Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 45, pp. 1–15. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu POSITIVE SOLUTIONS FOR A NONLINEAR SYSTEM OF FOURTH-ORDER ORDINARY DIFFERENTIAL EQUATIONS QIUYUE WANG, LU YANG Abstract. In this article, we consider the existence of positive solutions for a nonlinear system of fourth-order ordinary differential equations. By con- structing a single cone P in the product space C[0, 1] × C[0, 1] and applying fixed point theorem in cones, we establish the existence of positive solutions for a system in which the nonlinear terms are both superlinear or sublinear. In addition, by the construction of the product cone K1 ×K2 ⊂ C[0, 1] ×C[0, 1] along with the product formula of fixed point theory on a product cone, we investigate the existence of positive solutions involving nonlinear terms, one uniformly superlinear or sublinear, and the other locally uniformly sublinear or superlinear. 1. Introduction and main results In this article, we consider the existence of positive solutions for the nonlinear system of fourth-order ordinary differential equations u(4)(t) + β1u ′′(t)− α1u(t) = f1(t, u(t), v(t)), t ∈ (0, 1), v(4)(t) + β2v ′′(t)− α2v(t) = f2(t, u(t), v(t)), t ∈ (0, 1), u(0) = u(1) = u′′(0) = u′′(1) = 0, v(0) = v(1) = v′′(0) = v′′(1) = 0, (1.1) where fi ∈ C([0, 1]×R+ ×R+,R+), R+ = [0,+∞) and βi, αi ∈ R (i = 1, 2) satisfy the following conditions: βi < 2π2, −β2 i /4 6 αi, αi/π 4 + βi/π 2 < 1. These conditions involve a two-parameter nonresonance condition, see [5, 6]. In recent years, there have been extensive attention on the existence of positive solutions for second-order ordinary differential equations and systems, see [3, 4, 8] and the references therein. For example, in [8], by applying the product formula of fixed point theory on product cone and fixed point theory in cones, He-Yang discussed the existence and multiplicity of positive solutions for a system of non- linear Sturm-Liouville equations. Clearly, when β1 = β2 = β, α1 = α2 = α, f1(t, u, v) = f(t, u) and f2(t, u, v) = f(t, v), system (1.1) reduces to the following 2010 Mathematics Subject Classification. 34B18, 47H11, 47N20. Key words and phrases. Positive solution; fixed point theory; ordinary differential equation. c©2020 Texas State University. Submitted December 9, 2019. Published May 19, 2020. 1 2 Q. WANG, L. YANG EJDE-2020/45 fourth-order ordinary differential equation w(4)(t) + βw′′(t)− αw(t) = f(t, w(t)), t ∈ (0, 1), w(0) = w(1) = w′′(0) = w′′(1) = 0. (1.2) Equation (1.2) is used to describe the deformation of an elastic beam supported at the end points. Much has been studied for (1.2), see [9, 6, 1, 2] and references therein. For instance, in [9], by employing the fixed point theory in cones, Li presented the existence of positive solutions for (1.2) under the conditions that f(t, w) is either superlinear or sublinear with respect to w at w = 0 and w = +∞. Motivated by the works mentioned above, we shall deal with the existence of positive solutions for system (1.1). The purpose of this paper is to extend the result in [9] from two different aspects. Firstly, we consider the existence of positive solutions for system (1.1) with su- perlinear or sublinear nonlinearities. For this problem, we can change the existence of positive solutions for system (1.1) into that of nontrivial fixed points for the corresponding compactly continuous mappings on a single cone P (see (2.2)) in product space C[0, 1]× C[0, 1] and then choose proper open sets Pr ⊂ P in which the superlinearity or sublinearity can be applied directly. By using the fixed point theory in cones, we obtain the existence of positive solutions for system (1.1). Our main results are the following. Theorem 1.1. Assume that f1 and f2 satisfy the condition lim sup u+v→0+ max t∈[0,1] ∑2 i=1 fi(t, u, v) λ1u+ λ2v < 1 < lim inf u+v→+∞ min t∈[0,1] ∑2 i=1 fi(t, u, v) λ1u+ λ2v , (1.3) where λi = π4 − βiπ2 − αi (i = 1, 2). Then system (1.1) has at least one nonzero nonnegative