Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 86, pp. 1–21. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.86 EXISTENCE AND MULTIPLICITY OF SOLUTIONS TO TRIHARMONIC PROBLEMS QIFAN WEI, XUEMEI ZHANG Abstract. The authors consider the triharmonic equation (−∆)3u+ c1∆ 2u+ c2∆u = h(x)|u|p−2u+ g(x, u) in Ω, where p ∈ (1, 2), subject to Navier boundary conditions. Based on the least action principle, the Ekeland’s variational principle and a variant version of mountain pass lemma, we analyze the existence and multiplicity of nontrivial solutions to the above problem. In addition, we obtain the first eigenvalue of triharmonic operator and consider its structure. The conclusions are illustrated with several examples. 1. Introduction and main results Let Ω ⊂ RN (N ≥ 7) denote a smooth bounded domain. The purpose of this article is to study the sixth-order elliptic problem with combined nonlinearities (−∆)3u+ c1∆ 2u+ c2∆u = h(x)|u|p−2u+ g(x, u) in Ω, u = ∆u = ∆2u = 0 on ∂Ω, (1.1) where (−∆)3(·) = −∆((−∆)2(·)) stands for the triharmonic operator; c1, c2 ∈ R satisfying c1 ≥ 0 and c2 − c1µ1 < µ2 1 (µ1 is the first eigenvalue of (−∆, H1 0 (Ω)); h(x) is a weight function satisfying h(x) ∈ L∞(Ω) and there is a positive measure subset H ⊂ Ω satisfying h(x) > 0 in H; p ∈ (1, 2) and g(x, u) ∈ C(Ω̄× R,R). A function u ∈ H3 ϑ(Ω) is a weak solution to problem (1.1) if∫ Ω (∇∆u∇∆φ+ c1∆u∆φ− c2∇u∇φ) dx = ∫ Ω h(x)|u|p−1φdx+ ∫ Ω g(x, u)φdx for all φ ∈ H3 ϑ(Ω), where H3 ϑ(Ω) := {u ∈ H3(Ω) : ∆ju = 0 on ∂Ω for j < 3 2 } with the scalar product (u, v) = ∫ Ω (∇∆u∇∆v + c1∆u∆v − c2∇u∇v) dx and the equivalent suitable norm ∥u∥ϑ = (∫ Ω (|∇∆u|2 + c1|∆u|2 − c2|∇u|2) dx )1/2 . Thus solutions of (1.1) correspond to critical points of the energy functional E(u) = 1 2 ∫ Ω (|∇∆u|2 + c1|∆u|2 − c2|∇u|2) dx− 1 p ∫ Ω h(x)|u|p dx− ∫ Ω G(x, u) dx, (1.2) 2020 Mathematics Subject Classification. 35J30, 35J40. Key words and phrases. Triharmonic equation; Ekeland’s variational principle; mountain pass lemma; existence and multiplicity of solutions. ©2025. This work is licensed under a CC BY 4.0 license. Submitted June 23, 2025. Published August 18, 2025. 1 2 Q. WEI, X. ZHANG EJDE-2025/86 where G(x, u) is the primitive of g(x, u). Higher-order elliptic boundary problems have abundant applications in physics and engineering [36] and have also been studied in many areas of mathematics, including conformal geometry [9], some geometry invariants [7] and nonlinear elasticity [11]. Moreover, different higher order problems are associated with distinct applications. For instance, the fourth-order problems can describe static deflection of a bending beam [27] and traveling waves in suspension bridges [13]. The sixth-order problems appear in the study of the thin-film models [5, 32], the phase field crystal models [3, 10], fluid flows models [39] and geometric design models [30, 46]. The fourth-order problems have been extensively investigated in the previous four decades (see [6, 8, 14, 15, 21, 22, 25, 26, 28, 35, 38, 41, 43] and the references cited therein). In particular, by using a variant version of the mountain pass lemma, Hu-Wang [29] obtained the existence of nontrivial solutions for the following fourth-order problem ∆2u+ α∆u = f(x, u) in Ω, u = ∆u = 0 on ∂Ω, (1.3) where ∆2(u) = ∆(∆u) stands for the biharmonic operator, Ω ⊂ RN (N > 4) is a smooth bounded domain, and α < µ1 is a parameter, where µ1 is the first eigenvalue of (−∆) in H1 0 (Ω). Pu-Wu-Tang [37] studied the fourth-order problem ∆2u+ β∆u = a(x)|u|s−2u+ f(x, u) in Ω, u = ∆u = 0 on ∂Ω, (1.4) where Ω ⊂ RN (N > 4) is a smooth bounded domain, β < µ1, a(x) ∈ L∞(Ω), s ∈ (1, 2) and f ∈ C(Ω̄ × R,R). The authors established, by using the least action principle, the Ekeland’s variational principle and the mountain pass lemma, the existence and multiplicity of solutions for problem (1.4). On further results of biharmonic problems, we wish to bring the articles [2, 12, 18, 19, 20, 33, 40, 42, 43, 44, 45, 47] to the readers attention. On triharmonic problems, naturally hope that the excellent results of biharmonic problems can be generalized. However, such an extension will encounter essential difficulties. In fact, the triharmonic operator is negative, which fundamentally distinguishes it from the biharmonic operator and cannot be directly derived. In fact, unlike the biharmonic case, which often relies on established tools such as comparison principles and spectral theory, these methods are generally inapplicable to triharmonic equations. Moreover, the properties of the nonlinearities h(x)|u|p−2u and g(x, r) have a more significant impact on the nature of the solution. Comparing with the biharmonic problems, for which there have been a great many achievements, there seems only a few results on the triharmonic problems. For example, in [1], Abdrabou and El-Gamel developed a numerical scheme to