NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS

Transcription

1 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Abstract. We introduce a new Neumann problem for the fractional Laplacian arising from a simple probabilistic consideration, and we discuss the basic properties of this model. We can consider both elliptic and parabolic equations in any domain. In addition, we formulate problems with nonhomogeneous Neumann conditions, and also with mixed Dirichlet and Neumann conditions, all of them having a clear probabilistic interpretation. We prove that solutions to the fractional heat equation with homogeneous Neumann conditions have the following natural properties: conservation of mass inside, decreasing energy, and convergence to a constant as t. Moreover, for the elliptic case we give the variational formulation of the problem, and establish existence of solutions. We also stu the limit properties and the boundary behavior induced by this nonlocal Neumann condition. For concreteness, one may think that our nonlocal analogue of the classical Neumann condition νu = 0 on consists in the nonlocal prescription u(x) u(y) x y = 0 for x n+2s Rn \. 1. Introduction and results The aim of this paper is to introduce the following Neumann problem for the fractional Laplacian { ( ) s u = f in N s u = 0 in R n (1.1) \. Here, N s is a new nonlocal normal derivative, given by u(x) u(y) N s u(x) := c n,s, x Rn \. (1.2) The normalization constant c n,s is the one appearing in the definition the fractional Laplacian ( ) s u(x) u(y) u(x) = c n,s PV. (1.3) R n See [12, 20] for the basic properties of this operator (and for further details on the normalization constant c n,s, whose explicit value only plays a minor role in this paper) Mathematics Subject Classification. 35R11, 60G22. Key words and phrases. Nonlocal operators, fractional Laplacian, Neumann problem. The first author was supported by grant EPSRC EP/K024566/1 (Scotland). The second author was supported by grants MTM C04-01 (Spain) and 2009SGR345 (Catalunya). The third author was supported by grants ERC (Europe) and PRIN FYK7 (Italy). We thank Gerd Grubb for her very interesting comments on a previous version of this manuscript. 1

2 2 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI As we will see below, the corresponding heat equation with homogeneous Neumann conditions u t + ( ) s u = 0 in, t > 0 N s u = 0 in R n \, t > 0 (1.4) u(x, 0) = u 0 (x) in, t = 0 possesses natural properties like conservation of mass inside or convergence to a constant as t + (see Section 4). The probabilistic interpretation of the Neumann problem (1.4) may be summarized as follows: (1) u(x, t) is the probability distribution of the position of a particle moving randomly inside. (2) When the particle exits, it immediately comes back into. (3) The way in which it comes back inside is the following: If the particle has gone to x R n \, it may come back to any point y, the probability of jumping from x to y being proportional to x y n 2s. These three properties lead to the equation (1.4), being u 0 the initial probability distribution of the position of the particle. A variation of formula (1.2) consists in renormalizing N s u according to the underlying probability law induced by the Lévy process. This leads to the definition Ñ s u(x) := c n,s N s u(x). (1.5) x y n+2s Other Neumann problems for the fractional Laplacian (or other nonlocal operators) were introduced in [4, 8], [1, 3], [9, 10, 11], and [15]. All these different Neumann problems for nonlocal operators recover the classical Neumann problem as a limit case, and most of them has clear probabilistic interpretations as well. We postpone to Section 7 a comparison between these different models and ours. An advantage of our approach is that the problem has a variational structure. In particular, we show that the classical integration by parts formulae u = ν u and u v = v ( )u + v ν u are replaced in our setting by ( ) s u dx = and c n,s 2 R 2n \(C) 2 R n \ N s u dx ( )( ) u(x) u(y) v(x) v(y) dx = Also, the classical Neumann problem { u = f in ν u = g on v ( ) s u + v N s u. R n \ (1.6)

3 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 3 comes from critical points of the energy functional 1 u 2 fu g u, 2 without trace conditions. In analogy with this, we show that our nonlocal Neumann condition { ( ) s u = f in N s u = g in R n (1.7) \ follows from free critical points of the energy functional c R2n\(C)2 n,s u(x) u(y) 2 4 dx f u g u, R n \ see Proposition 3.7. Moreover, as well known, the theory of existence and uniqueness of solutions for the classical Neumann problem (1.6) relies on the compatibility condition f = g. We provide the analogue of this compatibility condition in our framework, that is f = g, see Theorem 3.9. Also, we give a description of the spectral properties of our nonlocal problem, which are in analogy with the classical case. The paper is organized in this way. In Section 2 we give a probabilistic interpretation of our Neumann condition, as a random reflection of a particle inside the domain, according to a Lévy flight. This also allows us to consider mixed Dirichlet and Neumann conditions and to get a suitable heat equation from the stochastic process. In Section 3 we consider the variational structure of the associated nonlocal elliptic problem, we show an existence and uniqueness result (namely Theorem 3.9), as follows: Let R n be a bounded Lipschitz domain, f L 2 (), and g L 1 (R n \ ). Suppose that there exists a C 2 function ψ such that N s ψ = g in R n \. Then, problem (1.7) admits a weak solution if and only if f = g. Moreover, if such a compatibility condition holds, the solution is unique up to an additive constant. Also, we give a description of a sort of generalized eigenvalues of ( ) s with zero Neumann boundary conditions (see Theorem 3.11): Let R n be a bounded Lipschitz domain. Then, there exist a sequence of nonnegative values 0 = λ 1 < λ 2 λ 3, and a sequence of functions u i : R n R such that { ( ) s u i (x) = λ i u i (x) for any x N s u i (x) = 0 for any x R n \. R n \ R n \

4 4 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Also, the functions u i (when restricted to ) provide a complete orthogonal system in L 2 (). By similarity with the classical case, we are tempted to consider the above λ i and u i as generalized eigenvalues and eigenfunctions. Though the word generalized will be omitted from now on for the sake of shortness, we remark that this spectral notion is not completely standard, since our eigenfunctions u i are defined in the whole of R n but satisfy the equation ( ) s u i = λ i u i only in the domain (indeed, outside they verify our nonlocal Neumann condition). Moreover, the orthogonality and density properties of u i also refer to their restriction in. In Section 4 we discuss the associated heat equation. As it happens in the classical case, we show that such equation preserves the mass, it has decreasing energy, and the solutions approach a constant as t +. In particular, by the results in Propositions 4.1, 4.2 and 4.3 we have: Assume that u(x, t) is a classical solution to u t + ( ) s u = 0 in, t > 0 N s u = 0 in R n \, t > 0 u(x, 0) = u 0 (x) in, t = 0. Then the total mass is conserved, i.e. for all t > 0 u(x, t) dx = u 0 (x)dx. Moreover, the energy E(t) = R2n\(C)2 u(x, t) u(y, t) 2 dx is decreasing in time t > 0. Finally, the solution approaches a constant for large times: more precisely u 1 u 0 in L 2 () as t +. In Section 5 we compute some limits when s 1, showing that we can recover the classical case. In particular, we show in Proposition 5.1 that: Let R n be any bounded Lipschitz domain. Let u and v be C0 2(Rn ) functions. Then, u lim N s u v = s 1 ν v. R n \ Also, we prove that nice functions can be extended continuously outside in order to satisfy a homogeneous nonlocal Neumann condition, and we characterize the boundary behavior of the nonlocal Neumann function. More precisely, in Proposition 5.2 we show that:

