# Uniqueness of Successive Positive Solution for Nonlocal Singular Higher-Order Fractional Differential Equations Involving Arbitrary Derivatives

Authors: Qiuyan Zhong; Xingqiu Zhang; Xinyi Lu; Zhengqing Fu
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

In this article, by means of fixed point theorem on mixed monotone operator, we establish the uniqueness of positive solution for some nonlocal singular higher-order fractional differential equations involving arbitrary derivatives. We also give iterative schemes for approximating this unique positive solution.


---

## Body

## 1. Introduction

We are interested in investigating the existence and iterative schemes of the unique positive solution for the following fractional differential equation (FDE):(1)D0+αut+ft,ut,D0+δut=0,0<t<1,D0+δu0=D0+δ+1u0=⋯=D0+δ+n-2u0=0,D0+ϑu1=λ∫0ηhtD0+ιutdt,where D0+α is the standard Riemann-Liouville derivative, n-1<α≤n,n≥3,ϑ-δ≥1,α-ϑ-1>0,0≤δ<1, 0<η≤1, λ is a positive parameter with 0≤λΓ(α-ϑ)∫0η‍h(t)tα-ι-1dt<Γ(α-ι), f∈C(J×R+′×R+′,R+),J=(0,1),R+′=(0,+∞),R+=[0,+∞),0<λΓ(α-ϑ)∫0η‍h(t)tα-ι-1dt<Γ(α-ι), h∈L1[0,1] is nonnegative, f(t,x1,x2) permits singularities at xi=0(i=1,2), and t=0,1.In recent years, fractional calculus and fractional models play more and more significant role in describing a wide spectrum of nonlinear phenomena in natural sciences, engineering, economics, biology, and signal and image processing; see books and monographs [1–3] and references [4–32] to name a few. More and more attention has been paid to nonlocal problem of fractional differential equation because of its wide applications to applied mathematics and physics such as chemical engineering, underground water flow, heat conduction, thermoelasticity, and plasma physics. Under different conjugate type integral conditions such as no parameters, only one or two parameters involved in boundary conditions, [8–16, 33, 34] investigate the existence, uniqueness, and multiplicity of positive solutions for FDEs when f is either continuous or singular. Very recently, in [16], we give two uniqueness results of solution for the following FDE: (2)D0+αut+ft,ut=0,0≤t≤1,u0=u′0=⋯=un-20=0,D0+βu1=λ∫0ηhtD0+γutdt,where f∈C(I×R,R), I=[0,1],R=(-∞,+∞), and h∈L1[0,1] is nonnegative. The whole discussion is based on the Banach contraction map principle and the theory of u0-positive linear operator.Motivated by the above papers, in this article we aim to obtain the uniqueness result of solution for BVP (1) by means of theory on mixed monotone operator. This article admits some new features. First, compared to [6–16] the problem considered in this paper performs more general form since another parameter δ is contained in boundary conditions. Second, the nonlinearity f is not only singular on space variable u but also relative with δ-order derivative of an unknown variable u. Finally, the method used in this paper is different from that in [16].


