Infinitely many radial positive solutions for nonlocal problems with lack of compactness
Fen Zhou
1, 2, Zifei Shen
1and Vicent
,iu D. R˘adulescu
B3, 41Department of Mathematics and Computer Science, Zhejiang Normal University, Jinhua 321004, P. R. China
2Department of Mathematics, Yunnan Normal University, Kunming 650500, P. R. China
3Faculty of Applied Mathematics, AGH University of Science and Technology, 30-059 Kraków, Poland
4Department of Mathematics, University of Craiova, 200585 Craiova, Romania
Received 14 January 2021, appeared 11 April 2021 Communicated by Dimitri Mugnai
Abstract. We are concerned with the qualitative and asymptotic analysis of solutions to the nonlocal equation
(−∆)su+V(|z|)u=Q(|z|)up inRN,
where N≥3, 0<s<1, and 1< p< N−2s2N . Asr→∞, we assume that the potentials V(r)andQ(r)behave as
V(r) =V0+a1 rα +O
1
rα+θ1
Q(r) =Q0+a2 rβ +O
1
rβ+θ2
where a1, a2∈R,α, β> N+2s+1N+2s , andθ1, θ2>0, V0, Q0>0. Under various hypothe- ses on a1,a2,α,β, we establish the existence of infinitely many radial solutions. A key role in our arguments is played by the Lyapunov–Schmidt reduction method.
Keywords: fractional Laplacian, radial solution, lack of compactness, Lyapunov–
Schmidt reduction method.
2020 Mathematics Subject Classification: 35R11, 35A15, 35B40, 47G20.
1 Introduction and the main result
We consider the following nonlocal equation driven by the fractional Laplace operator
(−∆)su+V(|z|)u=Q(|z|)up, inRN. (1.1)
BCorresponding author.
Emails: zhoufen_85@163.com (F. Zhou), szf@zjnu.edu.cn (Z. Shen), radulescu@inf.ucv.ro (V. D. R˘adulescu).
Fractional powers of the Laplacian arise in various equations in mathematical physics and related fields; see, e.g., [1], [9], and [14]. Numerous results related to equations with fractional Laplace operator sprout in literature. A characterization of the fractional Laplacian through Dirichlet–Neumann maps was given in [3]. Regularity for fractional elliptic equations was investigated in [4] and [17]. Existence of solutions was studied in many papers; see, e.g., [2,7,12].
Along with different results, there are various enlightening approaches. In [10], the author obtained some symmetry results for equations involving the fractional Laplacian inRNby the method of moving planes. In [2], symmetry results for nonlinear equations with fractional Laplacian were achieved by the sliding method. Geometric inequality was applied to inves- tigate symmetry properties for a boundary reaction problem in [18]. The method of moving planes and ABP (Aleksandrov–Bakelman–Pucci) estimates for fractional Laplacian were em- ployed in [6] to study radial symmetry and monotonicity properties for positive solutions of fractional Laplacian. We refer the readers to [7] and [11] for very recent new approaches dealing with fractional Laplacian equations, and to [15] for a comprehensive overview of variational methods for nonlocal fractional problems.
Inspired by [19], we obtain the existence of radial positive solutions to (1.1) by Lyapunov–
Schmidt reduction. To the best of our knowledge, this method has never been employed in investigating radial solutions to equations as (1.1).
We will use the radial solution of
(−∆)su+u=up in RN (1.2)
to build up the approximate solutions of problem (1.1). The uniqueness and nondegeneracy of the radial positive solution to problem (1.2) are established in [8].
Our result is based on the following growth assumptions forV(|z|)andQ(|z|)near infin- ity:
(V): there exist constants a1 ∈ R, α> 1, andθ1 > 0, such that V(r) = V0+ arα1 +O( 1
rα+θ1) as r→∞;
(Q): there exist constants a2 ∈ R, β> 1, and θ2 > 0, such that Q(r) =Q0+ a2
rβ +O( 1
rβ+θ2)as r→∞.
We assume throughout this paper thatV0 =1 andQ0 =1.
Let
xj =
rcos2(j−1)π
k ,rsin2(j−1)π k , 0
, j=1, . . . ,k, where 0 is the zero vector inRN−2,r ∈ r0kNN++2s2s−τ,r1kNN++2s2s−τ
,τ=min{α,β}, 0<r0 <r1, and k is the number of the bumps of the solution.
