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Generalized fractional-order hybrid of

block-pulse functions and Bernoulli

polynomials approach for solving fractional

delay diﬀerential equations

Octavian Postavaru1*

1*Center for Research and Training in Innovative Techniques of

Applied Mathematics in Engineering, University Politehnica of

Bucharest, Splaiul Independentei 313, Bucharest, 060042,

Romania.

Corresponding author(s). E-mail(s): opostavaru@linuxmail.org;

Abstract

In the current study, we propose a systematize technique for solving

fractional delay diﬀerential equations in the Caputo sense. There-

fore, we compute an exact Riemann-Liouville fractional integral oper-

ator for the generalized fractional-order hybrid of block-pulse func-

tions and Bernoulli polynomials, and we use it in order to reduce

the fractional delay diﬀerential equations into a system of alge-

braic equations. The last ones, are solved numerically using the

Newton’s iterative method. The relative simplicity of the procedure

together with accuracy of the results are the essential feature of

the present method. Finally, we conclude that for the same prob-

lems, our results are better than other results presented in literature.

Keywords: hybrid functions, block-pulse, Bernoulli polynomials, Caputo

derivative, fractional delay diﬀerential equations

1 Introduction

In the last decades, fractional diﬀerential equations began to play an increas-

ingly important role in science and engineering, and this is due to the fact that

1

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2Solving fractional delay diﬀerential equations

the dynamics of many systems in real world ask for using diﬀerential equations

of non-integer order [1–3]. Although one may recognize almost everywhere

around us delay problems, the study of fractional delay diﬀerential equations

is a quiet new topic. Among many examples of delay problems, we may enu-

merate applications in transportation systems [4], electromagnetism [5], ﬂuid

mechanics [6] or spherical ﬂames [7]. Besides, in the last period, delay mod-

els have become increasingly welcome in a large class of biology processes as

primary infection [8], immune response [9], tumor growth [10] and also in sta-

tistical analysis of ecology data [11,12]. The paper [13] is devoted to study

the stabilization problem of stochastic nonlinear delay systems with exoge-

nous disturbances and the event-triggered feedback control. In the article [14],

the authors studied stochastic highly nonlinear delay systems with neutral-

term, which does not satisfy the linear growth condition, and the paper [15] is

devoted to study the pth moment exponential stability problem for a class of

stochastic delay nonlinear systems driven by G-Brownian motion.

In many problems involving fractional calculus, it is worthwhile to use the

operational Riemann-Liouville integration matrix. Particularly, wavelets ψ(x)

like Chebyshev, Legendre, CAS, Haar and Bernoulli [16–19] received a peculiar

consideration in solving some fractional equations.

Recently, fractional order Legendre functions have been deﬁned [20] by

changing tto t=xα, (α > 0), to obtain an eﬃcient spectral technique of

solving fractional diﬀerential equations. In general, one looks for an operational

matrix Pα, which is obtained after some approximations, then, one looks for a

solution to the problem in the form of a linear combination of functions ψ(xα),

and also write all the polynomial components xνthat appear in the problem in

the same basis. It was found in the literature that, the best numerical results

are obtained when α=ν.

In science and engineering, one of the most eﬃcient methods used to solve

some smooth and non-smooth problems is obtained by introducing hybrid func-

tions, which are a mixture of block-pulse functions with polynomials [21–24].

To the extent of our knowledge, the ﬁrst derivation of the Riemann-Liouville

fractional integral operator for hybrid functions was published in [25–27]. The

considered hybrid functions were a combination of block-pulse functions and

Bernoulli polynomials.

In this paper, using the transformation t=xα, (α > 0), we deﬁne a new set

of fractional hybrid functions, consisting of block-pulse functions and Bernoulli

polynomials (GFOHBPB). In the calculation of the delay operational matrix,

we used the relation between the Taylor polynomials Tm(x) and Tm(x−τ), as

well as the relation between the Bernoulli and Taylor polynomials. This allows

us to reduce the Bernoulli polynomials B(x−τ) to B(x) and automatically the

hybrid functions Bα(x−τ) to Bα(x), getting rid of the binomial development.

Next, we deﬁne the Riemann-Liouville fractional integral operator attached

to GFOHBPB, and with its help together with the Newton-Cotes collocation

method, we reduce the fractional diﬀerential equations to systems of algebraic

equations, which we solve using the interactive Newton method. The theory is

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completed with error analysis, and we conclude the paper with four numerical

examples, to show the eﬃciency and simplicity of our method. Due to the

fact that our method is exact, we obtain better results than in Rahimkhani et

al. [28].

In this work we inspect two general types of fractional delay diﬀerential

equations [29], for 0 < τ < 1

Problem a

Dqf(x) = h(x, f (x), f(x−τ)) ,0≤x≤1, n −1< q ≤n ,

f(i)(0) = λi, i = 0 ,1, . . . , n −1, n ∈N,

f(x) = φ(x), x < 0.

(1)

Problem b

Dqf(x) = h(x, f (x), f(τ x)) ,0≤x≤1, n −1< q ≤n ,

f(i)(0) = λi, i = 0 ,1, . . . , n −1, n ∈N.(2)

In the above problems, we consider that his an analytic function, and we

denoted by τthe time delay, and by Dq,n−1< q ≤n, the fractional derivative

in the Caputo sense. λi,i= 0,1, . . . , n −1, and φare known real constants

and function, respectively, and fis the solution of the stated problems.

2 Preliminaries and notations

In this section, ﬁrst we introduce the deﬁnition of the Caputo’s fractional

derivative and the Riemann-Liouville fractional integration, and then we give

some basic deﬁnitions and properties of the fractional calculus theory which

we are using in the present paper.

