On the Gibbons' conjecture for equations involving the $p$-Laplacian

In this paper we prove the validity of Gibbons' conjecture for the quasilinear elliptic equation $ -\Delta_p u = f(u) $ on $\mathbb{R}^N.$ The result holds true for $(2N+2)/(N+2)<p<2$ and for a very general class of nonlinearity $f$.


Introduction
In this work we are concerned with the study of qualitative properties of weak solutions of class C 1 to the quasilinear elliptic equation where we denote a generic point of R N by (x , y) with x = (x 1 , x 2 , . . . , x N −1 ) ∈ R N −1 and y = x N ∈ R, p > 1 and N > 1. The nonlinear function f will be assumed to satisfy the following assumptions : A very special case covered by our assumptions is the well-known semilinear Allen-Cahn equation for which the following conjecture have been stated Gibbons' conjecture [5] Assume N > 1 and consider a bounded solution u ∈ C 2 (R N ) of (1.1) such that lim uniformly with respect to x . Then, is it true that for some α ∈ R?
Gibbons' conjecture was proven independently and with different methods by [2,3,10,11] (see also [12,13] for further results in the semilinear scalar case and [17] for a recent result concerning some related semilinear elliptic systems). Here we study Gibbons' conjecture for the quasilinear equation (P). To the best of our knowledge, there are no general results in this framework. This lack of results is mainly due to the fact that, unlike the semilinear case, when working with the singular operator −∆ p (·), both the weak and the strong comparison principles might fail. This (possible) failure being caused either by the presence of critical points or by the fact that the nonlinearity f changes sign. Those difficulties are even more magnified by the fact that we are facing a problem on an unbounded domain, the entire euclidean space R N . Also, in the pure quasilinear case, p = 2, we cannot exploit the usual arguments and tricks related to the linearity of the Laplace operator. Despite all those problems and difficulties, we are able to study and solve the quasilinear version of Gibbons' conjecture by making use of the the celebrated moving planes method which goes back to the papers of Alexandrov [1] and Serrin [25] (see also [4,19]).
Our main result is the following Theorem 1.1. Assume N > 1, (2N + 2)/(N + 2) < p < 2 and let u ∈ C 1 (R N ) be a weak solution of (P), such that |u| ≤ 1 on R N and To get our main result, we first prove a new weak comparison principle for quasilinear equations in half-spaces and then we exploit it to start the moving plane procedure at infinity in the y-direction. Then, by a delicate analysis based on the the use of the techniques developed in [7,8] and [14,15,16], the translation invariance of the considered problem and the method introduced in [10], we obtain the monotonicity of the solution in all the directions of the the upper hemi-sphere S N −1 + := {ν ∈ S N −1 + | (ν, e N )}. This result, in turn, will provide the desired one-dimensional symmetry result as well as the strict monotonicity.
The paper is organized as follows: In Section 2 we recall the definition of weak solution of (P), as well as some results about the strong maximum principle and the comparison principles for nonlinear equations involving the p-Laplace operator. In Section 3 we prove a new weak comparison principle in half-spaces. In Section 4 we prove the monotonicity of the solution in the y-direction, exploiting the moving plane procedure. In Section 5 we prove the the one-dimensional symmetry and the strict monotonicity of the solution.

Strong maximum principles and strong comparison principles for quasilinear elliptic equations
The aim of this section is to recall some results about the strong comparison principles and the strong maximum principles for quasilinear elliptic equations that will be used several times in the proof of our main theorem. To this end we first recall the definiton of weak solution for the quasilinear equation −∆ p u = f (u).
Let Ω be an open set of R N , N ≥ 1. We say that u ∈ C 1 (Ω) is a weak subsolution to Similarly, we say that u ∈ C 1 (Ω) is a weak supersolution to (2.4) if Finally, we say that u ∈ C 1 (Ω) is a weak solution of equation (2.4), if (2.5) and (2.6) hold. Sometimes for brevity, we shall use the term "solution" to indicate a weak solution to the considered problem.
