Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 48, pp. 1–12. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu 3 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS FOR SEMILINEAR ELLIPTIC EQUATIONS DUONG TRONG LUYEN, NGUYEN MINH TRI Abstract. In this article, we study the multiplicity of weak solutions to the boundary value problem −∆u = f(x, u) + g(x, u) in Ω, u = 0 on ∂Ω, where Ω is a bounded domain with smooth boundary in RN (N > 2), f(x, ξ) is odd in ξ and g is a perturbation term. Under some growth conditions on f and g, we show that there are infinitely many solutions. Here we do not require that f be continuous or satisfy the Ambrosetti-Rabinowitz (AR) condition. The conditions assumed here are not implied by the ones in [3, 15]. We use the perturbation method by Rabinowitz combined with estimating the asymptotic behavior of eigenvalues for Schrödinger’s equations. 1. Introduction In the previous decades, the boundary value problem for semilinear elliptic equa- tion −∆u = f(x, u) + g(x, u), u ∈ H1 0 (Ω) (1.1) has been studied by many authors, see for example [2, 14, 3] and the references therein. Here Ω is a bounded smooth domain of RN (N ≥ 2), f(x, ξ) is odd in ξ and g(x, ξ) is a non-odd perturbation term. The following condition was introduced in [1, 10] (AR) For some µ > 2, and R > 0, we have 0 < µF (x, ξ) ≤ f(x, ξ)ξ, ∀x ∈ Ω, ∀|ξ| ≥ R, where F (x, ξ) = ∫ ξ 0 f(x, τ) dτ . This condition plays an important role in the study of elliptic equations. Let us sketch some the results from the past 40 years. Bahri and Berestycki [2] proved that if f(x, ξ) ≡ |ξ|p−2ξ, g(x, ξ) ≡ g(x) ∈ L2(Ω), p ∈ (1, PN ), where PN is the largest root of the equation (2N − 2)P 2 − (N + 2)P −N = 0, N ≥ 2, then problem (1.1) has infinitely many solutions in H1 0 (Ω). This case was first studied by Bahri and Berestycki [2], and independently by Struwe [14]. 2010 Mathematics Subject Classification. 35J60, 35B33, 35J25, 35J70. Key words and phrases. Semilinear elliptic equations; multiple solutions; critical points; perturbation methods; boundary value problem. c©2021 Texas State University. Submitted September 18, 2019. Published May 28, 2021. 