Deﬁnition 1. The Caputo’s fractional derivative of order qis deﬁned

as [30]

(Dqf) (x) = 1

Γ(n−q)Zx

0

f(n)(s)

(x−s)q+1−nds , n −1< q ≤n , n ∈N,

where q > 0 is the order of the derivative and nis the smallest integer greater

than q.

The Caputo fractional derivative is a linear operation namely

Dq(λf(x) + µg(x)) = λDqf(x) + µDqg(x),

here λand µare constants, and has the property that

Dqxβ=(0, q ∈N0, β < q ,

Γ(β+1)

Γ(β−q+1) xβ−q,otherwise .

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Deﬁnition 2. The Riemann-Liouville fractional integral operator of order

qis deﬁned as [30]

Iqf(x) = (1

Γ(q)Rx

0

f(s)

(x−s)1−qds =1

Γ(q)xq−1∗f(x), q > 0,

f(x), q = 0 ,

where xq−1∗f(x) is the convolution product of xq−1and f(x). For the

Riemann-Liouville fractional integral we have

Iqxγ=Γ(γ+ 1)

Γ(γ+1+q)xγ+q, γ > −1.

The Riemann-Liouville fractional integral is also a linear operation.

The Caputo derivative and Riemann-Liouville integral satisfy the following

properties [30]

Iq(Dqf(x)) = f(x)−

n−1

X

k=0

f(k)(0)xk

k!.(3)

If q∈R,n−1< q ≤n,n∈Nthen

Dq(f(x)) = In−qDnf(x).

Theorem 1. (Generalized Taylor’s formula) [31] Suppose that Diαf(x)∈

C(0,1] for i= 0,1, . . . , m, then we have

f(x) =

m

X

i=0

xiα

Γ(iα + 1)Diα f(0+) + xmα+α

Γ(mα +α+ 1)Dmα+αf(ξ),

with 0 < ξ ≤x,∀x∈(0,1]. Also, one has

|f(x)−

m

X

i=0

xiα

Γ(iα + 1)Diα f(0+)|≤ Mα

xmα+α

Γ(mα +α+ 1) ,

where Mα= supξ∈(0,1] |Dmα+αf(ξ)|. It is worth mentioning that when α= 1,

the generalized Taylor’s formula scales down to the known Taylor’s formula.

3 Generalized fractional-order hybrid functions

In the beginning of this section, we recollect the deﬁnitions of hybrid functions

of block-pulse functions and Bernoulli polynomials, and then we construct

generalized fractional-order hybrid of block-pulse functions and Bernoulli

polynomials (GFOHBPB), suitable for delay problems.

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3.1 Hybrid of block-pulse functions and Bernoulli

polynomials

The hybrid functions bnm(t) are deﬁned on the interval [0,1) as [32]

bnm(t) = βm(N t −n+ 1) , t ∈n−1

N,n

N,

0,otherwise ,(4)

where n= 1,2, ..., N and m= 0,1, ..., M are the order of block-pulse functions

and Bernoulli polynomials, respectively. In Eq. (4), βm(x) are the Bernoulli

polynomials of order m, which are deﬁned by [33]

βm(t) =

m

X

k=0 m

kαm−ktk,

where αk,k= 1,2, ..., m, are the Bernoulli numbers [34]. The ﬁrst few Bernoulli

functions are

β0(x)=1, β1(x) = x−1

2, β2(x) = x2−x+1

6,....

3.2 Generalized fractional-order hybrid functions

We are introducing a new set of fractional-order functions entitled GFOHBPB,

bα;τ

nm (x), suitable for delay problems, which are constructed by changing the

variable tto (xα−τ), (α > 0, τ > 0) in the hybrid function of block-pulse

functions and Bernoulli polynomials

bα;τ

nm (x) = (βm(N(xα−τ)−n+ 1) , x ∈hn−1

N1/α −τ, n

N1/α −τ,

0,otherwise ,(5)

with n= 1,2, ..., N the order of block-pulse functions, and m= 0,1, ..., M the

order of the Bernoulli polynomials. The Bernoulli polynomials βm(x) of order

min Eq. (5), are deﬁned as

βα;τ

m(x)≡βm(xα−τ) =

m

X

k=0 m

kαm−k(xα−τ)k,0≤x≤1,

where αk,k= 1,2, ..., m, are the Bernoulli numbers.

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3.3 Function approximation

Let f∈L2[0,1]. We consider fˆm(x) to be the best approximation of f, and

write

f(x)≈fˆm(x) =

M

X

m=0

N

X

n=1

cnmbα;τ

nm (x) = CTBα;τ(x),(6)

where

C= [c10, . . . , cN0, c11 , . . . , cN1, . . . , c1M, . . . , cN M ]T,

and

Bα;τ(x) = [bα;τ

10 (x), . . . , bα;τ

N0(x), bα;τ

11 (x), . . . , bα;τ

N1(x), . . . , bα;τ

NM (x)]T,(7)

with ˆm=N(M+ 1).

3.4 Delay operational matrix

It is laborious to calculate the Riemann-Liouville fractional integral operator

for the hybrid function, and this is due to the binomial expansion (xα−τ)k.

It is more convenient to express the Bernoulli polynomials in Taylor series,

which is easier to handle.

Now, we derive a general delay operational matrix. One may write

βm(x)=ΞTm(x),(8)

where

B(x)=[β0(x), β1(x), . . . , βm(x)]T, Tm(x) = [1, x, . . . , xm]T,

and where

Ξ =

ξ00 0 0 ... 0

ξ10 ξ11 0... 0

ξ20 ξ21 ξ22 ... 0

... ... ... ... ...

