Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 88, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
INFINITELY MANY SOLUTIONS FOR FRACTIONAL
SCHRÖDINGER EQUATIONS IN RN
CAISHENG CHEN
Abstract. Using variational methods we prove the existence of infinitely
many solutions to the fractional Schrödinger equation
(−∆)s u + V (x)u = f (x, u),
x ∈ RN ,
where N ≥ 2, s ∈ (0, 1). (−∆)s stands for the fractional Laplacian. The potential function satisfies V (x) ≥ V0 > 0. The nonlinearity f (x, u) is superlinear,
has subcritical growth in u, and may or may not satisfy the (AR) condition.
1. Introduction and statement of main results
In this article, we investigate the existence of infinitely many solutions to the
fractional Schrödinger equation
(−∆)s u + V (x)u = f (x, u),
x ∈ RN ,
(1.1)
s
where N ≥ 2, s ∈ (0, 1). (−∆) stands for the fractional Laplacian. The function
f (x, u) is odd, sublinear or suplinear and subcritical in u, V (x) is positive and
bounded below in RN .
Equation (1.1) arises in the study of the fractional Schrödinger equation
∂ψ
+ (−∆)s ψ + V (x)ψ = f (x, ψ), x ∈ RN , t > 0,
(1.2)
∂t
when looking for standing waves, that is, solutions with the form ψ(x, t) = eiωt u(x),
where ω is a constant. This equation was introduced by Laskin [14, 15] and comes
from an expansion of the Feynman path integral and from Brownian-like to Lévylike quantum mechanical paths.
This equation is of particular interest in fractional quantum mechanics for the
study of particles on stochastic fields modelled by Lévy processes, which occur
widely in physics, chemistry and biology. The stable Lévy processes that gives rise
to equations with the fractional Laplacian have recently attracted much research
interest. For more details, we can see [5].
Nonlinear equations like (1.1) have recently been studied by Cabré and Roquejoffre [3], Cabré and Tan [4], Sire and Valdinoci [22], Iannizzotto et al. [13], Hua
i
2010 Mathematics Subject Classification. 35R11, 35A15, 35J60, 47G20.
Key words and phrases. Fractional Schrödinger equation; variational methods;
(PS) condition; (C)c condition.
c
2016
Texas State University.
Submitted January 31, 2016. Published March 30, 2016.
1
2
C. CHEN
EJDE-2016/88
and Yu [12]. A one-dimensional version of (1.1) has been studied in the context of
solitary waves by Weinstein [27].
Equations of the form (1.1) in the whole space RN were studied by a number
of authors; see for instance [6, 9, 19, 20, 21] and the references therein. Felmer
et al. [9] considered the existence and regularity of positive solution of (1.1) with
V (x) = 1 and s ∈ (0, 1) when f has subcritical growth and satisfies the AmbrosettiRabinowitz ((AR) for short) condition. Secchi [19] obtained the existence of ground
state solutions of (1.1) for s ∈ (0, 1) when V (x) → ∞ as |x| → ∞ and (AR)
condition holds. In [8], the authors proved the existence of infinitely many weak
solutions for (1.1) by variant fountain theorem under the assumption
0 < inf V (x) < lim inf V (x) = V∞ < ∞.
x∈RN
(1.3)
|x|→∞
Tang [25] studied (1.1) with a potential V (x) satisfying
0 < inf V (x),
x∈RN
meas({x ∈ RN |V (x) ≤ d}) < ∞,
∀d > 0.
(1.4)
Similar assumptions can be found in [10, 21, 23, 28]. Each of these conditions
ensures that the embedding W s,2 (RN ) ,→ Lq (RN ) is compact for 2 ≤ q < 2∗s =
2N
N −2s . On the other hand, Gou and Sun [11], Chang and Wang [7] investigated the
existence of radial solutions for (1.1).
In this article, we are interest in the existence of infinitely many solutions for
(1.1) under the assumptions (A3)–(A7) below. Our assumptions on f (x, u) are
different from that in the above papers. The weighted functions h1 (x), h2 (x) and
h3 (x) depend on the potential function V (x) and the nonlinear function f (x, u)
either satisfies (AR) condition or does not. Moreover, two cases that f (x, u) is
bounded and unbounded in x ∈ RN are considered. We note that, in [17, 23, 26],
f (x, u) is assumed to bounded in x ∈ RN
To state our main results, we recall some fractional Sobolev spaces and norms
[16]. Let V (x) satisfy (A1) below and
Z
Z
n
o
s,2
N
2s
2
E = u ∈ W (R ) :
|ξ| |û| dξ +
V (x)|u|2 dx < ∞
(1.5)
RN
RN
endowed with the norm
kukE =
Z
RN
where and in the sequel, kuk22,V =
transform of ω(x); that is,
|ξ|2s |û|2 dξ + kuk22,V
R
RN
1/2
,
(1.6)
V (x)|u|2 dx and ω̂ = ω̂(ξ) is the Fourier
Z
1
ω(x)e−iξ·x dx,
(2π)N/2 RN
Z
1
ω(x) = F −1 [ω̂] =
ω̂(ξ)eiξ·x dξ.
(2π)N/2 RN
ω̂ = F[ω(x)] =
(1.7)
In [16], the author shows that
((−∆)s u)(x) = F −1 [|ξ|2 û], ∀x ∈ RN ,
Z
2
2
[u]E =
|ξ|2 |û|2 dξ,
C(N, s) RN
(1.8)
(1.9)
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FRACTIONAL SCHRÖDINGER EQUATIONS
3
where
1/2
|u(x) − u(y)|2
dx
dy
(1.10)
N +2s
R2N |x − y|
is called the Gagliardo norm, and the constant C(N, s) depends only on the space
dimensional N and the order s, and it is explicitly given by the integral
Z
1
1 − cos(ζ1 )
dζ, ζ = (ζ1 , ζ2 , . . . , ζN ) ∈ RN .
