Existence of a positive bound state solution for the nonlinear Schrödinger–Bopp–Podolsky system

Existence of a positive bound state solution for the nonlinear Schrödinger–Bopp–Podolsky system

Kaimin Teng

^B

and Yunxia Yan

Department of Mathematics, Taiyuan University of Technology, Taiyuan, Shanxi 030024, P. R. China

Received 10 July 2020, appeared 11 January 2021

Communicated by Patrizia Pucci

Abstract. In this paper, we study a class of Schrödinger–Bopp–Podolsky system. Under some suitable assumptions for the potentials, by developing some calculations of sharp energy estimates and using a topological argument involving the barycenter function, we establish the existence of positive bound state solution.

Keywords: Schrödinger–Bopp–Podolsky system, variational approach, competing po- tentials, bound state solution.

2020 Mathematics Subject Classification: 35J48, 35Q60.

1 Introduction

In this paper, we consider the following Schrödinger–Bopp–Podolsky system

(

−

_∆

u + V ( x ) u + K ( x )

φu

= Q ( x )| u |

^p−1

u in

R³

,

−

_∆φ

+

_∆²φ

= K ( x ) u

²

in

R³

, (1.1)

where p ∈ ( _{3, 5} ) _, V ( x ) _, K ( x ) _and Q ( x ) are positive functions such that

|x

lim

|→_∞

V ( x ) = V

_∞

> 0, lim

|x|→_∞

Q ( x ) = Q

_∞

> 0, lim

|x|→_∞

K ( x ) = 0.

This system appears when a Schrödinger field

ψ

=

ψ

( t, x ) couple with its electromagnetic field in the Bopp–Podolsky electromagnetic theory. The Bopp–Podolsky theory, developed by Bopp [8], and independently by Podolsky [20], is a second order gauge theory for the electromagnetic field. As the Mie theory [19] and its generalizations given by Born and Infeld [7,

9], it was introduced to solve the so called infinity problem that appears in the classical

Maxwell theory. From the well known Gauss law (or Poisson equation), the electrostatic potential

φ

for a given charge distribution whose density is

ρ

satisfies the equation

−

_∆φ

=

ρ

in

R³

. (1.2)

BCorresponding author. Email: tengkaimin2013@163.com

If

ρ

= 4πδ

_x0

, with x

₀

∈

_R³

, the fundamental solution of (1.2) is E ( x − x

₀

) , where E ( x ) =

_|¹

x|

, and the electrostatic energy is

ε

( E ) =

¹₂

R

R³

|∇ E ( x )|

²

dx = +

∞. Thus, equation (1.2) was

replaced by

− div ∇

φ p

1 − |∇

φ

|

²

!

=

ρ

in

R³

in the Born–Infeld theory (see [2]) or replaced by

−

_∆φ

+ a

²∆²φ

=

ρ

in

R³

in the Bopp–Podolsky one, from the reason that, in both case if

ρ

= 4πδ

x0

, their energy is finite. In particular, when we consider the operator −

_∆

+

_∆²

, by [3], we know that K( x − x

₀

) , where K( x ) : =

¹⁻_|^e_x^−|_|^x|

, is the fundamental solution of the equation

−

_∆φ

+

_∆²φ

= 4πδ

_x0

,

it has no singularity in x

0

since it satisfies lim

x→x0

K( x − x

0

) = 1, and its energy satisfies

ε_BP

(K) = ¹

2

Z

R³

|∇K|

²

dx + ¹ 2

Z

R³

|

_∆

K|

²

dx < +

_∞.

In addition, the Bopp–Podolsky theory may be interpreted as an effective theory for short distances and for large distances it is indistinguishable from the Maxwell one. For more physical details we refer the reader to the recent paper[10,

11,14] and their references therein.

Indeed the operator −

_∆

+

_∆²

appears also in other different interesting mathematical and physical situations [5,

15].

Recently, P. d’Avenia and G. Siciliano in [3] introduced the following Schrödinger–Bopp–

Podolsky system

(

−

_∆u

+

ωu

+ q

²φu

= | u |

^p−2

u in

R³

,

−

_∆φ

+ a

²∆²φ

= 4πu

²

in

R³

, (1.3)

where a,

ω

> _0, q 6= 0, they developed the variational framework for system (1.3) and proved that when p ∈ ( 2, 6 ) , there exists q

^∗

> 0, for every q ∈ (− q

^∗

, q

^∗

) \

0 , problem (1.3) admits a nontrivial solution, when p ∈ ( 3, 6 ) , for q 6= 0, problem (1.3) admits a nontrivial solution.

In [22], G. Siciliano and K. Silva proved that the multiplicity and nonexistence of solutions for problem (1.3) by using the fibering method.

