# Existence and Uniqueness of the Solutions for Fractional Differential Equations with Nonlinear Boundary Conditions

Authors: Xiping Liu; Fanfan Li; Mei Jia; Ertao Zhi
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the existence and uniqueness of the solutions for the boundary value problem of fractional differential equations with nonlinear boundary conditions. By using the upper and lower solutions method in reverse order and monotone iterative techniques, we obtain the sufficient conditions of both the existence of the maximal and minimal solutions between an upper solution and a lower solution and the uniqueness of the solutions for the boundary value problem and present the iterative sequence for calculating the approximate analytical solutions of the boundary value problem and the error estimate. An example is also given to illustrate the main results.


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## Body

## 1. Introduction

In this paper, we study the following fractional differential equation with nonlinear boundary conditions:(1)CD0+δu(t)-Mu(t)=f(t,u(t)),t∈J,u′(0)=0,u(0)=g(u(T)),
where J=[0,T], 0<T<+∞. M≥0 is a constant. f∈C(J×R,R), g∈C1(R,R) are given functions. CD0+δ is the standard Caputo fractional derivative of order δ with 1<δ<2.With the development of mathematics, fractional derivative occurs more and more frequently in different research areas, such as physics, mechanics, electricity, and economics (see [1, 2]). At the same time, significant progress has also been made on the studies of fractional differential equations (see [3–12]).The upper and lower solutions method and monotone iterative techniques have been widely used in the studies for the boundary value problems of integer-order differential equations (see [13–15] and the references therein). In [13], the authors studied first order differential equation
(2)u′(t)=f(t,u(t),u(θ(t))),g(u(0))=u(T).
By using the upper and lower solutions method and the monotone iterative techniques, the authors concluded that (2) exists the maximal and minimal solutions in [β(t),α(t)], where α(t), β(t) are the lower and upper solutions of (2).As for the fractional differential problems, upper and lower solutions and monotone iterative techniques have also been widely used in the studies of boundary value problem (see [16–18]). For example, in [16], by using the method, the authors discussed the existence and uniqueness of solutions for the fractional differential equation with linear boundary condition
(3)Dδu(t)-Mu(t)=f(t,u(t)),t∈J,0<δ<1,u(0)=ru(T),
where M≥0, 0<r<1/Eδ,1(MTδ).In this paper, we study the existence and uniqueness of the solutions for the boundary value problem of the fractional differential equation (1), which has nonlinear boundary conditions. It is very difficult to get the iterative sequence which is used to obtain the solutions since the boundary value problem has nonlinear boundary conditions. By using the upper and lower solutions method in reverse order and monotone iterative techniques, we not only obtain the sufficient conditions that the boundary value problem (1) has the maximal and minimal solutions between an upper solution and a lower solution as well as having the unique solution, but also present the iterative sequence for calculating the approximate analytical solutions of the boundary value problem and the error estimate. An example is also given to illustrate the main results.The organization of this paper is as follows. In Section2, we provide the necessary background and comparison principles which are used to prove our main results. In Section 3, we consider a linear problem associated with (1). In Section 4, by using the method of upper and lower solutions and the monotone iterative techniques, we obtain the existence and uniqueness solutions of (1). In Section 5, the calculation method of the approximate analytical solutions for the boundary value problems (1) is obtained. In Section 6, an example is presented to illustrate our main results.


