Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 297, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SINGULAR CRITICAL ELLIPTIC PROBLEMS WITH
FRACTIONAL LAPLACIAN
XUEQIAO WANG, JIANFU YANG
Abstract. In this article, we consider the existence of solutions of the critical
problem with a Hardy term for fractional Laplacian
∗
u
(−∆)s u − µ 2s = u2s −1 in Ω,
|x|
u > 0 in Ω,
u=0
on ∂Ω,
where Ω ⊂ RN is a smooth bounded domain and 0 ∈ Ω, µ is a positive parameter, N > 2s and s ∈ (0, 1), 2∗s = N2N
is the critical exponent. (−∆)s stands
−2s
for the spectral fractional Laplacian. Assuming that Ω is non-contractible, we
show that there exists µ0 > 0 such that 0 < µ < µ0 , there exists a solution.
We also discuss a similar problem for the restricted fractional Laplacian.
1. Introduction
In this article, we consider the existence of solutions for the critical problem with
a Hardy term and fractional Laplacian
∗
u
(−∆)s u − µ 2s = u2s −1 in Ω,
|x|
(1.1)
u > 0 in Ω,
u=0
on ∂Ω
N
in a smooth bounded domain Ω ⊂ R and 0 ∈ Ω, where µ is a positive parameter,
s
N > 2s and s ∈ (0, 1), 2∗s = N2N
−2s is the critical exponent. The operator (−∆) is
the spectral Laplacian defined in section 2.
In the case s = 1, such a problem has been extensively studied, see [8, 11, 14, 16,
21, 27] etc. It is known that problem (1.1) with s = 1 has no nontrivial solutions
if µ ≥ 0 and Ω is star shaped [2]. However, the situation becomes different if
the domain Ω has nontrivial topology. In [17], a nontrivial solution was found for
problem (1.1) with s = 1 and µ = 0, if Ω is an annulus. Then it was shown in
[3] that there exists a nontrivial solution of (1.1) with s = 1 and µ = 0, if Ω has
nontrivial topology. If µ > 0, there is a Hardy term in (1.1) with s = 1. In [15],
it proved that problem (1.1) with s = 1 admits a solution in a non-contractible
2010 Mathematics Subject Classification. 35J20, 35J25, 35J61.
Key words and phrases. Fractional Laplacian; singular critical problem;
non-contractible domain.
c 2015 Texas State University.
Submitted September 6, 2015. Published December 3, 2015.
1
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X. WANG, J. YANG
EJDE-2015/297
domain. Since (1.1) with s = 1 is a critical problem, it involves the ground state
solution of the problem in the whole space
∗
u
−∆u − µ 2 = u2 −1 in RN ,
|x|
(1.2)
u > 0 in RN .
The ground state solutions of (1.2) are found in [25] for µ = 0 and in [27] for µ 6= 0.
Recently, Secchi et al [22] proved that Coron type problem admits a solution
for problem (1.1) with µ = 0 and the restricted fractional Laplacian; see section 2
for a definition. Similarly, the argument in [22] relies on, among other things, the
explicit form of the minimizer of the problem
R
|(−∆)s/2 u(x)|2 dx
RN
(1.3)
Λs =
inf
2/2∗s ,
R
∗
u∈Ḣ s (RN ),u6≡0
|u(x)|2s dx
RN
where the space Ḣ s (RN ) is defined as the completion of C0∞ (RN ) under the norm
Z
2
kukḢ s (RN ) =
|ξ|2s |û(ξ)|2 dξ,
(1.4)
RN
here û denotes the Fourier transform of u. In RN , the operator (−∆)s/2 , s ∈ R is
defined by the Fourier transform
\
s/2 u)(ξ) = |ξ|s û(ξ)
((−∆)
for u ∈ C0∞ (RN ). Therefore, for s > 0, we have
Z
k(−∆)s/2 uk2L2 (RN ) =
|ξ|2s |û(ξ)|2 dξ.
(1.5)
(1.6)
RN
For N > 2s, the minimizing problem Λs in (1.3) is related to the fractional
2N
Sobolev embedding Ḣ s (RN ) ,→ L N −2s (RN ). The continuity of this inclusion corresponds to the inequality
2
kuk2L2∗s (RN ) ≤ Λ−1
s kukḢ s (RN ) .
