ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
FRACTIONAL ELLIPTIC PROBLEMS WITH TWO CRITICAL SOBOLEV-HARDY EXPONENTS
WENJING CHEN
Communicated by Giovanni Molica Bisci
Abstract. By using the mountain pass lemma and a concentration compact- ness principle, we obtain the existence of positive solutions to the fractional elliptic problem with two critical Hardy-Sobolev exponents at the origin.
1. Introduction
In this article, we study the following doubly critical problem involving the frac- tional Laplacian
(−∆)su−γ u
|x|2s =|u|2∗s(α)−2u
|x|α +|u|2∗s(β)−2u
|x|β , u >0, in Rn, (1.1) wheres∈(0,1), 0< α, β <2s < nwithα6=β,γ < γH with
γH= 4sΓ2(n+2s4 ) Γ2(n−2s4 )
being the fractional best Hardy constant onRn, and 2∗s(α) = 2(n−α)/(n−2s) is the fractional critical Hardy-Sobolev exponent. The operator (−∆)s is the fractional Laplacian defined as
(−∆)su(x) =cn,spv Z
Rn
u(x)−u(y)
|x−y|n+2sdy, s∈(0,1), where pv stands for the Cauchy principle value and
cn,s= 22s−1π−n2Γ n+2s2
|Γ(−s)|
is the normalization constant so that the identity
(−∆)su=F−1(|ξ|2s(Fu)) ∀ξ∈Rn, s∈(0,1), u∈ S(Rn), holds, here Fu denotes the Fourier transform of u, Fu(ξ) = R
Rne−2πix·ξu(x)dx, andS(Rn) the Schwartz class, see [14] and references therein for the basics on the fractional Laplacian.
2010Mathematics Subject Classification. 35J20, 35J60, 47G20.
Key words and phrases. Fractional elliptic problems; mountain pass lemma;
critical fractional Hardy-Sobolev exponent; concentration compactness principle.
c
2018 Texas State University.
Submitted July 1, 2017. Published January 15, 2018.
1
In previous twenty years, the nonlocal elliptic problems have been investigated by many researchers, for example, [18, 27, 29, 30, 31] for the subcritical case, [3, 8, 23, 19, 28, 32, 33] for the critical case, [9, 10, 11] for the existence of solutions to fractional Laplacian system. Moreover, a great attention has been devoted to study the existence of solutions for the nonlocal problems with Hardy potential or nonlinearity term, we refer to see [1, 2, 4, 13, 15, 16, 34, 35, 36] and the references therein. In particular, the existence of solutions to the problem
(−∆)su−γ u
|x|2s = u2∗s(α)−1
|x|α , u >0 inRn, (1.2) corresponds to the minimization problem
µs,γ,α(Rn) = inf
u∈Hs(Rn)\{0}
R
Rn|(−∆)s/2u|2dx−γR
Rn
|u|2
|x|2sdx R
Rn
|u|2∗s(α)
|x|α dx2∗2 s(α)
. (1.3) Fall et al. [16] proved the existence of extremals for µs,0,α(Rn) in the case s= 12. Yang [35] proved that there exists a positive, radially symmetric and non-increasing extremal for µs,0,α(Rn) when s ∈ (0,1). Asymptotic properties of the positive solutions was given by Lei [24] and Yang-Yu [37]. The existence of extremals for µs,γ,α(Rn) in (1.3), when α∈ [0,2s) and γ ∈ (−∞, γH), was recently studied by Ghoussoub and Shakerian in [21]. Moreover, the authors in [21] used the mountain pass lemma to establish the existence of a nontrivial weak solution to the problem
(−∆)su−γ u
|x|2s =|u|2∗s−2u+|u|2∗s(α)−2u
|x|α , u >0, inRn.
Furthermore, the authors in [36] showed the existence of nontrivial solutions for fractional elliptic problem inRnwith the critical nonlocal Hartree term and critical fractional Hardy-Sobolev term.
It is worth pointing out that in the local case, i.e. s = 1, the existence and multiplicity of solutions for the Laplacian problems with Hardy terms have been extensively studied, we refer the reader to [5, 7, 12, 17, 22] and references therein.
