A critical elliptic problem for polyharmonic operators

Transcription

1 A critical elliptic problem for polyharmonic operators Yuxin Ge, Juncheng Wei and Feng Zhou Abstract In this paper, we study the existence of solutions for a critical elliptic problem for polyharmonic operators. We prove the existence result in some general domain by minimizing on some infinite-dimensional Finsler manifold for some suitable perturbation of the critical nonlinearity when the dimension of domain is larger than critical one. For the critical dimensions, we prove also the existence of solutions in domains perforated with the small holes. Some unstable solutions are obtained at higher level sets by Coron s topological method, provided that the minimizing solution does not exist. AMS classification scheme numbers: 35J35, 35J40, 35J60 1 Introduction This paper is a sequel to [21] on some semilinear critical problems for polyharmonic operators. Let K N and R N (N 2K + 1) be a smooth bounded domain in R N. We consider the semilinear polyharmonic problem with homogeneous Dirichlet boundary condition { ( ) K u = u s 2 u + f(x, u) in u = Du = = D K 1 (1) u = 0 on where s := 2N N 2K denotes the critical Sobolev exponent for ( )K and f(x, u) is a lowerorder perturbation of u s 2 u (see the assumption (H2) below). The equation (1) is of variational type: Solutions of (1) correspond to critical points of the energy functional E(u) = 1 2 u 2 K,2, 1 u s F (x, u), (2) s defined on the Hilbert space { } H0 K () = v H K () D i v = 0 on 0 i < K which is endowed with the scalar product (( ) M u)(( ) M v) (u, v) = ( ( ) M u)( ( ) M v) if K = 2M is even if K = 2M + 1 is odd The first author is supported by ANR project ANR-08-BLAN The second author is partially supported by a research Grant from GRF of Hong Kong and a Focused Research Scheme of CUHK. The third author is supported in part by NSFC No and by Shanghai Priority Academic Discipline. 1 (3)

2 u and K,2, is the corresponding norm, F (x, u) := f(x, t)dt is the primitive of f. 0 We assume that (H1) f(x, u) : R R is continuous and sup f(x, u) < for every M > 0; x, u M (H2) f(x, u) = a(x)u + g(x, u) with a(x) L () C (), g(x, u) = o(u) as u 0 uniformly in x and g(x, u) = o( u s 1 ) as u uniformly in x. From (H1) to (H2), it follows f(x, 0) = 0 and that f is a lower-order perturbation of u s 2 f(x, u) u at infinite in the sense that lim u u s 1 = 0 uniformly in x. Moreover, we assume that f(x, u) satisfies (H3) f u (x, u) is continuous on R; (H4) f u (x, u) C(1 + u s 2 ), u R uniformly in x ; (H5) f 1 (x, u) := f(x,u) u is non-decreasing in u > 0 and non-increasing in u < 0 for a.e. x. For K = 1, f(x, u) = λu and λ (0, λ 1 ) where λ 1 is the first eigenvalue of for Dirichlet boundary condition, the problem has a strong background from some variational problems in geometry and physics, such as the Yamabe s problem with lack of compactness. This was considered by Brezis and Nirenberg for positive solutions in their pioneer work in [5]. Then it has been studied extensively in the last three decades. We recall briefly some results about the existence and multiplicity of sign-changing solutions to the problem (1) for K = 1 and f(x, u) = λu. For any fixed λ > 0, the first multiplicity result was due to Cerami, Fortunato and Struwe [8]. They obtained the number of the solutions of (1) is bounded below by the number of the eigenvalues of lying in the open interval (λ, λ + S 2/N ), where S is the best constant for the Sobolev embedding D 1,2 (R N ) L 2 (R N ) (see the definition below) and denotes the Lebesgue measure of. Capozzi, Fortunato and Palmieri in [7] established the existence of a nontrivial solution for λ > 0 which is not an eigenvalue of when N 4 and for any λ > 0 when N 5 (see also [45]). In [11], Devillanova and Solimini proved that, if N 7, then (1) has infinitely many solutions for every λ > 0. They proved also in [12] that, if N 4 and λ (0, λ 1 ), then there exist at least N pairs of nontrivial solutions. Clapp and Weth [9] has extended this last result to all λ > 0 with N 4. In the same paper they also obtained some extensions to critical biharmonic problems for N 8. When the domain is a ball and N 4, Fortunato and Jannelli [15] proved there are infinitely many sign-changing solutions which are built using the symmetry of the domain. Schechter and Zou in [37] showed the same result for any domain when N 7. In particular, if λ λ 1, it has and only has infinitely many sign-changing solutions except zero. Their work is based on the estimates of Morse indices of nodal solutions. Concerning the polyharmonic case, Pucci and Serrin in [35] has studied the problem (1) for K = 2 and λ > 0 when is a ball. They proved that it admits nontrivial radial symmetric solutions for all λ (0, λ 1 ) if and only if N 8. If N = 5, 6, 7, then 2

