# Existence and Uniqueness of Positive Solutions for Singular Nonlinear Fractional Differential Equation via Mixed Monotone Operator Method

Authors: Tian Wang; Zhaocai Hao
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this article, we discuss the existence and uniqueness of positive solution for a class of singular fractional differential equations, where the nonlinear term contains fractional derivative and an operator. By applying the fixed point theorem in cone, we get the existence and uniqueness of positive solutions for the fractional differential equation. Moreover, we give an example to demonstrate our main result.


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## Body

## 1. Introduction

In this article, we consider the existence and uniqueness of solutions for the following singular fractional differential equations:(1)D0+αut+ptft,ut,D0+βut+qtgt,ut,Hut=0,0<t<1,u0=u′0=⋯=un−20=0,D0+γutt=1=ku1,where n−1<α≤n, n>3, 1≤β≤γ≤n−2, p,q∈C0,1,0,∞, pt and qt are allowed to be singular at t=0 or t=1, f:0,1×0,∞×0,∞⟶0,∞ is continuous and ft,u,v may be singular at t=0,1 and u=v=0, g:0,1×0,∞×0,∞⟶0,∞ is continuous and gt,u,v may be singular at t=0,1, and k:0,∞⟶0,∞ is continuous function.In recent years, we can see that differential equations have more and more attention in many fields, like electrical networks, quantum physics, and probability. There are a lot of interesting results from fractional differential equation, Schrodinger equation and k-hessian equations. In [1–4], the authors studied the solutions of fractional differential equations, and obtained some interesting results. Such as uniqueness of iterative positive solution, existence of multiple positive solutions or maximum and minimum solutions. In [5–9], the authors studied the solutions of Schrödinger equation. From these paper, there have several results about existence of infinitely solutions, existence and nonexistence of blow-up solutions, existence and asymptotic properties of solutions or existence and nonexistence of entire large solutions. In [10–12], the authors considered existence of blow-up radial solutions of the k-hessian equations or convergence analysis and uniqueness of blow-up solutions of the k-hessian equations. Inspired by these conclusions,we realize that the questions about differential equations are very interesting and varied. So we focus on fractional differential equation. The authors in [13] studied the following fractional differential equation:
(2)D0+αxt+ft,xt,xt+gt,xt=0,0<t<1,xi0=0,i=0,1,2,⋯,n−2,D0+βxtt=1=0,2≤β≤n−2,where n−1<α≤n, n>3, D0+α is the Riemann-Liouville fractional derivative, and f:0,1×0,∞×0,∞⟶0,∞ is a continuous function.In [14], by using cone expansion fixed point theorem, Li et al. studied the existence of positive solutions for the following fractional differential equation with Riemann-Stieltjes integral:
(3)D0+αxt+ptft,xt+qtgt,xt=0,0<t<1,x0=x′0=⋯=xn−20=0,x1=∫01ksxsdAs,where n−1<α≤n, n≥2, p,q:0,1⟶0,∞ are continuous, p,q∈L10,1, and allowed to be singular at t=0 or t=1, f,g:0,1×0,∞⟶0,∞ are continuous and ft,x,gt,x may be singular at x=0, k:0,1⟶0,∞ is continuous with k∈L10,1, where ∫01ksxsdAs is the Riemann-Stieltjes integral with a signed measure and A:0,1⟶ℝ is a bounded variation function, and D0+α is the standard Riemann-Liouville derivative of order α.In [15], Liu et al. investigated the iterative positive solutions for the following singular nonlinear fractional differential equation with integral boundary conditions.