solution. Moreover, if f1(t, 0, v(t)) 6≡ 0 and f2(t, u(t), 0) 6≡ 0 for all (u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. Theorem 1.2. Suppose that f1 and f2 satisfy the following assumptions: lim inf u+v→0+ min t∈[0,1] ∑2 i=1 fi(t, u, v) λ1u+ λ2v > 1 > lim sup u+v→+∞ max t∈[0,1] ∑2 i=1 fi(t, u, v) λ1u+ λ2v , (1.4) where λi = π4 − βiπ2 − αi (i = 1, 2). Then system (1.1) has at least one nonzero nonnegative solution. Moreover, if f1(t, 0, v(t)) 6≡ 0 and f2(t, u(t), 0) 6≡ 0 for all (u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. Secondly, we investigate the existence of positive solutions for (1.1) involving nonlinear terms in which one is uniformly superlinear or sublinear and the other is locally uniformly sublinear or superlinear. In this case, it is very difficult to directly construct proper open sets in the single cone P on product space owing to the different features of nonlinear terms. In order to overcome the obstacle, we will construct a product cone K1×K2 which is the Cartesian product of two cones {Ki}2i=1 ⊂ C[0, 1] (see (2.2)) and then choose proper open sets D = D1 × D2 ⊂ K1 ×K2, such that the different features of nonlinearities can be exploited better. Applying the product formula for the fixed point index on product cone and the fixed point index theory in cones, we establish the existence of positive solutions for system (1.1). Our main results are the following. EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 3 Theorem 1.3. Assume that f1 and f2 satisfy the following conditions: lim sup u→0+ max t∈[0,1] f1(t, u, v) u < λ1 < lim inf u→+∞ min t∈[0,1] f1(t, u, v) u (1.5) uniformly with respect to v ∈ R+; lim inf v→0+ min t∈[0,1] f2(t, u, v) v > λ2 > lim sup v→+∞ max t∈[0,1] f2(t, u, v) v (1.6) locally uniformly with respect to u ∈ [0,M ] for all M > 0, where λi = π4−βiπ2−αi (i = 1, 2). Then system (1.1) has at least one positive solution. Theorem 1.4. Suppose that f1 and f2 satisfy the following assumptions lim inf u→0+ min t∈[0,1] f1(t, u, v) u > λ1 > lim sup u→+∞ max t∈[0,1] f1(t, u, v) u (1.7) uniformly with respect to v ∈ R+; lim sup v→0+ max t∈[0,1] f2(t, u, v) v < λ2 < lim inf v→+∞ min t∈[0,1] f2(t, u, v) v (1.8) locally uniformly with respect to u ∈ [0,M ] for all M > 0, where λi = π4−βiπ2−αi (i = 1, 2). Moreover, if there exists a locally bounded function g = g(u) such that lim sup v→+∞ max t∈[0,1] f1(t, u, v) = g(u) (1.9) locally uniformly with respect to u ∈ [0,M ] for all M > 0, then system (1.1) has at least one positive solution. Remark 1.5. If one of the following two conditions is valid (i) f1 and f2 satisfy assumptions (1.5) and (1.8); or (ii) f2 and f1 satisfy assumptions (1.6), (1.7) and (1.9), then system (1.1) has at least one positive solution, which can be seen from the proofs of Theorems 1.3-1.4. In particular, when β1 = β2 = β, α1 = α2 = α, f1(t, u, v) = f(t, u) and f2(t, u, v) = f(t, v), equation (1.2) has at least one positive solution if f1 satisfies condition (1.5) or f2 satisfies condition (1.6), which is just the result in [9]. The rest of this article is organized as follows: in Section 2, we present some preliminaries; in Section 3, we prove Theorems 1.1-1.4; in Section 4, we give some applications. 2. Preliminaries In this section, we construct two classes of cones in which one is a single sub- cone P ⊂ C[0, 1] × C[0, 1], and the other is the Cartesian product K1 × K2 of subcones {Ki}2i=1 ⊂ C[0, 1]. And then we can change problem (1.1) into the fixed point problem in the constructed subcones. At the same time, we give some useful preliminaries for the proofs of our main results. Let µi,1 = 1 2 ( − βi + √ β2 i + 4αi ) , µi,2 = 1 2 ( − βi − √ β2 i + 4αi ) , i = 1, 2, andGi,j(t, s) (j = 1, 2) be the Green’s function of the linear boundary value problem −u′′(t) + µi,j u(t) = 0, t ∈ (0, 1), 4 Q. WANG, L. YANG EJDE-2020/45 u(0) = u(1) = 0, then for hi ∈ C[0, 1], the solution of linear boundary value problem u(4)(t) + βiu ′′(t)− αiu(t) = hi(t), t ∈ (0, 1), u(0) = u(1) = u′′(0) = u′′(1) = 