Expansion formulas in terms of integer-order derivatives for the ...

arXiv:1112.0693v1 [math.CA] 3 Dec 2011

Expansion formulas in terms of integer-order derivatives for the Hadamard fractional integral and derivative∗ Shakoor Pooseh [email protected]

Ricardo Almeida [email protected]

Delfim F. M. Torres [email protected]

Center for Research and Development in Mathematics and Applications Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal

Abstract We obtain series expansion formulas for the Hadamard fractional integral and fractional derivative of a smooth function. When considering finite sums only, an upper bound for the error is given. Numerical simulations show the efficiency of the approximation method.

MSC 2010: 26A33, 33F05. Keywords: Fractional Calculus, Hadamard fractional integrals, Hadamard fractional derivatives, numerical approximations.

1

Introduction

In general terms, Fractional Calculus allows to define integrals and derivatives of arbitrary real or complex order, and can be seen as a generalization of ordinary calculus. A fractional derivative of order α > 0, when α is integer, coincides with the classical derivative of order n ∈ N, while a fractional integral is an n-folded integral. Although there exist in the literature a large number of definitions for fractional operators (integrals and derivatives), the Riemann–Liouville and Caputo are the most common for fractional derivatives, and for fractional integrals the usual one is the Riemann–Liouville definition. In our work we consider the Hadamard fractional integral and fractional derivative. Although the definitions go back to the works of Hadamard in 1892 [5], this type of operators are not yet well studied and much exists to be done. Part of the first author’s Ph.D., which is carried out at the University of Aveiro under the Doctoral Program in Mathematics and Applications (PDMA) of Universities of Aveiro and Minho, supported by FCT fellowship SFRH/BD/33761/2009. Submitted 25-Jul-2011; revised 28-Nov-2011; accepted 02-Dec-2011; for publication in Numerical Functional Analysis and Optimization. ∗

1

As well-known, for most problems involving fractional operators, such as fractional differential equations or fractional control problems, one cannot provide methods to compute the exact solutions analytically. Therefore, numerical methods need to be employed. Typically, problems are addressed by looking to the fractional operators as special types of integrals and, using partitions of the domains, writing them as finite sums, with some error for the final result. Our approach is distinct from this, in the sense that we seek expansion formulas for the Hadamard fractional operators with integer-order derivatives. In this way we can rewrite the original problem, which depends on fractional operators, as a new one that involves integer derivatives only and then successfully apply the standard methods to obtain the desired solution. Also, in some cases, it can be easier to find the fractional derivative or integral using those expansions, instead of applying the direct definitions. In contrast with [2], where an expansion for the Riemann– Liouville fractional derivative is given, our expansions for the Hadamard fractional derivative and integral do not omit the first derivative, allowing one to obtain considerable better accuracy in computation. The paper is organized in the following way. In Section 2 we review some concepts of fractional calculus. Decomposition formulas for the left and right Hadamard fractional integrals are given in Section 3, together with approximation formulas and error estimations. Following the same approach, similar formulas are obtained for the left and right Hadamard fractional derivatives in Section 4. In Section 5 we test the efficiency of such approximations with some examples, comparing the analytical/exact solution with the numerical approximation.

2

Preliminaries

In this section we review some necessary definitions for our present work, namely the Hadamard fractional integral and derivative. For more on fractional calculus, we refer the interested reader to [8, 10, 11, 13]. For related work on Hadamard fractional operators, see [3, 4, 6, 7, 9, 12]. Let a, b be two reals with 0 < a < b and x : [a, b] → R be an integrable function. The left and right Hadamard fractional integrals of order α > 0 are defined by α a It x(t) =

1 Γ(α)

Z t

α t Ib x(t) =

1 Γ(α)

Z b

and

α−1

ln

t τ

x(τ ) dτ, τ

t ∈]a, b[,

ln

τ α−1 x(τ ) dτ, t τ

t ∈]a, b[,

a

t

respectively. These integrals were introduced by Hadamard in [5] in the special case a = 0. When α = m is an integer, these fractional integrals are m-folded integrals (see, e.g., [3, 6]): m a It x(t)

