# Analytical Solution for the Differential Equation Containing Generalized Fractional Derivative Operators and Mittag-Leffler-Type Function

Authors: V. B. L. Chaurasia; Ravi Shanker Dubey
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

We discuss and derive the analytical solution for the fractional partial differential equation with generalized Riemann-Liouville fractional operatorD0,tα,β of order α and β. Here, we derive the solution of the given differential equation with the help of Laplace and Hankel transform in terms of Fox's H-function as well as in terms of Fox-Wright function ψpq.


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## Body

## 1. Introduction, Definition, and Preliminaries

Applications of fractional calculus require fractional derivatives of different kinds [1–9].  Differentiation and integration of fractional order are traditionally defined by the right-sided Riemann-Liouville fractional integral operator Ia+P and the left-sided Riemann-Liouville fractional integral operator Ia-P, and the corresponding Riemann-Liouville fractional derivative operators Da+P and Da-P, as follows [10, 11]:(1.1)(Ia+μf)(x)=1Γ(μ)∫axf(t)(x-t)1-μdt(x>a;R(μ)>0),(1.2)(Ia-μf)(x)=1Γ(μ)∫xaf(t)(t-x)1-μdt(x<a;R(μ)>0),(1.3)(Da±μf)(x)=(±ddx)n(Ia±n-μf)(x)(R(μ)≥0;n=[R(μ)]+1),where the functionf is locally integrable, R(μ) denotes the real part of the complex number μ∈C and [R(μ)] means the greatest integer in R(μ).Recently, a remarkable large family of generalized Riemann-Liouville fractional derivatives of orderα(0<α<1) and type β(0≤β≤1) was introduced as follows [1–3, 5, 6, 8].Definition 1.1.
The right-sided fractional derivativeDa+α,β and the left-sided fractional derivative Da-α,β of order α(0<α<1) and type β(0≤β≤1) with respect to x are defined by
(1.4)(Da±α,βf)(x)=(±Ia±β(1-α)ddx(Ia±(1-β)(1-α)f))(x),
whenever the second number of (1.4) exists. This generalization (1.4) yields the classical Riemann-Liouville fractional derivative operator when β=0. Moreover, for β=1, it gives the fractional derivative operator introduced by Liouville [12] which is often attributed to Caputo now-a-days and which should more appropriately be referred to as the Liouville-Caputo fractional derivative. Several authors [7, 9] called the general operators in (1.4) the Hilfer fractional derivative operators. Applications of Da±α,β are given [3].
Using the formulas (1.1) and (1.2) in conjunction with (1.3) when n=1, the fractional derivative operator Da±α,β can be written in the following form:
(1.5)(Da±α,βf)(x)=(±Ia±β(1-α)(Da±α+β-αβf))(x).
The difference between fractional derivatives of different types becomes apparent from their Laplace transformations. For example, it is found for 0<α<1 that [1, 2, 9]
(1.6)L[(D0+α,βf)(x)](s)=sαL[f(x)](s)-sβ(α-1)(I0+(1-β)(1-α)f)(0+)(0<α<1),
where (I0+(1-β)(1-α)f)(0+) is the Riemann-Liouville fractional integral of order (1-β)(1-α) evaluated in the limit as t→0+, it being understood (as usual) that [13],
(1.7)L[f(x)](s)∶=∫0∞e-sxf(x)dx∶=F(s),
provided that the defining integral in (1.7) exists.
The familiar Mittag-Leffler functionsEμ(z) and Eμ,ν(z) are defined by the following series:
(1.8)Eμ(z)∶=∑n=0∞znΓ(μn+1)∶=Eμ,1(z)(z∈C;R(μ)>0),(1.9)Eμ,ν(z)∶=∑n=0∞znΓ(μn+ν)(z,ν∈C;R(μ)>0),
respectively. These functions are natural extensions of the exponential, hyperbolic, and trigonometric functions, since
(1.10)E1(z)=ez,E2(z2)=cosh⁡⁡z,E2(-z2)=cos⁡⁡z,E1,2(z)=ez-1z,E2,2(z2)=sinh⁡⁡zz.
For a detailed account of the various properties, generalizations, and applications of the Mittag-Leffler functions, the reader may refer to the recent works by, for example, Gorenflo et al. [14] and Kilbas et al. [15–17]. The Mittag-Leffler  function (1.1) and some of its various generalizations have only recently been calculated numerically in the whole complex plane [18, 19]. By means of the series representation, a generalization of the Mittag-Leffler function Eμ,ν(z) of (1.2) was introduced by Prabhakar [20] as follows:
(1.11)Eμ,νλ(z)=∑n=0∞(λ)nΓ(μn+ν)znn!(z,ν,λ∈C;R(μ)>0),
where (λ)ν denotes the familiar Pochhammer symbol, defined (for λ,ν∈C and in terms of the familiar Gamma function) by
(1.12)(λ)ν∶=Γ(λ+ν)Γ(λ)={1(ν=0;λ∈C∖{0})λ(λ+1)⋯(λ+n-1)(ν=n∈N;λ∈C).
Clearly, we have the following special cases:(1.13)Eμ,ν1(z)=Eμ,ν(z),Eμ,11(z)=Eμ(z).
Indeed, as already observed earlier by Srivastava and Saxena [21], the generalized Mittag-Leffler function Eμ,νλ(z) itself is actually a very specialized case of a rather extensively investigated function pΨq as indicated below [17]:
(1.14)Eu,νλ(z)=1Γ(λ)1Ψ1[(λ,1);(ν,u);z].
Here and in what follows, pΨq denotes the Wright (or more appropriately, the Fox-Wright) generalized of the hypergeometric pFq function, which is defined as follows [12]:
(1.15)pΨq=[(a1,A1),…,(ap,Ap);(b1,B1),…,(bq,Bq);z]=∑x=0∞Γ(a1+A1k)⋯Γ(ap+Apk)zkΓ(b1+B1k)⋯Γ(bq+Bqk)k!,(1.16)[R(Aj)>0(j=1,…,p);R(Bj)>0(j=1,…,q);1+R(∑j=1qBj-∑j=1pAj)≥0],
in which we assumed in general that
(1.17)aj,Aj∈C(j=1,…,p),bj,Bj∈C(j=1,…,q).
In application of Mittag-Leffler function, it is useful to have the following Laplace inverse transform formula:(1.18)L-1{Sγ-β(Sγ+A)k+1}=1k!tγk+β-1Eγ,βk(-Atγ),
where Eγ,βj(z)=(dj/dzj)Eγ,β(z).


