# Solution to the Linear Fractional Differential Equation Using Adomian Decomposition Method

Authors: Jin-Fa Cheng; Yu-Ming Chu
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

We obtain the analytical general solution of the linear fractional differential equations with constant coefficients by Adomian decomposition method under nonhomogeneous initial value condition, which is in the sense of the Caputo fractional derivative.


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## Body

## 1. Introduction

Fractional differential equations are hot topics both in mathematics and physics. Recently, the fractional differential equations have been the subject of intensive research. There are several methods to obtain the solution, such as the Laplace transform method, power series method, and Green function method. Many remarkable results for the fractional differential equations can be found in the literature [1–11]. In particular, the Adomian decomposition method has attracted the attention of many mathematicians [12–15].For a better understanding of the fractional derivatives and for a physical understanding of the fractional equations, the readers can refer to the recent publications in [16, 17]. Ebaid [18] suggested a modification of the Adomian method, and a few iterations lead to exact solution. Das [19] compared the variational iteration method with the Adomian method for fractional equations and found that the variational iteration method is much more effective. For other methods of the fractional differential equations, especially the homotopy perturbation method, variational iteration method and differential transform method were presented in [20, 21].Consider the followingn-term fractional differential equation with constant coefficients:(1.1)an[DCβn]y(t)+an-1[Dβn-1C]y(t)+⋯+a1[DCβ1]y(t)+a0[Dβ0C]y(t)=f(t),
where n+1>β≥n>βn-1>⋯>β1>β0 and ai(i=0,1,…,n) is a real constant. In [12], the authors obtain the particular solution of (1.1) of the homogeneous initial value problem of the form (1.2)yi(0)=0,i=0,1,…,n.However, it seems also more meaningful and more complicated for solving general solution of (1.1) under nonhomogeneous initial value condition. Therefore, in this paper, we will remove the restriction of the homogeneous initial value, consider the nonhomogeneous initial value problems of the form (1.3)yji(0)=ciji,i=0,1,…,n,ji=1,2,…,li,li-1⩽βi<li,
and obtain the analytical general solution of (1.1), which generalizes the result in [12].We organize the paper as follows. In Section2, we give some basic definitions and properties. In Section 3, we obtain the analytical general solution of the linear fractional differential equations by Adomian decomposition method.  Some explicit examples are given in Section 4.


## 2. Basic Definitions and Notations

Definition 2.1 (see [1]).
The Riemann-Liouville integral of orderp is defined by
(2.1)Dat-pf(t)=1Γ(p)∫at(t-τ)p-1f(τ)dτ,p>0.
From Definition 2.1, we clearly see that
(2.2)Dat-α((t-a)ν)=Γ(1+ν)Γ(1+ν+α)(t-a)ν+α,(2.3)Datα((t-a)ν)=Γ(1+ν)Γ(1+ν-α)(t-a)ν-α,
where α>0 and ν is a real number.Definition 2.2 (see [1]).
Forf(t)∈Cm(m∈N), the Caputo fractional derivative of f(t) is defined by
(2.4)DaCtpf(t)={Dm-pf(m)(t),m-1<p<m,d(m)dtmf(t),p=m,Dtpf(t)a,p⩽0.
Therefore,
(2.5)Dact-α((t-a)ν)=Dat-α((t-a)ν)=Γ(1+ν)Γ(1+ν+α)(t-a)ν+α.Lemma 2.3 (see [1]).
Iff(t) is continuous, then
(2.6)DaCt-nf(n)(t)=Dat-nf(n)(t)=f(t)-∑j=0n-1f(j)(a)(t-a)jΓ(j+1),n∈Z.Lemma 2.4 (see [1]).
Iff(t) is continuous, then
(2.7)DaCt-p(aCDt-q)f(t)=Dat-p(Dat-q)f(t)=Dat-p-qf(t)=DaCt-p-qf(t)=DaCt-q(aCDt-p)f(t).Lemma 2.5.
Iff(t) is continuous, then
(2.8)DaCt-p(aCDtq)f(t)=Datq-pf(t)-∑j=0m-1f(j)(a)(t-a)j-q+pΓ(j-q+p+1),
where m-1<q<m and q⩽p.Proof.
From Definition2.2 and Lemmas 2.3–2.4, we get
(2.9)aCDt-p(aCDtq)f(t)=Dat-p(Dtq-ma)f(m)(t)=Dat-p+q[Dt-qa(Dtq-ma)f(m)(t)]=Dat-p+q[Dt-ma(t)f(m)(t)]=Dat-p+q[f(t)-∑j=0m-1f(j)(a)(t-a)jΓ(j+1)]=Datq-pf(t)-∑j=0m-1f(j)(a)(t-a)j-q+pΓ(j-q+p+1).
It is easy to see that
(2.10)DaCt-p(aCDtp)f(t)=f(t)-∑j=0m-1f(j)(a)(t-a)jΓ(j+1).Proposition 2.6 (see [12]).
One has(2.11)∑k1=0∞⋯∑kn=0∞ak1,k2⋯kn-1kn=∑m=0∞‍∑k1,…,kn-1,kn⩾0k1+⋯+kn-1+kn=mak1,k2⋯kn-1kn.Proposition 2.7 (see [12]).
More over, one has(2.12)∑m=0∞∑k1,…,kn-1,kn⩾0k1+⋯+kn-1+kn=mak1,k2⋯kn-1kn=∑s=0∞∑k1,…,kn-1⩾0k1+⋯+kn-1=s∑kn=0∞ak1,k2⋯kn-1kn.


