Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 109, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
A SIGN-CHANGING SOLUTION FOR NONLINEAR PROBLEMS
INVOLVING THE FRACTIONAL LAPLACIAN
KAIMIN TENG, KUN WANG, RUIMENG WANG
Abstract. In this article, we establish the existence of a least energy signchanging solution for nonlinear problems involving the fractional Laplacian.
Our main tool is constrained minimization in a closed subset containing all
the sign-changing solutions of the equation.
1. Introduction
In this article, we establish the existence of least energy sign-changing solutions
for the nonlinear problem involving the fractional Laplacian,
(−∆)s u = f (x, u)
u=0
in Ω,
(1.1)
in RN \Ω,
where Ω ⊂ RN is a bounded domain with smooth boundary, s ∈ (0, 1), N > 2s,
(−∆)s stands for the fractional Laplacian, which is defined by
Z
u(x) − u(y)
(−∆)s u(x) = C(N, s) lim
dy, x ∈ RN ,
ε↓0 RN \B (x) |x − y|N +2s
ε
where C(N, s) is a suitable positive normalization constant. The nonlinearity f :
Ω̄ × R → R is of class C 1 and satisfies the growth conditions:
(F1) limt→0 f (x, t)/t = 0 uniformly for a.e. x ∈ Ω;
∗
(F2) lim|t|→+∞ f (x, t)/|t|2s −1 = 0 uniformly for a.e. x ∈ Ω;
(F3) There exist R > 0 and µ > 2 such that
0 < µF (t, x) ≤ tf (x, t),
∀x ∈ Ω, ∀|t| ≥ R;
(F4) f (x, t)/t is increasing in |t| > 0, for every x ∈ Ω.
Here 2∗s =
2N
N −2s
is the fractional critical exponent, and F (x, t) =
Rt
0
f (x, s) ds).
Remark 1.1. The conditions (F1) and (F4) imply that H(x, t) = tf (x, t)−2F (x, t)
is a nonnegative function, increasing in |s| with
tH 0 (x, t) = t2 f 0 (x, t) − f (x, t)t > 0
for any |t| > 0.
2010 Mathematics Subject Classification. 35B38, 35D05, 35J20.
Key words and phrases. Sign-changing solutions; fractional Laplacian; minimization;
Nehari manifold.
c 2015 Texas State University - San Marcos.
Submitted March 12, 2015. Published April 21, 2015.
1
2
K. TENG, K. WANG, R. WANG
EJDE-2015/109
Recently, a great attention has been focused on the study nonlinear problems
involving fractional Laplacian, because of its importance in the field of physics,
engineering, finance and so on. For example, it arises in phase transition, quantum mechanics, flame propagation, chemical reaction liquids American options in
finance, crystal dislocation, see [2, 11, 21] and so on. Owing to the pioneer work
Caffarelli and Silvestre [6], who developed the technique of harmonic extension to
define the fractional Laplace operator, many well-known results corresponding to
classical elliptic problems have been obtained, see [3, 5, 7, 20] and the references
therein. On the other hand, Servadei and Valdinoci [16, 17] who takes the fractional Laplace operator as singular integral operator, introduced a suitable Sobolev
space which can be to develop the variational formula, some existence and multiple
results for problem (1.1) have been obtained, one can see [9, 10, 14, 15, 18, 19]
and the references therein, they deal with another nonlocal operator, the spectral
fractional Laplacian, which is different from the operator considered in the present
paper defined by the singular integral kernel.
As we know, in the past decades, the existence and multiplicity of sign-changing
solutions for nonlinear elliptic problems have been intensively studied. There are
some powerful methods which have been developed, such as the descended flow
methods [12], constrained minimization methods [4], super and sub solution combining with truncation techniques [8] and so on. The existence and multiplicity of
sign-changing solutions for problem (1.1) has been investigated by the recent paper
of Chang and Wang [7], through using the descended flow methods and harmonic
extension techniques. In a recent paper [1], the authors proved the existence of a
least energy solution for Schrödinger-Poisson equations by the minimization methods, the problem is similar to ours, because of appearing the “nonlocal” term. In
this paper, we borrow some ideas from [1], we use the constrained minimization
methods to prove the existence of sign-changing solutions for problem (1.1).
For any measurable function u : RN → R we define the Gagliadro seminorm by
setting
Z
|u(x) − u(y)|2
dx dy,
[u]2s =
N +2s
R2N |x − y|
and we introduce the fractional Sobolev space
H s (RN ) = {u ∈ L2 (RN ) : [u]s < ∞},
which is a Hilbert space. We also define a closed subspace
X(Ω) = {u ∈ H s (RN ) : u = 0 a.e. in RN \Ω}.
