# A Numerical Algorithm for Solving a Four-Point Nonlinear Fractional Integro-Differential Equations

Authors: Er Gao; Songhe Song; Xinjian Zhang
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

We provide a new algorithm for a four-point nonlocal boundary value problem of nonlinear integro-differential equations of fractional orderq∈(1,2] based on reproducing kernel space method. According to our work, the analytical solution of the equations is represented in the reproducing kernel space which we construct and so the n-term approximation. At the same time, the n-term approximation is proved to converge to the analytical solution. An illustrative example is also presented, which shows that the new algorithm is efficient and accurate.


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## Body

## 1. Introduction

In recent years, differential equations of fractional order have been addressed by several researchers with the sphere of study ranging from the theoretical aspects of existence and uniqueness of solutions to the analytic and numerical methods for finding solutions. Several authors have used fixed point theory to show the existence of solution to differential equations of fractional order, see the monographs of Bai and Liu [1], Wu and Liu [2], Hamani et al. [3] and Ahmad and Sivasundaram [4]. At the same time, there may be several methods for solving differential equations of fractional order, such as the least squares finite-element method [5], collection method [6], fractional differential transform method [7], decomposition method [8], and variational iteration method [9]. Besides these cited works, few more contributions [10, 11] have been made to the analytical and numerical study of the solutions of fractional boundary value problems.Ahmad and Sivasundaram [4] proved the existence and uniqueness of solutions for a four-point nonlocal boundary value problem of nonlinear integro-differential equations of fractional order q∈(1,2] by applying some standard fixed point theorems:
(1.1)Dcqu(x)=f(x,u(x),(ϕu)(x),(ψu)(x)),1<q≤2,u′(0)+au(η1)=0,bu′(1)+u(η2)=0,0<η1≤η2<1,
where Dc is the Caputo's fractional derivative and f:[0,1]×X→X is continuous.In this paper, we consider the following nonlinear fractional integro-differential equation with four-point nonlocal boundary conditions:(1.2)Dqu(x)c+(ϕu)(x)+(ψu)(x)=f(x,u(x)),1<q≤2,u′(0)+au(η1)=0,bu′(1)+u(η2)=0,0<η1≤η2<1,
where cD is the Caputo's fractional derivative and f:[0,1]×X→X is continuous, for γ,δ:[0,1]×[0,1]→[0,+∞),
(1.3)(ϕu)(x)=∫0x‍γ(x,t)u(t)dt,(ψu)(x)=∫0x‍δ(x,t)u(t)dt,
and a,b∈(0,1). Here, (X,∥·∥) is a Banach space and C=C([0,1],X) denotes the Banach space of all continuous functions from [0,1]→X endowed with a topology of uniform convergence with the norm denoted by ∥·∥.Actually, we remark that the boundary conditions in (1.2) arise in the study of heat flow problems involving a bar of unit length with two controllers at t=0 and t=1 adding or removing heat according to the temperatures detected by two sensors at t=η1 and t=η2.The rest of the paper is organized as follows. We begin by introducing some necessary definitions and mathematical preliminaries of the fractional calculus theory which are required for establishing our results. Then we construct some special reproducing kernel spaces, and the new reproducing kernel method is introduced in Section3. In Section 4 we present one examples to demonstrate the efficiency of the method.


