Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 25, pp. 1–29. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu FRACTIONAL KIRCHHOFF HARDY PROBLEMS WITH WEIGHTED CHOQUARD AND SINGULAR NONLINEARITY SARIKA GOYAL, TARUN SHARMA Abstract. In this article, we study the existence and multiplicity of solutions to the fractional Kirchhoff Hardy problem involving weighted Choquard and singular nonlinearity M(‖u‖2)(−∆)su− γ u |x|2s = λ l(x) uq + 1 |x|α (∫ Ω r(y)|u(y)|p |y|α|x− y|µ dy ) r(x)|u|p−2u in Ω, u > 0 in Ω, u = 0 in RN \ Ω, where Ω ⊆ RN is an open bounded domain with smooth boundary containing 0 in its interior, N > 2s with s ∈ (0, 1), 0 < q < 1, 0 < µ < N , γ and λ are positive parameters, θ ∈ [1, p) with 1 < p < 2∗µ,s,α, where 2∗µ,s,α is the upper critical exponent in the sense of weighted Hardy-Littlewood-Sobolev inequality. Moreover M models a Kirchhoff coefficient, l is a positive weight and r is a sign-changing function. Under the suitable assumption on l and r, we established the existence of two positive solutions to the above problem by Nehari-manifold and fibering map analysis with respect to the parameters.The results obtained here are new even for s = 1. 1. Introduction Let Ω ⊆ RN be a bounded domain with smooth boundary containing 0 in its interior, N > 2s with s ∈ (0, 1). We consider the following problem with weighted Choquard and singular nonlinearity with weight functions M(‖u‖2)(−∆)su− γ u |x|2s = λl(x)u−q + 1 |x|α (∫ Ω r(y)|u(y)|p |y|α|x− y|µ dy ) r(x)|u|p−2u in Ω, u > 0 in Ω, u = 0 in RN \ Ω, (1.1) where M : R+ 0 → R+ 0 is a continuous function, which is defined as M(t) = c+ dtθ−1, with c > 0, d ≥ 0, 2020 Mathematics Subject Classification. 35A15, 35J75, 36B38. Key words and phrases. Fractional Kirchhoff Hardy operator; singular nonlinearity; weighted Choquard type nonlinearity; Nehari-manifold; fibering map. ©2022. This work is licensed under a CC BY 4.0 license. Submitted December 30, 2021. Published March 25, 2022. 1 2 S. GOYAL, T. SHARMA EJDE-2022/25 λ, γ are positive parameters, q ∈ (0, 1), µ ∈ (0, N), θ ∈ [1, p) with 1 < p < 2∗µ,s,α, where 2∗µ,s,α is upper critical exponent in the sense of weighted Hardy-Littlewood- Sobolev inequality. The fractional Laplacian operator (−∆)s is defined, up to a normalization constant, by the Riesz potential as (−∆)sξ(x) := ∫ RN 2ξ(x)− ξ(x+ y)− ξ(x− y) |y|N+2s dy where x ∈ RN and ξ ∈ C∞0 (RN ). We use the following assumptions on l and r; (A1) l : Ω→ R such that l > 0 a.e. in Ω and l ∈ Lm(Ω), where m = 2∗s 2∗s−1+q with 2∗s = 2N N−2s is the fractional critical Sobolev exponents. (A2) r : Ω → R such that r ∈ L 2∗s 2∗µ,s,α−p (Ω) is sign-changing function, where 2∗µ,s,α = (2N−2α−µ) (N−2s) is the upper critical exponent in the sense of weighted Hardy-Littlewood-Sobolev inequality. Problems of the type (1.1) are motivated by the weighted Hardy-Littlewood-Sobolev inequality proved by Stein and Weiss in [25]. Theorem 1.1. Let 1 < w, t < ∞, 0 < µ < N , α + β ≥ 0, α + β + µ ≤ N , and 1 w + 1 t + α+β+µ N = 2. Then there exists a constant C(α, β, µ,N, t, w) independent of f and h such that∫ RN ∫ RN f(x)h(y) |x|α|x− y|µ|y|β dx dy ≤ C(α, β, µ,N, t, w)‖f‖w‖h‖t (1.2) where 1− 1 w − µ N < α N < 1− 1 w . Currently, problems on nonlocal operators emerging an attractive research area, specifically the fractional Laplacian operator attracts a lot of interest in nonlinear analysis for references [5, 10, 15, 23]. The fractional Laplace operator (−∆)s is the infinitesimal generator of Lévy stable diffusion process which arises in anomalous diffusions in plasma, flames propagation, chemical reactions in liquids, geophysical fluid dynamics, and American options in finance, for references [2, 12]. There is a large amount of literature on the problems related to the fractional elliptic equations, so we refer here [19, 24] and references therein. Kirchhoff type problems, which arise in the modeling of various physical phenom- ena, specifically Kirchhoff in [17] extended the classical D’Alembert wave equation by considering the effects of the changes in the length of string during the vibrations and give a model which is governed by the equation ρ ∂2u ∂t2 −M (∫ L 0 ∣∣∂u ∂x ∣∣2 dx)∂2u ∂x2 = 0, where M (∫ L 0 ∣∣∂u ∂x ∣∣2 dx) = (ρ0 h + E 2L ∫ L 0 ∣∣∂u ∂x ∣∣2 dx) where ρ, ρ0, h, E, L are constants. For further study about the existence of solutions for fractional Kirchhoff problems, we refer here [11, 16, 18, 21]. Problem (1.1) is closely related to the nonlocal Choquard equation, which arise in the study of Hartree-Fock theory of one component plasma and in the model EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 3 of self-gravitating matter. Currently, lots of work has been done on the nonlocal Choquard equation, for the existence of solutions of such type of equations −∆u+ V (x)u = (Iα ∗ |u|p)|u|p−2u in RN , (1.3) where V (x) is an suitable function, Iα is Riesz potential and p > 1 is suitable constant, which is studied in [13, 20]. Moroz and Schaftingen in [20] studied the existence, qualitative properties and decay asymptotics of ground states for the problem (1.3) in case of V = 1. In [7], the authors study the critical nonlocal equation with weighted nonlocal term −∆u = 1 |x|α (∫ RN |u(y)|p |x− y|µ|y|α dy ) |u|p−2u in RN , where N ≥ 3, 0 < µ < N , α ≥ 0, 2α + µ ≤ N and 2 − 2α+µ N < p ≤ 2∗α,µ with 2∗α,µ = 2N−2α−µ N−2 . The critical exponent 2∗α,µ appears here from the weighted Hardy-Littlewood-Sobolev inequality. Here they prove the existence of a positive ground state solutions for subcritical cases by using Schwarz symmetrization and critical cases by a nonlocal version of the concentration-compactness principle. To manipulate the Hardy potential in (1.1), we recall the following fractional Hardy inequality γH ∫ Ω |φ(x)|2 |x|2s ≤ ∫∫ R2N |φ(x)− φ(y)|2 |x− y|N+2s dx dy (1.4) for any φ ∈ C∞0 (Ω), where γH is the sharp constant given in [6, 28] as γH := 22sΓ2(N+2s 4 ) Γ2(N−2s 4 ) , where Γ denotes the Gamma function. Abdellaoui et al. [1] studied the effect of the Hardy potential on the existence and summability of solutions to a class of nonlocal elliptic problems (−∆)su− λ u |x|2s = f(x, u) in Ω, u > 0 in Ω, u = 0 in RN \ Ω, for the case f(x, u) = h(x) uσ for t > 0. They prove that for σ ≥ 1, λ ≤ ΛN,s and h ∈ L1(Ω), this problem has a positive weak solution, where ΛN,s = 