5 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 5 Let R n be a domain with C 1 boundary. Let u be continuous in, with N s u = 0 in R n \. Then u is continuous in the whole of R n. The boundary behavior of the nonolcal Neumann condition is also addressed in Proposition 5.4: Let R n be a C 1 domain, and u C(R n ). Then, for all s (0, 1), lim Ñ s u(x) = 0, x x R n \ where Ñ is defined by (1.5). Also, if s > 1 2 and u C1,α (R n ) for some α > 0, then Ñ s u(x + ɛν) ν Ñ s u(x) := lim ɛ 0 + ɛ for some constant κ > 0. = κ ν u for any x, Later on, in Section 6 we deal with an overdetermined problem and we show that it is not possible to prescribe both nonlocal Neumann and Dirichlet conditions for a continuous function. Finally, in Section 7 we recall the various nonlocal Neumann conditions alrea appeared in the literature, and we compare them with our model. All the arguments presented are of elementary 1 nature. Moreover, all our considerations work for any general Lévy measure ν satisfying R n min(1, y 2 )dν(y) < +. However, for sake of clarity of presentation, we have done everything for the most canonical case of the fractional Laplacian. 2. Heuristic probabilistic interpretation Let us consider the Lévy process in R n whose infinitesimal generator is the fractional Laplacian ( ) s. Heuristically, we may think that this process represents the (random) movement of a particle along time t > 0. As it is well known, the probability density u(x, t) of the position of the particle solves the fractional heat equation u t + ( ) s u = 0 in R n ; see [21] for a simple illustration of this fact. Recall that when the particle is situated at x R n, it may jump to any other point y R n, the probability of jumping to y being proportional to x y n 2s. Similarly, one may consider the random movement of a particle inside a bounded domain R n, but in this case one has to decide what happens when the particle leaves. In the classical case s = 1 (when the Lévy process is the Brownian motion), we have the following: 1 To keep the notation as simple as possible, given functions f and g and an operator T, we will often write idendities like f = g in or T f = g in to mean f(x) = g(x) for every x or T f = g for every x, respectively. Also, if u : R n R, we often denote the restriction of u to again by u. We hope that this slight abuse of notation creates no problem to the reader, but for the sake of clarity we also include an appendix at the end of the paper in which Theorems 3.9 and 3.11 are proved using a functional analysis notation that distinguishes between a function and its restriction.

6 6 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI (1) If the particle is killed when it reaches the boundary, then the probability distribution solves the heat equation with homogeneous Dirichlet conditions. (2) If, instead, when the particle reaches the boundary it immediately comes back into (i.e., it bounces on ), then the probability distribution solves the heat equation with homogeneous Neumann conditions. In the nonlocal case s (0, 1), in which the process has jumps, case (1) corresponds to the following: The particle is killed when it exits. In this case, the probability distribution u of the process solves the heat equation with homogeneous Dirichlet conditions u = 0 in R n \, and solutions to this problem are well understood; see for example [18, 14, 13, 2]. The analogue of case (2) is the following: When the particle exits, it immediately comes back into. Of course, one has to decide how the particle comes back into the domain. In [1, 3] the idea was to find a deterministic reflection or projection which describes the way in which the particle comes back into. The alternative that we propose here is the following: If the particle has gone to x R n \, then it may come back to any point y, the probability of jumping from x to y being proportional to x y n 2s. Notice that this is exactly the (random) way as the particle is moving all the time, here we just add the restriction that it has to immediately come back into every time it goes outside. Let us finally illustrate how this random process leads to problems (1.1) or (1.4). In fact, to make the exposition easier, we will explain the case of mixed Neumann and Dirichlet conditions, which, we think, is very natural Mixed Dirichlet and Neumann conditions. Assume that we have some domain R n, and that its complement R n \ is splitted into two parts: N (with Neumann conditions), and D (with Dirichlet conditions). Consider a particle moving randomly, starting inside. When the particle reaches D, it obtains a payoff φ(x), which depends on the point x D where the particle arrived. Instead, when the particle reaches N it immediately comes back to as described before. If we denote u(x) the expected payoff, then we clearly have and ( ) s u = 0 u = φ in D, where φ : D R is a given function. Moreover, recall that when the particle is in x N then it goes back to some point y, with probability proportional to x y n 2s. Hence, we have that u(x) = κ in u(y) for x N, for some constant κ, possibly depending on the point x N, that has been fixed. In order to normalize the probability measure, the value of the constant κ is so that κ = 1.

7 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 7 Finally, the previous identity can be written as u(x) u(y) N s u(x) = c n,s = 0 for x N, and therefore u solves ( ) s u = 0 in N s u = 0 in N u = φ in D, which is a nonlocal problem with mixed Neumann and Dirichlet conditions. Note that the previous problem is the nonlocal analogue of u = 0 in ν u = 0 in Γ N u = φ in Γ D, being Γ D and Γ N two disjoint subsets of, in which classical Dirichlet and Neumann boundary conditions are prescribed. More generally, the classical Robin condition a ν u + bu = c on some Γ R may be replaced in our nonlocal framework by an s u + bu = c on some R R n \. Nonlinear boundary conditions may be considered in a similar way Fractional heat equation, nonhomogeneous Neumann conditions. Let us consider now the random movement of the particle inside, with our new Neumann conditions in R n \. Denoting u(x, t) the probability density of the position of the particle at time t > 0, with a similar discretization argument as in [21], one can see that u solves the fractional heat equation with u t + ( ) s u = 0 in for t > 0, N s u = 0 in R n \ for t > 0. Thus, if u 0 is the initial probability density, then u solves problem (1.4). Of course, one can now see that with this probabilistic interpretation there is no problem in considering a right hand side f or nonhomogeneous Neumann conditions { ut + ( ) s u = f(x, t, u) in N s u = g(x, t) in R n \. In this case, g represents a nonlocal flux of new particles coming from outside, and f would represent a reaction term. 3. The elliptic problem Given g L 1 (R n \ ) and measurable functions u, v : R n R, we set u H s := u 2L2() + g 1/2 u 2L2(Rn\) + u(x) u(y) 2 R 2n \(C) 2 dx (3.1)

8 8 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI and (u, v) H s := Then, we define the space u v dx + R n \ g u v dx (u(x) u(y))(v(x) v(y)) + R 2n \(C) 2 dx. H s := { u : R n R measurable : u H s < + }. (3.2) Notice that H s and its norm depend on g, but we omit this dependence in the notation for the sake of simplicity. Proposition 3.1. H s is a Hilbert space with the scalar product defined in (3.2). Proof. We point out that (3.2) is a bilinear form and u H s = (u, u) H s. Also, if u H s = 0, it follows that u L 2 () = 0, hence u = 0 a.e. in, and that R2n\(C)2 u(x) u(y) 2 dx = 0, which in turn implies that u(x) u(y) = 0 for any (x, y) R 2n \ (C) 2. In particular, a.e. x C and y we have that u(x) = u(x) u(y) = 0. This shows that u = 0 a.e. in R n, so it remains to prove that H s is complete. For this, we take a Cauchy sequence u k with respect to the norm in (3.1). In particular, u k is a Cauchy sequence in L 2 () and therefore, up to a subsequence, we suppose that u k converges to some u in L 2 () and a.e. in. More explicitly, there exists Z 1 R n such that Z 1 = 0 and u k (x) u(x) for every x \ Z 1. (3.3) Also, given any U : R n R, for any (x, y) R 2n we define ( ) U(x) U(y) χ R 2n \(C) 2(x, y) E U (x, y) :=. (3.4) 2 Notice that E uk (x, y) E uh (x, y) = ( ) u k (x) u h (x) u k (y) + u h (y) χ R 2n \(C) 2(x, y) 2 Accordingly, since u k is a Cauchy sequence in H s, for any ɛ > 0 there exists N ɛ N such that if h, k N ɛ then R2n\(C)2 ɛ 2 (u k u h )(x) (u k u h )(y) 2 dx = E uk E uh 2 L 2 (R 2n )..

