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Nontrivial solution for asymmetric -Laplacian Dirichlet problem
Boundary Value Problems volume 2014, Article number: 241 (2014)
We consider a class of particular -Laplacian Dirichlet problems with a right-hand side nonlinearity which exhibits an asymmetric growth at +∞ and −∞. Namely, it is linear at −∞ and superlinear at +∞. However, it need not satisfy the Ambrosetti-Rabinowitz condition on the positive semi-axis. Some existence results for a nontrivial solution are established by the mountain pass theorem and a variant version of the mountain pass theorem in the general case . Similar results are also established by combining the mountain pass theorem and a variant version of the mountain pass theorem with the Moser-Trudinger inequality in the case of .
Let Ω be a bounded domain in () with smooth boundary ∂ Ω. We consider the following quasilinear elliptic boundary problem:
where and , denotes the p-Laplacian operator defined by , is a real parameter, and .
It is known that the nontrivial solutions of problem (1.1) are equivalent to the corresponding nonzero critical points of the -energy functional
for all , where .
For the case of , and , there has been an increasing interest in looking for the existence of solutions of (1.1). Using the following conditions,
where and C is a constant, the authors in ,  prove that (1.1) has at least two nontrivial solutions by the three critical point theorems. Here and in the sequel, denote the eigenvalues of −△ in , and is the first eigenvalue of in (see ). For Eq. (1.1) with the right-hand side having p-linear growth at infinity, i.e., , the spectrum of in , the papers ,  get the existence of a nontrivial solution. In , the author extends the results in ,  under the general asymptotically linear condition.
The main purpose of this paper is to establish existence results of a nontrivial solution for problem (1.1) with when the nonlinearity term exhibits an asymmetric behavior as approaches +∞ and −∞. More precisely, we assume that for a.e. , grows superlinear at +∞, while at −∞ it has a linear growth. In case of , and , equations with nonlinearities which are superlinear in one direction and linear in the other were investigated by Arcoya and Villegas , de Figueiredo and Ruf , Perera . All three works express the superlinear growth at +∞ using the Ambrosetti-Rabinowitz condition ((AR)-condition, for short). Recall that a function is said to satisfy the (AR)-condition in the positive direction if there exist and such that
where . Since Ambrosetti and Rabinowitz proposed the mountain pass theorem in their celebrated paper , critical point theory has become one of the main tools for finding solutions to elliptic equations of variational type. When we apply the mountain pass theorem, the (AR)-condition usually plays an important role in verifying that the functional I has a ‘mountain pass’ geometry and showing that a related sequence is bounded.
By simple calculation, it is easy to see that the previous one side (AR)-condition implies that . That is, must be superlinear with respect to at positive infinity. Recently, Motreanu et al., Papageorgiou and Papageorgiou  and Papageorgiou and Smyrlis  studied asymmetric problem (1.1) with nonlinearity f not satisfying the (AR)-condition on the positive semi-axis when and . Nevertheless, all of the above-mentioned works involve the nonlinear term of a subcritical (polynomial) growth, say,
(SCP): there exist positive constants and and such that
where denotes the critical Sobolev exponent. One of the main reasons to assume this condition (SCP) is that they can use the Sobolev compact embedding , .
which is much weaker than (SCP). Note that in this case, we do not have the Sobolev compact embedding anymore. Our work again is to study asymmetric problem (1.1) without the (AR)-condition in the positive semi-axis. In fact, this condition was studied by Liu and Wang in  in the case of Laplacian (i.e., ) by the Nehari manifold approach. However, we will use the mountain pass theorem and a suitable version of the mountain pass theorem to get the nontrivial solution to problem (1.1) in the general case . Our results are different from those in – and our proof of the compactness condition is skillful.
Let us now state our results: Suppose that and satisfies:
(H1): uniformly for a.e. , where ;
(H2): uniformly for a.e. , where ;
(H3): uniformly for a.e. ;
(H4): is nonincreasing with respect to for a.e. .
