Abstract

We discuss and derive the analytical solution for the fractional partial differential equation with generalized Riemann-Liouville fractional operator 𝐷𝛼,𝛽0,𝑡 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 𝑝𝜓𝑞.

1. Introduction, Definition, and Preliminaries

Applications of fractional calculus require fractional derivatives of different kinds [19]. Differentiation and integration of fractional order are traditionally defined by the right-sided Riemann-Liouville fractional integral operator 𝐼𝑃𝑎+ and the left-sided Riemann-Liouville fractional integral operator 𝐼𝑃𝑎, and the corresponding Riemann-Liouville fractional derivative operators 𝐷𝑃𝑎+ and 𝐷𝑃𝑎, as follows [10, 11]:𝐼𝜇𝑎+𝑓1(𝑥)=Γ(𝜇)𝑥𝑎𝑓(𝑡)(𝑥𝑡)1𝜇𝐼𝑑𝑡(𝑥>𝑎;𝑅(𝜇)>0),(1.1)𝜇𝑎𝑓1(𝑥)=Γ(𝜇)𝑎𝑥𝑓(𝑡)(𝑡𝑥)1𝜇𝐷𝑑𝑡(𝑥<𝑎;𝑅(𝜇)>0),(1.2)𝜇𝑎±𝑓±𝑑(𝑥)=𝑑𝑥𝑛𝐼𝑛𝜇𝑎±𝑓[](𝑥)(𝑅(𝜇)0;𝑛=𝑅(𝜇)+1),(1.3)

where the function 𝑓 is locally integrable, 𝑅(𝜇) denotes the real part of the complex number 𝜇𝐶 and [𝑅(𝜇)] means the greatest integer in 𝑅(𝜇).

Recently, a remarkable large family of generalized Riemann-Liouville fractional derivatives of order 𝛼(0<𝛼<1) and type 𝛽(0𝛽1) was introduced as follows [13, 5, 6, 8].

