University of Mazandaran, Iran http://cjms.journals.umz.ac.ir ISSN: 1735-0611 CJMS
.
4(1)(2015), 61-75
Existence of infinitely many solutions for coupled system
of Schr¨
o
dinger-Maxwell’s equations
G.Karamali
1and M.K.Kalleji
21
PhD, Assistant Professor, Academic member of School of
Mathematics, Shahid Sattari Aeronautical University of Science and
Technology, South Mehrabad, Tehran, Iran.
2
School of Mathematics, Shahid Sattari Aeronautical University of
Science and Technology, South Mehrabad, Tehran, Iran.
Abstract. In this paper we study the existence of infinitely many large energy solutions for the coupled system of Schr¨odinger-Maxwell’s equations
−∆u+V(x)u+φu=Hv(x, u, v) inR3 −∆φ=u2 inR3
−∆v+V(x)v+ψv=Hu(x, u, v) inR3
−∆ψ=v2 in
R3,
via the Fountain theorem under Cerami condition. More precisely, we consider the More general case and weakenV1∗with respect to
the conditionV1 in [6].
Keywords: Schr¨odingerMaxwell system ,Cerami condition,Variational methods, Strongly indefinite functionals.
2000 Mathematics subject classification: 35Pxx; Secondary 46Txx.
1Corresponding author: m.kalleji@yahoo.com
Received: 30 November 2014 Revised: 15 March 2015 Accepted: 16 March 2015
1.
Introduction
In this paper, we study the nonlinear coupled system of Schr¨
o
dinger-Maxwell’s equations
−
∆
u
+
V
(
x
)
u
+
φu
=
H
v(
x, u, v
)
in
R
3−
∆
φ
=
u
2in
R
3−
∆
v
+
V
(
x
)
v
+
ψv
=
Hu
(
x, u, v
)
in
R
3−
∆
ψ
=
v
2in
R
3,
(1.1)
where
V
∈
C
(
R
3,
R
) and
H
∈
C
1(
R
3,
R
) which are satisfied in some
suitable conditions.
In the classical model, the interaction of a charge particle with an
electro-magnetic field can be described by the nonlinear Schr¨
o
dinger-Maxwell’s
equations.
In this article, we want to study the interaction of two
charge particles Simultaneously with same potential function
V
(
x
) and
different scalar potential
φ
and
ψ
which are satisfied in suitable
condi-tions. For more details on the physical aspects see [1] and the references
therein. More precisely, we have to solve the system 1.1 if we want to
find electrostatic-type solutions.
In [2] Zhang et al. considered the system 1.1 without the scalar potential
φ
and
ψ
, so-called the Hamiltonian elliptic system
−
∆
u
+
V
(
x
)
u
=
H
v(
x, u, v
)
in
R
3−
∆
v
+
V
(
x
)
v
=
H
u(
x, u, v
)
in
R
3,
(1.2)
They assumed the potential
V
(
x
) is non-periodic and sing changing,
and
H
(
x, z
) is non-periodic and asymptotically quadratic in
z
= (
u, v
) :
R
N→
R
×
R
.
For the case of a bounded domain the system 1.2 were
studied by many authors [3, 4, 5] and the references therein.
In [6] Li and Chen, studied the existence of infinitely many large energy
solutions for the superlinear Schr¨
o
dinger-Maxwell’equations
−
∆
u
+
V
(
x
)
u
+
φu
=
f
(
x, u
)
in
R
3−
∆
φ
=
u
2in
R
3,
(1.3)
They assumed on this paper that the potential function
V
(
x
) is bounded
from below with a positive constant and they are used to solve the
prob-lem 1.3, the Fountain theorem in critical point theory. In [7] Schaftingen
et al. studied positive bound states for the equation
−
2∆
u
+
V
(
x
)
u
=
K
(
x
)
f
(
u
)
in
R
N,
(1.4)
as nonlinear parametric Schr¨
o
dinger equation which were studied by
many authors see [8, 9, 10]. The infinitely many large solutions for the
equation 1.3 are obtained in [11] with the following variant
Ambrosetti-Rabinowitz type condition [6], there exist
µ >
4 such that for any
s
∈
R
and
x
∈
R
3µF
(
x, u
) :=
µ
Z
s0
f
(
x, t
)
dt
≤
sf
(
x, s
)
.
Existence of solutions are obtained via Fountain theorem in critical point
theory. More precisely, in this paper we consider the more general case
and weaken the condition of
V
1in [6] and we assume that the potential
V
is non-periodic and sing changing. We assume the following conditions
:
V
1∗)
V
∈
C
(
R
3,
R
) and there exists some
M >
0 such that the set
Ω
M=
{
x
∈
R
3;
V
(
x
)
≤
M
}
is not nonempty and has finite Lebesgue
measure.
