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Concentration Phenomena for Fractional Elliptic Equations Involving Exponential Critical Growth

  • Claudianor O. Alves , João Marcos do Ó EMAIL logo and Olímpio H. Miyagaki
Published/Copyright: October 11, 2016

Abstract

In this paper, we deal with the singular perturbed fractional elliptic problem ε(-Δ)1/2u+V(z)u=f(u) in , where (-Δ)1/2u is the square root of the Laplacian and f(s) has exponential critical growth. Under suitable conditions on f(s), we construct a localized bound state solution concentrating at an isolated component of the positive local minimum points of the potential of V as ε goes to 0.

1 Introduction

In this paper, we are concerned with existence and concentration of positive solutions for the following singular perturbed fractional elliptic problem:

(Pϵ) { ε ( - Δ ) 1 / 2 u + V ( z ) u = f ( u ) in , u H 1 / 2 ( ) , u > 0 on ,

where ε is a small positive parameter, the potential V is bounded away from zero, the non-linearity f(s) has exponential critical growth, and (-Δ)1/2u is the square root of the Laplacian, which may be defined for smooth functions as

( ( - Δ ) 1 / 2 u ) ( ξ ) = | ξ | ( u ) ( ξ ) ,

where is the Fourier transform, that is,

( ξ ) = 1 2 π e - i ξ x ϕ ( x ) 𝑑 x

for functions ϕ in the Schwartz class. Also, for sufficiently smooth u, (-Δ)1/2u can be equivalently represented, see [18, 24], as

( - Δ ) 1 / 2 u = - 1 2 π u ( x + y ) + u ( x - y ) - 2 u ( x ) | y | 2 𝑑 y ,

and, by [18, Propostion 3.6],

( - Δ ) 1 / 4 u L 2 2 := 1 2 π 2 ( u ( x ) - u ( y ) ) 2 | x - y | 2 𝑑 x 𝑑 y for all u H 1 / 2 ( ) .

Here H1/2() is the fractional Sobolev space

H 1 / 2 ( ) = { u L 2 ( ) : ( - Δ ) 1 / 4 u L 2 2 < } ,

endowed with the norm

u H 1 / 2 = ( u L 2 2 + ( - Δ ) 1 / 4 u L 2 2 ) 1 / 2 .

We suppose that the potential V: is bounded and satisfies the following hypotheses:

  1. V is locally Hölder continuous and there exists V0>0 such that

    V ( z ) V 0 for all z .

  2. There exists a bounded interval Λ such that

    V 0 inf Λ V ( z ) < min Λ V ( z ) .

The function f: satisfies the so-called Ambrosetti–Rabinowitz condition, introduced in [5], namely,

(AR) there exists θ > 2 with 0 < θ F ( s ) s f ( s ) for all s > 0 ,

where F(s)=0sf(t)𝑑t. In addition to the above condition we make the following assumptions on f:

  1. f:+ is a C1 function with f(s)=0 if s<0.

  2. f(s)=o(s) near the origin.

  3. f(s)/s is an increasing function in +.

  4. There exist constants p>2 and Cp>0 such that

    f ( s ) C p s p - 1 for all s > 0 ,

    where

    C p > [ β p ( 2 θ θ - 2 ) 1 min { 1 , V 0 } ] ( p - 2 ) / 2 ,

    with

    β p = inf 𝒩 0 J ~ 0 , 𝒩 0 = { v X 1 ( + 2 ) { 0 } : J ~ 0 ( v ) v = 0 }

    and

    J ~ 0 ( v ) = 1 2 + 2 | v | 2 𝑑 x 𝑑 y + 1 2 V 0 | v ( x , 0 ) | 2 𝑑 x - 1 p | v ( x , 0 ) | p 𝑑 x ,

    where X1(+2) is defined in (2.1).

We are interested in a bound state solution of () (solution with finite energy), when f has the maximal growth, which allows us to treat problem () variationally in the fractional Sobolev space H1/2() motivated by the following Trudinger–Moser type inequality due to T. Ozawa [26].

Theorem 1.1

There exists 0<ωπ such that, for all α(0,ω), there exists Hα>0 with

(1.1) ( e α u 2 - 1 ) 𝑑 x H α u L 2 2

for all uH1/2(R) with (-Δ)1/4uL221.

In view of (1.1), we say that f has exponential critical growth at +, if there exist ω(0,π) and α0(0,ω), such that

lim s + f ( s ) e α s 2 = 0 for all α > α 0 , and lim s + f ( s ) e α s 2 = + for all α < α 0 .

1.1 Statement of the Main Result

The following theorem contains our main result.

Theorem 1.2

Assume (V1), (V2), (AR), and (f1)(f4) hold. Then there exists ε0>0 such that for ε(0,ε0), problem () possesses a positive bound state solution uε(z) verifying the following conditions:

  1. uε has at most one local (hence global) maximum zε in and zεI.

  2. limε0+V(zε)=V1=infIV.

Theorem 1.2 may be considered as the extension of the main result for the Laplacian in [20] to the case of the square root of the Laplacian. The proof is made combining the Ozawa inequality [26] with the Del Pino and Felmer truncation argument [17] and a recent approach developed by Alves and Miyagaki [4]. In [16, 27, 29, 12], existence results in non-local situation were established, while in [21, 28, 13, 14], concentration phenomena were proved imposing a global condition in V. Recently, Felmer and Torres [23] studied a new class of the problem involving a non-linear Schrödinger equation with non-local regional diffusion, which generalizes, for instance, the fractional Laplacian operator. In [31], Zhang, do Ó and Squassina treated some problems involving fractional Schrödinger–Poisson systems.

