Nonlinear Analysis 164 (2017) 100–117
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Nonlinear Analysis www.elsevier.com/locate/na
Existence and multiplicity of solutions for fractional Choquard equations Pei Maa , Jihui Zhangb, * a
College of Science, Nanjing Forestry University, Nanjing 210037, PR China Jiangsu Key Laboratory for NSLSCS, School of Mathematical Sciences, Nanjing Normal University, Nanjing 210023, PR China b
article
info
Article history: Received 27 January 2017 Accepted 26 July 2017 Communicated by Enzo Mitidieri Keywords: Critical nonlinearity Ground state solutions Fractional Choquard equation Variational method Lusternik–Schnirelmann category theory
abstract We consider the following fractional order Choquard equation ∗
∗
(− △ )α/2 u(x) + (λV (x) − β)u = (|x|−µ ∗ |u|2µ )|u|2µ −2 u, x ∈ Rn , with the nonlinearity in the critical growth, where α ∈ (0, 2), n ≥ 3, λ, β ∈ R+ and 2∗µ = (2n − µ)/(n − α). Using the variational method, we establish the existence and multiplicity of weak solutions. © 2017 Elsevier Ltd. All rights reserved.
1. Introduction In this paper, we consider the following fractional Choquard equation { −µ p p−2 (−△)α/2 u + (λV (x) − β)u = (|x| ∗ |u| )|u| u, x ∈ Rn , α/2 n u ∈ H (R ), in the critical case p = 2∗µ = 2n−µ n−α , where α ∈ (0, 2) and µ ∈ (0, n). The fractional Laplacian in Rn is a nonlocal pseudo-differential operator taking the form ∫ u(x) − u(y) α/2 (−△) u(x) = Cn,α P V n+α dy Rn |x − y| ∫ u(x) − u(y) = Cn,α lim n+α dy, ε→0 Rn \Bε (x) |x − y|
(1)
(2)
where Cn,α is a normalization constant and P V is the Cauchy principal value. In this paper, we consider the fractional Laplacian in the weak sense.
*
Corresponding author. E-mail addresses:
[email protected] (P. Ma),
[email protected] (J. Zhang).
http://dx.doi.org/10.1016/j.na.2017.07.011 0362-546X/© 2017 Elsevier Ltd. All rights reserved.
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
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We denote by H α/2 (Rn ) the homogeneous fractional space. It is defined as the completion of C0∞ (Rn ) that } { ∫ ∫ (u(x) − u(y))2 α/2 n 2 n H (R ) = u ∈ L (R ) | n+α dxdy < ∞ . |x − y| Rn Rn This space is endowed with the norm (∫ ∥u∥H α/2 (Rn ) = α/2
Let H0
Rn
∫
(u(x) − u(y))2
Rn
n+α
|x − y|
) 12 |u| dx .
∫
2
dxdy + Rn
(Ω ) be the completion of C0∞ (Ω ) under the norm ∥ · ∥H α/2 . It is endowed with the norm defined as ∫ 2 |u(x) − u(y)| ∥u∥H α/2 (Ω) = n+α dxdy, 0 |x − y| Q
where Q := (Rn × Rn ) \ O and O := CΩ × CΩ ⊂ Rn × Rn . The homogeneous fractional Sobolev space α D 2 ,2 (Rn ) is defined as the completion of C0∞ (Rn ) } { |u(x) − u(y)| 2 n α/2,2 n 2 n ∈ L (R ) , D (R ) = u ∈ L (R ) | n+α |x − y| 2 and it is endowed with the norm ∥u∥2Dα/2,2 (Rn ) =
∫ R2n
2
|u(x) − u(y)| |x − y|
n+α
dxdy.
In recent years, the fractional Laplacian has attracted much attention. It appears in diverse physical phenomena, such as anomalous diffusion and quasi-geostrophic flows, turbulence and water waves, molecular dynamics, and relativistic quantum mechanics of stars. It also has various applications in probability and finance. In particular, the fractional Laplacian can be understood as the infinitesimal generator of a stable L´evy diffusion process and appear in anomalous diffusions in plasmas, flames propagation and chemical reactions in liquids, population dynamics, geographical fluid dynamics, and American options in finance. For readers who are interested in the applications of the fractional Laplacian, please refer to [1,4] and the references therein. When V (x) = 1, β = 0 and 2n−µ < p < 2n−µ n n−α , (1) becomes the following nonlocal problem −µ
(−△)α/2 u(x) + λu = (|x|
p
p−2
∗ |u| )|u|
u.
In [8], the authors obtained regularity of weak solutions, existence and properties of ground states, as well as multiplicity and nonexistence of solutions. When α = 1, the above problem has been used to model the dynamics of pseudo-relativistic boson stars. In particular, when λ = 1, µ = 1 and p = 2, (1) deduces to √ −1 2 (−△)u + u = (|x| ∗ |u| )u, and in [10], Frank and Lenzmann proved analyticity and radial symmetry of ground state solutions. Moreover, in [9], it was shown that the dynamical evolution of boson stars is effectively described by the nonlinear evolution equation with mass parameter m ≥ 0 √ −1 2 i∂t ψ = −△ + m2 ψ − (|x| ∗ |ψ| )ψ for the wave field ψ : [0, T ] × R3 → C. In fact, this dispersive nonlinear L2 critical PDE displays a rich variety of phenomena such as stable or unstable traveling solitary waves and finite-time blowup.
