Fractional Calculus and Fractional Differential Equations

Home , Integral transform, Laplace transform

CHAPTER 3

FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

[This chapter is based on the lectures of Professor R.K. Saxena of Jai Narain Vyas University, Jodhpur, Rajasthan, India. ]

3.0. Introduction

This section deals with certain properties of fractional calculus associated with Laplace and Mellin transforms. Composition relations between Riemann- Liouville fractional calculus and generalized Mittag-Leﬄer functions are pre- sented. Applications of fractional calculus in the solution of diﬀerential and integral equations of fractional order are demonstrated. This study will bring the reader to the research level.

3.1. Laplace Transform of the Fractional Integral

3.1.1. Laplace transform Notation 3.1.1. F(s) = L f (t); s = (L f )(s) : Laplace transform of f (t) with parameter s. { }

Notation 3.1.2. L 1 f (s); t : Inverse Laplace transform − { } Deﬁnition 3.1.1. The Laplace transform of a function f (t), denoted by F(s), is deﬁned by the equation

∞ st F(s) = (Lf )(s) = L f (t); s = e− f (t)dt, (3.1.1) { } Z0 175 176 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS where (s) > 0, which may be symbolically written as ℜ

1 F(s) = L f (t); s or f (t) = L− F(s); t , { } { } provided that the function f (t) is continuous for t 0, it being tacitly assumed that the integral in (3.1.1) exists. ≥

Example 3.1.1. Prove that

ρ 1 1 ρ t − L− s− = , (s) > 0, (ρ) > 0. (3.1.2) { } Γ(ρ) ℜ ℜ It follows from the Laplace integral

Γ ∞ st ρ 1 = (s) e− t − dt ρ , (s) > 0, (ρ) > 0. (3.1.3) Z0 s ℜ ℜ

F(s) Example 3.1.2. Find the inverse Laplace transform of a+sα ; a,α > 0; where (s) > 0, F(s) = L f (t); s . ℜ { } Solution 3.1.1. Let

1 ∞ r α αr a G(s) = = ( a) s− − , < 1. a + sα − | sα | Xr=0

Therefore,

1 1 ∞ r α αr L− G(s) = g(t) = L− ( a) s− − { }  −  Xr=0  α 1 α = t − E ( at ).  (3.1.4) α,α −  Application of convolution theorem of Laplace transform yields the result

F(s) x L 1 ; t = (x t)α 1E ( a(x t)α) f (t)dt (3.1.5) − + α − α,α (a s ) Z0 − − − where (α) > 0. ℜ 3.1.LAPLACETRANSFORMOFTHE FRACTIONALINTEGRAL 177 3.1.2. Laplace transform of the fractional integral

We have

x ν = 1 ν 1 0Ix− f (x) Γ (x t) − f (t)dt, (3.1.6) (ν) Z0 − where (ν) > 0. Applicationℜ of convolution theorem of the Laplace transform gives

ν 1 ν t − L 0I− f (x); s = L L f (t); s { x } (Γ(ν)) { } ν = s− F(s), (3.1.7) where (s) > 0, (ν) > 0. ℜ ℜ

3.1.3. Laplace transform of the fractional derivative

If n N, then by the theory of the Laplace transform, we know that ∈

n n 1 d n − n k 1 (k) L f ; s = s F(s) s − − f (0+) (3.1.8) (dxn ) − Xk=0 n 1 n − k (n k 1) = s F(s) s f − − (0+), (n 1 α < n) (3.1.9) − − ≤ Xk=0 where (s) > 0 and F(s) is the Laplace transform of f (t). By virtueℜ of the deﬁnition of the derivative, we ﬁnd that

n α d n α L 0D f ; s = L 0I − f ; s { x } (dxn x ) 178 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

n 1 n k 1 n n α − k d − − n α = s L I − f ; s s I − f (0+) 0 x − dxn k 1 0 x Xk=0 − − n 1 α − k α k 1 d = s F(s) s D − − f (0+), D = (3.1.10) − dx! Xk=0 n α k 1 α k = s F(s) s − D − f (0+) (3.1.11) − Xk=1 where (s) > 0. ℜ 3.1.4. Laplace transform of Caputo derivative α Notation 3.1.3. C aDx 0

Deﬁnition 3.1.2. The Caputo derivative of a casual function f (t) ( that is f (t) = 0 for t < 0 ) with α > 0 was deﬁned by Caputo (1969) in the form

n α = n α d = (n α) n C aDx f (x) aIx− f (x) aD−t − f (t) (3.1.12) 0 dxn x 1 n α 1 n = (x t) − − f (t)dt, (n 1 < α < n) (3.1.13) Γ(n α) Z − − − a where n N. From∈ (3.1.7) and (3.1.13), it follows that

α = (n α) (n) L C 0Dt f (t); s s− − L f (t) . (3.1.14) { 0 } { } On using (3.1.8), we see that

n 1 α = (n α) n − n k 1 (k) + L C 0Dt f (t); s s− − s F(s) s − − f (0 ) { 0 }  −  Xk=0    n 1  α − α k 1 (k) = s F(s) s − − f (0+), (n 1 < α n), (3.1.15) − − ≤ Xk=0 where (s) > 0 and (α) > 0. ℜ ℜ 3.1.LAPLACETRANSFORMOFTHE FRACTIONALINTEGRAL 179

Note 3.1.1. From (3.1.12), it can be seen that

α = C 0Dt A 0, where A is a constant, 0 whereas the Riemann-Liouville derivative At α DαA = − , (α , 1, 2, ), (3.1.16) 0 t Γ(1 α) ··· which is a remarkable result. −

Exercises 3.1.

3.1.1. Prove that

ν 1 ν ( I− f )(x) = L− x− L f (x) , (3.1.17) 0 x { } where (ν) > 0. ℜ 3.1.2. Prove that

ν ν 1 (xW Lf )(x) = Lx− L− f (x), (3.1.18) ∞ where (ν) > 0. ℜ 3.1.3. Prove that the solution of Abel integral equation of the second kind λ x φ(t)dt φ(x) = f (x), 0 < x < 1 Γ 1 α − (α) Z0 (x t) − α > 0, is given by −

x d α φ(x) = Eα[λ(x t) ] f (t)dt, (3.1.19) dx Z0 − where Eα(x) is the Mittag-Leﬄer function deﬁned by equation (3.5.1). 3.1.4. Show that

x α λ Eα(λt ) α dt = Eα(λx ) 1, α > 0. (3.1.20) Γ(α) Z (x t)1 α − 0 − − 180 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS 3.2. Mellin Transform of the Fractional Integrals and the Fractional Derivatives

3.2.1. Mellin transform Notation 3.2.1. m f (x); s , f (s), the Mellin transform { } ∗ Notation 3.2.2. m 1 f (s); x , Inverse Mellin transform − { ∗ }

Deﬁnition 3.2.1. The Mellin transform of a function f (x), denoted by f ∗(s), is deﬁned by

∞ s 1 f ∗(s) = m f (x); s = x − f (x)dx, x > 0. (3.2.1) { } Z0 The inverse Mellin transform is given by the contour integral

γ+i 1 1 ∞ s f (x) = m− f ∗(s); x = f ∗(s)x− ds, i = √ 1 (3.2.2) { } 2πi Zγ i − − ∞ where γ is real.

3.2.2. Mellin transform of the fractional integral Theorem 3.2.1. The following result holds true. Γ α (1 α s) m( I f )(s) = − − f ∗(s + α), (3.2.3) 0 x Γ(1 α) − where (α) > 0 and (α + s) < 1. ℜ ℜ Proof 3.2.1. We have

z α = ∞ s 1 1 α 1 m(0Ix f )(s) z − Γ (z t) − f (t)dtdz Z0 (α) Z0 − = 1 ∞ ∞ s 1 α 1 Γ f (t)dt z − (z t) − dz (3.2.4) (α) Z0 Zt − = t on setting z u , the z-integral becomes 3.2.MELLINTRANSFORMOF THE FRACTIONALINTEGRALS 181

1 α+s 1 α s α 1 α+s 1 t − u− − (1 u) − du = t − B(α, 1 α s), (3.2.5) Z0 − − − where (α) > 0, (α + s) < 1. Putting the above value of z-integral, the result follows.ℜ ℜ Similarly we can establish

Theorem 3.2.2. The following result holds true.

