Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 07, pp. 1–21. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu GROUND STATE AND MULTIPLE SOLUTIONS FOR CRITICAL FRACTIONAL SCHRÖDINGER-POISSON EQUATIONS WITH PERTURBATION TERMS LINTAO LIU, KAIMIN TENG Abstract. In this article, we study a class of critical fractional Schrödinger- Poisson system with two perturbation terms. By using variational methods and Lusternik-Schnirelman category theory, the existence of ground state and two nontrivial solutions are established. 1. Introduction In this article, we consider the nonlinear fractional Schrödinger-Poisson system with critical nonlinearity (−∆)su+ u+K(x)φu = a(x)|u|p−2u+ µb(x)|u|q−2u+ |u|2 ∗ s−2u, in R3, (−∆)tφ = K(x)u2, in R3, (1.1) where (−∆)α is the fractional Laplacian operator for α = s, t. p, q ∈ (4, 2∗s), s ∈ ( 3 4 , 1), 2s + 2t > 3, µ > 0 is a parameter, K(x), a(x) and b(x) satisfy the following conditions: (A1) K(x) ∈ C(R3), K(x) ≥ 0 and lim|x|→∞K(x) = K∞ > 0; (A2) there exist C0 > 0 and k > 0 such that K(x) ≤ K∞− C0 (1+|x|)k for all x ∈ R3; (A3) there exist C1 > 0 and d > 0 such that K(x) ≤ K∞+ C1 (1+|x|)d for all x ∈ R3; (A4) a(x) ∈ C(R3), a(x) ≥ 0 and lim|x|→∞ a(x) = a∞ > 0; (A5) there exist C2 > 0 and a > 0 such that a(x) ≥ a∞− C2 (1+|x|)a for all x ∈ R3; (A6) b(x) ∈ C(R3), b(x) ≥ 0 and lim|x|→∞ b(x) = 0; (A7) there exist C3 > 0 and b > 0 such that b(x) ≥ C3 (1+|x|)b for all x ∈ R3. Since the first equation in (1.1) is of fractional Schrödinger equation with a po- tential φ satisfying the fractional Poisson equation, we call system (1.1) a fractional Schrödinger-Poisson system. In recent years, equations or systems with fractional Laplace operators have been studied extensively because they are widely used in fractional quantum mechanics, physics, chemistry, obstacle problems, optimization and finance, we refer to see [12, 16, 20, 21, 23] and so on. It is also well applied in the mathematical theory of conformal geometry and minimal surface, see [9]. 2010 Mathematics Subject Classification. 35R11, 35B38. Key words and phrases. Fractional Schrödinger-Poisson system; variational methods; critical growth; Lusternik-Schnirelman category. c©2021 Texas State University. Submitted April 10, 2020. Published February 1, 2021. 1 2 L. LIU, K. TENG EJDE-2021/07 As far as we know, there are a few papers considering (1.1) after it was introduced in [15]. In [15], the author studied the local and global well-posedness of the Cauchy problem i∂tΨ + 1 2 ∆xΨ = A0Ψ + α|Ψ|γ−1Ψ, (t, x) ∈ R× R, (−∆)σ/2x A0 = |Ψ|2, Ψ(·, x) = f, where σ ∈ (0, 1), α = ±1, 1 < γ ≤ 5. Recently, Zhang, Do ó and Squassina [38] established the existence of radial ground state solution to the following fractional Schrödinger-Poisson system with a general subcritical or critical nonlinearity (−∆)su+ λφu = f(u), in R3, (−∆)tφ = λu2, in R3. Teng [33] studied the existence of a nontrivial ground state solution through using the method of Pohozaev-Nehari manifold, the monotonic trick and global compact- ness Lemma for the system (−∆)su+ V (x)u+ φu = |u|p−1u, in R3, (−∆)sφ = u2, in R3. Using a similar argument, Teng in [32] also studied the existence of ground state solutions for the critical problem with a perturbation term (−∆)su+ V (x)u+ φu = µ|u|q−1u+ |u|2 ∗ s−2u, in R3, (−∆)tφ = u2, in R3. For other related works, see [22, 28] and their references. On the other hand, when s = t = 1, system (1.1) reduces to classical Schrödinger- Poisson system written by a more general form −∆u+ V (x)u+K(x)φu = f(x, u), in R3, −∆φ = K(x)u2, in R3. (1.2) This is called the system of Schrödinger-Poisson equations because it consists of a Schrödinger equation coupled with a Poisson term. In the previous decades, there has been a lot of work dealing with the system (1.2) under different assumptions on V , K and f , see [2, 3, 4, 8, 10, 11, 15, 17, 19, 24, 27, 29, 35, 37, 39, 40] and the references therein. For example, in [3], the authors proved the existence of ground state solutions for the subcritical 3 < p < 6 and the critical case f = |u|p−2u+ u5 with 4 < p < 6. For the case p ≤ 2 or p ≥ 6, the reader may see [11] and for the case 2 < p < 6, can see [2, 3, 8, 10, 24]. In the case of V being non-radial, K ≡ 1 and f = |u|p−2u, the existence of ground state solution for system (1.2) was