Setz= (z0,z00),z0 ∈R2, z00 ∈RN−2 and define Hrs=
u:u∈ Hs(RN),uis even inzh, h=2, . . . ,N, u(rcosθ,rsinθ,z00) =u
rcos
θ+2πj k
,rsin
θ+ 2πj k
,z00
, whereHs(RN)represents the fractional Sobolev space
Hs(RN):= (
u∈ L2(RN): u(x)−u(y)
|x−y|N2+s ∈ L2(RN×RN) )
, 0<s<1.
Let W be the unique nondegenerate radial positive solution of problem (1.2), Then the result in [8] shows that there exist constants B1 >B2 >0, such that
B2
1+|z−xj|N+2s ≤Wxj(z)≤ B1
1+|z−xj|N+2s, whereWxj(z) =W(z−xj).
Set
Ur(z) =
∑
k j=1Wxj(z).
The main result of this paper establishes the following multiplicity property.
Theorem 1.1. Assume that V(r), Q(r) satisfy (V) and (Q), while a1, a2, α, β satisfy one of the following conditions:
(i) a1 >0, a2 =0, α< N+2s, and α≤ β;
(ii) a1 >0, a2 >0, α< N+2s, and β≥ N+2s;
(iii) a1 >0, a2 <0, α< N+2s, and α> β;
(iv) a1 =0, a2 <0, α≥ β, andβ< N+2s;
(v) a1 <0, a2 <0, α≥ N+2s, and β< N+2s.
Then there exists a positive integer k0such that for any k ≥k0, problem(1.1)has a solution Uk of the form
Uk(z) =Urk(z) +wk, where wk ∈ Hrs, rk ∈ r0kNN++2s2s−τ,r1kNN++2s2s−τ
, and as k→+∞,
Z
RN |(−∆)2swk|2+w2k
→0.
For some of the abstract methods used in this paper, we refer to the monographs by Molica Bisci and Pucci [13] and Papageorgiou, R˘adulescu and Repovš [16].
2 Reduction
Let
Pj = ∂Wxj
∂r , j=1, . . . ,k, where
xj =
rcos2(j−1)π
k ,rsin2(j−1)π k , 0
, j=1, . . . ,k and
r∈ S:=
"
N+2s τ −e
N+12s−τ
kNN++2s2s−τ,
N+2s τ +e
N+2s1−τ kNN++2s2s−τ
# .
We have denotedτ:=min{α,β}, whereαandβare the constants in the expansions ofVand Q, ande>0 is a small constant.
Define
H:=
u:u∈ Hrs, Z
RNWxPj−1Pju=0, j=1, . . . ,k.
. The norm and the inner product inHs(RN)are defined as
kuk= hu,ui12, u∈ Hs(RN), hu,vi=
Z
RN (−∆)2su(−∆)s2v+V(|z|)uv
, u,v∈ Hs(RN). We can easily check that
Z
RN (−∆)s2u(−∆)2sv+V(|z|)uv−pQ(|z|)Urp−1uv
, u,v∈ H
is a bounded bilinear functional inH. Thus, there exists a bounded linear operator Mfrom H toH satisfying
hMu,vi=
Z
RN (−∆)2su(−∆)s2v+V(|z|)uv−pQ(|z|)Urp−1uv
, u,v∈ H. (2.1) We now establish that M is invertible in H.
Lemma 2.1. There exists a constantρ >0, independent of k, such that for any r∈ S, kMuk ≥ρkuk, u∈ H.
Proof. We argue by contradiction. If the thesis does not hold, then for anyρk = 1k(k → +∞), there existsrk ∈S,uk ∈ H, such that
kMukk<ρkkukk. It follows that
kMukk= o(1)kukk. Then,
hMuk,ϕi=o(1)kukkkϕk, ∀ϕ∈H. (2.2) We can assumekukk2=k.