=
(1.11)
N +2s
C(N, s)
|ζ|
N
R
Moreover, by the Plancherel formula in Fourier analysis, we have
2
k(−∆)s/2 uk22 .
(1.12)
[u]2E =
C(N, s)
[u]E =
ZZ
Then, from (1.8)-(1.12), we obtain that the norm k · kE is equivalent to the norms
ZZ
Z
1/2
|u(x) − u(y)|2
dx
dy
kuk1 =
V (x)|u|2 dx +
,
N +2s
RN
R2N |x − y|
(1.13)
Z
1/2
2
s/2
2
kuk2 =
V (x)|u| dx + k(−∆) uk2
.
RN
In general, we define the fractional Sobolev space W s,p (RN )(0 < s < 1 < p, sp <
N ) as follows
o
n
|u(x) − u(y)|
p
2N
∈
L
(R
)
.
(1.14)
W s,p (RN ) = u ∈ Lp (RN ) :
N
|x − y| p +s
This space is endowed with the natural norm
ZZ
1/p
Z
|u(x) − u(y)|p
dx dy
kukW s,p =
|u|p dx +
,
(1.15)
N
+ps
RN
R2N |x − y|
while
ZZ
1/p
|u(x) − u(y)|p
(1.16)
[u]W s,p =
dx
dy
N +ps
R2N |x − y|
is called the Gagliardo norm. For the reader’s convenience, we recall the main
embedding results for W s,p (RN ).
Lemma 1.1 ([16]). Let s ∈ (0, 1) and p ≥ 1 such that sp < N . Then there exists
a positive constant S0 = S0 (N, p, s) such that, for any measurable and compactly
supported function u : RN → R, we have
kukp∗s ≤ S0 [u]W s,p ,
(1.17)
where p∗s =
s,p
N
pN/(N − ps) is the fractional critical exponent. Consequently, the space
W (R ) is continuously embedded in Lq (RN ) for any q ∈ [p, p∗s ]. Moreover, the
embedding W s,p (RN ) ,→ Lq (RN ) is locally compact whenever 1 < q < p∗s .
Remark 1.2. By the density of the compactly supported functions in W s,p (RN ),
we know that (1.17) holds for any u ∈ W s,p (RN ).
From the Hölder inequality and Lemma 1.1, we obtain the following lemma.
Lemma 1.3. Let s ∈ (0, 1), sp < N and p ≤ q ≤ p∗s . Then for any u ∈ X =
W s,p (RN ),
kukq ≤ Sq kukX
(1.18)
where Sq is a constant depending on s, q, p, N . In particular, we denote Sp∗s by S0 .
The inequality (1.18) shows that the embedding X ,→ Lq (RN ) is continuous.
4
C. CHEN
EJDE-2016/88
Proof. When q = p, inequality (1.18) is obvious. For q = p∗s , (1.18) can be obtained
from (1.17). Let p < q < p∗s . Then there exists t ∈ (0, 1) such that q = pt+p∗s (1−t).
It follows from the Hölder inequality and (1.17) that
Z
Z
Z
(1−t) Z
t
q
p∗
(1−t)
pt
p∗
s
s
|u| dx =
|u|
|u| dx ≤
|u| dx
|u|p dx
RN
RN
RN
RN
(1.19)
p∗
p∗
pt
q
1−t
pt
q
q
s (1−t)
s (1−t)
≤ S0 [u]X
kukp ≤ Sq kukX
kukX = Sq kukX ,
where Sqq = S01−t . This implies (1.18).
Similarly, for the Sobolev space E defined by (1.5), we have the following result.
Lemma 1.4. Let s ∈ (0, 1), 2s < N and 2 ≤ q ≤ 2∗s . Assume V (x) ≥ V0 > 0 in
RN . Then, for any u ∈ E,
kukq ≤ Sq kukE
(1.20)
where Sq is a constant depending on s, q, p, N and V0 . In particular, we denote S2∗s
by S0 .
Definition 1.5. A function u ∈ E is said to be a (weak) solution of (1.1) if for
any ϕ ∈ E, we have
Z
Z
Z
2s
|ξ| ûϕ̂dξ +
V (x)uϕdx =
f (x, u)ϕdx.
(1.21)
RN
RN
RN
Let J(u) : E → R be the energy functional associated with (1.1) defined by
Z
Z
Z
1
1
|ξ|2s |û|2 dξ +
V (x)|u|2 dx −
F (x, u)dx,
(1.22)
J(u) =
2 RN
2 RN
RN
Ru
where F (x, u) = 0 f (x, t)dt.
Using (1.18) and assumptions (A3)–(A7) below, we see that the functional J is
well defined and J ∈ C 1 (E, R) with
Z
Z
Z
0
2
J (u)ϕ =
|ξ| ûϕ̂dξ +
V (x)uϕdx −
f (x, u)ϕdx, ∀ϕ ∈ E.
(1.23)
RN
RN
RN
Throughout this article, the function f (x, u) ∈ C(RN × R) is odd in u. In
addition, we use the following assumptions.
(A1) The function V (x) ∈ C(RN ) satisfies inf x∈RN V (x) ≥ V0 > 0, where V0 is a
constant.
(A2) There exists a > 0 such that lim|y|→∞ meas({x ∈ Ba (y) : V (x) ≤ d}) = 0
for any d > 0, where “meas” denotes the Lebesgue measure on RN and
Br (x) denotes any open ball of RN centered at x and of radius r > 0, while
we simply write Br when x = 0.
