Existence and multiplicity of non-trivial solutions for a class of modified Schrödinger–Poisson systems

Accepted Manuscript Existence and multiplicity of non-trivial solutions for a class of modified Schr¨odinger–Poisson systems Jianjun Nie, Xian Wu PII: DOI: Reference:

S0022-247X(13)00555-6 http://dx.doi.org/10.1016/j.jmaa.2013.06.011 YJMAA 17673

To appear in:

Journal of Mathematical Analysis and Applications

Received date: 5 March 2012 Please cite this article as: J. Nie, X. Wu, Existence and multiplicity of non-trivial solutions for a class of modified Schr¨odinger–Poisson systems, J. Math. Anal. Appl. (2013), http://dx.doi.org/10.1016/j.jmaa.2013.06.011 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

Existence and Multiplicity of Non-trivial Solutions for a Class of Modified Schr¨odinger-Poisson Systems ∗ Jianjun Nie School of Mathematics and Computer Science, Guizhou Normal College, Guiyang, Guizhou 550018, P.R.China

E-mail: [email protected] Xian Wu

†

Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650092, P.R.China

E-mail: [email protected]

Abstract In the present paper, we study the following modified Schr¨ odinger-Poisson system

    −4u + V (x)u + φu − 21 u4u2 = |u|p−1 u + h(x), in R3 ,    −4φ = u2 ,

in

(P )

R3 ,

where 4 ≤ p + 1 < 12, h ∈ L2 (R3 ) and h 6= 0. This model describes the interaction of a charge particle with an electro-magnetic field. Under some hypotheses on V (x), we prove the existence of a non-trivial ground state solution and two non-trivial ground state solutions for the system (P ). Key Words: modified Schr¨odinger-Poisson system, Sobolev embedding, minimizer. ∗

This work was supported partially by the National Natural Science Foundation of China (10961028).

†

Corresponding Author: Tel./fax:+86 8715516877. [email protected]

1

1. Introduction and Preliminaries In this paper, we consider the following system     −4u + V (x)u + φu − 21 u4u2 = |u|p−1 u + h(x), in R3 ,    −4φ = u2 ,

in

(P )

R3 .

This system is a modified version of the Schr¨ odinger-Poisson systems, which arise in an interesting physical context. According to a classical model, the interaction of a charge particle with an electro-magnetic field can be described by coupling the nonlinear Schr¨ odinger’s and Poisson’s equations (we refer the reader to [1] and the references therein for more details on the physical aspects). In the recent years, there has been a lot of works dealing with the Schr¨ odinger-Poisson systems. Many interesting results were obtained (e.g. see [1-15]). Particularly, in [13], the authors study the following Schr¨ odinger-Poisson system     −4u + V (x)u + φu = f (x, u) + h(x), in R3 ,    −4φ = u2 ,

in

(1)

R3 .

Under some appropriate assumptions on V , f and h, they proved that system (1) has at least two different solutions by the Ekeland’s variational principle and the Mountain Pass Theorem. Moreover, some authors researched the following modified nonlinear Schr¨ odinger equation 1 −4u + V (x)u − u4u2 = |u|p−1 u, x ∈ RN 2

(2)

(e.g. see [16-32]). In [23], when p + 1 ∈ [4, N4N −2 ), the existence results of a positive ground state solution and a sign-changing ground state solution were given by using Nehari method. When p + 1 ∈ (2, N4N −2 ) and V satisfies stronger condition, the authors proved the existence of a positive ground state solution in [20]. They pointed out that the solutions of equation

2

(2) are the standing waves of the following modified nonlinear Schr¨ odinger equation iψt + 4ψ − V (x)ψ + k4(h(|ψ|2 ))h0 (|ψ|2 )ψ + g(x, ψ) = 0, x ∈ RN ,

(3)

where V (x) is a given potential, k is a real constant and g, h are real functions. This quasilinear version of the nonlinear Schr¨ odinger equation arises in several models of different physical phenomena, such as in the study of superfluid films in plasma physics, in condensed matter theory, etc (e.g. see [33,34,35]). These motivate us to consider the following modified Schr¨ odinger-Poisson system     −4u + V (x)u + φu − 21 u4u2 = |u|p−1 u + h(x), in R3 ,    −4φ = u2 ,

in

(P )

R3 ,

where 4 ≤ p + 1 < 12, h ∈ L2 (R3 ), h 6= 0 and V satisfies the following assumption: (V)

V ∈ C(R3 , R), inf V (x) ≥ V0 > 0 and for every M > 0, meas{x ∈ R3 : V (x) ≤ x∈R3

M } < ∞, where meas denote the Lebesgue measure in R3 . Because the term u4u2 appears in (P ), this system is new in the mathematical literature. Now we define the space Z

HV = {u ∈ H (R ) : 1

3

R3

V (x)|u|2 dx < ∞},

with the norm kukHV

Z 1 := ( (|∇u|2 + V (x)|u|2 )dx) 2 R3

and the space ∗

D1,2 (R3 ) = {u ∈ L2 (R3 ) : ∇u ∈ L2 (R3 )} with the norm Z kukD1,2 = (

R3

3

1

|∇u|2 dx) 2 .

Set X = {u ∈ HV :

Z

u2 |∇u|2 < +∞} = {u ∈ HV : u2 ∈ H 1 (R3 )}.

R3

Obviously, the embedding HV ,→ Ls (R3 ) is continuous for s ∈ [2, 6]. Then by using Sobolev embedding and interpolation, we conclude that X ⊂ Lq (R3 ) for q ∈ [2, 12]. Moreover, by Lemma 3.4 in [36], we know that the embedding HV ,→ Ls (R3 ) is compact for s ∈ [2, 6). It is well known that, for every u ∈ X ⊂ HV , the Lax-Milgram theorem implies that there exists a unique φu ∈ D1,2 (R3 ) such that −4φu = u2 . Moreover, φu has the following integral expression 1 φu (x) := 4π

Z

R3

u2 (y) dy. |x − y|

Furthermore, one has Z

x∈R3

φu u2 dx ≤ ckuk4L12/5 ≤ c1 kuk4HV ,

where c > 0 and c1 > 0. In form, the system (P ) has a variational structure, given by the functional 1 J(u) = 2

Z

1 (|∇u| +V (x)u +u |∇u| )dx+ 4 R3 2

2

2

2

Z

1 φu u dx− p+1 R3 2

Z

R3

p+1

|u|

Z dx−

R3

h(x)udx.

Given u ∈ X and ψ ∈ C0∞ (R3 ), the derivative of J in the direction ψ at u, denoted by hJ 0 (u), ψi. It is easy to check that 0

hJ (u), ψi =

Z

R3

[(1 + u2 )∇u · ∇ψ + u|∇u|2 ψ + V (x)uψ + φu uψ − |u|p−1 uψ − h(x)ψ]dx.

Hence, in a certain sense, (u, φu ) ∈ X × D1,2 (R3 ) is called a weak solution of (P ) if Z

R3

[(1+u2 )∇u·∇ψ +u|∇u|2 ψ +V (x)uψ +φu uψ −|u|p−1 uψ −h(x)ψ]dx = 0, ∀ψ ∈ C0∞ (R3 ).

