On the Approximation of Riemann-Liouville Integral by Fractional Nabla H-Sum and Applications L Khitri-Kazi-Tani, H Dib

On the approximation of Riemann-Liouville integral by fractional nabla h-sum and applications L Khitri-Kazi-Tani, H Dib

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L Khitri-Kazi-Tani, H Dib. On the approximation of Riemann-Liouville integral by fractional nabla h-sum and applications. 2016. hal-01315314

HAL Id: hal-01315314 https://hal.archives-ouvertes.fr/hal-01315314 Preprint submitted on 12 May 2016

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. On the approximation of Riemann-Liouville integral by fractional nabla h-sum and applications

L. Khitri-Kazi-Tani∗ H. Dib †

Abstract First of all, in this paper, we prove the convergence of the nabla h- sum to the Riemann-Liouville integral in the space of continuous functions and in some weighted spaces of continuous functions. The connection with time scales convergence is discussed. Secondly, the eﬃciency of this approximation is shown through some Cauchy fractional problems with singularity at the initial value. The fractional Brusselator system is solved as a practical case. Keywords. Riemann-Liouville integral, nabla h-sum, fractional derivatives, fractional diﬀerential equation, time scales convergence MSC (2010). 26A33, 34A08, 34K28, 34N05

1 Introduction

The numerous applications of fractional calculus in the mathematical modeling of physical, engineering, economic and chemical phenomena have contributed to the rapidly growing of this theory [12, 19, 21, 18], despite the discrepancy on deﬁnitions of the fractional operators leading to nonequivalent results [6]. Each deﬁnition has its own advantages for applications. Tacking into account some developments of fractional calculus theory, we can refer to [22, 17, 14, 20] and the references therein.

On the other side, the discrete fractional calculus is considered as new attempt. The fractional operators are obtained using the forward operators delta [16, 8] or backward operators nabla [11, 2] or combined delta and nabla fractional operators [3].

Besides that, since Hilger’s original concept of measure chain which provides a unifying framework to investigate both continuous and discrete time systems [13], many deﬁnitions of fractional integral/derivative in time scale arise to ex- tend and unify results on the real line. Invoke that any time scale is a closed non-empty subset of R (for more details see [1, 24, 4, 5]).

∗corresponding author: l [email protected] †h [email protected]

1 In this context, the h-fractional operators, when the time scale is chosen to be the subset (hZ) of real numbers, was our source of inspiration to consider an approximation of Riemann-Liouville fractional integral. Especially, the fractional summation obtained by the backward differences. We prove the convergence of this h-sum to the Riemann-Liouville integral firstly in the space of continuous functions, and then in more general weighted spaces. Moreover, we propose a numerical solution to the weighted fractional differential Cauchy problem.

The paper is organized as the following. Section 2 deals with some preliminary notions and results in the purpose of making this paper self-contained. Sec- tion 3 is about the convergence results. Section 4 treats the connection with convergence over time scales. Section 5 is extremely devoted to the approximation application to some fractional diﬀerential equations. First we show the accuracy of this approach for the Riemann-Liouville initial value problem in weighted spaces of continuous functions. Then the numerical solution for the non-linear fractional Brusselator system is presented as a practical case. Some new research directions are suggested in section 6.

2 Preliminaries

In this section we present some deﬁnitions and results which will be used in the sequel. For β > 0 the weighted space of continuous function is deﬁned by β Cβ [a, b] := f ∈ C (a, b] : lim (t − a) f(t) exists . t→a+ endowed with the norm

β kfkβ := max (t − a) |f(t)| t∈[a,b]

Cβ[a, b] is a Banach space. Remark that C0 [a, b] = C [a, b], the Banach space of all real continuous functions. We recall the expression of the norm of a continuous linear operator T from Cβ1 [a, b] to Cβ2 [a, b]: kT (f)k kT k = sup β2 β1,β2 kfk f6=0 β1

The following chain of inclusions holds C [a, b] ⊂ Cβ1 [a, b] ⊂ Cβ2 [a, b] if 0 < β1 < β2. Definition 1 (Riemann-liouville operators) Let f be a continuous function on [a, b], 0 < α < 1 and a ≤ x ≤ b. The fractional integral of order α is defined by 1 Z x Iαf(x) = (x − t)α−1f(t)dt Γ(α) a The Riemann-Liouville fractional derivative of order α is defined by 1 d Z x Dαf(x) = (x − t)α−1f(t)dt Γ(α) dx a

2 Deﬁnition 2 (Caputo fractional derivative) Let f ∈ C1 [a, b], 0 < α < 1 and a ≤ x ≤ b. The Caputo fractional derivative of order α is deﬁned by

1 Z x C Dαf(x) = (x − t)α−1f 0(t)dt Γ(α) a In the sequel, we will take [a, b] = [0, 1]. Some useful properties of the Riemann- Liouville operator are listed bellow.

