Symmetry and non-uniformly elliptic operators - CERMICS

Symmetry and non-uniformly elliptic operators Jean Dolbeault

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Ceremade, UMR no. 7534 CNRS, Université Paris IX-Dauphine, Place de Lattre de Tassigny, 75775 Paris Cédex 16, France E-mail: [email protected] Internet: http://www.ceremade.dauphine.fr/∼dolbeaul

Patricio Felmer

∗, †

Departamento de Ingenier´ıa Matem´ atica and Centro de Modelamiento Matem´ atico, UMR no. 2071 CNRS-UChile Universidad de Chile, Casilla 170 Correo 3, Santiago, Chile E-mail: [email protected]

Régis Monneau

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CERMICS, Ecole Nationale des Ponts et Chaussées, 6 et 8 avenue Blaise Pascal, Cité Descartes, Champs-sur-Marne, 77455 Marne-La-Vallée, France E-mail: [email protected] Internet: http://cermics.enpc.fr/∼monneau/home.html

Abstract. The goal of this paper is to study the symmetry properties of nonnegative solutions of elliptic equations involving a non uniformly elliptic operator. We consider on a ball the solutions of ∆p u + f (u) = 0 with zero Dirichlet boundary conditions, for p > 2, where ∆ p is the p-Laplace operator and f a continuous nonlinearity. The main tools are a comparison result for weak solutions and a local moving plane method which has been previously used in the p = 2 case. We prove local and global symmetry results when u is of class C 1,γ for γ large enough, under some additional technical assumptions. Keywords. Elliptic equations, non uniformly elliptic operators, p-Laplace operator, scalar field equations, monotonicity, symmetry, local symmetry, positivity, non Lipschitz nonlinearities, comparison techniques, weak solutions, maximum principle, Hopf’s lemma, local moving plane method, C 1,α regularity AMS classification (2000). 35B50, 35B65, 35B05

∗ †

Partially supported by ECOS-Conicyt. Partially supported by Fondecyt Grant # 1030929 and FONDAP de Matem´ aticas Aplicadas.

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2 March 21, 2004

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J. Dolbeault, P. Felmer & R. Monneau

Introduction and main results

The goal of this paper is to prove local and global solutions of   ∆p u + f (u) = 0 u≥0  u=0

symmetry results for the nonnegative in B in B on ∂B

(1)

where ∆p u = div(|∇u|p−2 ∇u) is the p-Laplace operator or p-Laplacian, and B = B(0, 1) is the unit ball in Rd , d ≥ 1. We will focus on the non uniformly elliptic case: p > 2, and consider nonlinearities f which are continuous on R + . In earlier works of two of the authors [14, 15, 16], it has been noted that for p = 2 the 2 C regularity of u is apparently a threshold in order to apply the moving plane technique. Here we will see that the threshold for p > 2 is rather C 1,1/(p−1) . As in [16], additional properties of f are required at least close to its zeros. The main tools of this paper are the maximum principle when the nonlinearity is nonincreasing or, based on a local inversion, when ∇u 6= 0, and a local moving plane technique. Our results are not entirely satisfactory, since the statements are rather technical, but on the other hand, they are a first achievement in obtaining symmetry results by comparison techniques in the case of non uniformly elliptic operators. Let us start with a simple case, where we assume the following additional conditions on f . (H1) Let f be a real function defined on R + such that: a) f ∈ C α (R+ ) for some α ∈ (0, 1), b) ∃ θ > 0 such that f > 0 on (0, θ), f < 0 on (θ, +∞), c) f is nonincreasing on (θ − η, θ + η) for some η > 0. Theorem 1 (Symmetry result for the p-Laplacian with p > 2) Let p > 2, γ ∈ (1/(p − 1), 1). Consider a function f satisfying (H1) and a solution u ∈ C 1,γ (B) of (1). Then u is positive and radially symmetric on B. Remark 1 A consequence of the C 1,γ regularity of the solution on the whole ball is that the maximum of this solution needs to be reached at the level u = θ (see Lemma 8). Does such a solution exist ? Given a function f and using a shooting method, we obtain a solution on a ball of radius large enough. Then a rescaling gives a solution on B for an appropriately rescaled nonlinearity f , which provides an example of existence of a solution. A result of symmetry similar to that of Theorem 1 also holds under weaker regularity assumptions on f . (H2) f ∈ C 0 (R+ ) and f > 0 on (0, +∞). Theorem 2 (Symmetry result with f only continuous) Let p > 2, γ ∈ (1/(p − 1), 1). Assume that (H2) holds and consider a solution u ∈ W 1,p (B) ∩ L∞ (B) of (1) such that, with M = kukL∞ (B) , u ∈ C 2 B \ (∇u)−1 (0) ∩ C 1,γ u−1 ([0, M )) . If f is nonincreasing in a neighborhood of M , then u is positive and radially symmetric on B.