Formally, seeking a weak solution (u, φu ) ∈ X × D1,2 (R3 ) of (P ) is equivalent to seek a critical point of the functional J. 4

It seems quite clear that X is the right space for studying the functional J. But the term

R

R3

u2 |∇u|2 dx is not convex and X is not even a vector space. So the usual min-max

techniques cannot be directly applied to J. Nevertheless X is a complete metric space with distance dX (u, v) = ku − vkHV + k∇u2 − ∇v 2 kL2 . Moreover, it is easy to check that J is continuous in X. In the latter parts of this paper, we use c to denote any positive constant.

2. Main Results We first give some notations. For u ∈ X we define β(u) := γ(u) :=

Z

R3

Z

R3

(|∇u|2 + V (x)u2 + u2 |∇u|2 )dx,

[(1 + 2u2 )|∇u|2 + V (x)u2 + φu u2 − |u|p+1 − h(x)u]dx.

Our main result is the following theorem.

Theorem

Assume (V ). Then there is a constant m0 > 0 such that whenever

khkL2 ≤ m0 , one has: (i) If p + 1 = 4, the system (P ) has at last one non-trivial weak solution. (ii) If 4 < p + 1 < 12, the system (P ) has at last two different non-trivial weak solutions. In order to prove our main result, we need the following facts.

Lemma 2.1.

Assume 4 ≤ p + 1 < 12. Then there exist m0 > 0, ρ > 0 and

α = α(ρ) > 0 such that whenever khkL2 ≤ m0 , one has: 5

(i) J(u) ≥ α for each u ∈ X\{0} with β(u) = ρ. (ii) For each u ∈ X\{0}, there exists a unique tu > 0 such that β(tu u) = ρ, and hence J(tu u) ≥ α. Proof. By the H¨older and Sobolev inequalities we have Z Z 11−p p−1 u2 dx) 10 ( |u|p+1 dx ≤ ( u12 dx) 10 3 R3 R3 Z ZR 11−p 2 2 ≤ c( (|∇u| + V (x)u )dx) 10 (

Z

R3

R3

≤ c(β(u)) If u 6= 0, then β(u) =

p+4 5

u2 |∇u|2 dx)

3(p−1) 10

(2.1)

.

R

R3 (|∇u|

2

+ V (x)u2 + u2 |∇u|2 )dx 6= 0. By the H¨ older inequality and

(2.1) there is a constant σ > 0 such that Z Z Z Z 1 1 1 (|∇u|2 + V (x)u2 + u2 |∇u|2 )dx + φu u2 dx − |u|p+1 dx − h(x)udx 2 R3 4 R3 p + 1 R3 R3 p+4 1 1 ≥ β(u) − σ[(β(u)) 5 + khkL2 (β(u)) 2 ] 2 2p+3 1 1 1 = (β(u)) 2 [ (β(u)) 2 − σ(β(u)) 10 − σkhkL2 ]. 2

J(u) =

1

Set g(s) = 21 s 2 − σs

2p+3 10

. Then there exists a constant ρ > 0 such that max g(s) = g(ρ) > 0. s≥0

Set m0 =

1 2σ g(ρ).

Then 1 1 J(u) ≥ ρ 2 g(ρ) := α > 0 2

(2.2)

whenever β(u) = ρ and khkL2 ≤ m0 . For each 0 6= u ∈ X, by solving the following equation β(tu) = t

2

Z

R3

(|∇u| + V (x)u )dx + t 2

2

6

4

Z

R3

u2 |∇u|2 dx = ρ,

we know that there exists a unique constant tu > 0 such that β(tu u) = ρ, and hence J(tu u) ≥ α whenever khkL2 ≤ m0 . This completes the proof. Throughout the latter parts of this paper, m0 , ρ and α mean these constants that appear in Lemma 2.1. For any u ∈ X \ {0}, we define a function fu (t) = J(tu) f or t ∈ (0, ∞), i.e. Z

Z

t4 (|∇u| +V (x)u )dx+ 2 R3

t2 fu (t) = 2

2

Z

t4 u |∇u| dx+ 4 R3

2

2

tp+1 φu u dx− p+1 R3

2

2

Z

R3

|u|

p+1

dx−t

Z

R3

h(x)udx.

Then fu0 (t)

=t

Z

(|∇u| +V (x)u )dx+2t 2

R3

2

3

Z

2

R3

2

3

u |∇u| dx+t

Z

R3

2

p

φu u dx−t

Z

R3

p+1

|u|

Z dx−

R3

h(x)udx

and fu00 (t) =

Z

R3

Remark 2.1. R

R3

Z

(|∇u|2 + V (x)u2 )dx + 6t2

R3

u2 |∇u|2 dx + 3t2

Z

R3

If p + 1 > 4, or p + 1 = 4 and 2

R

φu u2 dx − ptp−1

R3

u2 |∇u|2 dx +

Z

R3

R

R3

|u|p+1 dx.

φu u2 dx <

|u|p+1 dx, then there is a unique t0 > 0 such that fu00 (t0 ) = 0, fu00 (t) > 0 in (0, t0 ) and

fu00 (t) < 0 in (t0 , +∞). This implies that fu (t) has at most two critical points for t ∈ (0, +∞).

Lemma 2.2. p + 1 = 4 and 2

R

R3

Suppose that

u2 |∇u|2 dx +

R

R3

R

R3

h(x)udx > 0, khkL2 ≤ m0 and either p + 1 > 4, or

φu u2 dx <

R

R3

|u|p+1 dx. Then there exist bu > 0 with

J(bu u) > 0, t1 and t2 such that fu (t1 ) = fu (t2 ) = 0 and t1 ∈ (0, bu ) is the unique point such that fu (t1 ) = 0, t2 ∈ (bu , +∞) is the unique point such that fu (t2 ) = 0. Furthermore, we have fu0 (t1 ) > 0 and fu0 (t2 ) < 0. Proof.

Since

R

R3

h(x)udx > 0, fu (t) < 0 and fu0 (t) < 0 for small t. Moreover, by

Lemma 2.1, there exists a bu > 0 such that fu (bu ) > 0. Note that fu (t) → −∞ as t → +∞. We can find 0 < t1 < bu < t2 such that fu (t1 ) = fu (t2 ) = 0. If there exists another t3 > 0 7

such that fu (t3 ) = 0, which will contradict with Remark 2.1. Hence t1 ∈ (0, bu ) is the unique point such that fu (t1 ) = 0 and t2 ∈ (bu , +∞) is the unique point such that fu (t2 ) = 0. By Remark 2.1 fu (t) has at most two critical points for t ∈ (0, +∞), hence fu0 (t1 ) > 0 and fu0 (t2 ) < 0. This completes the proof. Now, we set M1 = {u ∈ X : γ(u) = 0, J(u) > 0},

m1 = inf J(u),

M2 = {u ∈ X : γ(u) = 0, J(u) < 0},

m2 = inf J(u).

