Neumann fractional $p-$Laplacian: eigenvalues and existence results

We develop some properties of the $p-$Neumann derivative for the fractional $p-$Laplacian in bounded domains with general $p>1$. In particular, we prove the existence of a diverging sequence of eigenvalues and we introduce the evolution problem associated to such operators, studying the basic properties of solutions. Finally, we study a nonlinear problem with source in absence of the Ambrosetti-Rabinowitz condition.


Introduction
Consider a bounded domain Ω of R N , N ≥ 1, with Lipschitz boundary. The aim of this paper is to investigate problems of the form is the nonlocal normal p−derivative, or p−Neumann boundary condition and describes the natural Neumann boundary condition in presence of the fractional p−Laplacian. It extends the notion of nonlocal normal derivative introduced in [7] for the fractional Laplacian, i.e. for p = 2. In our situation, p > 1, s ∈ (0, 1) and C N,s,p is the constant appearing in the definition of the fractional p−Laplacian; however, for the sake of simplicity, from now on, we will set C N,s,p = 1. The definition in (3) was introduced in [2], where basic integration by parts were given. Here, we present some further properties of the associated operator, following [7], where a detailed description of the case p = 2 was given. Indeed, we refer to [7] for several comments, justifications and reasons to consider such operators, and for this reason we shall skip these motivations; see also [14] for a general overview on fractional operators.
We shall also face the parabolic problem associated to this new class of operators, namely in Ω.
In this case, we will prove conservation of the mass and monotony of the associated energy, as in [7]. Investigations on parabolic equations in presence of the fraction p−Laplacian have started in recent years, but only in presence of Dirichlet boundary conditions, and there are not many contributions, yet, see for instance [1], [11], [20], [21]. On the other hand, [7] is the first paper where linear parabolic problems with the associated boundary condition are considered, and, in this direction, we intend to introduce the nonlinear case with the associated nonlinear Neumann conditions. We recall that Neumann boundary problems for the p−Laplacian were already introduced in [13], but the underlying operator was different from ours, since in their integral definition of fractional Laplacian only points in Ω were taken into account; more important, their Neumann boundary condition is a pointwise one, like that of [3], [4], [5], [15] and [19]. After these preliminary, but natural, properties, we will consider problem (1) first with a given source, just to treat the easy case. Then, we will study (1) in presence of a general nonlinear term which doesn't satisfy the usual Ambrosetti-Rabinowitz condition, showing the existence of two solutions, one being positive in the whole of R N , and the other being negative.
The paper is organized as follows. In Section 2 we consider the variational setting for the nonlocal elliptic problem associated to the p−Neumann boundary condition, recalling some properties from [2] and proving a maximum principle. In addition, we prove that the p−Neumann boundary condition is also valid pointwise (see Theorem 2.8).
In Section 3 we consider the associated eigenvalue problem. In particular, we prove the existence of an unbounded sequence of eigenvalues and we show that some classical properties of the set of eigenvalues for the p−Laplacian still hold true in this case. In particular, we show that any eigenfunction is bounded in the whole of R N .
In Section 4 we consider the associated parabolic problem and we show that, as in the classical case, the total mass is preserved and the energy is decreasing in time.
Finally, in Section 5, after treating the easy problem with an assigned source, we study a general problem where the right hand side function doesn't satisfy the Ambrosetti-Rabinowitz condition, and we show the the existence of two constant sign solutions by variational methods.

Functional setting for the normal p−derivative
In this section we follow the lines of [7], introducing the functional setting and the basic properties of the fractional p−Laplacian with associated p−Neumann boundary conditions.
To do that, fix a bounded domain with Lipschitz boundary Ω ⊂ R N , N ≥ 1, and for u : R N → R measurable, set Remark 2.1. It is clear that, Ω being "nice enough", in the previous setting we can equally write R N \Ω in place of R N \Ω. The abstract setting can be faced also for Ω less regular, replacing |g| , which is the natural norm in the general framework.
Though already stated in [2], we recall the following result, giving a detailed proof.
Proof. First, we show that · X is a norm. If u X = 0, we have u L p (Ω) = 0, so u = 0 a.e. in Ω. Moreover, we have In particular, we can take x ∈ CΩ and y ∈ Ω to obtain In this way, we have u = 0 a.e. in R N . Now, we prove that X is complete, and to do this we take a Cauchy sequence (u k ) k in X. In particular, u k is a Cauchy sequence in L p (Ω) and so (up to a subsequence) there exists u ∈ L p (Ω) such that u k converges to u in L p (Ω) and a.e. in Ω. This means that there exists Z 1 ⊂ Ω such that (4) |Z 1 | = 0 and u k (x) → u(x) for every x ∈ Ω \ Z 1 .
