Open Access Published by De Gruyter Open Access August 29, 2022

# Ground state solution for some new Kirchhoff-type equations with Hartree-type nonlinearities and critical or supercritical growth

• Li Zhou and Chuanxi Zhu
From the journal Open Mathematics

## Abstract

In this article, we study two classes of Kirchhoff-type equations as follows:

a + b R 3 u 2 d x Δ u + V ( x ) u = ( I α u p ) u p 2 u + f ( u ) , in R 3 , u H 1 ( R 3 ) ,

and

a + b R 3 u 2 d x Δ u + V ( x ) u = ( I α u p ) u p 2 u + m u l 2 u , in R 3 , u H 1 ( R 3 ) ,

where a > 0 , b 0 , α ( 0 , 3 ) , ( 3 + α ) / 3 < p < ( 3 + α ) , l 6 , m > 0 , V : R 3 R is a potential function and I α is a Riesz potential whose order is α ( 0 , 3 ) . Under some assumptions on V ( x ) and f ( u ) , we can prove that the equations have ground state solutions by variational methods.

MSC 2010: 35J60; 35J35; 35A15

## 1 Introduction

In this article, we study the following two classes of Kirchhoff-type equations:

(1) a + b R 3 u 2 d x Δ u + V ( x ) u = ( I α u p ) u p 2 u + f ( u ) , in R 3 , u H 1 ( R 3 ) ,

and

(2) a + b R 3 u 2 d x Δ u + V ( x ) u = ( I α u p ) u p 2 u + m u l 2 u , in R 3 , u H 1 ( R 3 ) ,

where a > 0 , b 0 , α ( 0 , 3 ) , ( 3 + α ) / 3 < p < ( 3 + α ) , l 6 , m > 0 , I α is a Riesz potential whose order is α ( 0 , 3 ) . Here, I α = Γ 3 α 2 Γ α 2 π 3 2 2 α x 3 α . Besides, V ( x ) : R 3 R is a potential function satisfying:

1. V C 1 ( R 3 ) L , and there exists a constant A ( 0 , a ) such that

( V ( x ) , x ) A 2 x 2 ,

for all x R 3 { 0 } ,

2. there is a constant V > 0 such that for all x R 3 ,

0 < V ( x ) liminf y + V ( y ) = V < + ,

3. inf x R 3 V ( x ) V 0 > 0 .

Furthermore, we suppose that the function f C 1 ( R , R ) satisfies:
1. there exists a constant C 0 > 0 and q ( 2 , 6 ) such that f ( t ) C 0 ( 1 + t q 1 ) , t R ,

2. f ( t ) = o ( t ) as t 0 ,

3. lim t + F ( t ) t 2 = , where F ( t ) = 0 t f ( s ) d s ,

4. f ( t ) t is increasing on ( , 0 ) ( 0 , + ) .

In the past decades, many scholars have studied the existence of nontrivial solutions for the Kirchhoff-type problem:

(3) a + b R 3 u 2 d x Δ u + V ( x ) u = g ( x , u ) , in R 3 , u H 1 ( R 3 ) ,

where a > 0 , b 0 , V : R 3 R is a potential function and g C ( R 3 × R , R ) . Problem (3) is a nonlocal problem because of the presence of the term b R 3 u 2 d x , which causes some mathematical difficulties, but at the same time makes the research of this problem particularly interesting. Besides, this problem has an interesting physical context. In fact, if we set V ( x ) = 0 and replace R 3 by a bounded domain Ω R 3 in (3), then we obtain the following Kirchhoff Dirichlet problem:

a + b Ω u 2 d x Δ u = g ( x , u ) , x Ω , u = 0 , x Ω .

