# Positive Solutions for a Fourth-Order Riemann–Stieltjes Integral Boundary Value Problem

Authors: Yujun Cui; Donal O’Regan; Jiafa Xu
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this paper, we use the fixed point index to study the existence of positive solutions for the fourth-order Riemann–Stieltjes integral boundary value problem−x4t=ft,xt,x′t,x″t,x″′t,t∈0,1x0=x′0=x″′1=0,x″0=αx″t, where f: 0,1×ℝ+×ℝ+×ℝ+×ℝ+⟶ℝ+ is a continuous function and αx″ denotes a linear function. Two existence theorems are obtained with some appropriate inequality conditions on the nonlinearity f, which involve the spectral radius of related linear operators. These conditions allow ft,z1,z2,z3,z4 to have superlinear or sublinear growth in zi,i=1,2,3,4.


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## Body

## 1. Introduction

In this paper, we investigate the existence of positive solutions for the following fourth-order Riemann–Stieltjes integral boundary value problem:(1)−x4t=ft,xt,x′t,x″t,x″′t,t∈0,1,x0=x′0=x″′1=0,x″0=αx″t,where αx″t=∫01x″tdβt denotes the Riemann–Stieltjes integral with a suitable function β of bounded variation and(2)H0α1∈0,1.The deformation of an elastic beam in an equilibrium state can be described by a fourth-order ordinary equation boundary value problem [1], and there are a large number of papers in the literature in this direction; for example, see [1–30] and the references therein. In [1], the author used Krasnosel’skii’s fixed point theorem to establish one or two positive solutions for the fourth-order boundary value problem:(3)u4t+βu″t=λft,ut,u″t,t∈0,1,u0=u1=∫01psusds,u″0=u″1=∫01qsu″sds.Whenf∈C0,1×ℝ+×ℝ−,ℝ+ and in [2], the authors studied the existence of positive solutions for the fourth-order m-point boundary value problem:(4)u4+αu″−βu=ft,u,t∈0,1,u0=∑i=1m−2aiuξi,u1=∑i=1m−2biuξi,u″0=∑i=1m−2aiu″ξi,u″1=∑i=1m−2biu″ξi,where f∈C0,1×ℝ+,ℝ+ satisfies superlinear and sublinear growth conditions:(5)superlinear:liminfu⟶+∞mint∈0,1ft,uu>λ∗,limsupu⟶0+maxt∈0,1ft,uu<λ∗,sublinear:liminfu⟶0+mint∈0,1ft,uu>λ∗,limsupu⟶+∞maxt∈0,1ft,uu<λ∗,where λ∗ is the first eigenvalue of the relevant linear operator.In [3], the authors studied the existence of an iterative solution for the fourth-order boundary value problem:(6)u4t=ft,ut,u′t,t∈0,1,u0=u′0=u′1=u″1=0,where f:0,1×ℝ2⟶ℝ is continuous and satisfies some appropriate Lipschitz condition, and in [4], the authors used the method of upper and lower solution to establish existence results for the fourth-order four-point boundary value problem on time scales:(7)uΔΔΔΔt=ft,uσt,uΔΔt,t∈0,1T,u0=uσ41=0,αuΔΔξ1−βuΔΔΔξ1=0,γuΔΔξ2+ηuΔΔΔξ2=0,where f:0,1×ℝ×ℝ⟶ℝ is a continuous function.There are only a few papers in the literature which consider general nonlinearities for fourth-order boundary value problems. The difficulty lies in a priori estimates for third-order derivatives, so some authors adopted a Nagumo-type growth condition (see (H7) in Section 3) to overcome this difficulty; for example, see [15, 16, 31–34] and the references therein. In [15], the author studied the existence of positive solutions for the fourth-order