22sΓ2(N+2s 4 ) Γ2(N−2s 4 ) . For the local case (s = 1) with λ = 0, in [3], the authors proved that for all h ∈ L1(Ω), there exists at least one distributional solution. Fiscella and Mishra [9] studied the following problem for singular and critical nonlinearities with Hardy term M (∫∫ R2N |u(x)− u(y)|2 |x− y|N+2s dx dy ) (−∆)su− µ u |x|2s = λf(x)u−γ + g(x)u2∗s−1 in Ω u > 0 in Ω, u = 0 in RN \ Ω, where Ω ⊂ RN , N > 2s with s ∈ (0, 1) and 2∗s = 2N N−2s , M(t) = a + bt2θ−2 is a Kirchhoff coefficients where θ ∈ [1, 2∗s/2), f is a positive weight while g is 4 S. GOYAL, T. SHARMA EJDE-2022/25 a sign-changing function. Using the Nehari-manifold technique, they showed the existence of two positive solutions for the combination of critical Sobolev and Hardy nonlinearities. Moreover in [8] author studied the above problem with µ = 0 and f = g = 1, here author prove the existence of two solutions by using variational methods with an appropriate truncation argument. Wang et al. [26] studied the Choquard-Kirchhoff type problem L(u) = λf(x)u−p + (∫ RN g(y)|u(y)|q |x− y|µ dy ) g(x)uq−1 in Ω u = 0 in RN \ Ω, where L(u) = ( a + b[u] (θ−1)p s,p ) (−∆)su, with s ∈ (0, 1), N ≥ 2, 1 < p < N s , 1 < q < p∗µ,s with a > 0, b ≥ 0, θ ∈ [1, 2q), 0 < β < 1 and µ ∈ (0, N), where λ, µ > 0. Under the above assumptions, the authors obtained the existence of two positive non-trivial solutions by using the Nehari-manifold approach. We also cite [14] where the author investigated the existence, non-existence, and multiplicity of positive solutions fractional problem with Hardy potential and singular nonlinearity. Before stating our main result for problem (1.1), it is worth noting that the nov- elty of this work is mainly to obtain the multiple positive solutions for the fractional Kirchhoff Hardy problem with weighted Choquard and singular nonlinearity using the Nehari-manifold and fibering map analysis. To the best of our knowledge, there is no article in the literature that deals with the fractional Kirchhoff Hardy prob- lem for subcritical weighted Choquard and singular nonlinearity. In fact, results are even new for the case s = 1 and α = 0. Because of the nonlocal behavior of the operator, the bounded support of the test function is not preserved which makes the analysis difficult. Also, because of the singular nature of the problem, the asso- ciated functional is not differentiable in the sense of Gâteaux. The results obtained here are somehow expected but we show how the results arise out of nature of the Nehari manifold. Now, we state our results. Theorem 1.2. Let N > 2s with s ∈ (0, 1), θ ∈ [1, p). c > 0, d ≥ 0 and the assumptions (A1), (A2) hold. Then (1.1) has at least two positive solutions for λ ∈ (0,Λ∗). In above theorem Λ∗ := min{Λ1,Λ2}, where Λ1 := ( 1 + q 2p− 2 ) 1+q 2p−2 (hc,γ(2p− 2) 2p+ q − 1 ) (2p+q−1) (2p−2) S (2p+q−1) 2(p−1) 1 ‖l‖m [ 1 Cr(α, µ,N) ] 1+q 2p−2 and Λ2 := S (θ+1)(2p+q−1) 2(2p−θ−1) [Cr(α, µ,N)] θ+q 2p−θ−1 [2 √ (1 + q)(2θ + q − 1)hc,γd 2p+ q − 1 ] θ+q 2p−θ−1 × [2√(2p− 2)(2p− 2θ)hc,γd (2p+ q − 1)‖l‖m ] . This article is organized as follows: In section 2, we recall some important results and useful inequalities. In section 3, we introduce the Nehari-manifold structure and the fiber map analysis related (1.1). Also, we show that the associated energy functional is bounded below and coercive. A crucial compactness property of the energy functional and some important Lemmas are proved in section 4. In section EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 5 5, we show the existence of a positive solution in N+ γ,λ and N−γ,λ respectively, which complete the proof of Theorem 1.2. To abbreviate notation, we denote ‖ · ‖X0 by ‖ · ‖, and ‖ · ‖Lp(Ω) by ‖ · ‖p. 2. Preliminaries In this section, we give the variational setting of problem (1.1) and state impor- tant results to be used later. Consider the fractional Sobolev space Hs(Ω) = { u ∈ L2(Ω) : u(x)− u(y) |x− y| N 2 +s ∈ L2(Q) } , where Q = R2N \ (CΩ× CΩ) and CΩ = RN \Ω. The norm for the space Hs(Ω) is ‖u‖X = ( ‖u‖22 + [u]2X )1/2 , where ‖u‖p is the norm for Lp(Ω) and [u]X = ( ∫ Q |u(x)−u(y)|2 |x−y|N+2s dx dy )1/2 . We define the function space X0 := {u ∈ Hs(RN ) : u = 0 a.e. in RN \ Ω} which is a Hilbert space with norm ‖u‖X0 = (∫∫ R2N |u(x)− u(y)|2 |x− y|N+2s dx dy )1/2 . The functional space X0 can be equivalently considered as the closure of C∞0 (Ω) under the norm ‖u‖ = [u]X . For further studies on X0, we refer the reader to [22]. Now if α = β and f(x) = |u(x)|p, h(y) = |u(y)|p, that is if w = t, then by Theorem 1.1, ∫ RN ∫ RN r(x)r(y)|u(x)|p|u(y)|p |x|α|x− y|µ|y|α dx dy is well defined if r|u|p ∈ Lt(RN ), where t = 2N 2N−2α−µ . Thus by fractional Sobolev embedding theorem, we have 2 ≤ tp ≤ 2∗s, which gives that 2∗µ,s,α = 2N − 2α− µ N ≤ p ≤ (2N − 2α− µ) (N − 2s) = 2∗µ,s,α, where 2∗µ,s,α is known as upper critical exponent and 2∗µ,s,α is lower critical expo- nent in the sense of weighted Hardy-Littlewood-Sobolev inequality. Thus from Theorem 1.1, r ∈ L 2∗s 2∗µ,s,α−p (RN ), and Hölder inequality, we obtain∫ RN ∫ RN r(x)r(y)|u(x)|p|u(y)|p |x|α|x− y|µ|y|α dx dy ≤ C(α, µ,N)‖r(x)|u|p‖2 2N 2N−2α−µ ≤ C(α, µ,N)‖r‖2 2∗s 2∗µ,s,α−p ‖u‖2p2∗s . Further, the fractional Sobolev inequality yields∫ RN ∫ RN r(x)r(y)|u(x)|p|u(y)|p |x|α|x− y|µ|y|α dx dy ≤ Cr(α, µ,N)S−p‖u‖2p, (2.1) 6 S. GOYAL, T. SHARMA EJDE-2022/25 where Cr(α, µ,N) = C(α, µ,N)‖r‖2 2∗s 2∗µ,s,α−p , and S denotes the best constant of the fractional Sobolev embedding theorem, is defined as S = inf v∈Ds,2(RN )\{0} ∫ R2N |v(x)−v(y)|2 |x−y|N+2s dx dy( ∫ RN |v(x)|2∗s dx )2/2∗s (2.2) whereDs,2(RN ) is the closure of C∞0 (RN ) under the norm (∫ R2N |v(x)−v(y)|2 |x−y|N+2s dx dy )1/2 . Lemma 2.1. Let N ≥ 3, q ∈ (1,∞) and {un} is a bounded sequence in Lq(RN ). If un → u a.e in RN as n→∞, then un ⇀ u weakly in Lq(RN ). A proof of the above lemma can be found in [27, Lemma 1.32]. Now we prove the following Brezis-Lieb type Lemma for singular Choquard in case of fractional. The idea of the proof comes from [7, Lemma 2.2] where is done in case of s = 1. Lemma 2.2. Let N ≥ 3, α ≥ 0, 0 < µ < N , 2α + µ ≤ N and 1 ≤ p ≤ 2N−2α−µ N−2s . If {un} is a bounded sequence in L 2Np 2N−2α−µ (RN ) such that un → u a.e. in RN as n→∞. Then∫ RN ∫ RN |un(x)|p|un(y)|p |x|α|x− y|µ|y|α dx dy − ∫ RN ∫ RN |(un − u)(x)|p|(un − u)(y)|p |x|α|x− y|µ|y|α dx dy → ∫ RN ∫ RN |u(x)|p|u(y)|p |x|α|x− y|µ|y|α dx dy. Proof. Since {un} is a bounded sequence in L 2Np 2N−2α−µ (RN ) such that un → u a.e. in RN as n→∞, then from the Brézis-Lieb Lemma [4], we know that |un − u|p − |un|p → |u|p in L 2N 2N−2α−µ (RN ). (2.3) The weighted Hardy-Littlewood-Sobolev inequality (1.2) implies that∫ RN |un|p |x|α|x− y|µ|y|α dy − ∫ RN |un − u|p |x|α|x− y|µ|y|α dy → ∫ RN |u|p |x|α|x− y|µ|y|α dy in L 2N 2α+µ (RN ). (2.4) Now we consider∫ RN ( |x|−µ ∗ ( 1 |x|α |un|2 ∗ µ,s,α )) |un|2∗µ,s,α |x|α dx − ∫ RN ( |x|−µ ∗ ( 1 |x|α |un − u|2 ∗ µ,s,α )) |un − u|2∗µ,s,α |x|α dx = ∫ RN ( |x|−µ ∗ ( 1 |x|α |un|2 ∗ µ,s,α − 1 |x|α |un − u|2 ∗ µ,s,α )) × ( |un|2 ∗ µ,s,α − |un − u|2 ∗ µ,s,α ) |x|α dx + 2 ∫ RN ( |x|−µ ∗ ( 1 |x|α |un|2 ∗ µ,s,α − 1 |x|α |un − u|2 ∗ µ,s,α )) |un − u|2∗µ,s,α |x|α dx. (2.5) By Lemma 2.1, we obtain |un − u|2 ∗ µ,s,α ⇀ 0 in L 2N 2N−2α−µ (RN ) (2.6) EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 7 Hence from (2.3), (2.4), (2.5) and (2.6), we obtain the required result. � Definition 2.3. We say u ∈ X0 is a weak solution of the problem (1.1), if l(x)(u)−qφ ∈ L1(Ω) and for any φ ∈ X0 it holds( c+ d‖u‖2θ−2 )∫∫ R2N ( u(x)− u(y) )( φ(x)− φ(y) ) |x− y|N+2s dx dy − γ ∫ Ω uφ |x|2s dx − λ ∫ Ω l(x)(u)−qφdx− ∫ Ω ∫ Ω r(x)r(y)(u(y))p(u(x))p−1φ(x) |x|α|x− y|µ|y|α dx dy = 0. To obtain a positive solution of (1.1), we consider the problem M(‖u‖2)(−∆) s u− γ u+ |x|2s = λl(x)(u+)−q + (∫ Ω r(y)(u+(y))p |y|α|x− y|µ dy )r(u+)p−1 |x|α in Ω, u > 0 in Ω, u = 0 in RN \ Ω, (2.7) where M(t) = c + dtθ−1 and u+ = max{u, 0}. Then the function u ∈ X0, u > 0 in Ω is a weak solution of the problem (2.7), if l(x)(u+)−qφ ∈ L1(Ω) and for any φ ∈ X0 it holds( c+ d‖u‖2θ−2 )∫∫ R2N (u(x)− u(y)) (φ(x)− φ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+φ |x|2s dx − λ ∫ Ω l(x)(u+)−qφdx− ∫ Ω ∫ Ω r(x)r(y)(u+(y))p(u+(x))p−1φ(x) |x|α|x− y|µ|y|α dx dy = 0. (2.8) One can easily see that if u > 0 is a solution to (2.7), then it is also a solution to (1.1). To find a solution to (2.7), we use a variational approach. So, we define the energy functional Jγ,λ : X0 → R corresponding to problem (2.7), as Jγ,λ(u) := c 2 ‖u‖2 + d 2θ ‖u‖2θ − γ 2 ‖u+‖2H − λ 1− q ∫ Ω l(x)(u+)1−q − 1 2p ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy. Here we denote ‖u‖2H := ∫ Ω |u(x)|2 |x|2s dx for all u ∈ X0. Then the functional Jγ,λ is well defined and continuous for any γ ∈ (0, cγH) by the fractional Hardy inequality (1.4). We note that the relation v+ ≤ |v| and (1.4) yield c ∫∫ R2N |v(x)− v(y)|2 |x− y|N+2s dx dy − γ ∫ Ω (v+)2 |x|2s dx ≥ hc,γ ∫∫ R2N |v(x)− v(y)|2 |x− y|N+2s dx dy (2.9) for any v ∈ X0 \ {0}, and hc,γ = c− γ γH > 0 for any γ ∈ (0, cγH). 8 S. GOYAL, T. SHARMA EJDE-2022/25 3. Fibering map analysis In this section, we show that N±γ,λ is non-empty and N 0 γ,λ = {0}. Moreover Jγ,λ is bounded below and coercive. Since the energy functional Jγ,λ is not bounded below on X0 but is bounded below on appropriate subset Nγ,λ of X0. Therefore in order to obtain the existence results, we introduce the Nehari set Nγ,λ := {u ∈ X0 : 〈Jγ,λ(u), u〉 = 0} = { u ∈ X0 : c‖u‖2 + d‖u‖2θ − γ‖u+‖2H − λ ∫ Ω l(x)(u+)1−q dx − ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy = 0 } . Here we define the fiber map φu : R+ → R as φu(t) := Jγ,λ(tu) for all t > 0. Then we have φu(t) = c 2 t2‖u‖2 + d 2θ t2θ‖u‖2θ − γ 2 t2‖u+‖2H − λt1−q 1− q ∫ Ω l(x)(u+)1−q dx − t2p 2p ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy, so φ′u(t) = ct‖u‖2 + t2θ−1d‖u‖2θ − tγ‖u+‖2H − λt −q ∫ Ω l(x)(u+)1−q dx − t2p−1 ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy and φ′′u(t) = c‖u‖2 + d(2θ − 1)t2θ−2‖u‖2θ − γ‖u+‖2H + λqt−q−1 ∫ Ω l(x)(u+)1−q dx − (2p− 1)t2p−2 ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy. One can easily see that the Nehari-manifold is closely related to the fibering map such that tu ∈ Nγ,λ if and only if φ′u(t) = 0. In particular, u ∈ Nγ,λ if and only if φ′u(1) = 0. So it is reasonable to split Nγ,λ into three parts corresponding to local maxima, local minima and point of inflection as N±γ,λ := {u ∈ Nγ,λ : φ′′u(1) ≷ 0}, N 0 γ,λ := {u ∈ Nγ,λ : φ′′u(1) = 0}. For the sign-changing function r, we denote R+ := { u ∈ X0 : ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy > 0 } , R− := { u ∈ X0 : ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy ≤ 0 } . Lemma 3.1. Let γ ∈ (0, cγH). Then the following holds: (1) For u ∈ R+, there exist Λ1 > 0 and a unique tmax := tmax(u) > 0, t+ = t+(u) > 0 and t− = t−(u) > 0 with t+ < tmax < t− such that t+u ∈ N+ γ,λ, t−u ∈ N−γ,λ, for any λ ∈ (0,Λ1). Also Jγ,λ(t+u) = min0≤t≤t− Jγ,λ(tu) and Jγ,λ(t−u) = maxt≥tmax Jγ,λ(tu). EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 9 (2) For u ∈ R−, γ ∈ (0, cγH) and λ > 0 there exists a unique t∗ > 0 such that t∗u ∈ N+ γ,λ and Jγ,λ(t∗u) = inft>0 Jγ,λ(tu). Proof. For fixed u ∈ X0, define Ψu : R+ → R as Ψu(t) := t2−2p[c‖u‖2−γ‖u+‖2H ]+dt2θ−2p‖u‖2θ−λt1−q−2p ∫ Ω l(x)(u+)1−q dx. (3.1) We observe that tu ∈ Nγ,λ if and only if t is the root of the equation, Ψu(t) = ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy. Using (3.1), we notice that Ψu(t) → −∞ as t → 0+ and Ψu(t) → 0 as t → ∞. Moreover, Ψ′u(t) = (2− 2p)t1−2p[c‖u‖2 − γ‖u+‖2H ] + d(2θ − 2p)t2θ−2p−1‖u‖2θ − λ(1− q − 2p)t−q−2p ∫ Ω l(x)(u+)1−q dx. Since, q < 1 < 2θ < 2p, we see that limt→0+ Ψ ′ u(t) > 0 and limt→∞Ψ′u(t) < 0. Then there exists a unique tmax = tmax(u) > 0 such that Ψu(t) is increasing in (0, tmax), decreasing in (tmax,∞) and Ψ′u(tmax) = 0. Now we estimate, Ψu(tmax) given as follows Ψu(tmax) = max t>0 Ψu(tmax) = max t>0 ( d‖u‖2θt2θ−2p + Ψu(t) ) > max t>0 Ψu(t), where Ψu(t) = [c‖u‖2 − γ‖u+‖2H ]t2−2p − λt1−q−2p ∫ Ω l(x)(u+)1−q dx. Now by (1.4), for γ ∈ (0, cγH), we can calculate max t>0 Ψu(t) ≥ φu(t), where φu(t) = hc,γ‖u‖2t2−2p − λt1−q−2p ∫ Ω l(x)(u+)1−q dx. We noticed that max t>0 φu(t) = ( q + 1 2p− 2 )( 2p− 2 2p+ q − 1 ) 2p+q−1 1+q ( hc,γ‖u‖2 ) 2p+q−1 1+q( λ ∫ Ω l(x)(u+)1−q dx ) 2p−2 1+q . Now using this, (A1), Sobolev inequality (2.2), and Hölder inequality, we obtain Ψu(tmax) ≥ ( 1 + q 2p− 2 )(hc,γ(2p− 2) 2p+ q − 1 ) 2p+q−1 1+q λ 2−2p 1+q (‖l‖m) 2−2p 1+q S (1−q)(2p−2) 2(1+q) ‖u‖2p > 0. Now, according to the behavior of r, we consider two cases case 1: Let u ∈ R+. Choose λ < Λ1 := ( 1 + q 2p− 2 ) 1+q 2p−2 (hc,γ(2p− 2) 2p+ q − 1 ) (2p+q−1) (2p−2) S (2p+q−1) 2(p−1) 1 ‖l‖m [ 1 Cr(α, µ,N) ] 1+q 2p−2 , 10 S. GOYAL, T. SHARMA