=

Z

a

t

dτ1 τ1

Z

a

τ1

dτ2 ··· τ2

2

Z

a

τm−1

x(τm ) dτm τm

and m t Ib x(t) =

Z

b

t

Z

dτ1 τ1

b

dτ2 ··· τ2

τ1

Z

b

τm−1

x(τm ) dτm . τm

For fractional derivatives, we also consider left and right operators. For α > 0, the left and right Hadamard fractional derivatives of order α are defined by

and α t Db x(t)



n

n−α−1

t ∈]a, b[,

  Z b d n τ n−α−1 x(τ ) 1 = −t ln dτ, dt Γ(n − α) t t τ

t ∈]a, b[,

=

d t dt

1 Γ(n − α)

Z t

x(τ ) dτ, τ

α a Dt x(t)

a

t ln τ

respectively, with n = [α] + 1. When α = m is an integer, we have (cf. [8])    m d m d m m x(t) and t Db x(t) = −t x(t). a Dt x(t) = t dt dt Hadamard’s fractional integrals and derivatives can be seen as inverse operations of each other (see Property 2.28 and Theorem 2.3 of [8]). When α ∈ (0, 1) and x ∈ AC[a, b], with AC[a, b] denoting the space of all absolutely continuous functions x : [a, b] → R, the Hadamard fractional derivatives may be expressed by α a Dt x(t)

x(a) = Γ(1 − α)



α t Db x(t)

x(b) = Γ(1 − α)



and

−α

1 + Γ(1 − α)

Z t

x(τ ˙ )dτ

(1)

−α

1 − Γ(1 − α)

Z b τ −α ln x(τ ˙ )dτ. t t

(2)

t ln a

b ln t

a

t ln τ

−α

For an arbitrary α > 0 we refer the reader to [7, Theorem 3.2]. If a function x admits derivatives of any order, then expansion formulas for the Hadamard fractional integrals and derivatives of x, in terms of its integer-order derivatives, are given in [4, Theorem 17]: α 0 It x(t)

=

∞ X k=0

and α 0 Dt x(t) =

where

S(−α, k)tk x(k) (t)

∞ X

S(α, k)tk x(k) (t),

k=0

  k 1 X k−j k (−1) jα S(α, k) = k! j j=1

is the Stirling function.

3

3

An expansion formula for the Hadamard fractional integral

In this section we consider the class of differentiable functions up to order n + 1, x ∈ C n+1 [a, b], and deduce expansion formulas for the Hadamard fractional integrals in terms of x(i) (·), for i ∈ {0, . . . , n}. Before presenting the result in its full extension, we briefly explain the techniques involved for the particular case n = 2. To that purpose, let x ∈ C 3 [a, b]. Integrating by parts three times, we obtain   Z t 1 t α−1 α x(t) = − 1 I x(τ )dτ − ln a t Γ(α) a τ  τ   Z t 1 1 1 t α t α x(a) − τ x(τ ˙ )dτ = − ln ln Γ(α + 1)  a  Γ(α + 1) a τ τ 1 1 t α t α+1 = x(a) + ax(a) ˙ ln ln Γ(α + 1) a Γ(α + 2) a   Z t 1 t α+1 1 (τ x(τ ˙ ) + τ 2x ¨(τ ))dτ − ln − Γ(α + 2) a τ τ   1 1 t α t α+1 x(a) + ax(a) ˙ = ln ln Γ(α + 1)  a  Γ(α + 2) a 1 t α+2 + (ax(a) ˙ + a2 x ¨(a)) ln Γ(α + 3) a  Z t 1 t α+2 ... + (x(τ ˙ ) + 3τ x ¨(τ ) + τ 2 x (τ ))dτ. ln Γ(α + 3) a τ On the other hand, using the binomial formula, we have       ln τ α+2 t α+2 t α+2 = ln 1 − at ln τ a ln a
p α+2 X  ∞ Γ(p − α − 2) ln τa t
. · = ln a Γ(−α − 2)p! ln at p p=0