Because of the fractional Sobolev inequality, X(Ω) is a Hilbert space with inner
product
Z
(u(x) − u(y))(v(x) − v(y))
hu, vi =
dx dy,
|x − y|N +2s
R2N
which induces a norm k · kX = [·]s . For u ∈ X(Ω), set
Z
1
J(u) = kukX −
F (x, u) dx.
2
Ω
Then, J ∈ C 1 (X(Ω)) and
Z
Z
(u(x) − u(y))(v(x) − v(y))
0
hJ (u), vi =
dx dy −
f (x, u)v dx,
|x − y|N +2s
R2N
Ω
∀v ∈ X(Ω).
EJDE-2015/109
A SIGN-CHANGING SOLUTION
3
Moreover, all its critical points are (up to a normalization constant depending on
s and N , which we will neglect henceforth) weak solutions of (1.1), namely they
satisfy
Z
hu, vi =
f (x, u)v dx, ∀v ∈ X(Ω).
Ω
Our main result is the following.
Theorem 1.2. Suppose that f satisfies (F1)–(F4). Then problem (1.1) possesses
a least energy sign-changing solution.
Remark 1.3. We observe that (F2) is weaker than the usual subcritical condition,
such as the (F2) of [7, Theorem 1.1].
Remark 1.4. Under the conditions (F1)–(F3), using the usual mountain pass
theorem, we can obtain two solutions: a positive solution and a negative solution.
Remark 1.5. In this framework, because there is a nonlocal term
Z
(u(x) − u(y))(v(x) − v(y))
dx dy,
|x − y|N +2s
R2N
we can not prove the exactly nodal domains corresponding to the sign-changing
solution obtained in Theorem 1.2.
In the proof of Theorem 1.2, we will prove that the functional J achieves a
minimum value on the nodal set
M = {u ∈ N : hJ 0 (u), u+ i = 0, hJ 0 (u), u− i = 0 and u± 6= 0}
where u+ (x) = max{u(x), 0}, u− (x) = min{u(x), 0} and
N = {u ∈ X(Ω)\{0} : hJ 0 (u), ui = 0}.
More precisely, we prove that there is w ∈ M such that
J(w) = inf J(u).
u∈M
2. Preliminary lemmas
Lemma 2.1. For any a, b ∈ R, we have:
(i) (ka)± = ka± , for all k ≥ 0, (a + b)± ≤ a± + b± ;
(ii) (a − b)(a+ − b+ ) ≥ (a+ − b+ )2 ;
(iii) (a − b)(a− − b− ) ≥ (a− − b− )2 .
Remark 2.2. The conclusion (ii) of Lemma 2.1 implies that (a+ −b+ )(a− −b− ) ≥ 0,
for any a, b ∈ R.
From Lemma 2.1, by simple computations, we deduce the following Lemma.
Lemma 2.3. For any u ∈ X(Ω), the following element facts hold.
(i) ku± kX ≤ kukX ;
(ii)
Z
Z
u+ (x)u− (y)
u− (x)u+ (y)
±
±
±
hu, u i = hu , u i −
dx
dy
−
dx dy;
N +2s
N +2s
R2N |x − y|
R2N |x − y|
(iii)
0
±
0
±
±
Z
hJ (u), u i = hJ (u ), u i −
R2N
u+ (x)u− (y)
dx dy −
|x − y|N +2s
Z
R2N
u− (x)u+ (y)
dx dy.
|x − y|N +2s
4
K. TENG, K. WANG, R. WANG
EJDE-2015/109
Lemma 2.4. Let {un } be a bounded sequence in X(Ω). Then up to a subsequence,
still denoted by {un }, there exists u ∈ X(Ω) such that
(i)
Z
Z
q
|u±
n | dx =
lim
n→∞
Ω
|u± |q dx;
Ω
(ii)
Z
±
u±
n f (x, un ) dx =
lim
n→∞
Ω
Z
u± f (x, u± ) dx;
Ω
(iii)
Z
F (x, u±
n ) dx =
lim
n→∞
Ω
Z
F (x, u± ) dx;
Ω
(iv)
h Z
lim inf −
n→∞
R2N
i
−
u+
n (x)un (y)
dx dy ≥ −
N
+2s
|x − y|
Z
R2N
u+ (x)u− (y)
dx dy.
|x − y|N +2s
Proof. From the boundedness of sequence {un } in X(Ω), it follows that up to a
subsequence, we may assume that there exists u ∈ X(Ω) such that
un * u,
un → u,
weakly in X(Ω),
strongly in Lq (RN ) with q ∈ [2, 2∗s ),
un (x) → u(x),
(2.1)
N
a.e. in R .