EJDE-2022/25 then there exists a unique t+ := t+(u) < tmax and t− := t−(u) > tmax, such that Ψu(t+) = ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy = Ψu(t−), that is t+u, t−u ∈ Nγ,λ. Now using the relation φ′′tu(1) = t2p+1Ψ′u(t), it follows that Ψ′u(t+) > 0 and Ψ′u(t−) < 0, which yield t+u ∈ N+ γ,λ and t−u ∈ N−γ,λ. Furthermore, assuming φ′u(t) = t2p−1 ( Ψu(t)− ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy ) , we have φ′u(t) < 0 for all t ∈ [0, t+) and φ′u(t) > 0 for all t ∈ (t+, t−). Therefore, Jγ,λ(t+u) = min 0≤t≤t− Jγ,λ(tu). Also, φ′u(t) > 0 for all t ∈ [t+, t−), φ′u(t−) = 0 and φ′u(t) < 0 for all t ∈ (t−,∞) yield that Jγ,λ(t−u) = max t≥tmax Jγ,λ(tu). case 2: Let u ∈ R−. Then using the fact Ψu(t) → −∞ as t → 0+, there exists a unique t∗ > 0 such that Ψu(t∗) = ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy for all λ > 0. As u ∈ R−, we have Ψu(t∗) < 0 and Ψ′u(t∗) > 0. Now by doing the similar calculations as in case(1), φ′′tu(1) = t2p+1Ψ ′ u(t) with Ψ′u(t∗) > 0, we deduce that t∗u ∈ N+ λ,γ . This completes the proof. � Lemma 3.2. Let γ ∈ (0, cγH), then there exists Λ2 > 0 such that N 0 γ,λ = {0}, for all λ ∈ (0,Λ2). Proof. Arguing by contradiction argument, we suppose that there exists u ∈ N 0 γ,λ \ {0} for all λ ∈ (0,Λ2). Then we have c‖u‖2 + d‖u‖2θ − γ‖u+‖2H − λ ∫ Ω l(x)(u+)1−q dx − ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α = 0, (3.2) c‖u‖2 + d(2θ − 1)‖u‖2θ − γ‖u+‖2H + λq ∫ Ω l(x)(u+)1−q dx − (2p− 1) ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α = 0. (3.3) Now we consider two cases: Case 1: ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x−y|µ|y|α dx dy = 0. Using this (3.2), (3.3), 2θ ≥ 1 > 1− q, and γ ∈ (0, cγH) with (2.9), we obtain 0 = (1 + q)[c‖u‖2 − γ‖u+‖2H ] + d(2θ + q − 1)‖u‖2θ ≥ hc,γ(1 + q)‖u‖2 + d(2θ + q − 1)‖u‖2θ > 0, which gives a contradiction. EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 11 Case 2: ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x−y|µ|y|α dx dy 6= 0. Equations (3.2) and (3.3) yield (2p− 2)[c‖u‖2 − γ‖u+‖2H ] + d(2p− 2θ)‖u‖2θ − λ(2p+ q − 1) ∫ Ω l(x)(u+)1−q dx = 0, (3.4) (1 + q) [ c‖u‖2 − γ‖u+‖2H ] + d(2θ + q − 1)‖u‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α = 0. (3.5) Let us define Eγ,λ : Nγ,λ → R as Eγ,λ := (1 + q)[c‖u‖2 − γ‖u+‖2H ] + d(2θ + q − 1)‖u‖2θ (2p+ q − 1) − ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy. Then from (3.5), it follows that Eγ,λ = 0 for all u ∈ N 0 γ,λ. Also using (2.1), (2.9), and (a+ b) ≥ 2 √ ab, for any a, b ≥ 0, we obtain Eγ,λ ≥ 2 √ (1 + q)(2θ + q − 1)hc,γd‖u‖θ+1 (2p+ q − 1) − Cr(α, µ,N)S−p‖u‖2p, = ‖u‖2p [2 √ (1 + q)(2θ + q − 1)hc,γd (2p+ q − 1)‖u‖2p−θ−1 − Cr(α, µ,N)S−p ] . Now using (2.9), Sobolev inequality (2.2), A.M ≥ G.M and Hölder inequality in (3.4), we have ‖u‖ ≤ [λ(2p+ q − 1)‖l‖mS −(1−q) 2 2 √ (2p− 2)(2p− 2θ)hc,γd ] 1 θ+q . Thus, λ < Λ2 := S (θ+1)(2p+q−1) 2(2p−θ−1) [Cr(α, µ,N)] θ+q 2p−θ−1 [2 √ (1 + q)(2θ + q − 1)hc,γd 2p+ q − 1 ] θ+q 2p−θ−1 × [2 √ (2p− 2)(2p− 2θ)hc,γd (2p+ q − 1)‖l‖m ] . Hence Eγ,λ(u) > 0 for all u ∈ N 0 γ,λ \ {0}, which gives a contradiction. � Lemma 3.3. Let γ ∈ (0, cγH) and λ ∈ (0,Λ2), then there exists a gap structure in Nγ,λ such that ‖U‖ > A0 > A1 > ‖u‖, for any u ∈ N+ γ,λ, U ∈ N−γ,λ, where A0 := [2 √ (1 + q)(2θ + q − 1)hc,γd (2p+ q − 1)Cr(α, µ,N)S−p ] 1 2p−θ−1 , A1 := [λ(2p+ q − 1)S −(1−q) 2 ‖l‖m 2 √ (2p− 2)(2p− 2θ)hc,γd ] 1 θ+q . 12 S. GOYAL, T. SHARMA EJDE-2022/25 Proof. If u ∈ N+ γ,λ ⊂ Nγ,λ, then (2p− 2)[c‖u‖2 − γ‖u+‖2H ] + d(2p− 2θ)‖u‖2θ − λ(2p+ q − 1) ∫ Ω l(x)(u+)1−qdx < 0. (3.6) Using (A1), (2.2), and Hölder inequality in (3.6), it follows that (2p− 2)[c‖u‖2 − γ‖u+‖2H ] + d(2p− 2θ)‖u‖2θ ≤ λ(2p+ q − 1)‖l‖mS −(1−q) 2 ‖u‖1−q. This and (2.9) yield (2p− 2)hc,γ‖u‖2 + d(2p− 2θ)‖u‖2θ < λ(2p+ q − 1)‖l‖mS −(1−q) 2 ‖u‖1−q. (3.7) At this moment, applying the condition AM ≥ GM in (3.7), we have ‖u‖ < [λ(2p+ q − 1)S −(1−q) 2 ‖l‖m 2 √ (2p− 2)(2p− 2θ)hc,γd ] 1 θ+q := A1. Now, if U ∈ N−λ,γ then using (2.1), we have (1 + q)[c‖U‖2 − γ‖U+‖2H ] + d(2θ + q − 1)‖U‖2θ < (2p+ q − 1)Cr(α, µ,N)S−p‖U‖2p. By this, (2.9), and AM ≥ GM , we obtain 2 √ (1 + q)(2θ + q − 1)hc,γd‖U‖θ+1 < (2p+ q − 1)Cr(α, µ,N)S−p‖U‖2p. Thus ‖U‖ > [2 √ (1 + q)(2θ + q − 1)hc,γd (2p+ q − 1)Cr(α, µ,N)S−p ] 1 2p−θ−1 := A0. It is easy to check that A0 > A1 for all λ ∈ (0,Λ2). Thus we conclude that ‖U‖ > A0 > A1 > ‖u‖ for all u ∈ N+ γ,λ, U ∈ N − γ,λ. This completes the proof. � Lemma 3.4. Let γ ∈ (0, cγH), then N−γ,λ is a closed set in X0 topology for all λ ∈ (0,Λ2). Proof. Assume {uk}k be any sequence in N−γ,λ, such that uk → u strongly in X0. Then u ∈ N−γ,λ ∪ {0}. Now by Lemma 3.3, we obtain ‖u‖ = lim k→∞ ‖uk‖ ≥ A0 > A1 > 0. Thus the above inequality, yields u 6≡ 0. Hence u ∈ N−γ,λ. � Lemma 3.5. Let u ∈ N±γ,λ with γ ∈ (0, cγH) and λ > 0. Then there is a ε > 0 and a continuous function ξ : Bε(0) → R+ such that ξ(v) > 0, ξ(0) = 1 and ξ(v)(u+ v) ∈ N±γ,λ for all v ∈ Bε(0), where Bε(0) = {v ∈ X0 : ‖v‖ < ε}. EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 13 Proof. Here we only give the proof for the case u ∈ N+ γ,λ, while the proof of the case N−γ,λ follows similarly. Define F : X0 × R+ → R as F (v, z) := z1+q ( c‖u+ v‖2 − γ‖(u+ v)+‖2H ) + z2θ−1+qd‖u+ v‖2θ − λ ∫ Ω l(x) ( (u+ v)+ )1−q dx − z2p+q−1 ∫ Ω ∫ Ω r(x)r(y) ( (u+ v)+(x) )p( (u+ v)+(y) )p |x|α|x− y|µ|y|α dx dy. Since u ∈ N+ γ,λ ⊂ Nγ,λ, it follows that F (0, 1) = c‖u‖2 + d‖u‖2θ − γ‖u+‖2H − λ ∫ Ω l(x)(u+)1−q dx − ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy = 0, (3.8) and ∂F ∂z (0, 1) = (1 + q) ( c‖u‖2 − γ‖u+‖2H ) + d(2θ − 1 + q)‖u‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(u+(x))p(u+(y))p |x|α|x− y|µ|y|α dx dy > 0. (3.9) Now applying the Implicit function theorem to the function F at (0, 1), there exists ε > 0 such that for every v ∈ X0 with ‖v‖ < ε, then F (v, z) = 0 has a unique solu- tion z = ξ(v) > 0. From (3.8), we have ξ(0) = 1. This together with F (v, ξ(v)) = 0 for any v ∈ X0 with ‖v‖ < ε yields that 0 = ξ(v) 1+q ( c‖u+ v‖2 − γ‖(u+ v)+‖2H ) + ξ(v) 2θ−1+q d‖u+ v‖2θ − λ ∫ Ω l(x) ( (u+ v)+ )1−q dx − ξ(v) 2p+q−1 ∫ Ω ∫ Ω r(x)r(y)((u+ v)+(x)) p ((u+ v)+(y)) p |x|α|x− y|µ|y|α dx dy = 1 ξ1−q(v) ( c‖ξ(v)(u+ v)‖2 − γ‖ξ(v)(u+ v)+‖2H − λ ∫ Ω l(x) ( ξ(v)(u+ v)+ )1−q dx + d‖ξ(v)(u+ v)‖2θ − ∫ Ω ∫ Ω r(x)r(y)(ξ(v)(u+ v)+(x)) p (ξ(v)(u+ v)+(y)) p |x|α|x− y|µ|y|α dx dy ) . This implies, ξ(v)(u+ v) ∈ Nγ,λ for any v ∈ X0 with ‖v‖ < ε. We notice that ∂F ∂z ∣∣∣ (v,ξ(v)) = 1 ξ2−q(v) [ (1 + q) ( c‖ξ(v)(u+ v)‖2 − γ‖ξ(v)(u+ v)+‖2H ) + (2θ − 1 + q)d‖ξ(v)(u+ v)‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(ξ(v)(u+ v)+(x)) p (ξ(v)(u+ v)+(y)) p |x|α|x− y|µ|y|α dx dy ] . 