We also define for every U : Since u k is a Cauchy sequence in X, for every ε > 0 there exists N ε > 0 such that for h, k ≥ N ε we have in particular So, T u k is a Cauchy sequence in L p (R 2N ), and up to a subsequence we can assume that T u k converges to some T in L p (R 2N ) and a.e. in R 2N . This means that there exists Z 2 ⊂ R 2N such that For any x ∈ Ω, we set From (6) and (5), we obtain |W | = 0, so by the Fubini's Theorem we have which implies that |R N \ S x | = 0 a.e. x ∈ Ω. It follows that |Ω \ V | = 0. This together with (4) implies that In particular, V \Z 1 = ∅ (nay, |V \Z 1 | = |Ω|), so we can take In addition, since x 0 ∈ V , we get |R N \ S x 0 | = 0. This means that for a.e. y ∈ R N , (x 0 , y) ∈ R 2N \ Z 2 and so lim k→∞ T u k (x 0 , y) = T (x 0 , y).
Moreover, since Ω × (CΩ) ⊆ R 2N \ (CΩ) 2 , we have for a.e. y ∈ CΩ. From this, we obtain for a.e. y ∈ CΩ. This and (4) imply that u k converges a.e. in R N , so we can say that u k converges a.e. to some u in R N . Now, since u k is a Cauchy sequence in X, for any ε > 0 there exists N ε > 0 such that, for any h ≥ N ε , where we used Fatou's Lemma. So u h converges to u in X. Starting this procedure with a generic subsequence, we can conclude that X is complete.
From the definition of X, it follows that X is embedded in L p (B(0, R)) for every R > 0. Indeed, by the convergence of the double integral, we get that for a.e. x ∈ Ω and so for every R > 0 In addition, we have hence the claim follows.
Remark 2.4. Under the previous setting, X is embedded continuously in W s,p (Ω). As a consequence, the standard compact embeddings in suitable L q (Ω) spaces hold true, see [8]. Now, we recall the analogous of the divergence theorem and of the integration by parts formula for the nonlocal case, see [2]: Proposition 2.6. Let u and v be bounded C 2 functions in R N . Then, The integration by parts formula in Proposition 2.6 leads to this natural definition: As a consequence of this definition, we have the following result Theorem 2.8. Let u be a weak solution of (7). Then, N s,p u = g a.e. in R N \ Ω.
Proof. First, we take v ∈ X such that v ≡ 0 in Ω as a test function in (8), obtaining Therefore, From the definition of weak solution, we have the following Proposition 2.9. Let f ∈ L p ′ (Ω) and g ∈ L 1 (R N \Ω). Let I g : X → R be the functional defined as for every u ∈ X. Then any critical point of I g is a weak solution of problem (7).
In addition, Then, if u ∈ X, we have The computation of the first variation of I g is standard.
The next result gives a sort of maximum principle.
Proposition 2.10. Let f ∈ L p ′ (Ω) and g ∈ L 1 (R N \ Ω). Let u ∈ X be a weak solution of (7) with f ≥ 0 and g ≥ 0. Then, u is constant.
Proof. First, we notice that v ≡ 1 belongs to X. So, using it as a test function in (8) we obtain Hence, f = 0 a.e. in Ω and g = 0 a.e. in R N \ Ω. Now, taking v = u as a test function again in (8), we get so u must be constant.
From now on, we concentrate on homogeneous boundary conditions, so that g ≡ 0.