3 there exists λ (0, λ 1 ) such that the problem admits no nontrivial radial symmetric solutions whenever λ (0, λ ]. Here λ 1 is understood as the first eigenvalue of 2 for Dirichlet boundary conditions. This is the counterpart of the well known result of [5] on the nonexistence for radial symmetric solutions for small λ in dimension N = 3 and K = 1 (where λ = λ 1 /4). They called these dimensions as critical dimensions. They conjectured that for general K 1, the critical dimensions are 2K + 1,, 4K 1. The conjecture is not completely solved for all K 1. Grunau [23] defined later the notion of weakly critical dimensions as the space dimensions for which a necessary condition for the existence of a positive radial solution of (1) in B 1 is λ (λ, λ 1 ) for some λ > 0. He proved that the conjecture is true in the weak sense. Gazzola, Grunau and Squassina [18] proved nonexistence of positive radial symmetric solutions for Navier boundary condition for small λ > 0. They established also some existence results for λ = 0. Their result strongly depends on the geometry of domains. For biharmonic operators, Bartsch, Weth and Willem in [3] and Ebobisse and Ahmedou in [13] have studied the problem (1) on domains with nontrivial topology under Dirichlet boundary condition and Navier boundary condition respectively. For related problems, we infer to [4], [14], [16], [17], [22], [24], [32] and the references therein. For general case K 1, Ge has studied in [21] the same type of equation (1) for Navier boundary condition when f(x, u) = λu with 0 λ < λ 1 and λ 1 the first eigenvalues of ( ) K. He established the existence of positive solutions in some general domain under the suitable assumptions. In particular unstable solutions in higher level set are obtained by Coron s topological method in domains perforated with the small holes. The purpose of this paper is to continue the study of the semilinear polyharmonic problem (1) to general K 1 with Dirichlet boundary condition for general domains. Let us denote the polyharmonic operator L := ( ) K a(x) and λ 1 () λ 2 () λ n () the eigenvalues of L under the homogeneous Dirichlet boundary condition. It is well known that each eigenvalue λ k (), k 1, can be described as the minimax value λ k () = min max V H0 K(),dimV =k v V vlv. v 2 It follows that λ k () is a non-increasing functional on the domains, that is, if 2 1, then λ k ( 2 ) λ k ( 1 ). Moreover, from the unique continuation principle, we have λ k ( 2 ) > λ k ( 1 ) for any k 1, provided 1 is connected (see [27, 34]). For the perforated domain := 1 \ 2 with the smooth bounded domains 2 1, with the help of the above description, we have lim λ k ( 1 \ 2 ) = λ k ( 1 ), where the limit is taken as the diameter of 2 goes to 0. To this aim, it suffices to consider 2 = B(x, ɛ) balls with small radius ɛ > 0 in the sequel. 3

4 Assume now λ n () 0 and λ n+1 () > 0 for some n 1. Let e i (x) be an eigenfunction associated to λ k () with e i K,2, = 1 for any 1 i n. Define M := {v H K 0 () \ {0} de(v)(w) = 0, w Span(v, e 1,, e n )}. We prove in Section 2 that under the hypothesis (H1) to (H5), M is then a complete C 1 Finsler manifold and it will be a C 1,1 Finsler manifold with additional assumptions (H6) to (H7) (see section 2). This permits to consider the following minimization problem We can prove then κ K N (S K()) N 2K S K () := κ := inf v M E(v). for any f satisfying (H1) to (H5), where we denote v 2 K,2, inf v H0 K()\{0} v 2 L s () the best constant for the embedding H K 0 () Ls (). Here, as for K = 1, it is well v 2 K,2,R N known that S K () is independent of and S K () = S K (R N ) := inf v H K (R N )\{0} v 2. L s (R N ) Therefore we denote it by S K in the sequel (see [17, 20]). We prove then that for non critical dimension case N 4K and f(x, u) = µu, the infimum above is achieved by some u M which is a solution of (1). This method is a generalization of the Nehari manifold method ([31]) and can be seen as an alternative approach to the linking method (see [39]). For the critical dimension 2K < N < 4K, the existence of solutions to (1) is a delicate issue. To our knowledge, there are few results on it. The reason is that the minimizing method fails, for example, for K = 1, when R 3 is a ball and when f(x, u) = λu with 0 < λ < λ 1 4. It is well known that there are no positive solutions. In Section 3, we study the existence of solutions for some perforated domains in such critical dimensions. We analyze the concentration phenomenon when κ equals to K N (S K) N 2K. Then following Coron s strategy of topological argument, we obtain the existence of unstable critical points in higher level sets for domains perforated with small holes. In all this paper, C, C and c denote generic positive constant independent of u, even their value could be changed from one line to another one. We give also some notations here. The space D K,2 (R N ) (resp. D K,2 (R N + )) is the completion of C0 (RN ) (resp. C0 (RN + )) for the norm K,2,R N (resp. K,2,R N ). + 2 Study of the energy functional E on M We begin this section by studying some properties of the set M. Observe that v M is equivalent to say v 0 and satisfying l 0 (v) := v 2 K,2, v s L s () f(x, v)v = 0 (4) l i (v) := (v, e i ) v s 2 ve i f(x, v)e i = 0, 1 i n. Let us denote V 0 := Span(e 1,, e n ) the n dimensional vector space spanned by e 1,, e n. We prove now the following proposition. 4

5 Proposition 1 Suppose (H1) to (H5) are satisfied. Then M is a complete C 1 Finsler manifold. Furthermore, suppose that (H6) 2 f u 2 (x, u) is continuous on R and u u s 2 u is C 2 on R; (H7) 2 u 2 f(x, u) C( u + 1) s 3, u R uniformly in x. Then M is a complete C 1,1 Finsler manifold. Proof. The proof is divided into several steps. Step 1. M is not empty. By the assumptions (H1)-(H2), E is a continuous functional on H K 0 (). Fixing v V 0 and let V := Span(v, e 1,, e n ). Clearly, for all w V, we have E(w) 1 2 w 2 K,2, 1 2 a(x)w 2 1 s w s, (5) since it follows from (H2) and (H5) that g(x,u) u 0 and F (x, u) 1 2 a(x)u2 for all u R\{0} and for a.e. x. As V is a finite dimensional vector space, all the norms on it are equivalent. In particular, the norms K,2, and L s () are equivalent on V. This implies lim E(w) =. (6) w V,w On the other hand, again from (H2), we infer for any given ε > 0, there exists C > 0 such that for all u R and for a.e. x so that for all w V g(x, u) ε u + C u s 1, F (x, u) 1 2 (a(x) + ε)u2 + C s u s, (7) E(w) 1 2 w 2 K,2, 1 2 (a(x) + ε)w C s Since v V 0, we can choose v V (V 0 ) such that 1 2 v 2 K,2, 1 2 taking a sufficiently small ε > 0, we have As a consequence, we obtain w s. a(x)(v ) 2 > 0. By 1 2 v 2 1 K,2, 2 (a(x) + ε)(v ) 2 ε v 2 K,2,. (8) sup E(w) > 0. (9) w V Together with (6), there exists ṽ V such that E(ṽ) = max w V E(w) since V is a finite dimensional vector space. Clearly, ṽ M. Step 2. M is closed. 5

