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

Authors: Dongliang Yan
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We show the existence of positive solutions for a singular superlinear fourth-order equation with nonlinear boundary conditions.u⁗x=λhxfux,x∈0,1,u0=u′0=0,u″1=0,u⁗1+cu1u1=0, where λ > 0 is a small positive parameter, f:0,∞⟶ℝ 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.


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## Body

## 1. Introduction

Consider the singular superlinear fourth-order nonlinear boundary value problem(1)u⁗x=λhxfux,x∈0,1,u0=u′0=0,u″1=0,u‴1+cu1u1=0,where λ > 0 is a small positive parameter, f:0,∞⟶ℝ 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 problem(2)u⁗x=ft,ux,Hux,x∈0,1,u0=u′0=0,u″1=0,u‴1+c˜u1=0,where 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 c˜s=css 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 partf 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 f:0,∞⟶ℝ 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 replaceh(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:(3)w⁗x=mx,x∈0,1,w0=w′0=0,w″1=0,w″′1+γw1=0.In detail, we construct Green’s functionG(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 constantβ with β<1/2 such that(4)limsupt⟶0+tβft<∞.(H4):f:0,∞⟶ℝ is continuous and limt⟶∞ft/t=∞.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 constantλ0 > 0 such that (1) has a positive solution uλ for λ < λ0 with uλ ⟶ ∞ 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 thatE 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.
LetE 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)
Ify ∈ P satisfies y = ζTy, θ ∈ [0, 1], then ‖y‖ ≠ r(b)
Ify ∈ P satisfies y = Ty + ξh, ξ ≥ 0, then ‖y‖ ≠ R
Then,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 ℬ=CI,I=0,1 coupled with the norm(5)u∞=maxt∈Iut,and the Banach space X = L1(0, 1) equipped with the norm ϕX=∫01ϕtdt.Lemma 2.
Letw satisfy(6)w⁗x=mx,x∈0,1,w0=w′0=0,w″1=0,w‴1+γw1=0,where m ∈ L1(0, 1), m(t) ≥ 0, for a.e. t ∈ (0, 1), γ ∈ [0, 2]. Then,(7)wx≥3−γ9−γw∞qx,x∈0,1,where(8)qx=minx2,x=x2.Proof of Lemma 2.
It is easy to verify that(9)wx=∫01Gx,smsds≔ℒwx.
Green’s functionG(x, s) is computed as follows:(10)Gx,s=16s23x−s+γ6−2γx2s2x−3s−3,0≤s≤x≤1,x23s−x+γ6−2γx2s2x−3s−3,0≤x≤s≤1.
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:(11)G0.587022479,0.999≈−0.0000059476,G0.5852750296,0.5852750296≈0.00979528106.
Letw∞=wx0 for some x0 ∈ (0, 1]. We verify that(12)Gx,sGx0,s≥qx,s,x,x0∈0,1.
Indeed, ifx, x0 ≤ s,(13)Gx,sGx0,s=x23s−x+γ/6−2γx2s2x−3s−3x023s−x0+γ/6−2γx02s2x0−3s−3≥x23−x/sx023−x0/s+γ/6−2γx02sx0−3s−3≥x23−x/s3s2+γ/6−2γs3s−32≥2x23+γ/6−2γ4≥3−γ9−γx2.
Ifx ≤ s ≤ x0,(14)Gx,sGx0,s=x23s−x+γ/6−2γx2s2x−3s−3s23x0−s+γ/6−2γx02s2x0−3s−3≥x23−x/ss3x0+γ/6−2γx02sx0−3s−3≥x23−x/s3x0+γ/6−2γs−32≥2x23+γ/6−2γ9≥3−γ9−γx2.
Ifs ≤ x ≤ x0,(15)Gx,sGx0,s=s23x−s+γ/6−2γx2s2x−3s−3s23x0−s+γ/6−2γx02s2x0−3s−3≥3x−s3x0−s+γ/6−2γx02x0−3s−3≥3x−s3x0+γ/6−2γx02s−32≥2x3+γ/6−2γ9≥3−γ9−γx.
Ifx0 ≤ s ≤ x,(16)Gx,sGx0,s=s23x−s+γ/6−2γx2s2x−3s−3x023s−x0+γ/6−2γx02s2x0−3s−3≥3x−sx0/s23s−x0+γ/6−2γx02x0−3s−3≥3x−ss/s23s+γ/6−2γx02x0−32≥2x3+γ/6−2γ4≥3−γ9−γx.
Thus,(17)wx=∫01Gx,sGx0,sGx0,swsds≥3−γ9−γw∞qx.Lemma 3.
Letk ∈ L1(0, 1) with k ≥ 0, and let u ∈ C3[0, 1] ∩ C 4(0, 1] satisfy(18)u⁗x≥−k,x∈0,1,u0≥0,u′0≥0,u″1≥0,u‴1+γu1≥0.
Supposeu∞>29−γ/3−γ2kX. Then, u(x) ≥ 0 and(19)ux≥3−γ29−γu∞qx,x∈τ,1.Proof.
Letw0x∈ℬ be the unique solution of the problem(20)w⁗x=−k,x∈0,1,w0=0,w′0=0,w″1=0,w‴1+γw1=0.
Then,(21)−w0x=16∫0xs23x−s+γ/6−2γx2s2x−3s−3ksds+∫x1x23s−x+γ/6−2γx2s2x−3s−3ksds,and accordingly,(22)−w0x≤∫0x3x2s+γ6−2γx2s2x−3s−3ksds+∫x13x2s+γ6−2γx2s2x−3s−3ksds≤∫013x2s+γ6−2γx2sx−3s−3ksds≤x2∫013s+γ6−2γ1−32sksds≤9−γ3−γkXqx.
Setyx=ux−w0x,y‴1+γy1−γy0+y′0/6+γ≤y″1≤y‴1+γy1−γy0+y′0,yτ=y∞. Then,(23)y⁗x≥0,x∈0,1,y0≥0,y′0≥0,y″1≥0,y‴1+γy1≥0,and by Lemma 2, we can get(24)yx≥3−γ9−γy∞qx,x∈τ,1.
It follows from (22) and (24) and the fact that y(x) = u(x) − w0x that(25)ux=yx+w0x≥3−γ9−γy∞qx−9−γ3−γkXqx=3−γ9−γu−w0∞qx−9−γ3−γkXqx≥3−γ9−γu∞qx−9−γ3−γw0∞qx−9−γ3−γkXqx≥3−γ9−γu∞qx+9−γ3−γkXqx2−9−γ3−γkXqx≥3−γ9−γu∞qx−9−γ3−γkXqx=3−γ9−γu∞−9−γ3−γkXqx,for x ∈ (τ, 1). This completes the Proof of Lemma 3.