Remark 2.2.

u∈M1

u∈M2

Under the conditions of Lemma 2.2, we know that we may take

bu = tu , where tu is the unique value such that β(tu u) = ρ in Lemma 2.1. Moreover, if u ∈ M1 we may take bu = 1.

Lemma 2.3.

Assume khkL2 ≤ m0 and 4 ≤ p + 1 < 12. Then

(i) for all u ∈ M1 , fu0 (t) < 0 for t > 1 and J(tu) < J(u) for t ∈ (0, ∞)\{1};

R

(ii) for any u ∈ X\{0} with either p + 1 = 4 and 2 R3

R

R3

u2 |∇u|2 dx +

|u|p+1 dx or p + 1 > 4, there exists a unique t > 0 such that tu ∈ M1 .

R

R3

φu u2 dx <

Proof. We first prove (i). Indeed, if u ∈ M1 , then J(u) > 0 and γ(u) = 0. Hence

1 0 > γ(u)−J(u) = 2

Z

3 (|∇u| +V (x)u )dx+ 2 R3 2

2

Z

3 u |∇u| dx+ 4 R3 2

2

Z

p φu u dx− p+1 R3 2

and hence Z

Z

4p 2 u |∇u| dx + φu u dx < 3(p + 1) R3 R3 2

2

2

Z

R3

|u|p+1 dx.

Particularly, when p + 1 = 4, one has 2

Z

R3

u |∇u| dx + 2

2

Z

2

R3

8

φu u dx <

Z

R3

|u|4 dx.

Z

R3

|u|p+1 dx,

Therefore, if

R

R3

h(x)udx ≤ 0, it is easy to see that fu0 (t) > 0 for 0 < t < 1 and fu0 (t) < 0

for t > 1. Hence fu (t) < fu (1), that is J(tu) < J(u) for t ∈ (0, ∞) and t 6= 1. Moreover, if

R

R3

h(x)udx > 0, by Remark 2.2 there is a unique 0 < t0 < 1 such that J(t0 u) = 0 and

fu0 (t0 ) > 0. Set k1 (t) =

t−1 tp −1 ,

k2 (t) =

t3 −1 tp −1

for 0 < t < 1 or t > 1. We claim that when

p + 1 ≥ 4, k1 (t) is a decreasing function and when p + 1 > 4, k2 (t) is also a decreasing function. We only give the proof for k2 (t). Indeed, when p + 1 > 4 k20 (t) =

3t2 (tp − 1) − ptp−1 (t3 − 1) ptp−1 − (p − 3)tp+2 − 3t2 = . (tp − 1)2 (tp − 1)2

Note that p ptp−1 = . (p − 3)tp+2 + 3t2 (p − 3)t3 + 3t3−p Set ϕ(t) = (p − 3)t3 + 3t3−p . Then ϕ0 (t) = 3(p − 3)t2 + 3(3 − p)t2−p = 3(p − 3)t2 (1 − t−p ). It is easy to see that ϕ0 (t) < 0 for 0 < t < 1 and ϕ0 (t) > 0 for t > 1. Note that ϕ(1) = p. Hence ϕ(t) > p for 0 < t < 1 or t > 1, and hence

ptp−1 (p−3)tp+2 +3t2

< 1 for 0 < t < 1 or t > 1.

Therefore, k20 (t) < 0 for 0 < t < 1 or t > 1, i.e. k2 (t) is a decreasing function for 0 < t < 1 or t > 1. Moreover, we have 1 lim k1 (t) = . p

t→1

Hence k1 (t) >

1 1 f or 0 < t < 1, k1 (t) < f or t > 1. p p

We also have 3 lim k2 (t) = . t→1 p Hence k2 (t) >

3 3 f or 0 < t < 1, k2 (t) < f or t > 1. p p

9

Since u ∈ M1 , fu0 (1) = 0. Hence 0 > −fu0 (t0 ) = fu0 (1) − fu0 (t0 ) Z Z Z = (1 − t0 ) (|∇u|2 + V (x)u2 )dx + 2(1 − t30 ) u2 |∇u|2 dx + (1 − t30 ) φu u2 dx 3 3 3 R R R Z −(1 − tp0 ) |u|p+1 dx R3 Z Z 1 − t0 2(1 − t30 ) 2 2 = (1 − tp0 )[ (|∇u| + V (x)u )dx + u2 |∇u|2 dx 1 − tp0 R3 1 − tp0 3 R Z Z 1 − t30 |u|p+1 dx]. + φu u2 dx − 1 − tp0 R3 3 R This implies that 1 − t0 1 − tp0

Z

Z

2(1 − t30 ) (|∇u| +V (x)u )dx+ 1 − tp0 R3 2

1 − t30 u |∇u| dx+ 1 − tp0 R3

2

2

2

Z

R3

Z φu u dx− 2

R3

|u|p+1 dx < 0. (2.3)

Hence, for t0 < t < 1, one has −fu0 (t)

=

fu0 (1)

fu0 (t)

3 p

2

2

3

R3

Recall that 0 < t0 < 1. We have k1 (t0 ) > k2 (t) ≤

Z

(|∇u| + V (x)u )dx + 2(1 − t ) u2 |∇u|2 dx R3 Z Z 3 2 p +(1 − t ) φu u dx − (1 − t ) |u|p+1 dx 3 3 R R Z Z 1 − t 2(1 − t3 ) p 2 2 = (1 − t )[ (|∇u| + V (x)u )dx + u2 |∇u|2 dx 1 − tp R3 1 − tp 3 R Z Z 1 − t3 2 p+1 + φu u dx − |u| dx] 1 − tp R3 R3 Z Z 2(1 − t30 ) p 1 − t0 2 2 < (1 − t )[ (|∇u| + V (x)u )dx + u2 |∇u|2 dx 1 − tp0 R3 1 − tp0 3 R Z Z 1 − t30 + φu u2 dx − |u|p+1 dx] < 0. 1 − tp0 R3 R3 −

= (1 − t)

Z

1 p

and k2 (t0 ) ≥ p3 . Moreover, since k1 (t) <

for all t > 1, k1 (t) < k1 (t0 ) and k2 (t) ≤ k2 (t0 )

for all t > 1, i.e. t−1 t0 − 1 1 − t0 t3 − 1 t30 − 1 1 − t30 < = , ≤ = tp − 1 tp − 1 tp0 − 1 1 − tp0 tp0 − 1 1 − tp0 10

1 p

and

for all t > 1. Hence, by (2.3), we have

<

Z t−1 (|∇u|2 + V (x)u2 )dx + tp − 1 R3 Z 1 − t0 (|∇u|2 + V (x)u2 )dx + 1 − tp0 R3

Z 2(t3 − 1) u2 |∇u|2 dx + tp − 1 R3 Z 2(1 − t30 ) u2 |∇u|2 dx + 1 − tp0 R3

Z Z t3 − 1 2 |u|p+1 dx φ u dx − u tp − 1 R3 3 ZR Z 1 − t30 |u|p+1 dx φu u2 dx − 1 − tp0 R3 R3