(4)D0+αxt+ptft,xt,xt+qtgt,xt=0,0<t<1,x0=x′0=⋯=xn−20=0,x1=∫01ksxsdAs,where n−1<α≤n, n≥2, p,q:0,1⟶0,∞ are continuous and allowed to be singular at t=0 or t=1, f:0,1×0,∞×0,∞⟶0,∞ is continuous and ft,x,y may be singular at t=0,1 and x=y=0, gt,x:0,1×0,∞⟶0,∞ is continuous and may be singular at t=0,1 and x=0, k:0,1⟶0,∞ is continuous with k∈L10,1, where ∫01ksxsdAs is the Riemann-Stieltjes integral with a signed measure and A:0,1⟶ℝ is a bounded variation function, and D0+α is the standard Riemann-Liouville derivative of order α.In [16], Zhang and Tian considered the following fractional differential equation:
(5)D0+νxt+ft,xt,D0+γxt+gt,xt=0,0<t<1,xi0=0,i=0,1,2,⋯,n−2,D0+βxtt=1=kx1,where n−1<ν<n, n>3, 1≤γ≤β≤n−2, and f:0,1×0,∞×0,∞⟶0,∞, g:0,1×0,∞⟶0,∞, and k:0,∞⟶0,∞ are continuous functions.Motivated by the abovementioned papers, the purpose of this article is to study the existence and uniqueness of positive solution for FBVP (1). Obviously, the problem in our article is more general. We not only consider the nonlinear term containing the derivative term, but also study the nonlinear term containing an operator. And we do not need the linearity of operator in nonlinear term. It may be a nonlinear operator.Comparing with the results in [17], boundary value problem (1) has a more general form. First, we discuss the singularity. It means that p,q are allowed to be singular at t=0,1, f may be singular at t=0,1 and x=y=0, and g may be singular at t=0. Second, we not only consider the derivative term but also consider the operator term, where the operator can be linear or nonlinear. Especially, when pt=qt=1 and Hu=0, we can see that the problem in [16] is a special case of problem (1). Comparing with [13], firstly, when ku=0, β=0, pt=qt=1, and Hut=ut, then our problem (1) reduced compared to the problem in [13]. Third, in [13], the author did not consider the singularity, but in problem (1), we consider the singularity. Compared with [14], first, our method is different from that of [14]. We use the fixed point theorem for the sum of operators rather than the fixed point theorem of cone expansion and compression. Second, ft,x,y not only has three variables but also is singular for both time and space variables, and gt,x,y also has three variables. Compared with [15], we consider the derivative term and the operator term. Our results extended and improved some existing results, such as [13–16].The rest of the paper is organized as follows: in Section2, we give some preliminaries and lemmas that will be used in our main result. In Section 3, we study the existence and uniqueness of a positive solution of the BVP (1). In Section 4, an example is provided to demonstrate our main results.


## 2. Preliminaries and Lemmas

In this section, we present some notations, definitions and lemmas that will be used in the paper.LetE,⋅ be a Banach space, and θ is defined as the zero element of E. A nonempty closed convex set P⊂E is a cone if it satisfies (1) x∈P,λ≥0⟹λx∈P and (2) x∈P,−x∈P⟹x=θ. Putting P∘=x∈PxisaninteriorpointofP, a cone P is said to be solid if P∘ is nonempty. In this paper, suppose that E,⋅ is a Banach space partially ordered by a cone P⊂E, that is, x≤y if and only if y−x∈P. A cone P called normal if there exists a constant N>0 such that for all x,y∈E,θ≤x≤y implies x≤Ny, and the smallest such N is called the normality constant of P. For all x,y∈P, we denote the notation x~y if there have constants λ,μ>0 such that λy≤x≤μy. Clearly, ~ is an equivalence relation. Given h>θ, we denote by Ph the set Ph=x∈Ex~h. And it is easy to know that Ph⊂P.Definition 1 (see [18]).
An operatorA:P×P⟶P is said to be a mixed monotone operator if Ax,y is increasing in the first component and decreasing in the second component, i.e., ui,vii=1,2∈P,u1≤u2,v1≥v2, implies Au1,v1≤Au2,v2. An element x∈P is called a fixed point of A if Ax,x=x.Definition 2 (see [19]).
LetD=P or D=P∘ and α be a real number with 0≤α≤1. An operator A:D⟶D is said to be α concave if it satisfies
(6)Atx≥tαAx,∀t∈0,1,x∈D.Definition 3 (see [19]).
An operatorB:P⟶P is said to be subhomogeneous if it satisfies
(7)Btx≥tBx,∀t∈0,1,x∈D.Definition 4 (see [20]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:0,∞⟶ℝ is given by
(8)I0+αyt=1Γα∫0tt−sα−1ysds,provided that the right-hand side is pointwise defined on 0,∞.Definition 5 (see [20]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function y:0,∞⟶ℝ is given by
(9)D0+αyt=1Γn−αddtn∫0tyst−sα−n+1ds,where n=α+1, where α denotes the integer part of the number α, provided that the right-hand side is pointwise defined on 0,∞.Lemma 6 (see [20]).