(i) From Lemma 2.1, one has
Z
Z
Z
q
± q
± q
|u±
|
dx
−
|u
|
dx
≤
|u±
n
n − u | dx
Ω
Ω
ZΩ
≤
|(un − u)± |q dx
Ω
Z
≤
|un − u|q dx.
Ω
Hence, the conclusion (i) can be deduced from (2.1).
(ii) By hypotheses (F1) and (F2), given ε > 0, there exists Cε > 0 such that
∗
|f (x, t)| ≤ ε|t| + Cε |t|q−1 + ε|t|2s −1
for all t ∈ R and almost all x ∈ Ω.
2∗
s −1
2∗
s −1
(2.2)
in L
(Ω), and so there exists v ∈ L
(Ω) such
Thus, {f (x, u±
n )} is bounded
2∗
s −1 (Ω). Because f (x, u± ) → f (x, u± ) a.e. x ∈ Ω, we obtain
that f (x, u±
n ) * v in L
n
±
2∗
s −1 (Ω). As a result, we deduce that
f (x, u±
n ) * f (x, u ) in L
Z
±
u± (f (x, u±
(2.3)
n ) − f (x, u )) dx → 0 as n → ∞.
Ω
By Hölder’s inequality and (2.2), we have
Z
±
±
±
[u±
n f (x, un ) − u f (x, u )]dx
Ω
Z
Z
±
±
±
≤
|u±
−
u
||f
(x,
u
)|
dx
+
|
u± (f (x, u±
n
n
n ) − f (x, u )) dx|
Ω
Ω
±
±
±
±
±
±
±
≤ εku±
ku±
n − u k2 kun k2 + Cε kun − u kq kun kq + εkun − u k2∗
n k2∗
s
s
Z
±
+|
u± (f (x, u±
n ) − f (x, u )) dx|
Ω
EJDE-2015/109
A SIGN-CHANGING SOLUTION
±
±
±
≤ C0 [ku±
n − u k2 + kun − u kq ] + C0 ε + |
Z
5
±
u± (f (x, u±
n ) − f (x, u )) dx|.
Ω
Taking the limit in the above inequality and using the arbitrariness of ε and (2.3),
conclusion (ii) follows.
(iii) By Hölder’s inequality and (2.3), we have
Z Z 1
Z
±
±
±
±
±
|u±
[F (x, un ) − F (x, u )] dx =
n − u ||f (x, θun + (1 − θ)u )| dθ dx
0
Ω
Ω
±
±
±
±
±
≤ εku±
n − u k2 kun k2 + Cε kun − u kq kun kq
±
+ εku±
ku±
n − u k2∗
n k2∗
s
s
±
±
±
≤ C0 [ku±
n − u k2 + kun − u kq ] + C0 ε.
Hence, taking the limit in the above inequality and using the arbitrariness of ε,
conclusion (iii) follows.
(iv) By Fatou’s Lemma, conclusion (iv) follows is trivially.
Lemma 2.5. There exists C1 > 0 and α > 0 such that
(i) J(u) ≥ ( µ2 − 1)kuk2X − C1 and kukX ≥ α, ∀u ∈ N ;
(ii) ku± kX ≥ α, ∀u ∈ M.
Proof. (i) For any u ∈ N , by (F3) and (2.2), we have
Z
µ
µJ(u) = µJ(u) − hJ 0 (u), ui = ( − 1)kuk2X + [uf (x, u) − µF (x, u)] dx
2
ZΩ
µ
2
= ( − 1)kukX +
[uf (x, u) − µF (x, u)] dx
2
{|u|≤R}
Z
+
[uf (x, u) − µF (x, u)] dx
{|u|≥R}
µ
≥ ( − 1)kuk2X − C1 .
2
From (F1) and (F2), for any ε > 0, there exists C2 > 0 such that
∗
f (x, t)t ≤ εt2 + C2 |t|2s ,
for all t ∈ R and a.e. x ∈ Ω.
(2.4)
Since hJ 0 (u), ui = 0, by Sobolev embedding and (2.4), we have
Z
Z
Z
∗
2∗
kuk2X =
uf (x, u) dx ≤ ε
u2 dx + C2
|u|2s dx ≤ C3 [εkuk2X + kukXs ],
Ω
Ω
Ω
∗
from this inequality, taking ε = 1/(2C3 ), α = (1/(2C3 ))2s −2 , it follows that
kukX ≥ α,
∀u ∈ N .
