Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 01, pp. 1–13. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu POSITIVE SOLUTIONS FOR A CLASS OF φ-LAPLACIAN DIFFERENTIAL SYSTEMS WITH MULTIPLE PARAMETERS XIAOZHU YU, SHIWEN JING, HAIRONG LIAN Abstract. In this article, we consider the double eigenvalue problem for a φ-Laplacian differential system. We prove the existence of positive solutions under the φ-super-linear condition by means of the Guo-Krasnosel’skii fixed point theorem and the topological degree. It is shown that there exists a continuous curve splitting R2 +\{(0, 0)} into disjoint subsets such that systems has at least two, at least one, or no positive solutions according to parameters in different subsets. 1. Introduction Eigenvalue problems have been studied for many years. There are lots of im- portant results on positive solutions of nonlinear problems. Among them, we refer to [3, 11, 18, 19, 22] for ordinary differential systems, to [4, 5, 7, 8, 9, 13, 14] for elliptic equations of partial differential systems, and to [1, 2, 6, 10, 15, 16, 17, 20, 21] for Laplacian systems. The technical methods are mainly the fixed point theory, Leray-Schauder degree theory, upper and lower solution method, and variational methods. Dunninger and Wang [8] considered the Dirichlet boundary value problem of the problem ∆u+ λk1(|x|)f(u, v) = 0, ∆v + µk2(|x|)g(u, v) = 0, where Ω = {x ∈ RN : R1 < |x| < R2, R1, R2 > 0} is an Annulus. They used the cone expansion and compression theorem to obtain sufficient conditions for the existence of the existence and multiplicity results. Later in [9], they extended the existence results to a more generalized boundary condition with f(0, 0) > 0, g(0, 0) > 0, α1u+ β1 ∂u ∂n = 0, α2u+ β2 ∂v ∂n = 0, |x| = R1, γ1u+ δ1 ∂u ∂n = 0, γ2u+ δ2 ∂v ∂n = 0, |x| = R2. They proved that superlinearity or sublinearity at zero or infinity of the nonlinearity can guarantee the existence results of differential systems. 2010 Mathematics Subject Classification. 34B16, 34L15. Key words and phrases. φ-Laplacian differential systems; eigenvalue; fixed point theorem; degree theory; positive solution. ©2022. This work is licensed under a CC BY 4.0 license. Submitted August 27, 2021. Published January 5, 2022. 1 2 X. YU, S. JING, H. LIAN EJDE-2022/01 Lee [14] considered the the radial solutions for semilinear elliptic systems on an annulus. By applying consecutive change of variables, the author transformed the partial differential equations into the ordinary differential equations u′′(t) + λh1(t)f(u(t), v(t)) = 0, v′′(t) + µh2(t)g(u(t), v(t)) = 0, with homogeneous and semi homogeneous Dirichlet boundary conditions. The au- thor proved the existence of one positive and multiple positive solutions based on the upper and lower solution technique, cone expansion and compression theorem and degree theory. For the multiple parameters of φ-Laplace systems, Wang [20, 21] considered the single parameter of the n-dimensional system (Φ(u′))′ + λh(t)f(u) = 0, 0 < t < 1, with one of the following conditions u(0) = u(1) = 0, u′(0) = u(1) = 0, u(0) = u′(1) = 0, where λ is a single parameter and Φ = (φ1, · · · , φn). Under