The aim of this paper is to consider the existence of nontrivial weak solutions of (1.1), which has a single pole with different powers of singularity and fractional critical Hardy-Sobolev exponents. We get the existence of nontrivial weak solutions of our problem by the Mountain Pass Lemma with concentration-compactness prin- ciple. Our result can be stated as follows.
Theorem 1.1. Let 0< s <1,0 < α, β <2s < nwith α6=β, and γ < γH. Then problem (1.1)admits a nontrivial solution.
This article is organized as follows: in Section 2, we give some preliminaries about fractional Laplacian harmonic extension and function space, and also the fractional Hardy-Sobolev inequality. We prove the compactness of the energy in Section 3. Section 4 is concerned with the proof of our main result.
2. Preliminary results
In this section, we first introduce suitable function spaces for the variational principles that will be needed in the sequel. Caffarelli and Silvestre in [6] showed that the fractional Laplacian operator can be realized in a local way by using one more variable and the so-called s−harmonic extension, that is, for a function
u ∈ Hs(Rn), we say that U = Es(u) is its s-harmonic extension to the upper half-space,Rn+1+ , i.e. it is a solution to the problem
div(y1−2s∇U) = 0 in Rn+1+ , U=u onRn× {y= 0}.
Define the spaceXs(Rn+1+ ) as the closure ofC0∞(Rn+1+ ) with the norm kUkXs(Rn+1+ ):=
ks Z
Rn+1+
y1−2s|∇U(x, y)|2dx dy1/2 ,
where ks = 21−2sΓ(s)Γ(1−s) is a normalization constant chosen in such a way that the extension operator U : Hs(Rn)→Xs(Rn+1+ ) is an isometry, that is, for any u∈Hs(Rn), we have
kUkXs(Rn+1+ )=kukHs(Rn)=k(−∆)s/2ukL2(Rn). (2.1) Conversely, for a function U ∈Xs(Rn+1+ ), we denote its trace onRn× {y= 0} as u= Tr(U) :=U(·,0). This trace operator is also well defined and satisfies
kukHs(Rn)=kU(·,0)kHs(Rn)≤ kUkXs(Rn+1+ ). (2.2) Caffarelli and Silvestre [6] showed that the extension functionU :=Es(u) is related to the fractional Laplacian of the original functionuin the following way:
(−∆)su(x) = ∂U
∂νs :=−ks lim
y→0+y1−2s∂U
∂y(x, y).
Thus, problem (1.1) can be written as the local problem
−div(y1−2s∇U) = 0 inRn+1+
∂U
∂νs =γ u
|x|2s +|u|2∗s(α)−2u
|x|α +|u|2∗s(β)−2u
|x|β onRn,
(2.3)
where and in the follows u = U(·,0). A function U ∈ Xs(Rn+1+ ) is said to be a weak solution to (2.3), if for all Ψ∈Xs(Rn+1+ ),
ks
Z
Rn+1+
y1−2sh∇U,∇Ψidx dy= Z
Rn
γ u
|x|2sψ dx+ Z
Rn
|u|2∗s(α)−2u
|x|α ψ dx +
Z
Rn
|u|2∗s(β)−2u
|x|β ψ dx, whereψ= Ψ(·,0). The energy functional corresponding to (2.3) is
J(U) =1 2kUk2Xs(
Rn+1+ )−γ 2
Z
Rn
|u|2
|x|2sdx− 1 2∗s(α)
Z
Rn
|u|2∗s(α)
|x|s dx
− 1
2∗s(β) Z
Rn
|u|2∗s(β)
|x|s dx.
We note that for any weak solutionU ∈Xs(Rn+1+ ) to (2.3), the functionu=U(·,0) is inHs(Rn) and is a weak solution to problem (1.1). Hence the associated trace of any critical pointU ofJ inXs(Rn+1+ ) is a weak solution for (1.1). Let us recall the following results.
Lemma 2.1. Assume that0< s <1.
(i) (The fractional Hardy inequality[20]) For all u∈Hs(Rn), we have γH
Z
Rn
|u|2
|x|2sdx≤ Z
Rn
|(−∆)s/2u|2dx, (2.4) whereγH= 4sΓ
2(n+2s4 )
Γ2(n−2s4 ) is the best constant in the above inequality onRn.