< 0 for all t > 1. Then, for t > 1, one has fu0 (t) = fu0 (t) − fu0 (1) Z Z 2 2 3 = (t − 1) (|∇u| + V (x)u )dx + 2(t − 1) u2 |∇u|2 dx R3 R3 Z Z 3 2 p +(t − 1) φu u dx − (t − 1) |u|p+1 dx 3 3 R R Z Z t − 1 2(t3 − 1) p 2 2 = (t − 1)[ p (|∇u| + V (x)u )dx + p u2 |∇u|2 dx t − 1 R3 t − 1 R3 Z Z t3 − 1 2 φu u dx − |u|p+1 dx] < 0. + p t − 1 R3 3 R

(2.4)

These show that fu0 (t) > 0 for t0 < t < 1 and fu0 (t) < 0 for t > 1, which implies J(tu) < J(u) f or t ∈ (t0 , +∞)\{1}. Since J(tu) = fu (t) < 0 for small t > 0 and t0 ∈ (0, 1) is the unique point such that fu (t0 ) = 0, J(tu) = fu (t) ≤ 0 < J(u) for 0 < t ≤ t0 . Now we prove (ii). Let u ∈ X\{0}. If

R

R3

h(x)udx ≤ 0, then J(tu) > 0 for small

t > 0, fu0 (t) > 0 for small t > 0 and fu0 (t) < 0 for large t > 0. Furthermore, we can deduce that there is t > 0 such that fu0 (t) = 0 and J(tu) > 0, which implies tu ∈ M1 . If R

R3

h(x)udx > 0, then by Lemma 2.2 there exist bu > 0 with J(bu u) > 0, t1 and t2 such that

fu (t1 ) = fu (t2 ) = 0 and t1 ∈ (0, bu ) is the unique point such that fu (t1 ) = 0, t2 ∈ (bu , +∞) is the unique point such that fu (t2 ) = 0. Furthermore, we have fu0 (t1 ) > 0 and fu0 (t2 ) < 0. Hence J(tu) > 0 for t1 < t < t2 and there is a t3 ∈ (t1 , t2 ) such that fu0 (t3 ) = 0, and hence 11

t3 u ∈ M1 . Moreover, the uniqueness follows from the fact that J(tu) < J(u) for all u ∈ M1 and t ∈ (0, ∞)\{1}. This completes the proof.

Lemma 2.4.

If khkL2 ≤ m0 , then m1 > 0.

Proof. By Lemma 2.1, we know that there exists a tu > 0 such that J(tu u) ≥ α for each u ∈ M1 . Consequently, by Lemma 2.3 (i), one has m1 ≥ α > 0. This completes the proof.

Lemma 2.5.

Suppose that khkL2 ≤ m0 and 4 ≤ p + 1 < 12. If u ∈ M2 , then

β(u) < ρ. Proof. If u ∈ M2 , we have fu (1) < 0 and fu0 (1) = 0. Hence, by Lemma 2.1, β(u) 6= ρ and there exists a unique tu > 0 such that β(tu u) = ρ and J(tu u) ≥ α > 0. We discuss two cases: Case1: If p + 1 > 4, or p + 1 = 4 and 2

R

R3

u2 |∇u|2 dx +

R

R3

φu u2 dx <

R

R3

|u|p+1 dx, then

by Lemma 2.3 (ii), there exists a unique δ > 0 such that δu ∈ M1 . Moreover, by Lemma 2.3 (i), one has fu0 (t) < 0 for all t > δ. If β(u) > ρ = β(tu u), then tu < 1. We claim that δ < 1. Otherwise, if δ ≥ 1, then by J(δu) > 0 > J(u) we deduce δ > 1. Hence tu < 1 < δ. Note that fu (tu ) > 0, fu (1) < 0, fu (δ) > 0. Consequently, there are two constants t1 and t2 such that tu < t1 < 1 < t2 < δ and fu (t1 ) = fu (t2 ) = 0. Since again fu (t) → −∞ as t → ∞, there exists t3 > δ such that fu (t3 ) = 0. This is not possible. Because if has only a solution for t > 0, and if

R

R3

R

R3

h(x)udx ≤ 0, it is easy to see that fu (t) = 0

h(x)udx > 0, then Lemma 2.2 implies that fu (t) = 0

has only two solutions for t > 0. This shows δ < 1. Consequently, fu0 (1) < 0 = fu0 (1), a contradiction. Hence β(u) < ρ.

12

Case 2:. If p + 1 = 4 and 2 fu0 (1)

=

Z

R3

R

R3

u2 |∇u|2 dx +

R

R3

φu u2 dx ≥

R

R3

|u|p+1 dx, then from

[(1 + 2u2 )|∇u|2 + V (x)u2 + φu u2 − |u|4 − h(x)u]dx = 0,

one has Z

(|∇u| + V (x)u )dx ≤ 2

R3

This implies that

R

R3

2

Z

R3

h(x)udx.

h(x)udx > 0. Hence fu0 (t) < 0 for 0 < t < 1 and fu (t) < 0 for small t.

Then J(tu) = fu (t) < 0 for 0 < t ≤ 1. Consequently, 1 < tu , and hence β(u) < β(tu u) = ρ. This completes the proof.

Lemma 2.6. R

β(u) < ρ, then

R3

Suppose that khkL2 ≤ m0 and 4 ≤ p + 1 < 12. If J(u) < 0 and

h(x)udx > 0.

Proof. Since J(u) < 0, u ∈ X\{0}. Hence, by Lemma 2.1(ii), there exists a unique tu > 0 such that β(tu u) = ρ and J(tu u) ≥ α > 0. Consequently, 0>

tp+1 u J(u)

Z Z 1 p+1 1 p+1 2 2 2 4 − J(tu u) = (t − tu ) (|∇u| + V (x)u )dx + (tu − tu ) u2 |∇u|2 dx 2 u 2 R3 R3 Z Z 1 p+1 + (tu − t4u ) φu u2 dx − (tp+1 − tu ) h(x)udx u 4 3 R R3

− t4u ≥ 0, tp+1 − tu > 0, − t2u > 0, tp+1 Since again β(u) < ρ = β(tu u), tu > 1. Hence tp+1 u u u and hence

R

R3

h(x)udx > 0. This completes the proof.

Lemma 2.7.

R R

Assume khkL2 ≤ m0 .

(i) For each u ∈ M2 , if p + 1 > 4 or p + 1 = 4 and 2 R3

R3

u2 |∇u|2 dx +

|u|p+1 dx, then tu > 1 and J(tu) > J(u) for all t ∈ (0, tu )\{1}; (ii) For each u ∈ X with

R3

R

φu u2 dx <

R

R3

R

R3

h(x)udx > 0, if p + 1 > 4 or p + 1 = 4 and 2

R

R3

|u|p+1 dx, then there exists a t ∈ (0, tu ) such that tu ∈ M2 .