102
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
In [5], the authors considered the following nonlinear fractional Choquard equation { −µ p p−2 (−△)α/2 u + u = (1 + a(x))(|x| ∗ |u| )|u| u in Rn , u(x) → 0 as |x| → ∞ in the subcritical cases (2n − µ)/n < p < (2n − µ)/(n − α). Under certain assumption on a(x) and lim|x|→∞ a(x) = 0, the authors proved the existence of ground state solutions. In [14], the authors studied the following nonlinear Choquard equation involving fractional Laplacian { −µ (−△)α/2 u + u = (|x| ∗ F (u))f (u) in Rn , α/2 n u(x) ∈ H (R ), where the nonlinearity satisfies the general Berestycki–Lions-type assumptions. They obtained the existence of ground states. In [13], the authors investigated the Br´ezis–Nirenberg type problem of nonlinear Choquard equation involving the fractional Laplacian { −µ 2∗ 2∗ −2 (−△)α/2 u − βu = (|x| ∗ |u| µ )|u| µ u in Ω , u=0 in Rn \ Ω , in a bounded domain Ω , where 2∗µ = (2n − µ)/(n − α) is the critical exponent in the sense of Hardy– Littlewood–Sobolev inequality. They obtained some existence, multiplicity, regularity and nonexistence results for solutions of the above equation using variational methods. For more details associated with the fractional Choquard equation, please refer to [7] and the references therein. In this paper, motivated by the works above, we consider more general equation in the whole space Rn and obtain the existence and multiplicity of the fractional Choquard equation (1) with critical nonlinearity. To the best of our knowledge, there has been any existence result in the critical case. Because of the lack of compactness for the corresponding variational functional in the critical case, it is very difficult for us to verify the (P S)c condition. So we use a new idea to overcome this difficulty and furthermore deduce the concentration of the weak solutions. And we also use the Lusternik–Schnirelmann theory to derive a multiplicity result. Throughout this paper, we assume that the potential V (x) satisfies (V1 ) V (x) ∈ C(Rn , R), V (x) ≥ 0, and Ω := intV −1 (0) is a nonempty bounded set with smooth boundary, ¯ = V −1 (0). 0 ∈ Ω and Ω (V2 ) There exists M0 > 0 such that L{x ∈ Rn : V (x) ≤ M0 } < ∞, where L denotes the Lebesgue measure in Rn . The following are our main theorems. Theorem 1.1. Let β1 be the first eigenvalue of (−△)α/2 on Ω with zero Dirichlet condition outside Ω . Suppose ( V1 ) and ( V2 ) hold, then for every 0 < β < β1 , there exists λ(β) > 0 such that for each λ ≥ λ(β), Eq. (1) has at least one ground state positive solution u. Theorem 1.2. Let {un } be a sequence of solutions for (1). If 2n−µ n−µ+α Sα,H n−µ+α , n → ∞ Jλn ,β (un ) → c < 4n − 2µ for 0 < β < β1 and λn → ∞, then {un } converges to a solution of { −µ 2∗ 2∗ −1 (−△)α/2 u − βu = (|x| ∗ |u| µ )|u| µ , x ∈ Ω , α/2 u ∈ H0 (Ω ). Here Sα,H is the best constant and will be defined precisely in Section 2.
(3)
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103
Theorem 1.3. Assume ( V1 ) and ( V2 ) hold, then there exists 0 < β ∗ < β1 and for each 0 < β ≤ β ∗ , 2n−µ n−µ+α such that if λ ≥ λ(β), then (1) has at least cat(Ω ) there exist λ(β) > 0 and 0 < c(β) < n−µ+α 4n−2µ Sα,H solutions with energy Jλ,β ≤ c(β), where cat(Ω ) is the category of the domain Ω .
2. Preliminaries and functional setting Proposition 2.1. (Hardy–Littlewood–Sobolev Inequality, [12]) Let t, r > 1 and 0 < µ < n with 1/r+1/t+µ/n = 2, f ∈ Lt (Rn ) and h ∈ Lr (Rn ). Then there exists a sharp constant C(n, µ, t, r) independent of f, h, such that ∫ ∫ f (x)h(y) (4) µ dxdy ≤ C(n, µ, t, r)|f |t |h|r . |x − y| n n R R Moreover, if t = r =
2n 2n−µ ,
then Γ ( n2 − µ2 ) . Γ (n − µ2 )
µ
C(n, µ, t, r) = C(n, µ) = π 2 In this case, (4) is an equality if and only if f ≡ Ch and
2n−µ 2
2
f = a(b2 + |x − x0 | )−
,
where a ∈ C, b ∈ R and b ̸= 0, x0 ∈ Rn . q
In particular, let f = h = |u| , then by (4), we know ∫ ∫ q q |u(x)| |u(y)| dxdy µ |x − y| Rn Rn is well defined if u ∈ Ltq (Rn ) for some t > 1 satisfying 2 µ + = 2. t n Thus, if u ∈ H α/2 (Rn ), by the Sobolev Embedding Theorem, 2n − µ 2n − µ ≤q≤ . n n−α In this paper, we study the fractional Choquard equation with the critical exponent the best constant ∫ |u(x)−u(y)|2 dxdy R2n |x−y|n+α Sα,H = inf n−α , ( ) 2n−µ α ,2 2∗ 2∗ |u(x)| µ |u(y)| µ u∈D 2 (Rn )\{0} ∫ dxdy |x−y|µ R2n
2n−µ n−α .
Let Sα,H be
(5)
and Sα be the best Sobolev constant ∫ Sα =
inf
α ,2
u∈D 2
(Rn )\{0}
R2n
(∫
|u(x)−u(y)|2 dxdy |x−y|n+α Rn
) 2∗ 2∗ |u| α dx 2α
,
2n where 2∗α = n−α is the fractional Sobolev critical exponent. It is well-known that Sα,H and Sα are both achieved if and only if
u=C
b
( b2
) n−α 2 2
+ |x − x0 |
, x ∈ Rn ,
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
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for some x0 ∈ Rn , C > 0 and t > 0 (see Theorem 2.15 of [8]). Moreover, Sα
Sα,H =
n−α
.
C(n, µ) 2n−µ Definition 1. We say that u ∈ H α/2 (Rn ) is a weak solution of (1) if ∫ ∫ (u(x) − u(y))(φ(x) − φ(y)) Cn,α dxdy + (λV (x) − β)uφdx n+α |x − y| Rn R2n ∫ −µ 2∗ 2∗ −2 = (|x| ∗ |u| µ )|u| µ uφdx
(6)
Rn
for all φ ∈ C0∞ (Rn ). The energy functional associated to Eq. (1) is ∫ ∫ Cn,α (u(x) − u(y))2 1 2 Jλ,β (u) = (λV (x) − β)|u| dx dxdy + n+α 2 2 2n n |x − y| R ∫R 1 −µ 2∗ 2∗ µ µ − (|x| ∗ |u| )|u| dx. 2 · 2∗µ Rn
(7)
This functional is well defined in H α/2 (Rn ), and the critical points of Jλ,β (u) are weak solutions of (1). Moreover, u0 is called a ground state of (1) if and only if u0 is a critical point of (7) which satisfies Jλ,β (u0 ) = c := inf{Jλ,β (u) : u ∈ H α/2 (Rn ) is a critical point of (7)}. Throughout this paper, we assume that (V1 ) and (V2 ) hold. We write |·|q for the Lq norm for q ∈ [1, ∞) and denote by β1 the first eigenvalue of (−△)α/2 on Ω with zero Dirichlet condition outside Ω . ∫ Let E = {u ∈ Dα/2,2 (Rn ) : Rn V (x)u2 dx < ∞} be the Hilbert space endowed with the norm ( ∥u∥ =
∥u∥2Dα/2,2
) 21 V (x)u dx .
∫
2
+ Rn
If λ > 0, it is equivalent to the norm ( ∥u∥λ =
∥u∥2Dα/2,2 + λ
∫
) 21 V (x)u2 dx .
