# Iterative Analysis of the Unique Positive Solution for a Class of Singular Nonlinear Boundary Value Problems Involving Two Types of Fractional Derivatives withp-Laplacian Operator

Authors: Fang Wang; Lishan Liu; Yonghong Wu; Yumei Zou
Journal: Complexity
 (2019)

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## Abstract

This article is concerned with a class of singular nonlinear fractional boundary value problems withp-Laplacian operator, which contains Riemann–Liouville fractional derivative and Caputo fractional derivative. The boundary conditions are made up of two kinds of Riemann–Stieltjes integral boundary conditions and nonlocal infinite-point boundary conditions, and different fractional orders are involved in the boundary conditions and the nonlinear term, respectively. Based on the method of reducing the order of fractional derivative, some properties of the corresponding Green’s function, and the fixed point theorem of mixed monotone operator, an interesting result on the iterative sequence of the uniqueness of positive solutions is obtained under the assumption that the nonlinear term may be singular in regard to both the time variable and space variables. And we obtain the dependence of solution upon parameter. Furthermore, two numerical examples are presented to illustrate the application of our main results.


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## Body

## 1. Introduction

In the past decades, fractional differential equations arise in many mathematical disciplines as the analogue modeling of systems and processes in many scientific fields, such as control theory and engineering. In fact, fractional-order models have proved to be valuable tools in modeling many physical phenomena (for details, see [1–3] and the references therein). Accordingly, there has been a substantial development in the research for fractional differential equations, the properties of solutions, especially. We refer the readers to the papers [4–54]. For instance, in [15], Xu and Wei investigated the positive solutions of the following fractional differential equations:(1)D0+αD0+αxt=ft,xt,x′t,−D0+αxt,t∈0,1,x0=x′0=x′1=0,D0+αx0=D0+α+1x0=D0+α+1x1=0,where D0+α is the Riemann–Liouville’s fractional derivative, 2<α≤3, and f∈C0,1×ℝ+3,ℝ+ ℝ+=0,+∞. The existence of positive solutions is obtained by the Krasnoselskii–Zabreiko fixed point theorem. And by using the method of lower and upper solutions, the authors discussed the uniqueness of positive solution of this problem.On the other hand, the fractional differential equations with thep-Laplacian operator can describe various phenomena, such as the flow of some specific fluid. Hence, the study of fractional differential equations with the p-Laplacian operator is gaining much significance and attention (see [16, 18–21, 37–39, 48–50]). For example, in [48], in order to research the mechanics phenomenon of turbulent flow in a porous medium, the author introduced the following type equation:(2)ϕpx′t′=ft,xt,x′t,where ϕpp>1 is the p-Laplacian operator, i.e., ϕps=sp−2sp>1. In the past thirty years, many papers discussed equation (2) with different boundary value conditions and then drew related conclusion in practice. We refer the reader to [49, 50] and the references cited therein. Recently, Liu et al. in [16] investigated the existence of positive solution of the following fractional differential equations with p-Laplacian operator:(3)D0+αϕpD0+βcxt=ft,xt,D0+βcxt,D0+βcx0=x′0=0,x1=r1xη,D0+βcx1=r2cD0+βxξ,where D0+α is the Riemann–Liouville’s fractional derivative and D0+βc is the Caputo fractional derivative, 1<α,β≤2, r1, r1≥0, ϕpp>1 is the p-Laplacian operator, and f∈C0,1×ℝ+×ℝ−,ℝ+ ℝ−=−∞,0. The existence of lower and upper solutions is obtained by using the monotone iterative technique.Based on the above illustration, it is of significance to study the fractional differential equations withp-Laplacian operator. In this article, we investigate the following fractional differential equation, which is a generalized form of the problem (2):(4)D0+αcϕpD0+βvt+ft,vt,D0+β1vt,…,D0+βn−2vt=0,0<t<1,with the boundary conditions(5)v0=D0+νiv0=D0+βv0=ϕpD0+βv0′=0,i=1,…,n−2,D0+r0v1=λ1∫01h1sD0+r1vsdA1s+λ2∫0ηh2sD0+r2vsdA2s+λ3∑j=1∞ρjD0+r3vηj,where D0+αc denotes the Caputo fractional derivative; D0+β denotes the Riemann–Liouville’s fractional derivative; ϕp is the p-Laplacian operator, i.e., ϕps=sp−2sp>1, 1<α≤2, n−1<β≤nn≥3, 1<β−βn−2≤2, i−1<βi, vi≤i, βn−2−vi≤n−2−ii=1,2,…,n−2, βn−2≤rk≤r0≤β−1k=1,2,3, 0<η≤1, 0<η1≤η2≤⋯≤1, ρj≥0j=1,2,…, λk≥0k=1,2,3; h1,h2:0,1⟶ℝ+=0,+∞ are continuous with h1,h2∈L10,1; ∫01h1svsdA1s, ∫01h2svsdA2s denote the Riemann–Stieltjes integral, in which A1,A2:0,1⟶ℝ are functions of bounded variation; and f:0,1×0,+∞n−1⟶ℝ+ is continuous. Furthermore, we also consider the following equation with a parameter:(6)D0+αcϕpD0+βvt+λft,vt,D0+β1vt,…,D0+βn−2vt=0,0<t<1,with the boundary conditions (5), where λ>0 is a parameter. In fact, we regard the problem (6)-(5) as a new problem for modeling the problem (4)-(5) involving a parameter. In this