6 We define the map L : H0 K () Rn+1 v (l 0 (v),, l n (v)). In view of the assumptions (H1)-(H2), L is continuous on H0 K(). Let (v k) M be a sequence in M such that v k v in H0 K (). Then we get L(v) = 0. Now it suffices to show v 0. First, we note v k V 0 for all k N. Indeed, we have v k 2 K,2, a(x)vk 2 = v k s L s () + g(x, v k )v k. (10) If we have v k V 0 for some k 1, the term on the left hand is non-positive. But that one on the right hand is non-negative. Thus, v k s L s () = 0 and the desired contradiction v k 0 gives the result. Now, we claim there exists some positive number c > 0 such that v k K,2, > c. We denote the orthogonal projection of v k on V 0 by v n k := (v k, e i ) e i and v k its orthogonal complementary As v k M, we obtain v k := v k v k. (v k, v k ) a(x)v k v k = ( v k s 2 + g(x, v k) )v k v k v. k Together with (10), we have vk 2 K,2, a(x)(vk ) 2 ( v k 2 a(x)(v k )2 ) = ( v k s 2 + g(x, v k) )((vk ) 2 (v k v )2 ) k which implies since vk 2 K,2, a(x)(vk ) 2 ( v k s 2 + g(x, v k) )(vk ) 2, (11) v k v k 2 K,2, a(x)(v k )2 0. Gathering (5), (8) and (11), we get 2ε vk 2 K,2, v k 2 (a(x) + ε)(vk ) 2 (1 + C) v k s 2 (vk ) 2 (1 + C) v k s 2 L s () v k 2 L s () C (1 + C) v k s 2 L s () v k 2 K,2, 6

7 Finally, v k L s () c > 0 and the desired claim follows. Step 3. dl(v) is surjective and its kernel splits for all v M. By (H3) and (H4), f(x, u)u and f(x, u) are C 1 on R and (f(x, u)u) (x, u) u C(1 + u s 1 ), uniformly in x and u R. (12) Therefore, L is C 1 on H0 K () provided the assumptions (H1)-(H4) hold. A direct calculation leads to dl 0 (v)(w) = 2(v, w) s v s 2 f(x, v) vw (f(x, v) + v )w, v dl i (v)(w) = (w, e i ) (s 1) v s 2 f(x, v) (13) we i we i, 1 i n. v We claim dl(v) V, the restriction on V of dl(v), is a bijective endomorphism from V on R n+1. As V and R n+1 have the same dimension, it suffices to prove Ker(dL(v) V ) = {0}. Let w Ker(dL(v) V ) and write w = µv+ n µ i e i where µ, µ i R for each i. Combining (4) and (13), we get dl 0 (v)(w) = (s 2) v s 2 f(x, v) vw ( f(x, v) + v )w = 0, dl i (v)(w) v f(x, v) f(x, v) = ( + )µve i (s 2) v s 2 µve i v v +( n µ j e j, e i ) (s 1) v s 2 n e i µ j e j for all 1 i n. On the other hand, we have Together with (14), we infer (s 2) + g(x, v) v µdl 0 (v)(w) + µ i dl i (v)(w) = 0. f(x, v) v n e i v s 2 w 2 + v s 2 ( µ j e j ) 2 f(x, v) f(x, v) + ( + )w 2 v v ( µ j e j ) 2 ( µ j e j, µ i e i ) + a(x)( µ j e j ) 2 = 0. We know from (H2) and (H5) that f(x,v) v ( µ j e j, µ i e i ) Finally, we deduce vw(x) = 0, + f(x,v) v 0, g(x,v) v 0 and a(x)( µ j e j ) 2 0. µ j e j = 0, (14) v µ j e j (x) = 0 for a.e. x (15) 7

8 and Thus we have ( µ j e j, µ i e i ) a(x)( µ j e j ) 2 = 0. (16) µv 2 = vw v µ j e j = 0 which yields µ = 0. Moreover, it follows from (16) that Lw = 0. By the unique continuation principle, we have either w 0 or w(x) 0 for a.e. x. Indeed, we state first w is regular. All the derivatives of w vanish a.e. on the set {x ; w(x) = 0} provided this set is not a negligible measurable set. Thus, w vanishes of infinite order at such points. By the strong unique continuation principle [27], w vanishes. Going back to (15), we have w 0 and the desired claim follows. As a consequence, for all v M, dl(v) is surjective and H0 K () = ker(dl(v)) V. M is thus a complete C1 Finsler manifold (see [26]). Furthermore, M is a complete C 1,1 Finsler manifold provided (H6) and (H7) are satisfied. For any v H K 0 () \ V 0, we denote by V + := {tv + µ i e i for all t > 0, µ i R}, the (n + 1) dimensional half space spanned by v and {e i } for all 1 i n. We have the following Lemma 1 Under the assumptions (H1) to (H5), then there exists an unique v 0 M such that M V + = {v 0 }. (17) Moreover we have E(v 0 ) = max E(w). (18) w V + Proof. Given v H K 0 () \ V 0, we define for any t > 0 the n dimensional affine vector space V t := tv + V 0. We divide the proof into several steps. Step 1. For any t > 0 there exists an unique v(t) V t such that E(v(t)) = max V t E. Moreover, {v(t), t > 0} is a C 1 curve in V +. From (H1) to (H4), it is known that E is C 2 on V +. Thanks to (6), we have lim E(w) = w V t,w 8

9 Thus there exists some v(t) V t such that E(v(t)) = max w Vt E(w). A direct calculation leads to d 2 E(v)(w, w) = w 2 K,2, a(x)w 2 ((s 1) v s 2 g(x, v) + )w 2 v By (H5), we infer g(x, v) v 0 and g(x, v) v g(x, v) v Hence, d 2 E(v) < 0 on V t, that is, the functional E is strictly concave on V t. This yields the uniqueness. We note {v(t), t > 0} = {w V + de(w) V0 = 0}. As the second variation d 2 E of E is negative define on V 0, it follows from the Implicit Function Theorem that {v(t), t > 0} is a C 1 curve in V + which finishes the proof of step 1. Step 2. For all w M V +, the restriction of E on V + has a strictly local maximum at w. Recall V := Span(v, e 1,, e n ). Let v 0 satisfying de(v) V = 0 and w = µv + n µ i e i V. As in the proof of Proposition 1, we have by (H2), 0. d 2 E(v)(w, w) = (s 2) v s 2 w 2 v s 2 ( µ j e j ) 2 f(x, v) f(x, v) ( + )w 2 g(x, v) ( µ j e j ) 2 v v v +( µ j e j, µ i e i ) a(x)( µ j e j ) 2 which implies from (H1) to (H5) Therefore, the desired claim follows. d 2 E(v)(w, w) < 0 provided w 0. Step 3. There exists an unique t 0 > 0 such that v(t 0 ) M. Moreover, de(v(t))(v(t)) > 0 for any 0 < t < t 0 and de(v(t))(v(t)) < 0 for any t > t 0. With the same arguments as in the proof of Proposition 1, we have On the other hand, it follows from (5) that w V 0 In particular, we obtain sup E(w) > 0. (19) w V + E(w) 0. (20) sup E(w) = sup E(w) = sup E(v(t)), w V + w V + t>0 9