boundary value problem:(8)u4=ft,u,u′,u″,u″′,t∈0,1,u0=u′0=u″1=u″′1=0,where f∈C0,1×ℝ+×ℝ+×ℝ+×ℝ−,ℝ+ satisfies some inequality conditions, where f grows both superlinearly and sublinearly about its variables u,u′,u″,andu″′. When f is superlinear, a Nagumo-type condition is used to restrict the growth of f on u″andu″′.Integral boundary conditions arise in thermal conduction problems [35], semiconductor problems [36], and hydrodynamic problems [37], and there are some papers in the literature devoted to this direction (see [1, 9, 19, 34, 38–48]). In [19], the authors studied p-Laplacian fourth-order differential equations with Riemann–Stieltjes integral boundary conditions:(9)ϕp1u″t″=λp1−1a1tf1t,ut,vt,0<t<1,ϕp1v″t″=μp2−1a2tf2t,ut,vt,u0=u1=∫01usdξ1s,ϕp1u″0=ϕp1u″1=∫01u″sdη1sϕp1,v0=v1=∫01vsdξ2s,ϕp2v″0=ϕp2v″1=∫01ϕp2v″sdη2s.The authors used fixed point theory in cones to obtain the existence of positive solutions for the above problem and provided the interval ranges of the parametersλandμ for these solutions.In [38], the authors studied the fractional differential equation with a singular decreasing nonlinearity and a p-Laplacian operator:(10)−D0+αφp−D0+γzx=fx,zx,0<x<1,z0=0,D0+γz0=D0+γz1=0,z1=∫01zxdχx.Using a double iterative technique, they showed that the above problem has a unique positive solution, and from an iterative technique, they established an appropriate sequence, which converges uniformly to the unique positive solution.Motivated by the aforementioned works, the aim of this paper is to study the existence of positive solutions for the fourth-order Riemann–Stieltjes integral boundary value problem (1). The novelty is of two folds: (1) we provide some useful inequality conditions on f involving the first eigenvalue of the relevant linear operator (these conditions imply that f grows superlinearly and sublinearly) and (2) for the superlinear case, an appropriate Nagumo-type condition is used to restrict the growth of f on x″′ in (1).


## 2. Preliminaries

In this section, we first transform (1) into an equivalent Hammerstein-type integral equation. For this, let x″t=yt, for t∈0,1. Then, from the conditions x0=x′0=0, we have(11)x′t=∫0tx″sds=∫0tysds,xt=∫0tx′sds=∫0t∫0syτdτds.Therefore, substituting (11) into (1) gives(12)−y″t=ft,∫0t∫0syτdτds,∫0tysds,yt,y′t,t∈0,1,y0=αyt,y′1=0.Lemma 1.
The problem (12) can be transformed into the Hammerstein-type integral equation:(13)yt=∫01G1t,sfs,∫0s∫0τyrdrdτ,∫0syτdτ,ys,y′sds,where G1t,s=1/1−α1∫01K1t,sdβt+K1t,s and K1t,s=mint,s, for t,s∈0,1.Proof.
Using the functiong on 0,1 to replaceft,∫0t∫0syτdτds,∫0tysds,yt,y′t in (12), we consider the following problem:(14)−y″t=gt,t∈0,1,y0=αyt,y′1=0.
From the differential equation in (14), we obtain(15)yt=∫0ts−tgsds+c1t+c2,for someci∈ℝ,i=1,2,and then(16)y′t=−∫0tgsds+c1,y″t=−gt.
The conditiony′1=0 implies that(17)c1=∫01gsds.
Using the conditiony0=αyt, it enables us to obtain(18)c2=∫01∫0ts−tgsds+∫01tgsds+c2dβt=∫01∫01K1t,sgsds+c2dβt.
Hence, we have(19)c2=11−α1∫01∫01K1t,sgsdsdβt.
As a result, substitutingc1andc2 into (15) gives(20)yt=∫0ts−tgsds+∫01tgsds+11−α1∫01∫01K1t,sgsdsdβt=∫01G1t,sgsds.