0, (2.1) can be expressed as u(t) = ∫ 1 0 ∫ 1 0 Gi,1(t, τ)Gi,2(τ, s)hi(s) dsdτ. Lemma 2.1 ([9]). The function Gi,j(t, s) (i = 1, 2 ; j = 1, 2) has the following properties: (i) Gi,j(t, s) > 0 for all t, s ∈ (0, 1); (ii) Gi,j(t, s) 6 Ci,jGi,j(s, s) for all t, s ∈ [0, 1], where Ci,j > 0 is a constant; (iii) Gi,j(t, s) > δi,jGi,j(t, t)Gi,j(s, s) for all t, s ∈ [0, 1], where δi,j > 0 is a constant. It is well known that C[0, 1] is a Banach space with the maximum norm ‖u‖ = maxt∈[0,1] |u(t)| and C+[0, 1] := {u ∈ C[0, 1]| u(t) > 0, ∀t ∈ [0, 1]} is a closed convex cone in C[0, 1]. In addition, let Mi,j = max t∈[0,1] Gi,j(t, t), mi,j = min t∈[1/4,3/4] Gi,j(t, t), Ci = ∫ 1 0 Gi,1(τ, τ)Gi,2(τ, τ) dτ, then Mi,j , mi,j , Ci > 0 (i = 1, 2; j = 1, 2). Lemma 2.2 ([9]). Let hi ∈ C+[0, 1] (i = 1, 2), then the solution of (2.1) satisfies u(t) > δi,1δi,2Ci Ci,1Ci,2Mi,1 Gi,1(t, t)‖u‖, ∀t ∈ [0, 1]. For λ ∈ [0, 1], u, v ∈ C+[0, 1], we define the mappings Tλ,1(·, ·), Tλ,2(·, ·) : C+[0, 1]×C+[0, 1]→ C+[0, 1] and Tλ(·, ·) : C+[0, 1]×C+[0, 1]→ C+[0, 1]×C+[0, 1] by Tλ,1(u, v)(t) = ∫ 1 0 ∫ 1 0 G1,1(t, τ)G1,2(τ, s)hλ,1(s, u(s), v(s)) dsdτ, Tλ,2(u, v)(t) = ∫ 1 0 ∫ 1 0 G2,1(t, τ)G2,2(τ, s)hλ,2(s, u(s), v(s)) dsdτ, Tλ(u, v)(t) = (Tλ,1(u, v)(t), Tλ,2(u, v)(t)), where hλ,1(s, u(s), v(s)) = λf1(s, u(s), v(s)) + (1− λ)f1(s, u(s), 0), hλ,2(s, u(s), v(s)) = λf2(s, u(s), v(s)) + (1− λ)f2(s, 0, v(s)). It is obvious that the existence of positive solutions of system (1.1) is equivalent to the existence of nontrivial fixed points of T1. We will find the nontrivial fixed point of T1 by using the fixed point theory in cones. On this purpose, we introduce the following sub-cones: Ki = {u ∈ C+[0, 1] : u(t) > σi‖u‖, ∀t ∈ [1/4, 3/4]} i = 1, 2; P = {(u, v) ∈ C+[0, 1]× C+[0, 1] : u(t) + v(t) > σ‖(u, v)‖, ∀t ∈ [1/4, 3/4]}, (2.2) EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 5 where σi = mi,1δi,1δi,2Ci/(Ci,1Ci,2Mi,1), σ = min{σ1, σ2} > 0 and ‖(u, v)‖ = ‖u‖+ ‖v‖. For simplicity and convenience, we will use the following notation: Kri = {u ∈ Ki : ‖u‖ < ri}, Pr = {(u, v) ∈ P : ‖(u, v)‖ < r}, ∂Kri = {u ∈ Ki : ‖u‖ = ri}, ∂Pr = {(u, v) ∈ P : ‖(u, v)‖ = r}, Kri = {u ∈ Ki : ‖u‖ 6 ri}, ∀ri > 0, Pr = {(u, v) ∈ P : ‖(u, v)‖ 6 r}, ∀r > 0. To calculate the fixed point index of T1, we need the following results. Lemma 2.3. (i) Tλ : K1 ×K2 → K1 ×K2 is completely continuous; (ii) T1 : P → P is completely continuous. Proof. (i) For (u, v) ∈ K1 ×K2, let h1(t) = hλ,1(t, u(t), v(t)), then Tλ,1(u, v)(t) is the solution of equation (2.1) with i = 1. By Lemma 2.2, we have Tλ,1(u, v)(t) = ∫ 1 0 ∫ 1 0 G1,1(t, τ)G1,2(τ, s)h1(s) dsdτ > δ1,1δ1,2C1 C1,1C1,2M1,1 G1,1(t, t)‖Tλ,1(u, v)‖, ∀t ∈ [0, 1], which implies that Tλ,1(u, v)(t) > σ1‖Tλ,1(u, v)‖, t ∈ [1/4, 3/4]. Similarly, Tλ, 2(u, v)(t) > σ2‖Tλ, 2(u, v)‖, t ∈ [1/4, 3/4]. Hence, Tλ(K1 × K2) ⊂ K1 × K2. By the Arzelà-Ascoli theorem, we know that Tλ : K1 ×K2 → K1 ×K2 is completely continuous. (ii) By (2.2) and the proof of Lemma 2.3 (i), we immediately obtain the desired conclusion. � Remark 2.4. In fact, let T (λ, u, v)(t) = Tλ(u, v)(t), then T ([0, 1]×Kr1 ×Kr2) is a compact set by the Arzelà-Ascoli theorem. Next, we will recall some concepts about the fixed point index, which will be used in the proofs of our theorems. Let E be a Banach space and K ⊂ E is a closed convex cone in E. Assume Ω is a bounded open subset of E with boundary ∂Ω and K ∩ ∂Ω 6= ∅. Let T : K ∩ Ω→ K be a completely continuous operator. If Tu 6= u for every u ∈ K ∩ ∂Ω, then the fixed point index i(T,K ∩Ω,K) is defined. One important fact is that if i(T,K ∩Ω,K) 6= 0 then T has a fixed point in K ∩Ω. The following results are useful in our proofs. Lemma 2.5 ([7]). Let E be a Banach space and K ⊂ E is a closed convex cone in E. For r > 0, denote Kr = {u ∈ K : ‖u‖ < r}, ∂Kr = {u ∈ K : ‖u‖ = r}. Let T : K → K be a complete continuous mapping, then the following conclusions are valid: (i) if µTu 6= u, ∀u ∈ ∂Kr and µ ∈ (0, 1], then i(T,Kr,K) = 1; (ii) if mapping T satisfies the following two