Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 118, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE GROUND STATE SOLUTIONS TO
SCHRÖDINGER-POISSON SYSTEMS WITH A NEGATIVE
NON-LOCAL TERM
YAN-PING GAO, SHENG-LONG YU, CHUN-LEI TANG
Abstract. In this article, we study the Schrödinger-Poisson system
−∆u + u − λK(x)φ(x)u = a(x)|u|p−1 u,
−∆φ = K(x)u2 ,
R3
x ∈ R3 ,
x ∈ R3 ,
R+
with p ∈ (1, 5). Assume that a :
→
and K : R3 → R+ are nonnegative
functions and satisfy suitable assumptions, but not requiring any symmetry
property on them, we prove the existence of a positive ground state solution
resolved by the variational methods.
1. Introduction and main results
In this article we study the Schrödinger-Poisson system
−∆u + V (x)u + λK(x)φ(x)u = f (x, u),
−∆φ = K(x)u2 ,
x ∈ R3 ,
x ∈ R3 ,
(1.1)
where V (x) = 1, λ < 0, f (x, s) = a(x)sp and a(x), K(x) satisfying some suitable
assumptions, we will prove problem (1.1) exists a positive ground state solution.
Similar problems have been widely investigated and it is well known they have
a strong physical meaning because they appear in quantum mechanics models (see
e.g. [9]) and in semiconductor theory [7, 8, 14, 15]. Variational methods and critical
point theory are always powerful tools in studying nonlinear differential equations.
In recent years, system (1.1) has been studied widely via modern variational methods under the various hypotheses, see [2, 4, 14, 19, 16] and the references therein.
Many researches have been devoted to the study of problem (1.1), but they mainly
concern either the autonomous case or, in the non-autonomous case, the search
of the so-called semi-classical states. We refer the reader interested in a detailed
bibliography to the survey paper [2]. All these works deal with systems like (1.1)
with λ > 0 and the nonlinearity f (x, s) = sp with p subcritical.
To the best of our knowledge, there are only a few article on the existence
of ground state solutions to (1.1) with the negative coefficient of the non-local
term. Recently, in [17], the author obtained a ground state solution. In [18], the
2010 Mathematics Subject Classification. 35J47, 35J50, 35J99.
Key words and phrases. Schrödinger-Poisson system; ground state solution;
variational methods.
c
2015
Texas State University - San Marcos.
Submitted January 20, 2015. Published April 30, 2015.
1
2
Y.-P. GAO, S.-L. YU, C.-L. TANG
EJDE-2015/118
author considered the nonlinearity f (x, s) = a(x)s2 and obtained a ground state
solution. In this article, we consider the nonlinearity f (x, s) = a(x)sp for following
Schrödinger-Poisson system
−∆u + u − λK(x)φ(x)u = a(x)|u|p−1 u,
−∆φ = K(x)u2 ,
x ∈ R3 ,
x ∈ R3 .
(1.2)
It is worth noticing that there are few works concerning on this case up to now.
As we shall see in Section 2, problem (1.2) can be easily transformed in a nonlinear Schrödinger equation with a non-local term. Briefly, the Poisson equation
is solved by using the Lax-Milgram theorem, then, for all u ∈ H 1 (R3 ), a unique
φu ∈ D1,2 (R3 ) is obtained, such that −∆φ = K(x)u2 and that, inserted into the
first equation, gives
−∆u + u − λK(x)φu (x)u2 = a(x)|u|p−1 u,
x ∈ R3 .
(1.3)
This problem is variational and its solutions are the critical points of the functional
defined in H 1 (R3 ) by
Z
Z
λ
1
1
K(x)φu (x)u2 dx −
a(x)|u|p+1 dx.
(1.4)
I(u) = kuk2 −
2
4 R3
p + 1 R3
In our research, we deal with the case in which p ∈ (1, 5), moreover we always
assume that a(x) and K(x) satisfy:
(A1) There exists a constant c > 0, such that a(x) > c for all x ∈ R3 and
6
a(x) − c ∈ L 5−p (R3 );
(K1) K ∈ L2 (R3 ).
Our main result reads as follows.
Theorem 1.1. Suppose a, K : R3 → R+ , λ > 0 and p ∈ (1, 5). Let (A1), (K1)
hold. Then problem (1.2) has a positive ground state solution.