ξm0ξm1ξm2... ξmm

,

with ξij =i

jαi−j.

Also, for Taylor polynomials, we have [35]

Tm(x−τ) = ΛTm(x),(9)

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where Λ is the following matrix

Λ =

λ00 0 0 ... 0

λ10 λ11 0... 0

λ20 λ21 λ22 ... 0

... ... ... ... ...

λm0λm1λm2... λmm

,

with λij =i

i−j(−τ)i−j.

Using Eq. (8) together with Eq. (9), one gets

βm(x−τ)=ΞTm(x−τ) = ΞΛTm(x) = ΞΛΞ−1βm(x)≡Mm

τβm(x),

or in matrix form

B(x−τ) = MrB(x).

Making the transformations x→xαand τ→Nτ ,Mrtransforms into Mα

r,

and we get

Bα(x−τ) = Mα

τBα(x),(10)

with

Bα(x, β)=[bα

10(x), . . . , bα

N0(x), bα

11(x), . . . , bα

N1(x), . . . , bα

NM (x)]T.

We may write the Riemann-Liouville delay fractional integral operator as

IβBα(x−τ)≡Bα(x−τ, β ),

and together with Eq. (10) gives

Bα(x−τ, β ) = Mα

τBα(x, β),(11)

with

Bα(x, β) = [Iβbα

10(x), . . . , I βbα

N0(x), Iβbα

11(x), . . . , I βbα

N1(x), . . . , Iβbα

NM (x)]T.

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4 Riemann-Liouville fractional integral

operator for hybrid of block-pulse functions

and Bernoulli polynomials

In this section, we calculate the Riemann-Liouville fractional integral operator

Iβfor Bα(x), i.e.,

IβBα(x)≡Bα(x, β).(12)

From Eq. (5), we have

bα

nm(x) = βm(N(x)α−n+ 1) µ(n−1

N)1/α (x)−µ(n

N)1/α (x),(13)

where µcis the unit step function deﬁned as

µc(x) = 1, x ≥c ,

0, x < c .

The hypergeometric function is a special function which has the following

integral representation [36]

2F1(a, b, c;z) = Γ(c)

Γ(b)Γ(c−b)Z1

0

tb−1(1 −t)c−b−1

(1 −tz)adt .

It is convenient to write this formula in the following form

2F1a, b, b + 1; c

x=b

cbZc

0

sb−11−s

x−a

ds . (14)

From the deﬁnition of the Riemann-Liouville fractional integral, together

with Eq. (14), for c,αand βpositive numbers, we obtain

Iβ(xαµc(x)) = xα+βΓ(α+ 1)

Γ(β+α+ 1)

−xβ−1

Γ(β)(α+ 1)cα+1 2F11−β, α + 1, α + 2; c

xµc(x).

Using, the deﬁnition of the Bernoulli polynomials in Eq. (13), one gets

Iβbα

nm(x) =

m

X

k=0

k

X

r=0 m

kk

rαm−kNr(1 −n)k−r

Iβxαrµ(n−1

N)1/α (x)−Iβxαrµ(n

N)1/α (x),

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or ﬁnally

Iβbα

nm(x) =

0, x ∈−∞,n−1

N1/α,

U(x), x ∈hn−1

N1/α ,n

N1/α,

U(x)−V(x), x ∈hn

N1/α ,∞,

with

U(x) =

m

X

k=0

k

X

r=0 m

kk

rαm−kNr(1 −n)k−rxαr+βΓ(αr + 1)

Γ(β+αr + 1)

−xβ−1

Γ(β)(αr + 1) n−1

N(αr+1)/α

×2F1 1−β, αr + 1, αr + 2; 1

xn−1

N1/α!# ,

and

V(x) =

m

X

k=0

k

X

r=0 m

kk

rαm−kNr(1 −n)k−rxαr+βΓ(αr + 1)

Γ(β+αr + 1)

−xβ−1

Γ(β)(αr + 1) n

N(αr+1)/α

2F11−β, αr + 1, αr + 2; 1

xn

N1/α .

5 Error bound and error estimation

In the following, we compute the error bound associated to the fractional

derivative established in Deﬁnition 1. We start the analysis with the Prob-

lem a).

Theorem 2. Let Diα f∈C(0,1] with i= 0,1, ..., M , ( ˆm=N(M+ 1)) and

Yα;τ

m={βα;τ

0(x), βα;τ

1(x),...βα;τ

M−1(x)}. If we denote by fM(x−τ) the best

approximation of f(x−τ)∈Yα;τ

mon the interval n−1

N1/α ,n

N1/αi, we

get for the approximate solution fˆm(x−τ) on the interval (0,1] the following

expression

kf(x−τ)−fˆm(x−τ)kL2[0,1]≤supx∈(0,1] |DM α+αf(x)|

Γ(Mα +α+ 1)√2M α + 2α+ 1 .(15)

Proof We deﬁne

f1(x−τ) =

M

X

i=0

(x−τ)iα

Γ(iα + 1) Diα f(0+).

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If x < τ ,f=fˆm. From Theorem 1, the generalized Taylor’s formula, we have

|f(x−τ)−f1(x−τ)|≤ (x−τ)Mα+α

Γ(Mα +α+ 1) sup

x∈Iα

n

|DMα+αf(x−τ)|,

where Iα

n=n−1

N1/α ,n

N1/αi.