1 2 D. T. LUYEN, N. M. TRI EJDE-2021/48 Rabinowitz [11, 12] studied problem (1.1), assuming that N ≥ 3 and f satisfies (AR), (R1), (R2), g(x, ξ) ≡ g(x) ∈ L2(Ω), and 2p N(p− 2) − 1 > µ µ− 1 , (1.2) where (R1) f(x, ξ) ∈ C(Ω× R,R), f(x,−ξ) = −f(x, ξ) for all (x, ξ) ∈ Ω× R. (R2) There exist 2 < p < 2∗ := 2N N−2 , C > 0 such that almost everywhere in Ω, |f(x, ξ)| ≤ C(1 + |ξ|p−1). He then proved that problem (1.1) has an unbounded sequence of solutions inH1 0 (Ω) (see [11, Theorem 1.5]). Assuming that f satisfies (AR), (R1), (R2), g(x, ξ) ∈ C(Ω× R,R) and |g(x, ξ)| ≤ C1 + C2|ξ|σ, 0 ≤ σ < µ− 1, 2p N(p− 2) − 1 > µ µ− σ − 1 , where C1, C2 are nonnegative real numbers, then he confirmed that the problem (1.1) has an unbounded sequence of solutions in H1 0 (Ω) (see [11, Remark 1.71]). Bahri and Lions [3] assumed that f(x, ξ) ≡ |ξ|p−2ξ, 2 < p < 2∗, (p < ∞, if N = 2) such that g : Ω× R→ R is a Carathéodory function satisfying |g(x, ξ)| ≤ g1(x) + C3|ξ| N+2 N−2 a.e. in Ω for some C3 ≥ 0, |G(x, ξ)| ≤ g2(x) + g3(x)|ξ|σ1 a.e. in Ω for some 0 ≤ σ1 < 2, where G(x, ξ) := ∫ ξ 0 g(x, τ) dτ , g1(x) ∈ L 2N/(N+2) + (Ω), g2(x) ∈ L1 +(Ω), N > 2, g3(x) ∈ Lβ+(Ω) ,with β > 1, β′ < 2N/(N − 2)(1/σ1), 1/β+ 1/β′ = 1, Lβ+(Ω) := {g : Ω→ R|g ∈ Lβ(Ω), g(x) ≥ 0 a.e. in Ω} and 2 < p < 2N − 2σ1 N − 2 . (1.3) Under the above assumptions, Bahri and Lions proved that problem (1.1) has in- finitely many solutions in H1 0 (Ω). Obviously, the assumption on p in (1.3) is weaker than the one in (1.2). Later Tanaka [15] obtained a similar existence result as in [3], assuming that f(x, ξ) ≡ f(ξ) satisfies (AR), (R1), (R2), g(x, ξ) ≡ g(x) ∈ L p p−1 (Ω), and 2p N(p− 2) > µ µ− 1 . (1.4) He then proved that problem (1.1) has an unbounded sequence of solutions inH1 0 (Ω) (see [15, Theorem 1]). The assumption on p in (1.4) is weaker than the one in (1.2). Tehrani [16] considered the case of a sign-changing potential. Bolle, Ghous- soub and Tehrani [4] also obtained some existence results on the perturbed elliptic equation −∆u = |u|p−2u+ g(x) in Ω, u = u0 on ∂Ω, where u0 ∈ C2(Ω,R) with ∆u0 = 0, 2 < p < 2∗. Long [8] considered a perturbed superquadratic second order Hamiltonian systems. Hirano and Zou [7] studied the elliptic boundary value problem −∆u = |u|p−2u+ βg(x, u), u ∈ H1 0 (Ω), (1.5) EJDE-2021/48 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS 3 where 2 < p < 2∗, (N ≥ 3) and g(x, ξ) ∈ C(Ω × R,R), g(x, ξ)ξ ≥ 0 for all x ∈ Ω, ξ ∈ R, limξ→0 g(x, ξ)/ξ = 0 uniformly in x ∈ Ω. Then they proved that for any m ∈ N, there is a βm > 0 such that for each β ∈ (0, βm), problem (1.5) has at least m distinct sign-changing solutions. Recently, Santos [13] using Leray-Schauder degree theory and the method of upper and lower solutions proved existence and multiplicity of solutions the problem (ϕ(u′))′ = f(t, u, u′) u(0) = u(T ) = u′(0), where ϕ is an