Remark 1.2. To the best of our knowledge, there are only two articles [17, 18] on
the existence of ground state solutions to (1.1) with the negative coefficient of the
non-local. In [17], the author discusses the negative coefficient of the non-local term
under symmetry assumption, but we get the positive ground state solution without
any symmetry assumption. Compared with the [18], we do not need conditions
lim
|x|→+∞
a(x) = a∞
and
lim
|x|→+∞
K(x) = K∞ .
The remainder of this paper is organized as follows. In Section 2, notation and
preliminaries are presented. In Section 3, we give the proof of Theorem 1.1.
2. Notation and preliminaries
Hereafter we use the following notation:
H 1 (R3 ) is the usual Sobolev space endowed with the standard scalar product
and norm
Z
Z
(u, v) =
(∇u · ∇v + uv)dx; kuk2 =
(|∇u|2 + u2 )dx.
R3
D
1,2
3
R3
(R ) is the completion of
C0∞ (R3 )
kukD1,2 =
Z
with respect to the norm
1/2
|∇u|2 dx
.
R3
EJDE-2015/118
POSITIVE GROUND STATE SOLUTIONS
3
Lp (Ω), 1 ≤ p ≤ +∞, Ω ⊆ R3 , denotes a Lebesgue space, the norm in Lp (Ω) is
denoted by kukLp (Ω) = |u|p,Ω when Ω is a proper subset of R3 , by kukLp (Ω) = | · |p
when Ω = R3 .
L∞ (Ω) is the space of measurable functions in Ω; that is,
ess supx∈Ω |u(x)| = inf{C > 0 : |u(x)| ≤ C a. e. in Ω} < +∞.
For any ρ > 0 and for any z ∈ R3 , Bρ (z) denotes the ball of radius ρ centered
at z, and |Bρ (z)| denotes its Lebesgue measure. C, C0 , C1 , C2 are various positive
constants which can change from line to line.
From the embeddings, H 1 (R3 ) ,→ L6 (R3 ) and D1,2 (R3 ) ,→ L6 (R3 ), we obtain
the inequalities
|u|6 ≤ C1 kuk
∀u ∈ H 1 (R3 )\{0},
|u|6 ≤ C2 kuk
∀u ∈ D1,2 (R3 )\{0}.
It is well known and easy to show that problem (1.2) can be reduced to a single
equation with a non-local term. Actually, considering for all u ∈ H 1 (R3 ), the linear
functional Lu defined in D1,2 (R3 ) by
Z
Lu (v) =
K(x)u2 v dx,
R3
the Hölder and Sobolev inequalities imply
Lu (v) ≤ |K|2 |u2 |3 |v|6 = |K|2 |u|26 |v|6 ≤ C2 |K|2 · |u|26 kvkD1,2 .
1
(2.1)
3
Hence, from the Lax-Milgram theorem, for every u ∈ H (R ), the Poisson equation
−∆φ = K(x)u2 exists a unique φu ∈ D1,2 (R3 ) such that
Z
Z
K(x)u2 v dx =
∇φu · ∇v dx,
(2.2)
R3
for any v ∈ D
1,2
R3
3
(R ). Using integration by parts, we get
Z
Z
∇φu · ∇v dx = −
v∆φu dx,
R3
R3
therefore,
−∆φu = K(x)u2 ,
in a weak sense and the representation formula
Z
1
K(y) 2
u (y)dy =
∗ Ku2
φu =
|x|
R3 |x − y|
(2.3)
holds. Moreover, φu > 0 when u 6= 0, because K does, and by (2.1), (2.2) and the
Sobolev inequality, the relations
Z
R3
Z
R3
kφu kD1,2 ≤ C2 C12 · |K|2 kuk2 , |φu |6 ≤ C2 kφu kD1,2 ,
Z
K(x)K(y) 2
u (x)u2 (y)dxdy =
K(x)φu u2 dx ≤ C22 C14 · |K|22 kuk4
|x − y|
3
R
(2.4)
(2.5)
hold. Substituting φu in problem (1.2), we are led to (1.3), whose solutions can be
obtained by looking for critical points of the functional I : H 1 (R3 ) → R where I is
defined in (1.4). Indeed, (2.4) and (2.5) imply that I is a well-defined C 2 functional,
and that
Z 0
hI (u), vi =
∇u · ∇v + uv − λK(x)φu uv − a(x)|u|p−1 uv dx.
