Existence of Nonradial Solutions for Hénon Type Biharmonic Equation Involving Critical Sobolev Exponents

and Applied Analysis 3 To avoid heavy notation from now on, we will write simply U j for U λ,x (j) , PU j for PU λ,x (j) , and φ j for φ λ,x (j) . We set d jl = x (l) − x (j) , d = 1 2 min j ̸ =l 󵄨󵄨󵄨󵄨 d jl 󵄨󵄨󵄨󵄨 = 1 2 󵄨󵄨󵄨󵄨 x (1) − x 󵄨󵄨󵄨󵄨 , (18) and we assume that 2d ≤ r and λd ≥ 1 for all λ under consideration. Note also that due to the definition of x, we have d = 2 (1 − r) sin π k ∼ C k (19) for all r small. Lemma 2. ∫ Ω U 2 ∗ −1 j U l dx = {{{{{{{{{ {{{{{{{{{ { S N/4 + O ((λr) −N ) , j = l,


Introduction
In this paper we consider the following Hénon type biharmonic problem: where  ≥ 0, 2 * = 2/(−4), Ω is the unit ball of R  ,  ≥ 5, and n denotes the unit outward normal at the boundary Ω.
We consider first the case where  = 0, namely, the equation It is well known that (2) admits no nontrivial radial solution (see [1], Theorem 3.11, or [2], Theorem 4).The nonexistence of any nontrivial solution to (2) seems to be still unknown; only more restricted results are available.In order to obtain existence results for (2), one should either add subcritical perturbations or modify the topology or the geometry of the domain.For subcritical perturbations, we refer to [2,3] and references therein.Domains with nontrivial topology are studied in [2,4].They demonstrated how domains with topology often carry solutions that cannot be present otherwise.The corresponding second order elliptic problem has been investigated by Bahri and Coron in [5].Berchio et al. [6], among other things, considered the minimization problem inf where  2 0,rad (Ω) denotes the subspace of radial functions in  2 0 (Ω).Actually, they treated general polyharmonic problem.They proved the infimum in (3) is attained.The minimizers of (3), after rescaling, are a solution of (1).It is natural to ask whether (1) has a nontrivial nonradial solution.We will answer this problem partially here.
Our main result is as follows.
Theorem 1.Let  ≥ 8 and let Ω be the unit ball in R  .Then, for every  > 0 large enough, problem (1) admits at least one nonradial solution.
The corresponding second order elliptic problem, namely, the Hénon equation, has been studied by many authors, where  > 1. Ni [7], among other things, proved the existence of radial positive is radial provided  is large enough.Serra [12] studied the case  = ( + 2)/( − 2) and proved the existence of nonradial positive solutions of (4) for  large.Theorem 1 can be regarded as an extension of Serra's result to biharmonic problem.
In order to outline the proof of Theorem 1, we introduce some notations.We write R  = R 2 × R −2 ≃ C × R −2 and  = (, ).For a given integer , let   be the group Z  × O( − 2).We consider the action of   on  2 0 (Ω) given by  () () =  () (, ) =  ( (2/) , ) , where  ∈ {0, 1, . . .,  − 1} and  ∈ O( − 2).Define It is easy to see that functions in   are radial in .Since both the numerator and the denominator of the functional   are invariant under the action of   , the functional   is invariant.So the critical points of   restricted to   are critical points of   .After scaling, these correspond to weak solution of (1), which are in fact classical solutions by standard elliptic theory (see [13,14]).Set () .
The paper is organized as follows.In Section 2, we establish some estimates we will need and investigate the compactness properties of Palais-Smale sequences for   .In Section 3, we prove Theorem 1.Throughout this paper, the constant  will denote various generic constants.
Case 1.Consider  = .Direct computations yield that where  denotes surface area of unit sphere in R  .Combining (21), we prove the first case of Lemma 2.

