Positive solutions for a nonhomogeneous Schrödinger-Poisson system

Jing Zhang; Rui Niu; Xiumei Han

doi:10.1515/anona-2022-0238

Open Access Published by De Gruyter March 9, 2022

Positive solutions for a nonhomogeneous Schrödinger-Poisson system

Jing Zhang , Rui Niu and Xiumei Han

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0238

Abstract

In this article, we consider the following Schrödinger-Poisson system:

− Δ u + u + k ( x ) ϕ ( x ) u = f ( x ) ∣ u ∣ p − 1 u + g ( x ) , x ∈ R 3 , − Δ ϕ = k ( x ) u 2 , x ∈ R 3 ,

with p ∈ ( 3 , 5 ) . Under suitable assumptions on potentials f ( x ) , g ( x ) and k ( x ) , then at least four positive solutions for the above system can be obtained for sufficiently small ‖ g ‖ H − 1 ( R 3 ) by taking advantage of variational methods and Lusternik-Schnirelman category.

Keywords: Schrödinger-Poisson system; Lusternik-Schnirelman category; multiple solutions

MSC 2010: 35A15; 35J10; 35B09

1 Introduction and main result

In this article, we are concerned with the existence of multiple positive solutions for the following nonlinear Schrödinger-Poisson system:

(1.1) − Δ u + u + k ( x ) ϕ ( x ) u = f ( x ) ∣ u ∣ p − 1 u + g ( x ) , x ∈ R 3 , − Δ ϕ = k ( x ) u 2 , x ∈ R 3 ,

which is a special case of the following generalized Schrödinger-Poisson system:

(1.2) − Δ u + V ( x ) u + k ( x ) ϕ ( x ) u = f ( x ) ∣ u ∣ p − 1 u + g ( x ) , x ∈ R 3 , − Δ ϕ = k ( x ) u 2 , x ∈ R 3 ,

where p ∈ ( 3 , 5 ) , 5 = 2 ∗ − 1 and 2 ∗ = 6 is the critical Sobolev exponent in R 3 . Systems like (1.1) and (1.2) have been extensively studied, and it is well known that they have a profound physical background since they arise in quantum mechanics models (see, e.g., [1,2]), semiconductor theory [3,4,5] and so on. The nonlinearity represents the particles interacting with each other. The term k ( x ) ϕ ( x ) u represents the interaction with the electric field. Historically, systems like (1.1) have been introduced in [1,3] as a model describing solitary waves. In system (1.1), the first equation is a nonlinear stationary Schrödinger equation in which the potential ϕ fulfills a Poisson equation, this is the reason that system (1.1) is usually known as Schrödinger-Poisson systems. We refer to [1,3,6] and references therein for more physical background.

The following assumptions are necessary for our main results.

f ( x ) ∈ C ( R 3 ) , f ( x ) ∈ ( 0 , 1 ] for all x ∈ R 3 , f ( x ) ≢ 1 .
f ( x ) → 1 as ∣ x ∣ → ∞ .
there exist δ > 0 and C > 0 such that
f ( x ) ≥ 1 − C e − ( 2 + δ ) ∣ x ∣ for all x ∈ R 3 .
k ∈ L 2 ( R 3 ) ∩ L ∞ ( R 3 ) , lim ∣ x ∣ → ∞ k ( x ) = 0 , k ( x ) ≥ 0 for all x ∈ R 3 , k ( x ) ≢ 0 .
g ( x ) ∈ H − 1 ( R 3 ) , g ( x ) ≥ 0 , g ( x ) ≢ 0 .

In the homogeneous case, g ≡ 0 , system (1.1) has been widely discussed. When k ( x ) = f ( x ) ≡ 1 , Coclite in [7] showed that system (1.1) has a nontrivial radial solution when 3 < p < 5 . The same result was obtained in [8] for 3 ≤ p < 5 . In [9], by means of a Pohozaev identity, D’Aprile and Mugnai proved that system (1.1) does not have any nontrivial solution for − 1 < p ≤ 1 or p ≥ 5 . In [10], Cerami and Vaira took into account system (1.1) with p ∈ ( 3 , 5 ) , and they obtained the existence of positive ground state solutions through minimization of the Nehari manifold method and the concentration compactness method. As for other existence or multiplicity results, we refer to [11,12, 13,14,15, 16,17] for various assumptions on the potential and nonlinearities. It is worth mentioning that the authors in [18] proved that the planar Schrödinger-Poisson system (1.2) has an axially symmetric solution, which has a special minimax characterization. We also refer to [19,20] for related results concerning ground state solution and semiclassical ground state in R 2 .

In the nonhomogeneous case, g ≢ 0 , this consideration about the existence of solutions for system (1.1) is relatively rare. By assuming that g ( x ) is radially symmetric and f ( x ) ≡ 1 , three radially symmetric solutions for system (1.1) were obtained for p ∈ ( 3 , 5 ) , see [21] for more details. The latest results about system (1.1) can be seen in [22,23, 24,25,26, 27,28] and references therein. For example, two positive solutions are obtained in [22] when the nonlinearity is asymptotically linear or asymptotically 3-linear at infinity. The existence results of two different solutions can be found in [23,26], one is a negative energy solution, and the other is a positive energy solution, based on the mountain pass theorem and the Ekeland variational principle. The authors in [25] proved the existence of two nontrivial radial solutions when f ( x ) ≡ 1 . Zhang et al. in [27] showed that system (1.1) has at least two positive radial solutions when the nonlinearity is asymptotically linear or superlinear at infinity. When the potential k ( x ) ≡ 0 , Adachi and Tanaka in [29] studied the corresponding equation (1.1) and obtained four positive solutions.

Inspired by the aforementioned works, we are interested in the multiplicity of positive solutions for system (1.1). For any u ∈ H 1 ( R 3 ) , there exists a unique ϕ u ∈ D 1 , 2 ( R 3 ) , whose expression will be given in Section 2, such that system (1.1) can be reduced to a nonlinear Schrödinger equation with a nonlocal term as follows:

− Δ u + u + k ( x ) ϕ u ( x ) u = f ( x ) ∣ u ∣ p − 1 u + g ( x ) .

When dealing with this problem, one has to face two main difficulties: one is the competing effect of the nonlocal term ϕ u ( x ) u with the nonlinear term, which leads to quite different situations as p varies in the interval ( 3 , 5 ) , another is the lack of compactness, that is, the embedding H 1 ( R 3 ) ↪ L q ( R 3 ) is not compact as q ∈ ( 2 , 6 ) , which makes us difficult to employ the variational techniques in a standard way.

Now we are in a position to state our main result as follows:

Theorem 1.1

Assume ( f 1 )–( f 3 ), ( k ) and ( g ) hold, and there exist d ¯ > 0 and δ ¯ > 0 such that g ( x ) and k ( x ) satisfy ‖ g ‖ H − 1 ( R 3 ) ≤ d ¯ , ‖ k ‖ L 2 ( R 3 ) ≤ δ ¯ , then system (1.1) has at least four positive solutions.

Finally, let us sketch the idea of the proof of Theorem 1.1 as follows: first, we try to get the existence of one positive solution, which is a local minimizer of energy functional near 0. Second, by using the Lusternik-Schnirelman category theory, one can obtain the other two positive solutions. Because of the lack of the compactness, which was caused by the nonlocal terms, the method in [29] could not be employed to verify the Palais-Smale condition. For this, we need to estimate the nonlocal term carefully. In the end, the fourth positive solution can be gained by getting through the energy difference.

The remainder of this article is organized as follows. In Section 2, we prove the existence of one positive solution as a local minimum of energy functional around zero. In Section 3, we verify the Palais-Smale compactness condition for energy functionals under some level. Section 4 is devoted to estimating the Lusternik-Schnirelman category of some level set. The proof of the main result is given in Section 5.

Notations:

H 1 ( R 3 ) is a Hilbert space endowed with the inner product and norm:
( u , v ) ≔ ∫ R 3 ( ∇ u ∇ v + u v ) d x , ‖ u ‖ = ( u , u ) 1 / 2 .
D 1 , 2 ( R 3 ) is a Hilbert space and is the closure of C 0 ∞ ( R 3 ) endowed with the norm
‖ u ‖ D 1 , 2 ( R 3 ) = ∫ R 3 ∣ ∇ u ∣ 2 d x 1 2 .
H − 1 ( R 3 ) denotes the dual space of H 1 ( R 3 ) .
L p ( R 3 ) , 1 ≤ p ≤ + ∞ , denotes a Lebesgue space, the norm in L p ( R 3 ) is denoted by ∣ ⋅ ∣ p .
C , C i , C ( i ) , C ′ are various positive constants.

2 Preliminaries and the first solution

It is well known that system (1.1) can be easily transformed into a single nonlinear Schrödinger equation with a nonlocal term. In short, the Poisson equation is solved by using the Lax-Milgram theorem, so for all u ∈ H 1 ( R 3 ) , a unique ϕ u ∈ D 1 , 2 ( R 3 ) is gained, such that − Δ ϕ = k ( x ) u 2 and that inserted into the first equation, then gives

− Δ u + u + k ( x ) ϕ u ( x ) u = f ( x ) ∣ u ∣ p − 1 u + g ( x ) .

