# Existence of Positive Solutions for a Class of Delay Fractional Differential Equations with Generalization to N-Term

Authors: Azizollah Babakhani; Dumitru Baleanu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We established the existence of a positive solution of nonlinear
fractional differential equations𝔏(D)[x(t)−x(0)]=f(t,xt), t∈(0,b] with finite delay x(t)=ω(t), t∈[−τ,0], where limt→0f(t,xt)=+∞, that is, f is singular at t=0 and xt∈C([−τ,0],ℝ≥0). The operator of 𝔏(D) involves the
Riemann-Liouville fractional derivatives. In this problem, the initial conditions with
fractional order and some relations among them were considered. The analysis rely
on the alternative of the Leray-Schauder fixed point theorem, the Banach fixed point
theorem, and the Arzela-Ascoli theorem in a cone.


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## Body

## 1. Introduction

Fractional differential equations have gained considerable importance due to their varied applications in viscoelasticity, electroanalytical chemistry, and many other physical problems [1–7]. So far there have been several fundamental works done on the fractional derivative and fractional differential equations [1–4, 6]. These works are in the introduction of the theory of the fractional derivative and fractional differential equations and provide a systematic understanding of the fractional calculus such as the existence and the uniqueness of some analytic methods for solving fractional differential equations, namely, the Green’s function method, the Mellin transform method, and the power series.The existence of positive solutions for fractional differential equations with delay has been carried out by various researchers [8–17]. In [8] the authors have investigated the following type of fractional differential equations:(1.1)L(D)[x(t)-x(0)]=f(t,xt),t∈(0,T],x(t)=ϕ(t)≥0,t∈[-w,0],
where(1.2)L(D)=Dαn-∑j=1n-1pj(t)Dαn-j,0<α1<⋯<αn<1,pj(t)=∑k=0Njajktk,pj(2m)(t)≥0,pj(2m+1)(t)≤0,m=0,1,…,[Nj2],j=1,2,…,n-1,
and Dαj is the standard Riemann-Liouville fractional derivative, T>0, w>0, ϕ∈C=C([-w,0],ℝ+) and f:I×C→ℝ+ is a given continuous function, I=[0,T].As a pursuit of this in this paper, we discuss the existence of positive solutions for initial nonlinear fractional differential equations with finite delay,(1.3)L(D)[x(t)-x(0)]=f(t,xt),t∈(0,b],x(t)=ω(t),t∈[-τ,0],
with initial conditions x(0)=x0, [Dα-n+1x(t)]t=0=xn-1, and  [Dα-n+jx(t)]t=0=xn-j, where ℒ(D)=Dα-∑j=1n-1ajDα-j, aj>0, for all j,j=1,2,…,n-1, n-1<α≤n, n∈ℕ, f:(0,b]×C→[0,+∞) is a given continuous function so that limt→0f(t,xt)=+∞ (i.e., f is singular at t=0), where C is the space of continuous functions from [-τ,0] to ℝ≥0 and xt∈C defined by xt(s)=x(t+s) for each s∈[-τ,0). In the initial conditions of (1.1), we also assume that(1.4)(i)xn-1≥0,(ii)xn-j≥∑k=1j-1akxk+n-j,∀j=2,3,…,n-1.
The paper is organized as follows.In Section2, we provide some definitions about the fractional derivatives and the fractional integrals as well as we list their basic properties. Required topics of functional analysis such as Banach fixed point theorem and Leray-Schauder Theorem were also introduced. Section 3 deals with existence of a positive solution theorem and it gives an explainable example. The unique positive solution theorem with an explainable example has been discussed in Section 4.


## 2. Preliminaries

Preliminaries from fractional calculus [1, 4, 6] and functional analysis [17] are outlined below.LetE be a real Banach space with a cone K.  K introduces a partial order ≤ in E as  x≤y if y-x∈K.Definition 2.1.
Forx,y∈E, the order interval 〈x,y〉 is defined as
(2.1)〈x,y〉={z∈E:x≤z≤y}.Theorem 2.2 (Leray-Schauder Theorem [17]).
