Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 223, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
GROUND STATE SOLUTIONS FOR NON-LOCAL FRACTIONAL
SCHRÖDINGER EQUATIONS
YANG PU, JIU LIU, CHUN-LEI TANG
Abstract. In this article, we study a time-independent fractional Schrödinger
equation with non-local (regional) diffusion
(−∆)α
ρ u + V (x)u = f (x, u)
in RN ,
where α ∈ (0, 1), N > 2α. We establish the existence of a non-negative ground
state solution by variational methods.
1. Introduction and statement of main results
Consider the fractional Schrödinger equation
(−∆)α
ρ u + V (x)u = f (x, u)
in RN ,
u ∈ H α (RN ),
(1.1)
where α ∈ (0, 1), N > 2α, (−∆)α
ρ denotes a non-local (regional) fractional Laplacian
operator with a range of scope determined by the positive function ρ ∈ C(RN , R+ ).
The fractional Schrödinger equation was firstly introduced by Laskin [1]. This
equation was of particular interest in fractional quantum mechanics in the study
of particles on stochastic fields modelled by Lévy processes that give rise to equations with the fractional Laplacian operator. Recently, the study on problems of
fractional Schrödinger equations has attracted much attention. Some existence and
nonexistence of Dirichlet problem involving the fractional Laplacian on bounded
domains have been established, see [2, 3] and their references. Using the equivalent
definition of the fractional operator, some authors introduced a variational principle
and studied the existence and multiplicity of solutions in RN . Cheng [4] considered
the equation
(−∆)α u + V (x)u = |u|p−1 u,
u ∈ H α (RN )
with unbounded potential V , he obtained the existence of ground state solution by
a Lagrange multiplier method and the Nehari manifold method. Dipierro, Palatucci
and Valdinoci in [5] proved existence and symmetry results for the solutions. In [6],
2010 Mathematics Subject Classification. 35J60, 35J20.
Key words and phrases. Nonlinear Schrödinger equation; fractional Laplacian;
non-local operator; ground state solution.
c
2015
Texas State University - San Marcos.
Submitted June 15, 2015. Published August 28, 2015.
1
2
Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
the authors studied the same equation with a more general right-hand side f (x, u).
Secchi [7] provided a generalization of the main result to equations of the form
(−∆)α u + V (x)u = f (x, u),
u ∈ H α (RN ).
(1.2)
The existence of positive solutions was obtained using the Nehari manifold method.
Chang [8] proved the existence of a positive ground state solution of (1.2) when
f (x, t) is asymptotically linear with respect to t at infinity.
There is another definition of regional fractional Laplacian in [10] and [11], the
α/2
authors introduced the regional fractional Laplacian operator ∆G in an arbitrary
open set G of RN , which is not the one used here. More recently, Felmer and Torres
[12] studied the regional fractional Laplacian equation
ε2α (−∆)α
ρ u + u = f (u),
u ∈ H α (RN ),
where α ∈ (0, 1), N > 2α. The operator (−∆)α
ρ was defined by
Z
Z Z
[u(x + z) − u(x)][v(x + z) − v(x)]
dz dx
(−∆)α
ρ u(x)v(x)dx =
|z|N +2α
RN
RN B(0,ρ(x))
for all u, v ∈ H α (RN ). They showed the existence of a ground state and analyzed
the behavior of the semi-classical solutions as ε → 0. Following some ideas in [12],
Felmer and Torres [13] proved that the ground state level is achieved by a radially
symmetry solution.
Up to now, no results for non-autonomous regional fractional Laplacian equations
with potential have appeared in the literature. In this note, we investigate the
existence of a non-negative ground state solution for such equations.
Throughout this article, we assume the following conditions:
(A1) ρ ∈ C(RN , R+ ), there exists a constant ρ0 > 0 such that ρ(x) ≥ ρ0 .
(A2) V ∈ C(RN , R+ ), inf x∈RN V (x) ≥ c > 0, there exists r0 > 0 such that, for
any M > 0,
lim meas {x ∈ RN : |x − y| ≤ r0 , V (x) ≤ M } = 0.
|y|→∞
(A3) f ∈ C(RN × R+ , R) and limt→0+ f (x,t)
= 0 uniformly in x ∈ RN .
t
f (x,t)
(A4) limt→+∞ t2∗α −1 = 0 uniformly in x ∈ RN , where 2∗α = N2N
−2α is the fractional
critical exponent.
(A5) limt→+∞ F (x,t)
= +∞ uniformly in x ∈ RN .
t2
The main results of this article are as follows:
Theorem 1.1. Assume (A1)–(A5) and
(A6) there exists T1 > 0 such that µF (x, t) ≤ f (x, t)t for t > T1 , where µ > 2 is
a constant.
