Radial and cylindrical symmetry of solutions to the Cahn–Hilliard equation

Abstract

The paper is devoted to the classification of entire solutions to the Cahn–Hilliard equation $-\Delta u=u-u^3-\delta$ in $\mathbb {R}^N$, with particular interest in those solutions whose nodal set is either bounded or contained in a cylinder. The aim is to prove either radial or cylindrical symmetry, under suitable hypothesis.

1 Introduction

We consider the entire equation

$$\begin{aligned} -\Delta u=f(u)-\delta&\text {in }\mathbb {R}^N, \end{aligned}$$

(1.1)

with $f(u):=u-u^3$ and $\delta \in \mathbb {R}$. This equation has a variational characterisation, indeed, if we consider it on a domain $\Omega \subset \mathbb {R}^N$, it arises as the Euler equation of the Ginzburg–Landau functional

$$\begin{aligned} E(u,\Omega ):=\int _\Omega \bigg (\frac{1}{2}|\nabla u|^2+W(u)\bigg ) dx, \quad W(t):=\frac{(1-t^2)^2}{4}, \end{aligned}$$

(1.2)

under the mass constraint

$$\begin{aligned} \frac{1}{|\Omega |}\int _\Omega u dx=m, \quad m\in (-1,1), \end{aligned}$$

(1.3)

which gives rise to the Lagrange multiplier $\delta$. The interest in the minimisers u of $E(\cdot p,\Omega )$ arises from the phase transitions theory. In other words, if two different fluids are mixed in a container $\Omega$, the number u(x) represents the density of one of the two at x, in an equilibrium configuration. Here we take $\delta \in (-\frac{2}{3\sqrt{3}},\frac{2}{3\sqrt{3}})$, so that the polynomial $f_\delta (t):=t-t^3-\delta$ admits exactly 3 real roots

$$\begin{aligned} z_1(\delta )<-1/\sqrt{3}<z_2(\delta )<1/\sqrt{3}<z_3(\delta ), \end{aligned}$$

with $z_2(\delta )$ satisfying $\delta z_2(\delta )\ge 0$. The main results of the paper deal with symmetry properties of entire solutions to the Cahn–Hilliard equation (1.1).

Theorem 1

Let $N\ge 2$, $\delta \in (-\frac{2}{3\sqrt{3}},\frac{2}{3\sqrt{3}})$ and let $u_\delta$ be a solution to (1.1) such that

$$\begin{aligned} u_\delta >z_2(\delta ) \quad \text {outside a ball } B_R\subset \mathbb {R}^N. \end{aligned}$$

(1.4)

• (1) If $\delta \in (-\frac{2}{3\sqrt{3}},0]$, then $u\equiv z_3(\delta )$.

• (2) If $\delta \in (0,\frac{2}{3\sqrt{3}})$, then $u_\delta$ is radially symmetric (not necessarily constant).

We note that, for $\delta >0$, nontrivial bubble solutions are known to exist. This is an important difference with the case $\delta \le 0$. Moreover, we will see that the zero level set of radial solutions is non empty. In particular, we have the following Corollary.

Corollary 2

Let $\delta \in (0,\frac{2}{3\sqrt{3}})$ and let $u_\delta$ be a non constant solution to (1.1) such that $u_\delta >z_2(\delta )$ outside a ball $B_R$. Then the nodal set of $u_\delta$ is a sphere.

This result agrees with the variational theory, which studies the asymptotic behaviour of the scaled functionals

$$\begin{aligned} E_\varepsilon (u,\Omega )=\int _\Omega \bigg (\frac{\varepsilon }{2}|\nabla u|^2+\frac{W(u)}{\varepsilon }\bigg ) dx \end{aligned}$$

(1.5)

as $\varepsilon \rightarrow 0$. For instance, Modica proved that, if $\varepsilon _k$ is a sequence of positive numbers tending to 0 and $u_{\varepsilon _k}$ is a sequence of minimisers of $E_{\varepsilon _k}(\cdot p,\Omega )$ under the constraint (1.3) such that $u_{\varepsilon _k}\rightarrow u_0$ in $L^1(\Omega )$, then $u_0(x)\in \{\pm 1\}$ for almost every $x\in \Omega$, and the boundary in $\Omega$ of the set $E:=\{x\in \Omega :\, u_0(x)=1\}$ has minimal perimeter among all subsets $F\subset \Omega$ such that $|F|=|E|$, where $|\cdot p|$ denotes the volume (see [15], Theorem 1). Further $\Gamma$-convergence results relating $E_\varepsilon (\cdot p,\Omega )$ to the perimeter can be found in [16]. Therefore, given a family $\{u_\varepsilon \}_{\varepsilon \in (0,\varepsilon _0)}$ of minimisers under the constraint (1.3), their nodal set is expected to be close to a compact Alexandrov-embedded constant mean curvature surface, at least for $\varepsilon$ small. Corollary 2, together with a scaling argument, shows that, for $\varepsilon$ small enough, the nodal set of any entire solution to

$$\begin{aligned} -\varepsilon \Delta u=\varepsilon ^{-1}(u-u^3)-\ell ,\quad \ell >0, \end{aligned}$$

(1.6)

in $\mathbb {R}^N$ such that $u>z_2(\varepsilon \ell )$ outside a ball is actually a sphere, which is known to be the unique compact Alexandrov-embedded constant mean surface in $\mathbb {R}^N$ (see [1]).