conditions: (a) infu∈∂Kr ‖Tu‖ > 0; (b) µTu 6= u, for all u ∈ ∂Kr and µ > 1, then i(T,Kr,K) = 0. Lemma 2.6 ([4]). Let X be a real Banach space, Pi ⊂ X be a closed convex cone, Wi be a bounded open subset of X with the boundary ∂Wi (i = 1, 2), P = P1 × P2 and W = W1 × W2. Assume that T : P ∩ W → P is completely continuous and there exist compactly continuous mappings Ai : Pi ∩Wi → Pi (i = 1, 2) and H : (P ∩W )× [0, 1]→ P such that 6 Q. WANG, L. YANG EJDE-2020/45 (a) H(·, 1) = T and H(·, 0) = A, where A(u, v) := (A1u,A2v) for all (u, v) ∈ P ∩W ; (b) Aiui 6= ui for all ui ∈ Pi ∩ ∂Wi; and (c) H(w, τ) 6= w for all (w, τ) ∈ (P ∩ ∂W )× (0, 1]. Then we have i(T, P ∩W,P ) = i(A1, P1 ∩W1, P1) · i(A2, P2 ∩W2, P2). 3. Proof of main results For Theorems 1.1 and 1.2, we choose a bounded open set PR\Pr in the single cone P on product space and calculate the fixed point index i(T1, PR\Pr, P ). Proof of Theorem 1.1. Firstly, by condition (1.3), there are ε ∈ (0, 1) and r > 0 such that 2∑ i=1 fi(t, u, v) 6 (1− ε)(λ1u+ λ2v), (3.1) for all t ∈ [0, 1], and all (u, v) ∈ {(u, v) ∈ (R+)2| u+ v ∈ [0, r]}. We claim that µT1(u, v) 6= (u, v), ∀µ ∈ (0, 1], ∀(u, v) ∈ ∂Pr. (3.2) In fact, if this is not true, then there exist µ0 ∈ (0, 1] and (u0, v0) ∈ ∂Pr, such that µ0T1(u0, v0) = (u0, v0); that is, (u0, v0) satisfies the system of differential equations u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) = µ0f1(t, u0(t), v0(t)), ∀t ∈ (0, 1), v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) = µ0f2(t, u0(t), v0(t)), ∀t ∈ (0, 1), u0(0) = u0(1) = u′′0(0) = u′′0(1) = 0, v0(0) = v0(1) = v′′0 (0) = v′′0 (1) = 0. (3.3) It follows from (3.1) and (3.3) that u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) + v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) 6 f1(t, u0(t), v0(t)) + f2(t, u0(t), v0(t)) 6 (1− ε)(λ1u0(t) + λ2v0(t)). Furthermore,∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt 6 (1− ε) ∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt. Noticing that ∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt > 0, we get 1 6 1 − ε, which is a contradiction. Secondly, from (1.3), there exist ε > 0 and m > 0 such that 2∑ i=1 fi(t, u, v) > (1 + ε)(λ1u+ λ2v), (3.4) for all t ∈ [0, 1] and all (u, v) ∈ {(u, v) ∈ (R+)2| u+ v > m}. Combining this with (3.4), it follows that 2∑ i=1 fi(t, u, v) > (1 + ε)(λ1u+ λ2v)− C, ∀t ∈ [0, 1], ∀(u, v) ∈ R+ × R+, (3.5) EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 7 where C = m(1 + ε)(λ1 + λ2). Now, we are in position to prove that there exists a R > r such that inf (u,v)∈∂PR ‖T1(u, v)‖ > 0 and µT1(u, v) 6= (u, v), ∀µ > 1, ∀(u, v) ∈ ∂PR. (3.6) In fact, if there are µ0 > 1 and (u0, v0) ∈ P such that µ0T1(u0, v0) = (u0, v0), then combining (3.3) with (3.5), we deduce that u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) + v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) > f1(t, u0(t), v0(t)) + f2(t, u0(t), v0(t)) > (1 + ε)(λ1u0(t) + λ2v0(t))− C . Then∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt > (1 + ε) ∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt− 2C π , which along with the definition of cone P implies that ‖(u0, v0)‖ 6 √ 2C σεmin{λ1, λ2} =: R. (3.7) Thus, for all µ > 1 and (u, v) ∈ ∂PR, µT1(u, v) 6= (u, v) as R > R. In addition, if R > m/σ, then by Lemma 2.1 and (3.4) we have that for all (u, v) ∈ ∂PR, ‖T1(u, v)‖ > (T1,1(u, v) + T1,2(u, v))(1/2) = ∫ 1 0 ∫ 1 0 2∑ i=1 Gi,1(1/2, τ)Gi,2(τ, s)fi(s, u(s), v(s)) dsdτ > 2∑ i=1 δi,1δi,2CiGi,1(1/2, 1/2) ∫ 1 0 Gi,2(s, s)fi(s, u(s), v(s)) ds > min i=1,2 {δi,1δi,2Cimi,1mi,2} ∫ 3/4 1/4 2∑ i=1 fi(s, u(s), v(s)) ds > min i=1,2 {δi,1δi,2Cimi,1mi,2}(1 + ε) ∫ 3/4 1/4 (λ1u(s) + λ2v(s)) ds > min i=1,2 {δi,1δi,2Cimi,1mi,2}min{λ1, λ2}(1 + ε)σR/2, which implies that inf(u,v)∈∂PR ‖T1(u, v)‖ > 0. So, we choose R > max{r,R,m/σ}. Finally, by (3.2), (3.6) and Lemmas 2.3-2.5, we have i(T1, PR\Pr, P ) = i(T1, PR, P )− i(T1, Pr, P ) = −1. Hence, system (1.1) has at least one solution in P\{(0, 0)}. Moreover, if f1(t, 0, v(t)) 6≡ 0 and f2(t, u(t), 0) 6≡ 