On the Allen-Cahn equation in the Grushin plane: a monotone entire solution that is not one-dimensional

We consider solutions of the Allen-Cahn equation in the whole Grushin plane and we show that if they are monotone in the vertical direction, then they are stable and they satisfy a good energy estimate. However, they are not necessarily one-dimensional, as a counter-example shows.


Introduction
We consider here the Grushin plane G (see [11]), that is R 2 endowed with the vector fields X = ∂ ∂x and Y = x ∂ ∂y . We also define T := [X, Y ] = ∂ ∂y . The Grushin gradient is then ∇ G := (X, Y ) (with the coordinates taken in the (X, Y )frame) and the Grushin Laplacian is ∆ G := X 2 + Y 2 . We denote by < ·, · > the standard scalar product (when the vectors are taken in the (X, Y )-frame), so that, for a smooth function v we have for any ζ ∈ R 2 . Moreover, the Grushin norm on G is defined as (x, y) := 4 |x| 4 + 4|y| 2 for any (x, y) ∈ R 2 , and then the Grushin ball of radius R > 0 centered at ζ ∈ R 2 is B R (ζ) := {η ∈ R 2 s.t. η − ζ < R}.
The main purpose of this paper is to study solutions of the Allen-Cahn equation in the Grushin plane, that is ∆ G u(ζ) + f (u(ζ)) = 0 (1.1) for any ζ ∈ R 2 . We take, for simplicity, f ∈ C 1 , though less regularity is also possible to be dealt with.
A particular case of interest is when f = −W ′ and W is a double-well potential. Namely, through this paper, we denote by W a function with the following properties: W ∈ C 2 (R) is an even function for which W (±1) = 0 ≤ W (r) for any r ∈ R, W ′′ (0) = 0, W ′′ (±1) = 0, and such that Inspired by a famous conjecture of De Giorgi (see [6]), one may wonder under which conditions the solutions of (1.1) are one-dimensional, i.e., their level sets are straight lines and so, up to rotation, they depend on only one variable (at least when f = −W ′ ). Natural requirements for such symmetry are monotonicity and stability conditions. Namely, if u is a solution of (1.1), we say that u is stable if for any φ ∈ C ∞ 0 (R 2 ). Stability is a natural condition in the calculus of variation, since it states that the energy functional associated to (1.1) has non-negative second derivative. The stability condition has thus been widely used in connection with the problems posed by [6] (see, for instance, [1,7] and references therein). Also, in the Euclidean setting, the stability condition holds true whenever u is monotone in some direction. The analogy in the Grushin setting is somehow more delicate, since the space is not homogeneous with respect to the choice of a particular direction. Thus, the monotonicity studied in this paper is the following. We are mostly concerned with solutions that are monotone in the y-direction, that is for which (1.4) We shall show that (1.4) implies (1.3) (see Proposition 3.1 below).
Symmetry properties for solutions of (1.1) in the Grushin plane have been recently studied in [8]. For instance, [8] pointed out the following result: Suppose that there exists C o ≥ 1 in such a way that Then, u depends only on the x-variable.
We observe that (1.6) is also a sort of monotonicity condition, while (1.5) is an energy growth requirement (and energy bounds are often needed in the Euclidean case too, see [1] The main purpose of this paper is in fact to show that the above question has a negative answer. This will be accomplished in Theorem 5.2, by constructing a counter-example which follows the lines of the one in [4]. The paper is organized in the following way. After gathering some elementary observations in Section 2, we point out in Section 3 that the monotonicity condition in (1.4) implies the stability condition in (1.3). Then, we develop in Section 4 the energy estimates which show that the monotonicity condition in (1.4) also implies the energy growth in (1.5). Finally, Section 5 contains the construction of the counter-example which shows that Question 1.2 has a negative answer.

Preliminaries
We collect in this section some elementary, but useful, observations. The expert reader may surely skip this section.

An integration by parts
We now point out a variation of Green formula, complicated here by the non-homogeneous Grushin scaling.
Then, by changing variable, we have Thence, by the standard Euclidean Divergence Theorem, where "·" denotes the standard Euclidean scalar product and "ν E " is the standard Euclidean outward normal of ∂B 1 (0). We write (2.2) as and so, since the Euclidean norm of ν E is 1, where | · | E is the Euclidean norm. We now observe that, for any (X, Y ) ∈ ∂B 1 (0), we have |X| ≤ 1 and From this and (2.3) we get (2.1).