(1.7)
The best constant Λs in (1.7) was computed in [10]. A minimizer u of Λs weakly
solves the problem
∗
(−∆)s u = |u|2s,α −2 u in RN
(1.8)
up to a multiplying constant. Using the moving plane method for integral equations,
Chen et al [9] classified the solutions of an integral equation, which is related to
problem (1.8). Positive regular solutions of (1.8), and then the minimizers of Λs
are precisely given by
N −2s
ε
2
Uε (x) = 2
(1.9)
ε + |x − x0 |2
for ε > 0 and x0 ∈ RN .
In this paper, we consider the existence of solutions of problem (1.1) with 0 <
µ < µH and s ∈ (0, 1) in a non-contractible domain Ω, where µH is the best constant
in the Hardy inequality. Problem (1.1) is related to the variational problem
R
2
k(−∆)s/2 uk2L2 (RN ) − µ RN |u(x)|
|x|2s dx
Λs,µ =
inf
.
(1.10)
2∗
R
2∗
u∈Ḣ s (RN ),u6≡0
s dx 2s
|u(x)|
N
R
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SINGULAR CRITICAL ELLIPTIC PROBLEMS
3
Although minimizers of Λ1,µ were found explicitly in [27] for s = 1, it is not the
case for s ∈ (0, 1). So in our argument, we need to avoid using it. Our main result
for the spectral Laplacian is as follows.
Theorem 1.1. Suppose Ω is not contractible. Then, there exists 0 < µ0 < µH
such that for each µ ∈ (0, µ0 ), there exists a solution of (Ps,µ ).
For the restricted fractional Laplacian (−∆|Ω )s , we consider the problem
∗
u
(−∆|Ω )s u − µ 2s = u2s −1 in Ω,
|x|
(1.11)
u > 0 in Ω,
u=0
on RN \ Ω.
Similarly, we have the following result.
Theorem 1.2. Suppose Ω is not contractible. Then, there exists 0 < µ0 < µH
such that for each µ ∈ (0, µ0 ), there exists a solution of (1.11).
The article is organized as follows. After some preparations in section 2, we
prove Theorems 1.1 and 1.2 in section 3.
2. Sobolev-Hardy inequality
In this section, we develop some properties of minimizers of Λs,µ , and give the
definition of fractional operator (−∆)s . First, we define for each s ≥ 0, the fractional Sobolev space
H s (RN ) = {u ∈ L2 (RN ) : |ξ|s û(ξ) ∈ L2 (RN )}
via the Fourier transform
1
û(ξ) =
(2π)N/2
Z
e−ix·ξ u(x) dx.
RN
For s ∈ (0, 1), it is known from [19] that there holds
Z
Z Z
|u(x) − u(y)|2
|ξ|2s |û(ξ)|2 dξ = Cs,N
dx dy,
N +2s
RN
RN RN |x − y|
where Cs,N is a positive constant. This provides an alternative norm
Z Z |u(x) − u(y)|2
1/2
kukḢ s (RN ) =
dx
dy
N +2s
RN RN |x − y|
(2.1)
on Ḣ s (RN ). If N > 2s, the optimal constant µs,N was found in [28] for the
fractional Hardy inequality
Z
Z
|u(x)|2
µH
dx ≤
|ξ|2s |û(ξ)|2 dξ,
(2.2)
2s
RN |x|
RN
where u ∈ C0∞ (RN ). By a denseness argument, we have
Z
Z
|u(x)|2
µH
dx
≤
|(−∆)s/2 u(x)|2 dx
2s
RN |x|
RN
Γ2 ( N +2s )
4
. If 0 < µ < µH , we may verify that
for u ∈ Ḣ s (RN ), where µH = 4s Γ2 ( N −2s
)
4
Z
|u(x)|2 1/2
s/2
2
|u|Ḣ s (RN ) := k(−∆) ukL2 (RN ) − µ
dx
2s
RN |x|
(2.3)
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X. WANG, J. YANG
EJDE-2015/297
defines an equivalent norm on Ḣ s (RN ). Therefore, there exists C > 0 such that for
any u ∈ Ḣ s (RN ),
Z
Z
hZ
22∗
∗
|u(x)|2 i
dx ,
(2.4)
|u(x)|2s dx s ≤ C
|(−∆)s/2 u(x)|2 dx − µ
2s
RN
RN
RN |x|
where 2∗s =
2N
N −2s .
Define
Λs,µ =
|u|2Ḣ s (RN )
inf
R
u∈Ḣ s (RN ),u6≡0
RN
2 .