Proposition 2.1 Let β ∈ ]0, 1[ , α − β ≥ 0 and f ∈ Cβ[0, 1] then

1. Iαf is continuous on [0, 1] . Γ(1 − β) 2. kIαk ≤ . β,0 Γ(α + 1 − β) Proof 1. Let 0 ≤ x < y ≤ 1 and ε > 0

1 Z 1 Iαf(y) − Iαf(x) = (1 − t)α−1 (yαf(yt) − xαf(xt)) dt Γ(α) 0 1 Z 1 = (1 − t)α−1 yα−β − xα−β)yβf(yt) + xα−β(yβf(yt) − xβf(xt) dt Γ(α) 0 Therefore,

Z 1 α α 1 α−1 −β α−β α−β β β |I f(y) − I f(x)| ≤ (1 − t) t (y − x )y t f(yt) dt Γ(α) 0 α−β Z 1 x α−1 −β β β β β + (1 − t) t (y t f(yt) − x t f(xt) dt Γ(α) 0

As f ∈ Cβ [0, 1] , there exists δ1 > 0, such that

α α |x − y| < δ1 ⇒ |tx − ty| < δ1, ∀t ∈ [0, 1] ⇒ |(yt) f(yt) − (xt) f(xt)| < ε

α−β The continuity of the function x , leads to the existence of δ2 > 0, such that α−β α−β |x − y| < δ2 ⇒ y − x < ε.

Thus, |x − y| < min(δ1, δ2) implies 1 ε kfkβ + 1 Z ε kfkβ + 1 Γ(1 − β) |Iαf(y) − Iαf(x)| ≤ (1−t)α−1t−βdt ≤ Γ(α) 0 Γ(1 + α − β)

2. ∀f ∈ Cβ[0, 1], ∀x ∈ [0, 1] 1 Z x |Iαf(x)| ≤ (x − t)α−1t−βtβ |f(t)| dt Γ(α) 0 Z x 1 α−1 −β Γ(1 − β) ≤ kfkβ (x − t) t dt ≤ kfkβ Γ(α) 0 Γ(α + 1 − β) and hence the result.

3 In order to use de nabla h-sum of order α, we start by deﬁning a uniform discretization of [0, 1] . Let 0 = t0 < t1 < ··· < tN = 1 be a mesh with step size h > 0. First we recall the deﬁnition of the fractional h-sum of order α at the point tn = nh by (see [15, formula (7)] )

n hα X Γ(n − i + α) ∇−αf(t ) = f(ih) h n Γ(α) Γ(n − i + 1) i=1 −α with, by convention, ∇h f(t0) = 0. This definition suggest to introduce the fractional h-sum operator by Definition 3 (Nabla h-sum operator) Let f be a continuous function, 0 ≤ x ≤ 1 and 1 ≤ n ≤ N, such that tn−1 < x ≤ tn. The fractional h-sum of order α is given by: 1 ∇−αf(x) = (t − x) ∇−αf(t ) + (x − t )∇−αf(t ) h h n h n−1 n−1 h n −α Note that ∇h f(.) is nothing but the broken continuous line joining the −α −α points (tn, ∇h f(tn)). Hence ∇h f(.) is a continuous function. The first prop- −α erty of the linear operator ∇h is given by the following −α Proposition 2.2 ∇h : C [0, 1] → C [0, 1] is a continuous linear operator with −α 1 ∇ ≤ . h 0,0 Γ(α + 1) For the proof the following lemma is needed Lemma 2.3 (See [10, Theorem 5] ) Let n ≥ 1 be an integer, and a, b > 0 be two real numbers. then n XΓ(k + a) 1 Γ(n + 1 + a) Γ(1 + a) = − . Γ(k + b) a − b + 1 Γ(n + b) Γ(b) k=1

Proof (of the prop 2.2) Let f ∈ C [0, 1] and tn = nh, n ∈ {0, 1,...,N} , then

α n α n−1 −α h XΓ(n − i + α) h X Γ(i + α) ∇ f(tn) ≤ |f(ih)| ≤ kfk (1) h Γ(α) Γ(n − i + 1) Γ(α) 0 Γ(i + 1) i=1 i=0 From lemma 2.3, we get

n−1 X Γ(i + α) 1 Γ(n + α) = Γ(i + 1) α Γ(n) i=0 Inequality (1) can be written as

α −α h Γ(n + α) ∇ f(tn) ≤ kfk h Γ(α + 1) 0 Γ(n) from Wendel’s inequality (see [23]) Γ(x + β) ≤ xβ, 0 < β < 1, x > 0 Γ(x)