The first result that we are going to present is the classical strong maximum principle due to J. L. Vazquez [27] (see also P. Pucci and J. Serrin book [22]) Theorem 2.2 (Strong Maximum Principle and Höpf's Lemma, [22,27]). Let u ∈ C 1 (Ω) be a non-negative weak solution to If u = 0, then u > 0 in Ω. Moreover for any point x 0 ∈ ∂Ω where the interior sphere condition is satisfied, and such that u ∈ C 1 (Ω) ∪ {x 0 } and u(x 0 ) = 0 we have that ∂ ν u > 0 for any inward directional derivative (this means that if y approaches x 0 in a ball B ⊆ Ω that has x 0 on its boundary, then lim y→x 0 It is very simple to guess that in the quasilinear case, maximum and comparison principles are not equivalent; for this reason we need also to recall the classical version of the strong comparison principle for quasilinear elliptic equations Theorem 2.3 (Classical Strong Comparison Principle, [6,22]). Let u, v ∈ C 1 (Ω) be two solutions to For the proof of this result we suggest [6]. The main feature of Theorem 2.3 is that it holds far from the critical set. Now we present a result which holds true, under stronger assumptions, on the entire domain Ω.
Theorem 2.4 (Strong Comparison Principle, [7]). Let u, v ∈ C 1 (Ω) be two solutions to where Ω is a bounded domain of R N and 2N +2 N +2 < p < +∞. Assume that at least one of the following two conditions (f u ),(f v ) holds: Suppose furthermore that Then u ≡ v in Ω unless Proof. The proof of this result follows by the same arguments in [7,15,23,24]. Note in fact that under the assumption (f u ) or (f v ), it follows that |∇u| −1 or |∇v| −1 has the summability properties exposed by Theorem 3.1 in [24]. Then the weighted Sobolev inequality is in force, see e.g. Theorem 8 in [15]. Now, it is sufficient to note that the Harnack comparison inequality given by Corollary 3.2 in [7] holds true, since the proof it is only based on the weighted Sobolev inequality.
Finally it is standard to see that the Strong Comparison Principle follows by the weak comparison Harnack inequality (that it is based on the Moser-iteration scheme [20,21]), see Theorem 1.4 in [7].
For future use we recall that, as it follows by the regularity results in [8,23,24], the directional derivatives of the solution ∂ η u (η ∈ S N −1 ) belong to the weighted Sobolev space H 1,2 ρ (Ω) and fulfils (2.15). In particular here below we recall two versions of the strong maximum principle for the linearized equation (2.15) that we shall use in our proofs. The first result holds far from the critical set: Theorem 2.5 (Classical Strong Maximum Principle for the Linearized Operator, [22]). Let u ∈ C 1 (Ω) be a solution to problem (2.7), with 1 < p < +∞. Let η ∈ S N −1 and let us assume that for any connected domain Ω ⊂ Ω \ {x ∈ Ω | |∇u(x)| = 0}.
Then ∂ η u ≡ 0 in Ω unless Next we recall a more general result which holds true on the entire domain Ω.
Theorem 2.6 (Strong Maximum Principle for the Linearized Operator, [7]). Let u ∈ C 1 (Ω) be a solution to problem (2.7), with 2N +2 N +2 < p < +∞. Assume that either We conclude this section by the following Remark 2.7. We want to point out the following properties satisfied by any weak solution to (P) such that |u| ≤ 1 on R N . They will be used several times throughout the paper.

Preliminary results
In this section we shall denote by Σ any (affine) open half-space of R N of the form where either a = −∞ and b ∈ R, or a ∈ R and b = +∞.
We also recall some known inequalities which will be used in this section. For any η, η ∈ R N with |η| + |η | > 0 there exists positive constants C 1 , C 2 , C 3 depending only on p such that (3.20) The first result that we need is a weak comparison principle between a subsolution and a supersolution to (P) ordered on the boundary of some open half-space Σ of R N . We prove the following Proposition 3.1. Assume N > 1, p > 1 and f ∈ C 1 (R). Let u, v ∈ C 1,α loc (Σ) such that |∇u|, |∇v| ∈ L ∞ (Σ) and . Moreover, let us assume that there are δ > 0, sufficiently small, and L > 0 such that The same result is true if Σ = R N −1 × (a, +∞) and (3.22) and (3.23) are replaced by Proof. We prove the result when (3.22) and (3.23) are in force. The other case is similar.
where the open set S (τ, ) ⊆ Σ (a,b) is such that and the open set I τ, The proof of this result is contained in [16,Theorem 1.6], where the authors proved the same result for a more general class of operators and nonlinearities and also in the presence of a first order term.