14 S. GOYAL, T. SHARMA EJDE-2022/25 Now using (3.9), take ε > 0 such that ε < ε, for any v ∈ X0 with ‖v‖ < ε, we obtain (1 + q) ( c‖ξ(v)(u+ v)‖2 − γ‖ξ(v)(u+ v)+‖2H ) + (2θ − 1 + q)d‖ξ(v)(u+ v)‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(ξ(v)(u+ v)+(x)) p (ξ(v)(u+ v)+(y)) p |x|α|x− y|µ|y|α dx dy > 0, that is ξ(v)(u+ v) ∈ N+ γ,λ for all v ∈ Bε(0). � Lemma 3.6. Let γ ∈ (0, cγH) and λ > 0, then the functional Jγ,λ is coercive and bounded from below on Nγ,λ. Proof. Let u ∈ Nγ,λ. Then this together with (2.9), (f1), (2.2), and Hölder inequal- ity with the condition 2θ < 2p, give Jγ,λ(u) := (1 2 − 1 2p )( c‖u‖2 − γ‖u+‖2H ) + ( 1 2θ − 1 2p ) d‖u‖2θ − λ ( 1 1− q − 1 2p )∫ Ω l(x)(u+)1−q dx ≥ (1 2 − 1 2p ) hc,γ‖u‖2 − λ ( 1 1− q − 1 2p ) ‖l‖mS −(1−q) 2 ‖u‖1−q, since 2 > 1− q. Hence Jγ,λ is coercive on Nγ,λ. let G(t) := (1 2 − 1 2p ) hc,γt 2 1−q − λ ( 1 1− q − 1 2p ) ‖l‖mS −(1−q) 2 t, which has minimum at tmin := (λ(2p+ q − 1)S −(1−q) 2 ‖l‖m (2p− 2)hc,γ ) 1−q 1+q . So, we have Jγ,λ(u) ≥ −(1 + q) 4p(1− q) ( λ(2p+ q − 1)S −(1−q) 2 ‖l‖m ) 2 (1+q) ((2p− 2)hc,γ) 1−q 1+q = −C(let), where C > 0. Hence Jγ,λ is bounded below on Nγ,λ. � 4. A compactness result for Jγ,λ By Lemma 3.4, N+ γ,λ∪{0} and N−γ,λ are two closed sets in X0 for λ < Λ2. Hence the Ekeland’s Variational principle can be applied to the problem of finding the infimum of Jγ,λ both on N+ γ,λ ∪ {0} and N−γ,λ. We define m+ γ,λ := inf u∈N+ γ,λ∪{0} Jγ,λ(u) and m−γ,λ := inf u∈N−γ,λ Jγ,λ(u). By Ekeland’s variational principle, we can find a minimizing sequence {uk} ⊂ N+ γ,λ ∪ {0}(N − γ,λ) for Jγ,λ such that m±γ,λ < Jγ,λ(uk) < m±γ,λ + 1 k , and Jγ,λ(u) ≥ Jγ,λ(uk) + 1 k ‖u− uk‖. (4.1) EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 15 In view of Lemma 3.6, we see that {uk}k is a bounded sequence in Nγ,λ with ‖uk‖ ≤ C1 with C1 > 0. Therefore up to a subsequence still denoted by {uk}, we may assume that there exists u0 ∈ X0 such that uk ⇀ u0 weakly in X0. (4.2) To prove our main result we need the following Lemmas. Lemma 4.1. Let {uk}k ⊂ N+ γ,λ satisfy (4.2) with γ ∈ (0, cγH) and λ ∈ (0,Λ1), where Λ1 is given in Lemma 3.1. Then there exists a constant C2 > 0 such that (1) if {uk} ⊂ N+ γ,λ for each k ∈ N, we have (1 + q)[c‖uk‖2 − γ‖u+ k ‖ 2 H ] + d(2θ + q − 1)‖uk‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy ≥ C2, (2) if {uk} ⊂ N−γ,λ for each k ∈ N, we have (1 + q)[c‖uk‖2 − γ‖u+ k ‖ 2 H ] + d(2θ + q − 1)‖uk‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy ≤ −C2. Proof. We present the proof of (1) only, the proof of (2) follows similarly. Since {uk}k ⊂ N+ γ,λ, it is enough to show that lim inf k→∞ [ (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ ] < λ(2p+ q − 1) ∫ Ω l(x)(u+ 0 )1−qdx. For this, we argue by contradiction, so we assume that lim inf k→∞ [ (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ ] = λ(2p+ q − 1) ∫ Ω l(x)(u+ 0 )1−q dx. Since {uk}k ∈ N+ γ,λ, we have (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ < λ(2p+ q − 1) ∫ Ω l(x)(u+ k )1−qdx. Now by l ∈ L 2∗s 2∗s−1+q (Ω) and by Vitali’s convergence theorem, we can show that lim k→∞ ∫ Ω l(x)(u+ k )1−qdx = ∫ Ω l(x)(u+ 0 )1−q dx. Then lim inf k→∞ [ (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ ] ≤ lim sup k→∞ [ (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ ] ≤ λ(2p+ q − 1) ∫ Ω l(x)(u+ 0 )1−qdx, 16 S. GOYAL, T. SHARMA EJDE-2022/25 which yields lim k→∞ [ (2p− 2) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2p− 2θ)‖uk‖2θ ] = λ(2p+ q − 1) ∫ Ω l(x)(u+ 0 )1−qdx. (4.3) From (4.3), it is clear that there exists A > 0 and Aγ > 0 with hc,γ ≤ Aγ ≤ cA for γ ∈ (0, cγH), by (2.9) such that c‖uk‖2 − γ‖u+ k ‖ 2 H → Aγ , ‖uk‖2 → A as k →∞. Then, we have (2p− 2)Aγ + d(2p− 2θ)Aθ = λ(2p+ q − 1) ∫ Ω l(x)(u+ 0 )1−qdx, which gives that λ ∫ Ω l(x)(u+ 0 )1−q dx = (2p− 2)Aγ (2p+ q − 1) + d(2p− 2θ)Aθ (2p+ q − 1) . (4.4) By Lemma 3.1, for λ ∈ (0,Λ1), we have 0 ≤ ( 1 + q 2p− 2 )( 2p− 2 2p+ q − 1 ) 2p+q−1 1+q (hc,γA) 2p+q−1 1+q ( λ ∫ Ω l(x)(u+ 0 )1−qdx ) 2−2p 1+q − lim k→∞ ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy. (4.5) Using {uk}k ⊂ N+ γ,λ ⊂ Nγ,λ and (4.4), we obtain lim k→∞ ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy = Aγ ( 1 + q 2p+ q − 1 ) + dAθ (2θ + q − 1 2p+ q − 1 ) . Now using (4.3), (4.5) in (4.4), together with hc,γA ≤ Aγ , we obtain dAθ (2θ + q − 1 2p+ q − 1 ) ≤ 0, which is a contradiction. Thus the proof is complete. � Fix ψ ∈ X0 with ψ ≥ 0. Consider the constants C1 > 0 with ‖uk‖ ≤ C1 and C2 > 0 given in the Lemma 4.1. Then for k ∈ N sufficiently large such that (1− q)C1 k < C2. (4.6) By Lemma 3.5, we can extract a sequence of functions (ξk)k such that ξk(0) = 1 and ξk(tψ)(uk + tψ) ∈ N±γ,λ for t > 0 sufficiently small. Using uk ∈ Nγ,λ and ξk(tψ)(uk + tψ) ∈ Nγ,λ, we have c‖uk‖2 + d‖uk‖2θ − γ‖u+ k ‖ 2 H − λ ∫ Ω l(x)(u+ k )1−q dx − ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy = 0 (4.7) EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 17 and ξ2 k(tψ) ( c‖uk + tψ‖2 − γ‖(uk + tψ)+‖2H ) − λξ1−q k (tψ) ∫ Ω l(x) ( (uk + tψ)+) )1−q dx+ dξ2θ k (tψ)‖uk + tψ‖2 − ξ1−q k (tψ) ∫ Ω ∫ Ω r(x)r(y) ((uk + tψ)+(x)) p ((uk + tψ)+(y)) p |x|α|x− y|µ|y|α dx dy = 0. (4.8) Now denote ξ′k(0) is the derivative of ξk at point 0 with 〈ξ′k(0), ψ〉 ∈ [−∞,∞] for any ψ ∈ X0. If it does not exist then ξ′k(0) can be replace by pk(0) = limk→∞ ξk(tψ)−1 tk for some (tk)k such that tk → 0 as k →∞ and tk > 0. Lemma 4.2. Let λ ∈ (0,Λ1), γ ∈ (0, cγH) and let {uk}k ⊂ N±γ,λ satisfying (4.1) and (4.2). Then 〈ξ′k(0), ψ〉 is uniformly bounded for every ψ ∈ X0 with ψ ≥ 0. Proof. Here we prove only for the case N+ γ,λ. The case N−γ,λ can be prove in a similar manner. Let {uk}k ⊂ N+ γ,λ and ξk(tψ)(uk + tψ) ∈ N+ γ,λ. Then in view of (4.7) and (4.8), we obtain 0 = [ξ2 k(tψ)− 1] ( c‖uk + tψ‖2 − γ‖(uk + tψ)+‖2H ) + ( c‖uk + tψ‖2 − γ‖(uk + tψ)+‖2H ) − ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d ( [ξ2θ k (tψ)− 1]‖uk + tψ‖2θ + ‖uk + tψ‖2θ − ‖uk‖2θ ) − λ ∫ Ω l(x) [( (uk + tψ)+ )1−q − (u+ k ) 1−q] dx − λ[ξ1−q k (tψ)− 1] ∫ Ω l(x) ( (uk + tψ)+ )1−q dx − [ξ2 k(tψ)− 1] ∫ Ω ∫ Ω r(x)r(y)(uk + tψ)+(x))p(uk + tψ)+(y))p |x|α|x− y|µ|y|α dx dy − ∫ Ω ∫ Ω