Then (1.1) has a non-negative ground state solution.
Theorem 1.2. Assume (A1)–(A5) and
(A7) f (x,t)
is increasing on (0, ∞).
t
Then (1.1) has a non-negative ground state solution.
Theorem 1.3. Assume that (A1)–(A5) and
(A8) There exist b > 0 and ν > 2 such that lim supt→+∞ F (x,t)
≤ b;
tν
f (x,t)t−2F (x,t)
N
(A9) lim inf t→+∞
≥ η > 0, where σ > 2α (ν − 2).
tσ
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GROUND STATE SOLUTIONS
3
Then (1.1) has a non-negative ground state solution.
Remark 1.4. Note that we impose the subcritical growth condition by a general
condition (A4). Also note that (A6), (A7) and (A9) almost cover all types of
superquadratic conditions.
When ρ ≡ +∞, the regional fractional operator becomes a common fractional
operator, our results are also new under the assumptions on f . Moreover, we can
find a positive ground state solution by the strong maximum principle [14].
2. Preliminaries
Firstly we give some basic notation. The fractional Sobolev space of order α on
RN is defined by
Z Z
o
n
|u(x) − u(y)|2
dx dy < ∞ ,
H α (RN ) = u ∈ L2 (RN ) :
N
+2α
RN RN |x − y|
endowed with the norm
Z
kukH α (RN ) =
Z
RN
RN
|u(x) − u(y)|2
dx dy +
|x − y|N +2α
Z
1/2
|u|2 dx
.
RN
In this article, we consider the space
Z Z
n
|u(x + z) − u(x)|2
2
N
E := u ∈ L (R ) :
dz dx
|z|N +2α
RN B(0,ρ(x))
Z
o
+
V (x)|u(x)|2 dx < ∞ ,
RN
equipped with the inner product
Z Z
[u(x + z) − u(x)][v(x + z) − v(x)]
hu, vi =
dz dx
|z|N +2α
RN B(0,ρ(x))
Z
+
V (x)u(x)v(x)dx
RN
and the norm
Z
kuk =
RN
Z
B(0,ρ(x))
|u(x + z) − u(x)|2
dz dx +
|z|N +2α
Z
V (x)|u(x)|2 dx
1/2
.
RN
We denote by k · kp the usual Lp -norm. We define
u± (x) = max{±u(x), 0}.
Obviously, u ∈ E implies that u+ , u− ∈ E.
Let I : E → R be the functional defined by
Z
1
I(u) = kuk2 −
F (x, u+ )dx
2
N
R
(2.1)
with F (x, t) being the primitive of f (x, t).
Next, we give some lemmas that play important roles in proving our main results.
Lemma 2.1. Suppose (A1) and (A2) hold. Then there exists a constant C0 > 0
such that
kuk22∗α ≤ C0 kuk2 .
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Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
Proof. From [9, Theorem 6.5], there exists C > 0 such that
Z Z
|u(x + z) − u(x)|2
2
dz dx.
kuk2∗α ≤ C
|z|N +2α
RN RN
As in the proof of [12, Proposition 2.1], we have
Z Z
|u(x + z) − u(x)|2
C
dz dx
|z|N +2α
RN RN
Z Z
|u(x + z) − u(x)|2
=C
dz dx
|z|N +2α
RN B(0,ρ0 )
Z Z
|u(x + z) − u(x)|2
+
dz
dx
|z|N +2α
RN B c (0,ρ0 )
Z Z
|u(x + z) − u(x)|2
2C|S n−1 |
≤C
dz dx +
kuk22
2α
N
+2α
|z|
αρ
N
R
B(0,ρ0 )
0
Z Z
Z
2C|S n−1 |
|u(x + z) − u(x)|2
dz dx +
≤C
V (x)|u(x)|2 dx
|z|N +2α
cαρ2α
RN B(0,ρ0 )
RN
0
≤ C0 kuk2 ,
n−1
|
where C0 = max{C, 2C|S
}. This completes the proof.
cαρ2α
0
Lemma 2.2. Suppose (A1)–(A2) hold, and that there exists a constant K > 0 such
that
kukq ≤ Kkuk,
where 2 ≤ q ≤ 2∗α . Then the embedding E ,→ Lq (RN ) is continuous for 2 ≤ q ≤ 2∗α
and E ,→ Lsloc (RN ) is compact for 2 ≤ s < 2∗α .
Proof. When q = 2, by the definition of k · k, there exists C1 > 0 such that
kuk2 ≤ C1 kuk.
When q = 2∗α , using Lemma 2.1, we obtain
1/2
kuk2∗α ≤ C0 kuk.