(2.6)
R3
4
Y.-P. GAO, S.-L. YU, C.-L. TANG
EJDE-2015/118
Hence, if u ∈ H 1 (R3 ) is a critical point of I, then the pair (u, φu ), with φu as in
(2.3), is a solution of (1.2).
Let us define the operator Φ:H 1 (R3 ) → D1,2 (R3 ) as
Φ[u] = φu .
In the next lemma we summarize some properties of Φ, useful for the study our
problem.
Lemma 2.1 ([11]).
(1) Φ is continuous;
(2) Φ maps bounded sets into bounded sets;
(3) if un * u in H 1 (R3 ) then Φ[un ] * Φ[u] in D1,2 (R3 );
(4) Φ[tu] = t2 Φ[u] for all t ∈ R.
Lemma 2.2 ([13]). Suppose r > 0, 2 < q < 2∗ (= 6). If {un } ⊂ H 1 (R3 ) is bounded
and
Z
sup
|un |q dx → 0, as n → +∞,
y∈R3
3
B(y,r)
then un → 0 in L (R ) for 2 < q < 2∗ .
q
3. Proof of main results
First wee give some properties of the nonlinear Schrödinger equation
− ∆u + u = c|u|p−1 u,
(3.1)
that has been broadly studied in [13, 12]. We set
N∞ := {u ∈ H 1 (R3 )\{0} : kuk2 = c|u|p+1
p+1 }.
Then for any u ∈ N∞ , we have
Z
1
c
1 1
2
|u|p+1 dx =
−
kuk2 ,
I∞ (u) = kuk −
2
p + 1 R3
2 p+1
(3.2)
(3.3)
and m∞ := inf{I∞ (u) : u ∈ N∞ }.
It is well known that (3.1) has at least a ground state solution which we denote
w∞ . By using (3.2) and (3.3), we know that
1
1 −
m∞ = I∞ (w∞ ) =
kw∞ k2
2 p+1
and
Z
kw∞ k2 = c
|w∞ |p+1 dx.
(3.4)
R3
For (1.2), it is not difficult to verify that the functional I is bounded either from
below or from above. So it is convenient to consider I restricted to a natural
constraint, the Nehari manifold, that contains all the nonzero critical points of I
and on which I turns out to be bounded from below. We set
N := {u ∈ H 1 (R3 )\{0} : G(u) = 0}
where
0
2
Z
G(u) = hI (u), ui = kuk − λ
2
Z
K(x)φu u dx −
R3
a(x)|u|p+1 dx.
R3
The following lemma states the main properties of N .
Lemma 3.1. I is bounded from below on N by a positive constant.
(3.5)
EJDE-2015/118
POSITIVE GROUND STATE SOLUTIONS
Proof. Let u ∈ N , from (A1) and Hölder’s inequality, we have
Z
Z
0 = kuk2 − λ
K(x)φu u2 dx −
a(x)|u|p+1 dx
R3
2
R3
4
≥ kuk − Ckuk − C0 kuk
5
(3.6)
p+1
from which we have
kuk ≥ C1 > 0,
∀u ∈ N
(3.7)
Using this inequality, λ > 0, K > 0, a > 0, when 1 < p < 3, we obtain
Z
1
1
1 1
2
I(u) =
kuk +
λ
K(x)φu u2 dx
−
−
2 p+1
p+1 4
R3
1
1 ≥
−
kuk2
2 p+1
1
1 2
≥
−
C > 0,
2 p+1 1
(3.8)
when 3 ≤ p < 5,
1
1 1
kuk2 +
−
4
4 p+1
1
≥ kuk2
4
1 2
≥ C1 > 0.
4
Z
I(u) =
a(x)|u|p+1 dx
R3
(3.9)
Setting m := inf{I(u) : u ∈ N }, as a consequence of Lemma 3.1, m turns out to
be a positive number. Then we obtain a sequence {un } ⊂ N , such that
lim I(un ) = m.
(3.10)
n→∞
Now we give the proof of our main result.
Proof of Theorem 1.1. First, we prove that
m < m∞ .