provide an approximate solution of the following triharmonic problem −∆3u = f(x, u) in Ω, u = ∂u ∂ν = ∂2u ∂ν2 = 0 on ∂Ω, where Ω stands for the rectangular domain denoted by Ω := {(x, y) : a1 < x < a2, a3 < y < a4}. Motivated by the above results, the main objective of this paper is to study the existence and multiplicity of nontrivial solutions to problem (1.1). We begin with demonstrating the existence of the first eigenvalue of ((−∆)3 + c1∆ 2 + c2∆, H 3 ϑ(Ω)) that will be used in the subsequent sections. For the eigenvalue problems, it is worth mentioning that Liu-Wang [34, Lemma 2.3] provided a detailed discussion of the first eigenvalue for the biharmonic problem ∆2u = f(x, u) in Ω, u = ∆u = 0 on ∂Ω. (1.5) Let λ1 = inf{ ∫ Ω |∆u|2 dx : u ∈ H2(Ω) ∩H1 0 (Ω), ∫ Ω |u|2 dx = 1}. Then they proved that λ1 is the first eigenvalue of (1.5) with a positive λ1- eigenfunction. Recently, Hu-Wang [29] obtained similar results for problem (1.3) by using the method of Liu-Wang [34]. While for sixth-order problems, EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 3 this trick fails since the triharmonic operator cannot be directly derived from the Laplace operator. For this reason, we need to introduce a new technique to study the structure of the first eigenvalue to problem (1.1) (see the demonstration of Lemma 2.1 for the details). In addition, comparing with [1], this article possesses the following features. Firstly, c1 ≥ 0 and c2 ∈ R are considered. Secondly, h(x) ̸= 0 is investigated, and the combined nonlinearities are studied in problem (1.1). Thirdly, we study the existence and multiplicity of nontrivial weak solutions to problem (1.1) via the least action principle, the Ekeland’s variational principle and a variant version of mountain pass lemma, which is not used in [1]. It is probably the first time that these techniques are to be used to deal with triharmonic problems. Our main results are stated in the following theorems, here we assume several hypotheses of g(x, r): (H1) lim|r|→0 g(x,r) r > −∞ uniformly in x ∈ Ω; (H2) lim|r|→∞ g(x,r) |r|s = 0 uniformly in x ∈ Ω, where s denotes a certain constant and s ∈ (1, N+6 N−6 ); (H3) lim|r|→0 g(x,r) r = k1 < µ1(µ 2 1 + c1µ1 − c2) uniformly in x ∈ Ω. Theorem 1.1. Let condition (H1) hold. In addition, we suppose that the function g(x, r) satisfies (H4) there is a constant k2 satisfying lim|r|→∞ sup g(x,r) r ≤ k2 < µ1(µ 2 1 + c1µ1 − c2) uniformly in x ∈ Ω. Then problem (1.1) admits a nontrivial solution. Theorem 1.2. Let condition (H1) hold. In addition, we suppose that the function g(x, r) satisfies (H5) lim|r|→∞ g(x,r) r = µ1(µ 2 1 + c1µ1 − c2) uniformly in x ∈ Ω; (H6) there is a constant k3 satisfying lim|r|→∞ sup 2G(x,r)−µ1(µ 2 1+c1µ1−c2)r 2 |r|p ≤ k3 < −∥h∥∞ p uniformly in x ∈ Ω. Then problem (1.1) admits a nontrivial solution. Theorem 1.3. Let conditions (H2) and (H3) hold. In addition, we suppose that the function g(x, r) satisfies (H7) lim|r|→∞ g(x,r) r = k4 ∈ (µ1(µ 2 1 + c1µ1 − c2),+∞) uniformly in x ∈ Ω and k4 ̸= µi(µ 2 i + c1µi − c2), where µi is the eigenvalue of (−∆, H1 0 (Ω)) and i is a positive integer. Then there is 0 < ᾱ ∈ R such that problem (1.1) admits two nontrivial solutions for ∥h∥∞ ≤ ᾱ. Theorem 1.4. Let conditions (H2) and (H3) hold. In addition, we suppose that the function g(x, r) satisfies (H8) lim|r|→∞ g(x,r) r = µi(µ 2 i + c1µi − c2)(i ̸= 1) uniformly in x ∈ Ω; (H9) there is a constant k5 satisfying lim|r|→∞ inf g(x,r)r−2G(x,r) |r|p ≥ k5 > ( 2p −1)∥h∥∞ uniformly in x ∈ Ω. Then there is 0 < ᾱ ∈ R such that problem (1.1) admits two nontrivial solutions for ∥h∥∞ ≤ ᾱ. Theorem 1.5. Let conditions (H2) and (H3) hold. In addition, we suppose that the function g(x, r) satisfies (H10) lim|r|→∞ g(x,r) r = +∞ uniformly in x ∈ Ω; (H11) there is a constant k6 satisfying lim|r|→∞ sup g(x,r)r−2G(x,r) |r|κ ≥ k6 > 0 uniformly in x ∈ Ω, where max{N 6 (s− 1), p} < κ < 2N N−6 . Then there is 0 < ᾱ ∈ R such that problem (1.1) admits two nontrivial solutions for ∥h∥∞ ≤ ᾱ. Remark 1.6. It is not hard to show that there are some elementary functions that satisfy the assumptions of Theorems 1.1–1.5. For instance, (1) g(x, r) = P1(x)r + Q1(x) r3 1+r2 , where P1(x) and Q1(x) are continuous functions with supx∈Ω(P1(x) +Q1(x)) < µ1(µ 2 1 + c1µ1 − c2); 4 Q. WEI, X. ZHANG EJDE-2025/86 (2) g(x, r) = µ1(µ 2 1 + c1µ1 − c2)r + pµ|r|p−2 + sν|r|s−2r, where µ < −∥h∥∞ 2p , ν > 0 and 0 < s < p; (3) g(x, r) = µr + k1−µ 1+r2 r, where µ > k1 and µ ∈ (µ1(µ 2 1 + c1µ1 − c2),+∞); (4) g(x, r) = µr + r2 1+r , where µ > 0 and µ ̸= µi; (5) g(x, r) = rsinr + r2. Throughout this paper, ∥ · ∥Lθ denotes the norm of Lθ(Ω). By using the Sobolev embedding theorem, there exists 0 < S ∈ R such that ∥u∥Lθ ≤ S∥u∥ϑ for 1 ≤ θ ≤ 2N N − 6 . Particularly, if θ = 2, then we have ∥u∥2L2 ≤ 1 λ1 ∥u∥2ϑ. The organization of the article is the following. In Section 2, we prove several preliminary results to be used in the subsequent sections. Section 3 will be devoted to the proof of Theorems 1.1–1.5. The main tools here are the least action principle, the Ekeland’s variational principle and a variant version of mountain pass lemma. 