Remark 1.3

(i) We recall that condition (AR) imposes some superquadratic growth condition on the non-linearity F.

(ii) Condition (f4) appeared first in [11], then for instance in [2, 20]. For the non-local situation it was used, e.g., in [19].

(iii) Critical growth of Trudinger–Moser type was used in [15] and also in [20, 1, 2]. The Ozawa inequality to discuss the non-local problem in bounded and unbounded domain was used in [25] and [19], respectively.

(iv) Notice that, if f(s) has exponential critical growth, instead of assumption (f4), it is enough to assume that there exist p>2 and μ>0 such that

lim inf s 0 + f ( s ) s p - 1 μ .

Throughout the paper, unless explicitly stated, the symbol C will denote a generic positive constant, which may vary from line to line.

1.2 Outline

The sequel of the paper is organized as follows. Section 2 contains some technical results, which are crucial tools to prove our main theorem. In Section 3, we adapt a method due to L. Caffarelli and L. Silvestre to obtain a local realization of the fractional Laplacian via a Dirichlet-to-Neumann operator. As a consequence of this argument we transform our non-local problem () into one local problem (LPϵ) defined on the upper half plane. Using variational techniques combined with Del Pino and Felmer’s truncation argument, we prove Theorem 1.2 in Section 4.

2 Preliminary Results

In this section we collect preliminary facts for future reference. First of all, let us set the standard notations to be used in the paper. We denote the upper half-space in 2 by +2={(x,y)2:y>0}. In the sequel, X1(+2) denotes the completion of C0(+2¯) with relation to the norm vε:

(2.1) X 1 ( + 2 ) := C 0 ( + 2 ¯ ) ¯ ε ,

where

v ε := ( + 2 | v ( x , y ) | 2 𝑑 x 𝑑 y + V ( ε x ) | v ( x , 0 ) | 2 𝑑 x ) 1 / 2 .

Moreover, we denote by the usual norm in X1(+2), that is,

v = ( + 2 | v ( x , y ) | 2 𝑑 x 𝑑 y + | v ( x , 0 ) | 2 𝑑 x ) 1 / 2 .

Since the potential V is bounded from above and below, it is easy to see that ε and are equivalent norms in X1(+2) with

min { 1 , V 0 } v v ε min { 1 , | V | } v for all v X 1 ( + 2 ) .

Using the above definition, we see that if vX1(+2), then u(x)=v(x,0) belongs to H1/2() and

v = u H 1 / 2 .

Since H1/2() is continuously embedded into Lq() for all q2, cf. [18, Theorem 6.9], it follows that X1(+2) is also continuously embedded into Lq() for all q2. Moreover, the embedded

X 1 ( + 2 ) L q ( A )

are compact for any bounded measurable set A, see [24, Proposition 3.6] and [19, Remark 2.1].

Our first lemma is an important Trudinger–Moser inequality on X1(+2), which was proved in [19, Lemma 2.4].

Lemma 2.1

Let (vn)X1(R+2) be a bounded sequence and assume supnNvn2=M. Then

sup n ( e α | v n ( x , 0 ) | 2 - 1 ) 𝑑 x < for every 0 < α < ω M 2 .

In particular, if M(0,1), then there exists αM<ω such that

sup n ( e α M | v n ( x , 0 ) | 2 - 1 ) 𝑑 x < .

Using the above lemma, we are able to prove some technical lemmas. The first of them is crucial in the study of the behavior of Palais–Smale sequences.

Lemma 2.2

Let (vn) be a sequence in X1(R+2) with

(2.2) lim sup n + v n 2 < 1 .

Then, there exist t>1 sufficiently close to 1 and C>0 satisfying

( e ω | v n ( x , 0 ) | 2 - 1 ) t 𝑑 x C for all n .

Proof.

Using (2.2), there are m>0 and n0 verifying

v n 2 < m < 1 for all n n 0 .

Fix t>1 sufficiently close to 1 and β>t satisfying βm<1. Then, there exists C=C(β)>0 such that

( e ω | v n ( x , 0 ) | 2 - 1 ) t 𝑑 x C ( e β m ω ( | v n ( x , 0 ) | / v n ) 2 - 1 ) 𝑑 x

for every nn0. Hence, by Lemma 2.1,

( e ω | v n ( x , 0 ) | 2 - 1 ) t 𝑑 x C 1 for all n n 0

for some positive constant C1. Now, the lemma follows fixing

C = max { C 1 , ( e ω | v 1 | 2 - 1 ) t 𝑑 x , , ( e ω | v n 0 | 2 - 1 ) t 𝑑 x } .

Corollary 2.3

Let (vn) be a sequence in X1(R+2) satisfying (2.2). If vnv weakly in X1(R+2) and vn(x,0)v(x,0) a.e in R, as n, then,

(2.3) F ( v n ( x , 0 ) ) F ( v ( x , 0 ) ) in L 1 ( - R , R ) ,
(2.4) f ( v n ( x , 0 ) ) v n ( x , 0 ) f ( v ( x , 0 ) ) v ( x , 0 ) in L 1 ( - R , R ) ,
(2.5) - R R f ( v n ( x , 0 ) ) ϕ ( x , 0 ) - R R f ( v ( x , 0 ) ) ϕ ( x , 0 ) ,

as n for all ϕX1(R+2) and R>0.

Proof.

By (f1), for each β>1 and α>α0, there is C>0 such that

| F ( s ) | C ( | s | 2 + ( e α β | s | 2 - 1 ) ) for all s ,

from where it follows that

(2.6) | F ( v n ( x , 0 ) ) | C ( | v n ( x , 0 ) | 2 + ( e α β | v n ( x , 0 ) | 2 - 1 ) ) for all n .