obtained in [3, 40] for 4 < p < 6 and 3 < p ≤ 4; In [5], the authors proved the existence of ground state and bound states for the case when V ≡ 1 and f = a(x)|u|p−2u with 4 < p < 6. In [37], the author considered a general critical situation with two perturbation term and obtained the existence and multiplicity of solutions via using Lusternik-Schnirelman category due to [1, 6]. EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 3 To the best of our knowledge, there are few papers on the multiplicity solutions for system (1.1). Inspired by [1, 6, 37], we construct two mappings: FR : S2 = {y ∈ R3 : |y| = 1} → {u ∈M : I(u) ≤ m∞ − ε(R)}, G : {u ∈M : I(u) < m∞} → S2, so that G◦FR homotopic to the identity. Using the theory of Lusternik-Schnirelman category, we will establish the existence of two nontrivial solutions for system (1.1). Our main results are stated as follows. Theorem 1.1. Assume that K, a and b satisfy (A1), (A2), (A4)–(A6) with 0 < k < α, where α = min{a, (3 + 2s)p}. Then problem (1.1) admits a positive ground state solution. Theorem 1.2. Suppose that (A1), (A3)–(A7) hold with b < min{α, β}, where α = min{a, (3 + 2s)p} and β = min{d, 6 + 4s}. Then problem (1.1) admits a positive ground state solution. Theorem 1.3. Assume that K ∈ C1(R3), a ∈ C1(R3) and b(x) satisfy (A1), (A3)– (A7) with K(x) ≥ K∞, a(x) ≤ a∞ and meas{x ∈ R3 : K(x) ≥ K∞} > 0. Then there exists µ0 > 0 small such that for any µ ∈ (0, µ0), problem (1.1) admits at least two nontrivial solutions. The rest of the paper is organized as follows: In Section 2, we give some prelim- inaries. In Section 3, we prove Theorem 1.1 and Theorem 1.2. Section 4 devotes to proving Theorem 1.3. 2. Preliminary lemmas In the sequel, we use the following notation: • Hs(R3) denotes the fractional sobolev space with norm ‖u‖2 := ∫ R3 (|(−∆) s 2u|2 + u2)dx and Ds,2(R3) := {u ∈ L2∗s (R3) : (−∆) s 2u ∈ L2(R3)} denotes the homogeneous fractional sobolev space with the norm ‖u‖2Ds,2 := ∫ R3 |(−∆) s 2u|2dx. • C denotes a universal positive constant (possibly different). • It is well known that Hα(R3) is continuously embedded into Lp(R3) for 2 ≤ p ≤ 2∗α(2∗α = 6 3−2α ), and for any α ∈ (0, 1), there exists a best constant Sα > 0 such that Sα = inf u∈Dα,2 ∫ R3 |(−∆) α 2 u|2dx ( ∫ R3 |u(x)|2∗αdx) 2 2∗α • For simplicity, we assume K∞ = 1 and a∞ = 1. Denote H = Hs(R3) and Ds,2 = Ds,2(R3). In this section, we assume (A1), (A4) and (A6) hold. Similar to the argument in [24], we know the function φtu has the following properties. Lemma 2.1. For any u ∈ H, we have 4 L. LIU, K. TENG EJDE-2021/07 (i) φtu ≥ 0; (ii) φt~u = ~2φtu, ∀~ > 0; (iii) ‖φtu‖Dt,2 ≤ C‖u‖2 12 3+2t ≤ C‖u‖2, ∫ R3 φ t uu 2dx ≤ C‖u‖4 12 3+2t ≤ C‖u‖4. By the Lax-Milgram theorem, there exists a unique φtu ∈ Dt,2(R3) such that (−∆)tφtu = K(x)u2. Thus, we can rewrite (1.1) as (−∆)su+ u+K(x)φtuu = a(x)|u|p−2u+ µb(x)|u|q−2u+ |u|2 ∗ s−2u. (2.1) To find weak solutions to (2.1), we look for critical points of the functional I(u) : H → R associated with (2.1) which is defined by I(u) = 1 2 ∫ R3 (|(−∆) s 2u|2 + u2)dx+ 1 4 ∫ R3 K(x)φtuu 2dx − 1 p ∫ R3 a(x)|u|pdx− µ q ∫ R3 b(x)|u|qdx− 1 2∗s ∫ R3 |u|2 ∗ sdx. To prove the compactness, we need to consider the following problem at infinity associated with (2.1): (−∆)su+ u+ φ̂tuu = |u|p−2u+ |u|2 ∗ s−2u, u > 0, (2.2) where φ̂tu ∈ Dt,2(R3) is the unique solution to problem (−∆)tφ = u2. The functional associated with (2.2) is I∞(u) = 1 2 ∫ R3 (|(−∆) s 2u|2 + u2)dx+ 1 4 ∫ R3 φ̂tuu 2dx − 1 p ∫ R3 |u|pdx− 1 2∗s ∫ R3 |u|2 ∗ sdx. Let m = inf u∈M I(u), m∞ = inf u∈M∞ I∞(u), where M = {u ∈ Hs(R3)\{0} : [I ′(u), u] = 0}, M∞ = {u ∈ Hs(R3)\{0} : [I ′∞(u), u] = 0} are Nehari manifolds correspond to the functionals I and I∞, respectively. Similar argument as [22, Proposition 3.4], we can obtain the following Lemma. Lemma 2.2. By using [22, Proposition 3.4], problem (2.2) has a positive ground state solution u∞ ∈ C1,2s+σ−1(R3) ∩ L∞(R3), where σ ∈ (0, 1) and 2s+ σ > 1. From u ∈ H ∩ C1,2s+σ−1(R3), we see that lim|x|→∞ u∞(x) = 0. Similar as the proof of [32, Proposition 3.8], we conclude that there exists C > 0 such that 0 < u∞(x) ≤ C (1 + |x|)3+2s , ∀x ∈ R3. Moreover, in [22], the authors showed that m∞ = c∞ = infγ∈Γ maxt∈[0,1] I∞(γ(t)), where Γ = {γ ∈ C([0, 1], H) : γ(0) = 0, I∞(γ(1)) < 0} and m∞ = c∞ = inf u∈H\{0} max