Positive Solutions for a Singular Superlinear Fourth-Order Equation with Nonlinear Boundary Conditions

Abstract

We show the existence of positive solutions for a singular superlinear fourth-order equation with nonlinear boundary conditions. where λ > 0 is a small positive parameter, is continuous, superlinear at ∞, and is allowed to be singular at 0, and h: [0, 1] ⟶ [0, ∞) is continuous. Our approach is based on the fixed-point result of Krasnoselskii type in a Banach space.

1. Introduction

Consider the singular superlinear fourth-order nonlinear boundary value problemwhere λ > 0 is a small positive parameter, is continuous, h: [0, 1] ⟶ [0, ∞) is continuous. If f(0) > 0, then (1) is called a positone problem; if f(0) < 0, then (1) is called a semipositone problem.

The study of semipositone problems was formally introduced by Castro and Shivaji [1]. From an application viewpoint, one is usually interested in the existence of positive solutions for semipositone problems. Significant processes on second-order semipositone problems have taken place in the last 10 years, see [1–5] and the references therein.

Fourth-order boundary value problems modeling bending equilibria of elastic beams have been considered in several papers [6–9]. Most of them are concerned with nonlinear equations with null boundary conditions. When the boundary conditions are nonzero or nonlinear, fourth-order equations can model beams resting on elastic bearings located in their extremities. See for instance, [10–12] and the references therein.

For instance, Cabrera et al. [10] studied the existence of positive solutions for the fourth-order positone problemwhere f(t, x, y): [0, 1] × [0, + ∞) × [0, + ∞) ⟶ [0, + ∞). f(t, x, y) is increasing in x and decreasing in y, for fixed t ∈ [0, + ∞]. For convenience, we denote in this article. In [10–12], the authors studied positive solutions of fourth-order nonlinear boundary value problems in the positone case based on a mixed monotone operator method and a well-known fixed-point theorem in cones.

It should be noted that nonlinear part f is either bounded or positive states in [1, 10–12]. In this paper, we prove the existence of positive solution to (1) by assuming that is continuous and is allowed to be singular at 0; in other words, f may be unbounded from below and satisfies the superlinear condition. Moreover, we prove a useful lemma (Lemma 3) in this paper which plays a key role to guarantee the positivity of solution. It can be obtained by concavity and convexity of solution or calculation for the second-order boundary value problem, but for the fourth-order boundary value problem, this becomes complicated.

In addition, we will replace h(x)f(u(x)) by m(x) and γ by c(s), and we perform a study of the sign of Green’s function of the corresponding linear problems:

In detail, we construct Green’s function G(x, s) by disconjugacy theory and give a sufficient condition to make ensure that G(x, s) is either positive or negative. This fact is crucial for our arguments.

Motivated by the above facts, in this paper, we shall obtain the existence of positive solutions for a singular superlinear fourth-order equation with nonlinear boundary conditions via the fixed-point result of Krasnoselskii type in a Banach space.

We shall make the following assumptions: (H1): h: [0, 1] ⟶ [0, ∞) is continuous. (H2): c: [0, ∞) ⟶ [0, ∞) is continuous. (H3): there exists a constant β with such that (H4): is continuous and .

By a positive solution of (1), we mean function u ∈ C³[0, 1] ∩ C ⁴(0, 1] with u > 0 on (0, 1) and satisfying (1).

Our main result is as follows.

Theorem 1. Let (H1)–(H4) hold. Then, there exists a constant λ₀ > 0 such that (1) has a positive solution u_λ for λ < λ₀ with u_λ ⟶ ∞ as λ ⟶ 0⁺ uniformly on compact subsets of (0, 1). The paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we prove the main result that the existence of a large positive solution to (1) for λ small.

2. Notation and Preliminaries

Suppose that E is a real Banach space which is partially ordered by a cone P ⊂ E, i.e., x ≤ y if and only if y − x ∈ P. If x ≤ y and x ≠ y, then we denote x < y or y > x. By θ, we denote the zero element of E. Recall that a nonempty closed convex set P ⊂ E is a cone if it satisfies (i) x ∈ P, λ ≥ 0 ⟹ λx ∈ P and (ii)  x ∈ P, − x ∈ P ⟹ x = θ.

We first recall the following fixed-point result of Krasnoselskii type in a Banach space (see e.g., [1], Theorem 12.3).