increasing homeomorphism such that ϕ(0) = 0, and f is a continuous function. In this article, we study the multiplicity of solutions to problem (1.1), using the following assumptions: f : Ω× R→ R is a Carathéodory function satisfying (A1) f(x,−ξ) = −f(x, ξ) for all (x, ξ) ∈ Ω× R. (A2) There exist 2 < p < 2∗, C1 > 0 such that |f(x, ξ)| ≤ C1(1 + |ξ|p−1) a.e. in Ω× R. (A3) There exists a positive constant r0 such that F (x, ξ) ≥ 0, (x, ξ) ∈ Ω× R and |ξ| ≥ r0, lim |ξ|→∞ F (x, ξ) ξ2 =∞ a.e. in Ω. (A4) There exist constants C2 > 0 and κ > N/2 such that |F (x, ξ)|κ ≤ C2|ξ|2κF̂ (x, ξ), (x, ξ) ∈ Ω× R and |ξ| ≥ r0, where F̂ (x, ξ) = 2−1ξf(x, ξ)− F (x, ξ). (A5) There exist a positive constant C3 > 0 and ρ1 ∈ [2, 2∗) such that F̂ (x, ξ) ≥ C3(|ξ|ρ1 − 1), for all (x, ξ) ∈ Ω× R. (A6) g : Ω × R → R is a Carathéodory function satisfying: There exist g1(x) ∈ Lp1(Ω), g2(x) ∈ Lp2(Ω), p1/(p1− 1) ≤ ρ1, p2 > 1, (σ1 + 1)p2/(p2− 1) < ρ1, σ1 ∈ [0, ρ1 − 1), p1 > max{1, 2∗p2 p2σ1+2∗ }, such that |g(x, ξ)| ≤ g1(x) + g2(x)|ξ|σ1 a.e. in Ω× R. The main results of this paper are the following theorems. Theorem 1.1. Suppose that (A1)–(A6) are satisfied, and 2p N(p− 2) > ρ1 ρ1 − σ1 − 1 . (1.6) Then problem (1.1) has an unbounded sequence of solutions in H1 0 (Ω). Remark 1.2. The result in Theorem 1.1 is not covered by the ones in [15]. For example, when N = 3, f(x, ξ) = 2ξ [ ln(1 + |ξ|1/3) + |ξ|1/3 6(1 + |ξ|1/3) ] , 4 D. T. LUYEN, N. M. TRI EJDE-2021/48 and g : Ω×R→ R is a Carathéodory function such that there exist g1(x) ∈ Lp1(Ω), g2(x) ∈ Lp2(Ω), (σ1 + 1)p2/(p2 − 1) < 2, σ1 ∈ [0, 4 7 ), p1 ≥ 6p2 p2σ1+6 , p2 ≥ 6 5−σ1 , such that |g(x, ξ)| ≤ g1(x) + g2(x)|ξ|σ1 a.e. in Ω× R, then f, g satisfies the conditions in Theorem 1.1, but f does not satisfy the condi- tions in [15, Theorem 1]. Theorem 1.3. Suppose that (AR), (A1), (A2) are satisfied, and g satisfies (A6’) there exist g3(x) ∈ Lp3(Ω), g4(x) ∈ Lp4(Ω), p3/(p3 − 1) ≤ µ, p4 > 1, (σ2 + 1)p4/(p4 − 1) < µ, σ2 ∈ [0, µ− 1), p3 > max{1, 2∗p4 p4σ2+2∗ }, such that |g(x, ξ)| ≤ g3(x) + g4(x)|ξ|σ2 a.e. in Ω× R, 2p N(p− 2) > µ µ− σ2 − 1 . Then problem (1.1) has an unbounded sequence of solutions in H1 0 (Ω). The proofs of Theorems 1.1 and 1.3 are quite long, but they contain several arguments similar to those in [9, 11]. Therefore sometimes, we will omit detailed discussions by referring to these papers. Remark 1.4. Theorem 1.3 generalizes results in Rabinowitz [11, 12] and in Tanaka [15], and it is not covered by [3] and [9, 11, 12, 14]. For example, when N = 