If a = 0 in problem (1.3), it reduces to the Schrödinger–Poisson system

(

−

_∆

u +

ωu

+ q

²φu

= | u |

^p−2

u in

R³

,

−

_∆φ

= 4πu

²

in

R³

. (1.4)

In the last decades, there are lots of results about the existence and multiplicity of solutions for system (1.4), we do not review the huge documents, just list some of them for interesting readers to see [1,

6,12,21] and the references therein.

The purpose of this paper is to describe some phenomena that can occur when the coef- ficients V ( x ) , K ( x ) and Q ( x ) are competing. In order to describe our main results, we first rewrite problem (1.1) in a more appropriate way to our aim. Let

V ( x ) = V

_∞

+ a ( x ) , Q ( x ) = Q

_∞

− b ( x ) ,

where a ( x ) _and b ( x ) satisfies the following assumptions:

( _H

₁

) _a ( _x ) ∈ L

³²

(

_R³

) _{, a} ( _x ) ≥ 0, a ( _x ) 6≡ 0, and lim

_|x|→∞

a ( _x ) = _0;

( H

₂

) b ( x ) ∈ L

^∞

(

_R³

) , 0 ≤ b ( x ) < Q

_∞

, b ( x ) 6≡ 0, and lim

_|x|→∞

b ( x ) = 0;

( H

₃

) K ∈ L

²

(

_R³

) and there exists R

₀

> 0 such that K ( x ) ≤ Ce

⁻²

√V_∞|x|

for all | x | ≥ R

₀

. Clearly, (1.1) becomes the following form

(

−

_∆u

+ ( V

_∞

+ a ( x )) u + K ( x )

φu

= ( Q

_∞

− b ( x ))| u |

^p−1

u in

R³

,

−

_∆φ

+

_∆²φ

= K ( x ) u

²

in

R³

. (1.5)

From the variational framework described in Section 2, we find that the difference be- tween problem (1.5) and system (1.4) is the nonlocal kernel K( x ) =

¹⁻_|^e_x^−|_|^x|

, comparing the Poisson kernel P ( x ) =

_|¹

x|

, K( x ) is nonhomogeneous and not singular at x = 0 because lim

_|x|→0

K( x ) = 1, which implies that K ∈ L

^∞

(

_R³

) . The non-homogeneity of K makes difficult the use of rescaling of type t → u ( t

^γ

·) and the non-singularity maybe weak some conditions in the estimates.

To the best of knowledge, the system (1.5) is a new one in the field of variational methods, there are only few works about the existence and multiplicity of solutions, such as the ground state. The purpose of this paper is to study the existence of bound state solution for system (1.5). The approach is inspired by the ideas in [4,

12], we explore some calculations of sharp

energy estimates and apply a topological argument involving the barycenter function to show that there exists a critical value of the energy functional, in a higher level of energy which can yield a solution of the problem (1.5). The main difficulties of this work are that the problem is given in the whole space, leading to the loss of compactness, and some sharp energy estimates. For dealing with these obstacles, we borrows a global compactness lemma to recover the compactness and use some careful computations to get the energy estimates.

Now we are ready to give the main results of the paper.

Theorem 1.1.

Suppose that ( H

₁

) – ( H

3

) hold and ( H

₄

) R

R³

a

³²

( x )| x |

²

e

²

√V_∞|x|

dx < +

_∞

and R

R³

b ( x )| x |

²

e

²

√V_∞|x|

dx < +

_∞.

Then (1.5) admits a positive bound state solution.

The paper is organised as follows. In Section 2, we give general preliminaries in order to attack our problem. In Section 3, we prove Theorem

1.1.

2 Preliminaries

In what follows, we will use the following notations:

• LetH¹(_R³)be the Sobolev space endowed with the inner product and norm (u,v):=

Z

R³(∇u∇v+uv)dx, kuk²:= Z

R³(|∇u|²+u²)dx.

• Dis the completion ofC^∞_c (_R³)with respect to the norm kukD=

Z

R³(|∇u|²+|∆u|²)_dx ¹₂

• L^q(O), 1≤q≤_∞,O ⊆_R³a measurable set, denotes the Lebesgue space, the norm inL^q(O)is denoted by| · |_Lq(O)whenO is a proper measurable subset ofR³and by| · |_q whenO=_R³.

• BR(y)denotes the ball of radiusRcentered aty, ify=0, we denote it byBR.

• c,c_i,C,C_i, . . . denote a number of positive constants.

In what follows, without any loss of generality we assumeV_∞=Q_∞=1.