Rn
We denote ⟨·, ·⟩ the L2 -inner product. Let Lλ := (−△)α/2 + λV (x), then for u, v ∈ E, ∫ ∫ (u(x) − u(y))(v(x) − v(y)) dxdy + λV (x)uvdx. ⟨Lλ u, v⟩ = Cn,α n+α |x − y| Rn R2n Obviously, aλ = inf{⟨Lλ u, u⟩ : u ∈ E, ∥u∥2 = 1} ≥ 0 and aλ is nondecreasing in λ. In the following, enlarging λ(β) if necessary, we assume λ ≥ β/M0 so that λM0 − β ≥ 0 f or all λ ≥ λ(β). ′ Recall that a sequence is called a (P S)c sequence for Jλ,β if Jλ,β (un ) → c and Jλ,β (un ) → 0 as n → ∞. Jλ,β satisfies the (P S)c condition if and only if every (P S)c sequence has a convergent subsequence. Before proving the theorems, we need the following two lemmas, whose proofs are similar to those in [6].
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
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α/2
Lemma 2.1. For λn ≥ 1 and λn → ∞ as n → ∞, if ∥un ∥2λn < C for un ∈ E, then there is a u ∈ H0 such that, up to a subsequence, un ⇀ u in E and un → u in L2 (Rn ).
(Ω )
Lemma 2.2. For every 0 < β < β1 , there exists λ(β) > 0 such that aλ ≥ (β + β1 )/2 for λ ≥ λ(β). Consequently, there exists a constant C > 0 such that C∥u∥2λ ≤ ⟨(Lλ − β)u, u⟩ for all u ∈ E, λ ≥ λ(β). Lemma 2.3. For each 0 < β < β1 and λ ≥ λ(β), every (P S)c sequence for Jλ,β is bounded in E. Proof . Let {un } be a Palais–Smale sequence such that Jλ,β (un ) → c, ′ Jλ,β (un ) → 0 in E ′ ,
where E ′ is dual space of E. By Definition 1, Lemma 2.2 and (7), we have 1 ⟨J ′ (un ), un ⟩ c + ∥un ∥λ ≥ Jλ,β (un ) − 2 · 2∗µ λ,β ( ) 1 1 = − ⟨(Lλ − β)un , un ⟩ 2 2 · 2∗µ ≥ C∥un ∥2λ . This shows that {un } is bounded in E. Proposition 2.2. 2n−µ
c<
For each 0 < β < β1 and λ ≥ λ(β), Jλ,β satisfies the (P S)c condition for all
n−µ+α n−µ+α . 4n−2µ Sα,H
Proof . By Lemma 2.3, {un } is bounded in E. Now up to a subsequence, still denoted by {un }, we may assume ⎧ in E, ⎨un ⇀ u un → u in L2loc (Rn ), ⎩ un → u a.e. in Rn . Then, 2∗ µ
|un |
⇀ |u|
2∗ µ
2n
in L 2n−µ (Rn ).
Using Hardy–Littlewood–Sobolev inequality, we derive −µ
|x|
2∗ µ
∗ |un |
⇀ |x|
−µ
∗ |u|
2∗ µ
in L
2n µ
.
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Moreover, |un |
2∗ µ −2
un ⇀ |u|
2∗ µ −2
2n
u in L n−µ+α (Rn ).
By H´ older inequality, −µ
(|x|
2∗
2∗ µ −2
∗ |un | µ )|un |
−µ
un ⇀ (|x|
2∗
2∗ µ −2
∗ |u| µ )|u|
2n
u in L n+α (Rn ).
By a standard argument, one can show that u is a weak solution of −µ
(−△)α/2 u + (λV (x) − β)u = (|x|
2∗
2∗ µ −2
∗ |u| µ )|u|
u,
hence by (6), ∫ R2n
2
|u(x) − u(y)| |x − y|
∫
(λV (x) − β)u2 dx =
dxdy +
n+α
Rn
∫
−µ
(|x|
2∗
2∗
∗ |u| µ )|u| µ dx.
Rn
Let wn = un − u, by the Br´ezis–Lieb lemma [11], ∫ −µ 2∗ 2∗ (|x| ∗ |un | µ )|un | µ dx ∫Rn ∫ −µ 2∗ 2∗ µ µ = (|x| ∗ |u| )|u| dx + Rn
−µ
(|x|
2∗
2∗
∗ |wn | µ )|wn | µ dx + o(1).
Rn
′ ′ Since ⟨Jλ,β (un ), un ⟩ → 0, ⟨Jλ,β (u), u⟩ = 0 and
∫ ∫ 2 |wn (x) − wn (y)| ′ ′ dxdy + (λV (x) − β)wn2 dx ⟨Jλ,β (un ), un ⟩ = ⟨Jλ,β (u), u⟩ + n+α n 2n |x − y| R R ∫ ∗ −µ 2∗ 2 − (|x| ∗ |wn | µ )|wn | µ dx + o(1), Rn
we can assume that ∫ R2n
2
|wn (x) − wn (y)| n+α
|x − y| ∫
(|x|
∫ dxdy + Rn
−µ
2∗
(λV (x) − β)wn2 dx → b, 2∗
∗ |wn | µ )|wn | µ dx → b.
Rn
Since, ∫ ∫ 2 1 |wn (x) − wn (y)| 1 c ← Jλ,β (un ) = Jλ,β (u) + (λV (x) − β)wn2 dx dxdy + n+α 2 R2n 2 Rn |x − y| ∫ 1 −µ 2∗ 2∗ − (|x| ∗ |wn | µ )|wn | µ dx, ∗ 2 · 2µ Rn and Jλ,β (u) ≥ 0, it is easy to deduce b≤
4n − 2µ c. n−µ+α
By Lemma 2.1 one can show that ∫ F
wn2 dx → 0, n → ∞,
(8)
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
107
where F = {x ∈ Rn : V (x) ≤ M0 }. Let F c = Rn \ F , then n−α (∫ ) 2n−µ −µ 2∗ 2∗ Sα,H (|x| ∗ |wn | µ )|wn | µ dx Rn 2
|wn (x) − wn (y)|
∫ ≤ R2n
n+α
|x − y|
dxdy
2
|wn (x) − wn (y)|
∫ dxdy + (λV (x) − β)wn2 dx n+α |x − y| Fc R2n ∫ ≤ ⟨(Lλ − β)wn , wn ⟩ + β wn2 dx ∫
≤
F
= ⟨(Lλ − β)wn , wn ⟩ + o(1). 2n−µ
n−α
2n−µ
Passing to the limits, it yields Sα,H b 2n−µ ≤ b. Hence either b = 0 or b ≥ Sα,H n−µ+α . If b ≥ Sα,H n−µ+α , then 2n−µ n−µ+α n−µ+α Sα,H n−µ+α ≤ b ≤ c, 4n − 2µ 4n − 2µ 2n−µ
n−µ+α . Therefore b = 0. This proves that w → 0 in E. Hence which is a contradiction with c < n−µ+α n 4n−2µ Sα,H un → u in E, which verifies the (P S)c condition.