2. Preliminaries In this section, we verify several preliminary conclusions. Lemma 2.1. Let λ1 = inf {∫ Ω (|∇∆u|2 + c1|∆u|2 − c2|∇u|2) dx : u ∈ H3 ϑ(Ω) and ∫ Ω |u|2 dx = 1 } be the first eigenvalue of the triharmonic eigenvalue problem (−∆)3u+ c1∆ 2u+ c2∆u = λu in Ω, u = ∆u = ∆2u = 0 on ∂Ω, and µ1 = inf {∫ Ω |∇u|2dx : u ∈ H1 0 (Ω) and ∫ Ω |u|2 dx = 1 } , where µ1 is the first eigenvalue of (−∆, H1 0 (Ω)), then λ1 = µ1(µ 2 1 + c1µ1 − c2). Proof. If φ1 > 0 attains µ1, then φ1(x) is a solution to −∆u = µ1u in Ω, u = 0 on ∂Ω. According to [29, Lemma 2.2], we know that φ1 ∈ H2(Ω) ∩H1 0 (Ω). As H3 ϑ(Ω) ⊂ H3(Ω) ⊂ H1(Ω) and C6 ⊂ C3, we obtain u ∈ H1(Ω) and ∂Ω ∈ C3. Since (−∆) is strictly elliptic in Ω and coefficients are constants, it follows from [24, Theorem 8.13] that φ1 ∈ H3(Ω) (i.e. k = 1 in Theorem 8.13). Then for every ψ ∈ H3 ϑ(Ω), we have∫ Ω (∇∆φ1∇∆ψ + c1∆φ1∆ψ − c2∇φ1∇ψ) dx = − ∫ Ω ∆φ1∆ 2ψ dx+ ∫ ∂Ω ∆φ1∇∆ψν dS + c1 ∫ Ω ∆φ1∆ψ dx− c2 ∫ Ω ∇φ1∇ψ dx = − ∫ Ω ∆φ1∆ 2ψ dx+ c1 ∫ Ω ∆φ1∆ψ dx− c2 ∫ Ω ∇φ1∇ψ dx = −(−µ1) ∫ Ω φ1∆ 2ψ dx+ c1 ∫ Ω ∆φ1∆ψ dx− c2 ∫ Ω ∇φ1∇ψ dx = µ1[ ∫ Ω ∆φ1∆ψ dx+ ∫ ∂Ω (φ1∇∆ψν −∇φ1∆ψν) dS] + c1 ∫ Ω ∆φ1∆ψ dx− c2 ∫ Ω ∇φ1∇ψ dx EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 5 = µ1 ∫ Ω ∆φ1∆ψ dx+ c1 ∫ Ω ∆φ1∆ψ dx− c2 ∫ Ω ∇φ1∇ψ dx = −µ2 1 ∫ Ω φ1∆ψ dx− c1µ1 ∫ Ω φ1∆ψ dx− c2µ1 ∫ Ω φ1ψ dx = −µ2 1[ ∫ Ω ∆φ1ψ dx+ ∫ ∂Ω (φ1∇ψν −∇φ1ψν) dS]− c1µ1[ ∫ Ω ∆φ1ψ dx + ∫ ∂Ω (φ1∇ψν −∇φ1ψν) dS]− c2µ1 ∫ Ω φ1ψ dx = −µ2 1 ∫ Ω ∆φ1ψ dx+ c1µ 2 1 ∫ Ω φ1ψ dx− c2µ1 ∫ Ω φ1ψ dx = µ3 1 ∫ Ω φ1ψ dx+ c1µ 2 1 ∫ Ω φ1ψ dx− c2µ1 ∫ Ω φ1ψ dx, which indicates∫ Ω (∇∆φ1∇∆ψ + c1∆φ1∆ψ − c2∇φ1∇ψ) dx = µ1(µ 2 1 + c1µ1 − c2) ∫ Ω φ1ψ dx. (2.1) Hence, considering the previous definition we obtain that φ1 ∈ H3 ϑ(Ω) is a solution to (−∆)3u+ c1∆ 2u+ c2∆u = µ(µ2 1 + c1µ1 − c2)u in Ω, u = ∆u = ∆2u = 0 on ∂Ω. For the purpose of proving the conclusion of this lemma, we only need to show that λ1 = µ2 1(µ 2 1 + c1µ1 − c2). Taking ψ(x) = φ1(x) in (2.1), we have∫ Ω (∇∆φ1∇∆φ1 + c1∆φ1∆φ1 − c2∇φ1∇φ1) dx = µ1(µ 2 1 + c1µ1 − c2) ∫ Ω φ2 1 dx. According to the definition of λ1 we deduce that λ1 ≤ ∫ Ω (|∇∆φ1|2 + c1|∆φ1|2 − c2|∇φ1|2) dx = µ1(µ 2 1 + c1µ1 − c2). In what follows we prove the other part by a similar method used in [31]. Assuming that v ∈ H3 ϑ(Ω) is the λ1-eigenfunction of ((−∆)3 + c1∆ 2 + c2∆, H 3 ϑ(Ω)) and ∫ Ω |v|2dx = 1. It is worth mentioning that λ1 and µ1 in this paper are essentially equivalent to the Γ(Ω) and λ(Ω) in reference [31], respectively. By making use of an integration by parts and two Cauchy-Schwarz inequalities, we find that for each 1 ≤ k < 3,(∫ Ω |∇kv|2 dx )2 ≤ ∫ Ω |∇k−1v|2 dx · ∫ Ω |∇k+1v|2 dx. (2.2) Letting vk = ∫ Ω |∇kv|2 dx, then we can rewrite the inequality (2.2) as v2k ≤ vk−1vk+1. Therefore, we use recursion to prove that the sequence (vk)k follows the rule vpk 3−k ≤ vqk2−kv3 for all 1 ≤ k < 3, where (pk)k and (qk)k are defined by (p1, q1) = (2, 1), and for every k ≥ 1, pk+1 = 2pk − qk, qk+1 = pk. The sequence (pk, qk)k≥1 forms a constant-recursive sequence of order 1, the solution of which is (pk, qk) = (k + 1, k). Take k = 1 and we can obtain that v31 ≤ v20v3, which indicates ( ∫ Ω |∇v|2dx )3 ≤ (∫ Ω |v|2dx )3(∫ Ω |∇∆v|2d ) . (2.3) 6 Q. WEI, X. ZHANG EJDE-2025/86 It follows from (2.3) that∫ Ω |∇∆v|2 dx+ c1 ∫ Ω |∆v|2dx− c2 ∫ Ω |∇v|2 dx ≥ (∫ Ω |∇v|2dx )3 + c1 (∫ Ω |∇v|2dx )2 − c2 ∫ Ω |∇v|2 dx. We define G = ∫ Ω |∇v|2 dx, then we have G ≥ µ1. Therefore, λ1 ≥ G3 + c1G2 − c2G ≥ µ3 1 + c1µ 2 1 − c2µ1 (c2 − c1µ1 < µ2 1). So the proof of this lemma is complete. □ Remark 2.2. Based on Lemma 2.1, we derive λi = µi(µ 2 i + c1µi − c2) for i = 1, 2, . . . , where 0 < µ1 < µ2 ≤ · · · ≤ µk ≤ . . . are the eigenvalues of (−∆, H1 0 (Ω)). Lemma 2.3. Let conditions (H2) and (H3) hold. Suppose that lim|r|→∞ g(x,r) r > λ1. Then there is 0 < ᾱ ∈ R such that E satisfies the following results provided ∥h∥∞ ≤ ᾱ: (a) there are positive constants σ and β satisfying E(u) ≥ β > 0 for all u ∈ H3 ϑ(Ω) with ∥u∥ϑ = σ; (b) there is a function γ ∈ H3 ϑ(Ω) with ∥γ∥ϑ > σ satisfying E(γ) ≤ 0. Proof. (a) It follows from g ∈ C(Ω̄×R,R) and conditions (H2) to (H3) that for τ1 = 1 2 (λ1−k1) > 0 there is α1 > 0 satisfying g(x, r)r ≤ (λ1 − τ1)r 2 + α1|r|s+1 for every (x, r) ∈ Ω× R. Thus, we obtain G(x, r) = ∫ 1 0 g(x, vr)r dv ≤ ∫ 1 0 ( (λ1 − τ1)vr 2 + α1v s|r|s+1 ) dv ≤ λ1 − τ1 2 r2 + α1 s+ 1 |r|s+1 (2.4) for every (x, r) ∈ Ω× R. We define α2 = α1 s+1 . Using the definition of E given in (1.2), together with (2.4) and the Sobolev inequality, we deduce that E(u) ≥ 1 2 ∥u∥2ϑ − ∥h∥∞ p ∫ Ω |u|p dx− ∫ Ω λ1 − τ1 2 u2 dx− ∫ Ω α2|u|s+1 dx ≥ 1 2 ( 1− λ1 − τ1 λ1 ) ∥u∥2ϑ − ∥h∥∞Sp p ∥u∥pϑ − α2Ss+1∥u∥s+1 ϑ ≥ ( τ1 2λ1 − ∥h∥∞Sp p ∥u∥p−2 ϑ − α2Ss+1∥u∥s−1 