Letα>0. If we assume u∈C0,1∩L10,1, then the fractional differential equation
(10)D0+αut=0has
(11)ut=C1tα−1+C2tα−2+⋯+CNtα−N,Ci∈ℝ,i=1,2,⋯,N,as the unique solution, where N=α+1.From the definition of the Riemann-Liouville derivative, we can obtain the statement.Lemma 7 (see [20]).
Assume thatu∈C0,1∩L10,1 with a fractional derivative of order α>0 that belongs to C0,1∩L10,1. Then,
(12)I0+αD0+αut=ut+C1tα−1+C2tα−2+⋯+CNtα−N,for some Ci∈ℝi=1,2,⋯,N, where N=α+1.In the following, we present the Green’s function of FBVP (1).Lemma 8 (see [16]).
Leth∈C0,1, then the unique solution of the linear problem
(13)D0+αut+ht=0,0<t<1,n−1<α≤n,u0=u′0=⋯=un−20=0,D0+γutt=1=ku1,1≤γ≤n−2is given by
(14)ut=∫01Gt,shsds+Γα−γΓαku1tα−1,where
(15)Gt,s=1Γαtα−11−sα−γ−1−t−sα−1,0≤s≤t≤1,tα−11−sα−γ−1,0≤t≤s≤1.Lemma 9 (see [16]).
The Green’s functionGt,s defined by (15) satisfies the following properties:
(1)
G:0,1×0,1⟶0,∞ is continuous(2)
For anyt,s∈0,1, there is
(16)0≤tα−11−sα−γ−11−1−sγ≤ΓαGt,s≤tα−11−sα−γ−1(3)
For anyt,s∈0,1, there is
(17)0≤tα−β−11−sα−γ−11−1−sγ−β≤Γα−βD0+βGt,s≤tα−β−11−sα−γ−1In [17], the operator equation Ax,x+Bx,x=x was studied, where A,B are mixed monotone operators, and for all t∈0,1,ψt∈t,1, there have Atx,t−1y≥ψtAx,y, Btx,t−1y≥tBx,y. Based on the result in [17], we get the following lemma:Lemma 10.
LetP be a normal cone in E, and let A,B:P×P⟶P be two mixed monotone operators and C:P⟶P is a decreasing operator, suppose that
(A1)
for allt∈0,1, there have ψt∈t,1 such that
(18)Atx,t−1y≥ψtAx,y,forallx,y∈P(A2)
for allt∈0,1,
(19)Btx,t−1y≥tBx,y,forallx,y∈P(A3)
for allt∈0,1,
(20)Ct−1y≥tCy,forally∈P(A4)
there existsh>θ and h∈Ph, such that Ah,h∈Ph,Bh,h∈Ph,Ch∈Ph(A5)
there exists a constantsδ>0, such that for all x,y∈P,
(21)Ax,y≥δBx,y+Cy
Then(1)
A:Ph×Ph⟶Ph,B:Ph×Ph⟶Ph,C:Ph⟶Ph(2)
There existu0,v0∈Ph and r∈0,1 such that
(22)rv0≤u0<v0,u0≤Au0,v0+Bu0,v0+Cv0≤Av0,u0+Bv0,u0+Cu0≤v0(3)
The operator equationAx,x+Bx,x+Cx=x has a unique solution x∗ in Ph(4)
For any initial valuesx0,y0∈Ph, constructing successively the sequences
(23)xn=Axn−1,yn−1+Bxn−1,yn−1+Cyn−1,n=1,2,⋯,yn=Ayn−1,xn−1+Byn−1,xn−1+Cxn−1,n=1,2,⋯,we have xn⟶x∗ and yn⟶x∗ as n⟶∞Proof.