9 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 9 That is, E uk is a Cauchy sequence in L 2 (R 2n ) and thus, up to a subsequence, we assume that E uk converges to some E in L 2 (R 2n ) and a.e. in R 2n. More explicitly, there exists Z 2 R 2n such that Now, for any x, we set We remark that Z 2 = 0 and E uk (x, y) E(x, y) for every (x, y) R 2n \ Z 2. (3.5) S x := {y R n : (x, y) R 2n \ Z 2 }, W := {(x, y) R 2n : x and y R n \ S x } and V := {x : R n \ S x = 0}. W Z 2. (3.6) Indeed, if (x, y) W we have that y R n \S x, that is (x, y) R 2n \Z 2, and so (x, y) Z 2, which gives (3.6). Using (3.5) and (3.6), we obtain that W = 0, hence by the Fubini s Theorem we have that 0 = W = R n \ S x dx, which implies that R n \ S x = 0 for a.e. x. As a consequence, we conclude that \ V = 0. This and (3.3) imply that \ (V \ Z 1 ) = ( \ V ) Z 1 \ V + Z 1 = 0. In particular, we have that V \ Z 1, so we can fix x 0 V \ Z 1. Since x 0 \ Z 1, we know from (3.3) that lim u k(x 0 ) = u(x 0 ). k + Furthermore, since x 0 V we have that R n \ S x0 = 0. That is, a.e. y R n (namely, for every y S x0 ), we have that (x 0, y) R 2n \ Z 2 and so lim E u k (x 0, y) = E(x 0, y), k + thanks to (3.5). Notice also that (C) R 2n \ (C) 2 and so, recalling (3.4), we have that E uk (x 0, y) := u k(x 0 ) u k (y), x 0 y n+2s 2 for a.e. y C. Thus, we obtain } lim u k(y) = lim {u k (x 0 ) x 0 y n+2s 2 E uk (x 0, y) k + k + = u(x 0 ) x 0 y n+2s 2 E(x 0, y), a.e. y C. This and (3.3) say that u k converges a.e. in R n. Up to a change of notation, we will say that u k converges a.e. in R n to some u. So, using that u k is a Cauchy sequence in H s,

10 10 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI fixed any ɛ > 0 there exists N ɛ N such that, for any h N ɛ, ɛ 2 lim inf u h u k 2 H k + s lim inf (u h u k ) 2 + lim inf g (u h u k ) 2 k + k + C R2n\(C)2 (u h u k )(x) (u h u k )(y) 2 + lim inf k + dx R2n\(C)2 (u h u) 2 + g (u h u) 2 (u h u)(x) (u h u)(y) 2 + dx = u h u 2 H s, C where Fatou s Lemma was used. This says that u h converges to u in H s, showing that Hs is complete Some integration by parts formulas. The following is a nonlocal analogue of the divergence theorem. Lemma 3.2. Let u be any bounded C 2 function in R n. Then, ( ) s u = N s u. R n \ Proof. Note that u(x) u(y) u(y) u(x) dx = dx = 0, since the role of x and y in the integrals above is symmetric. Hence, we have that ( ) s u(x) u(y) u dx = c n,s R n dx = c n,s u(x) u(y) = c n,s dx = as desired. R n \ R n \ R n \ u(x) u(y) dx N s u(y), More generally, we have the following integration by parts formula. Lemma 3.3. Let u and v be bounded C 2 functions in R n. Then, ( )( ) c n,s u(x) u(y) v(x) v(y) 2 dx = v ( ) s u + R 2n \(C) 2 where c n,s is the constant in (1.3). Proof. Notice that ( )( ) 1 u(x) u(y) v(x) v(y) 2 R 2n \(C) 2 dx u(x) u(y) = v(x) dx + R n R n \ Thus, using (1.3) and (1.2), the identity follows. v(x) R n \ v N s u, u(x) u(y) dx.

11 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 11 Remark 3.4. We recall that if one takes ν u = 1, then one can obtain the perimeter of by integrating this Neumann condition over. Indeed, = dx = ν u dx. (3.7) Analogously, we can define Ñsu, by renormalizing N s u by a factor w s, (x) := c n,s, that is Ñ s u(x) := N su(x) w s, (x) for x Rn \. (3.8) Now, we observe that if Ñsu(x) = 1 for any x R n \, then we find the fractional perimeter (see [6] where this object was introduced) by integrating such nonlocal Neumann condition over R n \, that is: Per s () := c n,s = = = R n \ R n \ R n \ R n \ w s, (x) dx dx w s, (x) Ñsu(x) dx N s u(x) dx, that can be seen as the nonlocal counterpart of (3.7). Remark 3.5. The renormalized Neumann condition in (3.8) can also be framed into the probabilistic interpretation of Section 2. Indeed suppose that C is partitioned into a Dirichlet part D and a Neumann part N and that: our Lévy process receives a final payoff φ(x) when it leaves the domain by landing at the point x in D, if the Lévy process leaves by landing at the point x in N, then it receives an additional payoff ψ(x) and is forced to come back to and keep running by following the same probability law (the case discussed in Section 2 is the model situation in which ψ 0). In this setting, the expected payoff u(x) obtained by starting the process at the point x satisfies ( ) s u = 0 in and u = φ in D. Also, for any x N, the expected payoff landing at x must be equal to the additional payoff ψ(x) plus the average payoff u(y) obtained by jumping from x to y, that is: u(y) for any x N, u(x) = ψ(x) +, which corresponds to Ñsu(x) = ψ(x).

