Entire Minimizers of Allen–Cahn Systems with Sub-Quadratic Potentials

Abstract

We study entire minimizers of the Allen–Cahn systems. The specific feature of our systems are potentials having a finite number of global minima, with sub-quadratic behaviour locally near their minima. The corresponding formal Euler–Lagrange equations are supplemented with free boundaries. We do not study regularity issues but focus on qualitative aspects. We show the existence of entire solutions in an equivariant setting connecting the minima of W at infinity, thus modeling many coexisting phases, possessing free boundaries and minimizing energy in the symmetry class. We also present a very modest result of existence of free boundaries under no symmetry hypotheses. The existence of a free boundary can be related to the existence of a specific sub-quadratic feature, a dead core, whose size is also quantified.

Appendices

Appendix

The Containment

The following result was established by the first author and P. Smyrnelis in unpublished work [9]. We reproduce it here for the convenience of the reader. For related applications of the method of proof we refer to [39].

Proposition 5

( [9]) Let $u : {{\mathbb {R}}}^n \rightarrow {{\mathbb {R}}}^m$ be a bounded ($| u(x) | < M$) critical point of the functional

$$\begin{aligned} J (u) = \int \left( \frac{1}{2} | \nabla u|^2 + W(u)\right) dx \end{aligned}$$

in the sense that $\forall \Omega \subset {{\mathbb {R}}}^n$, open, bounded,

$$\begin{aligned} \frac{d}{d \varepsilon }|_{ \varepsilon =0} J_{\Omega } (u + \varepsilon \phi ) = 0 ,\, \forall \, \phi \in C_0^1 (\Omega ) \end{aligned}$$

where

$$\begin{aligned} W(u) = {\left\{ \begin{array}{ll} W^{ \overline{\alpha }} (u) := \prod _{k=1}^{m+1} |u- a_k|^{{\alpha }_k} ,\, \overline{\alpha } = ({\alpha }_1,\ldots ,{\alpha }_{m+1}) , \, 0 < {\alpha }_k \le 2 \\ W^0 (u) := {\chi }_{\lbrace u \in S_A \rbrace } \end{array}\right. } \end{aligned}$$

(A.1)

and $S_A$ defined as the interior of the simplex with vertices $a_1,\ldots ,a_m,a_{m+1}$,

$$\begin{aligned} S_A := \left\{ \sum _{i=1}^{m+1} {\lambda }_i a_i \,\, ; \, {\lambda }_i \in [0,1) , \, \forall i=1,\ldots ,m+1 ,\, \sum _{i=1}^{m+1} {\lambda }_i =1 \right\} \end{aligned}$$

(A.2)

Then

$$\begin{aligned} u(x) \in {{\overline{S}}}_A ,\, x \in {{\mathbb {R}}}^n \end{aligned}$$

(A.3)

For $\alpha _k \in [0,1)$ we require that u in addition is a minimizer in the sense of (1.3), so that (A.5) is available.

Proof

Following an idea from [17] we introduce the set

• 1. ${\alpha }_k \in (0,1) ,\, k=1,\ldots ,m.$

  $$\begin{aligned} F_M := \lbrace u: {{\mathbb {R}}}^n \rightarrow {{\mathbb {R}}}^m ,\, u \,\, \text {minimizer of} \,\, J ,\, |u(x)| \le M \rbrace \end{aligned}$$

  (A.4)

  By Lemma 2.1 we have the uniform Hölder estimate

  $$\begin{aligned} | u |_{C^{\beta }( {{\mathbb {R}}}^n ; {{\mathbb {R}}}^m)} \le C(M) , u \in F_M \end{aligned}$$

  (A.5)

  Let $\Pi$ be the face of the simplex ${{\overline{S}}}_A$ defined by $a_2,\ldots ,a_{m+1}$, oppposite to $a_1$ and let $e \perp \Pi$. Set

  $$\begin{aligned} P(u;x) = \langle u(x) - a_2 , e \rangle \end{aligned}$$

  (A.6)

  where $\langle \cdot , \cdot \rangle$ is the inner product in ${{\mathbb {R}}}^m$ and the orientation of e is such that $\langle a_2-a_1 ,e \rangle >0$. Set

  $$\begin{aligned} P_M := \sup \lbrace P(u;x) \, : \, u( \cdot ) \in F_M ,\, x \in {{\mathbb {R}}}^n \rbrace \end{aligned}$$

Claim 2

$\, P_M \le 0$

Clearly the proposition follows from this claim. We proceed by contradiction. Suppose $P_M >0$. Thus there is $\lbrace u_k \rbrace \in F_M ,\, \lbrace x_k \rbrace \subset {{\mathbb {R}}}^n$, such that

$$\begin{aligned} P_M - \frac{1}{k} \le P(u_k, x_k) \le P_M. \end{aligned}$$

(A.7)

