LAZER-MCKENNA CONJECTURE FOR HIGHER ORDER ELLIPTIC PROBLEM WITH CRITICAL GROWTH

. This paper is concerned with the following problem involving critical Sobolev exponent and polyharmonic operator: (cid:40) where B 1 is the unit ball in R N , s 1 and λ are two positive parameters, ϕ 1 > 0 is the eigenfunction of (cid:16) ( − ∆) m , D m, 2 0 ( B 1 ) (cid:17) corresponding to the ﬁrst eigenvalue λ 1 with max y ∈ B 1 ϕ 1 ( y ) = 1, u + = max ( u, 0) and m ∗ = 2 N N − 2 m . By using the Lyapunov-Schmits reduction method, we prove that the number of solutions for ( P ) is unbounded as the parameter s 1 tends to inﬁnity, therefore proving the Lazer-McKenna conjecture for the higher order operator equation with critical growth.

where B 1 is the unit ball in R N , s 1 and λ are two positive parameters, ϕ 1 > 0 is the eigenfunction of (−∆) m , D m,2 0 (B 1 ) corresponding to the first eigenvalue λ 1 with max y∈B 1 ϕ 1 (y) = 1, u + = max (u, 0) and m * = 2N N −2m . By using the Lyapunov-Schmits reduction method, we prove that the number of solutions for (P ) is unbounded as the parameter s 1 tends to infinity, therefore proving the Lazer-McKenna conjecture for the higher order operator equation with critical growth.
1. Introduction. We consider the following polyharmonic elliptic problem with critical Sobolev exponent: where B 1 is the unit ball in R N , s 1 and λ > 0 are parameters, ϕ 1 > 0 is the eigenfunction of (−∆) m , D m,2 0 (B 1 ) corresponding to the first eigenvalue λ 1 with max y∈B1 ϕ 1 (y) = 1, u + = max (u, 0) and m * = 2N N −2m , N ≥ 2m + 1, m is positive integer and D m,2 0 (B 1 ) is the completion of C ∞ 0 (B 1 ) with respect to the norm induced by the scalar product: if m is even, Problem (1) is the higher order case of the Ambrosetti-Prodi type problem with critical exponent: where Ω is a bounded domain in R N , and g(t) satisfies lim t→−∞ g(t) t = ν < λ 1 , lim t→+∞ g(t) t = µ > λ 1 . Here µ = +∞ and ν = −∞ are allowed. It is known that the location of ν, µ with respect to the eigenvalues of −∆ with Dirichlet boundary condition in a bounded domain Ω plays an important role in the study of the existence and multiplicity of solutions for the problem (3). It has been a problem of considerable interest since the early 1970s, many interesting results have been achieved, see for example [1,4,5,8,9,10,11,14,15,16,20,21,22,23] and references therein. In particular, in the view of the results in [13,23], Lazer and McKenna [14] conjectured that if µ = +∞ and g(t) does not grow too fast at infinity, then (3) has an unbounded number of solutions ass → +∞. In case of m = 1, by using a partially numerical method, Breuer, McKenna and Plum [3] proved that if g(t) = t 2 and Ω ⊂ R 2 is the unit square, then (3) has at least four solutions; Dancer and Yan [6] proved that if g(t) = |t| p with p ∈ (1, +∞) for N = 2 and p ∈ (1, N +2 N −2 ) for N ≥ 3, then the Lazer-McKenna conjecture is true. While, in [7], they proved that the Lazer-McKenna conjecture is also true if g(t) = t p + + λt, λ ∈ (−∞, λ 1 ), N ≥ 3 and p ∈ (1, N +2 N −2 ). In the critical case, that is p = N +2 N −2 , Li, Yan and Yang [17] proved that if N ≥ 7 and λ ∈ (0, λ 1 ), s 1 > 0, the number of the solutions is unbounded as s 1 → +∞ (see also [18,24] for N ≥ 6). Note that the solutions constructed in [17,18] are different from those constructed in [24], where the previous one blows up near the maximum points of the function φ 1 (y) in Ω, while the solutions obtained in [24] have several peaks clustering near a boundary point which is a maximum of the function −∂φ1(z) ∂n . Motivated by the the above work, the aim of the present paper is going to discuss the Lazer-McKenna conjecture for the case of any m > 1 with critical nonlinearities. That is the following problem: As far as we know, no such type of results have been obtained for polyharmonic operator problems (4). However, we do not know whether the similar result is true for the general domain Ω due to the lack of positivity of the first eigenfunction. As a consequence, we prove the Lazer-McKenna conjecture for polyharmonic equation (4) in B 1 . Even this, due to the lack of of the comparision principle for polyharmonic operator, some new technique idea are needed. For example, instead of maximum principle, we turn to use Green's representation to obtain the essential estimate in the energy expansion.
Noting that if (Λ 1 ) or (Λ 2 ) holds. Then (4) has a non-negative solution u s1 = − s1 λ1−λ ϕ 1 . Moreover, if u s1 + u is a solution of (4), then u satisfies: where s = s1 λ1−λ > 0. Therefore in order to prove Theorem 1.1, it is sufficient to prove the number of solutions of (5) is unbounded as s → +∞. Let be the unique positive solution ( up to translation and scaling ) of the following equation where c 0 > 0 is a constant depending on m and N. Let P Ux ,μ be the projection of Ux ,μ on B 1 , i.e.
In the following of the paper, we will use the notations: We have the following existence result for problem (5).
Theorem 1.2. Assume that N ≥ 6m + 1, λ, s 1 satisfy either (Λ 1 ) or (Λ 2 ). Then for any integer k ≥ 1, there exists a s k > 0, depending on k, such that for any s ≥ s k , (5) has a solution of the form u s = k j=1 P U xs,j ,µs,j + ω s,k , satisfying as It is easy to see that Theorem 1.1 is a direct consequence of Theorem 1.2. Before the end of this introduction, let us outline the proof of Theorem 1.2 more precisely.
Recall that the functional corresponding to (5) is defined as To prove Theorem 1.2, follow the idea of [17], we first reduce the problem to a finite-dimensional problem. Then we prove that there exists a C 1 map ω s,x,µ from M s to D m,2 0 (B 1 ), such that ω s,x,µ ∈ E x,µ,k , and for some constants A j and B jh , where Note that in order to show that k j=1 P U xj ,µj + ω is a solution of (5), it is sufficient to find a suitable (x s , µ s ) ∈ M s such that (x s , µ s ) is a critical point of the function J s (x, µ, ω ( s, x, µ)), or the corresponding constants A j and B jh are all equal to zero. For this purposes, we turn to consider the function K(x, µ) = J s (x, µ, ω s,x,µ ), (x, µ) ∈ M s . And reduce the problem of finding the critical point of J s to the problem of finding the critical point of K. However, we may allow ( see Section 3 ) that K(x, µ) has a saddle point in M s . Hence the classical maximization procedure can not applied directly. To overcome this difficulty, follow the idea of [18], we first prove that there is a C 1 map µ(x) such that for each fixed x, ∂K(x,µ(x)) ∂µj = 0, j = 1, · · · , k. Then we prove that K(x, µ) has a critical point (x, µ(x)) by using a maximization procedure for the function K(x, µ(x)).
Our paper is organized as follows. In Section 2, we reduce the problem of finding solutions for (5) to a finite-dimensional problem. The proof of the main existence result will be given in Section 3. We put some essential estimates and the energy expansions in the Appendices A and B, respectively.

