Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 13, pp. 1–17. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.13 EXISTENCE OF SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS WITH ROBIN BOUNDARY CONDITIONS JUNHUI XIE, PENGFEI LI Abstract. This article studies the existence of solutions for the fractional p-Laplacian problem (−∆)spu = λ|u|q−2u+ |u|r−2u |x|α , in Ω, Ns,pu(x) + β(x)|u|p−2u = 0, in Rn\Ω, where Ω is a smooth bounded domain in Rn containing 0 with smooth bound- ary, (−∆)sp denotes the fractional p-Laplace operator and λ > 0, 1 < q < p < r < p∗α, p ∗ α is the fractional critical Hardy-Sobolev exponent for 0 ≤ α < ps < n and 0 < s < 1. By using fibering maps and Nehari manifold, we obtain the existence of solution for Hardy-Sobolev subcritical and critical cases. 1. Introduction Let Ω be a smooth bounded domain in Rn containing 0 with smooth boundary. We consider the fractional p-Laplacian Robin problem (−∆)spu = λ|u|q−2u+ |u|r−2u |x|α , in Ω, Ns,pu(x) + β(x)|u|p−2u = 0, in Rn\Ω, (1.1) where λ is a positive parameter, 0 < s < 1, 0 ≤ α < ps < n, 1 < q < p < r < p∗α and p∗α is the fractional critical Hardy-Sobolev exponent. The fractional p-Laplace operator (−∆)sp is defined by (−∆)spu(x) = cn,s,pP.V. ∫ Rn |u(y)− u(x)|p−2(u(y)− u(x)) |x− y|n+ps dy, where cn,s,p is a suitable positive normalization constant only depending on n, s and p, while Ns,pu(x) = cn,s,p ∫ Ω |u(y)− u(x)|p−2(u(y)− u(x)) |x− y|n+ps dy is the nonlocal normal derivative associated to (−∆)sp, see [6, 14] and [8] for its introduction in the case p = 2. Besides, β(x) is a given nonnegative function. We 2020 Mathematics Subject Classification. 35R11, 35S15, 35A15, 47G20. Key words and phrases. Fractional p-Laplacian; Nehari manifold; Robin boundary; Hardy-Sobolev exponent. ©2025. This work is licensed under a CC BY 4.0 license. Submitted September 28, 2024. Published February 18, 2025. 1 2 J. XIE, P. LI EJDE-2025/13 would like to point out that the Neumann operator Ns,2u(x) recovers the classical Neumann condition as a limit case, and has a clear probabilistic and variational interpretation a well, see [8] for the details. Recently, partial differential equations involving the fractional Laplacian oper- ator (−∆)s with s ∈ (0, 1) has received a special attention, because its arises in a quite natural way in many different contexts, such as, among the others, the thin obstacle problem, optimization, anomalous diffusion, ultra-relativistic limits of quantum mechanics, quasi-geostrophic flows, minimal surfaces, materials science and water waves, for more detail see [7]. In the framework of nonlocal problems, the following Brezis-Nirenberg type problem for the fractional p-Laplacian is considered (−∆)spu = λ|u|p−2u+ |u|p ∗ s−2u, in Ω, u = 0, in Rn\Ω, (1.2) where s ∈ (0, 1), n > sp, λ > 0 and p∗s = np n−sp is the fractional critical Sobolev exponent. In [12] the authors proved, among other results, that the above problem has a nontrivial weak solution for all λ > 0 provided that n3+s3p3 n(n+s) > sp2 and Ω is the domain of class C1,1. The fractional p-Laplace elliptic problems with Hardy term have also been stud- ied by many researchers. Chen-Mosconi-Squassina [5] studied the problem (−∆)spu = λ|u|q−2u+ |u|p∗ α−2u |x|α , in Ω, u = 0, in Rn\Ω, (1.3) where p ≤ q < np n−ps . By finding the minimizer of the corresponding energy func- tional on positive Nehari and sigh-changing Nehari sets, the existence of positive and sigh-changing least energy solutions for the above problem were established in [5]. Chen-Gui [4] studied the existence of multiple solutions for the fractional p- Kirchhoff problem M (∫ R2n |u(x)− u(y)|p |x− y|n+ps dx dy ) (−∆)spu = λ|u|q−2u+ |u|r−2u |x|α , in Ω, u = 0, in Rn\Ω. (1.4) It is worth