Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 214, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
GROUND STATE SOLUTIONS FOR SEMILINEAR PROBLEMS
WITH A SOBOLEV-HARDY TERM
XIAOLI CHEN, WEIYANG CHEN
Abstract. In this article, we study the existence of solutions to the problem
∗
|u|2s −2 u
, x ∈ Ω,
|y|s
u = 0, x ∈ ∂Ω,
−∆u = λu +
where Ω is a smooth bounded domain in RN (N ≥ 3). We show that there
is a ground state solution provided that N = 4 and λm < λ < λm+1 , or that
N ≥ 5 and λm ≤ λ < λm+1 , where λm is the m’th eigenvalue of −∆ with
Dirichlet boundary conditions.
1. Introduction
Let Ω be a smooth bounded domain of RN = Rk × RN −k , where 2 ≤ k < N ,
N ≥ 3. Suppose that a point (0, z0 ) ∈ Rk × RN −k and (0, z0 ) ∈ Ω. Without loss
of generality we assume that 0 ∈ Ω. In this article, we consider the existence of
solutions of the problem
∗
|u|2s −2 u
, x ∈ Ω,
−∆u = λu +
|y|s
u = 0, x ∈ ∂Ω,
(1.1)
where Ω is a smooth bounded domain in RN , and x = (y, z) ∈ Ω, 0 < s < 2, and
−s)
2∗s = 2(N
N −2 is the critical exponent related to the Hardy-Sobolev inequality
∗
2/2∗s Z
|u|2s
S
dy dz
≤
|∇u|2 dy dz, ∀u ∈ D1,2 (RN ),
(1.2)
s
RN |y|
RN
where S = S(N, k, t) is the best constant, see [3]. More general Hardy-Sobolev
inequalities are dealt in [4] and [5]. The minimizers of problem (1.2) are solutions
of the problem
Z
∗
− ∆u =
|u|2s −2 u
,
|y|s
u>0
in RN ,
u ∈ D1,2 (RN )
2000 Mathematics Subject Classification. 35J60, 35J65.
Key words and phrases. Existence; ground state; critical Hardy-Sobolev exponent;
semilinear Dirichlet problem.
c
2013
Texas State University - San Marcos.
Submitted April 19, 2013. Published September 26, 2013.
1
(1.3)
2
X. CHEN, W. CHEN
EJDE-2013/214
up to a constant. If s = 0, Equation (1.2) becomes the Sobolev inequality, for which
best constant was computed, and proved existence of minimizers in [2] and [16]. In
the case s = 2, (1.2) still holds true, it is an extension of the Hardy inequality. In
the more general case 0 ≤ s < 2 with k = N , the best constant was obtained in [8],
and minimizers were found in [10], which are radially symmetric. Therefore, it can
be shown by using ODEs, see [10], that up to dilations and translations, minimizers
take the form
1
N −2 .
(1 + |x|2−s ) 2−s
It is noted that equation (1.3) is invariant with respect to the scalings and ztranslations; that is, u is a solution of (1.3) if only if uα (x) = α(N −2)/2 u(αy, α(z −
z0 )), α > 0, satisfies the equation. Hence, problem (1.3) has lack of the compactness. In the case 0 < s < 2, 2 ≤ k < N , it was proved in [3] that the best constant
S > 0, and S is achieved by the concentration-compactness principle. So problem
(1.3) has a positive solution in D1,2 (RN ). Since the minimizer of problem (1.2)
can not be radially symmetric, they cannot be found among solutions of ODEs,
but of PDEs. This brings difficulties to find exact forms of the minimizer. In the
particular case s = 1, problem (1.3) becomes
N
u N −2
− ∆u =
,
|y|
u>0
in RN ,
u ∈ D1,2 (RN ).
(1.4)
By the moving plane method, it was proved in [7] that all solutions of (1.4) are
cylindrically symmetric. Thus, problem (1.4) can be reduced to an elliptic equation
in the positive cone in R2 , and it was shown in [7] that u is a solution of (1.4) if
and only if
u(y, z) = λ(N −2)/2 V (λy, λ(z + z0 ))
(1.5)
for some λ > 0 and z0 ∈ RN −k , where
V (x) = V (y, z) =
(N −2)/2
CN,k
((1 + |y|)2 + |z|2 )
(N −2)/2
=
((N − 2)(k − 1))
((1 + |y|)2 + |z|2 )
(N −2)/2
.