Γ α (s) α m(xI f )(s) = m t f (t); s ∞ Γ(s + α) { } Γ(s) = f ∗(s + α), (3.2.6) Γ(s + α) where (α) > 0, (s) > 0. ℜ ℜ α Note 3.2.1. If we set f (x) = x− φ(x), then using the property of the Mellin transform

α x φ(x) φ∗(s + α), (3.2.7) ↔ the results (3.2.3) and (3.2.6) become

Γ α α (1 α s) ( I x− f (x))(s) = − − f ∗(s), (3.2.8) 0 x Γ(1 s) − where (α) > 0, (α + s) < 1 and ℜ ℜ Γ α α (s) (xI x− f (x))(s) = f ∗(s), (3.2.9) ∞ Γ(s + α) where (α) > 0, and (s) > 0, respectively. ℜ ℜ 3.2.3. Mellin transform of the fractional derivative Theorem 3.2.3. If n N, then ∈ 182 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Γ(s) m f (n)(t); (s) = ( 1)n m f (t); s n , (3.2.10) { } − Γ(s n) { − } where (s) > 0, (s n) > 0. − ℜ ℜ − Proof 3.2.2. Integrate by parts and using the deﬁnition of the Mellin transform, the result follows.

Example 3.2.1. Find the Mellin transform of the fractional derivative.

Solution 3.2.1. We have

α = n α n = n n α 0Dx f 0Dx 0Dx− f 0Dx 0Ix− f. (3.2.11)

Therefore, n Γ α ( 1) (s) n α m( D f )(s) = − m I − f (s n), (n 1 (α) < n) (3.2.12) 0 x Γ(s n) 0 x − − ≤ ℜ ( 1)n−Γ(s)Γ(1 (s α )) = − − − m f (t); s α , (3.2.13) Γ(s n)Γ(1 s + n) { − } − − where (s) > 0, (s) < 1 + (α). ℜ ℜ ℜ Remark 3.2.1. An alternative form of ( 3.2.13 ) is given in Exercise 3.2.2.

Exercises 3.2.

3.2.1 Prove Theorem 3.2.2.

3.2.2 Prove that the Mellin transform of fractional derivative is given by

( 1)n Γ(s) sin[π(s n)] m( Dα f )(s) = − − m f (t); s α , (3.2.14) 0 x Γ(s α)sin[π(s α)] { − } where (s) > 0, (α s) > −1. − ℜ ℜ − − a b 3.2.3 Find the Mellin transform of (1 + x )− ; a, b > 0. 3.3. KOBER OPERATORS 183 3.3. Kober Operators

Kober operators are the generalization of Riemann - Liouville and Weyl operators. These operators have been used by many authors in deriving the solution of single, dual and triple integral equations possessing special functions of mathematical physics, as their kernels.

Notation 3.3.1. Kober operator of the ﬁrst kind

I I I α,η In,α [ f (x)], [α, η : f (x)], (α, η) f (x), E0,x f, x f.

Notation 3.3.2. Kober operator of the second kind

α,ζ ζ,α R[ f (x)], R[α, ζ : f (x)], R(α, ζ) f (x), Kx, f, Kx f. ∞ Deﬁnition 3.3.1. I = I = I = α,η [ f (x)] [α, η : f (x)] (α, η) f (x) E0,x f x η α x = Iη,α = − − α 1 η x f Γ (x t) − t f (t)dt, (3.3.1) (α) Z0 − where (α) > 0. ℜ Deﬁnition 3.3.2. R = R = R = α,ζ [ f (x)] [α, ζ : f (x)] (α, ζ) f (x) Kx, f ∞ ζ x ∞ = ζ,α = α 1 ζ α , Kx f Γ (t x) − t− − f (t)dt (3.3.2) (α) Zx − where (α) > 0. ℜ (3.3.1) and (3.3.2) hold true under the following conditions:

1 1 1 1 f L (0, ), (α) > 0, (η) > , (ζ) > , + = 1, p 1. ∈ p ∞ ℜ ℜ −q ℜ − p p q ≥ When η = 0, (3.3.1) reduces to Riemann - Liouville operator. That is,

0,α = α α Ix f x− 0Ix f. (3.3.3) α For ζ = 0, (3.3.2) yields the Weyl operator of t− f (t). That is, 184 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

0,α = α α Kx f xW t− f (t). (3.3.4) ∞ Theorem 3.3.1. [Kober (1940)].

If (α) > 0, (η s) > 1, f Lp(o, ), 1 p 2 (or f Mp(o, ), a subspaceℜ of L (ℜo, −) and p−> 2 ),∈ (η) >∞ 1 , 1≤+ 1≤= 1, then there∈ holds∞ the p ∞ ℜ − q p q formula Γ(1 + η s) m I(α, η) f (s) = − m f (x); s . (3.3.5) { } Γ(α + η + 1 s) { } − Proof 3.3.1. It is similar to the proof of Theorem 3.2.1.

In a similar manner, we can establish

Theorem 3.3.2. [Kober (1940)].

If (α) > 0, (s + ζ) > 0, f Lp(o, ), 1 p 2 (or f Mp(o, ), a subspaceℜ of L (oℜ, ) and p > 2 ) ∈ ∞ ≤ ≤ ∈ ∞ p ∞ 1 1 1 (ζ) > , + = 1, ℜ − p p q then, Γ(ζ + s) m (α, ζ) f (s) = m f (x); s . (3.3.6) {ℜ } Γ(α + ζ + s) { }

Semigroup property of the Kober operators has been given in the form of

Theorem 3.3.3. If f L (o, ),g L (o, ), 1 + 1 = 1, (η) > 1 , (ζ) > ∈ p ∞ ∈ q ∞ p q ℜ − q ℜ 1 , 1 p 2, (or f M (o, ), a subspace of L (o, ) and p > 2 ), then − p ≤ ≤ ∈ p ∞ p ∞ ∞ ∞ g(x)(I(α, η : f ))(x)dx = f (x)(R(α, ζ : g))(x)dx. (3.3.7) Z0 Z0

Proof 3.3.2. Interchange the order of integration. 3.3. KOBER OPERATORS 185

Remark 3.3.1. Operators deﬁned by (3.3.1.) and (3.3.2) are also called Erd´elyi- Kober operators.

Exercises 3.3.

3.3.1 Prove theorem 3.3.1.

3.3.2 For the modified Erdélyi-Kober operators, defined by the following equations for m > 0:

I(α, η : m) f (x) = I( f (x): α, η, m) x = m η mα+m 1 η m m α 1 Γ x− − − t (x t ) − f (t)dt, (3.3.8) (α) Zo − and R(α, ζ : m) f (x) = R( f (x): α,ζ, m) ζ = mx ∞ ζ mα+m 1 m m α 1 Γ t− − − (t x ) − f (t)dt, (3.3.9) (α) Zx − where f L (0, ), (α) > 0, (η) > 1 , (ζ) > 1 , 1 + 1 = 1, ﬁnd the ∈ p ∞ ℜ ℜ − q ℜ − p p q Mellin transforms of (i) I(α, η : m) f (x) and (ii) R(α, ζ : m) f (x) , giving the conditions of validity.

3.3.3 For the operators deﬁned by (3.3.8) and (3.3.9.), show that

∞ ∞ R( f (x): α, η, m)g(x)dx = f (x)I(g(x): α, η, m)dx, (3.3.10) Z0 Z0 where the parameters α, η, m are the same in both the operators I and R. Give conditions of validity of (3.3.10).

