Symmetry results for solutions of semilinear elliptic equations with convex nonlinearities

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1 Symmetry results for solutions of semilinear elliptic equations with convex nonlinearities Filomena Pacella Abstract In this paper we study the symmetry properties of the solutions of the semilinear elliptic problem { u = f(x, u) in Ω u = g(x) on Ω where Ω is a bounded symmetric domain in R N, N 2, and f : Ω R R is a continuous function of class C 1 in the second variable, g is continuous and f and g are somehow symmetric in x. Our main result is to show that all solutions of the above problem of index one are axially symmetric when Ω is an annulus or a ball, g 0 and f is strictly convex in the second variable. To do this we prove that the nonnegativity of the first eigenvalue of the linearized operator in the caps determined by the symmetry of Ω, is a sufficient condition for the symmetry of the solution, when f is a convex function. 1 Introduction In this paper we investigate the symmetry properties of classical C 2 (Ω) C(Ω) solutions of elliptic problems of the type { u = f(x, u) in Ω (1.1) u = g(x) on Ω Research supported by MURST, Project Metodi Variazionali ed Equazioni Differenziali non lineari. Dipartimento di Matematica, Università di Roma La Sapienza, P.le Aldo Moro, Roma, Italy. 1

2 where Ω is a bounded, somehow symmetric domain in R N, N 2, f : Ω R R is a continuous function of class C 1 with respect to the second variable, g is continuous and both f and g have some symmetry in x. It is well known that a classical tool to study this question is the moving planes method which goes back to Alexandrov and Serrin ([8]). Since it is essentially a monotonicity method it, usually, works very well when g 0, u > 0 in Ω and f has some monotonicity in x. In fact, under these hypotheses the moving planes method was successfully used by Gidas, Ni and Niremberg to prove, in the famous paper [6], the symmetry of the solutions of (1.1) when the domain Ω is symmetric with respect to a hyperplane T 0 and convex in the direction ν 0 orthogonal to T 0. However, when the domain is not convex in the direction ν 0 or some of the other hypotheses do not hold, in particular if f does not have the right monotonicity in x, the moving plane method cannot be applied to get the symmetry of the solutions. Indeed if some of these conditions fail there are examples of nonlinearities which give raise to nonsymmetric solutions of (1.1), such as in the case of an annulus and almost critical nonlinearities (see [2]) or when Ω is a ball and f(x) = x α u p, α > 0, p > 1 (see [9]). Nevertheless in some situations, or for a certain class of solutions, it is natural to expect that the solution inherits some or all symmetries of the domain, even if Ω is not convex in any direction, u changes sign and f may not have the right monotonicity in x. This is, for example, the case of ground states solutions in annuli or balls as we will show later. In this paper we use a new simple idea to study the symmetry of the solutions of (1.1) which works efficiently when f(x, s) is convex in the s variable. To be more precise we need some notations. Let us assume that Ω contains the origin and is symmetric with respect to the hyperplane T 0 = {x = (x 1,..., x N ) R N, x 1 = 0} and denote by Ω and Ω + the caps to the left and right of T 0, i.e. Ω = {x Ω, x 1 < 0}, Ω + = {x Ω, x 1 > 0}. Let u be a solution of (1.1) and let us consider the linearized operator at 2