March 21, 2004

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Symmetry and non-uniformly elliptic operators

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Remark 2 If we assume that u belongs to C 1,γ (B) with γ > 1/(p−1), then f (M ) = 0. See Lemma 8. Reciprocally if f (M ) 6= 0, as we shall see in Example 1, a solution like the one of Theorem 2 has a regularity which is not better than C 1,1/(p−1) at the points x such that u(x) = M . This is the reason why the C 1,γ regularity cannot be assumed at such points. The starting point of the moving plane technique is the celebrated paper by Gidas, Ni and Nirenberg [20] based on [1, 2, 26]. An improvement was made in [19, 5, 27] using the monotonicity properties of the nonlinearity. This lead to a local moving plane technique [14, 15, 16] in the case of the Laplacian, which is well suited for nonlinearities with low regularity. The corresponding notion of local symmetry also seems appropriate in the case of the p-Laplacian. Further interesting results were obtained in [17, 18]. For nonlinear elliptic operators, the story is shorter [8, 9, 10]. It strongly relies on comparison methods [29, 6, 7] but covers the case of the p-Laplace operator only when p < 2. For completeness, let us mention that related results have been obtained by rearrangement techniques, for instance in [23, 21], and that in [3, 4], F. Brock has been able to get symmetry results for the p-Laplace operator by a completely different approach based on a continuous Steiner symmetrization. Our goal is to get local symmetry results for p > 2 using an adapted version of the maximum principle [11, 24, 25] and regularity properties [12, 28, 13, 22], with a local moving plane method which is essentially adapted from [15, 16]. The paper is organized as follows. In the next section, we state a local symmetry result (Theorem 3). Theorem 2 is a simple corollary of Theorem 3. In the third section we start with two examples, which motivate the choice of the class of regularity of the solutions and then state three useful lemmas. The fourth section is devoted to the proof of Theorem 3 by the local moving plane method. Proofs of Theorems 1 and 2 are contained in the proof of Theorem 3.

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A general result

In this section we present a local symmetry result, generalizing Theorem 2 to sign-changing functions f . (H) The function f ∈ C 0 (R+ ) has a finite number of zeros and f (0) ≥ 0. Moreover, if f (u0 ) = 0 for some u0 ≥ 0, then there exists η > 0 such that: (a) either f is nonincreasing on (u0 − η, u0 + η) ∩ R+ , (b) or f (u) ≥ 0 on (u0 , u0 + η), We say that u0 ∈ f −1 (0) belongs to Za in case (a) and to Zb in case (b). Note that f −1 (0) = Za ∪ Zb . Theorem 3 (Local symmetry result with f only continuous) Let p > 2, γ ∈ (1/(p − 1), 1). Assume that f satisfies (H) and consider a C 1 (B) solution u of (1) such that, with M = kukL∞ (B) , u ∈ C 2 B \ (∇u)−1 (0) ∩ C 1,γ u−1 ([0, M )) .

If f is nonincreasing in a neighborhood of M and if the set J := u −1 (Zb ) ∩ (∇u)−1 (0) is empty, then u is locally radially symmetric and the set u −1 ([0, M ))∩(∇u)−1 (0) is contained in Za .

