Asymptotic symmetry for a class of nonlinear fractional reaction-diffusion equations

We study the nonlinear fractional reaction diffusion equation $\partial_{t}u + (-\Delta)^{s} u= f(t,x,u)$, $s\in(0,1)$ in a bounded domain $\Omega$ together with Dirichlet boundary conditions on $\R^N \setminus \Omega$. We prove asymptotic symmetry of nonnegative globally bounded solutions in the case where the underlying data obeys some symmetry and monotonicity assumptions. More precisely, we assume that $\Omega$ is symmetric with respect to reflection at a hyperplane, say ${x_1=0}$, and convex in the $x_1$-direction, and that the nonlinearity $f$ is even in $x_1$ and nonincreasing in $|x_1|$. Under rather weak additional technical assumptions, we then show that any nonzero element in the $\omega$-limit set of nonnegative globally bounded solution is even in $x_1$ and strictly decreasing in $|x_1|$. This result, which is obtained via a series of new estimates for antisymmetric supersolutions of a corresponding family of linear equations, implies a strong maximum type principle which is not available in the non-fractional case $s=1$.


Introduction
We consider the nonlinear fractional diffusion problem (P) ∂ t u + (−∆) s u = f (t, x, u) in (0, ∞) × Ω, where Ω is a bounded Lipschitz domain in R N , s ∈ (0, 1) and f is a nonlinearity defined on (0, ∞) × Ω × B. Here and in the following, B ⊂ R is an open interval (further assumptions on f are to be specified later). Moreover, (−∆) s denotes the fractional Laplacian, which for functions u ∈ H 2s (R N ) is defined via Fourier transform: Fueled by various applications in physics, biology or finance, linear and nonlinear equations of the form given in (P) or of similar type have received immensely growing attention recently. In particular, evolution equations involving the fractional Laplacian appear in the quasi-geostrophic equations (see e.g. [11,33]) and in the fractional porous medium equation (see [30]), while further applications in the context of stable processes are considered e.g. in [2,21]. Very recently, the fractional Laplacian has been studied in conformal geometry, see [14,22]. In order to incorporate the Dirichlet boundary condition on R N \ Ω in (P), the operator (−∆) s has to be replaced by the Friedrichs extension of the restriction of (−∆) s to the space C ∞ c (Ω) ⊂ L 2 (Ω) of test functions. Here and in the following, we identify L 2 (Ω) with the space {u ∈ L 2 (R N ) : u ≡ 0 on R N \ Ω}. This new operator, which we will also denote by (−∆) s in the following, has the form domain H s 0 (Ω) = {u ∈ H s (R N ) : u ≡ 0 on R N \ Ω}, and it is widely used in analysis and probability theory. In particular, it has recently been considered in the context of semilinear problems, see e.g. [3,5,25] and the references therein. In probabilistic terms, the operator coincides with the generator of the 2s-stable process in Ω killed upon leaving Ω. We note that for u ∈ C ∞ c (Ω) we have the representation for x ∈ Ω, where P.V. stands for the principal value integral and see e.g. [8,Remark 3.11].
For a solution u of (P), we define the ω-limit set (with respect to the norm · L ∞ ) as ω(u) := {z ∈ C 0 (Ω) : u(t k ) − z L ∞ → 0 for some t k → ∞} To state our main result, we introduce the following assumptions.
We note that (D2) is a technical assumption which is needed for some but not all of our results. The main result of this paper is the following.
Theorem 1.1. Let (D1), (F1), (F2) be satisfied, and let u be a nonnegative global solution of (P) satisfying the following conditions: (U1) There is c u > 0 such that u(t) L ∞ ≤ c u for every t > 0.
We immediately deduce the following corollary for equilibria and time-periodic solutions.

Remark 1.3. (i)
The nonnegativity assumption on u in Theorem 1.1 can be weakened in special cases. More precisely, if the other assumptions of Theorem 1.1 are satisfied, u(t 0 , ·) is nonnegative on Ω for some t 0 > 0 and f (t, ·, 0) ≥ 0 for all t ≥ t 0 , then u(t, ·) is nonnegative for t ≥ t 0 as a consequence of the weak maximum principle in the form discussed in Remark 2.6 below. Thus Theorem 1.1 applies to u after a time shift.
(iii) In the case where, in addition to the assumptions of Theorem 1.1, Ω ⊂ R N is a ball centered at zero and f is radially symmetric, i.e. f (t, x, u) =f (t, |x|, u), it follows -by the invariance of the equation under rotations -from Theorem 1.1 that every z ∈ ω(u) is radially symmetric as well. In the special case of equilibria, i.e., solutions of (6), this has been proved in [3] under more restrictive assumptions on the nonlinearity.