After that, we set

$$\begin{aligned} C_R:=\{(x',x_N)\in \mathbb {R}^N:\, |x'|<R\} \end{aligned}$$

and we consider solutions satisfying

$$\begin{aligned} u_\delta >z_2(\delta ) \quad \text {outside a cylinder } C_R\subset \mathbb {R}^N. \end{aligned}$$

(1.7)

The aim is to study their symmetry properties and their asymptotic behaviour as $\delta \rightarrow 0$, with particular interest in solutions which have one periodicity direction.

Theorem 3

Let $\{u_\delta \}_{\delta \in (0,\frac{2}{3\sqrt{3}})}$ be a family of non constant solutions to (1.1) in $\mathbb {R}^N$, with $N\ge 2$. Assume furthermore that $u_\delta$ is periodic in $x_N$ and, for any $\delta \in (0,\frac{2}{3\sqrt{3}})$, there exists $R(\delta )>0$ such that (1.7) is true. Then

• (1) $z_1(\delta )<u_\delta (x)<z_3(\delta )$, for any $x\in \mathbb {R}^N$.

• (2) $u_\delta$ is radially symmetric in $x'$.

• (3) $u_\delta \rightarrow -1$ as $\delta \rightarrow 0$ uniformly on compact subsets of $\mathbb {R}^N$.

In view of the aforementioned $\Gamma$-convergence results, given a solution u to (1.6) satisfying (1.7), with $\delta =\varepsilon \ell$, we expect its nodal set to be close to an Alexandrov-embedded constant mean curvature surface which is contained in a cylinder. This kind of surfaces are fully classified, at least the ones which are embedded in $\mathbb {R}^3$, in fact it is known that the unique examples are the sphere and Delaunay unduloids, that is a family of non compact revolution surfaces obtained by rotating a periodic curve around a fixed axis in $\mathbb {R}^3$, which can be taken to be the $x_3$-axis, parametrised by a real number $\tau \in (0,1)$. We will denote the period of $D_\tau$ by $T_\tau$. For a detailed introduction of Delaunay surfaces, we refer to [12, 14]. For any $\tau \in (0,1)$, Kowalczyk and Hernandez [11] constructed a family $\{u_{\tau ,\varepsilon }\}_{\varepsilon \in (0,\varepsilon _0)}$ of solutions to (1.6) in $\mathbb {R}^3$, with $\ell =\ell _\varepsilon$ depending on $\varepsilon$, such that

• (1) $\ell _\varepsilon$ is positive and bounded uniformly in $\varepsilon$.

• (2) $u_{\tau ,\varepsilon }$ is radially symmetric in $x'$.

• (3) $u_{\tau ,\varepsilon }(x)\rightarrow \pm 1$ as $\varepsilon \rightarrow 0$, uniformly on compact subsets of $\Omega _\tau ^\pm$, where $\Omega _\tau ^\pm$ denote the exterior and the interior of the Delaunay surface $D_\tau$, respectively.

• (4) $u_{\tau ,\varepsilon }(x',x_3)\rightarrow z_3(\varepsilon \ell _\varepsilon )$ as $|x'|\rightarrow \infty$, uniformly in $x_3$.

• (5) $u_{\tau ,\varepsilon }$ is periodic in $x_3$ of period $T_\tau$.

We observe that the solutions $u_{\varepsilon ,\tau }$ constructed in [11] are actually negative outside a cylinder, however, in order to obtain the aforementioned family, thanks to the oddness of f, it is enough to replace them with $-u_{\varepsilon ,\tau }$. An interesting question is uniqueness. In other words, we are interested in the following question.

Question 4

(Uniqueness) Let $\varepsilon _0>0$, $\tau \in (0,1)$ and let v be a non constant solution to (1.6) in $\mathbb {R}^3$ with $\ell =\ell _\varepsilon$, for $\varepsilon \in (0,\varepsilon _0)$. Assume in addition that

• $\ell _\varepsilon$ is bounded uniformly in $\varepsilon$.

• v is periodic in $x_3$, with period $T_\tau$.

• $v>z_2(\varepsilon \ell _\varepsilon )$ outside a ball $B_R$.

Is it true that $v=u_{\varepsilon ,\tau }$, at least if $\varepsilon _0$ is small enough?

This would be the counterpart of Corollary 2 for periodic solutions. For now we are not able to give a full answer to this question. However Theorem 3 is a first step in this direction, since it proves that any family $\{v_\varepsilon \}_{\varepsilon \in (0,\varepsilon _0)}$ of such solutions has to share many properties with the family $\{u_{\tau ,\varepsilon }\}_{\varepsilon \in (0,\varepsilon _0)}$ constructed by Hernandez and Kowalczyk. For instance, for $\varepsilon$ small, $v_\varepsilon$ has to satisfy (1), (2), (3) and the scaled functions $v_\varepsilon (\varepsilon x)$ tend to $-1$ uniformly on compact subsets of $\mathbb {R}^N$ as $\varepsilon \rightarrow 0$.

The plan of the paper is the following. In Sect. 2 we will state some quite general results, of which the Theorems stated in the introduction are consequences. Section 3 is devoted to the proofs. It is divided into three subsections, dedicated to prove global boundedness, radial symmetry and the asymptotic behaviour for $\delta$ small respectively.

2 Some relevant results

In this section we state some results that are proved in Sect. 3. First we prove boundedness of solutions, which holds irrespectively of the sign of $\delta$.

Proposition 5

Let $\delta \in (-\frac{2}{3\sqrt{3}},\frac{2}{3\sqrt{3}})$ and let $u_\delta \in L^3_{loc}(\mathbb {R}^N)$ be a distributional solution to the Cahn–Hilliard equation (1.1). Then

$$\begin{aligned} z_1(\delta )\le u_\delta (x)\le z_3(\delta ) \end{aligned}$$

a. e. in $\mathbb {R}^N$.