(A3) There exist 2 < α < β < 2∗s such that
|f (x, u)| ≤ h1 (x)|u|α−1 + h2 (x)|u|β−1 , ∀(x, u) ∈ RN × R,
(1.24)
where h1 (x), h2 (x) ∈ C(RN ) and
lim sup
r→∞ x∈B c
r
h1 (x)
= 0,
V t1 (x)
lim sup
r→∞ x∈B c
r
h2 (x)
=0
V t2 (x)
(1.25)
with t1 = (2∗s − α)/(2∗s − 2), t2 = (2∗s − β)/(2∗s − 2) and Brc = RN \ B r =
{x ∈ RN : |x| > r}.
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FRACTIONAL SCHRÖDINGER EQUATIONS
5
(A4) There exists µ > 2 such that
tf (x, t) − µF (x, t) ≥ 0,
∀(x, t) ∈ RN × R.
(1.26)
(A5) lim|t|→∞ (F (x, t)t−2 ) = ∞ for any x ∈ RN .
N
2k
(A6) There exist k > 2s
and 2 < α < β ≤ k−1
such that (1.24) and (1.25) hold.
Furthermore, there exist b, c1 ≥ 1 such that for x ∈ RN and |u| ≥ b,
1
uf (x, u) − F (x, u) ≥ 0,
2
|F (x, u)|k ≤ ck1 |u|2k |h3 (x)|2k G(x, u),
F (x, u) ≥ 0,
G(x, u) =
(1.27)
where h3 (x) ∈ C(RN ) satisfies
0
|h3 (x)|2k
= 0,
lim sup
r→∞ x∈B c V t3 (x)
r
with k 0 =
k
2∗ − 2k 0
, t3 = s ∗
.
k−1
2s − 2
(1.28)
(A7) There exist a constant C0 > 0 and 2 < α < β < 2∗s , such that
|f (x, u)| ≤ C0 (|u|α−1 + |u|β−1 ),
∀(x, u) ∈ RN × R.
(1.29)
Remark 1.6. Condition (A2), which is weaker than the coercivity assumption
V (x) → ∞ as |x| → ∞, was originally introduced by Bartsch and Wang in [1] to
overcome the lack of compactness. Clearly, if V (x) → ∞ as |x| → ∞, it is possible
that the functions h1 (x), h2 (x) and h3 (x) in (A3) and (A6) are unbounded on RN .
So, it is necessary to consider the condition (A7).
Our main results in this paper are as follows.
Theorem 1.7. Let s ∈ (0, 1), 2s < N . Assume (A1), (A2) and (A5) hold. In
addition, suppose that either (A3), (A4) or (A6) are satisfied. Then (1.1) admits
infinitely many solutions un ∈ E such that J(un ) → ∞ as n → ∞.
Theorem 1.8. Let s ∈ (0, 1), 2s < N . Assume (A1)–(A3) and (A7) hold. In
addition, suppose that either (A4) or (1.27) is satisfied with h3 (x) ≡ 1. Then (1.1)
admits infinitely many solutions un ∈ E such that J(un ) → ∞ as n → ∞.
Remark 1.9. Assumption (1.25) implies that the functions (h1 V −1 ), (h2 V −1 ),
(h1 V −t1 ), (h2 V −t2 ) belong to L∞ (RN ) and
lim sup [h1 (x)V −1 (x)] = lim sup [h2 (x)V −1 (x)] = 0.
r→∞ x∈B c
r→∞ x∈B c
r
Moreover, the condition k >
r
N
2s
in (A7) implies that
2k
k−1
< 2∗s .
Remark 1.10. Assumption (A4) isPcalled the (AR) condition. Obviously, the
n
power functions in u like f (x, u) = i=1 hi (x)|u|βi −2 u with 2 < βi < 2∗s satisfy
(A3) and (A4) for appropriate functions hi ∈ C(RN ). The functions like f (x, u) =
h(x)u log(1 + |u|) fails to satisfy condition (A4), but it satisfies (A6).
Teng [26] considered problem (1.1) under assumption (A4) with h1 (x), h2 (x) ∈
L∞ (RN ). Obviously, our assumptions on h, h1 and h2 are weaker than that in [26].
Without loss of generality, we let V0 = 1 in (A1).
6
C. CHEN
EJDE-2016/88
2. Proof of main results
To prove the main results, we recall some useful concepts and results.
Definition 2.1. Let E be a real Banach space and the functional J ∈ C 1 (E, R).
We say that J satisfies the (C)c condition if any (C)c sequence {un } ⊂ E:
J(un ) → c,
(1 + kun kE )kJ 0 (un )kE ∗ → 0
as n → ∞
(2.1)
has a convergent subsequence in E.
Lemma 2.2 ([18, 24]). Let E be an infinite dimensional real Banach space, the
functional J ∈ C 1 (E, R) be even and satisfy the (C)c condition for all c > 0 and
J(0) = 0. In addition, assume E = Y ⊕ Z, in which Y is finite dimensional, and
J satisfies
(A8) there exist constants ρ, α0 > 0 such that J(z) ≥ α0 on ∂Bρ ∩ Z;
(A9) for each finite dimensional subspace E0 ⊂ E, there is an R = R(E0 ) such
that J(z) ≤ 0 on E0 \ B R , where BR = {z ∈ E : kzkE < R}, ∂BR = {z ∈
E : kzkE = R}.
Then, J possesses an unbounded sequence of critical values, i.e. there exists a
sequence {un } ⊂ E such that J 0 (un ) = 0 and J(un ) → ∞ as n → ∞.