10 where V + is the closure of V +. Combining (6), (19) and (20) and using the continuity of E on V +, there exists some v 0 M V + such that We know E(v 0 ) = sup w V + E(w). M V + {w V + de(w) V0 = 0} = {v(t) t > 0} (21) so that there exists t 0 > 0 such that v(t 0 ) = v 0. Set α(t) := E(v(t)) then α (t) = de(v(t))(v (t)) = l 0(v(t)) t since v (t) v V 0 and de(v(t)) V0 = 0. We claim M V + = {v(t) α (t) = 0}. Obviously, M V + {v(t) α (t) = 0}. Conversely, for any v(t) with α (t) = 0, by the method of Lagrange multipliers, there exists µ 1,, µ n R such that de(v(t)) V + µ i dl i (v(t)) V = 0. Hence, we have on V 0, µ i d 2 E(v(t))(, e i ) = 0. By virtue of the fact d 2 E(v) V0 < 0 for all v V t, we infer µ 1 = µ n = 0 which proves the claim. Applying (6), we infer since lim α(t) =, t + lim inf w K,2, = +. t + w V t It follows from Step 2 that there exists only strictly local maximum points for α(t). Hence, t 0 is the only critical point of α(t). Moreover, α (t) > 0 for any 0 < t < t 0 and α (t) < 0 for any t > t 0. The lemma is proved. Now let us consider the minimization problem We have then Lemma 2 Under assumptions (H1) to (H5), there holds κ := inf E(v). (22) v M κ K N (S K) N 2K. (23) Proof. Let B(x 0, R) for some x 0 and R > 0. We consider for some small number ν > 0 and for all ɛ (0, ν), the function ɛ (N 2K)/2 u ɛ (x) := C N,K (ɛ 2 + x x 0 2 ) (N 2K)/2, 10

11 where the constant C N,K independent of ɛ is chosen such that u ɛ s L s (R N ) = u ɛ 2 K,2,R N = (S K ) N 2K. Let ξ C0 (B(x 0, R)) be a fixed cut-off function satisfying 0 ξ 1 and ξ 1 on B(x 0, R/2). Putting w ɛ := ξu ɛ C0 () as in [5] and [24], we obtain as ɛ 0 w ɛ s L s = (S K) N 2K + O(ɛ N ) and w ɛ 2 K,2, = (S K ) N 2K + O(ɛ N 2K ). (24) It is clear that as ɛ 0, we have w ɛ 0 weakly in H0 K(), w ɛ 0 weakly in L s (), strongly in L q () ( q < s) and a.e. in. Therefore, there holds f(x, w ɛ ) 0 strongly in L s s 1 (). (25) Indeed, for any M > 0, let f M (x, u) := { f(x, u), if u M 0, if u > M. From (H1) to (H2), it follows that δ > 0, there exists M > 0 such that Therefore, we have f M (x, u) f(x, u) δ u s 1 for a.e. x and u R. f(x, w ɛ ) L s s 1 f(x, w ɛ ) f M (x, w ɛ ) s + f L s 1 M(x, w ɛ ) s L s 1 δ w ɛ s 1 L s + f M (x, w ɛ ) s. (26) L s 1 Using Lesbegue s theorem, we infer that β > 0 Letting ɛ 0 in (26), we obtain Thus (25) is proved. Similarly, we have f M (x, w ɛ ) L β 0. lim sup f(x, w ɛ ) s ɛ 0 L s 1 lim F (x, w ɛ ) = 0. ɛ 0 2δC. Set e 0 = w ɛ. Clearly, e 0, e 1,, e n are linearly independent. Denote V ɛ the n + 1 dimensional vector space spanned by e 0,, e n and let w ɛ V ɛ M. We claim lim w ɛ w ɛ K,2, = 0. ɛ 0 11

12 For this purpose, fix some small number r > 0. For all (γ 0,, γ n ) R n+1 with n i=0 γ 2 i = r 2, with the same arguments as above, we have the following expansions: F (x, w ɛ + γ i e i ) = F (x, γ i e i ) + o(1), i=0 w ɛ + γ i e i 2 K,2, = (1 + γ 0 ) 2 w ɛ 2 K,2, + γ i e i 2 K,2, + o(1), i=0 w ɛ + γ i e i s = 1 + γ 0 s w ɛ s + γ i e i s + o(1) i=0 where o(1) tends to 0 uniformly with respect to (γ 0,, γ n ). As a consequence, we infer E(w ɛ + n i=0 γ i e i ) 1 2 (1 + γ 0) 2 w ɛ 2 K,2, 1 s 1 + γ 0 s w ɛ s L s () + 1 n 2 γ i e i 2 K,2, 1 s n γ i e i s L s () 1 a(x)( γ i e i ) 2 + o(1) (1 + γ 0) 2 w ɛ 2 K,2, 1 s 1 + γ 0 s w ɛ s L s () + 1 γi 2 λ i () 1 2 s n γ i e i s L s () + o(1), (27) since F (x, u) 1 2 a(x)u2 for a.e. x. Gathering (24) and (27), we deduce E(w ɛ + γ i e i ) < E(w ɛ ) i=0 provided ɛ is sufficiently small. On the other hand, E( w ɛ ) = sup v V ɛ E(v). Hence, we have w ɛ w ɛ = n i=0 γ i e i with Γ = (γ 0,, γ n ) R n+1 satisfying Γ 2 = n i=0 γ 2 i < r2, that is, the claim is proved. Now, applying (24) and (27), we infer This yields the desired result. lim E( w ɛ) = lim E(w ɛ ) = K ɛ 0 ɛ 0 N (S K) N 2K. The following lemma concerns the linear perturbation problem for the non critical dimensions case. Lemma 3 We suppose N 4K and f(x, u) = µu for some µ > 0. Then we have κ < K N (S K) N 2K. (28) Proof. We keep the same notations as in the proof of Lemma 2. Direct calculations lead to w ɛ s 1 L s 1 = O(ɛ (N 2K)/2 ), w ɛ L 1 = O(ɛ (N 2K)/2 ) and w ɛ 2 L 2 c 1ɛ 2K (29) 12