This series converges since τ ∈ [a, t] and α + 2 > 0. Combining these formulas, we get       1 1 t α t α+1 t α+2 1 2 α x(a)+ a x(a)+ ˙ (ax(a)+a ˙ x ¨(a)) ln ln ln I x(t) = a t Γ(α + 1) a Γ(α + 2) a Γ(α + 3) a   Z t ∞
1 τ p Γ(p − α − 2) t α+2 X 2 ...
x + ln (τ ) dτ. x(τ ˙ ) + 3τ x ¨ (τ ) + τ ln p t Γ(α + 3) a a Γ(−α − 2)p! ln a a p=0

Now, split the series into the two cases p = 0 and p = 1 . . . ∞, and integrate by parts the second one. We obtain     1 1 t α t α+1 α x(t) = I x(a) + ax(a) ˙ ln ln a t Γ(α + 1) a Γ(α + 2) a   α+2 ∞ X Γ(p − α − 2)  t 1 (tx(t) ˙ + t2 x ¨(t)) 1 + ln + Γ(α + 3) a Γ(−α − 2)p! p=1  α+2 X Z t ∞ τ p−1 Γ(p − α − 2) 1 t
ln (x(τ ˙ ) + τx ¨(τ ))dτ. + ln p Γ(α + 2) a a Γ(−α − 1)(p − 1)! ln at a p=1

4

Repeating this procedure two more times, we obtain the following:    α ∞ X 1 Γ(p − α − 2)  t α x(t) 1 + ln a It x(t) = Γ(α + 1) a Γ(−α)(p − 2)! p=3   α+1 ∞ X 1 Γ(p − α − 2)  t + tx(t) ˙ 1 + ln Γ(α + 2) a Γ(−α − 1)(p − 1)! p=2    α+2 ∞ X 1 Γ(p − α − 2)  t + (tx(t) ˙ + t2 x ¨(t)) 1 + ln Γ(α + 3) a Γ(−α − 2)p! p=1  α+2 X Z t ∞ τ p−3 x(τ ) Γ(p − α − 2) t 1
ln dτ, ln + p Γ(α) a a τ Γ(−α + 1)(p − 3)! ln at a p=3

or, in a more concise way, α a It x(t)

   t α t α+1 = A0 (α) ln x(t) + A1 (α) ln tx(t) ˙ a a  α+2   ∞ X t α+2−p t 2 B(α, p) ln (tx(t) ˙ +t x ¨(t)) + Vp (t) +A2 (α) ln a a 

p=3

with



 ∞ X 1 Γ(p − α − 2)  1 + A0 (α) = , Γ(α + 1) Γ(−α)(p − 2)! p=3   ∞ X Γ(p − α − 2)  1 1 + , A1 (α) = Γ(α + 2) Γ(−α − 1)(p − 1)! p=2   ∞ X 1 Γ(p − α − 2) 1 + , A2 (α) = Γ(α + 3) Γ(−α − 2)p! p=1

Γ(p − α − 2) Γ(α)Γ(1 − α)(p − 2)!

(3)

 τ p−3 x(τ ) (p − 2) ln dτ, a τ

(4)

B(α, p) = and Vp (t) =

Z

a

t

where we assume the series and the integral Vp to be convergent.

Remark 3.1. When useful, namely on fractional differential equations problems, we can define Vp as in (4) by the the solution of the system (
p−3 x(t) V˙p (t) = (p − 2) ln at t Vp (a) = 0 for all p = 3, 4, . . .

5

We now discuss the convergence of the series involved in the definitions of Ai (α), for i ∈ {0, 1, 2}. Simply observe that ∞ X

p=3−i

Γ(p − α − 2) = 1 F0 (−α − i, 1) − 1, Γ(−α − i)(p − 2 + i)!

and 1 F0 (a, x) converges absolutely when |x| = 1 if a < 0 ([1, Theorem 2.1.2]). For numerical purposes, only finite sums are considered, and thus the Hadamard left fractional integral is approximated by the decomposition     t α+1 t α α x(t) + A1 (α, N ) ln tx(t) ˙ a It x(t) ≈ A0 (α, N ) ln a a     N (5) X t α+2 t α+2−p +A2 (α, N ) ln B(α, p) ln (tx(t) ˙ + t2 x ¨(t)) + Vp (t) a a p=3

with

  N X Γ(p − α − 2) 1 1 + , A0 (α, N ) = Γ(α + 1) Γ(−α)(p − 2)! p=3   N X 1 Γ(p − α − 2)  1 + A1 (α, N ) = , Γ(α + 2) Γ(−α − 1)(p − 1)! p=2   N X Γ(p − α − 2)  1 1 + , A2 (α, N ) = Γ(α + 3) Γ(−α − 2)p! p=1