Actually, for each u ∈ H 1 ( R 3 ) , we define an operator T u on D 1 , 2 ( R 3 ) by

T u ( v ) = ∫ R 3 k ( x ) u 2 v d x .

The Hölder inequality and the fact k ∈ L 2 ( R 3 ) yield that there is a constant C > 0 such that for every v ∈ D 1 , 2 ( R 3 ) ,

∣ T u ( v ) ∣ ≤ C ∣ k ∣ 2 ‖ u ‖ 2 ‖ v ‖ D 1 , 2 ( R 3 ) .

Hence, by the Riesz representation theorem, there exists a unique ϕ u ∈ D 1 , 2 ( R 3 ) such that

∫ R 3 ∇ ϕ u ∇ v d x = ∫ R 3 k ( x ) u 2 v d x ,

thus ϕ u is a weak solution of − Δ ϕ u = k ( x ) u 2 and can be represented by

ϕ u ( x ) = 1 4 π ∫ R 3 k ( y ) ∣ x − y ∣ u 2 ( y ) d y .

Moreover, it is obvious that

‖ ϕ ‖ D 1 , 2 ( R 3 ) = ‖ T u ‖ ℒ ( D 1 , 2 , R ) ≤ C ∣ k ∣ 2 ‖ u ‖ 2 .

Also, there is

(2.1) ∫ R 3 k ( x ) ϕ u ( x ) u 2 d x ≤ C ∣ k ∣ 2 ‖ u ‖ 2 ‖ ϕ ‖ D 1 , 2 ( R 3 ) ≤ C ∣ k ∣ 2 2 ‖ u ‖ 4 .

Thus, substituting ϕ u in (1.1), we can prove that ( u , ϕ ) is a positive solution ( u > 0 ) of (1.1) if and only if u ∈ H 1 ( R 3 ) is a critical point of the functional I f , k , g : H 1 ( R 3 ) → R defined by

I f , k , g ( u ) = 1 2 ‖ u ‖ 2 + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x − 1 p + 1 ∫ R 3 f ( x ) ( u + ) p + 1 d x − ∫ R 3 g ( x ) u d x .

Consider the following nonlinear Schrödinger equation:

(2.2) − Δ u + u = ∣ u ∣ p − 2 u x ∈ R 3 .

It is well known that solutions of (2.2) are the critical points of

I ∞ ( u ) = 1 2 ∫ R 3 ( ∣ ∇ u ∣ 2 + ∣ u ∣ 2 ) d x − 1 p + 1 ∫ R 3 ∣ u ∣ p + 1 d x .

We know that (2.2) has a unique (up to translation) positive radially smooth solution w ( x ) , which satisfies (see for instance [30,31,32])

(2.3) w ( x ) ∣ x ∣ exp ( ∣ x ∣ ) → C ≥ 0 as ∣ x ∣ → + ∞ ,

(2.4) w ′ ( r ) r exp ( r ) → − C as r = ∣ x ∣ → + ∞ .

In particular, we have

C p + 1 − 1 = inf u ≠ 0 ‖ u ‖ ∣ u ∣ p + 1 = ‖ ω ‖ ∣ ω ∣ p + 1

with ω ( x ) critical point of I ∞ ( u ) corresponding to the mountain pass theorem, that is,

I ∞ ( ω ) = inf γ ∈ Γ max t ∈ [ 0 , 1 ] I ∞ ( γ ( t ) ) ,

where

Γ = { γ ∈ C ( [ 0 , 1 ] , H 1 ( R 3 ) ) : γ ( 0 ) = 0 , I ∞ ( γ ( 1 ) ) < 0 ) } .

Particularly,

(2.5) I ∞ ( ω ) = max t ≥ 0 I ∞ ( t ω ) .

In order to obtain the critical points of I f , k , g ( u ) , we need the condition that ‖ g ‖ H − 1 and ∣ k ∣ 2 are sufficiently small and the properties of I f , k , g are as follows:

I f , k , g ( u ) has a unique critical point u 1 in a neighborhood of 0 which is a local minimizer.
All of the critical points except u 1 can be gained by means of the following constraint problem:

J f , k , g ( v ) = max t > 0 I f , k , g ( t v ) : Σ + → R ,

where

Σ = { v ∈ H 1 ( R 3 ) : ‖ v ‖ = 1 } , Σ + = { v ∈ Σ : v + ≢ 0 } .

Observing that if k ≡ g ≡ 0 ,

(2.6) J f , 0 , 0 ( v ) = I f , 0 , 0 ∫ R 3 f ( x ) v + p + 1 d x − 1 p − 1 v = 1 2 − 1 p + 1 ∫ R 3 f ( x ) v + p + 1 d x − 2 p − 1 .

In particular,

J 1 , 0 , 0 ( v ) = 1 2 − 1 p + 1 ∫ R 3 v + p + 1 d x − 2 p − 1

and

inf v ∈ Σ + J 1 , 0 , 0 ( v ) = J 1 , 0 , 0 ω ‖ ω ‖ = 1 2 − 1 p + 1 ‖ ω ‖ ∣ ω ∣ p + 1 2 ( p + 1 ) p − 1 = 1 2 − 1 p + 1 C p + 1 − 2 ( p + 1 ) p − 1 .

Note that J 1 , 0 , 0 ω ‖ ω ‖ = I 1 , 0 , 0 ( ω ) . Thus,

inf v ∈ Σ + J 1 , 0 , 0 ( v ) = J 1 , 0 , 0 ω ‖ ω ‖ = I 1 , 0 , 0 ( ω ) .

Set

f ̲ = inf x ∈ R 3 f ( x ) > 0 , f ¯ = sup x ∈ R 3 f ( x ) ≥ 1 ,

it follows from (2.6) that for all v ∈ Σ +

f ¯ − 2 p − 1 J 1 , 0 , 0 ( v ) = J f ¯ , 0 , 0 ( v ) ≤ J f , 0 , 0 ( v ) ≤ J f ̲ , 0 , 0 ( v ) = f ̲ − 2 p − 1 J 1 , 0 , 0 ( v ) .

Thus,

(2.7) f ¯ − 2 p − 1 I 1 , 0 , 0 ( ω ) ≤ inf v ∈ Σ + J f , 0 , 0 ≤ f ̲ − 2 p − 1 I 1 , 0 , 0 ( ω ) .

In order to explore the properties of I f , k , g , we need the following technical lemmas.

Lemma 2.1

For u ∈ H 1 ( R 3 ) and ε ∈ ( 0 , 1 ) ,
(2.8) ( 1 − ε ) I f 1 − ε , 0 , 0 ( u ) + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x − 1 2 ε ‖ g ‖ H − 1 2 ≤ I f , k , g ( u ) ≤ ( 1 + ε ) I f 1 + ε , 0 , 0 ( u ) + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x + 1 2 ε ‖ g ‖ H − 1 2 .
For v ∈ Σ + and ε ∈ ( 0 , 1 ) , there exists a constant δ 1 > 0 such that if ∣ k ∣ 2 ≤ δ 1 , then
(2.9) ( 1 − ε ) p + 1 p − 1 J f , 0 , 0 ( v ) + o ( 1 ) − 1 2 ε ‖ g ‖ H − 1 2 ≤ J f , k , g ( v ) ≤ ( 1 + ε ) p + 1 p − 1 J f , 0 , 0 ( v ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 .
In particular, there exists a constant d 0 > 0 such that if ‖ g ‖ H − 1 ≤ d 0 , then
inf v ∈ Σ + J f , k , g ( v ) > 0 .

Proof

( i ) For ε ∈ ( 0 , 1 ) , the Hölder inequality and the Young inequality imply that

∫ R 3 g u d x ≤ ‖ g ‖ H − 1 ‖ u ‖ ≤ ε 2 ‖ u ‖ 2 + 1 2 ε ‖ g ‖ H − 1 2 .

Then we have

1 − ε 2 ‖ u ‖ 2 + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x − 1 p + 1 ∫ R 3 f ( x ) ( u + ) p + 1 d x − 1 2 ε ‖ g ‖ H − 1 2 ≤ I f , k , g ( u ) ≤ 1 + ε 2 ‖ u ‖ 2 + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x − 1 p + 1 ∫ R 3 f ( x ) ( u + ) p + 1 d x + 1 2 ε ‖ g ‖ H − 1 2 .

Thus, (2.8) follows.