## 3. The Analytical Solution of the Linear Constant Coefficient Fractional Differential Equation

For simplicity, ifa=0, then we denote aCDtp or aCDt-p by CDp or CD-p.In this section,  we use Adomian decomposition method to discuss the general form of the linear fractional differential equations with constant coefficients, and apply and some basic transformation and integration to obtain the solution of the equations.Let us consider the followingn-term linear fractional differential equations with constant coefficients:
(3.1)an[DβnC]y(t)+an-1[Dβn-1C]y(t)+⋯+a1[Dβ1C]y(t)+a0[Dβ0C]y(t)=f(t),y(ji)(0)=ciji,i=0,1,…,n,ji=1,2,…,li,li-1⩽βi<li,
where n+1>βn⩾n>βn-1⋯>β1>β0,aiandciji are real constants,  CDp=D0Ctp denotes Caputo fractional derivative of order α.ApplyingCD-βn to both sides of (1.1) and utilizing Lemma 2.5, we get
(3.2)y(t)+an-1anDCβn-1-βny(t)+⋯+a0anDCβ0-βny(t)=1anDC-βnf(t)+1an∑i=0nai∑ji=0li-1y(ji)(0)tβn-βi+jiΓ(1+βn-βi+ji).By the Adomian decomposition method, we obtain the recursive relationship as follows:(3.3)y0(t)=1anDC-βnf(t)+1an∑i=0nai∑ji=0li-1y(ji)(0)tβn-βi+jiΓ(1+βn-βi+ji),y1(t)=-(an-1anDCβn-1-βn+⋯+a0anDCβ0-βn)y0(t),y2(t)=(-1)2(an-1anDCβn-1-βn+⋯+a0anDCβ0-βn)2y0(t),⋮ys(t)=(-1)s(an-1anDCβn-1-βn+⋯+a0anDCβ0-βn)sy0(t),⋮By Adomian decomposition method, adding all terms of the recursion,  we obtain the solution of (1.1) as
(3.4)y(t)=∑s=0∞ys(t)=∑s=0∞(-1)s(an-1anDCβn-1-βn+⋯+a0anDCβ0-βn)sy0(t)=1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)sD-βnf(t)+1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)s∑i=0nai∑ji=0li-1y(ji)(0)tβn-βi+jiΓ(1+βn-βi+ji).
Let
(3.5)I1=1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)sD-βnf(t)I2=1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)s∑i=0nai∑ji=0li-1y(ji)(0)tβn-βi+jiΓ(1+βn-βi+ji).
Then,(3.6)y(t)=I1+I2.Next, we estimateI1 and I2, respectively.ForI1, by [12] we obtain
(3.7)1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)sD-βnf(t),=∫0t1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(m;k0,k1,…,kn-2)×∏p=0n-2(apan)kp(t-τ)(βn-βn-1)m+βn+∑j=0n-2(βn-1-βj)kj-1×Eβn-βn-1,βn+∑j=0n-2(βn-1-βj)kj(m)(-an-1anDβn-βn-1)f(τ)dτ.ForI2, by the initial conditions (2.10) we get
(3.8)1an∑s=0∞(-1)s(an-1anDβn-1-βn+⋯+a0anDβ0-βn)s∑i=0nai∑ji=0li-1y(ji)(0)tβn-βi+jiΓ(1+βn-βi+ji)=1an∑s=0∞(-1)s∑k0,k1,…,kn-1⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!(an-1an)kn-1(an-2an)kn-2⋯(a0an)k0×Dkn-1(βn-1-βn)+kn-2(βn-2-βn)+⋯+k0(β0-βn)∑i=0nai∑ji=0li-1cijitβn-βi+jiΓ(1+βn-βi+ji).Using formulas (2.2) and (2.3), the above expression can be written as(3.9)1an∑s=0∞(-1)s∑k0,k1,…,kn-1⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!