(iv) We point out that we do not require an a priori positivity assumption on elements in ω(u) in Theorem 1.1, and thus we also do not need to assume strict positivity of solutions of (6) in Corollary 1.2. This is a special feature of the nonlocal problems (P) and (6). The strong maximum principle given by Theorem 1.1 for elements z ∈ ω(u) and by Corollary 1.2 for nonnegative solutions of (6) is a consequence of the monotonicity of the nonlinearity, and it is derived as a byproduct of the method proving the symmetry results (see in particular Lemma 3.2 below). This contrasts with the local case s = 1, where counterexamples show that such a strong maximum principle is false, see [29,  The proof of Theorem 1.1 is based on a parabolic variant of the moving plane method. As far as the main structure of the argument is concerned, we follow the strategy elaborated by Poláčik [27,28] in the context of Dirichlet problems for fully nonlinear parabolic differential equations, but we need new and quite different tools. We recall that the moving plane method has its roots in a classical work of Alexandrov [1] on constant mean curvature surfaces and Serrin [32] on overdetermined boundary value problems, whereas Gidas,Ni and Nirenberg [20] provided the framework to consider Dirichlet problems for nonlinear elliptic differential equations. In the case where the underlying domain is R N , the method of moving plane has been applied in integral form in [16,18] to deduce symmetry and classification results for solutions of semilinear elliptic equations involving the fractional Laplacian. Birkner, Lopéz-Mimbela and Wakolbinger [3] used a variant of the moving plane method, paired with probabilistic methods, to prove radial symmetry of all equilibria of (P) in the case where the underlying domain is the unit ball B and the nonlinearity f is nonnegative, independent of t and x, and nondecreasing in u. Up to the authors' knowledge, our results are the first symmetry results for parabolic boundary value problems involving the fractional Laplacian and even for the elliptic problem if f depends on x or the domain is more general than a ball. We point out that -in comparison with the elliptic case -proving asymptotic symmetry in the parabolic setting with the moving plane approach requires much finer -time dependent -estimates. This is already evident from the seminal work of Poláčik [27,28] for the case of nonlinear differential equations. One key requirement is a special version of a parabolic Harnack inequality related to a linear fractional diffusion equation. Felsinger and Kassmann derived a parabolic Harnack inequality in [19], which requires nonnegativity of the solutions in the entire space. This global nonnegativity assumption is not technical since -already in the elliptic case -the Harnack inequality for the fractional Laplacian is not valid in a purely local form, see e.g. [23, Theorem 2.2] for a counterexample. However, since the moving plane method consists in studying the difference between the reflection of a solution of (P) at a hyperplane and the solution itself, we need to derive a corresponding Harnack inequality for antisymmetric (and therefore sign changing) supersolutions of a class of linear problems in the present paper. Another (closely related) problem in the fractional setting is the lack of local comparison principle to derive estimates via sub-or supersolutions. Here much finer quantitative arguments are needed to control the nonlocal effects and exclude the appearance of intersections in finite time. We will establish such estimates in two steps in Section 2.3 below, passing first to the Caffarelli-Silvestre extension of the solution u, which is defined, for each fixed time, on the half space R N+1 + (see [9]). It seems worthwhile to note that another type of Dirichlet boundary conditions has also been assigned to the fractional Laplacian in the literature. In [10,12,35]), the authors consider the s-th power of the Dirichlet Laplacian in spectral theoretic sense, which -in the case of a bounded domain Ω -is given by µ s k u k e k . Here µ k = µ k (Ω) are the eigenvalues of the Dirichlet Laplacian on Ω in increasing order (counted with multiplicity), e k , k ∈ N are the corresponding eigenfunctions and u k := Ω ue k dx the corresponding Fourier coefficients of u.
In order to explain the role of A s Ω in terms of stochastic processes, we recall that the 2s-stable process is constructed by subordinating Brownian motion with a s-stable subordinator, see [2,Chapter 1.3]. On the other hand, the process generated by A s Ω is obtained by first killing Brownian motion upon leaving Ω and then subordinating this process with a s-stable subordinator, see e.g. [34]. Hence the order of killing and subordination is reversed in this case. It is easy to see that the corresponding operators coincide only if Ω = R N (where the Dirichlet boundary conditions are not present). For more information related to these stochastic processes and their generators, we refer the reader to [4], [21] or [2,Chapter 3]. It is natural to ask whether a result similar to Theorem 1.1 is true for the corresponding problem with the operator A s Ω . For elliptic semilinear problems involving the operator A s Ω , symmetry and monotonicity results have been proved recently in special cases in [10,12] by applying the moving plane method to the Caffarelli-Silvestre extensions of the solutions.