12 12 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI 3.2. Weak solutions with Neumann conditions. The integration by parts formula from Lemma 3.3 leads to the following: Definition 3.6. Let f L 2 () and g L 1 (R n \ ). Let u H s. We say that u is a weak solution of { ( ) s u = f in N s u = g in R n (3.9) \ whenever c n,s 2 R 2n \(C) 2 for all test functions v H s. ( )( ) u(x) u(y) v(x) v(y) dx = With this definition, we can prove the following. f v + g v (3.10) R n \ Proposition 3.7. Let f L 2 () and g L 1 (R n \ ). Let I : H s R be the functional defined as I[u] := c R2n\(C)2 n,s u(x) u(y) 2 4 dx f u g u R n \ for every u H s. Then any critical point of I is a weak solution of (3.9). Proof. First of all, we observe that the functional I is well defined on H s. Indeed, if u Hs then f u f L 2 () u L 2 () C u H s, and g u g 1/2 g 1/2 u g 1/2 L 1 (R n \) g 1/2 u L 2 (R n \) C u H s. R n \ R n \ Therefore, if u H s we have that I[u] C u H s < +. Now, we compute the first variation of I. For this, we take ɛ < 1 and v H s. Then the function u + ɛv H s, and so we can compute I[u + ɛv] = c n,s 4 = I(u) + ɛ R2n\(C)2 (u + ɛv)(x) (u + ɛv)(y) 2 dx f(u + ɛv) g(u + ɛv) ( c n,s 2 R n \ R 2n \(C) 2 (u(x) u(y))(v(x) v(y)) dx + c n,s 4 ɛ2 R2n\(C)2 v(x) v(y) 2 dx. ) f v g v R n \

13 Hence, NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 13 I[u + ɛv] I[u] lim ɛ 0 ɛ = c n,s 2 which means that I [u](v) = c n,s 2 R 2n \(C) 2 (u(x) u(y))(v(x) v(y)) dx R 2n \(C) 2 (u(x) u(y))(v(x) v(y)) dx f v g v, R n \ f v g v. R n \ Therefore, if u is a critical point of I, then u is a weak solution to (3.9), according to Definition 3.6. Next result is a sort of maximum principle and it is auxiliary towards the existence and uniqueness theory provided in the subsequent Theorem 3.9. Lemma 3.8. Let f L 2 () and g L 1 (R n \ ). Let u be any H s function satisfying in the weak sense { ( ) s u = f in N s u = g in R n \, with f 0 and g 0. Then, u is constant. Proof. First, we observe that the function v 1 belongs to H s, and therefore we can use it as a test function in (3.10), obtaining that 0 f = g 0. This implies that R n \ f = 0 a.e. in and g = 0 a.e. in R n \. Therefore, taking v = u as a test function in (3.10), we deduce that and hence u must be constant. R2n\(C)2 u(x) u(y) 2 dx = 0, We can now give the following existence and uniqueness result (we observe that its statement is in complete analogy 2 with the classical case, see e.g. page 294 in [16]). Theorem 3.9. Let R n be a bounded Lipschitz domain, f L 2 (), and g L 1 (R n \). Suppose that there exists a C 2 function ψ such that N s ψ = g in R n \. Then, problem (3.9) admits a solution in H s if and only if f = g. (3.11) R n \ Moreover, in case that (3.11) holds, the solution is unique up to an additive constant. 2 The only difference with the classical case is that in Theorem 3.9 it is not necessary to suppose that the domain is connected in order to obtain the uniqueness result.

14 14 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Proof. Case 1. We do first the case g 0, i.e., with homogeneous nonlocal Neumann conditions. We also assume that f 0, otherwise there is nothing to prove. Given h L 2 (), we look for a solution v H s of the problem (v(x) v(y))(ϕ(x) ϕ(y)) v ϕ + R 2n \(C) 2 dx = h ϕ, (3.12) for any ϕ H s, with homogeneous Neumann conditions N sv = 0 in R n \. We consider the functional F : H s R defined as F(ϕ) := h ϕ for any ϕ H. s It is easy to see that F is linear. Moreover, it is continuous on H s : F(ϕ) h ϕ h L 2 () ϕ L 2 () h L 2 () ϕ H s. Therefore, from the Riesz representation theorem it follows that problem (3.12) admits a unique solution v H s for any given h L2 (). Furthermore, taking ϕ := v in (3.12), one obtain that v H s () C h L 2 (). (3.13) Now, we define the operator T o : L 2 () H s as T oh = v. We also define by T the restriction operator in, that is T h = T o h. That is, the function T o h is defined in the whole of R n, then we take T h to be its restriction in. In this way, we see that T : L 2 () L 2 (). We have that T is compact. Indeed, we take a sequence {h k } k N bounded in L 2 (). Hence, from (3.13) we deduce that the sequence of T h k is bounded in H s (), which is compactly embedded in L 2 () (see e.g. [12]). Therefore, there exists a subsequence that converges in L 2 (). Now, we show that T is self-adjoint. For this, we take h 1, h 2 C0 () and we use the weak formulation in (3.12) to say that, for every ϕ, φ H s, we have and T o h 1 ϕ + R 2n \(C) 2 (T o h 1 (x) T o h 1 (y))(ϕ(x) ϕ(y)) dx = h 1 ϕ, (3.14) (T o h 2 (x) T o h 2 (y))(φ(x) φ(y)) T o h 2 φ + R 2n \(C) 2 dx = h 2 φ, (3.15) Now we take ϕ := T o h 2 and φ := T o h 1 in (3.14) and (3.15) respectively and we obtain that h 1 T o h 2 = T o h 1 h 2 for any h 1, h 2 C0 (). Accordingly, since T oh 1 = T h 1 and T o h 2 = T h 2 in, we conclude that h 1 T h 2 = T h 1 h 2 (3.16)

15 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 15 for any h 1, h 2 C0 (). If h 1, h 2 L 2 (), there exist sequences of functions in C0 (), say h 1,k and h 2,k, such that h 1,k h 1 and h 2,k h 2 in L 2 () as k +. From (3.16) we have that h 1,k T h 2,k = T h 1,k h 2,k. (3.17) Moreover, from (3.13) we deduce that T h 1,k T h 1 and T h 2,k T h 2 in H s () as k +, and so h 1,k T h 2,k h 1 T h 2 as k + and T h 1,k h 2,k T h 1 h 2 as k +. The last two formulas and (3.17) imply that h 1 T h 2 = T h 1 h 2 for any h 1, h 2 L 2 (), which says that T is self-adjoint. Now we prove that Ker(Id T ) consists of constant functions. (3.18) First of all, we check that the constants are in Ker(Id T ). We take a function constantly equal to c and we observe that ( ) s c = 0 in (hence ( ) s c + c = c) and N s c = 0 in R n \. This shows that T o c = c in R n, and so T c = c in, which implies that c Ker(Id T ). Viceversa, now we show that if v Ker(Id T ) L 2 (), then v is constant. For this, we consider T o v H s. By construction ( ) s (T o v) + (T o v) = v in, (3.19) in the weak sense, and N s (T o v) = 0 in R n \. (3.20) On the other hand, since v Ker(Id T ), we have that Hence, by (3.19), we have that v = T v = T o v in. (3.21) ( ) s (T o v) = 0 in. Using this, (3.20) and Lemma 3.8, we obtain that T o v is constant. Thus, by (3.21), we obtain that v is constant in and this completes the proof of (3.18). From (3.18) and the Fredholm Alternative, we conclude that Im(Id T ) = Ker(Id T ) = {constant functions}, where the orthogonality notion is in L 2 (). More explicitly, { } Im(Id T ) = f L 2 () s.t. f = 0. (3.22) Now, let us take f such that f = 0. By (3.22), we know that there exists w L2 () such that f = w T w. Let us define u := T o w. By construction, we have that N s u = 0 in R n \, and that ( ) s (T o w) + (T o w) = w in.