In the proof of our results, we use the following lemma.
Lemma 2.3 ([17]). Let s ∈ (0, 1), 2s < N and 2 ≤ q < 2∗s =
and (A2). Then the embedding E ,→ Lq (RN ) is compact.
2N
N −2s .
Assume (A1)
For the prove Theorems 1.7 and 1.8, we need the following lemmas.
Lemma 2.4. Assume (A1) and (A2). If (A4) is satisfied, then any (C)c sequence
{un } is bounded in E.
Proof. Let the sequence {un } satisfy (2.1) and µ > 2. Then for large n, we have
1 0
J (un )un
µ
Z
1
1
1
= ( − )kun k2E +
f (x, un )un − µF (x, un ))dx.
2 µ
µ RN
c + 1 + kun kE ≥ J(un ) −
Then (A4) implies that {un } is bounded in E. The proof is complete.
(2.2)
Lemma 2.5. Assume (A1), (A2), (A5), (A6) hold. Then any (C)c sequence {un }
is bounded in E.
Proof. To prove the boundedness of {un }, arguing contradiction, we suppose that
un (x)
kun kE → ∞ as n → ∞. Let vn (x) = ku
. Then kvn kE = 1 for all n ≥ 1. By
n kE
Lemma 2.3, there exists a subsequence of {vn }, still denoted by {vn }, and v ∈ E
such that kvkE ≤ 1 and
vn * v weakly in E;
vn → v in Lq (RN ) (2 ≤ q < 2∗s );
vn (x) → v(x) a.e. in RN .
(2.3)
Clearly, it follows from (2.3) that there exists ω(x) ∈ Lq (RN )(2 ≤ q < 2∗s ) such
that |vn (x)| ≤ ω(x) a.e. in RN for all n ≥ 1.
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FRACTIONAL SCHRÖDINGER EQUATIONS
From (1.22), (1.23) and (2.1), it follows that for, n large,
Z
1 0
c + 1 ≥ J(un ) − J (un )un =
G(x, un )dx,
2
RN
where G(x, u) = 12 uf (x, u) − F (x, u), and
Z
1
|F (x, un )|
dx
≤ lim sup
2
kun k2E
n→∞
RN
Z
Z
|F (x, un )|
|F (x, un )|
2
|v
|
dx
+
lim
sup
|vn |2 dx,
≤ lim sup
n
2
|un |
|un |2
n→∞
n→∞
Brc
Br
7
(2.4)
(2.5)
for any r > 0. By (A6), we obtain, for any ε > 0, there exists δ > 0 such
(x,t)|
that |F|t|
≤ ε(h1 (x) + h2 (x)) for all 0 < |t| ≤ δ and all x ∈ RN . Denote
2
Xn = {x ∈ RN : |un (x)| ≤ δ}, Yn = {x ∈ RN : δ < |un (x)| ≤ b}, Zn = {x ∈ RN :
|un (x)| ≥ b}, where the constant b is given in (A6). Obviously, RN = Xn ∪ Yn ∪ Zn
and Brc = Brc ∩ (Xn ∪ Yn ∪ Zn ). Then
Z
Z
|F (x, un )|
2
(h1 (x) + h2 (x))|vn |2 dx
|v
|
dx
≤
ε
n
|un |2
Brc ∩Xn
Brc ∩Xn
Z
(2.6)
≤ ε(kh1 V −1 k∞ + kh2 V −1 k∞ )
V |vn |2 dx
RN
≤
2εM1 kvn k2E
= 2εM1 ,
where M1 = max{kh1 V −1 k∞ , kh2 V −1 k∞ }. Furthermore, by (1.24) and (1.25), one
sees that
Z
|F (x, un )|
|vn |2 dx
|un |2
c
Br ∩Yn
Z
β−2
≤b
(h1 (x) + h2 (x))|vn (x)|2 dx
Brc ∩Yn
Z
h2 (x) h1 (x)
+ sup
V (x)|vn |2 dx
x∈Brc V (x)
x∈Brc V (x)
Brc
h1 (x)
h2 (x) ≤ bβ−2 sup
+ sup
→ 0 as r = |x| → ∞.
x∈Brc V (x)
x∈Brc V (x)
≤ bβ−2
(2.7)
sup
On the other hand, from (1.27) and (2.4) it follows that
Z
|F (x, un )|
|vn |2 dx
|un |2
Brc ∩Zn
Z
1/k0
|F (x, u )| k 1/k Z
n
2k0
dx
≤
|h
v
|
dx
3 n
h23 |un |2
Brc ∩Zn
Brc ∩Zn
Z
Z
1/k 1/k0
0
≤ c1
G(x, un )dx
|h3 vn |2k dx
RN
≤ c1 (c + 1)1/k
Brc ∩Zn
Z
Brc
1/k0
|h3 vn |2k dx
.
0
(2.8)
8
C. CHEN
EJDE-2016/88
Note that 2 < 2k 0 < 2∗s . Let t3 = (2∗s − 2k 0 )/(2∗s − 2). By the Hölder inequality
and (1.20), we obtain
Z
Z
0
t3 Z
1−t3
0
|h3 (x)|2k 2
2∗
s dx
V
v
dx
|h3 vn |2k dx ≤ sup
|v
|
n
n
t
x∈Brc V 3 (x)
Brc
Brc
Brc
(2.9)
0
0
2k
|h3 (x)|2k
|h3 (x)|
2k0
≤ S0 sup
→0
≤ S0 kvn kE sup
t
t
x∈Brc V 3 (x)
x∈Brc V 3 (x)
as r = |x| → ∞, where S0 = S2∗s is the constant in (1.20). Then, and application
of (2.6)-(2.9) implies that for any ε > 0, there exist n0 , r0 ≥ 1, such that n ≥ n0 ,
r ≥ r0 ,
Z
|F (x, un )|
|vn |2 dx ≤ ε(2M1 + 1).
