Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 88, pp. 1–23. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.23 MULTIPLE SOLUTIONS FOR PARAMETRIC WEIGHTED (p, q)-EQUATIONS XIAOHUI ZHANG, XIAN XU Abstract. In this article, we prove that equations driven by a weighted (p, q)-Laplacian have at least two positive solutions, two negative solutions, and two sign-changing solutions. To obtain these result, we construct an operator that has invariant sets consisting of supersolustions and subsolutions. Then using this operator, we find a locally Lipschitz continuous operator and use it to construct a descending flow. Finally, by the method of invariant sets of descending flow, we obtain the 6 solutions stated above. 1. Introduction Let Ω ⊂ RN be a bounded domain with a C2-boundary ∂Ω. We study the parametric weighted (p, q)-equation −∆a1 p u(z)−∆a2 q u(z) = λ|u(z)|s−2u(z) + f(z, u(z)) in Ω, u ∣∣ ∂Ω = 0, 1 < s < q, 2 ⩽ q ⩽ p < p∗, λ > 0. (1.1) Given a a ∈ C0,1(Ω) and r ∈ (1,+∞), by ∆a ru(z), we denote the weighted r-Laplace differential operator ∆a ru(z) = div ( a(z)|Du|r−2Du ) ∀u ∈W 1,r 0 (Ω). In problem (1.1) we have the sum of two such operators. Many people have studied (p, q)-Laplacian equations (see [8, 14, 17, 18, 19, 20, 24, 25, 26]). Recently Papageorgiou and Scapellato [24] studied the positive and nodal solutions for weighted (p, q)-Laplacian equations. They proved global existence and multiplicity results. Papageorgiou, Qin and Rădulescu [18] proved the existence of infinitely many nodal solutions under symmetry conditions. Wu, Guo and Winkert [14] showed multiplicity of solutions for surperlinear (p, q)-equations in symmetrical domains by Lusternik- Schnirelmann category. The nonlinearity of (1.1) is a combination of convex and concave terms. Many people have studied this type of equations. For example, Ambrosetti, Brezisl and Cerami in their well known paper [1] considered the boundary value problem −∆u = λuq + up in Ω, u = 0 on ∂Ω. (1.2) They obtained the following results in [1], Theorem 1.1. Let 0 < q < 1 < p. Then there exists Λ > 0 such that (1) for all λ ∈ (0,Λ), (1.2) has two positive solutions; (2) for λ = Λ, (1.2) has at least one positive solution; (3) for all λ > Λ, (1.2) has no positive solutions. 2020 Mathematics Subject Classification. 35D30, 35J60, 35J92, 47K10, 58R05. Key words and phrases. Invariant sets with descending flow; parametric weighted (p, q)-equations multiple solutions. ©2025. This work is licensed under a CC BY 4.0 license. Submitted May 7, 2025. Published September 11, 2025. 1 2 X. ZHANG, X. XU EJDE-2025/23 Li and Wang [12] studied the multiple solutions of the boundary value problem −∆u = λ|u|q−2u+ g(u) in Ω, u = 0 on ∂Ω, (1.3) where g ∈ C1(R,R), g(u) = o(|u|) at 0 and g′(u) ⩾ −a for some a > 0. They obtained the existence of at least two positive solutions, at least two negative solutions and at least two sign-changing solutions. The main purpose of this paper is to investigate the solutions of equation (1.1) by the method of descending flow invariant sets. As far as we know, up to now, only few people have used the method of descending flow invariant sets to study the multiplicity of solutions of (p, q)-equation (see [17, 8, 26]). The main results of this paper generalize some results in [12]. A key challenge in this approach lies in identifying upper and lower solutions for equation (1.1) and constructing an appropriate pseudogradient vector field to ensure that certain sets related to upper and lower solutions are descending flow invariant. So we first find two supersolutions and two subsolutions for equation (1.1). Then we construct a compact operator, which ensures some sets with respect to supersolutions and subsolutions being invariant. Using this operator, we can obtain a locally Lipschitz continuous operator, which is used to construct a vector field. Finally, using the method of invariant sets of descending flow, we obtain the result of six solutions. 2. Main results Let X=W 1,p 0 (Ω) and Y=C1 0 (Ω), X ∗ be the topological dual of X and ⟨·, ·⟩X∗,X denote the duality pairing between X∗ and X. Let P be a closed convex cone of X, that is P = { u ∈W 1,p 0 (Ω) : u(z) ≥ 0 a.a. z ∈ Ω } . Let P1 = P ∩ Y and −P1 = −P ∩ Y . Then P1 = { u ∈ C1 0 (Ω) : u(z) ≥ 0 for all z ∈ Ω } . P1 has a nonempty interior in the Y topology and its interior in the Y topology is defined by intY P1 = { u ∈ C1 0 (Ω) : u(z) > 0 for all z ∈ Ω, ∂u ∂n ∣∣ ∂Ω < 0 } with ∂u ∂n = (Du, n)RN and n is the outward unit normal on ∂Ω. Let ∂YA be the boundary of A in Y if A ⊂ Y . We introduce the following conditions: (H0) a1, a2 ∈ C0,1(Ω), a1(z) ⩾ ĉ > 0 and a2(z) ⩾ 0 for all z ∈ Ω. (H1) f : Ω× R → R is a Carathéodory function such that f(z, 0) = 0 for a.a. z ∈ Ω and (i) |f(z, x)| ⩽ â(z)[1 + |x|r−1] for a.a. z ∈ Ω, all x ∈ R, with â ∈ L∞(Ω), p < r < p∗, (ii) let F (z, x) = ∫ x 0 f(z, s)ds and there exist m > p and M > 0 such that f(z, x)x ⩾ mF (z, x) for a.a. z ∈ Ω, all |x| ⩾M, (iii) f(z, x) is monotonically increasing in x ∈ R, for a.a. z ∈ Ω, (iv) there exist δ0 > 0 and τ ∈ (p, p∗) such that f(z, x)x ⩽ c0|x|τ for a.a. z ∈ Ω, all |x| ⩽ δ0, some c0 > 0. Remark 2.1. Condition (H1) (iii) can be replaced by a more general monotonicity condition, see [3, condition (H3”)]. Let ∥u∥X = (∫ Ω |Du|pdz )1/p , ∥u∥r = (∫ Ω |u|rdz )1/r be the standard norms of W 1,p 0 (Ω), and Lr(Ω) for r ⩾ 1, respectively. For λ > 0, we introduce the functional Jλ : X → R as Jλ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − λ s ∫ Ω |u|sdz − ∫ Ω F (z, u)dz. EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 3 Evidently, Jλ ∈ C1(X,R). Then we have ⟨J ′ λ(u), v⟩X∗,X = ∫ Ω a1(z)|Du|p−2⟨Du,Dv⟩RNdz + ∫ Ω a2(z)|Du|q−2⟨Du,Dv⟩RNdz − λ ∫ Ω |u|s−2uvdz − ∫ Ω f(z, u)vdz. (2.1) Let Aa1 p :W 1,p 0 (Ω) 7→W−1,p′ (Ω) and Aa2 q :W 1,q 0 (Ω) 7→W−1,q′(Ω) be defined by ⟨Aa1 p (u), v⟩X∗,X = ∫ Ω a1(z)|Du(z)|p−2⟨Du(z), Dv(z)⟩RNdz for u, v ∈W 1,p 0 (Ω), ⟨Aa2 q (u), v⟩X∗,X = ∫ Ω a2(z)|Du(z)|q−2⟨Du(z), Dv(z)⟩RNdz for u, v ∈W 1,q 0 (Ω), where 1 p + 1 p′ = 1 and 1 q + 1 q′ = 1. Let V (u) = Aa1 p (u) +Aa2 q (u) for all u ∈W 1,p 0 (Ω). Proposition 2.2 ([7]). The operator V (·) is bounded, continuous, strictly monotone and of type (S)+, i.e., if {un} is a sequence in W 1,p 0 (Ω) such that un ⇀ u in W 1,p 0 (Ω) and lim sup n→∞ ⟨V (un), un − u⟩X∗,X ⩽ 0, then un → u in W 1,p 0 (Ω). We introduce the following sets: L + = {λ > 0 : problem (1.1) has at least a positive solution}, L − = {λ > 0 : problem (1.1) has at least a negative solution}, S+ λ = {u : u is a positive solution of (1.1)}, S− λ = {u : u is a negative solution of (1.1)}. Then we have the following main result. Theorem 2.3. Suppose that (H0), (H1) hold. Then there exists λ > 0 such that for 0 < λ < λ, (1.1) has at least two positive solutions, two negative solutions, and two sign-changing solutions. Figure ?? shows the positions of six solutions. D1 D2 D3 G1 G2 u1 u2 u3 u4 u5 u6 O Figure 1. Corollary 2.4. If a2(z) = 0, then problem (1.1) is transformed into −∆a1 p u(z) = λ|u(z)|s−2u(z) + f(z, u(z)) in Ω, u ∣∣ ∂Ω = 0, 1 < s < q, 2 ⩽ q ⩽ p < p∗, λ > 0, (2.2) which is a p-Laplacian equation. Problem (2.2) has at least two positive solutions, two negative solutions, and two sign-changing solutions. 4 X. ZHANG, X. XU EJDE-2025/23 Corollary 2.5. If a2(z) = 0, a1(z) = 1 and p = 2, then problem (1.1) is transformed into −∆u(z) = λ|u(z)|s−2u(z) + f(z, u(z)) in Ω, u ∣∣ ∂Ω = 0, 1 < s < 2, λ > 0, (2.3) Then problem (2.3) has at least two positive solutions, two negative solutions, and two sign- changing solutions. To show Theorem 2.3 we need to give some Lemmas. Let’s recall some facts about the spectrum of the s-Laplacian with Dirichlet boundary condition. Consider the nonlinear eigenvalue problem −∆su(z) = λ|u(z)|s−2u(z) in Ω, u = 0 on ∂Ω. (2.4) We say that λ̂ is an eigenvalue of (−∆s,W 1,s 0 (Ω)) if the above problem has a nontrivial solution û, known as an eigenfunction corresponding to λ̂. Let λ̂1(s) be the smallest eigenvalue. Then it is positive, isolated, simple and satisfies λ̂1(s) = inf {∥Du∥ss ∥u∥ss : u ∈W 1,s 0 (Ω), u ̸= 0 } . By û1(s) we denote the positive, Ls-normalized eigenfunction corresponding to λ̂1(s). Lemma 2.6. If hypotheses (H0) and (H1) hold, then L + ̸= ∅ and, for any λ ∈ L +, S+ λ ⊂ intY P1. Proof. For each λ > 0, we consider the C1-functional ϑλ :W 1,p 0 (Ω) → R defined by ϑλ(u) = 1 p ∫ Ω a1(z)|Du|p dz + 1 q ∫ Ω a2(z)|Du|q dz − λ s ∫ Ω (u+)s dz − ∫ Ω F (z, u+) dz (2.5) for u ∈W 1,p 0 (Ω). Hypotheses (H1) (i) and (iv) imply that F (z, x) ⩽ d1(x τ + xr) for a.a. z ∈ Ω, ∀x ⩾ 0, some d1 > 0. (2.6) Using (2.6), we have ϑλ(u) ⩾ 1 p ĉ∥u∥pX − λ s ∫ Ω |u+|s dz − d1 ∫ Ω (|u+|τ + |u+|r) dz ⩾ 1 p ĉ∥u∥pX − λ s d2∥u∥sX − d3∥u∥τX − d4∥u∥rX , (2.7) for all u ∈W 1,p 0 (Ω), where d1, d2, d3, d4 > 0. Let ρ = ∥u∥ and choose α ∈ (0, 1 p−s ). We set ρ = λα and then from (2.7) we have ϑλ(u) ⩾ 1 p ĉλαp − d2 s λαs+1 − d3λ ατ − d4λ αr = [ ĉ p − d2 s λ1−α(p−s) − d3λ α(τ−p) − d4λ α(r−p) ] λαp. (2.8) Note that α(p − s) < 1 and so if λ → 0+, then d2 s λ 1−α(p−s) + d3λ α(τ−p) + d4λ α(r−p) → 0+. So, from (2.8) it follows that we can find λ0 > 0 such that ϑλ(u) ⩾ mλ > 0 for all ∥u∥X = ρλ = λα, λ ∈ (0, λ0). (2.9) On account of hypothesis (H1), we obtain F (z, x) ⩾ 0 for a.a. z ∈ Ω, all x ⩾ 0. Then we can find t ∈ (0, 1) small such that ϑ(tû1(s)) ⩽ tp p ∥a1∥∞∥Dû1(s)∥pp + tq q ∥a2∥∞∥Dû1(s)∥qq − λ s ts < 0. Using the Sobolev embedding theorem, we see that ϑλ(·) is sequentially weakly lower semicon- tinuous. Let Bλ = {u ∈ W 1,p 0 (Ω) : ∥u∥X ⩽ ρλ}, then from the reflexivity of W 1,p 0 (Ω) and using EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 5 the Eberlein-Smulian theorem [27] we have that Bλ is sequentially weakly compact. So, by the Weierstrass-Tonelli theorem, we can find uλ ∈ Bλ such that ϑλ(uλ) = inf[ϑλ(u) : u ∈ Bλ] =⇒ ϑλ(uλ) < 0 = ϑλ(0) =⇒ uλ ̸= 0. Then, from (2.9) we see that 0 < ∥uλ∥X < ρλ, =⇒ ϑ′λ(uλ) = 0, =⇒ ⟨V (uλ), h⟩X∗,X = λ ∫ Ω |u+λ | s−1h dz + ∫ Ω f(z, u+λ )h dz for all h ∈W 1,p 0 (Ω). (2.10) Here we choose the test function h = −u−λ ∈W 1,p 0 (Ω) and obtain ĉ∥Du−λ ∥ p p ⩽ 0 =⇒ uλ ⩾ 0, uλ ̸= 0. (2.11) So, uλ is a positive solution of (1.1), hence (0, λ0) ⊆ L + ̸= ∅. For u ∈ S+ λ , we have −∆a1 p u−∆a2 q u = λ|u|s−2u+ f(z, u) in Ω, u ∣∣ ∂Ω = 0. From [11, Theorem 7.1], we have that u ∈ L∞(Ω). Then, the nonlinear regularity theory of Lieberman [13] implies that u ∈ P1 \{0}. Invoking [24, Proposition 2.2], we have that u ∈ intY P1. Therefore S+ λ ⊆ intY P1. The proof is complete. □ Lemma 2.7. If (H0), (H1) hold, λ ∈ L +, uλ ∈ S+ λ , and µ ∈ (0, λ), then µ ∈ L + and there exists uµ ∈ S+ µ ⊆ intY P1 such that uµ ⩽ uλ. Proof. We introduce the Carathéodory function kµ(z, x) = { µ|x+|s−2x+ + f(z, x+) if x ⩽ uλ(z) µ|uλ(z)|s−2uλ(z) + f(z, uλ(z)) if uλ(z) < x, (2.12) where x+ = max{0, x}. Let Kµ(z, x) = ∫ x 0 kµ(z, s)ds and consider the C1-functional σµ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − ∫ Ω Kµ(z, u)dz for all u ∈W 1,p 0 (Ω). From (2.12) it is clear that σµ(·) is coercive. Also it is sequentially weakly lower semicontinuous. So we can find uµ ∈W 1,p 0 (Ω) such that σµ(uµ) = inf[σµ(u) : u ∈W 1,p 0 (Ω)]. (2.13) We see that for t ∈ (0, 1) small, we have σµ(tuλ(z)) ⩽ tp p ∥a1∥∞ ∫ Ω |Duλ(z)|pdz + tq q ∥a2∥∞ ∫ Ω |Duλ(z)|qdz − µ s ts ∫ Ω (uλ(z)) sdz − ∫ Ω F (z, tuλ(z))dz < 0. (2.14) So we have that σµ(uµ) < 0 = σµ(0) =⇒ uµ ̸= 0. (2.15) From (2.13), we have σ′ µ(uµ) = 0 =⇒ ⟨V (uµ), h⟩X∗,X = ∫ Ω kµ(z, uµ)hdz. (2.16) Choosing h = −u−µ ∈W 1,p 0 (Ω), we obtain ĉ∥Du−µ ∥pp ⩽ 0 =⇒ uµ ⩾ 0, uµ ̸= 0. (2.17) 6 X. ZHANG, X. XU EJDE-2025/23 In (2.16) we choose h = (uµ − uλ) + ∈W 1,p 0 (Ω). Then we have ⟨V (uµ), (uµ − uλ) +⟩X∗,X = ∫ Ω kµ(z, uµ)(uµ − uλ) +dz = ∫ Ω (µ|uλ|s−2uλ + f(z, uλ))(uµ − uλ) +dz ⩽ ∫ Ω (λ|uλ|s−2uλ + f(z, uλ))(uµ − uλ) +dz = ⟨V (uλ), (uµ − uλ) +⟩X∗,X , =⇒ uµ ⩽ uλ. (2.18) We have proved that uµ ∈ [0, uλ], uµ ̸= 0. (2.19) Then, (2.19), (2.12) and (2.16) imply that uµ ∈ S+ µ ⊆ intY P1. The proof is complete. □ Remark 2.8. Lemma 2.7 implies that L + is an interval. From (H1), we can see that f(z, x) ⩾ 0textfor a.a.z ∈ Ω, all x ⩾ 0. Based on this property of f(z, x), we consider the auxiliary Dirichlet problem −∆a1 p u(z)−∆a2 q u(z) = λ(u(z))s−1 