# Existence of Positive Solutions for Second-Order Third-Point Semipositive BVP

Authors: Hua Su; Jinmin Yu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study the existence of positive solutions for the following nonlinear second-order third-point semi-positive BVP. We derive an explicit interval of positive parameters, which for anyl,μ in this interval, the existence of positive solutions to the boundary value problem is guaranteed under the condition that at,x,bt,x are all superlinear (sublinear), or one is superlinear, the other is sublinear.


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## Body

## 1. Introduction

In the applied mathematical field, three-point BVP can describe many phenomena. Moshinsky [1] introduced the vibrations of a guy wire with a uniform cross-section and composed of N parts of different densities using a multipoint BVP. Timoshenko [2] also revealed that the theory of elastic stability can be used by the method of a three-point BVP. Il’in and Moviseev [3] were the first to study this aspect. Since then, more general nonlinear BVP have been studied by several authors [4–25].In their paper [7], Ma and Wang obtained the existence of positive solutions for a three-point BVP by Krasnoselskii’s fixed theorem:(1)u″t+atu′t+btut+htfu=0,0≤t≤1,u0=0,u1=αuη,where α is a positive constant, 0<η<1, at∈C0,1,R+, bt∈C0,1,R−, f∈CR+,R+,h∈C0,1,R+ and there exists x0∈0,+∞ such that hx0>0.In our paper, we study the existence of positive solutions of second-order third-point semipositive BVP:(2)Lxt+λat,x+μbt,x,0≤t≤1,x0=0,x1=αxξ,where Lut=u″t+ftu′t+gtut, λ,μ are positive parameters, 0<ξ<1, ft∈C0,1, and gt∈C0,1,−∞,0. And our paper also allows that at,x,bt,x are both semipositive and lower unbounded.Our main tool is the following fixed point index theory.Theorem 1 [4].
We suppose thatK⊂E is a cone in E, in which E is a real Banach space, the open bounded set Ω1,Ω2 is in E, θ∈Ω1, Ω¯1⊂Ω2, and T:K∩Ω¯2\Ω1⟶K. Suppose operator T can be completely continuous and satisfies one of the following conditions:(i)
Tx≤x, ∀x∈K∩∂Ω1; Tx≥x, ∀x∈K∩∂Ω2(ii)
Tx≥x, ∀x∈K∩∂Ω1; Tx≤x, ∀x∈K∩∂Ω2
Then, operatorT has at least one fixed point x∗ in K∩Ω¯2\Ω1.Theorem 2 [4].
We suppose thatP⊂E is a cone in E, in which E is a real Banach space, the open bounded set Ω1,Ω2,Ω3 is in E, θ∈Ω1, Ω¯1⊂Ω2, Ω¯2⊂Ω3, and T:P∩Ω3⟶P. Suppose operator A is completely continuous and satisfies the following conditions:(3)Tx≤x,∀x∈P∩∂Ω1,Tx≥x,Ax≠x,∀x∈P∩∂Ω2,Tx≤x,∀x∈P∩∂Ω3.Then, operatorT has at least two fixed points x∗ and x∗∗ in P∩Ω¯3/Ω1, and x∗∈P∩Ω2/Ω1 and x∗∗∈P∩Ω¯3/Ω¯2.