Definition 1.1. The right-sided fractional derivative 𝐷𝛼,𝛽𝑎+ and the left-sided fractional derivative 𝐷𝛼,𝛽𝑎 of order 𝛼(0<𝛼<1) and type 𝛽(0𝛽1) with respect to 𝑥 are defined by 𝐷𝛼,𝛽𝑎±𝑓(𝑥)=±𝐼𝛽(1𝛼)𝑎±𝑑𝐼𝑑𝑥(1𝛽)(1𝛼)𝑎±𝑓(𝑥),(1.4) 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 𝐷𝛼,𝛽𝑎± are given [3].
Using the formulas (1.1) and (1.2) in conjunction with (1.3) when 𝑛=1, the fractional derivative operator 𝐷𝛼,𝛽𝑎± can be written in the following form: 𝐷𝛼,𝛽𝑎±𝑓(𝑥)=±𝐼𝛽(1𝛼)𝑎±𝐷𝛼+𝛽𝛼𝛽𝑎±𝑓(𝑥).(1.5) 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] 𝐿𝐷𝛼,𝛽0+𝑓(𝑥)(𝑠)=𝑠𝛼𝐿[]𝑓(𝑥)(𝑠)𝑠𝛽(𝛼1)𝐼(1𝛽)(1𝛼)0+𝑓(0+)(0<𝛼<1),(1.6) where (𝐼(1𝛽)(1𝛼)0+𝑓)(0+) is the Riemann-Liouville fractional integral of order (1𝛽)(1𝛼) evaluated in the limit as 𝑡0+, it being understood (as usual) that [13], 𝐿[]𝑓(𝑥)(𝑠)=0𝑒𝑠𝑥𝑓(𝑥)𝑑𝑥=𝐹(𝑠),(1.7) provided that the defining integral in (1.7) exists.
The familiar Mittag-Leffler functions 𝐸𝜇(𝑧) and 𝐸𝜇,𝜈(𝑧) are defined by the following series: 𝐸𝜇(𝑧)=𝑛=0𝑧𝑛Γ(𝜇𝑛+1)=𝐸𝜇,1(𝐸𝑧)(𝑧𝐶;𝑅(𝜇)>0),(1.8)𝜇,𝜈(𝑧)=𝑛=0𝑧𝑛Γ(𝜇𝑛+𝜈)(𝑧,𝜈𝐶;𝑅(𝜇)>0),(1.9) respectively. These functions are natural extensions of the exponential, hyperbolic, and trigonometric functions, since 𝐸1(𝑧)=𝑒𝑧,𝐸2𝑧2=cosh𝑧,𝐸2𝑧2𝐸=cos𝑧,1,2𝑒(𝑧)=𝑧1𝑧,𝐸2,2𝑧2=sinh𝑧𝑧.(1.10)
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. [1517]. 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: 𝐸𝜆𝜇,𝜈(𝑧)=𝑛=0(𝜆)𝑛𝑧Γ(𝜇𝑛+𝜈)𝑛(𝑛!𝑧,𝜈,𝜆𝐶;𝑅(𝜇)>0),(1.11) where (𝜆)𝜈 denotes the familiar Pochhammer symbol, defined (for 𝜆,𝜈𝐶 and in terms of the familiar Gamma function) by (𝜆)𝜈Γ=(𝜆+𝜈)=Γ(𝜆)1(𝜈=0;𝜆𝐶{0})𝜆(𝜆+1)(𝜆+𝑛1)(𝜈=𝑛𝑁;𝜆𝐶).(1.12)
Clearly, we have the following special cases: 𝐸1𝜇,𝜈(𝑧)=𝐸𝜇,𝜈(𝑧),𝐸1𝜇,1(𝑧)=𝐸𝜇(𝑧).(1.13) Indeed, as already observed earlier by Srivastava and Saxena [21], the generalized Mittag-Leffler function 𝐸𝜆𝜇,𝜈(𝑧) itself is actually a very specialized case of a rather extensively investigated function 𝑝Ψ𝑞 as indicated below [17]: 𝐸𝜆𝑢,𝜈1(𝑧)=Γ(𝜆)1Ψ1(𝑧𝜆,1);(𝜈,u);.(1.14) 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]: 𝑝Ψ𝑞=𝑎1,𝐴1𝑎,,𝑝,𝐴𝑝;𝑏1,𝐵1𝑏,,𝑞,𝐵𝑞;𝑧=𝑥=0Γ𝑎1+𝐴1𝑘𝑎Γ𝑝+𝐴𝑝𝑘𝑧𝑘Γ𝑏1+𝐵1𝑘𝑏Γ𝑞+𝐵𝑞𝑘𝑅𝐴𝑘!,(1.15)𝑗𝐵>0(𝑗=1,,𝑝);𝑅𝑗>0(𝑗=1,,𝑞);1+𝑅𝑞𝑗=1𝐵𝑗𝑝𝑗=1𝐴𝑗0,(1.16) in which we assumed in general that 𝑎𝑗,𝐴𝑗𝐶(𝑗=1,,𝑝),𝑏𝑗,𝐵𝑗𝐶(𝑗=1,,𝑞).(1.17)
In application of Mittag-Leffler function, it is useful to have the following Laplace inverse transform formula: 𝐿1𝑆𝛾𝛽(𝑆𝛾+𝐴)𝑘+1=1𝑡𝑘!𝛾𝑘+𝛽1𝐸𝑘𝛾,𝛽(𝐴𝑡𝛾),(1.18) where 𝐸𝑗𝛾,𝛽(𝑧)=(𝑑𝑗/𝑑𝑧𝑗)𝐸𝛾,𝛽(𝑧).

2. Fox’s 𝐻-function

The Fox function, also referred as the Fox’s 𝐻-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]𝐻1,𝑝𝑝,𝑞+1|||||𝑥1𝑎1,𝐴1,,1𝑎𝑝,𝐴𝑝(0,1),1𝑏1,𝐵1,,1𝑏𝑞,𝐵𝑞=𝑘=0Γ𝑎1+𝐴1𝑘𝑎Γ𝑝+𝐴𝑝𝑘𝑏𝑘!Γ1+𝐵1𝑘𝑏Γ𝑝+𝐵𝑞𝑘𝑥𝑘.(2.1)

We need this relation𝐸𝑘𝛼,𝛽(𝑥)=𝑛=𝑘𝑛!𝑥𝑛𝑘=(𝑛𝑘)!Γ(𝛼𝑛+𝛽)𝑗=0Γ(𝑗+𝑘+1)𝑥𝑗𝑗!Γ(𝛼𝑗+𝛼𝑘+𝛽)=𝐻1,11,2||||.𝑥(𝑘,1)(0,1),(1𝛼𝑘𝛽,𝛼)(2.2)