This completes the proof.
LetE=C10,1, y=maxyC,y′C with yC=maxt∈0,1yt, and P=y∈E:yt≥0,y′t≥0,∀t∈0,1. Then, E,⋅ is a Banach space, and P is a cone on E. From Lemma 1, we can define an operator A:P⟶P as follows:(21)Ayt=∫01G1t,sfs,∫0s∫0τyrdrdτ,∫0syτdτ,ys,y′sds,fory∈E.
Then,A is a completely continuous operator from the Arzelà–Ascoli theorem (this argument is standard).Remark 1.
(i) In our work, we need the nonnegativity of Green’s functionG1, so we have the following assumption:
(22)
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1
∫
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1
K
1
t
,
s
d
β
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≥
0
,
for
s
∈
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.
(ii)
We need some inequality conditions on the nonlinearityft,z1,z2,z3,z4 with respect to the variables zi,i=1,2,3,4. We consider some useful linear operators:
(23)
L
1
y
t
=
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∫
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∫
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≔
∫
0
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2
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y
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d
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L
2
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=
∫
0
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1
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∫
0
1
∫
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d
τ
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∫
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3
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y
s
d
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,
for
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∈
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,
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.
If we know the functionβ, we can obtain the functions G2 and G3.Example 1.
Letβt=0 for t∈0,1. Then, G1t,s=K1t,s, for t,s∈0,1. Let(24)hs,τ=1,0≤τ≤s≤1,0,0≤s≤τ≤1,and then from (22), we find(25)G3t,s=∫01G1t,τhτ,sdτ,fort,s∈0,1.We consider two cases:(i)
Case 1: when0≤t≤s≤1, we have(26)G3t,s=∫0t0⋅τdτ+∫ts0⋅tdτ+∫s11⋅tdτ=t1−s.(ii)
Case 2: when0≤s≤t≤1, we have(27)G3t,s=∫0s0⋅τdτ+∫st1⋅τdτ+∫t11⋅tdτ=t−12t2−12s2.
We now calculateG2. For this, let ∫0τyrdr=jτ and kτ,r=0,0≤τ≤r≤1,1,0≤r≤τ≤1. Then, we have(28)∫01G1t,s∫0sjτdτds=∫01G3t,τjτdτ=∫01G3t,τ∫0τyrdrdτ=∫01G3t,τ∫01kτ,ryrdrdτ.
Therefore, from (22), we have(29)G2t,s=∫01G3t,τkτ,sdτ,fort,s∈0,1.
We consider two cases:(i)
Case 1: when0≤t≤s≤1, we have(30)G2t,s=∫0t0⋅t−12t2−12τ2dτ+∫ts0⋅t1−τdτ+∫s11⋅t1−τdτ=t12+12s2−s.(ii)
Case 2: when0≤s≤t≤1, we have(31)G2t,s=∫0s0⋅t−12t2−12τ2dτ+∫st1⋅t−12t2−12τ2dτ+∫t11⋅t1−τdτ=12t−ts+12t2s−16t3+16s3.Example 2.
Letβt=t/2 for t∈0,1. Then, we have(32)11−α1∫01K1t,sdβt=∫01K1t,sdt=∫0stdt+∫s1sdt=s−12s2,fors∈0,1.
Hence,G1t,s=s−1/2s2+K1t,s, for t,s∈0,1. Note (22) and Example 1, so we only need to calculate(33)∫01τ−12τ2hτ,sdτ=∫s1τ−12τ2dτ=13−12s2+16s3,∫0113−12τ2+16τ3kτ,sdτ=∫s113−12τ2+16τ3dτ=524−13s+16s3−124s4,fors∈0,1.
Therefore, we obtain(34)G3t,s=13−12s2+16s3+t1−s,0≤t≤s≤1,t−12t2−12s2,0≤s≤t≤1,G2t,s=524−13s+16s3−124s4+t12+12s2−s,0≤t≤s≤1,12t−ts+12t2s−16t3+16s3,0≤s≤t≤1.Lemma 2.