0
(ii) Assume that u ∈ M, we have that hJ (u), u± i = 0. From (iii) of Lemma 2.3,
we have
Z
Z
u+ (x)u− (y)
u− (x)u+ (y)
hJ 0 (u± ), u± i =
dx
dy
+
dx dy < 0.
N +2s
N +2s
R2N |x − y|
R2N |x − y|
Thus,
ku± k2X <
Z
u± f (x, u± ) dx.
Ω
As in the proof of (i), we deduce that ku± kX ≥ α.
6
K. TENG, K. WANG, R. WANG
EJDE-2015/109
Lemma 2.6. If {un } is a bounded sequence in M and p ∈ (2, 2∗s ), we have
Z
p
lim inf
|u±
n | dx > 0.
n→∞
Ω
Proof. By (F1) and (F2), for any ε > 0, there exists C4 > 0 such that
∗
tf (x, t) ≤ εt2 + C4 |t|p + ε|t|2s ,
for all t ∈ R and a.e. x ∈ Ω.
(2.5)
Since un ∈ M, by Lemma 2.5 and (2.5), we have
Z
Z
Z
Z
2
± 2
±
±
± 2
± p
2∗
s
α ≤ ku kX <
un f (x, un ) dx ≤ ε
|un | dx + C4
|un | dx + ε
|u±
n | dx.
Ω
Ω
Ω
Ω
From the boundedness of {un }, there is C5 > 0 such that
Z
2
p
α ≤ εC5 + C4
|u±
n | dx,
Ω
which implies
Z
p
|u±
n | dx > 0.
lim inf
n→∞
Ω
3. Proof of main result
In this section, we devote to proving Theorem 1.2. We denote by c0 the infimum
of J on M; that is,
c0 = inf J(u).
u∈M
To prove that this infimum can be attained, moreover, it is a critical point of
functional J, we need the following Lemmas.
Lemma 3.1. Let u ∈ X(Ω) with u± 6= 0. Then there are t, θ > 0 such that
hJ 0 (tu+ + θu− ), u+ i = 0 and hJ 0 (tu+ + θu− ), u− i = 0.
Proof. We define a function
γ(t, θ) = (hJ 0 (tu+ + θu− ), tu+ i, hJ 0 (tu+ + θu− ), θu− i),
obviously, γ ∈ C([0, +∞) × [0, +∞)). Since
Z
Z
u+ (x)u− (y)
dx
dy
−
tu+ f (x, tu+ )dx
hJ 0 (tu+ + θu− ), tu+ i = t2 ku+ k2X − 2ts
N +2s
|x
−
y|
2N
Ω
R
and
0
+
−
−
hJ (tu +θu ), θu i = θ
2
ku− k2X −2tθ
Z
R2N
u+ (x)u− (y)
dx dy−
|x − y|N +2s
Z
θu− f (x, θu− )dx,
Ω
from (2.4), by Sobolev embedding, we obtain
hJ 0 (tu+ + θu− ), tu+ i ≥ t2 ku+ k2X − 2tθ
2
− εt
ku+ k22
Z
R2N
∗
u+ (x)u− (y)
dx dy
|x − y|N +2s
2∗
− C2 t2s ku+ k2s∗s
2∗
∗
≥ t2 (1 − C6 ε)ku+ k2X − C7 t2s ku+ kXs .
Similarly,
∗
2∗
hJ 0 (tu+ + θu− ), θu− i ≥ θ2 (1 − C6 ε)ku− k2X − C7 θ2s ku− kXs .
EJDE-2015/109
A SIGN-CHANGING SOLUTION
7
Hence, There is r1 > 0 such that
hJ 0 (r1 u+ + θu− ), r1 u+ i > 0, ∀s > 0,
hJ 0 (tu+ + r1 u− ), r1 u− i > 0, ∀t > 0. (3.1)
From (F3), there are C8 , C9 > 0 such that
tf (x, t) ≥ C8 |t|µ − C9 , ,
for all t ∈ R and all x ∈ Ω̄.
(3.2)
Indeed, from (F3), we obtain
tf (x, t) ≥ µF (x, t) ≥ C8 |t|µ ,
∀|t| ≥ R and a.e. x ∈ Ω.
(3.3)
From the continuity of tf (x, t), tf (x, t) is bounded on Ω̄ × [−R, R]; that is, there
exists a constant k0 > 0 such that
tf (x, t) ≥ −k0 ≥ C8 |t|µ − C8 Rµ − k0 ,
∀|t| ≤ R.