suitable sufficient conditions, the author proved that when the nonlinearity is φ-suplinearity or sub- linearity at zero or infinity, there exists a λ0 such that the above n-dimensional system has at least two, at least one, or no positive radial solutions according to λ large enough or not. Lee et al [15] discussed a generalized Gelfand type Laplacian system with a vector parameter (Φ(u′))′ + λh(t)f(u) = 0, u(0) = u(1) = 0, where λ = diag(λ1, · · · , λn). By using the upper and lower solution method and fixed point index theory, they obtained a global multiplicity result with respect to the parameter. Recently, Lee and Xu [16] extended the existence result to a more general singular (p1, p2)-Laplacian system with multi-parameters. Inspired by the work listed above, we aim to study the double eigenvalue problem of φ-Laplace differential equations with mixed boundary condition (φ(u′(t)))′ + λh1(t)f(u(t), v(t)) = 0, 0 < t < 1, (φ(v′(t)))′ + µh2(t)g(u(t), v(t)) = 0, 0 < t < 1, u′(0) = v′(0) = u(1) = v(1) = 0, (1.1) where φ is an increasing and odd homemorphism. Throughout this article, we as- sume that the multiparameter λ, µ ∈ R2 +\{(0, 0)}, R+ = [0,+∞), hi ∈ C([0, 1],R+) is not always zero in any subinterval of [0, 1], i = 1, 2. f, g ∈ C(R2 +,R+). We define f∞ := lim |u|+|v|→∞ f(u, v) φ(|u|+ |v|) , g∞ := lim |u|+|v|→∞ g(u, v) φ(|u|+ |v|) . and use the assumptions (H1) f(u, v) and g(u, v) are quasi-nondecreasing on u, EJDE-2022/01 POSITIVE SOLUTIONS FOR φ-LAPLACIAN 3 (H2) f∞ = g∞ =∞. We summarize our main results as follows. Theorem 1.1. Suppose that (A1) and (A2) hold. Then there exists a continuous curve Γ separating R2 + \ {(0, 0)} into two disjoint subsets θ1 and θ2 such that the eigenvalue problem (1.1) has at least one positive solution for (λ, µ) ∈ θ1 ∪ Γ and has no solution for (λ, µ) ∈ θ2. The article is organized as follows: In Section 2, we present some lemmas and a theorem of upper and lower solutions method for our systems. In Section 3, we show that the parameters lying in different area make a difference for the existence and nonexistence results. In Section 4, we demonstrate the multiplicity results. 2. Preliminaries A cone and partial order are important definitions when the positive solution is discussed. After introducing the fixed point theorems on cones, we establish the upper and lower solution theory for our systems. Definition 2.1. Let X be a Banach space and K be a closed and convex subset of X. We call that K is a cone of X if and only if (1) If x ∈ K and λ ≥ 0, then λx ∈ K. (2) If x ∈ K and −x ∈ K, then x = θ, where θ is the zero element of X. Definition 2.2. Let Ω be a subset of R2. We call that P is a binary partial order of Ω if the following conditions hold. (1) For all (λ1, µ1) ∈ Ω, {(λ1, µ1), (λ1, µ1)} ∈ P . (2) If {(λ1, µ1), (λ2, µ2)} ∈ P, and {(λ2, µ2), (λ1, µ1)} ∈ P , then (λ1, µ1) = (λ2, µ2). (3) If {(λ1, µ1), (λ2, µ2)} ∈ P and {(λ2, µ2), (λ3, µ3)} ∈ P , then {(λ1, µ1), (λ3, µ3)} ∈ P . If {(λ1, µ1), (λ2, µ2)} ∈ P , we denote (λ1, µ1) ≤ (λ2, µ2). For convenience, we also use λ1 ≤ λ2, µ1 ≤ µ2 to mean (λ1, µ1) ≤ (λ2, µ2). The following lemmas are fixed point index theorems. We refer to Guo and