Where tu is the unique value such that β(tu u) = ρ in Lemma 2.1. 13

R

R3

φu u2 dx <

u2 |∇u|2 dx +

Proof. We first prove (i). If u ∈ M2 and either p+1 > 4, or p+1 = 4 and 2 R R

R

R

u2 |∇u|2 dx+

R3

|u|p+1 dx, then by Lemma 2.5 and Lemma 2.6, we have β(u) < ρ and

R3

φu u2 dx <

R3

h(x)udx > 0. Since β(tu u) = ρ > β(u), tu > 1. Moreover, by Remark 2.2, there is a

R3

unique θ ∈ (0, tu ) such that J(θu) = 0. Furthermore, fu0 (θ) > 0. Note that J(tu u) > 0 > J(u). There exists a t0 ∈ (1, tu ) such that J(t0 u) = 0. By the uniqueness of θ, we know θ = t0 ∈ (1, tu ). Since u ∈ M2 , fu0 (1) = 0. Hence 0 < fu0 (θ) = fu0 (θ) − fu0 (1) Z Z u2 |∇u|2 dx (|∇u|2 + V (x)u2 )dx + 2(θ3 − 1) = (θ − 1) 3 3 R R Z Z +(θ3 − 1) φu u2 dx − (θp − 1) |u|p+1 dx R3 R3 Z Z θ−1 2(θ3 − 1) = (θp − 1)[ p (|∇u|2 + V (x)u2 )dx + p u2 |∇u|2 dx θ − 1 R3 θ − 1 R3 Z Z θ3 − 1 2 + p φu u dx − |u|p+1 dx]. θ − 1 R3 R3

This implies that θ−1 θp − 1

Z

2(θ3 − 1) (|∇u| +V (x)u )dx+ p θ −1 R3 2

2

Z

θ3 − 1 u |∇u| dx+ p θ −1 R3 2

2

Z

R3

Z φu u dx− 2

R3

|u|p+1 dx > 0.

Note that when p + 1 ≥ 4, k1 (t) is a decreasing function for 0 < t < 1 or t > 1 and when p + 1 > 4, k2 (t) is also a decreasing function for 0 < t < 1 or t > 1. We have fu0 (t) = fu0 (t) − fu0 (1) Z Z 2 2 3 = (t − 1) (|∇u| + V (x)u )dx + 2(t − 1) u2 |∇u|2 dx R3 R3 Z Z +(t3 − 1) φu u2 dx − (tp − 1) |u|p+1 dx 3 3 R R Z Z 2(t3 − 1) t − 1 2 2 p (|∇u| + V (x)u )dx + p u2 |∇u|2 dx = (t − 1)[ p t − 1 R3 t − 1 R3 Z Z t3 − 1 + p φu u2 dx − |u|p+1 dx] t − 1 R3 3 R Z Z 3 θ − 1 2(θ − 1) > (tp − 1)[ p u2 |∇u|2 dx (|∇u|2 + V (x)u2 )dx + p θ − 1 R3 θ − 1 R3 14

3

+

θ −1 θp − 1

Z

R3

φu u2 dx −

Z

R3

|u|p+1 dx] > 0

for all 1 < t < θ. Moreover, note that k1 (t) > k1 (θ) and k2 (t) ≥ k2 (θ) for all 0 < t < 1. We have, for 0 < t < 1, −fu0 (t)

fu0 (1)

fu0 (t)

Z

Z

u2 |∇u|2 dx (|∇u| + V (x)u )dx + 2(1 − t ) R3 Z Z +(1 − t3 ) φu u2 dx − (1 − tp ) |u|p+1 dx 3 3 R R Z Z 1 − t 2(1 − t3 ) p 2 2 = (1 − t )[ (|∇u| + V (x)u )dx + u2 |∇u|2 dx 1 − tp R3 1 − tp R3 Z Z 1 − t3 2 + φ u dx − |u|p+1 dx] u 1 − tp R3 3 R Z Z θ − 1 2(θ3 − 1) p 2 2 > (1 − t )[ p (|∇u| + V (x)u )dx + p u2 |∇u|2 dx θ − 1 R3 θ − 1 R3 Z Z θ3 − 1 2 + p φu u dx − |u|p+1 dx] > 0. θ − 1 R3 3 R

=

−

= (1 − t)

2

2

3

R3

Therefore, fu0 (t) < 0 for all 0 < t < 1 and fu0 (t) > 0 for all 1 < t < θ. This shows J(tu) > J(u) for all t ∈ (0, θ)\{1}. Note that θ ∈ (0, tu ) is the unique value such that J(θu) = 0 and J(tu u) > 0. We have J(tu) ≥ 0 > J(u) for all t ∈ [θ, tu ), and hence J(tu) > J(u) for all t ∈ (0, tu )\{1}. This completes the proof of (i). Now we prove (ii): For each u ∈ X with

R

R3

h(x)udx > 0, by Remark 2.2 there is a

unique 0 < t1 < tu such that J(t1 u) = 0. Furthermore, fu0 (t1 ) > 0. Note that J(tu) < 0 and fu0 (t) < 0 for small t > 0. Hence J(tu) < 0 for 0 < t < t1 and there is a t0 ∈ (0, t1 ) such that fu0 (t0 ) = 0 , which implies that t0 u ∈ M2 . This completes the proof of (ii).

Lemma 2.8.

Assume p + 1 = 4.

(i) For each u ∈ M2 with 2 all t ∈ (0, ∞)\{1}.

R

R3

u2 |∇u|2 dx +

15

R

R3

φu u2 dx ≥

R

R3

|u|4 dx, J(tu) > J(u) for

(ii) For each u ∈ X with

R

R3

R

h(x)udx > 0 and 2

R3

u2 |∇u|2 dx+

R

R3

φu u2 dx ≥

R

R3

|u|4 dx,

u2 |∇u|2 dx+

R

R3

φu u2 dx ≥

R

R3

|u|4 dx,

there exists a unique 0 < t < ∞ such that tu ∈ M2 . R

Proof. We first prove (i). For each u ∈ M2 with 2 one has fu (1) =

1 2

Z

R3

(|∇u|2 +V (x)u2 +u2 |∇u|2 )dx+

1 4

Z

R3

φu u2 dx−

R3

1 4

Z

R3

|u|4 dx−

Z

R3

h(x)udx < 0

and fu0 (1) = γ(u) =

Z

R3

[(1 + 2u2 )|∇u|2 + V (x)u2 + φu u2 − |u|4 − h(x)u]dx = 0

and Z

R3

Hence

R

R3

(|∇u| + V (x)u )dx ≤ 2

2

Z

R3

h(x)udx.

h(x)udx > 0. Then we have fu0 (t) < 0 for 0 < t < 1 and fu0 (t) > 0 for t > 1. This

implies J(tu) > J(u) for all t ∈ (0, ∞)\{1}. The conclusion (i) is proved.

R

Now we prove (ii). For each u ∈ X with R3

φu u2 dx ≥

fu0 (t0 )

= t0

Z

R

R3

R3

R

R3

h(x)udx > 0 and 2

|u|4 dx, there is a unique t0 ∈ (0, ∞) such that

(|∇u| +V (x)u 2

2

)dx+2t30

Z

R3

2

u |∇u|

2

dx+t30

Z

R3

2

φu u

dx−t30

Z

R

R3

R3

u2 |∇u|2 dx +

Z |u| dx− 4

R3

h(x)udx = 0,

fu0 (t) < 0 for 0 < t < t0 and fu0 (t) > 0 for t0 < t. Note that fu (t) < 0 for small t. We have J(t0 u) < 0. Hence, t0 u ∈ M2 . The conclusion (ii) is proved.