3 u, that is L = f (x, u) where f denotes the derivative of f(x, s) with respect to s. We denote by λ 1 (L, D) the first eigenvalue of L in a subdomain D Ω with zero Dirichlet boundary conditions. Our symmetry results are based on the following proposition whose simple proof will be given in the next section. Proposition 1.1 If f(x, s) and g(x) are even in x 1, f is strictly convex in s and both λ 1 (L, Ω ) and λ 1 (L, Ω + ) are nonnegative then u is symmetric with respect to the x 1 variable, i.e. u(x 1,..., x N ) = u( x 1, x 2,..., x N ). The same result holds if f is only convex but λ 1 (L, Ω ) and λ 1 (L, Ω + ) are both positive. Remark 1.1 When f is convex but not strictly convex the previous result cannot be improved, i.e. the nonnegativity of the first eigenvalues in Ω and Ω + is not sufficient for the symmetry of the solution. This will be shown in section 2 with an easy example in the case when f(x, s) is linear with respect to s. Remark 1.2 A statement analogous to that of Proposition 1.1 holds also for semilinear problems with symmetric Neumann conditions on Ω. In this case the relevant eigenvalues to get the symmetry of the solutions are µ 1 (L, Ω ) and µ 1 (L, Ω + ) which denote the first eigenvalue of L in Ω (or Ω + ) with mixed boundary conditions, namely the normal derivative zero on Ω Ω (or Ω Ω + ) and a zero Dirichlet boundary condition on T 0 Ω (or T 0 Ω + ). This will be the subject of a further investigation. Having Proposition 1.1 in mind, now the question is how to prove the nonnegativity of the first eigenvalues of L in Ω and Ω + in order to get the symmetry result. A few easy cases will be considered in section 2, while in section 3 we apply Proposition 1.1 to get the main result of this paper which consists in proving the axial symmetry of solutions of index one. More precisely we will assume Ω to be either an annulus or a ball in R N, N 2 and u a solution of (1.1), with g(x) = 0, with index one, 3

4 i.e. such that the linearized operator L has only one negative eigenvalue in Ω (with homogeneous Dirichlet boundary conditions). This kind of solutions always exist for a large class of superlinear convex nonlinearities, as we can consider the so called least energy solutions or the ones of mountain pass type. In this case we can prove that if f(x, s) is strictly convex in s and radially symmetric in x then u is axially symmetric with respect to the axis passing through a maximum point. Moreover we are able to show that, if the solution u is not radial, all the critical points of u are located on the symmetry axis and we also describe some other qualitative properties of the solutions. For more general domains, symmetric with respect to the hyperplane T 0 = {x R N, x 1 = 0} we can prove the symmetry, with respect to x 1, of every solution of index one as soon as f(x, s) is even in x 1 and a maximum (or a minimum) point of u belongs to the hyperplane T 0. However let us remark immediately that this is not always the case as it can be shown with some counterexamples. All this will be described in detail in section 3. Finally let us note that even in nonsymmetric situations it is important to find out that λ 1 (L, D) is positive for some caps D Ω. In fact, if f is convex and D has the property that the reflected function v (with respect to some hyperplane) is well defined then, arguing as in the proof of Proposition 1.1 or Proposition 2 (see section 2), it is possible to show that u < v in D and this gives some information about the location of the maximum points of u. We feel that all this can help to understand the qualitative properties of the solutions of (1.1). 2 Proofs of Proposition 1.1 and applications We start with the proof of Proposition 1.1. Proof of Proposition 1.1. Let us denote by v and v + the reflected functions of u in the domains Ω and Ω + respectively: v (x) = u( x 1, x 2,..., x N ), x Ω 4

5 v + (x) = u( x 1, x 2,..., x N ), x Ω + We first assume f strictly convex. In this case we have f(x, v (x)) f(x, u(x)) f (x, u(x))(v (x) u(x)) in Ω f(x, v + (x)) f(x, u(x)) f (x, u(x))(v + (x) u(x)) in Ω + and the strict inequality holds whenever v (x) u(x) (resp. v + (x) u(x)). Hence by (1.1), using the symmetry of f and g in the x 1 variable and considering the functions w = v u and w + = v + u we have w f (x, u)w 0 in Ω (2.1) w + f (x, u)w + 0 in Ω + (2.2) with the strict inequality whenever w (x) 0 or w + (x) 0 and w = 0 (resp. w + = 0) on Ω (resp. Ω + ) (2.3) If w and w + are both nonnegative in the respective domains Ω and Ω + then w w + 0, by the very definition, and hence u is symmetric with respect to x 1. Therefore, arguing by contradiction, we can assume that one among the two functions, say w +, is negative somewhere in Ω +. Then, considering a connected component D in Ω + of the set where w + < 0, multiplying (2.2) by w +, integrating and using (2.3) and the strict convexity of f we get w + 2 f (x, u)(w + ) 2 < 0 (2.4) D D which implies that λ 1 (L, Ω + ) < 0 against the hypothesis. Hence u is symmetric. Now we assume that f is only convex but λ 1 (L, Ω ) > 0 and λ 1 (L, Ω + ) > 0. Then the maximum principle holds both in Ω and Ω +. Therefore by (2.1) (2.3) we get immediately w 0 and w + 0 which imply the symmetry of u. In particular if f is linear in u we get 5