(3.11)
We know that w∞ ∈ N∞ and I∞ (w∞ ) = m∞ . We claim that there exists t0 > 0
such that t0 w∞ ∈ N . Indeed, from (3.5), for all t ≥ 0 one has
Z
Z
2
G(tw∞ ) = t2 kw∞ k2 − λt4
K(x)φw∞ w∞
dx − tp+1
a(x)|w∞ |p+1 dx,
R3
R3
then G(0) = 0 and G(tw∞ ) → −∞ as t → +∞. Moreover,
Z
Z
0
2
2
2
p−1
Gt (tw∞ ) = t 2kw∞ k −4λt
K(x)φw∞ w∞ dx−(p+1)t
R3
a(x)|w∞ |p+1 dx ,
R3
then there exists tmax > 0 such that G0t (tw∞ ) > 0 for all 0 < t < tmax and
G0t (tw∞ ) < 0 for all t > tmax . Then G(tw∞ ) is increasing for all 0 < t < tmax
6
Y.-P. GAO, S.-L. YU, C.-L. TANG
EJDE-2015/118
and G(tw∞ ) decreasing for all t > tmax . Thus there exists t0 > 0 such that
G(t0 w∞ ) = 0. That is, t0 w∞ ∈ N . Our claim is true. It follows that
m ≤ I(t0 w∞ )
=
t20
t4
kw∞ k2 − 0 λ
2
4
Z
R3
2
K(x)φw∞ (x)w∞
dx −
tp+1
0
p+1
Z
a(x)|w∞ |p+1 dx
R3
Z
t20
tp+1
kw∞ k2 − 0
c|w∞ |p+1 dx
2
p + 1 R3
1
1 kw∞ k2
≤
−
2 p+1
= I∞ (w∞ ) = m∞ .
(3.12)
<
We assume that {un } is what obtained in (3.10). From (2.3), we can get {|un |} is
also a minimize sequence. Setting un (x) ≥ 0 in R3 a.e. by (3.8) and (3.9), we have
if p ∈ (1, 3), then
1
1 I(un ) ≥
−
kun k2 ,
2 p+1
and if p ∈ [3, 5), then
1
I(un ) ≥ kun k2 .
4
In both cases, being I(un ) is bounded, {un } is also bounded.
On the other hand, since {un } is bounded in H 1 (R3 ), there exists u ∈ H 1 (R3 )
such that, up to a subsequence,
in H 1 (R3 );
un * u,
un → u,
in
un (x) → u(x),
(3.13)
3
Lp+1
loc (R );
(3.14)
3
a.e. in R .
(3.15)
Setting
zn1 (x) = un (x) − u(x).
Obviously, zn1 * 0 in H 1 (R3 ), but not strongly. A direct computation gives
kun k2 = kzn1 + uk2 = kzn1 k2 + kuk2 + o(1).
(3.16)
According to the Brezis-Lieb Lemma [10], we deduce
p+1
1 p+1
|un |p+1
p+1 = |u|p+1 + |zn |p+1 + o(1).
1
Then, we claim that, for any h ∈ H (R ), we have
Z
Z
p−1
|un | un h dx →
|u|p−1 uh dx.
R3
(3.17)
3
(3.18)
R3
For every h ∈ C0∞ (R3 ), there exists a bounded open subset Ω ⊂ R3 , such that
supp h ⊂ Ω, where supp h = {x ∈ R3 : h(x) 6= 0}. From (3.14), we have
Z
|un |p−1 un h dx − |u|p−1 uh dx
R3
Z
|un |p−1 un h − |u|p−1 uhdx
<
3
ZR
≤
p(|un |p−1 + |u|p−1 )|un − u||h|dx
R3
EJDE-2015/118
POSITIVE GROUND STATE SOLUTIONS
Z
=
p|un |p−1 |un − u||h|dx +
R3
Z
7
p|u|p−1 |un − u||h|dx
R3
< p|un |p+1 |un − u|p+1,Ω |h|p+1 + p|u|p+1 |un − u|p+1,Ω |h|p+1 < ε
which proves (3.18). Let us show that
Z
Z
K(x)φun u2n dx =
K(x)φu u2 dx + o(1),
R3
R3
Z
Z
K(x)φun un h dx =
K(x)φu uh dx + o(1).
R3
(3.19)
(3.20)
R3
First let us observe that, in view of the Sobolev embedding theorem, (3.13) and (3)
of Lemma 2.1, un * ū in H 1 (R3 ) implies
un * u,
u2n
2
→u ,
φun * φu ,
in L6 (R3 );
in
(R );
(3.23)
L6loc (R3 ).