∗
|u(x)|2s dx 2∗s
(2.5)
It is proved in [13] that Λs,µ > 0 is achieved if µ ≥ 0.
We remark that any minimizer of Λs,µ does not change sign, and is radially
symmetric. Indeed, let u be a minimizer. By formula (A.11) in [23],
Z
Z
s/2
2
|(−∆) |u|| dx ≤
|(−∆)s/2 u|2 dx.
RN
RN
Hence, |u| is also a minimizer, and we have u > 0. Denote by u∗ the symmetricdecreasing rearrangement of u. By strict rearrangement inequalities in [12], u∗ is
also a minimizer of Λs,µ . Therefore, by strict rearrangement inequalities again,
u(x) = u∗ (x − A) for some A ∈ RN . Moreover, any minimizer u of Λs,µ weakly
solves (1.1) with Ω = RN up to multiplying a constant. In the case Ω = RN , since
both u(x) and u∗ (x) solve equation (1.1) , and (1.1) is not translation invariant, we
obtain that A = 0, that is u = u∗ .
Next, we define fractional Laplacians in a bounded domain. There are two
types of fractional Laplacians in bounded domainds, one is the spectral fractional
Laplacian, another one is the restricted fractional Laplacian.
In a bounded domain Ω ⊂ RN , we define the spectral fractional Laplacian (−∆)s
as follows. Let {λk , ϕk }∞
k=1 be the eigenvalues and corresponding eigenfunctions
of the Laplacian operator −∆ in Ω with zero Dirichlet boundary values on ∂Ω
normalized by kϕk kL2 (Ω) = 1, i.e.
−∆ϕk = λk ϕk
in Ω;
ϕk = 0
on ∂Ω.
2
For any u ∈ L (Ω), we may write
u=
∞
X
Z
uk ϕk ,
where
uk =
uϕk dx.
Ω
k=1
We define the space
H = {u =
∞
X
2
uk ϕk ∈ L (Ω) :
k=1
∞
X
2
λ2s
k uk < ∞},
(2.6)
k=1
which is equipped with the norm
kukH =
∞
X
2
λ2s
k uk
1/2
.
k=1
For any u ∈ H, the spectral fractional Laplacian (−∆)s , is defined by
(−∆)s u =
∞
X
k=1
λsk uk ϕk .
(2.7)
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SINGULAR CRITICAL ELLIPTIC PROBLEMS
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The space H defined in (2.6) is the interpolation space (H02 (Ω), L2 (Ω))s,2 , see
[1, 18, 26]. It was shown in [18] that (H02 (Ω), L2 (Ω))s,2 = H0s (Ω) if 0 < s < 1 and
1/2
s 6= 1/2; while (H02 (Ω), L2 (Ω)) 12 ,2 = H00 (Ω), where
Z 2
u (x)
1/2
H00 (Ω) = {u ∈ H 1/2 (Ω) :
dx < ∞},
Ω d(x)
and d(x) = dist(x, ∂Ω) for all x ∈ Ω.
Now, using the idea in [7], for any u ∈ H0s (Ω), we may define the extension
1
w = Es (u) of u as the solution w ∈ H0,L
(CΩ ) of the problem
− div(y 1−2s ∇w) = 0,
CΩ = Ω × (0, ∞),
∂L CΩ = ∂Ω × (0, ∞),
w = 0,
(2.8)
Ω × {0},
w = u,
where
Z
1
H0,L
(CΩ ) = w ∈ L2 (CΩ ) : w = 0 on ∂L CΩ , κs
y 1−2s |∇w|2 dx dy < ∞
CΩ
is a Hilbert space with the norm
kwk2H 1 (CΩ )
0,L
Z
y 1−2s |∇w|2 dx dy.
= κs
CΩ
1
(CΩ ). That is
The extension operator E is an isometry between H0s (Ω) and H0,L
1 (C ) = kukH s (Ω) .
kE(u)kH0,L
Ω
0
(2.9)
It was shown in [7], see also [5], that
− κs lim+ y 1−2s
y→0
∂w
= (−∆)s u,
∂y
(2.10)
where (−∆)s is the spectral fractional Laplacian. We remark that if Ω = RN , for
any u ∈ H s (RN ), the extension w = E(u) of u is given by
− div(y 1−2s ∇w) = 0,
+1
,
RN
+
RN ,
w = u,
(2.11)
then by [7], problem (2.11) corresponds to the fractional Laplacian given by the
Fourier transform in (1.5), which also satisfies (2.10). In this sense, the fractional
Laplacian given in (1.5) can be approached by the spectral fractional Laplacian
through extending the domain Ω in (2.8) to RN . In the sequel, we use the same
notation (−∆)s to denote these two operators.