Lakshmikantham [12] for proofs and further discussion of fixed point index. Lemma 2.3. Let X be a Banach space, and K be a cone in X. Ω is bounded open in X. Let 0 ∈ Ω and T : K ∩ Ω→ K be condensing. Suppose that Tx 6= νx for all x ∈ k ∩ ∂Ω and ν ≥ 1. Then i(T,K ∩ Ω,K) = 1. Lemma 2.4. Let X be a Banach space and K be a cone in X. For r > 0, define Kr = {x ∈ K : ‖x‖ < r}. Suppose that T : Kr → K is a compact operator such that Tx 6= x for x ∈ ∂Kr. If ‖x‖ ≤ ‖Tx‖ for x ∈ ∂Kr, then i(T,Kr,K) = 0. Consider the auxiliary boundary value problem (φ(u′(t)))′ + F1(t, u(t), v(t)) = 0, 0 < t < 1, (φ(v′(t)))′ + F2(t, u(t), v(t)) = 0, 0 < t < 1, u′(0) = v′(0) = a ≥ 0, u(1) = v(1) = b, (2.1) where Fi : [0, 1]× R2 → R are continuous functions for i = 1, 2. 4 X. YU, S. JING, H. LIAN EJDE-2022/01 Definition 2.5. We say a function F (t, u, v) is quasi-monotone nondecreasing with respect to (u, v), if for every fixed t ∈ [0, 1], F (t, u1, v1) ≤ F (t, u2, v2), whenever u1 ≤ u2, v1 ≤ v2. We say F (t, u, v) is quasi-monotone nondecreasing with respect to v(or u) if F (t, u, v1) ≤ F (t, u, v2) whenever v1 ≤ v2 . (or F (t, u1, v) ≤ F (t, u2, v) whenever u1 ≤ u2 .) Definition 2.6. The functions α1, α2 ∈ C([0, 1],R) are lower solutions of (2.1) if for all t ∈ (0, 1), it holds (φ(α′1(t)))′ + F1(t, α1(t), α2(t)) ≥ 0, 0 < t < 1, (φ(α′2(t)))′ + F2(t, α1(t), α2(t)) ≥ 0, 0 < t < 1, α′1(0) ≤ a, α1(1) ≤ b, α′2(0) ≤ a, α2(1) ≤ b. Similarly, we can define the upper solution (β1, β2) ∈ C([0, 1],R) × C([0, 1],R) of (2.1) if it satisfies the reverse inequality. Consider the Banach space X = C([0, 1],R)×C([0, 1],R). For u ∈ X, the norm is defined by ‖u‖ = maxt∈[0,1] |u(t)|, and ‖(u, v)‖ = ‖u‖+ ‖v‖. Here, we recall that the norm in R2 can be defined by |(u, v)| = |u| + |v|. Let α(t) = (α1(t), α2(t)), β(t) = (β1(t), β2(t)) and Dβα = {(t, u, v)|t ∈ [0, 1], α1(t) ≤ u(t) ≤ β1(t), α2(t) ≤ v(t) ≤ β2(t)}. For the next lemma we use the assumptions: (H3) (α1(t), α2(t)) ≤ (β1(t), β2(t)) for t ∈ [0, 1]; (H4) |Fi(t, u, v)| ≤ wi(t) for all (t, u, v) ∈ Dβα, where wi ∈ C([0, 1],R+) satisfy∫ 1 0 φ−1 (∫ s 0 wi(r)dr ) ds <∞, i = 1, 2; (H5) F1(t, u, v) is quasi-monotone nondecreasing on v and F2(t, u, v) is quasi- monotone nondecreasing on u. Lemma 2.7. Let (α1, α2), (β1, β2) be a pair of lower and upper solution of (2.1). Assume that (H3)–(H5) hold. Then problem (2.1) has at least one solution (u, v) in X satisfying (α1, α2) ≤ (u, v) ≤ (β1, β2). To prove the above lemma, we need to prove the following result first. Lemma 2.8. If there exist w1, w2 ∈ C((0, 1),R+) such that (H4) holds, then (2.1) is solvable. Proof. We define the operators q1 and q2 by q1(u, v) = b− ∫ 1 t φ−1 ( φ(a)− ∫ s 0 F1 ( r, u(r), v(r) ) dr ) ds, q2(u, v) = b− ∫ 1 t φ−1 ( φ(a)− ∫ s 0 F2 ( r, u(r), v(r) ) dr ) ds EJDE-2022/01 POSITIVE SOLUTIONS FOR φ-LAPLACIAN 5 and let T (u, v)(t) = ( q1(u, v)(t), q2(u, v)(t) ) . Clearly, T is well defined on X and the solution of (2.1) corresponds to the fixed point of T . From (H4), we can see that T is continuous. Furthermore, for any bounded set B ⊂ X, TB is uniformly bounded and if (u, v) ∈ B, t1, t2 ∈ [0, 1], it has |qi(u, v)(t1)− qi(u, v)(t2)| ≤ ∣∣∣ ∫ t2 t1 φ−1 (∫ s 0 φ−1 ( φ(a)− Fi(τ, u(τ), v(τ)) ) dτ ) ds ∣∣∣ ≤ ∣∣∣ ∫ t2 t1 φ−1 (∫ s 0 φ−1 ( φ(a) + ηkw1(τ) ) dτ ) ds ∣∣∣ → 0, uniformly as t1 → t2. Which implies that TB is equi-continuous. By Arezla-Ascoli Theorem, T is com- pletely continuous operator. Schauder fixed point theorem guarantees that T has a fixed point (u, v) in X which is the solution of (2.1). � Let Xi(t) = b− ∫ 1 t φ−1 ( φ(a)− ∫ s 0 wi(r)dr ) ds, i = 1, 2. Then X(t) = (X1(t), X2(t)) is the solution of the problem (Φ(u′))′ + w(t) = 0, 0 < t < 1, u′(0) = a, u(1) = b, where w = (w1, w2), w1, w2 defined in Lemma 2.7. Similarly, we define xi(t) = b− ∫ 1 t φ−1 ( φ(a) + ∫ s 0 wi(r)dr ) ds, i = 1, 2 . Then x(t) = (x1(t), x2(t)) is the solution of the problem (Φ(u′))′ −w(t) = 0, t ∈ (0, 1), u′(0) = a, u(1) = b. It is easy to prove that the solution of (2.1) satisfies x ≤ (u, v) ≤ X. The proof of Lemma 2.7. We consider the auxiliary boundary value problem (Φ(u′))′ + F ∗(t,u) = 0, 0 < t < 1, u′(0) = a, u(1) = b, (2.2) where u = (u1, u2), Φ(u) = (φ(u1), φ(u2)), F ∗ = (F ∗1 , F ∗ 2 ) and F ∗i : (0, 1)×R2 → R is defined as F ∗i (t, u1, u2) =  Fi(t, y1(t), y2(t))− wi(t) ui−βi(t) 1+ui−βi(t) , ui > βi(t), Fi(t, y1(t), y2(t)), αi(t) ≤ ui(t) ≤ βi(t), Fi(t, y1(t), y2(t)) + wi(t) αi(t)−ui 1+αi(t)−ui , ui < αi(t), where yi(t) =  βi(t), ui > βi(t), ui(t), αi(t) ≤ ui(t) ≤ βi(t), αi(t), ui < αi(t), 6 X. YU, S. JING, H. LIAN EJDE-2022/01 for i = 1, 2. Obviously, F ∗i is continuous and |F ∗i (t, u1, u2)| ≤ wi(t), i = 1, 2. From Lemma 2.8, we can see that (2.2) is solvable, its solution is denoted by u = (u1, u2). Let α(t) = (α1(t), α2(t)), β(t) = (β1(t), β2(t)) and α′(0) ≤ u′(0) ≤ β′(0), α(1) ≤ u(1) ≤ β(1). We will show that α(t) ≤ u(t) ≤ β(t), t ∈ [0, 1). On the contrary, suppose the right inequality does not hold. Then there exists j = 1 or 2 such that uj(t) 6≤ βj(t). Suppose there is t0 ∈ [0, 1) such that uj(t0)− βj(t0) = max t∈[0,1] (uj − βj)(t) > 0, u′j(t0)− β′j(t0) = 0, and for t ∈ (t0, t0 + δ], u′j(t)− β′j(t) < 0. (2.3) Note that yj(t0) = βj(t0) and yi(t0) ≤ βi(t0) for i 6= j. From (H5) it follows that F ∗j (t0, u1(t0), u2(t0)) = Fj(t0, y1(t0), y2(t0))− wj(t0) uj(t0)− βj(t0) 1 + uj(t0)− βj(t0) < Fj(t0, y1(t0), y2(t0)) ≤ Fj(t0, β1(t0), β2(t0)). Because F ∗j and Fj are continuous, we can choose δ0 ∈ (0, δ) such that F ∗j (t,u(t)) < Fj(t,β(t)) holds for t ∈ (t0, t0 + δ0). Notice that β is the upper solution of (2.1), when t ∈ (t0, t0 + δ0), it holds φ(u′j(t))− φ(u′j(t0)) = − ∫ t t0 F ∗j (s,u(s))ds > − ∫ t t0 Fj(s,β(s))ds > φ(β′j(t))− φ(β′j(t0)). Furthermore, because u′j(t0) = β′j(t0) and φ is increasing, it has u′j(t) > β′j(t), t ∈ (t0, t0 + δ0) ⊂ (t0, t0 + δ), which is an contradiction with (2.3). So u(t) ≤ β(t) holds for t ∈ [0, 1]. The left inequality can be similarly proved. � 3. Existence of positive solutions In this section, we prove Theorem 1.1 based on the upper and lower solution technique. We first prove some lemmas. We define three operators on X by Tλ(u, v)(t) = ∫ 1 t φ−1 (∫ s 0 λh1(τ)f ( u(τ), v(τ) ) dτ ) ds, Tµ(u, v)(t) = ∫ 1 t φ−1 (∫ s 0 µh2(τ)g ( u(τ), v(τ) ) dτ ) ds, T (u, v)(t) = ( Tλ(u, v)(t), Tµ(u, v)(t) ) for t ∈ [0, 1]. Clearly, T : X → X is continuous and the fixed points of