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Here locally radially symmetric [4] means that there exists an at most countable family (ui )i∈I of radial functions with supports in balls B i , i ∈ I, such that ui|Bi is a nonnegative radial nonincreasing solution of ∆ p ui +f (ui +Ci ) = 0 on Bi , where ui|(B\Bi ) ≡ 0 and Ci ≥ 0 is a constant satisfying f (Ci ) = 0, and such that X u= ui . i∈I

Remark 3 In the case of Theorem 3, I is finite. Monotonicity holds on balls or annuli. As a consequence, in Theorems 1 and 2, monotonicity holds along any radius. The monotonicity is even strict in case of Theorem 2, while there might be a plateau at level θ in case of Theorem 1. The regularity of locally radially symmetric solutions can be studied by elementary methods, which, under an additional condition on f in case (a) of (H), allows to give a sufficient condition under which all sufficiently smooth solutions are globally radially symmetric. Corollary 4 Let α ∈ (0, (p − 2)/2) and consider f such that for any u 0 ∈ Za , 0 < lim inf u→u0

|f (u)| < +∞ . |u − u0 |α

Under the same assumptions as in Theorem 3, if symmetric and decreasing along any radius.

α+1 p−α−1

< γ < 1, then u is globally radially

This corollary is a consequence of Proposition 5, which will be stated in the next section.

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Preliminaries

Let us start with two examples and a statement on the regularity of locally radially symmetric solutions of (1). Consider a radial solution in a ball centered at 0 given as a solution, with r = |x|, of the ordinary differential equation   

1 r d−1

0 r d−1 |u0 |p−2 u0 + f (u) = 0 ,

u(0) = u0 > 0

(2)

and u0 (0) = 0 .

Example 1 Let f (u0 ) 6= 0 and f (u) ≡ f (u0 ) on a neighborhood of u0 . Then u(r) = u0 − A sgn(f (u0 )) r β+1 is the solution of (2) in a neighborhood of r = 0, with β = 1/(p − 1) and A = (|f (u0 )|/d)β /(β + 1). The same formula for u is true up to lower order terms if f is smooth but not constant. In any case, u is exactly in C 1,1/(p−1) at r = 0, and this is why the level u = M has to be excluded in our results if f (M ) 6= 0. Example 2 Let f (u) = |u0 − u|α−1 (u0 − u). Then exactly as above, u(r) = u0 ± A r β+1 is a solution of (2) in a neighborhood of r = 0, with β = (α + 1)/(p − α − 1) and for some A > 0, which depends on d, α and p. In this case, u is exactly in C 1,β at r = 0 with β > 1/(p − 1) if α > 0. However, there is no uniqueness since u ≡ u 0 is also a solution. Actually we may find a continuum of solutions as follows: take u ≡ u 0 on (0, r0 ) for some r0 > 0, and u(r) = u0 ± A (r − r0 )β+1 (1 + o(r − r0 )), for r in a neighborhood of r0+ . This solution also has exactly a C 1,β regularity.

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Remark 4 If u0 ∈ Zb , then there is no radial solution with u(0) > u 0 and ∇u = 0 on {u = u0 }, because of Lemma 9 (strong maximum principle and Hopf’s lemma, see below). These two examples unveil regularity, not only for radially symmetric solutions of (1) but also for locally radially symmetric solutions, since (1) is invariant under translation. This can be stated as follows (the proof is left to the reader). Proposition 5 Assume that f is continuous. Any locally radially symmetric solution of (1) is of class C 2 outside of its critical set. At a critical point x 0 ∈ B, u is exactly in C 1,1/(p−1) if u0 = u(x0 ) 6∈ f −1 (0). If we further assume that f satisfies the hypotheses of Corollary 4 then u is either locally constant or at most of class C 1,β with β = (α + 1)/(p − α − 1). The proof of Corollary 4 is a straightforward consequence of Theorem 3 and Proposition 5. The next result relates the assumption u ≥ 0 with the fact that, in the definition of the local radial symmetry, ui ≥ 0 on Bi . Corollary 6 Assume that f is continuous and let u be a solution of ∆p u + f (u) = 0 in B , u=0 on ∂B , having the following property: there exists an at most countable Pfamily (u i )i∈I of radial monotone functions with supports in balls B i ⊂ B such that u = i∈I ui , where ui satisfies ∆p ui +f (ui +Ci ) = 0 on Bi , ui |∂Bi ≡ 0, the constants Ci are all non-negative and f (Ci ) = 0 for all i ∈ I. If d ≥ 2 and u ≥ 0, then ui > 0 in Bi for any i ∈ I, and u is locally radially symmetric in the sense of the definition given in Section 2. Moreover, for any d ≥ 1, if there exists an i ∈ I such that inf Bi ui < 0, then inf Bi u < 0. Proof. Under the assumptions of Theorem 3, this is a straightforward consequence of the proof. The general case is an extension to p > 2 of a result which has been proved for p = 2 in [14], Proposition 1 (also see [16]). Before proving Theorem 3, we shall state three lemmata. Let us start with a weak comparison result due to Montenegro [24] (also see [11, 25] for related results). Lemma 7 (Weak comparison principle) Let u and v be solutions in W 1,p ∩ L∞ (ω) of ∆p u + f (u) = 0