## 2. Preliminaries and Comparison Principles

In this section, we present some basic definitions, lemmas, and comparison principles which play important roles in our investigation.Definition 1 (see [3, 4]).
Letδ>0 and function u:(0,+∞)→R. The Riemann-Liouville fractional integral operator of order δ of u is defined by
(4)Iδu(t)=1Γ(δ)∫0t‍(t-s)δ-1u(s)ds,
provided the integral exists.
The Caputo derivative of orderδ for function u:(0,+∞)→R is given by
(5)CDδu(t)=1Γ(n-α)∫0t‍u(n)(s)(t-s)δ+1-nds,
provided the right side is pointwise defined on (0,+∞), where n∈N+ with n-1<δ<n.
Ifδ=n∈N+, then CDδu(t)=u(n)(t).Definition 2.
LetAC[0,1] be the space of the functions u which are absolutely continuous on [0,1]. We denote ACn[0,1] by the set of the functions u which have continuous derivatives up to order n-1 on [0,1] such that u(n-1)∈AC[0,1]. In particular, AC1[0,1]=AC[0,1].Lemma 3 (see [3]).
Ifu∈ACn[0,1], then the Caputo fractional derivative CDδu(t) exists almost everywhere on [0,1], where n is the smallest integer greater than or equal to δ.Lemma 4 (see [3]).
Supposeδ>0 and u∈ACn(J). Then
(6)IδCDδu(t)=u(t)+c0+c1t+c2t2+⋯+cn-1tn-1,ck=u(k)(0)k!,k=0,1,2,…,n-1,
where n is the smallest integer greater than or equal to δ.Definition 5.
One saysu is a solution of the boundary value problem (1) if u∈AC2(J) and CDδu∈C(J) and satisfies (1).Definition 6 (see [4]).
Let the realn>0. The function En is defined by
(7)En(z)=∑j=0∞‍zjΓ(jn+1),
whenever the series converges called the Mittag-Leffler function of order n.Definition 7 (see [4]).
Letn1,n2>0. The function En1,n2 is defined by
(8)En1,n2(z)=∑j=0∞‍zjΓ(jn1+n2),
whenever the series converges called the two-parameter Mittag-Leffler function with parameters n1 and n2.Remark 8 (see [4]).
It is evident that the one-parameter Mittag-Leffler functions may be defined in terms of their two-parameter counterparts via the relationEn(z)=En,1(z).Lemma 9 (see [4], Theorem 4.1).
The two-parameter Mittag-Leffler functionEn1,n2(z) for some n1,n2>0. The power series defining En1,n2(z) is convergent for all z∈R.Lemma 10 (see [3], Theorem 4.3).
Leth∈C(J). The Cauchy problem
(9)CD0+δu(t)-Mu(t)=h(t),t∈J,u(0)=b,u′(0)=d,b,d∈R,
with 1<δ<2 and M∈R has a unique solution
(10)u(t)=bEδ(Mtδ)+dEδ,2(Mtδ)t+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)h(s)ds.Lemma 11 (see [19]).
LetE be a partially ordered Banach space, xn≤yn(n=1,2,3,…), if xn→x*, yn→y0*. Then x*≤y*.In this paper, we assume the following conditions are satisfied.(H0)
A given functionc∈AC2(J) and CD0+δc∈C(J). c is monotone decreasing on J, and c(0)=c′(0)=0, c(T)≤-1.(H1)
g∈C1(R,R), g(0)≥0, and 0<g′(x)<1/Eδ(MTδ).Definition 12.
Letα∈AC2(J) and CD0+δα∈C(J); we say that α is a lower solution of the boundary value problem (1), if
(11)CD0+δα(t)-Mα(t)≥f(t,α(t))-aα(t),t∈J,α′(0)≥0,
where
(12)aα(t)={0,g(α(T))≤α(0),(CD0+δc(t)-Mc(t))×(α(T)-g-1(α(0))),g(α(T))>α(0).
Let β∈AC2(J) and CD0+δβ∈C(J); we say that β is an upper solution of the boundary value problem (1), if
(13)CD0+δβ(t)-Mβ(t)≤f(t,β(t))+bβ(t),t∈J,β′(0)≤0,
where
(14)bβ(t)={0,g(β(T))≥β(0),(CD0+δc(t)-Mc(t))×(g-1(β(0))-β(T)),g(β(T))<β(0),
and c(t) is defined in (H0).Lemma 13.
Letg1∈C1(R,R) be any function with g1(0)≥0 and 0<g1′(u)<1/Eδ(MTδ). If u∈AC2(J) and CD0+δu∈C(J) and satisfies
(15)CD0+δu(t)-Mu(t)≥0,t∈J,u′(0)≥0,u(0)≥g1(u(T)),
then u(t)≥0 for t∈J.Proof.
Ifu(0)<0, by the boundary conditions, we have
(16)0>u(0)≥g1(u(T))=g1(0)+g1′(ξ)u(T)≥g1′(ξ)u(T),
where ξ is a constant between 0 and u(T).
Becauseg1(0)≥0 and 0<g1′(u)<1/Eδ(MTδ), then u(T)<0 and u(0)>(1/Eδ(MTδ))u(T); that is,
(17)u(T)<u(0)Eδ(MTδ).
Leth(t)=CD0+δu(t)-Mu(t), u(0)=b, and u′(0)=d, and then h(t)≥0 for t∈J and d≥0.
By Lemma10, we can get that the Cauchy problem
(18)CD0+δu(t)-Mu(t)=h(t),t∈J,u(0)=b,u′(0)=d,
has a unique solution
(19)u(t)=bEδ(Mtδ)+dEδ,2(Mtδ)t+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)h(s)ds.
So we can obtain that
(20)u(T)=bEδ(MTδ)+dTEδ,2(MTδ)+∫0T(T-s)δ-1‍Eδ,δ(M(T-s)δ)h(s)ds≥bEδ(MTδ)=u(0)Eδ(MTδ),
which is contradictory to (17).
Hence,b=u(0)≥0. By (19), we can obtain that u(t)≥0 for t∈J.Let the functiong1(x)=r1x in Lemma 13; we can obtain Corollary 14.Corollary 14.
If there exists a constantr1 with 0<r1<1/Eδ(MTδ) such that u∈AC2(J) and CD0+δu∈C(J) and satisfies
(21)CD0+δu(t)≥Mu(t),t∈J,u′(0)≥0,u(0)≥r1u(T),
then u(t)≥0 on J.Corollary 15.
Suppose (H0) holds, and there exists a constantr2>Eδ(MTδ) such that u∈AC2(J) and CD0+δu∈C(J) and satisfies
(22)CD0+δu(t)≥Mu(t)-(CD0+δc(t)-Mc(t))×(u(T)-r2u(0)),t∈J,u′(0)≥0,u(T)≥r2u(0);
then u(t)≥0 for t∈J.Proof.
Letg1(x)=(1/r2)x and y(t)=u(t)+c(t)(u(T)-r2u(0)), t∈J.
Sincec(t) is monotone decreasing and c(0)=0, then c(t)≤0 on J. So y(t)≤u(t),t∈J.
By (22), we have
(23)CD0+δy(t)-My(t)=CD0+δu(t)-Mu(t)+(CD0+δc(t)-Mc(t))(u(T)-r2u(0))≥0,
and y′(0)=u′(0)≥0.
Becausec(0)=0 and c(T)≤-1, then y(T)=u(T)+c(T)(u(T)-r2u(0))≤r2u(0)=r2y(0).
Soy(0)≥(1/r2)y(T)=g1(u(T)).
By Lemma13, we can get that y(t)≥0, which implies that u(t)≥0 for t∈J.