## 2. Preliminaries and Lemmas

We set a Banach spaceE=C0,1,−∞,+∞ with norm x=maxt∈Ixt. We know of the following lemmas from Ref. [6].Lemma 3.
Settingξ1tas the positive solution of the equation, we have:(4)Lξ1t=0,0≤t≤1,ξ10=0,ξ11=1.Then,ξ1t∈0,1 is strictly increasing on 0,1, and ξ1′0>0.Lemma 4.
Settingξ2tas the positive solution of the equation, we have:(5)Lξ2t=0,0≤t≤1,ξ20=1,ξ21=0.
Then,ξ2t∈0,1 is strictly decreasing on 0,1.From Lemma3 and Lemma 4, we know that 0<ξ1t<1, 0<ξ2t<1. In the rest of our paper, the following condition is used:(C1)
0<αξ1η<1, whereξ1tis given by Lemma 3Throughout this paper, we shall use the following notation:(6)Gt,s=1ζξ1tξ2s,0≤t≤s≤1,ξ1sξ2t,0≤s≤t≤1,where ζ=ξ1′0ξ20.Obviously, from Ref. [6], we can be assured that when (C1) holds, the BVP(7)Lxt+yt=0,0≤t≤1,x0=0,x1=αuξ,is equivalent to the following integral equation:(8)xt=∫01Gt,sesysds+αξ1t1−αξ1ξ∫01Gξ,sesysds,where et=exp∫0tfsds.Setzt=minξ1t/ξ1,ξ2t/ξ2. From (6), for t∈0,1, we know that(9)ztGs,s≤Gt,s≤Gs,s.We present some other lemmas that are important to our main results.Lemma 5 [7].
Assume that for anyy∈C0,1,0,+∞, xt is the solution of the following BVP:(10)Lxt+yt=0,0<t<1,x0=0,x1=αxξ.
Then, we have(11)xt≥ztx,t∈0,1.Lemma 6.
Assume thatw¯ is a solution of the following BVP:(12)Lxt=−Bt,0<t<1,x0=0,x1=αxξ,where B∈C0,1,M>0. Then, there exists constant M>0 and satisfies(13)w¯t≤MBzt,t∈0,1.Proof.
Fort∈0,1, we can have(14)w¯t=∫01Gt,sesBsds+αξ1t1−αξ1ξ∫01Gξ,sesBsds.
Obviously, fort∈0,1, we have(15)∫01Gt,sesBsds=1ζ∫0tξ1sξ2tesBsds+∫t1ξ1tξ2sesMsds≤p1ζξ1tξ2t∫0tBsds+ξ1tξ2t∫t1Bsds=e1ξ1ξ2ζξ1tξ1ξ2tξ2∫0tMsds+ξ1tξ1ξ2tξ2∫t1Bsds≤e1ξ1ξ2ζzt∫01Bsds≤M1ztB,where M1=e1ξ1ξ2/ζ.
By the same method, we can know that(16)αξ1t1−αξ1ξ∫01Gξ,sesBsds≤M2ztB,where M2=αe1ξ1ξ2/1−αξ1ξ.
So, by choosing constantM≥M1+M2, we have(17)w¯t≤MBzt,0≤t≤1.
☐Lemma 7 [7].
Let0≤limx⟶∞¯bt,x/x≤L2, t∈0,1. Define the following function:(18)Gτ=max0≤t≤1,0≤x≤τbt,τ.
Then(i)
G is a nondecreasing function for τ(ii)
0≤limρ⟶∞¯Gτ/τ≤K2
For g assumptions:(C2)
at,x,bt,x∈C,0,1×0,+∞R
From (C2), there exists a functionBt∈C0,1,Bt>0, which satisfies(19)at,x≥−Bt,bt,x≥−Bt,∀t∈0,1,x≥0,where MB<1. M is given by Lemma 6.(C3)
B1≤a∞−≤∞,B2≤b∞−≤∞(C4)
0≤a∞+≤b1,0≤b∞+≤b2(C5)
K1≤a∞−≤∞,0≤b∞+≤K2where(20)minB1,B2≥2λ+μmin0≤t≤1∫01Gt,seszsds−1,b1+b2≤λ+μp1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sds−1,K1≥2λmin0≤t≤1∫01Gt,seszsds−1,a∞−=lim¯x⟶∞at,xx,a∞+=lim¯x⟶∞at,xx,b∞−=lim¯x⟶∞bt,xx,b∞+=lim¯x⟶∞bt,xx.
Letε=min0≤t≤1zt, and(21)H¯t,x=Ht,x,x≥0,Ft,0,x<0,Y¯t,x=Yt,x,x≥0,Gt,0,x<0,where Ht,x=at,x+Bt, Yt,x=bt,x+Bt.
For anyl>0, we set(22)Hl=max0≤t≤1,0≤x≤lH¯t,x,Yl=max0≤t≤1,0≤x≤lY¯t,x.From Lemma6, letting wt=w¯t, then xt is the positive solution of problem (2) if and only if x~t=xt+wt is the solution of the following problem:(23)Lxt+λH¯t,x−w+μY¯t,x−w=0,x0=0,x1=αuξ,and x~t>wt, 0<t<1; here, H¯, Y¯ is given by (21).Defining the coneP in E, we have(24)P=x∈E:xt≥xqt,t∈0,1.Obviously, problem (18) is equivalent to(25)xt=∫01Yt,sesλH¯s,x−w+μY¯s,x−wds+αξ1t1−αξ1ξ∫01Gξ,sesλH¯s,x−w+μY¯s,x−wds.Defining the operatorT:E⟶E, we have(26)Txt=∫01Gt,sesλH¯s,x−w+μY¯s,x−wds+αξ1t1−αξ1ξ∫01Gξ,sesλH¯s,x−w+μY¯s,x−wds.ObviouslyTP⊂P and T is completely continuous.