pointing out that Mugnai-Pinamonti-Vecchi [13] considered the bound- ary value problem driven by the p-fractional Laplacian with nonlocal Robin bound- ary conditions (−∆)spu = f(x, u), in Ω, Ns,pu(x) + β(x)|u|p−2u = 0, in Rn\Ω, (1.5) they provided necessary and sufficient conditions which ensure the existence of a unique positive solution for this problem. Recently, a wide interest arised in p-fractional Laplacian with nonlocal Robin boundary value problem, see [3, 5, 9, 11] and the references therein. In this article, we mainly focus on the existence of solution for fractional p- Laplacian Robin problem (1.1). To show our main result, we first give some nota- tion. For any couple of functions (u, v) and CΩ = Rn\Ω, we denote Hs,p(u, v) . = cn,s,p 2 ∫ ∫ R2n\(CΩ)2 |u(x)− u(y)|p−2(u(x)− u(y))(v(x)− v(y)) |x− y|n+sp dx dy. EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 3 Next, we define the fractional Sobolev space, which can be suitably modeled to deal with fractional Robin boundary conditions. Precisely, given β(x) ∈ L∞(Rn\Ω), we define the function space Xs,p β . = {u : Rn → R measurable: ∥u∥Xs,p β < +∞}, where ∥u∥p Xs,p β . = ∫ Ω |u|pdx+ ∫ ∫ R2n\(CΩ)2 |u(x)− u(y)|p |x− y|n+sp dx dy + ∫ Rn\Ω |β(x)||u|pdx = ∥u∥pLp(Ω) + [u]ps,p + ∥u∥pLp(β;Rn\Ω) . Observe that [u]s,p . = (∫ ∫ R2n\(CΩ)2 |u(x)− u(y)|p |x− y|n+sp dx dy )1/p is strictly related to the Gagliardo seminorm [u] = (∫ Ω×Ω |u(x)− u(y)|p |x− y|n+sp dx dy )1/p . We denote the fractional Hardy-Sobolev constant Sα by Sα = inf u∈W s,p(Ω)\{0} ∥u∥p ∥u∥p Lp∗α (Ω,|x|−αdx) and Lp∗ α(Ω, |x|−αdx) is the weighted Lp∗ α space with norm ∥u∥Lp∗α (Ω,|x|−αdx) = (∫ Ω |u|p∗ α |x|α dx )1/p∗ α , where p∗α = (n−α)p n−ps . When α = 0, S0 is the best fractional Sobolev constant. Moreover, p∗α = (n−α)p n−ps arises from the general fractional Hardy-Sobolev inequality(∫ Rn |u|p∗ α |x|α dx )1/p∗ α ≤ C(n, p, α) (∫ R2n |u(x)− u(y)|p |x− y|n+ps )1/p , for u ∈ W s,p 0 (Ω). Definition We say that u ∈ Xs,p β is a weak solution of (1.1) if Hs,p(u, φ) + ∫ Rn\Ω β(x)|u|p−2uφdx = λ ∫ Ω |u|q−2uφdx+ ∫ Ω |u|r−2uφ |x|α dx (1.6) for all φ ∈ Xs,p β . Formally, weak solutions of (1.1) coincide with the critical points of the functional Iλ(u) . = 1 p ∥u∥p Xs,p β + 1 p ∫ Rn\Ω β(x)|u|pdx− λ q ∫ Ω |u|qdx− 1 r ∫ Ω |u|r |x|α dx. (1.7) We can see that ⟨I ′λ(u), φ⟩ = ∥u∥p Xs,p β + ∫ Rn\Ω β(x)|u|pdx− λ ∫ Ω |u|qdx− ∫ Ω |u|r |x|α dx. (1.8) Now we state our main results. Theorem 1.1. Let 0 ≤ α < ps < n and 1 < q < p < r < p∗α. Then there exists Λ such that problem (1.1) has at least two solutions for λ ∈ (0, q pΛ). 4 J. XIE, P. LI EJDE-2025/13 Theorem 1.2. Let 0 ≤ α < ps < n, r = p∗α, q ≥ n(p−1) n−ps , then there exists λ∗ > 0 such that problem (1.1) has at least two solutions for λ ∈ (0, λ∗). This article organized as follows: we give some preliminary results in Section 2. In Section 3, we prove Theorem 1.1 by variational approach. Section 4 gives the proof of Theorem 1.2. 2. Preliminaries We want to collect several technical results needed in the upcoming sections and we will give some notations and properties of the Nehari manifold, which will be used to prove our main results. We define the manifold Nλ = {u ∈ X0\{0} : ⟨I ′λ(u), u⟩ = 0}. It is clear that all critical points of Iλ must lie on Nλ. We can see that u ∈ Nλ if and only if u ̸= 0 and ∥u∥p Xs,p β + ∫ Rn\Ω β(x)|u|pdx = λ ∫ Ω |u|qdx+ ∫ Ω |u|r |x|α dx. Set Ψλ(u) = ⟨I ′λ(u), u⟩. Then for u ∈ Nλ, we have ⟨Ψ′ λ(u), u⟩ = p∥u∥p Xs,p β + p ∫ Rn\Ω β(x)|u|pdx− λq ∫ Ω |u|qdx− r ∫ Ω |u|r |x|α dx = (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx− (r − q) ∫ Ω |u|r |x|α dx = (p− r)∥u∥p Xs,p β + (p− r) ∫ Rn\Ω β(x)|u|pdx− (q − r)λ ∫ Ω |u|qdx. Then Nλ can be divided into the following three parts N+ λ = {u ∈ Nλ|⟨Ψ′ λ(u), φ⟩ > 0}, N− λ = {u ∈ Nλ|⟨Ψ′ λ(u), φ⟩ < 0}, N0 λ = {u ∈ Nλ|⟨Ψ′ λ(u), φ⟩ = 0}. Lemma 2.1. The space Xs,p β is a reflexive Banach space for every 1 < p < ∞. Lemma 2.2. The embedding Xs,p β ↪→ Lq(Ω) is compact for every q ∈ [1, p∗), where p∗ = np n−ps if n < ps, p∗ = ∞ if n ≥ ps. The proofs of Lemmas 2.1 and 2.2 are the same as that [13, Lemmas 2.1 and 2.2] respectively. Lemma 2.3. Suppose u0 is a local minimizer of the functional Iλ on Nλ and u0 ̸∈ N0 λ. Then u0 is a critical point of Iλ. The proof of the above lemma is the same as that in Brown-Zhang [2, Theorem 2.3] and Chen-Gui [4, Theorem 2.1]. Let Λ = ( (p− q)ĈS r/p α r − q )(p−q)/(r−p) |Ω|−1/γSq/p α , EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 5 where |Ω| denotes the measure of Ω and γ = p∗ p∗ − q , Ĉ = (∫ Ω 1 |x| αp∗α r−p∗α ) r−p∗α p∗α . Lemma 2.4. If u0 ∈ Xs,p β \{0}, then there exists unique t0 > 0 such that for any λ ∈ (0,Λ), then there exist t+ > 0 and t− > 0 satisfying t+u ∈ N+ λ , t−u ∈ N− λ . Moreover, Iλ(t +u) = inf 0≤t≤t0 Iλ(tu), Iλ(t −u) = sup t≥0 Iλ(tu). Proof. Fix u0 ∈ Xs,p β \{0}, we consider the map ϕ : R+ → R defined by ϕ(t) = tp−q∥u∥p Xs,p β + tp−q ∫ Rn\Ω β(x)|u|pdx− tr−q ∫ Ω |u|r |x|α dx. Obviously, ϕ(0) = 0, limt→∞ ϕ(t) = −∞, ϕ′(t) = tp−q−1g(t), where g(t) = (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx− tr−p(r − q) ∫ Ω |u|r |x|α dx. Hence g′(t) = −tr−p−1(r − q)(r − p) ∫ Ω |u|r |x|α dx < 0. Then, we have g(t) is strictly decreasing on [0,+∞) and g(0) ≥ 0, limt→∞ g(t) = −∞, so there exists a unique t0 such that g(t0) = 0. Then ϕ(t) is strictly increasing on [0, t0] and strictly decreasing on (t0,+∞), which reaches the maximum at t0. Now ϕ(t0) = t−q 0 ( tp0∥u∥ p Xs,p β + tp0 ∫ Rn\Ω β(x)|u|pdx− tr0 ∫ Ω |u|r |x|α dx ) , where t0 = ( (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx (r − q) ∫ Ω |u|r |x|α dx )1/(r−p) . Using Hölder and Hardy-Sobolev inequalities, we obtain∫ Ω |u|qdx ≤ |Ω|1/γS−q/p 0 ∥u∥q Xs,p β (2.1)∫ Ω |u|r |x|α dx ≤ S−r/p α Ĉ−1∥u∥rXs,p β . (2.2) Combining the definition of t0 and (2.2), we have t0 > ( (p− q)∥u∥p Xs,p β (r − q) ∫ Ω |u(x)|r |x|α dx )1/(r−p) ≥ ( p− q (r − q)Ĉ−1S −r/p α )1/(r−p) ∥u∥−1 Xs,p β . = t′ ≥ 0, which implies that ϕ(t0) ≥ ϕ(t′) > t′p−q∥u∥p Xs,p β − t′r−q ∫ Ω |u|r |x|α dx ≥ ( (p− q)ĈS r/p α r − q )(p−q)/(r−p) ∥u∥q Xs,p β ≥ λ ∫ Ω |u|qdx (2.3) 6 J. XIE, P. LI EJDE-2025/13 for λ ∈ (0,Λ) by (2.1). Consequently, there exist t+ > 0 and t− > 0 with t+ < t0 < t− such that ϕ(t+) = ϕ(t−) = λ ∫ Ω |u(x)|qdx, which means t+u ∈ Nλ, t −u ∈ Nλ. If tu ∈ Nλ, then ⟨Ψ′ λ(tu), tu⟩ = tq+1ϕ′(t). According to t+u ∈ Nλ, t −u ∈ Nλ and ϕ′(t+) > 0, ϕ′(t−) < 0, then t+u ∈ N+ λ and t−u ∈ N− λ . Since ⟨I ′λ(tu), tu⟩ = tq(ϕ(t)− ∫ Ω |u(x)|qdx), we can see that Iλ(t −u) > Iλ(tu) > Iλ(t +u) for t ∈ [t+, t−], and Iλ(tu) > Iλ(t +u) for t ∈ [0, t+]. Thus Iλ(t +u) = inf 0≤t≤t0 Iλ(tu), Iλ(t −u) = sup t≥0 Iλ(tu). □ 3. Proof of Theorem 1.1 Definition We say that {uk} is a (PS)c sequence in Xs,p β for Iλ, if Iλ(uk) → c and I ′λ(uk) → 0 in X−s,p β as k → ∞. We say that Iλ satisfies the (PS)c condition if any (PS)c sequence {uk} in Xs,p β has a strongly convergent subsequence. Next, we prove some technical lemmas which will be very useful hereinafter. Lemma 3.1. If {uk} ⊂ Xs,p β is a (PS)c sequence for Iλ, then {uk} is bounded in Xs,p β . Proof. If {uk} ⊂ Xs,p β is a (PS)c sequence for Iλ, then Iλ(uk) → c, I ′λ(uk) → 0 in X−s,p β as k → ∞. Namely, 1 p ∥uk∥pXs,p β + 1 p ∫ Rn\Ω β(x)|uk|pdx− 1 q λ ∫ Ω |uk|qdx− 1 r ∫ Ω |uk|r |x|α dx = c+ o(1), (3.1) ∥uk∥pXs,p β + ∫ Rn\Ω β(x)|uk|pdx− λ ∫ Ω |uk|qdx− ∫ Ω |uk|r |x|α dx = o(∥uk∥Xs,p β ). (3.2) By (3.1), (3.2), Hölder inequality and Sobolev inequality, we obtain c+ o(∥uk∥Xs,p β ) = Iλ(uk)− 1 p ⟨I ′λ(uk), uk⟩ = ( 1 p − 1 q )λ ∫ Ω |uk|qdx+ ( 1 p − 1 r ) ∫ Ω |uk|r |x|α dx ≥ ( 1 p − 1 q )|Ω|1/γS−q/p 0 ∥uk∥qXs,p β . Hence {uk} is bounded in Xs,p β . □ Lemma 3.2. For any λ ∈ (0,Λ), we have N0 λ = ∅. Proof. On the contrary, ifN0 λ ̸= ∅, then there exists u ∈ N0 λ, this implies ⟨Ψ′(u), u⟩ = 0, we can deduce that (p− q)∥u∥p Xs,p β ≤ (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx = (r − q) ∫ Ω |u|r |x|α dx, (3.3) EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 7 and (r − p)∥u∥p Xs,p β ≤ (r − p)∥u∥p Xs,p β + (r − p) ∫ Rn\Ω β(x)|u|pdx = (r − q)λ ∫ Ω |u|qdx. (3.4) By (2.2), we obtain (r − q) ∫ Ω |u|r |x|α dx ≤ (r − q)S−r/p α Ĉ−1∥u∥rXs,p β . (3.5) From (3.3) and (3.5), we have ∥u∥Xs,p β ≥ ( (p− q)S r/p α Ĉ r − q )1/(r−p) . (3.6) By (2.1), we have (r − q)λ ∫ Ω |u|qdx ≤ (r − q)λ|Ω|1/γS−q/p 0 ∥u∥q Xs,p β . (3.7) Combining (3.4) and (3.6), it follows that ∥u∥Xs,p β ≤ ( (r − q)λ|Ω|1/γS−q/p 0 r − p )1/(p−q) . (3.8) Hence, by (3.6) and (3.8), we obtain λ ≥ Λ, which is a contradiction. □ Lemma 3.3. The energy functional Iλ is coercive and bounded from below on Nλ for all λ > 0. Proof. According to (2.1), for any λ > 0 and u ∈ Nλ, we can see that Iλ(u) = 1 p ∥u∥p Xs,p β + 1 p ∫ Rn\Ω β(x)|u|pdx− λ q ∫ Ω |u|qdx− 1 r ∫ Ω |u|r |x|α dx = ( 1 p − 1 r )∥u∥p Xs,p β + ( 1 p − 1 r ) ∫ Rn\Ω β(x)|u|pdx− ( 1 q − 1 r )λ ∫ Ω |u|qdx ≥ ( 1 p − 1 r )∥u∥p Xs,p β + ( 1 p − 1 r ) ∫ Rn\Ω β(x)|u|pdx− ( 1 q − 1 r )λ|Ω|1/γS−q/p 0 ∥u∥q Xs,p β . Then Iλ is coercive and bounded from below on Nλ for q < p < r. □ From Lemmas 3.2 and 3.3, for each λ ∈ (0,Λ), we know that Nλ = N+ λ ∪N− λ and Iλ is coercive and bounded from below on N+ λ and N− λ . Therefore we can define cλ = inf Nλ Iλ, c+λ = inf N+ λ Iλ, c−λ = inf N− λ Iλ. We have the following Lemma. Lemma 3.4. (1) If λ ∈ (0,Λ), then cλ ≤ c+λ < 0, (2) If λ ∈ (0, q pΛ), then c−λ > 0. Proof. (1) Let u ∈ N+ λ , then ⟨Ψ′ λ(u), u⟩ > 0, which means that p− q r − q ∥u∥p Xs,p β + p− q r − q ∫ Rn\Ω β(x)|u|pdx > ∫ Ω |u|r |x|α dx. 8 J. XIE, P. LI EJDE-2025/13 Then Iλ(u) = 1 p ∥u∥p Xs,p β + 1 p ∫ Rn\Ω β(x)|u|pdx− λ q ∫ Ω |u|qdx− 1 r ∫ Ω |u|r |x|α dx = ( 1 p − 1 q )∥u∥p Xs,p β + ( 1 p − 1 q ) ∫ Rn\Ω β(x)|u|pdx− ( 1 r − 1 q ) ∫ Ω |u|r |x|α dx < ( 1 p − 1 q )∥u∥p Xs,p β + ( 1 p − 1 q ) ∫ Rn\Ω β(x)|u|pdx + ( 1 q − 1 r ) (p− q r − q ∥u∥p Xs,p β + p− q r − q ∫ Rn\Ω β(x)|u|pdx ) = p− q q ( 1 r − 1 p ) ( ∥u∥p Xs,p β + ∫ Rn\Ω β(x)|u|pdx ) < 0. (3.9) Thus cλ ≤ c+λ < 0 follows from the definition of cλ and c+λ . (2) Similarly, we assume that u ∈ N− λ , then we can deduce that ⟨Ψ′ λ(u), u⟩ < 0, which implies that r − p r − q ∥u∥p Xs,p β + r − p r − q ∫ Rn\Ω β(x)|u|pdx > λ ∫ Ω |u|qdx, and p− q r − q ∥u∥p Xs,p β < p− q r − q ∥u∥p Xs,p β + p− q r − q ∫ Rn\Ω β(x)|u|pdx < ∫ Ω |u|r |x|α dx. By (2.2), we obtain ∥u∥Xs,p β ≥ ( (p− q)S r/p α Ĉ r − q )1/(r−p) . From (2.1), we find that Iλ(u) = 1 p ∥u∥p Xs,p β + 1 p ∫ Rn\Ω β(x)|u|pdx− λ q ∫ Ω |u|qdx− 1 r ∫ Ω |u|r |x|α dx = ( 1 p − 1 r )∥u∥p Xs,p β + ( 1 p − 1 r ) ∫ Rn\Ω β(x)|u|pdx− ( 1 q − 1 r )λ ∫ Ω |u|qdx ≥ ( 1 p − 1 r )∥u∥p Xs,p β − ( 1 q − 1 r )λ|Ω|1/γS−q/p 0 ∥u∥q Xs,p β = ∥u∥q Xs,p β ( ( 1 p − 1 r )∥u∥p−q Xs,p β − ( 1 q − 1 r )λ|Ω|1/γS−q/p 0 ) > ∥u∥q Xs,p β ( ( 1 p − 1 r ) ( (p− q)S r/p α Ĉ r − q )(p−q)/(r−p) − ( 1 q − 1 r )λ|Ω|1/γS−q/p 0 ) > 0 for λ ∈ (0, q pΛ), which implies that c−λ > 0. □ Lemma 3.5. Assume that λ ∈ (0,Λ). Then for each u ∈ Nλ, there exist ε > 0 and a differentiable map h : B(0, ε) ⊂ Xs,p β → R+, with h = 1 such that h(w)(u− w) ∈ EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 9 Nλ and ⟨h′(0), w⟩ = pΛ(u,w) + p ∫ Rn\Ω β(x)|u|p−2uw dx− q ∫ Ω |u|q−2uw dx− r ∫ Ω |u|r−2uw |x|α dx (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|p−2uw dx− (r − q) ∫ Ω |u|r−2uw |x|α dx , (3.10) where Λ(u,w) = ∫ ∫ R2n\(CΩ)2 |u(x)− u(y)|p−2(u(x)− u(y))(v(x)− v(y)) |x− y|n+sp dx dy for each w ∈ Xs,p β . Proof. For u ∈ Nλ, we define