Set

$$\begin{aligned} v_k (x):= u_k (x+ x_k), \end{aligned}$$

(A.8)

and note that $v_k \in F_M$ and

$$\begin{aligned} P_M - \frac{1}{k} \le P(v_k, 0) \le P_M \end{aligned}$$

(A.9)

By (A.5),

$$\begin{aligned} | v_k |_{C^{\beta }( {{\mathbb {R}}}^n ; {{\mathbb {R}}}^m)} \le C(M) \end{aligned}$$

hence by Arzela–Ascoli for a subsequence

$$\begin{aligned} v_k \xrightarrow {{C^{\beta }}} v ,\, \text {on compacts} \end{aligned}$$

(A.10)

We have

$$\begin{aligned} P(v; x) \le P_M = P(v; 0) > 0 ,\, x \in {{\mathbb {R}}}^n \end{aligned}$$

(A.11)

By the continuity of v there is $R >0$ such that

$$\begin{aligned}&\frac{P_M}{2} \le P(v; x) \le P_M ,\, x \in B(0;R) \end{aligned}$$

(A.12)

$$\begin{aligned}&P(v_k ;x) = \langle v_k(x) -a_2, e \rangle \ge \frac{P_M}{4} ,\, \text {on} \,\, B(0;R) \end{aligned}$$

(A.13)

for k large.

Thus $v_k (x)$ uniformly away from $a_1,\ldots ,a_m,a_{m+1}$, we have

$$\begin{aligned} \Delta v_k - W_u (v_k) = 0 ,\, \text {in} \,\, B(0;R) \end{aligned}$$

(A.14)

classically, since $W_u(u) \in C^1$ away from $a_1,\ldots ,a_m,a_{m+1}$ and $x \mapsto W_u(v_k(x))$ Holder by (A.10), thus $u \in C^{2+ \beta } (B(0;R))$.

We now calculate:

$$\begin{aligned} \Delta P&= \langle \Delta v, e \rangle = \langle W_u(u),e \rangle \\ \frac{\partial }{\partial v_j}W(v)&= \frac{\partial }{\partial v_j}\left( {\prod }_{ \nu =1}^{m+1} | v-a_{\nu } |^{{\alpha }_{\nu }} \right) = \sum _{i=1}^{m+1} \frac{\partial }{\partial v_j} \left( | v-a_i|^{{\alpha }_i}\right) {\prod }_{ \nu \ne i} | v-a_{\nu }|^{ {\alpha }_{\nu }} \end{aligned}$$

Notice that

$$\begin{aligned} \frac{\partial }{\partial v_j}\big (| v-a_i |^2\big )^{\frac{{\alpha }_i}{2}} = {\alpha }_i | v-a_i |^{{\alpha }_i -2} \cdot (v_j - a_i^j) \end{aligned}$$

where $a_i= \big (a_i^1,\ldots ,a_i^m\big )$

Hence

$$\begin{aligned} \begin{aligned} W_v(v)&= {\nabla }_v W(v) = \sum _{i=1}^{m+1} a_i \big (| v - a_i |^{ {\alpha }_i -2}\big ) (v-a_i) \prod _{ \nu \ne i} | v -a_{\nu } |^{{\alpha }_{\nu }} \\&= {\alpha }_2 | v- a_2 |^{{\alpha }_2 -2} (v-a_2) \prod _{\nu \ne 2} | v-a_{\nu } |^{{\alpha }_{\nu }} + \sum _{i \ne 2} {\alpha }_i | v-a_i |^{{\alpha }_i -2} (v-a_i) \prod _{\nu \ne i} | v- a_{\nu } |^{{\alpha }_{\nu }}. \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} \Delta P&= {\alpha }_2 | v- a_2 |^{{\alpha }_2 -2} \prod _{\nu \ne 2} | v-a_{\nu } |^{{\alpha }_{\nu }} \langle v-a_2 , e \rangle \\&\quad + \sum _{i \ne 2} {\alpha }_i | v-a_i |^{{\alpha }_i -2} \langle v-a_i, e\rangle \prod _{\nu \ne i} | v- a_{\nu } |^{{\alpha }_{\nu }} \end{aligned} \end{aligned}$$

Note that by the contradiction hypothesis, $\langle v(x) -a_i ,e \rangle >0$ (think of $a_2$ as the origin).

Hence $\Delta P >0$ on B(0; R) contradicting that P(v; x) takes its maximum at $x=0 .$

2. $\overline{\alpha } = 0$ For $W(u) = W^0 (u) := {\chi }_{\lbrace u \in S_A \rbrace }$, the proof proceeds similarly. The difference here is that $\Delta P = 0$, in B(0; R) which also leads to a contradiction by the maximum principle since P(v; x) takes its maximum at $x=0 .$

3. $\alpha _k \in [1,2],\, k=1,\ldots ,m$. In this case we define

$$\begin{aligned} F_M := \lbrace u : {{\mathbb {R}}}^n \rightarrow {{\mathbb {R}}}^m \, ,\, \Delta u - W_u(u) =0 ,\, | u(x) | \le M \rbrace \end{aligned}$$

u a weak $W^{1,2}$ solution. By linear elliptic theory we have the estimate (A.5). The rest of the argument is as before. The proof of the proposition is complete. $\square$

The Free Boundary

We follow closely the formal derivation from [1] p.140. We imbed the minimizer in a class of variations, $u(\tau ):=u(\cdot ,\tau )$, with u(0) corresponding to the minimizer, $u(\tau )=u(0)$ outside a ball B centered at some $x_0$ and quite arbitrary otherwise.