B(α, p) and Vp (t) as in (3)–(4), and N ≥ 3. We proceed with an estimation for the error on such approximation. We have proven before that       1 1 t α t α+1 t α+2 1 2 α x(a)+ ax(a)+ ˙ (ax(a)+a ˙ x ¨(a)) ln ln ln a It x(t) = Γ(α + 1) a Γ(α + 2) a Γ(α + 3) a
p   Z ∞ 1 t α+2 t X Γ(p − α − 2) ln τa ...
˙ ) + 3τ x¨(τ ) + τ 2 x (τ ))dτ. + ln t p (x(τ Γ(α + 3) a Γ(−α − 2)p! ln a p=0 a

When we consider finite sums up to order N , the error is given by   Z 1 t α+2 t ... |Etr (t)| = RN (τ )(x(τ ˙ ) + 3τ x ¨(τ ) + τ 2 x (τ ))dτ ln Γ(α + 3) a a with

Since τ ∈ [a, t], we have


p ∞ X Γ(p − α − 2) ln τa
. RN (τ ) = Γ(−α − 2)p! ln at p p=N +1

  2 ∞ ∞ X X α+2 e(α+2) +α+2 ≤ |RN (τ )| ≤ p pα+3 p=N +1 p=N +1 Z ∞ (α+2)2 +α+2 2 e e(α+2) +α+2 dp = . ≤ pα+3 (α + 2)N α+2 N 6

Therefore, |Etr (t)| ≤

1 Γ(α + 3)



ln

t a

α+2

where

2  e(α+2) +α+2  (t − a)L1 (t) + 3(t − a)2 L2 (t) + (t − a)3 L3 (t) , α+2 (α + 2)N

Li (t) = max |x(i) (τ )|, τ ∈[a,t]

i ∈ {1, 2, 3}.

Following similar arguments as done for n = 2, we can prove the general case with an expansion up to the derivative of order n. First, we introduce a notation. Given k ∈ N ∪ {0}, we define the sequences xk,0 (t) and xk,1 (t) recursively by the formulas x0,0 (t) = x(t) and xk+1,0 (t) = t and x0,1 (t) = x(t) ˙ and xk+1,1 (t) =

d xk,0 (t), for k ∈ N ∪ {0}, dt

d (txk,1 (t)), for k ∈ N ∪ {0}. dt

Theorem 3.2. Let n ∈ N, 0 < a < b and x : [a, b] → R be a function of class C n+1 . Then,     ∞ n X X t α+n−p t α+i α B(α, p) ln xi,0 (t) + Vp (t) Ai (α) ln a It x(t) = a a p=n+1 i=0

with Ai (α)

 1 1 + = Γ(α + i + 1)

∞ X

p=n−i+1

 Γ(p − α − n) , Γ(−α − i)(p − n + i)!

Γ(p − α − n) B(α, p) = , − α)(p
− n)! RΓ(α)Γ(1 p−n−1 x(τ ) t Vp (t) = a (p − n) ln τa τ dτ.

Moreover, if we consider the approximation α a It x(t)

≈

n X i=0



t Ai (α, N ) ln a

α+i

N X



t B(α, p) ln xi,0 (t) + a p=n+1

α+n−p

with N ≥ n + 1 and Ai (α, N ) =



1 1 + Γ(α + i + 1)

N X

p=n−i+1

then the error is bounded by the expression

Γ(p − α − n) , Γ(−α − i)(p − n + i)!