3.3.4 For the Erd´elyi-Kober operator, deﬁned by

2α 2η x = 2x− − 2 2 α 1 2η+1 Iη,α f (x) Γ (x t ) − t f (t)dt, (3.3.11) (α) Z0 − 186 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS where (α) > 0, establish the following results (Sneddon (1975)): ℜ

2β 2β (i) Iη,α x f (x) = x Iη+β,α f (x) (3.3.12)

(ii) Iη,αIη+α,β = Iη,α+β = Iη+α,β,Iη,α (3.3.13) 1 = (iii) Iη,α− Iη+α, α. (3.3.14) − Remark 3.3.2. The results of Exercise 3.3.4 also hold for the operator, defined by 2η 2x ∞ K = 2 2 α 1 2α 2η+1 η,α f (x) Γ (t x ) − t− − f (t)dt, (3.3.15) (α) Zx − where (α) > 0. ℜ Remark 3.3.3. Operators more general than the operators defined by (3.3.11) and (3.3.15) are recently defined by Galuéet al [Integral Transform & Spec. Funct. Vol. 9 (2000), No. 3, pp. 185-196] in the form

x η α x η,α = − − α 1 η , aIx f (x) Γ (x t) − t f (t)dt (3.3.16) (α) Za − where (α) > 0. ℜ

3.4. Generalized Kober Operators

Notation 3.4.1. I[α, β, γ : m, µ, η, a : f (x)], I[ f (x)]

Notation 3.4.2. I[α, β, γ : m,µ,δ, a : f (x)], I[ f (x)]

α,β,γ; Notation 3.4.3. R[ f (x)], R σ,ρ,a; : f (x) h i K K α,β,γ; Notation 3.4.4. [ f (x)], δ,ρ,a; : f (x) h i α,β,η; Notation 3.4.5. I0,x f (x) (Saigo, 1978)

α,β,η; Notation 3.4.6. Jx,α f (x) (Saigo, 1978) 3.4.GENERALIZEDKOBEROPERATORS 187 Deﬁnition 3.4.1. I[ f (x)] = I[α,β,γ : m, µ, η, a : f (x)] η 1 x µ µx− − at η = 2F1(α, β + m,γ; )t f (t)dt, (3.4.1) Γ(1 α) Z xµ − 0 where F ( ) is the Gauss hypergeometric function. 2 1 · Deﬁnition 3.4.2. I[ f (x)] = I[α,β,γ : m, µ, δ, a : f (x)] δ µ µx ∞ ax δ 1 = 2F1(α, β + m; γ; )t− − f (t)dt. (3.4.2) Γ(1 α) Z tµ − x

Operators deﬁned by (3.4.1) and (3.4.2) exist under the following conditions:

1 1 (i) 1 p, q < , p− + q− = 1, arg(1 a) < π (ii) ≤(1 α) > m∞, (η) > 1 , (δ) >| 1 , −(γ| α β m) > 1, m N ; γ , ℜ − ℜ − q ℜ − p ℜ − − − − ∈ 0 0, 1, 2, (iii)− f− L···(0, ) ∈ p ∞ Equations (3.4.1) and (3.4.2) are introduced by Kalla and Saxena (1969).

For γ = β, (3.4.1) and (3.4.2) reduce to generalized Kober operators, given by Saxena (1967).

Deﬁnition 3.4.3. α,β,γ; R[ f (x)] = R σ,ρ,a; f (x) hσ ρ x i = x− − σ ρ 1 t Γ t (x t) − 2F1[α, β; γ; a(1 )] f (t)dt. (3.4.3) (ρ) Z0 − − x Deﬁnition 3.4.4. K = K α,β,γ; [ f (x)] δ,ρ,a; f (x) hδ i = x ∞ δ ρ ρ 1 x Γ t− − (t x) − 2F1[α, β; γ; a(1 )] f (t)dt. (3.4.4) (ρ) Zx − − t 188 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

The conditions of validity of the operators (3.4.3) and (3.4.4) are given below:

1 1 (i) p 1, q < , p− + q− = 1, arg(1 a) < π. (ii) ≥(σ) > 1 ,∞ (δ) > 1 , (ρ) > 0.| − | ℜ − q ℜ − p ℜ (iii) γ , 0, 1, 2, ; (γ α β) > 0. (iv) f L (0−, −). ··· ℜ − − ∈ p ∞ The operators defined by (3.4.3) and (3.4.4) are given by Saxena and Kumb- a hat (1973). When a is replaced by α and α tends to infinity, the operators defined by (3.4.3) and (3.4.4) reduce to the following operators associated with confluent hypergeometric functions.

Definition 3.4.5. β,γ; = α,β,γ; R σ,ρ,a; f (x) lim R a f (x) α σ,ρ, α ; →∞ h i σ ρ x = x− − Φ t σ ρ 1 Γ [β,γ; a(1 )]t (x t) − f (t)dt. (3.4.5) (ρ) Z0 − x − Definition 3.4.6. K β,γ; = K α,β,γ; σ,ρ,a; f (x) lim a f (x) α δ,ρ, α ; →∞ h i δ = x ∞ Φ x δ ρ ρ 1 Γ [β,γ; a(1 )]t− − (t x) − f (t)dt, (3.4.6) (ρ) Zx − t − where (ρ) > 0, (δ) > 0. ℜ ℜ Remark 3.4.1. Many interesting and useful properties of the operators defined by (3.4.3) and (3.4.4) are investigated by Saxena and Kumbhat (1975), which deal with relations of these operators with well-known integral transforms, such as Laplace, Mellin and Hankel transforms. Equation (3.4.3) was first considered by Love (1967).

Remark 3.4.2. In the special case, when α is replaced by α + β,γ by α, σ by zero, ρ by α and β by η, then (3.4.3) reduces to the operator (3.4.7) considered by Saigo (1978). Similarly,− (3.4.4) reduces to another operator (3.4.9) introduced by Saigo (1978). 3.4.GENERALIZEDKOBEROPERATORS 189

Definition 3.4.7. Let α, β, η C, and let x R+ the fractional integral ( (α) > 0) and the fractional derivative∈ ( (α) < 0)∈ of the first kind of a function ℜ ℜ f (x) on R+ are defined by Saigo (1978) in the form

α β x α,β,η x− − α 1 I f (x) = (x t) − 0,x Γ (α) Z0 − t F (α + β, η; α;1 ) f (t)dt, (α) > 0 (3.4.7) × 2 1 − − x ℜ n d α+n,β n,η n = I − − f (x), 0 < (α) + n 1, (n N ). (3.4.8) dxn 0,x ℜ ≤ ∈ 0

Deﬁnition 3.4.8. The fractional integral ( (α) > 0) and fractional derivative ℜ ( (α) < 0) of the second kind of a function f (x)on R+ are given by Saigo (1978) inℜ the form

α,β,η = 1 ∞ α 1 α β Jx, f (x) Γ (t x) − t− − ∞ (α) Zx − x F (α + β, η; α;1 ) f (t)dt, (α) > 0 (3.4.9) × 2 1 − − t ℜ dn = n α+n,β n,η , < α + , N . ( 1) n Jx, − f (x) 0 ( ) n 1 (n 0) (3.4.10) − dx ∞ ℜ ≤ ∈