## 2. Preliminaries and Several Lemmas

Definition 1 (see [3]).
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,∞)→R is given by (3)I0+αxt=1Γα∫0tt-sα-1xsdsprovided the right-hand side is pointwise defined on (0,∞).Definition 2 (see [3]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function x:(0,∞)→R is given by (4)D0+αxt=1Γn-αddtn∫0txst-sα-n+1dswhere n=[α]+1,[α] denotes the integer part of the number α, provided that the right-hand side is pointwise defined on (0,∞).Lemma 3 (see [23]).
(1) Ifx∈L(0,1),ν>σ>0, then (5)I0+νI0+σxt=I0+ν+σxt,D0+σI0+νxt=I0+ν-σxt,D0+σI0+σxt=xt.
(2) Ifν>0,σ>0, then (6)D0+νtσ-1=ΓσΓσ-νtσ-ν-1.Letu(t)=I0+δx(t),x(t)∈C[0,1]. According to the definition of Riemann-Liouville derivative and Lemma 3, we have (7)D0+αut=dndtnI0+n-αut=dndtnI0+n-αI0+δxt=dndtnI0+n-α+δxt=D0+α-δxt;D0+δut=D0+δI0+δxt=xt;D0+δ+1ut=x′t,⋯,D0+δ+n-2ut=xn-2t;D0+β+δut=D0+βxt,D0+γ+δut=D0+γxt.Let β=ϑ-δ,γ=ι-δ. Then, by (7), BVP (1) can reduce to the following modified fractional boundary value problems (MFBVP): (8)D0+α-δxt+ft,I0+δxt,xt=0,0<t<1,x0=x′0=⋯=xn-20=0,D0+βx1=λ∫0ηhtD0+γxtdt.In a similar way, we can transform (8) into the form (1). Thus, MFBVP (8) is equivalent to BVP (1).Lemma 4.
Letu(t)=I0+δx(t),x(t)∈C[0,1]. Then BVP (1) can transform to (8). In addition, if x∈C[0,1] is a positive solution for (8), then I0+δx is a positive solution for BVP (1).Proof.
Substitutingu(t)=I0+δx(t) into (1), we know from Lemma 3 and (7) that (9)D0+αut=D0+α-δxt;D0+δut=xt;D0+δ+1ut=x′t,⋯,D0+δ+n-2ut=xn-2t;D0+ϑut=D0+β+δut=D0+βxt,D0+ιut=D0+γ+δut=D0+γxt.Considering this together with the boundary value conditions, we have (10)x0=D0+δu0=0,xit=D0+δ+iu0=0,i=1,2,⋯,n-2;D0+βx1=D0+ϑu1=λ∫0ηhtD0+ιutdt=λ∫0ηhtD0+γxtdt,and (11)D0+α-δxt=-ft,I0+δxt,xt.Thus, (1) is converted to (8).
Additionally, suppose thatx∈C[0,1] is a positive of (8). Let u(t)=I0+δx(t). Then, by Lemma 3, one gets (12)D0+αut=D0+αI0+δxt=D0+α-δxt=-ft,I0+δxt,xt=-ft,ut,D0+δut.The boundary condition x(0)=x′(0)=⋯=x(n-2)(0)=0, D0+βx(1)=λ∫0η‍h(t)D0+γx(t)dt together with (9) implies that (13)D0+δu0=D0+δ+1u0=⋯=D0+δ+n-2u0=0,D0+ϑu1=λ∫0ηhtD0+ιutdt.That is to say, I0+δx(t) is a positive solution of BVP (1).Remark 5.
Direct computation implies that(14)I0+δtα-δ-1=1Γδ∫0tt-sδ-1sα-δ-1ds=Bδ,α-δΓδtα-1=Γα-δΓαtα-1.The following two lemmas are isomorphic forms of those in [16].Lemma 6 (see [16]).