( i i ) Since

∣ ϕ u ( x ) ∣ = 1 4 π ∫ R 3 k ( y ) ∣ x − y ∣ u 2 ( y ) d y = 1 4 π ∫ B 1 ( 0 ) k ( y ) ∣ x − y ∣ u 2 ( y ) d y + ∫ B 1 c ( 0 ) k ( y ) ∣ x − y ∣ u 2 ( y ) d y ≤ 1 4 π ∣ k ∣ ∞ ∫ B 1 ( 0 ) 1 ∣ x − y ∣ 2 d y 1 2 ∣ u ∣ 4 2 + ∣ k ∣ 4 ∫ B 1 c ( 0 ) 1 ∣ x − y ∣ 4 d y 1 4 ∣ u ∣ 4 2 < C ( k ) ∣ u ∣ 4 2 ,

we obtain

∫ R 3 k ( x ) ϕ u u 2 d x ≤ ∣ ϕ u ∣ ∞ ∣ k ∣ 2 ∣ u ∣ 4 2 ≤ C ( k ) ∣ k ∣ 2 ∣ u ∣ 4 4 ≤ C ( k ) ∣ k ∣ 2 ( ∣ u ∣ 2 2 + ∣ u ∣ p + 1 p + 1 ) .

According to (2.8), for v ∈ Σ + ,

I f , k , g ≥ ( 1 − ε ) I f 1 − ε , 0 , 0 ( u ) − C ( k ) ∣ k ∣ 2 ( ∣ u ∣ 2 2 + ∣ u ∣ p + 1 p + 1 ) − 1 2 ε ‖ g ‖ H − 1 2 = 1 − ε 2 ∫ R 3 ∣ ∇ u ∣ 2 + 1 − 2 C ( k ) ∣ k ∣ 2 1 − ε u 2 d x − 1 p + 1 ∫ R 3 ( f ( x ) + ( p + 1 ) C ( k ) ∣ k ∣ 2 ) ∣ u + ∣ p + 2 d x − 1 2 ε ‖ g ‖ H − 1 2 .

In the same way, we have

I f , k g ≤ ( 1 + ε ) I f 1 + ε , 0 , 0 ( u ) + C ( k ) ∣ k ∣ 2 ( ∣ u ∣ 2 2 + ∣ u ∣ p + 1 p + 1 ) + 1 2 ε ‖ g ‖ H − 1 2 = 1 + ε 2 ∫ R 3 ∣ ∇ u ∣ 2 + 1 + 2 C ( k ) ∣ k ∣ 2 1 + ε u 2 d x − 1 p + 1 ∫ R 3 ( f ( x ) − ( p + 1 ) C ( k ) ∣ k ∣ 2 ) ∣ u + ∣ p + 2 d x + 1 2 ε ‖ g ‖ H − 1 2 .

When ∣ k ∣ 2 is sufficiently small, one has

( 1 − ε ) J f 1 − ε , 0 , 0 ( v ) + o ( 1 ) − 1 2 ε ‖ g ‖ H − 1 2 ≤ J f , k , g ( v ) ≤ ( 1 + ε ) J f 1 + ε , 0 , 0 ( v ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 .

By (2.6), we have

J f , k , g ( u ) ≤ ( 1 + ε ) J f 1 + ε , 0 , 0 ( v ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 = ( 1 + ε ) 1 2 − 1 p + 1 ∫ R 3 f ( x ) 1 + ε v + p + 1 d x − 2 p − 1 + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 = ( 1 + ε ) p + 1 p − 1 J f , 0 , 0 ( v ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 .

The same conclusion is also true for ( 1 − ε ) , then we obtain (2.9).

( i i i ) By (2.7) and (2.9)

inf v ∈ Σ + J f , k , g ( v ) ≥ ( 1 − ε ) p + 1 p − 1 inf v ∈ Σ + J f , 0 , 0 ( v ) − 1 2 ε ‖ g ‖ H − 1 2 ≥ ( 1 − ε ) p + 1 p − 1 f ¯ − 2 p − 1 I 1 , 0 , 0 ( ω ) − 1 2 ε ‖ g ‖ H − 1 2 .

Therefore, inf v ∈ Σ + J f , k , g ( v ) > 0 for sufficiently small ‖ g ‖ H − 1 .□

Now, we consider the properties of a function

I f , k , g : [ 0 , + ∞ ) → R : t → I f , k , g ( t v )

for v ∈ Σ + .

Lemma 2.2

For every v ∈ Σ + , the function t → I f , k , g ( t v ) has at most two critical points in [ 0 , + ∞ ) .
If ‖ g ‖ H − 1 ≤ d 0 ( d 0 is given in Lemma 2.1), then for any v ∈ Σ + there exists a unique t f , k , g ( v ) > 0 such that
J f , k , g ( v ) = I f , k , g ( t f , k , g ( v ) v ) .
Moreover, t f , k , g ( v ) > 0 satisfies
(2.10) t f , k , g ( v ) > 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 ≥ 1 p f ¯ C p + 1 p + 1 1 p − 1
and
(2.11) I f , k , g ″ ( t f , k , g ( v ) v ) ( v , v ) < 0 .
If ‖ g ‖ H − 1 ≤ d 1 and t → I f , k , g ( t v ) has a critical point different from t f , k , g ( v ) v , then it lies in 0 , 1 − 1 p − 1 ‖ g ‖ H − 1 .

Proof

Set

g ( t ) = I f , k , g ( t v ) = t 2 2 + t 4 4 ∫ R 3 k ( x ) ϕ v v 2 d x − t p + 1 p + 1 ∫ R 3 f ( x ) v + p + 1 d x − t ∫ R 3 g v d x .

( i ) We can see that

g ′ ( t ) = t + t 3 ∫ R 3 k ( x ) ϕ v v 2 d x − t p ∫ R 3 f ( x ) v + p + 1 d x − ∫ R 3 g v d x , g ″ ( t ) = 1 + 3 t 2 ∫ R 3 k ( x ) ϕ v v 2 d x − p t p − 1 ∫ R 3 f ( x ) v + p + 1 d x , g ‴ ( t ) = 6 t ∫ R 3 k ( x ) ϕ v v 2 d x − p ( p − 1 ) t p − 2 ∫ R 3 f ( x ) v + p + 1 d x .

Thus,

g ‴ ( t ) > 0 for t < t 0 = 6 ∫ R 3 k ( x ) ϕ v v 2 d x p ( p − 1 ) ∫ R 3 f ( x ) v + p + 1 d x 1 p − 3 , g ‴ ( t ) < 0 for t > t 0 = 6 ∫ R 3 k ( x ) ϕ v v 2 d x p ( p − 1 ) ∫ R 3 f ( x ) v + p + 1 d x 1 p − 3 .

This means that g ″ ( t ) is increasing for t ∈ [ 0 , t 0 ) and g ″ ( t ) is decreasing for t ∈ ( t 0 , + ∞ ) . Moreover, g ″ ( 0 ) = 1 , so that there exists a unique t 1 > t 0 such that g ″ ( t 1 ) = 0 and

(2.12) g ″ ( t ) > 0 when t ∈ [ 0 , t 1 ) , g ″ ( t ) < 0 when t ∈ ( t 1 , + ∞ ) .

For g ″ ( t 1 ) = 0 , that is,

g ″ ( t 1 ) = 1 + 3 t 1 2 ∫ R 3 k ( x ) ϕ v v 2 d x − p t 1 p − 1 ∫ R 3 f ( x ) v + p + 1 d x = 0 ,

then we have

t 1 ≥ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 .

In addition, when t → + ∞ , there is g ″ ( t ) → − ∞ , so that t 1 < + ∞ . Together with g ′ ( 0 ) < 0 and then g ′ ( t ) has at most two zeros t ̲ , t ¯ and they satisfy

0 ≤ t ̲ ≤ t 1 ≤ t ¯ .

( i i ) For the fact that g ( 0 ) = 0 , g ( t ) → − ∞ as t → + ∞ and by (iii) of Lemma 2.1 that sup t > 0 g ( t ) > 0 . As a consequence, there exists t f , k , g ( v ) ≥ t 1 such that I f , k , g ( t f , k , g ( v ) v ) = J f , k , g ( v ) . We claim that t f , k , g ( v ) ≠ t 1 . If it does, then by (2.12) that g ′ ( t ) ≤ 0 for all t > 0 . It contradicts that sup t > 0 g ( t ) > 0 , hence we obtain (2.10). Since g ″ ( t ) = I f , k , g ″ ( t v ) ( v , v ) , (2.11) follows from (2.10).

( i i i ) Assume that g ( t ) has a critical point t ˜ different from t f , k , g ( v ) . According to (2.10) and (2.12), it is easy to see that t ˜ ≤ t 1 .

Next we claim:

(2.13) t ˜ ≤ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 .

For the fact that g ′ ( t ) is increasing in [ 0 , t 1 ) , and g ′ ( t ˜ ) = 0 , if there exists g ′ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 ≥ 0 , then t ˜ ≤ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 is true. Indeed,

g ′ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 ≥ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 − 1 p ∫ R 3 f ( x ) v + p + 1 d x p p − 1 ∫ R 3 f ( x ) v + p + 1 d x − ∫ R 3 g v d x ≥ p − 1 p − 1 − p − p p − 1 1 ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 − ‖ g ‖ H − 1 ≥ p − 1 p − 1 − p − p p − 1 ( f ¯ C p + 1 p + 1 ) − 1 p − 1 − ‖ g ‖ H − 1 .

Therefore, if ‖ g ‖ H − 1 ≤ d 1 , small enough, then g ′ 1 p ∫ R 3 f ( x ) v + p + 1 d x 1 p − 1 ≥ 0 , so that (2.13) is obtained.