paper, a nonnegative function v∈C0,1 is called a positive solution of boundary value problem (4)-(5) (resp. (6)-(5)) if it satisfies the problem (4)-(5) (resp. (6)-(5)) for t∈0,1 and vt>0 for t∈0,1.In this paper, we make the following hypothesis:V1 f:0,1×0,+∞n−1⟶ℝ+ is continuous and satisfies(7)ft,u1,…,un−1=Ft,u1,…,un−1,u1,…,un−1+Gt,u1,…,un−1,u1,…,un−1,for allt,u1,…,un−1∈0,1×0,+∞n−1, where F,G:0,1×0,+∞2n−1⟶ℝ+ are continuous.V2 For any t∈0,1, v1,…,vn−1∈0,+∞n−1, Ft,u1,…,un−1,v1,…,vn−1, and Gt,u1,…,un−1,v1,…,vn−1 are nondecreasing in u1,…,un−1∈0,+∞n−1, and for any t∈0,1, u1,…,un−1∈0,+∞n−1, Ft,u1,…,un−1,v1,…,vn−1, and Gt,u1,…,un−1,v1,…,vn−1 are nonincreasing in v1,…,vn−1∈0,+∞n−1.V3 For any l∈0,1, there exists φl∈lq−1,1 such that for any t∈0,1, u1,…,un−1 and v1,…,vn−1∈0,+∞n−1:(8)Ft,lu1,…,lun−1,l−1v1,…,l−1vn−1≥φp−1lFt,u1,…,un−1,v1,…,vn−1,Gt,lu1,…,lun−1,l−1v1,…,l−1vn−1≥lp−1Gt,u1,…,un−1,v1,…,vn−1.V4 There exists ϱ>0 such that for any t∈0,1, u1,…,un−1 and v1,…,vn−1∈0,+∞n−1:(9)Ft,u1,…,un−1,v1,…,vn−1≥ϱp−1Gt,u1,…,un−1,v1,…,vn−1.V5(10)0<∫011−τα−1Fτ,1,1,…,1︸n−1times,τβ−1,τβ−1,…,τβ−1︸n−1timesdτ<+∞,0<∫011−τα−1Gτ,1,1,…,1︸n−1times,τβ−1,τβ−1,…,τβ−1︸n−1timesdτ<+∞.Our main results are as follows.Theorem 1.
Suppose thatV1−V5 hold, then the problem (4)-(5) has a unique solution v∗∈P, and there exists δ∈0,1 such that(11)Γβ−βn−2δΓβtβ−1≤v∗t≤Γβ−βn−2Γβδtβ−1,t∈0,1.
And for any initial valuesv0,v˜0∈Ph, by structuring the following sequences:(12)vnt=I0+βn−2∫01Jt,sϕq1Γα∫0ss−τα−1Fτ,I0+βn−2vn−1τ,I0+βn−2−β1vn−1τ,…,I0+βn−2−βn−3vn−1τ,vn−1τ,I0+βn−2v˜n−1τ,I0+βn−2−β1v˜n−1τ,…,I0+βn−2−βn−3v˜n−1τ,v˜n−1τdτds+∫01Jt,sϕq1Γα∫0ss−τα−1Gτ,I0+βn−2vn−1τ,I0+βn−2−β1vn−1τ,…,I0+βn−2−βn−3vn−1τ,vn−1τ,I0+βn−2v˜n−1τ,I0+βn−2−β1v˜n−1τ,…,I0+βn−2−βn−3v˜n−1τ,v˜n−1τdτds,v˜nt=I0+βn−2∫01Jt,sϕq1Γα∫0ss−τα−1Fτ,I0+βn−2v˜n−1τ,I0+βn−2−β1v˜n−1τ,…,I0+βn−2−βn−3v˜n−1τ,v˜n−1τ,I0+βn−2vn−1τ,I0+βn−2−β1vn−1τ,…,I0+βn−2−βn−3vn−1τ,vn−1τdτds+∫01Jt,sϕq1Γα∫0ss−τα−1Gτ,I0+βn−2v˜n−1τ,I0+βn−2−β1v˜n−1τ,…,I0+βn−2−βn−3v˜n−1τ,v˜n−1τ,I0+βn−2vn−1τ,I0+βn−2−β1vn−1τ,…,I0+βn−2−βn−3vn−1τ,vn−1τdτds,n=1,2,…,we have(13)supt∈0,1vnt−v∗t⟶0,supt∈0,1v˜nt−v∗t⟶0,asn⟶∞.Theorem 2.
Suppose thatV1−V5 hold, then for any λ>0, the problem (6)-(5) have a unique solution vλ∗∈P, and there exists δλ∈0,1 such that(14)Γβ−βn−2δλΓβtβ−1≤vλ∗t≤Γβ−βn−2Γβδλtβ−1,t∈0,1.
And moreover,(1)
If there exists γ∈0,1 such that(15)φl≥1ϱlγ−l+lγ,∀l∈0,1,then vλ∗ is continuous with respect to λ∈0,+∞. That is, for any λ0∈0,+∞,(16)supt∈0,1vλ∗t−vλ0∗t⟶0,asλ⟶λ0.(2)
If(17)φl≥1ϱl1/2−l+l1/2,∀l∈0,1,then 0<λ1<λ2 implies vλ1∗<vλ2∗.(3)
If there exists γ∈0,1/2 such that(18)φl≥1ϱlγ−l+lγ,∀l∈0,1,then(19)limλ⟶0+supt∈0,1vλ∗t=0,limλ⟶+∞supt∈0,1vλ∗t=+∞.The key argument of the problem (4)-(5) and (6)-(5) is the iterative positive solution by applying the method of reducing the order of fractional derivative and the fixed point theory of mixed monotone operator. The method of reducing the order of fractional derivative is based on certain semigroup properties of the Riemann–Liouville’s fractional integral and derivative. We refer the reader to [3, 24, 25, 34, 36, 46]. In this paper, different orders of Riemann–Liouville’s fractional derivative are involved in the nonlinearity f, which is solved in a more complex space, in most cases. By using the method of reduction, we transform the problem (4)-(5) (resp. (6)-(5)) into an equivalent and low-order problem, in which the nonlinearity f contains no fractional derivative. Therefore, the work could proceed in the space C0,1, which is more interesting and meaningful. On the other hand, by using the properties of relevant Green’s function and cone, the theory of mixed monotone operator could be applied on the research of fractional boundary value problems. We suggest that one refers to [13, 17, 24, 31, 32, 34, 47]. In this paper, by structuring a suitable mixed monotone operator, the problem (4)-(5) (resp. (6)-(5)), which contains the p-Laplacian operator, is solved. In [15, 16], the positive solutions are obtained under the assumption that the nonlinear term f is continuous. But, we obtain the uniqueness of positive solution with the assumption that the nonlinear term may be singular in regard to both the time variable and space variables. Compared with [15], our equation contains p-Laplacian operator, which is more general. Compared with [16], our boundary conditions are of general significance, which would be used to describe more phenomena in practice.The paper is organized as follows. In Section2, we present some relevant definitions and lemmas. And we transform the problem (4)-(5) (resp. (6)-(5)) into an equivalent and low-order problem (28)-(29) (resp. (44)-(29)) and obtain some properties of the corresponding Green’s function. In Section 3, we give the proof of our main results, as one researches the uniqueness of iterative positive solutions for the problem (4)-(5) and the other one analyses the dependence of solution upon parameter for the problem (6)-(5). In Section 4, we give two examples to illustrate our theoretical results. In Section 5, we give a conclusion of our main results.