Remark 6

• Using Proposition 5, standard elliptic estimates (see [10], Theorem 8.8 and Corollary 6.3) and a bootstrap argument, it is possible to show that any distributional solution $u\in L^3_{loc}(\mathbb {R}^3)$ is actually in $C^\infty (\mathbb {R}^N)$. This parallels the regularity result proved in [6] for the Allen–Cahn equation.

• It follows from the strong maximum principle that either $u_\delta$ is constant, and in this case it has to be either $z_1(\delta )$, or $z_2(\delta )$ or $z_3(\delta )$, or it satisfies $z_1(\delta )<u_\delta <z_3(\delta )$ in $\mathbb {R}^N$.

We observe that Proposition 5 and Remark 6 prove point (1) of Theorem 3, which is actually true for any non constant entire solution. After that, we rule out the case $\delta \le 0$, in which only constant solutions are allowed.

Proposition 7

Let $u_\delta$ be a solution to (1.1) in $\mathbb {R}^N$, with $-\frac{2}{3\sqrt{3}}<\delta \le 0$ such that $u_\delta >z_2(\delta )$ outside a stripe $\{x\in \mathbb {R}^N:\,|x_N|<L\}$. Then $u_\delta \equiv z_3(\delta )$.

We stress that the latter result proves point (1) of Theorem 1 and agrees with the sign of $\delta$ obtained by Hernández and Kowalczyk in [11]. Using boundedness and the famous result by Gidas et al. [9], or Theorem 2 of [7], which relies on the moving planes method, we can prove this symmetry result.

Proposition 8

Let $\delta \in (0,2/3\sqrt{3})$ and let $u_\delta$ be a non constant solution to (1.1) such that $u_\delta >z_2(\delta )$ outside a ball $B_R$, for some $R>0$. Then

• $u_\delta$ is radially symmetric, that is, up to a translation, $u_\delta (x)=w_\delta (|x|)$.

• $u_\delta$ is radially increasing, in the sense that $(\nabla u_\delta (x),x)> 0$, for any $x\in \mathbb {R}^N\backslash \{0\}$.

Proposition 8 proves point (2) of Theorem 1. More precisely, it is known that, for $\delta \in (0,\frac{2}{3\sqrt{3}})$, the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta v_\delta =v_\delta -v_\delta ^3-\delta &{}\quad \text {in }\mathbb {R}^N\\ v_\delta (0)=\min _{\mathbb {R}^N} v_\delta , \, v_\delta <z_3(\delta ), \, v_\delta (x)\rightarrow z_3(\delta ) &{}\quad \text {as }|x|\rightarrow \infty \end{array}\right. } \end{aligned}$$

(2.1)

admits a unique solution which is radially symmetric (see [4, 17, 18]), that is $v_\delta (x)=w_\delta (|x|)$. In view of this fact, we can actually prove the following classification result.

Proposition 9

Let $\delta \in (0,\frac{2}{3\sqrt{3}})$ and let $u_\delta$ be a non constant solution to (1.1) such that $u_\delta >z_2(\delta )$ outside a ball $B_R$. Then, up to a translation, $u_\delta =v_\delta$.

In the sequel, we will use the notation $W_\delta (t):=W(t)+\delta t$.

Remark 10

It is possible to see that, for any $\delta \in (0,\frac{2}{3\sqrt{3}})$, there exists $R(\delta )>0$ such that $w_\delta (R(\delta ))=0$. In fact, the energy

$$\begin{aligned} E_\delta (r):=\frac{1}{2}(w'_\delta )^2-W_\delta (w_\delta ) \end{aligned}$$

is strictly decreasing, since

$$\begin{aligned} \frac{d}{dr}E(r)=w'_\delta (w''_\delta -W'_\delta (w_\delta ))=-\frac{N-1}{r}(w'_\delta )^2<0, \quad \forall \,r>0. \end{aligned}$$

Thus, using that, by Proposition 8, $v_\delta$ is decreasing,

$$\begin{aligned} -W_\delta (w_\delta (0))=E_\delta (0)>0, \end{aligned}$$

(2.2)

which yields that $w_\delta (0)<0$.

In particular, in view of Remark 10, which yields that the nodal set of $v_\delta$ is neither empty nor a singleton, Corollary 2 is true.

Considering solutions that are approaching a positive limit just with respect to $N-1$ variables, we can prove the following.

Proposition 11

Let $\delta \in (0,2/3\sqrt{3})$ and let $u_\delta$ be a non constant solution to (1.1) such that $u_\delta >z_2(\delta )$ outside a cylinder $C_R$, for some $R>0$. If $u_\delta$ is periodic in $x_N$, then

• $u_\delta$ is radially symmetric in $x'$, that is, up to a translation, $u_\delta (x)=w_\delta (|x'|,x_N)$.

• $u_\delta$ is radially increasing, in the sense that $(\nabla u_\delta (x), (x',0))> 0$, for any $x=(x',x_N)\in \mathbb {R}^N\backslash \{0\}$.

We note that this proves point (2) of Theorem 3. Even in this case, our result agrees with the construction of [11], where the authors prove the existence of a family of solutions fulfilling the symmetries of the Delaunay surface $D_\tau$, hence, in particular they are periodic in $x_N$, radially symmetric and radially increasing in $x'$. Here we show that any periodic solution has to be radially symmetric and radially increasing in $x'$. Finally, in order to prove point (3) of Theorem 3, we need the following result, which shows that the phase transition has to be complete.