Denoting by X ′ the dual of X, we can define the operator A : X → X ′ such that In this way A is (p − 1)-homogeneous and odd, and such that By the uniform convexity of X, A satisfies the (S) property, that is,

The eigenvalue problem
In this section we consider the nonlinear eigenvalue problem depending on parameter λ ∈ R. If (9) admits a weak solution u ∈ X (notice that now g ≡ 0), that is for all v ∈ X, then we say that λ is an eigenvalue of (−∆) s p with p−Neumann boundary conditions and associated λ-eigenfunction u. As in the classical case, we call the set of all the eigenvalues the point spectrum of (−∆) s p in X and we denote it by σ(s, p). First of all we observe that for λ = 0 constant functions are all 0eigenfunctions. Since all the eigenvalues are obviously non negative, we have that λ 1 = 0 is the first eigenvalue. Moreover, for all v ∈ X implies u constant, so all the λ 1 -eigenfunctions are just constant functions. As usual, we can construct a sequence (λ k ) k of eigenvalues for problem (9), analogously to the Dirichlet case treated in [12], setting Here, if F is the family of all nonempty, closed, symmetric subsets of S = {u ∈ X : Ω |u| p = 1}, for all k ∈ N we have set is the cohomological index of Fadell and Rabinowitz [9].
In order to prove that λ k is an eigenvalue for every k ∈ N, we proceed in the standard way: set ϕ(u) =  Proof. Let (u n ) n ⊂ S and (µ n ) n ⊂ R be such that ϕ(u n ) → c as n → ∞ and ϕ ′ (u n ) − µ n I ′ (u n ) → 0 in X ′ . We have u n p X = 1 + ϕ(u n ) → 1 + c, so (u n ) n is bounded in X. Up to a subsequence, we have u n ⇀ u in X and u n → u in L p (Ω) for some u ∈ X as n → ∞, see Remark 2.4. In particular, u ∈ S. We also get that ϕ(u n ) − µ n → 0, and so µ n → c. Now, we have So, by the (S) property of A, we get that u n → u in X. Now we can give the desired result for the sequence (λ k ) k . Proposition 3.2. For all k ∈ N, λ k is an eigenvalue of (9). In addition, λ k → ∞.
The proof is standard, see for example the proof of [12, Proposition 2.2]. We also recall that in [6] a characterization of the second eigenvalue is given, together with the asymptotic for p → ∞. Now we show that every eigenfunction, except the ones corresponding to the first eigenvalue, changes sign.
Let v ∈ X be a solution to (9) such that v > 0 in Ω. Then λ = 0, hence v is constant.
Since all the eigenvalues are non negative, we have λ = 0 and so v belongs to the first eigenspace, as claimed. Now we want to prove the boundedness of eigenfunctions in the whole of R N , starting as in [10] to get the bound in Ω, and exploiting the p−Neumann condition to get the bound in the complementary set of Ω. More precisely, we have that the L ∞ −norm in Ω estimates the L ∞ −norm in the R N \ Ω. Proposition 3.4. Let s ∈ (0, 1), p > 1, and u ∈ X be a solution of (9) for some λ ≥ 0. Then u ∈ L ∞ (R N ) and Proof. First, we prove that u is bounded in Ω, concentrating on the case ps ≤ N, the case ps > N being trivial by the fractional Morrey-Sobolev embedding. As in [10], we only have to prove that u + is bounded in Ω, since both u ± are solutions, so we can get a bound for the negative part in the same way. To do that, it is enough to prove that (14) u where δ > 0 is still to be determined. Indeed, we can scale the function verifying (14), so there is no restriction in this. Now, for all k ≥ 0, we define the function see [10], also for the following facts: w k ∈ X and hold true for every k ≥ 0. Moreover, for every function v for all x, y ∈ R N . Now, we want to prove (14) using a standard argument relying on estimating the decay of U k := w k p L p (Ω) . First of all, using (16) with Taking w k+1 as a test function in (9) and then using (15), we get Using the fractional Sobolev embeddings, as in [10], we get N ps , where c > 0 depends on N, p, s. Proceeding as in [10], we get that u is bounded in Ω. Now, take x ∈ R N \ Ω. Since u is bounded in Ω, from (9) we get If u is constant, the result is trivial. On the other hand, if u is not constant, from Theorem 2.8 we have and so u L ∞ (R N \Ω) ≤ u L ∞ (Ω) , which concludes the proof.

The parabolic equation
In this section, we consider the problem in Ω.
We show that the solutions of (17) preserve their mass and have energy that decreases in time, as proved in [7] for p = 2. To do so, we assume that u is a classical solution of (17), so that (17) holds pointwise. In particular, we can differentiate with respect to time. which means that the total mass is preserved.
Proof. By the dominated convergence theorem and Proposition 2.5, we have So, Ω u dx does not depend on t, as desired.
Proof. From Proposition 2.6, we have since u is a solution of (17), and so the energy is decreasing.