Let be the first eigenvalue of and for every be the eigenfunction. Throughout this paper, we denote by the norm and the norm of u in will be defined by
Letand assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)) and satisfies (H1)-(H3). If, then there existssuch that for allwith, problem (1.1) has at least a nontrivial solution if l is not any of the eigenvalues ofon.
Letand assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)) and satisfies (H1)-(H3). Ifanduniformly for a.e. , then there existssuch that for allandwith, problem (1.1) has at least a nontrivial solution.
When , problem (1.1) is called resonant at negative infinity. This case is completely new. Here, we also give an example for . It satisfies our conditions (H1)-(H3) and (SCPI).
where , ; , ; ; ; ; . Moreover, there exists such that for all .
Letand assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)) and satisfies (H1)-(H4). Ifand, then there existssuch that for allwith, problem (1.1) has at least a nontrivial solution.
In case of , we have . In this case, every polynomial growth is admitted, but one knows easy examples that . Hence, one is led to look for a function with maximal growth such that
(SCE):f has subcritical (exponential) growth on Ω, i.e., uniformly on for all .
When and f has the subcritical (exponential) growth , our work is still to study asymmetric problem (1.1) without the (AR)-condition in the positive semi-axis. To our knowledge, this case is completely new. Our results are as follows.
Letand assume that f has the subcritical exponential growth on Ω (condition (SCE)) and satisfies (H1)-(H3). If, then there existssuch that for allwith, problem (1.1) has at least a nontrivial solution if l is not any of the eigenvalues ofon.
In view of conditions (H2), (H3) and (SCE), problem (1.1) is called asymmetric subcritical exponential problem. Hence, Theorem 1.4 is completely new.
Letand assume that f has the subcritical exponential growth on Ω (condition (SCE)) and satisfies (H1)-(H3). Ifanduniformly for a.e. , then there existssuch that for allandwith, problem (1.1) has at least one nontrivial solution.
Letand assume that f has the subcritical exponential growth on Ω (condition (SCE)) and satisfies (H1)-(H4). Ifand, then there existssuch that for allwith, problem (1.1) has at least one nontrivial solution.
2 Preliminaries and some lemmas
Let be a real Banach space with its dual space and . For , we say that I satisfies the condition if for any sequence with
there is a subsequence such that converges strongly in E. Also, we say that I satisfies the condition if for any sequence with
there is a subsequence such that converges strongly in E.
We have the following version of the mountain pass theorem (see ).
Let E be a real Banach space and suppose that satisfies the condition
for some, andwith. Letbe characterized by
whereis the set of continuous paths joining 0 and . Then there exists a sequencesuch that
Letandbe a-eigenfunction withand assume that (H1)-(H3) and (SCPI) hold. If, then there existssuch that for allwith, we have:
There exist such that for all with .
By (SCPI) and (H1)-(H3), if , for any , there exist , such that for all ,
Choose such that . By (2.1), the Poincaré inequality and the Sobolev inequality, ,
So, part (i) holds if we choose and .
On the other hand, if , take such that . By (2.2), we have
Since and , it is easy to see that
and part (ii) is proved. □
Let, thenfor all. Moreover,
The inequality is optimal: for any growthwith, the corresponding supremum is +∞.
Letandbe a-eigenfunction withand assume that (H1)-(H3) and (SCE) hold. If, then there existssuch that for allwith, we have:
There exist such that for all with .
By (SCE) and (H1)-(H3), if , for any , there exist , , and such that for all ,
Choose such that . By (2.3), the Holder inequality and the Moser-Trudinger embedding inequality, we get
where sufficiently close to 1, and . Set
So, part (i) holds if we choose and .