When 2 < q < 2∗α , according to the Hölder inequality and Lemma 2.1, we obtain
Z
2q−2
Z
22∗α∗ −q
∗ −2
−2
α
q
2∗
2
kukq ≤
|u| α dx
|u| dx α
RN
(q−2)2∗
α
2(2∗
α −2)
≤ C0
RN
kuk
(q−2)2∗
α
2∗
α −2
2(2∗
α −q)
2∗
α −2
C1
kuk
2(2∗
α −q)
2∗
α −2
≤ C2q kukq ,
where
∗
2(2∗
(q−2)2
α
α −q) 1/q
2(2∗ −2)
2∗ −2
C2 = C0 α C1 α
.
1/2
Let K = max{C0 , C1 , C2 }, we have
kukq ≤ Kkuk
for 2 ≤
2∗α and
q ≤ 2∗α . Therefore, the embedding E ,→ Lq (RN )
E ,→ Lsloc (RN ) is compact for 2 ≤ s < 2∗α .
is continuous for 2 ≤ q ≤
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GROUND STATE SOLUTIONS
5
Lemma 2.3. Suppose (A1)–(A2) hold. Then E ,→ Lq (RN ) is compact for any
2 ≤ q < 2∗α .
Proof. First, we show that E ,→ L2 (RN ) is compact. Let {un } ⊂ E with {un }
bounded in E. Then un * u weakly in E, by Lemma 2.2, which implies that
un → u strongly in L2loc (RN ).
Set δn = kun k2 , we may suppose that there is a subsequence of {un } and δ ∈ R
such that δn → δ. For any bounded domain Ω in RN , we have
Z
Z
Z
2
2
|u| dx = lim
|un | dx ≤ lim
|un |2 dx = δ 2 ,
n→∞
Ω
n→∞
Ω
RN
so kuk2 ≤ δ .
c
c
Let Ω1 := {x ∈ BR
|V (x) ≥ M }, Ω2 := {x ∈ BR
|V (x) < M }. On the basis of
Ω1 , we obtain
Z
Z
kun k2
V (x)
2
|un |2 dx ≤
.
|un | dx ≤
M
Ω1
RN M
S
∞
Let {yi } be a sequence satisfying RN ⊂ i=1 B(yi , r0 ) and each point x is contained
in at most 2N such balls B(yi , r0 ). By the Hölder inequality, we choose a positive
2∗
constant s ∈ (1, 2α ) such that
Z
∞ Z
X
|un |2 dx ≤
|un |2 dx
Ω2
i=1
≤
Ω2 ∩B(yi ,r0 )
∞ Z
X
1/s Z
|un | dx
2s
Ω2 ∩B(yi ,r0 )
i=1
1− 1s
.
1dx
Ω2 ∩B(yi ,r0 )
Define
kun k2B(yi ,r0 ) =
Z
Z
B(yi ,r0 )
B(0,ρ(x))∩B(yi ,r0 )
Z
|u(x + z) − u(x)|2
dz dx
|z|N +2α
V (x)|u(x)|2 dx.
+
B(yi ,r0 )
Using proof similar to the one of Lemmas 2.1 and 2.2, there exists C > 0 such that
Z
1s
Z
1s
2s
|un | dx
≤
|un |2s dx
≤ Ckun k2B(yi ,r0 ) .
Ω2 ∩B(yi ,r0 )
B(yi ,r0 )
Now, we can estimate
Z
∞
X
1
|un |2 dx ≤
Ckun k2B(yi ,r0 ) meas(Ω2 ∩ B(yi , r0 ))1− s ≤ 2N CεR kun k2 ,
Ω2
i=1
where εR = supyi meas (Ω2 ∩ B(yi , r0 ))
1− 1s
. Note that {un } is bounded in E, for
kun k2
M
M large enough, we have
→ 0. Since (A2) holds, for R large enough, we
obtain εR → 0. It is easy to check that, given any ε > 0, for sufficiently large R
and M ,
Z
Z
Z
2
2
|un | dx =
|un | dx +
|un |2 dx ≤ ε
c
BR
Ω1
Ω2
and
kuk22 =
Z
BR
|u|2 dx +
Z
c
BR
|u|2 dx
6
Y. PU, J. LIU, C.-L. TANG
Z
|un |2 dx
≥ lim
n→∞
= lim
EJDE-2015/223
BR
Z
n→∞
2
Z
|un |2 dx
|un | dx −
c
BR
RN
≥δ 2 − ε.
It means that δ ≤ kuk2 . Therefore, δ = kuk2 and it implies that E ,→ L2 (RN ) is
compact.