T corresponds to the solution of (1.1). Let P and K be defined by P = {(u, v) ∈ X : u, v ≥ 0}, K = {(u, v) ∈ P : u′(0) = v′(0) = 0 = u(1) = v(1), v are convex on (0, 1)}. Then P and K are both cones on X. Obviously, T (P) ⊂ K. EJDE-2022/01 POSITIVE SOLUTIONS FOR φ-LAPLACIAN 7 Lemma 3.1. If u ∈ C1([0, 1],R), φ(u′) ∈ C1((0, 1),R) and (φ(u′))′ ≤ 0, then u(t) ≥ min{t, 1− t} max t∈[0,1] u(t), t ∈ [0, 1]. Proof. Because (φ(u′))′ ≤ 0 and φ is increasing, u′ is increasing. For 0 ≤ t0 < t < t1 ≤ 1, we have u(t)− u(t0) = ∫ t t0 u′(s)ds ≥ (t− t0)u′(t), that is, u′(t) ≤ u(t)− u(t0) t− t0 . Similarly, we have u′(t) ≥ u(t1)− u(t) t1 − t . Therefore, u(t1)− u(t) t1 − t ≤ u(t)− u(t0) t− t0 . (3.1) Which is equivalent to u(t) ≥ (t− t0)u(t1) + (t1 − t)u(t0) t− t0 ≥ (t− t0)u(t1) + (t1 − t)u(t0). Let σ = {t∗|u(t∗) = maxt∈[0,1] u(t)}, t0 = 0, t1 = σ (or t0 = σ, t1 = 1). Then u(t) ≥ t max t∈[0,1] u(t), t ∈ [0, σ],( or u(t) ≥ (1− t) max t∈[0,1] u(t), t ∈ [σ, 1] ) . So the proof is complete. � Remark 3.2. If φ and u satisfy the conditions in Lemma 3.1, then min 1/4 0 such that for all (λ, µ) ∈ R∗ and all positive solutions (u, v) of (1.1) at (λ, µ), we have ‖(u, v)‖ < bR∗ . Proof. Suppose by contradiction that there is a sequence (un, vn), being positive solutions of (1.1) at (λn, µn) ∈ R∗(n = 1, 2, · · · ), such that ‖(un, vn)‖ → ∞ as n→∞. From Remark 3.2, min 1/4≤t≤3/4 (un(t) + vn(t)) ≥ 1 4 (‖un‖+ ‖vn‖). (3.2) Noticing that (λn, µn) ∈ R2 +\{(0, 0)}, without loss of generality, we suppose λn > 0. From condition (H2), we can choose Rf > 0 such that f(un, vn) ≥ ηφ(un + vn), ‖un‖+ ‖vn‖ ≥ Rf , (3.3) where η satisfies φ−1(λnη) 4 ∫ 3/4 t φ−1 (∫ s 1/4 h1(τ)dτ ) ds > 1, for t ∈ [1/4, 3/4). (3.4) 8 X. YU, S. JING, H. LIAN EJDE-2022/01 By inequalities (3.2)–(3.4), we have ‖un‖ ≥ un(t) = ∫ 1 t φ−1 (∫ s 0 λnh1(τ)f ( un(τ), vn(τ) ) dτ ) ds ≥ ∫ 3/4 t φ−1 (∫ s 1/4 λnh1(τ)f ( un(τ), vn(τ) ) dτ ) ds ≥ ∫ 3/4 t φ−1 (∫ s 1/4 λnh1(τ)ηφ ( un(τ) + vn(τ) ) dτ ) ds ≥ ∫ 3/4 t φ−1 (∫ s 1/4 h1(τ)dτ ) φ−1 ( λnη · φ (1 4 (‖un‖+ ‖vn‖) )) ds = ‖un‖+ ‖vn‖) φ−1(λ · η) 4 ∫ 3/4 t φ−1 (∫ s 1/4 h1(τ)dτ ) ds > ‖un‖+ ‖vn‖ > ‖un‖, for n sufficiently large and ‖un‖ + ‖vn‖ ≥ Rf , which is a contradiction. So u is bounded. For the case λn = 0, µn > 0, it can be proved similarly by using g∞ =∞. � Lemma 3.4. Suppose that (H1) holds. If (1.1) has a positive solution at (λ̄, µ̄), then (1.1) has a positive solution at (λ, µ) for all (λ, µ) ≤ (λ̄, µ̄). Proof. Let (ū, v̄) be a positive solution of (1.1) at (λ̄, µ̄) and (λ, µ) ∈ R2 0 \ {(0, 0)} satisfying (λ, µ) ≤ (λ̄, µ̄). Then (φ(ū′(t)))′ = −λ̄h1(t)f(ū(t), v̄(t)) ≤ λh1(t)f(ū(t), v̄(t)), (φ(v̄′(t)))′ = −µ̄h2(t)g(ū(t), v̄(t)) ≤ µh2(t)g(ū(t), v̄(t)), ū′(0) = v̄′(0) = 0 = ū(1) = v̄(1). So (ū, v̄) is an upper solution of (1.1) at (λ, µ). Similarly, (0, 0) is a lower solution of (1.1) at (λ, µ). Since (ū, v̄) 6= (0, 0), (ū, v̄) ≥ (0, 0) and (0, 0) is not a solution of (1.1) at (λ, µ), Lemma 2.7 implies that (1.1) has a positive solution at (λ, µ). � Lemma 3.5. Suppose that (H1) holds. Then there exists (λ0, µ0) > (0, 0) such that (1.1) has a positive solution for all (λ, µ) ≤ (λ0, µ0). Proof. Let βi(t) = ∫ 1 t φ−1 (∫ s 0 hi(τ)dτ ) ds for i = 1, 2. Then