It has relation to the stationary analogue of the equation:

ρ 2 u t 2 ρ 0 h + E 2 L 0 L u x d x 2 u x 2 = 0 ,

which was proposed by G. Kirchhoff as an extension of classical D’Alemberts wave equations for free vibration of elastic strings. Kirchhoff’s model considers the changes in length of the string, which were produced by transverse vibrations. Then J. L. Lions finished the previous work. He introduced a functional analysis approach. After that, more and more researchers have paid much attention to the problem (3). But most of their results need to assume:

1. V verifies (V): inf x R 3 V ( x ) V 0 > 0 and for each M > 0 .

meas { x R N : V ( x ) M } < + ,

2. g satisfies classical Ambrosetti-Rabinowitz condition, i.e., ( A R ) condition: there exists μ > 2 such that

0 < μ G ( x , s ) s g ( x , s )

for all s > 0 .

In fact, (V) is sufficient to ensure that the embedding

u H 1 ( R 3 ) : R 3 V ( x ) u 2 d x < + L p ( R 3 ) , 2 p < 6

is compact.

Unfortunately, there still are very few results of existence of ground state solution to (3) without (A-R) condition (see [1,2,3]).

In [3], Guo studied the following Kirchhoff-type problem:

(4) a + b R 3 u 2 d x Δ u + V ( x ) u = f ( u ) , in R 3 . u H 1 ( R 3 ) .

He proved the existence of positive ground states to (4), and in his paper, he did not use (A-R) type condition. He defined a new manifold:

= u H 1 ( R 3 ) : 1 2 Φ ( u ) , u + P ( u ) = 0 ,

which is named the Nehari-Pohozaev manifold. Here,

Φ ( u ) = 1 2 R 3 [ a u 2 + V u 2 ] d x + b 4 R 3 u 2 d x 2 R 3 F ( u ) d x

and

P ( u ) = 1 2 R 3 [ a u 2 + 3 V u 2 ] d x + b 2 R 3 u 2 d x 2 3 R 3 F ( u ) d x

are the energy functional and Pohozaev identity for the “limit problem” of problem (4), respectively. He first applied the result, which was obtained for the related “limit problem” of (4) to obtain a minimizer for problem (4) on the Nehari-Pohozaev manifold.

We must point out that f C 1 and the fourth assumption about f are very important in Guo [3]. Actually, only under aforementioned assumptions, is a C 1 manifold.

Most remarkably, as early as 2006, Ruiz [4] first proposed the prototype of this Nehari-Pohozaev manifold in his study of the Schrödinger-Poisson equation, which is a very great work.

On the other hand, when a = 1 , b = 0 , f = 0 , equation (1) becomes

(5) Δ u + V ( x ) u = ( I α u p ) u p 2 u .

We usually call it nonlinear Choquard-type equation. Its physical background can be found in [5], and the references therein. Besides, readers can see [6,7,8, 9,10,11, 12,13] for recent achievements.

Inspired by the aforementioned works, especially by [3,13, 14,15], we now research problem (3) with Hartree-type nonlinearities g ( x , u ) = ( I α u p ) u p 2 u + f ( u ) , which may be regarded as a Kirchhoff-type perturbation to (5). As we all know, there are very few results to (3) with Hartree-type nonlinearities and critical or supercritical growth.

The main outcomes of our investigation are as follows.

## Theorem 1.1

If V satisfies ( V 1)–( V 2), f C 1 ( R , R ) verifies ( f 1)–( f 4), then problem (1) has a ground state solution.

## Theorem 1.2

If V satisfies (V1)–(V3), then there exists some m 0 > 0 such that for m ( 0 , m 0 ] , problem (2) has a ground state solution.

For the convenience of expression, hereafter, we will use the following notations:

• X H 1 ( R 3 ) is a space in which an equivalent norm is defined as follows:

u = R 3 ( a u 2 + V ( x ) u 2 ) d x 1 2 ,

• L s ( R 3 ) ( 1 s ) denotes the Lebesgue space in which the norm is defined as follows:

u s = R 3 u s d x 1 / s ,

• For any u H 1 ( R 3 ) { 0 } , u t is denoted as follows:

u t = 0 , t = 0 , t u x t , t > 0 .