Lemma 1. Let E be a Banach space and T: P ⟶ P be a completely continuous operator. Suppose there exist h ∈ E, h₀ ≠ 0 and positive constants r, R with r ≠ R such that(a)If y ∈ P satisfies y = ζTy, θ ∈ [0, 1], then ‖y‖ ≠ r(b)If y ∈ P satisfies y = Ty + ξh, ξ ≥ 0, then ‖y‖ ≠ RThen, T has a fixed point y ∈ P with min{r, R} < ‖y‖ < max{r, R}. In the sequel, we describe the Banach space where the fixed points are found, as well as some notations which are used along the paper. Let us consider the Banach space coupled with the normand the Banach space X = L¹(0, 1) equipped with the norm .

Lemma 2. Let satisfywhere m ∈ L¹(0, 1), m(t) ≥ 0, for a.e. t ∈ (0, 1), γ ∈ [0, 2]. Then,where

Proof of Lemma 2. It is easy to verify thatGreen’s function G(x, s) is computed as follows:Obviously, G(x, s) is positive for any γ ∈ [0, 2], while G(x, s) changes sign for γ is large enough, and we can compute the value of G(x, s) when γ = 10000 by Mathematic 9.0:Let for some x₀ ∈ (0, 1]. We verify thatIndeed, if x, x₀ ≤ s,If x ≤ s ≤ x₀,If s ≤ x ≤ x₀,If x₀ ≤ s ≤ x,Thus,

Lemma 3. Let k ∈ L¹(0, 1) with k ≥ 0, and let u ∈ C³[0, 1] ∩ C ⁴(0, 1] satisfySuppose . Then, u(x) ≥ 0 and

Proof. Let be the unique solution of the problemThen,and accordingly,Set . Then,and by Lemma 2, we can getIt follows from (22) and (24) and the fact that y(x) = u(x) − thatfor x ∈ (τ, 1). This completes the Proof of Lemma 3.

3. Proof of the Main Results

Proof of Theorem 1. Let λ > 0. For , define , where u is the solution ofwhere = ,and q(x) is defined in Lemma 2. Then,where Green’s function is given byNote that ≤ 1 for all x, s. By (H3), there exists a constant M > 0 depending on such thatIt follows from the Lebesgue dominated convergence theorem thatDefine the cone P in byFor u ∈ P, it follows that, from (H3) and Lemma 3, we can get .
We next show that is a completely continuous operator.
Now, we show is continuous. To this end, let ∈ C[0, 1] be such that in P and letFix x, s ∈ (0, 1), and defineThen, , and there exists a constant N such that |H′(z)| ≤ N for 0 ≤ z ≤ 2, and the mean value theorem givesNotice thatand the factNow, these together with the Lebesgue dominated convergence theorem guarantee thatHence, is continuous.
Finally, we prove that is completely continuous. In fact, for x ∈ P,and for x, x′ ∈ [0, 1], we haveNow, the Arzela–Ascoli theorem guarantees that is compact.
Let a > 1 be such that f(z) > 0 for z ≥ a. Since , there exists a constant b > 0 such thatHence,for all z > 0, where for x ≥ a. Note that is nondecreasing.
Suppose , where and . Next, we shall verify the assumptions of Lemma 1 hold for .(a)There exists r_λ > 0 such that if u ∈ P satisfy , then ‖u‖_∞ ≠ r_λ. Indeed, let u ∈ P satisfy for some θ ∈ (0, 1]. Then, , and therefore, u satisfies for x ∈ I. Since q(s) ≤ 1 for all s ∈ I and a > 1, it follows from (43) that Hence, for x ∈ I, which implies Since and by (H4), there exists a constant r_λ > a such that Combining (47) and (48), we deduce that ‖u‖_∞ ≠ r_λ. Note that r_λ ⟶ ∞ as λ ⟶ 0.(b)There exists R_λ > r_λ such that if , then ‖u‖_∞ ≠ R_λ. Let u ∈ P satisfy for some ξ ≥ 0. Then, , and so, Let Then, k ∈ X. Since u satisfiesand by (41) and (42),it follows from Lemma 3 thatSuppose . Then,Sincefor s, x ∈Λ, it follows from (52), (54), and (55) thatfor x ∈Λ, where . Consequently,Since the left side of this inequality goes to ∞ as ‖u‖_∞ ⟶ ∞, it follows that ‖u‖_∞ < R_λ for R_λ ≫ 1, which establishes (b).
By Lemma 1, has a fixed point u_λ with . Since (19) holds (with ξ = 0) and r_λ ⟶ ∞ as λ ⟶ 0, and it follows that u_λ is a positive solution of (1) if λ is sufficiently small and u_λ(x) ⟶ ∞ as λ ⟶ 0 uniformly for x in compact subsets of (0, 1). This completes the Proof of Theorem 1.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Authors’ Contributions

The authors contributed equally to this paper. All authors read and approved the final manuscript.

Acknowledgments

This work was supported by the NSFC (No. 11671322).