ϑ ) ∥u∥2ϑ. (2.5) We define σ = ( τ1 4λ1α2Ss+1 ) 1 s−1 , ᾱ = τ1 8λ1Spσp−2 . So, if ∥h∥∞ ≤ ᾱ and ∥u∥ϑ = σ, then an application of (2.5) yields E(u) ≥ τ1 8λ1 σ2. Hence part (a) follows. (b) According to lim|r|→∞ g(x,r) r > λ1, there are constants δ > 0 and R > 0 satisfying g(x, r)r ≥ (λ1 + δ)r2 for every |r| ≥ R and a.e. x ∈ Ω, which indicates that G(x, r) = ∫ 1 R |r| g(x, vr)r dv +G ( x, Rr |r| ) EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 7 ≥ ∫ 1 R |r| (λ1 + δ)vr2 dv +G ( x, Rr |r| ) ≥ λ1 + δ 2 r2 − λ1 + δ 2 R2 +G ( x, Rr |r| ) . It follows from the continuity of G that there is α3 > 0 such that G(x, r) ≥ λ1 + δ 2 r2 − α3 (2.6) for each (x, r) ∈ Ω × R. Without loss of generality, we assume that φ1 > 0 denotes a λ1- eigenfunction and r > 0. Combined with (2.6), we can infer that E(rφ1) = r2 2 ∥φ1∥2ϑ − rp p ∫ Ω h(x)|φ1|p dx− ∫ Ω G(x, rφ1) dx ≤ r2 2 ∥φ1∥2ϑ − rp p ∫ Ω h(x)|φ1|p dx− r2 2 ∫ Ω (λ1 + δ)φ2 1 dx+ α3|Ω| = −r 2 2 ∫ Ω δφ2 1 dx− rp p ∫ Ω h(x)|φ1|p dx+ α3|Ω|. Therefore, taking r0 > 0 sufficiently large such that ∥γ∥ϑ = ∥r0φ1∥ϑ > σ, this gives the proof of part (b). □ For the sake of completeness, we present the (Ce)c condition and a variation of mountain pass lemma which we will be used. Definition 2.4 (Pu-Wu-Tang [37, Def. 2.1]). Let E ∈ C1(X,R). We say that E satisfies the Cerami condition at the level c ∈ R ((Ce)c for short) if any sequence {un} ⊂ X with E(un) → c and (1 + ∥un∥ϑ)E ′(un) → 0 possesses a convergent subsequence in X, E satisfies the (Ce) condition if E satisfies the (Ce)c condition for all c ∈ R. Lemma 2.5 (Costa-Miyagaki [23, Theorem 1]). Suppose that X is a real Banach space and E ∈ C1(X,R) satisfies max{E(0), E(u1)} ≤ a < b ≤ inf ∥u∥=ξ E(u) for some a < b, ξ > 0, and u1 ∈ X with ∥u1∥ ≥ ξ. Let c ≥ b be characterized by c = inf γ∈Γ max 0≤t≤1 E(γ(t)), where Γ = {γ ∈ C([0, 1], X) : γ(0) = 0, γ(1) = u1} denotes the set of continuous paths joining 0 and u1. Then, there is a sequence {un} ⊂ X such that E(un) n−→ c ≥ b, (1 + ∥un∥)∥E ′(un)∥X∗ n−→ 0, where X∗ denotes the dual space of X. Furthermore, if assume that E satisfies condition (Ce)c, then c is a critical value of E. In what follows, we introduce the Ekeland’s variational principle in order to find a local mini- mum. Proposition 2.6 (Ekeland [16, theorem 1]). Suppose that W denotes a complete metric space and F : W → R ∪ {+∞} is lower semicontinuous, bounded from below. For every ϵ > 0, there exists a certain point w ∈W with F (w) ≤ inf W F + ε and F (u) ≥ F (w) + εd (w, u) for all u ∈W. 8 Q. WEI, X. ZHANG EJDE-2025/86 3. Proof of main results 3.1. Proof of Theorem 1.1. From condition (H4), we obtain that for τ2 = 1 2 (λ1−k2) > 0, there is R > 0 satisfying g(x, r)r ≤ (λ1 − τ2)r 2 for every |r| ≥ R and a.e. x ∈ Ω. It follows from the continuity of G that there is α4 > 0 satisfying for every (x, r) ∈ Ω× R, G(x, r) = ∫ 1 R |r| g(x, vr)r dv +G ( x, Rr |r| ) ≤ λ1 − τ2 2 r2 + α4, which yields E(u) = 1 2 ∥u∥2ϑ − 1 p ∫ Ω h(x)|u|p dx− ∫ Ω G(x, u) dx ≥ 1 2 ∥u∥2ϑ − ∥h∥∞Sp p ∥u∥pϑ − ∫ Ω λ1 − τ2 2 u2 dx− α4|Ω| ≥ τ2 2λ1 ∥u∥2ϑ − ∥h∥∞Sp p ∥u∥pϑ − α4|Ω|. So, we know that E is coercive. Because E is coercive and weakly lower semicontinuous in H3 ϑ(Ω), we derive that u1 is a global minimum of E . In fact, combining the assumptions on h(x) and Lusin’s Theorem, there is a close subset H ⊂ K such that h(x) is continuous in H and meas(K \H) < 1 2 measK. Therefore, measH = measK −meas(K \H) > 1 2 measK > 0 and α̂ = inf x∈H h(x) > 0. For each δ > 0, there is an open set G satisfying meas(G \ H) < δ. Let η(x) ∈ C3 0 (Ω) be a function satisfying η(x) = 0 x ∈ Ω \G, 0 ≤ η(x) ≤ 1 x ∈ G \H, η(x) = 1 x ∈ H. It follows from h(x) ∈ L∞(Ω) that there is M0 > 0 satisfying h(x) > −M0 (x ∈ G \H). Therefore, ∫ Ω h(x)|η|p dx = ∫ G h(x)|η|p dx = ∫ H h(x)|η|p dx+ ∫ G\H h(x)|η|p dx ≥ α̂|H| −M0|G \H| ≥ α̂ 2 |K| −M0δ, which indicates that for δ > 0 small enough, we obtain∫ Ω h(x)|η|p dx > 0. According to condition (H1), we have that there is M > 0 such that g(x, r)r ≥ −Mr2 EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 9 for every |r| small enough and a.e. x ∈ Ω, which yields G(x, r) = ∫ 1 0 g(x, ηr)rdη ≥ −M 2 r2 for every |r| small enough and a.e. x ∈ Ω. Thus, we conclude that for r > 0, lim sup t→0 E(rη)r−p = lim sup r→0 (r2 2 ∥η∥2ϑ − rp p ∫ Ω h(x)|η|p dx− ∫ Ω G(x, rη) dx ) r−p ≤ lim sup r→0 (r2−p 2 ∥η∥2ϑ − 1 p ∫ Ω h(x)|η|p dx+ Mr2−p 2 ∫ Ω η2 dx ) < 0, which yields E(u1) < 0 and u1 is a nontrivial solution of problem (1.1). 