• For any x R 3 and r > 0 , B r ( x ) { y R 3 : y x < r } .

• C , C 1 , C 2 , denote positive constants, which are possibly different in different lines.

## 2 Preliminaries

Problem (1) has a variational structure, i.e., the critical points of the functional

(6) I ( u ) = 1 2 R 3 [ a u 2 + V ( x ) u 2 ] d x + b 4 R 3 u 2 d x 2 1 2 p R 3 ( I α u p ) u p d x R 3 F ( u ) d x

are weak solutions of problem (1).

## Lemma 2.1

Assume that ( f 1)–( f 4) hold, then we have

1. for all ε > 0 and q ( 2 , 6 ) , there is a C ε > 0 such that f ( t ) ε t + C ε t q 1 ,

2. for any s 0 , s f ( s ) > 2 F ( s ) and F ( s ) > 0 .

## Proof

We could easily obtain the results by elementary calculation.□

## Lemma 2.2

(Hardy-Littlewood-Sobolev inequality [16]). Let 0 < α < N , p , q > 1 and 1 r < s < be such that

1 p + 1 q = 1 + α N , 1 r 1 s = α N .

1. For any f L p ( R N ) and g L q ( R N ) , one has

R N R N f ( x ) g ( y ) x y N α d x d y C ( N , α , p ) f L p ( R N ) g L q ( R N ) .

2. For any f L r ( R N ) , one has

1 N α f L s ( R N ) C ( N , α , r ) f L r ( R N ) .

## Lemma 2.3

(Brezis-Lieb lemma [17]) Let s ( 1 , ) and { w n } be a bounded sequence in L s ( R N ) . If w n w almost everywhere on R N , then for any q [ 1 , s ] ,

(7) lim n R N w n q w n w q w q s q d x = 0

and

(8) lim n R N w n q 1 w n w n w q 1 ( w n w ) w q 1 w s q d x = 0 .

## Lemma 2.4

(Nonlocal Brezis lemma [6]) Let α ( 0 , N ) , N 3 , p 1 , 2 N N + α and { u n } be a bounded sequence in L 2 N p N + α ( R N ) . If u n u almost everywhere on R N , then

lim n R N ( I α u n p ) u n p d x R N ( I α u n u p ) u n u p d x = R N ( I α u p ) u p d x .

## Lemma 2.5

[13] Let α ( 0 , N ) , N 3 , p 1 , 2 N N + α and { u n } be a bounded sequence in L 2 N p N + α ( R N ) . If { u n } u almost everywhere on R N , then for any h L 2 N p N + α ( R N ) ,

lim n R N ( I α u n p ) u n p 2 u n h d x = R N ( I α u p ) u p 2 u h d x .

## Lemma 2.6

(Pohozaev identity [6,7,18,19]). Suppose V ( x ) satisfies ( V 1 )–( V 2 ) and let u X be a weak solution of Problem (1), then we have the following Pohozaev identity:

(9) 0 = P V ( u ) = a 2 R 3 u 2 d x + 3 2 R 3 V ( x ) u 2 d x + 1 2 R 3 ( V ( x ) , x ) x 2 d x + b 2 R 3 u 2 d x 2 3 + α 2 p R 3 ( I α u p ) u p d x 3 R 3 F ( u ) d x .

In particular, if V V , we have

(10) 0 = P ( u ) = a 2 R 3 u 2 d x + 3 2 R 3 V u 2 d x + b 2 R 3 u 2 d x 2 3 + α 2 p R 3 ( I α u p ) u p d x 3 R 3 F ( u ) d x .

## 3 Ground state solution for the “limit problem” of equation (1)

In this section, we will investigate the following limit problem that is associated with problem (1):

(11) ( a + b R 3 u 2 ) Δ u + V u = ( I α u p ) u p 2 u + f ( u ) , in R 3 , u H 1 ( R 3 ) .