Since fM(x−τ) is the best approximation of f(x−τ) out of Yα;τ

mon the interval

Iα

nand f1(x−τ)∈Yα;τ

m, we may write

kf−fˆmk2

L2[0,1] =

N

X

n=1

kf−fMk2

L2Iα

n≤

N

X

n=1

kf−f1k2

L2Iα

n

≤

N

X

n=1 ZIα

n"(x−τ)Mα+α

Γ(Mα +α+ 1) sup

x∈Iα

n

|DMα+αf(x−τ)|#2

dx

≤Z1

τ"(x−τ)Mα+α

Γ(Mα +α+ 1) sup

x∈(0,1]

|DMα+αf(x−τ)|#2

dx . (16)

Changing the variable x−τ→x, Eq. (16) becomes

kf−fˆmk2

L2[0,1] ≤Z1−τ

0"xMα+α

Γ(Mα +α+ 1) sup

x∈(0,1]

|DMα+αf(x)|#2

dx

≤(1 −τ)2Mα+2α+1

Γ(Mα +α+ 1)2(2M α + 2α+ 1) sup

x∈(0,1]

|DMα+αf(x)|!2

.

Counting that 1 −τ≤1, the theorem is proved by taking the square roots.

Corollary 1. We see from Eqs. (16) and (15) that if N→ ∞, respectively

M→ ∞, than kf−fˆmkL2[0,1]→0.

Theorem 3. Considering the problem a). If his Lipshitz with the Lipshitz

ηthen

kDqf−DqfˆmkL2[0,1]≤2ηsupx∈(0,1] |DM α+αf(x)|

Γ(Mα +α+ 1)√2M α + 2α+ 1 .

Proof

kDqf−DqfˆmkL2[0,1] =kh(x, IqDqf(x), I qDqf(x−τ))

−h(x, IqDqfˆm(x), I qDqfˆm(x−τ))kL2[0,1]

≤ηkIqDqf(x)−IqDqfˆm(x)kL2[0,1]

+ηkIqDqf(x−τ)−IqDqfˆm(x−τ)kL2[0,1]

=ηkf(x)−fˆm(x)kL2[0,1]

+ηkf(x−τ)−fˆm(x−τ)kL2[0,1] .

The Theorem 2 completes the proof.

Corollary 2. We see from Corollary 1 that if N→ ∞ or M→ ∞, than

kDqf−DqfˆmkL2[0,1]→0.

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We start the error analysis for the Problem b) with the following theorem:

Theorem 4. Let us assume that Diα f∈C(0,1] with i= 0,1, ..., M ,

( ˆm=N(M+ 1)) and let Yα

m={βα

0(xτ), βα

1(xτ), . . . , βα

M−1(xτ)}. If fM(xτ )

is the best approximation of f(xτ)∈Yα

mon the interval n−1

N1/α ,n

N1/αi,

then the error bound of fˆm(xτ ) on the interval (0,1] would satisfy the relation

kf(xτ)−fˆm(xτ )kL2[0,1] ≤supx∈(0,1] |DM α+αf(x)|

Γ(Mα +α+ 1)√2M α + 2α+ 1 .(17)

Proof We deﬁne

f1(xτ) =

M

X

i=0

(xτ)iα

Γ(iα + 1) Diα f(0+).

From Theorem 1, we have

|f(xτ)−f1(xτ )|≤ (xτ )M α+α

Γ(Mα +α+ 1) sup

x∈Iα

n

|DMα+αf(xτ )|,

where Iα

n=n−1

N1/α ,n

N1/αi.

Since fM(xτ) is the best approximation of f(xτ ) out of Yα

mon the interval Iα

n

and f1(xτ)∈Yα

m, we may write

kf−fˆmk2

L2[0,1] =

N

X

n=1

kf−fMk2

L2Iα

n≤

N

X

n=1

kf−f1k2

L2Iα

n

≤

N

X

n=1 ZIα

n"(xτ)M α+α

Γ(Mα +α+ 1) sup

x∈Iα

n

|DMα+αf(xτ )|#2

dx

≤Z1

0"(xτ)M α+α

Γ(Mα +α+ 1) sup

x∈(0,1]

|DMα+αf(xτ )|#2

dx . (18)

Changing the variable xτ →x, Eq. (18) becomes

kf−fˆmk2

L2[0,1] ≤Zτ

0"xMα+α

Γ(Mα +α+ 1) sup

x∈(0,1]

|DMα+αf(x)|#2

dx

≤τ2Mα+2α

Γ(Mα +α+ 1)2(2M α + 2α+ 1) sup

x∈(0,1]

|DMα+αf(x)|!2

.

Counting that τ≤1, the theorem is proven by taking the square roots.

Corollary 3. We see from Eqs. (18) and (17) that if N→ ∞, respectively

M→ ∞, than kf−fˆmkL2[0,1]→0.

Theorem 5. If his Lipshitz with the Lipshitz ηthen

kDqf−DqfˆmkL2[0,1]≤2ηsupx∈(0,1] |DM α+αf(x)|

Γ(Mα +α+ 1)√2M α + 2α+ 1 .

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Table 1: The absolute errors for N= 2, M= 3, q= 1 and various values of

τfor Example 1. Method 1 is given in [29], and Method 2 in [38].

xThis method Method 1 This method Method 1 This method Method 1 Method 2

τ= 0.0001 τ= 0.0001 τ= 0.001 τ= 0.001 τ= 0.01 τ= 0.01 τ= 0.01

0.2 0 8.33(−17) 0 1.94(−16) 0 0 2.77(−17)

0.4 0 2.22(−16) 0 3.33(−16) 0 1.11(−16) 2.77(−17)

0.6 0 1.47(−14) 0 8.60(−14) 0 3.15(−14) 5.55(−17)

0.8 0 1.57(−14) 0 8.57(−14) 0 3.23(−14) 0

1 0 0 0 0 0 0 0

Proof

kDqf−DqfˆmkL2[0,1] =kh(x, IqDqf(x), I qDqf(xτ))

−h(x, IqDqfˆm(x), I qDqfˆm(xτ))kL2[0,1]

≤ηkIqDqf(x)−IqDqfˆm(x)kL2[0,1]

+ηkIqDqf(xτ)−IqDqfˆm(xτ )kL2[0,1]

=ηkf(x)−fˆm(x)kL2[0,1] +ηkf(xτ)−fˆm(xτ )kL2[0,1] .