LetE be a Banach space with C⊆E closed and convex. Assume U is relatively open subset of C with 0∈U and F:U¯→C is a continuous, compact map. Then, either (i)
F has fixed point in U¯ or(ii)
there existsu∈∂U and γ∈(0,1) with u=γFu.Theorem 2.3 (Banach fixed point theorem [17]).
LetK be a closed subspace of a Banach space Ω. Let F be a contraction mapping with Lipschitz constant l(<1) from K to itself. Then, F has a unique fixed point x* in K. Moreover, if x0 is an arbitrary point in K and {xn} is defined by xn+1=Fxn,(n=0,1,2,…), then limn→∞xn=x*∈K and d(xn,x*)≤(kn/(1-k))d(x1,x0).The complete Gamma functionΓ(α) plays an important role in the theory of fractional integral and derivatives. A comprehensive definition of Γ(α) is provided by the Euler limit as (2.2)Γ(α)=lim⁡n→∞n!nαα(α+1)⋯(α+n).Ifα>0, then Γ(α) has the following familiar integral representation:(2.3)Γ(α)=∫0+∞tα-1e-tdt.
In this paper, the Beta function B(α,β) is used. We notice that B(α,β) is closely related to the Gamma function. If α,β>0, then it has the integral representation (2.4)B(α,β)=∫01tα-1(1-t)β-1dt=Γ(α)Γ(β)Γ(α+β).The definitions of Riemann-Liouville fractional derivative/integral and their properties are given bellow.Definition 2.4.
Letx:[a,b]→ℝ and x∈L1[a,b]; then the expression
(2.5)Ia+αx(t)=1Γ(α)∫at(t-s)α-1x(t)dt,x>a,
is called a left-sided factional integrals of order α.Definition 2.5.
Letn be a positive integer number and α∈(n-1,n]. Then the left-sided fractional derivative of a function x:[a,b]→ℝ is defined as
(2.6)Da+αx(t)=Dn(In-αx(t)),t∈[a,b].
We denote Da+αx(t) as Daαx(t) and Ia+αx(t) as Iaαx(t). Further, Dαx(t) and Iαx(t) are referred as Dαx(t) and Iαx(t), respectively.
If the fractional derivativeDaβx(t) is integrable; then
(2.7)Iaα(Daβx(t))=Iaα-βx(t)-{Ian-βx(t)}t=a(t-a)α-nΓ(α-n+1)-∑j=1n-1{Daβ-jx(t)}t=a(t-a)α-jΓ(α-j+1),
where n-1≤β<n,β≤α, and n∈ℕ [9]. Further, if x∈C[a,b], then {Ian-βx(t)}t=a=0  and (1.4) reduces to
(2.8)Iaα(Daβx(t))=Iaα-βx(t)-∑j=1n-1{Daβ-jx(t)}t=a(t-a)α-jΓ(α-j+1).


## 3. Existence Theorem

In this section, we show that the initial value problem (1.1) under the conditions among the initial value (i.e., (1.3)) has a positive solution.Lemma 3.1.
Letg:(0,b]→ℝ be a continuous function and limt→0+g(t)=+∞. If there exits σ∈(0,1) such that σ<α and by letting tσg(t) be a continuous function on [0,b], then H(t)=Iαtσg(t) is continuous [0,b], where n-1<α<n, n∈ℕ.Proof.
Let us considerL=maxtσg(t), t∈[0,b]. For all t∈[0,b) and for given all ϵ>0,
(3.1)|H(t+ϵ)-H(t)|=1Γ(α)|∫0t+ϵ(t+ϵ-s)α-1ds-∫0t(t-s)α-1ds|=1Γ(α)|∫0t+ϵ(t+ϵ-s)α-1s-σsσg(s)ds-∫0t(t-s)α-1s-σsσg(s)ds|≤LΓ(α)|∫0t+ϵ(t+ϵ-s)α-1s-σds-∫0t(t-s)α-1s-σds|=LΓ(α)B(1-σ,α)|(t+ϵ)α-σ-tα-σ|.
Hence we conclude that limϵ→0|H(t+ϵ)-H(t)|=0. We notice that a similar result is conclusion for |H(b-ϵ)-H(b)|.In the following, we want to show that (1.1) is equivalent to an integral equation.Theorem 3.2.