13 for some positive constant c 1 > 0. When N = 4K, c 1 could be any large constant as wanted. We have also for any i = 1,, n (w ɛ, e i ) = w ɛ ( ) K e i = O(ɛ (N 2K)/2 ). (30) Together with (24), we obtain, since lim ɛ 0 Γ = 0, w ɛ 2 K,2, = (1 + γ 0) 2 (S K ) N 2K + n γ i e i 2 K,2, + o(ɛ(n 2K)/2 ) w ɛ 2 L 2 () c 1(1 + γ 0 ) 2 ɛ 2K + n γ i e i 2 L 2 () + o(ɛ(n 2K)/2 ) On the other hand, using the fact that function s is convex on R, we have w ɛ s L s () (1 + γ 0 ) s e 0 s + (1 + γ 0 ) s (S K ) N 2K + o(ɛ (N 2K)/2 ). s(1 + γ 0 ) s 1 e 0 s 2 n e 0 γ i e i (31) (32) Gathering (31) and (32), there holds since and Finally, E( w ɛ ) 1 2 (1 + γ 0) 2 (S K ) N 1 2K s (1 + γ 0) s (S K ) N 1 n 2K + 2 γ i e i 2 K,2, µ 2 n γ i e i 2 L 2 () c 1µ 2 (1 + γ 0) 2 ɛ 2K + o(ɛ (N 2K)/2 ) K N (S K) N c 1 µ 2K 2 (1 + γ 0) 2 ɛ 2K + o(ɛ (N 2K)/2 ), 1 n 2 γ i e i 2 K,2, µ n 2 γ i e i 2 L 2 () 0 ( ) 1 sup t>0 2 t2 (S K ) N 1 2K s ts (S K ) N 2K = K N (S K) N 2K. E( w ɛ ) < K N (S K) N 2K provided ɛ is sufficiently small, which yields the desired result. Now we state our main result of this section. Theorem 1 Suppose (H1) to (H5) and (28) are satisfied. Then there exists u M such that E(u) = κ and u is a solution to (1). Proof. The strategy of the proof is standard. Let (u k ) M be a minimizing sequence for E. We prove first that (u k ) is bounded and then we can extract a subsequence, if necessary, which converges to some limit u. We prove then u 0, u M and u is a minimizer for κ. Step 1. (u k ) is a bounded sequence in H K 0 (). 13

14 Recall that (u k ) satisfies (4) and so that 1 2 u k 2 K,2, 1 s u k s L s () F (x, u k ) = κ + o(1) (33) ( ) K f(x, N u k s L s () + uk )u k F (x, u k ) = κ + o(1). (34) 2 From (H5), for a.e x and u R, we have F (x, u) 1 f(x, u)u, 2 which in turn (34) implies u k s L s () N κ + o(1). (35) K We infer from (H2) that for a.e x and u R, we have also thus Together with (33) and (35), u k 2 K,2, Hence Step 1 is proved. F (x, u) 1 2 a(x)u2 + 2ɛ u s s F (x, u k ) 2ɛ s u k s L s () + 1 a(x)u 2 k + C. 2 = 2 s u k s L s () C, (36) F (x, u k ) + 2κ + o(1) C( u k s L s () + u k 2 L s ()) + C + 2κ + o(1) C. Extracting a subsequence, there exists some u H K 0 () such that u k u weakly in H0 K(), u k u weakly in L s (), strongly in L q () ( q < s) and a.e. on, so that Setting v k = u k u, we have l i (u) = 0 1 i n. (37) u k 2 K,2, = v k 2 K,2, + u 2 K,2, + o(1) u k s L s () = u s L s () + v k s L s () + o(1). (38) Step 2. We have u 0. Suppose by contradiction that u = 0. As in the proof of Lemma 2, we have f(x, u k ) 0 in L s s 1 () (39) 14

15 and Combining (4), (33), (39) and (40), we deduce F (x, u k ) 0 in L 1 (). (40) u k s L s () = N K κ + o(1), u k 2 K,2, = N K κ + o(1) which yields u k 2 K,2, u k 2 L s () = ( N K κ ) s 2 s + o(1) < S K for sufficiently large k. This contradiction gives u 0. Consequently, we have u V 0 because of (37). Step 3. We have u M and E(u) = κ. We need to prove l 0 (u) = 0 to conclude that u M and E(u) = κ. So we should exclude two cases: (i) l 0 (u) < 0 and (ii) l 0 (u) > 0. First we suppose that the case (i) occurs. In this case there exists t (0, 1) such that u(t) M because of the Step 3 of Lemma1. Set v k := u k u as before and ũ k := tu k + u(t) tu = tv k + u(t). We define for all w H0 K(), E (w) := 1 2 w 2 K,2, 1 w s. s As v k 0 weakly in H K 0 (), we obtain E(ũ k ) = E (tv k ) + E(u(t)) + o(1). Suppose E(u(t)) > κ, otherwise E(u(t)) = κ and then we finish the proof. By Lemma 1 and the fact ũ k tu k V 0, we have E(ũ k ) E(u k ) = κ + o(1) which implies E (tv k ) < 0 for sufficiently large k. In particular, v k 0. Consequently, for sufficiently large k, tv k s L s () > s 2 tv k 2 K,2, s 2 S K tv k 2 L s () > S K tv k 2 L s () (41) so that On the other hand, we have v k s L s () > (S K) N 2K. (42) v k s L s () = u k s L s () u s L s () + o(1) N K κ u s L s () + o(1), (43) which contradicts (42) by using Lemma 2. Thus case (i) is impossible. Now we treat the case (ii). By the same arguments in the Step 2, we have f(x, u k )u k = f(x, u)u + o(1), (44) 15