Using this sort of extension, we may reformulate the nonlocal problem (1.1) in
a local way; that is,
− div(y 1−2s ∇w) = 0,
w = 0,
κs y
where
∂
∂ν
CΩ ,
∂L CΩ ,
1−2s ∂w
∗
u
= µ 2s + u2s −1
∂ν
|x|
is the outward normal derivative.
(2.12)
Ω × {0},
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X. WANG, J. YANG
EJDE-2015/297
The other type of fractional Laplacian is the restricted fractional Laplacian defined by the following formula:
Z
u(x) − u(y)
s
(−∆|Ω ) u(x) = cN,s P.V.
dy
|y
− x|N +2s
N
R
for all functions u which are zero outside Ω. Similarly, the extension problem is
given as follows.
div(y 1−2α ∇w) = 0 in RN +1 ,
w=0
y
in RN \ Ω,
∗
u
= µ 2s + u2s −1
∂ν
|x|
1−2α ∂w
(2.13)
on Ω.
3. Existence of solutions
In this section, we first show that problem (1.1) admits a nontrivial solution in
a non-contractible domain. Define the functional
Z
Z
∗
1
u2 1
Iµ (u) =
|(−∆)s/2 u|2 − µ 2s dx − ∗
|u|2s dx
2 Ω
|x|
2s Ω
on H0s (Ω), where (−∆)s is the spectral factional Laplacian defined in (2.7), and the
functional
Z
Z
∗
1
u2 1
Jµ (u) =
|(−∆)s/2 u|2 − µ 2s dx − ∗
|u|2s dx
2 RN
|x|
2s RN
on Ḣ s (RN ), where (−∆)s is given in (1.5). It is discussed in section 2 the relation
between the spectral factional Laplacian and the fractional Laplacian defined by
the Fourier transform.
We consider the minimizing problem
cµ = inf Jµ (u) : u ∈ Nµ,RN ,
where Nµ,Ω denotes the Nehari manifold
Z
Z
∗
u2 Nµ,Ω = u ∈ H0s (Ω) \ {0} :
|(−∆)s/2 u|2 − µ 2s dx =
|u|2s dx .
|x|
Ω
Ω
Obviously, we have
cµ = inf
s
N
Z
|(−∆)s/2 u|2 − µ
RN
u2 dx : u ∈ Nµ,RN
|x|2s
N −2s
We may verify that cµ = Ns Λs,µ . If u is a minimizer of Λs,µ , then v = Λs,µ4s u is
a minimizer of cµ , and vice verse. Since minimizers of Λs,µ are radially symmetric,
so are minimizers of cµ . The following result is proved in [20].
Lemma 3.1. Let 0 < s <
N
2
and {un } ⊂ Ḣ s (RN ) be a bounded sequence such that
inf kun kL2∗s ≥ C > 0.
n∈N
Then up to subsequence, there exist {xn } ⊂ RN , λn ∈ (0, ∞) such that
vn * v 6≡ 0
N −2s
2
where vn (x) := λn
un (xn + λn x).
in Ḣ s (RN ),
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SINGULAR CRITICAL ELLIPTIC PROBLEMS
7
Using Lemma 3.1, we have a description of (P S)c sequences. By a (P S)c sequences for Iµ we mean a sequence {un } ⊂ H0s (Ω) such that Iµ (un ) → c and
Iµ0 (un ) → 0.
Proposition 3.2. Let µ ∈ (0, µH ) and {un } ⊂ Nµ,Ω be a sequence such that for
0 < c < c0 ,
lim Iµ0 (un ) = 0.
lim Iµ (un ) = c,
n→∞
n→∞
(3.1)
Suppose problem (1.1) has no nontrivial solutions. Then, there exist {λn } ⊂ RN
+}
and {xn } ⊂ RN such that
lim λn = 0,
n→∞
lim xn = 0
n→∞
and
lim kun − uµλn ,xn kḢ s = 0,
n→∞
where uµ is a minimizer of cµ .