the map f : R+ ×Xs,p β → R as follows f(ξ, w) = ⟨I ′λ(ξ(u− w)), ξ(u− w)⟩ = ξp∥u− w∥p Xs,p β + ξp ∫ Rn\Ω β(x)|u− w|pdx −ξq ∫ Ω |u− w|qdx− ξr ∫ Ω |u− w|r |x|α dx (3.11) for ξ ∈ R+, w ∈ Xs,p β . Then we know f(1, 0) = ⟨I ′λ(u), u⟩. In addition combining with Lemma 3.2, we obtain df(1, 0) dξ = p∥u∥p Xs,p β + p ∫ Rn\Ω β(x)|u|pdx− λq ∫ Ω |u|qdx− r ∫ Ω |u|r |x|α dx = (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx− (r − q) ∫ Ω |u|r |x|α dx ̸= 0. (3.12) Using the Implicit Function Theorem, there exist ε > 0 and a C1 map h : B(0, ε) ⊂ Xs,p β → R+ with ξ = h(w) and h(0) = 1 such that ⟨h′(0), w⟩ = pΛ(u,w) + p ∫ Rn\Ω β(x)|u|p−2uw dx− q ∫ Ω |u|q−2uw dx− r ∫ Ω |u|r−2uw |x|α dx (p− q)∥u∥p Xs,p β + (p− q) ∫ Rn\Ω β(x)|u|pdx− (r − q) ∫ Ω |u|r |x|α dx , and f(h(w), w) = 0 for all w ∈ B(0, ε). Hence, ⟨I ′λ(h(w)(u− w)), h(w)(u− w)⟩ = 0. It implies that h(w)(u− w) ∈ Nλ. □ In Lemma 3.5, we replace u ∈ Nλ by u ∈ N− λ and ξ by ξ−, then the conclusion still holds. Moreover, the proof is similar to that in Lemma 3.5. Proposition 3.6. (1) If λ ∈ (0,Λ), then there exists a (PS)cλ sequence {uk} ⊂ Nλ for Iλ. (2) If λ ∈ (0, q pΛ), then there exists a (PS)c−λ sequence {uk} ⊂ N− λ for Iλ. 10 J. XIE, P. LI EJDE-2025/13 Proof. (1) By Ekeland’s Variational Principle, there exists a minimizing sequence {uk} ⊂ Nλ such that Iλ(uk) < cλ + 1 k , Iλ(uk) < Iλ(u) + 1 k ∥u− uk∥Xs,p β , ∀u ∈ Nλ. (3.13) Using that cλ < 0, we obtain Iλ(uk) = ( 1 p − 1 r )∥uk∥pXs,p β +( 1 p − 1 r ) ∫ Rn\Ω β(x)|uk|pdx− ( 1 q − 1 r )λ ∫ Ω |uk|qdx < cλ 2 . This yields cλqr 2(q − r) < λ ∫ Ω |uk|qdx < λ|Ω|1/γS−q/p 0 ∥uk∥qXs,p β , (3.14) ( 1 p − 1 r )∥uk∥pXs,p β < ( 1 q − 1 r )λ ∫ Ω |uk|qdx < ( 1 q − 1 r )λ|Ω|1/γS−q/p 0 ∥uk∥qXs,p β . (3.15) By (3.14) and (3.15), we have ∥uk∥Xs,p β > ( cλqrS q/p 0 2(q − r)λ|Ω|1/γ )1/q , ∥uk∥Xs,p β < ( (r − q)pλ|Ω|1/γ (r − p)qS q/p 0 )1/(p−q) . (3.16) Next we claim that ∥I ′λ(uk)∥X−s,p β → 0 as k → ∞. The proof of this claim is similar to [4, Proposition 3.1], hence we omit it here. From Lemma 3.5 (u ∈ N− λ ), using the same arguments, we obtain (2) of Proposition 3.6. □ Theorem 3.7. If λ ∈ (0,Λ), 1 < q < p < r < p∗α, then there exists u1 ∈ N+ λ and satisfies (1) Iλ(u1) = cλ = c+λ < 0, (2) u1 is a solution of the problem (1.1). Proof. (1) First, we prove Iλ(u1) = cλ. Since Iλ(u1) = 1 p ∥u1∥pXs,p β + 1 p ∫ Rn\Ω β(x)|u1|pdx− λ q ∫ Ω |u1|qdx− 1 r ∫ Ω |u1|r |x|α dx = ( 1 p − 1 r )∥u1∥pXs,p β + ( 1 p − 1 r ) ∫ Rn\Ω β(x)|u1|pdx− ( 1 q − 1 r )λ ∫ Ω |u1|qdx ≤ lim k→∞ inf ( ( 1 p − 1 r )∥uk∥pXs,p β + ( 1 p − 1 r ) ∫ Rn\Ω β(x)|uk|pdx − ( 1 q − 1 r )λ ∫ Ω |uk|qdx ) = lim k→∞ inf Iλ(uk) = cλ. It follows that Iλ(u1) = cλ. Then we claim that cλ = c+λ for u1 ∈ N+ λ . By I ′λ(u1) = 0 and Lemma 3.2, we have u1 ∈ N+ λ ∪ N− λ . Assume that u1 ∈ N− λ , and combining with Lemma 2.4, there exist t− > 0 and t+ > 0 with t− > t+ such that t−u1 ∈ N− λ , t+u1 ∈ N+ λ . In EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 11 particular t+ < t− = 1. Since dIλ(t +u1) dt = 0, d2Iλ(t +u1) dt2 > 0, there exists t ∈ (t+, 1] such that cλ ≤ Iλ(t +u1) < Iλ(tu1) = cλ, which is a contradiction, so u1 ∈ N+ λ . Then cλ = Iλ(u1) ≥ c+λ , this together with the definitions of cλ and we have cλ = c+λ . Hence we finish the proof of Iλ(u1) = cλ = c+λ . (2) By (1) of Proposition 3.6, there exists a bounded minimizing sequence {uk} ⊂ Nλ such that lim k→∞ Iλ(uk) = cλ ≤ c+λ < 0, I ′λ(uk) = ok(1). From Lemma 3.1, we know that {uk} is bounded in Xs,p β . Then there exists u1 ∈ Xs,p β such that, up to a subsequence, uk ⇀ u1 weakly in Xs,p β and uk → u1 strongly in Lθ(Ω, |X|−α) for any θ ∈ [1, p∗α) and 0 ≤ α < ps. In particular, we have λ ∫ Ω |uk|qdx → λ ∫ Ω |u1|qdx, ∫ Ω |uk|r |x|α dx → ∫ Ω |u1|r |x|α dx as k → ∞. Moreover, for all ϕ ∈ Xs,p β , o(1) = ⟨I ′λ(uk), ϕ⟩ = ⟨I ′λ(u1), ϕ⟩+ o(1). Thus, u1 ∈ Nλ is a nonzero solution of the problem (1.1) and Iλ(u1) ≥ cλ. □ Theorem 3.8. If λ ∈ (0, q pΛ), 1 < q < p < r < p∗α, then the functional Iλ has a minimizer u2 ∈ N− λ and satisfies (1) Iλ(u2) = c−λ , (2) u2 is a solution of the problem (1.1). Proof. By Proposition 3.6 (2), there exists a bounded minimizing sequence {uk} ⊂ N− λ such that lim k→∞ Iλ(uk) = c−λ , I ′λ(uk) = ok(1). As in the proof of Theorem 3.7, there exists u2 ∈ N− λ such that Iλ(u2) = c−λ and u2 is a solution of the problem (1.1). □ Proof of Theorem 1.1. By Theorems 3.7 and 3.8, we know that for 0 < λ < q pΛ, then problem (1.1) has two solutions u1 ∈ N+ λ and u2 ∈ N− λ in Xs,p β . Since N+ λ ∩N− λ = ∅, these two solutions are distinct. □ 4. Proof of Theorem 1.2 This section we consider the multiplicity of solutions for the critical case. We need the following lemmas. Lemma 4.1. Let r = p∗α, {uk} ⊂ Xs,p β be a sequence such that Iλ(uk) → c∗ with c∗ < cΛ = ( 1 p − 1 r )Sr/(r−p) α − c̄ r − q r (r − p pq )q/(q−r)( (p− q)λ pq )r/(r−q) and I ′λ(uk) → 0 in X−s,p β . Then there exists a strongly convergent subsequence. 12 J. XIE, P. LI EJDE-2025/13 Proof. By Lemma 3.1, we know that {uk} is bounder in Xs,p β , up to a subsequence, denote by itself, there exists u ∈ Xs,p β such that uk ⇀ u0 weakly in Xs,p β and uk → u0 strongly in Lγ(Ω, |x|−αdx) for any γ ∈ [1, p∗α) and 0 ≤ α < ps < n. Now from [9, Theorem 1.1], we can assume that there exist two positive measure µ, ν on Rn and at most countable set {xj}j∈J ⊆ Ω̄ such that∫ Rn |uk(x)− uk(y)|p |x− y|n+ps dy ⇀ µ, µ ≥ ∫ Rn |u(x)− u(y)|p |x− y|n+ps dy + ∑ j∈J µjδxj , (4.1) |uk|p ∗ α |x|α ⇀ ν, ν = |u|p∗ α |x|α νjδxj , (4.2) µj ≥ Sαν p/p∗ α j , ∀j ∈ J. (4.3) Next we claim that J = ∅. By contradiction, suppose that J ̸= ∅, then there exists j ∈ J , for this xj , define φδ,j(x) = φ( x−xj δ ), where x ∈ Rn, φ ∈ C∞ 0 (Rn) is a smooth cut off function, that is φ = 1 in B(0, 1) and φ = 0 in Rn\B(0, 2). Since ukφδ,j is bounded in Xs,p β , we have that ⟨I ′λ(uk), ukφδ,j⟩ → 0 as k → ∞. Then∫ R2n |uk(x)− uk(y)|p−2(uk(x)− uk(y))(uk(x)φδ,j(x)− uk(y)φδ,j(y)) |x− y|n+ps dx dy = ∫ R2n uk(x)|uk(x)− uk(y)|p−2(uk(x)− uk(y))(φδ,j(x)− φδ,j(y)) |x− y|n+ps dx dy + ∫ R2n φδ,j(y)|uk(x)− uk(y)|p |x− y|n+ps dx dy + ∫ Rn β(x)uk(x) pφδ,j(x)dx = λ ∫ Ω |uk(x)|qφδ,j(x)dx+ ∫ Ω |uk(x)|p ∗ αφδ,j(x) |x|α dx+ o(1). (4.4) Now using Hölder inequality and that uk is bounded in Xs,p β , we obtain∫ R2n uk(x)|uk(x)− uk(y)|p−2(uk(x)− uk(y))(φδ,j(x)− φδ,j(y)) |x− y|n+ps dx dy ≤ C (∫ R2n |uk(x)|p|φδ,j(x)− φδ,j(y)|p |x− y|n+ps dx dy )1/p , (4.5) where C is a positive constant. From [15, Lemma 2.3], it holds that lim δ→0 lim k→∞ ∫ R2n |uk(x)|p|φδ,j(x)− φδ,j(y)|p |x− y|n+ps dx dy = 0. (4.6) From (4.1) and (4.2), we have lim δ→0 lim k→∞ ∫ R2n φδ,j(y)|uk(x)− uk(y)|p |x− y|n+ps dx dy ≥ µj , (4.7) lim δ→0 lim k→∞ ∫ Ω |uk(x)|p ∗ αφδ,j(x) |x|α dx = νj , (4.8) lim δ→0 lim k→∞ λ ∫ Ω |uk(x)|qφδ,j(x)dx = 0. (4.9) From (4.4)-(4.9), we have νj ≥ µj . (4.10) EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 13 Combining (4.3) with (4.10), we obtain νj ≥ S p∗ α/(p∗ α−p) α . (4.11) and c∗ = lim k→∞ (Iλ(uk)− 1 p ⟨I ′λ(uk), uk⟩) = lim k→∞ ( ( 1 p − 1 q )λ ∫ Ω |uk(x)|qdx+ ( 1 p − 1 r ) ∫ Ω |uk(x)|p ∗ α |x|α dx ) ≥ ( 1 p − 1 q )λ ∫ Ω |u(x)|qdx+ ( 1 p − 1 r ) ∫ Ω |u(x)|p∗ α |x|α dx+ ( 1 p − 1 r )νj ≥ ( 1 p − 1 r )Sr/(r−p) α − c̄ r − q r (r − p pq )q/(q−r)( (p− q)λ pq )r/(r−q) , (4.12) where ( 1 q − 1 p )λ ∫ Ω |u(x)|qdx ≤ ( 1 q − 1 p )λ (∫ Ω ( |u(x)|q |x|αq/r )r/qdx )q/r(∫ Ω |x|αq/r·r/(r−q)dx )(r−q)/r = (r q ( 1 p − 1 r ) )q/r(∫ Ω |u(x)|r |x|α dx )q/r(r q ( 1 p − 1 r ) )−q/r × ( 1 q − 1 p )λ (∫ Ω |x|αq/(r−q)dx )(r−q)/r ≤ ( 1 p − 1 r ) ∫ Ω |u(x)|r |x|α dx+ c̄ r − q r (r q r − p pr )q/(q−r)( (p− q)λ pq )r/(r−q) (4.13) by Hölder