From [3], we know that for u ∈ H¹(_R³), there exists a unique φ ∈ D, denoted by φ_u^K, such that

−_∆φ+_∆²φ=K(x)u²and it possesses the explicit formula φ_u^K(x):=φ(x) =

Z

R³

(₁−e^−|x−z|)

4π|x−z| ^K(z)u²(z)dz, x∈R³. (2.1) Replacingφbyφ^K_u in the first equation in system (1.5), then this system reduces to a class of Schrödinger equation

−_∆u+ (V_∞+a(x))u+K(x)φ_u^Ku= (Q_∞−b(x))|u|^p−1u inR³. (2.2) The energy functionalI:H¹(_R³)→_Rcorresponding to equation (2.2) is defined as

I(u) =¹ 2 Z

R³|∇u|²+ (V_∞+a(x))u²dx+¹ 4 Z

R³K(x)φ^K_uu²dx− ¹ p+1

Z

R³(Q_∞−b(x))|u|^p+1dx, clearly, I ∈ C¹(H¹(_R³),R) and its critical points are weak solutions of problem (2.2). Therefore, in order to find the solutions of system (1.5), we only need to seek the critical points of functionalI. In other words, ifu∈H¹(_R³)is a critical point ofI, then(u,φu)is a weak solution for system (1.5).

Now, by Lemma 3.4 in [3] and applying a similar argument as in Proposition 2.2 of [12], we can show some properties ofφ_u.

Proposition 2.1. For each u∈H¹(R³), the following statements hold:

(i) φ_u^K∈ D,→L^∞(_R³); (ii) φ_u^K≥0;

(iii) For every s∈(3,+_∞],φ_u^K∈L^s(_R³)∩C0(_R³); (iv) For every s∈(³₂,+∞],∇φ^K_u ∈L^s(R³)∩C0(R³);

(v) φ_tu^K =t²φ_u^K; (vi) |φ_u^K|₆≤ckuk²;

(vii) For every y∈_R³,φ_u(·+y)^K =φ^K(·−y)u (·+y); (viii) If un*u in H¹(R³), then

φ^K_un→φ_u^K inD, Z

R³K(x)φ^K_unu²_ndx→ Z

R³K(x)φ_u^Ku²dx

and _Z

R³K(x)φ_u^K_nunϕdx→ Z

R³K(x)φ^K_uuϕdx ∀ϕ∈H¹(R³). Proof. We only need to verify that(iii),(iv)and(viii)hold true.

Observe thatKu²∈ L^r(_R³)forr∈ [1,³₂)owing toK∈ L²(_R³)andu∈ H¹(_R³). By(ii)of Lemma 3.3 in [3], we know thatφ_u^K∈ L^q(_R³)forq∈(_3−2r^3r ,+_∞]. From _3−2r^3r ∈[3,+_∞)and using Lemma 2.20 in [18], we see thatφ^K_u ∈L^s(_R³)∩C0(_R³). Similarly, we can get(iv).

(viii)For allu∈H¹(_R³), consider the linear functionalL_u :D →_R³defined by L_u(v) =

Z

R³K(x)u²vdx,

by the definition ofφ^K_u, we havekφ^K_uk_D =kL_uk_L(D,R). Therefore we intend to show that asn→_∞ kL_un− L_uk_L(D,R)→0.

Letε >0 be fixed, there exists a positive number Rε so large that |K|_L2(R³\B(0,R_ε)) < ε. Therefore forv∈ D(R³)_{, we have}

|Lu_n(v)− Lu(v)|= Z

R³K(u²_n−u²)vdx

≤ Z

R³\B(0,R_ε)K|u²_n−u²||v|dx+ Z

B(0,R_ε)K|u²_n−u²||v|dx

≤ |K|_L2(R³\B(0,R_ε))|u²_n−u²|₃|v|₆+ Z

B(0,Rε)K⁶⁵|u²_n−u²|⁶⁵dx ⁵₆

|v|₆

≤

"

εc+ _Z

B(0,R_ε)K⁶⁵|un+u|⁶⁵|un−u|⁶⁵dx ⁵₆#

kvk_D

Sinceun * u in H¹(_R³), we know thatun → u in L_loc¹²⁵ (_R³). Furthermore, set BM = x ∈ B(0,R_ε) : K(x) > M and remark being K ∈ L²(_R³)_, |B_M| → 0 as M → ∞. Therefore, for large M, R

B_MK²³₅

<ε. So we have Z

B(0,R_ε)K⁶⁵|un+u|⁶⁵|un−u|⁶⁵ dx

= Z

B_MK⁶⁵|un+u|⁶⁵|un−u|⁶⁵dx+ Z

B(0,R_ε)\B_MK⁶⁵|un+u|⁶⁵|un−u|⁶⁵ dx

≤ Z

B_M|K²|³⁵ Z

R³|un+u|^{6 1}₅ Z

R³|un−u|^{6 1}₅

dx +M⁶⁵

_Z

B(0,R_ε)|un+u|¹²⁵ dx ¹_{2 Z}

B(0,R_ε)|un−u|¹²⁵ dx ¹₂

≤cε+o(1). Thereforeφ^K_un →φ^K_u in D. And

Z

R³(K(x)φ_u^K_nu²_n−K(x)φ_u^Ku²)dx= Z

R³K(x)(u²_n−u²))φ_u^K_ndx+ Z

R³K(x)u²(φ_u^K_n−φ^K_u)dx.