Assume thatλΓ(α-δ-β)∫0η‍h(t)tα-δ-γ-1dt≠Γ(α-δ-γ). Then for any y∈L(0,1), the unique solution of the boundary value problems (15)D0+α-δxt+yt=0,0<t<1,x0=x′0=⋯=xn-20=0,D0+βx1=λ∫0ηhtD0+γxtdt,solves (16)xt=∫01Gt,sysds,t∈0,1,where (17)Gt,s=G1t,s+G2t,s,(18)G1t,s=tα-δ-11-sα-δ-β-1-t-sα-δ-1Γα-δ,0≤s≤t≤1,tα-δ-11-sα-δ-β-1Γα-δ,0≤t≤s≤1,(19)G2t,s=λΓα-δ-βtα-δ-1Γα-δ-γ-λΓα-δ-β∫0ηhttα-δ-γ-1dt∫0ηhtHt,sdt,(20)Ht,s=tα-δ-γ-11-sα-δ-β-1-t-sα-δ-γ-1Γα-δ,0≤s≤t≤1,tα-δ-γ-11-sα-δ-β-1Γα-δ,0≤t≤s≤1,Here, G(t,s) is called the Green function of BVP (15). Obviously, G(t,s) is continuous on [0,1]×[0,1].Clearly,x is a positive solution of BVP (8) if and only if x∈C[0,1] is a solution of the following nonlinear integral equation: (21)xt=∫01Gt,sfs,I0+δxs,xsds.Lemma 7 (see [16]).
The functionsG1(t,s) and G(t,s) given by (18) and (17), respectively, admit the following properties:
( a 1 )   G 1 ( t , s ) ≥ 1 / Γ ( α - δ ) t α - δ - 1 s ( 1 - s ) α - δ - β - 1 , ∀   t , s ∈ [ 0,1 ];
( a 2 )   G 1 ( t , s ) ≤ 1 / Γ ( α - δ ) ( α - δ - 1 ) s ( 1 - s ) α - δ - β - 1 , ∀   t , s ∈ [ 0,1 ];
( a 3 )   G ( t , s ) ≤ J ( s ) , J ( s ) = 1 / Γ ( α - δ ) ( α - δ - 1 ) s ( 1 - s ) α - δ - β - 1 + λ Γ ( α - δ - β ) / Γ ( α - δ - γ ) - λ Γ ( α - δ - β ) ∫ 0 η ‍ h ( t ) t α - δ - γ - 1 d t · ∫ 0 η ‍ h ( t ) H ( t , s ) d t ,   ∀   t , s ∈ [ 0,1 ];
( a 4 )   1 / ( α - δ - 1 ) t α - δ - 1 J ( s ) ≤ G ( t , s ) ≤ 1 / Γ ( α - δ ) Λ t α - δ - 1 ( 1 - s ) α - δ - β - 1, here Λ=(α-δ-1)+λΓ(α-δ-β)/Γ(α-δ-γ)-λΓ(α-δ-β)∫0η‍h(t)tα-δ-γ-1dt·∫0η‍h(t)tα-δ-γ-1dt, ∀t,s∈[0,1].LetP be a normal cone in a Banach space E and nonzero element e∈P with e≤1. A subset of cone P is given as follows: (22)Qe=x∈P∣thereexistconstantsm,M>0suchthatme≤x≤Me.Definition 8 (see [35]).
Assume thatA:Qe×Qe→Qe. A is said to be mixed monotone if A(x,y) is nondecreasing in x and nonincreasing in y, i.e., if x1≤x2(x1,x2∈Qe) implies A(x1,y)≤A(x2,y) for any y∈Qe, and y1≤y2(y1,y2∈Qe) implies A(x,y1)≤A(x,y2) for any x∈Qe. The element x∗∈Qe is said to be a fixed point of A if A(x∗,x∗)=x∗.Lemma 9 (see [17]).
Suppose thatA:Qe×Qe→Qe is a mixed monotone operator and there exists a constant σ, 0≤σ<1, such that (23)Atx,1ty≥tσAx,y,∀x,y∈Qe,0<t<1.Then A has a unique fixed point x∗∈Qe. Moreover, for any (x0,y0)∈Qe×Qe, (24)xk=Axk-1,yk-1,yk=Ayk-1,xk-1,k=1,2,⋯satisfy (25)xk→x∗,yk→x∗,where (26)xk-x∗=o1-rσk,yk-x∗=o1-rσk,where, r is a constant, 0<r<1, and dependent on x0,y0.