inequality, Young inequality, and c̄ = ∫ Ω |x|αq/(r−q)dx. According to the definition of cΛ, we have c∗ > cΛ, which is a contradiction. Hence J = ∅, which implies |uk|p ∗ α |x|α → |u|p ∗ α |x|α . Therefore, ⟨I ′λ(uk)− I ′λ(u), uk − u⟩ → 0 as k → ∞. By the well-known Simon inequalities: |α− β|m ≤ C ′ m(|α|m−2α− |β|m−2β)(α− β), for m ≥ 2, C ′′ m ( (|α|m−2α− |β|m−2β)(α− β) )m/2 (|α|m + |β|m)(2−m)/2, for 1 < m < 2, where α, β ∈ Rn, C ′ m, C ′′ m are positive constants depending only on m. Then, we obtain uk → u strongly in Xs,p β as k → ∞. □ In [1] the existence and properties of solutions for the minimization problem (1.6) when α = 0, were investigated. For 0 ≤ α < ps < n, from [10, Theorem 1.1], there exists a minimizer for Sα, for every minimizer Uα, there exist x0 ∈ Rn and a non-increasing u : R+ → R such that Uα(x) = u(|x − x0|). Next we fix a radially symmetric decreasing minimizer Uα = Uα(r) for Sα, multiplying Uα by a positive constant if necessary, we may assume that (−∆)spUα = U p∗ α−1 α |x|α , in Rn. 14 J. XIE, P. LI EJDE-2025/13 Lemma 4.2 ([10]). There exist c1, c2 > 0 and κ > 1 such that c1 r(n−ps)/(p−1) ≤ Uα(r) ≤ c2 r(n−ps)/(p−1) , Uα(κr) Uα(r) ≤ 1 2 for all r ≥ 1. For each δ ≥ ε > 0. Let mε,δ = Uα,ε(δ) Uα,ε(δ)− Uα,ε(κδ) , and gε,δ(t) =  0, if 0 ≤ t ≤ Uα,ε(κδ), mp ε,δ(t− Uα,ε(κδ)), if Uα,ε(κδ) ≤ t ≤ Uα,ε(δ), t+ Uα,ε(δ)(m p−1 ε,δ − 1), if t ≥ Uα,ε(δ). (4.14) The functions gε,δ and Gε,δ are nondecreasing and absolutely continuous. Consider now the radially symmetric nonincreasing function uα,ε,δ(r) = Gε,δ(Uα,ε(r)), which satisfies uα,ε,δ(r) = { Uα,ε(r), if r ≤ δ, 0, if r ≥ κδ. (4.15) Lemma 4.3 ([10]). There exists C̃ > 0 such that for any 0 < 2ε ≤ δ < κ−1δΩ, it holds ∫ R2n |uα,ε,δ(x)− uα,ε,δ(y)|p |x− y|α dx dy ≤ S(n−α)/(ps−α) α + C̃( ε δ )(n−ps)/(p−1), ∫ Rn |up∗ α α,ε,δ| |x|α dx ≥ S(n−α)/(ps−α) α − C̃( ε δ )(n−α)/(p−1). Moreover, for each 1 < q < p∗α, there exists Cq > 0 such that ∫ Rn uα,ε,δ(x) q ≥ Cq  εn− n−ps p q| log ε δ |, if q = n(p−1) n−ps , ε n−ps n(p−1) qδn− n−ps p−1 q, if q < n(p−1) n−ps , εn− n−ps p q, if q > n(p−1) n−ps . (4.16) Lemma 4.4. Assume that 0 ≤ α < ps < n and q ≥ n(p−1) (n−ps) . Then there exist λ̂ > 0 and u0 ∈ Xs,p β \{0} such that supt≥0 Iλ(tu0) < cΛ for all 0 < λ < λ̂, where cΛ is the constant given in Lemma 4.1. In particular, c−Λ < cΛ for all λ satisfying 0 < λ < λ̂. Proof. Let u0 = uα,ε,δ, which is defined in Lemma 4.2, we consider the function f(t) = Iλ(tu0) = 1 p tp∥u0∥pXs,p β + 1 p tp ∫ Rn\Ω β(x)|u0|pdx− λ q tq ∫ Ω |u0|qdx− 1 r tr ∫ Ω |u0|r |x|α dx, with f̃(t) = 1 p tp∥u0∥pXs,p β + 1 p tp ∫ Rn\Ω β(x)|u0|pdx− 1 r tr ∫ Ω |u0|r |x|α dx, for all t > 0, then there exists t∗ = (∥u0∥pXs,p β + ∫ Rn\Ω β(x)|u0|pdx∫ Ω |u0|r |x|α dx )1/(r−p) > 0 EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 15 such that f̃ ′(t∗) = 0 and f̃(t∗) ≥ f̃(t). Next we have sup t≥0 f̃(t) = f̃(t∗) = 1 p tp∗∥u0∥pXs,p β + 1 p tp∗ ∫ Rn\Ω β(x)|u0|pdx− 1 r tr∗ ∫ Ω |u0|r |x|α dx = ( 1 p − 1 r ) ( ∥u0∥pXs,p β + ∫ Rn\Ω β(x)|u0|pdx )r/(r−p) ( ∫ Ω |u0|r |x|α dx )p/(r−p) (4.17) ≤ ( 1 p − 1 r ) ( S (n−α)/(ps−α) α + C̃( εδ ) (n−ps)/(p−1) + ∫ Rn\Ω β(x)|u0|pdx )r/(r−p) ( S (n−α)/(ps−α) α − C̃( εδ ) (n−α)/(p−1) )p/(r−p) ≤ ( 1 p − 1 r )Sr/(r−p) α + C̃( ε δ )(n−ps)/(p−1). Then, we prove that supt≥0 Iλ(tu0) < cΛ in two cases 0 ≤ t ≤ τ1 and t ≥ τ1 for τ1 ∈ (0, 1). First, we have sup 0≥t≤τ1 Iλ(tu0) < cΛ. Then, from (4.17) and Lemma 4.3, we obtain sup t≥τ1 Iλ(tu0) = sup t≥τ1 ( f̃(t)− 1 q tqλ ∫ Ω |u0|qdx ) ≤ ( 1 p − 1 r )S r r−p α + C̃( ε δ ) n−ps p−1 − 1 q τ q1λ ∫ Ω |u0|qdx. (4.18) Hence, we cam compute that sup t≥τ1 Iλ(tu0) ≤ ( 1 p − 1 r )S r r−p α + C̃( ε δ ) n−ps p−1 − C̃λ  εn− n−ps p q|log ε δ |, if q = n(p−1) n−ps , ε n−ps n(p−1) qδn− n−ps p−1 q, if q < n(p−1) n−ps , εn− n−ps p q, if q > n(p−1) n−ps . Let ε = (λ p p−q ) p−1 n−ps ∈ (0, δ 2 )¿ Then we have sup t≥τ1 Iλ(tu0) ≤ ( 1 p − 1 r )S r r−p α + C̃λ p p−q − C̃λ (λ p p−q ) n(p−1) (n−ps)p |log(λ p p−q ) p−1 n−ps |, if q = n(p−1) n−ps ,( (λ p p−q ) p−1 n−ps )n−n−ps p q , if q > n(p−1) n−ps . If q > n(p−1) n−ps , then 1 + p p− q p− 1 n− ps ( n− n− ps p q ) < p p− q , hence, we can find δ2 > 0 such that for 0 < λ < δ2, C̃λ p p−q − C̃λ ( (λ p p−q ) p−1 n−ps )n−n−ps p q < −c̄ r − q r (r − p pq )q/(q−r)( (p− q)λ pq )r/(r−q) . 