(3.4)
Combining (3.3) and (3.4), we obtain that (3.2) holds. Hence, by 3.2), one has
Z
u+ (x)u− (y)
hJ 0 (tu+ + θu− ), tu+ i ≤ t2 ku+ k2X − 2tθ
dx dy
N +2s
R2N |x − y|
Z
− C8 tµ
|u+ |µ dx + C9 |Ω|
Ω
and
Z
u+ (x)u− (y)
hJ 0 (tu+ + θu− ), θu− i ≤ θ2 ku− k2X − 2tθ
dx dy
N +2s
R2N |x − y|
Z
− C8 θµ
|u− |µ dx + C9 |Ω|,
Ω
where |Ω| denotes the Lebesgue measure of Ω. Hence, there is r2 > 0 such that
hJ 0 (r2 u+ + θu− ), r2 u+ i < 0,
hJ 0 (tu+ + r2 u− ), r2 u− i > 0,
∀t, θ ∈ [r1 , r2 ]. (3.5)
From (3.1) and (3.5), this Lemma follows applying Miranda’s Theorem [13].
For u ∈ X(Ω) with u± 6= 0, we consider the function hu (t, θ) : [0, +∞) ×
[0, +∞) → R defined as
hu (t, θ) = J(tu+ + θu− )
and gu : [0, +∞) × [0, +∞) → R2 given by
gu (t, θ) = (
∂hu
∂hu
(t, θ),
(t, θ)) = (hJ 0 (tu+ + θu− ), u+ i, hJ 0 (tu+ + θu− ), u− i).
∂t
∂θ
By the assumption f ∈ C 1 , we can deduce that gu is also a C 1 map.
Lemma 3.2. Let u ∈ M,
(i) hu (t, θ) < hu (1, 1), for all θ, t ≥ 0 and (t, θ) 6= (1, 1);
(ii) det(gu )0 (1, 1) > 0.
Proof. (i) Since u ∈ M, then hJ 0 (u), u± i = 0; that is,
Z
Z
u+ (x)u− (y)
+ 2
ku kX − 2
dx dy =
u+ f (x, u+ ) dx,
N +2s
Ω
R2N |x − y|
Z
Z
u+ (x)u− (y)
− 2
ku kX − 2
dx dy =
u− f (x, u− ) dx .
N +2s
R2N |x − y|
Ω
8
K. TENG, K. WANG, R. WANG
EJDE-2015/109
The above two equalities imply that (1, 1) is a critical point of hu . By condition
(F3), it is easy to prove that
lim
|(t,θ)|→∞
hu (t, θ) = −∞.
By the property of continuous function, we can get that hu attains a global maximum in some (a, b) ∈ [0, +∞) × [0, +∞).
1. We claim that a, b > 0. If b = 0, obviously a 6= 0 and then hJ 0 (au+ ), au+ i = 0,
i.e.,
Z
a2 ku+ k2X =
au+ f (x, au+ ) dx.
(3.6)
Ω
Since
Z
u+ (x)u− (y)
dx dy
hJ 0 (u+ ), u+ i = hJ 0 (u), u+ i + 2
N +2s
R2N |x − y|
Z
u+ (x)u− (y)
=2
dx dy < 0,
N +2s
R2N |x − y|
we obtain
ku+ k2X <
Z
u+ f (x, u+ ) dx.
(3.7)
Ω
By (3.6) and (3.7), one has
1
(1 − 2 )ku+ k2X <
a
Z +
u f (x, u+ ) au+ f (x, au+ ) + 2
−
(u ) dx.
(u+ )2
(au+ )2
Ω
From (F4), we can infer that a ≤ 1. Thus, by Remark 1.1, we have
1
hu (a, 0) = J(au+ ) = J(au+ ) − hJ 0 (au+ ), au+ i
2
Z
1
+
=
[au f (x, au+ ) − 2F (x, au+ )] dx
2 Ω
Z
1
[u+ f (x, u+ ) − 2F (x, u+ )] dx
≤
2 Ω
Z
1
≤
[u+ f (x, u+ ) − 2F (x, u+ )] dx
2 Ω
Z
1
[u− f (x, u− ) − 2F (x, u− )] dx
+
2 Ω
Z
1
=
[uf (x, u) − 2F (x, u)]
2 Ω
1
= J(u) − hJ 0 (u), ui = hu (1, 1).
2
This contradicts tht (a, b) is a global maximum point for hu . Similarly, we can show
that a 6= 0, and the proof of claim is complete.
