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

Zhang, Yajing; Wang, Ningning; Lü, Yinmei; Hao, Jianghao

doi:https://doi.org/10.1155/2014/684679

Abstract and Applied Analysis

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Research Article | Open Access

Volume 2014 | Article ID 684679 | https://doi.org/10.1155/2014/684679

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

Yajing Zhang,¹Ningning Wang,¹Yinmei Lü,¹and Jianghao Hao¹

Academic Editor: Antonio Suárez

Received13 Jun 2014

Accepted19 Aug 2014

Published14 Oct 2014

Abstract

We prove the existence of nonradial solutions under some conditions for a semilinear biharmonic Dirichlet problem involving critical Sobolev exponents.

1. Introduction

In this paper we consider the following Hénon type biharmonic problem: where , is the unit ball of , and denotes the unit outward normal at the boundary .

We consider first the case where , 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 where denotes the subspace of radial functions in . 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 and let be the unit ball in . Then, for every 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 . Ni [7], among other things, proved the existence of radial positive solutions of (4) for all . For the case , there are many works concerning the limiting behavior of the ground state solutions of (4); see, for example, [8–11] and the references therein. In particular, Smets et al. [11] proved that, for every , no minimizer of is radial provided is large enough. Serra [12] studied the case 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 and . For a given integer , let be the group . We consider the action of on given by where and .

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 Notice that since in , we have , the best Sobolev constant for the embedding ; that is,

We now briefly outline the proof of Theorem 1. Firstly, we show that, for every , there exists such that for every integer . Next we prove that if , then is achieved. Finally, we obtain a bound from below for and then prove for all large enough, where denotes the space of radial functions in . Therefore, our solution cannot be radial.

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.

2. Asymptotic Estimates and Analysis of Palais-Smale Sequences

In this section, we first establish the estimate to prove that for suitable values of and .

In what follows, when we need to, we may define the trivial extension of functions in by zero; namely, for . For any fixed , the rescaling is defined by Notice that ; that is,

We choose and for every we define points in as Notice that the points are all in . Define where . We recall that are the unique positive solutions, radial about , of the equation in , and for all and . To fix ideas, we anticipate that we will let , and , with appropriate relations between , and . The functions are not in , so we will use instead their projections on defined by By Lemma 2.27 in [15], we have in . If we set , then To avoid heavy notation from now on, we will write simply for for , and for .

We set and we assume that and for all under consideration. Note also that due to the definition of , we have for all small.

Lemma 2. where .

Proof.
Case 1. Consider .
Direct computations yield that where denotes surface area of unit sphere in . Combining (21), we prove the first case of Lemma 2.
Case 2. Consider .
Writing with the same type of calculation as in the proof of the first case, we see that To estimate the integral over in (22), we follow exactly the calculation in [5] (see page 279-280). It is easy to see that We have also Hence for .
Direct calculations yield where , and From (27)–(29), we obtain Let We have From (30)–(33), it follows that Hence Combining (23) and (35), we prove

We denote by Green’s function of ; that is, where denotes the Dirac mass at , and is the outer unit normal at . We also denote by the regular part of ; that is, By Remark 1 in [16], we have

Lemma 3. Consider where .

Proof.
Case 1. Consider .
Set By the definition of and , we get For each , we have Thus, consequently, For each , we have Since , we have By [15] (page 155), we have the following explicit formula: where with . Using (45) and (47), we have, for all , Thus, We split the term to be estimated as and then for the last integral we have Concerning the integral over , 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., ). Since , using the explicit form of , it is not difficult to check that so that Note that . Using the symmetry of and the usual scaling arguments, we have 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.

Lemma 4. (i) Consider .(ii)Consider .

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 by definition, we have the decomposition Consequently, 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 .
Similar to (27), we have Now with elementary computations we have Inserting these in (64), we obtain Substituting (33), (32), and (66) into (63), we obtain By (61), (35), and the inequality above, we get the desired estimate.

