Symmetry of odd solutions to equations with fractional Laplacian

We present a symmetry result to solutions of equations involving the fractional Laplacian in a domain with at least two perpendicular symmetries. We show that if the solution is continuous, bounded, and odd in one direction such that it has a fixed sign on one side, then it will be symmetric in the perpendicular direction. Moreover, the solution will be monotonic in the part where it is of fixed sign. In addition, we present also a class of examples in which our result can be applied.


Introduction
In the following, we study symmetries of odd solutions to the nonlinear problem where Ω ⊂ ℝ N is an open set, f ∈ C(Ω × ℝ) , and (−Δ) s , s ∈ (0, 1) is the fractional Laplacian given for ∈ C ∞ c (ℝ N ) by with a normalization constant c N,s > 0. Symmetries of nonnegative solutions to problem (1.1) have been studied in detail by various authors (see [3,11,16,17]), where f satisfies some monotonicity and symmetry in x 1 and Ω is symmetric in x 1 . Here, we aim at investigating (1.1), where Ω has two perpendicular symmetries and the solution u is odd in one of these directions. For the variational framework, see also [1,2,24,25] and the references in there. To give a precise framework of our statements, we assume the following: (D) Ω ⊂ ℝ N with N ∈ ℕ , N ≥ 2 is open and bounded and, moreover, convex and symmetric in the directions x 1 and x N . That is, for every (x 1 , … , x N ) ∈ Ω , t, ∈ [−1, 1] we have (tx 1 , x 2 … , x N−1 , x N ) ∈ Ω. (F1) f ∈ C(Ω × ℝ) and for every bounded set K ⊂ ℝ there is L = L(K) > 0 such that (F2) f is symmetric in x 1 and monotone in |x 1 | . That is, for every u ∈ ℝ , x ∈ Ω , and t ∈ [−1, 1] we have f (tx 1 , In this work, we consider weak solutions of (1.1), i.e., u ∈ H s 0 (Ω) ∶= {v ∈ H s (ℝ N ) ∶ u = 0 onℝ N ⧵ Ω} is called a (weak) solution of (1.1), if whenever the right-hand side exists, where is the bilinearform associated to (−Δ) s . Here, H s (ℝ N ) = {u ∈ L 2 (ℝ N ) ∶ E s (u, u) < ∞} is the usual fractional Hilbert space of order s (see e.g. [1,5,7]). Denote e i ∶= ( ij ) 1≤j≤N ∈ ℝ N , where ij = 1 if j = i and 0 otherwise is the usual Kronecker Delta. Moreover, for ∈ ℝ , consider the halfspace and denote by the reflection of x at H i, ( ) . Note that r 1,0 (Ω) = r N,0 (Ω) = Ω , if assumption (D) is satisfied.
We call u ∶ ℝ N → ℝ symmetric with respect to H i, , if u•r i, = u and we call u antisymmetric with respect to H i, , if u•r i, = −u.
Symmetry of odd solutions to equations with fractional… and either u ≡ 0 in Ω or u| Ω∩H 1,0 ∩H N,0 is strictly decreasing in x 1 , that is, for every x, y ∈ Ω ∩ H 1,0 ∩ H N,0 with x 1 < y 1 we have u(x) > u(y).
We note that Theorem 1.1 is not surprising in the local case, where (−Δ) s is considered with s = 1 , if we have u > 0 in H N,0 ∩ Ω . In this case, the conclusion follows by simply considering the solution restricted to its part of nonnegativity and apply the usual symmetry result due to [13]. We emphasize however, that if this positivity assumption is reduced to a nonnegativity assumption, then in general the claimed monotonicity is not true in the local case and presents a purely nonlocal feature. Moreover, in the nonlocal setting, we are not able to simply restrict the solution to its set of nonnegativity. Due to this, we present in Sect. 2 below new maximum principles for doubly antisymmetric functions to certain linear problems, which we believe are of independent interest.
Let us emphasize that if u ∈ L ∞ (ℝ N ) is antisymmetric with respect to H N,0 , it follows that for any x ∈ H N,0 , such that u is regular enough at x, we have with a change of variables where (−Δ| H N,0 ) s denotes the so-called spectral fractional Laplacian (c.f. [6] for s = 1∕2 ). In particular, this difference of the kernel function does not meet the assumptions needed to conclude the symmetry result by a restriction to H N,0 and applying statements of [17].
In the particular case, where Ω = B 1 (0) is the unitary ball, it was shown in [10] that the second eigenfunction of the fractional Laplacian in B 1 (0) , denoted by 2 , is odd and can be chosen to be positive in {x N > 0} . Due to the regularity of 2 , Theorem 1.1 yields that for i = 1, … , N − 1 we have 1. 2 is symmetric with respect to H i,0 (see also [15]) and We emphasize that such a statement already follows due to [10] combined with [8], since thus the second eigenfunction can be written as a product of the first eigenfunction with a homogeneous function.
To give a more generalized application of our results to a class of nonlinear problems, we consider for 1 < p < 2N N−2s the minimization problem Clearly, the minimizer exists and is a solution of (1.1) with f (x, u) = |u| p−2 u (see e.g. [24,25]) and, since E s (|u|, |u|) ≤ E s (u, u) it can be chosen to be positive. For more information about the minimization problem (1.5) we refer to [21]. In the local case s = 1 , this is a well known problem, see e.g. [12,20]. If Ω satisfies (D), then it follows that this minimizer is also symmetric with respect to the symmetries of Ω (see [16]). In this case, we can also consider the minimizer in the set of H s 0 (Ω) -functions, which satisfy u = −u•r N,0 , that is In the next theorem, we prove that minimizers of (1.6) have constant sign in Ω ∩ H N,0 and in the particular case p = 2 we also prove a simplicity result for − 1,p (Ω). The existence, as mentioned above, follows immediately from a minimization problem. Moreover, by the known regularity theory it follows that indeed we have u ∈ C ∞ (Ω) ∩ C s (ℝ N ) , see e.g. [14,23]. We show here that this minimizer can actually be chosen to be nonnegative in Ω ∩ H N and thus the conclusion follows from Theorem 1.1.
This work is organized as follows. In Sect. 2 we give the framework for supersolutions and maximum principles used later on. Section 3 is devoted to prove Theorem 1.1 and in Sect. 4 we show Theorem 1.2.