The Theorem 4 completes the proof.

Corollary 4. We see from Corollary 3 that if N→ ∞ or M→ ∞, than

kDqf−DqfˆmkL2[0,1]→0.

Now, we give the error estimation of the numerical method presented in

Section 7. It is known that the error estimation by the residual error method

is eﬀective, therefore we use it to measure the error estimation in our method.

Deﬁnition 3. The error estimation of the numerical method is given by [37]

R(xi) =|Dqf(xi)−Dqfˆm(xi)|,

where xi∈[0,1].

6 Illustrative example

In this section, we give some numerical examples in order to demonstrate the

applicability and the accuracy of our method. All numerical calculations were

performed using Mathematica 10.

6.1 Example 1

Consider the fractional delay diﬀerential equation

Dqf(x) = f(x−τ)−f(x) + 2

Γ(3−q)t2−q−1

Γ(2−q)t1−q+ 2τx −τ2−τ ,

f(x)=0, x ≤0,(19)

where 0 < q ≤1. The exact solution when q= 1 is f(x) = x2−x.

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Solving fractional delay diﬀerential equations 13

0.2 0.4 0.6 0.8 1.0

-0.25

-0.20

-0.15

-0.10

-0.05

0.00

(a)

0.2 0.4 0.6 0.8 1.0

-0.25

-0.20

-0.15

-0.10

-0.05

(b)

Fig. 1: (a) Comparison of f(x) for N= 2, M= 4, τ= 0.01 and q=α=

0.7 (dashed, opal), 0.8 (dotted, brown), 0.9 (dashed-dotted, blue) and the

exact solution (continuous, purple). (b) Comparison between the exact and

the approximate solutions for N= 2, M= 4, τ= 0.01 and q=α= 1 for

example 1.

0.2 0.4 0.6 0.8 1.0

0.4

0.5

0.6

0.7

0.8

0.9

1.0

(a)

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

(b)

Fig. 2: (a) Comparison of f(x) for N= 2, M= 4 and q=α= 2.6 (dashed,

opal), 2.7 (dotted, brown), 2.8 (dashed-dotted, blue) and the exact solution

(continuous, purple). (b) Comparison between the exact and the approximate

solutions for N= 2, M= 4, α= 1 and q= 3 for example 1.

We solve this problem for τ= 0.01 by using the hybrid functions with

N= 2 and M= 1. Let

Dqf(x)'Dqfˆm(x) = CTBα(x)=[c10, c20 , c11, c21 ]

bα

10(x)

bα

20(x)

bα

11(x)

bα

21(x)

.(20)

Then, by using Eqs. (3), (12) and (20), we have

f(x)'fˆm(x) = Iq(Dqfˆm(x)) = CTBα(x, q).(21)

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14 Solving fractional delay diﬀerential equations

Table 2: Residual error for N= 2, M= 2, ( ˆm= 6) τ= 0.01 and diﬀerent

values of q=αfor Example 1. Method 1 is given in [29].

xThis method Method 1 This method Method 1 This method Method 1

γ= 0.7γ= 0.7γ= 0.8γ= 0.8γ= 0.9γ= 0.9

0.1 5.33(−3) 4.37(−1) 1.09(−2) 4.24(−1) 8.08(−3) 2.63(−1)

0.2 5.05(−3) 1.56(−1) 3.05(−3) 1.63(−1) 1.46(−3) 1.05(−1)

0.3 2.18(−3) 2.03(−2) 4.79(−3) 1.36(−2) 4.84(−3) 6.51(−3)

0.4 1.45(−3) 1.38(−1) 1.73(−2) 1.44(−1) 3.31(−3) 9.29(−2)

0.5 3.54(−4) 4.47(−2) 1.94(−4) 3.60(−2) 8.94(−4) 1.85(−2)

0.6 3.25(−4) 2.25(−2) 5.84(−4) 2.44(−2) 3.44(−4) 1.46(−2)

0.7 1.88(−4) 2.06(−2) 1.53(−5) 3.23(−3) 1.31(−4) 1.89(−3)

0.8 4.03(−4) 8.13(−2) 5.21(−3) 4.41(−2) 2.89(−4) 1.82(−2)

0.9 1.66(−4) 1.57(−1) 4.00(−5) 9.64(−2) 6.40(−5) 4.45(−2)

We may write f(x−τ) as

f(x−τ) = 0,0< x < τ ,

CTBα(x−τ, q ) = CTMα

τBα(x, q), τ ≤x≤1.(22)

Introducing Eqs. (20), (21) and (22) into (19), we can write

CTBα(x) = −CTBα(x, q) + 2

Γ(3−q)t2−q−1

Γ(2−q)t1−q+ 2τx −τ2−τ ,

0< x < τ ,

CTBα(x) = CTMα

τBα(x, q)−CTBα(x, q) + 2

Γ(3−q)t2−q−1

Γ(2−q)t1−q

+2τx −τ2−τ , τ ≤x≤1.

