Electronic Journal of Differential Equations, Vol. 2023 (2023), No. 52, pp. 1–16. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: https://doi.org/10.58997/ejde.2023.52 EXISTENCE AND NONEXISTENCE OF POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS MEIQIANG FENG, HAIPING CHEN Abstract. This article studies a fourth-order elliptic problem with and with- out an eigenvalue parameter. New criteria for the existence and nonexistence of positive solution are established under some sublinear conditions which involve the principal eigenvalues of the corresponding linear problems. The interesting point is that the nonlinear term f is involved in the second-order derivative explicitly. 1. Introduction Consider the fourth-order elliptic problem ∆2u = λf(u,−∆u) in Ω, u = ∆u = 0 on ∂Ω, (1.1) where (∆)2u = ∆(∆u) denotes the biharmonic operator, λ > 0 is a parameter, Ω is a smooth bounded domain in Rn (n ≥ 2), and the nonlinear term satisfies (A1) f : [0,+∞)× [0,+∞)→ [0,+∞) is continuous. Fourth-order elliptic problems belong to an open problem raised by Lion in [22, Section 4.2 (c)]. They are important questions for understanding related higher order problems, and they have important applications in the study of traveling waves in suspension bridges [6] and in static deflection of a bending beams [15]. Fourth-order elliptic problems have attracted the interest of many mathematicians; see for example [5, 7, 8, 10, 11, 12, 13, 16, 19, 20, 21, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38] and the references cited therein. In particular, Abid-Baraket [1] studied the existence of singular solution to the biharmonic elliptic problem ∆2u = up in Ω, u = ∆u = 0 on ∂Ω, (1.2) where Ω is a subset of Rn (n ≥ 5) with a smooth boundary. Let Σ be a compact submanifold of Ω without boundary of dimension (n −m) and 4 < m < n. When p > m m−4 and close enough to this value, the authors verified that problem (1.2) admits at least one solution which is singular on Σ. 2020 Mathematics Subject Classification. 35J60, 35J40. Key words and phrases. Fourth-order elliptic problems; Navier boundary conditions; positive solution; existence, uniqueness and nonexistence; principle eigenvalue. ©2023. This work is licensed under a CC BY 4.0 license. Submitted June 27, 2023. Published August 6, 2023. 1 2 M. FENG, H. CHEN EJDE-2023/52 Guo-Wei-Zhou [14] considered the existence, uniqueness, asymptotic behavior and further qualitative properties of singular radial solutions of the biharmonic equation ∆2u = up in Rn\{0}, u > 0, and lim |x|→0 u(x) = +∞, where n ≥ 5 and n n−4 < p < n+4 n−4 . In addition, the authors also constructed positive weak solutions with a prescribed singular set for problem (1.2). Let Ω be the unit ball in Rn (n ≥ 5) and ∂u ∂n denote the differentiation with respect to the exterior unit normal. Arioli-Gazzola-Grunau-Mitidieri [4] studied the fourth-order elliptic problem ∆2u = λeu in Ω, u = ∂u ∂n = 0 on ∂Ω, (1.3) where λ ≥ 0 is a parameter. For 5 ≤ n ≤ 16, the authors proved the existence of singular solutions for problem (1.3) by means of computer assistance. Liu-Wang [23] used a variant version of Mountain Pass Theorem to demonstrate the existence and nonexistence of positive solution for the fourth-order elliptic prob- lem ∆2u = f(x, u) in Ω, u = ∆u = 0 on ∂Ω, where Ω denotes a smooth bounded domain in Rn (n > 4). Recently, Feng [9] studied the Navier boundary value problem ∆2u = λf(x, u) in Ω, u = ∆u = 0 on ∂Ω, (1.4) where λ 6= 0 is a parameter, Ω is a smooth bounded domain in Rn (n ≥ 2). The author derived some criteria for the existence, multiplicity and nonexistence of positive solutions to (1.4) by applying fixed point theorems in a cone. However, to our best knowledge, there are almost no papers studying the fourth- order elliptic problem when the nonlinear term f is involved with the second-order derivative explicitly. In this article, we do some research on this problem. More precisely, this article has the following features. Firstly, comparing with [1, 4, 9, 14, 23], we discuss the fourth-order elliptic problem when the nonlinear term f is involved with the second-order derivative explicitly. Secondly, we find some new sublinear conditions, which involve the principle eigenvalues of the corresponding linear systems and do not appear in [1, 4, 9, 14, 23]. In addition, we are going to employ a simpler method, i.e. the theory of fixed points on cones to demonstrate the existence of positive solution for fourth-order elliptic problems, which is completely different from that used in [1, 4, 14, 23]. The article is organized as follows. In Section 2, we use a well-known fixed point theorem for completely continuous operators to prove existence results of positive solution to problem (1.1) without an eigenvalue parameter. Section 3 is devoted to analyzing the existence and nonexistence results of positive solution to problem (1.1) with an eigenvalue parameter. Finally, we give some comments on higher order elliptic problems and second-order elliptic systems in Section 4. EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 3 2. Existence of a positive solution without an eigenvalue parameter In this section, we demonstrate a general result of existence of positive solution to (1.1). Conclusions to be demonstrated in this section are true for all λ > 0. Hence, we may suppose that λ = 1 for simplicity and so study ∆2u = f(u,−∆u) in Ω, u = ∆u = 0 on ∂Ω. (2.1) Letting −∆u = v, one can transform the fourth-order elliptic problem (2.1) into the second-order elliptic system −∆u = v in Ω, −∆v = f(u, v) in Ω, u = 0 = v on ∂Ω. (2.2) From this system, we derive that u(x) = ∫ Ω̄ G(x, y)v(y) dy, (2.3) v(x) = ∫ Ω̄ G(x, y)f(u(y), v(y)) dy, (2.4) where G(x, y) denotes the Green’s function of −∆ on Ω, which satisfies 0 ≤ G(x, y) ≤ C|x− y|2−n, where n ≥ 3, and C is a constant, which depends only on Ω. In addition, for x, y ∈ Ω, x 6= y, we find that 0 ≤ G(x, y) ≤ 1 4π|x− y| , n = 3, 0 ≤ G(x, y) ≤ 1 2π ln d |x− y| , n = 2, where d is the diameter of Ω. Let the maximum norm in R2 be defined by |u| = max{|u|, |v|}, where u = (u, v) ∈ R2. We also let E = C(Ω̄)2 denote the real Banach space of continuous functions with the norm ‖u‖ = max{‖u‖0, ‖v‖0}, where ‖ · ‖0 denotes the supremum norm of the real Banach space C(Ω̄), and u(x) = (u(x), v(x)) for x ∈ Ω̄. We define a positive cone in E as K = {u ∈ E : u(x) ≥ 0, v(x) ≥ 0, x ∈ Ω̄}. (2.5) For % > 0, we also define K% = {u : u ∈ K, ‖u‖ < %}, ∂K% = {u : u ∈ K, ‖u‖ = %} K̄% = {u ∈ K : ‖u‖ ≤ %}. Let K1 = {w ∈ C(Ω̄) : w(x) ≥ 0, x ∈ Ω̄}. 