0, ∀(u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. � Proof of Theorem 1.2. Firstly, from hypothesis (1.4), there exist ε > 0 and r > 0 such that 2∑ i=1 fi(t, u, v) > (1 + ε)(λ1u+ λ2v), (3.8) 8 Q. WANG, L. YANG EJDE-2020/45 for all t ∈ [0, 1] and all (u, v) ∈ {(u, v) ∈ (R+)2| u+ v ∈ [0, r]}. Next, we can show that inf (u,v)∈∂Pr ‖T1(u, v)‖ > 0 and µT1(u, v) 6= (u, v), ∀µ > 1, ∀(u, v) ∈ ∂Pr. (3.9) In fact, if there are µ0 > 1 and (u0, v0) ∈ ∂Pr such that (u0, v0) = µ0T1(u0, v0), by (3.3) and (3.8) we obtain u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) + v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) > f1(t, u0(t), v0(t)) + f2(t, u0(t), v0(t)) > (1 + ε)(λ1u0(t) + λ2v0(t)) . Then∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt > (1 + ε) ∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt. Thus, 1 > 1 + ε, which is a contradiction. In addition, by Lemma 2.1 and (3.8) we obtain that for all (u, v) ∈ ∂Pr, ‖T1(u, v)‖ > (T1,1(u, v) + T1,2(u, v))(1/2) > min i=1,2 {δi,1δi,2Cimi,1mi,2}min{λ1, λ2}(1 + ε)σr/2, which implies that inf(u,v)∈∂Pr ‖T1(u, v)‖ > 0. Secondly, in view of hypothesis (1.4), there are ε ∈ (0, 1) and m > 0 such that 2∑ i=1 fi(t, u, v) 6 (1− ε)(λ1u+ λ2v), (3.10) for all t ∈ [0, 1] and all (u, v) ∈ {(u, v) ∈ (R+)2| u + v > m}. Furthermore, by the continuity of f1 and f2, there exists C > 0 such that 2∑ i=1 fi(t, u, v) 6 (1− ε)(λ1u+ λ2v) + C, ∀t ∈ [0, 1], ∀(u, v) ∈ R+ × R+. (3.11) Now, we can prove that there is a R > r such that µT1(u, v) 6= (u, v), ∀µ ∈ (0, 1], ∀(u, v) ∈ ∂PR. (3.12) In fact, if there exist µ0 ∈ (0, 1] and (u0, v0) ∈ P such that µ0T1(u0, v0) = (u0, v0), then by (3.3) and (3.11) we can obtain that u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) + v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) 6 f1(t, u0(t), v0(t)) + f2(t, u0(t), v0(t)) 6 (1− ε)(λ1u0(t) + λ2v0(t)) + C . Then∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt 6 (1− ε) ∫ 1 0 (λ1u0(t) + λ2v0(t)) sin(πt) dt+ 2C π . Combining this with the definition of cone P , we obtain ‖(u0, v0)‖ 6 √ 2C σεmin{λ1, λ2} =: R∗. (3.13) Therefore, for all µ ∈ (0, 1] and (u, v) ∈ ∂PR, µT1(u, v) 6= (u, v) as R > R∗. Hence, we choose R > max{r,R∗}. EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 9 Finally, from (3.9), (3.12) and Lemmas 2.3-2.5, we obtain i(T1, PR\Pr, P ) = i(T1, PR, P )− i(T1, Pr, P ) = 1. Thus, (1.1) has at least one solution in P\{(0, 0)}. Moreover, if f1(t, 0, v(t)) 6≡ 0 and f2(t, u(t), 0) 6≡ 0, for all (u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. � To prove Theorems 1.3 and 1.4, we choose a bounded open set D = (KR1\Kr1)× (KR2\Kr2) in product cone K1×K2 and verify that a family of operators {Tλ}λ∈[0,1] satisfy the sufficient conditions for the homotopy invariance of fixed point index on ∂D. Proof of Theorem 1.3. For convenience we present the proof four steps that deter- mine r1, R1, r2 and R2 in turn. Step 1. From assumption (1.5), there are ε ∈ (0, λ1) and r1 > 0 such that λf1(t, u, v)+(1−λ)f1(t, u, 0) 6 (λ1−ε)u, ∀t ∈ [0, 1], ∀(u, v) ∈ [0, r1]×R+. (3.14) We claim that µTλ,1(u, v) 6= u, ∀µ ∈ (0, 1], ∀(u, v) ∈ ∂Kr1 ×K. (3.15) In fact, if it is not valid, then there exist µ0 ∈ (0, 1] and (u0, v0) ∈ ∂Kr1 ×K, such that µ0Tλ,1(u0, v0) = u0, that is, (u0, v0) satisfies the differential equation u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) = µ0hλ,1(t, u0(t), v0(t)), t ∈ (0, 1), u0(0) = u0(1) = u′′0(0) = u′′0(1) = 0, (3.16) where hλ,1(t, u0(t), v0(t)) = λf1(t, u0(t), v0(t)) + (1− λ)f1(t, u0(t), 0). In combina- tion with (3.14), it follows that u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) 6 (λ1 − ε)u0(t), and then λ1 ∫ 1 0 u0(t) sin(πt) dt 6 (λ1 − ε) ∫ 1 0 u0(t) sin(πt) dt. Noticing that ∫ 1 0 u0(t) sin(πt) dt > 0, we have λ1 6 λ1− ε, which is a contradiction. Step 2. By hypothesis (1.5), there exist ε > 0 and m > 0 such that λf1(t, u, v) + (1− λ)f1(t, u, 0) > (λ1 + ε)u, (3.17) for all t ∈ [0, 1] and all (u, v) ∈ [m,+∞]× R+, which implies that λf1(t, u, v) + (1− λ)f1(t, u, 