Proposition 12

For any $\epsilon >0$ there exists $\delta _0\in (0,\frac{2}{3\sqrt{3}})$ such that, for any $\delta \in (0,\delta _0)$ and for any non constant solution $u_\delta$ to (1.1) satisfying $\sup _{\mathbb {R}^N}u_\delta =z_3(\delta )$, we have

$$\begin{aligned} \inf _{\mathbb {R}^N} u_\delta <-1+\epsilon . \end{aligned}$$

(2.3)

This result somehow parallels Lemma 2.5 of [8]. The proof relies on both the moving planes and the sliding method. For a detailed proof of point (3) of Theorem 3, we refer to Sect. 3.

3 The proofs

3.1 Boundedness

In order to prove boundedness for distributional solutions to (1.1), we will rely on a result proved by Brezis [2].

Lemma 13

(Brezis–Kato inequality) Let $p>1$ and assume that $v\in L^p_{loc}(\mathbb {R}^k)$ satisfies

$$\begin{aligned} -\Delta v+|v|^{p-1} v\le 0&\text {in }{\mathcal {D}}'(\mathbb {R}^N). \end{aligned}$$

Then $v\le 0$ a.e. in $\mathbb {R}^N$.

Now we prove Proposition 5.

Proof

Writing $-f_\delta (t)=(t-z_1(\delta ))(t-z_2(\delta ))(t-z_3(\delta ))$ and setting

$$\begin{aligned} \alpha&:=z_1(\delta )-z_2(\delta )<0,\\ \beta&:=z_3(\delta )-z_2(\delta )>0,\\ w&:=u_\delta -z_2(\delta ), \end{aligned}$$

we have

$$\begin{aligned} \Delta w=\Delta u_\delta =(u_\delta -z_1(\delta ))(u_\delta -z_2(\delta ))(u_\delta -z_3(\delta ))=w(w-\alpha )(w-\beta ), \end{aligned}$$

(3.1)

thus

$$\begin{aligned} \Delta (w-\beta )^+=\chi _{\{w>\beta \}}\Delta w=\chi _{\{w>\beta \}}w(w-\alpha )(w-\beta )\ge ((w-\beta )^+)^3, \end{aligned}$$

where $\chi _{\{w>\beta \}}$ denotes the characteristic function of the set $\{\in \mathbb {R}^N:\, w(x)>\beta \}$. By the Kato–Brezis inequality (see Lemma 13), we have $w\le \beta$. The same argument applied to $(\alpha -w)^+$ gives the lower bound $w\ge \alpha$. $\square$

Remark 14

A similar argument is used in [5] to prove boundedness for solutions to a class of vectorial equations of the form

$$\begin{aligned} \Delta u=u P'_n(|u|^2), \quad P_n(t):=\frac{1}{2}\Pi _{j=1}^n (t-k_j)^2, \end{aligned}$$

with $0<k_1<\cdots <k_n$. The scalar Allen–Cahn equation is included in this class. Here we prove that a similar result is true for a slightly different non linearity, due to the presence of $\delta$.

Now we can prove Proposition 7, using boundedness and a result of [6] where non-existence f ground states for some special non linearies is proved.

Proof

By Lemma 13, $z_1(\delta )\le u_\delta \le z_3(\delta )$, in particular, since $\delta \le 0$, $|z_1(\delta )|\le z_3(\delta )$, hence $|u_\delta |\le z_3(\delta )$. By Lemma 15, $u_\delta \rightarrow z_3(\delta )$ as $x_1\rightarrow \pm \infty$, the limit being uniform in $x'$. Moreover, setting $f_\delta (t):=f(t)-\delta$, we have

• $f_\delta (t)\ge 0$, $\forall \,t\in (0,z_3(\delta ))$,

• $f_\delta (t)+f_{-\delta }(t)=-2\delta \ge 0$, $\forall \,t\in (0,z_3(\delta ))$,

• $f_\delta (t)$ is non increasing in a left neighbourhood of $z_3(\delta )$.

Therefore, by Theorem 4.2 of [6], $u_\delta \equiv z_3(\delta )$. $\square$

3.2 Radial symmetry

The aim of this subsection is to prove Proposition 11. In order to do so, we need some decay at infinity of the solution. From now on, we denote the variables by $x:=(x_1,x'')\in \mathbb {R}\times \mathbb {R}^{N-1}$. For $\lambda \in \mathbb {R}$, we set

$$\begin{aligned} \Sigma _\lambda :=\{x\in \mathbb {R}^3:\, x_1<\lambda \}. \end{aligned}$$

(3.2)

This changing of notation is justified by the fact that several times this section $x_N$ is the periodicity variable, hence we are not allowed to start the moving planes in that direction.

Lemma 15

Let $u_\delta$ be a solution to (1.1). Assume furthermore that $u_\delta >z_2(\delta )$ in the half-space $\mathbb {R}^N\backslash \Sigma _\lambda$, for some $\lambda \in \mathbb {R}$. Then

$$\begin{aligned} u(x_1,x'')\rightarrow z_3(\delta ),&\text {as }x_1\rightarrow \infty ,\text { uniformly in }x''. \end{aligned}$$

(3.3)

Proof

The statement is trivial if $u_\delta$ is constant (see Remark 6), hence we can assume that it is non constant. We apply Lemma 2.3 of [6] to $w:=u_\delta -z_2(\delta )$ in the half space $\mathbb {R}^N\backslash \Sigma _\lambda$, where, by Lemma 13, $0<w<\beta$. This is possible since the non linearity $g(t):=-t(t-\alpha )(t-\beta )$ is positive in $(0,\beta )$ and $g'(0)>0$. We recall that the constants $\alpha$ and $\beta$ are defined in the Proof of Proposition 5. The conclusion is that

$$\begin{aligned} w(x_1,x'')\rightarrow \beta&\text {as }x_1\rightarrow \infty , \end{aligned}$$

and the limit is uniform in the other variables. $\square$

Using the fact that $f'(z_3(\delta ))<0$, we can actually prove a better result about the decay rate of $z_3(\delta )-u_\delta$.