2.
Finite-dimensional reduction. In this section, we will reduce the problem of finding a k−peak solution for (5) to a finite-dimension problem. We define , for any i = j.
It is easy to see that for ( For each (x, µ) ∈ M s , we expand J s (x, µ, ω) at ω = 0 as follows: where l s ∈ E x,µ,k satisfying and Q s is a bounded linear map from E x,µ,k to E x,µ,k satisfying and Lemma 2.1. We have where σ > 0 is a constant.
for some constants A i and B ih . Moreover, we have where l s , Q s , R s are defined respectively in (7), (8), (9). A direct calculation shows By Lemma 2.2, we obtain that Q s is invertible in E x,µ,k and there is a constant C > 0 such that ||Q −1 s || ≤ C. By implicit function theorem there is a ω s,x,µ ∈ E x,µ,k such that (18) holds. Moreover, ||ω s,x,µ || ≤ C||l s ||. On the other hand, by Lemma 2.1, we have Combining the above argument, we finish the proof.
where ω s,x,µ is the map obtained in Proposition 1.
From Proposition 1 and 4, we have For any x ∈ V k , noting that ϕ 1 ( It is easy to see that (21) has a solutionT = (T s,1 , · · · ,T s,k ) in By Lemma 9, we see that the matrix Thus (20) has a unique solution in W k and by the uniqueness of the solution of (20) in is the map obtained in Proposition 2. We consider the following problem: Let x s ∈ V k be a maximum point ofK(x) in V k . We will prove that x s is an interior point ofK(x), thus x s is a critical point ofK(x).
In the following, we proceed a contradiction argument.
We have From (19) and (23), we obtaiñ This is a contradiction. Thus we have proved that x s is an interior point of V k . As a result, x s is a critical point ofK(x).
Appendix A. Some basic estimates. Let x j ∈ Ω and µ j > 0, j = 1, · · · , k. In this section, we assume that d( We have the following estimates.
We have the following estimates for ψ xj ,µj : Proof. By Green's representation, we have where By Taylor's expansion, we have where d j = dist(x j , ∂B 1 ). On the one hand, we have By change of variables, we have Combining (26), (27) and (28), we obtain that

YUXIA GUO AND TING LIU
For ψ 2 , we have Thus, we obtain that Similarly, we have and Appendix B. Energy expansions. Proposition 3. Assume N ≥ 6m + 1. Then σ is some positive constant and 1172 YUXIA GUO AND TING LIU if m is even, Proof. Without loss of generality, we assume m is even.
By Lemma A.1, we have where we used for any positive numbers a and b, where σ > 0 is any fixed constant in (0, 1). Thus, The results follows from (34), (35) and (36).
Lemma B.2. Let A i and B ih be the constants obtained in Proposition 1. Then we have Proof. From Proposition 1 and Lemma B.1, we know that A i and B ih satisfy A direct calculation shows that (57) The results follow by solving (54) and (55).
The proof of Proposition 8. From Proposition 1, Lemma B.1 and Lemma B.2, we have To obtain the expansion of the second derivatives of K(x, µ), from (59), we find that In the following, we estimate each term of (60).
Proof. The proof is similar to [17], we just sketch it. Choose a jh and b j such that Noting that ω s,x,µ ∈ E x,µ,k , and we obtain that We know that ω s,x,µ satisfies Differentiating with respect to µ i , we obtain that A direct computation leads to for ∀η ∈ D m,2 0 (B 1 ), and Combining (65) and (66), from Lemma B.2, we obtain that for ∀η ∈ E x,µ,k , Combining (61) ∼ (68), we obtain that for ∀η ∈ E x,µ,k , Using Lemma 2.2, we have The result follows from (61) By Lemma B.2 and B.3, (60) leads to Let r s be the constant such that U 0,µi (r s ) = 2s. Then .

Next we have
Lemma B.5. If i = j, then Moreover, Proof. The proof is similar to [17], we sketch it. We only prove the case i = j, the other case can be proved similarly. We have where g s (y) By Hölder inequality, we have and Differentiating (54) and (55) The result follows by solving (82) and (83).
The proof of Proposition 9. From (72), Lemma B.5 ∼ B.7, we have The result follows from Proposition 7.