2

e(α+n) +α+n |Etr (t)| ≤ Ln (t) Γ(α + n + 1)(α + n)N α+n where Ln (t) = max |xn,1 (τ )|. τ ∈[a,t]

7



  t α+n ln (t − a), a

Vp (t)

Proof. Applying integration by parts repeatedly and the binomial formula, we arrive to α a It x(t)

n X

  1 t α+i = xi,0 (a) ln Γ(α + i + 1) a i=0   Z t ∞ Γ(p − α − n) t α+n X 1 τ p
xn,1 (τ )dτ. ln + ln p Γ(α + n + 1) a a Γ(−α − n)p! ln at a p=0

To achieve the expansion formula, we repeat the same procedure as for the case n = 2: we split the sum into two parts (the first term plus the remainings) and integrate by parts the second one. The convergence of the series Ai (α) is ensured by the relation ∞ X

p=n−i+1

Γ(p − α − n) = 1 F0 (−α − i, 1) − 1. Γ(−α − i)(p − n + i)!

The error on the approximation is given by  α+n Z t 1 t |Etr (t)| = RN (τ )xn,1 (τ )dτ ln Γ(α + n + 1) a a with


p ∞ X Γ(p − α − n) ln τa
. RN (τ ) = Γ(−α − n)p! ln at p p=N +1

Also, for τ ∈ [a, t],

|RN (τ )| ≤

  ∞ (α+n)2 +α+n X α+n ≤ e (α + n)N α+n . p

p=N +1

We remark that the error formula tends to zero as N increases. Similarly to what was done with the left fractional integral, we can also expand the right Hadamard fractional integral. Theorem 3.3. Let n ∈ N, 0 < a < b and x : [a, b] → R be a function of class C n+1 . Then, α t Ib x(t)

=

n X i=0



b Ai (α) ln t

α+i

with Ai (α)

=

(−1)i Γ(α + i + 1)

xi,0 (t) +



1 +

∞ X

p=n+1

∞ X

p=n−i+1

b B(α, p) ln t

α+n−p 

Wp (t)

Γ(p − α − n) , Γ(−α − i)(p − n + i)!

Γ(p − α − n) , B(α, p) = Γ(α)Γ(1 −α)(p − n)! p−n−1 Z b b x(τ ) (p − n) ln Wp (t) = dτ. τ τ t

8



Remark 3.4. Analogously to what was done for the left fractional integral, one can consider an approximation for the right fractional integral by considering finite sums in the expansion obtained in Theorem 3.3. In this case, the error is bounded by   2 e(α+n) +α+n b α+n |Etr (t)| ≤ Ln (t) (b − t), ln Γ(α + n + 1)(α + n)N α+n t where Ln (t) = max |xn,1 (τ )|. τ ∈[t,b]

4

An expansion formula for the Hadamard fractional derivative

Starting with formulas (1) and (2), and applying similar techniques as presented in Section 3, we are able to present expansion formulas, and respective approximation formulas with an error estimation, for the left and right Hadamard fractional derivatives. Due to restrictions on the number of pages, we omit the details here and just exhibit the results. Given n ∈ N and x ∈ C n+1 [a, b], we have     n X 1 t i−α t −α α Ai (α) ln x(t) + xi,0 (t) ln a Dt x(t) = Γ(1 − α) a a i=1 " #     ∞ X t n−α−p Γ(p + α − n) t −α B(α, p) ln + Vp (t) + x(t) ln a Γ(1 − α)Γ(α)(p − n)! a p=n+1

with Ai (α)

=



1 1 + Γ(i + 1 − α)

∞ X

p=n−i+1



Γ(p + α − n) , Γ(α − i)(p − n + i)!

i ∈ {1, . . . , n},

Γ(p + α − n) , p ∈ {n + 1, . . .}, + α)(p − n)!
RΓ(−α)Γ(1 p−n−1 x(τ ) t p ∈ {n + 1, . . .}. = a (p − n) ln τa τ dτ,

B(α, p) = Vp (t)

When we consider finite sums, α a Dt x(t)

≈

n X i=0



t Ai (α, N ) ln a

i−α

N X



n−α−p



i ∈ {0, . . . , n},

t xi,0 (t) + B(α, p) ln a p=n+1

Vp (t)

with Ai (α, N ) =



1 1 + Γ(i + 1 − α)

N X

p=n−i+1

and the error is bounded by

Γ(p + α − n) , Γ(α − i)(p − n + i)!