Lemma 16

Let $u_\delta$ be a solution to (1.1) such that $u_\delta >z_2(\delta )$ in the half space $\mathbb {R}^N\backslash \Sigma _\lambda$, for some $\lambda \in \mathbb {R}$. Then, for any $\gamma \in (0,\sqrt{-f'(z_3(\delta ))})$, there exists a constant $C(\gamma )>0$, depending on $\gamma$, such that

$$\begin{aligned} 0<z_3(\delta )-u_\delta (x_1,x'')\le C(\gamma ) e^{-\gamma x_1}, \quad \forall \, x=(x_1,x'')\in \mathbb {R}^N\backslash \Sigma _\lambda . \end{aligned}$$

(3.4)

Proof

We compare the bounded function $v:=z_3(\delta )-u_\delta$ with the barrier $\mu e^{-\gamma x_1}$, for $\gamma \in (0,\sqrt{-f'(z_3(\delta ))})$, in the half-space $\mathbb {R}^N\backslash \Sigma _M$, with $M>0$ large enough. In fact, on $\partial (\mathbb {R}^N\backslash \Sigma _M)$, we have

$$\begin{aligned} v(x)\le \Vert v\Vert _{L^\infty (\mathbb {R}^N)} \le \mu e^{-\gamma M}, \end{aligned}$$

provided $\mu \ge \Vert v\Vert _{L^\infty (\mathbb {R}^N)}e^{\gamma M}$. Note that here we use the fact that $v\in L^\infty$, which is true by Lemma 13. Moreover, setting $h_\delta (v):=-f_\delta (z_3(\delta )-v)$, we have $h_\delta (0)=-f_\delta (z_3(\delta ))=0$ and $h'_\delta (0)=f'(z_3(\delta ))<0$, thus

$$\begin{aligned} (-\Delta +\gamma ^2)(v-\mu e^{\gamma x_1})=h_\delta (v)+\gamma ^2 v\le 0 \end{aligned}$$

in $\mathbb {R}^N\backslash \Sigma _M$ if M is large enough, since, by Lemma 15, $z_3(\delta )-v$ is decaying as $x_1\rightarrow \infty$, uniformly with respect to $x''$. Thus, by the maximum principle for possibly unbounded domains (see Lemma 2.1 of [3]), we conclude that (3.4) is true in $\mathbb {R}^N\backslash \Sigma _M$. Changing, if necessary, the constant $C(\gamma )$, the required inequality is fulfilled in the whole space. $\square$

Now we prove Proposition 8

Proof

By Proposition 5, $z_1(\delta )<u_\delta <z_3(\delta )$ and, by Remark 6, $u_\delta$ is smooth. By Lemma 15, it converges to $z_3(\delta )$ as $|x|\rightarrow \infty$, therefore, by the famous symmetry result by [9], or by Theorem 2 of [7], we conclude that $u_\delta$ is radially symmetric and radially decreasing. $\square$

Now we prove Proposition 9.

Proof

Since, by Proposition 8, $u_\delta$ is radially symmetric and radially decreasing, then, up to translation, we have $u_\delta (0)=\min _{\mathbb {R}^N} u_\delta$. Since, by Lemma 15, $u_\delta (x)\rightarrow z_3(\delta )$ as $|x|\rightarrow \infty$, then it solves (2.1), therefore, by uniqueness, $u_\delta =v_\delta$. $\square$

In order to prove Proposition 11, we need to apply Theorem 2 of [7], which we recall, for the reader’s convenience.

Theorem 17

([7]) Let $v>0$ be a bounded entire solution to

$$\begin{aligned} -\Delta v=g(v) \end{aligned}$$

in $\mathbb {R}^N$, with $g\in C^1(\mathbb {R})$ such that $g'(s)\le 0$ in $(0,\eta )$, for some $\eta >0$. Writing $x=(y,z)\in \mathbb {R}^M\times \mathbb {R}^{N-M}$, we assume that

• $v(y,z)\rightarrow 0$ as $|y|\rightarrow \infty$, uniformly in z.

• v is periodic in z.

Then v is radially symmetric in y, that is, up to a translation, $v(y,z)=w(|y|,z)$, and radially decreasing in y, that is $\partial _{y_j} v(y,z)<0$ for any $x=(y,z)\in \mathbb {R}^M\times \mathbb {R}^{N-M}$ with $y\ne 0$.

Proof

By Proposition 5, $z_1(\delta )<u_\delta <z_3(\delta )$ and, by Remark 6, $u_\delta$ is smooth. By Lemma 15, it converges to $z_3(\delta )$ as $|x'|\rightarrow \infty$, uniformly in $x_N$. Since $u_\delta$ is periodic, in order to conclude that it is radially symmetric in $x'$ and radially decreasing, it is enough to apply Theorem 17 to $v:=z_3(\delta )-u_\delta$. $\square$

3.3 The asymptotic behaviour for $\delta$ small

First we show that if a solution lies between $1/\sqrt{3}$ and $z_3(\delta )$, then it is constant. This is proved by the moving planes method.

