Research

# Existence of positive solutions for a critical nonlinear Schrödinger equation with vanishing or coercive potentials

Shaowei Chen

Author Affiliations

School of Mathematical Sciences, Huaqiao University, Quanzhou, 362021, P.R. China

Boundary Value Problems 2013, 2013:201  doi:10.1186/1687-2770-2013-201

 Received: 2 May 2013 Accepted: 22 August 2013 Published: 8 September 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper we investigate the existence of positive solutions for the following nonlinear Schrödinger equation:

where and as with , , , and .

MSC: 35J20, 35J60.

##### Keywords:
semilinear Schrödinger equation; vanishing or coercive potentials

### 1 Introduction and statement of results

In this paper, we consider the following semilinear elliptic equation:

(1.1)

where . The exponent

(1.2)

with the real numbers b and s satisfying

(1.3)

By this definition, .

With respect to the functions V and K, we assume that

(A1) for every , and .

(A2) There exist and such that

(1.4)

A typical example for Eq. (1.1) with V and K satisfying (A1) and (A2) is the equation

(1.5)

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., [[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 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 [5] and [6]). 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

(1.6)

where

(1.7)

and

(1.8)

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

(1.9)

We denote this infimum by .

Our main result reads as follows.

Theorem 1.1Under assumptions (A1) and (A2), ifb, sandpsatisfy (1.3) and (1.2) and

(1.10)

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

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:

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

(2.1)

Lemma 2.1Under the above assumptions,

(2.2)

where

(2.3)

Proof Let . By direct computations,

(2.4)

and

Then

(2.5)

Since , we have and , . Then

(2.6)

Substituting (2.6) and into (2.5) results in

□

Let

(2.7)

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 ,

(2.8)

Theorem 2.2Ifis a weak solution of the equation

(2.9)

i.e., for every,

(2.10)

anduis defined by (2.1), thenand it is a weak solution of (1.1), i.e., for every,

(2.11)

Proof Using the spherical coordinates

where , , , we have

where . Recall that . Let . Then and

(2.12)

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.

Lemma 3.1If, then

(3.1)

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 ,

(3.2)

Lemma 3.2There exist constantsandsuch that for every,

Proof From conditions (A1) and (A2), we deduce that there exists a constant such that

(3.3)

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

(3.4)

If , then and

(3.5)

In this case, and

(3.6)

Conditions (A1) and (A2) imply that there exists a constant such that

(3.7)

Combining (3.5)-(3.7) yields that there exists a constant such that

(3.8)

If , (3.7) still holds. From Lemma 3.1 and (3.7), we deduce that there exists a constant such that for every ,

(3.9)

Then the desired result of this lemma follows from (3.4), (3.8) and (3.9) immediately. □

This lemma implies that

(3.10)

is equivalent to the standard norm in . We denote the inner product associated with by , i.e.,

(3.11)

By the Sobolev inequality, we have

(3.12)

and

(3.13)

By conditions (A1) and (A2), if , then is bounded in . Therefore, by (3.13), there exists such that

(3.14)

However, if , has a singularity at , i.e.,

(3.15)

Recall that and if . Then, by the Hardy-Sobolev inequality (see, for example, [[14], Lemma 3.2]), we deduce that there exists such that (3.14) still holds. Therefore, the functional

(3.16)

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

(4.1)

By this definition, we have, for ,

(4.2)

From

we deduce that the norm defined by

(4.3)

is equivalent to the standard norm in . The inner product corresponding to is

Lemma 4.1The infimum

(4.4)

is independent ofwith.

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

(4.5)

Note that

(4.6)

By , we have

It follows that

(4.7)

By (4.6) and , we get that

It follows that

(4.8)

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.

Proof Choosing in , we have

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

and is a solution of

(4.9)

In the next section, we shall show that Eq. (4.9) is the ‘limit’ equation of

(4.10)

It is easy to verify that

(4.11)

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

(4.12)

Proof Since u is a critical point of , we have

(4.13)

It follows that

(4.14)

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 5.1Under assumptions (A1) and (A2), letbe asequence ofJ. Then replacingif necessary by a subsequence, there exist a solutionof Eq. (4.10), a finite sequence, kfunctionsandksequencessatisfying:

(i) in,

(ii) , , , ,

(iii) ,

(iv) .

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,

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 [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,

It follows that . Now let .

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

(5.1)

thenin.

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

Therefore, . Now let . □

Lemma 5.4Let. Ifin, then

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 Brézis-Lieb lemma (see [15]) and its proof is similar to that of [[11], Lemma 1.32].

Lemma 5.5Letand. If

(a) is bounded in,

(b) a.e. on, then

Proof Let

Then j is a convex function. From [[15], Lemma 3], we have that for any , there exists such that for all ,

(5.2)

Hence

By Lemma 3.2 of [14], the map from is compact. We get that there exists such that for any n,

(5.3)

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

theninandis such that

Proof (1) Since in , we get that as ,

Therefore,

(5.4)

(2) Lemma 5.5 implies

(5.5)

By (5.4), (5.5) and the assumption , we get that

(3) Since in and , it is easy to verify that . For ,

(5.6)

By Lemma 5.4, we have

(5.7)

Combining (5.6) and (5.7) leads to . Then, by in and , we obtain that in . □

Lemma 5.7Ifand as,

then there existswithsuch thatandis such that

Proof We divide the proof into several steps.

(1) Since in , it is clear that

(2) For any ,

(5.8)

By the definition of the inner product (see (3.11)), we have

(5.9)

Since in , we have

(5.10)

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

(5.11)

By (5.11) and the Hölder inequality, we have

(5.12)

Since , for any , there exists such that

It follows that

(5.13)

Then

(5.14)

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 ,

(5.15)

Since in , we have in . It implies that

This together with (5.14), (5.15) and

yields that there exists such that, when ,

Thus

(5.16)

Combining (5.10), (5.12) and (5.16) leads to

(5.17)

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

(5.18)

Then, by the assumption in , we get , . Therefore, .

(3) From the definition of ,

(5.19)

By the definition of the norm (see (3.10)), we have

(5.20)

Since and a.e. on , using the Lebesgue convergence theorem, we get that

(5.21)

By (5.11), (5.20) and (5.21), we get that

(5.22)

Combining (5.19), (5.22) and (5.17) leads to

(5.23)

Note that

(5.24)

We obtain from Lemma 5.5 that

(5.25)

By Lemma 5.2,

(5.26)

Combining (5.24)-(5.26) yields

(5.27)

Combining (5.23), (5.27) and the assumption leads to

(4) For ,

(5.28)

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

Thus, as ,

holds uniformly for . Moreover, a similar argument as the proof of (5.16) yields that as ,

holds uniformly for . Therefore, as ,

(5.29)

holds uniformly for .

Second, from in and Lemma 5.4, we deduce that as ,

(5.30)

holds uniformly for . By the Hölder inequality, (3.14) and Lemma 5.2, we get that if , then

(5.31)

By Lemma 5.2, we get that for every , as ,

(5.32)

Combining (5.31) and (5.32) yields that

(5.33)

holds uniformly for . Then, by (5.30), (5.33) and

we get that as ,

(5.34)

holds uniformly for .

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

(5.35)

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,

It follows that is bounded.

(2) Assume that in and a.e. on . By Lemma 5.6, and is such that

(5.36)

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 (A1) and (A2), 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

(6.1)

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

(6.2)

Therefore, .

### 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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