# A Mixed Monotone Operator Method for the Existence and Uniqueness of Positive Solutions to Impulsive Caputo Fractional Differential Equations

Authors: Jieming Zhang; Chen Yang; Chengbo Zhai
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We establish some sufficient conditions for the existence and uniqueness of positive solutions to a class of initial value problem for impulsive fractional differential equations involving the Caputo fractional derivative. Our analysis relies on a fixed point theorem for mixed monotone operators. Our result can not only guarantee the existence of a unique positive solution but also be applied to construct an iterative scheme for approximating it. An example is given to illustrate our main result.


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## Body

## 1. Introduction

The purpose of this paper is to investigate the existence and uniqueness of positive solutions to the following impulsive initial value problem (IVP for short) for Caputo fractional-order differential equations:(1)Dαcy(t)=f(t,y(t),y(t))+g(t,y(t)),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiit∈[0,T],t≠tk,Δy|t=tk=Ik(y(tk-)),y(0)=y0,
where k=1,2,…,m, 0<α≤1, Dαc is the Caputo fractional derivative, f:[0,T]×R×R→R and g:[0,T]×R→R are given functions, Ik:R→R, y0∈R, 0=t0<t1<⋯<tm<tm+1=T, Δy|t=tk=y(tk+)-y(tk-), and y(tk+)=limh→0+y(tk+h) and y(tk-)=limh→0-y(tk+h) represent the right and left limits of y(t) at t=tk.Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, engineering, and biological sciences; see [1–6], for example. In the recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives; see the monographs of Miller and Ross [3], Podlubny [5], Kilbas et al. [6], and the papers [7–21] and the references therein. In these papers, many authors have investigated the existence of positive solutions for nonlinear fractional differential equation boundary value problems. On the other hand, the uniqueness of positive solutions for nonlinear fractional differential equation boundary value problems has been studied by some authors; see [10, 15, 19, 22, 23], for example.In [7], Ahmad and Sivasundaram considered the following impulsive hybrid boundary value problem for nonlinear fractional differential equations:
(2)Dqcx(t)=f(t,x(t)),1<q≤2,Dqcx(t)t∈[0,1]∖{t1,t2,…,tp},Δx(tk)=Ik(x(tk-)),Δx′(tk)=Jk(x(tk-)),tk∈(0,1),k=1,2,…,p,x(0)+x′(0)=0,x(1)+x′(1)=0,
where Dqc is the Caputo fractional derivative, f:[0,1]×R→R is a continuous function, Ik,Jk:R→R, Δx(tk)=x(tk+)-x(tk-) with x(tk+)=limh→0+x(tk+h), and x(tk-)=limh→0-x(tk+h), k=1,2,…,p for 0=t0<t1<t2<⋯<tp<tp+1=1. Based on contraction mapping principle and the Krasnoselskii fixed point theorem, they discussed some existence results for (2).In [24], Benchohra and Slimani concerned the existence and uniqueness of solutions for the following initial value problem for Caputo fractional-order differential equations:
(3)Dαcy(t)=f(t,y(t)),t∈J=[0,T],t≠tk,Δy|t=tk=Ik(y(tk-)),y(0)=y0,
where k=1,…,m, 0<α≤1, f:[0,T]×R→R is a given function, Ik:R→R, y0∈R, 0=t0<t1<⋯<tm<tm+1=T, Δy|t=tk=y(tk+)-y(tk-), and y(tk+)=limh→0+y(tk+h) and y(tk-)=limh→0-y(tk+h) represent the right and left limits of y(t) at t=tk. They gave an existence and uniqueness result for the IVP (3) which was based on the Banach fixed point theorem and also obtained two existence and uniqueness results; the first one was based on the Schaefer fixed point theorem and the second one was based on the nonlinear alternative of the Leray-Schauder type.Different from the above works [7, 24], in this paper, we will use a fixed point theorem for mixed monotone operators to study the existence and uniqueness of positive solutions for the IVP (1). Our result can not only guarantee the existence of unique positive solution but also be applied to construct iterative scheme for approximating it.With this context in mind, the outline of this paper is as follows. In Section2, we will recall certain results from the theory of fractional calculus and some definitions, notations, and results of mixed monotone operators. In Section 3, we will provide some conditions under which the IVP (1) will have a unique positive solution. Finally, in Section 4, we will provide one example, which explicates the applicability of our main result.