## 3. Main Result

Throughout this paper, we adopt the following assumptions:( H 1 )   f ( t , x 1 , x 2 ) = ϕ ( t , x 1 , x 2 ) + ψ ( t , x 1 , x 2 ), where, ϕ,ψ:J×R+′×R+′→R+ is continuous, ϕ(t,x1,x2) is nondecreasing on xi, and ψ(t,x1,x2) is nonincreasing on xi(i=1,2).( H 2 ) There exists σ∈J such that for xi>0(i=1,2),t,c∈J(27)ϕt,cx1,cx2≥cσϕt,x1,x2;ψt,c-1x1,c-1x2≥cσψt,x1,x2.( H 3 ) (28) 0 < ∫ 0 1 1 - s α - ϑ - 1 s - σ α - 1 ψ s , 1,1 d s < + ∞ ; 0 < ∫ 0 1 1 - s α - ϑ - 1 ϕ s , 1,1 d s < + ∞ .Remark 10.
According to(H2), for any t∈J,c≥1,xi>0(i=1,2), one has (29)ϕt,cx1,cx2≤cσϕt,x1,x2;ψt,c-1x1,c-1x2≤cσψt,x1,x2.Theorem 11.
Assume that(H1)−(H3) hold. Then, the BVP (1) has a unique solution u∗, and there exists a constant D>1 such that (30)Γα-δDΓαtα-1≤u∗t≤DΓα-δΓαtα-1,∀t∈0,1.Moreover, for any u0, we construct a successive sequence (31)uk+1t=I0+δ∫01Gt,sϕs,uks,D0+δuks+ψs,uks,D0+δuksds,k=1,2,⋯,and we have uk-u∗→0 as k→∞; the convergence rate is (32)uk-u∗=I0+δxk-I0+δx∗=o1-rσk,where r is a constant with 0<r<1 and is dependent on u0.Proof.
Lete(t)=tα-δ-1, and we define (33)Qe=x∈C0,1∣1Det≤xt≤Det,where (34)D>max⁡1Γα-δ∫01s1-sα-ϑ-1ρσsσα-1ϕs,1,1+2-σb-σψs,1,1ds1/σ-1,1,2ρ,ΛΓα-δ∫011-sα-ϑ-12σbσϕs,1,1+ρ-σs-σα-1ψs,1,1ds1/1-σ,where b=max{Γ(α-δ)/Γ(α),1},ρ=min{1,Γ(α-δ)/Γ(α)}. We consider the existence of positive solution for (8). For any x,y∈Qe, define an operator T as follows: (35)Tx,yt=∫01Gt,sϕs,I0+δxs,xs+ψs,I0+δys,ysds.It is clear that x is a positive solution of BVP (8) if and only if x∈C[0,1] is a fixed point of the operator T.
First, we are in position to show thatT:Qe×Qe→Qe is well defined. By (H2) and Remark 5, for any x,y∈Qe we have (36)ϕs,I0+δxs,xs≤ϕs,DΓα-δΓαsα-1,Dsα-δ-1≤ϕs,Db+1,Db+1≤Db+1σϕs,1,1≤2σbσDσϕs,1,1,s∈0,1,and (37)ψs,I0+δys,ys≤ψs,Γα-δDΓαsα-1,1Dsα-δ-1≤ψs,ρDsα-1,ρDsα-1≤ρDsα-1-σψs,1,1=ρ-σDσs-σα-1ψs,1,1,s∈0,1.
Considering the fact thatρ/Dsα-1<1, we have from (34), (H2), and Remark 5 that (38)ϕs,I0+δxs,xs≥ϕs,Γα-δDΓαsα-1,1Dsα-δ-1≥ϕs,ρDsα-1,ρDsα-δ-1≥ϕs,ρDsα-1,ρDsα-1≥ρDsα-1σϕs,1,1≥ρσD-σsσα-1ϕs,1,1,s∈0,1,and (39)ψs,I0+δys,ys≥ψs,DΓα-δΓαsα-1,Dsα-δ-1≥ψs,Dbsα-1,Dbsα-δ-1≥ψs,Db+1,Db+1≥Db+1-σψs,1,1≥2-σb-σD-σψs,1,1,s∈0,1.Thus, it follows from (36), (37), Lemma 7, and (H3) that (40)∫01Gt,sϕs,I0+δxs,xs+ψs,I0+δys,ysds≤1Γα-δΛtα-δ-1∫011-sα-δ-β-12σbσDσϕs,1,1+ρ-σDσs-σα-1ψs,1,1ds=1Γα-δΛtα-δ-1∫011-sα-ϑ-12σbσDσϕs,1,1+ρ-σDσs-σα-1ψs,1,1ds<+∞,∀t∈0,1.This means that T:Qe×Qe→P is well defined.
On the other hand, we can easily see from (34) and (40) that (41)Tx,yt≤Dtα-δ-1=Det,∀t∈0,1.At the same time, by (38), (39), and Lemma 7, we know that (42)Tx,yt=∫01Gt,sϕs,I0+δxs,xs+ψs,I0+δys,ysds≥1Γα-δtα-δ-1∫01s1-sα-δ-β-1ρσD-σsσα-1ϕs,1,1+2-σb-σD-σψs,1,1ds=1Γα-δtα-δ-1∫01s1-sα-ϑ-1ρσD-σsσα-1ϕs,1,1+2-σb-σD-σψs,1,1ds≥1Dtα-δ-1=1Det,∀t∈0,1.It follows from (40)-(42) that T:Qe×Qe→Qe is well defined.
Next, we shall prove thatT:Qe×Qe→Qe is a mixed monotone operator. To this end, let x1,x2∈Qe with x1≤x2. For any y∈Qe, it follows from (H1) together with the monotonicity of the operator I0+δ that (43)Tx1,yt=∫01Gt,sϕs,I0+δx1s,x1s+ψs,I0+δys,ysds≤∫01Gt,sϕs,I0+δx2s,x2s+ψs,I0+δys,ysds=Tx2,yt,which implies that T(x,y) is nondecreasing in x for any y∈Qe. In a similar manner, for any x,y1,y2∈Qe with y1≤y2, we have (44)Tx,y1t=∫01Gt,sϕs,I0+δxs,xs+ψs,I0+δy1s,y1sds≥∫01Gt,sϕs,I0+δxs,xs+ψs,I0+δy2s,y2sds=Tx,y2t.This is to say, T(x,y) is nonincreasing in y for any x∈Qe. Thus, T:Qe×Qe→Qe is a mixed monotone operator.
Finally, by(H2), one has (45)Tcx,c-1yt=∫01Gt,sϕs,cI0+δxs,cxs+ψs,c-1I0+δys,c-1ysds≥∫01Gt,scσϕs,I0+δxs,xs+ψs,I0+δys,ysds=cσTx,tt,∀t∈0,1,x,y∈Qe,which means that (23) in Lemma 9 is satisfied. Hence, Lemma 9 guarantees that there exists a unique positive solution x∗ for BVP (8). Let u∗(t)=I0+δx∗(t); then u∗ is the unique positive solution for BVP (1).
In addition, for anyu0(t)=I0+δx0(t)∈Qe, by Lemma 9, constructing a successive sequence (46)xk+1t=∫01Gt,sϕs,I0+δxks,xks+ψs,I0+δxks,xksds,k=1,2,⋯,and uk+1(t)=I0+δxk+1(t), then (47)uk+1t=I0+δ∫01Gt,sϕs,uks,D0+δuks+ψs,uks,D0+δuksds,k=1,2,⋯,and we have uk-u∗=I0+δxk-I0+δx∗→0 as k→∞; the convergence rate is (48)uk-u∗=I0+δxk-I0+δx∗=o1-rσk,where r is a constant with 0<r<1 and is dependent on u0. Moreover, by Remark 5, we get (49)Γα-δDΓαtα-1≤u∗t=I0+δx∗t≤DΓα-δΓαtα-1,∀t∈0,1.