The article is organized as follows. In Section 2, we develop the new tools we need to carry out the moving plane method for the fractional parabolic problem (P). We believe that the results of this Section could be of interest for other problems as well. Since, as already noted, the moving plane method consists in studying the difference between the reflection of a solution of (P) at a hyperplane and the solution itself, we are led to study antisymmetric supersolutions of linear problems here. Due to the nonlocality of the fractional Laplacian, it is important to estimate the influence of the negative part of these functions. This is one of the key differences in comparison with local problems involving classical differential operators. The first part of this Section is concerned with a parabolic small volume maximum principle. In Section 2.2 we establish, based on recent results in [19], a parabolic Harnack inequality for antisymmetric supersolutions of a class of linear fractional problems. Section 2.3 is devoted to a generalized subsolution estimate. The idea to control the positive part of the solution by comparing with suitable subsolutions is inspired by [27]. However, as mentioned above, the argument is essentially more involved in the present setting, and this is the only stage where we had to pass to the Caffarelli-Silvestre extension. In Section 2.4, we combine all estimates obtained so far to deduce our main result on antisymmetric supersolutions for a class of linear problems. This result should be seen as an analogue of [27, Theorem 3.7] for the fractional case. The moving plane argument is then carried out in Section 3. Here we follow the main structure of the argument in [27, Chapter 4], but we need to implement some new ideas at key points (see in particular the proof of Lemma 3.2) in the nonlocal setting. In the appendix, we present a sufficient condition for (U2), and we discuss a specific example to which Theorem 1.1 applies.

Notation
The following notation is used throughout the paper. For x ∈ R N and r > 0, B r (x) is the open ball centered at x with radius r and ω N will denote the volume of the N-dimensional ball with radius 1. For any subset M ⊂ R N , we denote by 1 M : R N → R the indicator function of M and diam(M) the diameter of M. Moreover, we let inrad(M) denote the supremum of all r > 0 such that every connected component of M contains a ball B r (x 0 ) with x 0 ∈ M. This notation -taken from [27] -differs slightly from the usual one but is very convenient in our setting. If T ⊂ R, Ω ⊂ R N are subsets and u : is a subset and w : M → R is a function, the inequalities w ≥ 0 and w > 0 are always understood in pointwise sense. Moreover, w + = max{w, 0} resp. w − = − min{w, 0} denote the positive and negative part of w, respectively. If M is measurable with |M| > 0 (where | · | always stands for Lebesgue measure) and w ∈ L 1 (M), we put to denote the mean of w over M. If D,U ⊂ R N are subsets, the notation D ⊂⊂ U means that D is compact and contained in the interior of U. Moreover, we set dist(D,U) := inf {|x − y| : x ∈ D, y ∈ U} , so this notation does not stand for the usual Hausdorff distance. If D = {x} is a singleton, we simply write dist(x,U) in place of dist({x},U). Finally, when we call an interval T ⊂ R a time interval, we assume that it consists of more than one point.

Antisymmetric supersolutions of a corresponding linear problem
Throughout this section, we consider a fixed open half space H and the reflection Q : R N → R N at ∂ H.
We will call a function w : for every x ∈ R N , i.e., w is antisymmetric with respect to Q. We first fix notions of supersolutions. For an open subset U ′ ⊂ R N , we introduce the function space endowed with the norm We note that if U ⊂⊂ U ′ is a pair of open sets and u ∈ V s (U ′ ), v ∈ H s 0 (U), then E (u, v) is well defined by (4).
for all ϕ ∈ H s 0 (U), ϕ ≥ 0 and a.e. t ∈ T . If, in addition, U ⊂ H and v is antisymmetric, we call v an antisymmetric supersolution. A supersolution of (8) on T × U will be called an entire supersolution if v ≥ 0 on T × (R N \U). If U ⊂ H, an antisymmetric supersolution of (8) on T ×U will be called an entire antisymmetric supersolution if v ≥ 0 on T × (H \ U).

Remark 2.2.
(i) Note that an entire antisymmetric supersolution v of (8) on T × U may take negative values in R N \ H, so in general it is not an entire supersolution of (8).
(ii) Let T,U and c be as in the definition above. We will mostly consider the case g ≡ 0 in the remainder of the paper, i.e., we consider supersolutions of on T ×U. We briefly explain the connection between (P) and (9). Suppose that (F1) is satisfied and that Let u be a nonnegative solution of (P), and let v(t, Then v is an entire antisymmetric supersolution of (9) with T = (0, ∞), U = H ∩ Ω and Indeed, by (10) where (11) was used in the last step.