16 16 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Consequently, in, f = w T w = w T o w = ( ) s (T o w) = ( ) s u, and we found the desired solution in this case. Viceversa, if we have a solution u H s of ( )s u = f in with N s u = 0 in R n \, we set w := f + u and we observe that ( ) s u + u = f + u = w in. Accordingly, we have that u = T o w in R n, hence u = T w in. This says that (Id T )w = w u = f in and so f Im(Id T ). Thus, by (3.22), we obtain that f = 0. This establishes the validity of Theorem 3.9 when g 0. Case 2. Let us now consider the nonhomogeneous case (3.9). By the hypotheses, there exists a C 2 function ψ satisfying N s ψ = g in R n \. Let ū = u ψ. Then, ū solves { ( )sū = f in with N s u = 0 in R n \, f = f ( ) s ψ. Then, as we alrea proved, this problem admits a solution if and only if f = 0, i.e., if 0 = f = f ( ) s ψ. (3.23) But, by Lemma 3.2, we have that ( ) s ψ = R n \ N s ψ = g. R n \ From this and (3.23) we conclude that a solution exists if and only if (3.11) holds. Finally, the solution is unique up to an additive constant thanks to Lemma Eigenvalues and eigenfunctions. Here we discuss the spectral properties of problem (1.1). For it, we will need the following classical tool. Lemma 3.10 (Poincaré inequality). Let R n be any bounded Lipschitz domain, and let s (0, 1). Then, for all functions u H s (), we have u 2 u u(x) u(y) 2 dx C dx, where the constant C > 0 depends only on and s. Proof. We give the details for the facility of the reader. We argue by contradiction and we assume that the inequality does not hold. Then, there exists a sequence of functions u k H s () satisfying u k = 0, u k L 2 () = 1, (3.24)

17 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 17 and u k (x) u k (y) 2 dx < 1 k. (3.25) In particular, the functions {u k } k 1 are bounded in H s (). Using now that the embedding H s () L 2 () is compact (see e.g. [12]), it follows that a subsequence {u kj } j 1 converges to a function ū L 2 (), i.e., u kj ū in L 2 (). Moreover, we deduce from (3.24) that ū = 0, and ū L 2 () = 1. (3.26) On the other hand, (3.25) implies that ū(x) ū(y) 2 dx = 0. Thus, ū is constant in, and this contradicts (3.26). We finally give the description of the eigenvalues of ( ) s with zero Neumann boundary conditions. Theorem Let R n be a bounded Lipschitz domain. Then, there exist a sequence of nonnegative values 0 = λ 1 < λ 2 λ 3, and a sequence of functions u i : R n R such that { ( ) s u i (x) = λ i u i (x) for any x N s u i (x) = 0 for any x R n \. Also, the functions u i (when restricted to ) provide a complete orthogonal system in L 2 (). Proof. We define L 2 0() := { u L 2 () : } u = 0. Let the operator T o be defined by T o f = u, where u is the unique solution of { ( ) s u = f in N s u = 0 in R n \ according to Definition 3.6. We remark that the existence and uniqueness of such solution is a consequence of the fact that f L 2 0 () and Theorem 3.9. We also define T to be the restriction of T o in, that is T f = T o f. In this way T : L 2 0 () L2 0 (). Also, we claim that the operator T is compact and self-adjoint. We first show that T is compact. Indeed, taking v = u = T o f in the weak formulation of the problem (3.10), we obtain c n,s 2 R2n\(C)2 u(x) u(y) 2 dx f L 2 () u L 2 (). (3.27)

18 18 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Now, using the Poincaré inequality in Lemma 3.10 (recall that u = 0), we deduce that ( u(x) u(y) 2 ) 1/2 u L 2 () C dx. (3.28) This and (3.27) give that ( u(x) u(y) 2 ) 1/2 dx C f L 2(). (3.29) Now, we take a sequence {f k } k N bounded in L 2 (). From (3.28) and (3.29) we obtain that u k = T f k is bounded in H s (). Hence, since the embedding H s () L 2 () is compact, there exists a subsequence that converges in L 2 (). Therefore, T is compact. Now we show that T is self-adjoint in L 2 0 (). For this, we take f 1 and f 2 in C0 (), with f 1 = f 2 = 0. Then from the weak formulation in (3.10) we have that, for every v, w H s, c n,s (T o f 1 (x) T o f 1 (y))(v(x) v(y)) 2 R 2n \(C) 2 dx = f 1 v (3.30) and c n,s 2 R 2n \(C) 2 (T o f 2 (x) T o f 2 (y))(w(x) w(y)) dx = f 2 w. (3.31) We observe that we can take v := T o f 2 in (3.30) and w := T o f 1 in (3.31) (and recall that T o f i = T f i in ), obtaining that f 1 T f 2 = f 2 T f 1, for any f 1, f 2 C0 (). (3.32) Now, if f 1, f 2 L 2 0 () we can find sequences of functions f 1,k, f 2,k C0 () such that f 1,k f 1 and f 2,k f 2 in L 2 () as k +. Therefore, from (3.32), we have f 1,k T f 2,k = f 2,k T f 1,k. (3.33) We notice that, thanks to (3.28) and (3.29), T f 1,k T f 1 and T f 2,k T f 2 in L 2 () as k +, and therefore, from (3.33), we obtain that f 1 T f 2 = f 2 T f 1, thus proving that T is self-adjoint in L 2 0 (). Thus, by the spectral theorem there exists a sequence of eigenvalues {µ i } i 2 of T, and its corresponding eigenfunctions {e i } i 2 are a complete orthogonal system in L 2 0 (). We remark that Indeed, suppose by contradiction that µ i = 0. Then µ i 0. (3.34) 0 = µ i e i = T e i = T o e i in. (3.35)

19 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 19 By construction, N s (T o e i ) = 0 in R n \. This and (3.35) give that T o e i (y) T o e i (x) = = 0 in R n \. Using this and (3.35) once again we conclude that T o e i 0 a.e. in R n. Therefore 0 = ( ) s (T o e i ) = e i in, which gives that e i 0 in, hence it is not an eigenfunction. This establishes (3.34). From (3.34), we can define λ i := µ 1 i. We also define u i := T o e i and we claim that u 2, u 3, is the desired system of eigenfunctions, with corresponding eigenvalues λ 2, λ 3, Indeed, u i = T o e i = T e i = µ i e i in, (3.36) hence the orthogonality and completeness properties of u 2, u 3, in L 2 0 () follow from those of e 2, e 3, Furthermore, in, we have that ( ) s u i = ( ) s (T o e i ) = e i = λ i u i, where (3.36) was used in the last step, and this proves the desired spectral property. Now, we notice that λ i > 0 for any i 2. (3.37) Indeed, its corresponding eigenfunction u i solves { ( ) s u i = λ i u i in N s u i = 0 in R n \, (3.38) Then, if we take u i as a test function in the weak formulation of (3.38), we obtain that c R2n\(C)2 n,s u i (x) u i (y) 2 2 dx = λ i u 2 i, which implies that λ i 0. Now, suppose by contradiction that λ i = 0. Then, from Lemma 3.8 we have that u i is constant. On the other hand, we know that u i L 2 0 (), and this implies that u i 0, which is a contradiction since u i is an eigenfunction. This establishes (3.37). From (3.37), up to reordering them, we can suppose that 0 < λ 2 λ 3 ; now, we notice that λ 1 := 0 is an eigenvalue (its eigenfunction is a constant), thanks to Lemma 3.8. Therefore, we have a sequence of eigenvalues 0 = λ 1 < λ 2 λ 3, and its corresponding eigenfunctions are a complete orthogonal system in L 2 (). This concludes the proof of Theorem Remark We point out that the notion of eigenfunctions in Theorem 3.11 is not completely standard. Indeed, the eigenfunctions u i corresponding to the eigenvalues λ i are defined in the whole of R n, but they satisfy an orthogonality conditions only in L 2 (). Alternatively, one can think that the natural domain of definition for u i is itself, since there the eigenvalue equation ( ) s u i = λ i u i takes place, together with the orthogonality condition, and then u i is naturally extended outside via the nonlocal