Let
Ωj =
z = (z0,z00)∈R2×RN−2: z0
|z0|, xj
|xj|
≥cosπ k
. By symmetry and the definition ofM, we conclude from (2.2) that for allϕ∈ H,
Z
Ω1
(−∆)s2uk(−∆)s2ϕ+V(|z|)ukϕ−pQ(|z|)Urpk−1ukϕ
= 1
khMuk,ϕi=o 1
√ k
kϕk. (2.3) Particularly,
Z
Ω1
|(−∆)s2uk|2+V(|z|)uk2−pQ(|z|)Urpk−1uk2
=o(1) and
Z
Ω1
|(−∆)s2uk|2+V(|z|)uk2
=1. (2.4)
Letuek(z) =uk(z+x1). Since
|x2−x1|=2rsinπ
k ≥2rπ 2k ≥
N+2s 2τ
N+12s−
τ
kN+τ2s−τπ,
it follows that for any R>0,BR(x1)⊂Ω1. Then from (2.4), we have for all R>0, Z
BR(0)
|(−∆)2suek|2+V(|z|)uek2
≤1.
So, we can assume that there existsu∈ Hs(RN), such that ask →+∞, uek *u, in Hlocs (RN),
and
uek →u, in L2loc(RN).
Sinceuek is even inzh, h=2, . . . .,N, thenuis even inzh, h=2, . . . ,N.
Besides, by
Z
RNWxP1−1P1uk =0, we know that
Z
RNWP−1∂W
∂x1uek =0.
So,usatisfies
Z
RNWP−1∂W
∂x1u=0. (2.5)
We prove in what follows thatusatisfies
(−∆)su+u−pWp−1u=0, in RN. (2.6) Define
He =
ϕ:ϕ∈ Hs(RN), Z
RNWP−1∂W
∂x1ϕ=0
.
For any R > 0, let ϕ ∈ C0∞(BR(0))∩He be any function which is even in zh, h = 2, . . . ,N.
Then ϕk(z):= ϕ(z−x1) ∈ C∞0 (BR(x1)). Substituting ϕin (2.3) with ϕk, then by Lemma A.1, we get
Z
RN (−∆)s2u(−∆)2sϕ+uϕ−pWp−1uϕ
=0. (2.7)
In addition, since u is even in zh, h = 2, . . . ,N, relation (2.7) holds for any ϕ ∈ C0∞(RN), which is odd in zh,h = 2, . . . ,N. Thus, relation (2.7) is true for any ϕ ∈ C0∞(BR(0))∩H. Bye the density ofC0∞(RN)in Hs(RN), we have
Z
RN (−∆)2su(−∆)s2ϕ+uϕ−pWp−1uϕ
=0, ∀ϕ∈ He (2.8)
Meanwhile, relation (2.8) holds for ϕ = ∂W∂x
1. Therefore, (2.8) holds for any ϕ ∈ Hs(RN). Substituting ϕin (2.8) with uyields (2.6).
Since W is non-degenerate, we have u = C∂W∂x
1 because u is even in zh, h = 2, . . . .,N. By (2.5), we know that
u=0, which implies
Z
BR(x1)uk2 =o(1), ∀R>0.
Besides, it follows from LemmaA.1that there existsC0 >0 such that Urk(x)≤C0, for all x∈ Ω1.
It follows that
o(1) =
Z
Ω1
|(−∆)s2uk|2+V(|z|)uk2−pQ(|z|)Urpk−1uk2
=
Z
Ω1
|(−∆)s2uk|2+V(|z|)uk2
+o(1)−C Z
Ω1
uk2
≥ 1 2
Z
Ω1
|(−∆)2suk|2+V(|z|)uk2
+o(1), which contradicts (2.4). The proof is now complete.
Define
I(u) = 1 2
Z
RN |(−∆)s2u|2+V(|z|)u2
− 1 p+1
Z
RNQ(|z|)|u|p+1. (2.9) Let
J(φ) = I(Ur+φ), φ∈ H.
Then, J(0) = 1
2 Z
RN |(−∆)2sUr|2+V(|z|)Ur2
− 1 p+1
Z
RNQ(|z|)|Ur|p+1.
= 1 2
Z
RNUr
∑
k j=1Wxpj+1 2
Z
RN(V(|z|)−1)Ur2− 1 p+1
Z
RN(Q(|z|)−1)Urp+1− 1 p+1
Z
RNUrp+1 becauseWxj solves (1.2).
Lemma 2.2. There exists a positive integer k0 such that for each k ≥ k0, there is a C1 map from S to Hrs: φk =φk(r), r=|x1|, satisfyingφk ∈ H, and
J0(φk)
H =0.