16 J. XIE, P. LI EJDE-2025/13 If q = n(p−1) n−ps , we can find δ3 > 0 such that for 0 < λ < δ3, C̃λ p p−q − C̃λ(λ p p−q ) n(p−1) (n−ps)p | log(λ p p−q ) p−1 n−ps | < −c̄ r − q r (r − p pq )q/(q−r)( (p− q)λ pq )r/(r−q) . Since |log(λ p p−q ) p−1 n−ps | → ∞ as λ → 0, and λ(λ p p−q ) n(p−1) (n−ps)p ∼ λ p p−q . Then taking δ̂ = min{δ1, δ2, δ3, ( δ 2 ) n−ps p−1 } > 0, we derive that sup t≥0 Iλ(tu0) < cΛ, for λ ∈ (0, δ̂). From the above inequality and Lemma 2.4, there exists t− > 0 such that t−u0 ∈ N− λ and c−λ ≤ Iλ(t −u0) ≤ sup t≥0 Iλ(tu0) < cΛ, for all λ ∈ (0, δ̂). □ Theorem 4.5. There exists Λ1 > 0 such that for 0 < λ < Λ1 and r = p∗α, the functional Iλ has a minimizer u3 ∈ N+ λ and satisfies (1) Iλ(u3) = cλ = c+λ < 0, (2) u3 is a solution of the problem (1.1). Proof. Set Λ1 = min{ q pΛ, δ̂}. Then cΛ > 0. From Lemma 3.4, we obtain cλ ≤ c+λ < 0, then cλ < cΛ. By Proposition 3.6 (1), for all 0 < λ < Λ1, there exists a bounded minimizing sequence {uk} ⊂ Nλ such that lim k→∞ Iλ(uk) = cλ ≤ c+λ , I ′ λ(uk) = o(1) in X−s,p β . Then there exists u3 ∈ Xs,p β such that, up to a subsequence, uk ⇀ u3 weakly in Xs,p β . By Lemma 4.1 and cλ < cΛ, we obtain uk → u3 strongly in Xs,p β . As in the proof of Theorem 3.7, we can obtain u3 ∈ N+ λ , Iλ(u3) = cλ = c+λ and u3 is a solution of the problem (1.1). □ Theorem 4.6. There exists Λ2 > 0 such that for 0 < λ < Λ2 and r = p∗α, the functional Iλ has a minimizer u4 ∈ N− λ and satisfies (1) Iλ(u4) = c−λ , (2) u4 is a solution of the problem (1.1). Proof. Set Λ2 = min{ q pΛ, δ̂}. By Lemma 4.4, it is easy to get c−λ < cΛ. By Proposition 3.6 (2), for all 0 < λ < Λ2, there exists a bounded minimizing sequence {uk} ⊂ N− λ such that lim k→∞ Iλ(uk) = c−λ , I ′ λ(uk) = o(1) in X−s,p β . By the same argument as in the proof of Theorem 4.5, there exists u4 ∈ N− λ such that Iλ(u4) = c−λ and u4 is a solution of problem (1.1). □ Proof of Theorem 1.2. Taking λ∗ = Λ2, by Theorems 4.5 and 4.6, for all λ ∈ (0, λ∗), problem (1.1) has two solutions u3 ∈ N+ λ and u4 ∈ N− λ in Xs,p β . In addition N+ λ ⋂ N− λ = ∅, then the two solutions u3 and u4 are distinct. □ EJDE-2025/13 SOLUTIONS TO FRACTIONAL P-LAPLACIAN PROBLEMS 17 Acknowledgments. 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Yang; The Brezis-Nirenberg problem for the frac- tional p-Laplacian, Calc. Var. Partial Differential Equations, 55 (2016), Art. 105. [13] D. Mugnai, A. Pinamonti, E. Vecchi; Towards a Brezis-Oswald-type result for fractional prob- lems with Robin boundary conditions, Calc. Var. Partial Differential Equations, 59 (2020), Paper No. 43. [14] D. Mugnai, E. Proietti Lippi; Neumann fractional p-Laplacian: eigenvalues and existence results, Nonlinear Anal., 188 (2019), 455–474. [15] M. Xiang, B. Zhang, X. Zhang; A nonhomogeneous fractional p-Kirchhoff type problem in- volving critical exponent in RN , Adv. Nonlinear Stud., 17 (2017), 611–640. Junhui Xie School of Mathematics and Statistics, Hubei University of Education, Wuhan, 430205, Hubei, China Email address: smilexiejunhui@hotmail.com Pengfei Li (corresponding author) School of Mathematics and Statistics, Fuzhou University, Fuzhou, 350108, Fujian, China Email address: pfliyou@163.com 1. Introduction 2. Preliminaries 3. Proof of Theorem 1.1 4. Proof of Theorem 1.2 Acknowledgments References