20 20 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Neumann condition. Notice indeed that the condition N s u i = 0 in R n \ is equivalent to prescribing u i outside from the values inside according to the formula u i (y) u i (x) = for any x R n \. In the following proposition we deal with the behavior of the solution of (1.1) at infinity. Proposition Let R n be a bounded domain and let u H s be a weak solution (according to Definition 3.6) of { ( ) s u = f in N s u = 0 in R n \. Then lim u(x) = 1 u uniformly in x. x Proof. First we observe that, since is bounded, there exists R > 0 such that B R. Hence, if y, we have that and so x R x y x + R, 1 R x y 1 + R x x x. Therefore, given ɛ > 0, there exists R > R such that, for any x R, we have x n+2s = 1 + γ(x, y), where γ(x, y) ɛ. Recalling the definition of N s u given in (1.2) and using the fact that N s u = 0 in R n \, we have that for any x R n \ u(y) x n+2s u(y) u(x) = = x n+2s (1 + γ(x, y)) u(y) = (1 + γ(x, y)) u(y) + γ(x, y) u(y) =. + γ(x, y) We set γ 1 (x) := γ(x, y) u(y) and γ 2 (x) := γ(x, y),

21 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 21 and we notice that γ 1 (x) Cɛ and γ 2 (x) ɛ, for some C > 0. Hence, we have that for any x R n \ u(x) u(y) + γ 1 (x) u(y) = u(y) 1 + γ 2 (x) γ 1(x) γ 2 (x) u(y) = 1 + γ 2 (x) C ɛ 1 ɛ. Therefore, sending ɛ 0 (that is, x + ), we obtain the desired result. Remark 3.14 (Interior regularity of solutions). We notice that, in particular, Proposition 3.13 implies that u is bounded at infinity. Thus, if solutions are locally bounded, then one could apply interior regularity results for solutions to ( ) s u = f in (see e.g. [17, 20, 7, 19]). 4. The heat equation Here we show that solutions of the nonlocal heat equation with zero Neumann datum preserve their mass and have energy that decreases in time. To avoid technicalities, we assume that u is a classical solution of problem (1.4), so that we can differentiate under the integral sign. Proposition 4.1. Assume that u(x, t) is a classical solution to (1.4), in the sense that u is bounded and u t + ( ) s u K for all t > 0. Then, for all t > 0, u(x, t) dx = u 0 (x)dx. In other words, the total mass is conserved. Proof. By the dominated convergence theorem, and using Lemma 3.2, we have d u = u t = ( ) s u = N s u = 0. dt R n \ Thus, the quantity u does not depend on t, and the result follows. Proposition 4.2. Assume that u(x, t) is a classical solution to (1.4), in the sense that u is bounded and u t + ( ) s u K for all t > 0. Then, the energy E(t) = is decreasing in time t > 0. R2n\(C)2 u(x, t) u(y, t) 2 dx

22 22 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Proof. Let us compute E (t), and we will see that it is negative. Indeed, using Lemma 3.3, E (t) = d R2n\(C)2 u(x, t) u(y, t) 2 dt dx ( )( 2 u(x, t) u(y, t) ut (x, t) u t (y, t) ) = R 2n \(C) 2 dx = 4 u t ( ) s u dx, c n,s where we have used that N s u = 0 in R n \. Thus, using now the equation u t + ( ) s u = 0 in, we find E (t) = 4 ( ) s u 2 dx 0, c n,s with strict inequality unless u is constant. Next we prove that solutions of the nonlocal heat equation with Neumann condition approach a constant as t + : Proposition 4.3. Assume that u(x, t) is a classical solution to (1.4), in the sense that u is bounded and u t + ( ) s u K for all t > 0. Then, u 1 u 0 in L 2 () as t +. Proof. Let m := 1 u 0 be the total mass of u. Define also A(t) := u m 2 dx. Notice that, by Proposition 4.1, we have ( A(t) = u 2 2mu + m 2) dx = Then, by Lemma 3.3, A (t) = 2 u t u dx = 2 u( ) s u dx = c n,s u 2 dx m 2. R 2n \(C) 2 u(x, t) u(y, t) 2 dx. Hence, A is decreasing. Moreover, using the Poincaré inequality in Lemma 3.10 and again Proposition 4.1, we deduce that A (t) c u m 2 dx = c A(t), for some c > 0. Thus, it follows that A(t) e ct A(0),

23 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 23 and thus lim u(x, t) m 2 dx = 0, t + i.e., u converges to m in L 2 (). Notice that, in fact, we have proved that the convergence is exponentially fast. 5. Limits In this section we stu the limits as s 1 and the continuity properties induced by the fractional Neumann condition Limit as s 1. Proposition 5.1. Let R n be any bounded Lipschitz domain. Let u and v be C0 2(Rn ) functions. Then, u lim N s u v = s 1 R n \ ν v. Proof. By Lemma 3.3, we have that N s u v = c n,s (u(x) u(y))(v(x) v(y)) 2 R 2n \(C) 2 dx R n \ Now, we claim that c n,s (u(x) u(y))(v(x) v(y)) lim s 1 2 R 2n \(C) 2 dx = v( ) s u. (5.1) u v. (5.2) We observe that to show (5.2), it is enough to prove that, for any u C0 2(Rn ), c R2n\(C)2 n,s u(x) u(y) 2 lim s 1 2 dx = u 2. (5.3) Indeed, R 2n \(C) 2 (u(x) u(y))(v(x) v(y)) dx = 1 2 R2n\(C)2 (u + v)(x) (u + v)(y) 2 dx R2n\(C)2 u(x) u(y) 2 dx R2n\(C)2 v(x) v(y) 2 dx. Now, we recall that c n,s lim s 1 1 s = 4n, ω n 1 (see Corollary 4.2 in [12]), and so we have to show that lim (1 s) u(x) u(y) 2 s 1 R 2n \(C) 2 dx = ω n 1 u 2. (5.4) 2n