t≥0 I∞(tu). (2.3) EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 5 To prove the (PS)c condition, we need the following function and its estimates (see [32]) vε(x) = ψ(x)Uε(x), x ∈ R3, where Uε(x) = ε− 3−2s 2 u∗(x/ε), u∗( x ε ) = ũ(x/S 1 2s s ) ‖ũ‖2∗s , κ ∈ R\{0}, µ > 0, and x0 ∈ R3 are fixed constants, ũ(x) = κ(µ2 + |x− x0|2)− 3−2s 2 , and ψ ∈ C∞(R3) such that 0 ≤ ψ ≤ 1 in R3, ψ(x) ≡ 1 in Bδ and ψ(x) ≡ 0 in R3\B2δ. We know that∫ R3 |(−∆) s 2 vε(x)|2dx ≤ S 3 2s s +O(ε3−2s), (2.4)∫ R3 |vε(x)|2 ∗ sdx = S 3 2s s +O(ε3), (2.5) ∫ R3 |vε(x)|pdx =  O(ε (2−p)3+2sp 2 ), p > 3 3−2s ; O(ε (2−p)3+2sp 2 | log ε|), p = 3 3−2s ; O(ε 3−2s 2 p), p < 3 3−2s . (2.6) Lemma 2.3. Let {un} ⊂ H be a bounded sequence such that I(un)→ c ∈ (0,m∞) and I ′(un)→ 0. Then {un} admits a strongly convergent subsequence in H. Proof. First we show that m∞ < s 3S 3 2s s . By (2.3), we see that c∞ ≤ supt≥0 I∞(tvε). Thus we only need to prove supt≥0 I∞(tvε) < s 3S 3 2s s for ε > 0 small. By Lemma 2.1, we have I∞(tvε) ≤ 1 2 t2‖vε‖2 + Ct4‖vε‖4 − 1 2∗s t2 ∗ s‖vε‖ 2∗s 2∗s . (2.7) Form (2.4)-(2.6), there exists ε1 > 0 small enough such that ‖vε‖2 := ∫ R3 (|(−∆) s 2 vε|2 + v2 ε)dx ≤ S 3 2s s +O(ε3−2s) +O(ε3−2s) ≤ 3 2 S 3 2s s , (2.8) ‖vε‖ 2∗s 2∗s = S 3 2s s +O(ε3) ≥ 1 2 S 3 2s s , (2.9) for ε ∈ (0, ε1). Thus, form (2.7)-(2.9), we have I∞(tvε) ≤ 3 4 t2S 3 2s s + C 9 4 t4S 3 s s − 1 2∗s t2 ∗ s 1 2 S 3 2s s . (2.10) By 2 < 4 < 2∗s, there exist a small t1 > 0 and a large t2 > 0 independent of ε ∈ (0, ε1) such that sup t∈[0,t1]∪[t2,+∞) I∞(tvε) < s 3 S 3 2s s . (2.11) Form Lemma 2.1 and (2.4)-(2.6), we obtain sup t∈[t1,t2] I∞(tvε) ≤ sup t≥0 [ 1 2 t2 ∫ R3 |(−∆) s 2 vε|2dx− 1 2∗s t2 ∗ s ∫ R3 |vε(x)|2 ∗ sdx] + C‖vε‖22 + C‖vε‖4 12 3+2t − C‖vε‖pp = s 3 S 3 2s s +O(ε3−2s)− Cε (2−p)3+2sp 2 . (2.12) 6 L. LIU, K. TENG EJDE-2021/07 In view of p ∈ (4, 2∗s), s ∈ ( 3 4 , 1), so we see that (2−p)3+2sp 2 < 3 − 2s. By choosing ε ∈ (0, ε1) small, we obtain sup t∈[t1,t2] I∞(tvε) < s 3 S 3 2s s . (2.13) By (2.11) and (2.13), we have m∞ < s 3 S 3 2s s . (2.14) Since {un} is bounded in H, up to a subsequence, we may assume that un ⇀ u weakly in H, un → u in Lrloc(R3) for 1 ≤ r < 2∗s and un → u a.e. R3. Thus by standard argument, we can show that I ′(u) = 0. Set vn = un − u. By the Brezis-Lieb Lemma in [36], we have that ‖vn‖2 = ‖un‖2 − ‖u‖2 + o(1), ‖vn‖ 2∗s 2∗s = ‖un‖ 2∗s 2∗s − ‖u‖2 ∗ s 2∗s + o(1), (2.15) and ∫ R3 a(x)|vn|pdx = ∫ R3 a(x)|un|p − ∫ R3 a(x)|u|p + o(1),∫ R3 b(x)|vn|qdx = ∫ R3 b(x)|un|q − ∫ R3 b(x)|u|q + o(1). From lim|x|→∞ a(x) = 1, lim|x|→∞ b(x) = 0, and vn → 0 in Lrloc(R3) for any r ∈ [1, 2∗s), we deduce that∫ R3 a(x)|un|p − ∫ R3 a(x)|u|p = ∫ R3 |vn|pdx+ o(1),∫ R3 b(x)|un|q − ∫ R3 b(x)|u|q = o(1). (2.16) By [33, Lemma 2.5], we can see that∫ R3 K(x)φtvnv 2 ndx = ∫ R3 K(x)φtunu 2 ndx− ∫ R3 K(x)φtuu 2dx+ o(1). From lim|x|→∞K(x) = 1 and Hölder’s inequality, it is easy to deduce that∫ R3 φ̂tvnv 2 n dx = ∫ R3 K(x)φtunu 2 ndx− ∫ R3 K(x)φtuu 2dx+ o(1). (2.17) Thus, from (2.15)-(2.17), it follows that c− I(u) = I∞(vn) + o(1). (2.18) By using [13, Proposition 5.1.1], we see that u ∈ L∞(R3). Then by [36, Lemmas 8.1 and 8.9], we have that | ∫ R3 (u 2∗s−1 n − u2∗s−1 − v2∗s−1 n )ϕdx| = o(1)‖ϕ‖, ∀ϕ ∈ H, | ∫ R3 a(x)(|un|p−2un − |u|p−2u− |vn|p−2vn)ϕdx| = o(1)‖ϕ‖, ∀ϕ ∈ H, | ∫ R3 b(x)(|un|q−2un − |u|q−2u− |vn|q−2vn)ϕdx| = o(1)‖ϕ‖, ∀ϕ ∈ H. (2.19) EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 7 Together with lim|x|→∞ a(x) = 1, lim|x|→∞ b(x) = 0, we deduce that∣∣ ∫ R3 [a(x)(|un|p−2un − |u|p−2u)− |vn|p−2vn]ϕdx ∣∣ = o(1)‖ϕ‖, ∀ϕ ∈ H,∣∣ ∫ R3 b(x)(|un|q−2un − |u|q−2u)ϕdx ∣∣ = o(1)‖ϕ‖, ∀ϕ ∈ H. (2.20) Using [33, Lemma 2.5], we have | ∫ R3 K(x)(φtunun − φ t uu− φtvnvn)ϕdx| = o(1)‖ϕ‖, ∀ϕ ∈ H. From lim|x|→∞K(x) = 1, and similar to the of proof of (2.17), we obtain∣∣ ∫ R3 K(x)(φtunun − φ t uu)ϕdx− ∫ R3 φ̂tvnvnϕdx ∣∣ = o(1)‖ϕ‖, ∀ϕ ∈ H. (2.21) Hence, by (2.19)-(2.21), it holds I ′∞(vn) = o(1). (2.22) We claim vn → 0 in H. Two cases occur: either lim n→∞ sup y∈R3 ∫ B1(y) |vn|2dx = 0, or there exists γ > 0 such that lim n→∞ sup y∈R3 ∫ B1(y) |vn|2dx ≥ γ. Thus, either ‖vn‖r → 0 for any r ∈ (2, 2∗s) through using vanishing Lemma, or there yn ∈ R3 with |yn| → ∞ such that vn(. + yn) ⇀ v 6= 0 weakly in H. If vn(.+ yn) ⇀ v 6= 0 weakly in H, from (2.18) and (2.22), it follows that c− I(u) = I∞(vn(.+ yn)) + o(1) and I ′∞(vn(.