Notation
The following notation is used. For subsets D,

A linear problem
For this section, we fix , ∈ ℝ and denote H 1 ∶= H 1, and H 2 ∶= H N, . Similarly, r 1 ∶= r 1, and r 2 ∶= r N, . We call w ∶ ℝ N → ℝ doubly antisymmetric (with respect to Symmetry of odd solutions to equations with fractional… In the following Lemma we collect some statements corresponding to the space V s (U) . The proofs can be found e.g. in [17][18][19].
The following Lemma gives an extension of Lemma 2.1. 3 to the case of doubly antisymmetric functions.
and we have Proof First note that since w is antisymmetric with respect to H i , i = 1, 2 , Lemma 2.1 and its proof imply Similarly, also w 2 is antisymmetric with respect to H 1 (resp. w 1 with respect to H 2 ) and thus also w 1,2 ∶= w 1 and it holds Similarly, we also have w r 1 ,2 = w 2 1 H c

It thus follows that
. Using the monotonicity of | ⋅ | and the antisymmetry of w and denoting r 1,2 ∶= r 1 •r 2 = r 2 •r 1 we have by several rearrangements and substitutions Symmetry of odd solutions to equations with fractional… |x − y| N+2s dxdy From here the statement of the Lemma follows, once we show the following claim: We write and In the following, fix x, y ∈ H 1 ∩ H 2 and without loss we may assume e 1 = (1, 0, … , 0) and e 2 = (0, 1, 0, … , 0) . Indeed, otherwise we may rotate the half spaces and since E s is invariant under rotations the situation remains the same. Then Thus with M ∶= |x − y| 2 we have Using the notation a = 4x 1 y 1 > 0 and b = 4x 2 y 2 > 0 , we may consider for fixed M > 0 the map it follows also that Fix such an open set U and let c ∈ L ∞ (U) . Then note that we may reflect c evenly across H 1 and denote V = U ∪ r 1 (U) . Then for any ∈ H s 0 (V) , which is antisymmetric with respect to H 1 and with ≥ 0 in U, we have Here, we have used the antisymmetry of w and with respect to H 1 and Lemma 2.1 to have 1 U ∈ H s 0 (U) , 1 r 1 (U) ∈ H s 0 (r 1 (U)) , and since we extended c evenly across H 1 .