4 M. FENG, H. CHEN EJDE-2023/52 For u, v ∈ K1, we define T1, T2 : K1 → C(Ω̄) as T1u(x) = ∫ Ω̄ G(x, y)v(y) dy, (2.6) T2v(x) = ∫ Ω̄ G(x, y)f(u(y), v(y)) dy. (2.7) From the continuity of f , it is clear that T1, T2 : K1 → K1 are completely continu- ous. On the other hand, it is well known that system (2.2) is equivalent to the fixed point equation u(x) = (T1u(x), T2v(x)) := Tu(x) for x ∈ Ω̄. (2.8) Since T1, T2 : K1 → K1 are completely continuous, so we derive that T : K → K is completely continuous. We define a linear integral operator G by Gu(x) = ∫ Ω̄ G(x, y)u(y) dy. (2.9) We also define a function e(x) on Ω̄ by e(x) = ∫ Ω̄ G(x, y) dy, ∀x ∈ Ω̄. Now we consider the integral operator G defined in (2.9). One can find in Kras- nosel’skii [17] and Amann [2] that the linear integral operator G is e-positive, i.e. for any v > θ, there exist ξ = ξ(v) > 0 and ζ = ζ(v) > 0 such that ξe ≤ Gv ≤ ζe. From this and the well-known Krein-Rutman theorem [18, Theorem 6.2], it is not difficult to see that µ1 ∈ (0,+∞) and that there exists ϕ1 ∈ K\{0} so that ϕ1 = µ1Gϕ1, (2.10) where µ1 = 1 r(G) and r(G) denotes the spectral radius of G. For a function f , we define f0 = lim |u|→0+ f(u) |u| , fα∞ = lim |u|→+∞ f(u) |u|α , where 0 < α < 1. Theorem 2.1. Let µ1 and ϕ1 be defined as in (2.10). Under assumption (A1), if f0 > µ1 and fα∞ < µ1, then system (2.1) admits at least one positive solution in K. Next, we shall apply the following fixed point theorem for completely continuous operators to demonstrate Theorem 2.1. Lemma 2.2 ([3, Theorem 12.3]). Let P be a cone in a real Banach space E. Assume Ω1,Ω2 are bounded open sets in E with θ ∈ Ω1, Ω̄1 ⊂ Ω2. If A : P ∩ (Ω̄2\Ω1)→ P is completely continuous such that either (i) there exists a u0 > 0 such that u − Au 6= tu0 for all u ∈ P ∩ ∂Ω2, t ≥ 0; Au 6= µu for all u ∈ P ∩ ∂Ω1, µ ≥ 1, or (ii) there exists a u0 > 0 such that u − Au 6= tu0 for all u ∈ P ∩ ∂Ω1, t ≥ 0; Au 6= µu for all u ∈ P ∩ ∂Ω2, µ ≥ 1. Then A has at least one fixed point in P ∩ (Ω2\Ω̄1). EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 5 Proof of Theorem 2.1. Assume that there is r1 > 0 such that u− Tu 6= θ, ∀u ∈ K, 0 < ‖u‖ ≤ r1. (2.11) If not, then there is u ∈ ∂Kr1 such that Tu = u. Considering f0 > µ1, there are ε > 0 and r∗ > 0 such that f(u) ≥ (ε+ µ1)|u|, ∀u ∈ R2 + with |u| ∈ [0, r∗]. (2.12) Suppose that ϕ = (ϕ1, ϕ1), where ϕ1 is defined as (2.10). We demonstrate that u− Tu 6= ζϕ ∀u ∈ ∂Kr, ζ ≥ 0, (2.13) where 0 < r < min{r1, r ∗}. If not, then there are u ∈ ∂Kr and ζ ≥ 0 so that u− Tu = ζϕ. Then (2.11) indicates that ζ > 0 and v0 = ζϕ1 + T2v0 ≥ ζϕ1. Set ζ∗ = sup{ζ|v0 ≥ ζϕ1}. (2.14) Then we obtain that 0 < ζ ≤ ζ∗ < +∞ and v0 ≥ ζ∗ϕ1. So, for each x ∈ Ω̄ and u ∈ ∂Kr, we derive from (2.7), (2.10), (2.12), and (2.14) that v0(x) = ∫ Ω̄ G(x, y)f(u(y)) dy + ζϕ1(x) ≥ ∫ Ω̄ G(x, y)(ε+ µ1)v0(y) dy ≥ ∫ Ω̄ G(x, y)(ε+ µ1)ζ∗ϕ1(y) dy = (ε+ µ1)ζ∗ ∫ Ω̄ G(x, y)ϕ1(y) dy = (ε+ µ1)ζ∗ ϕ1(x) µ1 . (2.15) So we derive that ζ∗ ≥ (ε+ µ1)ζ∗ 1 µ1 > ζ∗, which is a contradiction. Thus, (2.13) holds. Next, turning to fα∞ < µ1, there are ε1 > 0 and r2 > 0 so that f(u) ≤ (µ1 − ε1)|u|α, ∀u ∈ R2 + with |u| ≥ r2. From the continuity of f , there is L1 > 0 such that f(u) ≤ L1, ∀u ∈ R2 + with |u| ≤ r2. Hence we obtain that f(u) ≤ (µ1 − ε1)|u|α + L1, ∀u ∈ R2 +. (2.16) Moreover, it is not difficult to see that v ≤ |v|. Let L2 = max{1, |u| − (µ1 − ε1)|u|α}. Then v ≤ L2 + (µ1 − ε1)|u|α, ∀u, v ∈ R+. (2.17) Assume that R is large enough (R > r) so that L‖φ‖0 R + (µ1 − ε1)‖φ‖0 R1−α < 1, (2.18) 6 M. FENG, H. CHEN EJDE-2023/52 where L = max{L1, L2}, and φ ∈ C2(Ω̄) satisfies −∆φ = 1 in Ω, φ = 0 on ∂Ω. (2.19) We claim that for all u ∈ ∂KR, we have that µ ≥ 1⇒ Tu 6= µu. (2.20) As a matter of fact, if there are u ∈ ∂KR and µ ≥ 1 so that Tu = µu, then it follows from (2.7) and (2.16) that µv(x) = ∫ Ω̄ G(x, y)f(u(y)) dy ≤ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L1) dy = ((µ1 − ε1)‖u‖α + L1) ∫ Ω̄ G(x, y) dy ≤ ((µ1 − ε1)‖u‖α + L1)‖φ‖0. (2.21) Similarly, from (2.6) and (2.17) it follows that µu(x) = ∫ Ω̄ G(x, y)v(y)) dy ≤ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L2) dy = ((µ1 − ε1)‖u‖α + L2) ∫ Ω̄ G(x, y) dy ≤ ((µ1 − ε1)‖u‖α + L2)‖φ‖0. (2.22) Taking the maximum in (2.20) and (2.21), we have µ‖v‖0 ≤ ((µ1 − ε1)‖u‖α + L1)‖φ‖0, µ‖u‖0 ≤ ((µ1 − ε1)‖u‖α + L2)‖φ‖0. This indicates that µ‖u‖ ≤ ((µ1 − ε1)‖u‖α + L)‖φ‖0. Hence it follows from (2.18) that µ ≤ L‖φ‖0 R + (µ1 − ε1)‖φ‖0 R1−α < 1, which contradicts µ ≥ 1. So (2.20) holds. Applying (ii) of Lemma 2.2 to (2.13) and (2.20) yields that T possesses a fixed point u in KR\K̄r with r < ‖u‖ < R. It follows that system (2.1) admits at least one positive solution u with r < ‖u‖ < R. This completes the proof. � One of the contributions of Theorem 2.1 is to use a simple method, i.e. the theory of fixed points for completely continuous operators to prove the existence of positive solution for biharmonic problems. Lemma 2.3. Let ‖G‖0 denote the norm of the linear integral operator Gu(x) = ∫ Ω̄ G(x, y)u(y) dy. (2.23) Then G maps C(Ω̄) into C(Ω̄). EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 7 Proof. Let R be large enough (R > r) such that L‖G‖0 R + (µ1 − ε1)‖G‖0 R1−α < 1, (2.24) where L is defined as in (2.18). If there are u ∈ ∂KR and µ ≥ 1 so that Tu = µu, then it follows from (2.7) and (2.16) that µv(x) = ∫ Ω̄ G(x, y)f(u(y)) dy ≤ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L1) dy = ((µ1 − ε1)‖u‖α + L1) ∫ Ω̄ G(x, y) dy ≤ ((µ1 − ε1)‖u‖α + L1)‖G‖0. (2.25) Similarly, from (2.6) and (2.17) it follows that µu(x) = ∫ Ω̄ G(x, y)v(y)) dy ≤ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L2) dy = ((µ1 − ε1)‖u‖α + L2) ∫ Ω̄ G(x, y) dy ≤ ((µ1 − ε1)‖u‖α + L2)‖G‖0. (2.26) Taking the maximum in (2.25) and (2.26), we have µ‖v‖0 ≤ ((µ1 − ε1)‖u‖α + L1)‖G‖0, µ‖u‖0 ≤ ((µ1 − ε1)‖u‖α + L2)‖G‖0. This indicates that µ‖u‖ ≤ ((µ1 − ε1)‖u‖α + L)‖G‖0. Hence it follows from (2.24) that µ ≤ L‖G‖0 R + (µ1 − ε1)‖G‖0 R1−α < 1, which contradicts µ ≥ 1. So (2.20) holds. � Now we use the following assumptions: (A2) there exist ε2 > 0 and r3 > 0 such that f(u) ≥ (ε2 + µ1)v, ∀u ∈ R2 + with |u| ∈ [0, r3]; (A3) for 0 < α < 1, there exist ε3 > 0 and r4 > 0 such that f(u) ≤ (µ1 − ε3)uα, ∀u ∈ R2 + with |u| ≥ r4, From the proof of Theorem 2.1, we can derive the following results. Corollary 2.4. Under assumptions (A1)–(A3), system (2.1) admits at least one positive solution in K. Corollary 2.5. Under assumption (A1), if f0 > µ1 and fα∞ = 0, then system (2.1) admits at least one positive solution in K. 8 M. FENG, H. CHEN EJDE-2023/52 Corollary 2.6. Under assumption (A1), if f0 = ∞ and fα∞ < µ1, then system (2.1) admits at least one positive solution in K. Corollary 2.7. Under assumption (A1), if f0 =∞ and fα∞ = 0, then system (2.1) admits at least one positive solution in K. It is easy to see that the conditions in Corollary 2.7 does not depend on the µ1 defined in (2.10). 3. Existence of positive solution to (1.1) Letting −∆u = v, one can transform the biharmonic problem (1.1) into the second-order elliptic system −∆u = v in Ω, −∆v = λf(u, v) in Ω, u = 0 = v on ∂Ω. (3.1) From (3.1), we derive that u(x) = ∫ Ω̄ G(x, y)v(y) dy, (3.2) v(x) = λ ∫ Ω̄ G(x, y)f(u(y), v(y)) dy, (3.3) where G(x, y) denotes the Green’s function of −∆ on Ω. Because, for x, y ∈ Ω̄ ⊂ Rn (n ≥ 2), G(x, y) is nonnegative, continuous (when x 6= y) and symmetric, there must exist three points x0, y0 and z0 with x0 6= y0 6= z0, which are interior points of Ω, so that G(x0, y0) = G(y0, x0) > 0; G(x0, z0) = G(z0, x0) > 0; G(z0, y0) = G(y0, z0) > 0. So there exist τ1, τ2, τ3 > 0 and three disjoint small closed balls B1, B2, B3 ⊂ Ω such that G(x, y) ≥ τ1, ∀(x, y) ∈ (B1 ×B2) ∪ (B2 ×B1), G(y, z) ≥ τ2, ∀(y, z) ∈ (B2 ×B3) ∪ (B3 ×B2), G(x, z) ≥ τ3, ∀(x, z) ∈ (B1 ×B3) ∪ (B3 ×B1). (3.4) Here B1 = {x ∈ Ω : |x− x0| ≤ δ}, B2 = {x ∈ Ω : |x− y0| ≤ δ}, B3 = {x ∈ Ω : |x− z0| ≤ δ}, (3.5) where δ > 0 is small enough. It is not difficult to see that measB1 = measB2 = measB3. Let | · | denote the maximum norm in R2 defined by |u| = max{|u|, |v|}, where u = (u, v) ∈ R2. We also let E = C(Ω̄)2 denote the real Banach space of continuous functions with norm ‖u‖ = max{‖u‖0, ‖v‖0}, where ‖ · ‖0 denotes the supremum norm of the real Banach space C(Ω̄), and u(x) = (u(x), v(x)) for x ∈ Ω̄. EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 9 We define a positive cone in E as K = {u ∈ E : u(x) ≥ 0, v(x) ≥ 0, x ∈ Ω̄}. (3.6) For % > 0, we also define K% = {u ∈ K : ‖u‖ < %}, ∂K% = {u ∈ K : ‖u‖ = %}, K̄% = {u ∈ K : ‖u‖ ≤ %}. Let K1 = {u ∈ C(Ω̄) : u(x) ≥ 0, x ∈ Ω̄}. For u, v ∈ K1, we define T1, T3 : K1 → C(Ω̄) as T1u(x) = ∫ Ω̄ G(x, y)v(y) dy, (3.7) T3v(x) = λ ∫ Ω̄ G(x, y)f(u(y), v(y)) dy. (3.8) Under assumption (A1), it is clear that T1, T3 : K1 → K1 are completely continuous. On the other hand, it is well known that system (3.1) is equivalent to the following fixed point equation: u(x) = (T1u(x), T3v(x)) := T̄u(x) for x ∈ Ω̄. (3.9) Since T1, T3 : K1 → K1 are completely continuous, we derive that T̄ : K → K is completely continuous. Let G denote the