3, f(x, ξ) = ξ|ξ|1/4 − ξ|ξ|1/8, g(x, ξ) = |ξ|σ2 , 0 ≤ σ2 < 21 24 then on one hand f, g satisfy the conditions in Theorem 1.3, but f does not satisfy the conditions in [3, Theorem 1]. On the other hand, the function f satisfies the conditions in [9, Theorem 1.1], [11, 12, Theorem 1.5] and in [14, Theorem 3]. However, the function g may grow faster than the perturbation term in [9, 11, 12, 14]. 2. Proofs of the main results We define the Euler-Lagrange functional associated with problem (1.1) as follows Φ(u) = 1 2 ∫ Ω |∇u|2 dx− ∫ Ω F (x, u) dx− ∫ Ω G(x, u) dx. From [9, Proposition 2.2 ], (A2), and (A6), we have Φ is well defined on H1 0 (Ω) and Φ ∈ C1(H1 0 (Ω),R) with Φ′(u)(v) = ∫ Ω ∇u · ∇v dx− ∫ Ω f(x, u)v dx− ∫ Ω g(x, u)v dx for all v ∈ H1 0 (Ω). One can also check that the critical points of Φ are solutions of the problem (1.1). Lemma 2.1. Suppose that (A2), (A5), (A6) are satisfied, and u is a critical point of Φ. Then there is a constant C5 such that∫ Ω |u(x)|ρ1 dx ≤ C5(Φ2(u) + 1)1/2. (2.1) EJDE-2021/48 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS 5 Proof. Since u is a critical point of Φ, by (A2), (A5), and (A6), applying Hölder’s inequality, we obtain Φ(u) = Φ(u)− 1 2 Φ′(u)(u) ≥ ∫ Ω F̂ (x, u) dx− ∫ Ω (|2−1g(x, u)u|+ |G(x, u)|) dx ≥ C4 ∫ Ω |u|ρ dx− C6 (∫ Ω |u| (σ+1)p2 p2−1 dx ) p2−1 p2 − C7. (2.2) Then (2.1) follows from (2.2) and Young’s inequality. The proof is complete. � Next, we define a modified functional Φ(u). Let χ ∈ C∞(R,R) such that χ(t) = 1 for t ≤ 1, χ(t) = 0 for t > 2 and −2 < χ′ < 0 for t ∈ (1, 2). For u ∈ H1 0 (Ω), we put κ(u) = 2Θ ( (Φ(u))2 + 1 )1/2 , ψ(u) = χ ( κ(u)−1 ∫ Ω |u(x)|ρ1 dx ) , Φ(u) = ∫ Ω (1 2 |∇u|2 − F (x, u)− ψ(u)G(x, u) ) dx, (2.3) where Θ is a large enough positive constant, which will be chosen later in Lemma 2.3. Then, we obtain Φ ′ (u)(u) = (1 + T1(u)) (∫ Ω |∇u|2 dx− ∫ Ω f(x, u)udx ) − T2(u) ∫ Ω G(x, u) dx− (ψ(u) + T1(u)) ∫ Ω g(x, u)udx, (2.4) where T1(u) = χ′ ( κ(u)−1 ∫ Ω |u|ρ1 dx ) κ(u)−3(2Θ)2Φ(u) ∫ Ω |u|ρ1 dx ∫ Ω G(x, u) dx, T2(u) = ρ1χ ′ ( κ(u)−1 ∫ Ω |u|ρ1 dx ) κ(u)−1 ∫ Ω |u|ρ1 dx. Let supp(ψ) denote the support of ψ. Lemma 2.2. Suppose that (A1), (A2), (A5), (A6) are satisfied. (i) If u ∈ supp(ψ) then∣∣ ∫ Ω G(x, u) dx ∣∣ ≤ C8 ( |Φ(u)| σ1+1 ρ1 + |Φ(u)| 1 ρ1 + 1 ) . (ii) There is a constant C9, such that for any u ∈ H1 0 (Ω), |Φ(u)− Φ(−u)| ≤ C9 ( |Φ(u)| 1 ρ1 + |Φ(u)| σ1+1 ρ1 + 1 ) . (iii) There are constants M0, C10 > 0 such that whenever M ≥M0, Φ(u) ≥M , u ∈ supp(ψ) then Φ(u) ≥ C10M . (iv) For every δ > 0 small enough