## 2. Preliminaries and Lemmas

Definition 1.
Letα>0. The Riemann–Liouville fractional integral of order α of a function v:0,∞⟶ℝ is given by(20)I0+αvt=1Γα∫0tt−sα−1vsds,provided that the right-hand side is pointwise defined on 0,∞.Definition 2.
Letα>0. The Riemann–Liouville fractional derivative of order α of a continuous function v:0,∞⟶ℝ is given by(21)D0+αvt=ddtnI0+n−αvt=1Γn−αddtn∫0tvst−sα−n+1ds,where n=α+1, α denotes the integer part of the number α, provided that the right-hand side is pointwise defined on 0,∞. In particular, if α=n∈ℕ+, then D0+αvt=vnt.Definition 3.
Letα>0. The Caputo fractional derivative of order α of a continuous function v:0,∞⟶ℝ is given by(22)D0+αcvt=1Γn−α∫0tvnst−sα−n+1ds,where n=α+1, α denotes the integer part of the number α, provided that the right-hand side is pointwise defined on 0,∞. In particular, if α=n∈ℕ+, then D0+αcvt=vnt.Lemma 1 (see [7]).
Letα>0. If we assume v∈C0,1∩L10,1, then the equation(23)D0+αvt=0,has(24)vt=c1tα−1+c2tα−2+⋯+cntα−n,ci∈ℝ,i=1,2,…,n,as the unique solution, where n is the smallest integer greater than or equal to α.Lemma 2 (see [7]).
Letv∈C0,1∩L10,1 and D0+αv∈C0,1∩L10,1. Then,(25)I0+αD0+αvt=vt+c1tα−1+c2tα−2+⋯+cntα−n,ci∈ℝ,i=1,2,…,n,where n is the smallest integer greater than or equal to α.Lemma 3 (see [4]).
Letv∈ACn0,1. Then,(26)I0+αD0+αcvt=vt+c0+c1t+⋯+cn−1tn−1,ci∈ℝ,i=1,2,…,n,where n=α+1 for α∉ℕ+ and n=α for α∈ℕ+.Lemma 4 (see [4]).
(1)
If u∈L10,1, α≥β>0, then
(27)
D
0
+
α
I
0
+
α
u
t
=
u
t
,
D
0
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β
I
0
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α
u
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=
I
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+
α
−
β
u
t
.
(2)
Let α∈n−1,n. If u0=u′0=⋯=un−10, then D0+αcvt=D0+αvt.Letu=D0+βn−2v, then the problem (4)-(5) could be transformed to the following fractional differential equation:(28)D0+αcϕpD0+β−βn−2ut+ft,I0+βn−2ut,I0+βn−2−β1ut,…,I0+βn−2−βn−3ut,ut=0,0<t<1,with the boundary conditions(29)D0+vn−2−βn−2u0=D0+β−βn−2u0=ϕpD0+β−βn−2u0′=0,D0+r0−βn−2u1=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑j=1∞ρjD0+r3−βn−2uηj,where 1<β−βn−2≤2.Lemma 5.
If the problem (4)-(5) have a solution v∈C0,1, then u=D0+βn−2v is a solution of the problem (28)-(29). And in turn, if the problem (28)-(29) have a solution u∈C0,1, then v=I0+βn−2u is a solution of the problem (4)-(5).Proof.
Let(30)ut=D0+βn−2vt,t∈0,1,where v∈C0,1 is a solution of the problem (4)-(5). By using Lemma 2, we have(31)vt=I0+βn−2ut−c1tβn−2−1−⋯−cn−2tβn−2−n−2,where ci∈ℝi=1,2,…,n−2. By the boundary condition v0=0, we get cn−2=0. So, we have(32)vt=I0+βn−2ut−c1tβn−2−1−⋯−cn−3tβn−2−n−3.
Furthermore, we obtain(33)D0+v1vt=I0+βn−2−ν1ut−c1Γβn−2Γβn−2−ν1tβn−2−ν1−1−c2Γβn−2−1Γβn−2−ν1−1tβn−2−ν1−2−⋯−cn−3Γβn−2−n−4Γβn−2−ν1−n−4tβn−2−ν1−n−3,which combines with the boundary condition D0+ν1v0=0 meaning cn−3=0. In a similar manner, we obtain cn−4=⋯=c1=0 based on the boundary condition D0+ν2v0=⋯=D0+νn−3v0=0. Thus, we have(34)vt=I0+βn−2ut,t∈0,1.
Applying the operatorD0+β to both sides of this equality, we have(35)D0+βvt=D0+βI0+βn−2ut=dndtnI0+n−βI0+βn−2ut=dndtnI0+n−β−βn−2ut=d2dt2dn−2dtn−2I0+n−2I0+2−β−βn−2ut=d2dt2I0+2−β−βn−2ut=D0+β−βn−2ut.
Similarly, we know(36)D0+νn−2vt=D0+νn−2−βn−2ut,(37)D0+rivt=D0+riI0+βn−2ut=D0+ri−βn−2ut,i=0,1,2,3.
In addition, by means of Lemma 4, we can obtain(38)D0+βivt=D0+βiI0+βn−2ut=I0+βn−2−βiut,i=1,2,…,n−3.
According to (30), (34), (36), and (38), we have(39)D0+αcϕpD0+β−βn−2ut+ft,I0+βn−2ut,I0+βn−2−β1ut,…,I0+βn−2−βn−3ut,ut=cD0+αϕpD0+βvt+ft,vt,D0+β1vt,…,D0+βn−2vt=0.
It follows from (35)–(37) that(40)D0+β−βn−2u0=D0+βv0=0,(41)ϕpD0+β−βn−2u0′=ϕpD0+βv0′=0,(42)D0+νn−2−βn−2u0=D0+νn−2v0=0,(43)D0+r0−βn−2u1=D0+r0v1=λ1∫01h1sD0+r1vsdA1s+λ2∫0ηh2sD0+r2vsdA2s+λ3∑i=1∞ρiD0+r3vηi=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑i=1∞ρiD0+r3−βn−2uηi.