Moreover, there exists a constant C>0, independent of k, such that kφkk ≤ C
k
(N+2s)(τ−1)+τ 2(N+2s−τ) +δ
, (2.10)
whereδ >0is a small constant.
Proof. We expand J(φk)as
J(φk) = J(0) +l(φk) + 1
2hMφk,φki+R(φk), φk ∈ H, where
l(φk) =hI0(Ur),φki
=
Z
RN(V(|z|)−1)Urφk+
Z
RN
k
j
∑
=1Wxpj−Urp
φk−
Z
RN(Q(|z|)−1)Urpφk. Mis the bounded linear map defined in (2.1) and
R(φk) =− 1 p+1
Z
RNQ(|z|)
|Ur+φk|p+1−Urp+1−(p+1)UrPφk−1
2p(p+1)Urp−1φk2
. Sincel(φk)is a bounded linear functional in H, there existslk ∈ H, such that
l(φk) =hlk,φki. Then,φk being a critical point of J is equivalent to
lk+Mφk+R0(φk) =0. (2.11)
Since Mis invertible, we can infer from (2.11) that
φk = T(φk):=−M−1(lk+R0(φk)). Define
E= (
φk :φk ∈H, kφkk ≤ 1 k
(N+2s)(τ−1)+τ 2(N+2s−τ)
) . Next, we check thatTis a contraction map fromEto E.
Case 1: p≤2. It is easy to verify that
kR0(φk)k ≤Ckφkk|p. In fact,
|hR0(φk),vi|= Z
RNQ(|z|)(pUrp−1φkv+Urpv− |Ur+φk|p−1(Ur+φk)v)
= Z
RNQ(|z|)p(|Ur+θφk|p−1−Urp−1)φkv
≤C Z
RN|(|Ur+θφk|p−1−Urp−1)| |φk| |v|
≤C Z
RN|θφk|p−1|φk| |v|
≤Ckφkkpkvk where 0<θ <1.
Then, by the boundedness of Mand Lemma 2.3, kT(φk)k ≤C(klkk+kφkkp)≤ C
k
(N+2s)(τ−1)+τ 2(N+2s−τ) +δ
+ C
k
(N+2s)(τ−1)+τ
2(N+2s−τ) p ≤ 1
k
(N+2s)(τ−1)+τ 2(N+2s−τ)
(2.12) which implies thatTmaps EtoE.
In addition,
kR00(φk)k ≤Ckφkkp−1. In fact,
|hR00(φk)v,hi|= Z
RNQ(|z|)(pUrp−1vh−p|Ur+φk|p−1vh)
= p Z
RNQ(|z|)(Urp−1vh− |Ur+φk|p−1)vh
≤C Z
RN|φk|p−1|v||h|
≤Ckφkkp−1kvk khk. Thus, forφk1, φk2 ∈E,
kT(φk1)−T(φk2)k= kM−1R0(φk1)−M−1R0(φk2)k
≤ kM−1k kR0(φk1)−R0(φk2)k ≤ kM−1k kR00(φk1 +θ(φk2−φk1)k kφk2 −φk1k
≤C(kφk1kp−1+kφk2kp−1k)kφk1−φk2k ≤ 1
2kφk1 −φk2k.
Note that the last inequality holds only whenk is large enough, which implies the existence of k0 in Lemma 2.2. Therefore, T is a contraction map from E to E. Then the contraction mapping theorem implies the existence ofφk as a critical point of J restricted to H.
Case 2: p>2.
Settingh(t) =|Ur+tφk|p−1(Ur+tφk)v, then by Taylor’s formula,
|hR0(φk),vi|= Z
RNQ(|z|)(pUrp−1φkv+Urpv− |Ur+φk|p−1(Ur+φk)v)
= Z
RNQ(|z|)(−1 2h00(θ))
≤C Z
RN p(p−1)|Ur+θφk|p−2 Ur+θφk
|Ur+θφk|φ
2kv
≤C Z
RN(kφkk2+kφkkp)kvk
≤Ckφkk2kvk which implies thatkR0(φk)k ≤Ckφkk2.