## 3. Boundary Value Problems for the Linear Equation

In this section, we consider the existence and uniqueness of solutions for the linear fractional differential equation with nonlinear boundary conditions(24)CD0+δu(t)-Mu(t)=h(t),t∈J,u′(0)=0,u(0)=g(u(T)),
where h∈C(J).Theorem 16.
Assume that (H0) and (H1) hold; there exist a lower solutionα and an upper solution β of the boundary value problem (24) with α(t)≥β(t) on J. Then the boundary value problem (24) has a unique solution u. Moreover, α≥u≥β; that is, α(t)≥u(t)≥β(t) for t∈J.Proof.
(1) We show that the solution of (24) is unique if it exists.
Suppose thatu1, u2 are two solutions of (24) and let v=u1-u2. Then,
(25)CD0+δv(t)-Mv(t)=0,t∈J,v′(0)=0,v(0)=g(u1(T))-g(u2(T))=g′(τ1)v(T),
where τ1 is a constant between u1(T) and u2(T) and 0<g′(τ1)<1/Eδ(MTδ) is a constant.
By Corollary14, we have that v(t)≥0; that is, u1(t)≥u2(t) for t∈J.
Similarly, we can also obtain thatu2(t)≥u1(t), for t∈J.
Hence,u1=u2.
(2) We prove thatβ≤u≤α if u is a solution of the boundary value problem (24).
Letm=α-u.
Ifg(α(T))≤α(0), then aα(t)=0 on J. We have
(26)CD0+δm(t)-Mm(t)≥0,t∈J,m′(0)≥0,m(0)≥g(α(T))-g(u(T))=g′(τ2)m(T),
where τ2 is a constant between u(T) and α(T).
We denoter1=g′(τ2). Hence, 0<r1<1/Eδ(MTδ).
By Corollary14, we can obtain that m(t)=α(t)-u(t)≥0 on J.
Ifg(α(T))>α(0), then aα(t)=(CD0+δc(t)-Mc(t))(α(T)-g-1(α(0))). We have that
(27)CD0+δm(t)-Mm(t)=CD0+δα(t)-Mα(t)-CD0+δu(t)+Mu(t)≥-(CD0+δc(t)-Mc(t))(α(T)-g-1(α(0)))=-(CD0+δc(t)-Mc(t))×(α(T)-u(T)+g-1(u(0))-g-1(α(0)))=-(CD0+δc(t)-Mc(t))(m(T)-(g-1)′(τ3)m(0)),
where τ3 is a constant between α(0) and u(0).
We denoter2=(g-1)′(τ3). Hence, r2>Eδ(MTδ).
We can showm′(0)≥0, m(T)≥g-1(α(0))-g-1(u(0))=r2m(0).
By Corollary15, we have m(t)≥0; that is, α(t)≥u(t) on J. So α(t)≥u(t) on J.
Similarly, we can obtain thatu(t)≥β(t) for J. Therefore, β≤u≤α.
(3) We prove that the problem (24) has a unique solution.
Let(28)p(t)={α(t),g(α(T))≤α(0),α(t)+c(t)×(α(T)-g-1(α(0))),g(α(T))>α(0),(29)q(t)={β(t),g(β(T))≥β(0),β(t)-c(t)×(g-1(β(0))-β(T)),g(β(T))<β(0).
It is obvious thatp,q∈AC2(J), CD0+δp, CD0+δq∈C(J), and p(0)=α(0), q(0)=β(0). If g(α(T))≤α(0), we have p(T)=α(T). If g(α(T))>α(0), we have p(T)≤g-1(α(0)); that is,
(30)g(p(T))≤p(0).
Similarly, we can get
(31)g(q(T))≥q(0).
Ifg(α(T))≤α(0), we have
(32)CD0+δp(t)-Mp(t)≥h(t),t∈J.
If g(α(T))>α(0), since α is a lower solution of boundary value problem (24), then CD0+δα(t)-Mα(t)≥h(t)-aα(t). And by (12), we can obtain that
(33)CD0+δp(t)-Mp(t)=(CD0+δα(t)-Mα(t))+(CD0+δc(t)-Mc(t))(α(T)-g-1(α(0)))≥h(t)-aα(t)+(CD0+δc(t)-Mc(t))(α(T)-g-1(α(0)))=h(t),t∈J.
Hence, we obtain
(34)CD0+δp(t)-Mp(t)≥h(t),t∈J.
Using the same way as mentioned above, we can get that
(35)CD0+δq(t)-Mq(t)≤h(t),t∈J.
Since (H0) holds, it is easy to see that c(t)≤0 for t∈J. We can easily get that β(t)≤q(t) and p(t)≤α(t) for t∈J.
Letn(t)=p(t)-q(t). It follows from (34) and (35) that
(36)CD0+δn(t)-Mn(t)=CD0+δp(t)-Mp(t)-CD0+δq(t)+Mq(t)≥0,n′(0)=p′(0)-q′(0)=α′(0)-β′(0)≥0,n(0)=p(0)-q(0)≥g(p(T))-g(q(T))=g′(τ4)n(T),
where τ4 is a constant between p(T) and q(T) and 0<g′(τ4)<1/Eδ(MTδ) is a constant.
By Corollary14, we have n(t)≥0 for t∈J; that is,
(37)p(t)≥q(t)fort∈J.
According to Lemma10, for each λ∈R, the Cauchy problem
(38)CD0+δu(t)-Mu(t)=h(t),u′(0)=0,u(0)=λ,
has a unique solution
(39)u(t,λ)=λEδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)h(s)ds,t∈J.
In the following, we show thatq(T)≤u(T,λ)≤p(T) for each λ∈[g(q(T)),g(p(T))], where u(t,λ) is defined by (39).
Ifu(T,λ)<q(T), we denote l(t)=u(t,λ)-q(t), and then
(40)l(T)<0,(41)l(0)=u(0,λ)-q(0)≥λ-g(q(T))≥0.
So l(0)≥r1l(T), for any r1∈(0,1/Eδ(MTδ)).
It follows thatc′(0)=0 from (H0), and by (29), we have that q′(0)=β′(0). Then
(42)l′(0)=u′(0,λ)-q′(0)=-β′(0)≥0.
And from (35), we have
(43)CD0+δl(t)-Ml(t)=CD0+δu(t)-Mu(t)-(CD0+δq(t)-Mq(t))=h(t)-(CD0+δq(t)-Mq(t))≥0.
By Corollary 14, we can obtain that l(t)≥0, for all t∈J, which is a contradiction to (40). So
(44)u(T,λ)≥q(T).
Similarly, we can get
(45)u(T,λ)≤p(T).
LetH(λ)=g(u(T,λ))-λ. Since H(g(q(T)))H(g(p(T)))≤0 and
(46)H′(λ)=g′(u(T,λ))uλ′(T,λ)-1<1Eδ(MTδ)·Eδ(MTδ)-1=0,
then  H(λ) is strictly monotone decreasing and continuous for λ∈[g(q(T)),g(p(T))].
Hence, we can get thatH(λ)=0 has a unique solution λ0∈[g(q(T)),g(p(T))] with g(u(T,λ0))=λ0=u(0).
It is easy to see thatu(t,λ0) is the unique solution of the boundary value problem (24).