Lemma 18

Let $\delta \in [0,2/3\sqrt{3})$ and let $u_\delta$ be a solution to (1.1) in $\mathbb {R}^N$ such that $u_\delta (x)\ge 1/\sqrt{3}$, for any $x\in \mathbb {R}^N$. Then $u_\delta \equiv z_3(\delta )$.

Proof

We set $v:=z_3(\delta )-u_\delta$. Setting, for any $\lambda \in \mathbb {R}$, $v_\lambda (x):=v(2\lambda -x_1,x'')$, we have

$$\begin{aligned} v-v_\lambda \ge 0\text { in }\Sigma _\lambda ,\text { for any }\lambda \in \mathbb {R}. \end{aligned}$$

(3.5)

In order to prove this fact, we assume by contradiction that there exists $\lambda \in \mathbb {R}$ such that the open set $\Omega _\lambda :=\{x\in \Sigma _\lambda :v-v_\lambda <0\}$ is nonempty, and we observe that, in any connected component $\omega$ of $\Omega _\lambda$ we have

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta (v-v_\lambda )=h_\delta (v)-h_\delta (v_\lambda )<0 &{}\quad \text {in }\omega ,\\ v-v_\lambda =0 &{}\quad \text {on }\partial \omega , \end{array}\right. } \end{aligned}$$

due to the strict monotonicity of $f_\delta$ in $[1/\sqrt{3},1)$ (for the definition of $h_\delta$, see the Proof of Lemma 16). As a consequence, by the maximum principle for possibly unbounded domains, we have $v-v_\lambda \le 0$ in $\omega$, a contradiction.

By (3.5), we have $\partial _{x_1}v\le 0$ in $\mathbb {R}^N$. The same argument applied to ${\tilde{v}}(x):=v(-x_1,x')$ implies that also ${\tilde{v}}$ satisfies (3.5), hence $\partial _{x_1}v\ge 0$ in $\mathbb {R}^N$, thus $\partial _{x_1}v\equiv 0$. Composing v with any rotation of $\mathbb {R}^N$, we conclude that v is a constant solution to (1.1), thus $v\equiv 0$. $\square$

Given the double well potential $W(t)=\frac{(1-t^2)^2}{4}$, $0<\alpha <W(1/\sqrt{3})=\frac{1}{9}$ and $\delta \in (0,2/3\sqrt{3})$, we set

$$\begin{aligned} \mu (\delta ):=\max \{\mu <0:\, W_\delta (\mu )=\alpha \}. \end{aligned}$$

Moreover, we take a smooth cutoff function $\chi :\mathbb {R}\rightarrow [0,1]$ such that $\chi =1$ in $(-\infty ,-1)$ and $\chi =0$ in $(0,\infty )$ and we set

$$\begin{aligned} {\tilde{W}}_\delta :=\chi _\delta \alpha +(1-\chi _\delta )W_\delta , \quad \chi _\delta (t):=\chi (-t/\mu (\delta )). \end{aligned}$$

(3.6)

We will denote ${\tilde{W}}:={\tilde{W}}_0$. It is possible to see that ${\tilde{W}}_\delta$ enjoys the following properties:

$$\begin{aligned}&{\tilde{W}}_\delta \rightarrow {\tilde{W}}, \quad \text {as }\delta \rightarrow 0, \text { uniformly on compact subsets of }\mathbb {R}, \end{aligned}$$

(3.7)

$$\begin{aligned}&{\tilde{W}}_\delta (r)=W_\delta (r), \quad \text {for any }r\ge 0\text { and } \delta \in [0,2/3\sqrt{3}), \end{aligned}$$

(3.8)

and

$$\begin{aligned} \inf _{(-\infty ,0]}{\tilde{W}}= \alpha . \end{aligned}$$

(3.9)

In the sequel, we will be interested in a solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta \beta _{R,\delta }+{\tilde{W}}_\delta (\beta _{R,\delta })=0 &{}\quad \text {in }B_R,\\ \beta _{R,\delta }=z_1(\delta ) &{}\quad \text {on }\partial B_R, \end{array}\right. } \end{aligned}$$

(3.10)

for $\delta \ge 0$ small enough and R large. This will be used as a barrier in the Proof of Proposition 12, which relies on a sliding method. This can be obtained in a variational technique, by minimising the functional

$$\begin{aligned} J_{R,\delta }(v):=\int _{B_R}\bigg (\frac{1}{2}|\nabla v|^2+{\tilde{W}}_\delta (v)\bigg )dx. \end{aligned}$$

(3.11)

among all $H^1(B_R)$ functions with trace $z_1(\delta )$ on $\partial B_R$. The case $\delta =0$ is treated in Lemma 2.4 of [8].