Next, we prove that E ,→ Lq (RN ) is compact for 2 < q < 2∗α . Let r ∈ (0, 1) be
such that q = 2r + 2∗α (1 − r). Using the Hölder inequality again, we have
2∗ (1−r)
α
kun − ukqq ≤ kun − uk2r
2 kun − uk2∗
α
.
Since {un − u} is bounded in E, then un − u * 0, by Lemma 2.2, there exists C > 0
2∗ (1−r)
such that kun − uk2α∗α
≤ C. Thus,
kun − ukqq → 0.
Here, we can make a result of the proof of Lemma 2.3.
Lemma 2.4. Under the assumptions of Theorem 1.1 hold, the functional I satisfies
the (Ce)c condition for c > 0.
Proof. Assume that {un } ⊂ E is a (Ce)c sequence for c > 0,
(1 + kun k)I 0 (un ) → 0
I(un ) → c,
asn → ∞.
For any ϕ ∈ E, we obtain that
Z Z
|un (x + z) − un (x)||ϕ(x + z) − ϕ(x)|
dz dx
hI 0 (un ), ϕi =
|z|N +2α
RN B(0,ρ(x))
Z
Z
+
V (x)|un (x)||ϕ(x)|dx −
f (x, u+
n )ϕ(x)dx → 0
RN
(2.2)
RN
and
Z
1
+
kun k2 −
f (x, u+
(2.3)
n )un dx → 0.
2
RN
First we claim that {un } is bounded in E. In fact, if not, we may assume by the
contradiction that there exists a subsequence of {un } (still denoted by {un }) with
kun k → +∞, and we set
un
.
wn =
kun k
Clearly, {wn } is bounded in E. Going if necessary to a subsequence of {wn }, we
can assume that
hI 0 (un ), un i =
wn * w
wn → w
weakly in E,
strongly in Lq (RN )
wn → w
(2 ≤ q < 2∗α ),
a.e. x ∈ RN .
Similarly, we denote
wn+ =
u+
n
,
kun k
then
wn+ * w+
weakly in E,
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GROUND STATE SOLUTIONS
wn+ → w+
strongly in Lq (RN )
wn+ → w+
7
(2 ≤ q < 2∗α ),
a.e. x ∈ RN .
Next we claim that w 6= 0. Otherwise, if w ≡ 0, we know that wn+ → 0 strongly in
L2 (RN ). By (f1 ), for each ε > 0, there exists T0 > 0 such that
|f (x, t)t| < εt2 ,
for 0 < t < T0 . Through the continuity of f , there exists C > 0 such that
C
|f (x, t)t| ≤ C ≤ 2 t2
T0
for T0 ≤ t ≤ T1 . Hence, we have
|f (x, t)t| ≤
C 2
t
T02
(2.4)
for 0 < t ≤ T1 and
C 2
t
2T02
for 0 < t ≤ T1 . By (A6), it is easy to see that
|F (x, t)| ≤
(2.5)
1
f (x, t)t − F (x, t) ≥ 0
µ
for t > T1 . Combining (2.4) and (2.5), we have
1
1
1 C
f (x, t)t − F (x, t) ≥ −( − ) 2 t2
µ
2 µ T0
(2.6)
for all (x, t) ∈ RN × R+ . Under the definition of {un }, we see that
1 0
hI (un ), un i
µ
Z 1
1
1
+
=
−
kun k2 −
F (x, u+
f (x, u+
n) −
n )un dx
2 µ
µ
RN
= c − on (1).
I(un ) −
It follows from the preceding step and (2.6) that
0 = lim
n→∞
I(un ) − µ1 hI 0 (un ), un i
kun k2
Z F (x, u+ ) − 1 f (x, u+ )u+
n
n
n
µ
1
1
− − lim
2 µ n→∞ RN
kun k2
Z
2
1
1 C
1
1
(u+
n)
−
lim
≥ − −
dx
2 µ
2 µ T02 n→∞ RN kun k2
1
1
1
1 C
≥ − −
−
lim kw+ k2
2 µ
2 µ T02 n→∞ n 2
1
1
≥ − .
2 µ
Note that µ > 2, this is a contradiction, so w 6= 0.
Since {un } is (Ce)c sequence, by (2.2), we know that
=
dx
0
−
0
hI 0 (un ), u−
n i ≤ kI (un )kkun k ≤ kI (un )kkun k → 0.