The following observation will be useful in the sequel.
Proof. It is convenient to writex in place of Q(x) for x ∈ R N in the following. For ϕ ∈ H s 0 (H) and an antisymmetric v ∈ H s (R N ) we then have with J and κ H as defined above, as claimed. To see (13), let d > 0 and x, y as claimed in (13).

A small volume maximum principle
The main result of this subsection is the following.
In the proof we will use the following standard estimate, see e.g. [17, Lemma 6.1].

Lemma 2.5. For every measurable A ⊂ R N and every x ∈ R N we have
where K is given in Lemma 2.5.
By assumption and Lemma 2.5, we then have Without loss of generality, we may assume that For We first claim that Thus we find, using the symmetry of the kernel and the antisymmetry of u, ϕ is nonnegative. This shows (18). We now put Combining (17), (18) and (19), we get . This shows (16), as required.
Remark 2.6. For entire supersolutions v of (9), a corresponding small volume maximum principle can be derived in a similar but much easier way. More precisely, for every c ∞ , γ > 0 there exists δ > 0 such that for any bounded open subset U ⊂ R N with |U| ≤ δ , any time interval T : As a consequence, we may readily derive the following weak maximum principle: and v an entire supersolution of (9) on T × Ω such that v(t 0 , x) ≥ 0 for a.e. x ∈ Ω, then also v(t, x) ≥ 0 for all t ∈ T and almost every x ∈ Ω.

A Harnack inequality for antisymmetric supersolutions
In this part we state a Harnack inequality for antisymmetric supersolutions of (9). We will derive this inequality -via a reformulation of the problem -from a recent result in [19]. We need to introduce some notation.
We fix r 0 ∈ (0, 1] and C 1 ,C 2 > 0, and we consider a function k : The quadratic form corresponding to this kernel is given by on for ϕ ∈ H s 0 (U), ϕ ≥ 0 and a.e. t ∈ T . Next we introduce notation for parabolic cylinders. For t 0 ∈ R, In view of the scaling properties of (22), the following is a mere reformulation of a special case of [19, Theorem 1.1], see also [19, Remark after Theorem 1.2]. We point out that the notion of supersolution considered in [19] is weaker than the one considered here. Theorem 2.7. Let r 0 ∈ (0, 1] and ϑ ,C 1 ,C 2 > 0 be given. Then there are constants c i > 0, i = 1, 2 depending on N, s, r 0 , ϑ ,C 1 ,C 2 such that for any k : By an argument based on building chains of cylinders, we deduce the following Harnack inequality for general pairs of domains. We include the proof here since the argument is not completely standard. A similar argument has been detailed in [27, Appendix], but we need to argue somewhat differently since the triples of parabolic cylinders in Theorem 2.7 have a smaller overlap than the ones considered in [27]. Corollary 2.8. Let r 0 ∈ (0, 1], R, τ, ε > 0 and C 1 ,C 2 > 0 be given. Then there exist positive constants c i = c i (N, s, r 0 ,C 1 ,C 2 , R, ε, τ) > 0, i = 1, 2 with the following property: (21), and let D ⊂⊂ U ⊂ R N be a pair of bounded domains such that dist(D, ∂U) ≥ 2r 0 , |D| ≥ ε and diam(D) ≤ R. Moreover, let g ∈ L ∞ (T ×U) and a supersolution v of (22) Proof. We first note that there exist n = n(N, R, r 0 ) ∈ N and µ = µ(N, R, r 0 ) > 0 such that the following holds: For every subset D ⊂ R N with diam D ≤ R there exists a subset S D ⊂ D of n + 1 points such that D is covered by the balls B r 0 (x), x ∈ S D , and for every two points x * , x * ∈ S D there exists a finite sequence x j ∈ S D , j = 0, . . . , n such that We now fix D ⊂⊂ U ⊂ H as in the assertion, and we fix n, µ and a set S D with the property above. Next, we put ϑ = τ 7 min{ 1 17 , 1 n+3 }, and we claim the following: and The definition of ϑ and the restrictions on t * , t * then force (27), and (26) holds with s j

Since the cylinders
We now consider t * ∈ [t 0 + 3τ,t 0 + 4τ], x ∈ D arbitrary. Then we choose x * ∈ S D such that x ∈ B r 0 (x * ), and we choose s j , j = 0, . . . , m with the properties (26) and (27). Moreover, we fix a sequence of points x j ∈ S D , j = 0, . . . , m such that (25) holds with m in place of n. This may be done, since m ≥ n, by repeating some of the points in the chain if necessary. We now define Q j := Q(r 0 , ϑ , s j , x j ) and Q ± j := Q ± (r 0 , ϑ , s j , x j ) for j = 0, . . . , m.