H
1)
H
∈
C
1(
R
3×
R
2,
R
) and for some 2
< p <
2
∗= 6
,
and
M
1, M
2>
0
,
|
Hu
(
x, u, v
)
| ≤
M
1|
u
|
+
M
1|
u
|
p−1and
|
Hv
(
x, u, v
)
| ≤
M
2|
v
|
+
M
2|
v
|
p−1,
for a.e
x
∈
R
3and
u, v
:
R
3→
R
,
and also
lim
u→0Hu
(
x, u, v
)
u
= 0
and
ulim
→0Hv
(
x, u, v
)
v
= 0
,
uniformly for
x
∈
R
3and
u, v
∈
R
.
H
2)
lim
|(u,v)|→∞
H(x,u,v)
|(u,v)|4
= +
∞
,
uniformly in
x
∈
R
3and (
u, v
)
∈
R
2and
H
(
x,
0
,
0) = 0
, H
(
x, u, v
)
≥
0
for all (
x, u, v
)
∈
R
3×
R
×
R
.
H
3) There exists a constant
θ
≥
1 such that
θ
H
ˆ
(
x, u, v
)
≥
H
ˆ
(
x, su, sv
)
for all
x
∈
R
3,
(
u, v
)
∈
R
2and
t, s
∈
[0
,
1]
,
where
ˆ
H
(
x, u, v
) =
Hu
(
x, tu, v
)
tu
+
Hv
(
x, u, sv
)
sv
−
4
H
(
x, tu, sv
)
.
H
4)
H
(
x,
−
u, v
) =
H
(
x, u, v
) and
H
(
x, u,
−
v
) =
H
(
x, u, v
) for all
x
∈
R
3and (
u, v
)
∈
R
2.
Here, we express the Cerami condition which was established by G.
Cerami in [12]
Definition 1.1.
Suppose that functional
I
is
C
1and
c
∈
R
,
if any
sequence
{
u
n}
satisfying
I
(
u
n)
→
c
and (1 +
k
u
nk
)
I
0
To approach the main result, we need the following critical point
theorem.
Theorem 1.2.
(Fountain theorem under Cerami condition) Let
X
be a
Banach space with the norm
k
.
k
and let
X
jbe a sequence of subspace of
X
with
dimX
j<
∞
for any
j
∈
N
.
Further,
X
=
⊕
j∈NXj,
the closure
of the direct sum of all
X
j.
Set
W
k=
⊕
kj=0X
j, Z
k⊕
∞j=kX
j.
Consider
an even functional
I
∈
C
1(
X,
R
)
,
that is ,
I
(
−
u
) =
I
(
u
)
for any
u
∈
X.
Also suppose that for any
k
∈
N
,
there exist
ρ
k> r
k>
0
such that
I
1)
ak
:= max
u∈Wk,kuk=ρkI
(
u
)
≤
0
,
I
2)
bk
:= inf
u∈Zk,kuk=rkI
(
u
)
→
+
∞
as
k
→ ∞
,
I
3)
the Cerami condition holds at any level
c >
0
.
Then the functional
I
has an unbounded sequence of critical values.
Now, our main result is the following :
Theorem 1.3.
Let
V
1∗, H
1−
H
4be satisfied. Then the system 1.1 has
infinitely many solutions
{
((
u
k, φ
k)
,
(
v
k, ψ
k))
}
in product space
YHD
×
Y
HD(see section 2)which satisfies in
1
2
Z
R3
[
|∇
u
k|
2+
|∇
v
k|
2+
V
(
x
)(
u
2k+
v
k2) ]
dx
−
1
4
Z
R3
[
|∇
φ
k|
2+
|∇
ψ
k|
2]
dx
+
1
2
Z
R3
[
φ
ku
2k+
ψ
kv
k2]
dx
−
Z
R3
H
(
x, u, v
)
dx
→
+
∞
.
Remark
1.4
.
The assumption
V
1∗implies that the potential
V
is not
periodic and changes sing. This is different from condition
V
1in [6].
Also conditions
H
1and
H
3modified for the system 1.1 with respect to
the system 1
.
1 which considered by Li and Chen in [6].
2.
some auxiliary results and notations
In this section we give some notations and definitions on the function
product space. We set
H
1(
R
3) :=
{
u
∈
L
2(
R
3)
| |∇
u
| ∈
L
2(
R
3)
}
,
(2.1)
with the norm
k
u
k
H1:= (
Z
R3
|∇
u
|
2+
u
2dx
)
12,
(2.2)
and we consider the function space
D
1,2(
R
3) :=
{
u
∈
L
2∗
with the norm
k
u
k
D1,2:= (
Z
R3
|∇
u
|
2dx
)
12.