2

e(n−α) +n−α |Etr (t)| ≤ Ln (t) Γ(n + 1 − α)(n − α)N n−α 9

  t n−α (t − a), ln a

where Ln (t) = max |xn,1 (τ )|. τ ∈[a,t]

Remark 4.1. The series involved in the definition of Ai are convergent, for all i ∈ {1, . . . , n}. This is due to the fact that ∞ X

p=n−i+1

Γ(p + α − n) = 1 F0 (α − i, 1) − 1 Γ(α − i)(p − n + i)!

and 1 F0 (α − i, 1) converges ([1, Theorem 2.1.1]), since α − i < 0. Remark 4.2. For the right Hadamard fractional derivative, the expansion reads as α t Db x(t) =

    n X b i−α b −α 1 Ai (α) ln x(t) + xi,0 (t) ln Γ(1 − α) t t " #   i=1   ∞ X b n−α−p Γ(p + α − n) b −α B(α, p) ln + Wp (t) + x(t) ln t Γ(1 − α)Γ(α)(p − n)! t p=n+1

with Ai (α)

 (−1)i 1 + = Γ(i + 1 − α)

∞ X

p=n−i+1

5

i ∈ {1, . . . , n},

Γ(p + α − n) , p ∈ {n + 1, . . .}, Γ(−α)Γ(1  + α)(p− n)! Z b b p−n−1 x(τ ) dτ. = (p − n) ln τ τ t

B(α, p) = Wp (t)

 Γ(p + α − n) , Γ(α − i)(p − n + i)!

Examples

We obtained approximation formulas for the Hadamard fractional integrals and derivatives, and an upper bound for the error on such decompositions. In this section we study several cases, comparing the solution with the approximations. To gather more information on the accuracy, we evaluate the error using the distance s Z b 2 dt, ˜ dist = (X(t) − X(t)) a

˜ where X(t) is the exact formula and X(t) the approximation. To begin with, we consider α = 0.5 and functions x1 (t) = ln t and x2 (t) = 1 with t ∈ [1, 10]. Then, √ √ ln3 t ln t 0.5 0.5 and 1 It x2 (t) = 1 It x1 (t) = Γ(2.5) Γ(1.5)

10

(cf. [8, Property 2.24]). We consider the expansion formula for n = 2 as in (5) for both cases. We obtain then the approximations   N X p − 2 p 3 0.5 B(0.5, p) ln t 1 It x1 (t) ≈ A0 (0.5, N ) + A1 (0.5, N ) + p−1 p=3

and

0.5 1 It x2 (t)



≈ A0 (0.5, N ) +

N X p=3



B(0.5, p)

√ ln t.

The results are exemplified in Figures 1(a) and 1(b). As can be seen, the value N = 3 is enough in order to obtain a good accuracy in the sense of the error function. 1.8

3 Analytic N=3, E=4.1876e−016

Analytic N=3, E=4.1615e−016

1.6

2.5 1.4 2

1.2 1

1.5 0.8 1

0.6 0.4

0.5 0.2 0

1

2

3

4

5

6

7

8

9

0

10

t

1

2

3

4

5

6

7

8

9

10

t

(a) 1 It0.5 (ln t)

(b) 1 It0.5 (1)

Figure 1: Analytic vs. numerical approximation for n = 2. We now test the approximation on the power functions x3 (t) = t4 and x4 (t) = t9 , with t ∈ [1, 2]. Observe first that  Z t Z ln t tk 1 t −0.5 k−1 0.5 k τ dτ = ξ −0.5 e−ξk dξ ln 1 It (t ) = Γ(0.5) 1 τ Γ(0.5) 0 by the change of variables ξ = ln τt . In our cases, 0.5 4 1 It (t )

≈

√ √ 0.8862269255 4 0.5908179503 9 t erf(2 ln t) and 1 It0.5 (t9 ) ≈ t erf(3 ln t), Γ(0.5) Γ(0.5)

where erf(·) is the error function. In Figures 2(a) and 2(b) we show approximations for several values of N . We mention that, as N increases, the error decreases and thus we obtain a better approximation. Another way to obtain different expansion formulas is to vary n. To exemplify, we choose the previous test functions xi , for i = 1, 2, 3, 4, and consider the cases n = 2, 3, 4 with N = 5 fixed. 11