in Ω, u ∣∣ ∂Ω = 0, u ⩾ 0, λ > 0. (2.20) For this problem, we have the following existence and uniqueness result. Lemma 2.9. If hypotheses (H0), (H1) hold, then for every λ > 0 problem (2.20) has a unique positive solution uλ. Proof. First we show the existence of a positive solution. To this end, we consider the C1-functional φλ :W 1,p 0 (Ω) → R defined by φλ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − λ s ∫ Ω |u+|sdz for all u ∈W 1,p 0 (Ω). Since s < q ⩽ p, it is clear that φλ(·) is coercive. Using the Sobolev embedding theorem, we see that φλ(·) is weakly lower semicontinuous. So, we can find uλ ∈W 1,p 0 (Ω) such that φλ(uλ) = inf [ φλ(u) : u ∈W 1,p 0 (Ω) ] . (2.21) We can see that for t ∈ (0, 1) small, we have φλ(tû1(s)) ⩽ tp p ∥a1∥∞ ∫ Ω |Dû1(s)|pdz + tq q ∥a2∥∞ ∫ Ω |Dû1(s)|qdz − λ s ts < 0. So we have that φλ(uλ) < 0 = φλ(0) =⇒ uλ ̸= 0. (2.22) From (2.21), we have φ′ λ(uλ) = 0 =⇒ ⟨V (uλ), h⟩X∗,X = λ ∫ Ω |u+λ | s−1hdz. (2.23) Choosing h = −u−λ ∈W 1,p 0 (Ω), we obtain ĉ∥Du−λ ∥ p p ⩽ 0 =⇒ uλ ⩾ 0, uλ ̸= 0. (2.24) EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 7 Therefore uλ is a positive solution of (2.20). Then, the nonlinear regularity theory and Proposition 2.2 [24] imply uλ ∈ intY P1. Now we show the uniqueness of this positive solution. To this end, we introduce the integral functional j : L1(Ω) → R = R ∪ {+∞} defined by j(u) = { 1 p ∫ Ω a1(z)|Du1/q|pdz + 1 q ∫ Ω a2(z)|Du1/q|qdz if u ⩾ 0, u1/q ∈W 1,p 0 (Ω) +∞ otherwise (2.25) From Lemma 1[5], we know that j(·) is convex. Suppose that vλ ∈ W 1,p 0 (Ω) is an other positive solution of (2.20). Again we have vλ ∈ intY P1. So using [21, Proposition 4.1.22, p. 274], we have uλ vλ , vλ uλ ∈ L∞(Ω). If we let h = uqλ − vqλ, then for |t| < 1 small we have uqλ + th, vqλ + th ∈ dom j = {u ∈ L1(Ω) : j(u) <∞}. So, exploiting the convexity of j(·), we see that j(·) is Gâteaux differentiable at uqλ and at vqλ in the direction h. Using the nonlinear Green’s identity, we have j′(uqλ)(h) = 1 q ∫ Ω −∆a1 p uλ −∆a2 q uλ uq−1 λ hdz = 1 q ∫ Ω λus−q λ hdz, j′(vqλ)(h) = 1 q ∫ Ω −∆a1 p vλ −∆a2 q vλ vq−1 λ hdz = 1 q ∫ Ω λvs−q λ hdz. The convexity of j(·) implies the monotonicity of j′(·). Therefore, 0 ⩽ ⟨j′(uqλ)− j′(vqλ), u q λ − vqλ⟩ = 1 q ∫ Ω λ(us−q λ − vs−q λ )(uqλ − vqλ) ⩽ 0, =⇒ uλ = vλ. This proves the uniqueness of the positive solution uλ ∈ intY P1 of (2.20) for λ > 0. The proof is complete. □ Lemma 2.10. If hypotheses (H0) and (H1) hold for λ ∈ L +, then uλ ⩽ u for all u ∈ S+ λ . Proof. By Lemma 2.1, we can infer that S+ λ ⊆ intY P1. Let u ∈ S+ λ and consider the Carathéodory function lλ(z, x) = { λ|x+|s−1 if x ⩽ u(z) λu(z)s−1 if u(z) < x, (2.26) where x+ = max{0, x}. We set Lλ(z, x) = ∫ x 0 lλ(z, s)ds and consider the C1-functional ψλ : W 1,p 0 (Ω) → R defined by ψλ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − ∫ Ω Lλ(z, u)dz for u ∈W 1,p 0 (Ω). It is clear that ψλ(·) is coercive. Also, it is weakly lower semicontinuous. So, we can find ũλ ∈ W 1,p 0 (Ω) such that ψλ(ũλ) = inf[ψλ(u) : u ∈W 1,p 0 (Ω)]. (2.27) Since u ∈ intY P1, we can find t ∈ (0, 1) small such that tû1(s) ⩽ u. Also, as before, choosing t ∈ (0, 1) even smaller if necessary, we have ψλ(tû1(s)) < 0 =⇒ ψλ(ũλ) < 0 = ψλ(0) =⇒ ũλ ̸= 0. From (2.27), we have ψ′ λ(ũλ) = 0 =⇒ ⟨V (ũλ), h⟩X∗,X = ∫ Ω lλ(z, ũλ)hdz. (2.28) Choosing h = −ũ−λ ∈W 1,p 0 (Ω), we obtain ĉ∥Dũ−λ ∥ p p ⩽ 0 =⇒ ũλ ⩾ 0, ũλ ̸= 0. (2.29) 8 X. ZHANG, X. XU EJDE-2025/23 Next we choose h = (ũλ − u)+ ∈W 1,p 0 (Ω). We have ⟨V (ũλ), (ũλ − u)+⟩X∗,X = ∫ Ω lλ(z, ũλ)(ũλ − u)+dz = ∫ Ω λ|u(z)|s−1(ũλ − u)+dz ⩽ ∫ Ω λ|u(z)|s−1(ũλ − u)+dz + ∫ Ω f(z, u(z))(ũλ − u)+dz = ⟨V (u), (ũλ − u)+⟩X∗,X , =⇒ ũλ ⩽ u. So we have ũλ ∈ [0, u], ũλ ̸= 0. (2.30) Then (2.26), (2.28), and (2.30) imply that ũλ is a positive solution of (2.20). Hence ũλ = uλ =⇒ uλ ⩽ u for all u ∈ S+ λ . The proof is complete. □ This lower bound leads us to an existence of the smallest positive solution. Lemma 2.11. If hypotheses (H0), (H1) hold and λ ∈ L +, then problem (1.1) has a smallest positive solution u∗λ ∈ intY P1 (that is, u∗λ ∈ S+ λ , u∗λ ⩽ u for all u ∈ S+ λ ). Proof. From the proof of [22, Proposition 7], we know that S+ λ is downward directed (that is, if u1, u2 ∈ S+ λ , then there is u ∈ S+ λ such that u ⩽ u1, u ⩽ u2). Invoking [9, Lemma 3.10, p. 178], we can find a decreasing sequence {un}n∈N ⊂ S+ λ such that inf S+ λ = inf n∈N un. We have ⟨V (un), h⟩X∗,X = λ ∫ Ω (un) s−1hdz + ∫ Ω f(z, un)h dz for all h ∈W 1,p 0 (Ω), all n ∈ N, (2.31) uλ ⩽ un ⩽ u1 for all n ∈ N. (2.32) In (2.31) we use the test function h = un ∈ W 1,p 0 (Ω). Using (2.32) and hypotheses (H0), (H1)(i) we obtain ĉ∥un∥X ⩽ c for some c = c(λ) > 0 =⇒ {un}n∈N ⊂W 1,p 0 (Ω) is bounded. Then, for at least a subsequence, we have un ⇀ u∗λ in W 1,p 0 (Ω), un → u∗λ in Lk(Ω) for 1 ⩽ k < p∗, un(z) → u∗λ(z) a.e. on Ω, |un(z)| ⩽ h(z) a.e. on Ω, for all n ⩾ 1, with h ∈ Lk(Ω). (2.33) In (2.31) we choose h = un − u∗λ, pass to the limit as n→ ∞ and use (2.33). We obtain lim n→∞ ⟨V (un), un − u∗λ⟩X∗,X = 0. Then by Proposition 2.2 we obtain un → u∗λ in W 1,p 0 (Ω). (2.34) If in (2.31) we pass to the limit as n→ ∞ and use (2.34), then we obtain ⟨V (u∗λ), h⟩X∗,X = λ ∫ Ω |u∗λ|s−2u∗λhdz + ∫ Ω f(z, u∗λ)h dz for all h ∈W 1,p 0 (Ω). Also, for (2.32) we have uλ ⩽ u∗λ. Therefore, u∗λ ∈ S+ λ and u∗λ = inf S+ λ . The proof is complete. □ EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 9 Similarly, for S− λ , we have the following conclusion. Lemma 2.12. If hypotheses (H0), (H1) hold, λ ∈ L −, vλ ∈ S− λ , and µ ∈ (0, λ), then µ ∈ L − and there exists vµ ∈ S− µ ⊆ intY (−P1) such that vλ ⩽ vµ. Lemma 2.13. If hypotheses (H0), (H1) hold and λ ∈ L −, then problem (1.1) has a biggest negative solution v∗λ ∈ intY (−P1) (that is, v ∗ λ ∈ S− λ , v ⩽ v∗λ for all v ∈ S− λ ). The ideas for the proofs of Lemmas 2.6–2.13 come from [24]. Now we take λ ∈ L + ∩ L −. Lemma 2.14. For each 0 < λ < λ, there exist ũλ ∈ intY P1, v̂λ ∈ intY (−P1) such that ũλ ⩽ u∗λ, v∗λ ⩽ v̂λ, −∆a1 p ũλ −∆a2 q ũλ < λ|ũλ|s−2ũλ + f(z, ũλ) in Ω ũλ = 0 on ∂Ω (2.35) and −∆a1 p v̂λ −∆a2 q v̂λ > λ|v̂λ|s−2v̂λ + f(z, v̂λ) in Ω v̂λ = 0 on ∂Ω. (2.36) Proof. For