The local maximum principle implies $P^{f} \in L^{\infty }({\mathbb {R}})$ and then Schauder estimates give $P^{f} \in C^{2,\alpha }({\mathbb {R}})$. Since ${\dot{q}}(s) \rightarrow 0$ exponentially as $s \rightarrow \pm \infty$, a maximum principle argument shows that $P^{f}$ does also. $\square$

1.3 Penalized correctors

Finally, we prove the existence and regularity of the penalized correctors used in Sect. 4. Recall that these are the solutions $(P_{2}^{\delta })_{\delta > 0}$ of the penalized cell problem

$$\begin{aligned} {\tilde{m}}_{2}(s,y) {\dot{q}}(s) + \delta P_{2}^{\delta } + {\mathcal {D}}_{e}^{*} {\mathcal {D}}_{e} P^{\delta }_{2} + W''(q) P^{\delta }_{2} = 0 \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

(61)

We search for a solution of the form $P^{\delta }_{2}(s,y) = V^{\delta }_{2}(s,y) {\dot{q}}(s)$, employing the so-called Doob–Legendre transform. Plugging this ansatz into the equation, we find that $P^{\delta }_{2}$ solves (61) if and only if $V^{\delta }_{2}$ solves

$$\begin{aligned} {\tilde{m}}_{2}(s,y) + \delta V_{2}^{\delta } + {\mathcal {D}}_{e}^{*} {\mathcal {D}}_{e} V^{\delta }_{2} - \frac{2 \ddot{q}(s)}{{\dot{q}}(s)} \langle e, {\mathcal {D}}_{e} V^{\delta }_{2} \rangle = 0 \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

(62)

It is clear that (62) has a solution since it is degenerate elliptic and ${\tilde{m}}_{2}$ is bounded. However, we will not proceed in this way.

Instead, notice that $V^{\delta }_{2}$ is a classical solution of (62) if and only if the one-parameter family of functions $(v_{\zeta }^{\delta })_{\zeta \in {\mathbb {R}}}$ determined by

$$\begin{aligned} v_{\zeta }^{\delta }(x) = V^{\delta }_{2}(\langle x,e \rangle - \zeta , x) \end{aligned}$$

gives rise to solutions of the following family of equations:

$$\begin{aligned} {\tilde{m}}_{2}(\langle x,e \rangle - \zeta ,x) + \delta v^{\delta }_{\zeta } - \Delta v^{\delta }_{\zeta } - \frac{2 \ddot{q}(\langle x,e \rangle - \zeta )}{{\dot{q}}(\langle x,e \rangle - \zeta )} \langle e, Dv^{\delta }_{\zeta } \rangle = 0 \quad \text {in} \, \, {\mathbb {R}}^{d}. \end{aligned}$$

(63)

Where regularity considerations are concerned, it is convenient to construct the solution $V^{\delta }_{2}$ of (62) by studying the solutions of (63). Here is the main result in that regard:

Theorem 7

If ${\tilde{m}}$ satisfies (29), then, for each $\delta > 0$ and $\zeta \in {\mathbb {R}}$, there is a $v_{\zeta }^{\delta } \in C^{2,\alpha }({\mathbb {R}}^{d})$ solving (63). Furthermore, the map $(\zeta ,x) \mapsto v_{\zeta }^{\delta }(x)$ is twice continuously differentiable with respect to $\zeta$ and there is a constant $C > 0$ independent of $\delta$ such that

$$\begin{aligned} \Vert v_{\zeta }^{\delta }\Vert _{L^{\infty }({\mathbb {R}}^{d})} + \left\| \frac{\partial v_{\zeta }}{\partial \zeta } \right\| _{C^{1,\alpha }({\mathbb {R}}^{d})} + \left\| \frac{\partial ^{2} v_{\zeta }}{\partial \zeta ^{2}} \right\| _{C^{\alpha }({\mathbb {R}}^{d})} \le C (1 +\delta ^{-1}). \end{aligned}$$