It follows from g ′ ( t ˜ ) = 0 that

t ˜ + t ˜ 3 ∫ R 3 k ( x ) ϕ v v 2 d x − t ˜ p ∫ R 3 f ( x ) v + p + 1 d x − ∫ R 3 g v d x = 0 .

That means

t ˜ 1 − t ˜ p − 1 ∫ R 3 f ( x ) v + p + 1 d x ≤ ∫ R 3 g v d x .

By (2.13), we have

t ˜ ≤ 1 − 1 p − 1 ‖ g ‖ H − 1 .

This completes the proof.□

Lemma 2.3

There exists r 1 > 0 , 0 < δ 2 ≤ δ 1 and d 2 ∈ ( 0 , min { d 0 , d 1 } ] such that

If ∣ k ∣ 2 ≤ δ 2 , then I f , k , g ( u ) is strictly convex in B r 1 = { u ∈ H 1 ( R 3 ) : ‖ u ‖ < r 1 } .
If ‖ g ‖ H − 1 ≤ d 2 , then
inf ‖ u ‖ = r 1 I f , k , g ( u ) > 0 .
Moreover, I f , k , g ( u ) has a unique critical point u 1 in B r 1 and it satisfies
I f , k , g ( u 1 ) = inf u ∈ B r 1 I f , k , g ( u ) .

Proof

( i ) For h ∈ H 1 ( R 3 ) , we have

I f , k , g ″ ( u ) ( h , h ) = ‖ h ‖ 2 − p ∫ R 3 f ( x ) u + p − 1 h 2 d x + ∫ R 3 k ( x ) ϕ u h 2 + 2 ∫ R 3 k ( y ) ∣ x − y ∣ u ( y ) h ( y ) d y u h d x ≥ ‖ h ‖ 2 − p ∫ R 3 f ( x ) u + p − 1 h 2 d x + 2 ∫ R 3 k ( x ) ∫ R 3 k ( y ) ∣ x − y ∣ u ( y ) h ( y ) d y u h d x ≥ ‖ h ‖ 2 ( 1 − p f ¯ C p + 1 p + 1 ‖ u ‖ p − 1 ) + 2 ∫ R 3 k ( x ) ∫ R 3 k ( y ) ∣ x − y ∣ u ( y ) h ( y ) d y u h d x .

For the formula

(2.14) ∫ R 3 k ( y ) ∣ x − y ∣ u ( y ) h ( y ) d y ≤ ∣ k ∣ ∞ ∫ B 1 ( x ) u ( y ) h ( y ) ∣ x − y ∣ d y + ∫ B 1 c ( x ) k ( y ) u ( y ) h ( y ) ∣ x − y ∣ d y ≤ ∣ k ∣ ∞ ∫ B 1 ( x ) 1 ∣ x − y ∣ 2 d y 1 2 ∫ B 1 ( x ) u 4 d y 1 4 ∫ B 1 ( x ) h 4 d y 1 4 + ∣ k ∣ 4 ∫ B 1 c ( x ) 1 ∣ x − y ∣ 4 d y 1 4 ∫ B 1 c ( x ) u 4 d y 1 4 ∫ B 1 c ( x ) h 4 d y 1 4 ≤ C ( k ) ∣ h ∣ 4 ,

similarly that

2 ∫ R 3 k ( x ) ∫ R 3 k ( y ) ∣ x − y ∣ u ( y ) h ( y ) d y u h d x ≥ − C ( k ) ∣ k ∣ 2 ∣ h ∣ 4 2 ∣ u ∣ 4 ≥ − C ( k ) ∣ k ∣ 2 ‖ h ‖ 2 ,

then we have

I f , k , g ″ ( u ) ( h , h ) ≥ ‖ h ‖ 2 ( 1 − p f ¯ C p + 1 p + 1 ‖ u ‖ p − 1 − C ( k ) ∣ k ∣ 2 ) .

When ∣ k ∣ 2 is sufficiently small, for u ∈ B r 1 with r 1 satisfying 1 − p f ¯ C p + 1 p + 1 r 1 p − 1 = 0 , from the aforementioned inequality it follows that I f , k , g ″ ( u ) is positive definite for u ∈ B r 1 and I f , k , g ( u ) is strictly convex in B r 1 .

( i i ) For ‖ u ‖ = r 1 and 1 − p f ¯ C p + 1 p + 1 r 1 p − 1 = 0 , there holds

I f , k , g ( u ) = 1 2 ‖ u ‖ 2 + 1 4 ∫ R 3 k ( x ) ϕ u u 2 d x − 1 p + 1 ∫ R 3 f ( x ) ( u + ) p + 1 d x − ∫ R 3 g ( x ) u d x ≥ 1 2 − f ¯ p + 1 C p + 1 p + 1 r 1 p − 1 r 1 2 − ‖ g ‖ H − 1 r 1 ≥ 1 2 − 1 p ( p + 1 ) r 1 2 − ‖ g ‖ H − 1 r 1 .

Therefore, when ‖ g ‖ H − 1 ≤ d 1 is small enough, we have

inf ‖ u ‖ = r 1 I f , k , g ( u ) > 0 .

Since I f , k , g ( u ) is strictly convex in B r 1 and inf ‖ u ‖ = r 1 I f , k , g ( u ) > I f , k , g ( 0 ) = 0 , so that there exists a unique critical point u 1 of I f , k , g ( u ) in B r 1 and it satisfies

I f , k , g ( u 1 ) = inf ‖ u ‖ ≤ r 1 I f , k , g ( u ) ,

which implies that u 1 is a local minimum of I f , k , g ( u ) .□

Theorem 2.1

Let d 3 = min d 2 , 1 − 1 p r 1 > 0 and assume that ‖ g ‖ H − 1 < d 3 , then

J f , k , g ( v ) ∈ C 1 ( Σ + , R ) and
(2.15) J f , k , g ′ ( v ) h = t f , k , g ( v ) I f , k , g ′ ( t f , k , g ( v ) v ) h
for all h ∈ T v Σ + = { h ∈ H 1 ( R 3 ) : ⟨ h , v ⟩ = 0 } .
v ∈ Σ + is a critical point of J f , k , g ( v ) if and only if t f , k , g ( v ) v ∈ H 1 ( R 3 ) is a critical point of I f , k , g ( u ) .
The set of critical points of I a , K , f ( u ) has the following form:
(2.16) { t f , k , g ( v ) v : v ∈ Σ + , J f , k , g ′ ( v ) = 0 } ∪ { u 1 } .

Proof

( i ) By (2.11), we obtain

d 2 d t 2 t = t f , k , g ( v ) I f , k , g ( t v ) < 0 .

Then by the implicit function theorem, we have that t f , k , g ( v ) ∈ C 1 ( Σ + , ( 0 , + ∞ ) ) . Therefore,

J f , k , g ( v ) = I f , k , g ( t f , k , g ( v ) v ) ∈ C 1 ( Σ + , R ) .

Since

(2.17) I f , k , g ′ ( t f , k , g ( v ) v ) v = 0 ,

so that

J f , k , g ′ ( v ) h = I f , k , g ′ ( t f , k , g ( v ) v ) [ t f , k , g ( v ) h + ( t f , k , g ( v ) , h ) v ] = t f , k , g ( v ) I f , k , g ′ ( t f , k , g ( v ) v ) h

for h ∈ T v Σ + = { h ∈ H 1 ( R 3 ) , ⟨ h , v ⟩ = 0 } .

( i i ) By means of ( i ) , J f , k , g ′ ( v ) = 0 if and only if

I f , k , g ′ ( t f , k , g ( v ) v ) h = 0 for h ∈ T v Σ + .

Then by (2.17), it is equivalent to I f , k , g ′ ( t f , k , g ( v ) v ) = 0 .

( i i i ) Suppose that u ∈ H 1 ( R 3 ) is a critical point of I f , k , g ′ ( u ) . We write u = t v with v ∈ Σ + and t ≥ 0 . Because of Lemma 2.2( i i i ), we have

either t = t f , k , g ( v ) or t ≤ 1 − 1 p − 1 ‖ g ‖ H − 1 .

Hence, u ∈ H 1 ( R 3 ) is corresponding to a critical point of I f , k , g ( u ) or

‖ u ‖ = t ≤ 1 − 1 p − 1 ‖ g ‖ H − 1 ≤ r 1 .

By Lemma 2.3, I f , k , g ( u ) has a unique critical point in B r 1 and it is u 1 .□

3 The Palais-Smale condition for I f , k , g ( u ) and J f , k , g ( v )

Now, we study the break down of the Palais-Smale condition for I f , k , g ( u ) and J f , k , g ( v ) . The unique positive solution ω ( x ) of (2.2) plays an important role to describe the asymptotic behavior of the Palais-Smale sequence for I f , k , g ( u ) .