(Krein–Rutman, see([49] theorem 19.3)). Let P be a reproducing cone in a real Banach space E, and let L:E⟶E be a compact linear operator with LP⊂P. If rL>0, then there exists φ∈P,0 such that Lφ=rLφ, where rL is the spectral radius of L.Lemma 3.
For not all zero numbersa,b,c,andd≥0, we let(35)La,b,c,dyt=a∫01G2t,sysds+b∫01G3t,sysds+c∫01G1t,sysds+d∫01G1t,sy′sds,La,b,cyt=a∫01G2t,sysds+b∫01G3t,sysds+c∫01G1t,sysds,La,b,c∗ys=a∫01G2t,sytdt+b∫01G3t,sytdt+c∫01G1t,sytdt,fory∈P.Then,(36)130a+18b+13c+13dκ1≤rLa,b,c,d≤18a+13b+12c+12dκ2,130a+18b+13cκ1≤rLa,b,c≤18a+13b+12cκ2,130a+18b+13cκ1≤rLa,b,c∗≤18a+13b+12cκ2,where κ1≔αt−α1+1/1−α1 and κ2≔1/1−α1.Proof.
We first give some inequalities forGi,i=1,2,3. Note that ts≤K1t,s≤s, for t,s∈0,1, and from the definition of G1, we see that(37)κ1ts=αt−α1+11−α1ts≤G1t,s≤11−α1s=κ2s,fort,s∈0,1.With h and k as before, note(38)12κ1t1−s2=∫s1κ1tτdτ≤G3t,s=∫01G1t,τhτ,sdτ≤∫s1κ2τdτ=12κ21−s2fort,s∈0,1,16κ1t2+s3−3s=∫s112κ1t1−τ2dτ≤G2t,s=∫01G3t,τkτ,sdτ≤∫s112κ21−τ2dτ,=16κ22+s3−3s,fort,s∈0,1.
For convenience, letψ˜0t=t, ψ˜1t=1/62+t3−3t, and ψ˜2t=1/21−t2, for t∈0,1.
We only prove the inequalities in (35) about the spectral radius of La,b,c. For convenience, let Ga,b,ct,s=aG2t,s+bG3t,s+cG1t,s for t,s∈0,1. Then, we have(39)La,b,cyt=∫01Ga,b,ct,sysds,fory∈P.
Thus, we obtain(40)La,b,c=supy=1La,b,cy=supt∈0,1,y=1∫01Ga,b,ct,sysds≤maxt∈0,1∫01Ga,b,ct,sds≤∫01κ2aψ˜1s+bψ˜2s+cψ˜0sds≤18a+13b+12cκ2.
For alln∈ℕ+, we note that(41)La,b,cnyt=LLa,b,cn−1yt=∫01Ga,b,ct,sn−1La,b,cn−1ysn−1dsn−1=⋯=∫​01∫​01⋯∫​01︸nGa,b,ct,sn−1Ga,b,csn−1,sn−2⋯Ga,b,cs1,sysdsn−1dsn−2⋯ds,fory∈P.
Hence, we can obtain(42)La,b,cn≤maxt∈0,1∫​01∫​01⋯∫​01︸nGa,b,ct,sn−1Ga,b,csn−1,sn−2⋯Ga,b,cs1,sdsn−1dsn−2⋯ds≤κ2n∫01aψ˜1s+bψ˜2s+cψ˜0sdsn≤18a+13b+12cnκ2n.
Gelfand’s theorem implies that(43)rLa,b,c=limn⟶∞La,b,cnn≤18a+13b+12cκ2.
Next, we introduce a conclusion in ([50], problem 2.1.4). Let y∈C0,1, and a functional J on C0,1 be as(44)Jx=∫01xtytdt,∀x∈C0,1.