## 2. Preliminaries

Let us recall some basic definition and lemmas on fractional calculus.Definition 2.1.
For a functiong:[0,+∞)→ℝ, the Caputo derivative of fractional order q is defined as
(2.1)cDqg(t)=1Γ(n-q)∫0t‍(t-s)n-q-1g(n)(s)ds,n-1<q≤n,q>0,
where Γ denotes the gamma function.Definition 2.2.
The Riemann-Liouville fractional integral of orderq is defined as
(2.2)Iqg(t)=1Γ(q)∫0t‍g(s)(t-s)1-qds,q>0,
provided the integral exists.Lemma 2.3 (see [4]).
For a givenσ∈C[0,1], the unique solution of the boundary value problem
(2.3)Dqu(x)c=σ(x),0<x<1,1<q≤2,u′(0)+au(η1)=0,bu′(1)+u(η2)=0,0<η1≤η2<1,
is given by
(2.4)u(x)=∫0x‍(x-s)q-1Γ(q)σ(s)ds+a(b+η2-x)1+a(η1-η2-b)∫0η1‍(η1-s)q-1Γ(q)σ(s)ds+ax-(1+aη1)1+a(η1-η2-b)[b∫01‍(1-s)q-2Γ(q-1)σ(s)ds+∫0η2‍(η2-s)q-1Γ(q)σ(s)ds].To introduce the next lemma, we need the following assumptions.(A1)
There exist positive functionsL1(t),  L2(t),  L3(t) such that
(2.5)‖f(t,u(t),(ϕu)(t),(ψu)(t))-f(t,v(t),(ϕv)(t),(ψv)(t))‖≤L1(t)‖u-v‖+L2(t)‖ϕu-ϕv‖+L3(t)‖ψu-ψv‖,∀t∈[0,1],u,v∈X.Further,
(2.6)γ0=sup⁡t∈[0,1]|∫0t‍γ(t,s)ds|,δ0=sup⁡t∈[0,1]|∫0t‍δ(t,s)ds|,ILq=max⁡{sup⁡t∈[0,1]|IqL1(t)|,sup⁡t∈[0,1]|IqL2(t)|,sup⁡t∈[0,1]|IqL3(t)|},Iq-1L(1)=max⁡{|Iq-1L1(1)|,|Iq-1L2(1)|,|Iq-1L3(1)|},IqL(ηi)=max⁡{|IqL1(ηi)|,|IqL2(ηi)|,|IqL3(ηi)|},i=1,2.(A2)
There exist a numberκ such that Δ≤κ<1, t∈[0,1], where
(2.7)Δ=(1+γ0+δ0){ILq+λ1IqL(η1)+λ2(bIq-1L(1)+IqL(η2))},λ1=sup⁡t∈[0,1]|a(b+η2-t)1+a(η1-η2-b)|,λ2=sup⁡t∈[0,1]|at-(1+aη1)1+a(η1-η2-b)|.Lemma 2.4 (see [4]).
Assume thatf:[0,1]×X×X×X→X is a jointly continuous function and satisfies  assumption (A1). Then the boundary value problem (1.1) has an unique solution provided Δ<1, where Δ is given in assumption (A2).For more information on the mathematical properties of fractional derivatives and integrals one can consult the mentioned references.