each 0 < λ < λ, we take µ < λ. Let u∗λ ⊆ S+ λ be the smallest positive solution. Then by Lemma 2.7 there exists uµ ∈ S+ µ ⊆ intY P1 such that uµ ⩽ u∗λ. In Ω we have −∆a1 p uµ −∆a2 q uµ = µ|uµ|s−2uµ + f(z, uµ) < λ|uµ|s−2uµ + f(z, uµ). Let v∗λ ⊆ S− λ be the biggest negative solution. Then by Lemma 2.6 there exists vµ ∈ S− µ ⊆ intY (−P1) such that v∗λ ⩽ vµ. In Ω we have −∆a1 p vµ −∆a2 q vµ = µ|vµ|s−2vµ + f(z, vµ) > λ|vµ|s−2vµ + f(z, vµ). Let ũλ = uµ and v̂λ = vµ. The proof is complete. □ Lemma 2.15. If (H0) and (H1) hold, then for each 0 < λ < λ, there exist ûλ ∈ intY P1 and ṽλ ∈ intY (−P1) such that −∆a1 p ûλ −∆a2 q ûλ > λ|ûλ|s−2ûλ + f(z, ûλ) in Ω ûλ = 0 on ∂Ω (2.37) and −∆a1 p ṽλ −∆a2 q ṽλ < λ|ṽλ|s−2ṽλ + f(z, ṽλ) in Ω ṽλ = 0 on ∂Ω. (2.38) Proof. For any 0 < λ < λ, we take λ < λ0 < λ and uλ0 ∈ S+ λ0 ⊆ intY P1. In Ω we have −∆a1 p uλ0 −∆a2 q uλ0 = λ0|uλ0 |s−2uλ0 + f(z, uλ0) > λ|uλ0 |s−2uλ0 + f(z, uλ0). Taking vλ0 ∈ S− λ0 ⊆ intY (−P1) In Ω, we have −∆a1 p vλ0 −∆a2 q vλ0 = λ0|vλ0 |s−2vλ0 + f(z, vλ0 ) < λ|vλ0 |s−2vλ0 + f(z, vλ0 ). Let ûλ = uλ0 and ṽλ = vλ0 . The proof is complete. □ Now we introduce an auxiliary operator. For 0 < λ < +∞, v ∈ W 1,p 0 (Ω), define u = Aλ(v) ∈ W 1,p 0 (Ω) to be the unique weak solution of the equation −∆a1 p u(z)−∆a2 q u(z) = λ|v(z)|s−2v(z) + f(z, v(z)) in Ω. We can see that the set of fixed points of Aλ is exactly the set of critical points of Jλ. Lemma 2.16 ([3]). There exist positive constants c1, . . . , c4 such that for all ξ, η ∈ RN , ||ξ|p−2ξ − |η|p−2η| ⩽ c1(|ξ|+ |η|)p−2|ξ − η|, (|ξ|p−2ξ − |η|p−2η) · (ξ − η) ⩾ c2(|ξ|+ |η|)p−2|ξ − η|2, ||ξ|p−2ξ − |η|p−2η| ⩽ c3|ξ − η|p−1 if 1 < p ⩽ 2, (|ξ|p−2ξ − |η|p−2η) · (ξ − η) ⩾ c4|ξ − η|p if p > 2. 10 X. ZHANG, X. XU EJDE-2025/23 In the following, we introduce the famous Minty-Browder Theorem. Lemma 2.17 ([4]). Let X be a reflexive Banach space. Let A : X → X∗ be a continuous nonlinear map such that ⟨Av1 −Av2, v1 − v2⟩ > 0, ∀v1, v2 ∈ X, v1 ̸= v2, and lim ∥v∥→∞ ⟨Av, v⟩ ∥v∥ = ∞. Then for every f ∈ X∗ there exists a unique solution u ∈ X of the equation Au = f . Lemma 2.18. The operator Aλ :W 1,p 0 (Ω) →W 1,p 0 (Ω) is well defined, compact and continuous. Proof. We define Φ :W 1,p 0 (Ω) → R by Φ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz. Clearly Φ(u) is a C1 functional on W 1,p 0 (Ω). By Lemma 2.16, we infer that ⟨Φ′(u1)− Φ′(u2), u1 − u2⟩X∗,X = ∫ Ω a1(z)|Du1|p−2⟨Du1, Du1 −Du2⟩dz + ∫ Ω a2(z)|Du1|q−2⟨Du1, Du1 −Du2⟩dz − ∫ Ω a1(z)|Du2|p−2⟨Du2, Du1 −Du2⟩dz − ∫ Ω a2(z)|Du2|q−2⟨Du2, Du1 −Du2⟩dz. = ∫ Ω a1(z)⟨|Du1|p−2Du1 − |Du2|p−2Du2, Du1 −Du2⟩dz + ∫ Ω a2(z)⟨|Du1|q−2Du1 − |Du2|q−2Du2, Du1 −Du2⟩dz ⩾ ĉ ∫ Ω c4(p)|Du1 −Du2|pdz + c4(q) ∫ Ω a2(z)|Du1 −Du2|qdz ⩾ c∥u1 − u2∥p, which shows Φ′ is strictly monotone. Due to (H1)(i), we can see that for fixed v ∈ W 1,p 0 (Ω) the mapping φ → λ ∫ Ω |v|s−2vφdz + ∫ Ω f(z, v)φdz is a continuous linear functional on W 1,p 0 (Ω). According to Lemma 2.17, there exists a unique u = Aλ(v) satisfying ⟨Φ′(u), φ⟩X∗,X = λ ∫ Ω |v|s−2vφdz + ∫ Ω f(z, v)φdz, for all φ ∈W 1,p 0 (Ω). Therefore, Aλ is well defined. Next we prove that Aλ : X → X is compact. Take {vn} ⊂ X and ∥vn∥X ⩽ M1. So we can assume that, up to a subsequence, there exists v ∈ X such that vn ⇀ v in W 1,p 0 (Ω), vn → v in Lk(Ω) for 1 ⩽ k < p∗, vn(z) → v(z) a.e. on Ω and |vn(z)| ⩽ g(z) a.e. on Ω, for all n ⩾ 1, with g ∈ Lk(Ω). Let un = Aλ(vn). Then we have ⟨V (un), φ⟩X∗,X = λ ∫ Ω |vn|s−2vnφdz + ∫ Ω f(z, vn)φdz, for all φ ∈W 1,p 0 (Ω). We choose φ = un. Then by (H0), (H1)(i) and the Hölder inequality we have that ĉ∥un∥pX ⩽ λ ∫ Ω |vn|s−2vnundz + ∫ Ω f(z, vn)undz ⩽ λ ∫ Ω |vn|s−1|un|dz + d1 ∫ Ω (1 + |vn|r−1)|un|dz ⩽ d2 ∫ Ω (1 + |vn|p ∗−1)|un|dz ⩽ d3 (∫ Ω (1 + |vn|p ∗−1) p∗ p∗−1 dz ) p∗−1 p∗ ∥un∥p∗ EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 11 ⩽ d4 (∫ Ω (1 + |vn|p ∗ )dz ) p∗−1 p∗ ∥un∥p∗ ⩽ d5(1 + ∥vn∥p ∗−1 p∗ )∥un∥p∗ . So by the Sobolev inequality we have ∥un∥X ⩽ d6(1 + ∥vn∥ p∗−1 p−1 p∗ ) ⩽ d7(1 + ∥vn∥ p∗−1 p−1 X ). (2.39) This implies {un} is bounded in X. So there exist a subsequence{unk } ⊂ {un} and u ∈ X such that unk ⇀ u in W 1,p 0 (Ω), unk → u in Lt(Ω) for 1 ⩽ t < p∗, unk (z) → u(z) a.e. on Ω and |unk (z)| ⩽ h(z) a.e. on Ω, for all k ⩾ 1, with h ∈ Lt(Ω). From the above convergence properties, we obtain∫ Ω |vnk |s−2vnk (unk − u)dz → 0 and ∫ Ω f(z, vnk )(unk − u)dz → 0 as k → ∞. Then lim k→∞ ⟨V (unk ), unk − u⟩X∗,X = lim k→∞ (λ ∫ Ω |vnk |s−2vnk (unk − u)dz + ∫ Ω f(z, vnk )(unk − u)dz) = 0. It follows from Proposition 2.2 that unk → u in W 1,p 0 (Ω). In the following, we prove that Aλ is continuous. Assume that vn → v strongly in W 1,p 0 (Ω). Setting un = Aλ(vn) and u = Aλ(v), we need to show ∥un − u∥X → 0 as n → ∞. From (2.39), we can infer that {un} is bounded in X. Up to a subsequence, we may assume that un ⇀ u in W 1,p 0 (Ω), un → u in Lk(Ω) for 1 ⩽ k < p∗, un(z) → u(z) a.e. on Ω and |un(z)| ⩽ h(z) a.e. on Ω, for all n ⩾ 1, with h ∈ Lk(Ω). By (H1), we can infer that lim n→∞ ⟨Φ′(un)− Φ′(u), un − u⟩X∗,X = lim n→∞ ( λ ∫ Ω (|vn|s−2vn − |v|s−2v)(un − u)dz + ∫ Ω (f(z, vn)− f(z, v))(un − u)dz ) = 0. It is easy to show that ∥un − u∥pX ⩽ c′⟨Φ′(un)− Φ′(u), un − u⟩X∗,X , where c′ > 0. So limn→∞ ∥un − u∥pX = 0. Therefore, ∥un − u∥X → 0 as n → ∞. The proof is complete. □ Lemma 2.19. The operator Aλ : X → L∞(Ω) is bounded. Proof. Let D ⊂ X be a bounded set of X. Then there exists M1 such that for any v ∈ D, ∥v∥X ⩽M1. Set u = Aλ(v). From the proof of Lemma 2.18, we obtain ∥u∥X ⩽ d1(1 + |v| p∗−1 p−1 p∗ ) ⩽ d2(1 + ∥v∥ p∗−1 p−1 X ). (2.40) By Sobolev inequality we have |u|p∗ ⩽ d3(1 + ∥v∥ p∗−1 p−1 X ). From [11, Theorem 7.1, p. 286], we infer that Aλ : X → L∞(Ω) is bounded. The proof is complete. □ Lemma 2.20. There exists 0 < α < 1 such that Aλ : L∞(Ω) → C1,α 0 (Ω) is bounded. Proof. Let D ⊂ L∞(Ω) be a bounded set of L∞(Ω). Then there exists M1 > 0 such that for any v ∈ D, ∥v∥∞ ⩽M1. Take any v ∈ D. Let u = Aλ(v). By (2.40) and Sobolev inequality we have |u|p∗ ⩽ c(1 + ∥v∥ p∗−1 p−1 p∗ ) ⩽ c′(1 + ∥v∥ p∗−1 p−1 ∞ ) ⩽M2. 