Suppose thatf:(0,b]×C→[0,+∞) is a given continuous function so that limt→0f(t,xt)=+∞ (i.e., f is singular at t=0), where C is the space of continuous functions from [-τ,0] to ℝ≥0 and xt∈C defined by xt(s)=x(t+s) for each s∈[-τ,0]. If there exists σ∈(0,1) such that 0<σ<α∈(n-1,n) and tσf(t,xt) is a continuous function on [0,b], then the fractional differential equation
(3.2)(Dα-∑j=1n-1ajDα-j)[x(t)-x(0)]=f(t,xt),t∈(0,b],
is equivalent to the integral equation
(3.3)x(t)=x(0)+λ(t)+L(I)[x(t)-x(0)]+Iαf(t,xt),t∈[0,b],
where
(3.4)λ(t)=∑j=1n-1λjtα-j,λj=1Γ(α-j+1)(xj-∑k=1n-j-1akxk+j),j=1,…,n-2,λn-1=1Γ(α-n+1),L(I)=∑j=1n-1ajIjx(t).Proof.
From (3.2), we have
(3.5)Iα{Dα[x(t)-x(0)]}-∑j=1n-1ajIα{Dα-j[x(t)-x(0)]}=Iαf(t,xt),t∈(0,b].
By using (2.7), we conclude that
(3.6)Iα{Dα[x(t)-x(0)]}=[x(t)-x(0)]-∑j=1n-1xjtα-jΓ(α-j+1),
and for k=1,2,…,n-1,
(3.7)Iα{Dα-k[x(t)-x(0)]}=Ik[x(t)-x(0)]-∑j=1n-k-1xj+ktα-jΓ(α-j+1).
Note that, by Lemma 3.1, Iαf(t,xt)=Iα(tσt-σf(t,xt) exists and DαIαf(t,xt)=f(t,xt), as t-σf(t,xt) is continuous and Iαf(t,xt)∈C[0,b]. In view of (3.5), (3.6), and (3.7), (3.2) is equivalent to the following integral equation:
(3.8)x(t)=x(0)+λ(t)+L(I)[x(t)-x(0)]+Iαf(t,xt),t∈(0,b],
where
(3.9)λ(t)=∑j=1n-1λjtα-j,λj=1Γ(α-j+1)(xj-∑k=1n-j-1akxk+j),j=1,…,n-2,λn-1=1Γ(α-n+1),L(I)=∑j=1n-1ajIjx(t).
The proof is completed. Therefore, by Theorem 3.2, the other expression of (1.1) is given as follows:
(3.10)x(t)={x(0)+λ(t)+L(I)[x(t)-x(0)]+Iαf(t,xt),t∈(0,b],x(t)=ω(t),t∈[-τ,0],
where λ(t) and ℒ(I) are mentioned in above.
Lety:[-τ,b]→[0,+∞) be function defined by
(3.11)y(t)={ω(0),t∈[0,b],ω(t)≥0,t∈[-τ,0],
for each z∈C([0,b],ℝ) with z(0)=0; we denote by z¯ the function define by
(3.12)z¯(t)={z(t),t∈[0,b],0,t∈[-τ,0].
We can decompose x(·) as x(t)=z¯(t)+y(t),t∈[-τ,b], which implies that xt=z¯t+yt, t∈[0,b]. Hence by Theorem 3.2, (1.1) is equivalent to the following integral equation:
(3.13)z(t)=λ(t)+L(I)z(t)+Iαf(t,z¯t+yt),t∈[0,b].
Set Ω={z∈C([0,b],ℝ),z(0)=0}, and for each z∈Ω, let ∥z∥b be the seminorm in Ω defined by
(3.14)‖z‖b=∥z(0)∥+∥z∥=∥z∥=sup⁡{|z(t)|:t∈[0,b]}.Ω is a Banach space with norm ∥·∥b. Let K be a cone of Ω. K={z∈Ω:z(t)≥0,t∈[0,b]} and
(3.15)K*={x∈C([-τ,b],R≥0):x(t)=ω(t)≥0,t∈[-τ,0]}.