8
Y. PU, J. LIU, C.-L. TANG
Then wn− =
u−
n
kun k
EJDE-2015/223
→ 0, which implies that w− = 0 for a.e. x ∈ RN . We set
t
h(s) = sµ F (x, ),
s
then
t
t
t t
tt
µ−1
µF
x,
− sµ f x,
=
s
−
f
x,
.
s
s s2
s
s s
By (A5), we can find a T2 > 0 such that
h0 (s) = µsµ−1 F x,
inf
t≥T2 , x∈RN
F (x, t) > 0.
Denote T = max{T1 , T2 }. For s ∈ [1, Tt ], by (f4 ), we have
t
t t
µF (x, ) − f (x, ) ≤ 0.
s
s s
Hence, h0 (s) ≤ 0 and h(1) ≥ h( Tt ). This implies
F (x, t) ≥
1
F (x, T )tµ
T1µ
(2.7)
inf
(2.8)
for all (x, t) ∈ (RN , R+ ), where
c0 =
1
Tµ
t=T,x∈RN
F (x, t).
In accordance with (2.7) that
F (x, t) ≥ c0 tµ
for all (x, t) ∈ (RN , R+ ). Then, one has
f (x, t)t ≥ µc0 tµ
for all (x, t) ∈ RN × R+ . Related to (2.3), we obtain
Z
+
1
f (x, u+
hI 0 (un ), un i
n )un
=
−
dx = on (1).
kun kµ
kun kµ−2
kun kµ
RN
Therefore, we know
Z
+
f (x, u+
n )un
0 = lim
dx
n→∞ RN
kun kµ
≥ lim µc0 kwn+ kµµ
n→∞
≥ µc0 kw+ kµµ > 0.
However, this is a contradiction. So our assumption kun k → +∞ is false, and that
is to say, {un } is bounded in E.
Since {un } is bounded in E, there exists C1 > 0 such that
kun k ≤ C1 .
Moreover, there exists C2 > 0 such that
kun − uk2∗α ≤ C2 .
By the reflexivity of E, there exists a subsequence of {un } (which we also denote
by {un }) and u ∈ E such that
un * u weakly in E,
(2.9)
un → u strongly in Lq (RN ) (2 ≤ q < 2∗α ).
EJDE-2015/223
GROUND STATE SOLUTIONS
9
At the same time,
hI 0 (un ), un − ui → 0,
hI 0 (u), un − ui → 0.
Hence, we have
hI 0 (un ) − I 0 (u), un − ui
Z
+
+
+
= kun − uk2 −
f (x, u+
n ) − f (x, u ) (un − u )dx → 0.
(2.10)
Ω
By (A3), (A4) and f ∈ C(RN × R+ , R), for ε > 0, there exists Cε > 0 such that
∗
f (x, t) ≤ Cε t + εt2α −1
N
(2.11)
+
for all (x, t) ∈ R × R . By (2.11) and the Hölder inequality, we have
Z
Z
+
+
+
|f (x, u+
f (x, un )(un − u )dx ≤
n )||un − u|dx
N
RN
R
Z + 2∗
α −1 |u
Cε |u+
||u
−
u|
+
ε|u
|
−
u|
dx
≤
n
n
n
n
RN
2∗ −1
≤ Cε kun k2 kun − uk2 + εkun k2α∗α
≤ Cε KC1 kun − uk2 + εK
Related to (2.9), we obtain
Z
RN
2∗
α −1
kun − uk2∗α
2∗
α −1
C1
C2 .
+
+
f (x, u+
n )(un − u )dx → 0.
Consequently,
Z
RN
+
+
+
f (x, u+
n ) − f (x, u ) (un − u )dx → 0.
By (2.10), we have kun − uk2 → 0. So we derive that un → u strongly in E, we
conclude that the (Ce)c condition is satisfied.
Lemma 2.5. Under the assumptions of Theorem 1.2, the functional I satisfies the
(Ce)c condition for c > 0.
Proof. Assume that {un } ⊂ E is a (Ce)c sequence for c > 0,
I(un ) → c,
(1 + kun k)I 0 (un ) → 0
(n → ∞).
We first claim that the sequence {un } is bounded in E. Otherwise, there is a
subsequence, again denoted by {un }, such that kun k → +∞ as n → ∞. Set
wn =
un
,
kun k
wn+ =
u+
n
.
kun k
Clearly, {wn } is bounded in E. Going if necessary to a subsequence of {wn }, we
can assume that
wn * w
wn → w
weakly in E,
strongly in Lq (RN )
wn → w
(2 ≤ q < 2∗α ),
a.e. x ∈ RN .