## 2. Preliminaries

For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proof of our main theorem.Definition 1 (see [9]).
Letν>0 with ν∈R. Suppose that y:[a,∞)→R. Then the νth Riemann-Liouville fractional integral is defined to be
(4)Da+-νy(t):=1Γ(ν)∫aty(s)(t-s)ν-1ds,
whenever the right-hand side is defined. Similarly, with ν>0 and ν∈R, one defines the νth Riemann-Liouville fractional derivative to be
(5)Da+νy(t):=1Γ(n-ν)dndtn∫aty(s)(t-s)ν+1-nds,
where n∈N is the unique positive integer satisfying n-1≤ν<n and t>a.Definition 2 (see [11]).
For a functiony given on the interval [a,b], the Caputo fractional-order derivative of order ν of y is defined by
(6)Da+νcy(t):=1Γ(n-ν)∫aty(n)(s)(t-s)ν+1-nds,
where n=[ν]+1.
In the sequel, we present some basic concepts in the ordered Banach spaces for completeness and one fixed point theorem which will be used later. For convenience of readers, we suggest that one refers to [23, 25, 26] for details.
Suppose that(E,∥·∥) is a real Banach space which is partially ordered by a cone P⊂E. That is, x≤y if and only if y-x∈P. If x≤y and x≠y, then we denote that x<y or y>x. By θ we denote the zero element of E. Recall that a nonempty closed convex set P⊂E is a cone if it satisfies (i)  x∈P, λ≥0⇒λx∈P; (ii)  x∈P, -x∈P⇒x=θ.
P is called normal if there exists a constant N>0 such that, for all x, y∈E, θ≤x≤y implies that ∥x∥≤N∥y∥; in this case N is called the normality constant of P. If x1, x2∈E, the set [x1,x2]={x∈E∣x1≤x≤x2} is called the order interval between x1 and x2.Definition 3 (see [23, 25, 26]).
Consider thatA:P×P→P is said to be a mixed monotone operator if A(x,y) is increasing in x and decreasing in y; that is, ui, vi(i=1,2)∈P, u1≤u2, and v1≥v2 imply that A(u1,v1)≤A(u2,v2). Element x∈P is called a fixed point of A if A(x,x)=x.Lemma 4 (see [25]).
LetP be normal and let A:P×P→P be a mixed monotone operator. Suppose that(a)
there existv>θ and c>0, such that θ<A(v,θ)≤v and A(θ,v)≥cA(v,θ),(b)
for any0<a<b<1, there exists η=η(a,b)>0, such that
(7)A(tx,t-1y)≥t(1+η)A(x,y),∀a≤t≤b,θ≤y≤x≤v.Then A has a unique fixed point x* in [θ,v] with x*>θ. Moreover, for any initial values x0, y0∈[θ,v], constructing successively the sequences
(8)xn=A(xn-1,yn-1),yn=A(yn-1,xn-1),n=1,2,…,
one has xn→x* and yn→x* as n→∞.