The asscoiated energy function is given by:

(12) I ( u ) = 1 2 R 3 [ a u 2 + V u 2 ] d x + b 4 R 3 u 2 d x 2 1 2 p R 3 ( I α u p ) u p d x R 3 F ( u ) d x .

We prove the following results.

## Lemma 3.1

Let p 3 + α 3 , 3 + α , then I has no lower bounds.

## Proof

For u X \ { 0 } and t > 0 , we have

I ( u t ) = I ( t u ( t 1 x ) ) = a t 2 2 R 3 u 2 d x + t 4 2 R 3 V u 2 d x + b t 4 4 R 3 u 2 d x 2 t p + 3 + α 2 p R 3 ( I α u p ) u p d x t 3 R 3 F ( t u ) d x

as t , since p + 3 + α > 4 , and then we can obtain the conclusion.□

Next we define = { u X \ { 0 } : G ( u ) = 0 } , where

(13) G ( u ) = 1 2 I ( u ) , u + P ( u ) = a R 3 u 2 d x + 2 R 3 V u 2 d x + b R 3 u 2 d x 2 p + 3 + α 2 p R 3 ( I α u p ) u p d x 3 R 3 F ( u ) d x 1 2 R 3 f ( u ) u d x = d I ( u t ) d t t = 1 .

## Remark 3.2

For t > 0 , we set

(14) γ ( t ) = I ( u t ) = a t 2 2 R 3 u 2 d x + t 4 2 R 3 V u 2 d x + b t 4 4 R 3 u 2 d x 2 1 2 p t p + 3 + α R 3 ( I α u p ) u p d x t 3 R 3 F ( t u ) d x .

## Lemma 3.3

Let c 1 , c 2 , and c 3 be constants, which are positive and u X \ { 0 } . Then, the function

η ( t ) = c 1 t 2 + c 2 t 4 c 3 t ( p + 3 + α ) t 3 R 3 F ( t u ) d x f o r t 0

has only one positive critical point, which corresponds to its maximal value.

## Proof

One can obtain the result by elementary calculation.□

## Lemma 3.4

For any u X \ { 0 } , there exists only one t 0 > 0 such that u t 0 and I ( u t 0 ) = max t > 0 I ( u t ) .

## Proof

I ( u t ) has the form of the function η ( t ) defined earlier. Since by Lemma 3.3, η ( t ) has only one critical point t 0 > 0 , which corresponds to its maximal value. Thus, η ( t 0 ) = max t > 0 η ( t ) and η ( t 0 ) = 0 . It follows that G ( u t 0 ) = t 0 η ( t 0 ) = 0 . This implies u t 0 and I ( u t 0 ) = max t > 0 I ( u t ) .□

## Lemma 3.5

The functional I possesses the mountain-pass geometry, i.e.,

1. there exists ρ , δ > 0 such that I δ for all u = ρ ;

2. there exists e H 1 ( R 3 ) such that e > ρ and I ( e ) < 0 .

## Proof

(1) By Lemmas 2.1(1) and 2.2, we have

I ( u ) c 1 u 2 c 2 u 2 p C ε u q .

Thus, there exists ρ , δ > 0 such that I δ for all u = ρ > 0 small enough.

(2) For any u X \ { 0 } , by the definition of I ( u t ) , we see I ( u t ) < 0 for t > 0 large. Note that

u t 2 = a t 2 R 3 u 2 d x + t 4 R 3 V u 2 d x .

Taking e = u t 0 , with t 0 > 0 large, then we have e > ρ and I ( e ) < 0 .□

Now we can define the mountain-pass level of I :

c = inf γ Γ max t [ 0 , 1 ] I ( γ ( t ) ) > 0 ,

where Γ = { γ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = 0 , I ( γ ( 1 ) ) < 0 } .

Let

m = inf u I ( u ) ,

then for any u , we have

I ( u ) = I ( u ) 1 4 G ( u ) a 4 R 3 u 2 d x 0 .

Thus, m is well defined. In addition, by the similar argument as Chapter 4 [19], we have the following property:

c = inf u X { 0 } max t > 0 I ( u t ) = m = inf u I ( u ) .