(1.6)
This result allows one to obtain existence results for problem (1.1) in the case s = 1.
Denote by 0 < λ1 , . . . , λk , . . . the eigenvalues of −∆ with zero Dirichlet boundary
condition. When 0 < λ < λ1 and s = 1, it was proved in [1] and [6] that there exists
a solution of problem (1.1) by the mountain pass lemma and constrained variation
respectively.
In this article, we consider the existence of solutions to problem (1.1) for general
case 0 < s < 2 and λ in between λm and λm+1 for some m ∈ N. As far as we know,
the exact form of the minimizer of (1.2) is known, see Mancini and Sandeep [11].
However, even without knowing it, to control (P S)c sequences so that it may avoid
the energy levels where the compactness does not hold, we can always use the [6,
Lemma 3.4.2], if u is the solution of (1.3), then there exist C2 > C1 > 0 such that
C2
C1
≤ u(x) ≤
.
N
−2
1 + |x|
1 + |x|N −2
(1.7)
This estimate suffices to serve our purpose. Using the Nehari manifold method
introduced in [12], and developed in [14], we show the following result.
Theorem 1.1. Let N = 4 and λm < λ < λm+1 or N ≥ 5 and λm ≤ λ < λm+1 for
some m ∈ N, then there exists a ground state solution of problem (1.1).
EJDE-2013/214
GROUND STATE SOLUTIONS
3
In section 2, we describe a variational framework to study the ground state
solution of problem (1.1). We prove Theorem 1.1 in section 3.
2. Preliminaries
Denote by E =
H01 (Ω)
the Hilbert space with the scalar product
Z
hu, vi =
∇u∇v dx
Ω
and the induced norm k · k. Let (ϕj , λj ) be the eigenfunctions and eigenvalues of
−∆ in Ω with zero Dirichlet boundary condition. Suppose that m is a fixed positive
integer and λm ≤ λ < λm+1 , we define the subspaces E − = span{ϕ1 , . . . , ϕm } and
E + = span{ϕj , j ≥ m} of E, then E = E + ⊕ E − . The functional associated to
problem (1.1) is defined by
Z
Z
Z
∗
λ
1
|u|2s
1
|∇u|2 dx −
|u|2 dx − ∗
dx
J(u) =
2 Ω
2 Ω
2s Ω |y|s
for u ∈ H01 (Ω), which is C 1 and critical points of J are solutions of problem (1.1).
To find ground state solutions of (1.1), we introduce as [12] a submanifold of E.
Define
N = {u ∈ E \ {0} : h∇J(u), ui = 0, ∇J(u) ∈ E + }.
(2.1)
The set N is the intersection of the standard Nehari manifold {u ∈ E \ {0} :
h∇J(u), ui = 0} with the pre-image (∇J)−1 (E + ).
Proposition 2.1. The set N is a C 1 submanifold of E with codimension m + 1.
Moreover, every critical point of the restriction J|N is a nontrivial critical point of
the functional J.
Proof. The result can be proved as [15], see also [13]. We sketch the proof here for
reader’s convenience. Let F : E \ {0} → R × E − be a map defined by
F (u) = (h∇J(u), ui, Q∇J(u)),
where Q is the orthogonal projection of E onto E − , then N = F −1 (0). Consider
the inner product
(t1 , z1 ) · (t2 , z2 ) = t1 t2 + hz1 , z2 i for t1 , t2 ∈ R, z1 , z2 ∈ E − .
We claim that for every (t, z) ∈ R × E − , (t, z) 6= (0, 0), the inequality
(DF (u)(tu + z)) · (t, z) < 0
(2.2)
holds. This implies the first part of the proposition. Now, we prove the claim.
Indeed, for (t, z) 6= (0, 0), since
h∇J(u), ui = h∇J(u), zi = 0,
we deduce that
(DF (u)(tu + z)) · (t, z)
Z
Z
=
|∇z|2 dx − λ
|z|2 dx
Ω
−
Z (2.3)
Ω
(2∗s − 2)t2 |u|2 + 2(2∗s − 2)tzu + (2∗s − 1)|z|2
Ω
For λm ≤ λ < λm+1 , it is readily verified that (2.2) holds.
|u|2∗s −2
|y|s
dx.