Monotonicity with respect to x N
The purpose of this section consists in showing that all the non-trivial solutions u to (P) that satisfies (1.2) are increasing in the x N direction. Since in our problem the right hand side depends only on u, it is possible to define the following set Without any apriori assumption on the behaviour of ∇u, the set Z f (u) may be very wild, see Figure 1.
We start by proving a lemma that we will use repeatedly in the sequel of the work.
Let us define the upper hemisphere and let us assume that Proof. Using Theorem 2.6 we deduce that either ∂ η u > 0 in U or ∂ η u ≡ 0 in U. For contradiction, assume that ∂ η u ≡ 0 in U. Pick P 0 ∈ U and let us define We note that the infimum in (4.38) is well defined, since by definition the connected component U is an open set, and that t 0 ∈ [−∞, 0).
The proof is based on a nontrivial modification of the moving plane method. Let us recall some notations. We define the half-space Σ λ and the hyperplane T λ by We also define the critical set Z ∇u by (4.41) The first step in the proof of the monotonicity is to get a property concerning the local symmetry regions of the solution, namely any C ⊆ Σ λ such that u ≡ u λ in C.
Having in mind these notations we are able to prove the following: Under the assumption of Theorem 1.1, let us assume that u is a solution to (P) satisfying (1.2), such that Arguing by contradiction, let us assume that there exists We notice that, by construction,  and we observe that w 0 (x) = 0 for every x ∈ B ρ 0 (P 0 ), whereP 0 is the new centre of the slided ball. In fact, if this is not the case there would exist a pointz ∈ B ρ 0 (P 0 ) such that w 0 (z) = 0, but this is in contradiction with the fact that U 0 ∩ Z f (u) = ∅. We have to distinguish two cases. Since p < 2 and f is locally Lipschitz, we have that where C is a positive constant.
where C is a positive constant.
Using the Implicit Function Theorem we deduce that the set {u = u(z 0 )} is a smooth manifold near z 0 . Now we want to prove that u x N (z 0 ) > 0 and actually that the set {u = u(z 0 )} is a graph in the y-direction near the point z 0 . By our assumption we know that u x N (z 0 ) := u y (z 0 ) ≥ 0. According to [7,8] and (2.14), the linearized operator of (P) is well defined L u (u y , ϕ) = 0 ∀ϕ ∈ C 1 c (Σ λ ). Let us set z 0 = (z 0 , y 0 ). We have two possibilities: u y (z 0 ) = 0 or u y (z 0 ) > 0.
Claim: We show that the case u y (z 0 ) = 0 is not possible.
From what we have seen above, we have |∇u(z 0 )| = 0 and hence there exists a ball B r (z 0 ) where |∇u(x)| = 0 for every x ∈ B r (z 0 ). By Theorem 2.3 it follows that u ≡ u λ in B r (z 0 ) namely u ≡ u λ in a neighborhood of the point z 0 ∈ ∂U 0 . Since u y (z 0 ) > 0 and N f is finite and u y (x) > 0 in B r (z 0 ), as consequence, the set {u = u(z 0 )} is a graph in the y-direction in a neighborhood of the point z 0 . Now we have to distinguish two cases: Define the sets C 1 := x ∈ R N : x ∈ (B r (z 0 ) ∩ {y := y 0 }) and u(x) < u(z 0 ) We observe that C is an open unbounded path-connected set (actually a deformed cylinder), see Figure 4. Since f (u(z 0 )) has the right sign, by Theorem 2.4 it follows that u ≡ u λ in C and this in contradiction with the uniform limit conditions (1.2).
In this case the open ball B r (z 0 ) must intersect another connected component (i.e. ≡ U 0 ) of Σ λ \ Z f (u) , such that u ≡ u λ in a such component, see Figure 5. Here we used the fact that near the (new) first contact point, the corresponding level set is a graph in the y-direction. Now, it is clear that repeating a finite number of times the argument leading to the existence of the touching point z 0 , we can find a touching point z m such that The contradiction then follows exactly as in Case 1.