6 Corollary 2.1 Let A be a linear operator of the type A = c(x), with c(x) C(Ω), Ω as in the previous theorem and c(x) even in the x 1 variable. Then if λ 1 (A, Ω ) 0, any eigenfunction of A in Ω with homogeneous Dirichlet boundary conditions, corresponding to a negative eigenvalue µ j of A is even in x 1. The same statement holds if we assume λ 1 (A, Ω ) > 0 and µ j 0. Proof. If we consider an eigenfunction ϕ relative to an eigenvalue µ j, it solves the equation and the linearized operator at ϕ is ϕ = f(x, ϕ) = c(x)ϕ + µ j ϕ L = c(x) µ j. In both hypotheses we have that λ 1 (L, Ω ) > 0 and, since c(x) is even in x 1, also λ 1 (L, Ω + ) > 0. Hence, by Proposition 1.1, we get the symmetry of ϕ. Next we show with an example that if f is only convex, but not strictly convex, and λ 1 (L, Ω), λ 1 (L, Ω + ) are both zero then the solution u of (1.1) may not be symmetric. Let Ω be as in Proposition 1.1 and denote by µ 1 the first eigenvalue of the Laplace operator in Ω with homogeneous Dirichlet boundary conditions and by ϕ 1 the corresponding eigenfunction. By reflecting ϕ 1 in an odd way with respect to T 0 we get that the extended function ϕ 1 solves ϕ 1 = µ 1 ϕ 1 in Ω, ϕ 1 = 0 on Ω. Obviously the first eigenvalue of the linearized operator at ϕ 1 is zero in Ω or Ω +, while ϕ 1 is not even in x 1. Now we describe some situations when it is easy to prove the nonnegativity of the eigenvalues λ 1 (L, Ω ) and λ 1 (L, Ω + ), so to get the symmetry of the solution. A first trivial case is when the solution u of (1.1) is semi stable, i.e. λ 1 (L, Ω) 0. Then, obviously λ 1 (L, Ω ) > 0 and λ 1 (L, Ω + ) > 0; 6

7 therefore if f is convex we have that u is symmetric, in particular it is radial if Ω is an annulus or a ball. However, when f is strictly convex, it can be shown that the semistable solution is unique and from this, the symmetry follows easily. Another easy application of Proposition 1.1 is obtained by considering domains Ω which are also convex in the x 1 direction and assuming g = 0, u positive and f increasing in the x 1 variable in Ω. Under these hypotheses and keeping the same notations as in Proposition 1.1 we have Proposition 2.1 If f(x, s) is convex in s and u is a positive solution of (1.1) then λ 1 (L, Ω ) and λ 1 (L, Ω + ) are both nonnegative. Proof. Let us prove that λ 1 (L, Ω ) is nonnegative. Arguing by contradiction, we assume that λ 1 (L, Ω ) < 0. Then, by the continuity of the eigenvalues with respect to the domain, we have that for some µ < 0 in the cap Ω µ = {x Ω, x 1 < µ} the first eigenvalue λ 1 (L, Ω µ ) must be zero, because we know that when the measure of any cap is sufficiently small the first eigenvalue becomes positive. Since Ω is convex in the x 1 direction we can consider the function We have w µ = u(x) u(2µ x 1, x 2,..., x N ) in Ω µ. w µ = 0 on Ω µ Ω, w µ > 0 on Ω µ Ω (2.5) because µ < 0, u > 0 in Ω and u = 0 on Ω. By the convexity of f with respect to u and the monotonicity of f in x 1, we get, (as in (2.1)) w µ f (x, u)w µ 0 in Ω µ. (2.6) Therefore if wµ 0 in Ω µ, by the strong maximum principle and (2.5) we deduce wµ > 0 in Ω µ and this, together with (2.5) and (2.6) implies that λ 1 (L, Ω µ ) > 0 (see [4]) which gives a contradiction. Hence the function wµ is negative somewhere in Ω µ. Thus, considering a connected component D in Ω µ where wµ < 0, multiplying (2.6) by wµ and integrating we get wµ 2 f (u)(w ) 2 0 (2.7) D D 7