there exists M > 0 large enough such that for all u ∈ H1 0 (Ω),Φ(u) ≥M we have |T1(u)| ≤ δ, |T2(u)| ≤ 4ρ1. The proof of the above lemma is similar to the ones of [9, Lemmas 3.4, 3.5 , 3.6], so we omit it here. Now, we shall show that large critical values of Φ are critical values of Φ. 6 D. T. LUYEN, N. M. TRI EJDE-2021/48 Lemma 2.3. Suppose that (A1), (A2), (A5), (A6) are satisfied, and Θ is large enough. Then there exists M1 > 0 such that if u ∈ H1 0 (Ω) is a critical point of Φ and Φ(u) ≥M1, then u is a critical point of Φ and Φ(u) = Φ(u). Proof. Let u ∈ H1 0 (Ω) be such that Φ ′ (u) = 0. For M1 sufficiently large such that M1 > M0 then T1 is sufficiently small and T2 is bounded, with (2.4), we have Φ(u) = Φ(u)− Φ ′ (u)(u) 2(1 + T1(u)) ≥ C4 ∫ Ω |u|ρ dx− C11 (∫ Ω |u| (σ+1)p2 p2−1 dx ) p2−1 p2 − C11. Therefore, if we choose Θ large enough, κ(u)−1 ∫ Ω |u|ρ dx ≤ 1, it follows that ψ(u) = 1 and ψ′(u) = 0. � Definition 2.4. Let (V, ‖ · ‖V) be a real Banach space with its dual space V∗ and J ∈ C1(V,R). For c ∈ R we say that J satisfies condition (C)c if for each sequence {xm}∞m=1 ⊂ V with J(xm)→ c and (1 + ‖xm‖V)‖J ′(xm)‖V → 0 as m→∞, there exists a subsequence {xmk}∞k=1 that converges strongly in V. If J satisfies condition (Cc) for all c > 0, then we say that J satisfies the Cerami condition. Lemma 2.5. Suppose that (A1)–(A6) are satisfied. Then Φ ∈ C1(H1 0 (Ω),R) and there is a constant M2 > 0 such that Φ satisfies the (C)c condition for all c > M2. Proof. Since (A2), (A5), and (A6) are satisfied, and χ ∈ C∞(R,R), it follows that Φ ∈ C1(H1 0 (Ω),R). Let M0 be as in Lemma 2.2 and take M2 ≥ M0, c > M2. Let {um}∞m=1 ⊂ H1 0 (Ω) be a (C)c sequence, i.e., Φ(um)→ c as m→∞, lim m→∞ ( 1 + ‖um‖H1 0 (Ω) ) ‖Φ′(um)‖(H1 0 (Ω))∗ = 0. (2.5) Then Φ ′ (um)(um)→ 0, ‖um‖2H1 0 (Ω) − ∫ Ω 2F (x, um) dx− ∫ Ω 2ψ(um)G(x, um) dx→ 2c as m→∞. (2.6) We first show that {um}∞m=1 is bounded in H1 0 (Ω) by a contradiction argument. Indeed, we can (by passing to a subsequence if necessary) suppose that for any m, ‖um‖H1 0 (Ω) > 1 and ‖um‖H1 0 (Ω) →∞ as m→∞. (2.7) Setting wm = um ‖um‖H1 0 (Ω) , we have ‖wm‖H1 0 (Ω) = 1 and ‖wm‖Lν(Ω) ≤ τν‖wm‖H1 0 (Ω) = τν , 1 ≤ ν < 2∗. EJDE-2021/48 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS 7 Passing to a subsequence, assume that wm ⇀ w in H1 0 (Ω), then wm → w in Lν(Ω), 1 ≤ ν < 2∗. For 0 ≤ a < b, let Ωm(a, b) = {x ∈ Ω : a ≤ |um(x)| < b}. (2.8) In view of (A5) and (A6), for m large enough, we have c+1 ≥ Φ(um)− 1 2(1 + T1(um)) Φ ′ (um)(um) > 1 2 ∫ Ωm(r0,∞) F̂ (x, um) dx−C12, (2.9) where C12 is a positive constant independent of m. From (A6) and the definition of the functional ψ, for m large enough, we have∣∣ ∫ Ω ψ(um)G(x, um) dx ∣∣ ≤ C13. (2.10) By (A6), (2.6), (2.7) and (2.10), we obtain lim m→∞ ∫ Ω 2F (x, um) ‖um‖2H1 0 (Ω) dx = 1. (2.11) Now, we consider two possible cases: w = 0 and w 6= 0. Case 1: w = 0. Then wm → 0 in Lν(Ω), 1 ≤ ν < 2∗, and wm → 0 a.e. in Ω. From (A2) and Hölder’s inequality, we deduce that∫ Ωm(0,r0) F (x, um) ‖um‖2H1 0 (Ω) dx ≤ C14 ( 1 ‖um‖H1 0 (Ω) ‖wm‖L1(Ω) + ‖wm‖2L2(Ω) ) → 0 as m→∞, hence ∫ Ωm(0,r0) 2F (x, um) ‖um‖2H1 0 (Ω) dx→ 0 as m→∞. (2.12) Set q′ = q/(q−1) and q > N/2. Then 2q′ ∈ [2, 2∗). Therefore, from (A4) and (2.9), we have ∣∣ ∫ Ωm(r0,+∞) F (x, um) ‖um‖2H1 0 (Ω) dx| ≤ ∫ Ωm(r0,+∞) |F (x, um)| |um|2 |wm|2 dx ≤ [ ∫ Ωm(r0,+∞) ( |F (x, um)| |um|2 )q dx ]1/q[ ∫ Ωm(r0,+∞) |wm|2q ′ dx ]1/q ≤ C15 [ ∫ Ωm(r0,+∞) F̂ (x, um) dx ]1/q[ ∫ Ωm(r0,+∞) |wm|2q ′ dx ]1/q′ ≤ C16 [ ∫ Ωm(r0,+∞) |wm|2q ′ dx ]1/q′ dx→ 0, as m→∞. Then ∫ Ωm(r0,+∞) 2F (x, um) ‖um‖2H1 0 (Ω) dx→ 0 as m→∞. (2.13) In combination with (2.12), we obtain∫ Ω 2F (x, um) ‖um‖2H1 0 (Ω) dx→ 0 as m→∞, which contradicts (2.11). 8 D. T. LUYEN, N. M. TRI EJDE-2021/48 Case 2: w 6= 0. Setting Ω0 := {x ∈ Ω : w(x) 6= 0}, we have meas(Ω0) > 0 and lim m→∞ um(x) = lim m→∞ ‖um‖2H1 0 (Ω)wm(x) =∞, a.e. in Ω0. It follows from (A2), (A3), (A6), (2.5), (2.10) and Fatou’s lemma that 1 2 = lim m→∞ ∫ Ω F (x, um) ‖um‖2H1 0 (Ω) dx ≥ lim inf m→∞ ∫ Ω F (x, um) ‖um‖2H1 0 (Ω) dx ≥ lim inf m→∞ ∫ Ω0 F (x, um) ‖um‖2H1 0 (Ω) dx ≥ ∫ Ω0 lim inf m→∞ F (x, um) ‖um‖2H1 0 (Ω) dx = ∫ Ω0 lim inf m→∞ F (x, um) |um|2 w2 m dx = +∞, (2.14) which is a contradiction. Because of the above result, without loss of generality, we can assume that um ⇀ u weakly in H1 0 (Ω) as m→∞, um → u a.e. in Ω as m→∞, um → u strongly in Lν(Ω), 1 ≤ ν < 2∗ as m→∞. (2.15) Thus by (A2), (A6) and (2.15), we have∫ Ω (f(x, um)− f(x, u))(um − u) dx→ 0 as m→∞, (2.16)∫ Ω (g(x, um)− g(x, u))(um − u) dx→ 0 as m→∞. (2.17) If M2 is large enough, it follows from limm→∞ Φ ′ (um) = 0 and (2.15) that〈 (1 + T1(u))Φ ′ (um)− (1 + T1(um))Φ ′ (u), um − u 〉 → 0 as m→∞. (2.18) Moreover,〈 (1 + T1(u))Φ ′ (um)− (1 + T1(um))Φ ′ (u), um − u 〉 = (1 + T1(u))(1 + T1(um)) [ ∫ Ω ( |∇um −∇u|2 − f(x, um)(um − u) + f(x, u)(um − u) ) dx ] − (1 + T1(u))(ψ(um) + T1(um)) ∫ Ω g(x, um)(um − u) dx + (1 + T1(um))(ψ(u) + T1(u)) ∫ Ω g(x, u)(um − u) dx − (1 + T1(u))T3(um) ∫ Ω |um|ρ−1(um − u) dx + (1 + T1(um))T3(u) ∫ Ω |u|ρ−1(um − u) dx, (2.19) where