4
X. CHEN, W. CHEN
EJDE-2013/214
Next, we verify as in [15] that w ∈ E is a critical point of J if and only if u ∈ N
and DJ(u)|Tu N = 0. The proof is complete.
We recall that a ground state solution u to (1.1) is any element of N such that
DJ(u) vanishes on Tu N and J(u) = c, where
c = inf J.
(2.4)
N
By the argument in [14], for every v ∈ E + \ {0}, there is a unique continuous map
pair (f (v), g(v)) ∈ (0, ∞) × E − such that F (f (v)v + g(v)) = 0 and
J(f (v)v + g(v)) =
max
t>0,z∈E −
J(tv + z).
Hence,
c = inf J =
N
inf
v6=0,v∈E +
J(f (v)v + g(v)) =
inf
max
{v6=0,v∈E + } {t>0,z∈E − }
J(tv + z).
(2.5)
3. Existence results
In this section, we show that problem (2.4) is achieved. The minimizer of problem
(2.4) is actually a ground state solution of (1.1). Let
R
|∇u|2 dx
Ω
.
(3.1)
S = inf
R u2∗s
∗
u∈E,u6=0 (
dx)2/2s
Ω |y|s
We know from [3] that S can be achieved, which is independent of Ω and depends
only by N, k, s, moreover the infimum S is never achieved when Ω is a bounded
domain, we denote the minimizer by U (x) > 0. By (1.7),
C1
C2
≤ U (x) ≤
.
1 + |x|N −2
1 + |x|N −2
The following elementary lemma is readily verified.
Lemma 3.1. Suppose A > 0, B > 0. Then
∗
N −s
t2s
2 − s A 2−s
t2
max(A − B ∗ ) =
.
t>0
2
2s
2(N − s) B 2/2∗s
Lemma 3.2. Suppose that
c<
N −s
2−s
S 2−s ,
2(N − s)
(3.2)
then there exists v ∈ E + \ {0} such that
max
t>0,w∈E −
J(tv + w) = J(f (v)v + g(v)) = c.
Proof. Take any sequence {vn } in E + \ {0} such that kvn k = 1 and
max
t>0,w∈E −
J(tvn + w) → c.
Without loss of generality, we can assume that
vn * v
vn → v
vn → v
in E + ,
in L2 (Ω),
a.e. Ω.
(3.3)
EJDE-2013/214
GROUND STATE SOLUTIONS
5
Suppose
Z
|∇(vn − v)|2 dx ,
A = lim
n→∞
Z
B = lim
n→∞
Ω
Ω
∗
|vn − v|2s
dx .
|y|s
Using the Brezis-Lieb’s Lemma, from (3.3) we obtain
∗
1
1
J(tv + w) + At2 − ∗ Bt2s ≤ c,
2
2s
∀t > 0, ∀w ∈ E − .
(3.4)
If v = 0 and B = 0, from the assumption kvn k = 1, we deduce that A = 1.
Hence t2 ≤ 2c − 2J(w) for every t > 0 and every w ∈ E − , a contradiction.
Assume now B 6= 0. From Lemma 3.1, we obtain that
N −s
1
N −s
2 − s A 2−s
1
2−s
2
2∗
s .
Bt
S 2−s ≤
=
max
At
−
(3.5)
t>0
2(N − s)
2(N − s) B 2/2∗s
2
2∗s
If v = 0, we obtain from (3.2), (3.4) and (3.5) that
N −s
N −s
2−s
2−s
S 2−s ≤ c <
S 2−s ,
2(N − s)
2(N − s)
a contradiction. Thus v 6= 0.
Denote h = g(v)/f (v). It follows from the definition of c that
c ≤ J(f (v)(v + h)) = max J(t(v + h))
t>0
R
R
N −s
|∇(v + h)|2 dx − λ Ω |v + h|2 dx 2−s
2−s
Ω
=
.