Then, we have(45)J=∫01ytdt.
We note that there existst0∈0,1 such that maxt∈0,1∫01Ga,b,ct,sds=∫01Ga,b,ct0,sds. Then, in (38), for fixed t, we define a linear function:(46)La,b,c,ty=∫01Ga,b,ct,sysds,fory∈C0,1,and thus(47)La,b,c,t=∫01Ga,b,ct,sds.
Then, by the definition of the norm of linear function, we know that for allε>0, there exists yε,t∈C0,1 with yε,t=1 such that(48)∫01Ga,b,ct,sds=La,b,c,t=supy=1La,b,c,ty≥La,b,c,tyε,t≥∫01Ga,b,ct,sds−ε.
On the contrary, note from the definition of our norm we have(49)La,b,cy=supt∈0,1La,b,c,ty≥La,b,c,t0y.
Consequently, we have(50)La,b,c=supy=1La,b,cy≥La,b,c,t0yε,t0≥∫01Ga,b,ct0,sds−ε.
For the arbitrariness ofε, we have(51)La,b,c=∫01Ga,b,ct0,sds=maxt∈0,1∫01Ga,b,ct,sds.
Also, for alln∈ℕ+, we obtain(52)La,b,cn=maxt∈0,1∫01∫01⋯∫01︸nGa,b,ct,sn−1Ga,b,csn−1,sn−2⋯Ga,b,cs1,sdsn−1dsn−2⋯ds≥κ1nmaxt∈0,1t∫01∫01⋯∫01︸naψ˜1sn−1+bψ˜2sn−1+cψ˜0sn−1⋅sn−1aψ˜1sn−2+bψ˜2sn−2+cψ˜0sn−2⋯s1aψ˜1s+bψ˜2s+cψ˜0sdsn−1dsn−2⋯ds=κ1n∫01aψ˜1s+bψ˜2s+cψ˜0sds∫01asψ˜1s+bsψ˜2s+csψ˜0sdsn−1.
From Gelfand’s theorem, we have(53)rLa,b,c=limn⟶∞La,b,cnn≥κ1∫01asψ˜1s+bsψ˜2s+csψ˜0sds≥130a+18b+13cκ1.
This completes the proof.Lemma 4.
(see [51]). Let E be a real Banach space and P be a cone on E. Suppose that Ω⊂E is a bounded open set and that A:Ω¯∩P⟶P is a continuous compact operator. If there exists a ω0∈P,0 such that(54)ω−Aω≠λω0,∀λ≥0,ω∈∂Ω∩P,then iA,Ω∩P,P=0, where i denotes the fixed point index on P.Lemma 5.
(see [51]). Let E be a real Banach space and P be a cone on E. Suppose that Ω⊂E is a bounded open set with 0∈Ω and that A:Ω¯∩P⟶P is a continuous compact operator. If(55)ω−λAω≠0,∀λ∈0,1,ω∈∂Ω∩P,then iA,Ω∩P,P=1.