Lemma 19

Let $\delta _0>0$ be so small that $W_\delta (z_3(\delta ))<\alpha /2$, for any $\delta \in [0,\delta _0)$. Then, For any $R>0$ and $\delta \in [0,\delta _0)$, there exists a minimiser $\beta _{R,\delta }\in C^2(B_R)$ of (3.11) among all functions with trace $z_1(\delta )$ on $\partial B_R$. Moreover, there exists $R_0>0$ such that, for any $R\ge R_0$ and for any $\delta \in [0,\delta _0)$,

• $$\begin{aligned} z_1(\delta )<\beta _{R,\delta }(x)<z_3(\delta ), \quad \forall \, x\in B_R, \end{aligned}$$

  (3.12)
• $$\begin{aligned} \sup _{B_R}\beta _{R,\delta }>\frac{1}{\sqrt{3}}, \end{aligned}$$

  (3.13)

• there exists a solution $\beta _R$ of (3.10) with $\delta =0$ such that

  $$\begin{aligned} \sup _{B_R}\beta _{R,\delta }\rightarrow \sup _{B_R}\beta _R\in [\frac{1}{\sqrt{3}},1) \quad \text {as }\delta \rightarrow 0. \end{aligned}$$

  (3.14)

Proof

Existence follows from coercivity and weak lower semi continuity. By the fact that ${\tilde{W}}_\delta \equiv \alpha$ in $(-\infty ,\mu (\delta ))$ and (3.8), we can see the minimiser actually has to satisfy $z_1(\delta )\le \beta _{R,\delta }\le z_3(\delta )$, thus, due to the strong maximum principle, either (3.12) holds or $\beta _{R,\delta }\equiv z_1(\delta )$.

Now we prove (3.13), which, in particular, shows that $\beta _{R,\delta }> z_1(\delta )$ in $B_R$, at least for $R\ge R_0$. In order to do so, we assume that there exists a sequence $R_k\rightarrow \infty$ and a sequence $\delta _k\in [0,\delta _0)$ such that

$$\begin{aligned} \sup _{x\in \mathbb {R}^N}\beta _{R_k,\delta _k}\le \frac{1}{\sqrt{3}}. \end{aligned}$$

It follows that, on the one hand

$$\begin{aligned} J_{R_k,\delta _k}(\beta _{R_k,\delta _k})\ge \alpha \omega _N R_k^N, \end{aligned}$$

(3.15)

where $\omega _N$ denotes the surface of $S^{N-1}$. On the other hand, if, for $R>1$ and $\delta \in [0,\delta _0)$, we take $w_{R,\delta }$ to be equal to $z_1(\delta )$ on $\partial B_R$ and to $z_3(\delta )$ in $B_{R-1}$ with $|\nabla w_{R,\delta }|$ bounded uniformly in $\delta$, we can see that there exists a constant $C>0$ such that, for k large enough,

$$\begin{aligned} J_{R_k,\delta _k}(w_{R_k,\delta _k})\le C R_k^{N-1}+W_{\delta _k}(z_3(\delta _k))\omega _N R_k^N<\alpha \omega _N R_k^N, \end{aligned}$$

(3.16)

since $\delta _k\in [0,\delta _0)$, hence $W_{\delta _k}(z_3(\delta _k))<\alpha /2$. This contradicts the minimality of $\beta _{R_k,\delta _k}$.

Finally we prove (3.14). In the forthcoming argument, $R>0$ will always be arbitrary but fixed. We observe that, since $\beta _{R,\delta }$ is bounded uniformly in $R>0$ and $\delta >0$, then any sequence $\delta _k\rightarrow 0$ admits a subsequence, that we still denote by $\delta _k$, such that $\beta _{R,\delta _k}$ converges in $C^2(B_R)$ to a solution $\beta _R$ to

$$\begin{aligned} -\Delta \beta _R+{\tilde{W}}(\beta _R)=0\quad \text {in }B_R \end{aligned}$$

satisfying $\beta _R=-1$ on $\partial B_R$. Since the convergence is uniform and (3.12) holds, then

$$\begin{aligned} \sup _{B_R}\beta _{R,\delta }\rightarrow \sup _{B_R}\beta _R\in [-1,1] \end{aligned}$$

as $\delta \rightarrow 0$. Moreover, by (3.13) and the strong maximum principle, $\sup _{B_R}\beta _R\in [\frac{1}{\sqrt{3}},1)$. $\square$

Now we can prove Proposition 12.

Proof

It is enough to prove that, if there exists a sequence $\delta _k\rightarrow 0$, a sequence $u_{\delta _k}$ of solutions to (1.1) and $\nu >-1$ such that

$$\begin{aligned} \inf _{\mathbb {R}^N} u_{\delta _k}\ge \nu , \end{aligned}$$

(3.17)

then there exists a subsequence $\delta _{k'}$ such that $u_{\delta _{k'}}\equiv z_3(\delta _{k'})$.

Claim

For any $\varepsilon >0$ and $\rho >0$, there exists a subsequence, which we still denote by $u_{\delta _k}$, and a sequence $x^k\in \mathbb {R}^N$ such that

$$\begin{aligned} u_{\delta _k}(x)>1-\varepsilon , \quad \forall \,x\in B_\rho (x^k). \end{aligned}$$

(3.18)

Since $\sup _{\mathbb {R}^N}u_{\delta _k}=z_3(\delta _k)$, there exists $x^k\in \mathbb {R}^N$ such that

$$\begin{aligned} z_3(\delta _k)-u_{\delta _k}(x^k)<1/k. \end{aligned}$$

(3.19)

Therefore the sequence $u^k(x):=u_{\delta _k}(x+x^k)$ admits a subsequence converging, in $C^2_{loc}(\mathbb {R}^N)$, to a solution $u^\infty$ to the Allen–Cahn equation

$$\begin{aligned} -\Delta u^\infty =f(u^\infty ), \quad \text {in }\mathbb {R}^N. \end{aligned}$$

(3.20)

By (3.19), we can see that $u^\infty (0)=1$, thus $u^\infty \equiv 1$. As a consequence, for any $\varepsilon >0$ (small) and $\rho >0$, there exists a subsequence (still denoted by $u^k$) such that

$$\begin{aligned} \Vert u^k-1\Vert _{L^\infty (B_\rho )}<\varepsilon , \quad \forall k \end{aligned}$$

hence the claim is true.