Proof. We may assume without loss that H = {x ∈ R N : x 1 > 0}. For simplicity, we writex = Q(x) = (−x 1 , x 2 , . . . , x N ) for x ∈ R N . We consider J(x, y) as defined in Lemma 2.3. Obviously we have whereas, by Lemma 2.3 To define k with the asserted properties, we set Finally, we set k(x, y) := g(x + s(x, y)e 1 , y + s(x, y)e 1 ) for x, y ∈ R N , x = y. Then k is continuous, and properties (i) and (ii) follow directly by construction and (31). To see (iii), we note that if 0 < |x − y| ≤ β 2 then also |x −ỹ| ≤ β 2 , wherex = x + s(x, y)e 1 andỹ = y + s(x, y)e 1 . Furthermore we have that max{x 1 , y 1 } ≥ β and therefore min{x 1 , y 1 } ≥ β 2 . Consequently, by (32). It remains to show (iv): So let v ∈ H s (R N ) be antisymmetric, and let ϕ ∈ H s 0 (H β ). Then Lemma 2.3 gives If x ∈ H β , then for y ∈ H we have s(x, y) = 0 and thus J(x, y) = g(x, y) = k(x, y), while for y ∈ R N \ H we have that k(x, y) = 0. Hence we can rewrite the first integral of the RHS of (34) as Similarly, if y ∈ H β , then for x ∈ H \ H β we have s(x, y) = 0 and thus J(x, y) = g(x, y) = k(x, y), while for x ∈ R N \ H we have k(x, y) = 0. Hence we may rewrite the second integral of the RHS of (34) as where the last equality follows again since ϕ = 0 on R N \ H β . Combining these identities, we get and together with (33) it follows that as claimed in (30).
We may now complete the Proof of Theorem 2.9. Put β = 2r 0 , U 0 = {x ∈ U : dist(x, D) < β } ⊂⊂ U, and let k be the function given by Lemma 2.10 for this choice of β . Let v be an antisymmetric supersolution of (9) on T ×U, and considerṽ Since U 0 ⊂ H β , Lemma 2.10(iv) implies that Setting w(t) = w(t, ·) as usual, we observe that w(t) ≥ 0 on R N for all t ∈ T . Moreover, for any t ∈ T and any nonnegative ϕ ∈ H s 0 (U 0 ) we have Hence w is a nonnegative supersolution of (22) on T × U 0 with Applying Corollary 2.8 with U 0 in place of U (noting that dist(D, ∂U 0 ) = β = 2r 0 ) and using the properties of k given by Lemma 2.10, we find c i = c i (N, s, r 0 , R, ε, τ) > 0 such that Hence the assertion follows with K 1 = c 1 e −4τd and K 2 = c 2 d + 1. Note that both constants only depend on N, s, r 0 , c ∞ , ε, R and τ.

A lower bound based on a subsolution estimate
The aim of this subsection is to prove the following result.
Since v is antisymmetric, we have, by a simple change of variable, For x, z ∈ H and y > 0 we have Moreover, for x, z ∈ B ρ (x 0 ) we have |x − z| 2 ≤ 4ρ 2 and |x − Q(z)| 2 ≥ |x − z| 2 + 4ρ 2 , so that with Combining (40), (41) and (42) and using that v ≥ 0 on B 2ρ (x 0 ), we obtain the estimate for x, z ∈ B ρ and y ∈ (0, 1], we conclude that for x ∈ B 1 and y ∈ (0, 1]. Hence the claim follows withc 1 = let v be a supersolution of (9) on T × U such that v(t) ∈ H s (R N ) for all t ∈ T . Moreover, let w(t) ∈ H 1 (R N+1 + , y 1−2s ) be the L s -harmonic extension of v(t) given by (38) for each fixed time t ∈ T . Then for every nonnegative Φ ∈ H 1 (R N+1 + , y 1−2s ) with ϕ := tr(Φ) ∈ H s 0 (U) and every t ∈ T we have Proof. In case Φ is the L s -harmonic extension of ϕ, we have with d s as stated (see e.g. [9] or [8,Remark 3.11]) and therefore (44) is true. On the other hand, since w is L s -harmonic, Hence the assertion follows.