(ii) (The fractional Hardy-Sobolev inequality [21]) Assume 0 ≤ α ≤ 2s < n.
Then, there exist positive constantsc andC, such that for allu∈Hs(Rn), Z
Rn
|u|2∗s(α)
|x|α dx2∗2 s(α)
≤c Z
Rn
|(−∆)s/2u|2dx. (2.5) Moreover, if γ < γH, then
CZ
Rn
|u|2∗s(α)
|x|α dx2∗2 s(α)
≤ Z
Rn
|(−∆)s/2u|2dx−γ Z
Rn
|u|2
|x|2sdx, (2.6) for allu∈Hs(Rn).
Remark 2.2. One can use (2.1) to rewrite inequalities (2.4), (2.5) and (2.6) as the following trace class inequalities:
γH
Z
Rn
|u|2
|x|2sdx≤ kUk2Xs(
Rn+1+ ), (2.7)
Z
Rn
|u|2∗s(α)
|x|α dx2∗2 s(α)
≤ckUk2Xs(
Rn+1+ ), (2.8) CZ
Rn
|u|2∗s(α)
|x|α dx2∗2 s(α)
≤ kUk2Xs(
Rn+1+ )−γ Z
Rn
|u|2
|x|2sdx. (2.9) In what follows, we will denote by Xs(Rn+1+ ) the closure of C0∞(Rn+1+ ) for the following norm
kUk:=
ks
Z
Rn+1+
y1−2s|∇U|2dx dy−γ Z
Rn
|u|2
|x|2sdx1/2
for allγ < γH. (2.10) Note that inequality (2.7) asserts that Xs(Rn+1+ ) is embedded in the weighted space L2(Rn,|x|−2s) and this embedding is continuous. Set γ+ = max{γ,0} and γ−=−max{γ,0}. The following inequalities hold for anyu∈Xs(Rn+1+ ),
(1−γ+
γH)kUk2Xs(
Rn+1+ )≤ kUk2≤(1 + γ−
γH)kUk2Xs(
Rn+1+ ). (2.11) Thus,k · kis equivalent to the normk · kXs(Rn+1+ ).
The best constantµs,γ,α(Rn) in inequality (2.6) can be written as S(n, s, γ, α) = inf
U∈Xs(Rn+1+ )\{0}
Iγ,α(U), with
Iγ,α(U) = ksR
Rn+1+
y1−2s|∇U|2dx dy−γR
Rn
|u|2
|x|2sdx (R
Rn
|u|2∗s(α)
|x|α dx)2∗s2(α)
.
If S(n, s, γ, α) is attained at some function U ∈ Xs(Rn+1+ ), then u =U(.,0) will be a function inHs(Rn), where µs,γ,α(Rn) is attained. Recently, Ghoussoub and Shakerian [21] proved the extremal function ofS(n, s, γ, α) is attained as following.
Lemma 2.3 ([21]). Suppose0< s <1,0≤α <2s < n, andγ < γH. Then (1) If {α >0} or α= 0andγ≥0, then S(n, s, γ, α)is attained inXs(Rn+1+ )
byWγ,α.
(2) If α = 0 and γ < 0, then there are no extremals for S(n, s, γ, α) in Xs(Rn+1+ ).
3. Compactness lemmas
In this section, we study the compactness properties of the functional J(U) = 1
2kUk2− 1 2∗s(α)
Z
Rn
|u|2∗s(α)
|x|α dx− 1 2∗s(β)
Z
Rn
|u|2∗s(β)
|x|β dx (3.1) for U ∈ Xs(Rn+1+ ), where again u := U(·,0). From Lemma 2.1, we have that J ∈C1(Xs(Rn+1+ )).
Definition 3.1. Letc∈R,E be a Banach space andJ ∈C1(E,R).
(i){uk} is a (P S)c sequence inE for J ifJ(uk) =c+o(1) andJ0(uk) =o(1) strongly inE∗ ask→ ∞.
(ii) We say thatJ satisfies the (P S)c condition if any (P S)c sequence{uk} for J inE has a convergent subsequence.