(an-1an)kn-1(an-2an)kn-2⋯(a0an)k0×∑i=0nai∑ji=0li-1cijiΓ(𝒜)tβn-βi+ji+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0)Γ(𝒜)·Γ(𝒜+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0))=1an∑s=0∞(-1)s∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!(an-1an)kn-1(an-2an)kn-2⋯(a0an)k0×∑i=0nai∑ji=0li-1cijitβn-βi+ji+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0)Γ(𝒜+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0))=∑i=0nai∑ji=0li-1cijiD(βi-ji-1)[1an∑s=0∞(-1)s∑k0,k1,…,kn-1⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!∑k0,k1,⋯,kn-1⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!×(an-1an)kn-1(an-2an)kn-2⋯(a0an)k0×tβn-1+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0)Γ(βn+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0))∑k0,k1,⋯,kn-1⩾0k0+k1+⋯+kn-1=ss!k0!k1!⋯kn-1!],
where 𝒜 denotes 1+βn-βi+ji.Using Propositions2.6 and 2.7, the above solution is equivalent to the following form:
(3.10)∑i=0nai∑ji=0li-1cijiD(βi-ji-1)[1an∑kn-1=0∞∑m=0∞∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(-1)kn-1+m(kn-1+m)!k0!k1!⋯kn-1!×(an-1an)kn-1(an-2an)kn-2⋯(a0an)k0×tβn-1+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0)Γ(βn+kn-1(βn-βn-1)+kn-2(βn-βn-2)+⋯+k0(βn-β0))∑k0,k1,⋯,kn-2⩾0k0+k1+⋯+kn-2=mm!k0!k1!⋯kn-2!]=∑i=0nai∑ji=0li-1cijiD(βi-ji-1)[1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=mm!k0!k1!⋯kn-2!×∏r=0n-2(aran)krt(βn-βn-1)m+∑ξ=0n-2(βn-1-βξ)kξ+βn-1×∑kn-1=0∞(-1)kn-1(an-1an)kn-1(kn-1+m)!kn-1!×tkn-l(βn-βn-1)Γ(kn-1(βn-βn-1)+(βn-βn-1)m+∑δ=0n-2(βn-1-βδ)kδ+βn)∑k0,k1,⋯,kn-2⩾0k0+k1+⋯+kn-2=mm!k0!k1!⋯kn-2!]=∑i=0nai∑ji=0li-1cijiD(βi-ji-1)[1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(m;k0,k1,…,kn-2)×∏r=0n-2(aran)krt(βn-βn-1)m+∑ξ=0n-2(βn-1-βξ)kξ+βn-1×Eβn-βn-1,∑δ=0n-2(βn-1-βl)kδ+βn(m)(-an-1antβn-βn-1)∑k0,k1,⋯,kn-2⩾0k0+k1+⋯+kn-2=mm!k0!k1!⋯kn-2!].Therefore,(3.11)y(t)=I1+I2=∫0t1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(m;k0,k1,…,kn-2)×∏p=0n-2(apan)kp(t-τ)(βn-βn-1)m+βn+∑j=0n-2(βn-1-βj)kj-1×Eβn-βn-1,βn+∑j=0n-2(βn-1-βj)kj(m)(-an-1anDβn-βn-1)f(τ)dτ+∑i=0nai∑ji=0li-1cijiD(βi-ji-1)[1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(m;k0,k1,…,kn-2)×∏r=0n-2(aran)krt(βn-βn-1)m+∑ξ=0n-2(βn-1-βξ)kξ+βn-1×Eβn-βn-1,∑δ=0n-2(βn-1-βl)kδ+βn(m)(-an-1antβn-βn-1)∑k0,k1,⋯,kn-2⩾≥0k0+k1+⋯+kn-2=m(m;k0,k1,⋯,kn-2)],
where (3.12)(m;k0,k1,…,kn-2)=m!k0!k1!⋯kn-2!,
and Eλ,μ(i)(y) is the Mittag-Leffler function (3.13)Eλ,μ(i)(y)=didyiEλ,μ(y)=∑j=0∞(i+j)!yjj!Γ(λj+λi+μ).
Substituting the Green function
(3.14)Gn(t)=1an∑m=0∞(-1)mm!∑k0,k1,…,kn-2⩾0k0+k1+⋯+kn-2=m(m;k0,k1,…,kn-2)×∏p=0n-2(apan)kpt(βn-βn-1)m+βn+∑j=0n-2(βn-1-βj)kj-1×Eβn-βn-1,βn+∑j=0n-2(βn-1-βj)kj(m)(-an-1anDβn-βn-1)
into the above expression, we know that(3.15)y(t)=∫0tGn(t-τ)f(τ)dτ+∑i=0nai∑ji=0li-1cijiGn(βi-ji-1)(t)
is the analytical general solution of (1.1).