4 we get

α −α h α kfk0 ∇ f(tn) ≤ kfk n ≤ ∀n ∈ , 0 ≤ n ≤ N h Γ(α + 1) 0 Γ(α + 1) N

Furthermore, ∀x ∈ [0, 1]

−α 1 −α −α kfk0 ∇ f(x) = (tn − x)∇ f(tn−1) + (x − tn−1)∇ f(tn) ≤ h h h h Γ(α + 1)

−α Thus, ∇h f is a bounded operator on [0, 1] . The following technical lemmas play a key role in the proofs of ours main results. Lemma 2.4 There exists a function Φ, such that Γ(m + α) α ∀m ≥ 1 = mα−1 + Φ(m), with |Φ(m)| Γ(m + 1) Γ(1 − α) Γ(2 − α) ≤ mα−2. 2

Proof From the deﬁnition of the beta function we have

Γ(m + α) 1 Z 1 = t(m+α−1)(1 − t)−αdt (2) Γ(m + 1) Γ(1 − α) 0 Let t = e−u then the equality (2) becomes

Z +∞ Z +∞ Γ(m + α) 1 −mu u −α 1 −mu −α u α = e (e −1) du = e u ( u ) du Γ(m + 1) Γ(1 − α) 0 Γ(1 − α) 0 e − 1 Using the generating function of the Bernoulli numbers

+∞ u X uk G(u) = = B = 1 + ϕ(u) > 0 eu − 1 k k! k=0 where ϕ : [0, +∞[ → ]−1, 0] is a continuous function. We have

Γ(m + α) 1 Z +∞ = e−muu−α(1 + ϕ(u))αdu Γ(m + 1) Γ(1 − α) 0 Now, Taylor’s formula with integral remainder applied to the function (1 + ϕ(u))α gives

Z 1 (1 + ϕ(u))α = 1 + αϕ(u) (1 + ξϕ(u))α−1dξ 0 Therefore

Γ(m + α) 1 Z +∞ α Z +∞ Z 1 = e−muu−αdu+ e−muu−αϕ(u) (1+ξϕ(u))α−1dξdu Γ(m + 1) Γ(1 − α) 0 Γ(1 − α) 0 0

5 and then Γ(m + α) α Z +∞ Z 1 = mα−1 + e−muu−αϕ(u) (1+ξϕ(u))α−1dξdu (3) Γ(m + 1) Γ(1 − α) 0 0 Put Z +∞ Z 1 Φ(m) = e−muu−αϕ(u) (1 + ξϕ(u))α−1dξdu 0 0 From the identity 1+ξϕ(u) = 1+ξ −ξ(1−ϕ(u)) and the fact that 1−ϕ(u) > 0 for every u ≥ 0, we have

1 + ξϕ(u) ≥ 1 + ξ and (1 + ξϕ(u))α−1 ≤ (1 + ξ)α−1

Consequently

Z 1 Z 1 (1 + ξϕ(u))α−1dξ ≤ (1 + ξ)α−1dξ ≤ 1 0 0 Hence Z +∞ Z 1 Z +∞ −mu −α α−1 −mu −α e u ϕ(u) (1 + ξϕ(u)) dξdu ≤ e u |ϕ(u)| du (4) 0 0 0 ϕ(u) ϕ(u) 1 The function is strictly increasing on [0, +∞], lim = − and u u→0+ u 2 ϕ(u) lim = 0 . u→∞ u

ϕ(u) 1 So, ≤ , and the inequality (4) becomes u 2

1 Z +∞ |Φ(m)| ≤ e−muu1−αdu 2 0 and Γ(2 − α)mα−2 |Φ(m)| ≤ (5) 2 Lemma 2.5 Let 0 < α < 1, we have

1. For every m ≥ 1

1 mα−1 < ((m)α − (m − 1)α) < (m − 1)α−1 α ∞ 1 2. The series P (kα − (k − 1)α) − kα−1 is convergent. k=0 α

Proof 1. Let g(t) = tα−1 deﬁned on ]0, +∞[, g is a positive, convex, de- creasing and locally integrable function on ]0, +∞[ . Then

mα−1 < tα−1 < (m − 1)α−1

which leads, after integration from m − 1 to m, to the desired inequality.