R |v+h|2∗s 2/2∗s
2(N − s)
dx
s
|y|
Ω
By (3.4) and Lemma 3.1,
∗
1
1
c ≥ max J(t(v + h)) + At2 − ∗ Bt2s
t>0
2
2s
R
R
N −s
A + Ω |∇(v + h)|2 dx − λ Ω |v + h|2 dx 2−s
2−s
=
.
2/2∗s
R |v+h|2∗s
2(N − s)
B+
dx
s
Ω
(3.7)
|y|
Putting together (3.2), (3.5), (3.6) and (3.7), we obtain
Z
∗
2(N − s) N2−s
2/2∗s
|v + h|2s
−s
c
B+
dx
2−s
|y|s
Ω
2−s
2(N − s) N −s Z |v + h|2∗s
2/2∗s ∗
B 2/2s +
<
c
dx
2−s
|y|s
Z
Z Ω
<A+
|∇(v + h)|2 dx − λ
|v + h|2 dx
Ω
(3.6)
(3.8)
Ω
Z
∗
2/2∗s
2(N − s) N2−s
|v + h|2s
−s
c
B+
dx
,
≤
2−s
|y|s
Ω
a contradiction. Therefore, B = 0 and (3.4) yield
c ≤ J(f (v)v + g(v)) ≤ c.
The assertion follows.
6
X. CHEN, W. CHEN
EJDE-2013/214
From Lemma 3.2, we know that there exists a minimizer of problem (2.4) provided that (3.2) holds. By Proposition 2.1, such a minimizer is actually a solution
of problem (1.1). Therefore, to prove Theorem 1.1, it is sufficient to verify condition
(3.2). Choosing Bρ (0, z0 ) ⊂ Ω ⊂ BR (0, z0 ). Let ϕ ∈ C0∞ (Ω) be a cut-off function
satisfying
(
1, x ∈ B ρ2 (0, z0 )
ϕ(x) =
0, x ∈
/ Bρ (0, z0 ).
2−N
0)
For ε > 0, we define Uε (x) = ε 2 U ( x−(0,z
), uε = ϕ(x)Uε (x), Then uε ∈ E for
ε
ε > 0 small. We have following estimates for uε .
Lemma 3.3. Suppose N ≥ 3, we have
kuε k2 = kU k2 + O(εN −2 ) + O(εN −s ),
Z
Z
∗
∗
|uε |2s
|U |2s
dx
=
dx + O(εN −s ),
s
s
Ω |y|
RN |y|

2
N −2

),
N ≥ 5,
Z
Cε + O(ε
2
2
2
|uε (x)| dx ≥ Cε | ln ε| + O(ε ), N = 4,

Ω

Cε + O(ε2 ),
N = 3,
Z
uε (x)dx ≤ Cε(N −2)/2 ,
(3.9)
(3.10)
(3.11)
(3.12)
Ω
Z
Ω
∗
|uε |2s −1
dx ≤ Cε(N −2)/2 .
|y|s
(3.13)
Proof. First, we estimate (3.10). There holds
Z
Z
Z
Z
∗
∗
2∗
2∗
s
|uε |2s
|ϕUε |2s
Uε s
U
ε
2∗
dx =
dx =
dx − (1 − ϕ s ) s dx
s
s
s
|y|
|y|
Ω |y|
Ω
Ω |y|
Ω
Z
Z
Z
2∗
2∗
2∗
s
s
Uε
Uε
Uε s
2∗
s
=
dx −
dx − (1 − ϕ ) s dx
s
s
|y|
RN |y|
RN \Ω |y|
Ω
∗
Z
Z
Z
∗
2
2∗
U 2s
Uε s
Uε s
2∗
s
=
dx −
dx −
(1 − ϕ ) s dx.