0) > (λ1 + ε)u− C, (3.18) for all t ∈ [0, 1] and all (u, v) ∈ R+×R+, where C = (λ1 + ε)m. Now, we can prove that there exists a R1 > r1 such that µTλ,1(u, v) 6= u and inf u∈∂KR1 ‖Tλ,1(u, v)‖ > 0, (3.19) for all µ > 1 and all (u, v) ∈ ∂KR1 ×K. First, if there are (u0, v0) ∈ K ×K and µ0 > 1 such that µ0Tλ,1(u0, v0) = u0, then by (3.16) and (3.18), we obtain u (4) 0 (t) + β1u ′′ 0(t)− α1u0(t) > (λ1 + ε)u0(t)− C. 10 Q. WANG, L. YANG EJDE-2020/45 It follows that λ1 ∫ 1 0 u0(t) sin(πt) dt > (λ1 + ε) ∫ 1 0 u0(t) sin(πt) dt− 2C π ; that is, ∫ 1 0 u0(t) sin(πt) dt 6 2C πε . Furthermore, in view of the definition of cone K1, we know that ‖u0‖ 6 √ 2C σ 1 ε =: R̃. (3.20) Therefore, as R > R̃, µTλ,1(u, v) 6= u for all µ > 1 and (u, v) ∈ ∂KR ×K. In addition, if R > m/σ 1 , then by Lemma 2.1 and (3.17), for all (u, v) ∈ ∂KR×K, ‖Tλ,1(u, v)‖ > Tλ,1(u, v)(1/2) = ∫ 1 0 ∫ 1 0 G1,1(1/2, τ)G1,2(τ, s)[λf1(s, u(s), v(s)) + (1− λ)f1(s, u(s), 0)] dsdτ > δ1,1δ1,2C1G1,1( 1 2 , 1 2 ) ∫ 1 0 G1,2(s, s)[λf1(s, u(s), v(s)) + (1− λ)f1(s, u(s), 0)] ds > δ1,1δ1,2C1m1,1 m 1,2 ∫ 3/4 1/4 [λf1(s, u(s), v(s)) + (1− λ)f1(s, u(s), 0)] ds > δ1,1δ1,2C1m1,1 m1,2(λ1 + ε)σ1R/2, which implies infu∈∂KR ‖Tλ,1(u, v)‖ > 0. Hence, we chooseR1 > max{r1, R̃,m/σ1 }. Step 3. In view of condition (1.6), there exist ε > 0 and r2 > 0 such that λf2(t, u, v) + (1− λ)f2(t, 0, v) > (λ2 + ε)v, (3.21) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]× [0, r2]. We claim that µTλ,2(u, v) 6= v and inf v∈∂Kr2 ‖Tλ,2(u, v)‖ > 0, (3.22) for all µ > 1 and all (u, v) ∈ KR1 × ∂Kr2 . In fact, if there are µ0 > 1 and (u0, v0) ∈ KR1 × ∂Kr2 such that µ0Tλ,2(u0, v0) = v0, then (u0, v0) satisfies the differential equation v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) = µ0hλ,2(t, u0(t), v0(t)), t ∈ (0, 1), v(0) = v(1) = v′′(0) = v′′(1) = 0, (3.23) where hλ,2(t, u0(t), v0(t)) = λf2(t, u0(t), v0(t)) + (1 − λ)f2(t, 0, v0(t)). In combina- tion with (3.21), it follows that v (4) 0 (t) + β2v ′′ 0 (t)− α2v0(t) > (λ2 + ε)v0(t), which implies that λ2 ∫ 1 0 v0(t) sin(πt) dt > (λ2 + ε) ∫ 1 0 v0(t) sin(πt) dt. EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 11 By ∫ 1 0 v0(t) sin(πt) dt > 0, we get λ2 > λ2 +ε, which is a contradiction. In addition, by (3.21) and Lemma 2.1, we know that for all (u, v) ∈ KR1 × ∂Kr2 , ‖Tλ,2(u, v)‖ > Tλ,2(u, v)(1/2) > δ2,1δ2,2C2m2,1m2,2(λ2 + ε)σ2r2/2. So, infv∈∂Kr 2 ‖Tλ,2(u, v)‖ > 0. Step 4. By hypothesis (1.6) and the continuity of f2, there exist ε ∈ (0, λ2), n > 0 and C > 0 such that λf2(t, u, v) + (1− λ)f2(t, 0, v) 6 (λ2 − ε)v, (3.24) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]× [n,+∞); and λf2(t, u, v) + (1− λ)f2(t, 0, v) 6 (λ2 − ε)v + C, (3.25) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]× R+. Now, we are in a position to prove that there is a R2 > r2 such that µTλ,2(u, v) 6= v, ∀µ ∈ (0, 1], ∀(u, v) ∈ KR1 × ∂KR2 . (3.26) In fact, if there are µ0 ∈ (0, 1] and (u0, v0) ∈ KR1 ×K such that µ0Tλ,2(u0, v0) = v0, then by (3.23) and (3.25), similar to the proof of (3.20), we obtain ‖v0‖ 6 √ 2C σ 2 ε =: R′. (3.27) Thus, as R > R′, µTλ,2(u, v) 6= v for all µ ∈ (0, 1] and (u, v) ∈ KR1 × ∂KR. Hence, we choose R2 > max{r2, R′}. Let D = (KR1 \Kr1)×(KR2 \Kr2), then from the expressions (3.15), (3.19), (3.22) and (3.26), it is easy to verify that {Tλ}λ∈[0,1] satisfy the sufficient conditions for the homotopy invariance of fixed point index on ∂D. Moreover, by Remark 2.4 and Lemmas 2.5-2.6, we have i(T1, D,K1 ×K2) = i(T0,1,KR1 \Kr1 ,K1) · i(T0,2,KR2 \Kr2 ,K2) = −1. Hence, system (1.1) has at least one positive solution. � Proof of Theorem 1.4. Similar to the proof of Theorem 1.3, we will determine r1, R1, r2 and R2 in turn. Firstly, from assumption (1.7), there are ε > 0 and r1 > 0 such that λf1(t, u, v)+(1−λ)f1(t, u, 0) > (λ1+ε)u, ∀t ∈ [0, 1], ∀(u, v) ∈ [0, r1]×R+. (3.28) By (3.28) and the similar arguments in Step 3 of the proof for Theorem 1.3, we deduce that µTλ,1(u, v) 6= u, inf u∈∂Kr1 ‖Tλ,1(u, v)‖ > 0, ∀µ > 1, ∀(u, v) ∈ ∂Kr1 ×K. (3.29) Secondly, by hypotheses (1.7) and (1.9), there