Similar to the proof of the above, we can show that Z

R³K(x)(u²_n−u²))φ_u^K_n =o(1). Because whenn→∞,φ^K_un→φ_u^K, and by(H₃)we know that

Z

R³K(x)u²(φ_u^K_n−φ^K_u)dx≤ |K|₂|u²|₃|φ^K_un−φ_u^K|₆=o(1). Finally,

Z

R³(K(x)φ_u^K_nunϕ−K(x)φ_u^Kuϕ)dx = Z

R³(K(x)ϕun)(φ^K_un−φ_u^K)dx+ Z

R³(K(x)ϕφ^K_u)(un−u)dx.

This can be easily proved similar to the above.

It is not difficult to verify that the functional I is bounded neither from above nor from below in H¹(R³). Indeed, there existst∈R⁺ such thattu∈ H¹(R³)satisfies

I(tu) = ^t

2

2 Z

R³|∇u|²+ (V_∞+a(x))u²dx+ ^t

4

4 Z

R³K(x)φ^K_uu²dx− ^t

p+1

p+1 Z

R³(Q_∞−b(x))|u|^p+1dx.

Sincep>3, lim_t→+∞I(tu) =−∞. On the other hand, for someα,β∈_Rand for anyt>0, we have I(t^αu(t^βx)) = ¹

2 Z

R³t^2α+2β|∇u(t^βx)|²+ (V_∞+a(x))t^2α|u(t^βx)|²dx +^t

4α

4 Z

R³K(x)φ_u^Ku(t^βx)²dx−^t

(p+1)α

p+1 Z

R³(Q_∞−b(x))|u(t^βx)|^p+1dx

= ¹ 2 Z

R³t^{2α+2β−3β}|∇u|²+V_∞+ax t^β

t^2α−3β|u|²dx +^t

4α−3β

4 Z

R³Kx t^β

φ

K(^·

tβ)

u u²dx−^t

(p+1)α−3β

p+₁ Z

R³

Q_∞−bx t^β

u^p+1dx.

By(H₁)–(H3), choosingα,βsuch that 2α−β>(p+1)α−3β, that is 2β> (p−1)α. Particularly, we chose thatα=1,β= p, then lim_t→+∞I(tu) = +_∞.

Naturally, we consider that the functional I restricted in the Nehari manifoldN, that contains all the critical points ofI, is bounded from below, where

N :=ⁿu∈H¹(_R³)\{0}: I⁰(u)[u] =₀^o_. By using a standard argument, we can show the following lemma.

Lemma 2.2. Suppose that(H₁)–(H3)hold, the following statements are true:

(i) There exists a positive constant c>0such that for all u∈ N, there holds

|u|_p+1≥c>0.

(ii) N is a C¹regular manifold diffeomorphic to the sphere of H¹(R³). (iii) I is bounded from below onN by a positive constant.

(iv) u is a free critical point of I if and only if u is a critical point of I constrained onN. Proof. (i)Letu∈ N, by(H₁)–(H₃)and Sobolev’s embedding theorem, we have that

c₁|u|²_p+1≤ kuk²≤ kuk²+ Z

R³a(x)u²dx+ Z

R³K(x)φ_u^Ku²dx

= Z

R³(Q_∞−b(x))|u|^p+1dx≤c2|u|^p+1_p+1,

(2.3)

wherec₁,c2>0 independent ofu, and owing top>3, this estimate implies that

|u|_p+1≥c>0, kuk ≥c>0, ∀u∈ N, (2.4) wherec>0 independent ofu.

(ii) It suffices to show that G⁰(u)[u] < _{0 for} u ∈ N, where G(u) = I⁰(u)[u]_{. Clearly,} G ∈ C¹(H¹(_R³),R). By (2.4), for anyu∈ N, we deduce that

G⁰(u)[u] =2 Z

R³|∇u|²+ (V_∞+a(x))u²dx+4 Z

R³K(x)φ^K_uu²dx

−(_p+₁) Z

R³(_Q∞−b(_x))|u|^p+1dx

= −(p−1) Z

R³|∇u|²+ (V_∞+a(x))u²dx−(p−3) Z

R³K(x)φ^K_uu²dx

≤ −(p−3)C<0,

(2.5)

where C > 0 is dependent of u. By applying the implicit function theorem, we see that N is of C¹ manifold.

The remaining proofs of(ii)_,(iii)_and(iv)are standard, we omit them.