## 3. Proof of the Main Results

Proof of Theorem 1.
Letλ > 0. For v∈ℬ, define ℒλv=u, where u is the solution of(26)u⁗x=λhxfv˜,x∈0,1,u0=0,u′0=0,u″1=0,u‴1+γvu1=0,where(27)v˜x=maxvx,qx,γv = cv1,and q(x) is defined in Lemma 2. Then,(28)ux=λ∫01Gvx,shsfv˜sds,where Green’s function Gvx,s is given by(29)Gvx,s=16s23x−s+γv6−2γvx2s2x−3s−3,0≤s≤x≤1,x23s−x+γv6−2γvx2s2x−3s−3,0≤x≤s≤1.
Note thatGvx,s ≤ 1 for all x, s. By (H3), there exists a constant M > 0 depending on v∞ such that(30)fv˜x≤Mv˜β≤Mqβ∈L10,1.
It follows from the Lebesgue dominated convergence theorem that(31)u∈ℬi.e.ℒλ:ℬ⟶ℬ.
Define the coneP in ℬ by(32)P=u∈ℬux≥0,ux≥3−γ29−γu∞qx,x∈τ,1.
Foru ∈ P, it follows that, from (H3) and Lemma 3, we can get ℒλP⊂P.
We next show thatℒλ:P⟶P is a completely continuous operator.
Now, we showℒλ:P⟶P is continuous. To this end, let vn ∈ C[0, 1] be such that vn⟶v in P and let(33)un=ℒλvn,u=ℒλv.
Fixx, s ∈ (0, 1), and define(34)Hz=x23s−x+z6−2zx2s2x−3s−3,for0≤z≤2.
Then,H′z=1/6−2z2x2s2x−3s−3, and there exists a constant N such that |H′(z)| ≤ N for 0 ≤ z ≤ 2, and the mean value theorem gives(35)Gvnx,s−Gvx,s≤Nγvn−γv.
Notice that(36)σ˜ns=fv˜ns−fv˜s⟶0,n⟶∞fors∈0,1,and the fact(37)fv˜x≤Mqβ∈L10,1.
Now, these together with the Lebesgue dominated convergence theorem guarantee that(38)ℒλvn−ℒλv≤λ∫01Gvnx,s−Gvx,shsfv˜nds+∫01Gvx,shsfv˜n−fv˜ds≤λ∫01Nγvn−γvhsfv˜nds+∫01Gvx,shsσ˜nsds⟶0asn⟶∞.
Hence,ℒλ:P⟶P is continuous.
Finally, we prove thatℒλ:P⟶P is completely continuous. In fact, for x ∈ P,(39)ℒλv≤∫01Gγvx,shsMqβsds,and for x, x′ ∈ [0, 1], we have(40)ℒλvx−ℒλvx′≤∫01Gγvx,s−Gγvx′,shsMqβssds.
Now, the Arzela–Ascoli theorem guarantees thatℒλ:P⟶P is compact.
Leta > 1 be such that f(z) > 0 for z ≥ a. Since limsupx⟶0+xβfx<∞, there exists a constant b > 0 such that(41)fz≤bzβ,z∈0,a.
Hence,(42)fz≥−bzβ,(43)fz≤bzβ+f^maxz,afor all z > 0, where f^x=supa≤z≤xfz for x ≥ a. Note that f^ is nondecreasing.
