Abstract
Keywords:
semilinear Schrödinger equation; vanishing or coercive potentials1 Introduction and statement of results
In this paper, we consider the following semilinear elliptic equation:
with the real numbers b and s satisfying
With respect to the functions V and K, we assume that
(A_{2}) There exist and such that
A typical example for Eq. (1.1) with V and K satisfying (A_{1}) and (A_{2}) is the equation
When , the potentials are vanishing at infinity and when , the potentials are coercive.
Equation (1.1) arises in various applications, such as chemotaxis, population genetics, chemical reactor theory and the study of standing wave solutions of certain nonlinear Schrödinger equations. Therefore, they have received growing attention in recent years (one can see, e.g., [16] and [710] for reference).
Under the above assumptions, Eq. (1.1) has a natural variational structure. For an open subset Ω in , let be the collection of smooth functions with a compact support set in Ω. Let E be the completion of with respect to the inner product
From assumptions (A_{1}) and (A_{2}), we deduce that
are two equivalent norms in the space
Therefore, there exists such that
Moreover, assumptions (A_{1}) and (A_{2}) imply that there exists such that
Then, by the Hölder and Sobolev inequalities (see, e.g., [[11], Theorem 1.8]), we have, for every ,
where is a constant independent of u. It follows that there exists a constant such that
This implies that E can be embedded continuously into the weighted space
Then the functional
is well defined in E. And it is easy to check that Φ is a functional and the critical points of Φ are solutions of (1.1) in E.
In a recent paper [12], Alves and Souto proved that the space E can be embedded compactly into if and and Φ satisfies the PalaisSmale condition consequently. Then, by using the mountain pass theorem, they obtained a nontrivial solution for Eq. (1.1). Unfortunately, when , the embedding of E into is not compact and Φ no longer satisfies the PalaisSmale condition. Therefore, the ‘standard’ variational methods fail in this case. From this point of view, should be seen as a kind of critical exponent for Eq. (1.1). If the potentials V and K are restricted to the class of radially symmetric functions, ‘compactness’ of such a kind is regained and ‘standard’ variational approaches work (see [5] and [6]). However, this method does not seem to apply to the more general equation (1.1) where K and V are nonradially symmetric functions.
It is not easy to deal with Eq. (1.1) directly because there are no known approaches that can be used directly to overcome the difficulty brought by the loss of compactness. However, in this paper, through an interesting transformation, we find an equivalent equation for Eq. (1.1) (see Eq. (2.9) in Section 2). This equation has the advantages that its PalaisSmale sequence can be characterized precisely through the concentrationcompactness principle (see Theorem 5.1), and it possesses partial compactness (see Corollary 5.8). By means of these advantages, a positive solution for this equivalent equation and then a corresponding positive solution for Eq. (1.1) are obtained.
Before stating our main result, we need to give some definitions.
Let
where
and
Let be the Sobolev space endowed with the norm and the inner product
respectively, and let be the function space consisting of the functions on that are pintegrable. Since , can be embedded continuously into . Therefore, the infimum
Our main result reads as follows.
Theorem 1.1Under assumptions (A_{1}) and (A_{2}), ifb, sandpsatisfy (1.3) and (1.2) and
then Eq. (1.1) has a positive solution.
Remark 1.2 We should emphasize that condition (1.10) can be satisfied in many situations. For , let and be the closure of in . Under assumptions (A_{1}) and (A_{2}), we have
Then, for any , there exist and such that
It follows from this inequality and that if is small enough such that
then
This implies that (1.10) is satisfied if ϵ is chosen such that .
Notations Let X be a Banach space and . We denote the Fréchet derivative of φ at u by . The Gateaux derivative of φ is denoted by , . By → we denote the strong and by ⇀ the weak convergence. For a function u, denotes the functions . The symbol denotes the Kronecker symbol:
2 An equivalent equation for Eq. (1.1)
For , let . To u, a function in , we associate a function v, a function in , by the transformation
Lemma 2.1Under the above assumptions,
where
Proof Let . By direct computations,
and
Then
Substituting (2.6) and into (2.5) results in
□
Let
From the classical Hardy inequality (see, e.g., [[13], Lemma 2.1]), we deduce that for every bounded domain , there exists such that, for every ,
Theorem 2.2Ifis a weak solution of the equation
anduis defined by (2.1), thenand it is a weak solution of (1.1), i.e., for every,
Proof Using the spherical coordinates
where . Recall that . Let . Then and
Here, we used in the last inequality above. From (2.4), (2.12) and (2.8), we deduce that there exists such that for every bounded domain ,
Moreover,
Therefore, . Then, to prove that u satisfies (2.11) for every , it suffices to prove that (2.11) holds for every . For , let be such that
By using the divergence theorem and Lemma 2.1, we get that
Moreover,
and
Therefore,
This completes the proof. □
This theorem implies that the problem of looking for solutions of (1.1) can be reduced to a problem of looking for solutions of (2.9).