3. Proof of Theorems 1.1 and 1.2 First we prove the energy functional of problem (1) meets the geometry of the Mountain Pass Theorem. Lemma 3.1. For each 0 < β < β1 and λ ≥ λ(β), the functional Jλ,β satisfies the following conditions: (i) J(0) = 0 and there exist s, ρ > 0, such that Jλ,β (u) ≥ s > 0 for ∥u∥λ = ρ sufficiently small; (ii) There exists e ∈ E, such that Jλ,β (e) < 0 for ∥u∥λ > ρ. Proof . (i) J(0) = 0. For each 0 < β < β1 (Ω ), by the Sobolev Embedding and Hardy–Littlewood–Sobolev inequality, we have ∫ ∫ 1 (u(x) − u(y))2 1 2 Jλ,β (u) = dxdy + (λV (x) − β)|u| dx 2 R2n |x − y|n+α 2 Rn ∫ 1 −µ 2∗ 2∗ − (|x| ∗ |u| µ )|u| µ dx 2 · 2∗µ Rn ∫ ∫ (u(x) − u(y))2 1 1 2 dxdy + (λV (x) − β)|u| dx ≥ 2 R2n |x − y|n+α 2 Rn 2·2∗
−C1 ∥u∥2∗ µ α
2·2∗ µ
≥ C2 ∥u∥2λ − C1 ∥u∥λ
.
Then, we can choose ∥u∥λ = ρ sufficiently small, such that Jλ,β (u) ≥ s > 0. (ii) For any u ∈ E \ {0}, ∫ ∫ t2 (u(x) − u(y))2 t2 2 Jλ,β (tu) = dxdy + (λV (x) − β)|u| dx 2 R2n |x − y|n+α 2 Rn ∗ ∫ t2·2µ −µ 2∗ 2∗ − (|x| ∗ |u| µ )|u| µ dx. ∗ 2 · 2µ R n So we can choose t > 0 large enough, such that Jλ,β (tu) < 0. Taking e = tu, we complete the proof.
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
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By the Mountain Pass Theorem, we know that there exists a (P S)c sequence such that Jλ,β (un ) → c and → 0 at the minimax level
′ Jλ,β (un )
cλ,β = inf max Jλ,β (γ(t)) > 0, γ∈Γ t∈[0,1]
where Γ := {γ ∈ C([0, 1], E) : γ(0) = 0, Jλ,β (γ(1)) < 0}. Also note that the critical points of Jλ,β lie on the Nehari manifold ′ Mλ,β = {u ∈ E \ {0} : ⟨Jλ,β (u), u⟩ = 0} { ∫ = u ∈ E \ {0} : ⟨(Lλ − β)u, u⟩ =
(|x|
−µ
} 2∗ 2∗ ∗ |u| µ )|u| µ dx .
Rn
It thus implies that cλ,β := inf max Jλ,β (γ(t)) = γ∈Γ t∈[0,1]
inf
u∈Mλ,β
Jλ,β (u).
Mλ,β is radially diffeomorphic to { V=
∫
−µ
v∈E:
(|x|
} ∗ |u| )|u| dx = 1 . 2∗ µ
2∗ µ
Rn α/2
By our assumption V (x) = 0 on Ω . Hence, when u ∈ H0 (Ω ), Jλ,β (u) becomes ∫ ∫ ∫ 1 1 (u(x) − u(y))2 1 2 −µ 2∗ 2∗ Jβ,Ω (u) = dxdy − β|u| dx − (|x| ∗ |u| µ )|u| µ dx. 2 R2n |x − y|n+α 2 Ω 2 · 2∗µ Ω Its Nehari manifold is { Mβ,Ω :=
α/2
u ∈ H0
∫ (Ω ) \ {0} : ⟨(L0 − β)u, u⟩ =
(|x|
−µ
} 2∗ 2∗ ∗ |u| µ )|u| µ dx .
Ω
Mβ,Ω is radially diffeomorphic to { ∫ α/2 VΩ = v ∈ H0 (Ω ) :
(|x|
−µ
} ∗ |u| )|u| dx = 1 . 2∗ µ
2∗ µ
Rn
Set cβ,Ω =
inf
u∈Mβ,Ω
Jβ,Ω (u) = inf max Jβ,Ω (γ(t)), γ∈Γ t∈[0,1]
where α/2
Γ := {γ ∈ C([0, 1], H0
(Ω )) : γ(0) = 0, Jβ,Ω (γ(1)) < 0}.
Following the argument by Benci and Cerami [3] we can easily derive the similar result as in [6]. Proposition 3.1. If u ∈ Mλ,β is a critical point of Jλ,β such that Jλ,β (u) < 2cλ,β , then u does not change sign. Hence, |u| is a solution of (1). Lemma 3.2. If 0 < β < β1 and λ ≥ λ(β), then 0 < cλ,β < cβ,Ω <
2n−µ n+α−µ Sα,H n+α−µ . 4n − 2µ
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
109
Proof . From VΩ ⊂ V and ⟨Lλ v, v⟩ = ⟨L0 v, v⟩, It follows cλ,β ≤ cβ,Ω . In [13], the authors showed that for 2n−µ
n+α−µ 0 < β < β1 , cβ,Ω < n+α−µ and cβ,Ω is achieved at some u > 0. Therefore, cλ,β < cβ,Ω because 4n−2µ Sα otherwise cλ,β would be achieved at u which vanishes outside Ω . This contradicts the maximum principle.