## 3. Main Result

In this section, we apply Lemma4 to study the IVP (1) and then we obtain a new result on the existence and uniqueness of positive solutions. The existence and uniqueness result is relatively new to the fractional differential equations in this literature.In our considerations we will work in the Banach spaceC[0,T]={x:[0,T]→Riscontinuous} with the standard norm ∥x∥=sup{|x(t)|:t∈[0,T]}. Notice that this space can be equipped with a partial order given by
(9)x,y∈C[0,T],x≤y⟺x(t)≤y(t)fort∈[0,T].
Set P={x∈C[0,T]∣x(t)≥0,t∈[0,T]}, the standard cone. It is clear that P is a normal cone in C[0,T] and the normality constant is 1. Consider the following set of functions:
(10)PC[0,T]={y:[0,T]⟶R:y∈C(tk,tk+1],k=0,1,…,mandthereexisty(tk-)andy(tk+),k=1,2,…,mwithy(tk-)=y(tk)}.
Then PC[0,T] is a Banach space with the norm
(11)∥y∥PC=supt∈[0,T]|y(t)|.
Denote PC by PC={x∈PC[0,T]∣x(t)≥0,t∈[0,T]}; then PC is a standard and normal cone in PC[0,T].Lemma 5 (see [24]).
Let0<α≤1 and let h:J→R be continuous. A function y∈PC[0,T] is a solution of the fractional integral equation:
(12)y(t)={y0+1Γ(α)∫0t(t-s)α-1h(s)ds,ift∈[0,t1],y0+1Γ(α)∑i=1k∫ti-1ti(ti-s)α-1h(s)ds+1Γ(α)∫tkt(t-s)α-1h(s)ds+∑i=1kIi(y(ti-))ift∈(tk,tk+1],
where k=1,…,m, if and only if y is a solution of the fractional IVP
(13)Dαcy(t)=h(t),t∈[0,T],t≠tk,Δy|t=tk=Ik(y(tk-)),k=1,…,m,y(0)=y0.Remark 6.
The concept of solutions for fractional differential equations has been argued extensively; see [7, 18, 21, 24] and the references therein. In [18], the authors gave a counterexample to show that the result in Lemma 5 is not reasonable. However, the counterexample is inappropriate. From the very recent paper [21], the approach for finding the solution of impulsive fractional differential equations in [18] is inappropriate and some arguments like Lemma 5 are plausible.
For the convenience, we list the following conditions:(H1)
f:[0,T] × [0,∞) × [0,∞)→[0,∞) is continuous and g:[0,T] × [0,∞)→[0,∞) is continuous;(H2)
f(t,u,v) is increasing in u∈[0,∞) for fixed t∈[0,T] and v∈[0,∞) and decreasing in v∈[0,∞) for fixed t∈[0,T] and u∈[0,∞), and g(t,v) is decreasing in v∈[0,∞) for fixed t∈[0,T];(H3)
for any0<p<q<1, there exist β1, β2, and β3∈(0,∞) which depend on q, such that
(14)f(t,λu,λ-1v)≥λ1-λβ1f(t,u,v),∀t∈[0,T],p≤λ≤q,u,v∈[0,∞),iiiiiiiiiig(t,λ-1v)≥λ1-λβ2g(t,v),∀t∈[0,T],p≤λ≤q,v∈[0,∞),iiiiiiiiiiiiiiiiiiiiiiIk(λu)≥λ1-λβ3Ik(u),∀k=1,2,…m,p≤λ≤q,u∈[0,∞);(H4)
the functionsIk:[0,∞)→R are continuous and nondecreasing and there exists a constant M*>0 such that |Ik(u)|≤M* for each u∈[0,∞), k=1,…,m;(H5)
y0>mM*>0.Theorem 7.
Assume that(H1)–(H5) are satisfied and there exists R>0 such that
(15)y0+(m+1)(M1+M2)Γ(α+1)Tα+mM*≤R,
where M1=maxt∈[0,T]f(t,R,0), M2=maxt∈[0,T]g(t,0).
Then the IVP (1) has a unique positive solution u* in P[θ,R]={y∈PC∣0≤y(t)≤R,t∈[0,T]} with u*(t)≢0, t∈[0,T]. Moreover, for any initial values x0, y0∈PC, constructing successively the sequences
(16)xn+1(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,xn(s),yn(s))+g(s,yn(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,xn(s),yn(s))+g(s,yn(s))]ds+∑0<tk<tIk(xn(tk-)),yn+1(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,yn(s),xn(s))+g(s,xn(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,yn(s),xn(s))+g(s,xn(s))]ds+∑0<tk<tIk(yn(tk-)),
where n=0,1,2,…, then one has ∥xn-u*∥→0 and ∥yn-u*∥→0 as n→∞.Proof.
To begin with, from Lemma5, the IVP (1) has an integral formulation given by
(17)u(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,u(s),u(s))+g(s,u(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,u(s),u(s))+g(s,u(s))]ds+∑0<tk<tIk(u(tk-)).