Proposition 3.2. Suppose 0 < α, β < 2s and γ < γH, then the functional J defined in (3.1)satisfies the Palais-Smale condition (P S)c forc < c∗, where
c∗:= minn 2s−α
2(n−α)S(n, s, γ, α)2s−αn−α, 2s−β
2(n−β)S(n, s, γ, β)2s−βn−βo
. (3.2)
Proof. Let{Uk}k∈Nbe the Palais-Smale sequence of the functionalJ, i.e.
J(Uk)→c, J0(Uk)→0 in (Xs(Rn+1+ ))0 ask→ ∞.
Then
J(Uk) =1
2kUkk2− 1 2∗s(α)
Z
Rn
|uk|2∗s(α)
|x|α dx− 1 2∗s(β)
Z
Rn
|uk|2∗s(β)
|x|β dx
=c+ok(1),
(3.3) and
hJ0(Uk), Uki=kUkk2− Z
Rn
|uk|2∗s(α)
|x|α dx− Z
Rn
|uk|2∗s(β)
|x|β dx=ok(1)kwkk, (3.4) where againuk =Uk(·,0) andok(1)→0 ask→ ∞. From (3.3) and (3.4), we have
c+ok(1)kUkk=J(Uk)−1
2hJ0(Uk), Uki
=1 2 − 1
2∗s(α) Z
Rn
|uk|2∗s(α)
|x|α dx+1 2− 1
2∗s(β) Z
Rn
|uk|2∗s(β)
|x|β dx.
Since 2∗s(α)>2, 2∗s(β)>2, we have Z
Rn
|uk|2∗s(α)
|x|α dx≤C+ok(1)kUkk, Z
Rn
|uk|2∗s(β)
|x|β dx≤C+ok(1)kUkk. (3.5) By (3.4) and (3.5), we obtain
kUkk2+ok(1)kUkk= Z
Rn
|uk|2∗s(α)
|x|α dx+ Z
Rn
|uk|2∗s(β)
|x|β dx≤C+ok(1)kUkk, (3.6)
which implies that {Uk}k∈N is bounded in Xs(Rn+1+ ). It follows that there exists a subsequence, still denote by Uk, such that Uk * U in Xs(Rn+1+ ). For any Ψ∈ C0∞(Rn+1+ ), we have
ok(1)
=hJ0(Uk),Ψi
=ks Z
Rn+1+
y1−2sh∇Uk,∇Ψidx dy−γ Z
Rn
uk(x)
|x|2sψ(x)dx
− Z
Rn
|uk(x)|2∗s(α)−2uk(x)
|x|α ψ(x)dx− Z
Rn
|uk(x)|2∗s(β)−2uk(x)
|x|β ψ(x)dx.
(3.7)
SinceUk * U inXs(Rn+1+ ) ask→ ∞, we have that Z
Rn+1+
y1−2sh∇Uk,∇Ψidx dy−γ Z
Rn
uk(x)
|x|2s ψ(x)dx
→ Z
Rn+1+
y1−2sh∇U,∇Ψidx dy−γ Z
Rn
u(x)
|x|2sψ(x)dx, for all Ψ∈C0∞(Rn+1+ ), whereu=U(·,0).
Moreover, the boundedness of Uk in Xs(Rn+1+ ) implies that |uk|2∗s(α)−2uk and
|uk|2∗s(β)−2uk are bounded in L
2∗ s(α) 2∗
s(α)−1(Rn,|x|−α) and L
2∗ s(β) 2∗
s(β)−1(Rn,|x|−β) respec- tively. Therefore,
|uk|2∗s(α)−2uk *|u|2∗s(α)−2u inL
2∗ s(α) 2∗
s(α)−1(Rn,|x|−α),
|uk|2∗s(β)−2uk*|u|2∗s(β)−2u inL
2∗ s(β) 2∗
s(β)−1(Rn,|x|−β).
Thus, taking limits ask→ ∞in (3.7), we obtain 0 =hJ0(U),Ψi
=ks
Z
Rn+1+
y1−2sh∇U,∇Ψidx dy−γ Z
Rn
u(x)
|x|2sψ(x)dx
− Z
Rn
|u(x)|2∗s(α)−2u(x)
|x|α ψ(x)dx− Z
Rn
|u(x)|2∗s(β)−2u(x)
|x|β ψ(x)dx.
(3.8)
HenceU is a weak solution of (2.3).