Lemma 5. Consider

Proof.
Case 1. Consider .
Set For , we have By definition of , we get By (48), we obtain Therefore, we can write Since , 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 . Notice that depends on , and through the choice of the points .

Lemma 6. As (i.e., ), 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 is convergent. Recall that will be taken small so that we can always assume . By (80), we obtain from Lemma 2 Substituting (79) and (81) into (78) and recalling the definition of , we prove (76).
Having completed the estimate of the numerator of , we now go on to estimate the denominator, namely, . Recall that we denote and that we assume . We now set for . Then the ’s are positive disjoint and they are all contained in . Hence It is easy to see that where we have used the inequality for . We obtain therefore and we estimate the four integrals separately as above.
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 .
Finally, from Lemma 5, Substituting (86), (88), and (89) into (85) and recalling the definition of , we obtain the required estimate.

Proposition 7. Let . For every , there exists such that, for every integer ,

Proof. The function constructed in (75) depends on , and , and for each it belongs to . We show that, for appropriate choice of these parameters, there results . For simplicity, we set and we begin with an estimate of , noticing that we can write it as By definition of and since for all , we have Moreover, as in (87), so we obtain Note that , for all and small enough; we see that, from Lemma 6, Choose and with and small. It is easy to see that all the quantities depending on in the square brackets tend to zero as ; therefore, we obtain We must check that, for suitable values of the parameters, the right hand side is strictly less than . Direct computations show that it is enough to prove We take so large that and this is possible because and for all , as one immediately checks. Furthermore, noticing that , we see that since . Therefore, the third and the last big is unnecessary in the expression of . We are thus led to Since is fixed, we have for large (depending on ) if we take small enough (essentially ).

Next we show that if then is achieved. So we are led to analyze what happens if a minimizing sequence in tends weakly to zero in . Let be a minimizing sequence for problem (9) such that weakly in . Without loss of generality, we can assume that in by Ekeland’s variational principle.

Since is invariant under the action of , we also have in . By homogeneity of , we normalize to obtain a sequence (still denoted by ) such that as , The corresponding energy functional of problem (1) is defined by By (103), direct computation shows that Thus, is a Palais-Smale sequence for the functional at level .

Since for all by Lemma 10.1 in [17], there exists a sequence of rescaling such that Since supp, we get as and . We can also assume that .

Lemma 8. Let be the above sequence of rescaling satisfying , and weakly in . Then , and satisfies

Proof. We first prove as . Assume to the contrary that for all . Considering , we may regard . After a rotation of coordinates, we may assume that the sequence exhausts the half space . For any , we have for all large . Thus, the support of is contained in for all large .
Set The fact that is a Palais-Smale sequence for implies that We have used the fact that in and . Since this happens for all , we see that satisfies where for some constant . By Lemma 4 in [2](also see [4]), (111) has a unique solution , which contradicts . Therefore, we must have . In this case, we can repeat the above argument and take test function . The above computations imply that is a nontrivial solution of in , which also shows that since .

Remark 9. It is proved in [18] that the radial function solves in , where . So the function is nothing else than a multiple of . Precisely, .

Let be a cut-off function satisfying for all , for for , , and .

Since , we may take a sequence such that and as . Set . In this way, supp for all large and .

Note that Since , the first term tends to as . By Hölder inequality, the second term and the third term in inequality above, as . Thus, strongly in .

Lemma 10. Let be the sequence constructed above. Then, as , one has(i),(ii) in ,
where Moreover, the sequence is a Palais-Smale sequence for at level ; namely, it satisfies(i);(ii) in .

Proof. (i) Set ; then . Since and in , we have Set . Changing variables as in the first part, we have Since in , 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).
(ii) Without loss of generality, we assume that is nonnegative; otherwise, one replaces its th power by . Taking , Since in , we have By the Brézis-Lieb lemma, we have By (123), (124), and , we obtain Combing (122) and (125), we have since is a critical point of by Lemma 8. Thus, we prove (ii).
An immediate consequence of (i) and (ii) is the the sequence which is a Palais-Smale sequence for at level .