## 3. Reproducing Kernel Method

### 3.1. Some Reproducing Kernel Spaces

Firstly, inner spaceW21[0,1] is defined as W21[0,1]={u(x)∣u is absolutely continuous real-valued functions, u′∈L2[0,1]}. The inner product in W21[0,1] is given by
(3.1)(f,h)W21=f(0)h(0)+∫01‍f′(t)h′(t)dt,f,h∈W21[0,1],
and the norm ∥u∥W21 is denoted by ∥u∥W21=(u,u)W21. From [12], W21[0,1] is a reproducing kernel Hilbert space and the reproducing kernel is
(3.2)K1(t,s)=1+min⁡{t,s}.In order to solve (1.2) using RKM, we construct a reproducing kernel space H23[0,1] in which every function satisfies the boundary conditions of (1.2). Inner space H23[0,1] is defined as H23[0,1]={u(x)|u,u′,u′′ are absolutely continuous real valued functions, u′′′∈L2[0,1], and u′(0)+au(c)=0, bu′(1)+u(d)=0}, and the inner product is defined as follows:
(3.3)(f,h)H23=f(0)h(0)+∫01‍f′′′(t)h′′′(t)dt,f,h∈H23[0,1].Theorem 3.13.1.
H
2
3
[
0,1
] is a Hilbert reproducing kernel space.Proof.
Suppose{vn(x)}n=1∞ is a Cauchy sequence in H23[0,1], that means
(3.4)‖vn+p-vn‖2=(vn+p(0)-vn(0))2+∫01‍[vn+p(3)(x)-vn(3)(x)]2dx⟶0,n⟶∞.
Therefore, we have vn+p(0)-vn(0)→0 and ∫01‍[vn+p(3)(x)-vn(3)(x)]2dx→0, which shows that {vn(0)}n=1∞ is a Cauchy sequence in ℝ and {vn(3)(x)}n=1∞ is a Cauchy sequence in space L2[0,1]. So, we have
(3.5)lim⁡n→∞vn(0)⟶λ,∫01‍[vn(3)(x)-h(x)]2dx⟶0,n⟶∞,
where λ is a real constant and h(x)∈L2[0,1].
Let(3.6)g(x)=λ+12∫0x‍(x-t)2h(t)dt+a1x+a2x2,
where a1,a2 are determined by g′(0)+ag(c)=0, and bg′(1)+g(d)=0.
Fromh(x)∈L2[0,1], g′′(x)=∫0x‍h(t)dt+2a2 is absolutely continuous in [0,1] and g′′′(x)=h(x)∈L2[0,1] is almost true everywhere in [0,1]. Consequently, g(x)∈H23[0,1]. Moreover,
(3.7)‖vn-g(x)‖2=(vn(0)-λ)2+∫01‍[vn(3)(x)-h(x)]2dx⟶0,n⟶∞.
That means that, H23[0,1] is complete.
Similar to [13], we can prove that the point-evaluation functional x*(x*(x)=u(x), x∈[0,1]) of H23[0,1] is bounded. So H23[0,1] is a Hilbert reproducing kernel space.From [12, 14], we have the following.Theorem 3.2.
The reproducing kernel ofH23[0,1] is
(3.8)R(t,s)=1120R1(t,s)Δ2+R2(t,s)+R3(t,s)+R2(s,t)+R3(s,t)Δ+{1120s3(s2-5st+10t2),t≥s,1120t3(10s2-5st+t2),t<s,
where
(3.9)Δ=b(-2+a(-2+c)c)-d(d+ac(-c+d)),R1(t,s)=-s((5b(-4+d)d3-6d5)(s+ac(-c+s))-ss+ac3(c2-5cd+10d2)(b(-2+s)+d(-d+s))-ss-b((-40b+5(-4+d)d3)(ac(c-s)-s)=5+5a(-4+c)c3(b(-2+s)+d(-d+s))))t(t+ac(-c+t))+ac3s(-5b(-4+c)(ac(c-s)-s)+(c2-5cd+10d2)(ac(c-s)-s)+6ac2(b(-2+s)+d(-d+s)))t(b(-2+t)+d(-d+t))+120((d-s)(-d+a(c-d)(c-s)-s)+b(-2+a(-2c+c2-(-2+s)s)))×((d-t)(-d+a(c-d)(c-t)-t)+b(-2+a(-2c+c2-(-2+t)t))),R2(t,s)=t(t+ac(-c+t))(-124b(-4+s)s3+{1120s3(10d2-5ds+s2),d≥s1120d3(d2-5ds+10s2),d<s),R3(t,s)=-at(b(-2+t)+d(-d+t))={1120s3(10c2-5cs+s2),c≥s,1120c3(c2-5cs+10s2),c<s.Actually, it is easy to prove that for everyx∈[0,1] and u(y)∈H23[0,1], R(x,y)∈H23[0,1] and (u(y),R(x,y))=u(x) holds, that is, R(x,y) is the reproducing kernel of H23[0,1].