Then we claim that w 6= 0. Otherwise, if w ≡ 0, we obtain wn → 0 strongly in
Lq (RN ). Since I(tun )(t ∈ [0, 1]) is continuous, there exist tn ∈ [0, 1] such that
I(tn un ) = max I(tun ).
t∈[0,1]
10
Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
For tn ∈ (0, 1), we have
0
2
Z
hI (tn un ), tn un i = ktn un k −
RN
+
f (x, tn u+
n )tn un dx = tn
d
|t=tn I(tun ) = 0.
dt
It is obvious for tn = 0, we obtain
hI 0 (tn un ), tn un i = 0.
When tn = 1, we obtain
hI 0 (tn un ), tn un i = hI 0 (un ), un i → 0.
Hence, we have
hI 0 (tn un ), tn un i = ktn un k2 −
Z
RN
+
f (x, tn u+
n )tn un dx → 0
(2.12)
for tn ∈ [0, 1]. Set
vn = 2m1/2 wn
for each m > 0. By (2.11), we have
F (x, t) ≤
ε ∗
Cε 2
t + ∗ t2α
2
2α
(2.13)
for all (x, t) ∈ RN × R+ . Since vn → 0 strongly in L2 (RN ) and vn is bounded in
E, we come to a conclusion that
Z
F (x, vn+ )dx → 0.
RN
2m1/2
kun k
∈ (0, 1) for n large enough, so
Z
I(tn un ) ≥ I(vn ) = 2m −
F (x, vn+ ) ≥ m.
It is apparently showed that
RN
Then we have I(tn un ) → +∞. Denote H(x, t) = f (x, t)t − 2F (x, t), it follows
Z
0
H(x, u+
(2.14)
n )dx = 2I(un ) − hI (un ), un i → 2c.
RN
Assume 0 ≤ s1 ≤ s2 , by (f5 ), we have
H(x, s2 ) − H(x, s1 )
h1
i
= 2 (f (x, s2 )s2 − f (x, s1 )s1 ) − (F (x, s2 ) − F (x, s1 ))
Z s1 Z2 s2 f (x, s2 ) f (x, v) f (x, s2 ) f (x, s1 ) =2
−
vdv + 2
−
vdv
s2
v
s2
s1
s1
T2
f (x, s ) f (x, s ) 1
2
+ T22
−
≥ 0.
s2
s1
Thus, we have
Z
Z
H(x, u+
)dx
≥
H(x, tn u+
n
n )dx
RN
RN
= 2I(tn un ) − hI 0 (tn un ), tn un i → +∞.
This contradicts (2.14), so w 6= 0. Since {un } is (Ce)c sequence, we know that
0
−
0
hI 0 (un ), u−
n i ≤ kI (un )kkun k ≤ kI (un )kkun k → 0.
EJDE-2015/223
Thus, wn− =
GROUND STATE SOLUTIONS
u−
n
kun k
11
→ 0, which implies that w− = 0 for a.e. x ∈ RN . Then we set
Ω0 = {x ∈ RN |w(x) = 0},
Ω1 = {x ∈ RN |w(x) > 0}.
By (A5) and the Fatou Lemma, we have
Z
Z
F (x, u+
F (x, u+
n) 2
n) 2
lim inf
w
dx
≥
wn dx → +∞.
lim inf
n
2
2
n→∞
n→∞
u
u
Ω1
Ω1
n
n
In the meantime, by (A5), there exists T0 > 0, such that
F (x, t)
≥1
t2
for t > T0 . By the continuity of F , there exists C1 > 0 such that
|F (x, t)| ≤ C1
for 0 < t ≤ T0 . Combining the preceding two inequalities, there exists C2 > 0 such
that
Z
Z
Z
F (x, un )
F (x, un ) 2
F (x, u+
n) 2
w
dx
=
dx
+
wn dx
n
2
u2n
u2n
Ω0 (0<un (x)≤T0 ) kun k
Ω0 (un (x)>T0 )
Ω0
C3
meas Ω0 (0 < un (x)
≥−
kun k2
Z
wn2 dx > −C2 .
≤ T0 ) +
Ω0
2
Dividing (2.1) with kun k , we have
Z
I(un )
1
F (x, u+
n)
=
−
(wn+ )2 dx = on (1).
2
kun k2
2
u
RN
n
Hence, we obtain
F (x, u+
n) 2
wn dx
2
un
RN
Z
Z
F (x, u+
F (x, u+
n) 2
n) 2
w
dx
+
wn dx → +∞.
=
n
2
2
un
un
Ω1
Ω0
1
− on (1) =
2
Z
It is easy to see that it is a contradiction. So {un } is bounded in E.
By the standard processes similar to Lemma 2.4, we know that un → u strongly
in E. This completes the proof.
Lemma 2.6. Under the assumptions of Theorem 1.3, the functional I satisfies the
(Ce)c condition for c > 0.