Setting

h n ( x ) = C ( e α 0 β | v n ( x , 0 ) | 2 - 1 ) ,

we can fix β,q>1 sufficiently close to 1 and α sufficiently close to α0 such that

h n L q ( ) and sup n | h n | q < + ,

which is an immediate consequence of Lemma 2.2. Therefore, up to subsequence, we derive that

h n h = C ( e α 0 β | v ( x , 0 ) | 2 - 1 ) weakly in L q ( ) as n .

Since hn,h0, the last limit yields

h n h in L 1 ( - R , R ) for all R > 0 as n .

On the other hand, we know that

v n ( , 0 ) v ( , 0 ) in L 2 ( - R , R ) as n .

Gathering the above limits with (2.6), we get (2.3):

F ( v n ( x , 0 ) ) F ( v ( x , 0 ) ) in L 1 ( - R , R ) for all R > 0 as n ,

Limits (2.4) and (2.5) follow with the same type of arguments. ∎

The next lemma is a Lions type result, which is crucial in our approach. Since it follows with the same arguments found in [2, Proposition 2.3], we will omit its proof.

Lemma 2.4

Let (vn)X1(R+2) be a sequence with lim supn+vn2<1. If there is R>0 such that

lim n + sup z z - R z + R | v n ( x , 0 ) | 2 𝑑 x = 0 ,

then

lim n + F ( v n ( x , 0 ) ) 𝑑 x = lim n + f ( v n ( x , 0 ) ) v n ( x , 0 ) 𝑑 x = 0 .

3 Caffarelli and Silvestre’s Method

First of all, using the change of variable u(x)=v(εx), it is possible to prove that problem () is equivalent to the problem

(P’ϵ) { ( - Δ ) 1 / 2 u + V ( ε z ) u = f ( u ) in , u H 1 / 2 ( ) , u > 0 on .

Hereafter, to get a solution for (P’ϵ), we will use a version of the Caffarelli–Silvestre extension [10] due to Frank and Lenzmann [24] defined on the whole real line. In both papers, a local interpretation of the fractional Laplacian given in was developed considering a Dirichlet-to-Neumann type operator in the domain +2={(x,t)2:t>0}. For a similar extension in a bounded domain see, e.g., [6, 8, 9]. For uH1/2(), the solution wX1(+2) of

{ - div ( w ) = 0 in + 2 , w = u on × { 0 }

is called 1/2-harmonic extension w=E1/2(u) of u, and it is proved in [10] that

lim y 0 + w y ( x , y ) = - ( - Δ ) 1 / 2 u ( x ) .

To get a solution for the non-local problem (P’ϵ), we will study the existence of solutions for the local problem defined on the upper half plane:

(LPϵ) { - div ( w ) = 0 in + 2 , - w ν = - V ( ε x ) w + f ( w ) on × { 0 } ,

where

w ν = lim y 0 + w y ( x , y ) ,

since if w is a solution for (LPϵ), the function u(x)=w(x,0) is a solution for (P’ϵ).

Associated with (LPϵ), we have Jε:X1(+2) defined by

J ε ( v ) = 1 2 + 2 | v | 2 𝑑 x 𝑑 y + 1 2 V ( ε x ) | v ( x , 0 ) | 2 𝑑 x - F ( v ( x , 0 ) ) 𝑑 x ,

which is C1(X1(+2),) with derivative given by

J ε ( v ) ϕ = 1 2 + 2 v . ϕ d x d y + 1 2 V ( ε x ) v ( x , 0 ) ϕ ( x , 0 ) 𝑑 x - f ( v ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x for all ϕ X 1 ( + 2 ) .

We would like to point out that u is a solution of (P’ϵ) if, and only if, u=v(x,0) for all x, for some critical point v of Jε.

In what follows, we will not work directly with functional Jε, because we have some difficulties to prove that it verifies the Palais–Smale compactness condition. We recall that a C1-functional Ψ:X defined on a Banach space X satisfies the Palais–Smale condition at level c ((PS)c condition for short), if each sequence (un)X such that (i) Ψ(un)c and (ii) Ψ(un)0 in X* is relatively compact in X. Finally, any sequence (un) satisfying (i) and (ii) is called a Palais–Smale sequence at level c (a (PS)c for short), see [30].

Hereafter, we will use the same approach explored in [17], modifying the non-linearity of a suitable way. The idea is the following:

First of all, without loss of generality, we will assume that

0 Λ and V ( 0 ) = V 0 = inf x V ( x ) .

We recall that in assumption (f1) we imposed that f(t)=0 for all t0, because we are looking for positive solutions. Moreover, let us choose k>2θ/(θ-2) and a>0 verifying

f ( a ) a = V 0 k ,

where V0>0 was given in (V1). Using these numbers, we set the functions

f ~ ( t ) = { f ( t ) if t a , V 0 k t if t a

and

g ( x , t ) = χ Λ ( x ) f ( t ) + ( 1 - χ Λ ) f ~ ( t ) for all ( x , t ) 2 ,

where Λ was given in (V2) and χΛ denotes the characteristic function associated with Λ, that is,

χ ( x ) = { 1 if x Λ , 0 if x Λ c .

Using the above functions, we will study the existence of positive solutions for the following problem:

(AP) { ( - Δ ) 1 / 2 u + V ( ε x ) u = g ε ( x , u ) , x , u H 1 / 2 ( ) ,

where

g ε ( x , t ) = g ( ε x , t ) for all ( x , t ) 2 .

We recall from [10] that, to get a solution for problem (AP), it is enough to study the existence of solutions for the following problem:

(AP’) { - div ( w ) = 0 in + 2 , w ν = V ( ε x ) w - g ε ( x , w ) on × { 0 } ,

because if w is a solution of (AP’), the function u(x)=w(x,0) is a solution for (AP).