and the L s -harmonic extension w(t) of v(t) is nonnegative on B
Proof. Without loss of generality, we may assume that x 0 = 0, t 0 = 0 and σ = 1, and we put B ρ = B ρ (0). Let λ 1 > 0 be defined by (35), and let f : [0, ∞) → R denote the solution of the initial value problem with d s as in Lemma 2.13. We note that f is a scalar multiple of a rescaled MacDonald function (or modified Bessel function of the second kind), see e.g. [36]. We also note that f is strictly decreasing on [0, ∞). Moreover, the limit exists and only depends on s and ρ (via λ 1 ). We now put γ = c ∞ − κ 2 + 1, and we let Puttingw(t) =w(t, ·, ·) as usual, we then have for t ∈ [0,t 1 ]. Moreover, we havẽ by assumption and by construction ofw. In the following, we consider Moreover, we will write g resp.ṽ for the traces of h andw, respectively. Then, as a consequence of (37), (44), (47) and (49), we have Since furthermore B ρ g 2 (0, x) dx = 0 as a consequence of (48), we conclude that B ρ g 2 (t, x) dx = 0 and therefore g(t) ≡ 0 on B ρ for all t ∈ T .
We may now complete the Proof of Proposition 2.11. For given ρ, c ∞ > 0, letc i , i = 1, 2 be given by Lemma 2.12, and let γ be given by Lemma 2.14. Moreover, let Next, let T := [t 0 ,t 1 ] ⊂ R, σ 0 > 0 and σ 1 ≥ qσ 0 , and let v be an antisymmetric supersolution of (9) on T × B ρ (x 0 ) satisfying assumptions (i)-(iii). Suppose by contradiction that We may assume that t * > t 0 is chosen minimally with this property, so that Let w denote the L s -harmonic extension of v given by (40) for each fixed time t ∈ T . Then Lemma 2.12 implies that This contradicts (50), and thus the proof is finished.

Main Result on entire antisymmetric supersolutions
We recall from Section 1.1 that, for a subset D ⊂ R N , inrad(D) denote the supremum of all r > 0 such that every connected component of D contains a ball B r (x 0 ) with x 0 ∈ D. Note that inrad(D) ≥ ρ > 0 implies that every connected component of D has at least measure |B ρ (0)|, so in this case D has only finitely many connected components if it has finite measure. then: Proof. We let ρ, γ, q, δ , τ, R be given with the properties stated in the theorem, and we put ε = |B 2ρ (0)|. Let K 1 , K 2 -depending on these quantities -be given as in Theorem 2.9. We fix µ > 0 sufficiently small such that where Ψ is given in Proposition 2.11 depending on ρ. Next, we consider D ⊂⊂ U ⊂ H and an antisymmetric supersolution v of (9) on [t 0 , ∞) × U with the properties stated in the theorem, which implies in particular that ε ≤ |D * | ≤ (2R) N for every connected component D * of D. We put σ 0 = v − (t 0 ) L ∞ (U\D) and To prove (i), we suppose by contradiction that T 0 < ∞. Then there exists a connected component D * of Let U * be the connected component of U with D * ⊂ U * . Since v ≥ 0 on [t 0 ,t 0 + 8τ) × D * , we have, by Theorem 2.9, (51) and Proposition 2.4, We fix x 0 ∈ D * such that B 2ρ (x 0 ) ⊂ D * , which is possible by assumption. Since σ 1 ≥ qσ 0 by (52), the estimates (55) and (53) allow us to apply Proposition 2.14 with t 0 + 4τ in place of t 0 , which yields With the help of Theorem 2.9, (53) and (56), we find that by our choice of µ in (52), contradicting (54). We conclude that T 0 = ∞. In particular, (i) holds, and (ii) follows since, by (53),

Proof of the main symmetry result
In this section we complete the proof of Theorem 1.1. With the tools developed in Section 2, we may follow the main lines of the moving plane method as developed by Poláčik in [27], but some steps in the argument -in particular the proofs of Lemma 3.2 and Proposition 3.5 below -differ significantly from [27]. This is due to the fact that, contrary to [27], we do not a priori assume the existence of an element ϕ ∈ ω(u) with ϕ > 0. For λ ∈ R, we use the notations Moreover, we let Q λ : R N → R N denote the reflection at T λ given by Q λ (x) = (2λ − x 1 , x 2 , . . . , x N ). For a function z : R N → R, we put We now assume that the hypotheses (D1) and (F1), (F2) are satisfied, and we let u be a nonnegative global solution of (P) satisfying (U 1 ) and (U 2 ). We set l := max{x 1 : (x 1 , x ′ ) ∈ Ω for some x ′ ∈ R N−1 }, and we fix λ ∈ [0, l) for the moment. As discussed in Remark 2.2, the function v := V λ u is an entire antisymmetric supersolution of the problem Here the term entire antisymmetric supersolution refers to the notion defined in the beginning of Section 2 with respect to the half space H = H λ . Indeed, for λ ∈ [0, l) and this choice of H, (10) and (11) are satisfied as a consequence of assumptions (D1) and (F2). Moreover, as a consequence of (F1) and (U1), there exists c ∞ > 0 such that In the following, we fix c ∞ with this property. We also note that [V λ u](t) ∈ H s (R N ) for all t ∈ (0, ∞). For λ ∈ [0, l), we now consider the following statement: Our aim is to show, via the method of moving planes, that (S λ ) holds for every λ ∈ [0, l). We need the following lemmas.