3.2. Proof of Theorem 1.2. Let G(x, r) = λ1 2 r 2 +G(x, r) and g(x, r) = λ1r+ g(x, r). It follows from conditions (H5) and (H6) that lim |r|→∞ G(x, r) r2 = 0 and lim sup |r|→∞ G(x, r) |r|p = k3. Therefore, for 0 < δ < −∥h∥∞ p − k3, there is R > 0 satisfying G(x, r) ≤ (k3 + δ)|r|p (3.1) for |r| ≥ R and a.e. x ∈ Ω. Without loss of generality, we may suppose that {un} ⊂ H3 ϑ(Ω) is a sequence such that ∥un∥ϑ → +∞ as n→ ∞, E(un) ≤ α5 for some α5 ∈ R. We define Dn = {x ∈ Ω, |un(x)| ≥ R}, D̄n = {x ∈ Ω, |un(x)| ≤ R}. An application of the continuity of G yields that there is a constant α6 > 0 satisfying∫ D̄n G(x, r) dx ≤ α6. (3.2) In addition, it follows from the definition of D̄n and h(x) ∈ L∞(Ω) that there is a constant α7 > 0 satisfying 1 p ∫ D̄n h(x)|un|p dx ≤ α7. (3.3) We define wn = un ∥un∥ϑ , which yieldd that {wn} is bounded inH3 ϑ(Ω). By extracting a subsequence (still denoted by {wn}), we suppose that wn ⇀ w in H3 ϑ(Ω), wn → w in Lθ(Ω) ( 1 ≤ θ < 2N N − 6 ) , wn → w a.e. x ∈ Ω. From the definition of {un}, we deduce that α5 ∥un∥2ϑ ≥E(un) ∥un∥2ϑ ≥ 1 2 − 1 p∥un∥2−p ϑ ∫ Ω h(x)|wn|p dx− ∫ Ω λ1|wn|2 2 dx− ∫ Ω G(x, un) ∥un∥2ϑ dx ≥ 1 2 − ∥h∥∞ p∥un∥2−p ϑ ∫ Ω |wn|p dx− ∫ Ω λ1|wn|2 2 dx− ∫ Dn G(x, un) ∥un∥2ϑ dx− ∫ D̄n G(x, un) ∥un∥2ϑ dx. 10 Q. WEI, X. ZHANG EJDE-2025/86 Combining (3.1) and (3.2), we obtain α5 ∥un∥2ϑ ≥ 1 2 − ∥h∥∞ p∥un∥2−p ϑ ∫ Ω |wn|p dx− ∫ Ω λ1|wn|2 2 dx+ k3 − δ ∥un∥2−p ϑ ∫ Dn |wn|p dx− α6 ∥un∥2ϑ . Because 1 ∥un∥2−p ϑ → 0 as ∥un∥ϑ → +∞, we obtain − ∥h∥∞ p∥un∥2−p ϑ ∫ Ω |wn|p dx+ k3 − δ ∥un∥2−p ϑ ∫ Dn |wn|p dx→ 0. Therefore, ∫ Ω λ1|w|2 dx ≥ 1. It follows from the weak semicontinuity of norm and the Sobolev inequality that∫ Ω λ1|w|2 dx ≤ ∥w∥2ϑ ≤ 1 ≤ ∫ Ω λ1|w|2 dx. Obviously, the inequalities truly reduce to equalities. As the norm of w in H3 ϑ(Ω) is 1 and wn ⇀ w in H3 ϑ(Ω), we have {wn} → w in H3 ϑ(Ω),∫ Ω λ1|w|2 dx ≤ ∥w∥2ϑ. An application of the variational characterization of the first eigenvalue in [17] yields w = φ1 or w = −φ1. Thus, we find that |un(x)| = |wn(x)| · ∥un∥ϑ → ∞ a.e. x ∈ Ω. In what follows, we show that for all δ > 0, there is a subset Ωδ ⊂ Ω with meas(Ω \ Ωδ) < δ satisfying |un(x)| → ∞ uniformly as n→ ∞ for each x ∈ Ωδ. We may restrict our attention to the case in which |un(x)| → ∞ as n → ∞ for all x ∈ Ω. Given any K > 0 and every integer n > 0, let Ω[n,K] = ∩∞ k=n+1{x ∈ Ω, |un(x)| > K}, which implies Ω[n,K] is measurable and Ω[n,K] ⊂ Ω[k,K] when n < k. Since |un(x)| → ∞ for every x ∈ Ω, we conclude Ω = ∪∞ n=1Ω[n,K]. By using the properties of the Lebesgue measure, it follows that measΩ = lim n→∞ measΩ[n,K], which yields that lim n→∞ meas(Ω \ Ω[n,K]) = 0. Consequently, for each i, one can find an integer ni such that meas(Ω \ Ω[ni, i]) < δ 2i . Let Ωδ = ∩∞ i=1Ω[ni, i], which indicates that meas(Ω− Ωδ) = meas ( Ω \ ∩∞ i=1Ω[ni,i] ) = meas∪∞ i=1(Ω \ Ω[ni,i]) ≤ ∞∑ i=1 meas(Ω \ Ω[ni,i]) EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 11 < ∞∑ i=1 δ 2i = δ. Moreover, |un(x)| → ∞ as n → ∞ uniformly for every x ∈ Ωδ. For arbitrary K > 0, selecting i0 ≥ K yields Ωδ ⊂ Ω[ni0 , i0]. Consequently, |un(x)| ≥ i0 ≥ K for every n ≥ ni0 and x ∈ Ωδ. In view of (3.1) and (3.3), we conclude that α5 ≥ E(un) = ∥un∥2ϑ 2 − ∫ Ω λ1|un|2 2 dx− 1 p ∫ Ω h(x)|un|p dx− ∫ Ω G(x, un) dx ≥ −∥h∥∞ p ∫ Dn |un|p dx− α7 + (−k3 − δ) ∫ Dn |un|p dx− α6 ≥ ( − ∥h∥∞ p − k3 − δ )∫ Dn |un|p dx− α7 − α6. As −k3 < −∥h∥∞ p + δ, one can find a constant α8 > 0 such that∫ Dn |un|p dx ≤ α8. Furthermore, based on the definition of D̄n, we infer that∫ Ω |un|p dx = ∫ Dn |un|p dx+ ∫ D̄n |un|p dx ≤ α9 with some α9 > 0. This contradicts that |un(x)| → ∞ as n → ∞ uniformly for all x ∈ Ωδ, which confirms that E is coercive on H3 ϑ(Ω). Given that E is coercive and weakly lower semicontinuous in H3 ϑ(Ω), it achieves a global min- imum u1. Applying assumption (H1) and the argument used in Theorem 1.1, we obtain that E(u1) < 0. Hence, we obtain that u1 is a nontrivial solution of (1.1). 3.3. Proof of Theorem 1.3. In this subsection, we shall prove the multiplicity of solutions of (1.1) under the assumptions of Theorem 1.3 using the Ekeland’s variational principle and a variant version of mountain pass lemma. To this end, we first derive that E satisfies the (Ce) condition. Lemma 3.1. Under the hypotheses of Theorem 1.3, the functional E satisfies the (Ce) condition. Proof. Let un ⊂ H3 ϑ(Ω) be a (Ce)α sequence corresponding to some α > 0 such that E(un) → α and (1 + ∥un∥ϑ)E ′(un) → 0 as n→ ∞. Considering the definition of E , we obtain ⟨E ′(un), un⟩ = ∥un∥2ϑ − ∫ Ω h(x)|un|p dx− ∫ Ω g(x, un)un dx→ 0 (3.4) and ∫ Ω (∇∆un∇∆φ+ c1∆un∆φ− c2∇un · ∇φ) dx − ∫ Ω h(x)|un|p−1φdx− ∫ Ω g(x, un)φdx→ 0 (3.5) for every φ ∈ H3 ϑ(Ω). We begin by declaring that the sequence {un} is bounded inH3 ϑ(Ω). Suppose, to the contrary, that there is a subsequence {un} (also denoted as {un}) such that ∥un∥ϑ → ∞. We define wn = un ∥un∥ϑ . It is evident that the sequence {wn} is bounded in H3 ϑ(Ω). Thus, we may assume the existence of some w ∈ H3 ϑ(Ω) such that wn ⇀ w in H3 ϑ(Ω), 12 Q. WEI, X. ZHANG EJDE-2025/86 wn → w in Lθ(Ω) ( 1 ≤ θ < 2N N − 6 ) , wn → w a.e. x ∈ Ω. In addition, by applying [4, Theorem 1.2.7] again, we deduce that there is w0 ∈ Lθ(Ω) satisfying for each n, |wn| ≤ w0 a.e. in Ω. (3.6) It follows from conditions (H3) and (H7) that there is a constant M1 > 0 such that∣∣g(x, r) r ∣∣ ≤M1 (3.7) for every (x, r) ∈ Ω× R. In addition, we can find a constant M2 > 0 such that∣∣G(x, r) r2 ∣∣ ≤M2 (3.8) for every (x, r) ∈ Ω×R. Then we declare that w ̸= 0. Suppose to the contrary that w ≡ 0. Then it follows that wn → 0 in Lθ(Ω). By dividing (1.2) by ∥un∥2ϑ, we derive that E(un) ∥un∥2ϑ = 1 2 − 1 p∥un∥2−p ϑ ∫ Ω h(x)|wn|p dx− ∫ Ω G(x, un) u2n w2 n dx = on(1). From (3.8), we deduce that 1 2 = ∫ Ω G(x, un) u2n w2 n dx+ 1 p∥un∥2−p ϑ ∫ Ω h(x)|wn|p dx+ on(1) ≤M2 ∫ Ω w2 n dx+ ∥h∥∞ p∥un∥2−p ϑ ∫ Ω |wn|p dx+ on(1) → 0, leading to a contradiction. Thus, w ̸= 0 holds. We turn to prove that ∫ Ω g(x, un) un wnv dx→ ∫ Ω k4wv dx. (3.9) Set Φ = {x ∈ Ω|w(x) = 0}, Φ̃ = {x ∈ Ω|w(x) ̸= 0}. It follows from Hölder’s inequality and (3.7) that∣∣ ∫ Φ g(x, un) un wnv dx ∣∣ ≤ ∫ Φ ∣∣g(x, un) un ∣∣|wn||v|dx ≤M1 ∫ Φ |wn||v|dx ≤M1 (∫ Φ |wn|2 dx )1/2(∫ Φ |v|2 dx )1/2 . As wn → w in L2(Ω) and(∫ Φ |wn − w|2 dx )1/2 ≤ (∫ Ω |wn − w|2 dx )1/2 , it follows that wn → w in L2(Ω). Thus,∫ Φ g(x, un) un wnv dx→ 0 = ∫ Φ k4wv dx. Based on (3.6) and (3.7), we have ∣∣g(x, un) un wn ∣∣ ≤M1w0. An application of the dominated convergence theorem yields lim n→∞ ∫ Φ̃ g(x, un) un wnv dx = ∫ Φ̃ lim n→∞ g(x, un) un wnv dx. EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 13 Because ∥un∥ϑ → +∞, we obtain |un(x)| = |wn(x)| · ∥un∥ϑ → ∞ for x ∈ Φ̃. Consequently,∫ Φ̃ lim |un|→+∞ g(x, un) un wnv dx = ∫ Φ̃ k4wv dx, which leads to (3.9). Dividing both sides of (3.9) by ∥un∥ϑ yields∫ Ω (∇∆wn∇∆v + c1∆wn∆v − c2∇wn∇v) dx − 1 ∥un∥2−p ϑ ∫ Ω h(x)|wn|p−1v dx− ∫ Ω g(x, un) un wnv dx→ 0. (3.10) Given that 1 ∥un∥2−p ϑ → 0 as ∥un∥ϑ → +∞, we derive that 1 ∥un∥2−p ϑ ∫ Ω h(x)|wn|p−1v dx→ 0. It follows from (3.9) and (3.10) that∫ Ω (∇∆w∇∆φ+ c1∆w∆φ− c2∇w∇φ) dx = ∫ Ω k4wφdx. Hence, we conclude that w is a nontrivial solution to the problem (−∆)3w + c1∆ 2w + c2∆w = k4w. This contradicts that k4 ̸= µk(µ 2 k−c1µk−c2). Therefore, the assumption that ∥un∥ϑ → +∞ must be invalid. Consequently, the sequence {un} is bounded in H3 ϑ(Ω). Owing to the boundedness of un and the reflexivity of H3 ϑ(Ω), we can extract a subsequence {un} (still denoted by {un}) and a function u ∈ H3 ϑ(Ω) such that un ⇀ u in H3 ϑ(Ω), un → u in Lθ(Ω) ( 1 ≤ θ < 2N N − 6 ) . (3.11) In addition, one can find a constant M3 > 0 such that ∥un∥ϑ ≤M3. (3.12) In view of (3.4), we derive that ⟨E ′(un), un − u⟩ → 0 and ⟨E ′(u), un − u⟩ → 0. Therefore, ⟨E ′(un)− E ′(u), un − u⟩ = ∥un − u∥2ϑ − ∫ Ω h(x)|un − u|p dx− ∫ Ω (g(x, un)− g(x, u)) (un − u) dx→ 0. (3.13) It follows from (H2) and g ∈ C(Ω̄×R,R) that there are constants α10 > 0 and α11 > 0 satisfying |g(x, r)| ≤ α10|r|s + α11 for every (x, t) ∈ Ω× R. Applying Hölder’s inequality, we deduce that∣∣ ∫ Ω g(x, un)(un − u) dx ∣∣ ≤ ∫ Ω |g(x, un)||un − u|dx ≤ ∫ Ω (α10|un|s|un − u|+ α11|un − u|) dx ≤ ∫ Ω α10|un|s|un − u|dx+ ∫ Ω α11|un − u|dx ≤ α10∥un∥sLs+1∥un − u∥Ls+1 + α11|Ω| s s+1 ∥un − u∥Ls+1 ≤ α10Ss∥un∥sϑ∥un − u∥Ls+1 + α11|Ω| s s+1 ∥un − u∥Ls+1 . 14 Q. WEI, X. ZHANG EJDE-2025/86 Combining(3.11) and (3.12), we obtain∣∣∣∣∫ Ω g(x, un)(un − u) dx ∣∣∣∣ ≤ α10SsMs 3∥un − u∥Ls+1 + α11|Ω| s s+1 ∥un − u∥Ls+1 → 0, which indicates that ∣∣ ∫ Ω g(x, u)(un − u) dx ∣∣ → 0. Consequently, ∫ Ω (g(x, un)− g(x, u)) (un − u) dx→ 0. Analogously, it follows from Hölder’s inequality and (3.11) that∫ Ω h(x)|un − u|p dx→ 0. Furthermore, an application of (3.13) yields ∥un − u∥2ϑ → 0. Hence, we conclude that un → u in H3 ϑ(Ω), which implies that the (Ce) condition holds. □ Given σ > 0 as given in Lemma 2.3 (i), let Bσ = {u ∈ H3 ϑ(Ω), ∥u∥ϑ ≤ σ}, ∂Bσ = {u ∈ H3 ϑ(Ω), ∥u∥ϑ = σ}, which is a complete metric space endowed with the distance dist(u, v) = ∥u− v∥ϑ for u, v ∈ Bσ. According to Lemma 2.3, we have that E(u)|∂Bσ ≥ β > 0. It is obvious that E(u) ∈ C1(Bσ,R) and E(u) is lower semicontinuous. Hence, E(u) is bounded from below on Bσ. In fact, following the same approach as in the proof of Theorem 1.1, there is a function η(x) ∈ C3 0 (Ω) such that∫ Ω h(x)|η|p dx > 0. An application of (H3) yields g(x, r)r ≥ k1 2 r2 for every |r| small enough and a.e. x ∈ Ω, which implies G(x, r) = ∫ 1 0 g(x, vr)r dv ≥ k1 4 r2 for every |r| small enough and a.e. x ∈ Ω. Given r > 0, we infer that lim sup t→0 E(rη)r−p = lim sup r→0 (r2 2 ∥η∥2ϑ − rp p ∫ Ω h(x)|η|p dx− ∫ Ω G(x, rη) dx ) r−p ≤ lim sup r→0 ( t2−p 2 ∥η∥2ϑ − 1 p ∫ Ω h(x)|η|p dx− k1r 2−p 4 ∫ Ω |η|2 dx ) < 0. We define α̃ = inf{E(u)|u ∈ Bσ}, which implies α̃ < 0. It follows from Proposition 2.6 that for each k > 0, there exists a sequence {uk} such that α̃ ≤ E(uk) ≤ α̃+ 1 k . We now show that ∥uk∥ϑ < σ for k large enough. Suppose, on the contrary, that ∥uk∥ϑ = σ for infinitely many k. For simplicity, we take ∥uk∥ϑ = σ for every k ≥ 1. An application of Lemma 2.3 yields that E(uk) ≥ β > 0. EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 15 Taking k → ∞, we obtain the inequality 0 > α̃ ≥ β > 0, which leads to a contradiction. Indeed, for arbitrary u ∈ H3 ϑ(Ω) with ∥u∥ϑ = 1, we define wk = uk + ru, for each fixed k ≥ 1. Then it follows that ∥wk∥ϑ ≤ ∥uk∥ϑ + r, which yields that wk ∈ Bσ for r > 0 small enough. Hence, E(wk) = E(uk + ru) ≥ E(uk)− r k ∥u∥ϑ, which indicates that E ′(uk) = lim r→0+ E(uk + ru)− E(uk) r ≥ −1 k , E ′(uk) = lim r→0+ E(uk − ru)− E(uk) r ≤ 1 k . It follows that |E ′(uk)| ≤ 1 k → 0, E(uk) → α̃ as k → ∞, which implies that {uk} is a (Ce)α̃ sequence. According to Lemma 3.1, E fulfills the (Ce) condition. Therefore, there exists a function u1 ∈ H3 ϑ(Ω) satisfying E ′(u1) = 0. So we conclude that u1 is a nontrivial weak solution of (1.1) and E(u1) = α̃ < 0. Using Lemma 2.5, the second critical point u2 for E can be found satisfying E(u2) = inf ϕ∈B max r∈(0,1) E(ϕ(r)), where B = {ϕ ∈ C0([0, 1], H3 ϑ(Ω)), ϕ(0) = 0, ϕ(1) = e}. It follows that E(u2) ≥ β > 0, which completes the proof of Theorem 1.3. 3.4. Proof of Theorem 1.4. We start by showing that E fulfills the (Ce) condition under the assumptions of Theorem 1.3. Lemma 3.2. Under the hypotheses of Theorem 1.4, the functional E satisfies the (Ce) condition. Proof. Let {un} ⊂ H3 ϑ(Ω) be a sequence that satisfies the (Ce)α condition for α > 0, E(un) → α and (1 + ∥un∥ϑ)E ′(un) → 0 as n→ ∞. Thus, we obtain ⟨E ′(un), un⟩ = ∥un∥2ϑ − ∫ Ω h(x)|un|p dx− ∫ Ω g(x, un)un dx = on(1) (3.14) and E(un) = 1 2 ∥un∥2ϑ − 1 p ∫ Ω h(x)|un|p dx− ∫ Ω G(x, un) dx = c+ on(1). (3.15) Next, we demonstrate that the sequence {un} is bounded in H3 ϑ(Ω). If not, there exists a subse- quence (still denoted by {un}) satisfying ∥un∥ϑ → ∞ as n→ ∞. It follows from condition (H8) that, for arbitrary given ϵ > 0, there is a constant R1 > 0 satisfying g(x, r)r ≤ (λi + ε)r2 (3.16) for every |r| ≥ R1 and a.e. x ∈ Ω. An application of condition (H9) yields that for 0 < δ < k5 − ( 2p − 1)∥h∥∞, there is a constant R2 > 0 such that g(x, r)r − 2G(x, r) ≥ (k5 − δ)|r|p (3.17) 16 Q. WEI, X. ZHANG EJDE-2025/86 for every |r| ≥ R2 and a.e. x ∈ Ω. Let R = max{R1,R2}, and assume that Dn = {x ∈ Ω, |un(x)| ≥ R}, D̄n = {x ∈ Ω, |un(x)| ≤ R}. It follows from the definition of D̄n and h(x) ∈ L∞(Ω) that we can find a constant α12 > 0 satisfying ∫ D̄n h(x)|un|p dx ≤ α12. (3.18) In addition, employing the continuity of g and G, we obtain that there is a constant α13 > 0 such that ∣∣ ∫ D̄n (g(x, un)un − 2G(x, un)) dx ∣∣ ≤ α13. (3.19) According to (3.14) and (3.15), we infer that 2c+ on(1) = ( 1− 2 p ) ∫ Ω h(x)|un|p dx+ ∫ Ω (g(x, un)un − 2G(x, un)) dx. Based on (3.18) and (3.19), it follows that 2c+ on(1) ≥ ( 1− 2 p ) ∫ Dn h(x)|un|p dx+ ( 1− 2 p ) α12 + ∫ Dn (g(x, un)un − 2G(x, un)) dx− α13. Therefore, we can find a constant α14 > 0 satisfying α14 ≥ ( 1− 2 p ) ∫ Dn h(x)|un|p dx+ ∫ Dn (g(x, un)un − 2G(x, un)) dx. From (3.17), we deduce α14 ≥ ( 1− 2 p ) ∫ Dn h(x)|un|p dx+ ∫ Dn (k5 − δ)|un|p dx ≥ (( 1− 2 p ) ∥h∥∞ + k5 − δ )∫ Dn |un|p dx. As ( 1− 2 p ) ∥h∥∞ + k5 − δ > 0, we have that there is a constant α15 > 0 satisfying∫ Dn |un|p dx ≤ α15. (3.20) It follows from the definition of D̄n and g ∈ C(Ω̄ × R,R) that there is a constant α16 > 0 such that ∣∣ ∫ D̄n g(x, un)un dx ∣∣ ≤ α16. Owing to (3.14), we derive ∥un∥2ϑ − on(1) = ∫ Ω h(x)|un|p dx+ ∫ Ω g(x, un)un dx ≤ ∥h∥∞α15 + α12 + α16 + ∫ Dn g(x, un)un dx. (3.21) An application of (3.16) yields∫ Dn g(x, un)un dx ≤ (λi + ε) ∫ Dn u2n dx. Because p < 2, we can choose t = (p−2)N (N−6)p−2N ∈ (0, 1) satisfying 1 2 = 1−r p + (N−6)r 2N . Combining Hölder’s inequality and (3.20), we obtain∫ Dn u2n dx ≤ (∫ Dn |un|p dx ) 2(1−r) p (∫ Dn |un| 2N N−6 dx ) r(N−6) N ≤ α 2(1−r) p 15 ∥un∥2r L 2N N−6 . EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 17 Utilizing the Sobolev embedding theorem, we find that∫ Dn u2n dx ≤ S2rα 2(1−r) p 15 ∥un∥2rϑ . Letting α17 = (λi + ε)S2rα 2(1−r) p 15 , and based on (3.21), we conclude that ∥un∥2ϑ − on(1) ≤ ∥h∥∞α15 + α12 + α16 + α17∥un∥2rϑ . It follows from r < 1 that the assumption is invalid. Therefore, we obtain that {un} is bounded in H3 ϑ(Ω). Given that g(x, r) is subcritical according to condition (H2), and employing the compactness of the Sobolev embedding and the same approach as in the proof of Lemma 3.1, we have un → u in H3 ϑ(Ω). Therefore, E fulfills the (Ce) condition. □ It follows from Lemma 2.3 that the mountain geometrical structure exists. Then, an application of Lemma 3.2 yields that E satisfies the (Ce) condition. Consequently, by a similar method as in Theorem 1.3, we obtain two nontrivial solutions. 