(3.24)
in D
φun → φu ,
in
(3.21)
L3loc (R3 );
1,2
3
(3.22)
Furthermore, considering (3.22) and (3.24) respectively, we can assert that for any
choice of ε > 0 and ρ > 0, the relations
|u2n − u2 |3,Bρ (0) < ε,
(3.25)
|φun − φu |6,Bρ (0) < ε
(3.26)
hold for large n.
On the other hand, un being bounded in H 1 (R3 ), φun is bounded in D1,2 (R3 )
and in L6 (R3 ), because of (2) of Lemma 2.1 and the continuity of the Sobolev
embedding of D1,2 (R3 ) in L6 (R3 ). Moreover K ∈ L2 (R3 ), for any ε > 0, there
exists ρ = ρ(ε) such that
|K|2,R3 \Bρ (0) < ε,
∀ρ ≥ ρ.
(3.27)
Hence, by (3.25) and (3.27), for large n, we deduce that
Z
Z
K(x)φun u2n dx −
K(x)φu u2 dx
R3
R3
Z
Z
2
2
≤
K(x)φun (un − u )dx +
K(x)(φun − φu )u2 dx
R3
R3
Z
Z
2
2
≤
K(x)φun (un − u )dx + K(x)(φun − φu )u2 dx
3
3
R
R
Z
Z
2
2
≤
K(x)φun (un − u )dx + K(x)φun (u2n − u2 )dx
R3 \Bρ (0)
Z
+
Bρ (0)
K(x)(φun
R3 \Bρ (0)
≤
Z
− φu )u2 dx + Bρ (0)
|K|2,R3 \Bρ (0) |φun |6,R3 \Bρ (0) |u2n
+ |φun
2
− u |3,R3 \Bρ (0)
− φu |6,R3 \Bρ (0) |u2 |3,R3 \Bρ (0) + |K|2,Bρ (0) |φun |6,Bρ (0) |u2n − u2 |3,Bρ (0)
+ |K|2,Bρ (0) |φun − φu |6,Bρ (0) |u2 |3,Bρ (0)
≤ Cε
K(x)(φun − φu )u2 dx
8
Y.-P. GAO, S.-L. YU, C.-L. TANG
EJDE-2015/118
which proves (3.19).
Analogously, by (3.26) and (3.27), for large n, we infer that
Z
Z
K(x)φun un h dx −
K(x)φu uh dx ≤ ε
R3
R3
which proves (3.20). Therefore, by (3.16), (3.17) and (3.19) respectively, we obtain
Z
Z
1
λ
1
K(x)φun u2n dx −
a(x)|un |p+1 dx
I(un ) = kun k2 −
2
4 R3
p + 1 R3
Z
Z
1 1 2 1
λ
1
2
2
= kzn k + kuk −
K(x)φu u dx −
a(x)|u|p+1 dx
2
2
4 R3
p + 1 R3
(3.28)
Z
c
1 p+1
−
|z | dx + o(1)
p + 1 R3 n
= I(u) + I∞ (zn1 ) + o(1).
By (3.18) and (3.20) for any h ∈ C0∞ (R3 ),
Z
0
hI (un ), hi =
(∇un · ∇h + un h − λK(x)φun un h − a(x)|un |p−1 un h)dx
R3
Z
(3.29)
=
(∇u · ∇h + uh − λK(x)φu uh − a(x)|u|p−1 uh)dx + o(1)
R3
0
= hI (u), hi + o(1).
We now claim that
∇I(un ) → 0, in H 1 (R3 ).
(3.30)
By Lagrange’s multiplier theorem, we know that there exists λn ∈ R such that
o(1) = ∇I|N (un ) = ∇I(un ) − λn ∇G(un ).
(3.31)
So, taking the scalar product with un , we obtain
o(1) = (∇I(un ), un ) − λn (∇G(un ), un ).
G turns out to be a C 1 functional. Using (3.6) and λ > 0, K > 0, a > 0, when
1 < p ≤ 3, we deduce
Z
Z
0
2
2
hG (u), ui = 2kuk − 4λ
K(x)φu u dx − (p + 1)
a(x)|u|p+1 dx
R3
R3
Z
= (1 − p)kuk2 + λ(p − 3)
k(x)φu u2 dx
(3.32)
3
R
≤ (1 − p)kuk2
≤ −(p − 1)C1 < 0,
when 3 < p < 5,
hG0 (u), ui = 2kuk2 − 4λ
Z
K(x)φu u2 dx − (p + 1)
R3
Z
2
= −2kuk + (3 − p)
a(x)|u|p+1 dx
R3
Z
a(x)|u|p+1 dx
R3
(3.33)
≤ −2kuk2
≤ −2C2 < 0.