Proof. By (3.1) and the Hardy inequality, we know that {un } is bounded in H0s (Ω),
and
cN
> 0.
(3.2)
lim kun kL2∗s (Ω) =
n→∞
s
Therefore, we have
∗
un → u in L2s −1 (Ω),
un * u in H0s (Ω),
un → u a.e. Ω.
By the assumption that problem (1.1) does not have nontrivial solution, we have
u = 0. Extend un to be zero outside Ω, then the extension of un belongs to H s (RN ),
see [1, Theorem 7.40]. By (3.2) and Lemma 3.1, there exist {xn } ⊂ RN , λn ∈ (0, ∞)
such that
vn * v 6≡ 0
in Ḣ s (RN ),
N −2s
where vn (x) := λn 2 un (xn + λn x), and un has been extended to RN by setting
un = 0 outside Ω. We also have vn ∈ H0s (Ωn ), where Ωn = {x ∈ RN : xn + λn x ∈
Ω}. Moreover, vn satisfies
(−∆)s vn − µ
λ2s
2∗ −1
n vn
− vns → 0
2s
|xn + λn x|
(3.3)
and
Z
2
∗
λ2s
1
n vn
|(−∆) vn | − µ
dx − ∗
|vn |2s dx
2s
|x
+
λ
x|
2
n
n
Ωn
s Ωn
Z
Z
2 ∗
u
1
1
|un |2s dx → c
=
|(−∆)s/2 un |2 − µ n2s dx − ∗
2 Ω
|x|
2s Ω
1
2
Z
s/2
2
(3.4)
as n → ∞.
We may assume that λn → λ0 ≥ 0. If λ0 > 0, since un * 0 in Ḣ s (RN ), we have
vn * 0 in Ḣ s (RN ), which is a contradiction.
We may assume, up to a sequence, λxnn → x0 ∈ RN or | λxnn | → ∞ as n → ∞. If
xn
| λn | → ∞ as n → ∞, by (3.2), we see that the limit function v satisfies
∗
(−∆)s v = v 2s −1
in RN .
(3.5)
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X. WANG, J. YANG
EJDE-2015/297
Thus, equations (3.3) and (3.4) imply
Z
n1 Z
o
2
∗
λ2s
1
n vn
c0 > c = lim
|(−∆)s/2 vn |2 − µ
dx
−
|vn |2s dx
2s
∗
n→∞ 2 Ω
|xn + λn x|
2s Ωn
n
Z
2s 2
s
λn vn
=
lim
|(−∆)s/2 vn |2 − µ
dx
N n→∞ Ωn
|xn + λn x|2s
Z
s
|(−∆)s/2 v|2 dx ≥ c0 ,
≥
N Ωn
which is impossible.
So we have λxnn → x0 ∈ RN , which yields limn→∞ |xn | = 0. It follows that v
satisfies
∗
v
(−∆)s v − µ
= v 2s −1 in RN .
2s
|x0 + x|
By the translation, we may assume x0 = 0. Indeed, let ṽn (x) = vn (x −
ṽn * ṽ in Ḣ s (RN ). By (3.3), ṽn satisfies
(−∆)s ṽn − µ
ṽn
2∗ −1
− ṽns → 0
2s
|x|
xn
λn ).
Then
(3.6)
as n → ∞ and ṽ satisfies
(−∆)s ṽ − µ
∗
ṽ
= ṽ 2s −1
2s
|x|
in RN .
Now, we prove that
Z
|(−∆)s/2 (ṽn − ṽ)|2 dx = 0.
lim
n→0
RN
Suppose on the contrary that
Z
lim
n→∞
|(−∆)s/2 (ṽn − ṽ)|2 dx > 0.
RN
Noting
Z
|(−∆)
lim
n→∞
s/2
Z
2
(ṽn − ṽ)| dx +
RN
Z
= lim
n→∞
|(−∆)s/2 ṽ)|2 dx
RN
|(−∆)s/2 ṽn |2 dx,
RN
and by the Brézis-Lieb lemma [6],
∗
∗
∗
lim kṽn − ṽk2s + kṽk2s = lim kṽn k2s ,
n→∞
Z
Z
|ṽn − ṽ|2
|ṽ|2
|ṽn |2
dx
+
dx
=
lim
dx,
2s
n→∞ RN |x|2s
|x|2s
RN |x|
n→∞
Z
lim
n→∞
RN
we find from (3.6) that
Z
∗
|ṽn − ṽ|2
lim
|(−∆)s/2 (ṽn − ṽ)|2 − µ
− |ṽn − ṽ|2s dx = 0.
n→∞ RN
|x|2s
(3.7)
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SINGULAR CRITICAL ELLIPTIC PROBLEMS
9
That is, ṽn −ṽ is close to the manifold Nµ,RN . Let tn be such that tn (ṽn −ṽ) ∈ Nµ,RN .