T3(u) = ρχ′ ( κ(u)−1 ∫ Ω |u|ρ dx ) κ(u)−1 ∫ Ω G(x, u) dx. By (2.16), (2.17), (2.18) and (2.19) we obtain∫ Ω |∇um −∇u|2 dx→ 0 as m→∞. EJDE-2021/48 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS 9 Therefore, we conclude that um → u strongly in H1 0 (Ω). The proof is complete. � Lemma 2.6. Suppose that (A2), (A3), (A6) are satisfied. Then for any finite dimensional subspace X̂ ⊂ H1 0 (Ω), there is R = R(X̂) > 0 such that Φ(u) ≤ 0, ∀u ∈ X̂, ‖u‖H1 0 (Ω) ≥ R. Proof. Arguing by contradiction, suppose that for some sequence {um}∞m=1 ⊂ X̂ with ‖um‖H1 0 (Ω) > 0 for all m ∈ N and ‖um‖H1 0 (Ω) →∞ as m→∞, there is M > 0 such that Φ(um) ≥ −M for all m ∈ N. Setting wm = um ‖um‖H1 0 (Ω) , then ‖wm‖H1 0 (Ω) = 1. Therefore we can (by passing to a subsequence if necessary) suppose that wm ⇀ w weakly in H1 0 (Ω) as m→∞, wm → w a.e. in Ω as m→∞, wm → w strongly in Lν(Ω) as m→∞, 2 ≤ ν < 2∗. (2.20) Since X̂ is finite dimensional, it follows that wm → w strongly in X̂ as m→∞, and w ∈ X̂ with ‖w‖H1 0 (Ω) = 1. Therefore, from (2.13) we obtain 0 = lim m→∞ −M ‖um‖2H1 0 (Ω) ≤ lim m→∞ Φ(um) ‖um‖2H1 0 (Ω) = −∞. Hence we arrive at a contradiction. So, there is R = R(X̂) > 0 such that Φ(u) ≤ 0 for u ∈ X̂ and ‖u‖H1 0 (Ω) ≥ R. � Now, we show that Φ has an unbounded sequence of critical values. Let 0 < λ1 < λ2 ≤ λ3 ≤ · · · ≤ λk ≤ · · · denote the eigenvalues of the problem −∆u = λu in Ω, u = 0 on ∂Ω, (2.21) and e1, e2, . . . denote the corresponding eigenfunctions which normalized such that ‖ej‖H1 0 (Ω) = 1, for all j = 1, 2, . . . . For any k > 0, we put Vk = span{ej ; j ≤ k} in H1 0 (Ω), and V⊥k its orthogonal complement. Choose an increasing sequence Rk such that Φ(u) ≤ 0 if u ∈ Vk, ‖u‖H1 0 (Ω) ≥ Rk. Let BRk denote the closed ball of radius Rk in H1 0 (Ω),Wk ≡ BRk ⋂ Vk, and Γk = { h ∈ C(Wk, H 1 0 (Ω)) : h is odd and h(u) = u if ‖u‖H1 0 (Ω) = Rk } , Uk = { u = tek+1 + w : t ∈ [0, Rk+1], w ∈ BRk+1 ∩ Vk, ‖u‖H1 0 (Ω) ≤ Rk+1 } , Λk = { H ∈ C(Uk, H1 0 (Ω)) : H|Wk ∈ Γk and H(u) = u if ‖u‖H1 0 (Ω) = Rk+1 or u ∈ (BRk+1 \BRk) ∩ Vk } . (2.22) Now we define γk = inf H∈Λk max u∈Uk Φ(H(u)), k ∈ N, (2.23) βk = inf h∈Γk max u∈Wk Φ(h(u)), k ∈ N. (2.24) 10 D. T. LUYEN, N. M. TRI EJDE-2021/48 It is obvious that γk ≥ βk. We will give the lower bounds for βk in the next lemma. Lemma 2.7. Suppose that (A2), (A6) are satisfied. Then there are constants C17 > 0 and k0 ∈ N such that for all k ≥ k0, βk ≥ C17k 2p N(p−2) . (2.25) Proof. By (A2) and (A6) we obtain Φ(u) ≥ 1 2 ∫ Ω |∇u|2 dx− C18 ∫ Ω |u|p dx− C19. (2.26) Set