8 which implies that λ 1 (L, D) 0. Since, by (2.5), D is a proper subset of Ω µ we get λ 1 (L, Ω µ ) < 0 against what we assumed. In the same way it can be proved that also λ 1 (L, Ω + ) is nonnegative. Remark 2.1 From Proposition 1.1 and Proposition 2.1 we deduce the symmetry of the solutions of (1.1) (under the hypotheses of Proposition 2.1) with an approach slightly different from that used by Gidas, Ni and Nirenberg and later simplified by Berestycki and Niremberg ([3]). This is because we focus our attention on the first eigenvalue of the linearized operator rather than on the equation satisfied by the difference between u and its reflection. Of course this is possible because f is convex, therefore the symmetry result deduced by Proposition 2.1 is much weaker than that of [6]. Remark 2.2 In the classical paper [6], using the moving plane method, it is proved that u x 1 > 0 in Ω. Moreover, if Ω is smooth and f(0) 0, u x 1 is also positive on Ω Ω, by the Hopf s lemma. Then, if f does not depend on x, since the function u x 1 is a solution of the linearized equation, i.e. ( ) u L = 0 in Ω x 1 it follows that λ 1 (L, Ω ) > 0 and the same holds for λ 1 (L, Ω + ). Therefore Proposition 1.1 can be seen as a generalization of this result when Ω is not smooth or f(0) is not positive. It is also interesting to note that the nonnegativity of the first eigenvalue in Ω or Ω + can be deduced without knowing a priori that the solution is strictly monotone in the x 1 direction. Finally we recall that, using the sign of the eigenvalues λ 1 (L, Ω ), λ 1 (L, Ω + ), in the paper [4] it is shown that every solution of the linearized equation in Ω is symmetric when f(0) 0 and Ω is smooth. Let us observe that this result is a particular case of Corollary 2.1 (see Theorem 2.1 and Remark 2.1 in [4]). 8

9 3 Axial symmetry of solutions of index one Let us consider the semilinear problem { u = f( x, u) in A u = 0 on A (3.1) where A is either an annulus or a ball centered in the origin O of R N, N 2 and f has the same regularity as in (1.1). Let u be a solution of (3.1) which can be either positive or change sign and let P be a maximum point of u which lies in the interior of Ω, because of the boundary condition. We denote by r P the axis OP passing through the origin and P, by T any (n 1) dimensional hyperplane passing through the origin and by ν T the normal to T, directed towards the half space containing P, in case T does not pass through the axis r P. Our main result is the following Theorem 3.1 Let f( x, s) be strictly convex in s and u a solution of (3.1) of index one. Then i) u is axially symmetric with respect to the axis r P ii) if A is a ball and P is the origin then u is radially symmetric iii) if u is not radially symmetric then it is never symmetric with respect to any hyperplane T not passing through r P iv) if u is not radially symmetric then all the critical points of u belong to the symmetry axis r P ; in particular all the maximum points lie on the semi axis to which P belongs and u ν T (x) > 0 x T A (3.2) for every hyperplane T not passing through r P. Proof. i) Let us denote by T P any hyperplane passing through r P. Obviously T divides A in two open regions A P and A + P, i.e. A P A + P (T P A) = A. To show the symmetry of u with respect to T P we use Proposition 9