3.5. Proof of Theorem 1.5. First we prove a relevant lemma. Lemma 3.3. Under the hypotheses of Theorem 1.5, the functional E satisfies the (Ce) condition. Proof. Suppose that the sequence {un} ⊂ H3 ϑ(Ω) satisfies the (Ce)α condition for a positive constant α, E(un) → α and (1 + ∥un∥ϑ)E ′(un) → 0 as n→ ∞. It follows that ⟨E ′(un), un⟩ = ∥un∥2ϑ − ∫ Ω h(x)|un|p dx− ∫ Ω g(x, un)un dx = on(1) (3.22) and E(un) = 1 2 ∥un∥2ϑ − 1 p ∫ Ω h(x)|un|p dx− ∫ Ω G(x, un) dx = c+ on(1). (3.23) Next, we prove that the sequence {un} is bounded in H3 ϑ(Ω). Suppose, for contradiction, that there exists a subsequence (still denoted by un) such that ∥un∥ϑ → +∞ as n→ ∞. According to (H2), for every ϵ > 0, one can find R1 > 0 satisfying g(x, r) ≤ ϵ|r|s (3.24) for each |r| ≥ R1 and a.e. x ∈ Ω. In view of assumption (H11), for 0 < δ < k6, the re is a constant R2 > 0 such that g(x, r)r − 2G(x, r) ≥ (k6 − δ)|r|κ (3.25) for each |r| ≥ R2 and a.e. x ∈ Ω. Let R = max{R1,R2}, and define Dn = {x ∈ Ω, |un(x)| ≥ R}, D̄n = {x ∈ Ω, |un(x)| ≤ R}. It follows from the definition of D̄n and h(x) ∈ L∞(Ω) that there is a constant α18 > 0 such that∫ D̄n h(x)|un|p dx ≤ α18. (3.26) Furthermore, using the continuity of g and G, we can find a constant α19 > 0 such that∣∣∣ ∫ D̄n (g(x, un)un − 2G(x, un)) dx ∣∣∣ ≤ α19. (3.27) According to (3.22) and (3.23), we derive 2c+ on(1) = ( 1− 2 p ) ∫ Ω h(x)|un|p dx+ ∫ Ω (g(x, un)un − 2G(x, un)) dx. 18 Q. WEI, X. ZHANG EJDE-2025/86 Based on (3.26) and (3.27), it follows that 2c+ on(1) ≥ ( 1− 2 p ) ∥h(x)∥∞ ∫ Dn |un|p dx+ ( 1− 2 p ) α18 + ∫ Dn (g(x, un)un − 2G(x, un)) dx− α19. An application of (3.25) yields that there is a constant α20 > 0 such that α20 ≥ ( 1− 2 p ) ∥h(x)∥∞ ∫ Dn |un|p dx+ ∫ Dn (k6 − δ)|un|κ dx. It follows from Hölder’s inequality that α20 ≥ ( 1− 2 p ) ∥h∥∞|Dn|1− p κ (∫ Dn |un|κ dx )p/κ + ∫ Dn (k6 − δ)|un|κ dx. Owing to 2 p > 1, k6 − δ > 0 and p κ < 1, there is a constant α21 > 0 such that∫ Dn |un|κ dx ≤ α21. (3.28) Utilizing the definition of D̄n and g ∈ C(Ω× R,R), one can find a constant α22 > 0 satisfying∣∣ ∫ D̄n g(x, un)un dx ∣∣ ≤ α22. From (3.22), we deduce that ∥un∥2ϑ − on(1) = ∫ Ω h(x)|un|p dx+ ∫ D̄n g(x, un)un dx+ ∫ Dn g(x, un)un dx ≤ ∥h∥∞Sp∥un∥pϑ dx+ α22 + ∫ Dn g(x, un)un dx. Substituting ϵ < 1 into (3.24), it follows that∫ Dn g(x, un)un dx ≤ ∫ Dn |un|1+s dx. Considering that N 6 (s− 1) < 2N N+6s < s+1, we divide the discussion into the following two cases. Case 1: κ ≥ s+ 1. Applying Hölder’s inequality, we have∫ Dn |un|1+s dx ≤ (∫ Dn |un|κ dx )N+6 2N (∫ Dn |un| 2N(1+s)−(N+6)κ N−6 dx )N−6 2N . Owing to the fact that κ > 2N N+6s, we obtain 2N(1 + s)− (N + 6)κ N − 6 < 2N N − 6 , which implies∫ Dn |un|1+s dx ≤ (∫ Dn |un|κ dx )N+6 2N (∫ Dn |un| 2N N−6 dx )N−6 2N ≤ α N+6 2N 21 S∥un∥ϑ. Therefore, ∥un∥2ϑ − on(1) ≤ ∥h∥∞Sp∥un∥pϑ + C4 + α N+6 2N 21 S∥un∥ϑ, which yields that ∥un∥ϑ ≤ C. Case 2. N 6 (s− 1) < κ < s+ 1. As κ < s+ 1, one can choose r ∈ (0, 1) satisfying 1 s+ 1 = 1− s κ + (N − 6)r 2N . An application of Hölder’s inequality yields∫ Dn |un|1+s dx ≤ (∫ Dn |un|κ dx ) (1−r)(s+1) κ (∫ Dn |un| 2N N−6 dx ) r(s+1)(N−6) 2N EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 19 ≤ α (1−r)(s+1) κ 21 Sr(s+1)∥un∥r(s+1) ϑ . Then it follows that ∥un∥2ϑ − on(1) ≤ ∥h∥∞Sp∥un∥pϑ + C4 + α (1−r)(s+1) κ 21 Sr(s+1)∥un∥r(s+1) ϑ . Since κ > N 6 (s− 1) implies r(s+ 1) < 2, this contradicts our assumption. It follows that {un} is bounded in H3 ϑ(Ω). Given that g(x, r) is subcritical according to condition (H2), and employing the compactness of the Sobolev embedding and the same approach as in the proof of Lemma 3.1, we infer that un → u in H3 ϑ(Ω). Therefore, E satisfies the (Ce) condition. □ It follows from Lemma 2.3 the existence of the mountain geometrical structure. Hence, an application of Lemma 3.3 yields that E satisfies the (Ce) condition. Consequently, by a similar method to the one in Theorem 1.3, we can obtain two nontrivial solutions. Remark 3.4. We believe that the results in Theorems 1.1–1.5 are also valid for the higher-order elliptic equation (−∆)mu+ c1∆ m−1u+ · · ·+ cm−1∆u = h(x)|u|p−2u+ g(x, u) in Ω, u = ∆u = · · · = ∆m−1u = 0 on ∂Ω, where m ≥ 2 is a positive integer, (−∆)m(·) = −∆((−∆)m−1(·)) denotes the polyharmonic operator; Ω ⊂ RN (N ≥ 2m+1) stands for a smooth bounded domain; ci ∈ R for i ∈ {1, 2, . . . ,m− 1}. But we can not verify them right now. Acknowledgments. 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Zhang; Infinitely many nontrivial solutions for a class of biharmonic equations via variant fountain theo- rems. Electron. J. Qual. Theory Differ. Equ., 9 (2011), 1-14. [46] J. Zhang, L. You; Fast surface modelling using a 6th order pde. Comput. Graph. Forum, 23 (2004), 311-320 [47] J. Zhou, X. Wu; Sign-changing solutions for some fourth-order nonlinear elliptic problems. J. Math. Anal. Appl., 342 (2008), 542-558. Qifan Wei School of Mathematics and Physics, North China Electric Power University, Beijing 102206, China Email address: weiqifan0924@163.com EJDE-2025/86 SOLUTIONS TO TRIHARMONIC PROBLEMS 21 Xuemei Zhang School of Mathematics and Physics, North China Electric Power University, Beijing 102206, China Email address: zxm74@sina.com 1. Introduction and main results 2. Preliminaries 3. Proof of main results 3.1. Proof of Theorem ?? 3.2. Proof of Theorem ?? 3.3. Proof of Theorem ?? 3.4. Proof of Theorem ?? 3.5. Proof of Theorem ?? Acknowledgments References