Lemma 3.1. There is δ > 0 such that for each λ ∈ [0, l) the following statement holds. If K is a closed subset of Ω λ with |Ω λ \ K| < δ and there is t 0 ≥ 0 such that V λ u(t) ≥ 0 on K for all t ≥ t 0 , then for all t ≥ t 0 . In particular (S λ ) holds if λ < l is sufficiently close to l.
Proof. This follows immediately by applying Proposition 2.4 to γ = 1, c ∞ > 0 as fixed above, H = H λ and U = Ω λ \ K. Note that the number δ > 0 given by Proposition 2.4 in this case does not depend on λ and K. The second statement of the lemma follows since |Ω λ | < δ if λ is close to l.
Proof. (i) Since the set {u(t) : t ≥ 0} is relatively compact in C(Ω), the statement (S λ ) is equivalent to V λ z ≥ 0 on Γ λ for all z ∈ ω(u). Hence (S λ 0 ) holds by assumption and continuity of all z ∈ ω(u). (ii) Step one: We first claim that on each connected component U of Ω λ 0 we either have V λ 0 z > 0 on U or V λ 0 z ≡ 0 on U. To prove this, we fix z ∈ ω(u) and a connected component U of Ω λ 0 such that V λ 0 z ≡ 0 on Ω λ 0 . Since V λ 0 z ≥ 0, there exists x 0 ∈ U and ρ > 0 such that B := B ρ (x 0 ) ⊂⊂ Ω λ 0 and V λ 0 z > 0 on B. Since z ∈ ω(u), there exists a sequence of numbers t n > 0, such that t n → ∞ and u(t n ) → z in C(Ω), Consequently, there exists σ > 0 and n 0 ∈ N such that By the equicontinuity property (U2), there exists τ > 0 such that Now fix a subdomain D ⊂⊂ U. Applying Proposition 2.9 with U = Ω λ 0 , t 0 = t n − 4τ and using (60), we get Since D ⊂⊂ U was chosen arbitrarily, we conclude that V λ 0 z > 0 in U. This shows the claim.
Step two: Let z ∈ ω(u) be such that is nonempty. To finish the proof of (ii), we need to show that V λ 0 z ≡ 0 on R N . We suppose by contradiction that this is false; then there exists a compact set K ⊂ H λ 0 \ U z of positive measure such that By Step one above, U z is an open set. Hence we may fix a nonnegative function ϕ ∈ C ∞ c (U z ), ϕ ≡ 0, and we set D := supp ϕ. Moreover, we fix ρ > 0 with dist(D, ∂U z ) > 2ρ, and we note that there exists M > 0 such that To estimate the double integral on the right hand side of (63), we put where we used the fact that |x − Q λ 0 (y)| ≥ 2ρ for every x, y ∈ D ρ . To estimate the integral over H 2 , we first note that where in the last step we have set We now consider the function t → h(t) = U z v(t, x)ϕ(x) dx for t > 0. Combining the estimates above and using (58), we get . We now consider a sequence (t k ) k ⊂ (0, ∞) such that t k → ∞ and u(t k ) → z in L ∞ (Ω) as k → ∞, which yields in particular that h(t k ) → 0 as k → ∞. Using (61) and the equicontinuity property (U2), we find δ > 0 and Moreover, making δ > 0 smaller and k 0 ∈ N larger if necessary, we may assume that Combining (64), (65) and (66), we thus obtain This implies that contradicting the fact that v − (t) L ∞ (U z ) → 0 as t → ∞ and thus lim inf t→∞ h(t) ≥ 0. The proof of (ii) is finished.
(iii) Suppose that λ 0 > 0, and let z ∈ ω(u) such that V λ 0 z ≡ 0 on R N . In view of (ii), we need to show that z ≡ 0 on R N . For this we consider the reflected functions z : R N → R,z(x) = z(Q 0 (x)).