Here, we would like to point out that if vεX1(+2) is a solution of (AP’) with

v ε ( x , 0 ) < a for all x Λ ε c ,

where Λε=Λ/ε, then uε(x)=vε(x,0) is a solution of (P’ϵ).

Associated with (AP’), we have the energy functional Eε:X1(+2) given by

E ε ( v ) = 1 2 + 2 | v | 2 𝑑 x 𝑑 y + 1 2 V ( ε x ) | v ( x , 0 ) | 2 𝑑 x - G ε ( x , v ( x , 0 ) ) 𝑑 x ,

where

G ε ( x , t ) = 0 t g ε ( x , τ ) 𝑑 τ for all ( x , t ) 2 .

Using the definition of g, it follows that

(3.1) θ G ε ( x , t ) g ε ( x , t ) t for all ( x , t ) Λ ε × ,

and

(3.2) 2 G ε ( x , t ) g ε ( x , t ) t V 0 k | t | 2 for all ( x , t ) ( Λ ε ) c × .

From assumption (3.2), we deduce

(3.3) L ( x , t ) = V ( x ) - G ε ( x , t ) ( 1 - 1 2 k ) V ( x ) | t | 2 0 for all ( x , t ) ( Λ ε ) c × ,

and

(3.4) M ( x , t ) = V ( x ) - g ε ( x , t ) t ( 1 - 1 k ) V ( x ) | t | 2 0 for all ( x , t ) ( Λ ε ) c × .

Lemma 3.1

The functional Eε verifies the mountain pass geometry, that is,

  1. there are r , ρ > 0 such that E ε ( v ) ρ for v = r ;

  2. there is eX1(+2) with e>r and Eε(e)<0.

Proof.

By (3.1)–(3.4), there exist c1,c2>0 verifying

E ε ( v ) c 1 v 2 - c 2 v q for all v X 1 ( + 2 ) .

From the above inequality, we deduce that there are r,ρ>0 such that

E ε ( v ) ρ for v 1 , s = r ,

showing (i).

To prove (ii), fix φX1(+2) with suppφΛε×. Then, for t>0,

E ε ( t φ ) = t 2 2 φ 2 - F ( t φ ( x , 0 ) ) 𝑑 x .

From (f3), we know that there are c3,c40 verifying

F ( t ) c 1 | t | θ - c 2 for all t 0 .

Using the above inequality, we derive

lim t + E ε ( t φ ) = - .

Thereby, (ii) follows with e=tφ and t large enough. ∎

In what follows, we denote by cε the mountain pass level associated with Eε. Related to the case ε=0, it is possible to prove that there is w0X1(+2) such that

J 0 ( w 0 ) = c 0 and J 0 ( w 0 ) = 0 .

The existence of w0 can be obtained repeating the same approach explored in [2].

Lemma 3.2

The minimax level c0 verifies

0 < c 0 < min { 1 , V 0 } ( 1 2 - 1 θ ) .

Proof.

Consider w*X1(+2) verifying

J ~ 0 ( w * ) = β p and J ~ 0 ( w * ) = 0 .

By characterization of c0,

c 0 max t 0 J 0 ( t w * ) .

Consequently, by (f4),

c 0 max t 0 { t 2 2 + 2 | w * | 2 𝑑 x 𝑑 y + 1 2 V 0 | w * ( x , 0 ) | 2 𝑑 x - C p t p p | w * ( x , 0 ) | p 𝑑 x } ,

which implies that

c 0 C p 2 / ( 2 - p ) β p .

Hence, by (f4),

0 < c 0 < min { 1 , V 0 } ( 1 2 - 1 θ ) .

Hereafter, we will assume that k is large enough such that

0 < c 0 < min { 1 , V 0 } ( ( 1 2 - 1 θ ) - 1 k ) < min { 1 , V 0 } ( 1 2 - 1 θ ) .

The next lemma establishes an important relation between cε and c0.

Lemma 3.3

The numbers c0 and cε verify the equality

(3.5) lim ε 0 c ε = c 0 .

Hence, there is ε0>0 such that

(3.6) 0 < sup ε ( 0 , ε 0 ) c ε < min { 1 , V 0 } ( ( 1 2 - 1 θ ) - 1 k ) .

Proof.

By (V1), cεc0 for all ε0. Then,

(3.7) lim inf ε 0 c ε c 0 .

Next, fix tε>0 such that

t ε w ε = { v X 1 ( + 2 ) { 0 } : E ε ( v ) v = 0 } .

By definition of cε, we know that

c ε max t 0 E ε ( t w ) = E ε ( t ε w ) .

Using standard arguments as those in [20], we can prove

lim ε 0 t ε = 1 and lim ε 0 E ε ( t ε w ) = J 0 ( w ) .

Thus,

(3.8) lim sup ε 0 c ε J 0 ( w ) = c 0 .

By (3.7) and (3.8), we have

lim sup ε 0 c ε = c 0 ,

showing (3.5). Inequality (3.6) is an immediate consequence of (3.5) and Lemma 3.2. ∎

Lemma 3.4

Let ε(0,ε0) and let (vn)X1(R+2) be a (PS)cε sequence for Eε. Then,

(3.9) lim sup n + v n 2 < 1 .

Proof.