r(x)r(y) [ ((uk + tψ)+(x)) p ((uk + tψ)+(y)) p |x|α|x− y|µ|y|α − ( u+ k (x) )p( u+ k (y) )p |x|α|x− y|µ|y|α ] dx dy. Now dividing the above estimate by t > 0 and taking the limit t→ 0+, we obtain 0 = 〈ξ ′ k(0), ψ〉 [ 2 ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d2θ‖uk‖2θ − λ(1− γ) ∫ Ω l(x) ( (uk)+ )1−q − 2p ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy ] + ( 2c+ 2θd‖uk‖2θ−2) ∫ Ω ∫ Ω (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − 2γ ∫ Ω u+ k ψ |x|2s dx− 2p ∫ Ω ∫ Ω r(x)r(y)(u+ k (x)) p−1 ψ(x)(u+ k (y)) p |x|α|x− y|µ|y|α dx dy. 18 S. GOYAL, T. SHARMA EJDE-2022/25 This together with (4.7) yields 0 ≤ 〈ξ′k(0), ψ〉 [ (1 + q) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + d(2θ + q − 1)‖uk‖2θ − (2p− q + 1) ∫ Ω ∫ Ω r(x)r(y)(u+ k (x))p(u+ k (y))p |x|α|x− y|µ|y|α dx dy ] + ( 2c+ 2θd‖uk‖2θ−2) ∫ Ω ∫ Ω (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − 2γ ∫ Ω u+ k ψ |x|2s dx− 2p ∫ Ω ∫ Ω r(x)r(y)(u+ k (x)) p−1 ψ(x)(u+ k (y)) p |x|α|x− y|µ|y|α dx dy. Then by Lemma 4.1(1) and boundedness of the sequence {uk} in above, we obtain 〈ξ′k(0), ψ〉 is bounded from below for any ψ ∈ X0 with ψ ≥ 0. Now we show that 〈ξ′k(0), ψ〉 is bounded from above. Arguing by contradiction, we assume that 〈ξ′k(0), ψ〉 =∞. Note that |ξk(tψ)− 1|‖uk‖+ ξk(tψ)‖tψ‖ ≥ ‖ξk(tψ)(uk + tψ)− uk‖ (4.9) and ξk(tψ) > ξk(0) = 1 for sufficiently large k. Then by the definition of ξ′k(0) and (4.1) with u = ξk(tψ)(uk + tψ) ∈ N+ γ,λ, we obtain |ξk(tψ)− 1| ‖uk‖ k + ξk(tψ) ‖tψ‖ k ≥ 1 k ‖ξk(tψ)(uk + tψ)− uk‖ ≥ Jγ,λ(uk)− Jγ,λ (ξk(tψ)(uk + tψ)) = ( 1 1− q − 1 2 )[( c‖(uk + tψ)‖2 − γ‖ ( (uk + tψ)+ ) ‖2 H ) − ( c‖uk‖2 − γ‖u+ k ‖ 2 H )] + ( 1 1− q − 1 2θ ) d ( ‖uk + tψ‖2θ − ‖uk‖2θ ) + ( 1 1− q − 1 2θ ) d[ξ2θ k (tψ)− 1]‖uk + tψ‖2θ + ( 1 1− q − 1 2 ) [ξ2 k(tψ)− 1] ( c‖uk + tψ‖2 − γ‖(uk + tψ)+‖2H ) − ( 1 1− q − 1 2p ) [ξ2p k (tψ)− 1] ∫ Ω ∫ Ω r(x)r(y) ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy − ( 1 1− q − 1 2p ) ξ2p k (tψ) ∫ Ω ∫ Ω r(x)r(y) [ ((uk + tψ)+(x)) p ((uk + tψ)+(y)) p |x|α|x− y|µ|y|α − ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α ] dx dy. Dividing above estimate by t > 0 and passing the limit t→ 0, we obtain 〈ξ′k(0), ψ〉‖uk‖ k + ‖ψ‖ k ≥ 1 + q 1− q ( c ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx ) + 1 + q 1− q 〈ξ′k(0), ψ〉 ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + (2θ + q − 1 1− q ) d〈ξ′k(0), ψ〉‖uk‖2θ EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 19 + (2θ + q − 1 1− q ) d‖uk‖2θ−2 ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − (2p+ q − 1) (1− q) ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α − (2p+ q − 1) (1− q) 〈ξ′k(0), ψ〉 ∫ Ω ∫ Ω r(x)r(y) ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy. ≥ 〈ξ ′ k(0), ψ〉 (1− q) [ (1 + q) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + (2θ + q − 1)d‖uk‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y) ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy ] + (1 + q 1− q )( c ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx ) + (2θ + q − 1 1− q ) d‖uk‖2θ−2 ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − (2p+ q − 1) (1− q) ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α . This implies that ‖ψ‖ k ≥ 〈ξ ′ k(0), ψ〉 (1− q) [ (1 + q) ( c‖uk‖2 − µ‖u+ k ‖ 2 H ) + (2θ + q − 1)d‖uk‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y) ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy − (1− q)‖uk‖ k ] + (1 + q 1− q )( c ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx ) + (2θ + q − 1 1− q ) d‖uk‖2θ−2 ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − (2p+ q − 1) (1− q) ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α . which is impossible under the assumption that 〈ξ′k(0), ψ〉 = ∞. Now by Lemma 3.6(1), boundedness of {uk}k and (4.6), we have[ (1 + q) ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) + (2θ + q − 1)d‖uk‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy − (1− q)‖uk‖ k ] ≥ C2 − (1− q)C1 k > 0. Hence 〈ξ′k(0), ψ〉 is uniformly bounded for sufficiently large k with any ψ ∈ X0 and ψ ≥ 0. � 20 S. GOYAL, T. SHARMA EJDE-2022/25 Lemma 4.3. Let γ ∈ (0, cγH), λ ∈ (0,Λ1) and {uk}k ⊂ N±γ,λ verify (4.1) and (4.2). Then for every ψ ∈ X0, we have l(x)(u+ k )−qψ ∈ L1(Ω), and( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− λ ∫ Ω l(x)(u+ k )−qψdx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy = H(uk, ψ) = o(1) (4.10) as k →∞. Proof. Let ψ ∈ X0 with ψ ≥ 0. Then (4.1) and (4.9), we have [ξk(tψ)− 1] ‖uk‖ k + ξk(tψ) ‖tψ‖ k ≥ Jγ,λ(uk)− Jγ,λ (ξk(tψ)(uk + tψ)) = − ( ξ2 k(tψ)− 1 ) 2 ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) − ( ξ2 k(tψ)− 1 ) 2θ d‖uk‖2θ − ξ2 k(tψ) 2 [( c‖uk + tψ‖2 − γ‖(uk + tψ)+‖2H ) − ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) ] − ξ2θ k (tψ) 2θ d [ ‖uk + tψ‖2θ − ‖uk‖2θ ] + λ ( ξ1−q k (tψ)− 1 ) 1− q ∫ Ω l(x) ( (uk + tψ)+ )1−q dx + λ 1− q ∫ Ω l(x) [ ( (uk + tψ)+ )1−q − ((uk)+ )1−q ] dx + ξ2p k (tψ)− 1 2p ∫ Ω ∫ Ω r(x)r(y) ((uk + tψ)+(x)) p ((uk + tψ)+(y)) p |x|α|x− y|µ|y|α dx dy + 1 2p ∫ Ω ∫ Ω r(x)r(y) |x|α|x− y|µ|y|α [ ( (uk + tψ)+(x) )p ( (uk + tψ)+(y) )p − ( (uk)+(x) )p ( (uk)+(y) )p ] dx dy. Dividing by t > 0 and passing the limit t→ 0+, we obtain |〈ξ′k(0), ψ〉|‖uk‖ k + ‖ψ‖ k ≥ −〈ξ′k(0), ψ〉 [ ( c‖uk‖2 − γ‖u+ k ‖ 2 H ) − λ ∫ Ω l(x)(u+ k )1−qdx+ d‖uk‖2θ − ∫ Ω ∫ Ω r(x)r(y) ((uk)+(x)) p ((uk)+(y)) p |x|α|x− y|µ|y|α dx dy ] − ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy + γ ∫ Ω u+ k ψ |x|2s dx + ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 21 + lim t→0+ inf λ 1− q ∫ Ω l(x) [ ((uk + tψ)+) 1−q − (u+ k )1−q ] t dx. = − ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy + γ ∫ Ω u+ k ψ |x|2s dx + ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy + lim t→0+ inf λ 1− q ∫ Ω l(x)[((uk + tψ)+) 1−q − (u+ k )1−q] t dx, since {uk}k ∈ Nγ,λ. Applying the above inequality, we have lim inf t→0+ ∫ Ω l(x)[((uk + tψ)+) 1−q − (u+ k )1−q] t dx is finite. Now, it follows from l(x)[((uk + tψ)+) 1−q − (u+ k )1−q] ≥ 0 and {uk}k is bounded in X0, Fatou’s lemma and Lemma 4.1 that λ ∫ Ω l(x)(u+ k )−qψdx ≤ lim t→0+ inf λ 1− q ∫ Ω l(x)[((uk + tψ)+) 1−q − (u+ k )1−q] t dx ≤ 〈ξ ′ k(0), ψ〉‖uk‖+ ‖ψ‖ k − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy + ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx ≤ C1C3 + ‖ψ‖ k + ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy, where C3 > 0 is given for the boundedness of 〈ξ′k(0), ψ〉 and ‖uk‖ ≤ C1. At the moment, taking