Define the operator F:K→K by
(3.16)Fz(t)=λ(t)+L(I)z(t)+Iαf(t,z¯t+yt),t∈[0,b].Theorem 3.3.
Suppose thatf(t,xt), t∈(0,b], xt∈C, is a continuous function and limt→0+f(t,·)=+∞. If there exits σ∈(0,1) such that 0<σ<α∈(n-1,n) and tσf(t,xt) is a continuous function on [0,b], then the operator F, defined as (3.12), maps bounded set into bounded sets in K, continuous and completely continuous.Proof.
For allu∈K, since Fu(t)=λ(t)+ℒ(I)u(t)+Iαf(t,u¯t+yt) by Lemma 3.1 and the nonnegativeness of f, we obtain F:K→K.
Sincetσf(t,xt) is continuous on [0,b]×[0,+∞), there exists a positive constant N such that tσf(t,xt)≤N. Hence
(3.17)Iαf(t,z¯t+yt)=Iα(t-σtσf(t,xt))≤NIαt-σ=NΓ(α)∫0t(t-s)α-1s-σds=Ntα-σΓ(α)B(1-σ,α)≤Nbα-σΓ(α)B(1-σ,α).
Let G⊂K be bounded, that is, there exists a positive constant L such that ∥z∥b≤L, for all z∈C. In view of (3.13), for each z∈G, we obtain
(3.18)∥Fz(t)∥≤|λ(t)|+|L(I)z(t)|+|Iαf(t,z¯t+yt)|≤∥λ(t)∥+‖z‖b∑j=1najbjΓ(j+1)+Nbα-σΓ(α)B(1-σ,α)≤∥λ(t)∥+L∑j=1najbjΓ(j+1)+Nbα-σΓ(α)B(1-σ,α).
Thus, F(K) is bounded. In the following text, we would like to show that F is continuous. Let v0∈K and ∥v0∥=c0, if v∈K and ∥v-v0∥<l, then ∥v∥<l+c0; by the continuity of tσf(t,z¯t+yt), we know that tσf(t,z¯t+yt) is uniformly continuous on [0,b]×[0,c]. Thus, for all ϵ>0 there exists a δ>0 such that
(3.19)|tσf(t,u¯t+yt)-tσf(t,v¯t+yt)|<ϵ,
for all t∈[0,b] and u,v∈[0,c] with |u-v|<δ. Obviously, if ∥u-v∥<δ, then v0(t),v(t)∈[0,c] and |v(t)-v0(t)|<δ for each t∈[0,b]. Hence, we have
(3.20)|tσf(t,v¯t+yt)-tσf(t,v0¯t+yt)|<ϵ,
for all t∈[0,b] and v∈K with ∥v-v0∥<δ. For all t∈[0,b], let u,v∈K, and |u(t)-v(t)|<δ, we choose δ≤ϵ(∑j=1n-1(xjbj/Γ(j+1)))-1. By using (3.20), we get
(3.21)|Fu(t)-Fv(t)|≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+|Iα(f(t,u¯t+yt)-f(t,v¯t+yt))|=‖u-v‖b∑j=1n-1ajbjΓ(j+1)+|Iα(t-σtσf(t,u¯t+yt)-t-σtσf(t,v¯t+yt))|≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+1Γ(α)max⁡0≤t≤b∫0t(t-s)α-1s-σ|sσf(t,u¯t+yt)-sσf(t,v¯t+yt)|ds≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+ϵΓ(α)max⁡0≤t≤b∫0t(t-s)α-1s-σds=‖u-v‖b∑j=1n-1ajbjΓ(j+1)+ϵΓ(α)max⁡0≤t≤btα-σB(1-σ,α)≤(1+bα-σ)B(1-σ,α)ϵ.