By (3.7), we can show that tn → 1 as n → ∞. Hence, Jµ (ṽn −ṽ)−Jµ (tn (ṽn −ṽ)) → 0
as n → ∞. But Jµ (tn (ṽn − ṽ)) ≥ cµ , it yields
lim inf Jµ (ṽn − ṽ) ≥ cµ ,
n→∞
and then
lim inf Jµ (ṽn ) ≥ lim inf Jµ (ṽn − ṽ) + Jµ (ṽ) ≥ 2cµ .
n→∞
n→∞
This is a contradiction. Consequently, ṽn → ṽ strongly in Ḣ s (RN ) and Jµ (ṽ) = cµ .
The proof is complete.
In the case µ = 0, we have the following result, its proof is similar to that of
Proposition 3.2.
Proposition 3.3. Let{un } ⊂ N0,Ω be a sequence such that
lim I(un ) ≤ c0 ,
n→∞
lim I 0 (un ) = 0.
(3.8)
n→∞
N
such that
Then, there exist {λn } ⊂ RN
+ } and {xn } ⊂ R
lim λn = 0,
n→∞
lim xn = 0,
n→∞
lim kun − u0λn ,xn kḢ s = 0,
n→∞
where u0 is a minimizer of c0 .
For each set A ⊂ RN and each point x ∈ RN , d(x, A) denotes the distance
between x and A. For each d > 0, we denote Ωd = {x ∈ RN : d(x, Ω) < d} and
Ωid = {x ∈ Ω : d(x, ∂Ω) > d}. For subsets A, B ⊂ RN , A ∼
= B stands for that A
and B are homotopy equivalent.
Now, we choose d > 0 so that Ωd ∼
= Ω. Let
R
x|(−∆)s/2 u|2 dx
for u ∈ H0s (Ω) \ {0}.
β(u) = RΩ
s/2 u|2 dx
|(−∆)
Ω
Lemma 3.4. There exists µ0 ∈ (0, µH ) such that for each µ ∈ (0, µ0 ) and u ∈ Nµ,Ω
with I(u) < c0 , β(u) ∈ Ωd .
Proof. Suppose on the contrary that there exist µn ∈ R+ and un ∈ Nµ,Ω such that
limn→∞ µn = 0, Iµn (un ) < c0 and β(un ) 6∈ Ωd for all n ≥ 1. We may assume that
β(un ) → x0 ∈ RN \ Ωd . Since µn → 0, by the Hardy inequality,
Z
u2n
dx → 0
µn
2s
Ω |x|
as n → ∞. Hence,
Z
|(−∆)
lim
n→∞
s/2
Z
2
un | dx = lim
n→∞
Ω
∗
|un |2s dx,
Ω
lim Iµn (un ) ≤ c0 .
n→∞
By Proposition 3.3, there exist sequences {λn } and {xn } ⊂ RN such that
lim λn = 0
n→∞
and
lim kun − u0λn ,xn kḢ s = 0.
n→∞
By the assumption, we have limn→∞ xn = x0 . However, x0 6∈ Ωd , we have that
limn→∞ kun − u0λn ,xn kḢ s 6= 0. This is a contradiction. The assertion follows.
10
X. WANG, J. YANG
EJDE-2015/297
Now, we choose d1 > 0 such that Ω ∼
= Ωid1 . Let
µ
λ = inf{ 2s : x ∈ Ωd }.
|x|
Let ξ ∈ C ∞ (R+ ) be such that ξ(t) = 1 for t ∈ [0, d21 ] and ξ(t) = 0 for t ∈ [d1 , ∞).
For each (ε, z) ∈ R+ × RN , we define
wε,z (x) = τε,z ξ(x − z)Uε (x − z)
N
for x ∈ R , where Uε is given in (1.9) and τε,z is a positive constant such that wε,z
satisfying
Z
Z
∗
2
dx =
|(−∆)s/2 wε,z |2 − λwε,z
|wε,z |2s dx.