## 3. Main Results

In our paper, we letBρ=y∈P:y<ρ for ρ>0. Now, ∂Bρ=y∈P:y=ρ and B¯ρ=y∈P:y≤ρ. Now, we list our assumptions on the nonlinearity f.(56)H2f∈C0,1×ℝ+×ℝ+×ℝ+×ℝ+,ℝ+.(H3) There exist not all zero numbersa1,b1,c1,andd1≥0 and e1>0 such that rLa1,b1,c1,d1<1, and ft,z1,z2,z3,z4≤a1z1+b1z2+c1z3+d1z4+e1,fort,z1,z2,z3,z4∈0,1×ℝ+×ℝ+×ℝ+×ℝ+.(H4) There exist not all zero numbersa2,b2,c2,andd2≥0 and ρ1>0 such that rLa2,b2,c2,d2≥1, and ft,z1,z2,z3,z4≥a2z1+b2z2+c2z3+d2z4,fort,z1,z2,z3,z4∈0,1×0,ρ1×0,ρ1×0,ρ1×0,ρ1.(H5) There exist not all zero numbersa3,b3,c3,andd3≥0 and ρ2>0 such that rLa3,b3,c3,d3<1, and ft,z1,z2,z3,z4≤a3z1+b3z2+c3z3+d3z4,fort,z1,z2,z3,z4∈0,1×0,ρ2×0,ρ2×0,ρ2×0,ρ2.(H6) There exist not all zero numbersa4,b4,andc4≥0 and e2>0 such that rLa4,b4,c4∗>1, and ft,z1,z2,z3,z4≥a4z1+b4z2+c4z3−e2,fort,z1,z2,z3,z4∈0,1×ℝ+×ℝ+×ℝ+×ℝ+.(H7) For anyM>0, there exists a positive continuous function HMϱ on ℝ+ such that ∫0+∞ϱdϱ/HMϱ+δ0=+∞,∀δ0>0, and ft,z1,z2,z3,z4≤HMz4,fort,z1,z2,z3,z4∈0,1×0,M×0,M×0,M×ℝ+.Remark 2.
Considering Lemma3, one can adjust the coefficients ai,bi,ci,anddii=1,2,3,4 such that the spectral radii rLai,bi,ci,di,rLa4,b4,c4,rLa4,b4,c4∗i=1,2,3 satisfy their respective conditions in (H3)–(H6).Theorem 1.
Suppose that (H0–H4) hold. Then, (1) has at least one positive solution.Proof.
LetW=y∈P:y=λAy,λ∈0,1. Now, we prove that W is a bounded set in P. If y∈W, then from (H3), we have(57)yt=λAyt≤Ayt≤∫01G1t,sa1∫0s∫0τyrdrdτ+b1∫0syτdτ+c1ys+d1y′s+e1ds=∫01a1G2t,s+b1G3t,s+c1G1t,sysds+d1∫01G1t,sy′sds+e1ηt=La1,b1,c1,d1yt+e1ηt,where (58)ηt=∫0111−α1∫01K1t,sdβt+K1t,sds=11−α1∫01∫01K1t,sdβtds+t−12t2,for t∈0,1. This implies that(59)I−La1,b1,c1,d1yt≤e1ηt.
Note thatη∈Pη′t=1−t≥0,∀t∈0,1, y∈W, and we obtain(60)y′t=λAy′t=λ∫t1fs,∫0s∫0τyrdrdτ,∫0syτdτ,ys,y′sds≤∫01∂G1∂tt,sa1∫0s∫0τyrdrdτ+b1∫0syτdτ+c1ys+d1y′s+e1ds=∫01∂∂ta1G2t,s+b1G3t,s+c1G1t,sysds+d1∫01∂∂tG1t,sy′sds+e1η′t=La1,b1,c1,d1y′t+e1η′t.
Thus,(61)I−La1,b1,c1,d1y′t≤e1η′t.
SincerLa1,b1,c1,d1<1, we know that I−La1,b1,c1,d1 has a bounded inverse operator I−La1,b1,c1,d1−1, with(62)I−La1,b1,c1,d1−1=I+La1,b1,c1,d1+La1,b1,c1,d12+⋯+La1,b1,c1,d1n+⋯.
Note thatLa1,b1,c1,d1P⊂P, and we obtain I−La1,b1,c1,d1−1P⊂P. Therefore,(63)yt≤I−La1,b1,c1,d1−1e1ηt,y′t≤I−La1,b1,c1,d1−1e1η′t,fort∈0,1.
This implies that(64)yC≤I−La1,b1,c1,d1−1e1ηC,y′C≤I−La1,b1,c1,d1−1e1η′C.
Therefore, we have(65)y≤I−La1,b1,c1,d1−1e1η.