## 4. Existence and Uniqueness of the Solutions for Boundary Value Problem

In this section, we study the existence and uniqueness of the solutions for fractional differential equation with nonlinear boundary conditions (1).LetE=C(J) endowed with norm ∥u∥=max⁡t∈J|u(t)| for u∈E. Then E is a Banach space. Denote
(47)[β,α]={u∈C(J):β(t)≤u(t)≤α(t),t∈J},
and β≤α if and only if β(t)≤α(t) on J.Theorem 17.
Suppose that (H0) and (H1) hold; there existα0, β0 which are lower and upper solutions of the boundary value problem (1) with β0(t)≤α0(t) for t∈J. And f satisfies(H2)
f(t,x2)-f(t,x1)≥0 for any β0≤x1≤x2≤α0, t∈J.Then there exist monotone sequences {αn}, {βn} such that
(48)lim⁡n→∞αn(t)=α*(t),lim⁡n→∞βn(t)=β*(t),converging uniformly on J, and β*, α* are the minimal and the maximal solutions of (1) on [β0,α0].Proof.
We denoteD=[β0,α0]. For any γ∈D, we consider the following boundary value problem:
(49)CD0+δu(t)-Mu(t)=f(t,γ(t)),t∈J,u′(0)=0,u(0)=g(u(T)).
Since α0(t), β0(t) are lower and upper solutions of the boundary value problem (1), by (H2), we have
(50)CD0+δα0(t)-Mα0(t)≥f(t,α0(t))-aα(t)≥f(t,γ(t))-aα(t),CD0+δβ0(t)-Mβ0(t)≤f(t,β0(t))-bβ(t)≤f(t,γ(t))-bβ(t),
for t∈J.
Therefore,α0(t), β0(t) are also the lower and upper solutions of the boundary value problem (49), respectively.
In view of Theorem16, the boundary value problem (49) has a unique solution u and u∈D.
DefineP:[β0,α0]→[β0,α0] by Pγ=u. Hence, α0≥Pα0, β0≤Pβ0.
We will show thatPx1≤Px2 if β0≤x1≤x2≤α0.
Letm=Px2-Px1. By (H2) and (49), we have
(51)CD0+δm(t)-Mm(t)=f(t,x2(t))-f(t,x1(t))≥0,m′(0)=0,m(0)=g′(τ5)m(T),
where τ5 is a constant between Px1(T) and Px2(T), and denote r1=g′(τ5)<1/Eδ,1(MTδ). By Corollary 14, m(t)≥0, which implies Px1≤Px2. Hence, P is monotone increasing in [β0,α0].
Letαn=Pαn-1 and βn=Pβn-1 for n=1,2,…. We can get monotone iterative sequences
(52)β1≤β2≤⋯≤βn≤⋯≤αn≤⋯≤α2≤α1.
Therefore, there exist α*,β*∈E, such that
(53)lim⁡n→∞αn(t)=α*(t),lim⁡n→∞βn(t)=β*(t).
It is easy to see thatαn(t) satisfies
(54)CD0+δαn(t)-Mαn(t)=f(t,αn-1(t)),αn′(0)=0,αn(0)=g(αn(T)).
By Lemma 10, we have
(55)αn(t)=αn(0)Eδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)f(s,αn-1(s))ds.
Because f is continuous and αn-1∈D, there exists a constant L>0 such that
(56)|f(t,αn-1(t))|≤L,t∈J,n=1,2,…,f(t,αn-1(t))⟶f(t,α*(t)),t∈J,n⟶∞.
By Lebesgue dominated convergence theorem, we can get that
(57)lim⁡n→∞αn(t)=lim⁡n→∞αn(0)Eδ(Mtδ)+lim⁡n→∞∫0t‍(t-s)δ-1f(s,αn-1(s))ds=α*(0)Eδ(Mtδ)+∫0t‍(t-s)δ-1f(s,α*(s))ds.
That is,
(58)α*(t)=α*(0)Eδ(Mtδ)+∫0t‍(t-s)δ-1f(s,α*(s))ds.
Therefore,(59)α*(t)=Pα*(t)fort∈J.
It is similar to show that
(60)β*(t)=Pβ*(t)fort∈J.
It is clear thatα*, β* are solutions of the boundary value problem (1).
