Existence of positive solutions for a critical nonlinear Schrödinger equation with vanishing or coercive potentials
Boundary Value Problems volume 2013, Article number: 201 (2013)
In this paper we investigate the existence of positive solutions for the following nonlinear Schrödinger equation:
where and as with , , , and .
1 Introduction and statement of results
In this paper, we consider the following semilinear elliptic equation:
where . The exponent
with the real numbers b and s satisfying
By this definition, .
With respect to the functions V and K, we assume that
(A1) for every , and .
(A2) There exist and such that
A typical example for Eq. (1.1) with V and K satisfying (A1) and (A2) 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., [1–6] and [7–10] 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 (A1) and (A2), we deduce that
are two equivalent norms in the space
Therefore, there exists such that
Moreover, assumptions (A1) and (A2) imply that there exists such that
Then, by the Hölder and Sobolev inequalities (see, e.g., [, 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 , Alves and Souto proved that the space E can be embedded compactly into if and and Φ satisfies the Palais-Smale 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 Palais-Smale 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  and ). However, this method does not seem to apply to the more general equation (1.1) where K and V are non-radially 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 Palais-Smale sequence can be characterized precisely through the concentration-compactness 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 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 p-integrable. Since , can be embedded continuously into . Therefore, the infimum
We denote this infimum by .
Our main result reads as follows.
Theorem 1.1 Under assumptions (A1) and (A2), if b, s and p satisfy (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 (A1) and (A2), we have
Then, for any , there exist and such that
It follows from this inequality and that if is small enough such that
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:
We use to mean as .
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 2.1 Under the above assumptions,
Proof Let . By direct computations,
Since , we have and , . Then
Substituting (2.6) and into (2.5) results in
From the classical Hardy inequality (see, e.g., [, Lemma 2.1]), we deduce that for every bounded domain , there exists such that, for every ,
Theorem 2.2 If is a weak solution of the equation
i.e., for every ,
and u is defined by (2.1), then and it is a weak solution of (1.1), i.e., for every ,
Proof Using the spherical coordinates
where , , , we have
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 ,
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
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.
Lemma 3.1 If , then
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 3.2 There exist constants and such that for every ,
Proof From conditions (A1) and (A2), we deduce that there exists a constant such that
by (3.3) and the classical Hardy inequality (see, e.g., )
we deduce that there exists a constant such that
This together with the fact that yields that there exists a constant such that
If , then and
In this case, and
Conditions (A1) and (A2) 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
By conditions (A1) and (A2), 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 Hardy-Sobolev inequality (see, for example, [, 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
For with , let
By this definition, we have, for ,
we deduce that the norm defined by
is equivalent to the standard norm in . The inner product corresponding to is
Lemma 4.1 The infimum
is independent of with .
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
By , we have
It follows that
By (4.6) and , we get 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 4.2 Let be the infimum in (1.9). Then .
Proof Choosing in , we have
By Lemma 4.1, we have
It follows that
Since the functionals and are invariant by translations, the same argument as the proof of [, 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
and 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 4.3 Let satisfy . If is a critical point of , then
Proof Since u is a critical point of , we have
It follows that
Since , by and , we get that
This together with (4.14) yields the result of this lemma. □
5 The Palais-Smale 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 5.1 Under assumptions (A1) and (A2), let be a sequence of J. Then replacing if necessary by a subsequence, there exist a solution of Eq. (4.10), a finite sequence , k functions and k sequences satisfying:
, , , ,
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 [, Theorem 8.4].
Lemma 5.2 Let . Then, for any sequence ,
If , , then
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 , 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,
It follows that . Now let .
Using the same argument as above, for any , there exist and such that
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 5.3 Let . If is bounded in and
then in .
Proof Since , by Lemma 3.2 of , 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, [, Lemma 1.21]), we get that
Therefore, . Now let . □
Lemma 5.4 Let . If in , then
One can follow the proof of [, Lemma 8.1] step by step and use Lemma 5.2 to give the proof of this lemma.
Lemma 5.5 Let and . If
is bounded in ,
a.e. on , then
Then j is a convex function. From [, Lemma 3], we have that for any , there exists such that for all ,
By Lemma 3.2 of , 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 [, Lemma 1.32]. □
Lemma 5.6 If
then in and is such that
Proof (1) Since in , we get that as ,
Lemma 5.5 implies(5.5)
By (5.4), (5.5) and the assumption , we get that
Since in and , it is easy to verify that . For ,(5.6)
By Lemma 5.4, we have
Combining (5.6) and (5.7) leads to . Then, by in and , we obtain that in . □
Lemma 5.7 If and as ,
then there exists with such that and is such that
Proof We divide the proof into several steps.
Since in , it is clear that
For any ,(5.8)
By the definition of the inner product (see (3.11)), we have
Since in , we have
By assumption (A2) 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