By collocating this equation at the Newton-Cotes nodes, we get N(M+ 1) =

4 algebraic equations which can be solved for the unknown vector Cusing

Newton’s iterative method. We get the solutions c10 =−0.5, c20 = 0.5, c11 = 1

and c21 = 1, which together with the Eq. (21) gives f(x) = x2−x.

Table 1shows the absolute errors between the exact and approximate solu-

tions for q= 1 with N= 2, M= 1 ( ˆm= 4) and various choices of τand we

compare them to the Bernoulli wavelets method [29] with ˆm= 6 and to the

fractional-order Bernoulli method [38] with ˆm= 6. Due to the fact that the

best Bernoulli wavelets method is represented by [38], in our further compu-

tations one compare our method with this one. Beside, Fig. 1a displays the

approximate solutions obtained for diﬀerent values of q=α= 0.7,0.8,0.9,

for N= 2, M= 4 and τ= 0.01 and in Fig. 1b we compare the numerical

results for q=α= 1 to the exact solution. From these results, it is seen that

the approximate solutions converge to the exact solution.

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Solving fractional delay diﬀerential equations 15

Table 3: Comparison of the approximate solution for N= 2, M= 6, and

q= 1 with the exact solution for Example 2.

xExact solution Method [29] This method

0 1.0000 1.0000 1.0000

0.2 0.8187 0.8187 0.8187

0.4 0.6703 0.6703 0.6703

0.6 0.5488 0.5488 0.5488

0.8 0.4493 0.4494 0.4494

Due to the fact that the exact solution for the q6= 1 values is not known,

we measure the residual error deﬁned as

R(x) =

|Dqfˆm(x) + fˆm(x)−2

Γ(3−q)t2−q+1

Γ(2−q)t1−q−2τx +τ2+τ|,

0≤x≤τ ,

|Dqfˆm(x)−fˆm(x−τ) + fˆm(x)−2

Γ(3−q)t2−q+1

Γ(2−q)t1−q

−2τx +τ2+τ|, τ < x ≤1.

In the Table 2, we give the residual error for N= 2, M= 2 ( ˆm= 6) for

τ= 0.01 and q=α, and we compere them with the residual errors for

the method [29] with ˆm= 6. This table demonstrates the advantages and

the accuracy of the present method for solving fractional delay diﬀerential

equations.

6.2 Example 2

Consider the following fractional delay diﬀerential equation

Dqf(x) = −f(x)−f(x−0.3) + e(−t+0.3) ,0<x<1,2< q < 3,

f(0) = 1 , , f0(0) = −1, f 00(0) = 1 ,

f(x) = e−x, x < 0.

(23)

The exact solution when q= 3 is f(x) = e−x.

In order to solve this problem by using the present method, we let

Dqf(x)'Dqfˆm(x) = CTBα(x),(24)

and then, by using Eqs. (3), (12) and (24), we have

f(x)'fˆm(x) = CTBα(x, q)+1−x+1

2x2.(25)

We may write f(x−0.3) as

f(x−0.3) = e(−t+0.3) ,0≤x < 0.3,

CTMα

0.3Bα(x, q)+1.345 −1.3x+ 0.5x2,0.3≤x≤1.(26)

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Table 4: Residual error for N= 2, M= 6 ( ˆm= 14), α= 1 and diﬀerent

values of qfor Example 2. Method 1 is given in [29].

xThis method Method 1 This method Method 1 This method Method 1

γ= 2.6γ= 2.6γ= 2.7γ= 2.7γ= 2.8γ= 2.8

0.1 9.41(−8) 2.43(−1) 5.87(−8) 1.46(−1) 3.12(−8) 7.22(−2)

0.2 5.34(−8) 1.29(−1) 3.37(−8) 6.33(−2) 1.81(−8) 2.25(−2)

0.3 1.28(−5) 9.58(−2) 9.28(−6) 4.43(−2) 6.56(−6) 1.49(−2)

0.4 1.56(−6) 8.48(−2) 2.22(−6) 4.04(−2) 2.31(−6) 1.52(−2)

0.5 2.89(−6) 7.71(−2) 1.44(−6) 3.68(−2) 4.60(−7) 1.42(−2)

0.6 2.94(−6) 6.69(−2) 2.21(−6) 3.01(−2) 1.55(−6) 1.03(−2)

0.7 1.80(−6) 5.61(−2) 1.51(−6) 2.30(−2) 1.18(−6) 6.30(−3)

0.8 1.64(−7) 5.18(−2) 1.79(−7) 2.17(−2) 1.70(−7) 6.74(−3)

0.9 2.35(−8) 6.49(−2) 3.72(−8) 3.51(−2) 4.21(−8) 1.79(−2)

Introducing Eqs. (24), (25) and (26) into (23), we can write

CTBα(x) = −CTBα(x, q)−1 + x−1

2x2,0<x<0.3,

CTBα(x) = −CTBα(x, q)−CTMα

0.3Bα(x, q) + e(−t+0.3)

−2.345 + 2.3x−x2,0.3≤x≤1.

By collocating this equation at the Newton-Cotes nodes, we get N(M+1) alge-

braic equations which can be solved for the unknown vector Cusing Newton’s

iterative method.

Table 3displays the approximate solutions obtained for various values of x

by using the present method with N= 2 and M= 6 or ˆm= 14, the Bernoulli

wavelet method [29] for ˆm= 14, together with the exact solution. Numerical

results for values of q= 2.6,2.7,2.8 and the exact solution with N= 2, M= 4

are shown in Fig. 2a. In Fig. 2b we compare our numerical results for N= 2,

M= 4, α= 1 and q= 3 with the exact solution. From these results, it is seen

that, as qapproaches 3, the numerical solutions converge to the exact solution.