Finally, we want to prove that the operator F is equicontinuous. Let G⊆K be bounded, that is, three exits a positive constant l such that ∥u∥≤l, for all u∈G. Suppose, u∈K,t,r∈[0,b] and t<r. For a given ϵ>0, there exists δ>0, so that if |t-r|<δ, then
(3.22)|Fu(t)-Fu(r)|≤|λ(t)-λ(r)|+|L(I)u(t)-L(I)u(r)|+|Iαf(t,u¯t+yt)-Iαf(r,u¯r+yr)|≤∑j=1n-1λj|tα-j-rα-j|+∑j=1n-1lajΓ(j)|∫0t(t-s)j-1-(r-s)j-1ds|+∑j=1n-1lajΓ(j)∫tr(r-s)j-1ds+1Γ(α)|∫0t(t-s)α-1s-σsσf(s,u¯s+ys)ds-∫0r(r-s)α-1s-σsσf(s,u¯s+ys)ds|≤2∑j=1n-1λjbα-j+l∑j=1n-1ajtjΓ(j+1)+l∑j=1n-1ajΓ(j+1)[|r-t|j+|rj-tj|],+NB(1-σ,α)Γ(α)|tα-σ-rα-σ|≤2∑j=1n-1λjbα-j+l∑j=1n-1ajbjΓ(j+1)+2l∑j=1n-1ajΓ(j+1)+l∑j=1n-1ajΓ(j+1)|r-t|j+2NB(1-σ,α)Γ(α)bα-σ.
Set Δ=max{3Δ1/ϵ,3Δ2/ϵ,Δ3,Δ4,…,Δn+1}, where
(3.23)Δ1=∑j=1n-1(2λjbα-j+lajbjΓ(j+1)+2lajΓ(j+1)),Δ2=2NΓ(α)B(1-σ,α)bα-σ,Δj+2=l∑j=1n-1ajΓ(j+1),j=1,2,…,n-1.
Case 1.
If0<|r-t|<1<b, we choose δ=ϵ/(3(n-1)). Hence,
(3.24)|Fu(t)-Fu(r)|≤ϵ3Δ+ϵ3Δ+Δ∑j=1n-1|r-t|j≤ϵ3Δ+ϵ3Δ+(n-1)Δδ<Δϵ.Case 2.
If0<1≤|r-t|<b, we choose δ={ϵ/(3(n-1))}1/(n-1). Hence,
(3.25)|Fu(t)-Fu(r)|≤ϵ3Δ+ϵ3Δ+Δ∑j=1n-1|r-t|j≤ϵ3Δ+ϵ3Δ+(n-1)Δδn-1<Δϵ.
Therefore, F(G) is equicontinuous. The Arzela-Ascoli Theorem implies that F(G)¯ is compact. Thus, F:K→K is completely continuous.Theorem 3.4.
With all the details of  (1.1)-(1.3), further, if σ exists in (0,1) such that 0<σ<α∈(n-1,n), n∈ℕ, and tσf(t,xt) is a continuous function on [0,b], then (1.1) has at least one positive solution x*∈K*, satisfying ∥x*∥≤max{∥ω∥,h}, where h=2Λ/(1-Λ)+1 and Λ is  a positive constant which is observed in the proof of the theorem.Proof.
We know that the operatorF:K→K is continuous and it is completely continuous by Theorem 3.3. Here we show that there exists an open set U⊆K, with z≠γF(z) for γ∈(0,1) and z∈∂U. Let z∈K be any solution of z=γF(z), γ∈(0,1). According to the Theorem 3.3, since F:K→K is continuous and it is completely continuous, we have
(3.26)z(t)=γFz(t)=γ{λ(t)+L(I)z(t)+Iαf(t,z¯t+yt)},t∈[0,b]≤∑j=1n-1λjbj+∑j=1n-1aj|Ijz(t)|+|Iαf(t,z¯t+yt)|≤∑j=1n-1λjbj+∥z∥∑j=1n-1ajΓ(j+1)bj+1Γ(α)|∫0t(t-s)α-1s-σsσf(s,z¯s+ys)ds|.
Since there exists a positive constant N such that ∥sσf(s,z¯s+ys)∥≤N, as sσf(s,z¯s+ys) is continuous on [0,b], therefore
(3.27)z(t)≤∑j=1n-1λjbj+∥z∥∑j=1n-1ajΓ(j+1)bj+NΓ(α)B(1-σ,α).
Set Λ=max{Λ1,Λ2,Λ3}, where
(3.28)Λ1=∑j=1n-1λjbj,Λ2=∑j=1n-1ajΓ(j+1)bj,Λ3=NΓ(α)B(1-σ,α).