2. We claim that a, b ≤ 1. Since (hu )0 (a, b) = 0, we have
Z
Z
u+ (x)u− (y)
2
+ 2
dx dy =
au+ f (x, au+ ) dx,
a ku kX − 2ab
N +2s
Ω
R2N |x − y|
Z
Z
u+ (x)u− (y)
2
− 2
b ku kX − 2ab
dx dy =
bu− f (x, bu− ) dx.
N +2s
R2N |x − y|
Ω
EJDE-2015/109
A SIGN-CHANGING SOLUTION
9
Without loss of generality, we suppose that a ≥ b. Hence,
Z
Z
u+ (x)u− (y)
a2 ku+ k2X − 2a2
dx
dy
≥
au+ f (x, au+ ) dx.
N +2s
R2N |x − y|
Ω
Note that the assumption hJ 0 (u), u+ i = 0, we deduce that
Z +
au f (x, au+ ) u+ f (x, u+ ) + 2
0≥
−
(u ) dx.
(au+ )2
(u+ )2
Ω
The condition (F4) implies that a ≤ 1 and so the proof of the claim is complete.
To conclude the proof of (i), we only need to show that hu does not have a
global maximum in [0, 1] × [0, 1]\{(1, 1)}. If not, we assume that (a, b) is a global
maximum of hu in [0, 1] × [0, 1]\{(1, 1)}. From definition of hu and Remark 1.1, we
have
1
hu (a, b) = J(au+ + bu− ) − ((hu )0 (a, b), (a, b))
2
1
1
+
−
= J(au + bu ) − hJ 0 (au+ + bu− ), au+ i − hJ 0 (au+ + bu− ), bu− i
2
2
Z
a2 + 2
b2 − 2
u+ (x)u− (y)
= ku kX + ku kX − 2ab
dx dy
N +2s
2
2
R2N |x − y|
Z
Z
a2
b2
−
F (x, au+ ) dx −
F (x, bu− ) dx − ku+ k2X − ku− k2X
2
2
Ω
Ω
Z
Z
+
−
1
u (x)u (y)
dx dy +
au+ f (x, au+ ) dx
+ 2ab
N +2s
|x
−
y|
2
2N
Ω
Z R
1
−
−
+
bu f (x, bu ) dx
2
Z Ω
1
=
[au+ f (x, au+ ) − 2F (x, au+ ) dx
2 Ω
Z
1
+
[bu− f (x, bu− ) − 2F (x, bu− ) dx
2 Ω
Z
Z
1
1
<
[u+ f (x, u+ ) − 2F (x, u+ ) dx +
[u− f (x, u− ) − 2F (x, u− ) dx
2 Ω
2 Ω
Z
1
=
[uf (x, u) − 2F (x, u)] dx = hu (1, 1).
2 Ω
Hence, hu does not have a global maximum in [0, 1] × [0, 1]\{(1, 1)}. Therefore, the
proof of (i) is complete.
(ii) Since
Z
∂ 2 hu
+ 2
(t, θ) = ku kX −
f 0 (x, tu+ )(u+ )2 dx,
∂t2
Ω
Z
∂ 2 hu
− 2
(t, θ) = ku kX −
f 0 (x, θu− )(u− )2 dx,
∂θ2
Ω
Z
∂ 2 hu
u+ (x)u− (y)
(t, θ) = −2
dx dy,
N +2s
∂t∂θ
R2N |x − y|
by a simple computation, observing that u ∈ M and by Remark 1.1, we obtain
det(gu )0 (1, 1)
10
K. TENG, K. WANG, R. WANG
EJDE-2015/109
Z
Z
h
ih
i
= ku+ k2X −
f 0 (x, u+ )(u+ )2 dx ku− k2X −
f 0 (x, u− )(u− )2 dx
Ω
Ω
2
Z
u+ (x)u− (y)
dx dy
−4
N +2s
R2N |x − y|
Z
Z
i
h
u+ (x)u− (y)
+
+
0
+
+ 2
dx
dy
=
u f (x, u ) − f (x, u )(u ) dx + 2
N +2s
R2N |x − y|
Ω
Z
hZ i
+
u (x)u− (y)
×
dx
dy
u− f (x, u− ) − f 0 (x, u− )(u− )2 dx + 2
N +2s
R2N |x − y|
Ω
Z
+
−
2
u (x)u (y)
−4
dx dy > 0.
N
+2s
R2N |x − y|
Corollary 3.3. Suppose that u ∈ X(Ω) satisfies u± 6= 0 and hJ 0 (u), u± i ≤ 0. Then
there are t, s ∈ [0, 1] such that tu+ + θu− ∈ M.