6 2. Let n X 1 s = (kα − (k − 1)α) − kα−1 n α k=1 In view of 1 1 s − s = ((n + 1)α − (n)α) − (n + 1)α−1 > 0 n+1 n α

Then the sequence (sn)n∈N is increasing and n n X 1 1 X Z k 1 s = (kα−(k−1)α)−kα−1+ −1 = τ α−1 − kα−1 dτ+ −1 n α α α k=2 k=2 k−1 Moreover, k − 1 ≤ τ ≤ k implies kα−1 ≤ τ α−1 ≤ (k − 1)α−1 and 0 ≤ τ α−1 − kα−1 ≤ (k − 1)α−1 − kα−1 thus n X 1 1 1 s ≤ (k − 1)α−1 − kα−1 + − 1 = − nα−1 < n α α α k=2

Finally, the sequence (sn)n∈N is convergent.

3 Main results

−α α Now, our goal is to prove that ∇h f(.) converges uniformly to I f(.). In order to do so, we prove the following propositions. Proposition 3.1 Let f ∈ C([0, 1]) and

α n−1 h X α Sαf(t ) = [(n − i)α − (n − i − 1) ] f(t ) h n Γ(α + 1) i i=0

Then, there exists C1 > 0, such that α −α α Sh f(tn) − ∇h f(tn) ≤ C1h for all tn

Proof hα Sαf(t ) − ∇−αf(t ) = (nα − (n − 1)α)f(0) h n h n Γ(α + 1) n−1 hα X 1 Γ(n − i + α) + ((n − i)α − (n − i − 1)α) − f(ih) − hαf(t ) Γ(α) α Γ(n − i + 1) n i=1 From equality (3) we deduce hα Sαf(t ) − ∇−αf(t ) = (nα − (n − 1)α)f(0) − hαf(t )+ h n h n Γ(α + 1) n n−1 hα X 1 ((n − i)α − (n − i − 1)α) − (n − i)α−1 f(ih)+ Γ(α) α i=1 n−1 αhα X + [Φ(n − i)] f(ih) Γ(α)Γ(1 − α) i=1

7 Denote

n−1 X 1 S = ((n − i)α − (n − i − 1)α) − (n − i)α−1 f(ih) 1 α i=1 and n−1 X S2 = [Φ(n − i)] f(ih) i=1 From the lemma 2.5 we can assert that

|S1| ≤ kfk0 σ0

∞ P 1 α α α−1 where σ0 = (k − (k − 1) ) − k < +∞. k=0 α For S2 the inequality (5) gives

n−1 kfk Γ(2 − α) X kfk Γ(2 − α) |S | ≤ 0 (n − i)α−2 ≤ 0 σ 2 2 2 1 i=1

∞ P α−2 with σ1 = i < +∞. Consequently i=1

α α α −α h α α α h S f(tn) − ∇ f(tn) ≤ (n − (n − 1) ) kfk + h kfk + kfk σ0 h h Γ(α + 1) 0 0 Γ(α) 0 αhα + kfk Γ(2 − α)σ 2Γ(α)Γ(1 − α) 0 1 and then α −α 1 1 α α S f(tn) − ∇ f(tn) ≤ + 1 + σ0 + Γ(2 − α)σ1 kfk h h h Γ(α + 1) Γ(α) 2Γ(α)Γ(1 − α) 0 α ≤ C1h

2 Proposition 3.2 Assume that f ∈ C [0, 1] , then there exists C2 > 0 such that

α α |I f(tn) − Sh f(tn)| ≤ C2h for all tn

Proof An integration by parts gives

1 1 Z x Iαf(x) = xαf(0) + (x − t)αf 0(t)dt Γ(α + 1) Γ(α + 1) 0

α Let Rh be the associated Riemann sum

n 1 1 X Rαf(t ) = tαf(0) + h (t − t )αf 0(t ) h n Γ(α + 1) n Γ(α + 1) n i i i=1

8 We have " n # 1 Z tn X Iαf(t ) − Rαf(t ) = (t − t)αf 0(t)dt − h (t − t )αf 0(t ) n h n Γ(α + 1) n n i i 0 i=1 n 1 X Z ti = [(t − t)αf 0(t) − (t − t )αf 0(t )] dt Γ(α + 1) n n i i i=1 ti−1 n 1 X Z ti Z ti = ((t − t)α − (t − t )α) f 0(t)dt + (t − t )α (f 0(t) − f 0(t )) dt Γ(α + 1) n n i n i i i=1 ti−1 ti−1 then