Assumeu∈[β0,α0] is a solution of the boundary value problem (1). We can easily obtain that Pβ0(t)≤Pu(t)≤Pα0(t) by the fact that P is increasing in [β0,α0]. That is, β1(t)≤u(t)≤α1(t). Doing this repeatedly, we have βn(t)≤u(t)≤αn(t), for n=1,2,…. From Lemma 11, we obtain that β*(t)≤u(t)≤α*(t), as n→∞.
Hence,α*, β* are the maximal and the minimal solutions of the boundary value problem (1), respectively.Theorem 18.
Suppose that the conditions (H0) and (H1) hold; there existα0,β0 lower and upper solutions of the boundary value problem (1) with β0(t)≤α0(t) for t∈J, respectively. If there exists a constant k with 0<k<δ(1-aEδ(MTδ))/TδEδ,δ(MTδ), where a=max⁡u∈[β0,α0]g′(u), and f satisfies(H3)
f(t,x2)-f(t,x1)≤k(x2-x1) for any β0≤x1≤x2≤α0, t∈J,then the boundary value problem (1) has a unique solution u* on [β0,α0] and for any u0∈[β0,α0], the iterative sequence un=Pun-1, n=0,1,2,…, converging uniformly to u* on J and its error estimate is
(61)∥un-u*∥≤(kTδEδ,δ(MTδ)δ(1-aEδ(MTδ)))n∥α0-β0∥.Proof.
It is easy to check that the conditions of Theorem17 are satisfied. Then the boundary value problem (1) has the maximal solution and the minimal solution which are denoted by α*, β*, respectively.
Forβ0≤ω1≤ω2≤α0, t∈J, we have
(62)Pω2(t)-Pω1(t)=(Pω2(0)-Pω1(0))Eδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)×(f(s,ω2(s))-f(s,ω1(s)))ds,
and from (H3),
(63)Pω2(0)-Pω1(0)=g(Pω2(T))-g(Pω1(T))=g′(τ)(Pω2(T)-Pω1(T))=g′(τ)((Pω2(0)-Pω1(0))Eδ(MTδ)∫0T‍+∫0T‍(T-s)δ-1Eδ,δ(M(T-s)δ)×(f(s,ω2(s))-f(s,ω1(s)))ds∫0T‍)≤a((Pω2(0)-Pω1(0))Eδ(MTδ)∫0T‍+∫0T‍(T-s)δ-1Eδ,δ(M(T-s)δ)×(f(s,ω2(s))-f(s,ω1(s)))ds∫0T‍)≤a((Pω2(0)-Pω1(0))Eδ(MTδ)kTδEδ,δ(MTδ)δ+kTδEδ,δ(MTδ)δ∥ω2-ω1∥),
where τ∈[ω1(T),ω2(T)]⊂[β0(T),α0(T)] and P is defined in Theorem 17.
It follows that1-aEδ(MTδ)>0 from 0<g′(x)<1/Eδ(MTδ) for x∈R. Hence,
(64)0≤Pω2(0)-Pω1(0)≤kaTδEδ,δ(MTδ)δ(1-aEδ(MTδ))∥ω2-ω1∥,0≤Pω2(t)-Pω1(t)≤Eδ(MTδ)kaTδEδ,δ(MTδ)δ(1-aEδ(MTδ))∥ω2-ω1∥+kTδEδ,δ(MTδ)δ∥ω2-ω1∥.
We can easily get(65)∥Pω2-Pω1∥≤kTδEδ,δ(MTδ)δ(1-aEδ(MTδ))∥ω2-ω1∥.
Because 0<k<δ(1-aEδ(MTδ))/TδEδ,δ(MTδ), we have
(66)0<kTδEδ,δ(MTδ)δ(1-aEδ(MTδ))<1.
It implies that P is contraction mapping. By using the contraction mapping principle, the P has a unique fixed point.
Therefore, the boundary value problem (1) has a unique solution; that is, α*=β*, and we denote it by u*.
For anyu0∈[β0,α0], let the iterative sequence un=Pun-1, n=0,1,2,…. Similar to (65), we can get
(67)∥un-u*∥≤kTδEδ,δ(MTδ)δ(1-aEδ(MTδ))∥Pun-1-u*∥≤⋯≤(kTδEδ,δ(MTδ)δ(1-aEδ(MTδ)))n∥u0-u*∥.
It follows that {un} converging uniformly to u* on J from (66).
Sinceu0,u*∈[β0,α0], we can obtain that the error estimate of {un} is
(68)∥un-u*∥≤(kTδEδ,δ(MTδ)δ(1-aEδ(MTδ)))n∥α0-β0∥.