Ω
Ω
It is proved in [4] that if 0 ∈ Ω,
kξUε k2H0s (Ω) = kUε k2H0s (Ω) + O(εN −2s );
(
Cε2s + O(ε2s ),
if N > 4s,
2
kξUε kL2 (Ω) =
2s
2s
−Cε log ε + O(ε ), if N = 4s,
2∗ −1
kξUε kLs2∗s −1 (Ω) ≥ Cε
N −2s
2
Let
if N > 2s.
Z
1
|(−∆)s/2 u|2 − λu2 dx −
2 Ω
Then, we may verify that
(
c0 − λCε2s + O(ε2s ),
Q(wε,z ) =
c0 + Cε2s logε + O(ε2s ),
Q(u) =
1
2∗s
Z
∗
|u|2s dx.
Ω
if N > 4s,
if N = 4s,
for all z ∈ Ωid1 , it implies that for all z ∈ Ωid1 ,
Q(wε,z ) < c0
(3.9)
if ε > 0 small enough.
Lemma 3.5. Let µ ∈ (0, µ0 ). Then for ε > 0 small, there holds
sup{I(twε,z ,µ wε,z ) : z ∈ Ωid1 } < c0 .
Proof. Let t = twε,z . Since |x|µ2s > λ for x ∈ Ω,
Z
Z
2 wε,z
t2
t2
2
|(−∆)s/2 wε,z |2 − λwε,z
dx >
|(−∆)s/2 wε,z |2 − µ 2s dx
2 Ω
2 Ω
|x|
Z
2∗
∗
s
t
2s
=
wε,z
dx
2 Ω
∗ Z
t2s
2
=
|(−∆)s/2 wε,z |2 − λwε,z
dx.
2 Ω
Therefore, t < 1. It results from (3.9) that
Z
2 wε,z
st2
Iµ (twε,z ) =
|(−∆)s/2 wε,z |2 − µ 2s dx
N Ω
|x|
Z
s
2
≤
|(−∆)s/2 wε,z |2 − λwε,z
dx < c0 .
N Ω
The assertion follows.
EJDE-2015/297
SINGULAR CRITICAL ELLIPTIC PROBLEMS
11
Proof of Theorem 1.1. We argue by contradiction. For fixed µ ∈ (0, µ0 ), assume
problem (1.1) has no solutions. Hence, we know from [24] that there exists a pseudogradient flow η : [0, ∞) × Nµ,Ω → Nµ,Ω associated with Iµ , such that the function
η satisfies for s, t ∈ R+ with s > t and u ∈ Nµ,Ω that
Iµ (η(s, u)) < Iµ (η(t, u)),
lim Iµ (η(t, u)) > −∞ implies that
t→∞
lim Iµ0 (η(t, u)) = 0.
t→∞
For ε given in Lemma 3.5, we set ω = {twε,z ,µ wε,z : z ∈ ωdi 1 }. By the definition of wε,z , we have wε,z ∈ H0s (Ω) for each z ∈ ω. Therefore, ω ⊂ Nµ,Ω , and
sup{Iµ (η(t, u) : t ≥ 0)} is bounded from below for each u ∈ ω. By Proposition 3.2,
there exist {(εt , zt } ⊂ R+ × Ω such that limt→∞ zt = 0 and
lim kη(t, u) − uµεt ,zt kḢ s = 0.
t→∞
Since uµ is radially symmetric, we have
lim β(η(t, u)) = 0 ∈ Ω
t→∞
for all u ∈ ω. On the other hand, by Lemma 3.4,
{β(η(t, u)) : u ∈ ω} ⊂ Ωd .
Since {β(η(0, u)) : u ∈ ω} = Ωid1 , we see that Ωid1 is contractible in Ωd . This
contradicts to the assumption that Ωid1 ∼
=Ω∼
= Ωd and that Ω is not contractible.
This shows that problem (1.1) possesses a positive solution in Nµ,Ω .
Proof of Theorem 1.2. We remark that Lemma 3.1 can be applied in this case. The
proof of Theorem 1.2 is similar to that of Theorem 1.1 with minor changes, we omit
the details.
Acknowledgments. This work was supported by the NNSF of China: 11271170
and 11371254, and by the GAN PO 555 program of Jiangxi.
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Xueqiao Wang
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: wangxueqiao1989@126.com
Jianfu Yang
Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022,
China
E-mail address: jfyang 2000@yahoo.com