Now we define the operatorT:P×P⟶P by
(24)Tx,y=Ax,y+Bx,y+Cy,x,y∈P.So it is easy to know thatT is a mixed monotone operator. Next, we will prove that Th,h∈Ph and Ttx,t−1y≥φtTx,y, where φt∈t,1, Then according to the results of [21], the conclusions of Lemma 10 holds.Firstly, we will show thatTh,h∈Ph for h>θ and h∈Ph.From (18)–(20), for any t∈0,1, x,y∈P, there have
(25)At−1x,ty≤ψt−1Ax,y,(26)Bt−1x,ty≤t−1Bx,y,(27)Cty≤t−1Cy.According to (A4), we may assume that there exist constantsμi>0,νi>0i=1,2,3 such that
(28)μ1h≤Ah,h≤ν1h,μ2h≤Bh,h≤ν2h,μ3h≤Ch≤ν3h.Choose sufficiently small numbera∈0,1 such that ah≤x,y≤1/ah for any x,y∈Ph; then from (18)–(20) and (25)–(28), there have
(29)Ax,y≤A1ah,ah≤1ψaAh,h≤1ψaν1h,Ax,y≥Aah,1ah≥ψaAh,h≥ψaμ1h,and we know 1/ψaν1,ψaμ1>0. So Ax,y∈Ph and A:Ph×Ph⟶Ph. Similarly, we can easily get that Bx,y∈Ph, B:Ph×Ph⟶Ph, Cy∈Ph, and C:Ph⟶Ph. Then, we have Th,h=Ah,h+Bh,h+Ch∈Ph.Second, we will prove thatTtx,t−1y≥φtTx,y is satisfied for any x,y∈P,t∈0,1,φt∈t,1.For anyx,y∈P, from (A5), we have
(30)Ax,y+δAx,y≥δBx,y+Cy+δAx,y.Then1+δAx,y≥δAx,y+Bx,y+Cy=δTx,y. So
(31)Ax,y≥Tx,y1+δ−1.Thus, there have(32)Ttx,t−1y−tTx,y=Atx,t−1y+Btx,t−1y+Ct−1y−tAx,y+Bx,y+Cy≥ψt−tAx,y+Btx,t−1y−tBx,y+Ct−1y−tCy≥ψt−tAx,y≥ψt−t1+δ−1Tx,y.So(33)Ttx,t−1y≥t+ψt−t1+δ−1Tx,y.Now we noteφt=t+ψt−t/1+δ−1; it is easy to know that φt∈t,1, where t∈0,1. Then, Ttx,t−1y≥φtTx,y.Thus, by the results of Zhai and Zhang in [21], we know what the conclusions of Lemma 10 hold.


## 3. Main Results

For convenience, in this section, we set(34)E=xx∈C0,1,D0+βx∈C0,1.Clearly,E is a Banach space with the norm
(35)x=maxmaxt∈0,1xt,maxt∈0,1D0+βxt.Now, we let the coneP be defined by
(36)P=x∈Ext≥0,D0+βxt≥0,∀t∈0,1.There haveP⊂E, and P is a normal cone. And, the order relation in E is given by x≺¯y if xt≤yt,D0+βxt≤D0+βyt.From Lemma8, we get that the unique solution of the problem (1) satisfies the following equation:
(37)xt=∫01Gt,spsfs,xs,D0+βxsds+∫01Gt,sqsgs,xs,Hxsds+Γα−γΓαkx1tα−1,where Gt,s is the Green’s function given in (15).Theorem 11.