This leads immediately to a regularity result for $V^{\delta }$:

Corollary 2

If ${\tilde{m}}$ satisfies (29), then the unique viscosity solution $V_{2}^{\delta }$ of (62) is in $C^{2,\mu }({\mathbb {R}} \times {\mathbb {T}}^{d})$ and there is a $\delta$-independent constant $C > 0$ depending only on f such that

$$\begin{aligned} \left\| V_{2}^{\delta } \right\| _{C^{2,\alpha }({\mathbb {R}} \times {\mathbb {T}}^{d})} \le C(1+ \delta ^{-1}). \end{aligned}$$

Furthermore, making C larger if necessary, we also have:

$$\begin{aligned} \Vert P^{\delta }_{2}(s,\cdot )\Vert _{L^{\infty }({\mathbb {T}}^{d})} + \Vert \partial _{s} P_{2}^{\delta }(s,\cdot )\Vert _{L^{\infty }({\mathbb {T}}^{d})} \le C (1 + \delta ^{-1}) \exp \left( -C^{-1}|s|\right) . \end{aligned}$$

Proof of Theorem 7

Since the drift $\frac{\ddot{q}}{{\dot{q}}}$ is bounded and uniformly Lipschitz continuous by Proposition 15, (63) has a unique solution $v^{\delta }_{\zeta } \in C^{2,\mu }({\mathbb {R}}^{d})$ and Schauder estimates give

$$\begin{aligned} \Vert v^{\delta }_{\zeta }\Vert _{L^{\infty }({\mathbb {R}}^{d})}&\le \Vert f\Vert _{L^{\infty }({\mathbb {R}} \times {\mathbb {T}}^{d})} \delta ^{-1}, \\ \Vert v^{\delta }_{\zeta }\Vert _{C^{2,\mu }({\mathbb {R}}^{d})}&\le C' \left( \Vert v^{\delta }_{\zeta }\Vert _{L^{\infty }({\mathbb {R}}^{d})} + \Vert {\tilde{m}}\Vert _{C^{\mu }({\mathbb {R}} \times {\mathbb {T}}^{d})}\right) \le C' \Vert {\tilde{m}}\Vert _{C^{\mu }({\mathbb {R}} \times {\mathbb {T}}^{d})} (1 + \delta ^{-1}). \end{aligned}$$

By uniqueness, it is easy to see that $\zeta \mapsto v^{\delta }_{\zeta }$ is continuous with respect to the topology of local uniform convergence.

Recall the functions $({\tilde{v}}_{\zeta })_{\zeta \in {\mathbb {R}}}$ defined in (34) by translation. As pointed out above, these functions satisfy (35). Thus, employing the method of difference quotients, we deduce that the functions $(w_{\zeta })_{\zeta \in {\mathbb {R}}}$ defined by $w_{\zeta } = \frac{\partial {\tilde{v}}^{\delta }}{\partial \zeta }$ satisfy

$$\begin{aligned} \langle D_{y} {\tilde{m}}(\langle x,e \rangle ,x + \zeta e), e \rangle + \delta w_{\zeta } - \Delta w_{\zeta } + \frac{2 \ddot{q}(\langle x,e \rangle )}{{\dot{q}}(\langle x,e \rangle )} \langle e, Dw_{\zeta } \rangle = 0 \quad \text {in} \, \, {\mathbb {R}}^{d}. \end{aligned}$$

Similarly, the functions $(p_{\zeta })_{\zeta \in {\mathbb {R}}}$ given by $p_{\zeta } = \frac{\partial ^{2} {\tilde{v}}_{\zeta }}{\partial \zeta ^{2}}$ satisfy

$$\begin{aligned} \langle D^{2}_{y} {\tilde{m}}(\langle x,e \rangle , x + \zeta e)e ,e \rangle + \delta p_{\zeta } - \Delta p_{\zeta } + \frac{2 \ddot{q}(\langle x,e \rangle )}{{\dot{q}}(\langle x,e \rangle )} \langle e, Dp_{\zeta } \rangle = 0 \quad \text {in} \, \, {\mathbb {R}}^{d}. \end{aligned}$$