(2.10)
2
|u
|
c
n
Br
Set Tn = Xn ∪ Yn . Notice that for all x ∈ Br0 ∩ Tn ,
|F (x, un )|
|vn |2 ≤ bβ−2 (h1 (x) + h2 (x))|vn (x)|2
|un |2
β−2
≤b
2
(2.11)
1
M2 |ω(x)| ≡ d(x) ∈ L (Br0 ),
where
M2 = sup (h1 (x) + h2 (x)).
(2.12)
x∈Br0
If v(x) = 0 in Br0 , it follows from Fatou’s lemma that
Z
Z
|F (x, un )|
2
β−2
lim sup
|v
|
dx
≤
M
b
lim sup |vn |2 dx
n
2
2
|u
|
n→∞
n→∞
n
Br0 ∩Tn
Br
Z 0
= M2 bβ−2
|v|2 dx = 0.
(2.13)
B r0
Arguing as in (2.8) and (2.9), we obtain
Z
Z
1/k0
|F (x, un )|
2
2
2k0
|v
|
dx
≤
C
sup
|h
(x)|
|v
|
dx
n
1
3
n
|un |2
x∈Br0
B r0
Br0 ∩Zn
0
(2.14)
0
with C1 = c1 (c + 1)1/k . Similarly, since |vn (x)|2k ≤ |ω(x)|2k a.e. in RN and
0
|ω(x)|2k ∈ L1 (RN ), we obtain
Z
Z
Z
0
0
0
lim sup
|vn |2k dx ≤
lim sup |vn |2k dx =
|v|2k dx = 0.
n→∞
B r0
Br0
n→∞
Br0
Hence,
Z
lim sup
n→∞
Br0
|F (x, un )|
|vn |2 dx = 0.
|un |2
(2.15)
So, an application of (2.10) and (2.15) contradicts (2.5) and then meas(A) >
0, where A = {x ∈ RN : v(x) 6= 0}. Obviously, for a.e. x ∈ A, we have
limn→∞ |un (x)| = ∞. Hence, A ⊂ Zn for large n. Moreover, one sees that
Z
Z
|F (x, un )|
2
β−2
|vn | dx ≤ b
(h1 (x) + h2 (x))|vn |2 dx
|un |2
Tn
RN
Z
(2.16)
≤ bβ−2 (kh1 V −1 k∞ + kh2 V −1 k∞ )
V |vn |2 dx
RN
≤ 2b
β−2
M1 kvn k2E
= 2b
β−2
M1 .
EJDE-2016/88
FRACTIONAL SCHRÖDINGER EQUATIONS
9
Moreover, using assumption (A5) and Fatou’s lemma, it follows from J(un ) → c
that
h 1 Z F (x, u )
i
J(un )
c + o(1)
n
2
=
lim
≤
lim
|v
|
dx
0 = lim
−
n
n→∞ kun k2
n→∞ 2
n→∞ kun k2
|un |2
RN
E
Z E
1
F (x, un )
≤ + 2bβ−2 M1 − lim inf
|vn |2 dx
(2.17)
n→∞
2
|un |2
Zn
Z
1
F (x, un )
≤ + 2bβ−2 M1 −
χZn (x)|vn |2 dx = −∞,
lim inf
2
|un |2
RN n→∞
where χI denotes the characteristic function associated to the mensurable subset
I ⊂ RN . Clearly, (2.17) is impossible. Thus {un } is bounded in E and the proof of
Lemma 2.5 is finished.
Lemma 2.6. Assume (A1), (A2), (A5), (A7). In addition, suppose that (1.27) is
satisfied with h3 (x) ≡ 1. Then any (C)c sequence {un } is bounded in E.
Proof. Arguing as the proof of Lemma 2.5, we suppose that kun kE → ∞ as n → ∞.
un (x)
Let vn (x) = ku
. Then {vn } satisfies (2.3). For any ε > 0, we choose r1 > 0 such
n kE
R
2
that B c |v(x)| dx < ε when r ≥ r1 . Since vn (x) → v(x) in L2 (RN ), we obtain
r
Z
Z
Z
|vn (x)|2 dx ≤
lim sup |vn (x)|2 dx ≤
|v(x)|2 dx < ε.
(2.18)
lim sup
n→∞
Brc
Brc
n→∞
Brc
Then
Z
lim sup
n→∞
Brc ∩Tn
Z
|F (x, un )|
2
β−2
|v
|
dx
≤
2b
C
lim
sup
|vn |2 dx
n
0
|un |2
c
n→∞
Br
Z
2
≤ C2
|ω(x)| dx ≤ C2 ε,
(2.19)
Brc
where C2 = 2bβ−2 C0 , b is the constant in (A6) and C0 is given in (1.29).
On the other hand, from (1.27) with h3 (x) = 1 and (2.4) it follows that
Z
|F (x, un )|
|vn |2 dx
|un |2
c
Br ∩Zn
Z
|F (x, u )| k 1/k Z
1/k0
n
2k0
≤
dx
|v
|
dx
n
|un |2
Brc ∩Zn
Brc ∩Zn
(2.20)
Z
1/k Z
1/k0
0
≤ c1
G(x, un )dx
|vn |2k dx
RN
≤ c1 (c + 1)1/k
Brc ∩Zn
Z
0
|vn |2k dx
1/k0
.