Due to the fact that the exact solution for the q6= 3 values is not known,

we measure the residual error deﬁned as

R(x) = |Dqfˆm(x) + fˆm(x)|,0≤x < 0.3,

|Dqfˆm(x) + fˆm(x) + fˆm(x−0.3) −e(−t+0.3) |,0.3≤x≤1.

In the Table 4, we give the residual error for N= 2, M= 6 ( ˆm= 14), α= 1

and diﬀerent values of q, and we compere them with the residual errors for the

method [29] with ˆm= 14. The results reﬂect the superiority of our technique.

6.3 Example 3

Consider the fractional pantograph diﬀerential equation

Dqf(x) = af(x) + bf (τ x) + cos(x)−asin(x)−bsin(τ x),0< q ≤1,

f(0) = 0 .(27)

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Table 5: The maximum error Eˆmfor a=−1, b= 0.5 for Example 3. Method 1

is given in [39], and Method 2 is given in [29].

τPiecewise constant DG 1 Piecewise linear DG 1 Method 2 This method

ˆm= 64 ˆm= 64 ˆm= 12 ˆm= 12

0.5 1.4032(−2) 2.1429(−5) 6.1725(−6) 2.3022(−9)

Table 6: The absolute errors for a=−1, b=τ= 0.5, N= 2, q= 1 and

various values of Mfor Example 3. Method 1 is given in [29].

xThis method Method 1 This method Method 1 This method Method 1

ˆm= 4 ˆm= 4 ˆm= 8 ˆm= 8 ˆm= 12 ˆm= 12

0 0 6.71(−2) 0 4.51(−3) 0 5.93(−9)

0.2 840(−4) 3.51(−3) 7.16(−7) 1.54(−1) 7.82(−10) 2.27(−10)

0.4 652(−5) 2.67(−2) 4.97(−7) 1.29(−1) 7.01(−10) 1.22(−9)

0.6 196(−3) 7.68(−2) 2.72(−6) 3.23(−2) 1.93(−9) 5.57(−6)

0.8 945(−4) 1.17(−1) 2.27(−6) 5.22(−2) 1.64(−9) 4.56(−6)

The exact solution, when q= 1 is f(x) = sin(x) for any a , b ∈Rand 0 < τ < 1.

To solve this equation by using the present method, we let

Dqf(x) = CTBα(x).(28)

From the initial conditions we have

f(x) = CTB(x, q),(29)

and substituting Eqs. (28) and (29) into Eq. (27) one get

CTBα(x) = aCTB(x, q ) + bCTB(τ x, q) + cos(x)−asin(x)−bsin(τx).(30)

By collocating Eq. (34) at the Newton-Cotes nodes, we get N(M+ 1) nonlin-

ear algebraic equations, which can be solved for the unknown vector Cusing

Newton’s iterative method.

We deﬁne the maximum error for fˆm(x) as

Eˆm=kfˆm(x)−f(x)k∞= max{| fˆm(x)−f(x)|0≤x≤1},

where fˆm(x) is the approximate solutions and f(x) is the exact solution. In

the Table 5, we compare the maximum error of the present method for N=

2 and M= 5 with the discontinuous Galerkin (DG) method [39] and with

Bernoulli wavelet method [29]. Also, the absolute errors between the exact and

approximate solutions for N= 2 and diﬀerent values of Mare shown in the

Table 6. In addition, in the Table 6we compare our results with the results

obtained by the method [29]. Also, numerical results for diﬀerent choices of q

with N= 2, M= 4 are given in Fig. 3a. We see that, as qapproaches 1, the

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Table 7: Residual error for N= 2, M= 5, ( ˆm= 12) and diﬀerent values of

q=αfor Example 3. Method 1 is given in [29].

xThis method Method 1 This method Method 1 This method Method 1

γ= 0.7γ= 0.7γ= 0.8γ= 0.8γ= 0.9γ= 0.9

0.1 3.15(−6) 1.54(−3) 1.60(−5) 2.65(−3) 6.56(−7) 1.85(−3)

0.2 7.86(−6) 1.84(−3) 4.57(−6) 2.03(−4) 5.20(−6) 1.30(−3)

0.3 2.58(−5) 2.69(−3) 3.79(−6) 3.17(−3) 6.82(−6) 1.87(−3)

0.4 2.83(−6) 1.63(−5) 8.70(−5) 1.77(−3) 1.78(−5) 1.69(−3)

0.5 3.68(−6) 1.81(−5) 1.19(−5) 1.09(−5) 3.27(−5) 5.88(−6)

0.6 1.37(−6) 1.85(−5) 1.22(−5) 2.08(−5) 2.69(−5) 2.38(−5)

0.7 2.54(−6) 1.61(−5) 1.15(−5) 1.72(−6) 2.35(−5) 6.57(−6)

0.8 1.20(−6) 5.03(−5) 5.23(−6) 3.51(−5) 2.24(−5) 3.17(−5)

0.9 2.50(−6) 1.20(−5) 4.21(−6) 1.12(−5) 1.72(−5) 1.40(−5)

Table 8: The absolute errors for q= 1 and various values of xfor Example 4.

Method 1 is given in [39] and Method 2 is given in [29].

xMethod 1 Method 2 This method Method 2 This method

ˆm= 10 ˆm= 10 ˆm= 12 ˆm= 12

2−21.08(−5) 1.19(−8) 2.30(−8) 1.05(−8) 5.37(−10)

2−33.81(−5) 8.50(−8) 2.05(−8) 5.79(−9) 5.33(−10)

2−41.26(−5) 3.32(−7) 3.43(−8) 2.00(−8) 8.03(−10)

2−54.09(−5) 1.52(−7) 3.29(−8) 3.70(−9) 8.71(−10)

2−61.20(−5) 1.97(−7) 2.23(−8) 2.03(−8) 6.37(−10)

numerical solution converges to the solution of the integer order diﬀerential

equation. In Fig. 3b, we compare the numerical result for q=α= 1 and the

exact solution.