Assume that the following assumptions hold:(H1)
p,q:0,1⟶0,∞ are continuous, and pt,qt are allowed to be singular at t=0 or t=1. For all t∈0,1, there is pt≥qt≥m, where m is a constant(H2)
f:0,1×0,∞×0,∞⟶0,∞,g:0,1×0,∞×0,∞⟶0,∞,k:0,1⟶0,∞ are continuous, ft,u,v may be singular at t=0,1, and u=v=0,gt,u,v, may be singular at t=0,1(H3)
For fixedt∈0,1,v∈0,∞, ft,u,v is increasing in u∈0,∞ and decreasing in v∈0,∞; for fixed t∈0,1,v∈0,∞, gt,u,v is increasing in u∈0,∞ and decreasing in v∈0,∞; for fixed t∈0,1, kv is decreasing in v∈0,∞ with kv1≠0(H4)
There exists a constantϑ∈0,1 such that for all η,t∈0,1 and u,v∈0,∞,
(38)ft,ηu,η−1v≥ηϑft,u,v,gt,ηu,η−1v≥ηgt,u,v,kη−1v≥ηkv(H5)
gs,1,1≡0. There exists a constant δ1>0,δ2>0 such that for all t∈0,1 and u,v∈0,∞, ft,u,v≥δ1gt,u,0,ft,u,v≥δ2kv(H6)
∫011−sα−γ−1pssϑ1−αfs,1,1ds<∞ and ∫011−sα−1qss1−αgs,1,1ds<∞(H7)
H:C0,1⟶C0,1. Hu≥0 and Hu is increasing in u∈P. For λ∈0,1 and u∈P, there is Hλu≥λHu
Then problem (1) has a unique positive solution u∗∈Ph, where ht=tα−1. For any initial value u0,v0∈Ph, there are two iterative sequences un,vn for approximating u∗,v∗, that is, un⟶u∗,vn⟶v∗, as n⟶∞, where
(39)unt=∫01Gt,sfs,un−1s,D0+βvn−1sds+∫01Gt,sgs,un−1s,Hvn−1sds+Γα−γΓαkvn−11tα−1,n=1,2,⋯,vnt=∫01Gt,sfs,vn−1s,D0+βun−1sds+∫01Gt,sgs,vn−1s,Hun−1sds+Γα−γΓαkun−11tα−1,n=1,2,⋯.Proof.
We define operatorsA,B,C as follows:
(40)Au,vt=∫01Gt,spsfs,us,D0+βvsds,t∈0,1,(41)Bu,vt=∫01Gt,sqsgs,us,Hvsds,t∈0,1,(42)Cvt=Γα−γΓαkv1tα−1,t∈0,1.Furthermore, foru,v∈P, we obtain that
(43)D0+βAu,vt=∫01D0+βGt,spsfs,us,D0+βvsds,SSt∈0,1,(44)D0+βBu,vt=∫01D0+βGt,sqsgs,us,Hvsds,t∈0,1,(45)D0+βCvt=Γα−γΓα−βkv1tα−β−1,t∈0,1.Clearly,u is the solution of problem (1) if and only if u=Au,u+Bu,u+Cu. So, if we can prove that the operators A,B,C satisfy all conditions in Lemma 10, then we will obtain the conclusions in Theorem 11.First, we will prove thatA,B are well defined. Evidently, C is well defined, so we omit it.From (H4), for allt∈0,1,η∈0,1,u,v∈0,∞, we have
(46)ft,u,v=ft,ηη−1u,η−1ηv≥ηϑft,η−1u,ηv,gt,u,v=gt,ηη−1u,η−1ηv≥ηgt,η−1u,ηv.Furthermore, we have(47)ft,η−1u,ηv≤η−ϑft,u,v,gt,η−1u,ηv≤η−1gt,u,v.Takingu=v=1 in (H4) and (47), for η∈0,1,t∈0,1, we have
(48)ft,η,η−1≥ηϑft,1,1,ft,η−1,η≤η−ϑft,1,1,gt,η,η−1≥ηgt,1,1,gt,η−1,η≤η−1gt,1,1.For anyu,v∈Ph, there exists a constant b>0 such that bh≤u,v≤1/bh, where ht=tα−1,t∈0,1. So, from the above inequalities, we obtain