Thus, since $D_{y}{\tilde{m}}$ and $D^{2}_{y}{\tilde{m}}$ are just as regular as ${\tilde{m}}$, there is a $C > 0$ such that

$$\begin{aligned} \left\| \frac{\partial {\tilde{v}}_{\zeta }}{\partial \zeta } \right\| _{C^{2,\alpha }({\mathbb {R}}^{d})} + \left\| \frac{\partial ^{2} {\tilde{v}}_{\zeta }}{\partial \zeta ^{2}} \right\| _{C^{2,\alpha }({\mathbb {R}}^{d})} \le C (1 + \delta ^{-1}). \end{aligned}$$

Furthermore, if $\zeta , \zeta ' \in {\mathbb {R}}$, then the Hölder regularity of $D^{2}_{y}{\tilde{m}}$ yields

$$\begin{aligned} \left\| \frac{\partial ^{2} {\tilde{v}}_{\zeta }}{\partial \zeta ^{2}} - \frac{\partial ^{2} {\tilde{v}}_{\zeta '}}{\partial \zeta ^{2}} \right\| _{L^{\infty }({\mathbb {R}}^{d})}&\le \delta ^{-1} \Vert D^{2}{\tilde{m}}_{y}\Vert _{C^{\alpha }({\mathbb {R}} \times {\mathbb {T}}^{d})} |\zeta - \zeta '|^{\alpha }. \end{aligned}$$

These bounds readily carry over to $(v_{\zeta })_{\zeta \in {\mathbb {R}}}$, giving the desired estimates. $\square$

We proceed with the proof of Corollary 2:

Proof of Corollary 2

Define $V^{\delta }_{2} : {\mathbb {R}} \times {\mathbb {T}}^{d} \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} V^{\delta }_{2}(s,y) = v^{\delta }_{\langle y,e \rangle - s}(y). \end{aligned}$$

An exercise shows this is well-defined. Differentiating, we eventually find $V^{\delta }_{2} \in C^{2,\alpha }({\mathbb {R}} \times {\mathbb {T}}^{d})$. A calculus exercise shows that $V^{\delta }_{2}$ is a solution of (62).

The exponential convergence of $P^{\delta }_{2}$ follows from a maximum principle argument as in [32,  Proof of Proposition 30]. Using the fact that $\partial _{s} P^{\delta }_{2}$ satisfies a structurally similar linear PDE, we obtain a similar exponential estimate on $\partial _{s} P^{\delta }_{2}$. $\square$

Appendix D: Formal asymptotics

In this appendix, we review the approach of [9] as it pertains to (1). The aim is to show, in particular, how the assumptions made in that work arise naturally from a formal asymptotic expansion. While [9] only treats the case when $m \equiv 1$, it should be stressed the approach adapts readily to the case of a general m.

The key assumption in the analysis of (1) in [9] is the existence of smooth families of pulsating waves and correctors. Since there are currently no known examples of non-constant coefficients (a, m) for which this assumption holds, we will only present the formal asymptotics that underlie the arguments rather than a rigorous proof.