(2.4)
Now, we consider the function space
E
:=
{
u
∈
H
1(
R
3)
|
Z
R3
|∇
u
|
2+
u
2dx <
∞}
.
Then
E
is Hilbert space [13] with the inner product
(
u, v
)
E:=
Z
R3
(
∇
u.
∇
v
+
V
(
x
)
uv
)
dx
(2.5)
and the norm
k
u
k
E:= (
u, u
)
1 2E
.
We set
X
E:=
E
×
E, Y
HD:=
H
1(
R
3)
×
D
1,2(
R
3)
and Z
ED:=
E
×
D
1,2(
R
3)
.
Hence , we can define an inner product on
XE
as
((
u, v
)(
w, z
))
XE:= (
u, w
)
E+ (
v, z
)
E(2.6)
and the corresponding norm on
X
Eby this inner product as following
k
(
u, v
)
k
XE:= (
k
u
k
2E
+
k
v
k
2E)
12
= ((
u, u
)
E+ (
v, v
)
E)
12
.
(2.7)
Let
H
be a Hilbert space and
H
=
H
×
H
with the inner product
((
u, v
)
,
(
w, z
))
H= (
u, w
)
H+ (
v, z
)
Hand the corresponding generated norm by this inner product. Suppose
that
T
is operator on Hilbert space
H
. We assume that there is an
orthogonal decomposition
H
=
H
−⊕
H
+⊕
H
0such that
T
is negative definite (resp. positive definite) in
H
−(
resp.H
+)
and
H
0=
kerT.
Let
H
+=
H
+×
H
−,
H
−=
H
−×
H
+and
H
0=
H
0×
H
0.
Then for any
z
= (
u, v
)
∈ H
we have
z
=
z
−+
z
++
z
0,
where
z
+= (
u
+, v
−)
, z
−= (
u
−, v
+) and
z
0= (
u
0, v
0)
.
Therefore,
H
+,
H
−and
H
0are orthogonal. Hence,
H
=
H
+⊕ H
−⊕ H
0,
for details see [2].
Lemma 2.1.
[2]
If
V
1∗holds. Then
H
,
→
L
p(
R
N,
R
2)
is continuous for
p
∈
[2
,
2
∗]
and
H
,
→
L
ploc(
R
N,
R
2)
is compact for
p
∈
[2
,
2
∗)
.
Therefore, the system 1.1 is the Euler-Lagrange equations of the
func-tional
define by
J
((
u, φ
)
,
(
v, ψ
)) :=
1
2
k
(
u, v
)
k
2 XE
−
1
4
Z
R3
[
|∇
φ
|
2+
|∇
ψ
|
2]
dx
+
1
2
Z
R3
[
φu
2+
ψv
2]
dx
−
Z
R3
H
(
x, u, v
)
dx.
(2.8)
The functional
J
∈
C
1(
Z
ED
×
Z
ED,
R
) and its critical points are the
solutions of system 1.1.
Remark
2.2
.
[6] the functional
J
exhibits a strong indefiniteness, that
is, it is unbounded both from below and from above on infinitely
dimen-sional subspaces. This indefiniteness can be removed using the reduction
method described in [14], by which we are led to study a one variable
functional that does not present such a strongly indefiniteness nature.
For any
u
∈
E
the Lax-Milgram theorem [15] implies there exists a
unique
φu
∈
D
1,2(
R
3) such that
−
∆
φ
u=
u
2in a distributional (or weak) sense. Then we can obtain an integral
expression for
φu
:
φ
u=
1
4
π
Z
R3
u
2(
y
)
|
x
−
y
|
dy,
(2.9)
for any
u
∈
E.
Lemma 2.3.
[11]
For any
u
∈
E
i
)
k
φu
k
D1,2≤
M
3k
u
k
2L125
,
where
M
3is positive constant which does
not depend on
u.
In particular, there exists positive constant
M
4such
that
Z
R3
φ
uu
2dx
≤
M
4k
u
k
4E;
(2.10)
ii
)
φu
≥
0
.
Now, by the lemma 2.3 we define the functional
I
:
X
E→
R
by
I
(
u, v
) :=
J
((
u, φ
u)
,
(
v, ψ
v))
.
Remark
2.4
.
Using the relation
−
∆
φ
u=
u
2and integration by parts,
we can obtain
Z
R3
|∇
φ
u|
2dx
=
Z
R3
Then, we can consider the functional 2.8 as following
I
(
u, v
) =
1
2
k
(
u, v
)
k
2 XE
+
1
4
Z
R3
[
φuu
2+
ψvv
2]
dx
−
Z
R3
H
(
x, u, v
)
dx.