it is the unique solution of (φ(u′(t)))′ + hi(t) = 0, 0 < t < 1, u′(0) = 0 = u(1). Let Mf = maxt∈[0,1] f(β1(t), β2(t)) and Mg = maxt∈[0,1] g(β1(t), β2(t)). Then Mf > 0,Mg > 0, and when (λ0, µ0) = (1/Mf , 1/Mg), we have (φ(β′1(t)))′ + λ0h1(t)f(β1(t), β2(t)) = h1(t)[λ0f(β1(t), β2(t))− 1] ≤ 0, (φ(β′2(t)))′ + µ0h2(t)g(β1(t), β2(t)) = h2(t)[µ0g(β1(t), β2(t))− 1] ≤ 0. EJDE-2022/01 POSITIVE SOLUTIONS FOR φ-LAPLACIAN 9 This indicates that (β1, β2) is an upper solution of (1.1) at (λ0, µ0). Meanwhile, (0, 0) is a lower solution of (1.1) at (λ0, µ0) and (0, 0) < (β1, β2). Lemma 2.7 implies that (1.1) has a positive solution at (λ0, µ0). � We define ψ = {(λ, µ) ∈ R2 + \ {(0, 0)} : (1.1) has a positive solution at(λ, µ)}. From Lemma 3.5, we can see that ψ 6= ∅. (ψ,≤) is a partial ordered set. Lemma 3.6. Suppose that (H1) and (H2) hold. Then (ψ,≤) is bounded above. Proof. On the contrary, if (ψ,≤) is not bounded above, then there exists (λn, µn) such that (1.1) has solutions at (λn, µn) with λn → ∞ or µn → ∞ as n → ∞. Without loss of generality, we suppose λn → ∞ as n → ∞). Let (un, vn) are the positive solutions of (H3) at (λn, µn). From (H2), similarly to the proof of Lemma 3.3, when n is sufficient large and λn > δ, we can obtain ‖un‖ > ‖un‖, which is a contraction. So (ψ,≤) has an upper bound. � Lemma 3.7. Suppose that (H1, (H2) hold. Then every chain in ψ has a unique supremum in ψ. Proof. Let ψ̄ be a chain in ψ. Because ψ is bounded above, ψ̄ has an upper bound. Without loss of generality, we choose a distinct sequence {(λn, µn)} ⊂ ψ̄ with (λn, µn) ≤ (λn+1, µn+1), n = 1, 2, . . . . From Lemma 3.6, two sequences {λn} and {µn} are convergent to, say, λψ̄ and µψ̄, respectively. Next, we show that λψ̄, µψ̄ ∈ ψ. Because {(λn, µn)}∞n=1 has an upper bound, we can suppose that the sequence {(λn, µn)}∞n=1 belongs to a compact rectangle of R2 + ⊂ {(0, 0)}. Let (un, vn) be the positive solutions of (H3) at (λn, µn). Lemma 3.3 implies that {(un, vn)} are uniformly bounded on X. So the sequence {(un, vn)} has a conver- gent subsequence, denoted by {(un, vn)} → (uψ̄, vψ̄) ∈ X. By using the Lebesgue convergence theorem, we can prove that (uψ̄, vψ̄) is a solution of (1.1) at (λψ̄, µψ̄), that is, (λψ̄, µψ̄) ∈ ψ. Hence, the proof is complete. � Lemma 3.8. Suppose that (H1) and (H2) hold. Then there exists λ∗, µ∗ > 0 such that {(λ, 0)|λ ∈ [0, λ∗]} ∪ {(0, µ)|µ ∈ [0, µ∗]} ⊂ ∂(intψ) and intψ ⊂ [0, λ∗]× [0, µ∗]. Proof. Lemma 3.5 implies that intψ is nonempty and there exists (λ0, µ0) ∈ intψ ⊂ R2 +. Lemma 3.4 shows that (0, λ0]× (0, µ0] ⊂ intψ and {(λ0, 0), (0, µ0)} ⊂ ∂(intψ). So {λ > 0|(λ, 0) ∈ ∂(intψ)} and {µ > 0|(0, µ) ∈ ∂(intψ)} are nonempty and bounded above. Let λ∗ = sup{λ > 0|(λ, 0) ∈ ∂(intψ)}, µ∗ = sup{µ > 0|(0, µ) ∈ ∂(intψ)}, (3.5) then {(λ∗, 0), (0, µ∗)} ⊂ ∂(intψ), and {(λ, 0)|λ ∈ [0, λ∗]} ∪ {(0, µ)|µ ∈ [0, µ∗]} ⊂ ∂(intψ). (3.6) In fact, for all λ0 ∈ [0, λ∗), from Lemma 3.5, there exists (λ̄, µ̄) ∈ intψ ∩B((λ∗, 0), λ∗ − λ0) such that (0, λ̄) × (0, µ̄) ⊂ intψ, where B((λ∗, 0), λ∗ − λ0) is a circular area with (λ∗, 0) as its center and λ∗ − λ0 as its radius. In view of λ0 < λ̄, so (λ0, 0) ∈ ∂(intψ). From the selection of λ0, it holds {(λ, 0)|λ ∈ [0, λ∗)} ⊂ ∂(intψ). Similarly, {(0, µ)|µ ∈ [0, µ∗)} ⊂ ∂(intψ). 