We are now ready to describe the behavior of Palais-Smale sequence of .

Lemma 11. Let be a Palais-Smale sequence for at level and in . Then there is a positive (depending only on ) such that, for every , there exist sequences and , with and as , and there exists a nontrivial critical point of such that (up to subsequence) where .

Proof. It is clear that there exists a sequence of positive numbers , a sequence of points of with , and a nontrivial critical point of such that, setting , the sequence is a Palais-Smale sequence for at level .
We now iterate this scheme. If strongly in , then the fact that it is a Palais-Smale sequence implies that strongly in . Since also strongly in , we can write and the lemma is proved with . Otherwise, weakly in but not strongly. In this case, starting with Lemma 10.1 in [17], we can work on as we did for . So we can find sequences and a nontrivial critical point of such that the sequence is a Palais-Smale sequence for at level . Once again, if strongly in , then we obtain and the lemma is proved with . Otherwise, weakly in but not strongly, and we iterate the above argument. This procedure will end after a finite number of steps. Actually, notice that, by Remark 9, for all , by definition of , so that, after at most steps, the remainder will be a Palais-Smale sequence at level zero; namely, it will be in , obtaining the requested representation for and .

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 :

Lemma 13. Let be a minimizing sequence for problem (9) and weakly in . If , then is a minimum point and strongly in .

Proof. Notice first that since is weakly closed in . Weak convergence and the Brézis-Lieb Lemma yield Therefore, But is a minimizing sequence so that Since , we have Inserting (136) and (137) into (135), we obtain Since for large enough, it is easy to see that as . Otherwise, the right hand side of (138) will be strictly less than , which contradicts the definition of . Thus, we have strongly in .

The following proposition is the main result of this section.

Proposition 14. If , then is achieved.

Proof. Let be a (bounded) minimizing sequence for over . Then there exists a subsequence (still denoted by ) such that in . If , by Lemma 13, is a minimum point in , and the proof is completed. If, on the contrary, in , then, as we did above, without loss of generality we can assume that is a Palais-Smale sequence for at level . The behavior of such sequence is described in Lemma 11. In particular, Since is a multiple of the radial function , the cut-off function is also radial, and is a continuous group, we obtain that, for every as . Otherwise, (133) would be incompatible with the symmetry properties of . Thus, we can replace in (133) with its projection on so that we can assume that (133) holds in . This means that must be a multiple of ; say, , for some integer . But, from (139), namely, , which is impossible.

3. Proof of Theorem 1

This last section is devoted to the proof of Theorem 1.

Lemma 15. For any , there exists depending on and such that where denotes the space of radial functions in and

Proof. Let and define , where . Then It is easy to see that We obtain Set Direct computations yield , which shows is convex. Hence that is By Hardy’s inequality, we have consequently, Therefore, by (148) and (150), there exists depending only on such that By (143), we have For every , is achieved by standard arguments. Since is nondecreasing on , by (152), we obtain (141).

We are now ready for the main result of the paper.

Proof of Theorem 1. For every , problem (1) has a solution in some . Indeed, given , there exists such that, for , By Proposition 14, is achieved by a function . By invariance, is a critical point of on 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.
By Lemma 15, where the constant depends only on . We now show that the level of the solution we find is strictly below this threshold for large. To this aim, we must evaluate how large of Proposition 7 has to be in terms of . If we choose of the order of , we see that (102) essentially becomes Therefore, will be negative for all big enough when is sufficiently small, so we find a solution to (1) in the corresponding . Therefore, we find a solution at level for all large enough since for all . Therefore, our solution cannot be radial.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by NNSFC (no. 61374089), NSF of Shanxi Province (no. 2014011005-2), and International Cooperation Projects of Shanxi Province (no. 2014081026).

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Copyright © 2014 Yajing Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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