Proof. From (F3), it is easy to deduce that
lim
|(t,θ)|→∞
hu (t, θ) = −∞.
We assume that hu attain a global maximum in some (a, b) ∈ [0, +∞) × [0, +∞)
and then (hu )0 (a, b) = 0, we have
Z
Z
u+ (x)u− (y)
a2 ku+ k2X − 2ab
dx
dy
=
au+ f (x, au+ ) dx,
N +2s
R2N |x − y|
Ω
Z
Z
u+ (x)u− (y)
b2 ku− k2X − 2ab
dx
dy
=
bu− f (x, bu− ) dx.
N +2s
R2N |x − y|
Ω
If a ≥ b, then
a2 ku+ k2X − 2a2
Z
R2N
u+ (x)u− (y)
dx dy ≥
|x − y|N +2s
Z
au+ f (x, au+ ) dx.
Ω
Note that by the assumption hJ 0 (u), u+ i ≤ 0, we deduce
Z +
au f (x, au+ ) u+ f (x, u+ ) + 2
0≥
−
(u ) dx.
(au+ )2
(u+ )2
Ω
The condition (F4) implies that a ≤ 1 and so the proof is complete.
Proof of Theorem 1.2. Let {un } be a sequence in M such that
lim J(un ) = c0 .
n→∞
From Lemma 2.5 (i), {un } is bounded in X(Ω). From Lemma 2.6, it follows that
u± 6= 0. Then, by Lemma 3.1, there are t, θ > 0 such that
hJ 0 (tu+ + θu− ), u+ i = 0,
hJ 0 (tu+ + θu− ), u− i = 0.
Next, we show that t, θ ≤ 1. Since hJ 0 (un ), u±
n i = 0; that is,
Z
Z
+
−
un (x)un (y)
+
+ 2
kun kX − 2
dx dy =
u+
n f (x, un ) dx,
N +2s
R2N |x − y|
Ω
Z
Z
−
u+
n (x)un (y)
− 2
−
kun kX − 2
dx dy =
u−
n f (x, un ) dx .
N +2s
R2N |x − y|
Ω
(3.8)
EJDE-2015/109
A SIGN-CHANGING SOLUTION
11
From Lemma 2.4, by the weak semi-continuity of norm function in Banach space,
we obtain
Z
Z
u+ (x)u− (y)
ku+ k2X − 2
dx
dy
≤
u+ f (x, u+ ) dx,
N +2s
|x
−
y|
2N
R
Ω
and
ku− k2X − 2
Z
R2N
u+ (x)u− (y)
dx dy ≤
|x − y|N +2s
Z
u− f (x, u− ) dx;
(3.9)
Ω
that is,
hJ 0 (u), u+ i ≤ 0,
From (3.8), it follows that
Z
t2 ku+ k2X − 2tθ
R2N
hJ 0 (u), u− i ≤ 0.
u+ (x)u− (y)
dx dy =
|x − y|N +2s
Z
(3.10)
tu+ f (x, tu+ ) dx
Ω
and
θ2 ku− k2X − 2tθ
Z
R2N
u+ (x)u− (y)
dx dy =
|x − y|N +2s
Z
θu− f (x, θu− ) dx.
(3.11)
Ω
Without loss of generality, we can assume that θ ≥ t. From (3.9) and (3.11), we
deduce that
Z −
θu f (x, θu− ) u− f (x, u− ) − 2 0≥
−
(u ) dx
(θu− )2
(u− )2
Ω
which implies that θ ≤ 1 (using (F4)).
Next, we show that J(tu+ + θu− ) = c0 . By (3.8) and u± 6= 0, we see that
tu+ + θu− ∈ M. Thus, by Remark 1.1, one has
1
c0 ≤ J(tu+ + θu− ) = J(tu+ + θu− ) − hJ 0 (tu+ + θu− ), tu+ + θu− i
2
1
1
= J(tu+ ) − hJ 0 (tu+ ), tu+ i + J(θu− ) − hJ 0 (θu− ), θu− i
2
2
1
1
≤ J(u+ ) − hJ 0 (u+ ), u+ i + J(u− ) − hJ 0 (u− ), u− i
2
2
1 0
= lim J(un ) − hJ (un ), un i
n→∞
2
= lim J(un ) = c0 .
n→∞
Consequently, we have proved that there exists tu+ + θu− ∈ M such that J(tu+ +
θu− ) = c0 . Hereafter, we denote w = tu+ + θu− .
As in the proof of [1, Theorem 1.3], by using Lemma 3.2, we conclude that w is
a critical point of J on X(Ω).