Let

$$\begin{aligned} U(\tau ):=\{ |u(\cdot ,\tau )-a|>0\} \end{aligned}$$

(B.1)

for

$$\begin{aligned} a\in \{W=0\},\, u(\tau )=a\text { on }\partial U(\tau ) \end{aligned}$$

Set

$$\begin{aligned} \lambda (\tau ):=\frac{1}{2}\int _{U(\tau )}|\nabla u(\tau )|^2\,dx \, , \, \mu (\tau ):=\int _{U(\tau )} W(u(\tau ))\,dx \end{aligned}$$

(B.2)

We denote $V:=\frac{\partial X}{\partial \tau }$ where $X(s,\tau )$ is a parametrisation of $\partial U(\tau ), \,s\in \Omega \subset {\mathbb {R}}^{n-1}$.

Then we have:

$$\begin{aligned} {\dot{\lambda }}(\tau )=&\int _{U(\tau )} \nabla u(\tau )\nabla u_\tau (\tau )\,dx+\dfrac{1}{2}\int _{\partial U(\tau )}|\nabla u(\tau )|^2 V\cdot \nu dS\nonumber \\ =&\int _{U(\tau )} -\Delta u(\tau ) u_\tau (\tau )\,dx+\int _{\partial U(\tau )}\frac{\partial u}{\partial \nu }\cdot u_\tau \,dS+\frac{1}{2} \int _{\partial U(\tau )} |\nabla u(\tau )|^2 V\cdot \nu \,dS \end{aligned}$$

(B.3)

where $\nu$ is the unit outward normal to $\partial U( \tau )$ (pointing outside $U(\tau )$).

Now from $u(X(s,\tau ),\tau )=a$ we obtain:

$$\begin{aligned} 0=&\frac{\partial }{\partial \tau }[u(X(s,\tau ),\tau )]=\frac{\partial u}{\partial \tau }+\frac{\partial u}{\partial \nu }\frac{\partial X}{\partial \tau }\cdot \nu \nonumber \\ =&u_\tau +\frac{\partial u}{\partial \nu } V\cdot \nu \end{aligned}$$

(B.4)

Hence

$$\begin{aligned} u_\tau \cdot \frac{\partial u}{\partial \nu }=-|\frac{\partial u}{\partial \nu }|^2 V\cdot \nu \end{aligned}$$

(B.5)

Then from (B.3) and (B.5) and the equation $\Delta u = W_u(u)$ we get:

$$\begin{aligned} {\dot{\lambda }}(0)=\int _{U(0)} -W_u(u(0))u_\tau (0)\,dx-\frac{1}{2}\int _{\partial U(0)}|\nabla u(0)|^2 V\cdot \nu dS. \end{aligned}$$

(B.6)

On the other hand

$$\begin{aligned} {\dot{\mu }}(\tau )=\int _{\partial U(\tau )} W(u(\tau )) V\cdot \nu dS+\int _{U(\tau )} W_u(u(\tau ))u_{\tau }(\tau ) \,dx \end{aligned}$$

(B.7)

Here for $0<\alpha <2$ utilizing that $W(u(0))=0$ on $\partial U(0)$ we get:

$$\begin{aligned} 0=\,&{\dot{\mu }}(0)+{\dot{\lambda }} (0)\nonumber \\ =\,&-\frac{1}{2} \int _{\partial U(0)} |\nabla u(0)|^2 V\cdot \nu \,dS \end{aligned}$$

(B.8)

and since V is arbitrary

$$\begin{aligned} |\nabla u(0)|=0\,\text { on }\partial U(0)\text { for }\alpha \in (0,2). \end{aligned}$$

(B.9)

(we note that $u\in C^{1,\beta -1}$,$\beta =\frac{2}{2-\alpha }$ by [8]).

Now, for $\alpha =0$ we have $W(u(0))=1$ on $\partial U(0)$ and

$$\begin{aligned} 0=\,&{\dot{\mu }}(0)+{\dot{\lambda }} (0)\nonumber \\ =\,&\int _{\partial U(0)} V\cdot \nu \,dS-\frac{1}{2} \int _{\partial U(0)} |\nabla u(0)|^2 V\cdot \nu \,dS \end{aligned}$$

(B.10)

hence $\frac{1}{2}|\nabla _+ u(0)|^2=1$ (u is only Lipschitz, $\nabla _+$ is the one-sided gradient).