(49)ft,ut,vt≤ft,1btα−1,btα−1≤ft,1bt1−α,btα−1≤tϑ1−αft,1b,b≤tϑ1−α1bϑft,1,1,ft,ut,vt≥ft,btα−1,1btα−1≥ft,btα−1,1bt1−α≥tϑα−1ft,b,1b≥tϑα−1bϑft,1,1,gt,ut,vt≤gt,1btα−1,btα−1≤gt,1bt1−α,btα−1≤t1−αgt,1b,b≤t1−α1bgt,1,1,gt,ut,vt≥gt,btα−1,1btα−1≥gt,btα−1,1bt1−α≥tα−1gt,b,1b≥tα−1bgt,1,1.Then, combining assumption (H6), we get(50)∫01Gt,spsfs,us,D0+βvsds≤∫01Gt,spssϑ1−α1bϑfs,1,1ds≤tα−1Γα∫011−sα−γ−1pssϑ1−α1bϑfs,1,1ds<∞,∫01D0+βGt,spsfs,us,D0+βvsds≤∫01D0+βGt,spssϑ1−α1bϑfs,1,1ds≤tα−β−11−sα−γ−1Γα−β∫011−sα−γ−1pssϑ1−α1bϑfs,1,1ds<∞,∫01Gt,sqsgs,us,Hvsds≤∫01Gt,sqss1−α1bgs,1,1ds≤tα−1Γα∫011−sα−γ−1qss1−α1bgs,1,1ds<∞,∫01D0+βGt,sqsgs,us,Hvsds≤∫01D0+βGt,sqss1−α1bgs,1,1ds≤tα−β−11−sα−γ−1Γα−β∫011−sα−γ−1qss1−α1bgs,1,1ds<∞.Thus, we know thatA,B are well defined.Second, we will prove thatA,B,C satisfy conditions (A1)–(A3) of Lemma 10. From (H1)–(H3), for all u,v∈P,t∈0,1, we have Au,vt≥0,D0+βAu,vt≥0,Bu,vt≥0,D0+βBu,vt≥0,Cvt≥0,Cvt≥0. Therefore, A:P×P⟶P,B:P×P⟶P,C:P×P⟶P. And for all u,v∈P, t∈0,1, u≺¯v means ut≤vt,D0+βut≤D0+βvt. Thus, from assumptions (H2) and (H3), it is clear that operators A,B,C satisfy conditions (A1)–(A3) of Lemma 10. And A,B are mixed monotone operators and C is a decreasing operator.Third, we show thatAh,h∈Ph,Bh,h∈Ph,Ch∈Ph. From (H2) and (H3) and Lemma 9, we have
(51)Ah,h=∫01Gt,spsfs,sα−1,D0+βsα−1ds≤∫01tα−11−sα−γ−1Γαpssϑ1−α1bϑfs,1,1ds=ht∫011−sα−γ−1Γαpssϑ1−α1bϑfs,1,1ds,Ah,h=∫01Gt,spsfs,sα−1,D0+βsα−1ds≥∫01tα−11−sα−γ−11−1−sγΓαpssϑα−1bϑfs,1,1ds=ht∫011−sα−γ−11−1−sγΓαpssϑα−1bϑfs,1,1ds,D0+βAh,h=∫01D0+βGt,spsfs,sα−1,D0+βsα−1ds≤∫01tα−β−11−sα−γ−1Γα−βpssϑ1−α1bϑfs,1,1ds=D0+βht∫011−sα−γ−1Γαpssϑ1−α1bϑfs,1,1ds,D0+βAh,h=∫01D0+βGt,spsfs,sα−1,D0+βsα−1ds≥∫01tα−11−sα−γ−11−1−sγ−βΓαpssϑα−1bϑfs,1,1ds=D0+βht∫011−sα−γ−11−1−sγ−βΓαpssϑα−1bϑfs,1,1ds.Set(52)c1=∫011−sα−γ−11−1−sγ−βΓαpssϑα−1bϑfs,1,1ds,c2=∫011−sα−γ−1Γαpssϑ1−α1bϑfs,1,1ds.Now, in a similar way, for allt∈0,1, we obtain that
(53)Bh,h≤ht∫011−sα−γ−1Γαqss1−α1bgs,1,1ds,Bh,h≥ht∫011−sα−γ−11−1−sγΓαqssα−1bgs,1,1ds,D0+βBh,h≤D0+βht∫011−sα−γ−1Γαqss1−α1bgs,1,1ds,D0+βBh,h≥D0+βht∫011−sα−γ−11−1−sγ−βΓαqssα−1bgs,1,1ds.Set(54)c3=∫011−sα−γ−11−1−sγ−βΓαqssα−1bgs,1,1ds,c4=∫011−sα−γ−1Γαqss1−α1bgs,1,1ds.Because ofgt,1,1≡0, obviously, c4≥c3>0. Then c3h≺¯Bh,h≺¯c4h. And we have
(55)sϑα−1bϑfs,1,1≥sα−1bfs,1,1≥sα−1bδ1gs,1,1.Combining with the conditionpt≥qt, it is clear that 0<δ1c3≥c1≤c2. So, there is c1h≺¯Ah,h≺¯c2h. And from kv1≠0 and (42) and (45), obviously, we obtain that Ch∈Ph.Finally, we show that operatorsA,B,C satisfy assumption (A5) of Lemma 10.According to assumption (H6), fort∈0,1, we know ht∈0,1. So 0≥Hh≥H1. From assumption (H5), for all t∈0,1,u,v∈Ph, there is
(56)Au,v=∫01Gt,spsfs,us,D0+βvsds≥∫01Gt,sqsδ1gs,us,0ds≥∫01Gt,sqsδ1gs,us,Hvsds=δ1Bu,v,D0+βAu,v=∫01D0+βGt,spsfs,us,D0+βvsds≥∫01D0+βGt,sqsδ1gs,us,0ds≥∫01D0+βGt,sqsδ1gs,us,Hvsds=δ1D0+βBu,v.Then,Au,v≺¯δ1Bu,v.