linear integral operator defined as in (2.9). For each function f , we define f0 = lim |u|→0+ f(u) |u| , fα∞ = lim |u|→+∞ f(u) |u|α , where 0 < α < 1. Theorem 3.1. Let µ1 and ϕ1 be defined as in (2.10). Under assumption (A1), if f0 > µ1 and fα∞ < µ1, then system (3.1) admits at least one positive solution in K for each λ ≥ M τ1µ1 ∫ B2 ϕ1(z)dz , where M = max x∈B1 ϕ1(x). (3.10) Proof. Assume that there is r1 > 0 such that u− Tu 6= θ, ∀u ∈ K, 0 < ‖u‖ ≤ r1. (3.11) If not, then there is u ∈ ∂Kr1 so that Tu = u. Considering f0 > µ1, there are ε > 0 and r∗ > 0 such that f(u) ≥ (ε+ µ1)|u|, ∀u ∈ R2 + with |u| ∈ [0, r∗]. (3.12) Suppose that ϕ = (ϕ1, ϕ1), where ϕ1 is defined by (2.10). We demonstrate that u− Tu 6= ζϕ ∀u ∈ ∂Kr, ζ ≥ 0, (3.13) where 0 < r < min{r1, r ∗}. If not, there are u ∈ ∂Kr and ζ ≥ 0 so that u− Tu = ζϕ. Then (3.11) indicates that ζ > 0 and v0 = ζϕ1 + T1v0 ≥ ζϕ1. (3.14) 10 M. FENG, H. CHEN EJDE-2023/52 Set ζ∗ = sup{ζ|v0 ≥ ζϕ1}. Then we obtain that 0 < ζ ≤ ζ∗ < +∞ and v0 ≥ ζ∗ϕ1. So, for any x ∈ B1 and u ∈ ∂Kr, we derive from (3.4), (3.8), (3.9), (3.12), and (3.14) that v0(x) = λ ∫ Ω̄ G(x, y)f(u(y)) dy + ζϕ1(x) ≥ λ ∫ Ω̄ G(x, y)(ε+ µ1)v0(y) dy ≥ λ ∫ Ω̄ G(x, y)(ε+ µ1)ζ∗ϕ1(y) dy = λ(ε+ µ1)ζ∗ ∫ Ω̄ G(x, y)ϕ1(y) dy ≥ λ(ε+ µ1)ζ∗ ∫ B2 G(x, y)ϕ1(y) dy ≥ λτ1(ε+ µ1)ζ∗ ∫ B2 ϕ1(y) dy ≥ λτ1(ε+ µ1)ζ∗ ∫ B2 ϕ1(y) dy × ϕ1(x) M ≥ M τ1µ1 ∫ B2 ϕ1(z)dz τ1(ε+ µ1)ζ∗ ∫ B2 ϕ1(y) dy × ϕ1(x) M = (ε+ µ1)ζ∗ ϕ1(x) µ1 . (3.15) So we derive that ζ∗ ≥ (ε+ µ1)ζ∗ 1 µ1 > ζ∗, which is a contradiction. Thus, (3.13) holds. Next, turning to fα∞ < µ1, there are ε1 > 0 and r2 > 0 such that f(u) ≤ (µ1 − ε1)|u|α, ∀u ∈ R2 + with |u| ≥ r2. By the continuity of f , there is L1 > 0 such that f(u) ≤ L1, ∀u ∈ R2 + with |u| ≤ r2. Hence we obtain that f(u) ≤ (µ1 − ε1)|u|α + L1, ∀u ∈ R2 +. (3.16) Moreover, it is not difficult to see that u ≤ |u| < |u|+ (µ1 − ε1)‖u‖α. Therefore, there is a constant L2 > 0 small enough such that u ≤ L2 + (µ1 − ε1)‖u‖. (3.17) Assume that R is large enough (R > r) such that (λ+ 1)L‖φ‖0 R + (λ+ 1)(µ1 − ε1)‖φ‖0 R1−α < 1, (3.18) where φ satisfies (2.19). EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 11 We claim that for all u ∈ ∂KR, we have that µ ≥ 1⇒ Tu 6= µu. (3.19) As a matter of fact, if there are u ∈ ∂KR and µ ≥ 1 so that Tu = µu, then it follows from (3.8), (3.9) and (3.16) that µv(x) = λ ∫ Ω̄ G(x, y)f(u(y)) dy ≤ λ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L1) dy = λ((µ1 − ε1)‖u‖α + L1) ∫ Ω̄ G(x, y) dy ≤ λ((µ1 − ε1)‖u‖α + L1)‖φ‖0. (3.20) Similarly, from (3.7), (3.9) and (3.17) it follows that µu(x) = ∫ Ω̄ G(x, y)v(y)) dy ≤ ∫ Ω̄ G(x, y)((µ1 − ε1)‖u‖α + L2) dy = ((µ1 − ε1)‖u‖α + L2) ∫ Ω̄ G(x, y) dy ≤ ((µ1 − ε1)‖u‖α + L2)‖φ‖0. (3.21) Taking the maximum in (3.20) and (3.21), we have µ‖v‖0 ≤ λ((µ1 − ε1)‖u‖α + L1)‖φ‖0, µ‖u‖0 ≤ ((µ1 − ε1)‖u‖α + L2)‖φ‖0. This indicates that µ‖u‖ ≤ (λ+ 1)((µ1 − ε1)‖u‖α + L)‖φ‖0, where L = max{L1, L2}. Hence it follows from (3.18) that µ ≤ (λ+ 1)L‖φ‖0 R + (λ+ 1)(µ1 − ε1)‖φ‖0 R1−α < 1, which contradicts µ ≥ 1. So (3.19) holds. Applying (ii) of Lemma 2.2 to (3.13) and (3.19) yields that T possesses a fixed point u in KR\K̄r with r < ‖u‖ < R. It follows that system (3.1) admits at least one positive solution u withr < ‖u‖ < R. This completes the proof. � In the proof of Theorem 3.1, we use a new technique to prove that (3.15) holds, which is different from that used in proving (2.15). From the proof of Theorem 3.1, we can derive the following results. Corollary 3.2. If (A1)–(A3) hold, then system (3.1) admits at least one positive solution in K for each λ ≥ M τ1µ1 ∫ B2 ϕ1(z)dz , where M is defined in (3.10). 12 M. FENG, H. CHEN EJDE-2023/52 Corollary 3.3. Under assumption (̊A1), if f0 > µ1 and fα∞ = 0, then system (3.1) admits at least one positive solution in K for each λ ≥ M τ1µ1 ∫ B2 ϕ1(z)dz , where M is defined in (3.10). Corollary 3.4. Under assumption (A1), if f0 = ∞ and fα∞ < µ1, then system (3.1) admits at least one positive solution in K for each λ ≥ λ0, where λ0 > 0 is a finite number. Corollary 3.5. Under assumption (A1), if f0 =∞ and fα∞ = 0, then system (3.1) admits at least one positive solution in K for each λ ≥ λ0, where λ0 is defined in Corollary 3.4. Let f∞ = lim |u|→+∞ f(u) |u| . Theorem 3.6. Under assumption (A1), if f0 > µ1 and f∞ > µ1, then system (3.1) admits no positive solution for any λ > λ̄, where λ̄ > 0 is a finite number. Proof. Considering (A1), if f0 > µ1 and f∞ > µ1, then there are positive numbers ε3, ε4, h1, and h2 such that h1 < h2 and for u ∈ R2 + and |u| ≤ h1, we obtain f(u) ≥ (µ1 + ε3)|u|, (3.22) and for u ∈ R2 +, |u| ≥ h2, we obtain f(u) ≥ (µ1 + ε4)|u|. (3.23) Letting µ∗ = min { µ1 + ε3, µ1 + ε4, min {f(u) |u| : |u| ∈ [h1, h2] }} > 0, we have f(u) ≥ µ∗|u|, u ∈ R2 +. (3.24) Let u be a positive solution of system (3.1). We will demonstrate that this leads to a contradiction for λ > λ̄ = 1 µ∗τ1 measB1 . In deed, for x ∈ B1 and λ > λ̄, we derive from (3.4), (3.8), (3.9), and (3.24) that v(x) = λ ∫ Ω̄ G(x, y)f(u(y), v(y)) dy ≥ λ ∫ Ω̄ G(x, y)(µ∗ + ε5)‖u‖dy = λµ∗‖u‖ ∫ Ω̄ G(x, y) dy ≥ λµ∗‖u‖ ∫ B2 G(x, y) dy ≥ λµ∗‖u‖τ1 measB2 > λ̄µ∗‖u‖τ1 measB2 = ‖u‖. This indicates that ‖u‖ > ‖u‖, which is a contradiction. � EJDE-2023/52 POSITIVE SOLUTIONS FOR FOURTH-ORDER ELLIPTIC PROBLEMS 13 Corollary 3.7. Under assumption (A1), if f0 > 0 and f∞ > 0, then system (3.1) admits no positive solution for any λ > λ̄0, where λ̄0 > 0 is a finite number. 4. Comments on higher-order elliptic equations It is well known that one can change a higher-order elliptic equation into a second-order elliptic system. Thus we can employ some of the methods used for studying second-order elliptic systems. However, we find that it is more difficult to deal with higher-order elliptic equations than second-order elliptic systems. The main difficulty lies in the right-hand side of the equations. As an example, we consider the second-order elliptic system −∆u1 = λf1(u1, u2) in Ω, −∆u2 = λf2(u1, u2) in Ω, u1 = u2 = 0 on ∂Ω, (4.1) where Ω denotes a bounded domain in Rn (n ≥ 2) with smooth boundary ∂Ω, λ > 0 is a parameter, and for each i ∈ {1, 2}, fi satisfies: (A4) fi : R2 + → R+ is continuous. It follows from (4.1) that for i ∈ {1, 2}, ui(x) = λ ∫ Ω̄ G(x, y)fi(u1(y), u2(y)) dy, (4.2) where G(x, y) denotes