Equation (3.20) implies that
(3.29)∥z∥≤Λ1+∥z∥Λ2+Λ3≤2Λ+∥z∥Λ.
Therefore, we conclude that ∥z∥(1-Λ)≤2Λ. Then, as a result, any solution z=γF(z) satisfies ∥z∥≠h. Let U={z∈K:∥z∥<h}. Theorem 2.2 guarantees that F has a fixed point z∈U¯. By using Theorem 3.2, (1.1) under the conditions of (1.3) has a positive solution x*∈K satisfying ∥x*∥≤max{∥ω∥,h}.Example 3.5.
Consider the following nonlinear fractional differential equation:(3.30)L(D)[x(t)-x(0)]=t-1/2xt,t∈(0,1],x(t)=-sint≥0,t∈[-1,0],
where ℒ(D)=D5/2-D3/2-D1/2. Here, a1=a2=1, α=5/2∈(2,3), n=3 and f(t,xt)=t-1/2xt such that limt→0+f(t,·)=+∞. We select σ=11/12, 0<σ<α. Hence tσf(t,·)=t5/12xt is continuous on [0,1]. We consider the initial conditions
(3.31)x(0)=0,x1=D3/2x(t)|t=0=2,x2=D5/2x(t)|t=0=1,x1≥x2>0.
By using Theorem 3.2, the nonlinear fractional differential equation (3.20) is equivalent to an integral equation given below
(3.32)x(t)={λ(t)+L(I)x(t)=I5/2(t-1/2xt),t∈(0,1],x(t)=-sinπt,t∈[-1,0],
where in the view of the procedure proving Theorem 3.2, we have
(3.33)λ(t)=∑j=12λjt5/2-j=1Γ(5/2)t3/2+1Γ(1/2)t1/2,L(I)x(t)=∑j=12ajIjx(t)=∫0t(1+(t-s))x(s)ds.
In addition of that we have that Λ=max{Λ1,Λ2,Λ3}, where
(3.34)Λ1=∑j=12λjbj=λ1+λ2=1Γ(1/2)+1Γ(5/2)=73π,Λ2=∑j=12ajΓ(j+1)=1+12=32,Λ3=NΓ52B(1-1112,52)=4N3πB(112,52),N=∥t11/12xt∥.
Therefore, by using the Theorem 3.4, (3.30) has at least one positive solution x*∈K* satisfying ∥x*∥≤max{∥ω∥,h} where ∥ω∥=max-1≤t≤0|-sinπt|=1 and h=Λ/(1-Λ)+1,Λ=max{Λ1,Λ2,Λ3}.


## 4. Unique Existence of Solution

In this section, we give conditions onf and a1,a2,…,an, which render a unique positive solution to   (1.1).Theorem 4.1.
Letf:(0,1)×[0,∞)→[0,∞) be continuous and limt→0+f(t,·)=+∞. Suppose that there exists σ∈(0,1) so that 0<σ<α∈(n-1,n), n∈ℕ, and tσf(t,xt) is a continuous function on [0,b]. If further, the following conditions are satisfied (H1)
tσf(t,xt)  is  Lipschitz  with  respect  to  the  second  variable  with  Lipschitz constant  μ, that is,
(4.1)|tσf(t,xt)-tσf(t,zt)|≤μ∥u¯-v¯∥,∀u,v∈K,t∈(0,b],(H2)
(4.2)0<∑j=1n-1ajbjΓ(j+1)+μbα-σΓ(α)B(1-σ,α)<1,
where x(t)=u(t)+y(t) and z(t)=v(t)+y(t); then (1.1) under the conditions of  (1.3) has a unique positive solution.Proof.