Gathering Eε(un)-1θEε(un)un=cε+on(1) with the definition of g, we find

( 1 2 - 1 θ ) + 2 | v n | 2 𝑑 x 𝑑 y + ( ( 1 2 - 1 θ ) - 1 k ) V 0 | v n ( x , 0 ) | 2 𝑑 x c ε + o n ( 1 ) ,

from where it follows that

min { 1 , V 0 } ( ( 1 2 - 1 θ ) - 1 k ) lim sup n + v n 2 c ε < min { 1 , V 0 } ( ( 1 2 - 1 θ ) - 1 k ) ,

and thus (3.9). ∎

Lemma 3.5

For ε(0,ε0), the functional Eε verifies the (PS)cε condition.

Proof.

Let (vn)X1(+2) be a (PS)cε sequence for Eε, that is,

E ε ( v n ) c ε and E ε ( v n ) 0 as n .

By Lemma 3.4, (vn) is bounded in X1(+2) and lim supn+vn2<1. Since X1(+2) is reflexive, there are a subsequence of (vn), still denoted by itself, and vX1(+2) such that

v n v weakly in X 1 ( + 2 ) as n ,
v n v in L loc q ( ) for all q [ 2 , + ) as n ,
v n ( x , 0 ) v ( x , 0 ) a.e. in as n .

Moreover, by Corollary 2.3,

f ( v n ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x f ( v ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x

as n for all ϕC0(+2¯).

Using the above limits, it is possible to prove that v is a critical point for Eε, that is,

E ε ( v ) φ = 0 for all φ X 1 ( + 2 ) .

Considering φ=v, we have Eε(v)v=0, and so,

+ 2 | v | 2 𝑑 x 𝑑 y + Λ ε V ( ε x ) | v ( x , 0 ) | 2 𝑑 x + ( Λ ε ) c M ( x , v ( x , 0 ) ) 𝑑 x = Λ ε f ( v ( x , 0 ) ) v ( x , 0 ) 𝑑 x .

On the other hand, using the limit Eε(vn)vn=on(1), we derive that

+ 2 | v n | 2 𝑑 x 𝑑 y + Λ ε V ( ε x ) | v n ( x , 0 ) | 2 𝑑 x + ( Λ ε ) c M ( x , v n ( x , 0 ) ) 𝑑 x = Λ ε f ( v n ( x , 0 ) ) v n ( x , 0 ) 𝑑 x + o n ( 1 ) .

Since Λε is bounded, by the compactness of the Sobolev embedding and Corollary 2.3, we have

lim n + Λ ε f ( v n ( x , 0 ) ) v n ( x , 0 ) 𝑑 x = Λ ε f ( v ( x , 0 ) ) v ( x , 0 ) 𝑑 x

and

(3.10) lim n + Λ ε V ( ε x ) | v n ( x , 0 ) | 2 𝑑 x = Λ ε V ( ε x ) | v ( x , 0 ) | 2 𝑑 x .

Therefore,

lim sup n + ( + 2 | v n | 2 𝑑 x 𝑑 y + ( Λ ε ) c M ( x , v n ( x , 0 ) ) 𝑑 x ) = + 2 | v | 2 𝑑 x 𝑑 y + ( Λ ε ) c M ( x , v ( x , 0 ) ) 𝑑 x .

Now, recalling that M(x,t)0, Fatou’s lemma leads to

lim inf n + ( + 2 | v n | 2 𝑑 x 𝑑 y + ( Λ ε ) c M ( x , v n ( x , 0 ) ) 𝑑 x ) + 2 | v | 2 𝑑 x 𝑑 y + ( Λ ε ) c M ( x , v ( x , 0 ) ) 𝑑 x .

Hence,

(3.11) lim n + + 2 | v n | 2 𝑑 x 𝑑 y = + 2 | v | 2 𝑑 x 𝑑 y

and

lim n + ( Λ ε ) c M ( x , v ( x , 0 ) ) 𝑑 x = ( Λ ε ) c M ( x , v ( x , 0 ) ) 𝑑 x .

The last limit combined with the definition of function M gives

lim n + ( Λ ε ) c V ( ε x ) | v n ( x , 0 ) | 2 𝑑 x = ( Λ ε ) c V ( ε x ) | v ( x , 0 ) | 2 𝑑 x .

Gathering this limit with (3.10), we deduce that

(3.12) lim n + V ( ε x ) | v n ( x , 0 ) | 2 𝑑 x = V ( ε x ) | v ( x , 0 ) | 2 𝑑 x .

By (3.11) and (3.12),

lim n + v n ε 2 = v ε 2 .

Since X1(+2) is a Hilbert space and vnv weakly in X1(+2) as n, the above limit yields

v n v in X 1 ( + 2 ) as n ,

showing that Eε verifies the (PS)cε condition. ∎

Theorem 3.6

For ε(0,ε0), the functional Eε has a non-negative critical point vεX1(R+2) such that

(3.13) E ε ( v ε ) = c ε 𝑎𝑛𝑑 E ε ( v ε ) = 0 .

Proof.

By Lemma 3.3, there is ε0>0 such that Eε verifies the (PS)cε condition for ε(0,ε0). Then, the existence of vε is an immediate consequence of the Mountain Pass Theorem due to Ambrosetti and Rabinowitz (see, e.g., [30]). The function vε is non-negative, because Eε(vε)(vε-)=0 implies vε-=0, where vε-=min{vε,0}. ∎

Lemma 3.7

Decreasing ε0, if necessary, there are r,β>0 and (yε)R such that

(3.14) y ε - r y ε + r | v ε ( x , 0 ) | 2 𝑑 x β for all ε ( 0 , ε 0 ) .

Proof.

First of all, we recall that since (vε) satisfies (3.13), there is α>0, which is independent of ε, such that

(3.15) v ε ε 2 α for all ε > 0 .