Example 3.4.1. Find the value of

α,β,η σ 1 I x − F (a, b; c; a′ x) . 0,x 2 1 − n o Solution 3.4.1. We have

α,β,η σ 1 K = I x − F (a, b; c; ax) 0,x 2 1 − n r r o ∞ (a)r(b)r( 1) (a′) α,β,η r+σ 1 = − I x − . (c) r! 0,x Xr=0 r Applying the result of Exercise 3.4.1, we obtain 190 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Γ + Γ + + r σ β 1 ∞ r (a)r(b)r (σ r) (σ β η r)(a′) r K = x − − ( 1) − x − (c) r! Γ(σ β + r)Γ(α + η + σ + r) Xr=0 r − Γ Γ + σ β 1 (σ) (σ η β) = x − − − Γ(σ β)Γ(σ + α + η) − F (a, b,σ,σ + η β; c,σ β,σ + α + η; a′ x), × 4 3 − − − where (α) > 0, (σ) > 0, (σ + η β) > 0, c , 0, 1, 2, ; a′ x < 1. ℜ ℜ ℜ − − − ··· | | Example 3.4.2. Find the value of

α,β,η λ a′ Jx, (x 2F1(a, b; c; )). ∞ x

Solution 3.4.2. Following a similar procedure and using the result of Exercise 3.4.3, it gives

Γ Γ α,β,η λ a′ = (β λ) (η λ) λ β Jx, (x 2F1(a, b; c; )) − − x − ∞ x Γ( λ)Γ(α + β + η λ) − − a F (a, b,β λ, η λ; c, λ, α + β + η λ; ′ ), × 4 3 − − − − x where (α) > 0, (β λ) > 0, (η λ) > 0, x > 0, c , 0, 1, 2, ; x > a′ . ℜ ℜ − ℜ − − − ··· | | | | α,β,η α,β,η Remark 3.4.3. Special cases of the operators I0,x and Jx, are the operators of Riemann -Liouville: ∞

x α, α,η α 1 α 1 I − f (x) = D f (x) = (x t) f (t)dt, ( (α) > 0) (3.4.11) 0,x 0 −x Γ − (α) Z0 − ℜ the Weyl:

α, α,η = α = 1 ∞ α 1 Jx, − f (x) xW f (x) Γ (t x) − f (t)dt, ( (α) > 0) (3.4.12) ∞ ∞ (α) Zx − ℜ 3.4.GENERALIZEDKOBEROPERATORS 191 and the Erd´elyi - Kober operators: α η x α,0,η α,η x− − α 1 η I f (x) = E f (x) = (x t) − t f (t)dt, ( (α) > 0) (3.4.13) 0,x 0,x Γ (α) Z0 − ℜ and η α,0,η = α,η = x ∞ α 1 α η Jx, f (x) Kx, f (x) Γ (t x) − t− − f (t)dt, ( (α) > 0) (3.4.14) ∞ ∞ (α) Zx − ℜ Example 3.4.3. Prove the following theorem.

If (α) > 0 and (s) < 1 + min[0, (η β)], then the following formula holds for f (x) ℜL (0, ) withℜ 1 p 2 or f (x) ℜM −(0, ) with p > 2: ∈ p ∞ ≤ ≤ ∈ p ∞ Γ(1 s)Γ(η β + 1 s) m xβIα,β,η f = − − − m f (x) . (3.4.15) 0,x Γ(1 s β)Γ(α + η + 1 s) { } n o − − − Solution 3.4.3. Use the integral

Γ Γ Γ + ∞ σ γ γ 1 x = (γ) (σ) (γ σ α β) u− − (u x) − 2F1 (α, β; γ;1 )du Γ + Γ +− − , Zx − − u (γ σ α) (γ σ β) − −(3.4.16) where (γ) > 0, (σ) > 0, (γ + σ α β) > 0. ℜ ℜ ℜ − −

Exercises 3.4.

3.4.1. Prove that Γ + Γ + + α,β,η λ (1 λ) (1 λ η β) λ β I x = − x − , (3.4.17) 0,x Γ(1 + λ β)Γ(1 + λ + α + η) and give the conditions of validity. −

β α,β,η 3.4.2. Find the Mellin transform of x Jx, f (x), giving conditions of its validity. ∞ 3.4.3. Prove that Γ Γ α,β,η λ = (β λ) (η λ) λ β Jx, x − − x − (3.4.18) ∞ Γ( λ)Γ(α + β + η λ) − − 192 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS and give the conditions of validity.

3.4.4. Prove that

Γ + Γ + + α,β,η k λx (k 1) (η k β 1) k β I (x e− ) = − x − 0,x Γ(k β + 1)Γ(α + η + k + 1) − F (k + 1, η + k β + 1; k β + 1, α + η + k + 1; λx), (3.4.19) × 2 2 − − − and give the conditions of validity.

3.4.5. Prove that Γ β η α,β,η sx = η η β ( ) Φ + Jx, e− s x − − (1 α β, 1 η β; sx) ∞ Γ(α + β) − − − − Γ(η β) + sβ − Φ(1 α η, 1 + β η; sx), (3.4.20) Γ(α + η) − − − − α and give the conditions of its validity. Deduce the results for L[ xW f ](s) and α,η ∞ L[ Kx, f ](s). ∞ 3.4.6. Prove that [ Saxena and Nishimoto (2002) ]

Γ Γ + α,β,η σ 1 c c (σ) (σ η β) σ β 1 I [x − (a + bx) ] = a − x − − 0,x Γ(σ β)Γ(σ + α + η) − bx F (σ, σ + η β, c; σ β,σ + α + η; ), × 3 2 − − − − a (3.4.21) where (σ) > max[0, (β η)], bx < 1. ℜ ℜ − | a | 3.4.7. Evaluate

α,β,η σ 1 m,n λ (ap,Ap) I x − H ax , λ > 0, (3.4.22) 0,x p,q |(bq,Bq) and give the conditions of itsn validity.h io

3.4.8. Evaluate

α,β,η σ 1 m,n λ (ap,Ap) Jx, x − Hp,q ax− (b ,B ) , λ > 0, (3.4.23) ∞ | q q and give the conditions of itsn validity.h io 3.4.GENERALIZEDKOBEROPERATORS 193

3.4.9. Establish the following property of Saigo operators called “Integration by parts”.

∞ α,β,η = ∞ α,β,η f (x) I0,x g (x)dx g(x) Jx, f (x)dx. Z0 Z0 ∞ 3.4.10. From Exercise 3.4.6, deduce the formula for

α, α,η + c I0,x− (a bx) , (3.4.24) given by B. Ross (1993).

3.4.11. Prove that

Γ(k + 1) Rα xk = xk+α, (3.4.25) 0,x Γ(α + k + 1) where (α) > 0, (k) > 1, ℜ ℜ − 3.4.12. Prove that

Γ α α k = ( k) k+α Wx, x − − x , (3.4.26) ∞ Γ( k) − where (α) > 0, (k) < (α). ℜ ℜ −ℜ 3.4.13. Show that

α,β,η λ px = λ β 3,0 λ,α+β+η λ Jx, (x e− ) x − G2,3 px −0,β λ,η λ− , (3.4.27) ∞ | − − , h i where G3 0( ) is the Meijer’s G-function, (px) > 0, (α) > 0. 2,3 · ℜ ℜ Hint: Use the integral

px 1 s e− = Γ( s)(px) ds. (3.4.28) 2πi ZL −

3.4.14. Evaluate α,β,η σ 1 m,n λ (ap,Ap) I0,x x − Hp,q ax− (b ,B ) , λ > 0, (3.4.29) | q q giving the conditions of its validity. 194 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

3.4.15. Evaluate

α,β,η σ 1 m,n λ (ap,Ap) Jx, x − Hp,q ax (b ,B ) , λ > 0 (3.4.30) ∞ | q q and give the conditions of validity of the result.