10 1.1; so we need to prove that λ 1 (L, A P ) and λ 1 (L, A + P ) are both nonnegative, denoting, as in section 1, by λ 1 (L, A P ) and λ 1 (L, A + P ) the first eigenvalues of the linearized operator at u in A P and A + P, respectively, with homogeneous Dirichlet boundary conditions. Arguing by contradiction we can assume that one of these two numbers, say λ 1 (L, A P ), is negative. Since the solution is of index one, λ 2 (L, A) 0 and hence, by the variational characterization of the second eigenvalue, we have that λ 1 (L, A + P ) > 0. So in A + P the maximum principle holds for the operator L = f ( x, u). Then, considering in A + P the function w P + = v P + u, where v P + is the reflection of u with respect to T P and using the convexity of f, we have, as in (2.2) w + P f ( x, u)w + P 0 in A + P. (3.3) Since w P + 0 on A + P, from (3.3) we get, by the maximum principle, that w P + 0 in A + P. Thus, by the strong maximum principle either w P + 0 or w P + > 0 in A + P. The first case is not possible because it would imply that u is symmetric with respect to T P and hence the two eigenvalues λ 1 (L, A P ) and λ 1 (L, A + P ) should be equal while they have different sign. So, the only possibility is w P + > 0 in A + P. Then, by the Hopf s lemma we derive that w+ P < 0 on T ν P A, where ν is the outer normal to A + P and consequently u ν = 1 w P + 2 ν > 0 on T P A which is impossible because the maximum point P belongs to T P A. This contradiction shows that λ 1 (L, A P ) 0 and the same happens for λ 1 (L, A + P ). ii) It follows immediately from i) since the origin belongs to any symmetry hyperplane. iii) Arguing again by contradiction let us assume that u is symmetric with respect to a certain hyperplane T 1 not passing through r P. Since u is not radial, by ii) P is not the origin and hence P / T 1, so that, using the same notations as in i), P will belong to one of the two caps A 1, A + 1, created by T 1, say P A 1. Then, by symmetry, there exists P A + 1 such that u(p ) = u(p ) = max u. A 10

11 Now let us consider any hyperplane T close to T 1, i.e. denoting by ν 1 the unit normal to T 1, pointing towards the half space containing P, we consider, on the unit sphere, a neighborhood I(ν 1 ) of ν 1 and, for any ν I(ν 1 ) the hyperplane orthogonal to ν, passing through the origin. If the size of I(ν 1 ) is sufficiently small, we still have that P and P are on different caps, with respect to T : P Ã and P Ã+, for any ν I(ν 1 ). We claim that u is symmetric with respect to T. To prove this, we use again Proposition 1.1 so that we need to show that λ 1 (L, Ã ) and λ 1 (L, Ã+ ) are both nonnegative. If λ 1 (L, Ã ) < 0, then, since the solution u has index 1, we deduce that λ 1 (L, Ã+ ) > 0. Then, arguing exactly as in the proof of i) we have that the function w + = ṽ + u, where ṽ + is the reflected function of u in Ã+, is positive in Ã+. This means that u < ṽ + in Ã+, in particular u(p ) < ṽ + (P ) which is not possible since u(p ) is the maximum of u. This contradiction proves that λ 1 (L, Ã ) 0 and the same holds for λ 1 (L, Ã+ ). Hence u is symmetric with respect to the hyperplane T orthogonal to ν, for any direction ν in a suitable neighborhood of ν 1. Now we can think, without loss of generality, that all possible symmetry hyperplanes of A (i.e. all hyperplanes passing through the origin) correspond to unit vectors belonging to an hemisphere in R N (i.e. to half of thenunit sphere), having as boundary the vectors ν P orthogonal to the hyperplanes T P, passing through the axis r P. If we remove from this hemisphere all the vectors ν P we have an open connected set M of directions in R N. What we just proved is that the set S of directions ν which are orthogonal to hyperplanes of symmetry for the solution u is an open set in M. Since S is also, obviously, a closed subset of M we deduce that S = M and hence u is symmetric with respect to any hyperplane passing through the origin, i.e. u is a radial function. iv) Suppose that u is not radially symmetric and consider any hyperplane T not passing through r P. As usual we denote by A and A + the caps created by T and we assume that P belongs to A. Then, by iii) u is not symmetric with respect to T and hence, by Proposition 1.1, one among λ 1 (L, A ) and λ 1 (L, A + ) must be negative. Since P A it is 11