exist ε ∈ (0, λ1), m > 0 and C > 0 such that λf1(t, u, v) + (1− λ)f1(t, u, 0) 6 (λ1 − ε)u, (3.30) for all t ∈ [0, 1] and all (u, v) ∈ [m,+∞)× R+; and λf1(t, u, v) + (1− λ)f1(t, u, 0) 6 (λ1 − ε)u+ C, (3.31) for all t ∈ [0, 1] and all (u, v) ∈ R+ × R+. In view of (3.30)-(3.31) and similar arguments to those in Step 4 of the proof for Theorem 1.3, we can prove that there is a R1 > r1 such that µTλ,1(u, v) 6= u, ∀µ ∈ (0, 1], ∀(u, v) ∈ ∂KR1 ×K. (3.32) 12 Q. WANG, L. YANG EJDE-2020/45 Thirdly, by assumption (1.8), there are ε ∈ (0, λ2) and r2 > 0 such that λf2(t, u, v) + (1− λ)f2(t, 0, v) 6 (λ2 − ε)v, (3.33) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]× [0, r2]. From (3.33) and similar arguments as those in Step 1 of the proof for Theorem 1.3, we can show that µTλ,2(u, v) 6= v, ∀µ ∈ (0, 1], ∀(u, v) ∈ KR1 × ∂Kr2 . (3.34) Finally, in view of condition (1.8), there exist ε > 0, n > 0 and C > 0 such that λf2(t, u, v) + (1− λ)f2(t, 0, v) > (λ2 + ε)v, (3.35) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]× [n,+∞); and λf2(t, u, v) + (1− λ)f2(t, 0, v) > (λ2 + ε)v − C, (3.36) for all t ∈ [0, 1] and all (u, v) ∈ [0, R1]×R+. By using (3.35)-(3.36) and the similar arguments as those in Step 2 of the proof for Theorem 1.3, we can deduce that there exists a R2 > r2 such that µTλ,2(u, v) 6= v and inf v∈∂KR2 ‖Tλ,2(u, v)‖ > 0, (3.37) for all µ > 1 and all (u, v) ∈ KR1 × ∂KR2 . Let D = (KR1 \Kr1) × (KR2 \Kr2), then by (3.29), (3.32), (3.34) and (3.37), it is easy to verify that {Tλ}λ∈[0,1] satisfy the sufficient conditions for the homotopy invariance of fixed point index on ∂D. Furthermore, by Remark 2.4 and Lemmas 2.5-2.6, we have i(T1, D,K1 ×K2) = i(T0,1,KR1\Kr1 ,K1) · i(T0,2,KR2\Kr2 ,K2) = −1. Therefore, system (1.1) has at least one positive solution. � 4. Applications In this section, we present some corollaries to Theorems 1.1–1.4. Corollary 4.1. Assume fi ∈ C([0, 1]× R+ × R+,R+) and fi(t, u, v) = gi(t, λ1u+ λ2v), where gi ∈ C([0, 1] × R+,R+) (i = 1, 2) satisfying the following conditions: there exists a θ ∈ (0, 1) such that lim sup w→0+ max t∈[0,1] g1(t, w) w < θ and lim sup w→0+ max t∈[0,1] g2(t, w) w < 1− θ; (4.1) there exists a ϑ ∈ (0, 1) such that lim inf w→+∞ min t∈[0,1] g1(t, w) w > ϑ and lim inf w→+∞ min t∈[0,1] g2(t, w) w > 1− ϑ. (4.2) Then system (1.1) has at least one nonzero nonnegative solution. Moreover, if g1(t, λ2v(t)) 6≡ 0 and g2(t, λ1u(t)) 6≡ 0, ∀(u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. Proof. It is easy to verify that f1 and f2 satisfy condition (1.3). Hence, by Theorem 1.1, we immediately get the desired conclusion. � Similarly, applying Theorem 1.2 we have the following result. EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 13 Corollary 4.2. Suppose fi ∈ C([0, 1]×R+ ×R+,R+) and fi(t, u, v) = gi(t, λ1u+ λ2v), where gi ∈ C([0, 1] × R+,R+) (i = 1, 2) satisfying the following conditions: there is a θ ∈ (0, 1) such that lim inf w→0+ min t∈[0,1] g1(t, w) w > θ and lim inf w→0+ min t∈[0,1] g2(t, w) w > 1− θ; (4.3) there is a ϑ ∈ (0, 1) such that lim sup w→+∞ max t∈[0,1] g1(t, w) w < ϑ and lim sup w→+∞ max t∈[0,1] g2(t, w) w < 1− ϑ. (4.4) Then system (1.1) has at least one nonzero nonnegative solution. Furthermore, if g1(t, λ2v(t)) 6≡ 0 and g2(t, λ1u(t)) 6≡ 0, for all (u, v) ∈ P\{(0, 0)}, then system (1.1) has at least one positive solution. By Corollaries 4.1-4.2, it is easy to get the following result on the existence of positive solutions for equation (1.2) (see also [9, Theorem 1.1]). Corollary 4.3. Let λ = π4−βπ2−α. Assume that f ∈ C([0, 1]×R+,R+) satisfies one of the following conditions: lim sup w→0+ max t∈[0,1] f(t, w) w < λ < lim inf w→+∞ min t∈[0,1] f(t, w) w ; (4.5) or lim inf w→0+ min t∈[0,1] f(t, w) w > λ > lim sup w→+∞ max t∈[0,1] f(t, w) w . (4.6) Then (1.2) has at least one positive solution. Proof. Let β1 = β2 = β, α1 = α2 = α, θ = ϑ = 1/2 and g1(t, λ(u+ v)) = g2(t, λ(u+ v)) = 1 2 f(t, u+ v), then it is not difficult to verify that g1 and g2 satisfy conditions (4.1) and (4.2) (resp. (4.3) and (4.4)) as f satisfies condition (4.5) (resp. (4.6)). Therefore, by Corollaries 4.1-4.2, system (1.1) has at least one nonzero nonnegative solution denoted by (u 0 , v 0 ). Moreover, equation (1.2) has at least one positive solution w = u 0 +v 0 . � Next, we give corollaries related to Theorems 1.3 and 1.4. Corollary 4.4. Suppose that f1(t, u, v) = g1(t, u) + p1(u, v), f2(t, u, v) = g2(t, v) + p2(u, v), where gi ∈ C([0, 1]×R+,R+) and pi ∈ C(R+×R+,R+) (i = 1, 2) satisfying the following conditions: lim sup u→0+ max t∈[0,1] g1(t, u) u < λ1 < lim inf u→+∞ min t∈[0,1] g1(t, u) u ; (4.7) lim inf v→0+ min t∈[0,1] g2(t, v) v > λ2 > lim sup v→+∞ max t∈[0,1] g2(t, v) v ; (4.8) lim sup u→0+ p1(u, v) u = 0 uniformly w.r.t. v ∈ R+; (4.9) lim sup v→+∞ p2(u, v) v = 0 locally uniformly w.r.t. u ∈ [0,M ] (∀M > 0), (4.10) then system (1.1) has at least one positive solution. 14 Q. WANG, L. YANG EJDE-2020/45 Proof. It is clear that f1 and f2 satisfy all conditions in Theorem 1.3. Then, by Theorem 1.3, the desired conclusion yields. � Similarly, in view of Theorem 1.4 we obtain the following conclusion. Corollary 4.5. Assume that f1(t, u, v) = g1(t, u) + p1(u, v), f2(t, u, v) = g2(t, v) + p2(u, v), where gi ∈ C([0, 1]×R+,R+) and pi ∈ C(R+×R+,R+) (i = 1, 2) satisfying the following conditions: lim inf u→0+ min t∈[0,1] g1(t, u) u > λ1 > lim sup u→+∞ max t∈[0,1] g1(t, u) u ; (4.11) lim sup v→0+ max t∈[0,1] g2(t, v) v < λ2 < lim inf v→+∞ min t∈[0,1] g2(t, v) v ; (4.12) lim sup u→+∞ p1(u, v) u = 0 uniformly w.r.t. v ∈ R+, lim sup v→+∞ p1(u, v) = p(u) locally uniformly w.r.t. u ∈ [0,M ] (∀M > 0), where p(u) is a locally bounded function; (4.13) lim sup v→0+ p2(u, v) v = 0 locally uniformly w.r.t. u ∈ [0,M ] (∀M > 0), (4.14) then system (1.1) has at least one positive solution. Remark 4.6. In particular, when pi(u, v) ≡ 0 (i = 1, 2), Corollaries 4.4-4.5 reduce to the existence results of positive solutions to equation (1.2) (see Corollary 4.3). Acknowledgement. The authors want to express their gratitude to anonymous referee for the valuable comments and suggestions. This work was partially sup- ported by the Fundamental Research Funds for the Central Universities Grant (lzujbky-2018-112). References [1] A. R. Aftabizadeh; Existence and uniqueness theorems for fourth-order boundary value prob- lems, J. Math. Anal. Appl., 116 (1986), 415-426. [2] R. Agarwal; On fourth-order boundary value problems arising in beam analysis, Differential Integral Equations, 2 (1989), 91-110. [3] X. Cheng, Z. Feng; Existence and multiplicity of positive solutions to systems of nonlinear Hammerstein integral equations, Electron. J. Differential Equations, 2019 (2019) no. 52, 1-16. [4] X. Cheng, H. Lü; Multiplicity of positive solutions for a (p1, p2)-Laplacian system and its applications, Nonlinear Anal. RWA, 13 (2012), 2375-2390. [5] C. De Coster, C. Fabry, F. Munyamarere; Nonresonance conditions for fourth-order nonlinear boundary value problems, Internat. J. Math. Math. Sci., 17 (1994), 725-740. [6] M. A. Del Pino, R. F. Manasevich; Existence for a fourth-order boundary value problem under a two-parameter nonresonance condition, Proc. Amer. Math. Soc., 112 (1991), 81-86. [7] D. Guo, V. Lakshmikantham; Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988. [8] J. He, L. Yang; Existence of positive solutions for systems of nonlinear Sturm-Liouville differential equations with weight functions, Electron. J. Differential Equations, 2019 (2019) no. 111, 1-24. [9] Y. Li; Positive solutions of fourth-order boundary value problems with two parameters, J. Math. Anal. Appl., 281 (2003), 477-484. Qiuyue Wang School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, China Email address: wangqy17@lzu.edu.cn EJDE-2020/45 POSITIVE SOLUTIONS FOR A SYSTEM OF FOURTH-ORDER ODES 15 Lu Yang School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, China Email address: yanglu@lzu.edu.cn 1. Introduction and main results 2. Preliminaries 3. Proof of main results 4. Applications Acknowledgement References