### 3.2. The Reproducing Kernel Method

In recent years, there has been a growing interest in using a reproducing kernel to solve the operator equation. In this section, the representation of analytical solution of (1.2) is given in the reproducing kernel space H23[0,1].NoteLu=cDqu(x)+(ϕu)(x)+(ψu)(x)+β(x)u(x) and F(x,u(x))=f(x,u(x))+β(x)u(x). We can convert (1.2) into an equivalent equation Lu(x)=F(x,u(x)). It is clear that L:H23[0,1]→W21[0,1] is a bounded linear operator.Putφi(x)=K1(xi,x), Ψi(x)=L*φi(x), where L* is the adjoint operator of L. Then
(3.10)Ψi(x)=(L*φi(y),R(x,y))=(φi(y),LyR(x,y))=(LyR(x,y),φi(x))¯=LyR(x,y)|y=xi.Similar to [15], we can prove the following.Lemma 3.3.
Under the previous assumptions, if{xi}i=1∞ is dense on [0,1], then {Ψi(x)}i=1∞ is the complete basis of H23[0,1].The orthogonal system{Ψi¯(x)}i=1∞ of H23[0,1] can be derived from Gram-Schmidt orthogonalization process of {Ψi(x)}i=1∞, and
(3.11)Ψi¯(x)=∑j=1i‍βijΨj(x).We also can prove the following theorem.Theorem 3.4.
If{xi}i=1∞ is dense on [0,1] and the solution of (1.2) is unique, the solution can be expressed in the form
(3.12)u(x)=∑i=1∞‍∑k=1i‍βikF(xk,u(xk))Ψi¯(x).The approximate solution of the (1.2) is
(3.13)un(x)=∑i=1n‍∑k=1i‍βikF(xk,u(xk))Ψi¯(x).If (1.2) is linear, that is F(x,u(x))=F(x), then the approximate solution of (1.2) can be obtained directly from (3.13). Else, the approximate process could be modified into the following form:
(3.14)u0(x)=0,un+1(x)=∑i=1n+1‍BiΨi¯(x),
where Bi=∑k=1i‍βikF(xk,un(xk)).