Now, the limit equation corresponding to problem (2.2) is defined as

−_∆u+V_∞u=Q_∞|u|^p−1u inR³. (2.6) The energy functional I_∞:H¹(_R³)→_Rassociated to problem (2.6) given by

I_∞(u) = ¹ 2 Z

R³|∇u|²+V_∞|u|²dx− ¹ p+1

Z

R³Q_∞|u|^p+1dx

and it is easy to verify that I_∞∈C²(H¹(_R³),R). Denote the Nehari manifold of functionalI_∞by N_∞:=ⁿu∈ H¹(R³)\(0):I_∞⁰ (u)[u] =0o

. Set

m_∞:= inf

u∈N_∞I_∞(u).

m_∞ > 0 is achieved by a positive radially symmetric function ω, that is unique (up to translations) positive solution of (2.6) (see [17]), decreasing when the radial coordinate increases and such that

|x|→lim∞|D^jω(x)||x|e

√V_∞|x|

=dj >0, j=0, 1, dj∈R. (2.7) Moreover, for any sign-changing critical pointuofI_∞, by standard argument, the following inequality holds true

I_∞(u)≥2m_∞. (2.8)

Now we are ready to consider the constrained minimization problemm:=inf{I(u),u ∈ N }, we find that the relation between least energymandm_∞holds and it is not achieved, thus we can not look for the ground state.

Proposition 2.3. Suppose that(H₁)–(H₃)hold. Then there holds

m=m_∞ (2.9)

and the infimum is not achieved.

Proof. Letu∈ N, then there existstu >0 such thattuu∈ N_∞. Thus, we deduce that I(u)≥I(tuu)≥ ^t

2u

2 Z

R³(|∇u|²+V_∞u²)dx− ^t

p+1 u

p+1 Z

R³Q_∞|u|^p+1dx= I_∞(tuu)≥m_∞ from which, we getm≥m_∞.

Next, it suffices to find a sequence(un)_n,un ∈ N, such that limn→∞I(un) = m. For this purpose, let us consider(yn)_n, withyn ∈ _R³,|yn| → +_∞asn→_∞and setun = tnωy_n =tnω(x−yn), where tn=t(ωyn)is such thatun=tnωyn ∈ N. Since

I(un) = ^t

2n

2 Z

R³|∇ωy_n|²+ (V_∞+a(x))ω²_yndx+ ^t

4n

4 Z

R³K(x)φ^K_ωynω²_yndx

− ^t

p+1 n

p+1 Z

R³(Q_∞−b(x))|ω_yn|^p+1dx

= ^t

2n

2 Z

R³|∇ω|²+ (V_∞+a(x+yn))ω²dx+^t

4n

4 Z

R³K(x+yn)φ_ω^K(·+yn)ω²dx

− ^t

p+1 n

p+1 Z

R³(Q_∞−b(x+yn))|ω|^p+1dx,

and from(H₁),(H₂)and(H₃), by Lebesgue dominated theorem, we can deduce that

n→lim∞ Z

R³a(x+yn)ω²=0, lim

n→∞ Z

R³b(x+yn))|ω|^p+1=0,

and

n→lim∞ Z

R³K(x+yn)φ_ω^K(·+yn)ω²dx=0.

Combining withtnωyn ∈ N, we have that 1≤ tn ≤ C, whereC> 0 is a positive constant. Therefore, fromω∈ N_∞and p∈(3, 5), it follows thattn→1 and thus I(un)→m_∞.

To complete the proof, we argue by contradiction and we assume that v ∈ N exists such that I(v) =m=m_∞. Obviously, there existstv>0 such thattvv∈ N_∞, we have that

m_∞≤ I_∞(tvv) = 1

2− ¹ p+1

ktvvk²

≤ 1

2− ¹ p+1

ktvvk²+ 1

4− ¹ p+1

_Z

R³K(x)φ^K_tvv(tvv)²dx

≤ I(tvv)≤I(v) =m=m_∞ which implies thattv=1 and

Z

R³K(x)φ_v^Kv²dx=0. (2.10)

Hence,v ∈ N_∞ and I_∞(_v) =_m∞. By the uniqueness of solution of problem (2.6), there existsy ∈R³ such thatv(x) =ω(x−y)>0, for everyx∈_R³, which leads toR

R³K(x)φ^K_vv²dx >0, contradicts with (2.10).

In order to find a bound state in higher energy level in(m_∞, 2m_∞), the next results help us to recover the compactness of the bounded(PS)sequence in (m_∞, 2m_∞). Following the proof of Lemma 4.5 in [3], we can show the following splitting lemma.

Lemma 2.4(Splitting lemma). Suppose that(H₁)–(H₃)hold. Let(un)_n be a(PS)sequence of I constrained onN, i.e. un ∈ N, and I(un)is bounded,∇I|_N(un) →0 strongly in H¹(_R³). Then, up to a subsequence, there exist a solution u of (2.2), a number k ∈_N∪0 , k functions u¹, . . . ,u^k of H¹(_R³)and k sequences of points(y^j_n)_n, y_n^j ∈R³,0≤j≤k such that, as n→+_∞,

(i) un−_∑^k_j=1u^j(· −y^j_n)→u in H¹(_R³); (ii) I(u_n)→I(u) +_∑^k_j=1I_∞(u^j);

(iii) |yn^j| →+∞,|yⁱ_n−y^jn| →+∞if i6=j;

(iv) u^jare weak solution of (2.6).