That is,W is bounded. Now, we can select R1>maxsupW,ρ1 (ρ1 is defined in (H4)) such that(66)y≠λAy,fory∈∂BR1∩Pandλ∈0,1.
From Lemma5, we have(67)iA,BR1∩P,P=1.
On the contrary, sinceLa2,b2,c2,d2P⊂P and rLa2,b2,c2,d2≥1, it follows from Lemma 2 that there exists φ0∈P0 such that La2,b2,c2,d2φ0=rLa2,b2,c2,d2φ0 and φ0=rLa2,b2,c2,d2−1La2,b2,c2,d2φ0∈P. Now, we show that(68)y−Ay≠λφ0,fory∈∂Bρ1∩P,λ≥0.
If this claim is false, then there existy0∈∂Bρ1∩P and λ0≥0 such that y0−Ay0=λ0φ0. Note that λ0>0 (otherwise, the theorem is proved). Then, from (H4), we have(69)Ay0t=∫01G1t,sfs,∫0s∫0τy0rdrdτ,∫0sy0τdτ,y0s,y0′sds≥∫01G1t,sa2∫0s∫0τy0rdrdτ+b2∫0sy0τdτ+c2y0s+d2y0′sds=∫01a2G2t,s+b2G3t,s+c2G1t,sy0sds+d2∫01G1t,sy0′sds=La2,b2,c2,d2y0t,which implies that (70)y0=Ay0+λ0φ0≥La2,b2,c2,d2y0+λ0φ0≥λ0φ0.
Letλ∗=supλ>0:y0≥λφ0. Then, λ0∈λ>0:y0≥λφ0and y0≥λ∗φ0. However, we note that y0≥La2,b2,c2,d2y0+λ0φ0≥La2,b2,c2,d2λ∗φ0+λ0φ0=λ∗rLa2,b2,c2,d2+λ0φ0, and this contradicts the definition of λ∗ for rLa2,b2,c2,d2≥1. Therefore, (66) holds, as required. From Lemma 4, we have(71)iA,Bρ1∩P,P=0.
From (65) and (69), we have(72)iA,BR1∖B¯ρ1∩P,P=iA,BR1∩P,P−iA,Bρ1∩P,P=1,and hence A has at least one fixed point in BR1,B¯ρ1∩P, i.e., (1) has at least one positive solution. This completes the proof.Theorem 2.
Suppose that (H0–H2) and (H5–H7) hold. Then, (1) has at least one positive solution.Proof.
We show that(73)y≠λAy,fory∈∂Bρ2∩P,λ∈0,1.
If the claim is false, then there existy1∈∂Bρ2∩P and λ1∈0,1 such that y1t=λ1Ay1t, for t∈0,1. For t∈0,1, from (H5) we have(74)y1t≤∫01G1t,sfs,∫0s∫0τy1rdrdτ,∫0sy1τdτ,y1s,y1′sds≤∫01G1t,sa3∫0s∫0τy1rdrdτ+b3∫0sy1τdτ+c3y1s+d3y1′sds=La3,b3,c3,d3y1t.
Also,y1′t=λ1Ay1′t, for t∈0,1 implies that(75)y1′t=λ∫t1fs,∫0s∫0τy1rdrdτ,∫0sy1τdτ,y1s,y1′sds≤∫01∂G1∂tt,sfs,∫0s∫0τy1rdrdτ,∫0sy1τdτ,y1s,y1′sds≤∫01∂G1∂tt,sa3∫0s∫0τy1rdrdτ+b3∫0sy1τdτ+c3y1s+d3y1′sds=La3,b3,c3,d3y1′t.
Note thatLa3,b3,c3,d3P⊂P and rLa3,b3,c3,d3<1, and we have (76)I−La3,b3,c3,d3y1t≤0,I−La3,b3,c3,d3y1′t≤0,fort∈0,1.
Therefore,(77)y1t=0,y′t=0,fort∈0,1.