3.4.16. With the help of the following chain rules for Saigo operators ( Saigo, 1985)

α,β,η γ,δ,α+η = α+γ,β+δ,η I0,x I0,x f I0,x f, (3.4.31) and

α,β,η γ,δ,α+η = α+γ,β+δ,η Jx, Jx, f Jx, f, (3.4.32) ∞ ∞ ∞ derive the inverses

α,β,η 1 = α, β,α+η (I0,x )− I0−,x − . (3.4.33) and

α,β,η 1 = α, β,α+η (Jx, )− Jx−, − . (3.4.34) ∞ ∞

3.5. Compositions of Riemann-Liouville Fractional Calculus Operators and Generalized Mittag-Leﬄer Functions

In this section, composition relations between Riemann-Liouville fractional calculus operators and generalized Mittag-Leﬄer functions are derived. These relations may be useful in the solution of fractional diﬀerintegral equations. For details, one can refer to the work of Saxena and Saigo (2005). 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCULUS 195

δ 3.5.1. Composition relations between R-L operators and Eβ,γ(z)

Notation 3.5.1. Eα(x) : Mittag-Leﬄer function.

Notation 3.5.2. Eα,β(x) : Generalized Mittag-Leﬄer function.

α Notation 3.5.3. I0+ f : Riemann-Liouville left-sided integral. Notation 3.5.4. Iα f : Riemann-Liouville right-sided integral. − α Notation 3.5.5. D0+ f : Riemann-Liouville left-sided derivative. Notation 3.5.6. Dα f : Riemann-Liouville right-sided derivative. − δ ffl Notation 3.5.7. Eβ,γ(z) : Generalized Mittag-Le er function (Prabhakar, 1971). Definition 3.5.1. ∞ zk E (z):= , (α C, (α) > 0). (3.5.1) α Γ(αk + 1) ∈ ℜ Xk=0 Definition 3.5.2. ∞ zk E (z):= , (α, β C, (α) > 0, (β) > 0). (3.5.2) α,β Γ(αk + β) ∈ ℜ ℜ Xk=0 Definition 3.5.3. x α 1 f (t) (I + f )(x):= dt, (α) > 0. (3.5.3) 0 Γ(α) Z (x t)1 α ℜ 0 − − Definition 3.5.4. α = 1 ∞ f (t) (I f )(x): 1 α dt, (α) > 0. (3.5.4) − Γ(α) Z (t x) ℜ x − − Definition 3.5.5. [α]+1 α = d 1 α (D0+ f )(x): I0+−{ } (x); (α) > 0 (3.5.5) dx ℜ 1 d [α]+1 x f (t) = dt, (α) > 0. (3.5.6) Γ(1 α ) dx Z (x t) α ℜ −{ } 0 − { } 196 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS Definition 3.5.6. [α]+1 α d 1 α (D f )(x): = (I −{ } f )(x), (α) > 0 (3.5.7) − dx! − ℜ [α]+1 1 d ∞ f (t) = dt, (α) > 0. (3.5.8) Γ(1 α ) −dx! Z (t x) α ℜ −{ } x − { } Remark 3.5.1. Here [α] means the maximal integer not exceeding α and α is the fractional part of α. { }

Deﬁnition 3.5.7. ∞ (δ) zk Eδ (z):= k , (β,γ,δ C; (γ) > 0, (β) > 0). (3.5.9) β,γ Γ(βk + γ)k! ∈ ℜ ℜ Xk=0 For δ = 1, (3.5.9) reduces to (3.5.2).

R α Theorem 3.5.1. Let α > 0, β > 0, γ > 0 and α . Let I0+ be the left-sided operator of Riemann-Liouville fractional integral∈ (3.5.3). Then there holds the formula α γ 1 δ β = α+γ 1 δ β (I0+[t − Eβ,γ(at )])(x) x − Eβ,α+γ(ax ). (3.5.10)

Proof 3.5.1. By virtue of (3.5.3) and (3.5.9), we have

x n nβ+γ 1 α γ 1 δ β 1 α 1 ∞ (δ)na t − K (I +[t − E (at )])(x) = (x t) − dt. ≡ 0 β,γ Γ(α) − Γ(βn + γ)n! Z0 Xn=0 Interchanging the order of integration and summation and evaluating the inner integral by means of beta-function formula, it gives

β n α+γ 1 ∞ (δ)n(ax ) α+γ 1 δ β K x − = x − E + (ax ). ≡ Γ(α + βn + γ)(n)! β,α γ Xn=0 This completes the proof of Theorem 3.5.1.

Corollary 3.5.1. For α > 0, β > 0, γ > 0 and α R, there holds the formula ∈ α γ 1 β = α+γ 1 β (I0+[t − Eβ,γ(at )])(x) x − Eβ,α+γ(ax ). (3.5.11) 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCULUS 197

Remark 3.5.2. For β = α, (3.5.11) reduces to

γ 1 α γ 1 β x − α 1 (I +[t − E (at )])(x) = [E (ax ) ], (a , 0) (3.5.12) 0 α,γ a α,γ − Γ(γ) by virtue of the identity 1 E (x) = + xE + (x), (a , 0). (3.5.13) α,γ Γ(γ) α,α γ

R , α Theorem 3.5.2. Let α > 0, β > 0, γ > 0 and α , (a 0) and let I0+ be the left-sided operator of Riemann-Liouville fractiona∈ l integral (3.5.3). Then there holds the formula

α γ 1 δ β = 1 α+γ β 1 δ β δ 1 β (I0+[t − Eβ,γ(at )])(x) x − − [Eβ,α+γ β(ax ) Eβ,α− +γ β(ax )]. (3.5.14) a − − − Proof. Use Theorem 3.5.1.

The following two theorems can be established in the same way.

Theorem 3.5.3. Let α > 0, β > 0, γ > 0 and α R and let Iα be the right- sided operator of Riemann-Liouville fractional integral∈ (3.5.4). Then− we arrive at the following result:

α α γ δ β = γ δ β (I [t− − Eβ,γ(at− )])(x) x− [Eβ,α+γ(ax− )] (3.5.15) − Corollary 3.5.2. For α > 0, β > 0, γ > 0 and α R, there holds the formulas: ∈ α α γ β γ β (I [t− − Eβ,γ(at− )])(x) = x− [Eβ,α+γ(ax− )] (3.5.16) − and

α α 1 β 1 β (I t− − Eβ(at− ))(x) = x− [Eβ,α+1(ax− )]. (3.5.17) − Theorem 3.5.4. Let α > 0, β > 0, γ > 0, α R, (a , 0), α + γ>β and let Iα be the right-sided operator of Riemann-Liouville∈ fractional integral (3.5.4). Then− there holds the formula 198 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

α α γ δ β = 1 β γ δ β δ 1 β (I [t− − Eβ,γ(at− )])(x) x − [Eβ,α+γ β(ax− ) Eβ,α− +γ β(ax− )]. (3.5.18) − a − − −

Corollary 3.5.3. For α > 0, β > 0, γ > 0 with α + γ>β and for α R, (a , 0), there holds the formula ∈

α α γ β 1 β γ β 1 (I [t− − Eβ,γ(at− )])(x) = x − Eβ,α+γ β(ax− ) . (3.5.19) − a " − − Γ(α + γ β)# − Remark 3.5.3. (Kilbas and Saigo, (1998) ) α γ α α γ α x − α 1 (I [t− − Eα,γ(at− )])(x) = [Eα,γ(ax− ) ], (a , 0) (3.5.20) − a − Γ(γ) α 1 α α 1 α x − α (I [t− − Eα(at− )])(x) = Eα(ax− ) 1 , (a , 0). (3.5.21) − a − R α Theorem 3.5.5. Let α > 0, β > 0, γ > 0, γ > α, α and let D0+ be the left-sided operator of Riemann -Liouville fractional derivative∈ (3.5.6). Then there holds the formula.