The setRn∪ {∞}is compact for the standard topology which means that the measures can be identified as the dual spaceC(Rn∪ {∞}). For example,δ∞is well defined andδ∞=ϕ(∞). By the concentration compactness principle [25, 26], there exist a subsequence, still denoted by Uk and real numbers µ0, µ∞, ν0, ν∞, η0, η∞
andζ0, ζ∞ such that kUkk2Xs(
Rn+1+ )* dµ≥ kUk2Xs(
Rn+1+ )+µ0δ0+µ∞δ∞, (3.9)
|uk|2|x|−2s* dν =|u|2|x|−2s+ν0δ0+ν∞δ∞, (3.10)
|uk|2∗s(α)|x|−α* dη=|u|2∗s(α)|x|−α+η0δ0+η∞δ∞, (3.11)
|uk|2∗s(β)|x|−β * dζ=|u|2∗s(β)|x|−β+ζ0δ0+ζ∞δ∞, (3.12) whereδ0 andδ∞ are the Dirac mass at the origin and infinity respectively.
For% >0, defineB+% :={(x, y)∈Rn+1+ :|(x, y)|< %},B% :={x∈Rn :|x|< %}
and let Φ∈C0∞(Rn+1+ ) be a cut-off function such that Φ≡1 inB+1 2
and 0≤Φ≤1 inRn+1+ . We use ΦUk as test function, we have
hJ0(Uk),ΦUki
=ks Z
Rn+1+
y1−2sh∇Uk,∇(ΦUk)idx dy−γ Z
Rn
uk(x)2φ(x)
|x|2s dx
− Z
Rn
|uk(x)|2∗s(α)φ(x)
|x|α dx− Z
Rn
|uk(x)|2∗s(β)φ(x)
|x|β dx
=ks Z
Rn+1+
y1−2s|∇Uk|2Φ(x)dx dy−γ Z
Rn
uk(x)2φ(x)
|x|2s dx +ks
Z
Rn+1+
y1−2sUkh∇Uk,∇Φidx dy
− Z
Rn
|uk(x)|2∗s(α)φ(x)
|x|α dx− Z
Rn
|uk(x)|2∗s(β)φ(x)
|x|β dx,
(3.13)
whereφ= Φ(·,0). First, we have
%→0lim lim
k→∞
ks
Z
Rn+1+
y1−2sUkh∇Uk,∇Φidx dy
= 0.
Moreover, from (3.9)-(3.12), we obtain
%→0lim lim
k→∞
ks
Z
Rn+1+
y1−2s|∇Uk|2Φdx dy
≥µ0,
%→0lim lim
k→∞
Z
Rn
uk(x)2φ(x)
|x|2s dx=ν0,
%→0lim lim
k→∞
Z
Rn
|uk|2∗s(α)φ(x)
|x|α dx=η0, lim
%→0 lim
k→∞
Z
Rn
|uk|2∗s(β)φ(x)
|x|β dx=ζ0. Thus we obtain
%→0lim lim
k→∞hJ0(Uk),ΦUki ≥µ0−γν0−η0−ζ0. (3.14) By the fractional Hardy-Sobolev inequalities, we have
η
2 2∗
s(α)
0 S(n, s, γ, α)≤µ0−γν0, ζ
2 2∗
s(β)
0 S(n, s, γ, β)≤µ0−γν0. (3.15) By (3.14) and (3.15), we find
η
2 2∗
s(α)
0 S(n, s, γ, α)≤η0+ζ0, ζ
2 2∗
s(β)
0 S(n, s, γ, β)≤η0+ζ0. (3.16) So
η
2 2∗
s(α)
0
1−S(n, s, γ, α)−1η
2∗ s(α)−2
2∗ s(α)
0
≤S(n, s, γ, α)−1ζ0, ζ
2 2∗
s(β)
0
1−S(n, s, γ, β)−1ζ
2∗ s(β)−2 2∗
s(β)
0
≤S(n, s, γ, β)−1η0.
Since{Uk}k∈Nis bounded in Xs(Rn+1+ ), we have η0 ≤c1 andζ0≤c2 for positive constantsc1, c2, thus
η
2 2∗
s(α)
0
1−S(n, s, γ, α)−1c
2∗ s(α)−2 2∗
s(α)
1
≤S(n, s, γ, α)−1ζ0, ζ
2 2∗
s(β)
0
1−S(n, s, γ, β)−1c
2∗ s(β)−2
2∗ s(β)
2
≤S(n, s, γ, β)−1η0.