## 5. The Calculation Method of the Approximate Analytical Solutions

In this section, we give the calculation method of the approximate analytical solutions for the boundary value problem (1).Theorem 19.
Suppose (H0) and (H1) hold; there exist a lower solutionα0 and an upper solution β0 of the boundary value problem (1) with β0(t)≤α0(t) for t∈J. And there exists a constant κ with 0<κ<δ(1-bEδ(MTδ))/TδEδ,δ(MTδ), where b=sup⁡u∈Rg′(u)<1/Eδ(MTδ), such that f satisfies(H4)
0≤f(t,x2)-f(t,x1)≤κ(x2-x1) for any x1≤x2, t∈J.Then for any initial value y0∈[β0,α0], the sequence {yn} is defined by
(69)yn(t)=g(yn-1(T))Eδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)f(s,yn-1(s))ds,
converging uniformly to the unique solution u* of the boundary value problem (1) on the [β0,α0]. Furthermore, the error estimate is
(70)∥yn-u*∥≤ϱ0n-ϱ1n+11-ϱ1∥α0-β0∥,
where
(71)ϱ0=bEδ(MTδ)+κTδEδ,δ(MTδ)δ,ϱ1=κTδEδ,δ(MTδ)δ(1-bEδ(MTδ)).Proof.
Because0<κ<δ(1-bEδ(MTδ))/TδEδ,δ(MTδ), where b=sup⁡u∈Rg′(u)<1/Eδ(MTδ), then b≥a and κ<δ(1-aEδ(MTδ))/TδEδ,δ(MTδ), where a is defined by Theorem 18.
Hence, the conditions of Theorem18 hold, and we can get that the boundary value problem (1) has a unique solution u* on [β0,α0] and for any u0∈[β0,α0], the iterative sequence un=Pun-1, n=1,2,…, converging uniformly to u* on J. That is,
(72)∥un-u*∥⟶0,asn⟶∞.
By Lemma 10 and the definition of operator P, we can easily obtain that un=Pun-1 is equivalent to
(73)un(t)=un(0)Eδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)f(s,un-1(s))ds,
and un(0)=g(un(T)), n=0,1,2,….
Lety0=u0, n=1,2,3,…. By Lemma 10, the Cauchy problem
(74)CD0+δyn(t)-Myn(t)=f(t,yn-1(t)),yn′(0)=0,yn(0)=g(yn-1(T)),
has a unique solution
(75)yn(t)=g(yn-1(T))Eδ(Mtδ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)f(s,yn-1(s))ds,
which is (69).
In the following, we proveyn converging uniformly to u* on J.
Fort∈J, we have
(76)|yn(t)-un(t)|≤Eδ(MTδ)|g(yn-1(T))-g(un(T))|+∫0T‍(T-s)δ-1Eδ,δ(MTδ)×|f(s,yn-1(s))-f(s,un-1(s))|ds≤Eδ(MTδ)g′(ξn)|yn-1(T)-un(T)|+κTδEδ,δ(MTδ)δ∥yn-1-un-1∥≤bEδ(MTδ)(∥un-un-1∥+∥yn-1-un-1∥)+κTδEδ,δ(MTδ)δ∥yn-1-un-1∥≤bEδ(MTδ)∥un-un-1∥+(bEδ(MTδ)+κTδEδ,δ(MTδ)δ)∥yn-1-un-1∥,