Theorem 3.1

Let { u n } be a ( PS ) c sequence of I f , k , g ( u ) , i.e., u n ∈ H 1 ( R 3 ) satisfies

I f , k , g ( u n ) → c , I f , k , g ′ ( u n ) → 0 ,

as n → + ∞ . Then there exist a subsequence, still called { u n } , and a solution u ¯ of (1.1), a number k ∈ N ∪ { 0 } , k functions ω 1 , … , ω k of H 1 ( R 3 ) and k sequences of points ( y n j ) , y n j ∈ R 3 , 0 ≤ j ≤ k , such that

∣ y n j ∣ → + ∞ , ∣ y n j − y n i ∣ → + ∞ , if i ≠ j , n → + ∞ ;
u n − ∑ j = 1 k ω j ( ⋅ − y n j ) → u ¯ in H 1 ( R 3 ) ;
I f , k , g ( u n ) → I f , k , g ( u ¯ ) + ∑ j = 1 k I ∞ ( ω j ) ;
ω j are nontrivial weak solutions of (2.2).

Proof

This result can be proved by the argument in [10, Lemma 4.1], so we omit it here.□

Proposition 3.1

Suppose that ∣ k ∣ 2 ≤ δ 1 , where δ 1 is given in Lemma 2.1, ‖ g ‖ H − 1 ≤ d 3 , where d 3 is given in Lemma 2.1, if there holds

J f , k , g → ∞ as dist ( v , ∂ Σ + ) ≡ inf { ‖ v − u ‖ : u ∈ Σ , u + ≡ 0 } → 0 .
Assume that { v n } ⊂ Σ + satisfies as n → + ∞

(3.1) J f , k , g ( v n ) → c for some c > 0 ,

(3.2) ‖ J f , k , g ′ ( v n ) ‖ T v n ∗ Σ + ≡ sup { J f , k , g ′ ( v n ) h : h ∈ T v n Σ + , ‖ h ‖ = 1 } → 0 .

Then there exist a subsequence, still called { v n } , and a solution u ¯ of (1.1), a number k ∈ N ∪ { 0 } , k functions ω 1 , … , ω k of H 1 ( R 3 ) and k sequences of points ( y n j ) , y n j ∈ R 3 , 0 ≤ j ≤ k , such that

∣ y n j ∣ → + ∞ , ∣ y n j − y n i ∣ → + ∞ , if i ≠ j , n → + ∞ ;
v n − u ¯ + ∑ j = 1 k ω j ( ⋅ − y n j ) ‖ u ¯ + ∑ j = 1 k ω j ( ⋅ − y n j ) ‖ → 0 in H 1 ( R 3 ) ;
J f , k , g ( v n ) → I f , k , g ( u ¯ ) + ∑ j = 1 k I ∞ ( ω j ) ;
ω j are nontrivial weak solutions of (2.2).

Proof

( i ) By ( i i ) of Lemma 2.1 and (2.6), we have

J f , k , g ( v ) ≥ ( 1 − ε ) p + 1 p − 1 J f , 0 , 0 ( v ) + o ( 1 ) − 1 2 ε ‖ g ‖ H − 1 ≥ ( 1 − ε ) p + 1 p − 1 1 2 − 1 p + 1 ∫ R 3 f ( x ) v + p + 1 d x − 2 p − 1 + o ( 1 ) − 1 2 ε ‖ g ‖ H − 1 .

Since dist ( v , ∂ Σ + ) → 0 implies v + → 0 in H 1 ( R 3 ) , in particular,

∫ R 3 f ( x ) v + p + 1 d x → 0 ,

then ( i ) follows.

( i i ) According to (2.10) and (2.15), we have

‖ I f , k , g ′ ( t f , k , g ( v n ) v n ) ‖ = 1 t f , k , g ( v n ) ‖ J f , k , g ′ ( v n ) ‖ T v n ∗ Σ + ≤ ( p f ¯ C p + 1 p + 1 ) 1 p − 1 ‖ J f , k , g ′ ( v n ) ‖ T v n ∗ Σ + → 0 , as n → + ∞ .

We also have

I f , k , g ( t f , k , g ( v n ) v n ) = J f , k , g ( v n ) → c ,

as n → + ∞ . Then applying Theorem 3.1, we obtain conclusion ( i i ) .□

Corollary 3.2

Suppose that ‖ g ‖ H − 1 ≤ d 3 , ∣ k ∣ 2 ≤ δ 1 , then J f , k , g ( v ) satisfies the ( PS ) c condition for c < I f , k , g ( u 1 ) + I ∞ ( ω ) .

Proof

By Proposition 3.1, ( PS ) c breaks down only for

c = I f , k , g ( u ¯ ) + ∑ j = 1 k I ∞ ( ω j ) ,

where u ¯ ∈ H 1 ( R ) 3 is a critical point of I f , k , g ( u ) . By (2.16), (iii) of Lemma 2.1 and

I f , k , g ( u 1 ) = inf u ∈ B r 1 I f , k , g ( u ) ≤ I f , k , g ( 0 ) = 0 ,

it follows that

I f , k , g ( u 1 ) = inf { I f , k , g ( u ¯ ) : u ¯ ∈ H 1 ( R 3 ) is a critical point of I f , k , g ( u ) } .

Therefore, the lowest level of breaking down of ( PS ) c is I f , k , g ( u 1 ) + I ∞ ( ω ) .□

Later, we will find two critical points below the level

I f , k , g ( u 1 ) + I ∞ ( ω ) .

In order to get the fourth solution, we need some properties of the functional J f , 0 , 0 under ( f 1 )–( f 3 ). The next lemma is a rather standard result, which can be seen in [33,34].

Lemma 3.1

Assume ( f 1 )–( f 2 ) hold, then

inf v ∈ Σ + J f , 0 , 0 ( v ) = I ∞ ( ω ) .
inf v ∈ Σ + J f , 0 , 0 ( v ) is not attained.
J f , 0 , 0 ( v ) satisfies the, ( PS ) c condition for c ∈ ( − ∞ , I ∞ ( ω ) ) ∪ ( I ∞ ( ω ) , 2 I ∞ ( ω ) ) .

In order to be aware of the existence of a critical point of J f , 0 , 0 , the next lemma can be found in [29, Proposition 1.13] and it is essential to obtain our main result.

Lemma 3.2

Assume ( f 1 )–( f 3 ) hold, then J f , 0 , 0 ( v ) has at least one critical point v f ( x ) ∈ Σ + , which can be characterized as

J f , 0 , 0 ( v f ) = inf γ ∈ Γ sup y ∈ R 3 J f , 0 , 0 ( γ ( y ) ) ,

where

(3.3) Γ = γ ∈ C ( R 3 , Σ + ) : γ ( y ) = ω ( ⋅ − y ) ‖ ω ‖ for large ∣ y ∣ .

Moreover, v f ( x ) satisfies

J f , 0 , 0 ( v f ) ∈ ( I ∞ ( ω ) , 2 I ∞ ( ω ) ) .

Proposition 3.2

Assume ( f 1 )–( f 3 ) hold, then for any ε > 0 , there exists d ( ε ) ∈ ( 0 , d 3 ] such that ‖ g ‖ H − 1 ≤ d ( ε ) , also 0 < δ ( ε ) < δ 2 such that ∣ k ∣ 2 ≤ δ ( ε ) ,

inf v ∈ Σ + J f , k , g ( v ) ∈ [ I ∞ ( ω ) − ε , I ∞ ( ω ) + ε ] .
J f , k , g ( v ) satisfies the ( PS ) c condition for c ∈ ( − ∞ , I f , k , g ( u 1 ) + I ∞ ( ω ) ) ∪ ( I f , k , g ( u 1 ) + I ∞ ( ω ) , 2 I ∞ ( ω ) − ε ) .

Proof

( i ) According to Lemmas 3.1(i) and 2.1(ii), when ∣ k ∣ 2 and ‖ g ‖ H − 1 are sufficiently small, then ( i ) can be easily obtained.

( i i ) Similar to the statement in Corollary 3.2, ( PS ) c is collapsed only when c = I f , k , g ( u ¯ ) + ∑ j = 1 k I ∞ ( ω ) , where u ¯ is a critical point of I f , k , g ( u ) and k ∈ N . By Lemma 2.1(iii), u ¯ = u 1 , which is a local minimum or

I f , k , g ( u ¯ ) ≥ inf v ∈ Σ + J f , k , g ( v ) ≥ I ∞ ( ω ) − ε .

Therefore, the ( PS ) c condition is not established for

c = I f , k , g ( u 1 ) + I ∞ ( ω ) or c ≥ 2 I ∞ ( ω ) − ε .

As a consequence, ( i i ) is easily obtained.□

4 Category of [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ]

In order to find the second and third positive solutions, we use notation

[ J f , k , g ≤ c ] = { v ∈ Σ + : J f , k , g ( v ) ≤ c }

for c ∈ R . To find critical points of J f , k , g ( v ) , we will show that for sufficiently small ε > 0

[ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ]

is not empty and

(4.1) cat ( [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ] ) ≥ 2 ,

provided g ≥ 0 , g ≢ 0 and ‖ g ‖ H − 1 is small enough, where cat stands for the Lusternik-Schnirelman category. To prove (4.1), we need some preliminaries.