On the basis of (39)–(43), we know u=D0+βn−2v is a solution of the problem (28)-(29).
On the other hand, if the problem (28)-(29) has a solution u∈C0,1, then v=I0+βn−2u is a solution of the problem (4)-(5). Since the proof is similar to Lemma 3 in [44], we omit it here.Remark 5.
According to Lemma 5, we conclude that the work on considering the solution of the problem (4)-(5) is equivalent to the search for the solution of (28)-(29). Similarly, we can prove the work on the problem (6)-(5) is equivalent to the following equation:(44)D0+αcϕpD0+β−βn−2ut+λft,I0+βn−2ut,I0+βn−2−β1ut,…,I0+βn−2−βn−3ut,ut=0,0<t<1,with the boundary conditions (29).Lemma 6.
Letx∈C0,1∩L10,1. Then, the fractional differential equation(45)D0+β−βn−2ut+xt=0,0<t<1,1<β−βn−2≤2,with the boundary conditions(46)D0+νn−2−βn−2u0=0,D0+r0−βn−2u1=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑j=1∞ρjD0+r3−βn−2uηj,has a unique solution(47)ut=∫01Jt,sxsds,where(48)Jt,s=J0t,s+tβ−βn−2−1∫01J1τ,sh1τdA1τ+tβ−βn−2−1∫0ηJ2τ,sh2τdA2τ+tβ−βn−2−1∑j=1∞ρjJ3ηj,s,in which(49)J0t,s=1Γβ−βn−2tβ−βn−2−11−sβ−r0−1−t−sβ−βn−2−1,s≤t,tβ−βn−2−11−sβ−r0−1,t≤s,J1t,s=λ1σΓβ−βn−2Γβ−r1tβ−r1−11−sβ−r0−1−t−sβ−r1−1,s≤t,tβ−r1−11−sβ−r0−1,t≤s,J2t,s=λ2σΓβ−βn−2Γβ−r2tβ−r2−11−sβ−r0−1−t−sβ−r2−1,s≤t,tβ−r2−11−sβ−r0−1,t≤s,J3t,s=λ3σΓβ−βn−2Γβ−r3tβ−r3−11−sβ−r0−1−t−sβ−r3−1,s≤t,tβ−r3−11−sβ−r0−1,t≤s,σ=1Γβ−r0−λ1Γβ−r1∫01sβ−r1−1h1sdA1s−λ2Γβ−r2∫0ηsβ−r2−1h2sdA2s−λ3Γβ−r3∑j=1∞ρjηjβ−r3−1≠0.
Obviously,J:0,1×0,1⟶ℝ+ is continuous.Proof.
We may apply Lemma 2 to reduce equation (45) to an equivalent integral equation:(50)ut=c1tβ−βn−2−1+c2tβ−βn−2−2−I0+β−βn−2xt,where c1,c2∈ℝ. Since D0+νn−2−βn−2u0=0, we have c2=0. So(51)ut=c1tβ−βn−2−1−I0+β−βn−2xt.
Applying the operatorD0+ri−βn−2i=0,1,2,3 to both sides of (51), we have(52)D0+ri−βn−2ut=c1Γβ−βn−2Γβ−ritβ−ri−1−I0+β−rixt,i=0,1,2,3.
Hence, according to (52) and the boundary condition(53)D0+r0−βn−2u1=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑j=1∞ρjD0+r3−βn−2uηj,we can obtain(54)c1=1σΓβ−βn−21Γβ−r0∫011−sβ−r0−1xsds−λ1Γβ−r1∫01∫0ss−τβ−r1−1xτdτh1sdA1s−λ2Γβ−r2∫0η∫0ss−τβ−r2−1xτdτh2sdA2s−λ3Γβ−r3∑j=1∞ρj∫0ηjηj−sβ−r3−1xsds,where(55)σ=1Γβ−r0−λ1Γβ−r1∫01sβ−r1−1h1sdA1s−λ2Γβ−r2∫0ηsβ−r2−1h2sdA2s−λ3Γβ−r3∑j=1∞ρjηjβ−r3−1≠0.
Then(56)ut=−1Γβ−βn−2∫0tt−sβ−βn−2−1xsds+tβ−βn−2−1σΓβ−βn−21Γβ−r0∫011−sβ−r0−1xsds−λ1Γβ−r1∫01∫0ss−τβ−r1−1xτdτh1sdA1s−λ2Γβ−r2∫0η∫0ss−τβ−r2−1xτdτh2sdA2s−λ3Γβ−r3∑j=1∞ρj∫0ηjηj−sβ−r3−1xsds=−1Γβ−βn−2∫0tt−sβ−βn−2−1xsds+1Γβ−βn−2∫01tβ−βn−2−11−sβ−r0−1xsds+tβ−βn−2−1σΓβ−βn−2∫011−sβ−r0−1xsds×λ1Γβ−r1∫01sβ−r1−1h1sdA1s+λ2Γβ−r2∫0ηsβ−r2−1h2sdA2s+λ3Γβ−r3∑j=1∞ρjηjβ−r3−1−λ1Γβ−r1∫01∫0ss−τβ−r1−1xτdτh1sdA1s+λ2Γβ−r2∫0η∫0ss−τβ−r2−1xτdτh2sdA2s+λ3Γβ−r3∑j=1∞ρj∫0ηjηj−sβ−r3−1xsdstβ−βn−2−1σΓβ−βn−2=1Γβ−βn−2∫01tβ−βn−2−11−sβ−r0−1xsds−∫0tt−sβ−βn−2−1xsds+λ1tβ−βn−2−1σΓβ−βn−2Γβ−r1∫01∫01sβ−r1−11−τβ−r0−1xτdτ−∫0ss−τβ−r1−1xτdτh1sdA1s+λ2tβ−βn−2−1σΓβ−βn−2Γβ−r2∫0η∫01sβ−r2−11−τβ−r0−1xτdτ−∫0ss−τβ−r2−1xτdτh2sdA2s+λ3tβ−βn−2−1σΓβ−βn−2Γβ−r3∑j=1∞ρj∫01ηjβ−r3−11−sβ−r0−1xsds−∫0ηjηj−sβ−r3−1xsds=∫01J0t,sxsds+∫01tβ−βn−2−1∫01J1s,τxτdτh1sdA1s+∫0ηtβ−βn−2−1∫01J2s,τxτdτh2sdA2s+tβ−βn−2−1∫01∑j=1∞ρjJ3ηj,sxsds=∫01J0t,sxsds+∫01tβ−βn−2−1∫01J1τ,sh1τdA1τxsds+∫01tβ−βn−2−1∫0ηJ2τ,sh2τdA2τxsds+∫01tβ−βn−2−1∑j=1∞ρjJ3ηj,sxsds=∫01Jt,sxsds.
The proof is complete.Lemma 7.
Letx∈C0,1∩L10,1 be a given function. Then, the following equation(57)D0+αcϕpD0+β−βn−2ut+xt=0,0<t<1,with the boundary conditions(58)D0+νn−2−βn−2u0=D0+β−βn−2u0=ϕpD0+β−βn−2u0′=0,D0+r0−βn−2u1=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑j=1∞ρjD0+r3−βn−2uηj,has a unique solution(59)ut=∫01Jt,sϕq∫01Hs,τxτdτds,where ϕq=ϕp−1, Jt,s is defined in Lemma 6, and(60)Ht,s=1Γαt−sα−1,0≤s≤t≤1,0,0≤t≤s≤1.Proof.
LetϕpD0+β−βn−2ut=wt. Then, we can strip down the problem (57)‐(58) to the following two problems:(61)D0+αcwt+xt=0,0<t<1,w0=w′0=0,(62)D0+β−βn−2ut=ϕqwt,0<t<1,D0+νn−2−βn−2u0=0,D0+r0−βn−2u1=λ1∫01h1sD0+r1−βn−2usdA1s+λ2∫0ηh2sD0+r2−βn−2usdA2s+λ3∑j=1∞ρjD0+r3−βn−2uηj.