12 X. ZHANG, X. XU EJDE-2025/23 From [11, Theorem 7.1, p. 286,], we obtain ∥u∥∞ ⩽ M3. The nonlinear regularity theory of Liebermann [13] implies that there exists α ∈ (0, 1) and M4 > 0 such that u ∈ C1,α 0 (Ω) and ∥u∥C1,α 0 (Ω) ⩽M4, (2.41) for all u satisfying u = Aλ(v) and ∥v∥∞ ⩽M1. The proof is complete. □ Lemma 2.21. Let v ∈ L∞(Ω). Then v ⩾ ṽλ =⇒ Aλ(v) ≫ ṽλ, (2.42) v ⩽ ûλ =⇒ Aλ(v) ≪ ûλ, (2.43) v ⩽ v̂λ =⇒ Aλ(v) ≪ v̂λ, (2.44) v ⩾ ũλ =⇒ Aλ(v) ≫ ũλ. (2.45) Proof. Let u = Aλ(v) for any v ∈ L∞(Ω) and v ⩽ ûλ. Obviously, the operator Aλ is an increasing operator. Then we have u = Aλ(v) ⩽ Aλ(ûλ) ⩽ ûλ. Take ξ > 0. So we obtain that −∆a1 p ûλ −∆a2 q ûλ + ξ|ûλ|p−2ûλ > λ|ûλ|s−2ûλ + f(z, ûλ) + ξ|ûλ|p−2ûλ ⩾ λ|v|s−2v + f(z, v) + ξ|ûλ|p−2ûλ ⩾ −∆a1 p u−∆a2 q u+ ξ|u|p−2u. (2.46) Since ûλ ∈ intY P1, from (2.46) and [6, Proposition 3.2 ], we conclude that u = Aλ(v) ≪ ûλ. Similarly, we can infer that (2.42) (2.44) (2.45) hold. The proof is complete. □ Lemma 2.22. For v ∈W 1,p 0 (Ω), we have ⟨J ′ λ(v), v −Aλ(v)⟩X∗,X ⩾ c1∥v −Aλ(v)∥pX and ∥J ′ λ(v)∥X∗ ⩽ c2 ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 ) (∥v −Aλ(v)∥X), where c1, c2 are positive constants. Proof. Set u = Aλ(v). By the definition of Aλ and (H0) we obtain ⟨J ′ λ(v), v − u⟩X∗,X = ∫ Ω a1(z)|Dv|p−2⟨Dv,Dv −Du⟩dz + ∫ Ω a2(z)|Dv|q−2⟨Dv,Dv −Du⟩dz − λ ∫ Ω |v|s−2v(v − u)dz − ∫ Ω f(z, v)(v − u)dz. = ∫ Ω a1(z)|Dv|p−2⟨Dv,Dv −Du⟩dz + ∫ Ω a2(z)|Dv|q−2⟨Dv,Dv −Du⟩dz − ∫ Ω a1(z)|Du|p−2⟨Du,Dv −Du⟩dz − ∫ Ω a2(z)|Du|q−2⟨Du,Dv −Du⟩dz. = ∫ Ω a1(z)⟨|Dv|p−2Dv − |Du|p−2Du,Dv −Du⟩dz + ∫ Ω a2(z)⟨|Dv|q−2Dv − |Du|q−2Du,Dv −Du⟩dz ⩾ ĉ c2(p) ∫ Ω |Dv −Du|pdz + c2(q) ∫ Ω a2(z)|Dv −Du|qdz ⩾ c1∥v − u∥pX EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 13 and |⟨J ′ λ(v), φ⟩X∗,X | = ∣∣∣ ∫ Ω a1(z)|Dv|p−2⟨Dv,Dφ⟩dz + ∫ Ω a2(z)|Dv|q−2⟨Dv,Dφ⟩dz − λ ∫ Ω |v|s−2vφdz − ∫ Ω f(z, v)φdz ∣∣∣ = ∣∣∣ ∫ Ω a1(z)|Dv|p−2⟨Dv,Dφ⟩dz + ∫ Ω a2(z)|Dv|q−2⟨Dv,Dφ⟩dz − ∫ Ω a1(z)|Du|p−2⟨Du,Dφ⟩dz − ∫ Ω a2(z)|Du|q−2⟨Du,Dφ⟩dz ∣∣∣ = ∣∣∣ ∫ Ω a1(z)⟨|Dv|p−2Dv − |Du|p−2Du,Dφ⟩dz + ∫ Ω a2(z)⟨|Dv|q−2Dv − |Du|q−2Du,Dφ⟩dz ∣∣∣ ⩽ c3 ∫ Ω ∣∣|Dv|p−2Dv − |Du|p−2Du ∣∣|Dφ|dz + c4 ∫ Ω ∣∣|Dv|q−2Dv − |Du|q−2Du ∣∣|Dφ|dz ⩽ c3 (∫ Ω ∣∣|Dv|p−2Dv − |Du|p−2Du ∣∣ p p−1 dz ) p−1 p ∥φ∥X + c4( ∫ Ω ∣∣|Dv|q−2Dv − |Du|q−2Du ∣∣ q q−1 dz) q−1 q |Dφ|q (2.47) for any φ ∈W 1,p 0 (Ω), where ⟨·, ·⟩ denotes the inner product in RN . From Lemma 2.16, we obtain that (∫ Ω ∣∣|Dv|p−2Dv − |Du|p−2Du ∣∣ p p−1 dz ) p−1 p ⩽ c( ∫ Ω ( |Dv|+ |Du|) p(p−2) p−1 |Dv −Du| p p−1 dz ) p−1 p ⩽ c (∫ Ω (|Dv|+ |Du|)pdz ) p−2 p (∫ Ω |Dv −Du|pdz )1/p ⩽ c(p)(∥v∥X + ∥u∥X)p−2∥v − u∥X . (2.48) Similarly, we infer that(∫ Ω ∣∣|Dv|q−2Dv − |Du|q−2Du ∣∣ q q−1 dz ) q−1 q ⩽ c(q)(∥v∥X + ∥u∥X)q−2∥v − u∥X . (2.49) By (2.47), (2.48), (2.49), we can show that ∥J ′ λ(v)∥X∗ ⩽ c2((∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2)(∥v −Aλ(v)∥X). The proof is complete. □ Let Z=C1,α(Ω). We define D1 = {v ∈ Y : ṽλ ⩽ v ⩽ ûλ}, D2 = {v ∈ Y : v ⩾ ũλ}, D3 = {v ∈ Y : v ⩽ v̂λ}. Let D̃i be the closure of Di in X. Let Kλ = {x ∈ X : J ′ λ(u) = 0}. Note that Kλ ⊂ Z is the regularity results from Lemma 2.19 and Lemma 2.20. Lemma 2.23. For any λ ∈ (0, λ), there exists a locally Lipschitz continuous operator Bλ : X0 → Z with the following properties: (i) For v ∈ X0 and c1 from Lemma 2.22 we have 1 2 ∥v −Aλ(v)∥X ≤ ∥v −Bλ(v)∥X ≤ 2∥v −Aλ(v)∥X , ⟨J ′ λ(v), v −Bλ(v)⟩X∗,X ≥ c1 2 ∥v −Aλ(v)∥pX ; 14 X. ZHANG, X. XU EJDE-2025/23 (ii) Bλ(D̃i ∩X0) ⊂ intYDi for i = 1, 2, 3; (iii) Bλ(P ∩X0) ⊂ P1, Bλ(−P ∩X0) ⊂ (−P1); (iv) Bλ : X0 → L∞(Ω) is bounded; (v) Bλ : (X0 ∩ L∞(Ω), ∥ · ∥∞) → Z is bounded. Proof. For any v ∈ X0, setting ∆1(v) = 1 2 ∥v −Aλ(v)∥X (2.50) and setting ∆2(v) = c1 2c2 ∥v −Aλ(v)∥p−1 X ((∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2)−1, (2.51) we choose γ(v) ∈ (0, 1) such that ∥Aλ(u)−Aλ(w)∥X < min{∆1(u),∆1(w),∆2(u),∆2(w)} (2.52) holds for every u,w ∈ N(v) := {x ∈ W 1,p 0 (Ω) : ∥x − v∥X < γ(v)}. Let U be a locally finite open refinement of {N(v) : v ∈ X0}. We refine U in order to construct the required operator Bλ. For any U ∈ U , if U ∩ D̃2 ̸= ∅, U ∩ D̃3 ̸= ∅, then we replace U in the covering U by the two open sets U \ D̃2 and U \ D̃3. The new covering is U∗. We need to refine U∗ once more. For any U ⊂ U∗, if U ∩ D̃2 ̸= ∅, U ∩ D̃3 = ∅ and U ∩ (−P ) ̸= ∅, then we replace U in the covering U∗ by the two open sets U \ D̃2 and U \ (−P ). For any U ⊂ U∗, if U ∩ D̃3 ̸= ∅, U ∩ D̃2 = ∅ and U ∩P ̸= ∅, then we replace U in the covering U∗ by the two open sets U \ D̃3 and U \ P . The new covering is W. We refine W. For convenience, we set D̃4 = P and D̃5 = −P . For any V ⊂ X, we define IV = {i ∈ {1, 2, 3, 4, 5} : V ∩ D̃i ̸= ∅}. For any V ⊂ W, if IV = {1, 4, 5} and V ∩ D̃1 ∩ D̃4 ∩ D̃5 = ∅, then we replace V in the covering W by the two open sets U \ D̃4 and U \ D̃5. For any V ⊂ W, if IV = {1, 2, 4} and V ∩ D̃1 ∩ D̃2 = ∅, then we replace V in the covering W by the two open sets U \ D̃1 and U \ D̃2. If IV = {1, 3, 5} and V ∩ D̃1 ∩ D̃3 = ∅, then we replace V in the covering W by the two open sets U \ D̃1 and U \ D̃3. The new covering is W∗. We refine W∗ once more. For any V ⊂ W∗, if IV = {i, j} with i ̸= j and V ∩ D̃i ∩ D̃j = ∅, then we replace V in the covering W∗ by the two open sets U \ D̃i and U \ D̃j . The new covering is V∗. To make Bλ satisfy (iv) and (v), we need to refine V∗. If IV = {2, 4} and if inf v∈V ∩L∞(Ω) ∥v∥∞ < inf v∈V ∩D̃2∩L∞(Ω) ∥v∥∞ − 1 =: βV for some V ∈ V∗, then we replace V in the covering V∗ by the following two open subsets: V \ {v ∈ L∞(Ω) : ∥v∥∞ ≤ βV } and V \ D̃2; If IV = {3, 5} and if inf v∈V ∩L∞(Ω) ∥v∥∞ < inf v∈V ∩D̃3∩L∞(Ω) ∥v∥∞ − 1 =: βV for some V ∈ V∗, then we replace V in the covering V∗ by the following two open subsets: V \ {v ∈ L∞(Ω) : ∥v∥∞ ≤ βV } and V \ D̃3; We obtain a new covering V∗∗. We refine it. If IV = {i} with i = 4 or 5 and if inf v∈V ∩L∞(Ω) ∥v∥∞ < inf v∈V ∩D̃i∩L∞(Ω) ∥v∥∞ − 1 =: βV for some V ∈ V∗∗, then we replace V in the covering V∗∗ by the following two open subsets: V \ {v ∈ L∞(Ω) : ∥v∥∞ ≤ βV } and V \ D̃i; otherwise V is left unchanged. The new open covering V obtained in this way from V∗∗ is a locally finite open refinement of V∗∗, hence of {N(v) : v ∈ X0}, and in addition, any V ∈ V satisfies: V ∩ D̃i ̸= ∅ and V ∩ D̃j ̸= ∅ implies V ∩ D̃i ∩ D̃j ̸= ∅, inf v∈V ∩L∞(Ω) ∥v∥∞ ≥ inf