Case 2. Consider
with the same type of calculation as in the proof of the first case, we see that To estimate the integral over R  in (22), we follow exactly the calculation in [5] (see page 279-280).It is easy to see that We have also where We denote by  Green's function of Δ 2 ; that is, where   denotes the Dirac mass at , and n  is the outer unit normal at  ∈ Ω.We also denote by  the regular part of ; that is,  (, ) =      − where By the definition of   and ( () , ), we get For each  ∈ Ω, we have Thus, For each  ∈ Ω, we have By [15] (page 155), we have the following explicit formula: where with  ∈ Ω,  ∈ Ω.Using (45) and (47), we have, for all  ∈ Ω, Thus, We split the term to be estimated as and then for the last integral we have Concerning the integral over Ω\ /2 ( () ), we first notice that, by (39), Therefore, As in [5,12] we expand ( () , ⋅) up to the fifth order near  () , writing where   denotes the th order term (e.g., Note that Δ 2 ( () , ) = 0. Using the symmetry of   and the usual scaling arguments, we have (59) By ( 53)-(59), we obtain The proof of Lemma 3 is completed.
Case 2. Consider  ̸ = .Using the same argument similar to the ones in the proof of the first case, we get the desired result.
Proof.(i) The proof makes use of the same estimate as the one in the proof of Lemma 2, with  replaced by  this time.
(ii) We first write and notice that the first integral in the right hand side in (61) has been estimated in (35).Next, we treat the second integral.We will make use of notation and formulas already established in the proof of Lemma 2 to get estimate (35).Since /2 ≤ |  |/4 by definition, we have the decomposition Now we have to evaluate three integrals in the right hand side in (63).The first integral and the third integral have been estimated in (33) and (32), respectively.Finally, we deal with the second integral over Γ 2 \  /2 (0).
Case 1.Consider  = .Set For  ∈ Ω,  ∈ Ω, we have By definition of (, ), we get By (48), we obtain Therefore, we can write Since  ≤ (1/2), the last term can be estimated as in (59); namely, which gives the required estimate.
Case 2. Consider  ̸ = .The computation can be adapted from the ones in the proof of Case 1. Define Due to the definition of the points  () , we have ũ ∈   .Notice that ũ depends on , , and  through the choice of the points  () .Lemma 6.As  → ∞ (i.e.,  → 0), one has Proof.By definition of  (see (75)) and   , By Lemmas 2 and 3, we have By the symmetry of the points  () , we have since the series of  1− is convergent.Recall that  will be taken small so that we can always assume  ≤ 1/2.By (80), we obtain from Lemma 2 (81) Substituting ( 79) and ( 81) into (78) and recalling the definition of , we prove (76).
By the first part of Lemmas 4 and 5, and recalling that  >  ∼ / and The remainders generated by the second part of Lemma 4 can be dealt with as in (80).We obtain Therefore, since / → 0. Finally, from Lemma 5, (89) Substituting ( 86), (88), and (89) into (85) and recalling the definition of , we obtain the required estimate.Proposition 7. Let  ≥ 6.For every  > 0, there exists   > 0 such that, for every integer  ≥   , Σ  <  4/ . (90) Proof.The function ũ constructed in (75) depends on , , and , and for each  it belongs to   .We show that, for appropriate choice of these parameters, there results   (ũ) <  4/ .For simplicity, we set and we begin with an estimate of , noticing that we can write it as By definition of  and since | () −  () /| () | 2 | ≥  for all , , we have Moreover, as in (87), so we obtain Note that (1 − 2) 2/2 * ≥ 1 − 3, for all  ≥ 6,  > 0 and  small enough; we see that, from Lemma 6, Choose  =  −3(−5)/4(−2) and  =  1+ with  > 0 and small.It is easy to see that all the quantities depending on  in the square brackets tend to zero as / → ∞; therefore, we obtain   (ũ) We must check that, for suitable values of the parameters, the right hand side is strictly less than  4/ .Direct computations show that it is enough to prove We take  so large that and this is possible because and 2+(3/4)(−5) < −3 for all  ≥ 6, as one immediately checks.Furthermore, noticing that 3( − 5)/4( − 2) <  − 2, we see that since / → 0. Therefore, the third and the last big  is unnecessary in the expression of .We are thus led to Since  is fixed, we have  < 0 for  large (depending on ) if we take  small enough (essentially  < 3(−5)/4(−2)(− 4)).