limit k →∞, we obtain( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− ∫ Ω l(x)(u+ k )−qψdx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ≥ o(1). (4.11) Next, we prove that the equation (4.11) holds for any ψ ∈ X0. For this, in (4.11), we choose ψ = Ψ+ ε for ε > 0 as test function with Ψε = u+ k + εψ then as limit 22 S. GOYAL, T. SHARMA EJDE-2022/25 k →∞, we obtain o(1) ≤ ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (Ψ+ ε (x)−Ψ+ ε (y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k Ψ+ ε |x|2s dx− λ ∫ Ω l(x)(u+ k )−qΨ+ ε dx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 Ψ+ ε ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy = ( c+ d‖uk‖2θ−2 ) × ∫∫ R2N (uk(x)− uk(y)) ((Ψε + Ψ−ε )(x)− (Ψε + Ψ−ε )(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k (Ψε + Ψ−ε ) |x|2s dx− λ ∫ Ω l(x)(u+ k )−q(Ψε + Ψ−ε )dx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1( u+ k (y) )p (Ψε + Ψ−ε ) |x|α|x− y|µ|y|α dx dy. (4.12) We notice that for a.e. x, y ∈ RN , (uk(x)− uk(y)) ( u−k (x)− u−k (y) ) ≤ −|u−k (x)− u−k (y)|2, (4.13) (uk(x)− uk(y)) ( u+ k (x)− u+ k (y) ) ≤ |uk(x)− uk(y)|2. (4.14) Applying (4.14), we have∫∫ R2N (uk(x)− uk(y)) ((Ψε + Ψ−ε )(x)− (Ψε + Ψ−ε )(y)) |x− y|N+2s dx dy = ∫∫ R2N (uk(x)− uk(y)) ( u+ k (x)− u+ k (y) ) |x− y|N+2s dx dy + ε ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy + ∫∫ R2N (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy ≤ ∫∫ R2N |uk(x)− uk(y)|2 |x− y|N+2s dx dy + ε ∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy + ∫∫ R2N (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy. (4.15) Also,∫ Ω u+ k (Ψε + Ψ−ε ) |x|2s dx = ∫ Ω |u+ k |2 |x|2s dx+ ε ∫ Ω u+ k ψ |x|2s dx+ ∫ Ω u+ k Ψ−ε |x|2s dx ≥ ∫ Ω |u+ k |2 |x|2s dx+ ε ∫ Ω u+ k ψ |x|2s dx+ ε ∫ Ωε u+ k ψ |x|2s dx, (4.16) EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 23 where Ωε = {x ∈ RN : Ψε ≤ 0}. Now using (4.16), (4.15), and (4.12), yield o(1) ≤ [ ( c+ d‖uk‖2θ−2 ) ‖uk‖2 − γ‖u+ k ‖ 2 H − λ ∫ Ω l(x)(u+ k )1−q dx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ] + ε [ ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− λ ∫ Ω l(x)(u+ k )−qψdx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ] + ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy − εγ ∫ Ωε u+ k ψ |x|2s dx+ λ ∫ Ωε l(x)(u+ k )−q(u+ k + εψ)dx + ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 (u+ k + εψ)(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy. Using that {uk}k ∈ Nγ,λ, we deduce that o(1) ≤ ε [ ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− λ ∫ Ω l(x)(u+ k )−qψdx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ] + ( c+ d‖uk‖2θ−2 ) ∫∫ R2N (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy − εγ ∫ Ωε u+ k ψ |x|2s dx + ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 (u+ k + εψ)(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy. (4.17) Now the symmetry of the fractional kernel and (uk(x)− uk(y)) ( u+ k (x)− u+ k (y) ) ≥ |u+ k (x)− u+ k (y)|2, yield ∫∫ RN×RN (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy = ∫∫ Ωε×Ωε (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy + 2 ∫∫ Ωε×(RN\Ωε) (uk(x)− uk(y)) (Ψ−ε (x)−Ψ−ε (y)) |x− y|N+2s dx dy 24 S. GOYAL, T. SHARMA EJDE-2022/25 ≤ −ε (∫∫ Ωε×Ωε (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy + 2 ∫∫ Ωε×(RN\Ωε) (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy ) ≤ 2ε ∫∫ Ωε×RN ∣∣ (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s ∣∣ dx dy Now by Hölder inequality and boundedness of {uk}k in X0, we have∫∫ Ωε×RN ∣∣∣∣ (uk(x)− uk(y))(ψ(x)− ψ(y)) |x− y|N+2s ∣∣∣∣ dx dy ≤ C (∫∫ Ωε×RN ∣∣ (ψ(x)− ψ(y)) |x− y|(N+2s)/2 ∣∣2 dx dy)1/2 , (4.18) Here (ψ(x)−ψ(y)) |x−y|(N+2s)/2 ∈ L2(R2N ), Now for ν > 0 there exists Rν sufficiently large such that ∫∫ (suppψ)×[RN\BRν ] ∣∣ (ψ(x)− ψ(y)) |x− y|(N+2s)/2 ∣∣2 dx dy < ν 2 . Now by the definition of Ωε, we have Ωε ⊂ suppψ and |Ωε × BRν | → 0 as ε→ 0+. So by (ψ(x)−ψ(y)) |x−y|(N+2s)/2 ∈ L2(R2N ), there is a δν > 0 and εν > 0 such that for any ε ∈ (0, εν ], we have |Ωε×BRν | < δν and ∫∫ Ωε×BRν ∣∣ (ψ(x)−ψ(y)) |x−y|(N+2s)/2 ∣∣2 dx dy < ν 2 . Hence for ε ∈ (0, εν ], we obtain lim ε→0+ ∫∫ Ωε×RN ∣∣ (ψ(x)− ψ(y)) |x− y|(N+2s)/2 ∣∣2 dx dy = 0. (4.19) Hence by (4.18), lim ε→0+ ∫∫ Ωε×RN ∣∣ (uk(x)− uk(y))(ψ(x)− ψ(y)) |x− y|N+2s ∣∣ dx dy = 0. Now we claim that lim ε→0 ∫ Ωε u+ k ψ |x|2s dx = 0. (4.20) Using that for x ∈ Ωε, we have u+ k + εψ ≤ 0, which imply that ψ(x) ≤ 0. Hence, by (1.4) 0 ≤ ∣∣ ∫ Ωε u+ k ψ |x|2s dx ∣∣ ≤ ∫ Ωε |u+ k ||ψ| |x|2s dx ≤ ε ∫ Ωε |ψ|2 |x|2s dx ≤ ε‖ψ‖2H ≤ ε ‖ψ‖2 µH . As ε→ 0, we prove claim (4.20). Next, we show that lim ε→0+ 1 ε ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 (u+ k + εψ)(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy = 0 (4.21) For this, consider∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 (u+ k + εψ)(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ≤ ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p( u+ k (y) )p |x|α|x− y|µ|y|α dx dy EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 25 + ε ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy ≤ ∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p( u+ k (y) )p |x|α|x− y|µ|y|α dx dy + ε (∫ Ω ∫ Ωε r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p−1 ψ(y) |x|α|x− y|µ|y|α dx dy )1/2 × ( ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p( u+ k (y) )p |x|α|x− y|µ|y|α dx dy )1/2 ≤ CCr(α, µ,N) (∫ Ωε ( u+ k (x) )2∗s dx)p/2∗s + CεCr(α, µ,N) (∫ Ωε ( [u+ k (x)]p−1ψ(x) )2∗s/p dx)p/2∗s ≤ CCr(α, µ,N) (∫ Ωε ( u+ k (x) )2∗s dx)p/2∗s + CεCr(α, µ,N) (∫ Ωε ( u+ k (x) )2∗s dx)(p−1)/2∗s (∫ Ωε |ψ(x)|2 ∗ s dx )1/2∗s ≤ CCr(α, µ,N)εp (∫ Ωε |ψ(x)|2 ∗ s dx )p/2∗s + C̃εCr(α, µ,N)εp (∫ Ωε |ψ(x)|2 ∗ s dx )p/2∗s . Dividing the above estimate by ε and using |Ωε| → 0 as ε → 0+, we can easily deduce (4.21). Now dividing (4.17) by ε together with (4.19), (4.20), (4.21), and |Ωε| → 0 as ε→ 0+, we deduce that o(1) ≤ ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) (ψ(x)− ψ(y)) |x− y|N+2s dx dy − γ ∫ Ω u+ k ψ |x|2s dx− λ ∫ Ω l(x)(u+ k )−qψdx − ∫ Ω ∫ Ω r(x)r(y) ( u+ k (x) )p−1 ψ(x) ( u+ k (y) )p |x|α|x− y|µ|y|α dx dy. which prove (4.11). In consonance with the arbitrariness of ψ, we derive that (4.10) holds for any ψ ∈ X0. � Lemma 4.4. Let λ ∈ (0,Λ1), γ ∈ (0, cγH) and {uk}k ⊂ N±γ,λ with Jγ,λ(uk) → c, then the sequence {uk}k contains a subsequence strongly convergent to u0 in X0. Proof. By (4.2), {uk}k and {u−k }k are both bounded in X0. Taking ψ = u−k as k →∞ in (4.10), we have lim k→∞ ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) ( u−k (x)− u−k (y) ) |x− y|N+2s dx dy = 0. which together with c > 0 yields that ‖u−k ‖ → 0 as k → ∞. So, we can assume that {uk}k is a sequence of non-negative functions. Furthermore by Lemma 2.1 26 S. GOYAL, T. SHARMA EJDE-2022/25 and (1.4), we can extract a subsequence still denote by {uk}k such that uk ⇀ u0 weakly in L2∗s (Ω), ‖uk‖ → v uk → u0 in Lp(Ω) for any p ∈ (1, 2∗s) uk ⇀ u0 in L2∗s (Ω, |x|−2s), ‖uk‖H → l uk → u0 a.e. in Ω uk ≤ h a. e. in Ω, (4.22) as k →∞ with h ∈ Lp(Ω) for a fixed p ∈ [1, 2∗s). Now if v = 0 in (4.22), then we can easily see that uk → 0 strongly in X0 as k →∞. Hence we assume v > 0. Now by Brézis-Lieb Lemma [4] and Lemma 2.2, we obtain ‖uk‖2 = ‖uk − u0‖2 + ‖u0‖2 + o(1), (4.23) ‖uk‖2H = ‖uk − u0‖2H + ‖u0‖2H + o(1), (4.24)∫ Ω ∫ Ω r(x)r(y)(uk(x)) p (uk(y)) p |x|α|x− y|µ|y|α dx dy = ∫ Ω ∫ Ω r(x)r(y)((uk − u0)(x)) p (uk − u0)(y)) p |x|α|x− y|µ|y|α dx dy + ∫ Ω ∫ Ω r(x)r(y)(u0(x)) p (u0(y)) p |x|α|x− y|µ|y|α dx dy + o(1). (4.25) Now by (4.23), (4.24) and (4.25), we have o(1) = ( c+ d‖uk‖2θ−2 )∫∫ R2N (uk(x)− uk(y)) ((uk − u0)(x)− (uk − u0)(y)) |x− y|N+2s dx dy − γ ∫ Ω uk(uk − u0) |x|2s dx− λ ∫ Ω l(x)(uk)−q(uk − u0)dx − ∫ Ω ∫ Ω r(x)r(y)(uk(x))p−1(uk(y))p |x|α|x− y|µ|y|α (uk − u0) dx dy = ( c+ dv2θ−2 ) ( v2 − ‖u0‖2 ) − γ ( ‖uk‖2H − ‖u0‖2H ) − λ ∫ Ω l(x)(uk)−q(uk − u0)dx− ∫ Ω ∫ Ω r(x)r(y)(uk(x))p((uk(y))p |x− y|N+2s dx dy + ∫ Ω ∫ Ω r(x)r(y)(u0(x))p(u0(y))p |x− y|N+2s dx dy + o(1) = ( c+ dv2θ−2 ) ‖uk − u0‖2 − γ‖uk − u0‖2H − λ ∫ Ω l(x)(uk)−q(uk − u0)dx − ∫ Ω ∫ Ω r(x)r(y)((uk(x)− u0(x))) p ((uk(y)− u0(y))) p |x|α|x− y|µ|y|α dx dy + o(1). Taking limit as k →∞, we have( c+ dv2θ−2 ) lim k→∞ ‖uk − u0‖2 − γ lim k→∞ ‖uk − u0‖2H = λ lim k→∞ ∫ Ω l(x)(uk)−q(uk − u0)dx + lim k→∞ ∫ Ω ∫ Ω r(x)r(y)((uk(x)− u0(x))) p ((uk(y)− u0(y))) p |x|α|x− y|µ|y|α dx dy. (4.26) EJDE-2022/25 FRACTIONAL KIRCHHOFF HARDY PROBLEMS 27 Since q ∈ (0, 1) and l ∈ L 2∗s 2∗s−1+q (Ω), we conclude from Vitali’s convergence theorem that lim k→∞ ∫ Ω l(x)(u+ k )1−q dx = ∫ Ω l(x)(u+ 0 )1−q dx In view of Lemma 4.3, we obtain l(x)u−qk u0 ∈ L1(Ω) for each k ∈ N. It follows from Fatou’s lemma that ∫ Ω l(x)u1−q 0 dx ≤ lim inf k→∞ ∫ Ω l(x)u−qk u0dx. Now for r ∈ L 2∗s 2∗µ,s,α−p (Ω) using Vitali’s convergence theorem, one can noticed that r(x)|uk − u0|p → 0 strongly in L2∗µ,s,α(Ω). It follows from (2.1) that lim k→∞ ∫ Ω ∫ Ω r(x)r(y)((uk(x)− u0(x))) p ((uk(y)− u0(y))) p |x|α|x− y|µ|y|α dx dy = 0. (4.27) Using (4.27) in (4.26), we have 0 ≥ ( c+ dv2θ−2 ) lim k→∞ ‖uk − u0‖2 − γ lim k→∞ ‖uk − u0‖2H = c [ lim k→∞ ‖uk − u0‖2 − γ c lim k→∞ ‖uk − u0‖2H ] + dv2θ−2 lim k→∞ ‖uk − u0‖2. This and (2.9) yield 0 ≥ hc,γ lim k→∞ ‖uk − u0‖2 + dv2θ−2 lim k→∞ ‖uk − u0‖2 which implies that 0 ≥ lim k→∞ ‖uk − u0‖2 [ hc,γ + dv2θ−2 ] > 0, a contradiction. Hence v = 0. Thus uk → u0 strongly in X0. � 5. Proof of the main Theorem In this section we prove the existence of solutions in N+ γ,λ and N−γ,λ. Theorem 5.1. Let 0 < λ < Λ∗ = min(Λ1,Λ2). Assume f and g satisfies (A1) and (A2) respectively. Then the problem (2.7) has a positive solution in N+ γ,λ. Proof. We first show that m+ γ,λ = infu∈N+ γ,λ Jγ,λ(u) < 0. Since u ∈ N+ γ,λ ⊂ Nγ,λ we have Jγ,λ(u) = (1 2 − 1 1− q ) [c‖u‖2 − γ‖u+‖2H ] + ( 1 2θ − 1 1− q ) d‖u‖2θ − ( 1 2p − 1 1− q ) ∫ Ω ∫ Ω r(x)r(y)(u+(x)) p (u+(y)) p |x|α|x− y|µ|y|α dx dy = − 1 2p(1− q) [ (1 + q)[c‖u‖2 − γ‖u+‖2H ] + (2θ + q − 1)d‖u‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y)(u+(x)) p (u+(y)) p |x|α|x− y|µ|y|α dx dy ] < 0, since u ∈ N+ γ,λ. Hence m+ γ,λ < 0. 28 S. GOYAL, T. SHARMA EJDE-2022/25 Now for fix λ < Λ∗ = min (Λ1,Λ2). Then by Ekeland’s variational principle and Lemma 3.2, there exists a minimizing sequence {uk}k ⊂ N+ γ,λ∪{0}, satisfying (4.1) and (4.2). Hence Jγ,λ(uk)→ m+ γ,λ < 0 as n→∞, which gives that {uk}k ⊂ N+ γ,λ. Subsequently using Lemma 4.4 with c = m+ γ,λ, we know that uk → u0 in X0, upto a subsequence. Moreover, by Lemma 4.1 and (3.9), we have (1 + q) [ c‖u0‖2 − γ‖u0‖2H ] + d(2θ + q − 1)‖u0‖2θ − (2p+ q − 1) ∫ Ω ∫ Ω r(x)r(y) ( u+ 0 (x) )p( u+ 0 (y) )p |x|α|x− y|µ|y|α dx dy > 0, which implies that u0 ∈ N+ γ,λ, and m+ γ,λ is attained at u0 by Jγ,λ is continuous on X0. Take limit as k → ∞ together with Fatou’s lemma in (4.10) , we obtain H(u0, ψ) ≥ 0[where H is defined in (4.10)] for ψ ∈ X0 with ψ ≥ 0. Now letting a test function ψ = Ψ+ ε with Ψε = u+ 0 +εψ and ψ ∈ X0. Following the same process for u0 in place of uk from (4.10) to (4.21), we know that H(u0, ψ) ≥ 0 for ψ ∈ X0, which produce that λl(x)(u+ 0 )−qψ ∈ L1(Ω) and u0 ∈ N+ γ,λ. So accordingly Lemma 3.2, u0 6= 0. Furthermore, by (2.8) with ψ = u−0 together with (4.13), we obtain ‖u−0 ‖ = 0. Hence u0 is a positive solution of (1.1). � Theorem 5.2. Let 0 < λ < Λ∗ = min(Λ1,Λ2). Assume l satisfies (A1) and r satisfies (A2). Then problem (2.7) has a positive solution in N−γ,λ. Proof. Since N−γ,λ is a closed set in X0, we can extract a minimizing sequence {Uk}k ⊂ N−γ,λ satisfying the Ekeland’s variational principle for infu∈N−γ,λ Jγ,λ(u), since {Uk}k is bounded in X0, then suppose {Uk}k ⇀ U0 in X0, Now by Lemma 4.4, {Uk}k → U0 in X0 up to a subsequence because N−γ,λ is closed then U0 ∈ N−γ,λ with Jγ,λ(U0) = m−γ,λ, Now repeating the same argument as in Theorem 5.1, U0 verify H(U0, ψ) ≥ 0 (for H one can see (4.10)), so that λl(x)(U+ 0 )−qψ ∈ L1(Ω) for any ψ ∈ X0 and U0 in N−γ,λ. Binding this with Lemma 3.2, we obtain U0 is a positive solution to the problem (1.1). � Proof of Theorem 1.2. By above Theorems 5.1 and 5.2, we can see that problem (1.1) admits two positive solutions u0 and U0, since N+ γ,λ ∩ N − γ,λ = ∅. Hence these solutions are distinct. � Acknowledgements. 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Yafaev; Sharp constants in the Hardy-Rellich inequalities, Journal of Functional Analysis 168 (1999), no. 1, 121–144. Tarun Sharma Department of Mathematics, Bennett University, Greater Noida, Uttar Pradesh, India Email address: tarunsharma80065@gmail.com Sarika Goyal Department of Mathematics, Bennett University, Greater Noida, Uttar Pradesh, India Email address: sarika1.iitd@gmail.com 1. Introduction 2. Preliminaries 3. Fibering map analysis 4. A compactness result for J, 5. Proof of the main Theorem Acknowledgements References