Brc
From (2.9) and (2.18), for large n, we obtain
Z
0
(1−t3 )2∗
3
s
3
|vn |2k dx ≤ kvn k2t
≤ S0 kvn k2t
L2 (B c ) kvn kL2∗
L2 (B c ) ≤ S0 ε.
s (B c )
Brc
r
r
r
(2.21)
Then, an application of (2.19)-(2.21) gives that for any ε > 0 there exist n0 , r0 ≥ 1
such that n ≥ n0 , r ≥ r0 ,
Z
|F (x, un )|
|vn |2 dx ≤ ε(C2 + S0 C1 ).
(2.22)
|un |2
Brc
10
C. CHEN
EJDE-2016/88
Similar to (2.11), for a.e. x ∈ Br0 ∩ Tn and n ≥ 1, we obtain
|F (x, un )|
|vn |2 ≤ C2 |vn (x)|2 ≤ C2 |ω(x)|2 ≡ d(x) ∈ L1 (Br0 ).
|un |2
(2.23)
By Fatou’s lemma,
Z
lim sup
n→∞
Br0 ∩Tn
|F (x, un )|
|vn |2 ≤ C2
|un |2
Z
B r0
Z
= C2
lim sup |vn (x)|2
n→∞
(2.24)
|v(x)|2 dx.
B r0
Similar to (2.20), we derive
Z
Z
1/k0
|F (x, un )|
2
2k0
|v
|
≤
C
|v(x)|
lim sup
dx
.
n
1
|un |2
n→∞
B r0
Br0 ∩Zn
If v(x) = 0 in Br0 , an application of (2.24) and (2.25) gives that
Z
|F (x, un )|
|vn |2 = 0.
lim sup
2
|u
|
n→∞
n
B r0
(2.25)
(2.26)
Combining (2.22) with (2.26) contradicts (2.5). So, meas(A) > 0, where A = {x ∈
RN : v(x) 6= 0} and for a.e. x ∈ A, we have limn→∞ |un (x)| = ∞. Hence, A ⊂ Zn
for large n. Moreover, one sees that
Z
Z
Z
|F (x, un )|
2
2
|v
|
dx
≤
C
|v
|
dx
≤
C
V |vn |2 dx ≤ C1 kvn k2E = C1 .
n
1
n
1
2
|u
|
N
N
n
Tn
R
R
(2.27)
Moreover, using assumption (A5) and Fatou’s lemma, from J(un ) → c it follows
that
i
h 1 Z F (x, u )
J(un )
c + o(1)
n
2
=
lim
≤
lim
−
|v
|
dx
0 = lim
n
n→∞ kun k2
n→∞ 2
n→∞ kun k2
|un |2
RN
E
E
Z
1
F (x, un )
≤ + C1 − lim inf
|vn |2 dx
(2.28)
n→∞
2
|un |2
Zn
Z
F (x, un )
1
lim inf
χZn (x)|vn |2 dx = −∞.
≤ + C1 −
2
|un |2
RN n→∞
Clearly, the limit (2.28) is impossible. Thus {un } is bounded in E and the proof is
complete.
From Lemmas 2.4–2.6, we know that any (P S)c sequence and (C)c sequence
{un } of the functional J are bounded in E. Therefore, by Lemma 2.3, there exists
a subsequence of {un }, still denoted by {un }, and u ∈ E such that kun kE + kukE ≤
M (∀n ≥ 1) and
un * u weakly in E,
un → u in Lq (RN ) (2 ≤ q < 2∗s ),
un (x) → u(x) a.e. in RN
(2.29)
with some constant M > 0.
Lemma 2.7. Assume (A1)–(A6) hold. If the sequence {un } satisfies (2.29), then
Z
Z
α
α
lim
h1 (|un | − |u| )dx = 0,
lim
h2 (|un |β − |u|β )dx = 0,
(2.30)
n→∞
RN
n→∞
RN
EJDE-2016/88
Z
lim
n→∞
RN
FRACTIONAL SCHRÖDINGER EQUATIONS
Z
f (x, un )(un − u)dx = 0,
lim
f (x, u)(un − u)dx = 0,
n→∞ RN
Z
Z
lim
F (x, un )dx =
F (x, u)dx.
n→∞
RN
(2.31)
(2.32)
RN
Proof. First, we assume (A3), (A4). From (2.29), we obtain
Z
Z
lim
h1 (|un |α − |u|α )dx = 0,
lim
h2 (|un |β − |u|β )dx = 0
n→∞
11
n→∞
Br
(2.33)
Br
for any r > 0. On the other hand, we see from the Hölder inequality and (A3) that
Z
Z
t1 Z
1−t1
∗
h1 (x) V |un |2 dx
h1 |un |α dx ≤ sup t1
|un |2s dx
x∈Brc V (x)
Brc
Brc
Brc
≤ S0 sup
x∈Brc
h1 (x)
(1−t1 )2∗
1
s
kun k2t
E kun kE
V t1 (x)
h1 (x)
→ 0,
V t1 (x)
≤ S0 M α sup
x∈Brc
(2.34)
as r → ∞,
where t1 = (2∗s − α)/(2∗s − 2), S0 = S2∗s . Similarly, as r → ∞,
Z
h2 (x)
(1−t )2∗
2
kun k2t
kun kE 2 s
h2 |un |β dx ≤ S0 sup t2
E
x∈Brc V (x)
Brc
≤ S0 M β sup
x∈Brc
h2 (x)
→ 0,
V t2 (x)
(2.35)
where t2 = (2∗s − β)/(2∗s − 2). Then an application of (2.33), (2.34) and (2.35) gives
(2.30). Moreover, the limit (2.30) and Brezis-Lieb lemma [2] give that
Z
Z
α
lim
h1 |un − u| dx = 0,
lim
h2 |un − u|β dx = 0.