Obviously, the problem (61) can be written as(63)wt=c0+c1t−I0+αxt.
The boundary conditionsw0=w′0=0 imply that c0=c1=0. Thus,(64)wt=−I0+αxt=−∫01Ht,sxsds.
On the other hand, by Lemma 6, we know the problem (62) has a unique solution:(65)ut=−∫01Jt,sϕqwsds.
Therefore, the problem (57)-(58) has a unique solution:(66)ut=∫01Jt,sϕq∫01Hs,τxτdτds.Lemma 8.
Letσ>0 (defined in Lemma 6), λi≥0i=1,2,3, ∫01sβ−r1−1h1sdA1s≥0, ∫0ηsβ−r2−1h2sdA2s≥0, and 0≤∑j=1∞ρjηjβ−r3−1<+∞. Then, the following properties hold:(1)
tβ−βn−2−1g0s≤J0t,s≤Λ0tβ−βn−2−1, where Λ0=1/Γβ−βn−2,(67)g0s=1Γβ−βn−21−sβ−r0−11−1−sr0−βn−2.(2)
tβ−ri−1gis≤Jit,s≤Λitβ−ri−1i=1,2,3, where(68)gis=1Γβ−ri1−sβ−r0−11−1−sr0−ri,i=1,2,3,Λi=λiσΓβ−βn−2Γβ−ri,i=1,2,3.(3)
htgs≤Jt,s≤Λht, where ht=tβ−βn−2−1,(69)gs=g0s+∫01J1τ,sh1τdA1τ+∫0ηJ2τ,sh2τdA2τ+∑j=1∞J3ηj,s,Λ=Λ0+Λ1∫01τβ−r1−1h1τdA1τ+Λ2∫0ητβ−r2−1h2τdA2τ+Λ3∑j=1∞ρjηjβ−r3−1.Proof.
(1)
For0≤s≤t≤1,
(70)
J
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=
1
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.
For0≤t≤s≤1,(71)J0t,s=1Γβ−βn−2tβ−βn−2−11−sβ−r0−1≥1Γβ−βn−2tβ−βn−2−11−sβ−r0−11−1−sr0−βn−2=tβ−βn−2−1g0s,J0t,s=1Γβ−βn−2tβ−βn−2−11−sβ−r0−1≤Λ0tβ−βn−2−1.
(2)
Similarly as (1), we can obtain that(72)tβ−ri−1gis≤Jit,s≤Λitβ−ri−1,i=1,2,3.
(3)
In view of the conclusions of (1) and (2), we have(73)Jt,s=J0t,s+tβ−βn−2−1∫01J1τ,sh1τdA1τ+tβ−βn−2−1∫0ηJ2τ,sh2τdA2τ+tβ−βn−2−1∑j=1∞ρjJ3ηj,s≥tβ−βn−2−1g0s+tβ−βn−2−1∫01J1τ,sh1τdA1τ+tβ−βn−2−1∫0ηJ2τ,sh2τdA2τ+tβ−βn−2−1∑j=1∞ρjJ3ηj,s=htgs,Jt,s=J0t,s+tβ−βn−2−1∫01J1τ,sh1τdA1τ+tβ−βn−2−1∫0ηJ2τ,sh2τdA2τ+tβ−βn−2−1∑j=1∞ρjJ3ηj,s≤Λ0tβ−βn−2−1+Λ1tβ−βn−2−1∫01τβ−r1−1h1τdA1τ+Λ2tβ−βn−2−1∫0ητβ−r2−1h2τdA2τ+Λ3tβ−βn−2−1∑j=1∞ρjηjβ−r3−1=Λht.
The proof is complete.LetE,· be a Banach space, P be a cone in E, and θ be the zero element of E. P is said to be normal if there exists a constant N>0 such that(74)θ≤u≤v⟹u≤Nv,∀u,v∈P.The smallest constant, which satisfies the above inequality, is called the normality constant ofP. Then, E is a partially ordered Banach space by P; that is to say,(75)v≤u⟺u−v∈P.Foru,v∈E, the notation u∼v shows that there exist λ>0 and μ>0 such that λu≤v≤μu. Clearly, ∼ is an equivalence relation in E. For any h∈P\θ, let Ph=u∈Eu∼h=u∈Eμh≤u≤μ−1h,0<μ<1. Then, Ph⊂P is a component of P. For more details, we suggest readers to refer [10, 11, 13].Definition 4 (see [11]).
LetE be a Banach space and D⊂E. The operator T:D×D⟶E is called a mixed monotone operator if Tu,v is increasing in u∈D and decreasing in v∈D, i.e.,(76)u1≤u2,v1≥v2⟹Tu1,v1≤Tu2,v2,∀ui,vi∈D,i=1,2.Lemma 9.
(see [13, 17]). Let P be a normal cone in the Banach space E and T,S:Ph×Ph⟶Ph be mixed monotone operators which satisfy the following:(1)
For any l∈0,1, there exists φl∈l,1 such that(77)Tlu,l−1v≥φlTu,v,∀u,v∈Ph.(2)
or any l∈0,1, u,v∈Ph,(78)Slu,l−1v≥lSu,v.(3)
There exists a constant ϱ>0 such that for any u,v∈Ph, Tu,v≥ϱSu,v.
Then, the equationTu,u+Su,u=u has a unique fixed point u∗∈Ph. And for any initial values u0,v0∈Ph, by structuring the following sequences:(79)un=Tun−1,vn−1+Sun−1,vn−1,vn=Tvn−1,un−1+Svn−1,un−1,n=1,2,…,we have un−u∗⟶0 and vn−u∗⟶0 in E, as n⟶∞.Lemma 10.
(see [13, 17]). Assume T and S satisfy all the conditions of Lemma 9. Then, for any λ>0, the equation λTu,u+λSu,u=u has a unique solution uλ∗∈Ph, which satisfies the following:(1)
If there exists γ∈0,1 such that(80)φl≥1ϱlγ−l+lγ,∀l∈0,1,then uλ∗ is continuous with respect to λ∈0,+∞. That is, for any λ0∈0,+∞,(81)uλ∗−uλ0∗⟶0,asλ⟶λ0.(2)
If(82)φl≥1ϱl1/2−l+l1/2,∀l∈0,1,then 0<λ1<λ2 implies uλ1∗<uλ2∗.(3)
If there exists γ∈0,1/2 such that(83)φl≥1ϱlγ−l+lγ,∀l∈0,1,then(84)limλ⟶0+uλ∗=0,limλ⟶+∞uλ∗=+∞.
In this paper, we denoteE=C0,1 with the norm v=sup0≤t≤1vt. Then E,⋅ is a Banach space. Let P=v∈E:vt≥0,t∈0,1 be a cone in E. It is easy to check that P is normal in E with the normality constant N=1.