(2.11)
It is well-known that
I
is
C
1-functional with derivative given by
h
I
0(
u, v
)
,
(
u, v
)
i
=
Z
R3
[
∇
u.
∇
u
+
V
(
x
)
u
2+
φ
uu
2−
H
u(
x, u, v
) ]
dx
+
Z
R3
[
∇
v.
∇
v
+
V
(
x
)
v
2+
ψ
vv
2−
H
v(
x, u, v
) ]
dx.
(2.12)
Now, using the proposition 2.3 in [6] we can consider the following
proposition for our functional
J
:
Proposition 2.5.
The following statements are equivalent :
i
) ((
u, φu
)
,
(
v, ψv
))
∈
Z
ED×
Z
EDis a critical point of
J
;
ii
) (
u, v
)
is a critical point of functional
I
and
(
φ, ψ
) = (
φ
u, ψ
v)
.
Proof.
It follows using the remark 2.2 and theorem 2.3 in [14].
3.
proof of main theorem
We take an orthogonal basis
{
(
e
i, e
j)
}
of product space
X
:=
X
E=
E
×
E
and we define
W
k:=
span
{
(
e
i, e
j)
}
i,j=1,...,k, Z
k:=
W
k⊥.
Lemma 3.1.
[11]
for any
p
∈
[2
,
2
∗)
,
β
k:=
sup
u∈Zk,kuk=1k
u
k
Lp→
0
as
k
→ ∞
.
Now, we prove that the functional
I
:
X
E→
R
satisfies the Cerami
condition .
Proposition 3.2.
under the conditions
H
1−
H
3,
the functional
I
(
u, v
)
satisfies the Cerami condition at any positive level.
Proof.
Let
{
(
u
n, v
n)
}
be a sequence in
X
Esuch that for some
c
∈
R
,
I
(
un, vn
) =
1
2
k
(
un, vn
)
k
2 XE
+
1
4
Z
R3
[
φu
nu
2 n
+
ψv
nv
2 n
]
dx
−
Z
R3
H
(
x, un, vn
)
dx
→
c
(3.1)
and
(1 +
k
(
un, vn
)
k
XE)
I
0(
un, vn
)
→
0
, as n
→ ∞
.
(3.2)
From 3.1 and 3.2 for n large enough
1 +
c
≥
I
(
un, vn
)
−
1
4
h
I
0
1
2
k
(
u
n, v
n)
k
2 XE
+
1
4
Z
R3
[
φ
unu
2
n
+
ψ
vnv
2 n]
dx
−
Z
R3
H
(
x, u
n, v
n)
dx
−
1
4
[
Z
R3
[
∇
un.
∇
un
+
V
(
x
)
u
2n+
φu
nu
2n
−
Hu
n(
x, un, vn
)
un
]
dx
+
Z
R3
[
∇
vn.
∇
vn
+
V
(
x
)
v
n2+
ψv
nv
2n
−
Hv
n(
x, un, vn
)
vn
]
dx
]
1
2
k
(
u
n, v
n)
k
2 XE
−
Z
R3
H
(
x, u
n, v
n)
dx
−
1
4
[
Z
R3
[
∇
u
n.
∇
u
n+
V
(
x
)
u
2n]
dx
+
Z
R3
[
∇
v
n.
∇
v
n+
V
(
x
)
v
n2]
dx
]
+
1
4
[
Z
R3
Hu
n(
x, un, vn
)
undx
+
Z
R3
Hv
n(
x, un, vn
)
vndx
] =
1
4
k
(
un, vn
)
k
2 XE
+
1
4
[
Z
R3
Hu
n(
x, un, vn
)
undx
+
Z
R3
Hv
n(
x, un, vn
)
vndx
]
−
Z
R3
H
(
x, un, vn
)
dx.
(3.3)
Now, we shall show that the sequence
{
(
u
n, v
n)
}
is bounded. Suppose
that
k
(
un, vn
)
k
XE→ ∞
as
n
→ ∞
.
Then we consider
(
w
n, z
n) := (
u
nk
u
nk
,
v
nk
v
nk
)
∈
X
E,
so the sequence
{
(
wn, zn
)
}
is bounded.
For some (
w, z
)
∈
XE,
and
subsequence of (
w
n, z
n)
,
we can imply that
(
wn, zn
)
*
(
w, z
)
as n
→ ∞
in XE,
and
(
w
n, z
n)
→
(
w, z
)
in L
t(
R
3)
×
L
s(
R
3)
f or t, s
∈
[2
,
2
∗)
and
w
n(
x
)
→
w
(
x
)
and z
n(
x
)
→
z
(
x
)
(3.4)
for a.e.
x
∈
R
3.
Now, we consider two cases. In first case we suppose
that (
w, z
)
6
= (0
,
0) in
X
E.