We are now ready to prove Theorems 1.1 and 1.2. Proof of Theorem 1.1. Let {un } be the minimizing sequence for Jλ,β on Mλ,β . By Ekeland’s variational principle, we may assume that it is a PS sequence. It follows from Proposition 2.2 and Lemma 3.2 that a subsequence converges to a least energy solution of (1). Proof of Theorem 1.2. Let {un } be a sequence of solutions of (1) such that λn → ∞, ⟨Jλ′ n ,β (un ), un ⟩ → 0 and Jλn ,β (un ) → c <
2n−µ
n+α−µ n+α−µ 4n−2µ Sα,H
1 Jλn ,β (un ) = 2
∫ R2n
for 0 < β < β1 . Then
(un (x) − un (y))2 n+α
|x − y|
1 dxdy + 2
∫
2
(λn V (x) − β)|un | dx Rn
∫ 1 −µ 2∗ 2∗ − (|x| ∗ |un | µ )|un | µ dx 2 · 2∗µ Rn n+α−µ = ⟨(Lλn − β)un , un ⟩. 4n − 2µ
(9)
It follows from Lemma 2.2 that n+α−µ C∥un ∥2λn 4n − 2µ n+α−µ ≤ ⟨(Lλn − β)un , un ⟩ 4n − 2µ = Jλn ,β (un ) 2n−µ n+α−µ < Sα,H n+α−µ . 4n − 2µ α/2
Thus {un } is bounded in E. Lemma 2.1 shows that there exists a u ∈ H0 (Ω ) such that un ⇀ u in E and un → u in L2 (Rn ). Since {un } is a solution of (1), it holds ∫ ∫ (un (x) − un (y))(φ(x) − φ(y)) dxdy + (λn V (x) − β)un φdx n+α 2n |x − y| Rn ∫R −µ 2∗ 2∗ −1 = (|x| ∗ |un | µ )|un | µ φdx Rn
∫ α/2 for all φ ∈ E. If φ ∈ H0 (Ω ), then Rn λn V (x)un φdx = 0 for all n. Let n → ∞, we obtain ∫ ∫ ∫ (u(x) − u(y))(φ(x) − φ(y)) −µ 2∗ 2∗ −1 dxdy − βuφdx = (|x| ∗ |u| µ )|u| µ φdx n+α |x − y| R2n Ω Ω α/2
for all φ ∈ H0 (Ω ). This implies that u is a solution of (3). Next we want to show un → u in E. Let wn = un − u. Since V (x) = 0 for x ∈ Ω , it is easy to see that ⟨(Lλn − β)un , un ⟩ = ⟨(L0 − β)u, u⟩ + ⟨(Lλn − β)wn , wn ⟩ + o(1). By the Br´ezis–Lieb Lemma [11], ∫ −µ 2∗ 2∗ (|x| ∗ |un | µ )|un | µ dx ∫Ω ∫ −µ 2∗ 2∗ −µ 2∗ 2∗ = (|x| ∗ |u| µ )|u| µ dx + (|x| ∗ |wn | µ )|wn | µ dx + o(1). Ω
Ω
(10)
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
110
Since {un } is a sequence of solutions for (1) and u is a solution of (3), It holds ∫ −µ 2∗ 2∗ ⟨(Lλn − β)wn .wn ⟩ = (|x| ∗ |wn | µ )|wn | µ dx + o(1). Ω
∫
We can assume ⟨(Lλn − β)wn .wn ⟩ → b and for b > 0, we have (∫
Rn
−µ
Sα,H
(|x|
(|x|
−µ
2∗
2∗
∗ |wn | µ )|wn | µ dx → b. Next we prove b = 0. If not,
2∗ µ
2∗ µ
n−α ) 2n−µ
∗ |wn | )|wn | dx
Rn 2
|wn (x) − wn (y)|
∫ ≤
n+α
|x − y|
R2n
dxdy
2
|wn (x) − wn (y)|
∫ ≤ R2n
≤ ⟨(Lλn
∫ dxdy + (λn V (x) − β)wn2 dx n+α c |x − y| F ∫ 2 − β)wn , wn ⟩ + β wn dx F
= ⟨(Lλn − β)wn , wn ⟩ + o(1) ∫ −µ 2∗ 2∗ = (|x| ∗ |wn | µ )|wn | µ dx + o(1). Rn
It follows that (∫ Sα,H ≤
−µ
(|x|
2∗ µ
2∗ µ
) n+α−µ 2n−µ
∗ |wn | )|wn | dx
+ o(1)
Rn
(∫ ≤
−µ
(|x|
) n+α−µ 2n−µ ∗ |un | )|un | dx + o(1). 2∗ µ
2∗ µ
Rn 2n−µ
n+α−µ , we have Moreover, by (9) and the fact Jλn ,β (un ) → c < n+α−µ 4n−2µ Sα,H ∫ 2n−µ 2n−µ −µ 2∗ 2∗ n+α−µ Sα,H ≤ lim (|x| ∗ |un | µ )|un | µ dx < Sα,H n+α−µ ,
n→∞
Rn
which is a contradiction. Consequently, ∫ ⟨(Lλn − β)wn .wn ⟩ → 0,
(|x|
−µ
2∗
2∗
∗ |wn | µ )|wn | µ dx → 0.
Rn
Combining this with (10), it gives ⟨(Lλn − β)u, u⟩ = ⟨(L0 − β)un , un ⟩ + o(1).
(11)
Hence, ∫ Rn
V (x)u2n dx ≤
∫ Rn
λn V (x)u2n dx =
∫ Rn
λn V (x)wn2 dx ≤ ⟨(Lλn − β)wn , wn ⟩ + o(1),
here we use the fact un = wn in Rn \ Ω and V (x) = 0 in Ω . Therefore, have un → u in E.
∫
Rn
V (x)u2n dx → 0 and by (11) we
Corollary 3.1. For each 0 < β < β1 , limλ→∞ cλ,β = cβ,Ω . 2n−µ
n+α−µ . And from Theorem 1.1, we know Proof . By Lemma 3.2, we have cλ,β → c ≤ cβ,Ω < n+α−µ 4n−2µ Sα,H cλ,β is achieved for λ ≥ λ(β). Thus Theorem 1.2 implies that c is achieved by Jβ,Ω on Mβ,Ω . Hence, c ≥ cβ,Ω .
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
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4. Proof of Theorem 1.3 In this section, using the category theory introduced by Benci and Cerami [3], we prove the existence and multiplicity of weak solutions for problem (1). First we consider the equation { −µ 2∗ 2∗ −1 (−△)α/2 u = (|x| ∗ |u| µ )|u| µ , x ∈ Rn , (12) u ∈ Dα/2,2 (Rn ). Its energy functional is J∗ (u) =
1 2
2
|u(x) − u(y)|
∫ R2n
dxdy −
n+α
|x − y|
1 2 · 2∗µ
∫
−µ
(|x|
2∗
2∗
∗ |u| µ )|u| µ dx,
Rn
and its Nehari manifold is M∗ = {u ∈ Dα/2,2 (Rn ) \ {0} : ⟨J∗′ (u), u⟩ = 0}. Let c∗ := inf J∗ (u), u∈M∗
equivalently, we have c∗ =
max J∗ (tu).
inf
u∈D α/2,2 (Rn )\{0} t≥0
From the introduction, it is easy to see that c∗ =
2n−µ n+α−µ Sα,H n+α−µ . 4n − 2µ
Remark 4.1. From [13], we can see that cβ,Ω can be achieved by a function uβ ∈ Mβ,Ω . Lemma 4.1. For 0 < β < β1 and uβ ∈ Mβ,Ω , let tβ > 0 be the unique value such that tβ uβ ∈ M∗ . Then lim tβ = 1.