## 4. Illustrative Examples

In order to verify our conclusions,  we give some examples.(1). Consider an initial value problem for the relaxation-oscillation equation (see [1])
(4.1)DCαy(t)+Ay(t)=f(t),t>0,y(j)(0)=bj,j=1,2,…,m-1,m-1<α<m,
where bj are real constants.Utilizing Lemma2.5 and applying CD-α to both sides of (4.1), we obtain (4.2)y(t)+AD-αy(t)=D-αf(t)+∑j=0m-1y(j)(0)tjΓ(j+1).According to the above procedure of solving the linear fractional differential equations with constant coefficients and using the Adomian decomposition method, let(4.3)y0(t)=D-αf(t)+∑j=0m-1y(j)(0)tjΓ(j+1),y1(t)=-AD-αy0(t),y2(t)=-AD-αy1(t)=(-A)2D-2αy0(t),⋮ys(t)=-AD-αys-1(t)=(-A)sD-sαy0(t),⋮Adding all of the above terms, we obtain the solution of the equation by Adomian decomposition method as follows:(4.4)y(t)=∑s=0∞ys(t)=∑s=0∞(-A)sD-sαy0(t)=∑s=0∞(-A)sD(-s-1)αf(t)+∑s=0∞(-A)sD-sα∑j=0m-1y(j)(0)tjΓ(j+1)=∑s=0∞(-A)s1Γ((s+1)α)∫0t(t-τ)(s+1)α-1f(τ)dτ+∑j=0m-1bj∑s=0∞(-A)stj+sαΓ(j+sα+1)=∫0t∑s=0∞(t-τ)α-1·[-A(t-τ)α]sΓ(sα+α)f(τ)dτ+∑j=0m-1bjD(α-j-1)[∑s=0∞tα-1(-Atα)sΓ(sα+α)]=∫0t(t-τ)α-1Eα,α(-A(t-τ)α)f(τ)dτ+∑j=0m-1bjD(α-j-1)[tα-1Eα,α(-Atα)]=∫0tG2(t-τ)f(τ)dτ+∑j=0m-1bjD(α-j-1)G2(t),
where G2(t)=tα-1Eα,α(-Atα).It is easy to see that(4.5)y(t)=∫0tG2(t-τ)f(τ)dτ+b0D(α-1)G2(t),(0<α<1).(2). Consider an initial value problem for the nonhomogeneous Bagley-Torvik equation (see [5])
(4.6)Ay′′(t)+BCD3/2y(t)+Cy(t)=f(t),t>0,y(i)(0)=ai,i=0,1,
where aiare real constants.Utilizing Lemma2.5 and applying CD-2 to both sides of (4.6), we obtain
(4.7)y(t)+1A[BD-1/2y(t)+CD-2y(t)]=1AD-2f(t)+∑i=01y(i)(0)tiΓ(i+1)+BA∑i=01y(i)(0)t1/2+iΓ(3/2+i)=1AD-2f(t)+∑i=01y(i)(0)[tiΓ(i+1)+BA·t1/2+iΓ(3/2+i)]According to the above procedure of solving the linear fractional differential equation with constant coefficients and using the Adomian decomposition