3 The variational functional for Eq. (2.9)
The following inequality is a variant Hardy inequality.
Proof We only give the proof of (3.1) for since is dense in . For , we have the following identity:
By using the Hölder inequality, it follows that
Then we conclude that
□
From the definition of (see (2.3)), it is easy to verify that for ,
Lemma 3.2There exist constantsandsuch that for every,
Proof From conditions (A_{1}) and (A_{2}), we deduce that there exists a constant such that
Since
by (3.3) and the classical Hardy inequality (see, e.g., [13])
we deduce that there exists a constant such that
This together with the fact that yields that there exists a constant such that
Conditions (A_{1}) and (A_{2}) imply that there exists a constant such that
Combining (3.5)(3.7) yields that there exists a constant such that
If , (3.7) still holds. From Lemma 3.1 and (3.7), we deduce that there exists a constant such that for every ,
Then the desired result of this lemma follows from (3.4), (3.8) and (3.9) immediately. □
This lemma implies that
is equivalent to the standard norm in . We denote the inner product associated with by , i.e.,
By the Sobolev inequality, we have
and
By conditions (A_{1}) and (A_{2}), if , then is bounded in . Therefore, by (3.13), there exists such that
However, if , has a singularity at , i.e.,
Recall that and if . Then, by the HardySobolev inequality (see, for example, [[14], Lemma 3.2]), we deduce that there exists such that (3.14) still holds. Therefore, the functional
is a functional defined in . Moreover, it is easy to check that the Gateaux derivative of J is
and the critical points of J are nonnegative solutions of (2.9).
4 Some minimizing problems
By this definition, we have, for ,
From
we deduce that the norm defined by
is equivalent to the standard norm in . The inner product corresponding to is
Lemma 4.1The infimum
Proof In this proof, we always view a vector in as a matrix, and we use to denote the conjugate matrix of a matrix A.
For any with , let G be an orthogonal matrix such that . For any , let , . The assumption that G is an orthogonal matrix implies that , where I is the identity matrix. Then it is easy to check that
Note that
It follows that
It follows that
By (4.5), (4.7) and (4.8), we get that . This together with (4.5) leads to the result of this lemma. □
Since the infimum (4.4) is independent of with , we denote it by S.
Lemma 4.2Letbe the infimum in (1.9). Then.
By Lemma 4.1, we have
Let
Then
It follows that
□
Since the functionals and are invariant by translations, the same argument as the proof of [[11], Theorem 1.34] yields that there exists a positive minimizer for the infimum S. Moreover, from the Lagrange multiplier rule, it is a solution of
In the next section, we shall show that Eq. (4.9) is the ‘limit’ equation of
It is easy to verify that
is a functional defined in , the Gateaux derivative of is
and the critical points of this functional are solutions of (4.9).
Lemma 4.3Letsatisfy. Ifis a critical point of, then
Proof Since u is a critical point of , we have
It follows that
This together with (4.14) yields the result of this lemma. □
5 The PalaisSmale condition for the functional J
Recall that J is the functional defined by (3.16). By a sequence of J, we mean a sequence such that and in as , where denotes the dual space of . J is called satisfying the condition if every sequence of J contains a convergent subsequence in .
Our main result in this section reads as follows.
Theorem 5.1Under assumptions (A_{1}) and (A_{2}), letbe asequence ofJ. Then replacingif necessary by a subsequence, there exist a solutionof Eq. (4.10), a finite sequence, kfunctionsandksequencessatisfying:
This theorem gives a precise representation of the sequence for the functional J. Through it, partial compactness for J can be regained (see Corollary 5.8).
To prove this theorem, we need some lemmas. Our proof of this theorem is inspired by the proof of [[11], Theorem 8.4].
Lemma 5.2Let. Then, for any sequence,
Proof If , then is bounded in . In this case, the result of this lemma is obvious. If , then as . Since , by Lemma 3.2 of [14], the map from is compact. Therefore, for any , there exists such that
And there exists depending only on ϵ such that , . Then, for every n,
Using the same argument as above, for any , there exist and such that
and
Since , we have . Then, using the Lebesgue theorem and the above two inequalities, we get that
Let . Then we get the desired result of this lemma. □
Lemma 5.3Let. Ifis bounded inand
Proof Since , by Lemma 3.2 of [14], the map from is compact. Therefore, for any , there exists such that
And there exists depending only on ϵ such that , . By (5.1) and the Lions lemma (see, for example, [[11], Lemma 1.21]), we get that
One can follow the proof of [[11], Lemma 8.1] step by step and use Lemma 5.2 to give the proof of this lemma.
The following lemma is a variant BrézisLieb lemma (see [15]) and its proof is similar to that of [[11], Lemma 1.32].