Supposeλ<a/2c1+c2fa, where c1=b∫01hs/qβsds and c2=∫01hsds. Next, we shall verify the assumptions of Lemma 1 hold for ℒλ.(a)
There existsrλ > 0 such that if u ∈ P satisfy u=θℒλu,θ∈I, then ‖u‖∞ ≠ rλ.Indeed, letu ∈ P satisfy u=θℒλu for some θ ∈ (0, 1]. Then, u/θ=ℒλu, and therefore, u satisfies(44)ux=λθ∫01Gux,shsfu˜ds,forx ∈ I. Since q(s) ≤ 1 for all s ∈ I and a > 1, it follows from (43) that(45)fu˜≤bu˜β+f^maxu˜s,a≤bqβ+f^maxu˜s,a.Hence,(46)ux≤λb∫01hsqβds+f^maxu∞,a∫01hsds≤λc1+c2f^maxu∞,a,forx ∈ I, which implies(47)u∞c1+c2f^maxu∞,a≤λ.Sincea/c1+c2f^a>2λ and limz⟶∞z/c1+c2f^z=0 by (H4), there exists a constant rλ > a such that(48)rλc1+c2f^rλ=2λ.Combining (47) and (48), we deduce that ‖u‖∞ ≠ rλ. Note that rλ ⟶ ∞ as λ ⟶ 0.(b)
There existsRλ > rλ such that if u=ℒλu+ξ,ξ≥0, then ‖u‖∞ ≠ Rλ.Letu ∈ P satisfy u=ℒλu+ξ for some ξ ≥ 0. Then, u−ξ=ℒλu, and so,(49)ux−ξ=λ∫01Gux,shsfu˜ds.Let(50)kx=bhxqβx,x∈0,1.Then,k ∈ X. Since u satisfies(51)u⁗=λhxfu˜,0<x<1,u0=ξ≥0,u′0=u″1=0,u‴1+λvu1=λvξ≥0,and by (41) and (42),(52)hxfu˜x≥−bhxu˜βx≥−bhxqβx=−kx,it follows from Lemma 3 that(53)ux≥3−γ9−γu∞−9−γ3−γkXqx,x∈τ,1.
Supposeu∞>max9−γ/3−γ2kX,2/qσ. Then,(54)us≥29−γ2−9−γ229−γ2u∞qσ≥qσ2u∞,s∈τ+σ,1−σ≕Λ.
Since(55)Gux,s≥162x2s2,s≤x,2x2s2,x≤s.≥1256,for s, x ∈Λ, it follows from (52), (54), and (55) that(56)ux≥λ∫ΛGux,shsfu˜ds+∫ΛcGux,shsfu˜ds≥λ1/256f⌣qσu∞/2∫Λhsds−kX,for x ∈Λ, where f⌣x=infz≥xfz. Consequently,(57)1/256f⌣qσu∞/2∫Λhsds−kXu∞≤1λ.
Since the left side of this inequality goes to ∞ as ‖u‖∞ ⟶ ∞, it follows that ‖u‖∞ < Rλ for Rλ ≫ 1, which establishes (b).
By Lemma1, ℒλ has a fixed point uλ with uλ∞>rλ. Since (19) holds (with ξ = 0) and rλ ⟶ ∞ as λ ⟶ 0, and it follows that uλ is a positive solution of (1) if λ is sufficiently small and uλ(x) ⟶ ∞ as λ ⟶ 0 uniformly for x in compact subsets of (0, 1). This completes the Proof of Theorem 1.



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