Acknowledgments. The authors are grateful to the anonymous referee for his/her
suggestions. This research was supported by the National Natural Science Foundation of China (NSFC 11226117), and by the Shanxi Province Science Foundation
for Youths under grant 2013021001-3.
12
K. TENG, K. WANG, R. WANG
EJDE-2015/109
References
[1] C. O. Alves, M. A. S. Souto; Existence of least energy nodal solution for a Schrödinger-Poisson
system in bounded domains, Z. Angew. Math. Phys. 65: 1153–1166, 2014.
[2] D. Applebaum; Lévy process -from probability to finance and quantum groups, Notices Amer.
Math. Soc. 51: 1336–1347, 2004.
[3] B. Barrios, E. Colorado, A. de Pablo, U. Sánchez; On some critical problems for the fractional
Laplacian operator, J. Differential Equations 252: 6133–6162, 2012.
[4] T. Bartsch, T. Weth; Three nodal solutions of singularly perturbed elliptic equations on
domains without topology, Ann. I. H. Poincaré-AN, 22: 259–281, 2005.
[5] X. Cabré, J. Tan, Positive solutions of nonlinear problems involving the square root of the
Laplacian, Adv. Math. 224: 2052–2093, 2010.
[6] L. Caffarelli, L. Silvestre; An extension problem related to the fractional Laplacian, Comm.
Partial Differential Equations 32: 1245–1260, 2007.
[7] X. J. Chang, Z. Q. Wang; Nodal and multiple solutions of nonlinear problems involving the
fractional Laplacian, J. Differential Equations, 256: 2965–2992, 2014.
[8] E. N. Dancer, Y. H. Du; On sign-changing solutions of certain semilinear elliptic problems,
Appl. Anal. 56: 193–206, 1995.
[9] M. Ferrara, G. Molica Bisci, B. Zhang; Existence of weak solutions for non-local fractional
problems via Morse theory, Discrete and Continuous Dynamical Systems Series B, 19: 2493–
2499, 2014.
[10] A. Iannizzotto, M. Squassina; 1/2-Lapalcain problems with exponential nonlinearity, J. math.
Anal. Appl. 414: 372–385, 2014.
[11] N. Laskin; Fractional quantum mechanics and Lévy path integrals, Physics Letters A 268:
298–305, 2000.
[12] Z. L. Liu, J. X. Sun; Invariant sets of descending flow in critical point theory with applications
to nonlinear differential equations, J. Differential Equations, 172: 257–299, 2001.
[13] C. Miranda; Un’osservazione su un teorema di Brouwer, Bol. Un. Mat. Ital. 3: 5–7, 1940.
[14] G. Molica Bisci; Sequence of weak solutions for fractional equations, Mathematical Research
Letters, 21: 241–253, 2014.
[15] G. Molica Bisci, D. Repovs; Fractional nonlocal problems involving nonlinearities with
bounded primitive, J. Math. Anal. Appl. 420: 167–176, 2014.
[16] R. Servadei, E. Valdinoci; Mountain Pass solutions for non-local elliptic operators, J. Math.
Anal. Appl. 389: 887–898, 2012.
[17] R. Servadei and E. Valdinoci, Variational methods for non-local operators of elliptic type,
Discrete Contin. Dyn. Syst. 33: 2105–2137, 2013.
[18] R. Servadei, E. Valdinoci; Lewy-Stampacchia type estimates for variational inequalities driven
by (non)local operators, Rev. Mat. Iberoam. 29: 1091–1126, 2013.
[19] R. Servadei, E. Valdinoci; A Brezis-Nirenberg result fornon-local critical equations in low
dimension, Commu. Pure App. Anal. 12: 2445–2464, 2013.
[20] J. Tan; The Brezis-Nirenberg type problem involving the square root of the Laplacian, Calc.
Var. 42: 21–41, 2011.
[21] E. Valdinoci; From the long jump random walk to the fractional Laplacian, Bol. Soc. Esp.
Mat. Apl. SMA 49: 33–44, 2009.
Kaimin Teng
Department of Mathematics, Taiyuan University of Technology, Taiyuan,
Shanxi 030024, China
E-mail address: tengkaimin2013@163.com
Kun Wang
Department of Mathematics, Taiyuan University of Technology, Taiyuan,
Shanxi 030024, China
E-mail address: 1090417245@qq.com
Ruimeng Wang
Department of Mathematics, Taiyuan University of Technology, Taiyuan,
Shanxi 030024, China
E-mail address: 1042583643@qq.com