K(u) = 1 2 ‖u‖2H1 0 (Ω) − C18‖u‖pLp(Ω) ∈ C 2(H1 0 (Ω),R). (2.27) Then we can see that Φ(u) ≥ K(u)− C19, (2.28) K ′′(u)(h, h) = ( (−∆− C18p(p− 1)|u|p−2)h, h ) for all u ∈ H1 0 (Ω) (2.29) and that the functional K(u) satisfies the following assumptions: (A7) K(0) = 0; (A8) K(−u) = K(u) for all u ∈ H1 0 (Ω); (A9) for each finite dimensional subspace E ⊂ H1 0 (Ω), there is an R = R(E) > 0 such that K(u) < 0, for all u ∈ E with ‖u‖H1 0 (Ω) ≥ R(E); (A10) K ′(u) = u + κ(u) for u ∈ H1 0 (Ω), where κ : H1 0 (Ω) → H1 0 (Ω) is a compact operator; (A11) If for some M > 0, {uj}∞j=1 ⊂ H1 0 (Ω) satisfies K(uj) ≤ M for all j, and ‖K ′(uj)‖(H1 0 (Ω))∗ → 0 as j →∞, then there exists a subsequence {ujk}∞k=1 which converges strongly in H1 0 (Ω); (A12) If for some M > 0, {uj}∞j=1 ⊂ Vm satisfies K(uj) ≤ M for all j and ‖(K|Vm)′(uj)‖(Vm)∗ → 0 as j →∞, then there exists a subsequence {ujk}∞k=1 which converges strongly in Vm; (A13) If for some M > 0, {uj}∞j=1 ⊂ H1 0 (Ω) satisfies uj ∈ Vj , K(uj) ≤ M for all j and ‖(K|Vj )′(uj)‖(Vj)∗ → 0 as j → ∞, then there exists a subsequence {ujk}∞k=1 which converges strongly in H1 0 (Ω). Next, we define minimax values ωk = inf h∈Γk max u∈Wk K(h(u)), k ∈ N. (2.30) From (2.28), we obtain βk ≥ ωk − C19. (2.31) From [15, Theorem B], there is a vk ∈ H1 0 (Ω) such that K(vk) ≤ ωk, (2.32) K ′(vk) = 0, (2.33) index0K ′′(vk) ≥ k, (2.34) where index0K ′′(vk) := max { dimE : E ⊂ H1 0 (Ω) is a subspace such that K ′′(vk)(h, h) ≤ 0, for h ∈ E } . EJDE-2021/48 MULTIPLE SOLUTIONS TO BOUNDARY VALUE PROBLEMS 11 Therefore, from (2.29) and (2.34), we obtain that −∆− C18p(p− 1)|vk|p−2 possesses at least k non-positive eigenvalues. (2.35) Let N (V ) denote the number of non-positive eigenvalues (multiplicities counted) of the problem −∆u− V (x)u = λu in Ω, u = 0 on ∂Ω, where V (x) ∈ LN/2(Ω). Then from [15, Lemma 2.3] (or [5, 6]) there is a constant CN > 0 such that N (V ) ≤ CN‖V (x)‖ N 2 L N 2 (Ω) . (2.36) From (2.35) and (2.36), we obtain C20k ≤ ‖|vk|p−2‖ N 2 L N 2 (Ω) . 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Duong Trong Luyen Division of Computational Mathematics and Engineering, Institute for Computational Science, and Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam Email address: duongtrongluyen@tdtu.edu.vn Nguyen Minh Tri Institute of Mathematics, Vietnam Academy of Science and Technology, 18 Hoang Quoc Viet, 10307 Cau Giay, Hanoi, Vietnam Email address: triminh@math.ac.vn 1. Introduction 2. Proofs of the main results Acknowledgments References