Proof Let
Then j is a convex function. From [[15], Lemma 3], we have that for any , there exists such that for all ,
Hence
By Lemma 3.2 of [14], the map from is compact. We get that there exists such that for any n,
And there exists depending only on ϵ such that , . Then
By the Lebesgue theorem, , . This together with (5.3) yields
The left proof is the same as the proof of [[11], Lemma 1.32]. □
Lemma 5.6If
Proof (1) Since in , we get that as ,
Therefore,
(2) Lemma 5.5 implies
By (5.4), (5.5) and the assumption , we get that
(3) Since in and , it is easy to verify that . For ,
By Lemma 5.4, we have
Combining (5.6) and (5.7) leads to . Then, by in and , we obtain that in . □
then there existswithsuch thatandis such that
Proof We divide the proof into several steps.
(1) Since in , it is clear that
By the definition of the inner product (see (3.11)), we have
By assumption (A_{2}) and the definition of , we have . This yields
Moreover, together with (2.8) and the fact that yields that for any fixed ,
Combining the above two limits leads to
By (5.11) and the Hölder inequality, we have
Since , for any , there exists such that
It follows that
Then
where the constant C is independent of ϵ and n. There exists a subsequence of , denoted by itself for convenience, and with such that as . Then, by , we get that as ,
and converges to θ uniformly for . Therefore, there exists such that, when ,
Since in , we have in . It implies that
This together with (5.14), (5.15) and
yields that there exists such that, when ,
Thus
Combining (5.10), (5.12) and (5.16) leads to
We obtain, by the Hölder inequality and Lemma 5.2, that as ,
where and C are positive constants independent of n and h. This together with (5.8) and (5.17) yields
Then, by the assumption in , we get , . Therefore, .
By the definition of the norm (see (3.10)), we have
Since and a.e. on , using the Lebesgue convergence theorem, we get that
By (5.11), (5.20) and (5.21), we get that
Combining (5.19), (5.22) and (5.17) leads to
Note that
We obtain from Lemma 5.5 that
By Lemma 5.2,
Combining (5.24)(5.26) yields
Combining (5.23), (5.27) and the assumption leads to
We shall give the limits for and as .
First, as (5.9), we have
By the Hölder inequality and (5.11), we get that if , then
holds uniformly for . Moreover, a similar argument as the proof of (5.16) yields that as ,
holds uniformly for . Therefore, as ,
Second, from in and Lemma 5.4, we deduce that as ,
holds uniformly for . By the Hölder inequality, (3.14) and Lemma 5.2, we get that if , then
By Lemma 5.2, we get that for every , as ,
Combining (5.31) and (5.32) yields that
holds uniformly for . Then, by (5.30), (5.33) and
Finally, combining (5.28), (5.29) and (5.34) leads to
holds uniformly for . This together with the fact that and in yields in . □
Proof of Theorem 5.1 We divide the proof into two steps.
(1) For n big enough, we have
As mentioned in Section 3, the norm is equivalent to the norm . Therefore, there exists a constant such that , . Then by (5.35) there exists a constant such that for n big enough,
(2) Assume that in and a.e. on . By Lemma 5.6, and is such that
Let us define
If , Lemma 5.3 implies that in . Since in , it follows that
and the proof is complete. If , we may assume the existence of such that
Let us define . We may assume that in and a.e. on . Since
it follows from the Rellich theorem that
and . But in , so that is unbounded. We may assume that . Finally, by (5.36) and Lemma 5.7, there exists with such that and satisfies
Moreover, Lemma 4.3 implies that
Iterating the above procedure, we construct sequences , and . Since for every l, , the iteration must terminate at some finite index k. This finishes the proof of this theorem. □
The following corollary is a direct consequence of Theorem 5.1 and Lemma 4.3. It implies that the functional J satisfies the condition if .
Corollary 5.8Under assumptions (A_{1}) and (A_{2}), any sequencesuch that
contains a convergent subsequence.
6 Proof of Theorem 1.1
Recall that the critical points of J are nonnegative solutions of (2.9). By Corollary 2.2, to prove that Eq. (1.1) has a positive solution, it suffices to prove that J has a nontrivial critical point. Moreover, by Corollary 5.8, it suffices to apply the classical mountain pass theorem (see, e.g., [[11], Theorem 1.15]) to J with the mountain pass value .
By assumption (1.10) and Lemma 4.2, there exists a nonnegative such that
We obtain
By (3.14), we get
Therefore, there exists such that
Moreover, there exists such that and . It follows from (6.1) that
By Corollary 5.8 and the mountain pass theorem (see [[11], Theorem 1.15]), J has a critical value c such that and Eq. (2.9) has a positive solution . Then, by Theorem 2.2, the function u defined by (2.1) is a positive solution of (1.1). To complete the proof, it suffices to prove that . Using the divergence theorem, Lemma 2.1 and (2.12), we get that
Moreover, by Lemma 2.1 and (2.12), we get that
Competing interests
The author declares that they have no competing interests.
Acknowledgements
Shaowei Chen was supported by Science Foundation of Huaqiao University.
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