+ yn)) = o(1). Thus I ′∞(v) = 0 and c− I(u) = I∞(vn(.+ yn))− 1 4 [I ′∞(vn(.+ yn)), vn(.+ yn)] = 1 4 ‖(vn(.+ yn)‖2 + ( 1 4 − 1 p ) ∫ R3 |(vn(.+ yn)|pdx + ( 1 4 − 1 2∗s ) ∫ R3 |(vn(·+ yn)|2 ∗ sdx+ o(1), form which we obtain c ≥ I(u) + 1 4 ‖v‖2 + ( 1 4 − 1 p ) ∫ R3 |v|pdx+ ( 1 4 − 1 2∗s ) ∫ R3 |v|2 ∗ sdx = I(u) + I∞(v)− 1 4 [I ′∞(v), v)] = I(u) + I∞(v). By the definition of m∞, we have I∞(v) ≥ m∞. Since I ′(u) = 0, we have I(u) = I(u)− 1 4 [I ′(u), u] = 1 4 ‖u‖2 + ( 1 4 − 1 p ) ∫ R3 a(x)|u|pdx + ( 1 4 − 1 q )µ ∫ R3 b(x)|u|qdx+ ( 1 4 − 1 2∗s ) ∫ R3 |u|2 ∗ sdx ≥ 0, 8 L. LIU, K. TENG EJDE-2021/07 which leads to a contradiction with c < m∞. Thus ‖vn‖Lr → 0 for any r ∈ (2, 2∗s). By (2.18) and (2.22), we have c− I(u) = 1 2 ‖vn‖2 − 1 2∗s ‖vn‖ 2∗s 2∗s + o(1), ‖vn‖2 − ‖vn‖ 2∗s 2∗s = o(1). Up to a subsequence, we may assume that ‖vn‖2 → l. Thus ‖vn‖ 2∗s 2∗s → l. If l > 0, by the definition of Ss, we obtain l ≥ (Ss) 3 2s . Hence, c = I(u) + 1 2 ‖vn‖2 − 1 2∗s ‖vn‖ 2∗s 2∗s = I(u) + ( 1 2 − 1 2∗s )l ≥ s 3 S 3 2s s , which contradicts with c < m∞ < s 3S 3 2s s . Thus l = 0 and we complete the proof. � Lemma 2.4. Suppose that α, β > n, f, g ∈ L∞(Rn) and f(x) ≤ C1 (1 + |x|)α , g(x) ≤ C2 (1 + |x|)β . Then there exits C > 0 such that |f ∗ g(x)| ≤ C (1 + |x|)γ , where γ = min{α, β}. Proof. By direct computations, |f ∗ g(x)| = ∣∣∣ ∫ Rn C1 (1 + |x− y|)α C2 (1 + |y|)β dy ∣∣∣ = ∫ |x−y|≥ |x|2 C1 (1 + |x− y|)α C2 (1 + |y|)β dy + ∫ |x−y|< |x|2 C1 (1 + |x− y|)α C2 (1 + |y|)β dy ≤ C1 (1 + |x| 2 )α ∫ |x−y|≥ |x|2 C2 (1 + |y|)β dy + ∫ |x| 2 <|y|< 3 2 |x| C1 (1 + |x− y|)α C2 (1 + |y|)β dy ≤ C1 (1 + |x| 2 )α ∫ Rn C2 (1 + |y|)β dy + C2 (1 + |x| 2 )β ∫ |x| 2 <|y|< 3 2 |x| C1 (1 + |x− y|)α dy ≤ C1C22α (2 + |x|)α ∫ Rn 1 (1 + |y|)β dy + C1C22β (2 + |x|)β ∫ |x−y|< 5 2 |x| 1 (1 + |x− y|)α dy ≤ C1C22α (2 + |x|)α ∫ Rn 1 (1 + |y|)β dy + C1C22β (2 + |x|)β ∫ Rn 1 (1 + |x− y|)α dy ≤ C( 1 (2 + |x|)α + 1 (2 + |x|)β ) ≤ C (1 + |x|)γ , where γ = min{α, β}. � Now we recall the definition of Lusternik-Schnirelman category. EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 9 Definition 2.5. (i) For a topological space X, we say a non-empty, closed subset A ⊂ X is contractible to a point in X if and only if there exist a continuous mapping η : [0, 1]×A→ X such that for some x0 ∈ X, (a) η(0, x) = x for all x ∈ A, (b) η(1, x) = x0 for all x ∈ A. (ii) We define cat(X) = min { k ∈ N : there exist closed subsets A1, . . . , Ak ⊂ X such that Ai is contractible to a point in X for all i and ∪ki=1 Ai = X } . We say cat(X) = ∞ if do not exist finitely many closed subsets A1, . . . , Ak ⊂ X such that Ai is contractible to a point in X for all i and ∪ki=1Ai = X. We need the following two important lemmas. See [1, Proposition 2.4 and Lemma 2.5]. Lemma 2.6. Suppose that M is a Hilbert manifold and Ψ ∈ C1(M,R). Assume that there exist c0 ∈ R and k ∈ N such that Ψ(u) satisfies the Palais-Smale condition for c ≤ c0 and cat({u ∈ M : Ψ(u) ≤ c0}) ≥ k. Then Ψ(u) has at least k critical points in {u ∈M : Ψ(u) ≤ c0}. Lemma 2.7. Let X be a topological space. Suppose that there exist two continuous mappings F : S2 = {y ∈ R3 : |y| = 1} → X and G : X → S2, such that G ◦ F is homotopic to identity id : S2 → S2, that is, there is a continuous mapping ζ : [0, 1] × S2 → S2 such that ζ(0, x) = (G ◦ F )(x) for all x ∈ S2 and ζ(1, x) = x for all x ∈ S2. Then cat(X) ≥ 2. 3. Proof of main results Proof of Theorem 1.1. Let {un} ⊂ M be a minimizing sequence for functional I, that is, {un} ⊂M and I(un)→ m, where M = {u ∈ H\{0} : G(u) = [I ′(u), u] = 0}. We claim I ′(un) → 0. By the Lagrange multiplier Theorem, there exists λn ∈ R such that I ′(un)− λnG′(un)→ 0. Since un ⊂M , we have m+ o(1) = I(un)− 1 4 (I ′(un), un) ≥ 1 4 ‖un‖2, which implies that {un} is bounded in H. Hence λn[G′(un), un]→ 0. (3.1) By (A4) and (A6), for any ε > 0, there exists Cε > 0 such that a(x)|u|p + b(x)|u|q + |u|2 ∗ s ≤ ε|u|2 + Cε|u|2 ∗ s . Taking ε = 1/2 and recalling the definition of Ss, we have ‖un‖2 ≤ ∫ R3 a(x)|un|pdx+ ∫ R3 µb(x)|un|qdx+ ∫ R3 |un|2 ∗ sdx 10 L. LIU, K. TENG EJDE-2021/07 ≤ 1 2 ∫ R3 |un|2dx+ C1/2 ∫ R3 |un|2 ∗ sdx ≤ 1 2 ∫ R3 |un|2dx+ C1/2 ‖un‖2 ∗ s S 3 3−2s s , which implies that ‖un‖2 ≥ S 3 2s s (2C1/2) 3−2s 2s . (3.2) By (3.2), we obtain [G′(un), un] = [G′(un), un]− 4[I ′(un), un] = 2‖un‖2 + 4 ∫ R3 K(x)φtunu 2 ndx− p ∫ R3 