β→0
Proof . Recall that M∗ = {u ∈ H α/2 (Rn ) \ {0} : ⟨J∗′ (u)u⟩ = 0}, and the fact tβ uβ ∈ M∗ , we have ∫ ∫ 2 |uβ (x) − uβ (y)| 2·2∗ −µ 2∗ 2∗ µ t2β dxdy = t (|x| ∗ |uβ | µ )|uβ | µ dx. β n+α |x − y| R2n Rn Since uβ ∈ Mβ,Ω , it holds ∫ 2 |uβ (x) − uβ (y)| R2n
n+α
|x − y|
∫ dxdy − β Rn
u2β dx =
∫
−µ
(|x|
2∗
2∗
∗ |uβ | µ )|uβ | µ dx.
Rn
For 0 < β < β1 , (
1 1 − 2 2 · 2∗µ
= Jβ,Ω (uβ ) − ≤
) (∫
2
|uβ (x) − uβ (y)|
Rn
|x − y|
1 J ′ (uβ )uβ 2 · 2∗µ β,Ω
2n−µ n+α−µ Sα,H n+α−µ . 4n − 2µ
n+α
∫ dxdy − β Rn
) u2β dx
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
112
Thus,
∫
R2n
|u(x)−u(y)|2 dxdy |x−y|n+α
is uniformly bounded by β, and 2
|uβ (x) − uβ (y)|
∫ R2n
|x − y|
n+α
∫ dxdy =
(|x|
−µ
2∗
2∗
∗ |uβ | µ )|uβ | µ dx + o(1).
Rn
We obtain lim tβ = 1.
β→0
Lemma 4.2. limβ→0 cβ,Ω = c∗ , where c∗ and cβ,Ω are defined as before. Proof . From Lemma 3.2, we have cβ,Ω < c∗ , thus lim cβ,Ω ≤ c∗ .
β→0
To prove the Lemma, it is sufficient to show that lim cβ,Ω ≥ c∗ .
β→0 α/2
For any u0 ∈ H0
(Ω ), max Jβ,Ω (tu0 ) t≥0 { ∫ ∫ 2 |u0 (x) − u0 (y)| t2 t2 dxdy − β u20 dx = max n+α t≥0 2 R2n 2 |x − y| Ω } ∫ 2·2∗ µ t −µ 2∗ 2∗ − (|x| ∗ |u0 | µ )|u0 | µ dx 2 · 2∗µ Ω ∫ 2 ) 2n−µ |u0 (x)−u0 (y)|2 (∫ n + α − µ R2n |x−y|n+α dxdy − β Ω u0 dx n+α−µ . = n−α ∫ −µ 2∗ 2∗ 4n − 2µ ( (|x| ∗ |u | µ )|u | µ dx) 2n−µ 0
Ω
0
Then, ∫ 2n−µ |u0 (x)−u0 (y)|2 ( ) n+α−µ dxdy R2n n+α−µ |x−y|n+α lim max Jβ,Ω (tu0 ) = n−α β→0 t≥0 4n − 2µ (∫ (|x|−µ ∗ |u |2∗µ )|u |2∗µ dx) 2n−µ 0 0 Ω 2n−µ n+α−µ ≥ Sα,H n+α−µ . 4n − 2µ Because u0 is chosen arbitrarily, we have lim cβ,Ω ≥ c∗ .
β→0
This completes the proof. Next, we introduce some notations for the proof of Theorem 1.3. Since Ω is a bounded smooth domain, we may fix r > 0 small enough such that + Ω2r = {x ∈ Rn : dist(x, Ω ) < 2r}
and Ωr− = {x ∈ Ω : dist(x, ∂Ω ) > r}
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
113
are homotopically equivalent to Ω . Moreover, we assume Br = {x ∈ Rn : |x| < r} ⊂ Ω . We define cβ,r = cβ,Br . Then as in the proof of Lemma 3.2, we have cβ,Ω < cβ,r <
2n−µ n+α−µ Sα,H n+α−µ , 4n − 2µ
for 0 < β < β1 . α/2 For 0 ̸= u ∈ H0 (Ω ) we consider the corresponding center of mass ϕ(u) = (ϕ1 (u), . . . , ϕn (u)), where ∫ ϕi (u) =
xi ·
Q
∫ Q
|u(x)−u(y)|2 dxdy |x−y|n+α
|u(x)−u(y)|2 dxdy |x−y|n+α
, i = 1, 2, . . . , n.
Lemma 4.3. There is a β∗ = β∗ (r) ∈ (0, β1 ) such that for 0 < β ≤ β∗ and every u ∈ Mβ,Ω with Jβ,Ω (u) ≤ cβ,r , we have ϕ(u) ∈ Ωr+ . Proof . We argue by contradiction. Assume that there exists εn → 0, βn → 0 and un ∈ Mβn ,Ω such that Jβn ,Ω (un ) < cβn ,r + εn and ϕ(un ) ̸∈ Ωr+ . α/2
It is easy to see {un } is bounded in H0 (Ω ). By Lemma 4.2, limβn →0 Jβn ,Ω = c∗ . Let tn be the unique value such that tn un ∈ M∗ , from Lemma 4.1, we have tn → 1. Since Jβn ,Ω (un ) − J∗ (tn un ) ( )∫ ∫ 2 |un (x) − un (y)| βn βn n+α−µ 2 (1 − t2n ) − = dxdy − |un | dx n+α ∗ 4n − 2µ 2 2 · 2 2n |x − y| R Ω µ = o(1). we have J∗ (tn un ) → c∗ . Therefore, {tn un } is a (PS) sequence of J∗ at level c∗ . From Theorem 2.15 in [8], the minimum of J∗ on M∗ is achieved by the function of the form ( Ubn (x − x0 ) = C
) n−α 2
bn b2n + |x − x0 |
2
,
for some x0 ∈ Rn , C > 0 and bn > 0. It is easy to prove that tn un − Ubn (x − x0 ) → 0 in Dα/2,2 (Rn ), for some bn ∈ Rn \ {0} and x0 ∈ Ω . Then, we can write tn un = Ubn (· − x0 ) − vn , ∫ (x)−vn (y)|2 where vn is a function satisfying Q |vn|x−y| dxdy → 0 and Ubn (· − x0 ) = vn on Rn \ Ω . We write n+α n x = (x1 , x2 , . . . , xn ) ∈ R . The ith coordinate of the barycenter of un satisfies ∫ 2 |tn un (x) − tn un (y)| ϕ(un )i dxdy n+α |x − y| Q
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
114
∫ xi ·
=
|tn un (x) − tn un (y)| |x − y|
Q
2
dxdy
n+α
∫ 2 2 |vn (x) − vn (y)| |Ubn (x − x0 ) − Ubn (y − x0 )| dxdy + dxdy xi · xi · n+α n+α |x − y| |x − y| Q Q ∫ 2(Ubn (x − x0 ) − Ubn (y − x0 ))(vn (x) − vn (y)) − xi · dxdy n+α |x − y| Q ∫ ∫ 2 2 |vn (x) − vn (y)| |Ubn (x − x0 ) − Ubn (y − x0 )| dxdy + dxdy = x · xi · i n+α n+α |x − y| |x − y| Q Q ∫ 2(Ubn (x − x0 ) − Ubn (y − x0 ))(vn (x) − vn (y)) − dxdy xi · n+α |x − y| Q = An + Bn − 2Cn , ∫
=
0 0 here we use the fact Ubn (· − x0 ) = vn on Rn \ Ω . Let z = x−x and ξ = y−x bn bn , by simple computations, we have ∫ ∫ 2 2 |U1 (z) − U1 (ξ)| |U1 (z) − U1 (ξ)| dzdξ + x dzdξ, An = bn zi · 0(i) n+α n+α |z − ξ| |z − ξ| Q′ Q′
where Q′ = {(z, ξ) : z =
x−x0 bn , ξ
=
y−x0 bn , x
∈ Ω , y ∈ Ω }. Let n → ∞, we know bn → 0. Hence,
∫ bn
Q′
zi ·
|U1 (z) − U1 (ξ)| |z − ξ|
n+α
2
dzdξ → 0.