To prove Proposition 4.2 we need of the following result: Arguing by contradiction let us assume that there exists a sequence of point P n = (x n , x N,n ), with −M − 1 < x N,n < −M + 1 for every n ∈ N, such that ∂ x N u(P n ) → 0 as n → +∞ in Up to subsequences, let us assume that Let us now defineũ n (x , x N ) := u(x + x n , x N ) so that ũ n ∞ = u ∞ ≤ 1. By standard regularity theory, see [9,26], we have that ũ n C 1,α loc (R N ) ≤ C for some 0 < α < 1. By Ascoli's Theorem we havẽ −→ũ up to subsequences, for α < α. By construction ∂ x Nũ ≥ 0 and ∂ x Nũ (0,x N ) = 0, hence by Theorem 2.5 it follows that ∂ This gives a contradiction (by Theorem 2.5) with the fact that lim , see Remark 2.7.
With the notation introduced above, we set Note that, by Proposition 3.1 (with v = u t ), it follows that Λ = ∅, hence we can define (4.48)λ := sup Λ.
Moreover it is important to say that by the continuity of u and u λ , it follows that u ≤ uλ in Σλ.
The proof of the fact that u(x , x N ) is monotone increasing in the x N -direction in the entire space R N is done once show thatλ = +∞. To do this we assume by contradiction that λ < +∞, and we prove a crucial result, which allows us to localize the support of (u − uλ) + . This localization, that we are going to obtain, will be useful to apply the weak comparison principle given by Proposition 3.1 and Theorem 3.2.
In particular, to get (4.51), we choose κ in Proposition 4.5 such that 2κ = C * . Then we deduce that Using (4.52), we can apply Proposition 3.1 in {x N < −4M − 1} and therefore, together Lemma 4.4 and Proposition 4.5, we actually deduce In particular, if we look to (4.49), we deduce that supp W + ε must belong to the set We now apply Theorem 3.2 in the set A. Let us choose (in Theorem 3.2) and take τ 0 = τ 0 (p,λ,M , N, L 0 ) > 0 and 0 = 0 (p,λ,M , N, L 0 ) > 0 as in Theorem 3.2. Let µ, ε in Proposition 4.5 such that 2(µ + ε) < τ 0 and let us redefine κ eventually such that κ := min{C * /2, 0 }. We finally apply Theorem 3.2 concluding that actually W + ε = 0 in the set A. This gives a contradiction, in view of the definition (4.48) ofλ. Consequently we deduce thatλ = +∞. This implies the monotonicity of u, that is ∂ x N u ≥ 0 in R N . By Theorem 2.6, it follows that since by Lemma 4.1, the case ∂ x N u ≡ 0 in some connected component, say U, of R N \ Z f (u) can not hold.

1-Dimensional Symmetry
In this section we pass from the monotonicity in x N to the monotonicity in all the directions of the upper hemisphere S N −1 + defined in (4.37). We refer to [10] for the case of the Laplacian operator, where in the proof the linearity of the operator was crucial. Here we have to take into account the singular nature and the nonlinearity of the operator p-Laplacian.
Lemma 5.1. Under the same assumption of Theorem 1.1, given ρ > 0 and k > 0, we define Then, there exists an open neighbourhood O η of η in S N −1 + , such that Proof. Arguing by contradiction let us assume that there exist two sequences {P m } ∈ R N and {ν m } ∈ S N −1 + such that, for every m ∈ N we have that P m = (x m , x N,m ) ∈ Σ ρ k , |(ν m , η) − 1| < 1/m and ∂ νm u(P m ) ≤ 0. Since −k < x N,m < k for every m ∈ N, then up to subsequences x N,m →x N . Now, let us definẽ so that ũ m ∞ = u ∞ ≤ 1. By standard regularity theory, see [9,26], we have that ũ m C 1,α loc (R N ) ≤ C. By Ascoli's Theorem, via a standard diagonal process, we have, up to subsequences for some 0 < α < α.
• If P 0 ∈ Σ ρ k \Z f (ũ) by Theorem 2.6 it follows that ∂ ηũ > 0 in the connected component of R N \ Z f (ũ) containing the point P 0 . Indeed the case ∂ ηũ ≡ 0 in the connected component of R N \ Z f (ũ) containing P 0 can not hold since Lemma 4.1.