## Lemma 3.6

Assume that ( f 1)–( f 4) hold, then m is obtained.

## Proof

Let { u n } be such that I ( u n ) m . Since G ( u n ) = 0 , we have:

1 + m > I ( u n ) = I ( u n ) 1 4 G ( u n ) = a 4 R 3 u n 2 d x + 1 8 R 3 [ f ( u n ) u n 2 F ( u n ) ] d x + p 1 + α 8 p R 3 ( I α u p ) u p d x > a 4 R 3 u n 2 d x

for n large enough. Therefore, { u n 2 2 } is bounded. So there exists A 0 such that R 3 u n 2 d x A . We claim that { u n 2 2 } is also bounded. By Lemmas 2.1(1) and 2.2(1), the definition of G ( u n ) and Sobolev inequality, we have

2 R 3 V u n 2 d x = 1 2 R 3 [ 6 F ( u n ) d x + f ( u n ) u n ] d x + p + 3 + α 2 p R 3 ( I α u n p ) u n p d x a R 3 u n 2 d x b R 3 u n 2 d x 2 < 1 2 R 3 [ 6 F ( u n ) + f ( u n ) u n ] d x + p + 3 + α 2 p R 3 ( I α u n p ) u n p d x < ε u n 2 2 + C ε u n q q + C u n 6 p 3 + α 2 p < ε u n 2 2 + C ε u n 2 q + C u n 2 2 p ,

which indicates { u n 2 2 } is bounded. Then let δ = lim n sup y R 3 B 1 ( y ) u n 2 d x . Next we prove δ > 0 . If δ = 0 , then by Lions’ concentration compactness principle [19], we have u n 0 in L q ( R 3 ) for q ( 2 , 6 ) . From Lemma 2.2, we can obtain that

R 3 ( I α u n p ) u n p d x C u n 6 p 3 + α 2 p 0 ,

since 2 < 6 p 3 + α < 6 . Together with G ( u n ) = 0 , we have u n 0 in X . This conflicts with the fact that c > 0 . So δ > 0 and there exists { y n } R 3 such that B 1 ( y n ) u n 2 d x δ 2 > 0 . We set v n ( x ) = u n ( x + y n ) , then

u n = v n , B 1 ( 0 ) v n 2 d x > δ 2

and

I ( v n ) m , G ( v n ) = 0 .

Therefore, there exists v X \ { 0 } such that

v n v in X , v n v in L l o c s ( R 3 ) , s [ 1 , 6 ) v n v a . e . on R 3

and R 3 v n 2 d x A . Then, we set

I A , ( u ) = a + b A 2 R 3 u 2 d x + 1 2 R 3 V u 2 d x 1 2 p R 3 ( I α u p ) u p d x R 3 F ( u ) d x .

Since I ( v n ) 0 and v n v in X , we have I A , ( v ) = 0 . Besides v satisfied the following Pohozaev identity:

P A , = a + b A 2 R 3 u 2 d x + 3 2 R 3 V u 2 d x 3 + α 2 p R 3 ( I α u p ) u p d x 3 R 3 F ( u ) d x = 0 .

Suppose that R 3 v 2 d x < A , then we have

0 = I A , ( v ) , v 2 + P A , ( v ) > G ( v ) .

Thus, by Lemma 3.4, there exists t 0 ( 0 , 1 ) such that v t 0 . Hence, we have

c I ( v t 0 ) = I ( v t 0 ) 1 4 G ( v t 0 ) = a t 0 2 4 R 3 v 2 d x + t 0 3 8 R 3 [ f ( t v ) t v 2 F ( t v ) ] d x + p + α 1 8 p t 0 p + 3 + α R 3 ( I α v p ) v p d x < a 4 R 3 v 2 d x + 1 8 R 3 [ f ( v ) v 2 F ( v ) ] d x + p + α 1 8 p R 3 ( I α v p ) v p d x lim n a 4 R 3 v n 2 d x + 1 8 R 3 [ f ( v n ) v n 2 F ( v n ) ] d x + p + α 1 8 p R 3 ( I α v n p ) v n p d x = lim n I ( v n ) 1 4 G ( v n ) = m .