## 3. Proof of Main Results

In this section, letPh=u∈E:u∼h, where ht=tβ−βn−2−1,t∈0,1 (defined in Lemma 8). Then, Ph is a component of P. Let us define three operators T:Ph⟶P and T1,T2:Ph×Ph⟶P as follows:(85)Tut=∫01Jt,sϕq∫01Hs,τfτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτdτds,T1u,vt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds,T2u,vt=∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds.It is easy to check thatu∈C0,1 is a positive solution of the problem (28)-(29) if it is a fixed point of T in Ph.Proof of Theorem 1.
Obviously, byV1, Tu=T1u,u+T2u,u. And it is easy to check that u∈C0,1 is a solution of the problem (28)-(29) if it satisfies u=Tu=T1u,u+T2u,u.
The first work is to proveT1,T2:Ph×Ph⟶P are well defined. For any u,v∈Ph, there exists a constant μ∈0,1 such that(86)μht≤ut≤μ−1ht,μht≤vt≤μ−1ht,t∈0,1.
On the other hand, it is easy to check that(87)I0+βn−2ht=1Γβn−2∫0tt−sβn−2−1sβ−βn−2−1ds=Γβ−βn−2Γβtβ−1≤1,t∈0,1,(88)I0+βn−2−βiht=1Γβn−2−βi∫0tt−sβn−2−βi−1sβ−βn−2−1ds=Γβ−βn−2Γβ−βitβ−βi−1≤1,i=1,2,…,n−3,t∈0,1.
Let(89)r=minΓβ−βn−2Γβ,Γβ−βn−2Γβ−β1,…,Γβ−βn−2Γβ−βn−3,1.
Then,0<r≤1. According to V1–V3 and (87)–(89), we have(90)Fτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτ≤Fτ,I0+βn−2μ−1hτ,I0+βn−2−β1μ−1hτ,…,I0+βn−2−βn−3μ−1hτ,μ−1hτ,I0+βn−2μhτ,I0+βn−2−β1μhτ,…,I0+βn−2−βn−3μhτ,μhτ≤Fτ,μ−1,μ−1,…,μ−1,μ−1,μΓβ−βn−2Γβτβ−1,μΓβ−βn−2Γβ−β1τβ−β1−1,…,μΓβ−βn−2Γβ−βn−3τβ−βn−3−1,τβ−βn−2−1≤Fτ,μr−1,μr−1,…,μr−1,μr−1,μrτβ−1,μrτβ−β1−1,…,μrτβ−βn−3−1,μrτβ−βn−2−1≤1φp−1μrFτ,1,1,…,1,1,τβ−1,τβ−β1−1,…,τβ−βn−3−1,τβ−βn−2−1≤1φp−1μrFτ,1,…,1,τβ−1,…,τβ−1,(91)Fτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτ≥Fτ,I0+βn−2μhτ,I0+βn−2−β1μhτ,…,I0+βn−2−βn−3μhτ,μhτ,I0+βn−2μ−1hτ,I0+βn−2−β1μ−1hτ,…,I0+βn−2−βn−3μ−1hτ,μ−1hτ≥Fτ,μΓβ−βn−2Γβτβ−1,μΓβ−βn−2Γβ−β1τβ−β1−1,…,μΓβ−βn−2Γβ−βn−3τβ−βn−3−1,μτβ−βn−2−1,μ−1,μ−1,…,μ−1,μ−1≥Fτ,μrτβ−1,μrτβ−β1−1,…,μrτβ−βn−3−1,μrτβ−βn−2−1,μr−1,μr−1,…,μr−1,μr−1≥φp−1μrFτ,τβ−1,τβ−β1−1,…,τβ−βn−3−1,τβ−βn−2−1,1,1,…,1,1≥φp−1μrFτ,τβ−1,…,τβ−1,1,…,1.
Similarly, we have(92)Gτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτ≤1μrp−1Gτ,1,…,1,τβ−1,…,τβ−1,(93)Gτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτ≥μrp−1Gτ,τβ−1,…,τβ−1,1,…,1.
On the basis of (90), (92), and V5, we know that for any t∈0,1,(94)T1u,vt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds≤ΛhtΓαq−1φμr∫01ϕq∫011−τα−1Fτ,1,…,1,τβ−1,…,τβ−1dτds<+∞,(95)T2u,vt≤ΛhtΓαq−1μr∫01ϕq∫011−τα−1Gτ,1,…,1,τβ−1,…,τβ−1dτds<+∞.
That is,T1,T2:Ph×Ph⟶P are well defined. On the other hand, by means of V2, for any u1,v1, u2,v2 satisfying u1≤u2 and v1≥v2, we have(96)T1u1,v1t=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2u1τ,I0+βn−2−β1u1τ,…,I0+βn−2−βn−3u1τ,u1τ,I0+βn−2v1τ,I0+βn−2−β1v1τ,…,I0+βn−2−βn−3v1τ,v1τdτds≤∫01Jt,sϕq∫​01Hs,τFτ,I0+βn−2u2τ,I0+βn−2−β1u2τ,…,I0+βn−2−βn−3u2τ,u2τ,I0+βn−2v2τ,I0+βn−2−β1v2τ,…,I0+βn−2−βn−3v2τ,v2τdτds=T1u2,v2t,t∈0,1,(97)T2u1,v1t=∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2u1τ,I0+βn−2−β1u1τ,…,I0+βn−2−βn−3u1τ,u1τ,I0+βn−2v1τ,I0+βn−2−β1v1τ,…,I0+βn−2−βn−3v1τ,v1τdτds≤∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2u2τ,I0+βn−2−β1u2τ,…,I0+βn−2−βn−3u2τ,u2τ,I0+βn−2v2τ,I0+βn−2−β1v2τ,…,I0+βn−2−βn−3v2τ,v2τdτds=T2u2,v2t,t∈0,1.
Hence,T1,T2:Ph×Ph⟶P are mixed monotone operators.
In the following, we prove thatT1,T2:Ph×Ph⟶Ph. It follows from (91) and (93) that for any u, v∈Ph and t∈0,1,(98)T1u,vt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds≥∫01Jt,sϕqφp−1μr∫01Hs,τFτ,τβ−1,…,τβ−1,1,…,1dτds≥φμrht∫01gsϕq∫01Hs,τFτ,τβ−1,…,τβ−1,1,…,1dτds≥M−1ht,(99)T2u,vt≥μrht∫01gsϕq∫01Hs,τGτ,τβ−1,…,τβ−1,1,…,1dτds≥M−1ht,where M is a constant, which satisfies(100)maxΛΓαq−1φμr∫01ϕq∫011−τα−1Fτ,1,…,1,τβ−1,…,τβ−1dτds,ΛΓαq−1μr∫01ϕq∫​011−τα−1Gτ,1,…,1,τβ−1,…,τβ−1dτds,φμr∫01gsϕq∫01Hs,τFτ,τβ−1,…,τβ−1,1,…,1dτds−1,μr∫01gsϕq∫01Hs,τGτ,τβ−1,…,τβ−1,1,…,1dτds−1<M.
According to (94) and (95), we have(101)T1u,vt≤ΛhtΓαq−1φμr∫01ϕq∫011−τα−1Fτ,1,…,1,τβ−1,…,τβ−1dτds≤Mht,(102)T2u,vt≤ΛhtΓαq−1μr∫01ϕq∫011−τα−1Gτ,1,…,1,τβ−1,…,τβ−1dτds≤Mht.
On the basis of (98), (99), (101), and (102) we infer that T1,T2:Ph×Ph⟶Ph.