Proof of Theorem.

The proof proceeds in two steps.

Step 1: Let 4 ≤ p + 1 < 12. Set {vn } ⊂ M2 be such that J(vn ) → m2 . By Lemma 2.5, we know that β(vn ) ≤ ρ. Hence {vn } is bounded in HV and {vn2 } is bounded in D1,2 (R3 ), and hence both {vn } and {vn2 } are bounded in H 1 (R3 ). By the Sobolev embedding, passing to a subsequence, we can assume that vn * v in HV and H 1 (R3 ), vn2 * v 2 in H 1 (R3 ) 16

and D1,2 (R3 ), vn → v in Ls (R3 ) for all s ∈ [2, 6) and vn (x) → v(x) for a.e. x ∈ R3 . For 6 ≤ s < 12, from Z

|vn − v| dx = s

R3

Z

|vn − v|

6(s−2) 5

12−s

· |vn − v| 5 dx R3 Z Z 12−s s−2 |vn − v|2 dx) 10 ≤ ( (|vn − v|2 )6 dx) 10 ( 3 R3 ZR 12−s |vn − v|2 dx) 10 ≤ c( R3

we know that vn → v in Ls (R3 ) for all s ∈ [6, 12). Since vn * v in H 1 (R3 ), φvn * φv in D1,2 (R3 ). Note that Z

φvn vn2 dx =

R3

Z

R3

−φvn ∆φvn dx =

Z

R3

|∇φvn |2 dx = kφvn k2D1,2 .

Hence Z

R3

φv v dx = 2

kφv k2D1,2

≤ lim inf n→∞

Z

R3

φvn vn2 dx.

Obviously, one has Z

R3

(|∇v|2 + V (x)v 2 )dx ≤ lim inf n→∞

Z

R3

(|∇vn |2 + V (x)vn2 )dx

and Z

1 1 v 2 |∇v|2 dx = kv 2 k2D1,2 ≤ lim inf kvn2 k2D1,2 = lim inf n→∞ n→∞ 4 4 3 R

Z

R3

vn2 |∇vn |2 dx.

Consequently, γ(v) ≤ lim inf γ(vn ) = 0,

β(v) ≤ lim inf β(vn ) ≤ ρ,

n→∞

n→∞

and J(v) ≤ lim inf J(vn ) = lim J(vn ) = m2 < 0. n→∞

n→∞

Since J(v) < 0, β(v) < ρ. Then by Lemma 2.6, one has

R

R3

h(x)v > 0. Moreover, Lemma

2.1 implies that there exists a unique tv > 0 such that β(tv v) = ρ and J(tv v) ≥ α > 0. 17

We discuss two cases to prove v ∈ M2 and J(v) = m2 . Case 1: If p + 1 > 4 or p + 1 = 4 and 2

R

R3

v 2 |∇v|2 dx +

R

R3

φv v 2 dx <

R

R3

|v|4 dx, then by

Lemma 2.7(ii), there exists 0 < t1 < tv such that t1 v ∈ M2 , and hence fv0 (t1 ) = 0. Moreover, if γ(v) 6= 0, then γ(v) < 0, i.e. fv0 (1) < 0. By Remark 2.2 there exist a and b such that fv (a) = fv (b) = 0 and a ∈ (0, tv ) is the unique point such that fv (a) = 0, b ∈ (tv , +∞) is the unique point such that fv (b) = 0. Furthermore, fv0 (b) < 0 < fv0 (a). Hence there is an e ∈ (a, b) such that fv0 (e) = 0. Note that fv (1) = J(v) < 0, fv (t1 ) = J(t1 v) < 0 and fv (tv ) = J(tv v) > 0. The uniqueness implies that a ∈ (1, tv ) and t1 < a. Hence there exists t2 ∈ (1, a) such that fv0 (t2 ) = 0. By Remark 2.1, we know that fv (t) has at most two critical points for t ∈ (0, +∞). Hence t1 = t2 ∈ (1, a). Set v∗ = t1 v ∈ M2 . By Lemma 2.7(i), we have tv∗ > 1. Hence

∈ (0, tv∗ )\{1}. Consequently, Lemma 2.7(i) implies that

1 t1

m2 ≤ J(t1 v) < J(

1 · t1 v) = J(v) ≤ m2 , t1

a contradiction. Hence γ(v) = 0. This implies that v ∈ M2 and J(v) = m2 . Case 2: If p+1 = 4 and 2

R

R3

v 2 |∇v|2 dx+

R

R3

φv v 2 dx ≥

R

R3

|v|4 dx, then by Lemma 2.8(ii),

there exists a unique 0 < t3 < ∞ such that t3 v ∈ M2 . Note that 2 R

R3

φt3 v (t3 v)2 dx ≥

R

R3

R

2 2 R3 (t3 v) |∇(t3 v)| dx

+

|t3 v|4 dx. We claim t3 = 1. Otherwise, by Lemma 2.8(i), one has m2 ≤ J(t3 v) < J(

1 · t3 v) = J(v) ≤ m2 , t3

a contradiction. This also implies that v ∈ M2 and J(v) = m2 . Now we prove that (v, φv ) is a weak solution of (P ). Otherwise, we can choose a ψ ∈ C0∞ (R3 ) such that 0

hJ (v), ψi =

Z

R3

[(1 + v 2 )∇v · ∇ψ + v|∇v|2 ψ + V (x)vψ + φv vψ − |v|p−1 vψ − h(x)ψ]dx < −1. 18

Choose a small δ > 0 and a small ε1 ∈ (0, 1) be such that whenever |t−1| ≤ ε1 and |σ| ≤ 3δ, 1 hJ 0 (tv + σψ), ψi ≤ − . 2 Since fv (t) is continuous in t, there exists an 0 < ε < min{ε1 , δ} such that J(tv) ≤ J(v) + δ = m2 + δ, ∀ t ∈ (1 − ε, 1 + ε). Take a C ∞ -cut-off function 0 ≤ η ≤

3δ ε

be such that η(t) =

3δ ε

(2.5)

for |t − 1| ≤ ξ < ε and

η(t) = 0 for |t − 1| ≥ ε. For t ∈ [1 − ε, 1 + ε], we define ω(t) = tv + εη(t)ψ. It is easy to see that ω(t) is continuous curve in the metric space (X, d). When |t − 1| < ε, by using the mean value theorem to the C 1 map [0, ε] 3 σ 7−→ J(tv + ση(t)ψ) ∈ R, there is a σ ∈ (0, ε) such that 1 J(ω(t)) = J(tv) + hJ 0 (tv + ση(t)ψ), εη(t)ψi ≤ J(tv) − η(t)ε. 2 Especially for |t − 1| ≤ ξ, by (2.5), we have 1 3 J(ω(t)) ≤ J(tv) − η(t)ε ≤ m2 + δ − δ < m2 . 2 2

(2.6)