where ξn is between un(T) and yn-1(T).
Therefore,(77)∥yn-un∥≤bEδ(MTδ)∥un-un-1∥+ϱ0∥yn-1-un-1∥.
Because0<κ<δ(1-bEδ(MTδ))/TδEδ,δ(MTδ), we can show
(78)0<ϱ0=bEδ(MTδ)+κTδEδ,δ(MTδ)δ<1.
On the other hand, by (73), we have
(79)∥un-un-1∥=max⁡t∈J⁡|(un(0)-un-1(0))Eδ(Mtδ)∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)+∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)×(f(s,un-1(s))-f(s,un-2(s)))ds∫0t‍(t-s)δ-1Eδ,δ(M(t-s)δ)|≤Eδ(MTδ)g′(ηn)∥un-un-1∥+∫0T‍(T-s)δ-1Eδ,δ(MTδ)×|f(s,un-1(s))-f(s,un-2(s))|ds≤bEδ(MTδ)∥un-un-1∥+κTδEδ,δ(MTδ)δ∥un-1-un-2∥,
where ηn is between un(0) and un-1(0).
In view ofb<1/Eδ(MTδ), we can get(80)∥un-un-1∥≤κTδEδ,δ(MTδ)δ(1-bEδ(MTδ))∥un-1-un-2∥=ϱ1∥un-1-un-2∥.It follows that 0<κ<δ(1-bEδ(MTδ))/TδEδ,δ(MTδ) from 0<ϱ1<1.
By (80), we can easily get
(81)∥un-un-1∥≤ϱ1∥un-1-un-2∥≤ϱ12∥un-2-un-2∥≤⋯≤ϱ1n-1∥u1-u0∥.
Substituting (81) into (77), we have
(82)∥yn-un∥≤bEδ(MTδ)ϱ1n-1∥u1-u0∥+ϱ0∥yn-1-un-1∥.
It is easy to show that
(83)∥yn-un∥≤bEδ(MTδ)(ϱ1n-1+ϱ0ϱ1n-2)∥u1-u0∥+ϱ02∥yn-2-un-2∥≤bEδ(MTδ)(ϱ1n-1+ϱ0ϱ1n-2+ϱ02ϱ1n-3)∥u1-u0∥+ϱ03∥yn-3-un-3∥≤⋯≤bEδ(MTδ)(ϱ1n-1+ϱ0ϱ1n-2+ϱ02ϱ1n-3+⋯+ϱ0n-1)×∥u1-u0∥+ϱ0n∥y0-u0∥=bEδ(MTδ)(ϱ1n-1+ϱ0ϱ1n-2+ϱ02ϱ1n-3+⋯+ϱ0n-1)×∥u1-u0∥.
We claimϱ0>ϱ1 from (71).
Ifϱ0≤ϱ1, we have
(84)bEδ(MTδ)+κTδEδ,δ(MTδ)δ≤κTδEδ,δ(MTδ)δ(1-bEδ(MTδ)).
So
(85)κ≥δ(1-bEδ(MTδ))TδEδ,δ(MTδ),
which is a contradiction to the hypothesis of the theorem.
Therefore, it follows from (78) that
(86)∥yn-un∥≤bEδ(MTδ)(ϱ0n-ϱ1n)ϱ0-ϱ1∥u1-u0∥⟶0,asn⟶∞.
By (72) and (86), we can show that
(87)∥yn-u*∥≤∥yn-un∥+∥un-u*∥⟶0,asn⟶∞.
Then for any initial value u0∈[β0,α0], the sequence {yn} is defined by (69) converging uniformly to the unique solution u* of the boundary value problem (1) on the [β0,α0].
In view of (61), we have
(88)∥un-u*∥≤(κTδEδ,δ(MTδ)δ(1-aEδ(MTδ)))n∥α0-β0∥=ϱ1n∥α0-β0∥.
According to (86), we can show
(89)∥yn-un∥≤bEδ(MTδ)(ϱ0n-ϱ1n)ϱ0-ϱ1∥α0-β0∥.
It is easy to see that(90)ϱ0-ϱ1=bEδ(MTδ)(1-ϱ1).
By (88) and (89), we can obtain
(91)∥yn-u*∥≤ϱ1n∥α0-β0∥+bEδ(MTδ)(ϱ0n-ϱ1n)ϱ0-ϱ1∥α0-β0∥=(ϱ1n+ϱ0n-ϱ1n1-ϱ1)∥α0-β0∥=ϱ0n-ϱ1n+11-ϱ1∥α0-β0∥.
Hence, the error estimate of {yn} is
(92)∥yn-u*∥≤ϱ0n-ϱ1n+11-ϱ1∥α0-β0∥.