α γ 1 δ β = γ α 1 δ β (D0+[t − Eβ,γ(at )])(x) x − − Eβ,γ α(ax ). (3.5.22) − Proof 3.5.2. By virtue of (3.5.9) and (3.5.6), we have

[α]+1 α γ 1 δ β d 1 α γ 1 δ β K (D +[t − E (at )])(x) = I −{ } t − E (at ) (x) ≡ 0 β,γ dx 0+ β,γ h i n [α]+1 x ∞ a (δ)n d nβ+γ 1 α = t − (x t)−{ }dt Γ(γ + nβ)Γ(1 a )n! dx Z − Xn=0 −{ } 0 n [α]+1 ∞ a (δ)n d nβ+γ α = x −{ } Γ(γ + nβ + 1 α )n! dx Xn=0 −{ } + ∞ n δ γ nβ α 1 = a ( )n x − − = γ α 1 δ β x − − Eβ,γ α(ax ), Γ(nβ + γ α)n! − Xn=0 − which proves the theorem. 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCULUS 199

By using a similar procedure, we arrive at the following theorem.

Theorem 3.5.6. Let α > 0,γ>β> 0, α R, (a , 0), γ>α + β and let α ∈ D0+ be the left-sided operator of Riemann-Liouville fractional derivative (3.5.6). Then there holds the formula

α γ 1 δ β = 1 γ α β 1 δ β δ 1 β D0+[t − Eβ,γ(at )] (x) x − − − Eβ,γ α β(ax ) Eβ,γ− α β(ax ) . (3.5.23) a − − − − − h i Corollary 3.5.4. Let α > 0,γ>β> 0, α R, (a , 0), γ>α + β, then there holds the formula. ∈

α γ 1 β = 1 γ α β 1 β 1 D0+[t − Eβ,γ(at )] (x) x − − − Eβ,γ α β(ax ) . (3.5.24) a " − − − Γ(γ α β)# − − Theorem 3.5.7. Let α > 0, γ > 0,γ α > 0 with γ α + α > 1, α R, and let Dα be the right-sided operator of Riemann-Liouville− − fraction{ }al derivative∈ (3.5.8). Then− there holds the formula.

α α γ δ β = γ δ β D [t − Eβ,γ(at− )] (x) x− Eβ,γ α(ax− ). (3.5.25) − − Theorem 3.5.8. Let α > 0, β > 0 with γ α > 1, α R, γ>α+β, (a , 0) and let Dα be the right-sided operator of Riemann-Liouville−{ } ∈ fractional derivative (3.5.8). Then− there holds the formula

β γ α α γ δ β = x − δ β δ 1 β D [t − Eβ,γ(at− )] (x) Eβ,γ α β(ax− ) Eβ,γ− α β(ax− ) . (3.5.26) − a − − − − − h i

Exercises 3.5.

3.5.1. Show that

β δ β = δ β δ 1 β , ax Eβ,γ(ax ) Eβ,γ β(ax ) Eβ,γ− β(ax ), (a 0) (3.5.27) − − − 3.5.2. Show that 200 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

γ 1 α γ 1 α x − α 1 I [t − E (at )] (x) = E (ax ) , (a , 0). (3.5.28) 0+ α,γ a α,γ Γ(γ) " − # 3.5.3. Prove Theorem 3.5.3.

3.5.4. Prove Theorem 3.5.4.

3.5.5. Prove Theorem 3.5.6.

3.5.6. Prove Theorem 3.5.7.

3.5.7. Prove Theorem 3.5.8.

3.5.8. Prove that

α ω m,n σ (ap,Ap) ω+α m,n+1 σ ( ω,σ),(ap,Ap) = − I0+t Hp,q t (bq,Bq) (x) x Hp+1,q+1 x (bq,Bq),( ω α,σ) , (3.5.29) | | − − giving conditionsh of validity.i h i

3.5.9. Evaluate

α ω m,n σ (ap,Ap) I t Hp,q t (bq,Bq) (x), (3.5.30) − | and give the conditions of validity. h i

3.6. Fractional Diﬀerential Equations

Differential equations contain integer order derivatives, whereas fractional ff dα di erential equations involve fractional derivatives, like dxα , which are defined for α > 0. Here α is not necessarily an integer and can be rational, irrational or even complex-valued. Today, fractional calculus models find applications in physical, biological, engineering, biomedical and earth sciences. Most of the problems discussed involve relaxation and diffusion models in the so called complex or disordered systems. Thus, it gives rise to the generalization of initial value problems involving ordinary differential equations to generalized fractional-order differential equations and Cauchy problems involving 3.6.FRACTIONALDIFFERENTIALEQUATIONS 201 partial differential equations to fractional reaction, fractional diffusion and fractional reaction-diffusion equations. Fractional calculus plays a dominant role in the solution of all these physical problems.

3.6.1. Fractional relaxation In order to formulate a relaxation process, we require a physical law, say the relaxation equation d 1 f (t) + f (t) = 0, t > 0, c > 0, (3.6.1) dt c to be solved for the initial value f (t = 0) = f0. The unique solution of (3.6.1) is given by

t f (t) = f e− c , t 0, c > 0. (3.6.2) 0 ≥ Now the problem is as to how we can generalize the initial-value problem (3.6.1) into a fractional value problem with physical motivation. If we incorporate the initial value f0 into the integrated relaxation equation (3.6.1), we ﬁnd that

1 1 f (t) f = D− f (t), (3.6.3) − 0 − c 0 t 1 where 0Dt− is the standard Riemann integral of f (t). 1 1 1 α On replacing c 0Dt− f (t) by cα 0Dt− f (t), it yields the fractional integral equation

1 α f (t) f = D− f (t), α > 0 (3.6.4) − 0 − cα ! 0 t with initial value

f0 = f (t = 0). ﬀ α Applying the Riemann-Liouville di erential operator 0Dt from the left and mak- ing use of the formula (3.1.16), we arrive at

α α t− α D f (t) = f = c− f (t), α > 0, c > 0, (3.6.5) 0 t 0 Γ(1 α) − − with initial condition f0 = f (t = 0). 202 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Theorem 3.6.1. The solution of the fractional diﬀerential equation (3.6.4) is given by

α t (0,1) f (t) = f H1,1 , (3.6.6) 0 1,2 c (0,1),(0,α) where α > 0, c > 0.

Proof 3.6.1. If we apply the Laplace transform to equation (3.6.4), it gives

1 1 α F(s) f s− = s− F(s), (3.6.7) − 0 −cα where we have used the result (3.1.7) and F(s) is the Laplace transform of f (t). Solving for F(s), we have

s 1 F(s) = L f (t) = f − . (3.6.8) 0 + α { } "1 (cs)− # Taking inverse Laplace transform, (3.6.8) gives

s 1 f (t) = L 1 F(s) = f L 1 − − 0 − + α { } "1 (cs)− #

1 ∞ k αk αk 1 = f L− ( 1) c− s− − 0  −  Xk=0   k t αk  ∞ ( 1) ( )  = f − c 0 Γ(αk + 1) Xk=0 t α = f0Eα ( ) , (3.6.9) − c where Eα( ) is the Mittag-Leﬄer function. (3.6.9) can be written in terms of the H-function· as

α t (0,1) f (t) = f H1,1 , (3.6.10) 0 1,2 c (0,1),(0,α) where c > 0, α > 0. This completes the proof of the Theorem 3.6.1. Alternative form of the solution. By virtue of the identity

Ap (a ,A ) 1 (ap, ) m,n µ p p = m,n µ H x H x B , (µ > 0) (3.6.11) p,q (bq,Bq) p,q q µ " (bq, µ ) #

3.6.FRACTIONALDIFFERENTIALEQUATIONS 203 the solution (3.6.10) can be written as

f t (0, 1 ) f (t) = 0 H1,1 α , (3.6.12) 1,2 (0, 1 ),(0,1) α c α where α > 0, c > 0.

Remark 3.6.1. In the limit as α 1, one recovers the result (3.6.2) → t t f (t) = f exp( ) = f E ( ). (3.6.13) 0 −c 0 1 c Remark 3.6.2. In terms of Wright’s function, the solution (3.6.10) can be ex- pressed in the form t f (t) = f ψ (1,1) ( )α , (3.6.14) 0 1 1 (1,α) − c where α > 0, c > 0. In a similar manner, we can establish Theorems 3.6.2 and 3.6.3 given below.