1.1 Preliminaries

By analogy with the well-studied case where $(a,m) \equiv (\text {Id},1)$, we expect that there are open sets $\{E_{t}\}_{t \ge 0}$ such that

$$\begin{aligned} \{u^{\epsilon }(\cdot ,t) \approx 1\} \rightarrow E_{t} \quad \text {and} \quad \{u^{\epsilon }(\cdot ,t) \approx -1\} \rightarrow {\mathbb {R}}^{d} \setminus {\overline{E}}_{t} \quad \text {as} \, \, \epsilon \rightarrow 0^{+}. \end{aligned}$$

The question is the identification of $\{E_{t}\}_{t \ge 0}$. Let us hypothesize that it is governed by a geometric flow of the following form:

$$\begin{aligned} {\bar{M}}^{(a,m)}(n_{\partial E_{t}}) V_{\partial E_{t}} = \text {tr} \left( \bar{{\mathcal {S}}}^{a}(n_{\partial E_{t}}) A_{\partial E_{t}} \right) . \end{aligned}$$

(64)

Here $\bar{{\mathcal {S}}}^{a}$ and ${\bar{M}}^{(a,m)}$ are effective coefficients, which we expect to appear in the sharp interface limit due to averaging. The formal arguments below suggest that the influence of m only appears in the effective mobility ${\bar{M}}^{(a,m)}$, hence the notation.

Since we are arguing formally, we will assume the sets $\{E_{t}\}_{t \ge 0}$ have smooth boundaries, which vary smoothly as functions of t.

To relate the solution $u^{\epsilon }$ of (1) to the macroscopic interface $\partial E_{t}$, we will use the signed distance function $d : {\mathbb {R}}^{d} \times (0,\infty ) \rightarrow {\mathbb {R}}$ to $\{E_{t}\}_{t \ge 0}$, defined by

$$\begin{aligned} d(x,t) = \left\{ \begin{array}{ll} \text {dist}(x,\partial E_{t}), &{}\quad \text {if} \, \, x \in E_{t}, \\ -\text {dist}(x,\partial E_{t}), &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Note that the smoothness of the evolution $\{E_{t}\}_{t \ge 0}$ implies the smoothness of d locally near points on the interface, and (64) holds if and only if d satisfies

$$\begin{aligned} {\bar{M}}^{(a,m)}(Dd) d_{t} - \text {tr} \left( {\bar{S}}^{a}(Dd) D^{2} d \right) = 0 \quad \text {on} \, \, \bigcup _{t \ge 0} \partial E_{t} \times \{t\}. \end{aligned}$$

(65)

1.2 Asymptotic expansion

Let us rewrite the ansatz (7) already introduced in the introduction in a more suggestive form:

$$\begin{aligned} u^{\epsilon }(x,t)&= U_{Dd(x,t)}\left( \epsilon ^{-1} d(x,t), \epsilon ^{-1}x\right) + \epsilon Q^{D^{2}d(x,t)}_{Dd(x,t)} \left( \epsilon ^{-1} d(x,t), \epsilon ^{-1} x \right) \nonumber \\&\quad + \epsilon P_{Dd(x,t)}^{d_{t}(x,t)}(\epsilon ^{-1}d(x,t),\epsilon ^{-1}x) + \cdots \end{aligned}$$

(66)

Here $\{U_{e}\}_{e \in S^{d-1}}$, $\{Q^{X}_{e}\}_{(e,X) \in S^{d-1} \times {\mathcal {S}}_{d}}$ , and $\{P^{q}_{e}\}_{(e,q) \in S^{d-1} \times {\mathbb {R}}}$ are functions to be determined and d is the signed distance to $\{E_{t}\}_{t \ge 0}$ as above.

We search for $\{U_{e}\}$, $\{Q^{X}_{e}\}$, and $\{P^{q}_{e}\}$ among functions that are periodic in the second variable, reflecting the periodicity of a and m. Hence we fix the domain of these functions to be ${\mathbb {R}} \times {\mathbb {T}}^{d}$.

In what follows, we will assume everything is smooth, including not only the evolving sets $\{E_{t}\}_{t \ge 0}$, but also the functions $\{U_{e}\}$, $\{Q^{X}_{e}\}$, and $\{P^{q}_{e}\}$ and the dependence of these functions on the parameters (e, X, q).