9

350 Analytic N=3, E=0.21341 N=4, E=0.067535 N=5, E=0.029254

8 7

Analytic N=3, E=45.4709 N=4, E=15.0158 N=5, E=6.7188

300

250 6 5

200

4

150

3 100 2 50

1 0

1

1.2

1.4

1.6

1.8

0

2

t

1

1.2

1.4

1.6

1.8

2

t

(a) 1 It0.5 (t4 )

(b) 1 It0.5 (t9 )

Figure 2: Analytic vs. numerical approximation for n = 2. The results are shown in Figures 3(a), 3(b), 3(c) and 3(d). Observe that as n increases, the error may increase. This can be easily explained by analysis of the error formula, and the values of the sequence x(k,0) involved. For example, for x4 we have x(k,0) (t) = 9k t9 , for k = 0 . . . , n. This suggests that, when we increase the value of n and the function grows fast, in order to obtain a better accuracy on the method, the value of N should also increase. We now proceed with some examples for the Hadamard fractional derivatives. The test functions are the same as before, and in Figures 4(a), 4(b), 4(c) and 4(d) we exemplify the results. In this case, √ ln t 0.5 , 1 Dt x1 (t) = Γ(1.5) 1 0.5 √ , 1 Dt x2 (t) = Γ(0.5) ln t √ 1 0.8862269255 4 0.5 √ 4t erf(2 ln t), + 1 Dt x3 (t) ≈ Γ(0.5) Γ(0.5) ln t √ 0.5908179503 1 0.5 √ + 9t9 erf(3 ln t). 1 Dt x4 (t) ≈ Γ(0.5) Γ(0.5) ln t One main advantage of this method is that we can replace fractional integrals and fractional derivatives as a sum of integer/classical derivatives, and by doing this we are rewriting the original problem, that falls in the theory of fractional calculus, into a new one where we can apply the already known techniques (analytical or numerical) and thus solving it. For example, when in presence of a fractional integral or a fractional derivative, with a number of initial conditions, replace the fractional operator by the appropriate approximation, with the value of n given by the number of initial conditions.

12

1.8

3 Analytic n=2, E=5.5435e−016 n=3, E=4.5003e−016 n=4, E=3.7701e−016

2.5

Analytic n=2, E=5.2015e−016 n=3, E=5.1896e−016 n=4, E=4.1615e−016

1.6 1.4

2

1.2 1

1.5 0.8 1

0.6 0.4

0.5 0.2 0

1

2

3

4

5

6

7

8

9

0

10

1

2

3

4

5

t

6

7

8

9

10

t

(a) 1 It0.5 (ln t)

(b) 1 It0.5 (1)

8

350 Analytic n=2, E=0.029254 n=3, E=0.02319 n=4, E=0.032662

7

Analytic n=2, E=6.7188 n=3, E=12.1029 n=4, E=38.7158

300

6

250

5 200 4 150 3 100

2

50

1 0

1

1.2

1.4

1.6

1.8

0

2

1

1.2

t

1.4

1.6

1.8

2

t

(c) 1 It0.5 (t4 )

(d) 1 It0.5 (t9 )

Figure 3: Analytic vs. numerical approximation for n = 2, 3, 4 and N = 5. For example, consider the problem  p   D 0.5 x(t) + x(t) = x(t) + ln t 1 t Γ(1.5)   x(1) = 0.

(6)

Obviously, x(t) = ln t is a solution for (6). Since we have only one initial condition, we replace the operator 1 Dt0.5 (·) by the expansion with n = 1 and thus obtaining  p N  X   x(t)  0.5−p −0.5 0.5   B(0.5, p)(ln t) Vp (t) = 1 + A0 (0.5, N )(ln t) x(t) + A1 (0.5, N )(ln t) tx(t) ˙ + + ln t,   Γ(1.5)   p=2  x(t) V˙p (t) = (p − 1)(ln t)p−2 , p = 2, 3, . . . , N,   t    x(1) = 0,     V (1) = 0, p = 2, 3, . . . , N. p

(7)

13

1.8

6 Analytic N=3, E=7.1715e−016

1.6

Analytic N=3, E=2.9216e−016 5

1.4 1.2

4

1 3 0.8 0.6

2

0.4 1 0.2 0

1

2

3

4

5

6

7

8

9

0

10

1

2

3

4

5

t

6

7

8

9

10

t

(a) 1 Dt0.5 (ln t)