(2.36)
n→∞
n→∞
RN
Thus, from (2.36), it follows that
Z
Z
h1 |un |α−1 |un − u|dx ≤
RN
RN
h1 |un |α dx
(α−1)/α Z
RN
≤ (S0 M α kh1 V −t1 k∞ )(α−1)/α
1/α
h1 |un − u|α dx
RN
Z
h1 |un − u|α dx
1/α
→ 0,
as n → ∞
RN
(2.37)
and
Z
h2 |un |β−1 |un − u|dx
Z
1−1/β Z
≤
h2 |un |β dx
RN
RN
1/β
h2 |un − u|β dx
(2.38)
RN
≤ (S0 M β kh2 V −t2 k∞ )1−1/β
Z
h2 |un − u|β dx
1/β
→ 0,
as n → ∞.
RN
Hence,
Z
Z
|f (x, un )(un − u)|dx ≤
RN
(h1 |un |α−1 + h2 |un |β−1 )|un − u|dx → 0, (2.39)
RN
as n → ∞. This proves the first limit of (2.31). The second limit of (2.31) can be
obtained in a similar way.
12
C. CHEN
EJDE-2016/88
To prove the limit (2.32), we use (1.24) and derive that
|F (x, un ) − F (x, v)|
h
i
≤ C h1 (x)(|un |α−1 + |u|α−1 ) + h2 (x)(|un |β−1 + |u|β−1 ) |un − u|.
(2.40)
Then an application of (2.37) and (2.38) yields that the limit (2.32). The proof is
complete.
Lemma 2.8. Let the assumptions in Theorem 1.8 hold. If the sequence {un }
satisfies (2.29), then the limits (2.31) and (2.32) hold.
Proof. Choose ψ ∈ C0∞ (R) such that suppψ ⊂ [−2, 2] and ψ(t) = 1 on [−1, 1].
Denote g(x, t) = ψ(t)f (x, t), H(x, t) = (1 − ψ(t))f (x, t). Then f (x, t) = g(x, t) +
H(x, t). Furthermore, from (1.29), there exist the constants a1 , b1 > 0 such that
|g(x, t)| ≤ a1 |t|α−1 ,
|H(x, t)| ≤ b1 |t|β−1 ,
∀(x, t) ∈ RN × R.
(2.41)
Denote
Z
0
|g(x, un ) − g(x, u)|α dx,
An =
Z
0
|H(x, un ) − H(x, u)|β dx, (2.42)
Dn =
RN
RN
0
where t = t/(t − 1). By (2.29), there exist ω1 (x) ∈ Lα (RN ) and ω2 (x) ∈ Lβ (RN )
such that |un (x)| ≤ ω1 (x) and |un (x)| ≤ ω2 (x) a.e. in RN for all n ≥ 1. Note that
0
|g(x, un ) − g(x, u)|α ≤ C3 (|un (x)|α + |u(x)|α )
≤ C3 (|ω1 (x)|α + |u(x)|α ) ≡ d1 (x) ∈ L1 (RN )
(2.43)
and
0
|H(x, un ) − H(x, u)|β ≤ C3 (|un (x)|β + |u(x)|β )
≤ C3 (|ω2 (x)|β + |u(x)|β ) ≡ d2 (x) ∈ L1 (RN ),
(2.44)
where C3 is a constant independent of n. By the Lebesgue dominated convergence
theorem and (2.29), we have
Z
0
lim An =
lim |g(x, un ) − g(x, u)|α dx = 0,
n→∞
N n→∞
ZR
(2.45)
β0
lim Dn =
lim |H(x, un ) − H(x, u)| dx = 0.
n→∞
RN n→∞
Therefore, by the Hölder inequality,
Z
|f (x, un ) − f (x, u)||un − u|dx
RN
Z
≤
(|g(x, un ) − g(x, u)| + |H(x, un ) − H(x, u)|)|un − u|dx
≤
RN
0
A1/α
kun
n
0
(2.46)
0
− ukα + Dn1/β kun − ukβ
0
0
0
≤ (A1/α
+ Dn1/β )kun − ukE ≤ M (A1/α
+ Dn1/β ).
n
n
An application of (2.45) and (2.46) gives
Z
Z
lim
f (x, un )(un − u)dx = lim
n→∞
RN
n→∞
RN
f (x, u)(un − u)dx.
(2.47)
EJDE-2016/88
FRACTIONAL SCHRÖDINGER EQUATIONS
13
Similarly, from (2.41), (2.43) and (2.44), we can derive that
Z
Z
f (x, u)(un − u)dx =
lim f (x, u)(un − u)dx = 0.
lim
n→∞
RN n→∞
RN
(2.48)
Consequently, the limit (2.31) is given. Similarly, (2.32) can be proved and the
proof is complete.
Lemma 2.9. Let the assumptions in Theorems 1.7 and 1.8 hold. Let {un } be the
sequence in Lemmas 2.4–2.6 satisfying (2.29). Then u is a critical point of the
functional J and un → u in E.
Proof. First, we show that J 0 (u) = 0 in E ∗ . By Lemmas 2.4–2.6, the sequence
{un } is bounded in E. So, there exists a subsequence, still denoted by {un }, such
that {un } satisfies (2.29). Moreover, one sees that for all ϕ ∈ C0∞ (RN ),
Z
Z
lim
|ξ|2 (uˆn − û)ϕ̂dx +
V (x)(un − u)ϕdx = 0.