a(x)|un|pdx− q ∫ R3 µb(x)|un|qdx − 2∗s ∫ R3 |un|2 ∗ sdx− 4[‖un‖2 + ∫ R3 K(x)φtunu 2 ndx− ∫ R3 a(x)|un|pdx − ∫ R3 µb(x)|un|qdx− ∫ R3 |un|2 ∗ sdx] = −2‖un‖2 + (4− p) ∫ R3 a(x)|un|pdx + (4− q) ∫ R3 µb(x)|un|qdx+ (4− 2∗s) ∫ R3 |un|2 ∗ sdx ≤ −2‖un‖2 ≤ −2 S 3 2s s (2C1/2) 3−2s 2s . From (3.1), we have λn → 0. Thus I ′(un)→ 0. This means that {un} is a (PS)m sequence for I, that is, I(un)→ m and I ′(un)→ 0. By Lemma 2.2, if m ∈ (0,m∞), then un → u in H and thus I(u) = m and I ′(u) = 0. Hence, m is attained by u ∈ H\{0}. For this purpose, it is sufficient to prove m < m∞. Similar argument as (2.3), we can obtain the equivalent characterization of the least energy m: m = inf u∈H\{0} max t≥0 I(tu). (3.3) Let R > 0 and γ ∈ R3 with |γ| = 1. By (3.3), clearly, we have m ≤ sup t≥0 I(tu∞(x−Rγ)), where u∞ is a positive ground state solution for limit problem (2.2). Since I(tu∞(x−Rγ)) ≤ t2 2 ‖u∞(x−Rγ)‖2 + Ct4‖u∞(x−Rγ)‖4 − t2 ∗ s 2∗s ‖u∞(x−Rγ)‖2 ∗ s 2∗s = t2 2 ‖u∞‖2 + Ct4‖u∞‖4 − t2 ∗ s 2∗s ‖u∞‖ 2∗s 2∗s , there exist a small t′ > 0 and a large t′′ > 0 independent of R and γ such that sup t∈[0,t′]∪[t′′,+∞) I(tu∞(x−Rγ)) < m∞. (3.4) EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 11 On the other hand, by (A6), for any u ∈ H, we have I(tu) ≤ I∞(tu) + t4 4 ∫ R3 (K(x)− 1)φtuu 2dx− 1 p tp ∫ R3 (a(x)− 1)|u|pdx + t4 4 ∫ R3 (φtu − φ̂tu)u2 dx = I∞(tu) + t4 4 ∫ R3 (K(x)− 1)φtuu 2dx+ t4 4 ∫ R3 (K(x)− 1)φ̂tuu 2 dx − 1 p tp ∫ R3 (a(x)− 1)|u|pdx. Thus, choosing u = u∞(x − Rγ) in the inequality above and using (K2), (a1), we obtain I(tu∞(x−Rγ)) ≤ I∞(tu∞)− t4 4 C0 ∫ R3 1 (1 + |x+Rγ|)k ∫ R3 K(y +Rγ)u2 ∞(y) |x− y|3−2t dy|u∞(x)|2 dx dy − t4 4 C0 ∫ R3 1 (1 + |x+Rγ|)k φ̂tu∞(x)|u∞(x)|2dx + 1 p tpC2 ∫ R3 1 (1 + |x+Rγ|)a |u∞(x)|pdx ≤ I∞(tu∞)− t4 4 C0 ∫ R3 1 (1 + |x+Rγ|)k ∫ R3 K(y +Rγ)u2 ∞(y) |x− y|3−2t |u∞(x)|2 dx dy − t4 4 C0 ∫ R3 1 (1 + |x+Rγ|)k φ̂tu∞(x)|u∞(x)|2dx + 1 p tpC2C p σ ∫ R3 1 (1 + |x+Rγ|)a 1 (1 + |x|)(3+2s)p dx ≤ I∞(tu∞)− t4 4 C0 ∫ R3 1 (1 + |x+Rγ|)k φ̂tu∞(x)|u∞(x)|2dx + 1 p tpC2C p σ ∫ R3 1 (1 + |x+Rγ|)a 1 (1 + |x|)(3+2s)p dx. Set l(t) = I∞(tu∞), t ∈ (0,∞). It is easy to verify that supt≥0 l(t) = I∞(u∞) = m∞. Moreover, we have∫ R3 1 (1 + |x+Rγ|)k φ̂tu∞(x)|u∞(x)|2dx ≥ ∫ |x|≤1 1 (1 + |x+Rγ|)k φ̂tu∞(x)|u∞(x)|2dx ≥ C ∫ |x|≤1 1 (2 +R)k φ̂tu∞(x)|u∞(x)|2dx ≥ C̃ (2 +R)k . By Lemma 2.4, we have∫ R3 1 (1 + |x+Rγ|)a 1 (1 + |x|)(3+2s)p dx ≤ C (1 +R)α , 12 L. LIU, K. TENG EJDE-2021/07 where α = min{a, (3 + 2s)p}. Thus sup t′≤t≤t′′ I(tu∞(x−Rγ)) ≤ m∞ − (t′)4 4 C0C̃ 1 (2 +R)k + 1 p (t′′)pC2C p σC 1 (1 +R)α . By 0 < k < α, there exists R̂ > 0 large such that for R > R̂, sup t′≤t≤t′′ I(tu∞(x−Rγ)) < m∞, ∀ |γ| = 1. Thus, combing with (3.4), for R > R̂, we have sup t≥0 I(tu∞(x−Rγ)) < m∞, ∀ |γ| = 1, which yields m < m∞. The remaining of the proof of Theorem 1.1 is to show that the solution u ∈ H is positive. � Proof of Theorem 1.2. The argument is similar to the on in Theorem 1.1, we only need to prove for R > 0 large, supt≥0 I(tu∞(x−Rγ)) < m∞ uniformly in γ. Clearly, there exist 0 < t′ < t′′ independent of R and γ such that sup t∈[0,t′]∪[t′′,+∞) I(tu∞(x−Rγ)) < m∞. On the other hand, by (A3), (A5), (A7), and we have for any σ > 0, there exist Cσ > 0 such that sup t∈[t′,t′′] I(tu∞(x−Rγ)) ≤ sup t≥0 I∞(tu∞) + C1(t′′)4 4 ∫ R3 1 (1 + |x+Rγ|)d ∫ R3 K(y +Rγ)u2 ∞(y) |x− y|3−2t dy|u∞(x)|2dx + C1(t′′)4 4 ∫ R3 1 (1 + |x+Rγ|)d φ̂tu∞(x)|u∞(x)|2dx + 1 p (t′′)pC2C P σ ∫ R3 1 (1 + |x+Rγ|)a 1 (1 + |x|)(3+2s)p dx − µC3(t′)q q ∫ R3 1 (1 + |x+Rγ|)b |u∞(x)|qdx. By calculations, we have∫ R3 1 (1 + |x+Rγ|)b |u∞(x)|qdx ≥ ∫ |x|≤1 1 (1 + |x+Rγ|)b |u∞(x)|qdx ≥ ∫ |x|≤1 1 |2 +R|b |u∞(x)|qdx ≥ C 1 (2 +R)b . (3.5) From Lemma 2.4, we obtain∫ R3 1 (1 + |x+Rγ|)a 1 (1 + |x|)(3+2s)p dx ≤ α (1 +R)α , (3.6) where α = min{a, (3 + 2s)p}. EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 13 By Hölder’s inequality, (A3) and (3.6), we have∫ R3 1 (1 + |x+Rγ|)d ∫ R3 K(y +Rγ)u2 ∞(y) |x− y|3−2t dy|u∞(x)|2dx ≤ (1 + C1) ∫ R3 1 (1 + |x+Rγ|)d ∫ R3 u2 ∞(y) |x− y|3−2t dy|u∞(x)|2dx = (1 + C1) ∫ R3 1 (1 + |x+Rγ|)d φ̂tu∞(x)|u∞(x)|2dx ≤ C‖φ̂tu∞(x)‖2∗s [ ∫ R3 ( 1 (1 + |x+Rγ|)d 1 (1 + |x|)6+4s ) 6 3+2s dx ] 3+2s 6 ≤ C [ 1 (1 +R)m ] 3+2s 6 ≤ C 1 (1 +R)β , (3.7) where m = min{ 6d 3+2s , 12}. Similar as the above argument, we obtain∫ R3 1 (1 + |x+Rγ|)d φ̂tu∞(x)|u∞(x)|2dx ≤ C 1 (1 +R)β , (3.8) where β = min{d, 6 + 4s}. By (3.5)-(3.8), we have sup