For the ansatz (66) to produce a solution of (1), we require that

$$\begin{aligned} 0 = m(\epsilon ^{-1} x, \epsilon Du^{\epsilon }) u_{t}^{\epsilon } - \text {div}(a(\epsilon ^{-1}x)Du^{\epsilon }) + \epsilon ^{-2} W'(u^{\epsilon }) = \epsilon ^{-2} A_{1} + \epsilon ^{-1} A_{2} + \cdots , \end{aligned}$$

(67)

where the neglected terms are of lower order in $\epsilon$.

1.3 Vanishing to order $\epsilon ^{-2}$

Setting $A_{1} = 0$ and substituting $y = \epsilon ^{-1} x$ and $e = Dd(x,t)$ leads to the following equations for $\{U_{e}\}$:

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}U_{e}) + W'(U_{e}) = 0 \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

(68)

At the same time, away from the interface $\partial E_{t}$, we know that $u^{\epsilon } \approx 1$ in $E_{t}$ and $u^{\epsilon } \approx -1$ outside. That suggests the limiting condition

$$\begin{aligned} \lim _{s \rightarrow \pm \infty } U_{e}(s,y) = \pm 1. \end{aligned}$$

Finally, it is convenient to add the monotonicity assumption $\partial _{s}U_{e} > 0$.

Next, we set $A_{2} = 0$ and proceed similarly. At this stage, derivatives of the map $e \mapsto U_{e}$ appear. In order to avoid calculus on manifolds, it is convenient to extend $\{U_{e}\}_{e \in S^{d-1}}$ to $\{U_{v}\}_{v \in {\mathbb {R}}^{d} \setminus \{0\}}$ according to the rule (cf. [32,  Remark 2])

$$\begin{aligned} U_{v}(s,y) = U_{e}(\Vert v\Vert s,y) \quad \text {for} \, \, e = \frac{v}{\Vert v\Vert }. \end{aligned}$$

Next, to ease the notation, we define the vector-valued functions $\{R_{v}\}_{v \in {\mathbb {R}}^{d} \setminus \{0\}}$ to be the derivative of $v \mapsto U_{v}$, hence

$$\begin{aligned} \langle R_{v}(s,y), \xi \rangle = \lim _{h \rightarrow 0} \frac{U_{v + h \xi }(s,y) - U_{v}(s,y)}{h} \quad \text {for} \, \, (s,y) \in {\mathbb {R}} \times {\mathbb {T}}^{d}, \, \, \xi \in {\mathbb {R}}^{d}. \end{aligned}$$

(69)

1.4 Vanishing to order $\epsilon ^{-1}$

We now proceed to investigate the consequences of the identity $A_{2} = 0$. Making the substitutions

$$\begin{aligned} y = \epsilon ^{-1}x, \quad e = Dd(x,t), \quad X = D^{2}d(x,t), \quad q = d_{t}(x,t), \end{aligned}$$

we derive the equation

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}(Q^{X}_{e} + P^{q}_{e}))+ W'(U_{e}) (Q^{X}_{e} + P^{q}_{e}) = G(s,y,e,X,q) \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}, \end{aligned}$$

(70)

where G is given by

$$\begin{aligned} G(s,y,e,X,q)&= G_{1}(s,y,e,X) - G_{2}(s,y,e,q), \\ G_{1}(s,y,e,X)&= \text {tr} \left( a(y) X \right) \partial _{s}U_{e} + 2 \langle a(y) e, X \partial _{s} R_{e}(s,y) \rangle \\&\quad + 2 \text {tr}(a(y) D_{y}R_{e}(s,y) X) + \langle (\text {div} \, a)(y), X R_{e}(s,y) \rangle , \\ G_{2}(s,y,e,q)&= q m(y,{\mathcal {D}}_{e}U_{e})\partial _{s}U_{e}. \end{aligned}$$

The question now is the solvability of the linear equation (70).