16 F (x, u k ) = Thus, according to (34), (44) and (45), we have Similarly, we have v k s L s () = N K κ + N K v k 2 K,2, = N K κ + N K F (x, u) + o(1). (45) (F (x, u) 1 2 f(x, u)u) u s L s () + o(1). (46) F (x, u) + (1 N 2K ) f(x, u)u u 2 K,2, + o(1). (47) Combining (45), (47), we see that l 0 (u) > 0 implies for sufficiently large k v k s L s () > v k 2 K,2,. Consequently, by the definition of S K, we obtain v k s L s () > (S K) N 2K for sufficiently large k. This is (42) and as before, we conclude that (ii) does not occur and thus u M. Moreover E(u k ) = E(u) + E (v k ) + o(1) and v k s L s () = v k 2 K,2, + o(1). Thus E(u) = E(u k ) K N v k 2 K,2, + o(1). Finally, we deduce v k 2 K,2, = o(1) and therefore E(u) = κ. Step 4. u is a solution to (1). In fact u is a critical point of E on M. By the method of Lagrange multipliers, there exists µ, µ 1,, µ n R such that de(u) + µdl 0 (u) + µ i dl i (u) = 0. We consider its restriction on V, this means ( ) µdl 0 (u) + µ i dl i (u) V = 0 since de(u) V = 0. On the other hand, we have seen from Proposition 1 that dl(u) V is an isomorphism from V on R n+1. Consequently, µ = µ 1 = = µ n = 0, that is, de(u) = 0. Finally, u solves the problem (1) which finishes the proof. 16

17 3 Existence of solutions for some perforated domains In this section, we analyze first the concentration phenomenon for the problem (1). For this purpose, set (( ) M v) 2 if K = 2M F K (v) := ( ) M v 2 if K = 2M + 1. Similarly to Theorem 6 of [21], we have the following theorem and here we just give a sketch of the proof. Theorem 2 Suppose the assumptions (H1) to (H5) are satisfied. Moreover, suppose that and κ = K N (S K) N 2K (48) E(v) > κ, v M. (49) Let (u k ) M be a minimizing sequence for κ, that is, lim n E(u k ) = κ. Then there exists x 0 such that and µ k := ζ F K (u k ) dx S K δ x0 weakly in R(R N ) ν k := ζ u k s dx S K δ x0 weakly in R(R N ), where R(R N ) denotes the space of non-negative Radon measures on R N with finite mass, δ x0 denotes the Dirac measure concentrated at x 0 with mass equal to 1 and ζ designates the characteristic function of. Proof. As in the proof of Theorem 1, we see that (u k ) is bounded in H0 K a subsequence, there exists some u H0 K () such that (). Extracting u k u weakly in H0 K(), u k u weakly in L s () and a.e. on. Moreover, for all 1 j n, we have l j (u) = 0. Furthermore, we have u = 0. Otherwise, with the same arguments as in Theorem 1, we infer u M and E(u) = κ which contradicts (49). Now the rest of proof is just a consequence of concentration compactness principle (for details cf [28, 19, 21]). In the following, we give some classification result. First we recall a basic fact for non-existence result on the half space R N +. It can be stated as follows: Lemma 4 Let u D K,2 (R N + ) be a weak positive solution of the problem ( ) K u = u s 2 u in R N + u = Du = = D K 1 u = 0 on R N +. (50) Then u 0. 17

18 A stronger result have been obtained by Reichel and Weth in[36] very recently. Here we give a proof based on the Pohozaev formula (see [29]). Proof. It follows from the Pohozaev formula D K u = 0 on R N + (see the details cf [21] for the Navier boundary conditions). Now, ( ) K 1 (( )u) = u s > 0 in R N + verifying Dirichlet boundary condition ( )u = = D K 2 ( )u = 0 on R N +. Thanks to the Boggio s result, we know the Green function for the operator ( ) K 1 on the half space with Dirichlet boundary condition is positive. Thus, ( )u > 0 in R N +. From Hopf s Maximum principle, u n > 0 on RN +. This contradiction finishes the proof of Lemma. A similar problem in the whole space can be stated as follows: Lemma 5 Let u D K,2 (R N ) be a weak positive solution of the problem ( ) K u = u s 2 u in R N. (51) Then there exists a constant λ 0 and a point x 0 R N such that ( u(x) = 2λ 1 + λ 2 x x 0 2 ) N 2K 2 This result has been proved by Wei-Xu (Theorem 1.3 in [44]).. (52) Lemma 6 Let u D K,2 (R N + ) (resp. u D K,2 (R N ) ) be a weak sign changing solution of the problem (50) (resp. (51)). Then E (u) 2K N (S K) N 2K. (53) Proof. Our proof is an adaptation of Gazzola-Grunau-Squassina s approach [18]. We consider the closed convex cone and its dual cone C 1 = {v D K,2 (R N + ) v 0 a.e. in R N + } C 2 = {w D K,2 (R N + ) (w, v) R N + 0 v C 1 }. We claim that C 2 C 1. Given h C 0 (RN + ) C 1, let v be the solution to the problem ( ) K v = h in R N +. Again from the Boggio s result, we have v 0 since the Green function for the operator ( ) K on the half space with Dirichlet boundary condition is positive. Consequently, for all w C 2, we have hw = ( ) K vw = (v, w) R N 0. + R N + R N + This implies w 0 a.e. in R N +. Hence the claim is proved. Using a result of Moreau [30], for any u D K,2 (R N + ), there exists an unique pair (u 1, u 2 ) C 1 C 2 such that u = u 1 + u 2 with (u 1, u 2 ) R N + = 0. 18