Proposition 4.1

Assume ( f 1 )–( f 3 ) hold, suppose that ‖ g ‖ H − 1 ≤ d 3 , g ≥ 0 , g ≢ 0 , also suppose k ≥ 0 , k ≢ 0 , 0 < δ 3 < δ 2 such that ∣ k ∣ 2 ≤ δ 3 , then there exists λ 0 > 0 such that

(4.2) I f , k , g ( u 1 + t ω λ ) < I f , k , g ( u 1 ) + I ∞ ( ω )

for all ∣ λ ∣ > λ 0 and t ≥ 0 . Here ω λ ( ⋅ ) = ω ( ⋅ + λ e ) and e is a unit vector in R 3 .

Proof

A straightforward computation gives that

I f , k , g ( u 1 + t ω λ ) = 1 2 ∫ R 3 ∣ ∇ ( u 1 + t ω λ ) ∣ 2 + ∣ u 1 + t ω λ ∣ 2 d x + 1 4 ∫ R 3 k ( x ) ϕ u 1 + t ω λ ( u 1 + t ω λ ) 2 d x − 1 p + 1 ∫ R 3 f ( x ) ( u 1 + t ω λ ) p + 1 d x − ∫ R 3 g ( x ) ( u 1 + t ω λ ) d x = 1 2 ∫ R 3 ∣ ∇ u 1 ∣ 2 + u 1 2 d x + 1 4 ∫ R 3 k ( x ) ϕ u 1 u 1 2 d x − 1 p + 1 ∫ R 3 f ( x ) u 1 p + 1 d x − ∫ R 3 f ( x ) u 1 d x + 1 2 ∫ R 3 ∣ ∇ t ω λ ∣ 2 + ( t ω λ ) 2 d x − 1 p + 1 ∫ R 3 ( t ω λ ) p + 1 d x + ∫ R 3 ∇ u 1 ∇ t ω λ + u 1 t ω λ d x + ∫ R 3 k ( x ) ϕ u 1 u 1 t ω λ d x − ∫ R 3 f ( x ) u 1 p t ω λ d x − ∫ R 3 g ( x ) t ω λ d x + 1 p + 1 ∫ R 3 ( t ω λ ) p + 1 d x − 1 p + 1 ∫ R 3 f ( x ) [ ( u 1 + t ω λ ) p + 1 − u 1 p + 1 − ( p + 1 ) u 1 p t ω λ ] d x − 1 4 ∫ R 3 k ( x ) ϕ u 1 u 1 2 d x + 1 4 ∫ R 3 k ( x ) ϕ u 1 + t ω λ ( u 1 + t ω λ ) 2 d x − ∫ R 3 k ( x ) ϕ u 1 u 1 t ω λ d x .

Since I f , k , g ( u 1 + t w λ ) → − ∞ as t → + ∞ , and I f , k , g ( u 1 + t w λ ) → I f , k , g ( u 1 ) < 0 as t → 0 , we only consider t ∈ [ t 2 , t 1 ] .

(4.3) I f , k , g ( u 1 + t ω λ ) = I f , k , g ( u 1 ) + I ∞ ( t ω λ ) + ⟨ I f , k , g ′ ( u 1 ) , t ω λ ⟩ + 1 p + 1 ∫ R 3 ( 1 − a ( x ) ) ( t ω λ ) p + 1 − 1 p + 1 ∫ R 3 f ( x ) [ ( u 1 + t ω λ ) p + 1 − u 1 p + 1 − ( t ω λ ) p + 1 − ( p + 1 ) u 1 p t ω λ ] d x + 1 4 ∫ R 3 ( k ( x ) ϕ u 1 + t ω λ ( u 1 + t ω λ ) 2 − k ( x ) ϕ u 1 u 1 2 ) d x − ∫ R 3 k ( x ) ϕ u 1 u 1 t ω λ d x = I f , k , g ( u 1 ) + I ∞ ( t ω λ ) + ( I ) − ( I I ) + 1 4 ∫ R 3 k ( x ) ϕ u 1 ( 2 t u 1 ω λ + t 2 ω λ 2 ) d x + 1 4 ∫ R 3 k ( x ) t φ ( x ) ( u 1 2 + 2 t u 1 ω λ + t 2 ω λ 2 ) + k ( x ) ϕ t ω λ ( u 1 2 + 2 t u 1 ω λ + t 2 ω λ 2 ) − ∫ R 3 k ( x ) ϕ u 1 u 1 t ω λ d x ,

where

φ ( x ) = 1 t [ ϕ u 1 + t ω λ ( x ) − ϕ u 1 ( x ) − ϕ t ω λ ( x ) ] = 1 4 π ∫ R 3 2 k ( y ) ∣ x − y ∣ u 1 ( y ) ω λ ( y ) d y .

Similar to (2.14), we can obtain that there is a positive constant C which is independent of ∣ λ ∣ such that ∣ φ ∣ ∞ ≤ C ∣ k ∣ ∞ ‖ u ‖ ‖ ω ‖ . Moreover, by Fubini’s theorem, we have

(4.4) 1 4 ∫ R 3 k ( x ) φ u 1 2 d x = 1 2 ∫ R 3 ∫ R 3 k ( x ) k ( y ) u 1 ( y ) ω λ ( y ) u 1 2 ( x ) 4 π ∣ y − x ∣ d y d x = 1 2 ∫ R 3 ∫ R 3 k ( x ) k ( y ) u 1 ( y ) ω λ ( y ) u 1 2 ( x ) 4 π ∣ y − x ∣ d x d y = 1 2 ∫ R 3 k ( x ) ϕ u 1 u 1 ω λ d x .

Similarly, we obtain

(4.5) 1 4 ∫ R 3 k ( x ) ϕ t w λ u 1 2 d x = t 2 4 ∫ R 3 k ( x ) ϕ u 1 w λ 2 d x

and

(4.6) 1 4 ∫ R 3 k ( x ) ϕ t w λ u 1 t w λ d x = t 3 4 ∫ R 3 k ( x ) φ w λ 2 d x .

By (2.5) (3.3), (4.4), (4.5) and (4.6), we obtain

(4.7) I f , k , g ( u 1 + t ω λ ) ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) + ( I ) − ( I I ) + t 1 2 ∫ R 3 k ( x ) ϕ u 1 ω λ 2 d x + t 1 2 2 ∫ R 3 k ( x ) φ u 1 ω λ d x + t 1 3 4 ∫ R 3 k ( x ) φ ω λ 2 d x + t 1 2 4 ∫ R 3 k ( x ) ϕ u 1 ω λ 2 d x + t 1 3 2 ∫ R 3 k ( x ) φ ω λ 2 d x + t 1 4 4 ∫ R 3 k ( x ) ϕ ω λ ω λ 2 d x .

Recalling the fact that for some c > 0 ,

ω ( ∣ x ∣ ) ∣ x ∣ N − 1 2 e ∣ x ∣ → c , as ∣ x ∣ → + ∞

(see [30,31, 32,35]). In particular, we have

there exists a constant C 0 > 0 such that
ω ( x ) ≤ C 0 e − ∣ x ∣ for all x ∈ R N ;
for any ε > 0 there exists a constant C ε > 0 such that
ω ( x ) ≥ C ε e − ( 1 + ε ) ∣ x ∣ for all x ∈ R N .

Also note that

for all ( s , t ) ∈ [ 0 , + ∞ ) × [ 0 , + ∞ ) ,
( s + t ) p + 1 − s p + 1 − t p + 1 − ( p + 1 ) s p t ≥ 0 .
for any τ > 0 one can find a constant A ( τ ) > 0 such that
( s + t ) p + 1 − s p + 1 − t p + 1 − ( p + 1 ) s p t ≥ A ( τ ) t 2
for all ( s , t ) ∈ [ τ , + ∞ ) × [ 0 , + ∞ ) .

Thus, set A = A ( min ∣ x ∣ ≤ 1 u 1 ) > 0 , then we have

(4.8) ( II ) ≥ ∫ ∣ x ∣ ≤ 1 1 p + 1 ∫ R 3 f ( x ) [ ( u 1 + t w λ ) p + 1 − u 1 p + 1 − ( t w λ ) p + 1 − ( p + 1 ) u 1 p t w λ ] d x ≥ f ̲ A t 2 2 p + 1 ∫ ∣ x ∣ ≤ 1 ω ( x + λ e ) 2 d x ≥ C ε ′ e − 2 ( 1 + ε ) ∣ λ ∣ .

Also by the condition ( f 3 ) , there holds

(4.9) ( I ) ≤ 1 p + 1 ∫ R 3 C e − ( 2 + δ ) ∣ x ∣ C 0 p + 1 e − ( p + 1 ) ∣ x + λ e ∣ d x ≤ C ′ e − min { p + 1 , 2 + δ } ∣ λ ∣ .