α α |I f(tn) − Rh (tn)| ( n n ) 1 X Z ti X Z ti ≤ ((t − t)α − (t − t )α) |f 0(t)| dt + (t − t )α (f 0(t) − f 0(t )) dt Γ(α + 1) n n i n i i i=1 ti−1 i=1 ti−1 ( n n ) 1 X Z ti X Z ti ≤ kf 0k ((t − t)α − (t − t )α) dt + (t − t )α f 0(t) − f 0(t )dt Γ(α + 1) 0 n n i n i i i=1 ti−1 i=1 ti−1 the Taylor formula with integral remainder allows us to write

α α |I f(tn) − Rh (tn)| ( n n ) 1 X Z ti X Z ti Z ti ≤ kf 0k ((t − t )α − (t − t )α) dt + (t − t )α f 00(θ)dθ dt Γ(α + 1) 0 n i−1 n i n i i=1 ti−1 i=1 ti−1 t According to Fubini’s theorem we have

( n ! ) 1 X Z ti Z θ |Iαf(t ) − Rα(t )| ≤ kf 0k h(t − t )α + f 00(θ) dt dθ n h n Γ(α + 1) 0 n 0 i=1 ti−1 ti−1 ( n ) 1 X ≤ kf 0k h + h |f 00(θ )(θ − t )| , for some θ ∈ ]t , t [ Γ(α + 1) 0 i i i−1 i i−1 i i=1 1 ≤ (kf 0k + kf 00k ) h Γ(α + 1) 0 0 Finally α α |I f(tn) − Rh f(tn)| ≤ C3h (6) 1 with C = (kf 0k + kf 00k ) 3 Γ(α + 1) 0 0 On the other hand n 1 X Rαf(t ) − Sαf(t ) = (t − t )α [hf 0(t ) − f(t ) + f(t )] h n h n Γ(α + 1) n i i i i−1 i=1 From Taylor-Lagrange formula we have

n h2 X Rαf(t ) − Sαf(t ) = (t − t )αf 00(θ ), t < θ < t h n h n 2Γ(α + 1) n i i i−1 i i i=1

9 In addition, the intermediate value theorem ensures the existence of θ ∈ [θ1, θn] such that n h2 X Rαf(t ) − Sαf(t ) = f 00(θ) (t − t )α h n h n 2Γ(α + 1) n i i=1 That we can write

n n−1 h2+α X h2+α X Rαf(t ) − Sαf(t ) = f 00(θ) (n − i)α = f 00(θ) iα h n h n 2Γ(α + 1) 2Γ(α + 1) i=1 i=0 We have then the inequality

h2+α h |Rαf(t ) − Sαf(t )| ≤ |f 00(θ)| nα+1 ≤ |f 00(θ)| h n h n 2Γ(α + 1) 2Γ(α + 1) then kf 00k |Rαf(t ) − Sαf(t )| ≤ 0 h h n h n 2Γ(α + 1) From

Finally, an essential result is the following corollary Corollary 3.3 Under the assumption that f ∈ C2 [0, 1], there exists C > 0 such that α −α α I f(tn) − ∇h f(tn) ≤ Ch for all tn

Proof As

α −α α α α −α I f(tn) − ∇h f(tn) ≤ |I f(tn) − Sh f(tn)| + Sh f(tn) − ∇h f(tn) The conclusion follows from Propositions 3.1 and 3.2.

Now, we are able to state and prove the following convergence result.

2 −α α Corollary 3.4 For every function f ∈ C [0, 1] we have ∇h f − I f 0 tends to 0 when h tends to 0.

Proof let tn−1 < x ≤ tn

−α α 1 −α −α α ∇ f(x) − I f(x) = (tn − x)∇ f(tn−1) + (x − tn−1)∇ f(tn) − I f(x) h h h h

1 −α α 1 −α α ≤ (tn − x)∇ f(tn−1) − (tn − x)I f(tn−1) + (x − tn−1)∇ f(tn) − (x − tn−1)I f(tn) h h h h

1 α α α + [(tn − x)I f(tn−1) + (x − tn−1)I f(tn)] − I f(x) h

10 First, estimate the interpolation error of Iαf(x) expressed with the third term of the previous inequality 1 [(t − x)Iαf(t ) + (x − t )Iαf(t )] − Iαf(x) = h n n−1 n−1 n α Z 1 α−1 α Z 1 α−1 (tn − x)tn−1 (1 − s) (x − tn−1)tn (1 − s) = f(tn−1s)ds + f(tns)ds h 0 Γ(α) h 0 Γ(α) Z 1 (1 − s)α−1 − xα f(xs)ds 0 Γ(α) Z 1 α−1 α (1 − s) 1 1 = x (tn − x)f(tn−1s) + (x − tn−1)f(tns) − f(xs) ds 0 Γ(α) h h Z 1 α−1 Z 1 α−1 1 α α (1 − s) 1 α α (1 − s) + (tn − x) tn−1 − x f(tn−1s)ds + (x − tn−1)(tn − x ) f(tns)ds h 0 Γ(α) h 0 Γ(α) From the error involved in the interpolation of the function f, there exists ξ between tn and tn−1 such that