Moreover, in the case k=0, the above holds without u^j.

In the end of this section, we recall a technical result for some estimates in the next section, its proof is found in [4,12].

Lemma 2.5. If g∈ L^∞(R³)and h∈ L¹(R³)are such that, for someα≥0, b≥0,γ∈R

|x|→+lim∞g(x)e^α|x||x|^b =γ and Z

R³|h(x)|e^α|x||x|^bdx<+_∞, then, for every z∈_R³\ {0},

ρ→+∞lim Z

R³g(_x+_ρz)_h(_x)_dx

e^α|ρz||ρz|^b=γ Z

R³h(_x)_e^{−α(x·z)/|z|}_dx.

3 Proof of Theorem 1.1

Now, we turn to build tools and topological techniques to prove the existence of an higher energy solution when (1.5) has no ground state solution. First we recall the definition of barycenter β : H¹(_R³)\ {0} →_R³of a functionu∈H¹(R³),u6=0, given in [13], set

µ(u)(x) = ¹

|B1(0)|

Z

B₁(x)|u(y)|dy, thenµ(u)∈L^∞(_R³)and is continuous inH¹(_R³)_{. Let}

ˆ u(x) =

µ(u)(x)−¹

2maxµ(u)(x) +

,

it is easy to check that ˆu∈C0(R³)and then defineβ:H¹(R³)\ {0} →R³as follows β(u) = ¹

|uˆ|₁ Z

R³xuˆ(x)dx∈_R³.

Since ˆuhas compact support,βis well defined and it is easy to verify the following properties:

(1) βis continuous in H¹(R³)\ {0}; (2) ifuis a radial function,β(u) =0;

(3) for allt6=0 and for allu∈H¹(_R³)\ {0},β(tu) =β(u); (4) givenz∈R³and settinguz(x) =u(x−z),β(uz) =β(u) +z.

By Proposition2.3, we see thatmcan not be achieved, with the help of the barycenter mappingβ, we can add some refined constraint in the Nehari manifoldN. For this purpose, define the following minimization problem

B₀:=inf

I(u):u∈ N,β(u) =0 .

Clearly, we havem=m_∞≤ B₀. Furthermore, the strict inequality holds true.

Lemma 3.1. Suppose that(H1)–(H3)hold. Then

m=m_∞<B₀

Proof. By contradiction, we assume that B₀ = m_∞, then there exists un ∈ N such that β(un) = 0, I(un)→m_∞ and ∇I|_N(un) →0. By the Ekeland’s variational principle (see Theorem 8.15 in [24]), a sequence of(vn)_n∈ H¹(_R³)exists so that

vn ∈ N, I(vn) =m_∞+on(1) and |β(vn)−β(un)|=o(1). (3.1) By Lemma2.4, we have that

m_∞= I(un) +o(1) =I(u) +_Σ^k_j=1I_∞(u^j) +o(1).

Owing to I(u)≥m=m_∞andI_∞(u^j)≥m_∞, we have thatk=0. Thus,vn →u. Sinceuis a nontrivial solution of (1.5), we deduce that

I(u) =m_∞, I⁰(u) =0, β(u) =0,

which means thatm=m_∞is achieved, contradicts with Proposition2.3. The proof is completed.

Lemma 3.2. The functional I constrained onN satisfies the Palais–Smale condition in(m_∞, 2m_∞).

Proof. Let{un}be a Palais–Smale sequence of I|_N such that I(un) →c∈ (m_∞, 2m_∞). By Lemma2.4, we have that

c=I(un) +o(1) =I(u) +Σ^k_j=1I_∞(u^j) +o(1).

The conclusion follows observing that any critical point ¯u of I is such that I(u¯) ≥ m = m_∞, any solution of (2.6) verifies thatI_∞(u)≥m_∞and if it changes sign,I_∞(u)≥2m_∞. Whatever any case, we can obtain the sequence{un}strongly convergence inH¹(_R³). The compactness is proved.

Letξ∈_R³with|ξ|=1 andΣ=z∈_R³:|z−ξ|=2 . Forρ>0 and(z,s)∈_Σ×[0, 1], define ψ_ρ[z,s](x):= (1−s)ωρz(x) +sω_ρξ(x) = (1−s)ω(x−ρz) +sω(x−ρξ), x∈R³,

wherewis a unique radically symmetric positive solution of problem (2.6), then by virtue of standard argument, there exist positive numberstρ,z,s:=t_ψ

ρ[z,s]andτ_ρ,z,s :=τ_ψ

ρ[z,s] such that

ψ_ρ[z,s] =tρ,z,sψ_ρ[z,s]∈ N, ψ_∞,ρ[z,s] =τ_ρ,z,sψ_ρ[z,s]∈ N_∞. (3.2) Remark 3.3. Note that ψ_ρ[z,s] → ω(x−ρz)as s→ 0 and ψ_ρ[z,s] →ω(x−ρξ) ass → 1, moreover, τρ,z,s→1 ass→0 ors→1 due toω(x−ρz)∈ N_∞andω(x−ρξ)∈ N_∞.