To show (3.14), it is enough to prove that, for any sequence (εn)(0,+) with εn0, the limit

lim n + sup y y - r y + r | v ε n ( x , 0 ) | 2 𝑑 x = 0 ,

does not hold for any r>0. Otherwise, if it holds for some r>0, by Lemma 2.4,

f ( v ε n ( , 0 ) ) v n ( x , 0 ) 𝑑 x 0 as n ,

implying that vεnε20 as n+, which contradicts (3.15). ∎

Lemma 3.8

For any εn0, consider the sequence (yεn)R given in Lemma 3.7 and ψn(x,y)=vεn(x+yεn,y). Then, up to subsequence, there is ψX1(R+2) such that

(3.16) ψ n ψ in X 1 ( + 2 ) as n .

Moreover, there is x0Λ such that

(3.17) lim n 0 ε n y ε n = x 0 𝑎𝑛𝑑 V ( x 0 ) = V 0 .

Proof.

We begin the proof showing that (εnyεn) is a bounded sequence. Hereafter, we denote by (yn) and (vn) the sequences (yεn) and (vεn), respectively.

Since Eεn(vn)ϕ=0 for all ϕX1(+2), we have that

+ 2 v n ϕ d x d y + V ( ε n x ) v n ( x , 0 ) ϕ ( x , 0 ) 𝑑 x - g ε ( x , v n ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x = 0 .

Then,

+ 2 | v n | 2 𝑑 x 𝑑 y + V ( ε n x ) | v n ( x , 0 ) | 2 𝑑 x - g ε ( x , v n ( x , 0 ) ) v n ( x , 0 ) 𝑑 x = 0 .

From the definition of g, we see that

g ε ( x , t ) f ( t ) for all t 0 .

Recalling that vn0, we infer that

+ 2 | v n | 2 𝑑 x 𝑑 y + V 0 | v n ( x , 0 ) | 2 𝑑 x - f ( v n ( x , 0 ) ) v n ( x , 0 ) 𝑑 x 0 .

Therefore, there is sn(0,1) such that

s n v n 0 = { v X 1 ( + 2 ) { 0 } : J 0 ( v ) v = 0 } .

Using the characterization of c0, we know that

c 0 J 0 ( s n v n ) for all n .

As

J 0 ( w ) E ε ( w ) for all w X 1 ( + 2 ) , ε > 0 ,

it follows that

c 0 J 0 ( s n v n ) E ε n ( s n v n ) max s 0 E ε n ( s v n ) = E ε n ( v n ) = c ε n .

Recalling that cεnc0 as n, the last inequality gives

( s n v n ) 0 for all n , and J 0 ( s n v n ) c 0 as n .

By change of variable, we also have

( s n ψ n ) 0 for all n , and J 0 ( s n ψ n ) c 0 as n .

Using Ekeland’s variational principle, we can assume that (snvn) is a (PS)c0 sequence, that is,

( s n ψ n ) 0 for all n , J 0 ( s n ψ n ) c 0 and J 0 ( s n ψ n ) 0 as n .

A direct computation shows that (sn) is a bounded sequence with

lim inf n + s n > 0 .

Thus, in what follows, we can assume that for some subsequence, there is s0>0 such that sns0 as n. From definition of yn and ψn, we know that ψX1(+2){0}. Moreover, since J0(snψn)0, we also have J0(s0ψ)=0. Thereby, by definition of c0, we obtain

c 0 J 0 ( s 0 ψ ) .

On the other hand, by Fatou’s lemma we obtain

lim inf n + J 0 ( s n ψ n ) J 0 ( s 0 ψ )

which implies

J 0 ( s 0 ψ ) = c 0 and J 0 ( s 0 ψ ) = 0 .

The above equalities combined with Fatou’s lemma, up to a subsequence, give

s n ψ n s 0 ψ in X 1 ( + 2 ) as n .

Recalling that sns0>0 as n, we can conclude that

ψ n ψ in X 1 ( + 2 ) as n ,

showing (3.16).

Using the last limit, we are able to prove (3.17). First, we prove the following fact:

Claim 3.9

limn+dist(εnyn,Λ¯)=0.

Indeed, if the claim does not hold, there is δ>0 and a subsequence of (εnyn), still denoted by itself, such that

dist ( ε n y n , Λ ¯ ) δ for all n .

Consequently, there is r>0 such that

( ε n y n - r , ε n y n + r ) Λ c for all n .

From definition of ψn, we have that

+ 2 | ψ n | 2 𝑑 x 𝑑 y + V ( ε n x + ε n y n ) | ψ n ( x , 0 ) | 2 𝑑 x = g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ψ n ( x , 0 ) 𝑑 x .

Note that

g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ψ n ( x , 0 ) 𝑑 x
- r / ε n r / ε n g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ψ n ( x , 0 ) 𝑑 x + ( - - r / ε n + r / ε n + ) g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ψ n ( x , 0 ) d x ,

and so,

g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ψ n ( x , 0 ) 𝑑 x
V 0 k - r / ε n r / ε n | ψ n ( x , 0 ) | 2 𝑑 x + ( - - r / ε n + r / ε n + ) f ( ψ n ( x , 0 ) ) ψ n ( x , 0 ) d x .