and

∆p v + f (v) = 0 ,

respectively, where ω is a bounded open connected set in R d with a C 1 by parts boundary. Assume that f is a nonincreasing function on u(ω) ∪ v(ω). If u ≥ v on ∂ω a.e., then u ≥ v in ω a.e. Lemma 8 (Characterization of the critical set) Consider a domain ω in R d . Let u ∈ C 1,γ (ω) be a solution of ∆p u + f (u) = 0 in ω and consider x0 ∈ ω such that ∇u(x0 ) = 0. If x0 ∈ ∂ω, assume moreover that u ∈ C 1,γ (ω), ∂ω is Lipschitz and A(x0 ) = lim→0 −d Vol({x ∈ ω : |x − x0 | < }) > 0. If γ > 1/(p − 1), then f (u(x0 )) = 0.

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Proof. Assume for simplicity that x 0 = 0 and u(x0 ) = 0. Consider then u (x) = −1 u(x). A straightforward computation gives −1 ∆p u + f ( u ) = 0 . Because of the C 1,γ regularity of u, |∇u (x)|p−1 ≤ C |x|γ(p−1) γ(p−1) for some nonnegative constant C. Let φ be a nonnegative nonzero radial test function, then R R f (0) A(x0 ) Rd φ dx = lim→0 −1 ωRf (u ) φ dx |p−2 ∇u · ∇φ dx = lim→0 −1 −1 ω |∇u R ≤ C lim→0 γ(p−1)−1 Rd |x|γ(p−1) |∇φ| dx = 0 , which proves that f (0) = 0. Here we assume A(x 0 ) = |S d−1 |/d if x0 ∈ ω.

For the completeness of this paper, we finally recall a result due to Vázquez [29]. Lemma 9 (Strong maximum principle and Hopf ’s lemma) Let ω be a domain in Rd , d ≥ 1, and let u ∈ C 1 (ω) be such that ∆p u ∈ L2loc (ω), u ≥ 0 a.e. in ω. Assume that −∆p u + β(u) ≥ 0

a.e.

in

ω

with β : [0, +∞) → R continuous, nondecreasing, β(0) = 0 and such there exists R 1 that either −1/p an > 0 for which β ≡ 0 on (0, ) or β(s) > 0 for any s > 0 and 0 (s β(s)) ds = +∞. Then if u does not vanish identically on ω, it is positive everywhere in ω. Moreover, if u ∈ C 1 (ω ∪ {x0 }) for an x0 ∈ ∂ω that satisfies an interior sphere condition and u(x 0 ) = 0, then the derivative along the unit outgoing normal vector at x 0 satisfies: ∂u ∂ν (x0 ) > 0.