Proof. Assume that {un } ⊂ E is a (Ce)c sequence for c > 0,
I(un ) → c,
(1 + kun k)I 0 (un ) → 0 as n → ∞.
We argue that the sequence {un } is bounded in E. Otherwise, there is a subsequence, again denoted by {un }, such that kun k → +∞ as n → ∞. By (f6 ), we see
that, there exists R1 > 0 such that
f (x, t)t ≤ νbtν
(2.15)
for t ≥ R1 . By (A9), for 0 < δ < η, there exists R2 > 0 such that
f (x, t)t − 2F (x, t) ≥ (η − δ)tσ
(2.16)
12
Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
for t ≥ R2 . Set T = max{R1 , R2 }, let
Ω1 = {x ∈ RN : un (x) ≥ T },
Ω2 = {x ∈ RN : 0 < un (x) < T }.
Moreover, by the continuity of f and F , there exists C1 > 0 such that
Z
(f (x, un )un − 2F (x, un )) dx ≤ C1 .
Ω2
Since {un } ⊂ E is a (Ce)c sequence, we obtain
Z
+
+
2c + on (1) =
f (x, u+
n )un − 2F (x, un ) dx
N
ZR
≥
(f (x, un )un − 2F (x, un ))dx − C1 .
Ω1
Thus,
Z
(f (x, un )un − 2F (x, un ))dx ≤ 2c + C1 + on (1).
Ω1
It follows from (2.16) that there exists C2 > 0 such that
Z
uσn dx ≤ C2 .
Ω1
Using the continuity of f again, there exists C3 > 0 such that
Z
f (x, un )un dx ≤ C3 .
Ω2
Through the definition of {un }, it reaches
Z
2
+
kun k − on (1) =
f (x, u+
n )un dx
RN
Z
Z
=
f (x, un )un dx +
f (x, un )un dx
Ω1
Ω2
Z
≤
f (x, un )un dx + C3 .
Ω1
From (2.15), we obtain
Z
Z
f (x, un )un dx ≤ νb
Ω1
uνn dx.
Ω1
Then, we will consider two cases.
t
Case 1. σ ≥ ν, there exists t ∈ (0, 1) such that ν1 = 1−t
σ + 2 . By the Hölder
inequality,
Z
Z
(1−t)ν
Z
tν2
σ
uνn dx ≤
uσn dx
u2n dx
Ω1
Ω1
(1−t)ν
σ
≤ C2
Ω1
K tν kun ktν .
(N −2α)t
Case 2. σ < ν, there exists t ∈ (0, 1) such that ν1 = 1−t
. By the Hölder
σ +
2N
inequality,
Z
Z
(1−t)ν
Z
tν(N2N−2α)
2N
σ
uνn dx ≤
uσn dx
unN −2α dx
Ω1
Ω1
Ω1
EJDE-2015/223
GROUND STATE SOLUTIONS
(1−t)ν
σ
13
K tν kun ktν .
≤ C2
Hence, one has
(1−t)ν
σ
kun k2 − on (1) ≤ C3 + νbC2
K tν kun ktν .
N
It is easily observed that σ > 2α
(ν − 2) is equivalent to tν < 2, it is easy to concluce
that the assumption is false. So {un } is bounded in E.
Since (A3) and (A4) hold, we know that un → u strongly in E by the standard
processes similar to Lemma 2.4. So I satisfies the (Ce)c condition.
Lemma 2.7. Assume that (A1)–(A4) hold. Then I satisfies the the following
conditions:
(1) There exist θ > 0, ξ > 0 such that I(u) ≥ ξ > 0 for all u ∈ E with kuk = θ;
(2) There exists e ∈ E with kek > θ such that I(e) ≤ 0.
Proof. (1) By (A3), (A4) and f ∈ C(RN × R+ , R), for 0 < ε <
Cε > 0 such that
∗
f (x, t)t ≤ εt2 + Cε t2α
N
1
2K 2 ,
there exists
(2.17)
+
for all (x, t) ∈ R × R . This implies that
Z 1
F (x, t) =
f (x, st)tds
0
Z
≤
1
∗
∗
εst2 + Cε s2α −1 t2α ds
(2.18)
0
≤
ε 2 Cε 2∗α
t + ∗t
2
2α
for all (x, t) ∈ RN ×R+ . Pay attention to the definition of I given in (2.1), it follows
from (2.18) and Lemma 2.2 that
Z
1
2
F (x, u+ )dx
I(u) = kuk −
2
RN
Z
Z
ε + 2
Cε + 2∗α
1
2
(u ) dx −
(u ) dx
≥ kuk −
∗
2
2
N
N
R
R 2α
∗
(2.19)
1
Cε K 2α
2∗
α
≥
kuk
1 − εK 2 kuk2 −
2
2∗α
1 C K 2∗α
∗
ε
≥
−
kuk2α −2 kuk2 .