## 3.1. Some Reproducing Kernel Spaces

Firstly, inner spaceW21[0,1] is defined as W21[0,1]={u(x)∣u is absolutely continuous real-valued functions, u′∈L2[0,1]}. The inner product in W21[0,1] is given by
(3.1)(f,h)W21=f(0)h(0)+∫01‍f′(t)h′(t)dt,f,h∈W21[0,1],
and the norm ∥u∥W21 is denoted by ∥u∥W21=(u,u)W21. From [12], W21[0,1] is a reproducing kernel Hilbert space and the reproducing kernel is
(3.2)K1(t,s)=1+min⁡{t,s}.In order to solve (1.2) using RKM, we construct a reproducing kernel space H23[0,1] in which every function satisfies the boundary conditions of (1.2). Inner space H23[0,1] is defined as H23[0,1]={u(x)|u,u′,u′′ are absolutely continuous real valued functions, u′′′∈L2[0,1], and u′(0)+au(c)=0, bu′(1)+u(d)=0}, and the inner product is defined as follows:
(3.3)(f,h)H23=f(0)h(0)+∫01‍f′′′(t)h′′′(t)dt,f,h∈H23[0,1].Theorem 3.13.1.
H
2
3
[
0,1
] is a Hilbert reproducing kernel space.Proof.
Suppose{vn(x)}n=1∞ is a Cauchy sequence in H23[0,1], that means
(3.4)‖vn+p-vn‖2=(vn+p(0)-vn(0))2+∫01‍[vn+p(3)(x)-vn(3)(x)]2dx⟶0,n⟶∞.
Therefore, we have vn+p(0)-vn(0)→0 and ∫01‍[vn+p(3)(x)-vn(3)(x)]2dx→0, which shows that {vn(0)}n=1∞ is a Cauchy sequence in ℝ and {vn(3)(x)}n=1∞ is a Cauchy sequence in space L2[0,1]. So, we have
(3.5)lim⁡n→∞vn(0)⟶λ,∫01‍[vn(3)(x)-h(x)]2dx⟶0,n⟶∞,
where λ is a real constant and h(x)∈L2[0,1].
Let(3.6)g(x)=λ+12∫0x‍(x-t)2h(t)dt+a1x+a2x2,
where a1,a2 are determined by g′(0)+ag(c)=0, and bg′(1)+g(d)=0.
Fromh(x)∈L2[0,1], g′′(x)=∫0x‍h(t)dt+2a2 is absolutely continuous in [0,1] and g′′′(x)=h(x)∈L2[0,1] is almost true everywhere in [0,1]. Consequently, g(x)∈H23[0,1]. Moreover,
(3.7)‖vn-g(x)‖2=(vn(0)-λ)2+∫01‍[vn(3)(x)-h(x)]2dx⟶0,n⟶∞.
That means that, H23[0,1] is complete.
Similar to [13], we can prove that the point-evaluation functional x*(x*(x)=u(x), x∈[0,1]) of H23[0,1] is bounded. So H23[0,1] is a Hilbert reproducing kernel space.From [12, 14], we have the following.Theorem 3.2.
The reproducing kernel ofH23[0,1] is
(3.8)R(t,s)=1120R1(t,s)Δ2+R2(t,s)+R3(t,s)+R2(s,t)+R3(s,t)Δ+{1120s3(s2-5st+10t2),t≥s,1120t3(10s2-5st+t2),t<s,
where
(3.9)Δ=b(-2+a(-2+c)c)-d(d+ac(-c+d)),R1(t,s)=-s((5b(-4+d)d3-6d5)(s+ac(-c+s))-ss+ac3(c2-5cd+10d2)(b(-2+s)+d(-d+s))-ss-b((-40b+5(-4+d)d3)(ac(c-s)-s)=5+5a(-4+c)c3(b(-2+s)+d(-d+s))))t(t+ac(-c+t))+ac3s(-5b(-4+c)(ac(c-s)-s)+(c2-5cd+10d2)(ac(c-s)-s)+6ac2(b(-2+s)+d(-d+s)))t(b(-2+t)+d(-d+t))+120((d-s)(-d+a(c-d)(c-s)-s)+b(-2+a(-2c+c2-(-2+s)s)))×((d-t)(-d+a(c-d)(c-t)-t)+b(-2+a(-2c+c2-(-2+t)t))),R2(t,s)=t(t+ac(-c+t))(-124b(-4+s)s3+{1120s3(10d2-5ds+s2),d≥s1120d3(d2-5ds+10s2),d<s),R3(t,s)=-at(b(-2+t)+d(-d+t))={1120s3(10c2-5cs+s2),c≥s,1120c3(c2-5cs+10s2),c<s.Actually, it is easy to prove that for everyx∈[0,1] and u(y)∈H23[0,1], R(x,y)∈H23[0,1] and (u(y),R(x,y))=u(x) holds, that is, R(x,y) is the reproducing kernel of H23[0,1].


## 3.2. The Reproducing Kernel Method

In recent years, there has been a growing interest in using a reproducing kernel to solve the operator equation. In this section, the representation of analytical solution of (1.2) is given in the reproducing kernel space H23[0,1].NoteLu=cDqu(x)+(ϕu)(x)+(ψu)(x)+β(x)u(x) and F(x,u(x))=f(x,u(x))+β(x)u(x). We can convert (1.2) into an equivalent equation Lu(x)=F(x,u(x)). It is clear that L:H23[0,1]→W21[0,1] is a bounded linear operator.Putφi(x)=K1(xi,x), Ψi(x)=L*φi(x), where L* is the adjoint operator of L. Then
(3.10)Ψi(x)=(L*φi(y),R(x,y))=(φi(y),LyR(x,y))=(LyR(x,y),φi(x))¯=LyR(x,y)|y=xi.Similar to [15], we can prove the following.Lemma 3.3.
Under the previous assumptions, if{xi}i=1∞ is dense on [0,1], then {Ψi(x)}i=1∞ is the complete basis of H23[0,1].The orthogonal system{Ψi¯(x)}i=1∞ of H23[0,1] can be derived from Gram-Schmidt orthogonalization process of {Ψi(x)}i=1∞, and
(3.11)Ψi¯(x)=∑j=1i‍βijΨj(x).We also can prove the following theorem.Theorem 3.4.
If{xi}i=1∞ is dense on [0,1] and the solution of (1.2) is unique, the solution can be expressed in the form
(3.12)u(x)=∑i=1∞‍∑k=1i‍βikF(xk,u(xk))Ψi¯(x).The approximate solution of the (1.2) is
(3.13)un(x)=∑i=1n‍∑k=1i‍βikF(xk,u(xk))Ψi¯(x).If (1.2) is linear, that is F(x,u(x))=F(x), then the approximate solution of (1.2) can be obtained directly from (3.13). Else, the approximate process could be modified into the following form:
(3.14)u0(x)=0,un+1(x)=∑i=1n+1‍BiΨi¯(x),
where Bi=∑k=1i‍βikF(xk,un(xk)).