As it was pointed out in the previous section, (1.1) is equivalent to (3.13). Thus, for u,v∈K we obtain
(4.3)|Fu(t)-Fv(t)|=|L(I)u(t)-L(I)v(r)|+|Iαf(t,u¯t+yt)-Iαf(r,v¯r+yr)|≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+|Iα(f(t,u¯t+yt)-f(t,v¯t+yt))|=‖u-v‖b∑j=1n-1ajbjΓ(j+1)+|Iα(t-σtσf(t,u¯t+yt)-t-σtσf(t,v¯t+yt))|≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+1Γ(α)max⁡0≤t≤b∫0t(t-s)α-1s-σ|sσf(t,u¯t+yt)-sσf(t,v¯t+yt)|ds≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+μΓ(α)max⁡0≤t≤b∫0t(t-s)α-1s-σds≤‖u-v‖b∑j=1n-1ajbjΓ(j+1)+μ‖u-v‖bΓ(α)max⁡0≤t≤btα-σB(1-σ,α)≤‖u-v‖b{∑j=1n-1ajbjΓ(j+1)+μbα-σΓ(α)B(1-σ,α)},
where F is given in (3.16). Hence, in view of the Theorem 2.3, F will have a unique fixed point in K, which is the unique positive solution of (1.1).Example 4.2.
Consider the nonlinear fractional differential equation given below(4.4)L(D)[x(t)-x(0)]=ηt∫-π-t1+xt(s)1+xt2(s)ds,t∈(0,π],η>0,x(t)=-sint≥0,t∈[-π,0],
where ℒ(D)=D5/2-a1D3/2-a2D1/2 with initial conditions x(0)=0, x1=D3/2x(t)|t=0=2, x2=D5/2x(t)|t=0=1, x1≥x2>0, 0<a1≤1/(4π), 0<a2≤8/π2, and η=3π-3/7)/16B(1/12,5/12). Here limt→0+f(t,xt)=+∞. We select σ=11/12; then 0<σ<α and tσf(t,xt)=t5/12f(t,xt) is continuous on [0,π]. Now, we review the Lipschitz condition with respect to the second variable. Note that from equations (3.11) and (3.12), for xt,zt∈C([-π,0],ℝ≥0), we have xt=u¯t+yt,zt=v¯t+yt,t∈[-π,π] and xt=u(t)+y(t), zt=v(t)+y(t), t∈(0,π]) where u,v∈K. Without loss of generality, we suppose that xt(s)≥yt(s), for all s∈[-π,0]. Then, for each t∈(0,π],
(4.5)|t11/12f(t,xt)-t11/12f(t,zt)|=ηt11/12∫-π-t|1+x(t+s)1+x2(t+s)-1+z(t+s)1+z2(t+s)|ds=ηt11/12∫t-π0|1+x(s)1+x2(s)-1+z(s)1+z2(s)|ds≤ηt11/12∫t-π0|1+x(s)1+x2(s)-1+z(s)1+x2(s)|ds≤ηmax⁡0≤t≤πt11/12∫t-π0|x(s)-z(s)|1+x2(s)ds≤η∥x(t)-z(t)∥max⁡0≤t≤πt11/12∫t-π0ds=η∥u-v∥max⁡0≤t≤πt11/12(π-t)=ηπ23/12∥u-v∥.
Hence, the condition of (H1) holds with μ=π23/12η. On other hand, we have
(4.6)a1bΓ(2)+a2b2Γ(3)+μπ11/12Γ(5/2)B(112,512)∈(0,1).
Thus, (4.4) satisfies the conditions required by the Theorem 4.1. This Theorem implies that the nonlinear equation (4.4) has a unique solution in K. By using the Theorem 3.2, (4.4) is equivalent to the following integral equation:
(4.7)x(t)=1Γ(5/2)(2-14π)t3/2+1Γ(3/2)t1/2+a1Ix(t)+a2I2x(t)+I5/2f(t,xt).
The solution of (4.4) is x(t)=limn→+∞xn(t), where xn+1(t)=Fxn(t). Therefore, the iterated sequence is given by
(4.8)x1(t)=a1t3/2+a2t1/2+ηe-ttI5/2[∫-π-t1+x(s)1+x2(s)ds]=a1t3/2+a2t1/2+ηe-ttI5/2[12ln⁡(1+t21+π2)-tan-1t],x2(t)=(a1I+a2I2+I5/2)x1(t)+x1(t),⋮xn+1(t)=∑i=0n(a1I+a2I2+I5/2)n-ix1(t),n=1,2,….



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*Source: 391971-2011-06-13.xml*