And it is clear that 2
∫ xi ·
|vn (x) − vn (y)| n+α
|x − y|
Q
By H´ older inequality, ∫ xi ·
2(Ubn (x − x0 ) − Ubn (y − x0 ))(vn (x) − vn (y)) n+α
|x − y|
Q
(∫ ≤ C
(Ubn (x − x0 ) − Ubn (y − x0 ))2 n+α
|x − y|
Q
We also know
→ 0. ∫ |tn un (x)−tn un (y)|2 Q
dzdξ → 0.
|x−y|n+α
dxdy =
∫ Q
x0(i) ϕ(un )i = ∫
) 12 (∫ dxdy Q
|U1 (z)−U1 (ξ)|2 dzdξ |z−ξ|n+α
∫
Q′
Q′
dxdy
(vn (x) − vn (y))2 |x − y|
n+α
) 12 dxdy
+ o(1). Then ,
|U1 (z)−U1 (ξ)|2 dzdξ |z−ξ|n+α
|U1 (z)−U1 (ξ)|2 dzdξ |z−ξ|n+α
+ o(1) .
+ o(1)
¯ , which is a contradiction. We complete the proof. Moreover, ϕ(un ) ∈ Ω ¯ ⊂ BR and let As in [2], we choose R > 0 with Ω { 1, 0 ≤ t ≤ R, ξ(t) = R/t, R ≤ t. Define ∫ ϕ0 (u) =
2
ξ(|x|)x · |u(x)−u(y)| dydx |x−y|n+α for u ∈ Dα/2,2 (Rn ) \ {0}. ∫ 2 |u(x)−u(y)| dydx R2n |x−y|n+α
R2n
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
115
Lemma 4.4. There exists β∗ = β∗ (r) ∈ (0, β1 ) and for each 0 < β ≤ β∗ , there exist Λ(β) ≥ λ(β) such that + for all λ ≥ Λ(β) and all u ∈ Mλ,β with Jλ,β ≤ cβ,r . ϕ0 (u) ∈ Ω2r Proof . We prove by contradiction. Assume that for β arbitrarily small, there is a sequence {un } such + . Then, by Lemma 2.1, there is a that un ∈ Mλn ,β with λn → ∞, Jλn ,β (un ) → c ≤ cβ,r and ϕ0 (un ) ̸∈ Ω2r α/2 2 n uβ ∈ H0 (Ω ) such that un ⇀ uβ in E and un → uβ in L (R ). We prove in two cases: ∫ −µ 2∗ 2∗ Case i: Rn (|x| ∗ |uβ | µ )|uβ | µ dx ≤ ⟨(L0 − β)uβ , uβ ⟩. Let wn = un − uβ . Since V (x) = 0 in Ω , ⟨(Lλn − β)un , un ⟩ = ⟨(L0 − β)uβ , uβ ⟩ + ⟨(Lλn − β)wn , wn ⟩ + o(1). Moreover, by the Br´ezis–Lieb Lemma [11], ∫ −µ 2∗ 2∗ (|x| ∗ |un | µ )|un | µ dx ∫Rn ∫ −µ 2∗ 2∗ = (|x| ∗ |uβ | µ )|uβ | µ dx + Rn
(|x|
2∗
−µ
(13)
2∗
∗ |wn | µ )|wn | µ dx + o(1).
Rn
Since un ∈ Mλn ,β , ∫ ⟨(Lλn − β)wn , wn ⟩ ≤ We claim that since
∫
Rn
(|x|
−µ
2∗ µ
−µ
(|x|
2∗
Rn
2∗ µ
∗ |wn | )|wn | dx → 0. Otherwise (∫
−µ
Sα,H
(|x|
2∗
∗ |wn | µ )|wn | µ dx + o(1).
2∗
∫
Rn
(|x|
−µ
2∗
2∗
∗ |wn | µ )|wn | µ dx → b > 0. Then,
n−α ) 2n−µ
2∗
∗ |wn | µ )|wn | µ dx
Rn
|wn (x) − wn (y)|
∫ ≤
|wn (x) − wn (y)|
∫
dxdy
|x − y|
R2n
≤
2
n+α
R2n
≤ ⟨(Lλn
2
∫ dxdy + (λn V (x) − β)wn2 dx n+α |x − y| Fc ∫ − β)wn , wn ⟩ + β wn2 dx F
= ⟨(Lλn − β)wn , wn ⟩ + o(1) ∫ −µ 2∗ 2∗ = (|x| ∗ |wn | µ )|wn | µ dx + o(1), Rn
it follows that (∫ Sα,H ≤
(|x|
−µ
(|x|
−µ
) n+α−µ 2n−µ 2∗ 2∗ + o(1) ∗ |wn | µ )|wn | µ dx
Rn
(∫ ≤
2∗
2∗
) n+α−µ 2n−µ
∗ |un | µ )|un | µ dx
+ o(1).