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Proof of Theorem 3

We will split the proof of Theorem 3 in five steps, which are more or less classical in moving plane techniques. In the first one, we prove that one can start the method. Then we characterize the obstructions and get the global symmetry result when there is no obstruction. In the third and fourth steps, we prove that obstructions are due only to balls on which the solution is radially symmetric. Finally in the fifth step we establish the local symmetry result. We also indicate how the special cases of Theorems 1 and 2 can be treated. Consider e1 ∈ S d−1 and denote by x1 the coordinate along the direction given by e 1 , and by x0 the coordinate in the orthogonal hyperplane identified with R d−1 . Let λ > 0 and Tλ = {x ∈ B : x = (x1 , x0 ) ∈ R × Rd−1 , x1 = λ} , Σλ = {x ∈ B : x = (x1 , x0 ) ∈ R × Rd−1 , x1 > λ} , xλ = (2λ − x1 , x0 ) if x = (x1 , x0 ) , uλ (x) = u(xλ )

and wλ = uλ − u .

Step 1: Starting the moving plane method Consider λ0 = sup{x1 ∈ (0, 1] : ∃ x0 such that (x1 , x0 ) ∈ B and u(x1 , x0 ) > 0}. Up to a change of coordinates, we may therefore assume that λ 0 > 0. We claim that the moving plane method can be started, that is, there is a λ < λ 0 such that wλ ≥ 0 in Σλ for any λ ∈ (λ, λ0 ).

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Let us assume first that 0 ∈ Zb or f (0) > 0. In this case we notice that for some η > 0, −∆p u ≥ 0

on {x ∈ B : 0 < u(x) < η} .

∂u (1, 0) < 0. Thus λ0 = 1 and the We can use Lemma 9 to conclude that u > 0 in B, and ∂x 1 claim is proved. Next we assume that 0 ∈ Za , that is, f is nonincreasing on (0, η). Assume by contradiction that there exists a sequence λn % 1, such that for each n ∈ N, there exists x n ∈ Σλn such that wλn (xn ) < 0 .

For n large, we can certainly have that η > supx∈Σλn u(x). Noticing that on ∂Σλn we have uλn ≥ u, we may apply Lemma 7 with ω = Σλn and get a contradiction that proves the claim.

Step 2: Characterizing the obstructions Let λ = inf{λ > 0 : wλ (x) ≥ 0 ∀ x ∈ Σλ }. We assume that λ > 0: there exists a sequence λn % λ and a sequence xn ∈ Σλn such that wλn (xn ) < 0. We may assume that xn is a minimum point of wλn , so that ∇wλn (xn ) = 0. Up to the extraction of a subsequence, xn converges as n → +∞ to some x ∈ Σλ , which satisfies wλ (x) = 0 and ∇wλ (x) = 0. We call this point x an obstruction. We observe that in case 0 ∈ Zb or f (0) > 0, u is decreasing in the direction x 1 near ∂B ∩ Σλ for λ close to λ and so that x 6∈ ∂B. Then one of the following cases holds: 1) x ∈ Σλ . At this point we may have ∇u(x) = 0, which implies that f (u(x)) = 0 by Lemma 8 if u(x) < M , or wλ ≡ 0 on B(x, ) for some > 0. In fact, if ∇u(x) 6= 0, we can use a local inversion method and the strong maximum principle as in [15] to conclude that wλ ≡ 0 near x, since wλ has a local minimum at x. ∂w

∂u 2) x ∈ Tλ ∩B. Since x ∈ B it is standard that ∂x (x) = − 12 ∂x1λ (x) = 0, which implies that 1 ∇wλ (x) = ∇u(x) = 0. Actually, if ∇u(x) 6= 0, we can use a local inversion method as in [15] and get a contradiction with Hopf’s lemma for usual elliptic operators. Now we have two possibilities: either u(x) = M , or we can use Lemma 8 to find that f (u(x)) = 0. In summary, either u(x) = M or f (u(x)) = 0, and ∇u(x) = 0.

3) In case 0 ∈ Za it may occur that x ∈ ∂B∩Σλ . If x ∈ ∂B∩Σλ then uλ (x) = u(x) = 0 and so ∇uλ (x) = 0, since xλ ∈ B. Consequently, if ∇u(x) 6= 0 we have that ∇wλ (x) · e1 < 0, which is a contradiction. Thus we must have ∇u(x) = 0. If x ∈ T λ ∩ B then ∂wλ /∂x1 (x) = −2∂u/∂x1 (x) = 0 and we also have ∇u(x) = 0.