12 easy to see, arguing as in i) or iii) that λ 1 (L, A ) < 0 and λ 1 (L, A + ) > 0. Thus in Ω + the maximum principle holds for the operator L and this implies that the function w + = v + u (same notations as before) satisfying { w + f ( x, u)w + 0 in A + (3.4) w + = 0 on A is nonnegative in A + and, actually, w + > 0, by the strong maximum principle. This means that u < v + in A + and hence u cannot have any maximum point in A +. Letting T vary and, in particular, taking T as the hyperplane orthogonal to r P we deduce that all maximum points belong to the radius joining the origin with P. Moreover, applying, as before, the Hopf s lemma to the function w + in A + we get (3.2) which implies that all critical points of u belong to the symmetry axis r P. Remark 3.1 The previous theorem gives the precise location of all maximum points of the solution. However it is natural to expect that solutions of index one, in particular least energy solutions, have only one maximum point, at least for a large class of nonlinearities f independent of x. This can be proved for solutions of index one of some asymptotic nonlinear problems, using a blow up argument, in a more general context (see [7]). Theorem 3.1 applies to a very large class of nonlinear problems since we only require f(x, s) to be radial in x and strictly convex in s. Also the existence of solutions of (3.1) of index one is well known in many cases, at least for positive solutions. Though changing sign solutions of index one exist, they do not occur frequently since the nonlinearity f must satisfy some peculiar hypotheses (see [1]). Hence we do not think that Theorem 3.1 is easily applicable to solutions which change sign. On the contrary when u is positive there are many interesting applications of Theorem 3.1. We would like to single out two of them. I) Let f(x, u) = f(u) = u P + λu, 1 < p < if N = 2, 1 < p < N+2 N 2 if N 3, λ < λ 1 (A), where λ 1 (A) is the first eigenvalue of the Laplace operator in A with zero Dirichlet boundary conditions. If λ > 0 we can also take p = N+2 N 2. 12

13 In this case a positive solution of (3.1) of index one can be either found using the famous mountain pass lemma or by a constrained minimization procedure. If A is an annulus it can be proved that this positive solution is, in general, not radial (see [2]); in fact the Gidas Ni Nirenberg symmetry result does not apply since A is not convex in any direction. II) Let f(x, u) = f( x, u) = x α u P where α > 0, 1 < p < if N = 2, 1 < p < N+2 if N 3. N 2 The corresponding equation is called the Hénon equation. In this case the symmetry result of Theorem 3.1 is interesting both in the annulus and in the ball, since there are results which assert that ground states solutions of (3.1) are not always radially symmetric if A is a ball (see [9]). We recall that also for this nonlinearity the Gidas Ni Nirenberg theorem does not apply since f is not decreasing with respect to x. Finally consider a general domain Ω R N, N 2, containing the origin and symmetric with respect to the hyperplane T = {x R N, x 1 = 0}. In Ω we define the usual problem { u = f(x, u) in Ω (3.5) u = 0 on Ω where f has the usual regularity and f is even in x 1. We have the following result. Theorem 3.2 Let f be strictly convex in the second variable and u a solution of (3.5) with index one. If a maximum point P of u belongs to the symmetry hyperplane T, then u is even in x 1. Proof. It is the same as for i) of Theorem 3.1. In view of Theorem 3.2 a natural question is whether a maximum point of a solution (in particular of index one) belongs to the symmetry hyperplane T 0. The answer is that this is not, in general, true since it is possible to find some counterexamples. One of these could be constructed by taking Ω as a dumbbell and f(u) = u P. Then in the paper [5] of Dancer it is shown that there exists a positive solution of (3.5) which has only one maximum point near the center of one of the two kind of balls which form the dumbbell. 13

14 References [1] T. Bartsch, K.C. Chang, Z.Q. Wang, On the Morse indices of sign changing solutions of nonlinear elliptic problems, Math. Z., 233 (2000), [2] H. Brezis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math., 36 (1983), [3] H. Berestycki, L. Nirenberg, On the method of moving planes and the sliding method, Bol. Soc. Bras. Mat., 22 (1991), [4] L. Damascelli, M. Grossi, F. Pacella, Qualitative properties of positive solutions of semilinear elliptic equations in symmetric domains via the maximum principle, Ann. Inst. H. Poincaré, 16 (1999), [5] E.N. Dancer, The effect of domain shape on the number of positive solutions of certain nonlinear equations, Journ. Diff. Eq., 74 (1988), [6] B. Gidas, W.M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phis., 68 (1979), [7] K. El Mehdi, F. Pacella, Morse index and blow up points of solutions of some nonlinear problems, (to appear). [8] J. Serrin, A symmetry problem in potential theory, Arch. Rat. Mech. Anal., 43 (1971), [9] D. Smets, J. Su, M. Willem, Nonradial ground states for the Hénon equation, (to appear). 14