This is a contradiction. Consequently, we obtain R 3 v 2 d x = A = lim n R 3 v n 2 d x . So G ( v ) = 0 , I ( v ) = m . This completes the proof.□

## Lemma 3.7

If m is obtained at some u ˜ , then u ˜ is a critical point of I ( u ) .

## Proof

It is obvious that u ˜ 0 since u ˜ . Then we claim that for every fixed v H 1 ( R 3 ) , there exists ε > 0 such that u ˜ + s v 0 for all s ( ε , ε ) . In fact, by contradiction, there exists a sequence { s i } i = 1 such that lim i + s i = 0 and u ˜ + s i v = 0 a.e. on R 3 . Letting i + , we have u ˜ = 0 a.e. on R 3 , which is a contradiction with u ˜ 0 . Then by Lemma 3.3, there exists only one t 0 > 0 such that ( u ˜ + s v ) t ( s ) . Now consider the function Φ ( t , s ) = G ( ( u ˜ + s v ) t ) defined for ( t , s ) ( 0 , + ) × ( ε , ε ) . Since u ˜ , one has Φ ( 1 , 0 ) = G ( u ˜ ) = 0 . Moreover, Φ is a C 1 function and

Φ ( t , s ) t ( t , s ) = ( 1 , 0 ) < 0 .

By the implicit function theorem, the function t ( s ) is C 1 and being t ( 0 ) = 1 , then one can know t ( s ) 0 near 0. By letting γ ( s ) = I ( ( u ˜ + s v ) t ( s ) ) , one has γ is differentiable for all small s and attains its minimum at s = 0 . Therefore, we can deduce

0 = γ ( 0 ) = d I ( ( u ˜ + s v ) t ( s ) ) d s s = 0 = I ( ( u ˜ + s v ) t ) t ( t , s ) = ( 1 , 0 ) d t d s t = 0 + I ( ( u ˜ + s v ) t ) s ( t , s ) = ( 1 , 0 ) = G ( u ˜ ) t ( 0 ) + I ( u ˜ ) , v = I ( u ˜ ) , v .

Since v X is arbitrary, we deduce that I ( u ˜ ) = 0 .□

Then by Lemmas 3.6 and 3.7, we can have the following result.

## Theorem 3.8

Under assumptions ( f 1 )–( f 4 ), Problem (11) has a ground state solution.

## Proposition 4.1

(See [20]) Let ( X , ) be a Banach space and T R + be an interval. Φ τ ( u ) is a family of C 1 functions on X of the following form:

Φ τ ( u ) = A ( u ) τ B ( u ) , τ T ,

with B ( u ) 0 , u X and either A ( u ) + or B ( u ) + as u . Suppose that there are two points v 1 , v 2 X such that

c τ = inf γ Γ max t [ 0 , 1 ] Φ τ ( γ ( t ) ) > max { Φ τ ( v 1 ) , Φ τ ( v 2 ) } , τ T ,

where Γ = { γ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = v 1 , γ ( 1 ) = v 2 } , then for almost every τ T , there is a bounded ( P S ) c τ sequence in X.

Set T = [ δ , 1 ] , where δ > 0 . We investigate a family of functionals on X with the following form:

I V , τ ( u ) = 1 2 R 3 [ a u 2 + V ( x ) u 2 ] d x + b 4 R 3 u 2 d x 2 τ 1 2 p R 3 ( I α u p ) u p d x + R 3 F ( u ) d x , τ [ δ , 1 ] .

Then we can set I V , τ ( u ) = A ( u ) τ B ( u ) . Here,

A ( u ) = 1 2 R 3 [ a u 2 + V ( x ) u 2 ] d x + b 4 R 3 u 2 d x 2 + ,

as u + , and

B ( u ) = 1 2 p R 3 ( I α u p ) u p d x + R 3 F ( u ) d x 0 .