24 24 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI For this, we first show that lim (1 s) u(x) u(y) s 1 (C) 2 dx = 0. (5.5) Without loss of generality, we can suppose that B r B R, for some 0 < r < R. Since u C 2 0 (Rn ), then (C) u(x) u(y) 2 dx 4 u 2 L (R n ) 4 u 2 L (R n ) which implies (5.5). Hence, (C) 4 u 2 L (R n ) ω n 1 = 4 u 2 L (R n ) ω n 1 B R (CB r) B R B R 1 dx 1 dx dx dx = 4 u 2 ω n 1 r 2s L (R n ) 2s + r + B R = 4 u 2 ωn 1 2 Rn r 2s L (R n ), 2s r dx ρ n 1 ρ n 2s dρ ρ 1 2s dρ lim (1 s) u(x) u(y) 2 s 1 R 2n \(C) 2 dx = lim s 1 (1 s) u(x) u(y) 2 dx = C n u 2, (5.6) where C n > 0 depends only on the dimension, see [5]. In order to determine the constant C n, we take a C 2 -function u supported in. In this case, we have u 2 dx = u 2 dx = R n ξ 2 û(ξ) 2 dξ, R n (5.7) where û is the Fourier transform of u. Moreover, R2n\(C)2 u(x) u(y) 2 dx = R2n u(x) u(y) 2 dx ξ 2s û(ξ) 2 dx, R n = 2 c 1 n,s

25 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 25 thanks to Proposition 3.4 in [12]. Therefore, using Corollary 4.2 in [12] and (5.7), we have lim (1 s) u(x) u(y) 2 s 1 R 2n \(C) 2 dx 2(1 s) = lim ξ 2s û(ξ) 2 dx s 1 c n,s R n = ω n 1 ξ 2 û(ξ) 2 dx 2n R n = ω n 1 u 2 dx. 2n Hence, the constant in (5.6) is C n = ω n 1 2n. This concludes the proof of (5.4), and in turn of (5.2). On the other hand, ( ) s u u uniformly in R n, (see Proposition 4.4 in [12]). This, (5.1) and (5.2) give N s u v = u v + v u = as desired. lim s 1 R n \ u ν v, 5.2. Continuity properties. Following is a continuity result for functions satisfying the nonlocal Neumann condition: Proposition 5.2. Let R n be a domain with C 1 boundary. Let u be continuous in, with N s u = 0 in R n \. Then u is continuous in the whole of R n. Proof. First, let us fix x 0 R n \. Since the latter is an open set, there exists ρ > 0 such that x 0 y ρ for any y. Thus, if x B ρ/2 (x 0 ), we have that x y x 0 y x 0 x ρ/2. Moreover, if x B ρ/2 (x 0 ), we have that x y y x 0 x 0 x y 2 + ( ) ( y y 4 x ρ ) y 2 2, provided that y R := 4 x 0 + 2ρ. As a consequence, for any x B ρ/2 (x 0 ), we have that ( u(y) + 1 2n+2s ( u L () + 1) χbr (y) ρ n+2s + χ ) Rn \B R (y) y n+2s =: ψ(y) and the function ψ belongs to L 1 (R n ). Thus, by the Neumann condition and the Dominated Convergence Theorem, we obtain that u(y) u(y) lim u(x) = lim x 0 y n+2s x x 0 x x 0 = This proves that u is continuous at any points of R n \. = u(x 0 ). x 0 y n+2s

26 26 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Now we show the continuity at a point p. We take a sequence p k p as k +. We let q k be the projection of p k to. Since p, we have from the minimizing property of the projection that p k q k = inf p k ξ p k p, ξ and so q k p q k p k + p k p 2 p k p 0 as k +. Therefore, since we alrea know from the assumptions the continuity of u at, we obtain that lim u(q k) = u(p). (5.8) k + Now we claim that lim u(p k) u(q k ) = 0. (5.9) k + To prove it, it is enough to consider the points of the sequence p k that belong to R n \ (since, of course, the points p k belonging to satisfy p k = q k and for them (5.9) is obvious). We define ν k := (p k q k )/ p k q k. Notice that ν k is the exterior normal of at q k. We consider a rigid motion R k such that R k q k = 0 and R k ν k = e n = (0,, 0, 1). Let also h k := p k q k. Notice that h 1 k R kp k = h 1 k R k(p k q k ) = R k ν k = e n. (5.10) Then, the domain k := h 1 k R k has vertical exterior normal at 0 and approaches the halfspace Π := {x n < 0} as k +. Now, we use the Neumann condition at p k and we obtain that u(y) p u(p k ) u(q k ) = k y n+2s u(q k ) p k y n+2s u(y) u(q k ) p = k y n+2s = I 1 + I 2, p k y n+2s with I 1 := and I 2 := B h k (q k ) \B h k (q k ) u(y) u(q k ) p k y n+2s p k y n+2s u(y) u(q k ) p k y n+2s. p k y n+2s

27 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 27 We observe that the uniform continuity of u in gives that As a consequence lim sup k + y B (q h k ) k u(y) u(q k ) = 0. I 1 sup u(y) u(q k ) 0 (5.11) y B (q h k ) k as k +. Moreover, exploiting the change of variable η := h 1 k R ky and recalling (5.10), we obtain that u(y) u(q k ) \B h (q k ) p k y n+2s I 2 k p k y n+2s Notice that, if η k \ B 1/ hk 2 u L () = 2 u L () then \B h k (q k ) k \B 1/ hk k p k y n+2s p k y n+2s e n η n+2s. e n η n+2s e n η n+2s = e n η n+s e n η s e n η n+s( η 1 ) s for large k. Therefore Since e n η n+s( ) h 1/2 s k 1 en η n+s h s/4 I 2 2h s/4 k lim k + k k u L () k k e n η n+s = e n η n+2s e n η n+s. e n η n+2s Π Π k e n η n+s, e n η n+2s we conlude that I 2 0 as k +. This and (5.11) imply (5.9). From (5.8) and (5.9), we conclude that hence u is continuous at p. lim u(p k) = u(p), k +

28 28 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI As a direct consequence of Proposition 5.2 we obtain: Corollary 5.3. Let R n be a domain with C 1 boundary. Let v 0 C(R n ). Let v 0 (x) if x, v 0 (y) v(x) := x y n+2s if x Rn \. Then v C(R n ) and it satisfies v = v 0 in and N s v = 0 in R n \. Proof. By construction, v = v 0 in and N s v = 0 in R n \. Then we can use Proposition 5.2 and obtain that v C(R n ). Now we stu the boundary behavior of the nonlocal Neumann function Ñsu. Proposition 5.4. Let R n be a C 1 domain, and u C(R n ). Then, for all s (0, 1), lim Ñ s u(x) = 0. (5.12) x x R n \ Also, if s > 1 2 and u C1,α (R n ) for some α (0, 2s 1), then Ñ s u(x + ɛν) ν Ñ s u(x) := lim ɛ 0 + ɛ for some constant κ > 0. = κ ν u for any x, (5.13) Proof. Let x k be a sequence in R n \ such that x k x as k +. By Corollary 5.3 (applied here with v 0 := u), there exists v C(R n ) such that v = u in and N s v = 0 in R n \. By the continuity of u and v we have that Moreover This and (5.14) imply that that is (5.12). lim u(x k) v(x k ) = u(x ) v(x ) = 0. (5.14) k + Ñ s u(x k ) = Ñsu(x k ) Ñsv(x k ) u(x k ) u(y) v(x k ) v(y) x = k y n+2s x k y n+2s x k y n+2s u(x k ) v(x k ) x = k y n+2s x k y n+2s = u(x k ) v(x k ). lim Ñ s u(x k ) = 0, k +