Since Ω and the nonlinearity f are symmetric in the x 1 -variable,ũ is also a solution of (P) satisfying the same hypotheses as u. Moreover,z ∈ ω(ũ). Putting λ * := l − 2λ 0 ∈ (−l, l), it follows from V λ 0 z ≡ 0 on R N thatz ≡ 0 on Ω λ * and therefore For λ ∈ ( λ * +l 2 , l) sufficiently close to l, it also follows from Lemma 3.1 that (S λ ) holds forũ in place of u, so that (68) and (ii) imply that V λz ≡ 0 on R N for λ < l sufficiently close to l.
(69) ¿From this we easily conclude thatz ≡ 0 and therefore z ≡ 0 on R N , as claimed.
For the proof of this lemma, the following observation is useful. Proof. We first note that since Ω λ ∩ T λ = ∅ by assumption (D1).
Since Ω λ 0 ⊂ Ω λ for λ < λ 0 and V λ z → V λ 0 z uniformly on Ω λ 0 for every z ∈ M, we have lim sup . Now suppose by contradiction that there exists sequences of numbers λ n ∈ (0, λ 0 ), of functions u n ∈ M and of points x n ∈ Ω λ n such that λ n → λ and V λ n u n (x n ) → c < I λ 0 (M) for n → ∞.
The following Proposition evidently completes the Proof of Theorem 1.1.
Proposition 3.5. Suppose that (D2) holds or that z ≡ 0 on Ω for all z ∈ ω(u). Then we have: (ii) For every z ∈ ω(u), we either have the following alternative. Either z ≡ 0 on Ω, or z is strictly decreasing in |x 1 | and therefore strictly positive in Ω.
(iii) Let z ∈ ω(u) be given such that z is not strictly decreasing in |x 1 |. Then there exists λ > 0 such that V λ z is not strictly positive in Ω λ . By Lemma 3.2(ii), applied to λ in place of λ 0 , we then have that V λ z ≡ 0 on R N . By (ii), z therefore has two different parallel symmetry hyperplanes. This implies that z ≡ 0, since z vanishes outside a bounded subset of R N .
In the special case f ≡ 0, the interior regularity estimate (80) is an immediate consequence of [19, Theorem 1.2], but we could not find any reference where the case f ≡ 0 is considered. Before giving the proof of this proposition, we discuss an example.
(84) satisfying 0 ≤ u(t, x) ≤ 1 for all t ∈ (0, ∞), x ∈ Ω, so that condition (U1) is satisfied for u. Furthermore, if ϕ(x) ≤ C 1 dist(x, ∂ Ω) s for x ∈ Ω with some constant C 1 > 0, then (U2) is also satisfied by Proposition 4.1(ii). We remark that the solution u can be found as a the unique mild solution of (84), i.e., the unique solution of the nonlinear integral equation where dom(A) is the space of all functions u ∈ H s 0 (Ω) ∩ C 0 (Ω) such that (−∆) s u, defined in distributional sense, is contained in C 0 (Ω). Moreover, F : [0, ∞) × C 0 (Ω) → C 0 (Ω) is the substitution operator given by [F(t, w)](x) = f (t, x, w(x)) for t ∈ [0, ∞), x ∈ Ω. The m-dissipativity of the operator A in C 0 (Ω) is essentially a consequence of the following recent regularity result given in [31, Proposition 1.1]: If Ω ⊂ R N is a bounded domain satisfying the exterior sphere condition and w ∈ L ∞ (Ω), then the unique weak solution u ∈ H s 0 (Ω) of the equation −∆u = w belongs to C 0 (Ω). Another important fact needed for the local existence and uniqueness of solutions of (P) is the local uniform (in time) Lipschitz continuity of F : [0, ∞) × C 0 (Ω) → C 0 (Ω), which follows since f satisfies (F2). In order to show solutions of (85) are also solutions of (84), one may essentially argue as in [13] for the semilinear heat equation, noting the following additional useful property of the substitution operator F: If M ⊂ C 0 (Ω) ∩ H s 0 (Ω) is bounded with respect to · ∞ , then F(M) ⊂ H s 0 (Ω), and there exists L = L(M) > 0 such that E (F(t, u), F(t, u)) ≤ LE (u, u) for all u ∈ M, t > 0.
This property can be checked immediately by using (F2) and the definition of the quadratic form E . Note that (83) is just a particular example of a nonlinearity which admits an ordered pair of a bounded subsolution ϕ * and a bounded supersolution ϕ * and which satisfies (F1) with B = R. In such a setting, an initial condition ϕ ∈ C 0 (Ω) ∩ H s 0 (Ω) always gives rise to a global bounded solution of (P).
The remainder of this appendix is devoted to the proof of Proposition 4.1. The assertion (80) on interior regularity will be deduced from the Harnack inequality of Felsinger and Kassmann [19]. More precisely, we will use the following rescaled variant of a special case of [19,Corollary 5.2].