(2.49)
n→∞
RN
RN
Under assumptions (A3)–(A7), we obtain
Z
lim
(f (x, un ) − f (x, u))ϕdx = 0.
n→∞
(2.50)
RN
Furthermore, from (2.49), (2.50) and the assumption J 0 (un ) → 0 in E ∗ , we have
0 = lim J 0 (un )ϕ = J 0 (u)ϕ,
n→∞
∀ϕ ∈ C0∞ (RN ).
(2.51)
By the denseness of C0∞ (RN ) in E, it follows that J 0 (u)ϕ = 0, ∀ϕ ∈ E. Hence, u is
a critical point of J in E. On the other hand, from (2.29) it follows that
Z
Z
Rn =
|ξ|2 û(ûn − û)dξ +
V (x)u(un − u)dx → 0, as n → ∞.
(2.52)
RN
RN
Set
Z
Z
f (x, un )(un − u)dx,
Wn :=
RN
f (x, u)(un − u)dx,
Sn :=
∀n ∈ N. (2.53)
RN
From (2.31), it follows that Wn , Sn → 0 as n → ∞ and so
J 0 (u)(un − u) = Rn − Sn → 0.
(2.54)
Similarly, we set
Qn := (J 0 (un ) − J 0 (u))(un − u) = kun − uk2E − Wn + Sn ,
∀n ∈ N.
(2.55)
Obviously, relation (2.55) can be reduced to the form
kun − uk2E = Wn + Qn − Sn ,
∀n ∈ N.
(2.56)
0
From (2.53), (2.54) and J (un ) → 0, we find Qn → 0 and kun −ukE → 0 as n → ∞.
Thus un → u in E as n → ∞ under assumptions (A3)–(A7). Therefore, J satisfies
the (C)c condition in E and the proof is complete.
Proof of Theorem 1.7. Clearly, the functional J defined by (1.22) is even. By
Lemma 2.9, the functional J satisfies the (C)c condition. Next, we prove that
J satisfies (A8) and (A9) in Lemma 2.2. From (A3), we have
Z
Z
1
1
2
2
F (x, u)dx ≥ kukE −
(h1 |u|α + h2 |u|β )dx.
(2.57)
J(u) = kukE −
2
2
RN
RN
14
C. CHEN
EJDE-2016/88
Arguing as in the proof of (2.34) and (2.35), we obtain
J(u) ≥
1
ρ2
β
2 1
α−2
kuk2E −S0 M3 (kukα
)≥
≡ α0 > 0, (2.58)
E +kukE ) ≥ ρ ( −2S0 M3 ρ
2
2
4
1
where M3 = max{kh1 V −t1 k∞ , kh2 V −t2 k∞ } and kukE = ρ = min{1, (8S0 M3 ) 2−α }.
Thus, by (2.58), condition (A8) is satisfied. We now satisfy condition (A9). For
any finite dimensional subspace E0 ⊂ E, we assert that there holds J(un ) → −∞
when un ∈ E0 and kun kE → ∞. Arguing by contradiction, suppose that for some
sequence {un } ⊂ E0 with kun kE → ∞, there is M4 > 0 such that J(un ) ≥ −M4 , for
un (x)
, then kvn kE = 1. Passing to a subsequence, we may
all n ≥ 1. Set vn (x) = ku
n kE
assume that vn * v in E, vn (x) → v(x) a.e on RN . Since E0 is finite dimensional,
then vn → v in E0 and so v 6= 0 a.e.in RN . Set Ω = {x ∈ RN : v(x) 6= 0}, then
meas(Ω) > 0. For x ∈ Ω, we have limn→∞ |un (x)| = ∞.
Then, from (A5) it follows that
i
h 1 Z F (x, u )
−M
J(un )
n
2
0 = lim sup
≤
lim
sup
=
lim
sup
−
|v
|
dx
n
2
2
2
|un |2
n→∞ kun kE
n→∞ kun kE
n→∞
RN
(2.59)
Z
1
F (x, un )
2
≤ −
lim inf
χ
(x)|v
|
dx
=
−∞,
Ω
n
2
|un |2
RN n→∞
and we have a contradiction. So, there exists R = R(E0 ) > 0 such that J(u) <
0 for u ∈ E0 and kukE ≥ R. Therefore, condition (A9) is satisfied. Then an
application of Lemma 2.2 shows that (1.1) admits infinitely many solutions un ∈ E
with J(un ) → ∞ as n → ∞. This completes the proof.
Proof of Theorem 1.8. Clearly, the functional J defined by (1.22) is even. By
Lemma 2.9, the functional J satisfies the (C)c condition. Next, we prove that
J satisfies (A8) and (A9) in Lemma 2.2. From (A7), we have
Z
Z
1
1
J(u) = kuk2E −
F (x, u)dx ≥ kuk2E − C0
(|u|α + |u|β )dx.
(2.60)
2
2
RN
RN
Furthermore, from (1.20) it follows that
J(u) ≥
ρ2
1
β
2 1
α−2
kuk2E − C4 (kukα
)≥
≡ α1 > 0,
E + kukE ) ≥ ρ ( − C4 ρ
2
2
4
(2.61)
1
with kukE = ρ = min{1, (4C4 ) 2−α } and C4 = max{Sαα , Sββ }. Thus, by (2.61),
condition (A8) is satisfied. Similarly, we can derive (2.59) and the verification of
condition (A9) is finished. Again, using Lemma 2.2, we complete the proof of
Theorem 1.8.
Acknowledgments. This work is supported by the Fundamental Research Funds
for the Central Universities of China (2015B31014) and by the National Natural
Science Foundation of China (No.11571092).
The author would like to express his sincere gratitude to the reviewers for their
valuable comments and suggestions.
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Caisheng Chen
College of Science, Hohai University, Nanjing 210098, China
E-mail address: cshengchen@hhu.edu.cn