## 2. Fox’sH-function

The Fox function, also referred as the Fox’sH-function, generalizes the Mellin-Barnes function. The importance of the Fox function lies in the fact that it includes nearly all special functions occurring in applied mathematics and statistics as special cases. Fox H-function is defined as [22](2.1)Hp,q+11,p[-x|(1-a1,A1),…,(1-ap,Ap)(0,1),(1-b1,B1),…,(1-bq,Bq)]=∑k=0∞Γ(a1+A1k)⋯Γ(ap+Apk)k!Γ(b1+B1k)⋯Γ(bp+Bqk)xk.We need this relation(2.2)Eα,βk(x)=∑n=k∞n!xn-k(n-k)!Γ(αn+β)=∑j=0∞Γ(j+k+1)xjj!Γ(αj+αk+β)=H1,21,1[-x|(-k,1)(0,1),(1-αk-β,α)].


## 3. Finite Hankel Transform

Iff(r) satisfies Dirichlet conditions in closed interval (0,a) and if its finite Hankel transform is defined to be [23](3.1)H[f(r)]=f¯(λn)=∫0arf(r)J0(rλn)dr,whereλn are the roots of the equation J0(r)=0. Then at each point of the interval at which f(r) is continuous:(3.2)f(r)=2a2∑n=1∞f¯(λn)J0(λnr)J12(λna),where the sum is taken over all positive roots ofJ0(r)=0, J0 and J1 are Bessel functions of first kind.In application of the finite Hankel transform to physical problems, it is useful to have the following formula [23](3.3)H[d2fdr2+1rdfdr]=-λn2f¯(r)+aλnf(a)J1(λna).Example 3.1.
Solve the differential equation(3.4)D0,t2α,βu(r,t)+aD0,tα,βu(r,t)=d(∂2u(r,t)∂r2+1ru(r,t))+f(t),
where 0<α≤1/2 and 0≤β≤1 with initial condition
(3.5)It(1-β)(1-2α)u(r,0)=ϕ1(r),It(1-β)(1-α)u(r,0)=ϕ2(r),u(r,t)=0everywherefort<0,u(r,t)=0forr=1,t>0,u(r,t)=finiteatr=0,t>0.Solution 1.
Taking Laplace transform of (3.4), we get
(3.6)s2αũ(r,s)-sβ(2α-1)ϕ1(r)+asαũ(r,s)-asβ(α-1)ϕ2(r)=d[∂2ũ(r,s)∂r2+1rũ(r,s)]+f̃(s).
Taking Hankel transform on both side of the above equation, we get
(3.7)s2αũ̃(r,s)-sβ(2α-1)ϕ̃1(r)+asαũ̃(r,s)-asβ(α-1)ϕ̃2(r)=d[-λn2ũ̃(r,s)]+f̃(s)J1(λn)λn,
then we get
(3.8)ũ̃(r,s)=sβ(2α-1)ϕ̃1(r)(s2α+asα+dλn2)+asβ(α-1)ϕ̃2(r)(s2α+asα+dλn2)+f̃(s)(s2α+asα+dλn2)J1(λn)λn,(3.9)ũ̃(r,s)=G̃̃1ϕ̃1(r)+aG̃̃2ϕ̃2(r)+G̃̃3f̃(s)J1(λn)λn,
where
(3.10)G̃̃1=sβ(2α-1)(s2α+asα+dλn2),(3.11)G̃̃2=sβ(α-1)(s2α+asα+dλn2),(3.12)G̃̃3=1(s2α+asα+dλn2).
On taking Laplace inverse of (3.10), (3.11), and (3.12), respectively,