Define an operator A:PC×PC→PC[0,T] by
(18)A(u,v)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,u(s),v(s))+g(s,v(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,u(s),v(s))+g(s,v(s))]ds+∑0<tk<tIk(u(tk-)).
It is easy to prove that u is the solution of the IVP (1) if and only if u is the fixed point of A. From (H1), (H4), and (H5), we know that
(19)A(u,v)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,u(s),v(s))+g(s,v(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,u(s),v(s))+g(s,v(s))]ds+∑0<tk<tIk(u(tk-)),≥y0-mM*>0,
for any u,v∈PC, t∈[0,T]. So, A:PC×PC→PC. In the sequel we check that A satisfies all assumptions of Lemma 4.
Firstly, we prove thatA is a mixed monotone operator. In fact, for ui,vi∈PC, i=1,2 with u1≥u2, v1≤v2, we know that u1(t)≥u2(t), v1(t)≤v2(t), and t∈[0,T], and by (H2) and Definition 3, we have that
(20)A(u1,v1)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,u1(s),v1(s))+g(s,v1(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,u1(s),v1(s))+g(s,v1(s))]ds+∑0<tk<tIk(u1(tk-))≥y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,u2(s),v2(s))+g(s,v2(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,u2(s),v2(s))+g(s,v2(s))]ds+∑0<tk<tIk(u2(tk-))=A(u2,v2)(t).
That is, A(u1,v1)≥A(u2,v2).
Next, we show thatA satisfies the condition (a) of Lemma 4. Now, we set v(t)≡R(t∈[0,T]). Then, v>θ.
On the one hand, from(H1) and (H5), we have
(21)A(v,θ)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,R,0)+g(s,0)]ds+1Γ(α)∫tkt(t-s)α-1[f(s,R,0)+g(s,0)]ds+∑0<tk<tIk(R),≥y0-mM*>0.
That is to say, A(v,θ)>θ. From (15), for any t∈[0,T], we obtain
(22)A(v,θ)(t)≤y0+m(M1+M2)Γ(α+1)Tα+(M1+M2)Γ(α+1)Tα+mM*≤R=v(t),
That is, A(v,θ)≤v.
On the other hand, we takec∈(0,1) such that
(23)c≤min{m1+m2M1+M2,m3M*},
where m1=mint∈[0,T]f(t,0,R), m2=mint∈[0,T]g(t,R), and m3=mink∈{1,…,m}Ik(0). For any t∈[0,T],
(24)A(θ,v)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,0,R)+g(s,R)]ds+1Γ(α)∫tkt(t-s)α-1[f(s,0,R)+g(s,R)]ds+∑0<tk<tIk(0)≥y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1(m1+m2)ds+1Γ(α)∫tkt‍(t-s)α-1(m1+m2)ds+∑0<tk<tm3≥y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1c(M1+M2)ds+1Γ(α)∫tkt(t-s)α-1c(M1+M2)ds+∑0<tk<tcM*≥cy0+cΓ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,R,0)+g(s,0)]ds+cΓ(α)∫tkt‍(t-s)α-1[f(s,R,0)+g(s,0)]ds+c∑0<tk<tIk(R)=cA(v,θ)(t).
That is to say, A(θ,v)≥cA(v,θ). Therefore, condition (a) of Lemma 4 holds.
Finally, we show thatA satisfies condition (b) of Lemma 4. Let 0<a<b<1 and
(25)η=min{1b-1,11-aβ1-1,11-aβ2-1,11-aβ3-1}>0.
For any a≤λ≤b, x,y∈PC, and t∈[0,T], by (H3) we have
(26)A(λx,λ-1y)(t)=y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[f(s,λx(s),λ-1y(s))+g(s,λ-1y(s))]ds+1Γ(α)∫tkt(t-s)α-1[f(s,λx(s),λ-1y(s))+g(s,λ-1y(s))]ds+∑0<tk<tIk(λx(tk-))≥y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[λ1-λβ1f(s,x(s),y(s))+λ1-λβ2g(s,y(s))]ds+1Γ(α)∫tkt(t-s)α-1[λ1-λβ1f(s,x(s),y(s))+λ1-λβ2g(s,y(s))]ds+∑0<tk<tλ1-λβ3Ik(x(tk-))≥y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[λ1-aβ1f(s,x(s),y(s))+λ1-aβ2g(s,y(s))]ds+1Γ(α)∫tkt(t-s)α-1[λ1-aβ1f(s,x(s),y(s))+λ1-aβ2g(s,y(s))]ds+∑0<tk<tλ1-aβ3Ik(x(tk-))≥b(1+η)y0+1Γ(α)×∑0<tk<t∫tk-1tk(tk-s)α-1[λ(1+η)f(s,x(s),y(s))+λ(1+η)g(s,y(s))]ds+1Γ(α)∫tkt(t-s)α-1[λ(1+η)f(s,x(s),y(s))+λ(1+η)g(s,y(s))]ds+∑0<tk<tλ(1+η)Ik(x(tk-))≥λ(1+η)A(x,y)(t).
That is to say, A(λx,λ-1y)≥λ(1+η)A(x,y), for all a≤λ≤b, x,y∈PC. Hence, condition (b) of Lemma 4 is satisfied. So, the conclusion of Theorem 7 follows from Lemma 4.