s
s
|y|
RN |y|
RN \Ω |y|
Ω\B ρ (0,z0 )
2
Since
2∗
Z
RN \BR (0,z0 )
Uε s
dx ≤
|y|s
2∗
Z
RN \Ω
Uε s
dx ≤
|y|s
2∗
Z
RN \Bρ (0,z0 )
Uε s
dx,
|y|s
while
Z
RN \BR (0,z0 )
2∗
Uε s
dx =
|y|s
∗
Z
εs−N
RN \BR (0,z0 )
∗
Z
=
ε
RN \BR (0)
≤ Cεs−N
0 ) 2s
U ( x−(0,z
)
ε
dx
s
|y|
Z
s−N
U ( xε )2s
dx
|y|s
1
RN \BR (0)
2∗s 1
dx
1 + | xε |N −2
|y|s
EJDE-2013/214
GROUND STATE SOLUTIONS
Z
= CεN −s
RN \BR (0)
N −s
= O(ε
7
∗
2s 1
1
dx
εN −2 + |x|N −2
|y|s
),
and similarly,
2∗
Z
Uε s
dx = O(εN −s ).
|y|s
RN \Bρ (0,z0 )
Thus, we obtain
∗
|uε |2s
dx =
|y|s
Z
Ω
∗
Z
RN
U 2s
dx + O(εN −s ).
|y|s
That is, (3.10) holds.
Next, we estimate (3.11). In fact,
Z
Z
Z
|uε |2 dx =
ϕ2 |Uε |2 dx ≤
Ω
Ω
2
|Uε | dx
Bρ (0,z0 )
Z
x 2
=ε
U ( ) dx
ε
B (0)
Z ρ
C
≤ ε2−N
dx
x N −2 2
Bρ (0) 1 + | |
ε
Z
C
= εN −2
2 dx
Bρ (0) (εN −2 + |x|N −2 )
Z
Z
C
C
N −2
≤ εN −2
dx
+
ε
dx
2(N −2)
2(N −2)
ε
|x|
Bε (0)
Bρ (0)\Bε (0)

2
N −2

),
N ≥ 5,
Cε + O(ε
2
= Cε | ln ε| + O(ε2 ), N = 4,


Cε + O(ε2 ),
N = 3.
2−N
Now, we estimate (3.9). Observe that
Z
Z
Z
|∇uε |2 dx =
Uε2 |∇ϕ|2 dx +
∇Uε ∇(ϕ2 Uε ) dx
Ω
and −∆Uε =
Ω
Uε2∗−1 /|y|s ,
Ω
we find
Z
∇Uε ∇(ϕ2 Uε ) dx =
Z
Ω
ϕ2
Ω
Uε2∗
dx
|y|s
and
Z
2
Z
|∇uε | dx =
Ω
|∇ϕ|
2
Uε2
Z
dx +
Ω
Ω
Since ∇ϕ = 0 in Bρ (0, z0 ), we have
Z
Z
|∇ϕ|2 Uε2 dx =
|∇ϕ|2 Uε2 dx
Ω
Ω\Bρ (0,z0 )
Z
|∇ϕ|2 Uε2 dx
≤
BR (0,z0 )\Bρ (0,z0 )
Z
≤C
BR (0,z0 )\Bρ (0,z0 )
Uε2 dx
ϕ2
Uε2∗
.
|y|s
8
X. CHEN, W. CHEN
Z
EJDE-2013/214
ε2−N
≤C
BR (0)\Bρ (0)
1
(1 +
| xε |N −2 )2
dx = O(εN −2 ).
On the other hand, we can show that
Z
Z
Z
∗
2∗
s
U 2s
2 |Uε |
N −s
ϕ
dx =
dx + O(ε
)=
|∇U |2 dx + O(εN −s ).
s
|y|s
Ω
RN |y|
RN
Therefore,
Z
|∇uε |2 dx = k∇U k2 + O(εN −2 ) + O(εN −s ).
Ω
Now, we estimate (3.12).
Z
Z
uε dx =
Ω
ε
2−N
2
Bρ (0,z0 )
Z
≤C
ε
2−N
2
Bρ (0)
=ε
Z
N −2
2
Bρ (0)
≤ Cε
N −2
2
≤ Cε
x − (0, z0 )
)
ε
1
dx
1 + | xε |N −2
1
dx
εN −2 + |x|N −2
Z
1
Bε (0)
N −2
2
U(
εN −2
dx + Cε
N −2
2
Z
Bρ (0)\Bε (0)
1
dx
|x|N −2
.