An interpolation inequality
We point out the following elementary estimate: Proof. We may assume that both h L ∞ (R) and h ′′ L ∞ (R) are finite, otherwise (2.4) is void. First, we observe that, for any j ∈ Z, there exists t j ∈ [j, j + 1] in such a way that We assume that the second possibility holds (the first case is analogous). Then, This contradiction proves (2.5). Then, making use of (2.5), given any j ∈ Z and any t ∈ [j, j + 1],

ODE analysis
The scope of this section is an elementary analysis of the solutions h ∈ C 2 (R) of Recall that for any any C 2 solution of (2.6) and any s, t ∈ R, for a suitable C > 0, possibly depending on h L ∞ (R) . Also, for any t ∈ R, Proof. By construction, |W ′ | and so, by (2.4), we get (2.8).
We take Let also t n be a sequence for which Let w n (t) := h(t + t n ). From (2.8), we have that w n converges, up to subsequence, in C 1 loc (R) to some function w ∈ C 1 (R).
We now suppose that σ = inf R h (for this argument, the case σ = sup R h is completely analogous). Then, for any t ∈ R, so w ′ (0) = 0 and therefore, by (2.7), for any t ∈ R, Lemma 2.5. If h has two or more critical points, then it is periodic.
for any t ∈ R, and so h has period 2T . Proof.
Proof. If h is constant but not zero, then either h is constantly equal to −1 or it is constantly equal to +1, because of (1.2). Since in such cases (2.11) is obvious, we focus on the case in which h is not periodic. Then, by Lemma 2.5, h has at most one critical point. (2.12) In particular, h attains either its sup or its inf at either +∞ or −∞. So, let us assume, for definiteness that sup the other cases being analogous. By (2.8), we obtain that the limit in (2.13) holds in C 1 , therefore Thus, by (1.2), we would have that h(t o ) = 0. That is, in contradiction with our assumptions.
Proof. By Lemma 2.5, we see that only two cases hold: either h ′ never vanish or h ′ has only one zero. In any case, there exists c ∈ R in such a way that h ′ (t) = 0 for any t ∈ (−∞, c) ∪ (c, +∞). Consequently, by Lemma 2.8, for any a < c < Then, the desired result follows by sending a → −∞ and b → +∞.

A compactness result
We now point out a useful compactness criterion: Lemma 2.11. Let u k be a sequence of solutions of (1.1) in the whole R 2 . Then, up to subsequence, u k converges locally uniformly to some u which is also a solution of (1.1) in the whole R 2 .
Moreover, for any a ∈ (0, 1), the Grushin operator ∆ G is uniformly elliptic in D a := {|x| ∈ (a, 1/a)}, therefore standard elliptic estimates give that for any x ∈ D a . Since a can be taken arbitrarily small, we have that (2.19) holds for any x ∈ R 2 \ {0}. But then, since the map x → W ′ (u(x)) is continuous, by means of (2.18), it follows that ∆ G u is continuous too and so (2.19) holds for any x ∈ R 2 .

Then, there exists λ ∈ R for which there exists a non-trivial solution φ of
Moreover, we can take for a suitable C > 0.

Extension of bounded harmonic functions
Then, it may be extended to a ∆ G -harmonic in B r .

Monotonicity and stability
We show that (1.4) is sufficient for stability. This is in analogy with the fact that monotonicity in any direction implies stability in the Euclidean setting (see [1]) -but in the Grushin plane the directions do not play the same role, thus (1.4) somehow selects the good direction for stability. a solution of (1.1) satisfying (1.4). Then, u is stable.
Proof. The argument we present here is a modification of a classical one (see [1] and also Section 7 in [7] for a general result). We recall that we need to prove that for any smooth φ, compactly supported For any ϕ smooth and compactly supported, we have Therefore, by taking ϕ := φ 2 /(T u), and making use of the Cauchy-Schwarz inequality,

Energy estimates
We follow here some ideas of [1] to estimate the energy For this, for any t ∈ R, we define the translation u t (x, y) := u(x, y + t) and the translated energy Of course, E R (0) = F R (u).
Then, there exists a structural constant C in such a way that for any t ∈ R and any R > 0.
Proof. We prove (4.1) for t > 0 (this is enough, since u(x, −y) is also a solution).
Assume that (1.4) holds true. Then, there exists a structural constant C in such a way that for any R > 0. As a consequence, (1.5) holds true.
Proof. We have that u is bounded and monotone in y, thanks to (1.4). Thus, we may define u ± (x) := lim y→±∞ u(x, y). Then, from Lemma 2.11, we have that In fact, since u does not depend on y, we may write (4.3) as and so we may apply to u ± the ODE analysis developed in Section 2.3. For this, we observe that at least one between u + and u − is either constant or non-periodic. (4.5) To prove (4.5), we argue by contradiction, supposing that u + and u − are both periodic and non-constant. In particular, by Cauchy Uniqueness Theorem, |u ± | < 1 and then, by Lemma 2.6, we would have that But from (1.4), we know that and so, if we set x ± min , x ± max be such that we deduce from (4.6) and (4.7) that This contradiction proves (4.5). We now claim that either u + or u − is non-periodic or constant but not zero. (4.8) To prove this, we argue by contradiction. Suppose (4.8) is false. Then, both u − and u + are periodic. Then, at least one, say u + is constant, because of (4.5). If u + were not equal to zero, then (4.8) would be true, thus we have to say that u + is constantly equal to zero and that u − is periodic. But then u − cannot be constant, otherwise (4.4), ( This would say that u − is constant, while we know it is not the case. This contradiction proves (4.8).
By means of (4.8), up to a sign change, we may suppose that u + is either constant but not zero or it is non-periodic. Consequently, by Lemma 2.10, for a suitable C + > 0, with In fact, (2.11) and (4.9) give that Moreover, by (4.1), Thus, by (4.10), 5 The counter-example
That is, Ω is T -convex when vertical segments joining two points of Ω lie in Ω.
The proof of this result is done through the sliding method introduced in [2] for uniformly elliptic equations. This method uses two fundamental ingredients: the Maximum Principle in small domains and the invariance of the operator with respect to "sliding". In [5] the equivalent of Theorem 5.1 was proved for sub-elliptic equations in nilpotent Lie groups. There, the key "new" ingredient being a Hölder estimate for Hörmander type operators proved in [13] that allowed to prove the Maximum Principle in small domains. The operator is invariant by T s translations and our equation satisfies the hypotheses of [13], hence the proof of Theorem 5.1 proceeds exactly like the one given in [5], and we omit it.