19 Now let u be a sign-changing solution of the problem (50). Then u i 0 for all = 1, 2. From the above claim, we see u 1 0 and u 2 0 so that u(x) s 2 u(x)u i (x) u i (x) s for i = 1, 2. Applying the Sobolev inequality for u i (i = 1, 2), we obtain S K u i 2 L s u i 2 K,2,R N + = (u, u i ) R N = + R N + ( ) K uu i R N + u i (x) s = u i s L s so that u i s L s () (S K) N 2K. Consequently, using the fact u 2 K,2,R N + = u s Ls, we infer E (u) = K N u 2 K,2,R N + = K N ( u 1 2 K,2,R N + + u 2 2 K,2,R N + ) K N S K( u 1 2 L s + u 2 2 L s) 2K N (S K) N 2K. Similarly, we have the same result for u D K,2 (R N ). Theorem 3 Assume (H1), (H2), (H5), (48) and (49) are satisfied. Let (u k ) H K 0 () be a (P.S.) β sequence such that Then (u k ) is precompact in H K 0 (). E(u k ) β ( K N (S K) N 2K 2K, N (S K) N 2K ) (54) de(u k ) 0 in (H0 K ()). (55) Proof. The blow up analysis for (P.S.) β sequences is more or less standard. Its proof follows from the P. Lions concentration compactness principle and it is close to one in [21]. The only difference is that we need Lemma 6 to rule out sign changing bubbles. We leave this part to interested readers. As a consequence, we have Corollary 1 Under the assumptions (H1) to (H5), (48) and (49), assume moreover (H8) e n () < 0. Let (u k ) M be a (P.S.) β sequence for E on M such that Then (u k ) is precompact in M. E(u k ) β ( K N (S K) N 2K 2K, N (S K) N 2K ), (56) de(u k ) (Tuk M) 0. (57) 19

20 Proof. As in the proof of Theorem 1, (u k ) is a bounded sequence in H0 K (). On the other hand, using (33), (34) and (H2), we infer that(u k ) is bounded from below by some positive constant in H0 K() and also in Ls (). Set V k the n + 1 dimensional vector space spanned by u k, e 1,, e n. If there is no confusion, we drop the index k. We claim there exists some positive constant c > 0 independent of k such that k N, w H0 K (), we can decompose w = w 1 + w 2 (58) where w 1 V k and w 2 T uk M satisfying w 1 K,2, c w K,2,, w 2 K,2, c w K,2,. Set e 0 = u k and θ i = dl i (u k )(w) R for all i = 0,, n. Using (13) and the fact that (u k ) is a bounded sequence in H K 0 (), the vector Θ = (θ 0,, θ n ) T is bounded in R n+1 with respect to k. Moreover, we can estimate Θ c w K,2,. Define (n + 1) (n + 1) symmetric matrix M(k) = (m ij ) 0 i,j n by We write m ij = d 2 E(u k )(e i, e j ). w 1 = ψ i e i i=0 where ψ i R. Denote the vector Ψ = (ψ 0,, ψ n ) T R n+1. Again from (13), the decomposition (58) is equivalent to solve d 2 E(u k )(w 1, e i ) = dl i (u k )(w) 0 i n, that is, M(k)Ψ = Θ. As in the proof of Lemma 1, the matrix is negative definite. Clearly, the matrix M(k) is uniformly bounded. We show there exists c > 0 independent of k such that M(k) ci where I is the identity matrix. For this purpose, for any vector Γ T = (γ 0,, γ n ) R n+1, denote ξ = n i=0 γ i e i we have Γ T M(k)Γ = d 2 E(u k )(ξ, ξ) (s 2) u k s 2 ξ 2 u k s 2 ( γ j e j ) 2 +( γ j e j, γ i e i ) a(x)( γ j e j ) 2 s 2 u k s γ0 2 + γj 2 λ j (). s 1 Thus, the desired result follows. sequence. More precisely, we infer As a consequence, (Ψ = (M(k)) 1 Θ) k is a bounded w 1 K,2, c w K,2,. 20

21 Therefore, that is, the claim is proved. Hence, w 2 K,2, ( w K,2, + w 1 K,2, ) c w K,2,, de(u k )(w) = de(u k )(w 2 ) c de(u k ) (Tuk M) w K,2,. Thus, there holds so that de(u k ) (H K 0 ()) c de(u k) (Tuk M) lim de(u k) n (H K 0 ()) = 0. Finally, applying Theorem 3, we finish the proof. Now, we can prove the main result for domains with the small holes. Recall that = 1 \ 2 is a bounded domain satisfying 2 B(0, ɛ) and 1 is fixed. To search solutions of (1) in such, we minimize the energy functional E on the Finsler manifold M. We see that the concentration phenomenon occurs if E can not reach the minimum. In this case, we will employ Coron s strategy to search unstable critical points in higher level sets. Theorem 4 Let be a bounded domain satisfying the above assumption. Assume (H1) to (H7) hold. Then there exists η > 0 such that for all ɛ < η, the problem (1) admits a non trivial solution in. Proof. Thanks to Lemma 2, we have κ K N (S K) N 2K. In the case κ < K N (S K) N 2K, the desired result follows from Theorem 1. So we suppose κ = K N (S K) N 2K. If there exists u M such that E(u) = κ, we finish the proof by Step 4 in the proof of Theorem 1. Hence, we assume v M there holds E(v) > κ. From the properties of eigenvalues λ i () described in the previous sections, (H8) is always satisfied for the perforated domain, provided ɛ is sufficiently small. In fact, in case λ i ( 1 ) 0 for all i N, it follows from the continuity of λ i (). In the case λ n ( 1 ) = = λ n+k ( 1 ) = 0, we have λ n () > 0. We devide the proof into several steps. Step 1. We choose a radially symmetric function ϕ C0 (RN ) such that 0 ϕ 1, ϕ 1 on the annulus {x R N 1/2 < x < 1} and ϕ 0 outside the annulus {x R N 1/4 < x < 2}. For any R 1, define ϕ(rx) if 0 x < 1/R ϕ R (x) = 1 if 1/R x < R ϕ(x/r) if x R. Denote the unit sphere S N 1 = {x R N [ u σ 1 t t (x) = C N,K (1 t) 2 + x tσ 2 x = 1}. For σ S N 1, 0 t < 1, we set ] N 2K 2 H K (R N ), 21