In order to prove our result, we first observe that, by (3.13), for $\delta _0$ small as in Lemma 19 and $\delta \in (0,\delta _0)$, there exists $R>0$ and a solution $\beta _{R,\delta }$ to (3.10) such that

$$\begin{aligned} \sup _{B_R}\beta _{R,\delta }>\frac{1}{\sqrt{3}}, \quad \forall \,\delta \in (0,\delta _0). \end{aligned}$$

(3.21)

Moreover, by (3.14), there exists a solution $\beta _R$ to

$$\begin{aligned} -\Delta \beta _R+{\tilde{W}}(\beta _R)=0\text { in }B_R, \quad \beta _R=-1 \text { on }\partial B_R \end{aligned}$$

and $\delta _1=\delta _1(R)>0$ such that, for any $\delta \in (0,\delta _1)$, we have

$$\begin{aligned} \sup _{B_R}\beta _{R,\delta }<\frac{\sup _{B_R}\beta _R +1}{2}<1, \quad \forall \,\delta \in (0,\delta _1). \end{aligned}$$

(3.22)

As a consequence, for any $\delta \in (0,{\bar{\delta }})$, where ${\bar{\delta }}={\bar{\delta }}(R):=\min \{\delta _0,\delta _1(R)\}$, we get

$$\begin{aligned} \frac{1}{\sqrt{3}}<\sup _{B_R}\beta _{R,\delta }<\frac{\sup _{B_R}\beta _R +1}{2}<1. \end{aligned}$$

(3.23)

Now, applying the claim with $\rho =R$ and

$$\begin{aligned} \varepsilon :=1-\frac{\sup _{B_R}\beta _R +1}{2}, \end{aligned}$$

we can prove the existence of a subsequence, still denoted by $u_{\delta _k}$, and a sequence $x^k$ in $\mathbb {R}^N$ such that

$$\begin{aligned} u_{\delta _k}(x)>1-\varepsilon >\sup _{B_R}\beta _{R,\delta _k}\ge \beta _{R,\delta _k}(x-x^k), \quad \forall \, x\in B_R(x^k),\, \forall \, k. \end{aligned}$$

Sliding $\beta _{R,\delta _k}$, with $k\ge k_0$ fixed, we get the lower bound

$$\begin{aligned} u_{\delta _k}(x)> 1-\varepsilon > \frac{1}{\sqrt{3}}, \quad \forall \, x\in \mathbb {R}^N,\, \forall \, k\ge k_0. \end{aligned}$$

In conclusion, by Lemma 18, $u_{\delta _k}\equiv z_3(\delta _k)$. $\square$

Proposition 20

Let $\delta \in (0,2/3\sqrt{3})$ and let $\{u_\delta \}_{\delta \in (0,\frac{2}{3\sqrt{3}})}$ be a family of non constant solutions to (1.1) in $\mathbb {R}^N$ such that

• for any $\delta \in (0,2/3\sqrt{3})$ there exists $R(\delta )>0$ such that $u_\delta >z_2(\delta )$ outside the cylinder $C_{R(\delta )}$.

• $u_\delta$ is periodic in $x_N$.

Then

$$\begin{aligned} u_\delta \rightarrow -1 \quad \text {as }\delta \rightarrow 0,\text { uniformly on compact subsets of }\mathbb {R}^N. \end{aligned}$$

(3.24)

and

$$\begin{aligned} R(\delta )\rightarrow \infty \quad \text {as }\delta \rightarrow 0. \end{aligned}$$

(3.25)

Remark 21

We note that point (3) of Theorem 3 is a consequence of Proposition 20.

Proof

By Lemma 13, the family $u_\delta$ is uniformly bounded, hence any sequence $\delta _k\rightarrow 0$ admits a subsequence, that we still denote by $\delta _k$, such that $u_{\delta _k}$ converges in $C^2_{loc}(\mathbb {R}^N)$ to a solution $u^\infty$ to the Allen–Cahn equation (3.20). Since $u_\delta$ are all non constant solutions, then, by Proposition 12, we have

$$\begin{aligned} \inf _{\mathbb {R}^N}u_\delta \rightarrow -1, \quad \text {as }\delta \rightarrow 0. \end{aligned}$$

(3.26)

By periodicity and Theorem 11, we know that, for $\delta$ small, $u_\delta$ is radially symmetric in $x'$ and, up to a translation,

$$\begin{aligned} \inf _{\mathbb {R}^N}u_\delta =u_\delta (0), \end{aligned}$$

hence, passing to the limit, we get

$$\begin{aligned} u^\infty (0)=\lim _{k\rightarrow \infty }u_{\delta _k}(0)=\lim _{k\rightarrow \infty }\inf _{\mathbb {R}^N}u_{\delta _k}=-1, \end{aligned}$$

which yields that $u^\infty \equiv -1$, thus (3.24) holds.

In order to prove (3.25), we assume by contradiction that there exists ${\bar{R}}>0$ and a sequence $\delta _k\rightarrow 0$ such that $R(\delta _k)\le {\bar{R}}$. By (3.24), $u_{\delta _k}\rightarrow -1$ uniformly in $B^{N-1}_{2{\bar{R}}}\times [-1,1]$, thus, for k large enough,

$$\begin{aligned} u_{\delta _k}(x'_k,0)<-\frac{1}{2}<z_2(\delta _k) \end{aligned}$$

if, for instance, $x'_k=(2R(\delta _k),0)\in \mathbb {R}\times \mathbb {R}^{N-2}$, which contradicts the fact that $u_{\delta _k}$ is radially increasing. $\square$