For the form

(4.10) ∣ ϕ u ( x ) ∣ = 1 4 π ∫ R 3 k ( y ) ∣ x − y ∣ u 2 ( y ) d y ≤ 1 4 π ∣ k ∣ ∞ ∫ B 1 ( x ) u 2 ( y ) ∣ x − y ∣ d y + ∫ B 1 c ( x ) u 2 ( y ) ∣ x − y ∣ d y ≤ 1 4 π ∣ k ∣ ∞ ∫ B 1 ( x ) 1 ∣ x − y ∣ 2 d y 1 / 2 ∫ B 1 ( x ) u 4 d y 1 / 2 + ∫ B 1 c ( x ) 1 ∣ x − y ∣ 4 d y 1 / 4 ∫ B 1 c ( x ) ∣ u ∣ 8 / 3 d y 3 / 4 ≤ C ∣ k ∣ ∞ ‖ u ‖ 2 ,

we have

∫ R 3 k ( x ) ϕ u 1 ω λ 2 d x ≤ ∣ ϕ u 1 ∣ ∞ ∫ R 3 k ( x ) ω λ 2 d x ≤ C ∣ k ∣ 2 ∣ ω λ ∣ 4 2 .

Similarly, we can obtain

∫ R 3 k ( x ) ϕ ω λ ω λ 2 d x ≤ ∣ ϕ ω λ ∣ ∞ ∣ k ∣ 2 ∣ ω λ ∣ 4 2 , ∫ R 3 k ( x ) φ ω λ 2 d x ≤ ∣ φ ∣ ∞ ∣ k ∣ 2 ∣ ω λ ∣ 4 2 , ∫ R 3 k ( x ) φ u 1 ω λ d x ≤ ∣ φ ∣ ∞ ∣ k ∣ 2 ∣ u 1 ∣ 4 ∣ ω λ ∣ 4 .

Choosing 2 ε < δ for any ε > 0 , we can find λ 0 > 0 when ∣ λ ∣ ≥ λ 0 , there is ( I ) < ( II ) . Also when ∣ k ∣ 2 is sufficiently small, then we obtain

I f , k , g ( u 1 + t ω λ ) < I f , k , g ( u 1 ) + I ∞ ( ω λ ) for ∣ λ ∣ > λ 0 .

Thus, the proof is complete.□

Now, we are going to show (4.1). First, we recall the definition of the Lusternik-Schnirelman category.

Definition 4.1

For a topological space X , we say that a nonempty, closed subset A ⊂ X is contractible to a point in X if and only if there exists a continuous mapping
η : [ 0 , 1 ] × A → X ,
such that for some x 0 ∈ X
1 ∘ η ( 0 , x ) = x for all x ∈ A , 2 ∘ η ( 1 , x ) = x 0 for all x ∈ A .
Define
cat ( X ) = min { k ∈ N : there exist closed subsets A 1 , … , A k ⊂ X such that 1 ∘ A j is contractible to a point in X for all j , 2 ∘ ∪ j = 1 k A j = X } .
If there do not exist finitely many closed subset A 1 , … , A k ⊂ X such that 1 ∘ and 2 ∘ hold, then it is said that cat ( X ) = + ∞ .

For fundamental properties of Lusternik-Schnirelman category, we refer to Ambrosetti [36] and Schwartz [37]. Here we use the following property:

Proposition 4.2

Suppose that M is a Hilbert manifold and Ψ ∈ C 1 ( M , R ) . Assume that for c 0 ∈ R and k ∈ N

1 ∘ Ψ ( x ) satisfies the ( PS ) c condition for c ≤ c 0 , 2 ∘ cat ( { x ∈ M : Ψ ( x ) ≤ c 0 } ) ≥ k .

Then Ψ ( x ) has at least k critical points in { x ∈ M : Ψ ( x ) ≤ c 0 } .

To estimate the Lusternik-Schnirelman category, the following lemma is essential, its proof can be found in [29, Lemma 2.5].

Lemma 4.1

Let N ≥ 1 and X be a topological space. Suppose that there exist two continuous mappings

F : S N − 1 = { y ∈ R N : ∣ y ∣ = 1 } → X , G : X → S N − 1 ,

such that G ∘ F is homotopic to identity: S N − 1 → S N − 1 ; x ↦ x , that is, there exists a continuous mapping h : [ 0 , 1 ] × S N − 1 → S N − 1 such that

h ( 0 , x ) = ( G ∘ F ) ( x ) for all x ∈ S N − 1 , h ( 1 , x ) = x for all x ∈ S N − 1 .

Then

cat ( X ) ≥ 2 .

From now on, we construct two mappings

F : S 2 → [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ] ,

G : [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ] → S 2 ,

so that G ∘ F is homotopic to the identity.

Define a mapping F λ : S 2 → Σ + in the following way: according to Proposition 4.1, we know that for ∣ λ ∣ ≥ λ 0 and for all t ≥ 0 ,

I f , k , g ( u 1 + t ω ( x + λ e ) ) < I f , k , g ( u 1 ) + I ∞ ( ω ) .

For ∣ λ ∣ ≥ λ 0 , we find s = s ( k , g , λ ) > 0 such that

u 1 + s ω ( x + λ e ) = t f , k , g u 1 + s ω ( x + λ e ) ‖ u 1 + s ω ( x + λ e ) ‖ u 1 + s ω ( x + λ e ) ‖ u 1 + s ω ( x + λ e ) ‖ ,

that is

(4.11) t f , k , g u 1 + s ω ( x + λ e ) ‖ u 1 + s ω ( x + λ e ) ‖ = ‖ u 1 + s ω ( x + λ e ) ‖ .

This implies that

J f , k , g u 1 + s ω ( x + λ e ) ‖ u 1 + s ω ( x + λ e ) ‖ = I f , k , g ( u 1 + s ω ( x + λ e ) ) < I f , k , g ( u 1 ) + I ∞ ( ω ) .

Proposition 4.3

Assume ( f 1 )–( f 3 ) hold, then there exist d 4 ∈ ( 0 , d 3 ] and λ 1 > λ 0 such that for any ‖ g ‖ H − 1 ≤ d 4 and for any ∣ λ ∣ ≥ λ 1 there exists a unique s = s ( k , g , λ ) > 0 in a neighborhood of 1 satisfying (4.11). Moreover,

{ λ ∈ R : ∣ λ ∣ ≥ λ 1 } → ( 0 , + ∞ ) ; λ ↦ s ( k , g , λ )

is continuous.

Proof

Set

Φ ( s , k , g , λ ) = ⟨ I f , k , g ′ ( u 1 + s ω ( x + λ e ) ) , u 1 + s ω ( x + λ e ) ⟩ = ‖ u 1 + s ω ( x + λ e ) ‖ 2 + ∫ R 3 k ( x ) ϕ u 1 + s ω ( x + λ e ) ( u 1 + s ω ( x + λ e ) ) 2 d x − ∫ R 3 f ( x ) ( u 1 + s ω ( x + λ e ) ) p + 1 d x − ∫ R 3 g ( x ) ( u 1 + s ω ( x + λ e ) ) d x .

Then (4.11) holds if and only if Φ ( s , k , g , λ ) = 0 . Since ω ( x ) is a unique positive radial solution of (2.2), that is,

‖ ω ‖ 2 − ∫ R 3 ω p + 1 d x = 0 .

Moreover, by ( f 2 ) and ( k ) , one has

Φ ( s , k , 0 , λ ) = ⟨ I f , k , 0 ′ ( s ω ( x + λ e ) ) , s ω ( x + λ e ) ⟩ = ‖ s ω ‖ 2 + ∫ R 3 k ( x ) ϕ s ω ( x + λ e ) ( s ω ( x + λ e ) ) 2 d x − ∫ R 3 f ( x ) ( s ω ( x + λ e ) ) p + 1 d x → 0 ,

as ∣ λ ∣ → + ∞ and s → 1 .

∂ ∂ s s = 1 Φ ( s , k , 0 , λ ) = 2 ‖ ω ‖ 2 + 4 ∫ R 3 k ( x ) ϕ ω ( x + λ e ) ω ( x + λ e ) 2 d x − ( p + 1 ) ∫ R 3 f ( x ) ω ( x + λ e ) p + 1 d x → − ( p − 1 ) ‖ ω ‖ 2 < 0 ,

as ∣ λ ∣ → + ∞ .

Therefore, by the implicit function theorem, one can find a unique s = s ( k , g , λ ) in a neighborhood of 1 such that Φ ( s , k , g , λ ) = 0 . The continuity of λ ↦ s ( k , g , λ ) is also clear.□

Now, define a function F λ : S 2 → Σ + by

F λ ( y ) = u 1 + s ( k , g , λ y ) ω ( x + λ e ) ‖ u 1 + s ( k , g , λ y ) ω ( x + λ e ) ‖

for ‖ g ‖ H − 1 ≤ d 4 and ∣ λ ∣ ≥ λ 1 .

Proposition 4.4

For 0 < ‖ g ‖ H − 1 ≤ d 4 and ∣ λ ∣ ≥ λ 1 , there exists ε 0 ( λ ) > 0 such that

F λ ( S 2 ) ⊂ [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε 0 ( λ ) ] .

Proof

By Proposition 4.3, it follows that

F λ ( S 2 ) ⊂ [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) ] .