10 X. YU, S. JING, H. LIAN EJDE-2022/01 Next, we prove intψ ⊂ [0, λ∗]× [0, µ∗]. Suppose, on the contrary that (λ̄, µ̄) ∈ intψ and (λ̄, µ̄) /∈ [0, λ∗]× [0, µ∗] . Then λ̄ > λ∗ or µ̄ > µ∗. Without loss of generality, we suppose λ̄ > λ∗. From Lemma 3.4, [0, λ̄] × [0, µ̄] ⊂ ψ. So (λ̄, 0) ∈ ∂(intψ). While, (3.5) implies λ̄ ≤ λ∗, which is a contradiction. � We define the line cluster L(t) = {(λ, µ) ∈ R2 + \ {(0, 0)} : µ = λ− t}, where t ∈ [−µ∗, λ∗]. We define λ∗(t) = sup{λ | (λ, µ) ∈ L(t) ∩ intψ}, µ∗(t) = λ∗(t)− t, and Γ(t) = (λ∗(t), µ∗(t)). Lemma 3.9. Suppose that (H1) and (H2) hold. For any given t ∈ [−µ∗, λ∗], Γ(t) ∈ L(t) ∩ ∂(intψ). Proof. For any t ∈ [−µ∗, λ∗] and λ∗(t), there exists {(λn, µn)}∞n=1 ⊂ L(t) ∩ intψ with λn converging to λ∗(t). Notice µ∗(t) = λ∗(t)− t and (λ∗(t), µ∗(t)) ∈ L(t). So (λ∗(t), µ∗(t)) = limn→∞(λn, µn) and (λ∗(t), µ∗(t)) ∈ L(t) ∩ intψ. � Lemma 3.10. Suppose that (H1) and (H2) hold. Then (I) {Γ(t) | t ∈ [−µ∗, λ∗]} is continuous; (II) {Γ(t) | t ∈ [−µ∗, λ∗]} ∩ {(λ, µ)|λµ = 0} = {(λ∗, 0), (0, µ∗)}. Proof. (I) Obviously, λ∗(t) is increasing and µ∗(t) is decreasing. Let λ∗(tn) = λ∗n, µ ∗(tn) = µ∗n for any tn ∈ [−µ∗, λ∗]. For any given t1, t2 ∈ [−µ∗, λ∗], t1 < t2, we have |Γ(t1)− Γ(t2)| = |(λ∗1, µ∗1)− (λ∗2, µ ∗ 2)| = |(λ∗1 − λ∗2, µ∗1 − µ∗2)| = |λ∗1 − λ∗2|+ |µ∗1 − µ∗2| = (λ∗2 − λ∗1) + (µ∗1 − µ∗2) = (λ∗2 − µ∗2)− (λ∗2 − µ∗2) = t2 − t1 = |t2 − t1|, where | · | denotes the norm of R2. Therefore, Γ is equi-continuous. (II) It suffices to prove that {t | t ∈ [−µ∗, λ∗], λ∗(t) = 0} = {−µ∗} and {t | t ∈ [−µ∗, λ∗], µ∗(t) = 0} = {λ∗}. Let t ∈ (−µ∗, 0]. By Lemma 3.8, we have (0, µ∗) ∈ ∂(intψ). And there exists (λ0, µ0) ∈ intψ ∩ B((0, µ∗), ε/3) ⊂ R+ × R+, where ε = t+ µ∗. It is easy to see that λ0 − t < µ0. Furthermore, by Lemma 3.5, (λ0, λ0 − t) ∈ L(t) ∩ intψ, t ∈ (−µ∗, 0]. Therefore, λ∗(t) ≥ λ0 > 0. Meanwhile, because λ∗(t) is increasing on [−µ∗, λ∗], it holds λ∗(t) ≥ λ∗(0) > 0 for all t ∈ [0, λ∗]. So {t ∈ [−µ∗, λ∗], λ∗(t) = 0} = {−µ∗}. Similarly, we can prove that when {t ∈ [−µ∗, λ∗], it holds µ∗(t) = 0} = {λ∗}. � Proof of Theorem 1.1. From Lemmas 3.4–3.10, we can see that Theorem 1.1 holds. The curve Γ divides the set R2 + \ {(0, 0)} into two disjoint subsets θ1 and θ2, where θ1 is bounded and θ2 is unbounded. Lemma 3.5 shows that (1.1) has a positive solution at Γ(t), t ∈ [−µ∗, λ∗]. If (λ, µ) ∈ θ1, then (λ, µ) ∈ L(t), t ∈ [−µ∗, λ∗], and (λ, µ) < Γ(t), Lemma 3.8 shows that (1.1) has a solution at (λ, µ). If (λ, µ) ∈ θ2, then either t /∈ [−µ∗, λ∗] or (λ, µ) > Γ(t), t ∈ [−µ∗, λ∗]. In both cases, (1.1) has no solution at (λ, µ). � EJDE-2022/01 POSITIVE SOLUTIONS FOR φ-LAPLACIAN 11 4. Multiplicity of positive solutions In this section, we prove that (1.1) has another positive solution when (λ, µ) ∈ θ1. Choose t0 ∈ [−µ∗, λ∗] with (λ, µ) ∈ L(t0). From Lemma 3.5, it is known that (1.1) has a positive solution at Γ(t0), denoted by (u∗0, v ∗ 0), and set Γ(t0) = (λ∗0, µ ∗ 0). Obviously, (λ, µ) < (λ∗0, µ ∗ 0). For the convenience, set u∗ε (t) = u∗0(t) + ε, v∗ε (t) = v∗0(t) + ε for ε > 0. Lemma 4.1. If (λ, µ) ∈ θ1, then there exists ε0 > 0 such that for any ε ∈ (0, ε0], (u∗ε , v ∗ ε ) is an upper solution of (1.1) at (λ, µ). Proof. Choose a constant H > 0 such that f(u∗0(t), v∗0(t)) ≥ H, g(u∗0(t), v∗0(t)) ≥ H, 0 < t < 1. Because f and g are continuous, there exists ε0 > 0 such that∣∣f(u∗ε (t), v∗ε (t) ) − f ( u∗0(t), v∗0(t) )∣∣ < H(λ∗0 − λ) λ ,∣∣g(u∗ε (t), v∗ε (t) ) − g ( u∗0(t), v∗0(t) )∣∣ < H(µ∗0 − µ) µ hold for all t ∈ [0, 1] and 0 < ε ≤ ε0. Obviously, u′∗ε (0) = v′∗ε (0) = 0 and u∗ε (1) = v∗ε (1) = ε > 0. For 0 < λ < λ∗0, we have( φ ( u∗ε ′(t) ))′ + λh1(t)f(u∗ε (t), v ∗ ε (t)) = −λ∗0h1(t)f(u∗0(t), v∗0(t)) + λh1(t)f(u∗ε (t), v ∗ ε (t)) = λh1(t)[f(u∗ε (t), v ∗ ε (t))− f(u∗0(t), v∗0(t))]− (λ∗0 − λ)h1(t)f(u∗0(t), v∗0(t)) < H(λ∗0 − λ)h1(t)− (λ∗0 − λ)h1(t)f(u∗0(t), v∗0(t)) = (λ∗0 − λ)h1(t)(H − f(u∗0(t), v∗0(t)) ≤ 0, 0 < t < 1. Similarly, ( φ ( v∗ε ′(t) ))′ + µh2(t)g(u∗ε (t), v ∗ ε (t)) ≤ 0, 0 < t < 1. So (u∗ε , v ∗ ε ) is an upper solution of (1.1) at (λ, µ). � In addition, for 0 < ε ≤ ε0 and (u∗ε , v ∗ ε ), we have u∗ε (t) > ∫ 1 t φ−1 (∫ s 0 λh1(τ)f(u∗ε (τ), v∗ε (τ))dτ ) ds, (4.1) v∗ε (t) > ∫ 1 t φ−1 (∫ s 0 µh2(τ)g(u∗ε (τ), v∗ε (τ))dτ ) ds. (4.2) Theorem 4.2. Suppose that (H1) and (H2) hold. Then there exists a continuous curve Γ splitting R2 0 \ {(0, 0)} into two disjoint subsets θ1 and θ2 such that the eigenvalue problem (1.1) has at least two positive solutions on θ1, at least one solution on Γ, and no solution on θ2. Proof. From Theorem 1.1, suffices to prove that (1.1) has a second positive solution for (λ, µ) ∈ θ1. Let (λ, µ) ∈ θ1 and (λ, µ) ∈ L(t0). Let Ω = {(u, v) ∈ X : −ε < u(t) < u∗ε (t),−ε < v(t) < v∗ε (t), t ∈ [0, 1]}. Clearly, 0 ∈ Ω, Ω ⊂ X and Ω is bounded open. T : K ∩ Ω̄ → K is completely continuous. Suppose (u, v) ∈ K ∩ ∂Ω. There exists t0 ∈ [0, 1] such that either 12 X. YU, S. JING, H. LIAN EJDE-2022/01 u(t0) = u∗ε (t0) or v(t0) = v∗ε (t0). Without loss of generality, we suppose u(t0) = u∗ε (t0). Then by (H1) and (4.1), we arrive at Tλ(u, v)(t0) = ∫ 1 t0 φ−1 (∫ s 0 λh1(τ)f(u(τ), v(τ))dτ ) ds ≤ ∫ 1 t0 φ−1 (∫ s 0 λh1(τ)f(u∗ε (τ), v∗ε (τ))dτ ) ds < u∗ε (t0) = u(t0) ≤ νu(t0) for all ν ≥ 1. Thus for all (u, v) ∈ K ∩ ∂Ω and ν ≥ 1, T (u, v) 6= ν(u, v). If v(t0) = v∗ε (t0), by (H1) and (4.2), the prior inequality also holds. From Lemma 2.3, i(T,K ∩ Ω,K) = 1. From (H2), there exists Rf > 0 such that f(u, v) ≥ ηφ(u+ v), ‖u‖+ ‖v‖ ≥ Rf , (4.3) with η satisfies (3.4) with λn replacing with λ. Let R = max{bR∗ , 4Rf , ‖(u∗ε , v∗ε )‖}, where bR∗ is given in Lemma 3.3 with R∗ a rectangle in R2 + \ {(0, 0)} containing (λ, µ). Let KR = {(u, v) ∈ K : ‖(u, v)‖ < R}. Then ‖T (u, v)‖ ≥ ‖Tλ(u, v)‖ > ‖(u, v)‖. From Lemma 2.4, we have i(T,KR,K) = 0. 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Xiaozhu Yu School of Science, China University of Geosciences, Beijing 100083, China Email address: yuxiaozhu@cugb.edu.cn Shiwen Jing School of Science, China University of Geosciences, Beijing 100083, China Email address: 2019210011@email.cugb.edu.cn Hairong Lian (corresponding author) School of Science, China University of Geosciences, Beijing 100083, China Email address: lianhr@126.com 1. Introduction 2. Preliminaries 3. Existence of positive solutions 4. Multiplicity of positive solutions References