## 3. Our Main Three Results

Theorem 8.
Suppose condition (C1), condition (C2), and condition (C3) hold. Then, for the small numberλ,μ, problem (2) has at least one positive solution.Proof.
Firstly, we choose sufficiently smallλ,μ which satisfies the following:(27)λ+μ<H1+Y1p1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sds−1.
LettingΩ1=x∈E:x<1, for any x∈P∩∂Ω1, t∈0,1, by the definition of operator T, we have(28)Txt≤∫01Gs,sλH1+μY1ds+αξ1t1−αξ1ξ∫01Gξ,sesλH1+μY1ds≤λ+μH1+Y1e1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sds<1=x.
Thus, we have(29)Tx≤x,∀x∈P∩∂Ω1.
Secondly, by (C3), we know that there exists constantl1>0 which satisfies(30)at,x≥B1x,bt,x≥B2x,∀x≥l1,t∈0,1.
Lettingr=max2MB,2l1/ε,2, then r>1. Set Ω2=x∈E:x<r, for any x∈P∩∂Ω2, t∈0,1, we have(31)xt−wt≥xt−MBzt≥xt−MB1rxt≥12xt.
Therefore, we havext−wt≥1/2xt≥x/2zt≥εr/2≥l1.
Thus, by the definition ofH¯,Y¯ and (30), we can have(32)λH¯s,xs−ws+μY¯s,xs−ws≥B1λxs−ws+B2μxs−ws≥minB1,B2λ+μxs−ws.
We have(33)Txt≥∫01Gt,sesλH¯s,x−w+μY¯s,x−wds≥min0≤t≤1∫01Gt,sminB1,B2λ+μxs−wsds≥12λ+μminB1,B2min0≤t≤1∫01Gt,sxsds.
Then, by Lemma5, we have(34)Txt≥12λ+μminB1,B2min0≤t≤1∫01Gt,seszsdsxt.
Therefore, by the definition ofB1,B2, we have(35)Tx≥x,∀x∈K∩∂Ω2.
Then, by (29), (35) and Theorem 1, operator T has at last one fixed point x~t∈P∩Ω2¯/Ω1, i.e., x~t is the solution of problem (2), and it is easy to know x~≥1.
Finally, by (C2) and Lemma3, we have(36)x~t≥x~zt≥zt>MBzt≥w¯t=wt.
Thus,x=x~−w is the positive solution of problem (2).☐Theorem 9.
We suppose that condition (C1), (C2), and (C4) hold, and the following condition also holds:(C6)
There exist constantD>0,ρ>0, and we have(37)at,x≥ρ,bt,x≥ρ,x∈D,∞,t∈0,1.Then, for the small numberλ,μ, problem (2) has at least one positive solution.Proof.
Firstly, letr=max2MB,2D/ε,2, and(38)1=2rmin0≤t≤1∫01Yt,sesλ+μρds−1.
SetΩ1=x∈E:x<r, for any x∈P∩∂Ω1,s∈0,1, we have(39)xs−ws≥xs−MBzs≥xs−MBrxs≥12xs.
Thus,xs−ws≥1/2xs≥x/2zs≥εr/2≥D. Therefore, by (C6) and the definition of operator T, we have(40)Txt=∫01Gt,sesλH¯s,x−w+μY¯s,x−wds+αξ1t1−αξ1ξ∫01Gξ,sesλH¯s,x−w+μY¯s,x−wds.
ForBt>0,t∈0,1, we have(41)Txt≥∫01Gt,sesλH¯s,x−w+μY¯s,x−wds≥∫01Gt,sesλρ+Bs+μρ+Bsds≥12min0≤t≤1∫01Gt,sesλ+μρds=r=x.
We can know that by the above discussion, we have(42)Tx≥x,∀x∈P∩∂Ω1.