Theorem 3.6.2. The solution of the fractional integral equation

µ 1 ν ν N(t) N t − = c D− N(t), (3.6.15) − 0 − 0 t is given by µ 1 ν µ N(t) = N Γ(µ)t − E ( c t ), (3.6.16) 0 ν,µ − where E ( ) is the generalized Mittag-Leﬄer function (3.5.2), ν > 0, µ > 0. ν,µ · Remark 3.6.3. When µ = 1, we obtain the result given by Haubold and Mathai (2000).

Theorem 3.6.3. If c > 0, ν > 0, µ > 0, then for the solution of the integral equation

µ 1 γ ν ν ν N(t) N t − E [ (ct) ] = c D− N(t), (3.6.17) − 0 ν,µ − − 0 t there holds the formula

µ 1 γ+1 ν N(t) = N t − E [ (ct) ]. (3.6.18) 0 ν,µ − 204 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Hint: Use the formula

1 β α γ β 1 γ α L− s− (1 as− )− = t − E (at ), (3.6.19) − α,β n 1 o where (α) > 0, (β) > 0, (s) > a (α) , (s) > 0. ℜ ℜ ℜ | | ℜ ℜ Corollary 3.6.1. If c > 0, µ > 0, ν > 0, then for the solution of

µ 1 ν ν ν ν N(t) N t − E [ c t ] = c D− N(t), (3.6.20) − 0 ν,µ − − 0 t there holds the relation

N0 µ 1 ν ν ν ν N(t) = t − Eν,µ 1( c t ) + (1 + ν µ)Eν,µ( c t ) . (3.6.21) ν − − − − h i Theorem 3.6.4. The Cauchy problem for the integro-diﬀerential equation

µ ν D f (x) + λ D− f (x) = h(x), (λ,µ,ν C) (3.6.22) 0 x 0 x ∈ with the initial condition

µ k 1 D − − f (0) = a , k = 0, 1, , [µ], (3.6.23) x k ··· where (ν) > 0, (µ) > 0 and h(x) is any integrable function on the ﬁnite intervalℜ [0,b] has theℜ unique solution, given by

x µ 1 µ+ν f (x) = (x t) − Eµ+ν,µ[ λ(x t) ]h(t)dt Z0 − − − n 1 − α k 1 µ+ν + ak x − − Eµ+ν,µ k( λx ) (3.6.24) − − Xk=0

Proof 3.6.2. Exercise.

Theorem 3.6.5. The solution of the equation

1 1 2 + = − 2 = 0Dt f (t) bf (t) t > 0; 0Dt f (t) C, (3.6.25) t=0 where C is a constant is given by 3.6.FRACTIONALDIFFERENTIALEQUATIONS 205

1 1 = 2 2 f (t) C t− E 1 , 1 ( bt ), (3.6.26) 2 2 − where E 1 , 1 ( ) is the Mittag-Leﬄer function. 2 2 · Proof 3.6.3. Exercise.

Remark 3.6.4. Theorem 3.6.5 gives the generalized form of the equation solved by Oldham and Spanier (1974).

Exercises 3.6.

3.6.1. Prove that if c > 0, ν > 0, µ > 0, then the solution of µ 1 2 ν ν ν ν N(t) N t − E (c t ) = c D− N(t), (3.6.27) − 0 ν,µ − 0 t is given by µ 1 = µ 1 3 ν ν = N0t − ν ν N(t) N0t − Eν,µ( c t ) Eν,µ 2( c t ) − 2ν2 − − ν ν + 3(ν + 1) 2µ Eν,µ 1( c t ) { − } − − 2 2 ν ν + 2ν + µ + 3ν 2µ 3νµ + 1 Eν,µ( c t ) , (3.6.28) n − − o − where (ν) > 0, (µ) > 2. ℜ ℜ 3.6.2. Prove that if ν > 0, c > 0, d > 0, µ > 0, c , d, then for the solution of the equation

µ 1 ν ν ν ν N(t) N t − E ( d t ) = c D− N(t), (3.6.29) − 0 ν,µ − − 0 t there holds the formula.

µ ν 1 t − − ν ν ν ν N(t) = N0 Eν,µ ν( d t ) Eν,µ ν( c t ) . (3.6.30) cν dν − − h − − − i 3.6.3. Prove that if c > 0, ν >−0, µ > 0, then for the solution of the equation

µ 1 ν ν ν ν N(t) N t − E ( c t ) = c D− N(t), (3.6.31) − 0 ν,µ − − 0 t the following result holds: 206 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

N0 µ 1 ν ν ν ν N(t) = t − Eν,µ 1( c t ) + (1 + ν µ)Eν,µ( c t ) . (3.6.32) ν − − − − h i 3.6.4. Solve the equation

Q + q = 0Dt f (t) 0Dt f (t) g(t), where q Q is not an integer or a half integer and the initial condition is −

q 1 + Q 1 = 0Dt − f (t) 0Dt − f (t) C (3.6.33) t=0 where C is a constant. 3.6.5. Solve the equation

Dα x(t) λx(t) = h(t), (t > 0), (3.6.34) 0 t − subject to the initial conditions

α k 0D − h(t) = bk, (k = 1, , n) (3.6.35) t t=0 ··· h i where n 1 < α < n. − 3.6.6 Prove Theorem 3.6.4

3.6.6 Prove Theorem 3.6.5.

3.6.2. Fractional diﬀusion Theorem 3.6.6. The solution of the following initial value problem for the fractional diﬀusion equation in one dimension

∂2U(x, t) Dα U(x, t) = λ2 , (t > 0, < x < ) (3.6.36) 0 t ∂x2 −∞ ∞ 3.6.FRACTIONALDIFFERENTIALEQUATIONS 207 with initial conditions :

= α 1 = lim U(x, t) 0; 0Dt − U(x, t) φ(x) (3.6.37) x t=0 →±∞ h i is given by

∞ U(x, t) = G(x ζ, t)φ(ζ)dζ, (3.6.38) Z − −∞ where

1 ∞ α 1 2 2 α G(x, t) = t − Eα,α( k λ t )cos kx dk. (3.6.39) π Z0 −

Solution 3.6.1. Let 0 < α < 1. Using the boundary conditions (3.6.37), the Fourier transform of (3.6.36) with respect to variable x gives

α ¯ + 2 2 ¯ = 0Dx U(k, t) λ k U(k, t) 0 (3.6.40) α 1 0D − U¯ (k, t) = φ¯(k), (3.6.41) t t=0 h i where k is a Fourier transform parameter and ‘ ’ indicates Fourier transform. Applying the Laplace transform to (3.6.40) and using− (3.6.41), it gives

φ¯(k) U≃ (k, s) = , (3.6.42) sα + k2λ2 where ‘ ’ indicates Laplace transform. The inverse Laplace transform of (3.6.42) yields∼

α 1 2 2 2 U¯ (k, t) = t − φ¯(k)E ( λ k t ), (3.6.43) α,α − and then the solution is obtained by taking inverse Fourier transform. By taking inverse Fourier transform of (3.6.43) and using the formula

1 ∞ ikx 1 ∞ e− f (k)dk = f (k) cos(kx)dk (3.6.44) 2π Z π Z0 −∞ we have 208 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

∞ U(x, t) = G(x ζ, t)φ(ζ)dζ, (3.6.45) Z − −∞ where

1 ∞ α 1 2 2 α G(x, t) = t − Eα,α( k λ t ) cos(kx)dk (3.6.46) π Z0 − with (α) > 0, k > 0. ℜ

Exercise 3.6.

3.6.8 Evaluate the integral in (3.6.46).

3.6.9 Find the solution of the Fick’s diﬀusion equation

∂ ∂2 P(x, t) = λ P(x, t), ∂t ∂x2 with the initial condition P(x, t = 0) = δ(x), where δ(x) is the Dirac delta function.