Symmetry of odd solutions to equations with fractional…
In the next statement, we give a Hopf type lemma for Eq. (2.2) similar to [9, Proposition 3.3]. U ⊂ H 1 ∩ H 2 open. Furthermore, let c ∈ L ∞ (U) and let u ∈ V s (U) be a doubly antisymmetric supersolution of (2.2). Assume u ≥ 0 in It is clear that w•r 1 = −w = w•r 2 , that is, w is doubly antisymmetric. Let ∈ H s 0 (B) with ≥ 0 . Then, we have  there is a sign-changing solution to this problem. Assuming that Ω is bounded and starshaped with C 1,1 boundary and there exists a bounded solution of (1.1) with f (x, u) = |u| 2 * s −2 u , it first follows that u ∈ C s (ℝ N ) ∩ C ∞ (Ω) (see e.g. [23]) and the fractional Pohozaev identity from [22] implies

3
However, by [9,Proposition 3.3] it then follows that if Ω has additionally a symmetry hyperplane T and u is odd with respect to reflections across this hyperplane and of one sign on one side of the hyperplane, then there cannot be such an odd solution of the problem. Similarly, using instead Proposition 2.4, it follows that there can also be no doubly antisymmetric solution of this problem if Ω satisfies (D).

Symmetry of solutions
In the following, we use the notation from Sect. 2 and assume Ω ⊂ ℝ N satisfies (D). Moreover, f ∈ C(Ω × ℝ) satisfies (F1) and (F2) and let u ∈ L ∞ (U) ∩ H s 0 (Ω) be a solution of problem (1.1) which satisfies u•r N,0 = −u . Note that by (F1) and [23] it follows that u ∈ C s (ℝ N ) . For ∈ ℝ we may than define Then it follows that v is antisymmetric with respect to H N,0 and H 1, , hence doubly antisymmetric, and it satisfies due to (F2) where Note that by assumption (F1) we have Finally, let 1 ∶= sup x∈Ω x 1 .
Proof of Theorem 1.1 Assume that u is nontrivial. We apply the moving plane method to then prove that u is symmetric with respect to H 1,0 and decreasing in x 1 . For this let

3
Symmetry of odd solutions to equations with fractional… Next note that by (D) and Proposition 2.3 it follows that there is > 0 such that v ≥ 0 for all ∈ ( 1 − , 1 ) and thus by Proposition 2.4 we have 0 ≤ 1 − . Assume next by contradiction that 0 > 0 . Then by continuity v 0 ≥ 0 in H N,0 ∩ H 1, 0 . By Proposition 2.4 it follows that either v 0 ≡ 0 or v 0 > 0. If v 0 ≡ 0 , this implies that we have u ≡ 0 in Ω ⧵ H 1, 0 − 1 . But then, we can also start moving the hyperplane from the left (working instead with ℝ N ⧵ H 1, ), up to the same 0 . It then follows that u has two different parallel symmetry hyperplanes, but since u vanishes outside of Ω , this implies u ≡ 0 , which cannot be the case.
If v 0 > 0 , let > 0 be given by Proposition 2.3 according to c ∞ . Then by continuity there is > 0 such that and a compact set K ⊂ Ω 0 such that |Ω 0 ⧵ K| ≤ 2 and v 0 ≥ 2 in K. Again, by continuity, we can find ∈ (0, Whence, 0 > 0 is not possible. Thus 0 = 0 and this finishes the proof. ◻

A symmetric sign-changing solution
Let Ω ⊂ ℝ N open and bounded and consider the functional Then by a constraint minimization argument using the framework as explained e.g. in [24,25], see also [4], it follows that there exists such a minimizer u of J| M . That is, the minimum is attained. Similar to [4,Theorem 3.1], it can be shown that this minimizer is bounded and then, by an iteration of the results of [14,23], we have u ∈ C ∞ (Ω) . If in addition Ω is of class C 1,1 , then [14,23] also imply that u ∈ C s (ℝ N ).

Proof of Theorem 1.2 Let −
1,p be as in (4.1) and let u be the minimizer as explained in the above remarks. In view of Theorem 1.1 it remains to show that u can be chosen of one sign in Ω + ∶= Ω ∩ H N,0 . In the following Ω − = Ω ⧵ Ω + . Assume by contradiction that u changes sign in Ω + and let Ω + 1 ∶= {x ∈ Ω + ∶ u(x) > 0} and Ω + 2 ∶= {x ∈ Ω + ∶ u(x) ≤ 0} . We also let Ω − 1 = r N,0 (Ω + 1 ) , and Ω − 2 = r N,0 (Ω + 2 ) . By the property of u, it is clear that u < 0 in Ω − 1 and u ≥ 0 in Ω − 2 . Now let u be defined by Then u ∈ M , that is u ∈ H s 0 (Ω) satisfies u•r N,0 = −u and Moreover, we have Using the notation above, we rewrite  That is Whence u ≡ 0 in Ω + 2 and therefore u ≥ 0 in Ω + . This is in contradiction with the hypothesis. It follows that u does not change sign in Ω + and, without loss of generality, we may assume u ≥ 0 in Ω + . By the strong maximum principle [9, Corollary 3.4] we have u > 0 in Ω + .   ◻