On the other hand, if , taking such that and using (2.4), we have
Thus part (ii) is proved. By exactly slight modification to the proof above, we can prove (ii) if . □
For the functional I defined by (1.2), ifa.e. , and
then there exists a subsequence, still denoted by, such that
By as , for a suitable subsequence, we may assume that
for all n, where denotes the norm of .
We claim that for any and ,
Indeed, for any , at fixed and , if we set
and our claim (2.6) is proved.
On the other hand,
3 Proofs of the main results
Here, we only prove Theorems 1.1-1.4. Others follow these results.
Proof of Theorem 1.1
By Lemma 2.1, the geometry conditions of mountain pass theorem hold. So, we only need to verify condition . Let be a sequence such that for every ,
for all , where is a positive constant and is a sequence which converges to zero.
Step 1. In order to prove that has a convergent subsequence, we first show that it is a bounded sequence. To do this, we argue by contradiction assuming that for a subsequence, which we denote by , we have
Without loss of generality, we can assume for all and define . Obviously, , , and then it is possible to extract a subsequence (denoted also by ) such that
where and . Dividing both sides of (3.2) by , we obtain
for all . Passing to the limit we deduce from (3.3) that
for all .
Now we claim that for a.e. . To verify this, let us observe that by choosing in (3.7), we have
where . On the other hand, from (H2) and (H3), it implies
for some positive constant . Moreover, using for a.e. , (3.5) and the superlinearity of f (see (H3)), we also deduce
Therefore, if , by Fatou’s lemma, we will obtain that
which contradicts (3.8). Thus and the claim is proved.
Clearly, . By (H2), there exists such that for a.e. . By using the Lebesgue dominated convergence theorem in (3.7), we have
for all . This contradicts our assumption, i.e., l is not any of the eigenvalues of on .
Step 2. Now, we prove that has a convergent subsequence. In fact, we can suppose that
Now, since f has the subcritical growth on Ω, for every , we can find a constant such that
Similarly, since in , . Since is arbitrary, we can conclude that
By (3.2), we have
Using an elementary inequality
we can imply that
So we have in , which means that I satisfies . □
Proof of Theorem 1.2
Since , obviously, Lemma 2.1(i) holds. We only need to show that Lemma 2.1(ii) holds. Let . Using condition (H3), we have
as , where M is a positive constant large enough. By Proposition 2.1, there exists a sequence such that
Clearly, (3.13) implies that
To complete our proof, we first need to verify that is bounded in . Similar to the proof of Theorem 1.1, we have , , and
for all . By the maximum principle (see ), is an eigenfunction of , then for a.e. . By our assumptions, we have
uniformly in , which implies that
On the other hand, (3.14) implies that
which contradicts (3.15). Hence is bounded. According to Step 2 of the proof of Theorem 1.1, we have in , which means that I satisfies . □
Proof of Theorem 1.3
Since is bounded in , it is possible to extract a subsequence (denoted also by ) such that
where and .
If as , then . In fact, letting and noticing , from (H3) we have that
So for a.e. . But, if , then . Hence
On the other hand, by as , we have as . From Lemma 2.4 and (3.12), we get
Obviously, it contradicts (3.17). So is bounded in . According to Step 2 of the proof of Theorem 1.1, we have in , which means that I satisfies . □
Proof of Theorem 1.4
By Lemma 2.3, the geometry conditions of mountain pass theorem hold. So, we only need to verify condition . Similar to Step 1 of the proof of Theorem 1.1, we easily know that the sequence is bounded in . Next, we prove that has a convergent subsequence. Without loss of generality, suppose that
Now, since f has the subcritical exponential growth (SCE) on Ω, we can find a constant such that
Thus, by the Moser-Trudinger inequality (see Lemma 2.2),
Similar to the last proof of Theorem 1.1, we have in , which means that I satisfies . □
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This work was supported by the National NSF (Grant No. 11101319) of China and Planned Projects for Postdoctoral Research Funds of Jiangsu Province (Grant No. 1301038C).
The authors declare that they have no competing interests.
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.