Therefore,

+ 2 | ψ n | 2 𝑑 x 𝑑 y + V ( ε n x + ε n y n ) | ψ n ( x , 0 ) | 2 𝑑 x
V 0 k - r / ε n r / ε n | ψ n ( x , 0 ) | 2 𝑑 x + ( - - r / ε n + r / ε n + ) f ( ψ n ( x , 0 ) ) ψ n ( x , 0 ) d x ,

implying that

(3.18) + 2 | ψ n | 2 𝑑 x 𝑑 y + A N | ψ n ( x , 0 ) | 2 𝑑 x ( - - r / ε n + r / ε n + ) f ( ψ n ( x , 0 ) ) ψ n ( x , 0 ) d x ,

where A=V0(1-1k). By (3.16), we have

( - - r / ε n + r / ε n + ) f ( ψ n ( x , 0 ) ) ψ n ( x , 0 ) d x 0 as n ,

and

+ 2 | ψ n | 2 𝑑 x 𝑑 y + A | ψ n ( x , 0 ) | 2 𝑑 x + 2 | ψ | 2 𝑑 x 𝑑 y + A | ψ ( x , 0 ) | 2 𝑑 x > 0 as n ,

which contradicts (3.18). This proves Claim 3.9.

By Claim 3.9, there are a subsequence of (εnyn) and x0Λ¯ such that

lim n + ε n y n = x 0 .

Claim 3.10

x0Λ.

Indeed, from definition of ψn, we get

+ 2 | ψ n | 2 𝑑 x 𝑑 y + V ( ε n x + ε n y n ) | ψ n ( x , 0 ) | 2 𝑑 x f ( ψ n ( x , 0 ) ) ψ n ( x , 0 ) 𝑑 x .

Then, by (3.16),

+ 2 | ψ | 2 𝑑 x 𝑑 y + V ( x 0 ) | ψ ( x , 0 ) | 2 𝑑 x f ( ψ ( x , 0 ) ) ψ ( x , 0 ) 𝑑 x .

Hence, there is s1(0,1) such that

s 1 ψ V ( x 0 ) = { v X 1 ( + 2 ) { 0 } : J ~ V ( x 0 ) v = v } ,

where J~V(x0):X1(+2) is given by

J ~ V ( x 0 ) ( v ) = 1 2 + 2 | v | 2 𝑑 x 𝑑 y + 1 2 V ( x 0 ) | v ( x , 0 ) | 2 𝑑 x - F ( v ( x , 0 ) ) 𝑑 x .

If c~V(x0) denotes the mountain pass level associated with J~V(x0), we must have

c ~ V ( x 0 ) J ~ V ( x 0 ) ( s 1 ψ ) lim inf n + E ε n ( v n ) = lim inf n + c ε n = c 0 = c ~ V ( 0 ) .

Hence,

c ~ V ( x 0 ) c ~ V ( 0 ) ,

from where it follows that

V ( x 0 ) V ( 0 ) .

As V0=infxV(x), the above inequality implies that

V ( x 0 ) = V ( 0 ) = V 0 .

Moreover, by (V2), we have x0Λ. Then, x0Λ, finishing the proof. ∎

Corollary 3.11

Let (ψn) be the sequence given in Lemma 3.8. Then, ψn(,0)L(R) and there is K>0 such that

| ψ n ( , 0 ) | K for all n

and

(3.19) ψ n ( , 0 ) ψ ( , 0 ) in L p ( ) for all p ( 2 , + ) as n .

As an immediate consequence, the sequence hn(x)=g(εnx+εnyn,ψn(x,0)) must verify

(3.20) h n f ( ψ ( , 0 ) ) in L p ( ) for all p ( 2 , + ) as n .

Proof.

In what follows, for each L>0, we set

ψ n , L ( x , y ) = { ψ n ( x , y ) if ψ n ( x , y ) L , L if ψ n ( x , y ) L , and z n , L = ψ n , L 2 ( β - 1 ) ψ n

with β>1 to be determined later. Since

+ 2 ψ n ϕ d x d y + V ( ε n x + ε n y n ) ψ n ( x , 0 ) ϕ ( x , 0 ) 𝑑 x
- g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x = 0 for all ϕ X 1 ( + 2 ) , n ,

adapting the same approach explored in [3, Lemma 4.1] and using the fact that (ψn) is bounded in X1(+2), we conclude that there is K>0 such that

| ψ n ( , 0 ) | K for all n .

Now, the limit (3.19) is obtained by interpolation on the Lp spaces, while (3.20) follows combining the growth condition on g with (3.19). ∎

In what follows, we denote by (wn)H1/2() the sequence (ψn(,0)), that is,

w n ( x ) = ψ n ( x , 0 ) for all x .

Since

+ 2 ψ n ϕ d x d y + V ( ε n x + ε n y n ) ψ n ( x , 0 ) ϕ ( x , 0 ) 𝑑 x
- g ( ε n x + ε n y n , ψ n ( x , 0 ) ) ϕ ( x , 0 ) 𝑑 x = 0 for all ϕ X 1 ( + 2 ) ,

we have that wn is a solution of the problem

( - Δ ) 1 / 2 w n + V ( ε n x + ε n y n ) w n = g ( ε n x + ε n y n , w n ) in ,

or equivalently

(3.21) ( - Δ ) 1 / 2 u + w n = χ n in ,

where

χ n ( x ) = w n ( x ) + g ( ε n x + ε n y n x , w n ( x ) ) - V ( ε n x + ε n y n ) w n ( x ) , x .

Denoting χ(x)=w(x)+f(w(x))-V(x0)w(x), we deduce from Corollary 3.11 that

χ n χ in L p ( ) for all p [ 2 , + ) as n ,

and that there is k1>0 such that

| χ n | k 1 for all n .

Motivated by some results found in [7] (see also [22]), which hold for the whole line, we deduce that

w n ( x ) = ( 𝒦 χ n ) ( x ) = 𝒦 ( x - y ) χ n ( y ) 𝑑 y ,

where 𝒦 is the Bessel kernel which verifies:

  1. 𝒦 is positive and even on {0}.

  2. There is C>0 such that 𝒦(x)C/|x|2 for all x{0}.

  3. 𝒦Lq() for all q[1,].