3. Finite Hankel Transform

If 𝑓(𝑟) satisfies Dirichlet conditions in closed interval (0,𝑎) and if its finite Hankel transform is defined to be [23]𝐻[]=𝑓(𝑟)𝑓𝜆𝑛=𝑎0𝑟𝑓(𝑟)𝐽0𝑟𝜆𝑛𝑑𝑟,(3.1)

where 𝜆𝑛 are the roots of the equation 𝐽0(𝑟)=0. Then at each point of the interval at which 𝑓(𝑟) is continuous:2𝑓(𝑟)=𝑎2𝑛=1𝑓𝜆𝑛𝐽0𝜆𝑛𝑟𝐽21𝜆𝑛𝑎,(3.2)

where the sum is taken over all positive roots of 𝐽0(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]𝐻𝑑2𝑓𝑑𝑟2+1𝑟𝑑𝑓𝑑𝑟=𝜆2𝑛𝑓(𝑟)+𝑎𝜆𝑛𝑓(𝑎)𝐽1𝜆𝑛𝑎.(3.3)

Example 3.1. Solve the differential equation 𝐷2𝛼,𝛽0,𝑡𝑢(𝑟,𝑡)+𝑎𝐷𝛼,𝛽0,𝑡𝜕𝑢(𝑟,𝑡)=𝑑2𝑢(𝑟,𝑡)𝜕𝑟2+1𝑟𝑢(𝑟,𝑡)+𝑓(𝑡),(3.4) where 0<𝛼1/2 and 0𝛽1
with initial condition 𝐼𝑡(1𝛽)(12𝛼)𝑢(𝑟,0)=𝜙1𝐼(𝑟),𝑡(1𝛽)(1𝛼)𝑢(𝑟,0)=𝜙2(𝑟),𝑢(𝑟,𝑡)=0everywherefor𝑡<0,𝑢(𝑟,𝑡)=0for𝑟=1,𝑡>0,𝑢(𝑟,𝑡)=niteat𝑟=0,𝑡>0.(3.5)

Solution 1. Taking Laplace transform of (3.4), we get 𝑠2𝛼̃𝑢(𝑟,𝑠)𝑠𝛽(2𝛼1)𝜙1(𝑟)+𝑎𝑠𝛼̃𝑢(𝑟,𝑠)𝑎𝑠𝛽(𝛼1)𝜙2𝜕(𝑟)=𝑑2̃𝑢(𝑟,𝑠)𝜕𝑟2+1𝑟+̃𝑢(𝑟,𝑠)𝑓(𝑠).(3.6) Taking Hankel transform on both side of the above equation, we get 𝑠2𝛼̃̃𝑢(𝑟,𝑠)𝑠𝛽(2𝛼1)𝜙1(𝑟)+𝑎𝑠𝛼̃̃𝑢(𝑟,𝑠)𝑎𝑠𝛽(𝛼1)𝜙2(𝑟)=𝑑𝜆2𝑛̃+𝐽̃𝑢(𝑟,𝑠)𝑓(𝑠)1𝜆𝑛𝜆𝑛,(3.7) then we get ̃𝑠̃𝑢(𝑟,𝑠)=𝛽(2𝛼1)𝜙1(𝑟)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛+𝑎𝑠𝛽(𝛼1)𝜙2(𝑟)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛+𝑓(𝑠)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛𝐽1𝜆𝑛𝜆𝑛̃𝐺,(3.8)̃𝑢(𝑟,𝑠)=1𝜙1𝐺(𝑟)+𝑎2𝜙2𝐺(𝑟)+3𝐽𝑓(𝑠)1𝜆𝑛𝜆𝑛,(3.9) where 𝐺1=𝑠𝛽(2𝛼1)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛,𝐺(3.10)2=𝑠𝛽(𝛼1)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛𝐺,(3.11)3=1𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛.(3.12)
On taking Laplace inverse of (3.10), (3.11), and (3.12), respectively, 𝐿1𝑠𝛽(2𝛼1)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛=𝑚=0(1)𝑚𝑎𝑚+1𝑡𝛼+𝛽2𝛼𝛽𝑚𝛼1𝐸𝑚!𝑚𝛼,𝛼+𝛽2𝛼𝛽2𝑚𝛼𝑑𝜆2𝑛𝑎𝑡𝛼,𝐿(3.13)1𝑠𝛽(𝛼1)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛=𝑚=0(1)𝑚𝑎𝑚+1𝑡𝛼+𝛽𝛼𝛽𝑚𝛼1𝐸𝑚!𝑚𝛼,𝛼+𝛽𝛼𝛽2𝑚𝛼𝑑𝜆2𝑛𝑎𝑡𝛼𝐿,(3.14)11𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛=𝑚=0(1)𝑚𝑎𝑚+1𝑡𝛼𝑚𝛼1𝐸𝑚!𝑚𝛼,𝛼2𝑚𝛼𝑑𝜆2𝑛𝑎𝑡𝛼.(3.15) After taking Inverse Laplace and Hankel transform of (3.9) put the value (3.13) through (3.15) in (3.9), we get 𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝑚!𝐽21𝜆𝑛𝜙1(𝑟)𝑡2𝛼𝛽𝑚𝛼+𝛼+𝛽1𝑗=0(𝑗+𝑚+1)!𝑑𝜆2𝑛𝑡𝛼/𝑎𝑗𝑗!Γ(𝑗𝛼+𝛼+𝛽2𝛼𝛽2𝑚𝛼)+2𝑎𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝐽21𝜆𝑛𝜙2(𝑟)𝑡𝛼𝛽𝑚𝛼+𝛼+𝛽1𝑗=0(𝑗+𝑚+1)!𝑑𝜆2𝑛𝑡𝛼/𝑎𝑗𝑗!Γ(𝑗𝛼+𝛼+𝛽𝛼𝛽𝑚𝛼)+2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼1𝑗=0(𝑗+𝑚+1)!𝑑𝜆2𝑛𝑢𝛼/𝑎𝑗𝑗!Γ(𝛼𝑗+𝛼𝑚𝛼)𝑓(𝑡𝑢)𝑑𝑢.(3.16)𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝐽21𝜆𝑛𝜙1(𝑟)𝑡2𝛼𝛽𝑚𝛼+𝛼+𝛽1𝐻1,11,2𝑑𝜆2𝑛𝑡𝛼𝑎||||(𝑚1,1)(0,1),(1𝛼𝛽+2𝛼𝛽+2𝑚𝛼,𝛼)+2𝑎𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝐽21𝜆𝑛𝜙2(𝑟)𝑡𝛼𝛽𝑚𝛼+𝛼+𝛽1𝐻1,11,2𝑑𝜆2𝑛𝑡𝛼𝑎||||(𝑚1,1)(0,1),(1𝛼𝛽+𝛼𝛽+𝑚𝛼,𝛼)+2𝑛=0𝑚=0(1)𝑛𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼1𝐻1,11,2𝑑𝜆2𝑛𝑢𝛼𝑎||||𝑓(𝑚1,1)(0,1),(1𝛼+𝑚𝛼,𝛼)(𝑡𝑢)𝑑𝑢.(3.17)