By dividing relation 3.1 with
k
(
un, vn
)
k
4XEand lemma 2.3 we get that
Z
R3
H
(
x, u
n, v
n)
k
(
un, vn
)
k
4 XEdx
=
1
2
k
(
un, vn
)
k
2 XE+
Z
R3
(
φ
unu
2n
+
ψ
vnv
2n
)
dx
−
c
4
k
(
un, vn
)
k
4XE
+
O
(
k
(
un, vn
)
k
−4XE
)
≤
M
5<
∞
,
(3.5)
where
M
5is a positive constant. we consider Ω :=
{
x
∈
R
3|
w
(
x
)
, z
(
x
)
6
=
0
}
.
By condition
H
2for all
x
∈
Ω
,
H
(
x, u
n, v
n)
k
(
un, vn
)
k
4 XE=
H
(
x, u
n, v
n)
|
(
un, vn
)
|
4(
w
4Since
|
Ω
|
>
0
,
by Fatou’s lemma,
Z
R3
H
(
x, un, vn
)
k
(
u
n, v
n)
k
4XEdx
→
+
∞
as n
→ ∞
.
This contradicts with 3.5. In second case, we assume that (
wnzn
) =
(0
,
0) then we define sequence (
t
n, s
n)
∈
R
2by
I
(
tnun, snvn
)
max
(t,s)∈[0,1]×[0,1]
I
(
tu, sv
)
.
Let
m >
0 be fixed and let
( ¯
w,
z
¯
) := (
√
4
m
un
k
un
k
E,
√
4
m
un
k
un
k
E) = 2
√
m
(
wn, zn
)
∈
XE
.
By condition
H
1we have
|
Hu
(
x, u, v
)
| ≤
M
1|
u
|
2+
M
1|
u
|
p−1and
|
Hv
(
x, u, v
)
| ≤
M
2|
v
|
2+
M
2|
u
|
p−1for any
x
∈
R
3and (
u, v
)
∈
R
2,
and
H
(
x, u, v
) =
Z
10
Hu
(
x, tu, v
)
tudt
+
Z
10
Hv
(
x, u, sv
)
svds.
Therefore,
H
(
x, u, v
)
≤
M
12
|
u
|
2
+
M
1p
|
u
|
p
+
M
22
|
v
|
2
+
M
2p
|
v
|
p
.
(3.6)
Using the relation 3.5 we get that
lim
n→∞Z
R3
H
(
x,
w
¯
n,
z
¯
n)
dx
≤
lim
n→∞[
Z
R3
M
12
|
w
¯
n|
2
+
M
1p
|
w
¯
n|
p
+
M
22
|
z
¯
n|
2
+
M
2p
|
z
¯
n|
p
]
dx
≤
lim
n→∞
M
6Z
R3
[
|
w
¯
n|
2+
|
z
¯
n|
2]
dx
+
M
7Z
R3
[
|
w
¯
n|
p+
|
z
¯
n|
p]
dx,
where
M
6:= max
{
M22,
M23}
and
M
7:= max
{
Mp2,
Mp3}
.
Hence,
lim
n→∞Z
R3
H
(
x,
w
¯
n,
z
¯
n)
dx
≤
lim
n→∞M
6Z
R3
[
|
w
¯
n|
2+
|
z
¯
n|
2]
dx
+ lim
n→∞M
7Z
R3
[ (
|
w
¯
n|
2+
|
z
¯
n|
2)
p]
dx
lim
n→∞M
6Z
R3
|
( ¯
wn,
zn
¯
)
|
2dx
+
M
7Z
R3
(
|
( ¯
wn,
zn
¯
)
|
2)
pdx
= 0
.
(3.7)
Therefore, for
n
large enough,
I
((
t
nu
n, s
nv
n))
≥
I
(( ¯
w
n,
z
¯
n)) = 2
m
+
1
4
Z
R3
[
φ
w¯nw
¯
2n
+
ψ
z¯nz
¯
2 n]
dx
−
Z
R3
By relation 3.8, lim
n→∞
I
(
tnun, snvn
) = +
∞
.
Since
I
(0
,
0) = 0 and
I
(
un, vn
)
→
c
then
t
ns
n∈
(0
,
1)
.
Hence, for
t
n, s
n∈
(0
,
1) and
n
large enough we
ob-tain
Z
R3
∇
t
nu
n.
∇
t
nu
n+
V
(
x
)
t
nu
nt
nu
n+
φ
tnunt
nu
nt
nu
n−
H
tnun(
x, t
nu
n, v
n)
t
nu
ndx
+
Z
R3
∇
snvn.