Moreover, it follows fromV3, for any l∈0,1 and u, v∈Ph and t∈0,1, we have(103)T1lu,l−1vt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2luτ,I0+βn−2−β1luτ,…,I0+βn−2−βn−3luτ,luτ,I0+βn−2l−1vτ,I0+βn−2−β1l−1vτ,…,I0+βn−2−βn−3l−1vτ,l−1vτdτds=∫01Jt,sϕq∫01Hs,τFτ,lI0+βn−2uτ,lI0+βn−2−β1uτ,…,lI0+βn−2−βn−3uτ,luτ,l−1I0+βn−2vτ,l−1I0+βn−2−β1vτ,…,l−1I0+βn−2−βn−3vτ,l−1vτdτds≥∫01Jt,sϕqφp−1l∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds=φl∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds=φlT1u,vt,(104)T2lu,l−1vt=∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2luτ,I0+βn−2−β1luτ,…,I0+βn−2−βn−3luτ,luτ,I0+βn−2l−1vτ,I0+βn−2−β1l−1vτ,…,I0+βn−2−βn−3l−1vτ,l−1vτdτds≥l∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds=lT2u,vt.
FromV4, for any u, v∈Ph and t∈0,1, we can obtain(105)T1u,vt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds≥∫01Jt,sϕqϱp−1∫01Hs,τGτ,I0+βn−2uτ,I0+βn−2−β1uτ,…,I0+βn−2−βn−3uτ,uτ,I0+βn−2vτ,I0+βn−2−β1vτ,…,I0+βn−2−βn−3vτ,vτdτds=ϱT2u,vt.
So, based on (103)–(105) and Lemma 9, we know the equation T1u,u+T2u,u=u has a unique fixed point u∗∈Ph, which means u∗ is the unique positive solution of the problem (28)-(29), and there exists δ∈0,1 such that(106)δtβ−βn−2−1=δht≤u∗t≤δ−1ht=δ−1tβ−βn−2−1,t∈0,1.
And for any initial valuesu0,u˜0∈Ph, we can construct the following sequences:(107)unt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2un−1τ,I0+βn−2−β1un−1τ,…,I0+βn−2−βn−3un−1τ,un−1τ,I0+βn−2u˜n−1τ,I0+βn−2−β1u˜n−1τ,…,I0+βn−2−βn−3u˜n−1τ,u˜n−1τdτds+∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2un−1τ,I0+βn−2−β1un−1τ,…,I0+βn−2−βn−3un−1τ,un−1τ,I0+βn−2u˜n−1τ,I0+βn−2−β1u˜n−1τ,…,I0+βn−2−βn−3u˜n−1τ,u˜n−1τdτds,u˜nt=∫01Jt,sϕq∫01Hs,τFτ,I0+βn−2u˜n−1τ,I0+βn−2−β1u˜n−1τ,…,I0+βn−2−βn−3u˜n−1τ,u˜n−1τ,I0+βn−2un−1τ,I0+βn−2−β1un−1τ,…,I0+βn−2−βn−3un−1τ,un−1τdτds+∫01Jt,sϕq∫01Hs,τGτ,I0+βn−2u˜n−1τ,I0+βn−2−β1u˜n−1τ,…,I0+βn−2−βn−3u˜n−1τ,u˜n−1τ,I0+βn−2un−1τ,I0+βn−2−β1un−1τ,…,I0+βn−2−βn−3un−1τ,un−1τdτds,t∈0,1,n=1,2,…,such that(108)supt∈0,1unt−u∗t=un−u∗⟶0,supt∈0,1u˜nt−u∗t=u˜n−u∗⟶0,asn⟶∞.
Finally, on the basis of Lemma 5, we knowv∗=Iβn−2u∗ is the unique positive solution of problem (4)-(5). By (106), we have(109)Γβ−βn−2δΓβtβ−1≤v∗t≤Γβ−βn−2Γβδtβ−1,t∈0,1.
Letvn=I0+βn−2un and v˜n=I0+βn−2u˜n; by using the monotonicity and continuity of fractional integral, we have(110)supt∈0,1vnt−v∗t=vn−v∗⟶0,supt∈0,1v˜nt−v∗t=v˜n−v∗⟶0,asn⟶∞.Proof of Theorem 2.
For anyλ>0, let us define three operators T˜,T˜1,T˜2:Ph×Ph⟶P by T˜=λq−1T, T˜1=λq−1T1, and T˜2=λq−1T2, respectively. It is easy to check that u∈C0,1 is a positive solution of the problem (44)–(29) if it is a fixed point of T˜ in Ph, i.e., u=T˜u=T˜1u,u+T˜2u,u. Considering the results of Theorem 1 and Lemma 10 together, we know the following equation(111)T˜1u,u+T˜2u,u=λq−1T1u,u+λq−1T2u,u=u,has a unique solution uλ∗∈Ph, which implies uλ∗ is the unique positive solution of the problem (44)–(29), and there exists δλ∈0,1 such that(112)δλtβ−βn−2−1=δλht≤uλ∗t≤δλ−1ht=δλ−1tβ−βn−2−1,t∈0,1.
Furthermore,uλ∗ satisfies the following conclusion:(1)
If there existsγ∈0,1 such that(113)φl≥1ϱlγ−l+lγ,∀l∈0,1,then uλ∗ is continuous with respect to λ∈0,+∞.That is, for anyλ0∈0,+∞,(114)supt∈0,1uλ∗t−uλ0∗t=uλ∗−uλ0∗⟶0,asλ⟶λ0.(2)
If(115)φl≥1ϱl1/2−l+l1/2,∀l∈0,1,then0<λ1<λ2 implies uλ1∗<uλ2∗.(3)
If there existsγ∈0,1/2 such that(116)φl≥1ϱlγ−l+lγ,∀l∈0,1,then(117)limλ⟶0+supt∈0,1uλ∗t=limλ⟶0+uλ∗=0,limλ⟶+∞supt∈0,1uλ∗t=limλ⟶+∞uλ∗=+∞.
Finally, by Lemma 5, we deducevλ∗=I0+βn−2uλ∗ is a unique positive solution of the problem (6)-(5), which satisfies(118)Γβ−βn−2δλΓβtβ−1≤vλ∗t≤Γβ−βn−2Γβδλtβ−1,t∈0,1.
And by using the monotonicity and continuity of fractional integral, we have the following:(1)
If there existsγ∈0,1 such that(119)φl≥1ϱlγ−l+lγ,∀l∈0,1,thenvλ∗ is continuous with respect to λ∈0,+∞. That is, for any λ0∈0,+∞,(120)supt∈0,1vλ∗t−vλ0∗t=vλ∗−vλ0∗⟶0,asλ⟶λ0.(2)
If(121)φl≥1ϱl1/2−l+l1/2,∀l∈0,1,then0<λ1<λ2 implies vλ1∗<vλ2∗.(3)
If there existsγ∈0,1/2 such that(122)φl≥1ϱlγ−l+lγ,∀l∈0,1,then(123)limλ⟶0+supt∈0,1vλ∗t=limλ⟶0+vλ∗=0,limλ⟶+∞supt∈0,1vλ∗t=limλ⟶+∞vλ∗=+∞.