Two p−Neumann problems with source
In this section we consider two problems in presence of the p−Neumann condition: the first one is the easy case of a given source term, which we study for completeness of the subject, while the second one takes into account a source not satisfying the Ambrosetti-Rabinowitz condition or some of its standard generalizations (see [16]).
Let us start with with f ∈ L p ′ (Ω).
Definition 5.1. We say that u ∈ X is a weak solution of problem (18) if For the sake of simplicity, in this section we replace the usual norm in X with the equivalent one Hence, as usual, we can define the functional so that every critical point of J is a weak solution of (18). Not surprisingly, we have the following existence result: Proposition 5.2. Let f ∈ L p ′ (Ω), s ∈ (0, 1) and p > 1. Then (18) admits a unique solution.
Proof. First of all, the functional J is coercive, in fact Moreover, J is also strictly convex, hence by the Weierstrass Theorem it has a global minimum, which is a critical point of J. Uniqueness follows by strict convexity. Now, we consider the problem where f : Ω×R → R a Carathéodory function such that f (x, 0) = 0 for almost every x ∈ Ω. In addition, we assume the following hypotheses: (f 1 ) there exists a ∈ L q (Ω), a ≥ 0, with q ∈ ((p * s ) ′ , p), c > 0 and r ∈ (p, p * s ) such that |f (x, t)| ≤ a(x) + c|t| r−1 for a.e. x ∈ Ω and for all t ∈ R; uniformly for a.e. x ∈ Ω; (f 3 ) if σ(x, t) = f (x, t)t − pF (x, t), then there exist ϑ ≥ 1 and β * ∈ L 1 (Ω), β * ≥ 0, such that for a.e. x ∈ Ω and all 0 ≤ t 1 ≤ t 2 or t 2 ≤ t 1 ≤ 0; uniformly for a.e. x ∈ Ω. As usual, in (f 1 ) we have denoted by p * s the fractional Sobolev exponent of order s, that is so that the embedding in L q (Ω) of W s,p (Ω) (and thus of X) is compact for every q < p * s . Remark 5.3. A few comments on (f 3 ) are mandatory. Such a condition was introduced in [17] with ϑ = 1. However, it is clear that assuming ϑ ≥ 1 enlarges the set of admissible positive (or definitely positive) functions σ's considered in [17] (as it happens for the model case f (x, t) = |t| r−2 r). On the other hand, if σ were negative, admitting ϑ < 1 would make the situation more general. However, if (f 1 ) − (f 4 ) hold for some ϑ > 0, then σ(x, t) > 0 for a.e. x ∈ Ω and all t, at least for |t| large, that is there existst ≥ 0 such that σ(x, t) > 0 for a.e. x ∈ Ω and all |t| >t. Indeed, reasoning with t positive, if for every t > 0 there exists τ > t such that σ(x, τ ) ≤ 0, we get for a.e. x ∈ Ω and all t. As a consequence, ( t p for every t < s. Letting s → +∞, we get a contradiction with (f 2 ).
As a consequence, in (f 3 ) the requirement ϑ ≥ 1 is the most general one. Now we are ready to give the definition of a weak solution of our problem.
Definition 5.4. Let u ∈ X. With the same assumption on f as above, we say that u is a weak solution of (19) if With this definition, we have that any critical point of the functional E : X → R given by is a weak solution of (19). Our main result is the following First, we introduce the functionals where u + and u − are the classical positive part and negative part of u.
We want to prove that both E ± satisfies the Cerami condition, (C) for short, which states that any sequence (u n ) n in X such that (E ± (u n )) n is bounded and (1 + u n )E ′ ± (u n ) → 0 as n → ∞ admits a convergent subsequence.
We will also use the following inequality: for any x, y ∈ R.
Proposition 5.6. Under the assumptions of Theorem 5.5, E ± satisfies the (C) condition.