v∈V ∩D̃i∩L∞(Ω) ∥v∥∞ − 1 if IV = {i} with i = 4, or 5, EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 15 inf v∈V ∩L∞(Ω) ∥v∥∞ ≥ inf v∈V ∩D̃2∩L∞(Ω) ∥v∥∞ − 1 if IV = {2, 4}, inf v∈V ∩L∞(Ω) ∥v∥∞ ≥ inf v∈V ∩D̃3∩L∞(Ω) ∥v∥∞ − 1 if IV = {3, 5}. Now we are ready to construct the operator Bλ. Let {πV : V ∈ V} be the standard partition of unity subordinated to V defined by πV (v) = ( ∑ U∈V αU (v) )−1 αV (v), where αV (v) = dist(v,X0 \ V ). For each V ∈ V choose aV ∈ V such that aV ∈ V ∩ L∞(Ω) and ∥aV ∥∞ ≤ inf v∈V ∩L∞(Ω) ∥v∥∞ + 1 (2.53) if IV = ∅, and aV ∈ V ∩i∈IV D̃i ∩ L∞(Ω) and ∥aV ∥∞ ≤ inf v∈V ⋂ i∈IV D̃i∩L∞(Ω) ∥v∥∞ + 1 (2.54) if IV ̸= ∅. Now we define Bλ : X0 → X by Bλ(v) = ∑ V ∈V πV (v)Aλ(aV ). As a consequence of the Lipschitz continuity of πV , the locally finiteness of the covering V and the fact Aλ(L ∞(Ω)) ⊂ Z, Bλ : X0 → Z is locally Lipschitz continuous. For each v ∈ X0, we have ∥Bλ(v)−Aλ(v)∥X = ∥∥∥ ∑ V ∈V πV (v)Aλ(aV )− ∑ V ∈V πV (v)Aλ(v) ∥∥∥ X ≤ ∑ V ∈V πV (v) ∥∥Aλ(aV )−Aλ(v) ∥∥ X . (2.55) By (2.50), (2.51), (2.52), and (2.55), we infer that ∥Bλ(v)−Aλ(v)∥X ≤ 1 2 ∥v −Aλ(v)∥X , ∥Bλ(v)−Aλ(v)∥X ≤ c1 2c2 ∥v −Aλ(v)∥p−1 X ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )−1 . Thus, we have ∥v −Bλ(v)∥X ≤ ∥v −Aλ(v)∥X + ∥Aλ(v)−Bλ(v)∥X ≤ ∥v −Aλ(v)∥X + 1 2 ∥v −Aλ(v)∥X ≤ 2∥v −Aλ(v)∥X and ∥v −Aλ(v)∥X ≤ ∥v −Bλ(v)∥X + ∥Bλ(v)−Aλ(v)∥X ≤ ∥v −Bλ(v)∥X + 1 2 ∥v −Aλ(v)∥X . So 1 2 ∥v −Aλ(v)∥X ≤ ∥v −Bλ(v)∥X ≤ 2∥v −Aλ(v)∥X . And by Lemma 2.22, we obtain ⟨J ′ λ(v), v −Bλ(v)⟩X∗,X ≥ ⟨J ′ λ(v), v −Aλ(v)⟩X∗,X − ∥J ′ λ(v)∥X∗∥Bλ(v)−Aλ(v)∥X ≥ c1∥v −Aλ(v)∥pX − 1 2 c1∥v −Aλ(v)∥pX = c1 2 ∥v −Aλ(v)∥pX . (2.56) Thus (i) is proved. 16 X. ZHANG, X. XU EJDE-2025/23 If v ∈ D̃i ∩ X0 with i = 1, 2, 3, then v ∈ D̃i ∩ V for any V with πV (v) ̸= 0. From the construction it follows that aV ∈ V ∩ D̃i ∩ L∞(Ω) for any V with πV (v) ̸= 0. This implies Bλ(v) ∈ conv Aλ(D̃i ∩ L∞(Ω)). So by Lemma 2.21 (ii) is proved. If v ∈ P ∩ X0, then v ∈ P ∩ V for any V with πV (v) ̸= 0. From the construction it follows that aV ∈ V ∩ P ∩ L∞(Ω) for any V with πV (v) ̸= 0. Since Aλ is an increasing operator and Aλ : L∞(Ω) → Z, Aλ(P ∩L∞(Ω)) ⊂ P1. This implies Bλ(v) ∈ P1. So Bλ(P ∩X0) ⊂ P1. Similarly, we can infer that Bλ(−P ∩X0) ⊂ (−P1). Thus (iii) is proved. To prove (iv), we suppose v ∈ X0 and ∥v∥X ≤ d1. If πV (v) ̸= 0, then v ∈ V . Since v, aV ∈ V ⊂ N(u) for some u ∈ X0, and since γ(u) < 1 we have ∥aV − v∥X ≤ ∥aV − u∥X + ∥u− v∥X < 2 and thus ∥aV ∥X ≤ d1 + 2. So ∥Aλ(aV )∥∞ ≤ d2 for all V ∈ V with πV (v) ̸= 0 by Lemma 2.13. Then we have ∥Bλ(v)∥∞ ≤ ∑ V ∈V πV (v)∥Aλ(aV )∥∞ ≤ d2. It remains to prove (v). Suppose v ∈ X0 ∩ L∞(Ω) and ∥v∥∞ ≤ d3. If πV (v) ̸= 0 then v ∈ V ∩ L∞(Ω). If IV = ∅, then ∥aV ∥∞ ≤ inf v∈V ∩L∞(Ω) ∥v∥∞ + 1 ≤ d3 + 1. If IV = {i}, i = 4, 5, then ∥aV ∥∞ ≤ inf v∈V ∩D̃i∩L∞(Ω) ∥v∥∞ + 1 ≤ inf v∈V ∩L∞(Ω) ∥v∥∞ + 2 ≤ d3 + 2. If IV = {2, 4}, then ∥aV ∥∞ ≤ inf v∈V ∩D̃2∩L∞(Ω) ∥v∥∞ + 1 ≤ inf v∈V ∩L∞(Ω) ∥v∥∞ + 2 ≤ d3 + 2. If IV = {3, 5}, then ∥aV ∥∞ ≤ inf v∈V ∩D̃3∩L∞(Ω) ∥v∥∞ + 1 ≤ inf v∈V ∩L∞(Ω) ∥v∥∞ + 2 ≤ d3 + 2. If IV = {4, 5}, then aV = 0. Obviously, ∥aV ∥∞ ≤ d3. If IV = {1}, {1, 4}, {1, 5}, {1, 4, 5}, {1, 2, 4}, {1, 3, 5}, since D̃1 ∩ L∞(Ω) is bounded in L∞(Ω), we have ∥aV ∥∞ ≤ d4. In any case, by Lemma 2.20 we have ∥Aλ(aV )∥Z ≤ d5 for any V ∈ V with πV (v) ̸= 0 and ∥Bλ(v)∥Z ≤ ∑ V ∈V πV (v)∥Aλ(aV )∥Z ≤ d5. The proof is complete. □ For v ∈ Y0 = Y \Kλ ⊂ X0 we consider the initial value problem, both in X0 and in Y0, dσ dt = −σ(t, v) +Bλ(σ(t, v)), σ(0, v) = v. (2.57) Since Bλ : X0 → Z is locally Lipschitz continuous, the solution of (2.57) considered in X0 and the solution of (2.57) considered in Y0 are exactly the same. Let σ(t, v) be the unique solution of (2.57) with its right maximal existence interval [0, τ(v)). From Lemma 2.23, we infer that Jλ(σ(t, v)) is strictly decreasing in t ∈ [0, τ(v)). Lemma 2.24. If v ∈ Di \Kλ, then σ(t, v) ∈ intY Di for 0 < t < τ(v). A proof of the above lemma can be found in [10]. Lemma 2.25. For each b ∈ R, there exists a constant c3 = c3(b) > 0 such that ∥v∥X + ∥Aλ(v)∥X ⩽ c3(1 + ∥v −Aλ(v)∥X) holds for every v ∈ X with Jλ(v) ⩽ b. EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 17 Proof. For v ∈ X, we have Jλ(v)− 1 m ⟨J ′ λ(v), v⟩X∗,X = (1 p − 1 m ) ∫ Ω a1(z)|Dv|pdz + (1 q − 1 m ) ∫ Ω a2(z)|Dv|qdz + ( λ m − λ s ) ∫ Ω |v|sdz + ∫ Ω ( 1 m f(z, v)v − F (z, v) ) dz. (2.58) If Jλ(v) ⩽ b, then (H1)(ii) implies ∥v∥pX ⩽ d1(1 + ∥v∥sX + ∥J ′ λ(v)∥X∗∥v∥X). Then using Lemma 2.22 we obtain ∥v∥pX ⩽ d2 ( 1 + ∥v∥sX + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 ) ∥v −Aλ(v)∥X∥v∥X ) . (2.59) Then Young’s inequality gives ∥v∥pX ⩽ d2 ( 1 + ∥v∥sX + C(ε) ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′ ∥v −Aλ(v)∥p ′ X + ε∥v∥pX ) , (2.60) where 1 p + 1 p′ = 1, ε > 0 is arbitrarily small, C(ε) = (εp)−p′/p/p′. We take ε > 0 sufficiently small such that d2ε < 1. Then we have ∥v∥pX ⩽ d3 ( 1 + ∥v∥sX + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′ ∥v −Aλ(v)∥p ′ X ) . Obviously, there exist some positive constant d4, d5 such that d4∥v∥pX − d5 ⩽ ∥v∥pX − d3∥v∥sX . Thus, ∥v∥pX ⩽ d6 ( 1 + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′ ∥v −Aλ(v)∥p ′ X ) and so ∥v∥X ⩽ d7 ( 1 + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′/p ∥v −Aλ(v)∥p ′/p X ) . (2.61) Using (2.61), We infer that ∥Aλ(v)∥X ⩽ ∥v −Aλ(v)∥X + ∥v∥X = ∥v −Aλ(v)∥p ′/p X ∥v −Aλ(v)∥ 1− p′ p X + ∥v∥X ⩽ ∥v −Aλ(v)∥p ′/p X (∥v∥X + ∥Aλ(v)∥X)1− p′ p + ∥v∥X = ∥v −Aλ(v)∥p ′/p X ( (∥v∥X + ∥Aλ(v)∥X)p−2 )p′/p + ∥v∥X ⩽ ∥v −Aλ(v)∥p ′/p X ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′/p + ∥v∥X ⩽ d8 ( 1 + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′/p ∥v −Aλ(v)∥p ′/p X ) . Thus, we have ∥v∥X + ∥Aλ(v)∥X ⩽ d9 ( 1 + ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 )p′/p ∥v −Aλ(v)∥p ′/p X ) . Using Young’s inequality again, we have ∥v∥X + ∥Aλ(v)∥X 18 X. ZHANG, X. XU EJDE-2025/23 ⩽ d9 ( 1 + ε′ ( (∥v∥X + ∥Aλ(v)∥X)p−2 + (∥v∥X + ∥Aλ(v)∥X)q−2 ) 1 p−2 + C(ε′)∥v −Aλ(v)∥X ) ⩽ d9 ( 1 + c(p)ε′ ( (∥v∥X + ∥Aλ(v)∥X) + (∥v∥X + ∥Aλ(v)∥X) q−2 p−2 ) + C(ε′)∥v −Aλ(v)∥X ) We take ε′ > 0 is sufficiently small such that d9c(p)ε ′ < 1. Then ∥v∥X + ∥Aλ(v)∥X ⩽ d10 ( 1 + ∥v −Aλ(v)∥X + (∥v∥X + ∥Aλ(v)∥X) q−2 p−2 ) . Obviously, there exist some positive constant d11, d12 such that d11(∥v∥X + ∥Aλ(v)∥X)− d12 ⩽ ∥v∥X + ∥Aλ(v)∥X − d10(∥v∥X + ∥Aλ(v)∥X) q−2 p−2 . So there exists a constant c3 = c3(b) > 0 such that ∥v∥X + ∥Aλ(v)∥X ⩽ c3(1 + ∥v −Aλ(v)∥X). The proof is complete. □ Lemma 2.26. For each λ > 0, Jλ satisfies the Palais-Smale condition in X. Proof. Let {vn} ⊂ X be such that |Jλ(vn)| ⩽M1 for someM1 > 0 and J ′ λ(vn) → 0 inW−1,p′ 0 (Ω) = W 1,p 0 (Ω)∗ ( 1p + 1 p′ = 1) as n→ ∞. We first claim that {vn} is bounded. From |Jλ(vn)| ⩽M1, we have 1 p ∫ Ω a1(z)|Dvn|pdz + 1 q ∫ Ω a2(z)|Dvn|qdz − λ s ∫ Ω |vn|sdz − ∫ Ω F (z, vn)dz ⩽M1. (2.62) From J ′ λ(vn) → 0, we have |⟨J ′ λ(vn), vn⟩X∗,X | ⩽ c∥vn∥X for some c > 0, namely − ∫ Ω a1(z)|Dvn|pdz − ∫ Ω a2(z)|Dvn|qdz + λ ∫ Ω |vn|sdz + ∫ Ω f(z, vn)vndz ⩽ c∥vn∥X . (2.63) By (H1) (ii), for some c1 > 0 we have∫ Ω (f(z, vn)vn −mF (z, vn))dz ⩾ −c1. (2.64) Then by (H0), (2.62), (2.63), and (2.64), we have(m p − 1 ) ĉ∥vn∥pX ⩽ mM1 + c1 + c∥vn∥X + λc2∥vn∥sX . Since m > p > s, we infer that {vn} ⊂W 1,p 0 (Ω) is bounded. Going if necessary to a subsequence, we assume that vn ⇀ v in W 1,p 0 (Ω), vn → v in Lk(Ω) for 1 ⩽ k < p∗, vn(z) → v(z) a.e. on Ω and |vn(z)| ⩽ g(z) a.e. on Ω, for all n ⩾ 1, with g ∈ Lk(Ω). We can deduce from ∥J ′ λ(vn)∥X∗ → 0 and vn ⇀ v that |⟨J ′ λ(vn), vn − v⟩X∗,X | → 0 as n→ +∞. This reads∣∣∣⟨V (vn), vn − v⟩X∗,X − λ ∫ Ω |vn|s−2vn(vn − v)dz − ∫ Ω f(z, vn)(vn − v)dz ∣∣∣ → 0 as n→ +∞. We have λ ∫ Ω |vn|s−2vn(vn − v)dz + ∫ Ω f(z, vn)(vn − v)dz → 0 as n→ ∞, and so lim n→∞ ⟨V (vn), vn − v⟩X∗,X = 0. From Proposition 2.2, we can deduce that vn → v in W 1,p 0 (Ω). The proof is complete. □ Definition 2.27 ([15]). A nonempty subset D of Y is called an invariant set of descending flow of (2.57) if o(v0) ⊂ D for all v0 ∈ D, where o(v0) = {σ(t, v0) ⊂ Y : t ∈ [0, τ(v0))}. EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 19 Definition 2.28 ([15]). LetM,D ⊂ Y be invariant sets of descending flow of (2.57) with D ⊂M . Denote CM (D) = {v0 : v0 ∈ D or v0 ∈M\D and there exists t′ ∈ (0, τ(v0)) such that σ(t′, v0) ∈ D}. If D = CM (D), then D is called a complete invariant set of descending flow of (2.57). Lemma 2.29 ([15]). Let G ⊂ Y be a connected and invariant set of (2.57) and D be an open invariant subset of G. Then the following assertions hold: (1) CG(D) is an open subset of G; (2) ∂GCG(D) is a complete invariant set of descending flow of (2.57). Lemma 2.30 ([28]). Assume U is bounded connected open set of R2 and (0, 0) ∈ U , then there exists a connected component Γ′ of the boundary of U , and each one sided ray l through the origin satisfies l ∩ Γ′ ̸= ∅. We set G1 = {u ∈ Y : ũλ ≪ u≪ ûλ}, G2 = {u ∈ Y : ṽλ ≪ u≪ v̂λ}. Proof of Theorem 2.3. For convenience, we divide the proof into four steps. Step 1. There exist a positive solution u1 ∈ G1 and a negative solution u2 ∈ G2. For each λ ∈ (0, λ), we introduce the Carathéodory function kλ(z, x) = { λ|x+|s−2x+ + f(z, x+), x ⩽ ûλ(z), λ|ûλ(z)|s−2ûλ(z) + f(z, ûλ(z)), ûλ(z) < x, (2.65) where x+ = max{0, x}. Let Kλ(z, x) = ∫ x 0 kλ(z, s)ds and consider the C1 functional ζλ : W 1,p 0 (Ω) → R defined by ζλ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − ∫ Ω Kλ(z, u)dz, ∀u ∈W 1,p 0 (Ω). It is clear that ζλ(·) is coercive. Also it is sequentially weakly lower semicontinuous. So, we can find uλ ∈W 1,p 0 (Ω) such that ζλ(uλ) = inf [ ζλ(u) : u ∈W 1,p 0 (Ω) ] . (2.66) We see that for t ∈ (0, 1) small, we have ζλ(tû1(s)) < 0 =⇒ ζλ(uλ) < 0 = ζλ(0) =⇒ uλ ̸= 0. From (2.66), we have ζ ′λ(uλ) = 0 =⇒ ⟨V (uλ), h⟩X∗,X = ∫ Ω kλ(z, uλ)hdz. (2.67) In (2.66) first we choose h = −u−λ ∈W 1,p 0 (Ω). We obtain ĉ∥Du−λ ∥ p p ⩽ 0 =⇒ uλ ⩾ 0, uλ ̸= 0. Next we choose h = (uλ − ûλ) + ∈W 1,p 0 (Ω). We have ⟨V (uλ), (uλ − ûλ) +⟩ = ∫ Ω kλ(z, uλ)(uλ − ûλ) +dz = ∫ Ω λ|ûλ(z)|s−1(uλ − ûλ) +dz + ∫ Ω f(z, ûλ(z))(uλ − ûλ) +dz ⩽ ⟨V (ûλ), (uλ − ûλ) +⟩ =⇒ uλ ⩽ ûλ. We have proved that uλ ∈ [0, ûλ], uλ ̸= 0. (2.68) Then, (2.65), (2.67), and (2.68) imply that uλ is a positive solution, and by Lemma 2.21 we have uλ ∈ G1. We denote this solution as u1. Similarly, for G2 we obtain a negative solution u2. 20 X. ZHANG, X. XU EJDE-2025/23 Step 2. There exist a positive solution u3 ∈ ∂D2 CD2 ((intY D1) ∩ D2) and a negative solution u4 ∈ ∂D3CD3((intY D1) ∩D3). First we consider the set CD2((intY D1) ∩D2), which is an open subset of D2. Let e1, e2 ∈ Y be linearly independent and denote Y1 = span{e1, e2}. Clearly, (H1) (ii) implies F (z, x) ⩾ c4|x|m − c5 for x ∈ R for some positive constants c4 and c5. So we have Jλ(u) = 1 p ∫ Ω a1(z)|Du|pdz + 1 q ∫ Ω a2(z)|Du|qdz − λ s ∫ Ω |u|sdz − ∫ Ω F (z, u)dz. ⩽ c3 p ∫ Ω |Du|pdz + c4 q ∥Du∥qq − λ s ∫ Ω |u|sdz − ∫ Ω c5|u|mdz + c6. (2.69) This implies that if u ∈ Y1 and ∥u∥Y → +∞, then Jλ(u) → −∞. Therefore, CD2 ((intY D1)∩D2) ̸= D2 and ∂D2 CD2 ((intY D1) ∩ D2) ̸= ∅. By Lemma 2.29, we have ∂D2 CD2 ((intY D1) ∩ D2) is an invariant set of descending flow of (2.57). In addition, inf u∈∂D2 CD2 ((intY D1)∩D2) Jλ(u) ⩾ d := inf u∈(intY D1)∩D2 Jλ(u) > −∞. Take u0 ∈ ∂D2 CD2 ((intY D1) ∩D2). So we have σ(t, u0) ∈ ∂D2 CD2 ((intY D1) ∩D2) for 0 ⩽ t ⩽ τ(u0). Since Jλ(σ(t, u)) is strictly decreasing in t ∈ [0, τ(u)), we obtain d ⩽ Jλ(σ(t, u0)) ⩽ Jλ(u0) for 0 ⩽ t ⩽ τ(u0). (2.70) Next we prove that there exists u3 ∈ Kλ and an increasing sequence (tn)n with tn → τ(u0) such that limn→∞ ∥σ(tn, u0)− u3∥X = 0. By (2.57) and Lemma 2.23, for 0 < t1 < t2, we have ∥σ(t2, u0)− σ(t1, u0)∥X ⩽ ∫ t2 t1 ∥σ(s, u0)−Bλ(σ(s, u0))∥Xds ⩽ 2 ∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥Xds. (2.71) At first, we assume τ(u0) < +∞. As a consequence of the Hölder inequality we have∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥Xds ⩽ (∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥pXds )1/p (t2 − t1) 1− 1 p . Thus, (2.57) and Lemma 2.23 imply∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥Xds ⩽ d1(Jλ(σ(t1, u0)− Jλ(σ(t2, u0))) 1/p(t2 − t1) 1− 1 p . In view of (2.70) and the finiteness of τ(u0), we see that lim t1,t2→τ(u0)−0 ∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥Xds = 0. So by (2.71) there exists