## 6. Example

We consider the boundary value problem(93)CD0+3/2u(t)-14u(t)=t10arctanu(t)+t210,u′(0)=0,u(0)=15arctanu(1).Letf(t,u)=(t/10)arctanu+t2/10 and g(u)=(1/5)arctanu.It is obvious thatM=1/4, T=1, δ=3/2, and b=sup⁡u∈Rg′(u)=1/5<1/Eδ(MTδ)(≈0.834179). Then (H1) holds.We takeβ(t)=0 for t∈J.Hence,β(0)=0, g(β(1))=0, and bβ(t)=0. CD0+δβ(t)=0, f(t,β(t))=0, and then β(t) is an upper solution of the boundary value problem (93).Letα(t)=5t3/2/6π+5t/8π+35/48π≤35/16π for t∈J.We haveα′(0)=5/8π, α(0)=35/48π≈0.411388, g(α(1))=(1/5)arctan(35/16π)≈0.177966, and aα(t)=0. Consider
(94)CD0+δα(t)-14α(t)≥58-354×16π≈0.316459>0.188983≈f(1,3516π)≥f(t,α(t)),
for t∈J. Then α(t) is a lower solution of the boundary value problem (93). And β(t)≤α(t).It is easy to verify that the (H2) holds. By using Theorem17, we know the problem (93) has minimal and maximal solutions in [β,α].Letκ=1/10; we know κ<δ(1-bEδ(MTδ))/TδEδ,δ(MTδ)≈0.905853.By calculating, we have(95)ϱ0≈0.323682,ϱ1≈0.110393.It is obvious that(96)0≤f(t,x2)-f(t,x1)≤110(x2-x1)=κ(x2-x1),foranyx1≤x2,t∈J.
By Theorem 19, we can get the problem (93) has a unique solution on the [β0,α0].For any initial valueu0∈[β0,α0], the sequence {yn} defined by
(97)yn(t)=15arctan(yn-1(1))E3/2(14t3/2)+∫0t‍(t-s)1/2E3/2,3/2(14(t-s)3/2)×(s10arctanyn-1(s)+s210)ds,
converging uniformly to the unique solution u* of the boundary value problem (93) on the [β0,α0], and its error estimate is
(98)∥yn-u*∥≤ϱ0n-ϱ1n+11-ϱ1∥α0-β0∥=35(ϱ0n-ϱ1n+1)16π(1-ϱ1),
where y0=u0.We takeu0=0. For n=3, its error is not more than 0.046841; for n=5, its error is not more than 0.00492662; for n=8, its error is not more than 0.000167155; for n=10, its error is not more than 0.0000175132.



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*Source: 758390-2014-06-19.xml*