This contradicts the fact thaty1∈∂Bρ2∩P. Hence, (71) is true, as required. From Lemma 5, we have(78)iA,Bρ2∩P,P=1.
On the contrary, from Lemma2 there exists ψ1∈P0 such that La4,b4,c4∗ψ1s=rLa4,b4,c4∗ψ1s, for s∈0,1. Let U=y∈P:y−Ay=λφ2,forλ≥0, where φ2t=t−1/2t2∈P, for t∈0,1. Note that λ>0 (otherwise, the theorem is proved). We shall show that U is a bounded set in P. If y∈P, then from (H6), we have(79)yt=Ayt+λφ2t≥Ayt≥∫01G1t,sa4∫0s∫0τyrdrdτ+b4∫0syτdτ+c4ys−e2ds≥∫01a4G2t,s+b4G3t,s+c4G1t,sysds−e2∫01G1t,sds≥∫01a4G2t,s+b4G3t,s+c4G1t,sysds−12κ2e2.
Multiplying both sides of the above inequality byψ1t and integrating from 0 to 1 yields (80)∫01ytψ1tdt≥∫01ψ1t∫01a4G2t,s+b4G3t,s+c4G1t,sysdsdt−12κ2e2∫01ψ1tdt≥rLa4,b4,c4∗∫01ytψ1tdt−12κ2e2∫01ψ1tdt.
This, together withrLa4,b4,c4∗>1, implies that (81)∫01ytψ1tdt≤κ2e2∫01ψ1tdt2rLa4,b4,c4∗−2.
Note thaty∈U, and we have(82)y′t=Ay′t+λφ2′t=∫t1fs,∫0s∫0τyrdrdτ,∫0syτdτ,ys,y′sds+λ1−t≥0,y″t=Ay″t+λφ2″t=−ft,∫0t∫0τyrdrdτ,∫0tyτdτ,yt,y′t−λ≤0.
Then,y is a concave and increasing function on 0,1. Hence, (83)∫01ytψ1tdt=∫01yt1⋅1+1−t1⋅0ψ1tdt≥∫01ty1+1−ty0ψ1tdt≥∫01ty1ψ1tdt.
This enables us to obtain(84)yC=y1≤κ2e2∫01ψ1tdt2rLa4,b4,c4∗−2∫01tψ1tdt−1.
Now, note (82), and we see there is an M>0 such that(85)∫0s∫0τyrdrdτ,∫0syτdτ,ys≤M,fors∈0,1.
This, together with (H7), implies that(86)fs,∫0s∫0τyrdrdτ,∫0syτdτ,ys,y′s≤HMy′s,fors∈0,1.
Note thaty∈U, and we obtain(87)−y″t=−Ay″t−λφ2″t=ft,∫0t∫0τyrdrdτ,∫0tyτdτ,yt,y′t+λ≤HMy′t+λ.
This implies that(88)−y″ty′tHMy′t+λ≤y′t,and then if we let ϱ=y′, we have(89)∫0y′CϱdϱHMϱ+λ≤∫01∫01y′tdt=y1−y0≤κ2e2∫01ψ1tdt2rLa4,b4,c4∗−2∫01tψ1tdt−1.
Therefore, combining this and (H7), there existsN∗>0 such that(90)y′C≤N∗.
Thus,U is bounded (see (82) and (88)). Taking R2>maxsupU,ρ2, we have(91)y−Ay≠λφ2,fory∈∂BR2∩P,λ≥0.
From Lemma4, we have(92)iA,BR2∩P,P=0.
From (76) and (90), we have(93)iA,BR2∖B¯ρ2∩P,P=iA,BR2∩P,P−iA,Bρ2∩P,P=−1,and hence A has at least one fixed point in BR2,B¯ρ2∩P, i.e., (1) has at least one positive solution. This completes the proof.



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*Source: 3748631-2019-12-18.xml*