Finally, there holds
Z
Z
∗
∗
|uε |2s −1
|Uε |2s −1
dx
≤
dx
|y|s
|y|s
Ω
Bρ (0,z0 )
Z
2∗s −1 dx
N +2−2s
1
2
≤ε
N −2 + |x|N −2
|y|s
Bρ (0) ε
Z
∗
2s −1 dx
N +2−2s
1
=ε 2
N
−2
N
−2
+ |x|
|y|s
Bε (0) ε
Z
2∗s −1 dx
N +2−2s
1
+ε 2
N −2 + |x|N −2
|y|s
Bρ (0)\Bε (0) ε
Z
N +2−2s
1
dx
≤ Cε(N −2)/2 + ε 2
N +2−2s
s
2
2
|y|
2
Bρ (0)\Bε (0) (|y| + |z| )
and
ε
N +2−2s
2
Z
1
Bρ (0)\Bε (0)
=ε
N +2−2s
2
+ε
≤ Cε
(|y|2 + |z|2 )
N +2−2s
2
dx
|y|s
Z
1
(Bρ (0)\Bε (0))∩{x=(y,z):|y|≥|z|} (|y|2
N +2−2s
2
+
Z
Z
+ Cε
+
dx
|z|N +2−2s |y|s
|z|2 )
1
{x=(y,z): √ε2 <|y|,|z|<ρ}
N +2−2s
2
N +2−2s
2
dx
|y|s
1
(Bρ (0)\Bε (0))∩{x=(y,z):|y|<|z|} (|y|2
N +2−2s
2
|z|2 )
Z
{x=(y,z): √ε2 <|y|,|z|<ρ}
1
dx
|y|N +2−2s |y|s
N +2−2s
2
dx
|y|s
EJDE-2013/214
GROUND STATE SOLUTIONS
9
≤ Cε(N −2)/2 ,
which implies (3.13).
Proposition 3.4. There holds
c<
N −s
2−s
S 2−s .
2(N − s)
Proof. We will check that
max
t>0,v∈E −
J(tuε + v) <
N −s
2−s
S 2−s .
2(N − s)
(3.14)
Let ω = Ω \ suppϕ. By [15, Lemma 3.3], v 7→ kvkL2∗s (ω) defines a norm on E − .
Since dim E − = m < +∞, all the norms are equivalent on E − . For every t > 0
and every v ∈ E − , by convexity we deduce
∗
|tuε (x) + v(x)|2s
dx
|y|s
Ω
Z
Z
∗
∗
|tuε (x) + v(x)|2s
|v(x)|2s
=
dx
+
dx
|y|s
|y|s
Ω\ω
ω
Z
Z
∗
∗
∗
|uε (x)|2s −1 v(x)
|uε (x)|2s
∗ 2∗
−1
2∗
s
s
dx + 2s t
dx + 2∗s C1 kvk2s .
≥t
s
s
|y|
|y|
Ω
Ω
Z
(3.15)
It follows that
Z
Z
1
J(tuε + v) ≤ J(tuε ) + t
∇uε ∇v +
|∇v|2 dx
2 Ω
Ω
Z
Z
λ
− λt
uε (x)v(x) dx −
|v(x)|2 dx
2 Ω
Ω
Z
∗
∗
∗
|uε (x)|2s −1 v(x)
− t2s −1
dx − C1 kvk2s .
s
|y|
Ω
By the assumption λm ≤ λ < λm+1 ,
Z
Z
|∇v|2 dx − λ
|v(x)|2 dx ≤ (λm − λ)kvk2 ≤ 0.
Ω
(3.16)
(3.17)
Ω
In particular, we can write
∗
∗
∗
J(tuε + z) ≤ A(t2 + tkvk + t2s −1 kvk) − B(t2s + kvk2s )
for suitable constants A > 0 and B > 0. Hence there exists R > 0 such that, for ε
small, t > R and v ∈ E − there holds J(tuε + v) ≤ 0. On the other hand, whenever
t ≤ R,
(N −2)(N −s) ∗
J(tuε +v) ≤ J(tuε )+O ε(N −2)/2 kvk−C1 kvk2s ≤ J(tuε )+O ε N +2−2s . (3.18)
10
X. CHEN, W. CHEN
EJDE-2013/214
Indeed, integrating by parts and using the definition of E − , we obtain
Z
Z
∇uε ∇v dx − λ
uε (x)v(x) dx
Ω
Ω
Z
Z
= (−∆v)uε dx − λ
uε (x)v(x) dx
Ω
Ω
Z
Z
≤ |λm − λ|
|uε (x)v(x)| dx ≤ |λm − λ|kv(·)kL∞
|uε (x)| dx
Ω
Ω
Z
≤ C|λm − λ|kvk
|uε (x)| dx
(3.19)
Ω
and
Z
Ω
∗
|uε (x)|2s −1 v dx ≤ Ckvk
|y|s
By (3.12) and (3.13) , we get
Z
|uε (x)| dx ≤ Cε(N −2)/2 ,
Ω
Z
Ω
Z
Ω
∗
|uε (x)|2s −1
dx.