29 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 29 Now, we prove (5.13). For this, we suppose that s > 1 2, that 0 and that the exterior normal ν coincides with e n = (0,, 0, 1); then we use (5.12) and the change of variable η := ɛ 1 y in the following computation: ɛ 1( ) Ñ s u(ɛe n ) Ñsu(0) = ɛ 1 Ñ s u(ɛe n ) ɛ 1 u(ɛe n ) u(y) ɛe = n y n+2s ɛe n y n+2s ɛ 1 u(ɛe n ) u(ɛη) 1 ɛ = e n η n+2s = I 1 + I 2, e n η n+2s 1 ɛ where I 1 := and I 2 := u(ɛe n ) (e n η) 1 ɛ e n η n+2s e n η n+2s ɛ 1 1 ɛ u(ɛe n ) u(ɛη) ɛ u(ɛe n ) (e n η) 1 ɛ e n η n+2s. e n η n+2s 1 ɛ So, if Π := {x n < 0}, we have that lim I 1 = ɛ 0 + = Π Π u(0) (e n η) e n η n+2s Π e n η n+2s n u(0)(1 η n ) e n η n+2s, e n η n+2s Π where we have used that, for any i {1,, n 1} the map η its integral averages to zero. So, we can write (1 η n ) lim I Π e 1 = κ n u(0) with κ := n η ɛ 0 + Π n+2s iu(0) η i e n η n+2s is odd and so. (5.15) e n η n+2s

30 30 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI We remark that κ is finite, since s > 1 2. Moreover ɛ 1 u(ɛen ) u(ɛη) ɛ u(ɛe n ) (e n η) 1 ( ) = u(tɛe n + (1 t)ɛη) u(ɛe n ) (e n η) dt As a consequence ɛ α I 2 0 u C 1,α(R n ) e n η 1 u C 1,α(R n )ɛ α e n η 1+α. u C 1,α(R n ) 1 ɛ 1 ɛ 0 e n η n+2s 1 α e n η n+2s tɛe n + (1 t)ɛη ɛe n α dt u C 1,α(R n ) Π Π e n η n+2s 1 α e n η n+2s as ɛ 0, which is finite, thanks to our assumptions on α. This shows that I 2 0 as ɛ 0. Hence, recalling (5.15), we get that ) lim ɛ 1( Ñ s u(ɛe n ) Ñsu(0) = κ n u(0), ɛ 0 + which establishes (5.13). 6. An overdetermined problem In this section we consider an overdetemined problem. For this, we will use the renormalized nonlocal Neumann condition that has been introduced in Remark 3.4. Indeed, as we pointed out in Remark 3.5, this is natural if one considers nonhomogeneous Neumann conditions. Theorem 6.1. Let R n be a bounded and Lipschitz domain. Then there exists no function u C(R n ) satisfying { u(x) = 0 for any x R n \ Ñ s u(x) = 1 for any x R n (6.1) \. Remark 6.2. We notice that u = χ satisfies (6.1), but it is a discontinuous function. Proof. Without loss of generality, we can suppose that 0. We argue by contradiction and we assume that there exists a continuous function u that satisfies (6.1). Therefore, there exists δ > 0 such that u 1/2 in B δ. (6.2) Since is Lipschitz, up to choosing δ small enough, we have that B δ = B δ, where := {x = (x, x n ) R n 1 R s.t. x n < γ(x )} for a suitable Lipschitz function γ : R n 1 R such that γ(0) = 0 and x γ(0) = 0. Now we let x := ɛ e n R n \, for suitable ɛ > 0 sufficiently small. We observe that u(ɛ e n ) = 0. (6.3)

31 NONLOCAL PROBLEMS WITH NEUMANN BOUNDARY CONDITIONS 31 Moreover we consider the set { 1 ɛ = y = (y, y n ) R n 1 R s.t. y n < 1 } ɛ γ(ɛy ). We also define K := { y = (y, y n ) R n 1 R s.t. y n < L y }, where L is the Lipschitz constant of γ. We claim that K ɛ 1. (6.4) Indeed, since γ is Lipschitz and 0, we have that and so, if y K, γ(ɛy ) = γ(ɛy ) + γ(0) L ɛ y, y n L y 1 ɛ γ(ɛy ), which implies that y ɛ 1. This shows (6.4). Now we define Σ ɛ := ɛe n y n+2s, and we observe that B δ B δ u(y) u(ɛ e n ) ɛe n y n+2s 1 2 Σ ɛ, (6.5) thanks to (6.3) and (6.2). Furthermore, if y R n \ B δ and ɛ δ/2, we have which implies that u(y) u(ɛ e n ) C \B δ ɛe n y n+2s y ɛe n y ɛ y 2, \B δ up to renaming the constants. On the othe hand, we have that ɛe n y n+2s Finally, we observe that ɛ 2s Σ ɛ =ɛ 2s = =:κ, ɛe n y n+2s C R n \B δ B δ B δ B δ/ɛ (ɛ 1 ) B δ/ɛ K y n+2s = C δ 2s, (6.6) ɛe n y n+2s = Σ ɛ. (6.7) ɛe n y n+2s dz e n z n+2s dz e n z n+2s where we have used the change of variable y = ɛz and (6.4). (6.8)

32 32 SERENA DIPIERRO, XAVIER ROS-OTON, AND ENRICO VALDINOCI Hence, using the second condition in (6.1) and putting together (6.5), (6.6), (6.7) and (6.8), we obtain 0 = ɛe n y n+2s u(ɛe n ) u(x) ɛe n y n+2s = ɛe n y n+2s Σ ɛ 1 2 Σ ɛ C δ 2s u(ɛe n ) u(x) B δ ɛe n y n+2s u(ɛe n ) u(x) \B δ ɛe n y n+2s = 1 2 Σ ɛ C δ 2s ( ) ɛ = ɛ 2s 2s 2 Σ ɛ C ɛ 2s δ 2s ( ɛ 2s κ 2 C ɛ2s δ 2s) > 0 if ɛ is sufficiently small. This gives a contradiction and concludes the proof. 7. Comparison with previous works In this last section we compare our new Neumann nonlocal conditions with the previous works in the literature that also deal with Neumann-type conditions for the fractional Laplacian ( ) s. The idea of [4, 8] (and also [9, 10, 11]) is to consider the regional fractional Laplacian, associated to the Dirichlet form ( )( ) u(x) u(y) v(x) v(y) c n,s dx. (7.1) This operator corresponds to a censored process, i.e., a process whose jumps are restricted to be in. The operator can be defined in general domains, and seems to give a natural analogue of homogeneous Neumann condition. However, no nonhomogeneous Neumann conditions can be considered with this model, and the operator depends on the domain. On the other hand, in [1, 3] the usual diffusion associated to the fractional Laplacian (1.3) was considered inside, and thus the particle can jump outside. When it jumps outside, then it is reflected or projected inside in a deterministic way. Of course, different types of reflections or projections lead to different Neumann conditions. To appropriately define these reflections, some assumptions on the domain (like smoothness or convexity) need to be done. In contrast with the regional fractional Laplacian, this problem does not have a variational formulation and everything is done in the context of viscosity solutions. Finally, in [15] a different Neumann problem for the fractional Laplacian was considered. Solutions to this type of Neumann problems are large solutions, in the sense that they are not bounded in a neighborhood of. More precisely, it is proved in [15] that the following problem is well-posed ( ) s u = f in u = 0 in R n \ ν ( u/d s 1 ) = g on,