Lemma 3.4. For allρ>0, we have

B₀≤ T_ρ:= max

Σ×[0,1]I(ψ_ρ[z,s]).

Proof. Observing thatβ(ψρ[z, 0]) =β(tρ,z,0ψ_ρ[z, 0]) =β(tρ,z,0ωρz) =β(ωρz) =ρzand β(ψρ[z, 1]) =ρξ.

Let

G(z,s) =sρξ+ (1−s)ρz, (3.3)

thenG(z,s)∈C(_Σ×(0, 1]). Define a mapping by

h(t,z,s) =tG(z,s) + (1−t)β(ψ_ρ[z,s]), ∀t∈[0, 1], (3.4) thenh(t,z,s)∈C([0, 1]×_Σ×(0, 1])is continuous and

h(t,z, 0) =tρz+ (1−t)β(ψ_ρ[z, 0]) =tρz+ (1−t)ρz=ρz6=0, ∀z∈Σ and

h(t,z, 1) =tρξ+ (1−t)β(ψ_ρ[z, 1]) =tρξ+ (1−t)ρξ =ρξ6=0, ∀z∈Σ

which implies that 06∈h(t,∂(_Σ×(0, 1])), for everyt∈[0, 1]. Therefore, by the homotopical invariance of Brouwer degree, we get deg(h(t),Σ×(0, 1], 0) =constant. Thus

deg(h(0),Σ×(0, 1], 0) =deg(h(1),Σ×(0, 1], 0), that is

deg(β(ψ_ρ[z,s]),Σ×(0, 1], 0) =deg(G(s,z),Σ×(0, 1], 0).

Clearly, deg(G(z,s),Σ×(0, 1], 0) 6=0. Thus, it follows from the solvable property of Brouwer degree that there exists(z,s)∈Σ×(0, 1]such thatβ(ψ_ρ[z,s]) =0. Therefore, by the definition ofB₀, we have that

B₀≤I(ψ_ρ[z,s])≤ T_ρ.

In order to showT_ρ< 2m_∞, we have to give some estimates from the decay ofωand coefficients a(x),K(x)andb(x).

Lemma 3.5. Suppose that(H3)holds. There exists c>0such that for allρ>3R0, the following holds Z

R³K(x)φ^K_ω

ρζω²_ρζdx≤ce⁻⁴³

√V_∞ρ, for allζ∈R³ with|ζ| ≥1. (3.5)

Proof. Let ρ > 3R0, then ¹₃ρ > R0and if |y| ≤ ¹₃ρ and |ζ| ≥ 1, we have that|y−ρζ| ≥ ρ|ζ| − |y| ≥ ρ−¹₃ρ = ²₃ρ. Thus, by the exponential decay (2.7) of ω, ^1−e_|x|^−|x| ∈ L^s(_R³) for all s ∈ (3,+_∞] (see Lemma 3.3 in [3]), Hölder’s inequality and(H3), we deduce that

φ^K_ω

ρζ(x) = ¹ 4π

Z

R³

1−e^−|x−y|

|x−y| ^K(y)ω²(y−ρζ)dy

= ¹ 4π

Z

|y|≤¹₃ρ

1−e^−|x−y|

|x−y| ^K(y)ω²(y−ρζ)dy+ Z

|y|>¹₃ρ

1−e^−|x−y|

|x−y| ^K(y)ω²(y−ρζ)dy

!

≤ ¹ 4π





 Z

|y|≤¹₃ρ

K²(y)ω²(y−ρζ)dy ¹₂



 Z

|y|≤¹₃ρ

1−e^−|x−y|

|x−y|

!4

dy





1 4Z

|y|≤¹₃ρ

ω⁴(y−ρζ)dy ¹₄

+ Z

|y|>¹₃ρ

1−e^−|x−y|

|x−y| ^K(y)ω²(y−ρζ)dy







≤c





e⁻²³

√V_∞ρ

Z

|y|≤(¹₂−q)ρK²(y)dy ¹₂



 Z

R³

1−e^−|x−y|

|x−y|

!4

dy





14

Z

R³ω⁴(y−ρζ)dy ¹₄

+e⁻²³

√V_∞ρ Z

|y|>¹₃ρ

1−e^−|x−y|

|x−y| ^ω

2 ρζdy







≤ce⁻²³

√V_∞ρ C+ Z

|y|>¹₃ρ

1−e^−|x−y|

|x−y| ^ω

2 ρζdy

!

≤ce⁻²³

√V_∞ρ.

Thus, a similar computation gives Z

R³K(x)φ_ω^K

ρζ(x)ω²_ρζdx≤ce⁻²³

√V_∞ρ Z

R³K(x)ω²_ρζdx

=ce⁻²³

√V_∞ρ Z

|x|≤¹₃ρ

K(x)ω²_ρζdx+ Z

|x|>¹₃ρ

K(x)ω²_ρζdx

!