1 1 f 00(ξs) (t − x)f(t s) + (x − t )f(t s) − f(xs) = s2(t − x)(x − t ) h n n−1 h n−1 n 2 n n−1 Therefore

Z 1 α−1 00 (1 − s) 1 1 kf k0 2 (tn − x)f(tn−1s) + (x − tn−1)f(tns) − f(xs) ds ≤ h 0 Γ(α) h h Γ(α + 3) Otherwise Z 1 α−1 Z 1 α−1 1 α α (1 − s) 1 α α (1 − s) (tn − x) tn−1 − x f(tn−1s)ds + (x − tn−1)(tn − x ) f(tns)ds h 0 Γ(α) h 0 Γ(α) kfk tα − tα ≤ 0 ( n n−1 ) Γ(α) α Then from lemma 2.5, we have

Z 1 α−1 Z 1 α−1 1 α α (1 − s) 1 α α (1 − s) (tn − x) tn−1 − x f(tn−1s)ds + (x − tn−1)(tn − x ) f(tns)ds h 0 Γ(α) h 0 Γ(α) kfk ≤ 0 (n − 1)α−1 hα Γ(α)

Consequently for all x between tn−1 and tn with n > 1 the following inequality holds

00 1 α α α kf k0 kfk0 α [(tn − x)I f(tn−1) + (tn−1 − x)I f(tn)] − I f(x) ≤ + h h Γ(α + 3) Γ(α) As a result, and after using the corollary 3.3, for all x ≥ h

00 −α α kf k0 kfk0 α ∇ f(x) − I f(x) ≤ 2C + + h h Γ(α + 3) Γ(α)

11 Now, for x between 0 and h

Z x α−1 −α α x −α α α−1 (x − t) ∇h f(x) − I f(x) = ∇h f(h) − I f(x) = xh f(h) − f(t)dt h 0 Γ(α) Hence

Z x α−1 −α α α−1 (x − t) ∇h f(x) − I f(x) ≤ xh |f(h)| + |f(t)| dt 0 Γ(α) xα 1 ≤ hα kfk + kfk ≤ 1 + kfk hα 0 Γ(α + 1) 0 Γ(α + 1) 0

This proves the corollary.

Our ﬁrst main result is

−α α Theorem 3.5 (Uniform convergence of ∇h f to I f) −α α For every f ∈ C [0, 1] , we have ∇h f − I f 0 tends to 0 when h goes to 0.

Proof For every ε > 0, Weiestrass approximation theorem, ensures the existence of a polynomial p such that, |f(x) − p(x)| ≤ ε, ∀x ∈ [0, 1] . As the previous −α α corollary applies to p, the boundedness of operators ∇h and I implies

−α α α −α α −α ∇h f(x) − I f(x) ≤ |I (f − p)(x)| + ∇h p(x) − I p(x) + ∇h (f − p)(x) ε ≤ 2 + Chα Γ(α + 1)

By taking ε = hα, we get

−α α α ∇h f(x) − I f(x) ≤ Kh hence the conclusion.

Now, we enunciate the convergence result in weighted spaces.

Theorem 3.6 (Convergence in weighted spaces) For every function f in Cβ [0, 1] −α α with 0 < β < 1 and α −β ≥ 0 we have ∇h f − I f β tends to 0 when h tends to 0.

Proof Since tβf(t) is continuous on [0, 1] , then |f(t)| ≤ Mt−β, β where M = sup t f(t) = kfkβ . t∈[0,1] f(t) h ≤ t ≤ 1 Let F (t) := . It is a continuous function on [0, 1] . f(h) 0 ≤ t ≤ h It is obvious that

−α −α ∇h F (x) = ∇h f(x), for every x ≥ h If x < h, x ∇−αf(x) = ∇−αf(h) h h h

12 and x x ∇−αF (x) = ∇−αF (h) = ∇−αf(h) = ∇−αf(x) h h h h h h Then, ∀x ∈ [0, 1] we have

|f(t) − f(h)| ≤ Mt−β + Mh−β ≤ 2Mt−β, then when x ≥ h

1 Z h 2M Z h α α α−1 α−1 −β |I F (x) − I f(x)| = (x − t) (f(h) − f(t))dt ≤ (x−t) t dt Γ(α) 0 Γ(α) 0 It is easy to see that

h Z h Z x Z 1 α−1 −β α−β α−1 −β α−β α−1 −β (x−t) t dt = x (1−t) t dt = x (1−t) t χ[0, h ](t)dt 0 0 0 x 1 1 1 1 Let 1 < p < min( , ) and q the conjugate exponent i.e, + = 1. The 1 − α β p q H¨olderinequality gives