Proof. We do the proof for E + , the proof for E − being analogous. Let (u n ) n in X be such that for some M 1 > 0 and all n ≥ 1, and for every h ∈ X and with ε n → 0 as n → ∞, that is (23) Taking h = u − n in (23), we obtain By (20), we have So, we have that From (21) we have for M 1 > 0 and all n ≥ 1, and since for some M 2 > 0 and all n ≥ 1. Adding (26) to (25) we obtain for some M 3 > 0 and all n ≥ 1, that is Now we want to prove that (u + n ) n is bounded in X, and to do this we argue by contradiction. Passing to a subsequence if necessary, we assume that u + n → ∞ as n → ∞. Defining y n = u + n / u + n , we can assume (28) y n ⇀ y in X and y n → y in L q (Ω) for every q ∈ (p, p * s ) and y ≥ 0. First we consider the case y = 0. We define Z(y) = {x ∈ Ω : y(x) = 0}, and so we have |Ω \ Z(y)| > 0 and u + n → ∞ for almost every x ∈ Ω \ Z(y) as n → ∞. By hypothesis (f 2 ), we have n (x) p y n (x) p → ∞ for almost every x ∈ Ω \ Z(y). By Fatou's Lemma, we have and so As before, from (21) and (24) we have for some M 4 > 0 and n ≥ 1. Since u n p ≤ 2 p−1 ( u + n p + u − n p ), we obtain for some M 5 > 0, and so Passing to the limit, we have lim sup n→∞ Ω for some M 6 , which is in contradiction with (29), and this concludes the case y = 0. Now,we deal with the case y ≡ 0. We consider the continuous functions γ n : [0, 1] → R, defined as γ n (t) := E + (tu + n ) with t ∈ [0, 1] and n ≥ 1. So, we can define t n such that Now we define v n := (pλ) 1 p y n ∈ X for λ > 0. From (28), it follows that v n → 0 in L q (Ω) for all q ∈ (p, p * s ). Starting from (f 1 ) and performing some integration, we have as n → ∞. Since u + n → ∞, there exists n 0 ≥ 1 such that (pλ) 1 p / u + n ∈ (0, 1) for all n ≥ n 0 . Then, from (30), we have γ n (t n ) ≥ γ n (pλ) 1 p u + n for all n ≥ n 0 . It follows that From (31), we have E + (t n u + n ) ≥ λ + o(1), and since λ is arbitrary we have for all n ≥ 1. In addition, we have E + (0) = 0, and from (21), (24) and (20), we have E + (u + n ) ≤ M 7 for some M 7 > 0. Together with (32), this implies that t n ∈ (0, 1) for all n ≥ n 1 ≥ n 0 . Since t n is a maximum point, we also have 0 = t n γ ′ n (t n ) = R 2N \(CΩ) 2 J p (t n u n (x) − t n u n (y))(t n u + n (x) − t n u + n (y)) |x − y| N +ps dxdy + Ω |t n u + n | p dx − Ω f (x, t n u + n (x))t n u + n (x) dx, and so, from (20), (34) t n u + n p − Ω f (x, t n u + n (x))t n u + n (x) dx ≤ 0.
We have proved that (u + n ) n is bounded in X, so from (24) we have that (u n ) n is bounded in X. Hence, we can assume (36) u n ⇀ u in X and u n → u in L q (Ω) with q ∈ (p, p * so from (f 2 ) we have Ω F (x, tu) (tu) p u p dx → ∞ as n → ∞. It follows that E + (tu) → −∞ as t → ∞, and so there exists e ∈ X such that e ≥ ρ and E + (e) > 0. Now, we can apply the Mountain Pass Theorem to E + and obtain a non-trivial critical point u. In particular, we have 0 = R 2N \(CΩ) 2 J p (u n (x) − u n (y))(u − (x) − u − (y)) |x − y| N +ps dxdy From (20), we get 0 ≥ u − p , and so u − ≡ 0. As a consequence, we have E + (u) = E (u), and so u ≥ 0 is a solution of (19). Suppose that there exists x 0 ∈ R N \ Ω such that u(x 0 ) = 0. Then, from Theorem 2.8 we would get Ω u p−1 (y) |x − y| N +sp dy = 0, so that u = 0 in Ω and thus, using u as test function in the equation, u = 0 in R N , while u is non-trivial. Now, assume that the equation i (19) holds pointwise and suppose by contradiction that there exists x ∈ Ω such that u(x) = 0. From the equation we would get R N u(y) p−1 |x − y| N +ps dy = 0. This would imply that u = 0 a.e. in R N , which is a contradiction since the solution is non-trivial. It follows that u > 0 in R N .
Arguing in the same way for E − , we can find a non-trivial negative solution for (19).

Some open questions.
(1) Is any solution of problem (19) continuous in R N ? In the Dirichlet case "u = 0 on R N \Ω", this last condition helps significantly in obtaining the desired regularity. In our case, we believe this result is true, but at the moment we are not able to prove it.