Rn
Moreover, by (9) and Jλn ,β (un ) → c <
2n−µ
n+α−µ n+α−µ , 4n−2µ Sα,H
∫
2n−µ
−µ
Sα,H n+α−µ ≤ lim
n→∞
which is a contradiction. As a result,
∫
(|x|
2∗
2∗
2n−µ
∗ |un | µ )|un | µ dx < Sα,H n+α−µ ,
Rn −µ
Rn
we have
(|x|
2∗
2∗
∗ |wn | µ )|wn | µ dx → 0. Combining this with (13), it gives
⟨(Lλn − β)uβ , uβ ⟩ = ⟨(L0 − β)un , un ⟩ + o(1).
(14)
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
116
Hence, ∫ ∫Rn ≤ n ∫R
= Rn
V (x)u2n dx λn V (x)u2n dx λn V (x)wn2 dx
≤ ⟨(Lλn − β)wn , wn ⟩ + o(1), ∫ here we use the fact un = wn in Rn \ Ω and V (x) = 0 in Ω . Therefore, Rn V (x)u2n dx → 0 and by (14) we know ∫ ∫ 2 2 |uβ (x) − uβ (y)| |un (x) − un (y)| dxdy = dxdy. lim n+α n+α n→∞ R2n |x − y| |x − y| R2n Therefore, ϕ0 (un ) → ϕ(uβ ). However since Jβ,Ω (uβ ) ≤ limn→∞ Jλn ,β (un ) ≤ cβ,r , it follows from Lemma 4.3 + . that ϕ(uβ ) ∈ Ωr+ . This contradicts our assumption that ϕ0 (un ) ̸∈ Ω2r ∫ ∗ −µ 2∗ 2 µ µ Case ii: Rn (|x| ∗ |uβ | )|uβ | dx > ⟨(L0 − β)uβ , uβ ⟩. In this case tuβ ∈ Mβ,Ω for some t ∈ (0, 1), then Jβ,Ω (tuβ ) =
n+α−µ 2 t 4n − 2µ
|uβ (x) − uβ (y)|
∫ R2n
|x − y|
2
dxdy −
n+α
n+α−µ 2 βt 4n − 2µ
∫
2
|uβ | dx. Rn
Recall that un ∈ Mλn ,β , we have Jλn ,β (un ) =
∫ ∫ 2 n+α−µ n+α−µ |un (x) − un (y)| 2 dxdy + λn V (x)|un | dx n+α 4n − 2µ R2n 4n − 2µ Rn |x − y| ∫ n+α−µ 2 − β|un | dx. 4n − 2µ Rn
Therefore, cβ,Ω ≤ Jβ,Ω (tuβ ) n+α−µ 2 = t 4n − 2µ
|uβ (x) − uβ (y)|
∫
n+α
|x − y|
R2n
n+α−µ n→∞ 4n − 2µ
∫
≤ lim
R2n
2
n+α−µ 2 dxdy − βt 4n − 2µ
∫
2
|uβ | dx Rn
2
∫ n+α−µ 2 2 βt |uβ | dx n+α 4n − 2µ n |x − y| R ∫ 2 |un (x) − un (y)| 2 dxdy + λn V (x)|un | dx n+α n |x − y| R |un (x) − un (y)|
dxdy −
n+α−µ 4n − 2µ R2n ∫ n+α−µ 2 2 − βt |uβ | dx 4n − 2µ n R ∫ ∫ n+α−µ 2 n+α−µ 2 2 ≤ cβ,r + β|un | dx − βt |uβ | dx. 4n − 2µ Rn 4n − 2µ Rn ∫
≤ lim
n→∞
It follows that when n → ∞, ⏐∫ ⏐ ∫ 2 2⏐ ⏐ |un (x) − un (y)| |u (x) − u (y)| 4n − 2µ ⏐ ⏐ β β dxdy − t2 (cβ,r − cβ,Ω ). ⏐ ⏐ dxdy ≤ n+α n+α ⏐ R2n ⏐ n +α−µ 2n |x − y| |x − y| R Since |cβ,r − cβ,Ω | → 0 as β → 0, this implies |ϕ0 (un(β) ) − ϕ(tuβ )| < r for all β sufficiently small. But by + Lemma 4.3, ϕ(tuβ ) ∈ Ωr+ , whereas ϕ0 (un(β) ) ̸∈ Ω2r .
P. Ma, J. Zhang / Nonlinear Analysis 164 (2017) 100–117
117
For a given function J : M → R, we set J ≤b = {z ∈ M : J(z) ≤ b}. We will use the following result of the Lusternik–Schnirelmann theory to prove Theorem 1.3. Proposition 4.1 ([6]). Let J ∈ C 1 (M, R) be an even functional on a complete symmetric C 1,1 submanifold M ⊂ V \ {0} of some Banach space V and J is bounded below and satisfies the (P S)c condition for all c ≤ b. Moreover, define the following two maps l
ϕ
X→ − J ≤b − → Y, where ϕ ◦ l is a homotopy equivalence, and ϕ(z) = ϕ(−z) for all z ∈ M ∩ J ≤b . Then J has at least cat(X) pairs of critical points with J(z) = J(−z) ≤ b. Proof of Theorem 1.3. For 0 < β ≤ β∗ and λ ≥ Λ(β), we define two maps l
≤c
ϕ
0 + Ωr− → − Mλ,β ∩ Jλ,ββ,r −→ Ω2r
as follows: The map ϕ0 is the one defined above. Lemma 4.4 shows that ϕ0 is well defined. Let ur ∈ α/2 H0 (Br (x)) ⊂ E be the minimizer of Jλ,β on Mβ,Br with ur > 0 and set l(x) = ur (· − x). Obviously, ϕ0 ◦ l is a homotopy equivalence. Since l(x) = 0 in Rn \ Ω for every x ∈ Ωr− , it follows that l(x) ∈ Mλ,β + and that Jλ,β (l(x)) = Jβ,Br (l(x)) = cβ,r . Since ur is radially symmetric, ϕ0 (l(x)) = x ∈ Ω2r for every − x ∈ Ωr . Clearly, Jλ,β (u) = Jλ,β (−u) and ϕ0 (u) = ϕ0 (−u) for every u ∈ E \ {0}. On the other hand, 2n−µ
n+α−µ . Then by Proposition 2.2, J cβ,r < n+α−µ λ,β satisfies the (P S)c condition for all c ≤ cβ,r . It 4n−2µ Sα,H follows from the Lusternik–Schnirelmann theory, Proposition 4.1 and Lemma 4.4 that (1) has at least cat(Ω ) positive solutions.
Acknowledgment The authors were partially supported by NSFC (NO. 11571176). References
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