Since un ∈ N , we have (∇I(un ), un ) = 0; by inequalities (3.32) and (3.33), we have
(∇G(un ), un ) < C < 0. Thus λn → 0 for n → +∞. Moreover, by the boundedness
EJDE-2015/118
POSITIVE GROUND STATE SOLUTIONS
9
of {un }, ∇G(un ) is bounded and this implies λn ∇G(un ) → 0, so (3.31) follows from
(3.30). By (3.29) and (3.30), we have hI 0 (u), hi = 0, so u is a solution of problem
(1.2). By (3.19), we have
Z
Z
0
2
2
hI (un ), un i = kun k − λ
K(x)φun un dx −
a(x)|un |p+1 )dx
3
3
R
R
Z
Z
2
1 2
2
K(x)φu u dx −
a(x)|u|p+1 dx
= kuk + kzn k − λ
R3
R3
Z
−c
|zn1 |p+1 dx + o(1)
R3
0
(zn1 ), zn1 i + o(1),
= hI 0 (u), ui + hI∞
which implies that
0
(zn1 ), zn1 i = kzn1 k2 − c|zn1 |p+1
o(1) = hI∞
p+1 .
(3.34)
Setting
δ := lim sup
n→+∞
Z
|zn1 |p+1 dx .
sup
y∈R3
B1 (y)
We claim δ = 0. By [19, Lemma 1.21], one has
zn1 → 0,
in Lp+1 (R3 ).
(3.35)
From (3.34) and (3.35), we obtain
0
o(1) = hI∞
(zn1 ), zn1 i
= kzn1 k2 − c|zn1 |p+1
p+1
= kzn1 k2 + o(1)
= kun − uk2 + o(1),
so un → u in H 1 (R3 ). Let u = u, so I(u) = m, I 0 (u) = 0 and u(x) > 0 a.e. in R3 .
Let us prove δ = 0. Actually, if δ > 0, there exists sequence {yn1 } ⊂ R3 , such
that
Z
δ
|zn1 |p+1 dx > .
2
1
B1 (yn )
Let us now consider zn1 (· + yn1 ). We assume that zn1 (· + yn1 ) * u1 in H 1 (R3 ) and,
then, zn1 (x + yn1 ) → u1 (x) a.e. on R3 . Since
Z
δ
|zn1 (x + yn1 )|p+1 dx > ,
2
B1 (0)
from the Rellich theorem it follows that
Z
δ
|u1 (x)|p+1 dx ≥ ,
2
B1 (0)
and thus u1 6= 0. Finally, let us set
zn2 (x) = zn1 (x + yn1 ) − u1 (x).
Then, using (3.16), (3.17) and the Brezis-Lieb Lemma, we have
kzn2 k2 = kzn1 k2 − ku1 k2 + o(1),
|zn2 |p+1
p+1
=
|un |p+1
p+1
−
|u|p+1
p+1
−
|u1 |p+1
p+1
+ o(1).
(3.36)
(3.37)
10
Y.-P. GAO, S.-L. YU, C.-L. TANG
EJDE-2015/118
These equalities imply
I∞ (zn2 ) = I∞ (zn1 ) − I∞ (u1 ) + o(1),
hence, by using (3.28), we obtain
I(un ) = I(u) + I∞ (zn1 ) + o(1)
= I(u) + I∞ (u1 ) + I∞ (zn2 ) + o(1).
(3.38)
Using (3.34), (3.36) and (3.37), we obtain
0
hI∞
(zn1 ), zn1 i = kzn1 k2 − c|zn1 |p+1
p+1
2 2
2 p+1
= ku1 k2 − c|u1 |p+1
p+1 + kzn k − c|zn |p+1 + o(1)
0
0
= hI∞
(u1 ), u1 i + hI∞
(zn2 ), zn2 i + o(1),
which implies
0
o(1) = hI∞
(zn2 ), zn2 i = kzn2 k2 − c|zn2 |p+1
p+1 .
Moreover, we obtain
1
c
1 2 2
1
|zn2 |p+1
−
kzn k + o(1).