## 4. Numerical Examples

In this section, we give two simple theoretical numerical examples which justify Theorems 1 and 2.Example 1.
We consider the following equation:(124)D0+3/2cϕ3/2D0+29/10vt+t−19/20v1/2t+t−9/20v1/2t+2t−29/40v1/4tD0+9/10vt1/4+2t19/20v−1/2t+2t1/2D0+9/10vt−1/2=0,0<t<1,with the boundary conditions(125)v0=D0+10/11v0=D0+29/10v0=ϕ3/2D0+29/10v0′=0,D0+19/10v1=1100Γ32∫01s−1/2D0+7/5vsdA1s+1100∫01s−1D0+9/10vsdA2s+1100∑i=1∞1ii+1D0+9/10v1,where ϕ3/2s=s−1/2s and(126)A1t=211,t∈0,11311,t=1,A2t=113,t∈0,1,1413,t=1.Proof.
Let(127)ft,u1,u2=t−19/20u11/2+t−9/20u11/2+2t−29/40u11/4u21/4+2t19/20u1−1/2+2t1/2u2−1/2,ϕps=ϕ3/2s, α=3/2, β=29/10, β1=9/10, ν1=10/11, r0=19/10, r1=7/5, r2=9/10, r3=9/10, ρi=1/ii+1, ηi=1, i=1,2,…, η=1, λ1=1/100Γ3/2, λ2=1/100, λ3=1/100, h1t=t−1/2, and h2t=t−1. Then, the problem (124)-(125) can be expressed as the problem (4)-(5). Notice that(128)∫01sβ−r1−1h1sdA1s=1>0,∫0ηsβ−r2−1h2sdA2s=1>0,∑i=1∞ρiηiβ−r3−1=1,σ=1Γβ−r0−λ1Γβ−r1∫01sβ−r1−1h1sdA1s−λ2Γβ−r2∫0ηsβ−r2−1h2sdA2s−λ3Γβ−r3∑i=1∞ρiηiβ−r3−1=0.97>0.
We infer that the properties of Green’s function in Lemma 8 are achieved. Let(129)Ft,u1,u2,v1,v2=t−19/20u11/2+t−29/40u11/4u21/4+t19/20v1−1/2+t1/2v2−1/2,Gt,u1,u2,v1,v2=t−9/20u11/2+t−29/40u11/4u21/4+t19/20v1−1/2+t1/2v2−1/2.
It is easy to check the following conditions:V1 ft,u1,u2=Ft,u1,u2,u1,u2+Gt,u1,u2,u1,u2.V2 For fixed t∈0,1, and v1,v2∈0,+∞2, Ft,u1,u2,v1,v2, Gt,u1,u2,v1,v2 are increasing in u1,u2∈0,+∞2; for fixed t∈0,1 and u1,u2 ∈0,+∞2, Ft,u1,u2,v1,v2 and Gt,u1,u2,v1,v2 are decreasing in v1,v2∈0,+∞2.V3 Let φl=l. Then, for l∈0,1, t∈0,1, and u1,u2,v1,v2∈0,+∞4,(130)Ft,lu1,lu2,l−1v1,l−1v2=t−19/20lu11/2+t−29/40lu11/4lu21/4+t19/20l−1v1−1/2+t1/2l−1v2−1/2≥l1/2Ft,u1,u2,v1,v2=φ1/2lFt,u1,u2,v1,v2=φp−1lFt,u1,u2,v1,v2,Gt,lu1,lu2,l−1v1,l−1v2=t−9/20lu11/2+t−29/40lu11/4lu21/4+t19/20l−1v1−1/2+t1/2l−1v2−1/2≥l1/2Gt,u1,u2,v1,v2=lp−1Gt,u1,u2,v1,v2.V4 Let ϱ=1. Then, for all u1,u2,v1,v2∈0,+∞4,(131)Ft,u1,u2,v1,v2≥ϱGt,u1,u2,v1,v2.V5 The functions F and G satisfy(132)0<∫011−τα−1Fτ,1,1,τβ−1,τβ−1dτ=B1.5,0.05+B1.5,0.275+B1.5,1+B1.5,0.55<+∞,0<∫011−τα−1Gτ,1,1,τβ−1,τβ−1dτ=B1.5,0.55+B1.5,0.275+B1.5,1+B1.5,0.55<+∞.
Thus, the assumptions of Theorem 1 are satisfied. By calculation, we obtain the approximate solution of the problem (124)-(125) is v∗t≈34.315t1.9. On the other hand, let u0t=t and u˜0=10t. We take v0t=Iβ1u0t and v˜0=Iβ1u˜0t. By iterating the sequences, the numerical results of the iterative process are shown as follows (Figure 1 and Tables 1 and 2).Figure 1
Iterative process ofvi and v˜i.
(a)
(b)Table 1
Iterative process ofvi.
t
v
0
v
1
v
2
v
3
v
4
v
5
v
∗
0.0
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.1
0.0069
0.0094
0.0106
0.0113
0.0116
0.0118
0.4320
0.2
0.0257
0.0352
0.0397
0.0421
0.0433
0.0440
1.6123
0.3
0.0556
0.0759
0.0857
0.0908
0.0935
0.0950
3.4835
0.4
0.0960
0.1307
0.1476
0.1563
0.1610
0.1635
6.0172
0.5
0.1466
0.1983
0.2239
0.2373
0.2443
0.2481
9.1945
0.6
0.2073
0.2769
0.3126
0.3312
0.3411
0.3464
13.0008
0.7
0.2779
0.3631
0.4100
0.4344
0.4474
0.4543
17.4249
0.8
0.3581
0.4520
0.5104
0.5408
0.5570
0.5657
22.4572
0.9
0.4480
0.5361
0.6054
0.6415
0.6606
0.6710
28.0895
1.0
0.5472
0.6045
0.6828
0.7234
0.7450
0.7567
34.3150Table 2
Iterative process ofv˜i.
t
v
˜
0
v
˜
1
v
˜
2
v
˜
3
v
˜
4
v
˜
5
v
∗
0.0
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.1
0.0689
0.0943
0.1064
0.1128
0.1161
0.1179
0.4320
0.2
0.2571
0.3518
0.3971
0.4207
0.4333
0.4400
1.6123
0.3
0.5555
0.7591
0.8570
0.9079
0.9350
0.9496
3.4835
0.4
0.9596
1.3071
1.4757
1.5633
1.6100
1.6351
6.0172
0.5
1.4663
1.9835
2.2394
2.3724
2.4433
2.4814
9.1945
0.6
2.0733
2.7688
3.1262
3.3120
3.4109
3.4641
13.0008
0.7
2.7788
3.6311
4.1000
4.3438
4.4735
4.5433
17.4249
0.8
3.5814
4.5203
5.1043
5.4079
5.5695
5.6565
22.4572
0.9
4.4796
5.3612
6.0542
6.4144
6.6062
6.7093
28.0895
1.0
5.4724
6.0459
6.8277
7.2340
7.4502
7.5666
34.3150Example 2.
We consider the equation(133)D0+3/2cϕ3/2D0+35/12vt+λ2t−35/144v1/12tD0+11/12vt1/12+2t23/72v−1/6t+2t1/6D0+11/12vt−1/6+1=0,0<t<1,with the boundary conditions(134)v0=D0+12/13v0=D0+35/12v0=ϕ3/2D0+35/12v0′=0,D0+23/12v1=1100Γ32∫01s−1/2D0+17/12vsdA1s+1100∫01s−1D0+11/12vsdA2s+1100∑i=1∞1ii+1D0+11/12v1,where ϕ3/2s=s−1/2s and(135)A1t=211,t∈0,1,1311,t=1,A2t=113,t∈0,1,1413,t=1.Proof.
Let(136)ft,u1,u2=λ2t−35/144u11/12u21/12+2t23/72u1−1/6+2t1/6u2−1/6+1,ϕps=ϕ3/2s, α=3/2, β=35/12, β1=11/12, ν1=12/13, r0=23/12, r1=17/12, r2=11/12, r3=11/12, ρi=1/ii+1, ηi=1, i=1,2,…, η=1, λ1=1/100Γ3/2, λ2=1/100, λ3=1/100, h1t=t−1/2, and h2t=t−1. Then, the problems (133)-(134) can be expressed as the problems (6)-(5). Notice that(137)σ=1Γβ−r0−λ1Γβ−r1∫01sβ−r1−1h1sdA1s−λ2Γβ−r2∫0ηsβ−r2−1h2sdA2s−λ3Γβ−r3∑i=1∞ρiηiβ−r3−1=0.97>0.
We infer that the properties of Green’s function in Lemma 8 are achieved. Let(138)Ft,u1,u2,v1,v2=λt−35/144u11/12u21/12+t23/72v1−1/6+t1/6v2−1/6+1,Gt,u1,u2,v1,v2=λt−35/144u11/12u21/12+t23/72v1−1/6+t1/6v2−1/6.
It is easy to check the conditionsV1–V5 are all satisfied with φl=l1/3 and ϱ=1. On the other hand, let γ=4/9<1/2. Then,(1)
φl=l1/3≥1/ϱlγ−l+lγ=6/5l4/9−29/40l,∀l∈0,1(2)
φl=l1/3≥1/ϱl1/2−l+l1/2=6/5l1/2−29/40l,∀l∈0,1
So, the assumptions of Theorem 2 are satisfied. In the following, we give the graphical simulations and table of the solutionvλ∗ with respect to λ=0.5, 1, 1.5, 2, 3, 10, respectively (Figure 2 and Table 3).Figure 2
Dependence of the solutionvλ∗ upon λ.Table 3
Numerical value of the solutionvλ∗.
t
λ = 0.5
λ = 1
λ = 1.5
λ = 2
λ = 3
λ = 10
0.0
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.1
0.0093
0.0342
0.0759
0.1362
0.3198
5.0454
0.2
0.0351
0.1289
0.2867
0.5140
1.2073
19.0488
0.3
0.0763
0.2805
0.6237
1.1182
2.6262
41.4358
0.4
0.1324
0.4868
1.0825
1.9407
4.5582
71.9187
0.5
0.2030
0.7466
1.6603
2.9765
6.9909
110.3027
0.6
0.2880
1.0589
2.3548
4.2216
9.9151
156.4408
0.7
0.3870
1.4229
3.1642
5.6727
13.3233
210.2155
0.8
0.4998
1.8379
4.0871
7.3272
17.2093
271.5289
0.9
0.6264
2.3034
5.1222
9.1830
21.5678
340.2972
1.0
0.7666
2.8188
6.2684
11.2379
26.3942
416.4475Based on the graphical simulations and table, we obtain the following conclusions:(1)
vλ∗ is continuous with respect to λ∈0,+∞. That is, for any λ0∈0,+∞,(139)supt∈0,1vλ∗t−vλ0∗t=vλ∗−vλ0∗⟶0,asλ⟶λ0.(2)
0<λ1<λ2 implies vλ1∗<vλ2∗.(3)
limλ⟶0+supt∈0,1vλ∗t=limλ⟶0+vλ∗=0,limλ⟶+∞supt∈0,1vλ∗t=limλ⟶+∞vλ∗=+∞.


## 5. Conclusion

In this paper, we introduce the fixed point theorem of mixed monotone operator for finding the uniqueness of positive solution of a class of fractional boundary value problems, which is a generalized form of turbulent flow problem in a porous medium. Two theoretical numerical examples are given to illustrate Theorems 1 and 2; the results then bring us a step closer to research the characters of solutions. Furthermore, as the application of mixed monotone operator operator, further research and discussion are required in practice.



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