∗
4
2α
Set
2∗α 2∗α1−2
.
∗
8Cε K 2α
Taking kuk = θ, it follows from (2.19) that
θ=
I(u) ≥
1 2
θ > 0,
8
then (1) is proved.
(2) By (A5), for any M > 0, there exists a T0 > 0 such that
F (x, t) ≥ M t2 > 0
14
Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
for all t ≥ T0 . By the continuity of F , there exists C > 0 such that
F (x, t) ≥ M t2 − C
for all (x, t) ∈ RN × R+ . We choose ϕ ∈ C0∞ (RN ) with
R ϕ ≥ 0, kϕk = 1 and
supp(ϕ) ⊂ B(0, R) for some R > 0. we have that for M B(0,R) ϕ2 dx > 1/2.
For t > 0 large enough, it follows from (A5) that F (x, tϕ) ≥ 0. Hence,
Z
t2
I(tϕ) = kϕk2 −
F (x, tϕ)dx
2
RN
Z
t2
−
F (x, tϕ)dx
≤
2
B(0,R)
Z
1
≤ t2
−M
ϕ2 dx + C|B(0, R)|.
2
B(0,R)
Choosing kek = ktϕk > ρ, we have I(e) < 0, then (2) is proved.
3. Proof of main results
Proof of Theorem 1.1. By Lemmas 2.4 and 2.7, it is easy to obtain a nontrivial
critical point u0 of I by the mountain pass theorem, which implies that
c = inf max I(γ(t)) = I(u0 ),
γ∈Γ t∈(0,1)
where Γ = {γ ∈ C([0, 1], E), γ(0) = 0, γ(1) = e} and I(u0 ) ≥ θ > 0. Let
N = {u ∈ E\{0} : I 0 (u) = 0},
since u0 ∈ N , N 6= ∅. Then we claim that I is bound from below on N , moreover,
there exists d > 0 such that I(u) > d for every u ∈ N . If not, there exists a
sequence {un } ⊂ N such that
I(un ) <
1
,
n
∀n ∈ N+ .
Using (2.18), we know that
∗
∗
1
Cε K 2α
1
> I(un ) ≥
1 − εK 2 kun k2 −
kun k2α .
n
2
2∗α
Since un ∈ N , I 0 (un ) = 0, it follows from (2.17) that
Z
+
2
2
2∗
2∗
α
α
kun k2 =
f (x, u+
n )un dx ≤ εK kun k + Cε K kun k .
(3.1)
RN
Thus,
1
1
1 >
− ∗ (1 − εK 2 )kun k2 .
n
2 2α
By the definition of ε <
1
2K 2 ,
we obtain
1
2 > 1 − ∗ kun k2 .
n
2α
Since 2 < 2∗α , it is easy to see that kun k → 0. From (3.1) and ε <
kun k2 <
∗
∗
1
kun k2 + Cε K 2α kun k2α ,
2
1
2K 2 ,
we know
EJDE-2015/223
GROUND STATE SOLUTIONS
15
so there exists C such that kun k > C. It is a contradiction. Hence, there exists
c1 > 0 such that
c1 = inf I(u).
u∈N
Clearly, c1 ≤ c. Let {vn } ⊂ N be a minimizing sequence for c1 , so {vn } is a
(Ce)c1 sequence. By Lemma 2.4, {vn } is bounded in E and it has a convergence
subsequence {vn } such that vn → u2 in E. Since u2 ∈ N , we know that
− 2
hI(u2 ), u−
2 i = −ku2 k = 0,
therefore, u2 is a non-negative ground state solution.
Proof of Theorem 1.2. Since the (Ce)c condition is satisfied by Lemma 2.5 and the
mountain geometrical structure is proved by Lemma 2.7. So we can get a ground
state solution by using the same method as that of Theorem 1.1.
Proof of Theorem 1.3. Since the (Ce)c condition is satisfied by Lemma 2.6 and the
mountain geometrical structure is proved by Lemma 2.7. So we can get a ground
state solution by using the same method as that of Theorem 1.1.
Acknowledgments. This research was supported by National Natural Science
Foundation of China (No. 11471267). The authors want to thank the anonymous
referees for their valuable comments.
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16
Y. PU, J. LIU, C.-L. TANG
EJDE-2015/223
Yang Pu
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China.
School of Mathematics and Information College, China West Normal University, Nanchong, Sichuan 637002, China
E-mail address: pt8820@swu.edu.cn
Jiu Liu
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: jiuliu2011@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