Since F λ is continuous and then F λ ( S 2 ) is compact, the conclusion is obtained.□

Thus, we construct a mapping

F λ : S 2 → [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε 0 ( ∣ λ ∣ ) ] .

By Proposition 3.1 and Lemma 3.1, the next lemma is similar to the one in [29].

Lemma 4.2

Assume ( f 1 )–( f 2 ) hold, then there exists a constant δ 0 > 0 such that if J f , 0 , 0 ( v ) ≤ I ∞ ( ω ) + δ 0 , then

∫ R 3 x ∣ x ∣ ( ∣ ∇ v ∣ 2 + ∣ v ∣ 2 ) d x ≠ 0 ,

where v ∈ Σ + .

Lemma 4.3

There exists the constant d 5 such that 0 < d 5 ≤ d 4 . If ‖ g ‖ H − 1 ≤ d 5 , ∣ k ∣ 2 ≤ δ 1 , then

(4.12) [ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] ⊂ [ J f , 0 , 0 < I ∞ ( ω ) + δ 0 ] ,

where δ 0 > 0 is defined in Lemma 4.2.

Proof

By (2.9), there is for any ε ∈ ( 0 , 1 )

(4.13) J f , 0 , 0 ( v ) ≤ ( 1 − ε ) − p + 1 p − 1 J f , k , g ( v ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 ,

for all v ∈ Σ + . Note that I f , k , g ( u 1 ) ≤ 0 and for all v ∈ [ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] , then we have J f , k , g ( v ) < I ∞ ( ω ) . Then by (4.13) we have

J f , 0 , 0 ( v ) ≤ ( 1 − ε ) − p + 1 p − 1 I ∞ ( ω ) + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2

for all v ∈ [ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] . Since ε ∈ ( 0 , 1 ) is arbitrary, the desired result is obtained for sufficiently small ‖ g ‖ H − 1 .□

Now we can define

G : [ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] → S 2

G ( v ) = ∫ R 3 x ∣ x ∣ ( ∣ ∇ v ∣ 2 + ∣ v ∣ 2 ) d x ∫ R 3 x ∣ x ∣ ( ∣ ∇ v ∣ 2 + ∣ v ∣ 2 ) d x .

By Lemmas 4.2 and 4.3,

∫ R 3 x ∣ x ∣ ( ∣ ∇ v ∣ 2 + ∣ v ∣ 2 ) d x ≠ 0

for all v ∈ [ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] and G ( v ) is well defined.

Proposition 4.5

For sufficiently large ∣ λ ∣ > λ 1 and for sufficiently small ‖ g ‖ H − 1 > 0 and ∣ k ∣ 2 > 0 ,

G ∘ F λ : S 2 → S 2 ; y ↦ G ( F λ ( y ) )

is homotopic to the identity.

Proof

Define

ψ ( θ , y ) : [ 0 , 1 ] × S 2 → S 2

ψ ( θ , y ) = G ∘ F λ 1 − θ ( y ) , θ ∈ [ 0 , 1 ) , y , θ = 1 .

It is easy to see that ψ ( θ , y ) ∈ C ( [ 0 , 1 ] × S 2 → S 2 ) and

ψ ( 0 , y ) = G ∘ F λ ( y ) for all y ∈ S 2 , ψ ( 1 , y ) = y for all y ∈ S 2 ,

for large enough ∣ λ ∣ and sufficiently small ‖ g ‖ H − 1 > 0 and ∣ k ∣ 2 > 0 .□

By Lemma 4.1, we have

Proposition 4.6

For sufficiently large ∣ λ ∣ ≥ λ 1 ,

cat ( [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε 0 ( ∣ λ ∣ ) ] ) ≥ 2 .

Theorem 4.1

Assume ( f 1 )–( f 2 ) hold, then there exists d 6 > 0 , δ 3 > 0 , such that if ‖ g ‖ H − 1 ≤ d 6 , g ≥ 0 , g ≢ 0 , ∣ k ∣ 2 ≤ δ 3 , then J f , k , g ( v ) has at least two critical points in

[ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) ] .

Proof

Since the ( PS ) c condition holds for J f , k , g ( v ) as c ∈ ( − ∞ , I f , k , g ( u 1 ) + I ∞ ( ω ) ) , by Propositions 4.2 and 4.6, the desired result can be easily obtained.□

5 Proof of main result

Define

b f , k , g = inf γ ∈ Γ sup y ∈ R 3 J f , k , g ( γ ( y ) ) ,

where Γ is defined in (3.3). Owing to (2.9) and for any ε ∈ ( 0 , 1 ) , when ∣ k ∣ 2 is small enough, then we acquire that

( 1 − ε ) p + 1 p − 1 b f + o ( 1 ) − 1 2 ε ‖ g ‖ H − 1 2 ≤ b f , k , g ≤ ( 1 + ε ) p + 1 p − 1 b f + o ( 1 ) + 1 2 ε ‖ g ‖ H − 1 2 ,

where b f = inf γ ∈ Γ sup y ∈ R 3 J f , 0 , 0 ( γ ( y ) ) . Thus, the next lemma is easily gained.

Lemma 5.1

For any δ ˜ > 0 there exist d 7 > 0 such that ‖ g ‖ H − 1 ≤ d 7 , also δ 4 > 0 such that ∣ k ∣ 2 ≤ δ 4 , then

b f , k , g ∈ ( b f − δ ˜ , b f + δ ˜ ) .

Together with Lemma 3.2, we have the next result.

Theorem 5.1

There exist d 8 > 0 such that ‖ g ‖ H − 1 ≤ d 8 , also δ 5 > 0 such that ∣ k ∣ 2 ≤ δ 5 , then there is a critical point v f , k , g of J f , k , g ( v ) , which satisfies

J f , k , g ( v f , k , g ) = b f , k , g ≥ I f , k , g ( u 1 ) + I ∞ ( ω ) .

Proof

Since b f ∈ ( I ∞ ( ω ) , 2 I ∞ ( ω ) ) , by Lemma 5.1, letting ‖ g ‖ H − 1 and ∣ k ∣ 2 be sufficiently small, we have

b f , k , g ∈ ( I ∞ ( ω ) + ε , 2 I ∞ ( ω ) − ε )

for some ε > 0 . According to Proposition 3.2(ii), J f , k , g ( v ) satisfies the ( PS ) c condition for c = b f , k , g so that there is a critical point which corresponds to b f , k , g .□

Proof of Theorem 1.1

The first solution is u 1 , which is a local minimum of I f , k , g and satisfies

I f , k , g ( u 1 ) < 0 .

By Theorem 4.1, there exist two critical points v 2 and v 3 in

[ J f , k , g < I f , k , g ( u 1 ) + I ∞ ( ω ) ] .

Let u 2 = t f , k , g ( v 2 ) v 2 and u 3 = t f , k , g ( v 3 ) v 3 be the corresponding solutions satisfying

0 < I f , k , g ( u k ) = J f , k , g ( v k ) < I f , k , g ( u 1 ) + I ∞ ( ω ) for k = 2 , 3 .

Finally, the fourth solution is not difficult to get through Theorem 5.1. Indeed, there is a fourth positive solution v 4 ( x ) . Let u 4 ( x ) be the corresponding positive solution which satisfies

I f , k , g ( u 4 ) = J f , k , g ( v 4 ) ≥ I f , k , g ( u 1 ) + I ∞ ( ω ) .

In conclusion, u 1 , u 2 , u 3 , u 4 are different from each other, thus the main result is obtained. This completes the proof of Theorem 1.1.□

Funding information: J. Zhang was supported by the National Natural Science Foundation of China (No. 11962025). R. Niu was supported by the National Natural Science Foundation of China (No. 11871199) and Doctoral Research Fund of Heilongjiang Institute of Technology (No. 2018BJ05). X. Han was supported by the Natural Science Foundation of Shandong Province (No. ZR2019PA020).
Conflict of interest: The authors state no conflict of interest.

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Received: 2021-08-15

Revised: 2021-09-26

Accepted: 2022-01-15

Published Online: 2022-03-09

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

Positive solutions for a nonhomogeneous Schrödinger-Poisson system

Abstract

1 Introduction and main result

Theorem 1.1

2 Preliminaries and the first solution

Lemma 2.1

Proof

Lemma 2.2

Proof

Lemma 2.3

Proof

Theorem 2.1

Proof

3 The Palais-Smale condition for I f , k , g ( u ) and J f , k , g ( v )

Theorem 3.1

Proof

Proposition 3.1

Proof

Corollary 3.2

Proof

Lemma 3.1

Lemma 3.2

Proposition 3.2

Proof

4 Category of [ J f , k , g ≤ I f , k , g ( u 1 ) + I ∞ ( ω ) − ε ]

Proposition 4.1

Proof

Definition 4.1

Proposition 4.2

Lemma 4.1

Proposition 4.3

Proof

Proposition 4.4

Proof

Lemma 4.2

Lemma 4.3

Proof

Proposition 4.5

Proof

Proposition 4.6

Theorem 4.1

Proof

5 Proof of main result

Lemma 5.1

Theorem 5.1

Proof

Proof of Theorem 1.1

References

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