## 4. An Example

We present one example to illustrate Theorem7.Example 1.
Consider the following fractional IVP:(27)cD1/5y(t)=τ(t)+δ(t)+y(t)1+y(t)+21+y(t)+2,t∈[0,1],t≠13,Δy|t=(1/3)=y(1/3-)4+y(1/3-)+1,y(0)=2,
where τ, δ∈C([0,1,0,∞)).
In this example,(28)α=15,T=1,t1=13,m=1,y0=2,f(t,x,y)=τ(t)+1+x1+x+11+y,g(t,y)=δ(t)+1+11+y,I1(x)=1+x4+x.
Then, f(t,x,y)∈C([0,1]×[0,∞)×[0,∞), [0,∞)) is continuous and increasing in x∈[0,∞) for fixed t∈[0,1] and y∈[0,∞) and decreasing in y∈[0,∞) for fixed t∈[0,1] and x∈[0,∞). And g(t,y)∈C([0,1]×[0,∞), [0,∞)) is continuous and decreasing in y∈[0,∞) for fixed t∈[0,1]. Moreover, for any 0<p<q<1, we take
(29)β1≥logq(1-3q1+2q),β2≥logq(1-2q1+q),β3≥logq(1-2q1+q).
For any p≤λ≤q, t∈[0,1], and x,y∈[0,∞), we have
(30)f(t,λx,λ-1y)=τ(t)+1+λx1+λx+λλ+y≥λq(τ(t)+1)+λ(x1+x+11+y).
We show that
(31)f(t,λx,λ-1y)≥λ1-λβ1f(t,x,y),
if and only if we can prove that
(32)1q(τ(t)+1)+x1+x+11+y≥11-λβ1(τ(t)+1+x1+x+11+y).
That is to say,
(33)(τ(t)+1)(1q-11-λβ1)≥(11-λβ1-1)(x1+x+11+y).
From (29), we get
(34)λβ1≤qβ1≤1-3q1+2q,
and then
(35)1q-11-λβ1≥2(11-λβ1-1).
Hence, (33) is satisfied. So, (31) holds.
Similarly, we obtain that(36)g(t,λ-1y)=δ(t)+1+λλ+y≥λq(δ(t)+1)+λ1+y≥λ1-λβ2(δ(t)+1+11+y)=λ1-λβ2g(t,y),I1(λx)=1+λx4+λx≥λq+λx4+x≥λ1-λβ3(1+x4+x)=λ1-λβ3I1(x).
Further, I1(x) is continuous, increasing in x∈[0,∞), and
(37)|I1(x)|=|x4+x|≤1≡M*,∀x∈[0,∞).
And, we also have y0=2>1=mM*>0.
Finally, owing toτ,δ∈C([0,1],[0,∞)), we have that
(38)τ*=maxt∈[0,1]τ(t),δ*=maxt∈[0,1]δ(t).
We take R>0 such that
(39)3+2(5+τ*+δ*)Γ((1/5)+1)≤R.
And thus,
(40)M1=maxt∈[0,1]f(t,R,0)=maxt∈[0,1](τ(t)+1+R1+R+1)≤3+τ*,M2=maxt∈[0,1]g(t,0)=maxt∈[0,1](δ(t)+1+1)=δ*+2.
Then,
(41)y0+(m+1)(M1+M2)Γ(α+1)Tα+mM*≤2+2Γ((1/5)+1)(3+τ*+2+δ*)+1=3+2(5+τ*+δ*)Γ((1/5)+1)≤R.
Hence, (15) holds. Therefore, all the conditions of Theorem 7 are satisfied. An application of Theorem 7 implies that problem (27) has a unique positive solution.



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