Secondly, by (C4), we can have(43)0≤limx⟶∞H¯s,x−wu≤b1,0≤limx⟶∞Y¯s,x−wu≤b2,s∈0,1.
Then, there exists constantl2>0 which satisfies(44)H¯s,x−w≤b1x,Y¯s,x−w≤b2x,∀x≥l2,s∈0,1.
LettingR=max2l2,2r, then r<R. Set Ω2=x∈E:x<R, for any x∈P∩∂Ω2,t∈0,1, we have(45)Txt≤∫01Gs,sλb1xs+μb2xsds+αξ1t1−αξ1ξ∫01Gξ,sesλb1xs+μb2xsds.
Thus, we have(46)Txt≤λ+μb1+b2e1∫01Gs,sxsds+αξ111−αξ1ξ∫01Gξ,sxsds.
So, we have(47)Txt≤λ+μb1+b2e1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sdsx.
Then, we can have by the definition ofb1,b2(48)Tx≤x,∀x∈K∩∂Ω2.
☐Then, similar to the proof of heorem8, we have that result of heorem 9 by Theorem 1.Theorem 10.
Suppose condition (C1), condition (C2), and condition (C5) hold. Then, for sufficiently smallλ,μ, problem (2) has at least two positive solutions.Proof.
Firstly, by Lemma7, there exists constant τ>0 which satisfies(49)Gτ≤K2τ.
Therefore, settingΩ1=x∈E:x<τ, for any x∈P∩∂Ω1,t∈0,1, by the above discussion, for the quite small λ,μ, we have(50)λHτ+μYτe1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sds≤τ.
We have(51)Txt≤λHτ+μYτ∫01Gt,sesds+αξ111−αξ1ξ∫01Gξ,sesds≤λHτ+μYτe1∫01Gs,sds+αξ111−αξ1ξ∫01Gξ,sds≤τ=x.
Then, we have(52)Tx≤x,∀x∈P∩∂Ω1.
Secondly, by (C5), there exists a constantl3>1, which satisfies(53)at,x≥K1x,∀x≥l3.
Lettingr=max2MB,2l3/ε,2τ, and Ω2=x∈E:x<r, for any x∈P∩∂Ω2,t∈0,1, we have(54)xt−wt≥xt−MBzt≥xt−M∥B∥rxt≥12xt.
Then,xt−wt≥1/2xt≥x/2zt≥rε/2≥l3.
Therefore, by the definitions ofH¯,Y¯ and the above discussion, we have(55)Txt≥∫01Gt,sesλH¯s,x−w+μY¯s,x−wds≥∫01Gt,sesλK1x−wds≥K12λmin0≤t≤1∫01Gt,seszsds≥r=x.
Thus, we have(56)Tx≥x,∀x∈P∩∂Ω2.
Finally, letting(57)R=maxλHR+μYR∫01Gs,sesds+αξ111−αξ1ξ∫01Gξ,sesds,2r,then, τ<r<R. Set Ω3=x∈E:x<R, for any x∈P∩∂Ω3,t∈0,1, by the definition of operator T, we have(58)Txt≤λHR+μYR∫01Gs,sesds+αξ111−αξ1ξ∫01Gξ,sesds.
Thus, we have(59)Tx≤x,∀x∈P∩∂Ω3.
Then, similar to the proof of heorem8, we have the result of heorem 10 by Theorem 2.☐Remark 11.
The results of these three theorems in our paper also hold under the condition in which nonlinearat,x,bt,x are both lower semicontinuous.Remark 12.
We can obtain the results of Theorem10 if we replace condition (C5) with (C6)K1≤b∞−≤∞, 0≤a∞+≤K2.



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