t∈[t′,t′′] I(tu∞(x−Rγ)) ≤ m∞ − C1 1 (2 +R)b + C2 1 (1 +R)α + C3 1 (1 +R)β , where C1, C2, C3 are positive constants. Since b < min{α, β}, we obtain that there exists R0 > 0 such that for R > R0, supt≥0 I(tu∞(x−Rγ)) < m∞ uniformly in γ. The proof is complete. � 4. Proof of Theorem 1.3 Let h(t) = I(tu∞(x − Rγ)), t ∈ (0,∞), γ ∈ R3 with |γ| = 1. Form the proof of Theorem 1.2, we know there exists R0 > 0 such that for R > R0, there exists ε(R) > 0 satisfying sup t≥0 h(t) ≤ m∞ − ε(R) < m∞ uniformly in γ. For any fixing R and γ, it is easy to check that h(t) attains its maximum at a unique point t = t∞. Hence, we define a mapping FR : S2 = {γ ∈ R3 : |γ| = 1} →M by FR(γ) = t∞u∞(x−Rγ). Immediately we have the following Lemma. Lemma 4.1. There exists R0 > 0 such that for R > R0, there exists ε(R) > 0 satisfying FR(S2) ⊂ {u ∈M : I(u) ≤ m∞ − ε(R)} uniformly in γ ∈ S2. For u ∈ H, we define a map Φ : H → H by Φ(u)(x) := 1 |B1(x)| ∫ B1(x) |u(y)|dy, ∀x ∈ R3, where |B1(x)| is the Lebesgue measure of B1(x). Let û(x) = [Φ(u)(x)− 1 2 max x∈R3 Φ(u)(x)]+, 14 L. LIU, K. TENG EJDE-2021/07 and β : H\{0} → R3 given by β(u) = 1 ‖û‖1 ∫ R3 xû(x)dx. Obviously, β(u) is well defined for all u ∈ H\{0} and β(u) has a compact support in R3. Moreover, β(u) is continuous in H\{0} and satisfies the following properties. Lemma 4.2. (i) For any t 6= 0 and u ∈ H\{0}, β(tu) = β(u). (ii) For any z ∈ R3 and u ∈ H\{0}, β(u(x− z)) = β(u) + z. Define a functional J : H → R given as follows J(u) = 1 2 ‖u‖2 + 1 4 ∫ R3 K(x)φtuu 2dx− 1 p ∫ R3 a(x)|u|pdx− 1 2∗s ∫ R3 |u|2 ∗ sdx, u ∈ H. Lemma 4.3. m0 := infM0 J(u) = m∞ is not attained, where M0 = {u ∈ H\{0} : [J ′(u), u] = 0}. Proof. First, we show that for any u ∈ M0, there exists a unique 0 < τ ≤ 1 such that τu ∈M∞. Indeed, by u ∈M0 and τu ∈M∞, we have ‖u‖2 + ∫ R3 K(x)φtuu 2dx = ∫ R3 a(x)|u|pdx+ ∫ R3 |u|2 ∗ sdx, (4.1) and then τp ∫ R3 a(x)|u|pdx+ τ2∗s ∫ R3 |u|2 ∗ sdx ≤ τp ∫ R3 |u|pdx+ τ2∗s ∫ R3 |u|2 ∗ sdx = τ2‖u‖2 + τ4 ∫ R3 φ̂tuu 2dx. (4.2) From (A3) and K(x) ≥ 1 for any x ∈ R3, it follows that∫ R3 φ̂tuu 2dx ≤ ∫ R3 K(x)φ̂tuu 2dx ≤ ∫ R3 K(x)φtuu 2dx. (4.3) If τ > 1, by (4.1), (4.2) and (4.3), we deduce that τ4(‖u‖2 + ∫ R3 K(x)φtuu 2dx) ≥ τ4(‖u‖2 + ∫ R3 φ̂tuu 2dx) ≥ τp( ∫ R3 a(x)|u|pdx+ ∫ R3 |u|2 ∗ sdx) = τp(‖u‖2 + ∫ R3 K(x)φtuu 2dx), which yields τ ≤ 1, this achieves a contradiction. Hence τ ≤ 1 and the claim is true. For u ∈M0, using (4.3), we have J(u) = J(u)− 1 p [J ′(u), u] = (1 2 − 1 p ) ‖u‖2 + ( 1 4 − 1 p ) ∫ R3 K(x)φtuu 2dx+ ( 1 p − 1 2∗s ) ∫ R3 |u|2 ∗ sdx ≥ (1 2 − 1 p ) ‖τu‖2 + ( 1 4 − 1 p )τ4 ∫ R3 φ̂tuu 2dx+ ( 1 p − 1 2∗s ) ∫ R3 |τu|2 ∗ sdx = I∞(τu)− 1 p [I ′∞(τu), τu] EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 15 = I∞(τu) ≥ m∞, which implies that m0 ≥ m∞. Next we prove m0 ≤ m∞. Let wn = u∞(.− zn), where zn ∈ R3 with |zn| → ∞. We claim that for wn ∈ M∞, there exists τn ≥ 1 such that τnwn ∈ M0. In fact, from wn ∈M∞ and τnwn ∈M0, it holds ‖wn‖2 + ∫ R3 φ̂twnw 2 ndx = ∫ R3 |wn|pdx+ ∫ R3 |wn|2 ∗ sdx, and then τpn ∫ R3 |wn|pdx+ τ 2∗s n ∫ R3 |wn|2 ∗ sdx ≥ τpn ∫ R3 a(x)|wn|pdx+ τ 2∗s n ∫ R3 |wn|2 ∗ sdx = τ2 n‖wn‖2 + τ4 n ∫ R3 K(x)φtwnw 2 ndx. If τn < 1, then τpn( ∫ R3 |wn|pdx+ ∫ R3 |wn|2 ∗ sdx) ≥ τ4 n(‖wn‖2 + ∫ R3 K(x)φtwnw 2 ndx) ≥ τ4 n(‖wn‖2 + ∫ R3 φ̂twnw 2 ndx) = τ4 n( ∫ R3 |wn|pdx+ ∫ R3 |wn|2 ∗ sdx), which leads to a contradiction with τn < 1. Hence τn ≥ 1 and the claim holds. By the definition of m0 and τnun ∈M0, we have m0 ≤ J(τnwn) = 1 2 τ2 n‖u∞‖2 + 1 4 τ4 n ∫ R3 K(x)φtwnw 2 ndx − 1 p τpn ∫ R3 a(x)|wn|pdx− 1 2∗s τ 2∗s n ∫ R3 |u∞(x)|2 ∗ sdx. By Lebesgue dominated convergence Theorem, we deduce that lim n→∞ ∫ R3 K(x)φtwnw 2 ndx = lim n→∞ ∫ R3 K(x+ zn) ∫ R3 K(y + zn)u2 ∞(y) |x− y|3−2t dy|u∞(x)|2 dx dy = ∫ R3 φ̂tu∞(x)|u∞(x)|2dx, lim n→∞ ∫ R3 a(x)|wn|pdx = lim n→∞ ∫ R3 a(x+ zn)|u∞(x)|pdx = ∫ R3 |u∞(x)|pdx. (4.4) If τn → 1, we obtain m0 ≤ limn→∞ J(tnwn) = I∞(u∞) = m∞, form which we see that m0 = m∞. Thus we only need to prove τn → 1. By τnwn ∈M0, with τn ≥ 1, we have τ4 n(‖wn‖2 + ∫ R3 K(x)φtwnw 2 ndx) 16 L. LIU, K. TENG EJDE-2021/07 ≥ τ2 n‖wn‖2 + τ4 n ∫ R3 K(x)φtwnw 2 ndx = τpn ∫ R3 a(x)|wn|pdx+ τ 2∗s n ∫ R3 |wn|2 ∗ sdx ≥ τpn( ∫ R3 a(x)|wn|pdx+ ∫ R3 |wn|2 ∗ sdx). Thus, by (4.4), we deduce that 1 ≤ τp−4 n ≤ ‖wn‖2 + ∫ R3 K(x)φtwnw 2 ndx∫ R3 a(x)|wn|pdx+ ∫ R3 |wn|2∗sdx = ‖u∞‖2 + ∫ R3 φ̂ t u∞(x)|u∞(x)|2dx+ o(1)∫ R3 |u∞(x)|pdx+ o(1) + ∫ R3 |u∞(x)|2∗sdx , which yields τn → 1 by using u∞ ∈M∞. Next we prove m0 is not attained. Assume by contradiction that there exists u0 ∈M0 such that m0 = J(u0). We claim J ′(u0) = 0. Set G̃(u) = [J ′(u), u], By the Lagrange multipliers Theorem, we obtain λ ∈ R such that J ′(u0) − λG̃′(u0) → 0, similar to the of proof of Theorem 1.1, we have J ′(u0) = 0. Note that if u0 is sing- changing, by Remark 5.6 in [34], we see that J(u0) ≥ 2m0, a contradiction. Thus we may assume that u0 ≥ 0 in H and u0 6≡ 0, we claim u0 > 0, by the definition of φtu0 (x), there exists C > 0 such that φtu0 (x) = ∫ |x−y|≥1 K(y)u2 0(y) |x− y|3−2t dy + ∫ |x−y|<1 K(y)u2 0(y) |x− y|3−2t dy ≤ C‖u0‖22 + C ∫ |x−y|<1 1 |x− y|3−2t dy < +∞, and |g| ≤ C(|u0|+ |u0|q−1), where g(x) = a(x)|u0(x)|p−2u0(x) + |u0(x)|2∗s−2u0(x)− u0(x) − K(x)φtu0 (x)u0(x). Then it follows from [22, Proposition 3.4] that there exists σ ∈ (0, 1) such that u0 ∈ C0,σ. Let ω satisfy −∆ω = −u0 − K(x)φtu0 u + a(x)|u0|p−2u0 + |u0|2 ∗ s−2u0 ∈ C0,σ. By the Hölder regularity theory for the Lapla- cian, we have ω ∈ C2,σ. It follows from 2s + σ > 1 that (−∆)1−sω ∈ C1,2s+σ−1. Then, since (−∆)s(u0 − (−∆)1−sω) = 0, the function u − (−∆)1−sω is harmonic and we obtain u0 has the same regularity as (−∆)1−sω. That is, u0 ∈ C1,2s+σ−1. The regularity obtained above implies that (−∆)su0 = − ∫ R3 u0(x+ y) + u0(x− y)− 2u0(x) |y|3+2s dy. Assume that there exists x0 ∈ R3 such that u0(x0) = 0, then by u0 6≡ 0 and u0 ≥ 0, (−∆)su(x0) = − ∫ R3 u(x0 + y) + u(x0 − y) |y|3+2s dy < 0. However, noting that −∆u0 = −u0 −K(x)φtu0 u0 + a(x)|u0|p−2u0 + |u0|2 ∗ s−2u0 we obtain −∆u0(x0) = 0, which is a contradiction. Therefore, u0 > 0. From the above proof, we see that for u0 ∈M0, there exists a unique τ0 ≤ 1 such that τ0u0 ∈M∞. Thus, m∞ ≤ I∞(τ0u0) = I∞(τ0u0)− 1 p [I ′∞(τ0u0), τ0u0] EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 17 = (1 2 − 1 p ) ‖τ0u0‖2 + ( 1 4 − 1 p )τ4 0 ∫ R3 φ̂tu0 u2 0dx+ ( 1 p − 1 2∗s ) ∫ R3 |τ0u0|2 ∗ sdx ≤ (1 2 − 1 p ) ‖u0‖2 + ( 1 4 − 1 p ) ∫ R3 K(x)φtu0 u2 0dx+ ( 1 p − 1 2∗s ) ∫ R3 |u0|2 ∗ sdx = J(u0)− 1 p [J ′(u0), u0] = J(u0) = m0. From m0 = m∞, it follows that(1 2 − 1 p ) ‖τ0u0‖2 + ( 1 4 − 1 p )τ4 0 ∫ R3 φ̂tu0 u2 0dx+ ( 1 p − 1 2∗s ) ∫ R3 |τ0u0|2 ∗ sdx = (1 2 − 1 p ) ‖u0‖2 + ( 1 4 − 1 p ) ∫ R3 K(x)φtu0 u2 0dx+ ( 1 p − 1 2∗s ) ∫ R3 |u0|2 ∗ sdx, that is τ2 0 ‖u0‖2 + τ4 0 ∫ R3 φ̂tu0 u2 0dx+ τ 2∗s 0 ∫ R3 |u0|2 ∗ sdx = ‖u0‖2 + ∫ R3 K(x)φtu0 u2 0dx+ ∫ R3 |u0|2 ∗ sdx. Thus (1− τ2 0 )‖u0‖2 + ∫ R3 (K(x)− 1)φtu0 u2 0dx+ ∫ R3 (φtu0 − φ̂tu0 )u2 0dx + (1− τ4 0 ) ∫ R3 φ̂tu0 u2 0dx+ (1− τ2∗s 0 ) ∫ R3 |u0|2 ∗ sdx = (1− τ2 0 )‖u0‖2 + ∫ R3 (K(x)− 1)(φtu0 + φ̂tu0 )u2 0dx + (1− τ4 0 ) ∫ R3 φ̂tu0 u2 0dx+ (1− τ2∗s 0 ) ∫ R3 |u0|2 ∗ sdx = 0, by τ0 ≤ 1, so ∫ R3 (K(x)− 1)(φtu0 + φ̂tu0 )u2 0dx = 0, this contradicts u0 being positive, K(x) ≥ 1 and meas{x ∈ R3 : K(x) > 1} > 0. � Lemma 4.4. There exists ρ0 > 0 such that for u ∈M0 satisfying J(u) ≤ m∞+ρ0, it holds |β(u)| > 0. Proof. Assume by the contrary that there exists {un} ⊂ M0 such that J(un) → m∞ = m0 and |β(u)| = 0. Similar to the proof Theorem 1.1, we can derive by the Lagrange multipliers Theorem that J ′(un) → 0. We omit the proof here. Similar to the proof Lemma 2.3, we obtain un ⇀ u weakly in H, J ′(u) = 0, and m∞ − J(u) = I∞(vn) + o(1) and I ′∞(vn) = o(1), (4.5) where vn = un − u. For the sequence {vn}, two cases may occur: ‖vn‖r → 0 for any r ∈ (2, 2∗s), or there yn ∈ R3 with |yn| → ∞ such that vn(. + yn) ⇀ v 6= 0 weakly in H. By virtue of J ′(u) = 0, we can deduce that J(u) ≥ 0. From Lemma 2.3, we see that m∞ < s 3S 3 2s s . Thus m∞ − J(u) < s 3S 3 2s s . If ‖vn‖r → 0 for any r ∈ (2, 2∗s), by (4.6), we have m∞ − J(u) = 1 2‖vn‖ 2 − 1 2∗s ‖vn‖ 2∗s 2∗s + o(1) and ‖vn‖2 − ‖vn‖ 2∗s 2∗s = o(1). Up to a subsequence, we may assume 18 L. LIU, K. TENG EJDE-2021/07 that ‖vn‖2 → l and then ‖vn‖ 2∗s 2∗s → l. If l > 0, by the definition of Ss, we obtain l ≥ S 3 2s s . So m∞ − J(u) = 1 2‖vn‖ 2 − 1 2∗s ‖vn‖ 2∗s 2∗s = s 3 l ≥ s 3S 3 2s s , a contradiction with m∞ − J(u) < s 3S 3 2s s . Thus, l = 0 and then un → u in H, we obtain m0 = J(u), a contradiction with m0 is not attained. Therefore, vn(.+ yn) ⇀ v 6= 0 weakly in H. Similar to the proof Lemma 2.3, we can deduce that m∞ − J(u) = I∞(vn(.+ yn)) + o(1), I ′∞(vn(.+ yn)) = o(1). Hence, I ′∞(v) = 0 and by using Fatou’s Lemma, we have m∞ − J(u) = I∞(vn(.