## 4. An Example

Example 1.
Consider the following fractional differential equation integral boundary value problems:(50)D0+9/2ut+ft,ut,D0+1/4ut=0,0<t<1,D0+1/4u0=D0+5/4u0=D0+9/4u0=D0+13/4u0=0,D0+7/4u1=23∫04/5t-3/4D0+3/2utdt,where α=9/2,n=5,δ=1/4,ϑ=7/4,ι=3/2,λ=2/3,h(t)=t-3/4, (51)ft,ut,D0+1/4ut=t-2/3+cos⁡tu1/8t+t-1/4u-1/9t+6t4/5D0+1/4ut1/7+3-2tD0+1/4ut-1/7.By simple computation, we have λΓ(α-ϑ)∫0η‍h(t)tα-ι-1dt=2/3×Γ(11/4)×∫04/5‍t-3/4·t2dt=2/3×1.6084×0.2690=0.2884<2=Γ(α-ι). It is easy to know that (H1) holds for (52)ϕt,x1,x2=t-2/3+cos⁡tx11/8+6t4/5x21/7,ψt,x1,x2=t-1/4x1-1/9+3-2tx2-1/7.At the same time, for any (t,x1,x2)∈J×R+′×R+′ and c∈J, one has (53)ϕt,cx1,cx2=c1/8t-2/3+cos⁡tx11/8+c1/76t4/5x21/7≥c1/7ϕt,x1,x2,ψt,c-1x1,c-1x2=c1/9t-1/4x1-1/9+c1/73-2tx2-1/7≥c1/7ψt,x1,x2. Thus, (H2) is valid for σ=1/7. Notice that ϕ(s,1,1)=s-2/3+cos⁡s+6s4/5,ψ(s,1,1)=s-1/4+(3-2s), and one gets (54)∫011-sα-ϑ-1s-σα-1ψs,1,1ds=∫011-s7/4s-1/2s-1/4+3-2sds≈5.9264<+∞,and (55)∫011-sα-ϑ-1ϕs,1,1ds=∫011-s7/4s-2/3+cos⁡s+6s4/5ds<∫011-s7/4s-2/3+1+6s4/5ds≈3.0765<+∞,which implies that (H3) is also satisfied. Thus, by Theorem 11 we know that BVP (50) has a unique positive solution.
In addition, for any initialu0=I0+1/4x0∈Qe, we construct a successive sequence (56)xk+1t=∫01Gt,sϕs,I0+1/4xks,xks+ψs,I0+1/4xks,xksds,k=1,2,⋯,and uk+1(t)=I0+1/4xk+1(t); then (57)uk+1t=I0+1/4∫01Gt,sϕs,uks,D0+1/4uks+ψs,uks,D0+1/4uksds,k=1,2,⋯,and we have uk-u∗=I0+1/4xk-I0+1/4x∗→0 as k→∞; the convergence rate is (58)uk-u∗=I0+δxk-I0+1/4x∗=o1-r1/7k,where r is a constant with 0<r<1 and is dependent on u0.



---

*Source: 6207682-2018-09-06.xml*