|y|s
∗
|uε (x)|2s −1
dx ≤ Cε(N −2)/2 .
|y|s
By the Young inequality,
(N −2)(N −s) ∗
O ε(N −2)/2 kvk ≤ O ε N +2−2s
+ C1 kvk2s .
Therefore, together with (3.17), we see that (3.18) holds.
−s)
> 2. By Lemma 3.2, for ε > 0 small enough,
Since N ≥ 5 implies (NN−2)(N
+2−2s
J(tuε + v)
(N −2)(N −s) ≤ max J(tuε ) + O ε N +2−2s
max
t>0,v∈E −
t>0
N −s
(N −2)(N −s) kuε k2 − λkuε (x)k2L2 (Ω) 2−s
2−s
=
+
O
ε N +2−2s
∗
R
∗
(x)|2s
2(N − s)
( Ω |uε|y|
dx)2/2s
s
−s
(N −2)(N −s) N2−s
2−s ≤
+ O ε N +2−2s
S − Cλε2 + O(εN −2 )
2(N − s)
N −s
2−s
<
S 2−s .
2(N − s)
Assume now that N = 4. From (3.12) and (3.13) , we obtain
Z
Z
∗
|uε (x)|2s −1
|uε (x)| dx ≤ Cε,
dx ≤ Cε.
|y|s
Ω
Ω
By the assumption λm < λ < λm+1 ,
Z
Z
|∇v|2 dx − λ
|v(x)|2 dx ≤ (λm − λ)kvk2 = −C2 kvk2 .
Ω
Ω
Inequality (3.16), (3.19) and (3.20) imply that, for t ≤ R,
J(tuε + v) ≤ J(tuε ) + O(ε)kvk − C2 kvk2 ≤ J(tuε ) + O(ε2 ).
From Lemma 3.2, for ε > 0 small enough, we obtain
max
t>0,v∈E −
J(tuε + v)
(3.20)
EJDE-2013/214
GROUND STATE SOLUTIONS
11
4−s
kuε k2 − λkuε k2L2 (Ω) 2−s
2−s
≤
+ O(ε2 )
R |uε |2∗s
∗
2(4 − s)
2/2
s
( Ω |y|s dx)
4−s
kU k2 + O ε2 − λ Cε2 |ln ε| + O(ε2 ) 2−s
2−s
+ O(ε2 )
≤
∗
R
|U |2s
2(4 − s)
4−s ) 2/(4−s)
dx
+
O
(ε
s
RN |y|
4−s
2−s
≤
S − Cλε2 |ln ε| + O ε2 2−s + O(ε2 )
2(4 − s)
4−s
2−s
<
S 2−s .
2(4 − s)
Proof of Theorem 1.1. By Lemma 3.1 and Proposition 3.4, there exists u ∈ N such
that J(u) = c and DJ(u)|Tu N = 0. It follows from Proposition 2.1 that DJ(u) = 0
on X.
Acknowledgments. Xiaoli Chen is supported by NNSF of China (grant 11261023),
the foundation of teacher’s division of Jiangxi Province (grant GJJ12203), Startup
Foundation for Doctors of Jiangxi Normal University. WeiYang Chen is supported
by NNSF of China (grant 11271170), GAN PO 555 program of Jiangxi, NSF of
Jiangxi Province (grant 20122BAB201008)
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12
X. CHEN, W. CHEN
EJDE-2013/214
Xiaoli Chen
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: littleli chen@163.com
Weiyang Chen (corresponding author)
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: xiaowei19901207@126.com