By Lemma 2.7(i) and fv0 (t) = 1t γ(tv), we can choose a smaller ε such that γ(ω(1 − ε)) < 0 and γ(ω(1 + ε)) > 0. Since the map t 7→ γ(ω(t)) is continuous, we can choose a 0 < ξ < ε such that γ(ω(1−ξ)) < 0 and γ(ω(1 +ξ)) > 0. Hence there exits s0 ∈ (1−ξ, 1+ξ) such that γ(ω(s0 )) = 0. Moreover, by (2.6), we have J(ω(s0 )) < 0, and hence ω(s0 ) = s0 v + εη(s0 )ψ ∈ M2 . But by (2.6), one has J(ω(s0 )) < m2 , a contradiction. This shows that (v, φv ) is a weak solution of the system (P ). Now, we have completed the proof of (i). 19

Step 2: Let 4 < p + 1 < 12. Set {un } ⊂ M1 be such that J(un ) → m1 . By γ(un ) = 0 and the H¨older inequality, we have 1 2

Z

R3

u2n |∇un |2 )dx

1 + 4

Z

(|∇un | + V + φun u2n dx R3 Z Z 1 − h(x)un dx |un |p+1 dx − p + 1 R3 R3 Z Z 1 1 1 2 2 2 = ( − (|∇un | + V (x)un )dx + ( − u2 |∇un |2 dx ) ) 2 p + 1 R3 2 p + 1 R3 n Z Z p 1 1 2 +( − φu u dx − h(x)un dx ) 4 p + 1 R3 n n p + 1 R3 Z 1 1 2 ≥ ( − )β(un ) − c( u2n dx) 2 2 p+1 R3 1 1 2 ≥ ( − )β(un ) − c(β(un )) 2 . 2 p+1

J(un ) =

(x)u2n

2

This implies that {β(un )} is bounded. Hence, passing to a subsequence, we can assume that un * u in HV and H 1 (R3 ), u2n * u2 in H 1 (R3 ) and D1,2 (R3 ), un → u in Ls (R3 ) for s ∈ [2, 12). For each w ∈ M1 , we have 1 J(w) = 2

Z

1 (|∇w| +V (x)w +w |∇w| )dx+ 4 R3 2

2

2

2

Z

1 φw w dx− p+1 R3 2

Z

R3

|w|

p+1

Z dx−

R3

h(x)wdx > 0

and γ(w) =

Z

R3

[(1 + 2w2 )|∇w|2 + V (x)w2 + φw w2 − |w|p+1 − h(x)w]dx = 0.

Hence γ(w) − J(w) < 0, and hence 1 2

Z

3 (|∇w| + V (x)w )dx + 2 R3 2

2

Z

3 w |∇w| dx + 4 R3 2

2

Z

p φw w dx < p+1 R3 2

Z

R3

|w|p+1 dx.

Consequently, by (2.1), 1 1 β(w) ≤ 2 2

Z

3 (|∇w| + V (x)w )dx + 2 R3 2

2

Z

3 w |∇w| dx + 4 R3 2

2

Z

R3

φw w2 dx < c(β(w))

p+4 5

.

This shows that β(w) is bounded away from zero on M1 . Hence {β(un )} is bounded away from zero, and hence

R

R3

|un |p+1 dx+

R

R3

h(x)un dx is bounded away from zero. Consequently, 20

u 6= 0. Therefore, by Lemma 2.3(ii), there is a t4 > 0 such that t4 u ∈ M1 . From the fact that un * u in HV and H 1 (R3 ), u2n * u2 in H 1 (R3 ) and D1,2 (R3 ), we deduce Z

R3

Z

φu u dx ≤ lim inf n→∞

φun u2n dx,

R3

Z

R3

(|∇un |2 + V (x)u2n )dx

u |∇u| dx ≤ lim inf

Z

R3

u2n |∇un |2 dx.

φt4 u (t4 u)2 dx ≤ lim inf

Z

φt4 un (t4 un )2 dx,

(|∇u| + V (x)u )dx ≤ lim inf 2

R3

Z

2

2

n→∞

and Z

R3

2

2

n→∞

Therefore, one has

Z

R3

Z

R3

n→∞

[|∇(t4 u)| + V (x)(t4 u) ]dx ≤ lim inf 2

2

n→∞

R3

Z

[|∇(t4 un )|2 + V (x)(t4 un )2 ]dx

Z

(t4 un )2 |∇(t4 un )|2 dx.

R3

and Z

(t4 u) |∇(t4 u)| dx ≤ lim inf 2

R3

2

n→∞

R3

Then together with the fact that un → u in Ls (R3 ) for s ∈ [2, 12), one has J(t4 u) ≤ lim inf J(t4 un ). n→∞

Moreover, by Lemma 2.3(i), we have J(t4 un ) ≤ J(un ) for all n. Now we obtain m1 ≤ J(t4 u) ≤ lim inf J(t4 un ) ≤ lim J(un ) = m1 . n→∞

n→∞

This implies J(t4 u) = m1 . Set u1 = t4 u. We prove that (u1 , φu1 ) is also a weak solution of the system (P ). Otherwise, we can choose a ψ ∈ C0∞ (R3 ) such that hJ 0 (u1 ), ψi =

Z

R3

[(1+u21 )∇u1 ·∇ψ+u1 |∇u1 |2 ψ+V (x)u1 ψ+φu1 u1 ψ−|u1 |p−1 u1 ψ−h(x)ψ]dx < −1.

Choose a small 0 < ε < 1 such that whenever |t − 1| ≤ ε and |σ| ≤ ε, one has 1 hJ 0 (tu1 + σψ), ψi ≤ − . 2 21

Choose a C ∞ -cut-off function 0 ≤ η ≤ 1 such that η(t) = 1 for |t − 1| ≤ 2ε , η(t) = 0 for |t − 1| ≥ ε and η(t) > 0 for

ε 2

< |t − 1| < ε. For |t − 1| ≤ ε, we define ω(t) = tu1 + εη(t)ψ.

Note that J is continuous in the metric space (X, d) and ω(t) is a continuous curve in (X, d). Then, by Lemma 2.4, we can choose a small 0 < ε < 1 such that J(ω(t)) > 0 for |t − 1| < ε. When |t − 1| < ε, by using the mean value theorem to the C 1 map [0, ε] 3 σ 7−→ J(tu1 + ση(t)ψ) ∈ R, there is a σ ∈ (0, ε) such that 1 J(ω(t)) = J(tu1 ) + hJ 0 (tu1 + σεη(t)ψ), εη(t)ψi ≤ J(tu1 ) − η(t)ε < m1 . 2

(2.7)

By Lemma 2.3(i) and fu0 1 (t) = 1t γ(tu1 ), we can choose a smaller ε such that γ(ω(1 − ε)) > 0 and γ(ω(1 + ε)) < 0. Furthermore, by the continuity of the map t 7−→ γ(ω(t)), there exists t0 ∈ (1 − ε, 1 + ε) such that γ(ω(t0 )) = 0. Hence ω(t0 ) = t0 u1 + εη(t0 )ψ ∈ M1 . But, by (2.7), one has J(ω(t0 )) < m1 , a contradiction. This shows that (u1 , φu1 ) is a weak solution of system (P ). Moreover, by step 1, we know that there exists u2 ∈ M2 such that (u2 , φu2 ) is also a weak solution of the system (P ). Finally, by m2 < 0 < m1 , we know that u1 6= u2 , and hence (u1 , φu1 ) and (u2 , φu2 ) are two different weak solutions of (P ). The conclusion (ii) is proved.