method, let(4.8)y0(t)=1AD-2f(t)+∑i=01y(i)(0)[tiΓ(i+1)+BA·t1/2+iΓ(3/2+i)],y1(t)=-1A[BD-1/2y0(t)+CD-2y0(t)]=-CA(BCI+D-3/2)D-1/2y0(t),y2(t)=-CA(BCI+D-3/2)D-1/2y1(t)=(-CA)2(BCI+D-3/2)2D-2/2y0(t),⋮yn(t)=-CA(BCI+D-3/2)D-1/2yn-1(t)=(-CA)n(BCI+D-3/2)nD-n/2y0(t),⋮Adding all of the above terms, we obtain the solution of the equation by Adomian  decomposition method as follows:(4.9)y(t)=∑n=0∞yn(t)=∑n=0∞(-CA)n(BCI+D-3/2)nD-n/2y0(t)=∑n=0∞(-CA)n(BCI+D-3/2)nD-n/2[1AD-2f(t)]+∑n=0∞(-CA)n(BCI+D-3/2)nD-n/2∑i=01y(i)(0)[tiΓ(i+1)+BA·t1/2+iΓ(3/2+i)]=1A∑n=0∞(-CA)n∑k=0nCnn-k(BCI)n-kD-3k/2D-n/2D-2f(t)+∑i=01ai∑n=0∞(-CA)n∑k=0nCnn-k(BCI)n-kD-3k/2D-n/2[tiΓ(i+1)+BA·t1/2+iΓ(3/2+i)]=1A∑n=0∞(-CA)n∑k=0nn!(n-k)!k!(BCI)n-kD-(3k/2+n/2+2)f(t)+∑i=01ai∑n=0∞(-CA)n∑k=0nn!(n-k)!k!(BCI)n-kD-(3k/2+n/2)[tiΓ(i+1)+BA·t1/2+iΓ(3/2+i)]=1A∫0t∑n=0∞(-CA)n∑k=0nn!(n-k)!k!(BCI)n-k1Γ(3k/2+n/2+2)(t-τ)3k/2+n/2+2-1f(τ)dτ+∑i=01ai∑n=0∞(-CA)n∑k=0nn!(n-k)!k!(BCI)n-k[ti+3k/2+n/2Γ(i+3k/2+n/2+1)+BA·t1/2+i+(3k/2)+(n/2)Γ(i+3k/2+n/2+3/2)]=1A∫0t∑k=0∞∑m=0∞(-CA)k+m(k+m)!m!k!(BCI)m1Γ(3/2k+(k+m/2)+2)×(t-τ)3k/2+(k+m)/2+1f(τ)dτ+∑i=01aiD(1-i)[∑k=0∞∑m=0∞(-CA)k+m(k+m)!m!k!(BCI)mt3k/2+(k+m)/2+1Γ(3/2k+(k+m)/2+2)]+∑i=01aiD(1/2-i)[∑k=0∞‍∑m=0∞(-CA)k+m(k+m)!m!k!(BCI)mt3k/2+(k+m)/2+1Γ(3/2k+(k+m)/2+2)]=1A∫0t∑k=0∞(-C/A)kk!(t-τ)2k+1∑m=0∞(k+m)!m!·(-(B/A)(t-τ)1/2)mΓ(k/2+m/2+3k/2+2)f(τ)dτ+A∑i=01aiD(1-i)[1A∑k=0∞(-C/A)kk!(t-τ)2k+1∑m=0∞(k+m)!m!·(-(B/A)(t-τ)1/2)mΓ(k/2+m/2+3k/2+2)]+A∑i=01aiD(1/2-i)[1A∑k=0∞(-C/A)kk!(t-τ)2k+1×∑m=0∞(k+m)!m!·(-(B/A)(t-τ)1/2)mΓ((k/2)+(m/2)+(3k/2)+2)]=1A∫0t∑k=0∞(-C/A)kk!(t-τ)2k+1E1/2,3k/2+2(k)(-BA(t-τ)1/2)f(τ)dτ+A∑i=01aiD(1-i)[1A∑k=0∞(-C/A)kk!t2k+1E1/2,3k/2+2(k)(-BAt1/2)]+A∑i=01aiD(1/2-i)[1A∑k=0∞(-C/A)kk!t2k+1E1/2,3k/2+2(k)(-BAt1/2)]=∫0tG3(t-τ)f(τ)dτ+A∑i=01ai[D(1-i)G3(t)+D(1/2-i)G3(t)],
where G3(t)=(1/A)∑k=0∞((-C/A)k/k!)t2k+1E1/2,3k/2+2(k)(-(B/A)t1/2).



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