## 4. Convergent Theorem of the Numerical Method

In this section, we will give the following convergent theorem of our algorithm.Lemma 4.1.
There exists a constantM, satisfied |u(x)|≤M∥u∥H23, for all u(x)∈H23[0,1].Proof.
For all thex∈[0,1] and u∈H23[0,1], there are
(4.1)|u(x)|=|(u(⋅),K3(⋅,x))|≤‖K3(⋅,x)‖H23⋅‖u‖H23K3(·,x)∈H23[0,1], and note that
(4.2)M=max⁡x∈[0,1]‖K3(⋅,x)‖H23.
That is, |u(x)|≤M∥u∥H23.By Lemma4.1, it is easy to obtain the following lemma.Lemma 4.2.
Ifun→∥·∥u- (n→∞), ∥un∥ is bounded, xn→y  (n→∞) and F(x,u(x)) is continuous, then F(xn,un-1(xn))→F(y,u-(y)).Theorem 4.3.
Suppose that∥un∥ is bounded in (3.13) and (1.2) has a unique solution. If {xi}i=1∞ is dense on [0,1], then the n-term approximate solution un(x) derived from the above method converges to the analytical solution u(x) of (1.2).Proof.
First, we will prove the convergence ofun(x).
From (3.14), we infer that
(4.3)un+1(x)=un(x)+Bn+1Ψn+1¯(x).
The orthonormality of {Ψi¯}i=1∞ yields that
(4.4)‖un+1‖2=‖un‖2+(Bn+1)2=⋯=∑i=1n+1‍(Bi)2.
That means ∥un+1∥≥∥un∥. Due to the condition that ∥un∥ is bounded, ∥un∥ is convergent and there exists a constant ℓ such that
(4.5)∑i=1∞‍(Bi)2=l.
If m>n, then
(4.6)‖um-un‖2=‖um-um-1+um-1-um-2+⋯+un+1-un‖2.
In view of (um-um-1)⊥(um-1-um-2)⊥⋯⊥(un+1-un), it follows that
(4.7)‖um-un‖2=‖um-um-1‖2+‖um-1-um-2‖2+⋯+‖un+1-un‖2=∑i=n+1m‍(Bi)2⟶0asn⟶∞.
The completeness of H23[0,1] shows that un→u- as n→∞ in the sense of ∥·∥H23.
Secondly, we will prove thatu- is the solution of (1.2).
Taking limits in (3.12), we get
(4.8)u-(x)=∑i=1∞‍BiΨi¯(x).
So
(4.9)Lu-(x)=∑i=1∞‍BiLΨi¯(x),(Lu-)(xn)=∑i=1∞‍Bi(LΨi¯,φn)=∑i=1∞‍Bi(Ψi¯,L*φn)=∑i=1∞‍Bi(Ψi¯,Ψn).
Therefore,
(4.10)∑i=1n‍βnj(Lu-)(xn)=∑i=1∞‍Bi(Ψi¯,∑j=1n‍βnjΨj)=∑i=1∞‍Bi(Ψi¯,Ψn¯)=Bn.
If n=1, then
(4.11)Lu-(x1)=F(x1,u0(x1)).
If n=2, then
(4.12)β21Lu-(x1)+β22Lu-(x2)=β21F(x1,u0(x1))+β22F(x2,u1(x2)).
It is clear that
(4.13)(Lu-)(x2)=F(x2,u1(x2)).
Moreover, it is easy to see by induction that
(4.14)(Lu-)(xj)=F(xj,uj-1(xj)),j=1,2,….
Since {xi}i=1∞ is dense on [0,1], for all Y∈[0,1], there exists a subsequence {xnj}j=1∞ such that
(4.15)xnj⟶Yasj⟶∞.
It is easy to see that (Lu-)(xnj)=F(xnj,unj-1(xnj)). Let j→∞; by the continuity of F(x,u(x)) and Lemma 4.2, we have
(4.16)(Lu-)(Y)=F(Y,u-(Y)).
At the same time, u-∈H23[0,1]; clearly, u satisfies the boundary conditions of (1.2).
That is,u- is the solution of (1.2).
The proof is complete.In fact,un(x) is just the orthogonal projection of exact solution u-(x) onto the space Span{Ψi¯}i=1n¯.