∇
snvn
+
V
(
x
)
snvnsnvn
+
φs
nvnsnvnsnvn
−
Hs
nvn(
x, un, snvn
)
snvndx
=
h
I
0((
tnun, snvn
))
,
(
tnun, snvn
)
i
= 0
.
Then by
H
3we can get that
I
((
un, vn
))
−
1
4
h
I
0
(
un, vn
)
,
(
un, vn
)
i
=
1
4
k
(
u
n, v
n)
k
2 XE
+
1
4
Z
R3
[
H
un(
x, u
n, v
n)
u
n+
H
vn(
x, u
n, v
n)]
dx
−
Z
R3
H
(
x, u
n, v
n)
dx
=
1
4
k
(
u
n, v
n)
k
2 XE
+
1
4
Z
R3
ˆ
H
(
x, u
n, v
n)
dx
≥
1
4
θ
k
(
u
n, v
n)
k
2 XE+
1
4
θ
Z
R3
[
H
tnun(
x, t
nu
n, v
n)
t
nu
n+
H
snvn(
x, u
n, s
nv
n)
s
nv
n]
dx
−
Z
R3
H
(
x, t
nu
n, s
nv
n)
dx
1
θ
I
((
tnun, snvn
))
−
1
4
θ
h
I
0
((
tnun, snvn
))
,
(
tnun, snvn
)
i →
+
∞
.
This is contradiction with 3.3. Hence, the sequence
{
(
u
n, v
n)
}
is bounded
in
XE.
Since (
un, vn
)
,
→
(
u, v
) in
XE
,
so by lemma 2.1, (
un, vn
)
→
(
u, v
)
in for any
s, t
∈
[2
,
2
∗)
.
By 2.12 we can get that
k
(
u
n, v
n)
−
(
u, v
)
k
X2E=
k
(
u
n−
u, v
n−
v
)
k
2XE
= ((
u
n−
u, v
n−
v
)
,
(
u
n−
u, v
n−
v
))
XE=
Z
R3
∇
(
u
n−
u
)
.
∇
(
u
n−
u
)
V
(
x
)(
u
n−
u
)
2dx
+
Z
R3
∇
(
v
n−
v
)
.
∇
(
v
n−
v
)
V
(
x
)(
v
n−
v
)
2dx
=
h
I
0(
u
n, v
n)
−
I
0
(
u, v
)
,
(
u
n, v
n)
−
(
u, v
)
i −
Z
R3
[ (
φ
unu
n−
φ
uu
)(
u
n−
u
) ]
dx
+
Z
R3
[ (
ψ
vnv
n−
ψ
vv
)(
v
n−
v
) ]
dx
+
Z
R3
[ (
H
un(
x, u
n, v
n)
−
H
u(
x, u, v
))(
u
n−
u
) ]
dx
+
Z
R3
Since (
un, vn
)
,
→
(
u, v
) in
X
E,
and (
un, vn
)
→
(
u, v
) in for any
s, t
∈
[2
,
2
∗)
,
we can imply
h
I
0(
u
n, v
n)
−
I
0
(
u, v
)
,
(
u
n, v
n)
−
(
u, v
)
i →
0
,
as
n
→ ∞
.
On the other hand,
|
Z
R3
[ (
φu
nun
−
φuu
)(
un
−
u
) ]
dx
+
Z
R3
[ (
ψv
nvn
−
ψv
v
)(
vn
−
v
) ]
dx
| ≤
|
Z
R3
[ (
φu
nun
−
φuu
)(
un
−
u
) ]
dx
|
+
|
Z
R3
[ (
ψv
nvn
−
ψvv
)(
vn
−
v
) ]
dx
|
.
By H¨
o
der inequality and 2.3,
|
φ
unu
n(
u
n−
u
)
dx
| ≤ k
φ
unu
nk
L2k
u
n−
u
k
L2≤ k
φ
unk
L6k
u
nk
L3k
u
n−
u
k
L2M
8k
φu
nk
D1,2k
un
k
L3k
un
−
u
k
L2≤
M
3M
8k
un
k
2L125
k
un
k
L3
k
un
−
u
k
L2,
(3.9)
where
M
8is positive constant. We can similarity conclude the following
inequalities :
|
φuu
(
un
−
u
)
dx
| ≤ k
φuu
k
L2k
un
−
u
k
L2≤ k
φu
k
L6k
un
k
L3k
un
−
u
k
L2M
9k
φ
uk
D1,2k
u
nk
L3k
u
n−
u
k
L2≤
M
3M
9k
u
k
2 L125k
u
k
L3k
u
n−
u
k
L2,
(3.10)
and
|
ψ
vnv
n(
v
n−
v
)
dx
| ≤ k
ψ
vnv
nk
L2k
v
n−
v
k
L2≤ k
ψ
vnk
L6k
v
nk
L3k
v
n−
v
k
L2M
10k
ψv
nk
D1,2k
vn
k
L3k
vn
−
v
k
L2≤
M
3M
10k
vn
k
2L125k
vn
k
L3k
vn
−
v
k
L2,
(3.11)
and
|
ψ
vv
(
v
n−
v
)
dx
| ≤ k
ψ
vv
k
L2k
v
n−
v
k
L2≤ k
ψ
vk
L6k
v
nk
L3k
v
n−
v
k
L2M
11k
ψ
vk
D1,2k
v
nk
L3k
v
n−
v
k
L2≤
M
3M
11k
v
k
2L125
k
v
k
L3
k
v
n−
v
k
L2,
(3.12)
where
M
9, M
10and
M
11are positive constants.