Step 3: The obstruction set and the critical points of u Let λ be the first value of λ for which an obstruction point x appears in applying the moving plane method. Remark that all possible obstruction points x are contained in the set K := x ∈ ∂Σλ : ∇u(x) = 0 ∪ x ∈ Σλ \ T λ : u(x) = uλ (x) .

We claim that the set K1 = {x ∈ K : ∇u(x) 6= 0} is not empty. Assume by contradiction that K1 = ∅: ∇u(x) = 0 for all x ∈ K. Let us denote by Qε (x) the set {y ∈ Rd : maxi=1,2,...d |yi − xi | < ε}, where ε > 0 is taken small. From the covering ∪x∈K Qε (x) of the compact set K if 0 ∈ Zb or f (0) > 0, and of the compact set

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K ∪ K ∩ T λ ∩ ∂B if 0 ∈ Za , we can extract a finite subcovering corresponding to points xj ∈ K, j = 1, 2,... L for some number L, and define the open set ωε :=

L [

Qε (xj ) ⊃ K .

j=1

If 0 ∈ Zb or by Lemma 9 if f (0) > 0, we can choose ε small enough so that ω ε ⊂ B and observe that by construction ωε has a boundary of class C 1 by parts. Next we define Γ = ∂ωε . We can find δ1 > 0 and δ > 0 such that uλ ≥ u + δ Next we claim that

Σλ+δ1 ∩ Γ .

on

∂u < −δ ∂x1

on

Tλ ∩ Γ ,

for a possibly smaller δ > 0. If the claim was false, then there would be a point x 0 ∈ Tλ ∩ Γ ∂u with ∂x (x0 ) = 0. But then, since x0 ∈ Γ and uλ (x0 ) = u(x0 ), we have ∇u(x0 ) 6= 0. Thus, 1 we can use a local inversion method and the strong maximum principle as in [15] to prove that uλ ≡ u in a neighborhood of x0 in Σλ and then K1 6= ∅, which is a contradiction with our assumption. If 0 ∈ Za , we take δ1 smaller than dist T λ ∩ ∂B, ∂ωε , and we replace ωε by ωε ∩ B and Γ by ∂ωε ∩ B. Let us consider η > 0 small. First we analyze the behavior of u and uλ in A1 = (B \ B(0, 1 − η)) ∩ Σλ+δ1 . For any x0 ∈ Γ ∩ ∂B ∩ Σλ+δ1 , by definition of K, uλ (x0 ) = uλ (x0 ) − u(x0 ) ≥ δ0 > 0. By continuity, for x in a neighborhood of x 0 and for λ − λ > 0 small enough, have uλ (x0 ) − u(x0 ) ≥ δ0 /2 > 0. Taking η smaller if necessary, this implies that uλ ≥ u in Γ ∩ A1 , for λ − λ > 0 small enough. Next we consider A 2 = B(0, 1 − η) ∩ Σλ+δ1 . Then that by continuity, the property uλ ≥ u + δ on Γ ∩ A2 can be extended to uλ ≥ u + δ/2

on

Γ ∩ A2 ,

for λ − λ > 0 small enough. The property ∂u < −δ ∂x1

on Tλ ∩ Γ

holds for the same reasons as in the first case. Thus, in both cases, using continuity again, for possibly smaller δ and δ 1 , we see that ∂u < −δ ∂x1

on

Σλ \Σλ+δ1 ∩ Γ

and, for η > 0 small enough, we also have uλ ≥ u + δ/2 on Σλ+δ1 ∩ Γ , for λ ∈ (λ − η, λ) , ∂u ≤ −δ/2 on Σλ \Σλ+δ1 ∩ Γ , for λ ∈ (λ − η, λ) . ∂x1

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From the last inequality, we get uλ ≥ u

on

Σλ \Σλ+δ1 ∩ Γ ,

from where it follows that uλ ≥ u

on

Γ ∩ Σλ .