Here is where ${\bar{M}}^{(a,m)}$ and $\bar{{\mathcal {S}}}^{a}$ come into the picture. There is a natural solvability condition associated with equations of the form

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}P) = F(s,y) \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

To see this, first, notice that differentiating (68) with respect to s shows that the function $V_{e} := \partial _{s} U_{e}$ solves the linear PDE

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}V_{e}) + W''(U_{e}) V_{e} = 0 \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

Hence, multiplying the previous equation by $V_{e}$ and integrating by parts, we obtain

$$\begin{aligned} 0 = \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} \left( {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}V_{e}) + W''(U_{e}) V_{e} \right) P \, dy \, ds = \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} F(s,y) V_{e} \, dy \, ds. \end{aligned}$$

Due to this solvability condition, we are led to the following equations for $Q^{X}_{e}$ and $P^{q}_{e}$:

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e} Q^{X}_{e}) + W''(U_{e}) Q_{e}^{X}&= G_{1}(s,y,e,X) - {\overline{G}}_{1}(e,X) \partial _{s}U_{e} \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}, \end{aligned}$$

(71)

$$\begin{aligned} {\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}P^{q}_{e}) + W''(U_{e}) P_{e}^{q}&= -\left[ G_{2}(s,y,e,q) - {\overline{G}}_{2}(e,q) \partial _{s}U_{e}\right] \quad \text {in} \, \, {\mathbb {R}} \times {\mathbb {T}}^{d}. \end{aligned}$$

(72)

Notice these equations are of the form (9). Invoking the solvability condition, we compute the necessary equations for ${\overline{G}}_{1}$ and ${\overline{G}}_{2}$:

$$\begin{aligned} {\overline{G}}_{1}(e,X)&= \Vert V_{e}\Vert _{L^{2}({\mathbb {R}} \times {\mathbb {T}}^{d})}^{-2} \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} V_{e}\left( \text {tr}(a(y)X)V_{e} + 2 \langle a(y) e, X \partial _{s} R_{e} \rangle + 2 \text {tr}(a(y) D_{y} R_{e} X) \right. \\&\quad \left. + \langle (\text {div} \, a)(y), X R_{e} \rangle \right) \, dy \, ds, \\ {\overline{G}}_{2}(e,q)&= q \Vert V_{e}\Vert _{L^{2}({\mathbb {R}} \times {\mathbb {T}}^{d})}^{-2} \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} m(y,{\mathcal {D}}_{e}U_{e}) |V_{e}|^{2} \, dy \, ds. \end{aligned}$$

By linearity in X and q, we can fix a symmetric matrix-valued function $\bar{{\mathcal {S}}}^{a}$ and a positive function ${\bar{M}}^{(a,m)}$ such that

$$\begin{aligned} \text {tr}(\bar{{\mathcal {S}}}^{a}(e) X) = {\overline{G}}_{1}(e,X) \Vert V_{e}\Vert ^{2}_{L^{2}({\mathbb {R}} \times {\mathbb {T}}^{d})}, \quad {\bar{M}}^{(a,m)}(e) q = {\overline{G}}_{2}(e,q) \Vert V_{e}\Vert ^{2}_{L^{2}({\mathbb {R}} \times {\mathbb {T}}^{d})}. \end{aligned}$$

Finally, in order for the sum $P^{X}_{e} + P^{q}_{e}$ to solve (70), we require that ${\overline{G}}_{1}(e,X) - {\overline{G}}_{2}(e,q) = 0$, which, rewritten in terms of d, yields the equation

$$\begin{aligned} {\bar{M}}^{(a,m)}(Dd(x,t)) d_{t}(x,t) - \text {tr}(\bar{{\mathcal {S}}^{a}}(Dd(x,t)) D^{2}d(x,t)) = 0, \end{aligned}$$

which is precisely the PDE we sought to derive.

Remark 5

As shown in [9], it is possible to build a proof from the previous arguments provided that solutions of (68), (71), and (72) do, in fact, exist and are regular enough as functions of (s, y, e, X, q). (See also Theorem 1 above and [32,  Theorem 8] for rigorous proofs in special cases.) Of course, as mentioned already in the introduction, in general, existence and smoothness need not hold.