1 1 Z 1 h q α−1 −β p (1 − t) t χ[0, h ](t)dt ≤ (B(1 − (1 − α)p, 1 − βp)) 0 x x where B(., .) is the Euler beta function. Consequently, ∀x ≥ h

β α α α− 1 1 1 x |I F (x) − I f(x)| ≤ K1x q h q ≤ K1h q .

Now if x < h

1 Z h α α α−1 |I F (x) − I f(x)| = (x − t) (f(t) − f(h))dt Γ(α) 0 2M Z x ≤ (x − t)α−1t−βdt Γ(α) 0 2MΓ(1 − β) ≤ xα−β Γ(1 − β + α) and then,

2MΓ(1 − β) 2MΓ(1 − β) xβ |IαF (x) − Iαf(x)| ≤ xα ≤ hα Γ(1 − β + α) Γ(1 − β + α) 1 Because α > , we get q

α α α 1 x |I F (x) − I f(x)| ≤ K2h q , ∀x ∈ [0, 1]

13 Since F is continuous on [0, 1], from the theorem 3.5 we can state that, there exists K > 0 such that,

−α α α ∇h F (x) − I F (x) ≤ Kh , ∀x ∈ [0, 1] (7) leading to β −α α α x ∇h F (x) − I F (x) ≤ Kh , ∀x ∈ [0, 1] Therefore

1 1 β −α α α q q x ∇h f(x) − I f(x) ≤ Kh + K2h ≤ K3h , ∀x ∈ [0, 1] hence −α α lim ∇h f − I f = 0. h→0 β 4 Connection with time scale theory

∗ Let T = [0, 1] and Th = hZ ∩ [0, 1] be time scales for each h ∈ R+. We consider Hd(A, B) the Hausdorﬀ distance between two sets deﬁned by Hd(A, B) = max sup d(a, B), sup d(b, A) a∈A b∈B where, as usual, d(a, B) = inf |a − b|. b∈B Remark 1 It is easy to see that

Hd(Th, T) = h/2 and n ˜ o Hd(Th, Th˜ ) = max h/2, h/2

We can reformulate the result in theorem 3.5 in term of convergence over time scales. Corollary 4.1 (Convergence on time scales) Let f be any continuous function on [0, 1], then there exists a constant K˜ > 0, such that

−α α ˜ α ∇h f − I f 0 ≤ K.Hd(Th, T) (8)

It is interesting to see that the inequality (8) can be seen as continuity of an operator, mapping the space of time scales into the space of continuous function. Indeed, let (τ, Hd) be the metric space of all time scales on [0, 1] endowed with the Hausdorff metric. Let f be a fixed continuous function. Assume that we can define a mapping from (τ, Hd) to (C [0, 1] , k.k0):

Jf :(τ, Hd) → (C [0, 1] , k.k0) S7→Jf (S)

14 α which can be called integral with respect to S and such that Jf (T) =I f −α and Jf (Th) = ∇h f. The corollary 4.1 says that

lim Jf ( h) =Jf ( ) τ T T Th→T So will conjecture the following,

Conjecture 4.2 Assume that Jf is well deﬁned. Then Jf is continuous.

Note that in the subset {Th, h > 0} ∪ {T}, the time scales Th are isolated ”points”. We have then shown that the conjecture is true in the previous subset.

5 Applications

In this section, we use our proposed approximation and show how it is numeri- cally efficient through some problems of fractional differential equations. The proposed procedure for numerical solution of Cauchy fractional differential equations consists of two steps. First, initial conditions are used to reduce a given initial-value problem to an equivalent Volterra integral equation. Then a system of algebraic equations is obtained by replacing fractional integral by the nabla fractional h-sum. In case of linear fractional differential equations the resulting algebraic system is a lower triangular. For the non linear ones, the corresponding algebraic systems is non linear too, and must be solved by adequate methods. In the presence of singularity at the origin for the Riemann-Liouville initial value problem, it is important to note that this procedure is appropriate. Example 1 Consider the linear fractional initial value problem

( Dαy(t) = λy(t) t ∈ (0, 1] lim t1−αy(t) = c (9) t→0+ This problem has a known analytical solution [14]:

α−1 α y(t) = cΓ(α)t Eα,α(λt ) where Eµ,ν (t) is a Mittag-Leﬄer type function [14]. The problem (9) is reduced to the following Volterra integral equation

y(t) = ctα−1 + Iα(λy(t))

In Figure 1, we show accuracy of the method with the next parameters, α = 0.75, c = 1, λ = 1. 1−α 4 −2 For h = 0.002, we obtain max ti |y(ti) − yi| , i = 1,..., 10 = 2.9 × 10 , where yi is the approximation value of y(ti) at the point ti = ih, derived from the recurrence formula

 i−1  1 XΓ(i − j + α) y = λhα y + ctα−1 , i = 1,..., 104 i Γ(α)(1 − λhα)  Γ(i − j + 1) j i  j=1

15 (a) Exact solution in the continuous line, numerical one in dotted line. (b) Error with respect step size h.

Figure 1: Comparison between analytical solution and numerical solution for the linear fractional initial value problem.

Example 2 Consider the following nonhomogeneous diﬀerential equation with Riemann-Liouville fractional derivative (see [14] (4.2.101) with our notation),

Dαy(t) = λy(t) + ctµ, t ∈ (0,T ] (10) D1−αy(0+) = η where λ, c, η ∈ R, 0 < α < 1, µ > −1.The solution of the problem is given by α−1 α α+µ α y(t) = T ηt Eα,α(λt ) + Γ(µ + 1)ct Eα,α+µ+1(λt ) Since lim t1−αy(t)exists then by [14] lemma 3.2 the problem (10) can be written t→0+ as ( Dαy(t) = λy(t) + ctµ, t ∈ (0,T ] η lim t1−αy(t) = t→0+ Γ(α) which is reduced to the following Volterra integral equation η y(t) = tα−1 + Iα(λy(t) + ctµ) Γ(α) In ﬁgure 2, we show accuracy of the method with the next parameters, α = 0.5, η = 1, c = 2, mu = 2, λ = 1,T = 1.

Example 3 Consider the fractional-order Brusselator system [25],[18]. It’s an example of autocatalytic chemical reaction which is mathematically described as follows: C Dαx(t) = a − (µ + 1)x(t) + x2(t)y(t) C Dαy(t) = µx(t) − x2(t)y(t) with initial conditions x(0) = x0, y(0) = y0

16 Figure 2: Numerical solutions for the nonhomogeneous fractional initial value problem with α = 0.5. Exact solution in the continuous line, the numerical solutions: dotted line ( h = 10−2), dashed line (h = 10−3). and where C Dα(.) is the fractional derivative in Caputo sense. by means of [9, lemma 6.2], The formulation of the Brusselator problem in Volterra integral form is: x(t) = x(0) + Iα a − (µ + 1)x(t) + x2(t)y(t) y(t) = y(0) + Iα µx(t) − x2(t)y(t) We show eﬃciency of the numerical process by setting parameters given in the ﬁrst simulation in [18]. a = 1, µ = 3, α = 0.7, x0 = 1.1, y0 = 2.9,T = 80. with h = 0.001, we obtain the graphical representation of Brusselator dynamics below (see Figure 3).

(a) The components x(t) in the continuous line and y(t) in the dotted line. (b) The trajectories in phase plan.

Figure 3: Numerical solution of the fractional order Brusselator system.

17 We have taken as reference the numerical solution obtained in [18] using Adams-Bashford-Moulton method [9]. The error between our approach and the reference is of the order of 10−2, when h = 0.005.

6 Conclusion

Using standard techniques from functional analysis we have proved the convergence of nabla h-sum operator to the Riemann-Liouville integral. It is a powerful tool of approximation which allowed us to obtain the numerical solution of some Cauchy fractional problems even with singularity at the initial point. Otherwise, the previous analysis can be the starting point to write an Euler-like formula extending the classical theory to the fractional case. Unfortunately we were not able to ﬁnd a satisfactory such formula. This will be done in further work. Another interesting problem consist to investigate the convergence of the (q, h)− ∇−α operator to the Riemann-Liouville integral. This will be a new step toward the achievement of our conjecture.

7 Acknowledgments

The numerical results are obtained by the Python programming language and special function for SciPy. The Mittag-Leﬄer functions are not available from standard libraries, there were generously provided by Zulﬁkar Moinuddin Ahmed [26], we are especially grateful to him.

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Laboratoire de Syst`emesDynamiques et Applications Department of Mathematics, Abou Bekr Belkaid University, Tlemcen, Algeria