Using the above information, we are able to prove the following result.

Lemma 3.12

The sequence (wn) verifies wn(x)0 as |x|+, uniformly in nN.

Proof.

Given δ>0, we have

0 w n ( x ) 𝒦 ( x - y ) | χ n | ( y ) 𝑑 y = ( - x - 1 / δ + x + 1 / δ + ) 𝒦 ( x - y ) | χ n | ( y ) d y + x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ n | ( y ) 𝑑 y .

By (K2), we have that, for all n,

( - x - 1 / δ + x + 1 / δ + ) 𝒦 ( x - y ) | χ n | ( y ) d y C δ 1 / 2 | χ n | ( - x - 1 / δ + x + 1 / δ + ) d y | x - y | 3 / 2
(3.22) C δ 1 / 2 k 1 ( - x - 1 + x + 1 + ) d y | x - y | 3 / 2 = C 1 δ 1 / 2 .

On the other hand,

x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ n | ( y ) 𝑑 y x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ n - χ | ( y ) 𝑑 y + x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ | ( y ) 𝑑 y .

Fix q>1 with q sufficiently close to 1 and q>2 such that 1/q+1/q=1. By (K2) and (3.21),

x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ n | ( y ) 𝑑 y | 𝒦 | q | χ n - χ | q + | 𝒦 | q | χ | L q ( x - 1 / δ , x + 1 / δ ) .

Since

| χ n - χ | q 0 as n +

and

| χ | L q ( x - 1 / δ , x + 1 / δ ) 0 as | x | + ,

we deduce that there are R>0 and n0 such that

(3.23) x - 1 / δ x + 1 / δ 𝒦 ( x - y ) | χ n | ( y ) 𝑑 y δ for all n n 0 , | x | R .

By (3.22) and (3.23),

𝒦 ( x - y ) | χ n | ( y ) 𝑑 y C 1 δ d + δ for all n n 0 , | x | R .

The same approach can be used to prove that for each n{1,,n0-1}, there is Rn>0 such that

𝒦 ( x - y ) | χ n | ( y ) 𝑑 y C 1 δ d + δ , | x | R n .

Hence, increasing R, if necessary, we must have

𝒦 ( x - y ) | χ n | ( y ) 𝑑 y C 1 δ d + δ for | x | R , uniformly in n .

Since δ is arbitrary, the proof is finished. ∎

Corollary 3.13

There is n0N such that

v n ( x , 0 ) < a for all n n 0 , x Λ ε n c .

Hence, un(x)=vn(x,0) is a solution of (Pεn) for nn0.

Proof.

By Lemma 3.8, we know that εnynx0 for some x0Λ. Therefore, there is r>0 such that up to subsequence we have

( r - ε n y n , r + ε n y n ) Λ for all n .

Hence,

( y n - r / ε n , y n + r / ε n ) Λ ε n for all n ,

or equivalently

Λ ε n c ( - , y n - r / ε n ) ( y n + r / ε n , + ) for all n .

Now, by Lemma 3.12, there is R>0 such that

w n ( x ) < a for all | x | R , n ,

from where it follows that

v n ( x , 0 ) = ψ n ( x - y n , 0 ) = w n ( x - y n ) < a for all x ( - , y n - R ) ( y n + R , + ) , n .

On the other hand, we have that

Λ ε n c ( - , y n - r / ε n ) ( y n + r / ε n , + ) for all n .

Thus, there is n0 such that

( - , y n - r / ε n ) ( y n + r / ε n , + ) ( - , y n - R ) ( y n + R , + ) for all n n 0 ,

implying that

v n ( x , 0 ) < a for all x Λ ε n c , n n 0 ,

finishing the proof. ∎

4 Proof of Theorem 1.2

By Theorem 3.6, we know that problem (AP) has a non-negative solution vε for all ε>0. Applying Corollary 3.13, there is ε0 such that

v ε ( x , 0 ) < a for all x Λ ε c , ε ( 0 , ε 0 ) ,

that is, vε(,0) is a solution of (P’ϵ) for ε(0,ε0). Thus,

u ε ( x ) = v ε ( x / ε , 0 ) for ε ( 0 , ε 0 )

is a solution for original problem ().

If xε denotes a global maximum point of uε, it is easy to see that there is τ0>0 such that

u ε ( x ε ) τ 0 for all ε > 0 .

In what follows, setting zε=(xε-εyε)ε-1, we have that zε is a global maximum point of wε and

w ε ( z ε ) τ 0 for all ε > 0 .

Now, we claim that

(4.1) lim ε 0 V ( x ε ) = V 0 .

Indeed, by Lemma 3.12, we know that

w ε n ( x ) 0 as | x | + uniformly in n .

Therefore, (zε) is a bounded sequence. Moreover, for some subsequence, we also know that there is x0Λ satisfying V(x0)=V0 and

ε n y ε n x 0 as n .

Hence,

x ε n = ε n z ε n + ε n y ε n x 0 as n ,

implying that

V ( x ε n ) V 0 as n ,

showing that (4.1) holds.


Communicated by Patrizia Pucci


Funding statement: The authors were supported by INCTMAT/CNPq/Brazil. C. O. Alves was supported by CNPq/Brazil Proc. 304036/2013-7. J. M. do Ó was supported by CNPq/Brazil Proc. 407099/2013-1. O. H. Miyagaki was supported by CNPq/Brazil Proc. 304015/2014-8 and CAPES/Brazil Proc. 2531/14-3.

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Received: 2016-07-09
Revised: 2016-08-26
Accepted: 2016-08-28
Published Online: 2016-10-11
Published in Print: 2016-11-01

© 2016 by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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