=
I∞ (zn2 ) = kzn2 k2 −
p+1
2
p+1
2 p+1
(3.39)
Since zn1 * u1 in H 1 (R3 ) and u1 6= 0, according to (3.34), one has u1 ∈ N∞ .
Because of u ∈ N , from Lemma 3.1, we obtain I(u) > 0. Thus, using (3.38) and
(3.39), we obtain
m = lim inf I(un )
n→∞
≥ I(u) + I∞ (u1 ) + lim inf I∞ (zn2 )
n→∞
1
≥ I∞ (u ) ≥ m∞
which contradicts with (3.11).
Acknowledgments. This research was supported by National Natural Science
Foundation of China (No.11471267).
The authors would like to thank the anonymous referees for their valuable suggestions.
References
[1] C. O. Alves; Schrödinger-Possion equations without Ambrosetti-Rabinowitz condition, J.
Math. Anal. Appl. 377 (2011) 584-592.
[2] A. Ambrosetti; On Schrödinger-Poisson systems, Milan J. Math. 76 (2008) 257-274.
[3] A. Azzollini; Concentration and compactness in nonlinear Schrödinger-Poisson system with
a general nonlinearity, J. Differential Equations 249 (2010) 1746-1763.
[4] A. Ambrosetti, D. Ruiz; Multiple bound states for the Schrödinger-Poisson problem, Commun. Contemp. Math. 10 (2008) 391-404.
[5] A. Ambrosetti, P. Rabinowitz; Dual variational methods in critical point theory and applications, J. Funct. Anal. 14 (1973) 349-381.
[6] A. Azzollini, A. Pomponio; Ground state solutions for the nonlinear Schrödinger-maxwell
equations, J. Math. Anal. Appl. 345 (2008) 90-108.
[7] V. Benci, D. Fortunato; An eigenvalue problem for the Schröinger-Maxwell equations, Topol.
Methods Nonlinear Anal. 11 (1998) 283-293.
[8] V. Benci, D. Fortunato; Solitary waves of the nonlinear Klein-Gordon equation coupled with
Maxwell equations, Rev. Math. Phys. 14 (2002) 409-420.
[9] R. Benguria, H. Brezis, E.H. Lieb; The Thomas-Fermi-von Weizsäcker theory of atoms and
molecules, Comm. Math. Phys. 79 (1981) 167-80.
EJDE-2015/118
POSITIVE GROUND STATE SOLUTIONS
11
[10] H. Brezis, E. Lieb; A relation between pointwise convergence of functions and convergence
of functionals, Proc. Amer. Math. Soc. 88 (1983) 486-490.
[11] G. Cerami, Giusi Vaira; Positive solutions for some non-autonomous Schrödinger-Poisson
systems, J. Differential Equations 248 (2010) 521-543.
[12] M. K. Kwong; Uniqueness of positve solutions of ∆u − u + up = 0 in RN , Arch. Ration.
Mech. Anal. 105 (1989) 243-266.
[13] P. L. Lions; The concentration-compactness principle in the calculus of variations. The locally
compact case, part 2, Anal. Nonlinéaire I (1984) 223-283.
[14] P. L. Lions; Solutions of Hartree-Fock equations for Coulomb systems, Comm. Math. Phys.
109 (1984) 33-97.
[15] P. Markowich, C. Ringhofer, C. Schmeiser; Semiconductor Equations, Springer-Verlag, New
York, 1990.
[16] D. Ruiz; The Schrödinger-Poisson equation under the effect of a nonlinear local term, J.
Funct. Anal. 237 (2006) 655-674.
[17] G. Vaira; Ground states for Schrödinger-Poisson type systems, Ricerche mat. 60 (2011) 263297.
[18] G. Vaira; Existence of bounded states for Schrödinger-Poisson type systems, S. I. S. S. A.
251 (2012) 112-146.
[19] M. Willem; Minimax Theorems, Progr. Nonlinear Differential Equations Appl., vol. 24,
Birkhäuser, 1996.
[20] L. Zhao, F. Zhao; On the existence of solutions for the Schrödinger-Poisson equations, J.
Math. Anal. Appl. 346 (2008) 155-169.
Yan-Ping Gao
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: gao0807@swu.edu.cn
Sheng-Long Yu
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: ysl345827434@163.com
Chun-Lei Tang (corresponding author)
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: tangcl@swu.edu.cn, Phone +86 23 68253135, Fax +86 23 68253135