Therefore, by relations 3.9, 3.10, 3.11 and 3.12 we obtain that
|
Z
R3
[ (
φ
unu
n−
φ
uu
)(
u
n−
u
) ]
dx
+
Z
R3
[ (
ψ
vnv
n−
ψ
vv
)(
v
n−
v
) ]
dx
| ≤
[
M
3M
8k
un
k
2L125
k
un
k
L3
+
M
3M
9k
u
k
2L125
k
u
k
L3
]
k
un
−
u
k
L2+[
M
3M
10k
v
nk
2L125k
v
nk
L3+
M
3M
11k
v
k
2
L125
k
v
k
L3
]
k
v
n−
v
k
L2.
Then
Z
R3
[ (
φ
unu
n−
φ
uu
)(
u
n−
u
) ]
dx
+
Z
R3
Now, by condition
H
1and H¨
o
der inequality we obtain that
|
Z
R3
H
un(
x, u
n, v
n)(
u
n−
u
)
dx
| ≤
Z
R3
|
H
un(
x, u
n, v
n)
| |
u
n−
u
|
dx
≤
Z
R3
[
M
12
|
u
n|
+
M
1p
|
u
n|
p−1
]
|
u
n
−
u
|
dx
≤
[
M
12
k
un
k
2 L2
+
M
1p
k
un
k
p−1
LP
]
k
un
−
u
k
LP.
(3.13)
Similarly, we can get the following inequalities by condition
H
1and
H¨
o
der inequality:
|
Z
R3
Hu
(
x, u, v
)(
un
−
u
)
dx
| ≤
Z
R3
|
Hu
(
x, u, v
)
| |
un
−
u
|
dx
≤
Z
R3
[
M
12
|
u
|
+
M
1p
|
u
|
p−1]
|
un
−
u
|
dx
≤
[
M
12
k
u
k
2 L2
+
M
1p
k
u
k
p−1
LP
]
k
u
n−
u
k
LP(3.14)
and
|
Z
R3
Hv
n(
x, un, vn
)(
vn
−
v
)
dx
| ≤
Z
R3
|
Hv
n(
x, un, vn
)
| |
vn
−
v
|
dx
≤
Z
R3
[
M
22
|
vn
|
+
M
2p
|
vn
|
p−1]
|
vn
−
v
|
dx
≤
[
M
22
k
v
nk
2 L2
+
M
2p
k
v
nk
p−1
LP
]
k
v
n−
v
k
LP(3.15)
and
|
Z
R3
H
v(
x, u, v
)(
v
n−
v
)
dx
| ≤
Z
R3
|
H
v(
x, u, v
)
| |
v
n−
v
|
dx
≤
Z
R3
[
M
22
|
v
|
+
M
2p
|
v
|
p−1]
|
vn
−
v
|
dx
≤
[
M
22
k
v
k
2 L2
+
M
2p
k
v
k
p−1
LP
]
k
v
n−
v
k
LP.
(3.16)
Hence, by relations 3.13, 3.14, 3.15 and 3.16 we obtain
Z
R3
[ (
H
un(
x, u
n, v
n)
−
H
u(
x, u, v
))(
u
n−
u
) ]
dx
+
Z
R3
≤
2[
M
12
k
u
nk
2 L2
+
M
1p
k
u
nk
p−1
LP
]
k
u
n−
u
k
LP+2[
M
22
k
v
nk
2 L2
+
M
2p
k
v
nk
p−1
LP
]
k
v
n−
v
k
LP.