Remark 2.3.

When h = 0, |u1 | ∈ M1 is also a minimizer of J|M1 , and hence

(|u1 |, φ|u1 | ) is a solution of system (P ).

22

References [1] V. Benci, D. Fortunato, An eigenvalue problem for the Schr¨ odinger-Maxwell equations, Topol. Methods Nonlinear Anal. 11(1998), 283-293. [2] A. Azzollini, A. Pomponio, Ground state solutions for the nonlinear Schr¨ odingerMaxwell equations, J. Math. Anal. Appl. 345(2008), 90-108. [3] A. Ambrosetti, D. Ruiz, Multiple bound states for the Schr¨ odinger-Poisson problem, Commun. Contemp. Math. 10(2008), 391-404. [4] A. Salvatore, Multiple solitary waves for a non-homogeneous Schr¨ odinger-Maxwell system in R3 , Adv. Nonlinear Stud. 6(2006), 157-169. [5] C. Mercuri, Positive solutions of nonlinear Schr¨ odinger-Poisson systems with radial potentials vanishing at infinity, (English summary) Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl. 19(2008), 211-227. [6] D. Ruiz, The Schr¨odinger-Poisson equation under the effect of a nonlinear local term, J. Funct. Anal. 237(2006), 655-674. [7] G. M. Coclite, A multiplicity result for the nonlinear Schr¨ odinger-Maxwell equations, Commun. Appl. Anal. 7(2003), 417-423. [8] G. M. Coclite, A multiplicity result for the Schr¨ odinger-Maxwell equations with negative potential, Ann. Polon. Math. 79(2002), 21-30. [9] H. Kikuchi, On the existence of a solution for elliptic system related to the MaxwellSchr¨ odinger equations, Nonlinear Anal. 67(2007),1445-1456.

23

[10] L. G. Zhao, F. K. Zhao, Positive solutions for Schr¨ odinger-Poisson equations with a critical exponent, Nonlinear Anal. 70(2009), 2150-2164. [11] L. G. Zhao, F. K. Zhao, On the existence of solutions for the Schr¨ odinger-Poisson equations, J. Math. Anal. Appl. 346(2008), 155-169. [12] S. J. Chen, C. L. Tang, High energy solutions for the superlinear Schr¨ odinger-Maxwell equations, Nonlinear Anal. 71(2009), 4927-4934. [13] S. J. Chen, C. L. Tang, Multiple solutions for nonhomogeneous Schr¨ odinger-Maxwell and Klein-Gordon-Maxwell equations on R3 , Nonlinear Differ. Equ. Appl. 17(2010), 559-574. [14] T. D’Aprile, D. Mugnai, Solitary waves for nonlinear Klein-Gordon-Maxwell and Schr¨ odinger-Maxwell equations, Proc. R. Soc. Edinb. Sect. A 134(2004), 893-906. [15] Z. P. Wang, H. S. Zhou, Positive solution for a nonlinear stationary Schr¨ odinger-Poisson system in R3 , Discrete Contin. Dyn. Syst. 18(2007), 809-816. [16] A. Moameni, Existence of soliton solutions for a quasilinear Schr¨ odinger equation involving critical exponent in RN , J. Diff. Eqns. 229(2006), 570-587. [17] A. Moameni, On a class of periodic quasilinear Schr¨ odinger equations involving critical growth in R2 , J. Math. Anal. Appl. 335(2007), 775-786. [18] A. Moameni, On the existence of standing wave solutions to quasilinear Schr¨ odinger equations, Nonlinearity. 19(2006), 937-957. [19] B. Guo, J. Chen, F. Su, The blow-up problem for a quasilinear Schr¨ odinger equation, J. Math. Phys. 46(2005), 073510-073519. 24

[20] David Ruiz, Gaetano Siciliano, Existence of ground states for a modified nonlinear Schr¨ odinger equation, Nonlinearity. 23(2010), 1221-1233. [21] J. Liu, I. Sim, Z. Q. Wang, Bifurcations for quasilinear Schr¨ odinger equations, Nonlinear Anal. 67(2007), 3152-3166. [22] J. Liu, Y. Wang, Z. Q. Wang, Soliton solutions for quasilinear Schr¨ odinger equations: II, J. Diff. Eqns. 187(2003), 473-493. [23] J. Liu, Y. Wang, Z. Q. Wang, Solutions for quasilinear Schr¨ odinger equations via the Nehari method Commun, Partial Diff. Eqns. 29(2004), 879-901. [24] J. Liu, Z. Q. Wang, Soliton solutions for quasilinear Schr¨ odinger equations: I, Proc. Amer. Math. Soc. 131(2003), 441-448. [25] J. Liu, Z. Q. Wang, Symmetric solutions to a modified nonlinear Schr¨ odinger equation, Nonlinearity. 21(2008), 121-133. [26] M. Colin, L. Jeanjean, Solutions for a quasilinear Schr¨ odinger equation: a dual approach, Nonlinear Anal. 56(2004), 213-226. [27] M. J. Alves, P. C. Carri˜ ao, O. H. Miyagaki, Soliton solutions for a class of quasilinear elliptic equations on R, Adv. Nonlinear Stud. 7(2007), 579-598. [28] M. Poppenberg, K. Schmitt, Z. Q. Wang, On the existence of soliton solutions to quasilinear Schr¨odinger equations, Calc. Var. Partial Diff. Eqns. 14(2002), 329-344. [29] M. Poppenberg, On the local well posedness of quasi-linear Schr¨ oodinger equations in arbitrary space dimension, J. Diff. Eqns. 172(2001), 83-115.

25

´ J. M. do, O. H. Miyagaki, S. H. M. Soares, Soliton solutions for quasilinear [30] O. Schr¨ odinger equations: the critical exponential case, Nonlinear Anal. 67(2007), 33573372. [31] U. B. Severo, Symmetric and nonsymmetric solutions for a class of quasilinear Schr¨ odinger equations, Adv. Nonlinear Stud. 8(2008), 375-389. [32] U. Severo, Existence of weak solutions for quasilinear elliptic equations involving the p-Laplacian, Electro. J. Diff. Eqns. 56(2008), (16pp). [33] A. Nakamura, Damping and modification of exciton solitary waves, J. Phys. Soc. Japan. 42(1977), 1824-1835. [34] R. W. Hasse, A general method for the solution of nonlinear soliton and Kink Schr¨ odinger equations, Z. Physik B. 37(1980), 83-87. [35] S. Kurihura, Large-amplitude quasi-solitons in superfluid films, J. Phys. Soc. Japan. 50(1981), 3262-3267. [36] W. M. Zou, M. Schechter, Critical point theory and its applications, Springer, New York, 2006.

26