By the mean value theorem we obtain
|hR00(φk)v,hi|= p Z
RNQ(|z|)(Urp−1vh− |Ur+φk|p−1)vh
= p(p−1) Z
RNQ(|z|)|Ur+θφk|p−3(Ur+θφk)φkvh
≤C Z
RN(Urp−2+|φk|p−2)|φk||v||h|
where 0<θ <1.
By Hölder’s inequality, Z
RNUrp−2|φk||v||h| ≤C Z
RNUrp+1
pp−+21 Z
RN|φk|p+1
p+11 Z
RN|v|p+1
p+11 Z
RN|h|p+1 p+11
≤Ckφkk kvk khk and
Z
RN|φk|p−1|v||h| ≤ Z
RN|φk|p+1
pp−+11 Z
RN|v|p+1
p+11 Z
RN|h|p+1 p+11
≤Ckφkkp−1kvkkhk. Therefore,kR00(φk)k ≤Ckφkk.
Arguing similarly as in case 1, we have,
kT(φk)k ≤C(klkk+kφkk2)≤ 1 k
(N+2s)(τ−1)+τ 2(N+2s−τ)
. (2.13)
and T is a contraction map from E to E. The existence of φk follows from the contraction mapping theorem, and (2.10) follows from (2.12) and (2.13).
Following the argument employed in [5] to prove Lemma 4.4, we conclude that φk(r) is continuously differentiable in r.
Lemma 2.3. Ifτ=min{α,β}< N+2s, there exists a small constantδ>0such that klkk ≤ C
k
(N+2s)(τ−1)+τ 2(N+2s−τ) +δ
. Proof. We have
hlk,φki= l(φk)
=
Z
RN(V(|z|)−1)Urφk+
Z
RN
k
j
∑
=1Wxpj−Urp
φk−
Z
RN(Q(|z| −1)Urpφk. (2.14) By symmetry,
Z
RN(V(|z|)−1)Urφk =
Z
RN(V(|z|)−1) k
j
∑
=1Wxj
φk
=
∑
k j=1Z
RN(V(|z|)−1)Wxjφk =k Z
RN(V(|z|)−1)Wx1φk (2.15) and
Z
RN(V(|z|)−1)Wx1φk =
Z
Br 2(0)
(V(|z|)−1)Wx1φk+
Z
RN\Br 2(0)
(V(|z|)−1)Wx1φk. For z∈RN\Br
2(0),
V(|z|)−1= a1
|z|α +O 1
|z|α+θ1
≤ C
|z|α ≤ 2
αC
|r|α,
Z
RN\Br 2(0)
(V(|z|)−1)Wx1φk = 2
αC
|r|α
Z
RN\Br
2(0)Wx1φk
≤ 2
αC
|r|α Z
RN\Br 2(0)Wx21
12Z
RN\Br 2(0)
φk2 12
=O 1
rα
kφkk, Z
Br 2(0)
(V(|z|)−1)Wx1φk ≤ C Z
Br 2(0)Wx21
12Z
Br 2(0)
φ2k 12
≤C 1
rN2+2skφkk. Then we conclude that
Z
RN(V(|z|)−1)Wx1φk ≤O 1
rα
kφkk+C 1
rN2+2skφkk. (2.16) By the mean value theorem and LemmaA.1,
Z
RN
k
∑
j=1Wxpj −Urp
φk
=k Z
Ω1
k
j
∑
=1Wxpj−Urp
φk
≤Ck Z
Ω1
Wxp1−1 k
j
∑
=2Wxj
φk
≤Ck 1 (k−1r)N+2s
Z
Ω1
Wxp1 p
−1
p Z
Ω1
|φk|p 1p
≤Ck 1
(k−1r)N+2skφkk. (2.17)
By the boundedness ofUrwe have Z
RN(Q(|z| −1)Urpφk =
Z
RN(Q(|z| −1)Urp−1Urφk
≤C Z
RN(Q(|z| −1)Urφk ≤Ck Z
RN(Q(|z| −1)Wx1φk
≤Ck O 1
rβ
+ 1 rN2+2s
!
kφkk. (2.18)
Combining relations (2.14)–(2.18), we obtain hlk,φki=
Z
RN(V(|z|)−1)Urφk+
Z
RN
∑
k i=1Wxpj−Urp
! φk−
Z
RN(Q(|z| −1)Urpφk
≤ k O 1
rα
+O 1
rβ
+C 1 rN2+2s
+ C
(k−1r)N+2s
! kφkk
≤ k O 1
rτ
+ C
rN2+2s + C (k−1r)N+2s
!
kφkk. (2.19)
Byr∈S, it holds thatr ∼kNN++2s2s−τ, C
(k−1r)N+2s ∼O r1τ
,
kO 1
rτ
<C 1
k(NN++2s2s−)ττ−1
≤ C
k
(N+2s)(τ−1)+τ 2(N+2s−τ) +δ
, andk 1
rN2+2s < 1
k
(N+2s)(τ−1)+τ 2(N+2s−τ)
, forτ< N+2s.