## Lemma 4.2

Assume (V2) holds, then we have

1. there exists a v X \ { 0 } such that I V , τ ( v ) 0 for all τ [ δ , 1 ] ,

2. c τ = inf γ Γ max t [ 0 , 1 ] I V , τ ( γ ( t ) ) > max { I V , τ ( 0 ) , I V , τ ( v ) } for all τ [ δ , 1 ] , where

Γ = { γ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = 0 , γ ( 1 ) = v } .

## Proof

(1) Fix u X \ { 0 } , then for τ [ δ , 1 ] and t > 0 , we have

I V , τ ( u t ) I V , δ ( u t ) = a t 2 2 R 3 u 2 d x + t 4 2 R 3 V u 2 d x + b t 4 4 R 3 u 2 d x 2 δ t p + 3 + α 2 p R 3 ( I α u p ) u p d x + t 3 R 3 F ( t u ) d x

as t + . Just by taking v = u t with t large, one can have

I V , τ ( u t ) I V , δ ( u t ) < 0 .

(2) By Lemmas 2.1(1) and 2.2, we have

I V , τ ( u ) c 1 u 2 c 2 u 2 p C ε u q .

Since p > 1 , I V , τ ( u ) has a strictly local minima at 0, i.e., there exists r > 0 such that

b = inf u = r I V , τ ( u ) > 0 = I V , τ ( 0 ) I V , τ ( v ) ,

and hence taking u t = v , we obtain c τ > max { I V , τ ( 0 ) , I V , τ ( u t ) } = 0 .□

## Lemma 4.3

(See [20]) Assume the conditions of Proposition 4.1 hold, the map τ c τ is nonincreasing and left continuous.

By Theorem 3.8, we conclude that for τ [ δ , 1 ] , the “limit problem” of the following type

(15) ( a + b R 3 u 2 ) Δ u + V u = τ [ ( I α u p ) u p 2 u + f ( u ) ] , in R 3 , u H 1 ( R 3 )

has a ground state solution u τ H 1 ( R 3 ) , i.e., for τ [ δ , 1 ] , there exists u τ τ = { u X \ { 0 } : G , τ ( u ) = 0 } such that I V , τ ( u τ ) = 0 and I V , τ ( u τ ) = m τ = inf u τ I V , τ ( u ) . Here,

I V , τ ( u ) = 1 2 R 3 [ a u 2 + V u 2 ] d x + b 4 R 3 u 2 d x 2 τ 1 2 p R 3 ( I α u p ) u p d x + R 3 F ( u ) d x ,

G , τ ( u ) = a R 3 u 2 d x + 2 R 3 V u 2 d x + b R 3 u 2 d x 2 p + 3 + α 2 p τ R 3 ( I α u p ) u p d x 3 τ R 3 F ( u ) d x τ 2 R 3 f ( u ) u d x .

## Lemma 4.4

Suppose that ( V 1 )–( V 2 ) hold, and V ( x ) V , then c τ < m τ for τ [ δ , 1 ] .

## Proof

Let u τ be the minimizer of m τ . By Lemma 4.2, there exists t ˜ ( 0 , t 0 ) such that

c τ = inf γ Γ max t [ 0 , 1 ] I V , τ ( γ ( t ) ) max 0 < t < t 0 I V , τ t u τ x t = I V , τ t ˜ u τ x t ˜ < I V , τ t ˜ u τ x t ˜ max t > 0 I V , τ t u τ x t = I V , τ ( u τ ) = m τ .

Next we provide the following global compactness lemma.

## Lemma 4.5

(See [3,13]) Suppose that ( V 1 )–( V 2 ) and ( f 1 )–( f 4 ) hold. For c > 0 and τ [ δ , 1 ] , let { u n } X be a bounded ( P S ) c sequence for I V , τ . Then there exists a