the Green’s function of −∆ on Ω. Now we define a positive cone in E = C(Ω̄)2 as (3.6). Let K1 = {u ∈ C(Ω̄) : u(x) ≥ 0, x ∈ Ω̄}. For u1, u2 ∈ K1 and i ∈ {1, 2}, we define T̄i : K1 → C(Ω̄) as T̄iui(x) = λ ∫ Ω̄ G(x, y)fi(u1(y), u2(y)) dy. (4.3) Under assumption (A4), it is clear that T̄i : K1 → K1 is completely continuous for i ∈ {1, 2}. On the other hand, it is well known that system (4.1) is equivalent to the fixed point equation u(x) = (T̄1u(x), T̄2v(x)) := T̂u(x) for x ∈ Ω̄, (4.4) where u(x) = (u1(x), u2(x)). Since T̄1, T̄2 : K1 → K1 are completely continuous, we derive that T̂ : K → K is completely continuous. For each i ∈ {1, 2}, let (fi)∞ = lim |u|→+∞ fi(u) |u| , (fi)0 = lim |u|→0+ fi(u) |u| , where u = (u1, u2). Theorem 4.1. Under assumption (A4), if (fi)0 < µ1 and (fi)∞ < µ1 for i ∈ {1, 2}, then system (4.1) admits no positive solution for any λ < λ∗, where λ∗ > 0 is a finite number. Proof. Considering (A4), if (fi)0 < µ1 and (fi)∞ < µ1 for each i ∈ {1, 2}, then there are positive numbers ε6, ε7, h3 and h4 such that h3 < h4 and for u ∈ R2 + and |u| ≤ h3, we derive fi(u) ≤ (µ1 − ε6)|u|, (4.5) 14 M. FENG, H. CHEN EJDE-2023/52 and for u ∈ R2 + and |u| ≥ h2, we derive fi(u) ≤ (µ1 − ε7)|u|. (4.6) Letting µ∗∗ = max { µ1 − ε6, µ1 − ε7, max {fi(u) |u| : |u| ∈ [h1, h2] }} > 0, we have fi(u) ≤ µ∗∗|u|, u ∈ R2 +. (4.7) Let u be a positive solution of system (4.1). We will demonstrate that this leads to a contradiction for λ < λ̂ = 1 µ∗∗‖φ‖0 . Indeed for x ∈ Ω̄ and λ < λ̂, we derive from (4.3), (4.4) and (4.7) that ui(x) = λ ∫ Ω̄ G(x, y)fi(u1(y), u2(y)) dy ≤ λ ∫ Ω̄ G(x, y)µ∗∗‖u‖dy = λµ∗∗‖u‖ ∫ Ω̄ G(x, y) dy ≤ λµ∗∗‖u‖‖φ‖0 < λ̂µ∗∗‖u‖‖φ‖0 = ‖u‖. This indicates that ‖u‖ < ‖u‖, which is a contradiction. � Similar to the proof of Theorem 4.1, we can prove the following result. Theorem 4.2. Under assumption (A4), if (fi)0 > µ1 and (fi)∞ > µ1 for i ∈ {1, 2}, then system (4.1) admits no positive solution for any λ > λ∗∗, where λ∗∗ > 0 is a finite number. For each i ∈ {1, 2}, let (fi) α ∞ = lim |u|→+∞ fi(u) |u|α , where u = (u1, u2). Then, similar to the proof of Theorem 3.1, we can prove the following result. Theorem 4.3. Let µ1 and ϕ1 be defined as in (2.10). Under assumption (A4), if there exists i0 ∈ {1, 2} such that (fi0)0 > µ1, and (fi) α ∞ < µ1 for each i ∈ {1, 2}, then system (4.1) admits at least one positive solution in K for each λ ≥ M τ1µ1 ∫ B2 ϕ1(z)dz , where τ1, B2 and M are respectively defined in (3.4), (3.5), and (3.10). We believe the following conclusion also holds, but we do not have a proof right now. Theorem 4.4. 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Meiqiang Feng School of Applied Science, Beijing Information Science & Technology University, Bei- jing, 100192, China Email address: meiqiangfeng@sina.com Haiping Chen School of Applied Science, Beijing Information Science & Technology University, Bei- jing, 100192, China Email address: chpmath@sina.com 1. Introduction 2. Existence of a positive solution without an eigenvalue parameter 3. Existence of positive solution to (??) 4. Comments on higher-order elliptic equations Acknowledgments References