(3.13)L-1{sβ(2α-1)(s2α+asα+dλn2)}=∑m=0∞(-1)mam+1tα+β-2αβ-mα-1m!Eα,α+β-2αβ-2mαm(-dλn2atα),(3.14)L-1{sβ(α-1)(s2α+asα+dλn2)}=∑m=0∞(-1)mam+1tα+β-αβ-mα-1m!Eα,α+β-αβ-2mαm(-dλn2atα),(3.15)L-1{1(s2α+asα+dλn2)}=∑m=0∞(-1)mam+1tα-mα-1m!Eα,α-2mαm(-dλn2atα).
After taking Inverse Laplace and Hankel transform of (3.9) put the value (3.13) through (3.15) in (3.9), we get
(3.16)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)m!J12(λn)ϕ1(r)t-2αβ-mα+α+β-1∑j=0∞(j+m+1)!(-dλn2tα/a)jj!Γ(jα+α+β-2αβ-2mα)+2a∑n=0∞∑m=0∞(-1)mam+1J0(λnr)J12(λn)ϕ2(r)t-αβ-mα+α+β-1∑j=0∞(j+m+1)!(-dλn2tα/a)jj!Γ(jα+α+β-αβ-mα)+2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)λnJ1(λn)∫0tuα-mα-1∑j=0∞(j+m+1)!(-dλn2uα/a)jj!Γ(αj+α-mα)f(t-u)du.(3.17)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)J12(λn)ϕ1(r)t-2αβ-mα+α+β-1⋅H1,21,1[dλn2tαa|(-m-1,1)(0,1),(1-α-β+2αβ+2mα,α)]+2a∑n=0∞∑m=0∞(-1)mam+1J0(λnr)J12(λn)ϕ2(r)t-αβ-mα+α+β-1⋅H1,21,1[dλn2tαa|(-m-1,1)(0,1),(1-α-β+αβ+mα,α)]+2∑n=0∞∑m=0∞(-1)nJ0(λnr)λnJ1(λn)∫0tuα-mα-1H1,21,1[dλn2uαa|(-m-1,1)(0,1),(1-α+mα,α)]f(t-u)du.Example 3.2.
Solve the differential equation (3.4) with initial condition
(3.18)It(1-β)(1-2α)u(r,0)=0,It(1-β)(1-α)u(r,0)=0,u(r,t)=0everywherefort≤0,u(r,t)=0forr=1,t>0,u(r,t)=finiteatr=0,t>0.Solution 2.
Taking Laplace and Hankel transform of (3.4), we get
(3.19)ũ̃(r,s)=J1(λn)λnf̃(s)(s2α+asα+dλn2),
on taking Inverse Laplace transform of equation (3.19), we get
(3.20)ũ(r,t)=L-1{f̃(s)J1(λn)λn}L-1{(1s2α+asα+dλn2)}.
By using convolution theorem for Laplace transform and taking inverse Hankel transform, we get(3.21)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)λnJ1(λn)∫0tuα-mα-1Eα,α-2mαm(-dλn2uαa)f(t-u)du,
or
(3.22)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)λnJ1(λn)∫0tuα-mα-1∑j=0∞(j+m+1)!(j)!(-dλn2uα/a)jΓ(αj+α-mα).
By using the relation (2.2)
(3.23)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)λnJ1(λn)∫0tuα-mα-1H1,21,1[dλn2uαa|(-m-1,1)(0,1),(1-α+mα,α)]f(t-u)du,
or
(3.24)u(r,t)=2∑n=0∞∑m=0∞(-1)mam+1J0(λnr)λnJ1(λn)∫0tuα-mα-11Γ(m)1Ψ1[(m,1);(α-2mα,α);-dλn2uαa],
which is the required solution.



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