≤ce⁻²³

√V_∞ρ

"

Z

|x|≤¹₃ρ

K²(x)ω_ρζ² dx

!¹₂ Z

|x|≤¹₃ρ

ω²_ρζdx

!¹₂ +

Z

|x|>¹₃ρ

K(x)ω²_ρζdx

!

≤ce⁻²³

√V_∞ρe⁻²³

√V_∞ρ=ce⁻⁴³

√V_∞ρ. From (3.5), it is easy to very that

Z

R³K(x)φ_ω^K

ρξω²_ρξdx=o(ρ⁻¹e⁻

√V_∞ρ), Z

R³K(x)φ_ω^K_ρzω²_ρzdx=o(ρ⁻¹e⁻

√V_∞ρ) (3.6)

and we denoteε_ρ=ρ⁻¹e⁻

√V_∞ρfor convenience. Moreover, we can obtain the following estimate:

Lemma 3.6.

Z

R³K(x)φ_ψ^K

ρ[z,s]ψ²_ρ[z,s]dx=o(ε_ρ), ∀s∈[0, 1] and z∈_Σ. (3.7) Proof. Since

φ^K

ψ_ρ[z,s] = ¹ 4π

Z

R³

1−e^−|x−y|

|x−y| ^K(y)ψ²_ρ[z,s](y)dy≤2φ^K_ωρz+2φ^K_ω

ρξ

and then Z

R³K(x)φ_ψ^K

ρ[z,s]ψ²_ρ[z,s]dx≤2 Z

R³K(x)φ_ψ^K

ρ[z,s](ω²_ρξ+ω_ρz²)dx

≤4 Z

R³K(x)(φ^K_ωρz+φ^K_ω

ρξ)(ω²_ρξ+ω²_ρz)dx

=4 Z

R³K(x)(φ^K_ωρzω²_ρz+φ_ω^K

ρξω²_ρz+φ_ω^K_ρzω²_ρξ+φ^K_ω

ρξω²_ρz)dx.

Now, similar to the proof of Lemma3.5, we have that Z

R³K(x)φ_ω^K_ρzω²_ρξ(x)dx ≤ce⁻²³

√V_∞ρ Z

R³K(x)ω²_ρξ(x)dx

=ce⁻²³

√V_∞ρ Z

|x|≤¹₃ρ

K(x)ω²_ρξ(x)dx+ Z

|x|>¹₃ρ

K(x)ω²_ρξ(x)dx

!

≤ce⁻²³

√V_∞ρ

Z

|x|≤¹₃ρ

K²(x)ω²_ρξ(x)dx

!¹₂ Z

|x|≤¹₃ρ

ω_ρξ² (x)dx

!¹₂ +

Z

|x|>¹₃ρ

K(x)ω²_ρξ(x)dx

!

≤ce⁻²³

√V_∞ρe⁻²³

√V_∞ρ=ce⁻⁴³

√V_∞ρ. Thus,

Z

R³K(x)φ^K_ωρzω_ρξ² (x)dx=o(ε_ρ), and the same argument leads to

Z

R³K(x)φ^K_ω

ρξω²_ρz(x)dx=o(ε_ρ). Therefore, from (3.6), the estimate (3.7) follows.

Next, we give some estimates which are used in the sequel.

Lemma 3.7. The following estimates hold:

Z

R³ωρz^pω_ρξdx=O(ε_eρ) =o(ε_ρ), Z

R³ω_ρξ^p ωρzdx=O(ε_eρ) =o(ε_ρ), (3.8) Z

R³a(x)ω²_ρzdx=o(ε_ρ), Z

R³a(x)ω²_ρξdx=o(ε_ρ), Z

R³a(x)ψ²_ρ[z,s]dx=o(ε_ρ), (3.9) Z

R³b(x)|ψ_ρ[z,s]|^p+1dx=o(ε_ρ)_{, (3.10)} whereεe_ρ=ρ⁻²e⁻²

√V_∞ρ.

Proof. (i)By (2.7), we can deduce that Z

R³ω_ρξ^p ωρzdx= Z

R³ω^p(x−ρξ)ω(x−ρz)dx= Z

R³ω^p(y)ω(y+ρ(ξ−z))dy

∼c Z

R³|y+ρ(ξ−z)|⁻¹e⁻

√V_∞|y+ρ(ξ−z)|

ω^p(y)dy In order to apply Lemma2.5, let us seth(x) =ω^p(x)∈L¹(_R³),g(x) =|x|⁻¹e⁻

√V_∞|x|, takingα=√ V_∞ andb=1, clearly,

|x|→+∞lim g(x)|x|e

√V_∞|x|=1,

Z

R³ω^p(x)e

√V_∞|x||x|dx≤c Z

R³e^−(p−1)

√V_∞|x||x|^−(p−1)dx<+∞.