22 where the choice of C N,K is such that u σ t 2 K,2,R N = u σ t s L s (R N ) = (S K) N 2K. Let w σ t,r (x) = u σ t (x)ϕ R (x) and wt,r σ N 2K (x) = (4R) 2 w t,r σ (4Rx). Hence wσ t,r HK 0 (B(0, 1/2)\B(0, 1/16R2 )), σ S N 1 and t [0, 1). Clearly, A direct computation leads to R > 1 and Consequently w σ t,r L s (R N ) = w σ t,r L s (R N ) (59) w σ t,r K,2,R N = w σ t,r K,2,R N. (60) w σ t,r u σ t 2 K,2,R N C(1 t) N 2K R 2K N (61) w σ t,r u σ t s L s (R N ) CR N (1 t) N. (62) lim R wσ t,r 2 K,2,R N = lim R wσ t,r s L s (R N ) = (S K) N 2K uniformly for t [0, 1) and σ S N 1. Set w σ t,r M Vect{e 1(),, e n (), w σ t,r } where = 1 \ 2, B(0, 1/2) 1 and 2 B(0, 1/16R 2 ). Thanks to the Implicit Function Theorem, the continuous map yields a continuous map w R : S N 1 [0, 1) H K 0 () (σ, t) w σ t,r w R : S N 1 [0, 1) M (σ, t) w t,r σ. Recall 1 is fixed. A basic observation is that e i () e i ( 1 ) for all i = 1,, n in Cloc ( 1 \ {0}) away from 0 and strongly in H0 K( 1) as R +. We remark that E(u) 1 2 u 2 K,2, 1 s u s L s () 1 2 a(x)u 2. In the following, we consider the simple case F (x, u) = 1 2 a(x)u2 (we can treat the general case with the same argumens). Fix some small number r > 0. As in the proof of Lemma 2, for all Γ = (γ 0,, γ n ) R n+1 with n i=0 γ 2 i r2, we infer sup E(wt,R σ + t,σ, 2 i=0 γ i e i ) 1 2 (1 + γ 0) 2 (S K ) N 1 2K s 1 + γ 0 s (S K ) N 2K + 1 γi 2 λ i ( 1 ) 1 n 2 s (63) γ i e i ( 1 ) s L s ( 1 ) + o(1), where o(1) is uniformly with respect to Γ as R. Consequently, we deduce sup E(wt,R σ + t,σ, 2 γ i e i ) < E(wt,R) σ i=0 for γi 2 = r 2 i=0 22

23 provided R is sufficiently large. This implies w t,r σ wt,r σ = γ i e i () for some Γ < r, i=0 so that lim sup E( w t,r) σ = K R t,σ, 2 N (S K) N 2K. Hence, we can choose R 0 > 0 such that for any R R 0 sup E( w t,r) σ < 2K t [0,1), σ S N 1, 2 B(0,1/16R 2 ) N (S K) N 2K. (64) Thus we can define a map α : B(0, 1) M (t, σ) w t,r σ 0. Step 2. Set η := 1/16R 2 0 and fix 2 B(0, η). From (59) to (62), we infer that lim t 1 wσ t,r 0 2 K,2, = lim w t,r σ t 1 0 s L s () = (S K) N 2K uniformly for σ S N 1 which implies for any σ S N 1 Step 3. For any v M, let γ(v) = lim E(α(t, σ)) = K t 1 N (S K) N 2K. x v(x) s dx R N denotes its center mass. We claim there exists δ > 0 such that for any v M satisfying E(v) K N (S K) N 2K + δ, we have γ(v) R N \ B(0, ɛ 2 (S K ) N 2K /2) (65) where B(0, ɛ 2 ) 2. Otherwise, we can find a sequence (v n ) M satisfying Applying Theorem 2, there exists x 0 such that lim E(v n) = K n N (S K) N 2K, (66) γ(v n ) B(0, ɛ 2 (S K ) N 2K /2). (67) ζ v n (x) s dx (S K ) N 2K δx0. Consequently, γ(v n ) (S K ) N 2K x0 B(0, ɛ 2 (S K ) N 2K ) 23

24 which contradicts (67). Thus, the desired claim yields. Choosing t 0 [0, 1) such that σ S N 1 and t [t 0, 1), we have E(α(t, σ)) < K N (S K) N 2K + δ, we set β := min max f H (t,σ) (0,t 0 ] S N 1 E(f(t, σ)), where H is the set of any function homotopic to α on B(0, t 0 ) with the fixed boundary data, that is, H = {f f : B(0, t 0 ) M is continuous, f B(0,t0 ) = α B(0,t0 ) and f is homotopic to α}. We see that f H, γ f : B(0, t 0 ) R N is a contraction of the loop γ α B(0,t0 ) R N \ B(0, ɛ 2 (S K ) N 2K /2). On the other hand, it follows from Steps 1 and 2 lim γ α(t, σ) = (S K) N σ 2K uniformly in σ S N 1. t 1 4R 0 Thus, γ α B(0,t0 ) is a non-trivial loop in R N \ B(0, ɛ 2 (S K ) N 2K /2). Using (67), we obtain which implies sup E(f(t, σ)) K (t,σ) B(0,t 0 N (S K) N 2K ) β K N (S K) N 2K On the other hand, it follows from Step 1 β + δ > K N (S K) N 2K. + δ, sup E(α(t, σ)) < 2K (t,σ) B(0,t 0 N (S K) N 2K. ) Recalling Theorem 1 and Corollary 1 and using the deformation lemma, we infer β is a critical value. Finally, the problem (1) admits a non trivial critical point u such that E(u) = β. Remark 0 The condition a L () C () could be weakened. Remark 1 We can use the above strategy to treat also the Navier boundary conditions. References [1] A. Ambrosetti and P. Rabinowitz, Dual variational methods in critical point theory and applications, J. Funct. Anal. 14, (1973). [2] T. Aubin, Equations differentielles non lineaires et probleme de Yamabe concernant la courbure scalaire, J. Math. Pure Appl. 55, (1976). [3] T. Bartsch, T. Weth and M. Willem, A Sobolev inequality with remainder term and critical equations on domanins with topology for the polyharmonic operator, Calc. Var. and PDE. 18, (2003). 24

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