1.5 Interpretation of $\bar{{\mathcal {S}}}^{a}$

It is worth noting that the coefficient $\bar{{\mathcal {S}}}^{a}$ computed above has a variational interpretation.

Recall that the equation (1) is the gradient flow of the energy ${\mathcal {F}}^{a}$ given by (5) with respect to the metric determined by m. The energy ${\mathcal {F}}^{a}$ has been studied in its own right. More precisely, in [2], Ansini, Braides, and Chiadò Piat prove that the large scale behavior of ${\mathcal {F}}^{a}$ is an anisotropic perimeter functional $\bar{{\mathcal {F}}}^{a}$ of the form

$$\begin{aligned} \bar{{\mathcal {F}}}^{a}(F; \text{\O}mega ) = \int _{\partial F \cap \text{\O}mega } {\bar{\sigma }}(n_{\partial E}(\xi )) \, {\mathcal {H}}^{d-1}(d\xi ). \end{aligned}$$

(See also [17, 18].) We refer to ${\bar{\sigma }}$ as the surface tension.

It has been shown in [32,  Proposition 1] that if $U_{e}$ is a smooth solution of (68) with $\lim _{s \rightarrow \pm \infty } U_{e}(s,y) = \pm 1$ and $\partial _{s} U_{e} \ge 0$, then

$$\begin{aligned} {\bar{\sigma }}(e) = \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} \left( \frac{1}{2} \langle a(y) {\mathcal {D}}_{e}U_{e}, {\mathcal {D}}_{e}U_{e} \rangle + W(U_{e}) \right) \, dy \, ds. \end{aligned}$$

Furthermore, $U_{e}$ generates a continuous one-parameter family of minimizers of ${\mathcal {F}}^{a}$, the graphs of which foliate ${\mathbb {R}}^{d} \times (-1,1)$. This fact is precisely the obstruction to the existence of smooth solutions of (68) alluded to in the introduction: [32] provides examples where such foliations do not exist (cf. [24]).

It turns out that the surface tension reappears in the macroscopic velocity law ${\bar{M}}^{(a,m)}(n) V = \text {tr}(\bar{{\mathcal {S}}}^{a}(n) A)$. Specifically, in the (smooth) context of the preceding discussion, it is possible to prove ${\bar{\sigma }}$ is smooth away from the origin and $\bar{{\mathcal {S}}}^{a}$ is given by

$$\begin{aligned} \bar{{\mathcal {S}}}^{a}(e) = D^{2} {\bar{\sigma }}(e). \end{aligned}$$

(73)

Indeed, as long as everything is smooth, this follows from manipulations of the integral representation of $\bar{{\mathcal {S}}}^{a}$ computed above, exactly as in [32,  Proposition 39].

The appearance of the surface tension in (73) shows that, at least according to this formal analysis, the gradient flow and homogenization “commute.” For more on this, see [32] and the references therein.

1.6 Special case $a \equiv \text {Id}$

Let us show, for the sake of completeness, that the formal derivation above coincides with the approach taken in the paper when a is constant.

When a is constant, say, $a \equiv \text {Id}$, one can prove that the only possible choices of $\{U_{e}\}_{e \in S^{d-1}}$ that depend continuously on e are necessarily given by

$$\begin{aligned} U_{e}(s,y) = q(s + s_{0}(e)), \end{aligned}$$

where, as above, q is the standing wave solution of the spatially homogeneous Allen–Cahn equation with $q(0) = 0$ and $s_{0} : S^{d-1} \rightarrow {\mathbb {R}}$ is an arbitrary continuous function. This claim follows from [32,  Sections 6.3–6.5], the key point being that solutions of (68) inherit symmetry from a when $e \notin {\mathbb {R}} {\mathbb {Z}}^{d}$.