The exact solution for the values of q6= 1 is not known. Therefore, in order

to show the eﬃciency of our method for this problem, we compute the residual

error

R(x) =|Dqfˆm(x)−af ˆm(x)−bf ˆm(τx)−cos(x) + asin(x) + bsin(τ x)|,

for N= 2, M= 5, ( ˆm= 12) and diﬀerent values of q=αand we compare

them to the residual errors obtained by method [29] with ˆm= 12. The results

are presented in Table 7.

6.4 Example 4

Consider the fractional pantograph diﬀerential equation

Dqf(x) = −f(x) + τ

2f(τx)−τ

2e−τx ,0<x<1,0< q ≤1,

f(0) = 1 ,(31)

with τ= 0.2. The exact solution, when q= 1 is f(x) = e−x.

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0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

(a)

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

(b)

Fig. 3: (a) Comparison of f(x) for N= 2, M= 4, and q=α= 0.7 (dashed,

opal), 0.8 (dotted, brown), 0.9 (dashed-dotted, blue) and the exact solution

(continuous, purple). (b) Comparison between the exact and the approximate

solutions for N= 2, M= 4, and q=α= 1 for example 3.

0.2 0.4 0.6 0.8 1.0

0.4

0.5

0.6

0.7

0.8

0.9

1.0

(a)

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

(b)

Fig. 4: (a) Comparison of f(x) for N= 2, M= 4, and q=α= 0.7 (dashed,

opal), 0.8 (dotted, brown), 0.9 (dashed-dotted, blue) and the exact solution

(continuous, purple). (b) Comparison between the exact and the approximate

solutions for N= 2, M= 4, and q=α= 1 for example 4.

In order to solve this equation, we let

Dqf(x) = CTBα(x).(32)

From the initial conditions we have

f(x) = CTB(x, q)+1,(33)

and substituting Eqs. (32) and (33) into Eq. (31) one get

CTBα(x) = −CTB(x, q) + τ

2CTB(τ x, q) + τ

2−1−τ

2e−τx .(34)

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Table 9: Residual error for N= 2, M= 5, ( ˆm= 12) and diﬀerent values of

q=αfor Example 4. Method 1 is given in [39].

xThis method Method 1 This method Method 1 This method Method 1

γ= 0.7γ= 0.7γ= 0.8γ= 0.8γ= 0.9γ= 0.9

0.1 7.47(−8) 2.24(−5) 3.94(−7) 5.22(−5) 1.60(−8) 3.86(−5)

0.2 1.79(−7) 2.67(−5) 1.10(−7) 3.83(−6) 1.25(−7) 2.76(−5)

0.3 5.73(−7) 3.76(−5) 9.01(−8) 6.29(−5) 1.63(−7) 3.95(−5)

0.4 4.46(−6) 9.50(−6) 2.04(−6) 3.50(−5) 4.23(−7) 3.53(−5)

0.5 4.56(−6) 1.05(−5) 1.64(−6) 5.17(−6) 4.43(−7) 1.59(−6)

0.6 4.59(−6) 1.10(−5) 1.67(−6) 9.96(−6) 5.02(−7) 6.56(−6)

0.7 4.71(−6) 9.40(−6) 1.69(−6) 8.74(−7) 5.44(−7) 1.79(−6)

0.8 4.96(−6) 2.98(−5) 1.67(−6) 1.67(−5) 5.70(−7) 8.71(−6)

0.9 5.45(−6) 7.02(−6) 1.65(−6) 5.38(−6) 5.94(−7) 3.85(−6)

By collocating Eq. (34) at the Newton-Cotes nodes, we get N(M+ 1)

unknown coeﬃcients, which can be determined by using any standard iterative

technique, like Newton’s iterative method.

In Table 8, we compare the absolute errors between the exact and approx-

imate solutions using the present method for q= 1, N= 2, M= 4 and 5,

with Ref. [39] and Ref. [29]. This table shows the superiority of our technique.

Furthermore, numerical results for diﬀerent choices of qwith N= 2, M= 4,

are given in Fig. 4a. From Fig. 4a, we see that, as qapproaches 1, the numer-

ical solutions converges to the solution of integer order diﬀerential equation.

In Fig. 4b we compare the numerical and exact solution for q=α= 1.

In order to show the eﬃciency of our method in the case q6= 1 for this

problem, we compute the residual error

R(x) =|Dqfˆm(x) + fˆm(x)−τ

2fˆm(τx) + τ

2e−τx |,

for N= 2, M= 5, ( ˆm= 12) and diﬀerent values of q=αand we compare

them to the residual errors obtained by method [29] with ˆm= 12. The results

are presented in the Table 9. Again the results show the primacy of the present

technique.

7 Conclusions

In this article, we compute an exact Riemann-Liouville fractional integral oper-

ator for the generalized fractional-order hybrid of block-pulse functions and

Bernoulli polynomials, and we use it in order to reduce the fractional delay

diﬀerential equations into a system of algebraic equations. In order to solve

numerically this system of equations, we use the collocation method. In the

section Illustrative example, we show that this method is eﬃcient even with a

relatively small basis.

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Solving fractional delay diﬀerential equations 21

Declarations

No funds, grants, or other support was received. The authors have no ﬁnancial

or proprietary interests in any material discussed in this article. Data sharing

not applicable to this article as no datasets were generated or analysed during

the current study.

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