Example 3.2. Solve the differential equation (3.4) with initial condition 𝐼𝑡(1𝛽)(12𝛼)𝐼𝑢(𝑟,0)=0,𝑡(1𝛽)(1𝛼)𝑢(𝑟,0)=0,𝑢(𝑟,𝑡)=0everywherefor𝑡0,𝑢(𝑟,𝑡)=0for𝑟=1,𝑡>0,𝑢(𝑟,𝑡)=niteat𝑟=0,𝑡>0.(3.18)

Solution 2. Taking Laplace and Hankel transform of (3.4), we get ̃𝐽̃𝑢(𝑟,𝑠)=1𝜆𝑛𝜆𝑛𝑓(𝑠)𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛,(3.19) on taking Inverse Laplace transform of equation (3.19), we get ̃𝑢(𝑟,𝑡)=𝐿1𝐽𝑓(𝑠)1𝜆𝑛𝜆𝑛𝐿11𝑠2𝛼+𝑎𝑠𝛼+𝑑𝜆2𝑛.(3.20)
By using convolution theorem for Laplace transform and taking inverse Hankel transform, we get 𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼1𝐸𝑚𝛼,𝛼2𝑚𝛼𝑑𝜆2𝑛𝑢𝛼𝑎𝑓(𝑡𝑢)𝑑𝑢,(3.21) or 𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼1𝑗=0(𝑗+𝑚+1)!(𝑗)!𝑑𝜆2𝑛𝑢𝛼/𝑎𝑗Γ.(𝛼𝑗+𝛼𝑚𝛼)(3.22) By using the relation (2.2) 𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼1𝐻1,11,2𝑑𝜆2𝑛𝑢𝛼𝑎||||(𝑚1,1)(0,1),(1𝛼+𝑚𝛼,𝛼)𝑓(𝑡𝑢)𝑑𝑢,(3.23) or 𝑢(𝑟,𝑡)=2𝑛=0𝑚=0(1)𝑚𝑎𝑚+1𝐽0𝜆𝑛𝑟𝜆𝑛𝐽1𝜆𝑛𝑡0𝑢𝛼𝑚𝛼11Γ(𝑚)1Ψ1(𝑚,1);(𝛼2𝑚𝛼,𝛼);𝑑𝜆2𝑛𝑢𝛼𝑎,(3.24) which is the required solution.