Next we show that if Σ  <  ( The corresponding energy functional of problem ( 1) is defined by by Lemma 10.1 in [17], there exists a sequence of rescaling Since supp  ⊂ Ω, we get   → ∞ as  → ∞ and   ∈ Ω.
We can also assume that   →  ∈ Ω.
Lemma 10.Let T    be the sequence constructed above.Then, as  → ∞, one has Moreover, the sequence Set   () = (/  +   ).Changing variables as in the first part, we have Since    →  in  2,2 (R  ), we get by the Brézis-Lieb lemma [19]: By changing variables, we obtain Inserting these into (118), we get which, combined with (116), yields (i).
Proof.It is clear that there exists a sequence of positive numbers   → ∞, a sequence  1 of points of Ω with  1 →  1 ∈ Ω \ {0}, and a nontrivial critical point V 1 of  , 1 such that, setting T 1 = T( 1 ,  1 ), the sequence is a Palais-Smale sequence for   at level (2/)Σ /4 −   (V 1 ).We now iterate this scheme.If  1 → 0 strongly in  2 * (R  ), then the fact that it is a Palais-Smale sequence implies that  1 → 0 strongly in  2,2 (R  ).Since also and the lemma is proved with  = 1.Otherwise,  1 ⇀ 0 weakly in  2 * (R  ) but not strongly.In this case, starting with Lemma 10.1 in [17], we can work on  1 as we did for   .So we can find sequences  2 → ∞,  2 →  2 ∈ Ω \ {0} and a nontrivial critical point V 2 of  , 2 such that the sequence is a Palais-Smale sequence for   at level (R  ), then we obtain and the lemma is proved with  = 2. Otherwise,  2 ⇀ 0 weakly in  2 * (R  ) but not strongly, and we iterate the above argument.This procedure will end after a finite number of steps.Actually, notice that, by Remark

Proof of Theorem 1
This last section is devoted to the proof of Theorem 1.We are now ready for the main result of the paper.
Proof of Theorem 1.For every  > 0, problem (1) has a solution in some   .Indeed, given  > 0, there exists   > 0 such that, for  ≥   , By Proposition 14, Σ  is achieved by a function  ∈   .By invariance,  is a critical point of   on  2 0 (Ω) which, after scaling, gives rise to a weak solution of (1).By [13],  is a classical solution.We have to show that, at least for  large,  is not radial.
4/  for suitable values of  and .
4/  then Σ  is achieved.So we are led to analyze what happens if a minimizing sequence in   tends weakly to zero in  2 0 (Ω).Let   ∈   be a minimizing sequence for problem (9) such that   ⇀ 0 weakly in  2 0 (Ω).Without loss of generality, we can assume that    (  ) → 0 in   by Ekeland's variational principle.Since   is invariant under the action of   , we also have    (  ) → 0 in  2 0 (Ω).By homogeneity of   (), we normalize   to obtain a sequence (still denoted by   ) such that as  → ∞,   (  ) → Σ  , T    +  (1) ∫ We are now ready to describe the behavior of Palais-Smale sequence of   .Let {  } be a Palais-Smale sequence for   at level (2/)Σ /4 and   ⇀ 0 in  2 0 (Ω).Then there is a positive  (depending only on Σ  ) such that, for every  = 1, 2, . . ., , there exist sequences {  } ⊂ R + and {  } ⊂ Ω, with   → ∞ and   →   ∈ Ω \ {0} as  → ∞, and there exists a nontrivial critical point V  ∈  2,2 (R  ) of  ,  such that (up to subsequence) Remark 12. Checking the process of the proof of Lemma 11, it is easy to see that if one does not suppress the cut-off functions   , one can obtain the following representation of   : by definition of , so that, after at most  := [(Σ  /) /4 ] steps, the remainder will be a Palais-Smale sequence at level zero; namely, it will be (1) in  2,2 (R  ), obtaining the requested representation for   and   (  ).