(b) 1 Dt0.5 (1)

35

1600 Analytic N=3, E=0.72399 N=4, E=0.38 N=5, E=0.22964

30

Analytic N=3, E=153.6572 N=4, E=83.9273 N=5, E=52.2765

1400 1200

25

1000 20 800 15 600 10

400

5

0

200

1

1.2

1.4

1.6

1.8

0

2

t

1

1.2

1.4

1.6

1.8

2

t

(c) 1 Dt0.5 (t4 )

(d) 1 Dt0.5 (t9 )

Figure 4: Analytic vs. numerical approximation for n = 2. In Figure 5 we compare the analytical solution of the FDE (6) with the numerical result for N = 2 in (7).

Acknowledgments Work supported by FEDER funds through COMPETE — Operational Programme Factors of Competitiveness (“Programa Operacional Factores de Competitividade”) — and by Portuguese funds through the Center for Research and Development in Mathematics and Applications (University of Aveiro) and the Portuguese Foundation for Science and Technology (“FCT – Funda¸c˜ ao para a Ciˆencia e a Tecnologia”), within project PEst-C/MAT/UI4106/2011 with COMPETE number FCOMP-01-0124-FEDER-022690. Shakoor Pooseh was also supported by the Ph.D. fellowship SFRH/BD/33761/2009. The authors are grateful to three referees for their constructive and helpful comments and valuable suggestions. 14

0.7 Analytic Approximation: n=1, N=2, E=3.9231e−005 0.6

0.5

x(t)

0.4

0.3

0.2

0.1

0

1

1.2

1.4

1.6

1.8

2

t

Figure 5: Analytic vs. numerical approximation for the FDE (6) with one initial condition.

References [1] G. E. Andrews, R. Askey and R. Roy, Special functions, Encyclopedia of Mathematics and its Applications, 71, Cambridge Univ. Press, Cambridge, 1999. [2] T. M. Atanackovic and B. Stankovic, On a numerical scheme for solving differential equations of fractional order, Mech. Res. Comm. 35 (2008), no. 7, 429–438. [3] P. L. Butzer, A. A. Kilbas and J. J. Trujillo, Mellin transform analysis and integration by parts for Hadamard-type fractional integrals, J. Math. Anal. Appl. 270 (2002), no. 1, 1–15. [4] P. L. Butzer, A. A. Kilbas and J. J. Trujillo, Stirling functions of the second kind in the setting of difference and fractional calculus, Numer. Funct. Anal. Optim. 24 (2003), no. 7-8, 673–711. [5] J. Hadamard, Essai sur l’´etude des fonctions donn´ees par leur d´eveloppement de Taylor, Journ. de Math. 4 (1892), no. 8 , 101-186. [6] U. N. Katugampola, New approach to a generalized fractional integral, Appl. Math. Comput. 218 (2011), no. 3, 860–865. [7] A. A. Kilbas, Hadamard-type fractional calculus, J. Korean Math. Soc. 38 (2001), no. 6, 1191–1204. [8] A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Theory and applications of fractional differential equations, North-Holland Mathematics Studies, 204, Elsevier, Amsterdam, 2006. 15

[9] A. A. Kilbas and A. A. Titioura, Nonlinear differential equations with MarchaudHadamard-type fractional derivative in the weighted space of summable functions, Math. Model. Anal. 12 (2007), no. 3, 343–356. [10] K. S. Miller and B. Ross, An introduction to the fractional calculus and fractional differential equations, A Wiley-Interscience Publication, Wiley, New York, 1993. [11] I. Podlubny, Fractional differential equations, Mathematics in Science and Engineering, 198, Academic Press, San Diego, CA, 1999. [12] D. Qian, Z. Gong and C. Li, A generalized Gronwall inequality and its application to fractional differential equations with Hadamard derivatives, 3rd Conference on Nonlinear Science and Complexity (NSC10), Cankaya University, 1–4, Ankara, Turkey, 28-31 July, 2010. [13] S. G. Samko, A. A. Kilbas and O. I. Marichev, Fractional integrals and derivatives, translated from the 1987 Russian original, Gordon and Breach, Yverdon, 1993.

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