+ yn))− 1 4 [I ′∞(vn(.+ yn)), vn(.+ yn)] + o(1) = 1 4 ‖(vn(.+ yn)‖2 + ( 1 4 − 1 p ) ∫ R3 |(vn(.+ yn)|pdx + ( 1 4 − 1 2∗s ) ∫ R3 |(vn(.+ yn)|2 ∗ sdx+ o(1) ≥ 1 4 ‖v‖2 + ( 1 4 − 1 p ) ∫ R3 |v|pdx+ ( 1 4 − 1 2∗s ) ∫ R3 |v|2 ∗ sdx = I∞(v)− 1 4 (I ′∞(v), v)) = I∞(v) ≥ m∞. Combining with J(u) ≥ 0, we obtain J(u) = 0 and then vn(.+yn) = un(.+yn)→ v in H. By Lemma 4.2, we have β(v(x)) + o(1) = β(un(x+ yn)) = β(un)− yn = −yn. Which yields |β(v(x))| =∞, this leads to a contradiction. � Lemma 4.5. There exists µ0 > 0 small such that for µ ∈ (0, µ0), we have |β(u)| > 0 for u ∈ {u ∈M : I(u) < m∞}. Proof. Let u ∈M be such that I(u) < m∞, then we have m∞ > I(u) = I(u)− 1 4 [I ′(u), u] ≥ 1 4 ‖u‖2. (4.6) Using the conditions (A4), (A6) and u ∈ M , we obtain for any ε > 0, there exists Cε > 0 such that ‖u‖2 ≤ ∫ R3 a(x)|u|pdx+ µ ∫ R3 b(x)|u|qdx+ ∫ R3 |u|2 ∗ sdx ≤ (1 + µ) [ ε ∫ R3 |u|2dx+ Cε ∫ R3 |u|2 ∗ sdx ] . (4.7) Choose ε ∈ (0, 1/4), we have 1 2‖u‖ 2 ≤ (1+µ)Cε ∫ R3 |u|2 ∗ sdx for µ ∈ (0, 1). In fact, if ε ∈ (0, 1 4 ) and µ ∈ (0, 1), we obtain 0 < (1 + µ)ε < 1/2, and then 1 2 − (1 + µ)ε > 0. Thus, it holds 1 2 ∫ R3 |∇u|2dx+ ( 1 2 − (1 + µ)ε) ∫ R3 |u|2dx ≥ 0, that is 1 2 ‖u‖2 ≤ ‖u‖2 − (1 + µ)ε ∫ R3 |u|2dx, EJDE-2021/07 GROUND STATE AND MULTIPLE SOLUTIONS 19 by (4.7), we have ‖u‖2 − (1 + µ)ε ∫ R3 |u|2dx ≤ (1 + µ)Cε ∫ R3 |u|2 ∗ sdx, so 1 2 ‖u‖2 ≤ (1 + µ)Cε ∫ R3 |u|2 ∗ sdx. Thus, by the definition of Ss, there exists L0 > 0 independent of µ ∈ (0, 1) such that ∫ R3 |u|2 ∗ sdx ≥ L0 (1 + µ) 3 2s . (4.8) Similar to the argument of Lemma 4.3, we can deduce that for any u ∈ M , there exists a unique τ(u) ≥ 1 such that τ(u)u ∈M0. Then τ4(u)(‖u‖2 + ∫ R3 K(x)φtuu 2dx) ≥ τ2(u)‖u‖2 + τ4(u) ∫ R3 K(x)φtuu 2dx = τp(u) ∫ R3 a(x)|u|pdx+ τ2∗s (u) ∫ R3 |u|2 ∗ sdx ≥ τ2∗s (u) ∫ R3 |u|2 ∗ sdx, which implies that τ2∗s−4(u) ≤ ‖u‖2 + ∫ R3 K(x)φtuu 2dx∫ R3 |u|2∗sdx . Together with (4.6) and (4.8), we derive there exists C > 0 independent of µ ∈ (0, 1) such that 1 ≤ τ2∗s−4(u) ≤ C(1 + µ) 3 2s . (4.9) Note that for u ∈M with I(u) < m∞, thus m∞ > I(u) = sup t≥0 I(tu) ≥ I(t(u)u) = J(t(u)u)− µt q(u) q ∫ R3 b(x)|u|qdx. By (4.6) and (4.9), there exists a small µ0 ∈ (0, 1) such that µ ∈ (0, µ0), J(t(u)u) < m∞ + µ tq(u) q ∫ R3 b(x)|u|qdx ≤ m∞ + ρ0. Form Lemma 4.4, we have |β(t(u)u)| > 0. Hence, Lemma 4.2 implies |β(u)| > 0. � Lemma 4.6. For µ ∈ (0, µ0), define G : {u ∈ M : I(u) < m∞} → S2 by G(u) = β(u) |β(u)| . Then for R > R0 and µ ∈ (0, µ0), the map G ◦ FR : S2 → S2; y → G ◦ (FR(y)) is homotopic to the identity. Proof. Similar to the argument of [1, 37, Proposition 2.9], we define the map ζ(θ, y) : [0, 1]× S2 → S2 by ζ(θ, y) =  G((1− 2θ)FR(y) + 2θu∞(x−Ry)), θ ∈ [0, 1/2), G(u∞(x− R 2(1−θ)y), θ ∈ [1/2, 1), y, θ = 1. 20 L. LIU, K. TENG EJDE-2021/07 By the definition of G and Lemma 2.7, tt is not difficult to check that ζ(θ, y) ∈ C([0, 1]× S2, S2), ζ(0, y) = G ◦ (FR(y)) for y ∈ S2 and ζ(1, y) = y for y ∈ S2. The proof is complete. � Proof of Theorem 1.3. Form Lemma 2.7, Lemma 4.1 and Lemma 4.6, we have that for R > R0 and µ ∈ (0, µ0), it holds cat({u ∈M : I(u) ≤ m∞ − ε(R)}) ≥ 2. Then by Lemma 2.3 and Lemma 2.6, we see that I admits at least two nontrivial critical point in {u ∈M : I(u) < m∞}. � Acknowledgments. 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Zhang, Ground state and multiple solutions for SchrödingeršCPoisson equations with crit- ical nonlinearity, J. Math. Anal. Appl., 440 (2016), 466-482. [38] J. J. Zhang, J. M. DO Ó, M. Squassina, Fractional Schrödinger-Poisson systems with a general subcritical or critical nonlinearity, Adv. Nonlinear Stud., 16 (2016), 15-30. [39] L. Zhao, H. Liu, F. Zhao, Existence and concentration of solutions for the Schrödinger-Poisson equations with steep well potential, J. Differential Equations., 255 (2013), 1-23. [40] L. Zhao, F. Zhao, On the existence of solutions for the Schrödinger-Poisson equations, J. Math. Anal. Appl., 346 (2008), 155-169. Lintao Liu Department of Mathematics, Taiyuan University of Technology, Taiyuan, Shanxi 030024, China Email address: 956484600@qq.com Kaimin Teng Department of Mathematics, Taiyuan University of Technology, Taiyuan, Shanxi 030024, China Email address: tengkaimin2013@163.com 1. Introduction 2. Preliminary lemmas 3. Proof of main results 4. Proof of Theorem 1.3 Acknowledgments References