Existence of monotone solutions that are not one-dimensional
The following result shows that Question 1.2 has a negative answer: There exists a solution of in R 2 such that T u = ∂ y u > 0. Also, such u is not one-dimensional.
Proof. We follow the two steps of [4].

Construction of a monotone solution in a bounded set.
Let M > 0 be greater then the Lipschitz constant of f , let g(u) := f (u) + Mu, . We consider the operator T on C α such that T v = u is the classical solution of where 0 ≤ ψ ≤ 1 with ψ(0) = 0 and ψ(R 2 ) = 1.
This is just the Maximum Principle, because with our choice of M we get that We remark that we can take such a λ o in the light of Lemma 2.12. Let ϕ o be the corresponding eigenfunction normalized by sup ϕ o = 1. Our choice of λ o implies that there exists ε > 0 such that Observe that, in B, the Grushin operator ∆ G is uniformly elliptic, and so by standard estimates we know that v ∈ C α (Q + R ). Using (5.5), (P2) and (P3), we get that So, iteratively, T k (v o ) ≥ v o for any k ∈ N. This proves (P4). By Lemma 2.11, we may and do suppose that u k := T k (v o ) converges to a solutioñ u =ũ R of ∆ Gũ − W ′ (ũ) = 0 in Q + R u(x, 0) = 0,ũ(x, R 2 ) = 1ũ(−R, y) = ψ(y),ũ(R, y) = ψ(y) Note thatũ satisfies 0 ≤ũ ≤ 1. Hence, using Theorem 5.1, we know that Finally, we extend the solution to Q R = Q + R ∪ Q − R by taking v R (x, y) := ũ(x, y) for (x, y) ∈ Q + R −ũ(x, −y) for (x, y) ∈ Q − R Clearly, v R is a solution in Q + R ∪ Q − R . Also the solution u is C 2 up to the boundary for x = 0. Hence we get that v R is a solution in Q R \ {(0, 0)}. To check that v R is a solution in all of Q R . Observe that the map ζ → W ′ (v R (ζ)) is in C α (Q R ), hence there exists w ∈ C 2,α loc (Q R ) ∩ C(Q R ) solution of Then w − v R is ∆ G -harmonic in Q R \ {(0, 0)} and it is bounded. Thus, by Lemma 2.13, it is ∆ G -harmonic in Q R , and so v R is a solution in all of Q R .
Furthermore v R is monotone in T , in the sense that T v R = ∂ y v R > 0, because of (5.7).
Step 2. Let R → ∞. Then, by Lemma 2.11, v R locally uniformly converges to some u, which is a solution of ∆ G u − W ′ (u) = 0 in R 2 .
Furthermore, in Q + R , v R =ũ = lim k→+∞ u k ≥ v 0 , due to (P4) and so, by (5.6), u ≡ 0 in Q + R . Then, u is monotone i.e. ∂ y u > 0, and it is therefore the counter-example we are looking for. Indeed, u is not one-dimensional; suppose, by contradiction, that there exists a function g such that u(x, y) = g(ax + by), for any (x, y) ∈ R 2 . Then, the strict monotonicity in T of u implies that b = 0. (5.8) Clearly g would be a solution of (a 2 + b 2 x 2 )g ′′ (ax + by) − W ′ (g(ax + by)) = 0, (5.9) for any (x, y) ∈ R 2 . This implies that for any t along the lines ax + by = t, (a 2 + b 2 x 2 )g ′′ (t) − W ′ (g(t)) = 0.
Since b = 0, this implies that g ′′ ≡ 0. Hence W ′ (g(t)) = 0 for any t and so g would be constant, in contradiction with the fact that T u > 0.