Therefore, there exist constants A=A(α,2∗s(α), c1) and B =B(β,2∗s(β), c2) such that
η
2 2∗
s(α)
0 ≤Aζ0, and ζ
2 2∗
s(β)
0 ≤Bη0. In particular, we have that eitherη0= 0 and ζ0= 0, or
η0≥S(n, s, γ, α)2s−αn−α, ζ0≥S(n, s, γ, β)2s−βn−β. On the other hand, we know that
c=J(Uk)−1
2hJ0(Uk), Uki+ok(1)
≥ 2s−α 2(n−α)
Z
Rn
|uk(x)|2∗s(α)
|x|α dx+η0 + 2s−β
2(n−β) Z
Rn
|uk(x)|2∗s(β)
|x|β dx+ζ0
≥ 2s−α
2(n−α)η0+ 2s−β 2(n−β)ζ0.
(3.17)
By the assumption thatc < c∗, we obtain thatη0= 0,ζ0= 0.
For the concentration at infinity, we defineBR+:={(x, y)∈Rn+1+ :|(x, y)|< R}, BR :={x∈Rn :|x|< R} and let Ψ∈C0∞(Rn+1+ ) be a cut-off function such that Ψ = 0 inBR+and Ψ≡1 inRn+1+ \B2R+ and 0≤Ψ≤1 inRn+1+ . Consider
µ∞= lim
R→∞ lim
k→∞sup ks
Z
Rn+1+ \B2R+
y1−2s|∇Uk|2Ψdx dy , ν∞= lim
R→∞ lim
k→∞sup Z
Rn\B2R
uk(x)2ψ(x)
|x|2s dx, η∞= lim
R→∞ lim
k→∞sup Z
Rn\B2R
|uk(x)|2∗s(α)ψ(x)
|x|α dx, ζ∞= lim
R→∞ lim
k→∞sup Z
Rn\B2R
|uk(x)|2∗s(β)ψ(x)
|x|β dx.
By the same arguments as the concentration at the origin, we can get the following facts: eitherη∞= 0 andζ∞= 0, or
η∞≥S(n, s, γ, α)2s−αn−α, ζ∞≥S(n, s, γ, β)2s−βn−β. As for (3.17), we obtain
c≥ 2s−α
2(n−α)η∞+ 2s−β
2(n−β)ζ∞. (3.18)
By the assumption that c < c∗, we obtain that η∞= 0,ζ∞= 0. Therefore, up to a subsequence{Uk}k converges strongly toU inXs(Rn+1+ ).
LetWγ,α be the extremal function ofS(n, s, γ, α) inXs(Rn+1+ ), whose existence was obtained by Ghoussoub and Shakerian in [21] forα >0 orα= 0 and 0≤γ <
γH.
Lemma 3.3. Let 0< s <1,0< α, β <2s < n, andγ < γH. Then sup
t≥0
J(tWγ,ϑ)< c∗ forϑ=α, β, wherec∗ is defined in Proposition 3.2.
Proof. Forϑ=α, we have J(tWγ,α) = t2
2kWγ,αk2−t2∗s(α) 2∗s(α)
Z
Rn
|wγ,α|2∗s(α)
|x|α dx−t2∗s(β) 2∗s(β)
Z
Rn
|wγ,α|2∗s(β)
|x|β dx.
wherewγ,α:=T r(Wγ,α) =Wγ,α(·,0). By construction, we have that J(tWγ,α)≤fα(t) := t2
2kWγ,αk2−t2∗s(α) 2∗s(α)
Z
Rn
|wγ,α|2∗s(α)
|x|α dx
Straightforward computations yield thatfα(t) attains its maximum at the point
˜t= kWγ,αk2 R
Rn
|wγ,α|2∗s(α)
|x|α dx 2∗ 1
s(α)−2
.
It follows that sup
t≥0
fα(t) = 2s−α 2(n−α)
kWγ,αk2 (R
Rn|wγ,α|2∗s(α)/|x|αdx)2∗s2(α) 2s−αn−α
.