We conclude that ifτ<N+2s, then
klkk ≤ C k
(N+2s)(τ−1)+τ 2(N+2s−τ) +δ
. The proof is now complete.
3 Proof of the main result
Define
G(r) =I(Ur+φk), ∀r ∈S, whereφk =φk(r)is the map obtained in Lemma2.2.
According to Lemma 6.1 in [5], ifris a critical point of G(r), thenUr+φk(r)is a solution of (1.1).
From the energy expansion in the Appendix, we have J(0) = I(Ur)
=k D+ a1A1
rα − a2A2
rβ − B
(k−1r)N+2s +O 1
rα+τ1
+O 1
rβ+τ2
+O
1 (k−1r)N+2s+σ
! . Set
H(r) = a1A1
rα − a2A2
rβ − B
(k−1r)N+2s.
We prove in what follows that in any of the cases in Theorem 1.1, H(r) has a maximum pointrk.
For case (i): ifa1 >0, a2=0, α<N+2s, andα≤ βthen H0(r) =−αa1A1
rα+1 + B(N+2s)kN+2s rN+2s+1 andrk satisfies
αa1A1
rαk+1 = B(N+2s)kN+2s rkN+2s+1 .
Actually, calculating the maximum points in these cases can be summed up as τC
rτk+1 = C
0(N+2s)kN+2s rNk+2s+1 , whereτ=min{α,β}. ThenH(r)has a maximum point
rk =
(N+2s)C0 τC
N+12s−τ
kNN++2s2s−τ,
which is an interior point ofS. Then, there exists a small constant δsuch that J(0) =k D+ a1A1
rα − a2A2
rβ − B
(k−1r)N+2s +O 1
k(NN++2s2s−)ττ+δ !
.
Consequently,
G(r) =I(Wr+φk) = I(Wr) +l(φk) + 1
2hMφk,φki+R(φk)
= J(0) +O(klkkkφkk+kφkk2)
= J(0) +O
1 k(NN++2s2s−)ττ−1+δ
=k D+ a1A1
rα − a2A2
rβ − B
(k−1r)N+2s+O 1
k(NN++2s2s−)ττ+δ !
.
Since H(r)has a maximum point rk which is an interior point ofS in any of the cases listed, thenG(r)has a critical pointrek in the interior ofS. This means that the function
Urek+φk(rek)
is a solution of problem (1.1). The proof is now complete.
A Appendix. Energy expansions
In this section, we obtain some energy estimates for the approximate solutions. Recall that Ωj =
z = (z0,z00)∈R2×RN−2: z0
|z0|, xj
|xj|
≥cosπ k
,
xj =
rcos2(j−1)π
k ,rsin2(j−1)π k , 0
, j=1, . . . ,k,
r∈ S:=
"
N+2s τ
−e N+12s−τ
kNN++2s2s−τ,
N+2s τ
+e N+2s1−τ
kNN++2s2s−τ
# ,
I(u) = 1 2
Z
RN |(−∆)s2u|2+V(|z|)u2
− 1 p+1
Z
RNQ(|z|)|u|p+1, and
I(Ur) = 1 2
Z
RN |(−∆)2sUr|2+V(|z|)Ur2
− 1 p+1
Z
RNQ(|z|)|Ur|p+1
= 1 2
Z
RNUr
∑
k j=1Wxpj +1 2
Z
RN(V(|z|)−1)Ur2
− 1 p+1
Z
RN(Q(|z|)−1)Urp+1− 1 p+1
Z
RNUrp+1. Lemma A.1. For any z∈ Ω1, there exists C >0, such that
∑
k j=2Wxj(z)≤ C (k−1r)N+2s