References

Caputo, M. (1969). Elasticit´ae Dissipazione, Zanichelli, Bologna.

Dzherbashyan, M.M. (1966). Integral Transforms and Representation of Func- tions in Complex Domain (in Russian), Nauka, Moscow.

Erd´elyi, A. (1950-51). On some functional transformations, Univ. Politec. Torino, Rend. Sem. Mat. 10, 217-234.

Erd´elyi, A., Magnus, W., Oberhettinger, F. and Tricomi, F. G. (1953). Higher Transcendental Functions, Vol. 1, McGraw - Hill, New York, Toronto and Lon- don. 3.6.FRACTIONALDIFFERENTIALEQUATIONS 209

Erd´elyi, A., Magnus, W., Oberhettinger, F. and Tricomi, F. G. (1954). Tables of Integral Transforms, Vol. 1, McGraw - Hill, New York, Toronto and London.

Erd´elyi, A., Magnus, W., Oberhettinger, F. and Tricomi, F. G. (1954a). Tables of Integral Transforms, Vol. 2, McGraw - Hill, New York, Toronto and London.

Erd´elyi, A., Magnus, W., Oberhettinger, F. and Tricomi, F. G. (1955). Higher Transcendental Functions, Vol. 3, McGraw - Hill, New York, Toronto and Lon- don.

Fox, C. (1963). Integral transforms based upon fractional integration, Proc. Cambridge Philos. Soc., 59, 63-71.

Haubold, H. J. and Mathai, A. M. (2000). The fractional kinetic equation and thermonuc1ear functions, Astrophysics and Space Science, 273, 53-63.

Hilfer, R., (Ed.). (2000). Applications of Fractional Calculus in Physics, World Scientiﬁc, Singapore.

Kalla, S. L. and Saxena, R. K. (1969). Integral operators involving hypergeometric functions, Math. Zeitschr., 108, 231-234.

Kilbas. A. A. and Saigo, M. (1998). Fractional calculus of the H-function, Fukuoka Univ. Science Reports, 28, 41-51.

Kilbas, A. A. and Saigo, M. (1996). On Mittag- Leﬄer type function, fractional calculus operators and solutions of integral equations, Integral Transforms and Special Functions, 4, 355-370.

Kilbas, A. A, Saigo, M. and Saxena, R. K. (2002). Solution of Volterra integrod- iﬀerential equations with generalized Mittag-Leﬄer function in the kernels, J. Integral Equations and Applications, 14, 377-396

Kilbas, A. A., Saigo, M. and Saxena, R. K. (2004). Generalized Mittag-Leﬄer function and generalized fractional calculus operators, Integral Transforms and Special Functions, 15, 31-49.

Kober, H. (1940). On fractional integrals and derivatives, Quart. J. Math. Ox- ford, Ser. ll, 193 -211. 210 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Love, E. R (1967). Some integral equations involving hypergeometric functions, Proc. Edin. Math. Soc., 15(2), 169-198.

Mathai, A. M. and Saxena, R. K. (1973). Generalized Hypergeometric Functions with Applications in Statistics and Physical Sciences, Lecture Notes in Mathe- matics, 348, Springer- Verlag, Berlin, Heidelberg.

Mathai, A. M. and Saxena, R. K. (1978). The H-function with Applications in Statistics and Other Disciplines, John Wiley and Sons, New York - London - Sydney.

Miller, K. S. and Ross, B. (1993). An Introduction to the Fractional Calculus and Fractional Diﬀerential Equations, Wiley, New York.

Mittag-Leﬄer, G. M. (1903). Sur la nouvelle fonction Eα(x), C. R. Acad. Sci. Paris, (Ser. II) 137, 554-558.

Oldham, K. B. and Spanier, J. (1974). The Fractional Calculus: Theory and Applications of Diﬀerentiation and Integration to Arbitrary Order, Academic Press, New York.

Podlubny, 1. (1999). Fractional Diﬀerential Equations, Academic Press, San Diego.

Podlubny, 1. (2002). Geometric and physical interpretations of fractional integration and fractional diﬀerentiation, Frac. Calc. Appl. Anal., 5(4), 367-386.

Prabhakar, T. R. (1971). A singular integral equation with a generalized Mittag- Leﬄer function in the kernel, Yokohama Math. J., 19, 7-15.

Ross, B. (1994). A formula for the fractional integration and diﬀerentiation of (a + bx)c, J. Fract. Calc., 5, 87-89.

Saigo, M. (1978). A remark on integral operators involving the Gauss hypergeometric function, Math. Reports of College of Gen. Edu. , Kyushu University, 11, 135 -143.

Saigo, M. and Raina, R. K. (1988). Fractional calculus operators associated with a general class of polynomials, Fukuoka Univ. Science Reports, 18,15-22. 3.6.FRACTIONALDIFFERENTIALEQUATIONS 211

Samko, S. G. , Kilbas, A. A. and Marichev, 0. 1. (1993). Fractional Integrals and Derivatives, Theory and Applications, Gordon and Breach, Reading.

Saxena, R. K. (1967). On fractional integration operators, Math. Zeitsch., 96, 288-291.

Saxena, R. K. (2002). Certain properties of generalized Mittag-Leﬄer function, Proceedings of the Third Annual Conference of the Society for Special Functions and Their Applications, Varanasi, March 4-6, 75-81.

Saxena, R. K. (2003). Alternative derivation of the solution of certain integro- diﬀerential equations of Volterra-type, Ganita Sandesh, 17(1), 51-56.

Saxena,R. K. (2004). On a uniﬁed fractional generalization of free electron laser equation, Vijnana Parishad Anusandhan Patrika, 47(l), 17-27.

Saxena, R. K. and Kumbhat, R. K. (1973). A generalization of Kober operators, Vijnana Parishad Anusandhan Patrika, 16, 31-36.

Saxena, R. K. and Kumbhat, R. K. (1974). Integral operators involving H- function, Indian J. Pure appl. Math., 5, 1-6.

Saxena, R. K. and Kumbhat, R. K. (1975). Some properties of generalized Kober operators, Vijnana Parishad Anusandhan Patrika, 18, 139-150.

Saxena, R. K, Mathai, A. M and Haubold, H. J. (2002). On fractional kinetic equations, Astrophysics and Space Science, 282, 281-287.

Saxena, R. K, Mathai, A. M and Haubold, H. J. (2004). On generalized fractional kinetic equations, Physica A , 344, 657-664.

Saxena, R. K, Mathai, A. M. and Haubold, H. J. (2004). Uniﬁed fractional kinetic equations and a fractional diﬀusion equation, Astrophysics and Space Sci- ence, 290, 241-245.

Saxena, R. K. and Nishimoto, K. ( 2002). On a fractional integral formula of Saigo operator, J. Fract. Calc., 22, 57-58. 212 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQUATIONS

Saxena, R. K. and Saigo, M. (2005). Certain properties of fractional calculus operators associated with generalized Mittag-Leffler function. Frac. Calc. Appl. Ana1., 8(2), 141-154. Sneddon, I. N. (1975). The Use in Mathematical Physics of Erdélyi- Kober Op- erators and Some of Their Applications, Lecture Notes in Mathematics (Edited by B. Ross), 457, 37-79. Srivastava, H. M. and Saxena, R. K. (2001). Operators of fractional integration and their applications, Appl. Math. Comput., 118, 1- 52. Stein, E. M. (1970). Singular Integrals and Differential Properties of Functions, Princeton University Press, New Jersey. Wiman, A. (1905). Uber den Fundamental satz in der Theorie de Funktionen Eα(x). Acta Math., 29, 191-201 Weyl, H. (1917). Bemerkungen zum Begriff des Differentialquotienten gebroch- ener Ordnung, Vierteljahresschr. Naturforsch. Gen. Zurich, 62, 296-302.