u3 ∈ X such that limt→τ(u0)−0 ∥σ(t, u0) − u3∥X = 0. Since [0, τ(u0)) is the maximal interval of existence of σ(t, u0) in X0, we have u3 ∈ Kλ. It remains to consider the case τ(u0) = +∞. By (2.57) and Lemma 2.17 there exists an increasing sequence {tn} with tn → +∞ such that 0 ⩽ d2∥σ(tn, u0)−Aλ(σ(tn, u0))∥pX ⩽ − d dt Jλ(σ(t, u0)) ∣∣∣ t=tn → 0. Now {σ(tn, u0)} is bounded in X by Lemma 2.19. Since Aλ : X → X is a compact operator it follows that lim n→∞ ∥σ(tn, u0)− u3∥X = lim n→∞ ∥Aλ(σ(tn, u0))− u3∥X = 0 EJDE-2025/23 PARAMETRIC WEIGHTED (p, q)-EQUATIONS 21 for some u3 ∈ Kλ. Next we prove that limn→∞ ∥σ(tn, u0) − u3∥Y = 0. We first claim that {σ(t, u0) : 0 ⩽ t < τ(u0)} is bounded in X. Suppose that ∥σ(t, u0)∥X ⩾ 2c3 for t ∈ [t1, t2] ⊂ [0, τ(u0)) with c3 from Lemma 2.25. Then we have ∥σ(t, u0)−Aλ(σ(t, u0))∥X ⩾ 1 for t ∈ [t1, t2]. (2.72) So using (2.57), Lemma 2.23, (2.72), (2.70), we have ∥σ(t2, u0)− σ(t1, u0)∥X ⩽ 2 ∫ t2 t1 ∥σ(s, u0)−Aλ(σ(s, u0))∥pXds. ⩽ d2(Jλ(σ(t 1, u0)− Jλ(σ(t 2, u0))) ⩽ d3. (2.73) Then we obtain {σ(t, u0) : 0 ⩽ t < τ(u0)} is bounded in X. It follows from (2.57) that σ(t, u0) = e−tu0 + e−t ∫ t 0 esBλ(σ(s, u0))ds for 0 ⩽ t < τ(u0). (2.74) As a consequence of Lemma 2.23, Bλ(σ(s, u0)) : [0, τ(u0)) → Z is continuous. Since u0 ∈ Y ⊂ L∞(Ω) and since Bλ : X0 → L∞(Ω) and Bλ : (X0 ∩ L∞(Ω), ∥ · ∥∞) → Z is bounded, {e−t ∫ t 0 esBλ(σ(s, u0))ds : 0 ⩽ t < τ(u0)} is bounded in Z and relatively compact in Y . This fact and (2.74) imply that {σ(t, u0) : 0 ⩽ t < τ(u0)} is relatively compact in Y . So limn→∞ ∥σ(tn, u0)− u3∥Y = 0 and u3 ∈ ∂D2CD2((intY D1) ∩ D2). Since {σ(tn, u0)} ⊂ ∂D2CD2 ( (intY D1) ∩ D2 ) , we have u3 /∈ (intY D1) ∩ D2. So u3 /∈ intY D1. Thus, u3 = Aλ(u3) and Aλ(D1) ⊂ intY D1 imply u3 /∈ D1. So u3 /∈ G1 and u3 ̸= u1. It follows from u3 ∈ D2 that u3 is the other positive solution. Similarly, we can find a negative solution u4 ∈ ∂D3 CD3 ((intY D1) ∩D3) and u4 ̸= u2. Step 3. There exists a sign-changing solution u5 ∈ ∂Y CY (intY D1). By (2.69), we can infer that CY (intY D1) ̸= Y and so ∂Y CY (intY D1) ̸= ∅. It follows from Lemma 2.29 that CY (intY D1) is an open invariant set of (2.57). Let Y2 ⊂ Y be a two-dimensional subspace of Y . So CY (intY D1)∩Y2 is an open subset of Y2. According to Lemma 2.30, there exists a connected component Γ′ of ∂Y2(CY (intY D1)∩Y2), and each one sided ray l through the origin of Y2 satisfies l∩Γ′ ̸= ∅. Let Γ be the connected component of ∂Y CY (intY D1) containing Γ′. Then Γ is an invariant set of (2.57). Obviously, CΓ(intY D2 ∩Γ) and CΓ(intY D3 ∩Γ) are two open subsets of Γ. By the connectedness of Γ we see that Λ := Γ \ (CΓ(intY D2 ∩ Γ) ∪ CΓ(intY D3 ∩ Γ)) ̸= ∅. Since Γ is an invariant set of (2.57), CΓ(intY D2 ∩ Γ) and CΓ(intY D3 ∩ Γ) are two complete invariant set of (2.57) in Γ, Λ is closed invariant set of (2.57) in Γ. Moreover, c = inf u∈Λ Jλ(u) ⩾ inf u∈∂Y CY (intY D1) Jλ(u) > −∞. Then for any u0 ∈ Λ, we obtain {σ(t, u0) : 0 ⩽ t < τ(u0)} ⊆ Λ and c ⩽ Jλ(σ(t, u0)) ⩽ Jλ(u0) for 0 ⩽ t < τ(u0). Using a similar argument as before we find an increasing sequence {tn} with tn → τ(u0) and u5 ∈ Kλ such that lim n→∞ ∥σ(tn, u0)− u5∥Y = 0. Since u5 ∈ Λ, we obtain u5 /∈ intY D2 and u5 /∈ intY D3. By Lemma 2.21, we can infer that u5 /∈ D2 and u5 /∈ D3. Thus, u5 is a sign-changing solution. Step 4. There exists a sign-changing solution u6 ∈ ∂D1CD1((intY D2) ∩D1). Since ((intY D2) ∩ D1) ∩ ((intY D3) ∩D1) = ∅ and ((intY D3) ∩D1) is an invariant set of descending flow of (2.57), we have CD1 ((intY D2) ∩ D1) ̸= D1. Then ∂D1 CD1 ((intY D2) ∩ D1) ̸= ∅. By Lemma 2.29, we have ∂D1 CD1 ((intY D2) ∩ D1) is a closed invariant set of descending flow of (2.57). Let c̃ = infu∈∂D1 CD1 ((intY D2)∩D1) Jλ(u) > −∞. So there exists {vn} ⊂ ∂D1 CD1 ((intY D2) ∩D1) such that c̃ ⩽ Jλ(vn) ⩽ c̃+ 1 n ⩽ c̃+ 1. 22 X. ZHANG, X. XU EJDE-2025/23 Using a similar argument as before we find an increasing sequence {tm} with tm → τ(vn) and ṽn ∈ Kλ such that lim m→∞ ∥σ(tm, vn)− ṽn∥X = 0 for all n ∈ N, lim m→∞ ∥σ(tm, vn)− ṽn∥Y = 0 for all n ∈ N. Since J ∈ C1(X,R), we obtain J ′ λ(ṽn) = 0 and c̃ ⩽ Jλ(ṽn) ⩽ Jλ(vn) ⩽ c̃+ 1 n ⩽ c̃+ 1 for any n ∈ N. Since Jλ satisfies the Palais-Smale condition, there exists a subsequence {ṽnk } and v0 ∈ Kλ such that lim k→∞ ∥ṽnk − v0∥X = 0. This implies {ṽnk } is bounded in X. By Lemma 2.19, there exists M1 > 0 such that ∥ṽnk ∥∞ ⩽M1 for all k ∈ N. By the nonlinear regularity theory of Liebermann [13], we can find β ∈ (0, 1) and M2 > 0 such that ṽnk ∈ C1,β 0 (Ω) and ∥ṽnk ∥C1,β 0 (Ω) ⩽M2 for all k ∈ N. (2.75) The compact embedding of C1,β 0 (Ω) into C1 0 (Ω) and (2.75) imply at least for a subsequence we have ṽnk → v0 in C1 0 (Ω). This implies v0 ∈ ∂D1 CD1 ((intY D2) ∩ D1) and Jλ(v0) = c̃. Since ∂D1 CD1 ((intY D2) ∩ D1) ∩ (intY D3) ∩D1) = ∅, v0 /∈ (intY D3) ∩D1. Since Aλ(D3) ⊂ intY D3, v0 /∈ D3 . This shows that v0 is not a negative solution. Since v0 ∈ ∂D1CD1((intY D2) ∩ D1), v0 /∈ D2. This shows that v0 is not a positive solution. Thus, setting v0 = u6, we have that u6 is either a trivial solution or a sign-changing solution. Next, we show that u6 is a sign-changing solution. Let e3, e4 ∈ Y be linearly independent with e3 ∈ (intY D2) ∩D1, e4 ∈ Y \(P1 ∪ (−P1)) and denote Y3 = span{e3, e4}. Since Jλ(u) ⩽ c3 p ∫ Ω |Du|pdz + c4 q ∥Du∥qq − λ s ∫ Ω |u|sdz, there exists ε > 0 such that sup u∈Y3∩Sε Jλ(u) < 0, where Sε = {u ∈ Y : ∥u∥Y = ε}. By Lemma 2.23, if ε > 0 is small enough, we can choose w1 ∈ ( intY P1 ∩ Sε ) such that w1 ∈ CD1 ((intY D2) ∩D1). Choose a w2 ∈ int Y (−P1) ∩ Sε. By the connectedness of Sε, it follows that Sε ∩ ∂D1CD1((intY D2) ∩D1) ∩ Y3 ̸= ∅. Then we have Jλ(v0) = c̃ ⩽ inf u∈∂D1 CD1 ( (intY D2)∩D1 ) ∩Sε∩Y3 Jλ(u) < 0. This implies v0 ̸= 0, and so v0 = u6 is a sign-changing solution. The proof is complete. □ Acknowledgments. This research was supported by the National Scince Foundation of China (11871250). References [1] A. Ambrosetti, H. Brezis, G. Cerami; Combined effects of concave and convex nonlinearities in some elliptic problems. J. Funct. Anal., 122 (1994), no. 2, 519-543. [2] M. Badiale, E. 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