Having in mind the previous lemma, now we are able to prove the monotonicity in a small cone of direction around η in the entire space.
Proof. We fixδ > 0 and let k = k(δ) > 0 be such that u < −1 +δ in {x N < −k}, u > 1 −δ in {x N > k} and (3.22) holds in {|x N | > k}. By Lemma 5.1 it follows that for all ρ > 0 one has For simplicity of exposition we set Our claim is to show that u − ν = 0 in A ∪ D. In order to do this we split the proof in two part.
Step 1. We show that u − ν = 0 in A. We set where α > 1, R > 0 large, A(2R) := A ∩ B 2R and ϕ R is a standard cutoff function such that First of all we notice that ϕ belongs to W 1,p 0 (A(2R)). To see this, use the definition of ϕ R and note that by Lemma 4.4 and Lemma 5.1, it follows that u − ν = 0 on the hyperplanes |x N | = k, namely on ∂A.
Having in mind all these fixed parameters let us define It is easy to see that L(R) ≤ CR N −1 . By (5.73) (up to a redefining of the constant involved) we deduce that (5.74) L(R) ≤ ϑL(2R) + C R β−(N −2) holds for every R > 0. By applying Lemma 2.1 in [14] it follows that L(R) = 0 for all R > 0. Since p < 2, passing to the limit in (5.74), we deduce that for a.e. x ∈ D This actually implies that u − ν (x) = 0 in D. Indeed let us suppose that would exist a point P ∈ D such that u − ν (P ) = 0. Let us consider the connected component U of D \ {x ∈ D : u − ν (x) = 0} containing P . By the continuity of u − ν , it follows that u − ν = 0 on the boundary ∂U. On the other hand u − ν must be constant in U (since by (5.75) |∇u − ν | = 0 there) .This is a contradiction.
By this two step we deduce that u ν ≥ 0 in R N . Finally by Lemma 4.1 we get (5.55).
Proof of Theorem 1.1. Using Proposition 4.2 we get that the solution is monotone increasing in the y-direction and this implies that ∂ y u ≥ 0 in R N . In particular we have ∂ y u > 0 in R N \ Z f (u) by (4.39). By Proposition 5.2, actually we obtain that the solution is increasing in a cone of directions close to the y-direction. This allows us to show that in fact, for i = 1, 2, · · · , N − 1, ∂ x i u = 0 in R N , just exploiting the arguments in [10] (see also [17,18]). We provide the details for the sake completeness. Let Ω be the set of the directions η ∈ S N −1 + for which there exists an open neighborhood O η ⊂ S N −1 + such that ∂ ν u = u ν ≥ 0 in R N and ∂ ν u = u ν > 0 in R N \ Z f (u) , for every ν ∈ O η . The set Ω is non-empty, since e N ∈ Ω, and it is also open by Proposition 5.2. Now we want to show that it is also closed. Letη ∈ S N −1 + and let us consider the sequence {η n } in Ω such that η n →η as n → +∞ in the topology of S N −1 + . Since by our assumptions ∂ ηn u ≥ 0 in R N , passing to the limit we obtain that ∂ηu ≥ 0 in R N . By Lemma 4.1 it follows that ∂ηu > 0 in R N \ Z f (u) . By Proposition 5.2 there exists an open neighborhood Oη such that (5.55) is true for every ν ∈ Oη; henceη ∈ Ω and this implies that Ω is also closed. Now, since S N −1 + is a path-connected set, we have that Ω = S N −1 + . Then there exists v ∈ C 1,α loc (R) such that u(x , y) = v(y). Now, let us assume that there exists b ∈ Z f (u) \{−1, 1} such that v (b) = 0. Then, by u y ≥ 0, the level set {v = v(b)} must be a bounded closed interval (possibly reduced to a single point), i.e., there exist α, β ∈ R with α ≤ β such that Therefore, by Höpf's Lemma, we have that v (β) > 0. The latter clearly implies that {v = v(b)} = {β} = {b} and so v (b) > 0, which is in contradiction with our initial assumption. Hence we deduce that ∂ y u > 0 in R N , concluding the proof.