The choice of $s_{0}$ is irrelevant in what follows. Hence let us set $s_{0} \equiv 0$. Notice this implies $U_{e}$ does not depend on e and, thus, a direct computation shows that the functions $\{R_{e}\}$ of (69) are given by

$$\begin{aligned} R_{e}(s,y) = s {\dot{q}}(s) e. \end{aligned}$$

Using the shorthand $c_{W} = \int _{-\infty }^{\infty } {\dot{q}}(s)^{2} \, ds$, let us compute $\bar{{\mathcal {S}}}^{\text {Id}}$:

$$\begin{aligned} \langle \bar{{\mathcal {S}}}^{\text {Id}}(e) v,v \rangle&= \text {tr}(\bar{{\mathcal {S}}}^{\text {Id}}(e) (v \otimes v)) \\&= \text {tr}(v \otimes v) \int _{-\infty }^{\infty } {\dot{q}}(s)^{2} \, ds + 2 |\langle e, v \rangle |^{2} \int _{-\infty }^{\infty } \frac{d}{ds}\{s {\dot{q}}(s)\} \, ds \\&= c_{W} \Vert v\Vert ^{2}. \end{aligned}$$

This proves $\bar{{\mathcal {S}}}^{\text {Id}}(e) = c_{W} \text {Id}$. We similarly compute ${\bar{M}}^{(\text {Id},m)}$:

$$\begin{aligned} {\bar{M}}^{(\text {Id},m)}(e) = \int _{{\mathbb {R}} \times {\mathbb {T}}^{d}} m(y,{\dot{q}}(s) e) {\dot{q}}(s)^{2} \, dy \, ds. \end{aligned}$$

In particular, ${\bar{M}}^{(\text {Id},m)} \equiv c_{W} {\overline{m}}$, where ${\overline{m}}$ is defined in (17). Thus, after factoring out $c_{W}$, the limiting motion is precisely the one obtained in Theorem 1.

Finally, let us show how one arrives at the simplified ansatz (19) starting from (66). To start with, notice that the function $G_{2}$ in the cell problem (71) depends linearly on q. Thus, if $P_{e}^{1}$ is a solution when $q = 1$, then $P_{e}^{q} \equiv q P_{e}^{1}$ defines a solution for all $q \in {\mathbb {R}}$. This gives rise to the the linear dependence on the time derivative $d_{t}$ in (19).

Where $G_{1}$ is concerned, recall that since d is a (signed) distance function, the following identity holds at any point (x, t) where d is smooth:

$$\begin{aligned} D^{2}d(x,t) Dd(x,t) = 0. \end{aligned}$$

Hence $G_{1}(s,y,Dd(x,t),D^{2}d(x,t)) = \text {tr}(a(y)X) V_{e}(s,y)$ at points (x, t) near the interface. Our computation above shows that ${\overline{G}}_{1}(e,X) = \text {tr}(a(y)X)$, and, therefore, for the values of $(e,X) = (Dd(x,t),D^{2}d(x,t))$ appearing in (66), the cell problem (71) for $Q_{e}^{X}$ is the trivial equation ${\mathcal {D}}_{e}^{*}(a(y) {\mathcal {D}}_{e}Q^{X}_{e}) + W''(U_{e}) Q^{X}_{e} = 0$. Of course, the zero function $Q_{e}^{X} \equiv 0$ gives a solution of this PDE, and, thus, the term $Q^{X}_{e}$ can be dropped from the ansatz (66).

Combining all these observations, the expansion in (66) simplifies to the following

$$\begin{aligned} u^{\epsilon }(x,t) = q\left( \frac{d(x,t)}{\epsilon }\right) + \epsilon d_{t}(x,t) P_{e}^{1}(\epsilon ^{-1}d(x,t),\epsilon ^{-1}x) + \cdots , \end{aligned}$$

which is precisely the form proposed in (19).