But we also have uλ = u on Tλ ∩ ωε and uλ ≥ u in ∂B ∩ Γ . As a consequence, for ωελ = ωε ∩ Σλ , we get uλ ≥ u on ∂ωελ . By Lemma 8 we have f (u(K) \ {M }) ⊂ {0}, and then, by the assumptions of Theorem 3, u(K) ⊂ Za ∪ {M }. We recall that in a neighborhood of every point in Z a ∪ {M }, f is nonincreasing,by hypothesis. Thus, for ε, η small enough, this guarantees that f is nonincreasing λ λ on u ωε ∪ uλ ωε . Then Lemma 7 applies and gives that uλ ≥ u

on ωελ ,

for λ ∈ (λ − η, λ). This proves that there is no obstruction in ω ελ , a contradiction: K1 6= ∅.

Step 4: Getting a local symmetry result Let us notice that K1 ⊂ Σλ . Because of the local inversion method and the strong maximum principle as in [15] again, K 1 is an open set. Let C + be one of its connected components: ∇u = 0 on Γ+ = ∂C + ∩ Σλ \Tλ .

By Lemma 8 and the assumptions of Theorem 3, u (Γ + ) ⊂ Za ∪ {M } and u is equal to a constant on each component of ∂C + ∩ Σλ . This in particular implies that C + ∩ Tλ 6= ∅, otherwise we would have C + ⊂ Σλ \Tλ : the fact that u is constant in each connected component of ∂C + would contradict the fact that u is nondecreasing and ∇u 6= 0 in C + . Define C = C+ ∪ C− , where C − is the reflexion of C + with respect to Tλ . The set C is connected and symmetric with respect to Tλ , and by construction the function u is symmetric on C with respect to Tλ .

Step 5: Getting a global symmetry result Let us define u1 = inf C u and u2 = supC u. By monotonicity of u in the x1 direction, we know that u takes either the value u 1 > 0 on the boundary of C, where ∇u = 0, or u1 = 0 on ∂C ∩ ∂B. Consequently, by Lemma 8, we have f (u 1 ) = 0 if u1 6= 0. Similarly, either u takes the value u2 on another connected component of the boundary of C and then f (u2 ) = 0, or u reaches its maximum in an interior point x ∈ C that we may take in Tλ ∩ C. Because u1 and u2 belong to {0} ∪ Za ∪ {M }, which is a finite set, and |∇u| is uniformly bounded in B, we deduce that C contains a ball of some radius r 0 > 0, which depends only on f and u. There are therefore only a finite number of such components C. Moreover, if we consider all possible sets C, symmetric with respect to a certain hyperplane, along which u is increasing and where ∇u = 0 on ∂C (except maybe for u 1 = 0), we still have a finite number, say N0 .

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Now we consider N = d N0 + 1 directions γi ∈ S d−1 , i = 1, 2, ..., N , such that the angle (γi , γj ) is 2π-irrational for any (i, j) with i 6= j, and such that any family of d such unit vectors generates Rd . It is not hard to prove that a set which is symmetric with respect to those directions is actually radially symmetric, with respect to some point in it. See [16] for more details. Thus, if we apply the moving plane procedure in all the directions given above, we find a connected component C which is radially symmetric, i.e. an annulus (or a ball) C = Br1 (x0 )\Br2 (x0 ) with r1 > r2 ≥ 0, and such that u = u1 on ∂Br1 (x0 ) and u = u2 on ∂Br2 (x0 ). Then we can consider the following modifications of the function u. First we consider v1 defined in B as v1 ≡ u1 in Br1 (x0 ) and v1 = u in B \ Br1 (x0 ). Second, in case r2 > 0, we define v2 = u in Br2 (x0 ). At this point, under the hypothesis of Theorem 1, the function u is radially symmetric, and M = θ. We observe that the set {x : u(x) = M } is a closed ball of radius r ≥ 0. Under the hypothesis of Theorem 2, we also see that the function u is radially symmetric. In the case of Theorem 3, we reapply Step 1-5 to the function x 7→ v 1 (x) in B, and the function x 7→ v2 (x) − u2 in Br2 (x0 ). We remark that u ≥ u2 on Br2 (x0 ) because the monotonicity of the solution in x 1 . Thus all assumptions of Theorem 3 hold with replaced by Br2 (x0 ). A finite iteration provides the result.

to of B

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