## 5. Numerical Example

To give a clear overview of the methodology as a numerical tool, we consider one example in this section. We apply the reproducing kernel method and results obtained by the method are compared with the analytical solution of each example and are found to be in good agreement with each other. Also, the numerical results obtained are compared with the corresponding experimental results obtained by the methods presented in [8, 9].Example 5.1.
Consider the following boundary value problem:(5.1)D3/2cu(t)=15∫0te-(s-t)+e-(s-t)/25u(s)ds+u2(t)-2t2u(t)+20t17u(t)+454153u(t)+f(t),t∈[0,1],u′(0)+12u(13)=0,14u′(1)+u(23)=0,
where f(t)=1674244/585225-2354et/2/3825-169et/3825+4t/π-3316t/2601-162647t2/65025+20t3/17+t4. According to Lemma 2.4, the boundary value problem (5.1) has a unique solution on [0,1]. u(t)=t2+10t/17-227/153 is the solution of (5.1), so it is the one and the only one solution. Using our method, taking xi=(i-1)/(N-1),  i=1,2,…,N,N=11,101, the numerical results are given in Table 1.Table 1
Absolute errors for Example5.1.
x
True solution
DM [8]
VIM [9]
RKM (u511)
RKM (u5101)
0
0.05566
5.61689
E
-
5
1.48366
E
-
4
2.16993
E
-
5
2.11552
E
-
7
0.1
0.19798
5.47421
E
-
5
1.46246
E
-
4
1.92463
E
-
5
1.90591
E
-
7
0.2
0.39473
5.43973
E
-
5
1.49244
E
-
4
1.70817
E
-
5
1.59788
E
-
7
0.3
0.60560
5.47579
E
-
5
1.55807
E
-
4
1.49046
E
-
5
1.36933
E
-
7
0.4
0.77891
5.56624
E
-
5
1.65449
E
-
4
1.26031
E
-
5
1.14128
E
-
7
0.5
0.83516
5.68976
E
-
5
1.78225
E
-
4
1.02477
E
-
5
9.25561
E
-
8
0.6
0.70036
5.80389
E
-
5
1.94554
E
-
4
7.88466
E
-
6
6.84821
E
-
8
0.7
0.50144
5.82441
E
-
5
2.1518
E
-
4
5.65515
E
-
6
4.82468
E
-
8
0.8
0.29175
5.5968
E
-
5
2.41183
E
-
4
3.54481
E
-
6
2.9536
E
-
8
0.9
0.11797
4.85764
E
-
5
2.74062
E
-
4
1.50572
E
-
6
9.8474
E
-
9
1.0
0.01485
3.18397
E
-
5
3.15881
E
-
4
9.79078
E
-
7
1.3896
E
-
8


## 6. Conclusion

In this paper, RKM is presented to solve four-point nonlocal boundary value problem of nonlinear integro-differential equations of fractional orderq∈(1,2]. The results of numerical examples demonstrate that the present method is more accurate than the existing methods.



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*Source: 329575-2012-09-05.xml*