(57)Au,v=∫01Gt,spsfs,us,D0+βvsds≥∫01Gt,smδ2ds≥mδ2tα−1Γα1α−γ−1α≥mkv1tα−1Γα1α−γ−1α−β=mΓα−γ1α−γ−1α−βCv,D0+βAu,v=∫01D0+βGt,spsfs,us,D0+βvsds≥∫01D0+βGt,smδ2kv1ds≥mδ2kv1tα−β−1Γα−β1α−γ−1α−β=mδ2Γα−β1α−γ−1α−βD0+βCv.Then,Au,v≻¯δ′Cv, where δ′=mδ2/Γα−β1/α−γ−1/α−β. Now, choose δ=minδ1,δ′, then Au,v≻¯δBu,v+Cv.So, applying Lemma10, the proof of Theorem 11 is completed.Remark 12.
In Lemma10, if we assume that A,B:Ph×Ph⟶Ph and C:Ph×Ph, then it is easy to prove the results in Lemma 10 are also satisfied. So, we also can use the changed lemma to prove Theorem 11.


## 4. Example

Example 13.
We consider the following FBVP:(58)D0+9/2ut+2tt2+ut+1tD0+3/2ut+1+1t∫0tutdt=0,0<t<1,u0=u′0=u″0=u‴0=0,D0+5/2utt=1=1u1+1.
Letpt=qt=1/t,ft,u,v=t2+u+1/v+1,gt,u,v=t2+u+v,kv=1/v+1. Clearly, p,q,f,g,k satisfied conditions (H1)–(H3). And for Hut=∫0tutdt, we can see that assumption (H7) is satisfied. For η∈0,1, we have
(59)ft,ηu,η−1v=t2+ηu+1η−1v+1=t2ηu+ηv+η≥ηt2+u+1v+1=η1/2ft,u,v,gt,ηu,η−1v=t2+ηu+η−1v≥ηt2+ηu+ηv=ηt2+u+v=ηgt,u,v,kη−1v=1η−1v+1≥η1v+1=ηkv.
Then, we know that assumption (H4) of Theorem11 is satisfied.
Clearly,gt,1,1≡0. Moreover, if we let δ1=δ2=1, then we have
(60)ft,u,v=t2+u+1v+1≥t2+u=gt,u,0,ft,u,v=t2+u+1v+1≥1v+1=kv.
Then, assumption (H5) of Theorem11 is satisfied.
Finally, we show that assumption (H6) of Theorem11 is also satisfied. It is easy to see from the following inequalities:
(61)∫011−sα−γ−1pssϑ1−αfs,1,1ds=∫011−ss−1s−7/4s2+1+22<∞,∫011−sα−1qss1−αgs,1,1ds=∫011−s7/4s−1s−7/4s2+1+sds<∞.
Therefore the assumptions of Theorem11 are satisfied. Thus, problem (58) possesses the uniqueness solution.Remark 14.
It is clear thatHut=∫01utdt is linear in this example. If we let operator H be a nonlinear operator, for example, Hut=uat, where 0<a<1, it is obvious that assumption (H7) of Theorem 11 is also satisfied. So using different operators, we also can obtain that BVP (58) has a unique positive solution from the same discussion process. This fact means that our theorem has a wide range of applications.



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*Source: 2354927-2020-07-10.xml*