Since, (
u
n, v
n)
→
(
u, v
) in
L
t(
R
3)
×
L
s(
R
3) for any
t, s
∈
[2
,
2
∗) then
Z
R3
[ (
Hu
n(
x, un, vn
)
−
Hu
(
x, u, v
))(
un
−
u
) ]
dx
+
Z
R3
[ (
Hv
n(
x, un, vn
)
−
Hv
(
x, u, v
)(
vn
−
v
)) ]
dx
→
0
,
as
n
→ ∞
.
proof of theorem 1.3
By proposition 3.2 the functional
I
(
u, v
) satisfies
in Cerami condition. Now, we show that
I
(
u, v
) satisfies in condition
I
1and
I
2in theorem 1.2. From
H
2,
lim
|(u,v)|→∞
H
(
x, u, v
)
|
(
u, v
)
|
4= +
∞
,
for any
M >
0 there exists
N >
0 such that
∀
x
∈
R
3,
|
(
u, v
)
| ≥
N,
H
(
x, u, v
)
≥
1
4
M
|
(
u, v
)
|
4
,
for any
x
∈
R
3and (
u, v
)
∈
R
2.
Therefore,
I
(
u, v
) =
1
2
k
(
u, v
)
k
2 XE
+
1
4
Z
R3
[
φuu
2+
ψvv
2]
dx
−
Z
R3
H
(
x, u, v
)
dx
≤
1
2
k
(
u, v
)
k
2 XE
+ [
M
44
(
k
u
k
4
E
+
k
v
k
4E) ]
−
M
4
k
(
u, v
)
k
4
XE
+ ˜
M
k
(
u, v
)
k
2 XE,
where ˜
M
:=
sup
|(u,v)|<N
(
M4|
(
u, v
)
|
4−
H(x,u,v)|(u,v)|2
)
.
Then
H
(
x, u, v
)
≥
M
4
k
(
u, v
)
k
4
XE
−
M
˜
k
(
u, v
)
k
2 XE.
Hence,
I
(
u, v
)
≤
1
2
k
(
u, v
)
k
2 XE
+
M
44
(
k
u
k
4
E
+
k
v
k
4E)
−
M
12M4
[
k
u
k
4
E
+
k
v
k
4E]
+
M
12M
˜
(
k
u
k
2E+
k
v
k
2E)
,
where
M
12is positive constant. Since,
M44−
M124M<
0
,
so for
M
large
enough, it follows that
a
k:=
max
(u,v)∈Wk,k(u,v)kXE=ρk
I
(
u, v
)
≤
0
,
for some positive constant large enough. Now, using the lemma 2.3 and
2.1 we show that
I
(
u, v
) satisfies in condition
I
2.
I
(
u, v
) =
1
2
k
(
u, v
)
k
2 XE
+
1
4
Z
R3
[
φ
uu
2+
ψ
vv
2]
dx
−
Z
R3
≥
1
2
k
(
u, v
)
k
2 XE
−
(
M
12
k
u
k
2 L2
+
M
1p
k
u
k
p Lp
+
M
22
k
v
k
2 L2
+
M
2p
k
v
k
p Lp
)
≥
1
2
[
k
u
k
2
E
+
k
v
k
2E]
−
M
1C
emb2
k
u
k
2 E
−
β
kpM
1C
embp
k
u
k
p E
−
M
2C
emb2
k
v
k
2 E
−
M
2C
embα
pkp
k
v
k
p E
= (
1
2
−
M
1C
emb2
)
k
u
k
2 E
+ (
1
2
−
M
2C
emb2
)
k
v
k
2 E
−
[
M
1C
embβ
kpp
k
u
k
p E
+
M
2C
embα
pkp
k
v
k
p E
]
.
We choose
r
k:= (
M1βCpemb k+
M2Cembαpk
)
1p−2
.
Hence,
b
k:=
inf
(u,v)∈Zk,k(u,v)kXE=rk
I
(
u, v
)
≥
inf
(u,v)∈Zk,k(u,v)kXE=rk
[ (
1
2
−
M
1C
emb2
)
k
u
k
2 E
−
M
1Cembβ
kpp
k
u
k
p E
]
+(
1
2
−
M
2C
emb2
)
k
v
k
2 E
−
M
2Cembα
pkp
k
v
k
p E
}
≥
inf
(u,v)∈Zk,k(u,v)kXE=rk
M
13(
k
u
k
2E+
k
v
k
2E)
−
M
14(
k
u
k
pE+
k
v
k
pE)
,
where
M
13:=
min
{
1−M12Cemb,
1−M22Cemb}
and
M
14:=
min
{
1−M1βpkCemb
p
,
1−M2αpkCemb
p
}
are positive constants. Therefore,
I
(
u, v
)
≥
inf
(u,v)∈Zk,k(u,v)kXE=rk
M
13k
(
u, v
)
k
2XE−
M
14k
(
u, v
)
k
p
XE
≥
M
15r
2 k(1
−
r
p−2 k
)
,
where
M
15:=
min
{
M
13, M
14}
is positive constant. Now, if
k
→ ∞
,
then by lemma 3.1 we have
rk
→
+
∞
.
Therefore, by theorem 1.2 the
system 1.1 has infinitely many solutions.
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