SinceWγ,α is an extremal forS(n, s, γ, α) onXs(Rn+1+ ), we obtain that sup
t≥0
J(tWγ,α)≤sup
t≥0
fα(t) = 2s−α
2(n−α)S(n, s, γ, α)2s−αn−α. (3.19) We now need to show that equality does not hold in (3.19). Indeed, otherwise we would have that supt≥0J(tWγ,α) = supt≥0fα(t). Considert1(resp. t2>0) where supt≥0J(tWγ,α) (resp. supt≥0fα(t)) is attained. We obtain
fα(t1)−t2
∗ s(β) 1
2∗s(β) Z
Rn
|wγ,α|2∗s(β)
|x|β dx=fα(t2),
which means thatfα(t1)> fα(t2) since t1>0. This contradicts the fact thatt2 is a maximum point offα(t), hence the strict inequality holds in (3.19).
Similarly, forϑ=β, we obtain sup
t≥0
J(tWγ,β)<sup
t≥0
fβ(t) = 2s−β
2(n−β)S(n, s, γ, β)2s−βn−β.
This completes the proof.
4. Proof of main result
Proof of Theorem 1.1. For any U ∈ Xs(Rn+1+ ), the energy functional to problem (2.3) is
J(U) = 1
2kUk2− 1 2∗s(α)
Z
Rn
|u|2∗s(α)
|x|α dx− 1 2∗s(β)
Z
Rn
|u|2∗s(β)
|x|β dx,
where againu:=T r(U) =U(·,0). By fractional Hardy-Sobolev inequality, we have J(U)≥ 1
2kUk2− 1
2∗s(α)S(n, s, γ, α)−2
∗ s(α)
2 kUk2∗s(α)
− 1
2∗s(β)S(n, s, γ, β)−
2∗ s(β)
2 kUk2∗s(β)
=1 2− 1
2∗s(α)S(n, s, γ, α)−2
∗s(α)
2 kUk2∗s(α)−2
− 1
2∗s(β)S(n, s, γ, β)−2
∗ s(β)
2 kUk2∗s(β)−2 kUk2.
Sinceα, β∈(0,2s), we have that 2∗s(α)>2,2∗s(β)>2. By (2.11), we then get that there existsR >0 such thatJ(U)≥ρfor allU ∈Xs(Rn+1+ ) withkUkXα(Rn+1+ )=R.
Moreover, forϑ=αorϑ=β, J(tWγ,ϑ) = t2
2kWγ,ϑk2−t2∗s(α) 2∗s(α)
Z
Rn
|wγ,ϑ|2∗s(α)
|x|α dx−t2∗s(β) 2∗s(β) Z
Rn
|wγ,ϑ|2∗s(β)
|x|α dx, hence limt→+∞J(tWγ,ϑ) =−∞, then there existst0>0 such thatkt0Wγ,ϑk> R andJ(t0Wγ,ϑ)<0. Set
cϑ:= inf
g∈Γϑ max
t∈[0,1]J(g(t)), where
Γϑ:=
g∈C0([0,1], Xs(Rn+1+ )) :g(0) = 0, g(1) =t0Wγ,ϑ .
Thus by Mountain Pass Lemma, there exists a sequence {Uk} in Xs(Rn+1+ ) such that
J(Uk)→c, J0(Uk)→0 in (Xs(Rn+1+ ))0 ask→ ∞.
By Lemma 3.3, we have 0< c≤ sup
t∈[0,1]
J(tt0Wγ,ϑ)≤sup
t>0
J(tWγ,ϑ)< c∗.
By Proposition 3.2, we deduce that{Uk}has a subsequence, still denote by{Uk}, such thatUk→Ustrongly inXs(Rn+1+ ). ThusUis a nontrivial solution of problem (2.3), andu:=T r(U) =U(·,0) is a nontrivial solution of problem (1.1).
Acknowledgments. This research was supported by the NSFC No. 11501468, by the Chongqing Research Program of Basic Research and Frontier Technology cstc2016jcyjA0323, by the Fundamental Research Funds for the Central Universities XDJK2017C049, and by the innovation support program of Chongqing cx2017070.
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Wenjing Chen
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China E-mail address:[email protected]
