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Generalized fractional-order hybrid of block-pulse functions and Bernoulli polynomials approach for solving fractional delay differential equations

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In the current study, we propose a systematize technique for solving fractional delay differential equations in the Caputo sense. Therefore , we compute an exact Riemann-Liouville fractional integral operator 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. 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 problems , our results are better than other results presented in literature.
a Comparison of f(x) for N=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N = 2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, τ=0.01\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau =0.01$$\end{document} and q=α=0.7\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =0.7$$\end{document} (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\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N =2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, τ=0.01\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau =0.01$$\end{document} and q=α=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =1$$\end{document} for Example 1
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a Comparison of f(x) for N=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N = 2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document} and q=α=2.6\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =2.6$$\end{document} (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\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N =2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, α=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha =1$$\end{document} and q=3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=3$$\end{document} for Example 1
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a Comparison of f(x) for N=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N = 2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, and q=α=0.7\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =0.7$$\end{document} (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\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N =2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, and q=α=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =1$$\end{document} for example 3
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a Comparison of f(x) for N=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N = 2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, and q=α=0.7\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =0.7$$\end{document} (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\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N =2$$\end{document}, M=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M =4$$\end{document}, and q=α=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q=\alpha =1$$\end{document} for example 4
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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 [13]. 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 [1619] 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 [2124].
To the extent of our knowledge, the first derivation of the Riemann-Liouville
fractional integral operator for hybrid functions was published in [2527]. 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(xτ), as
well as the relation between the Bernoulli and Taylor polynomials. This allows
us to reduce the Bernoulli polynomials B(xτ) to B(x) and automatically the
hybrid functions Bα(xτ) 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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Solving fractional delay differential equations 3
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(xτ)) ,0x1, 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)) ,0x1, 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,n1< 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(nq)Zx
0
f(n)(s)
(xs)q+1nds , 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(βq+1) xβq,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)
(xs)1qds =1
Γ(q)xq1f(x), q > 0,
f(x), q = 0 ,
where xq1f(x) is the convolution product of xq1and 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)
n1
X
k=0
f(k)(0)xk
k!.(3)
If qR,n1< q n,nNthen
Dq(f(x)) = InqDnf(x).
Theorem 1. (Generalized Taylor’s formula) [31] Suppose that Df(x)
C(0,1] for i= 0,1, . . . , m, then we have
f(x) =
m
X
i=0
x
Γ( + 1)D f(0+) + x+α
Γ( +α+ 1)D+αf(ξ),
with 0 < ξ x,x(0,1]. Also, one has
|f(x)
m
X
i=0
x
Γ( + 1)D f(0+)|≤ Mα
x+α
Γ( +α+ 1) ,
where Mα= supξ(0,1] |D+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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Solving fractional delay differential equations 5
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 n1
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
kαmktk,
where αk,k= 1,2, ..., m, are the Bernoulli numbers [34]. The first few Bernoulli
functions are
β0(x)=1, β1(x) = x1
2, β2(x) = x2x+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 (xατ), (α > 0, τ > 0) in the hybrid function of block-pulse
functions and Bernoulli polynomials
bα;τ
nm (x) = (βm(N(xατ)n+ 1) , x hn1
N1 τ, n
N1 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(xατ) =
m
X
k=0 m
kαmk(xατ)k,0x1,
where αk,k= 1,2, ..., m, are the Bernoulli numbers.
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3.3 Function approximation
Let fL2[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 (xατ)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
jαij.
Also, for Taylor polynomials, we have [35]
Tm(xτ) = ΛTm(x),(9)
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Solving fractional delay differential equations 7
where Λ is the following matrix
Λ =
λ00 0 0 ... 0
λ10 λ11 0... 0
λ20 λ21 λ22 ... 0
... ... ... ... ...
λm0λm1λm2... λmm
,
with λij =i
ij(τ)ij.
Using Eq. (8) together with Eq. (9), one gets
βm(xτ)=ΞTm(xτ) = ΞΛTm(x) = ΞΛΞ1βm(x)Mm
τβm(x),
or in matrix form
B(xτ) = MrB(x).
Making the transformations xxαand τNτ ,Mrtransforms into Mα
r,
and we get
Bα(xτ) = 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(xτ)Bα(xτ, β ),
and together with Eq. (10) gives
Bα(xτ, 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)αn+ 1) µ(n1
N)1 (x)µ(n
N)1 (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(cb)Z1
0
tb1(1 t)cb1
(1 tz)adt .
It is convenient to write this formula in the following form
2F1a, b, b + 1; c
x=b
cbZc
0
sb11s
xa
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)
xβ1
Γ(β)(α+ 1)cα+1 2F11β, 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
rαmkNr(1 n)kr
Iβxαrµ(n1
N)1 (x)Iβxαrµ(n
N)1 (x),
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or finally
Iβbα
nm(x) =
0, x −∞,n1
N1,
U(x), x hn1
N1 ,n
N1,
U(x)V(x), x hn
N1 ,,
with
U(x) =
m
X
k=0
k
X
r=0 m
kk
rαmkNr(1 n)krxαr+βΓ(αr + 1)
Γ(β+αr + 1)
xβ1
Γ(β)(αr + 1) n1
N(αr+1)
×2F1 1β, αr + 1, αr + 2; 1
xn1
N1!# ,
and
V(x) =
m
X
k=0
k
X
r=0 m
kk
rαmkNr(1 n)krxαr+βΓ(αr + 1)
Γ(β+αr + 1)
xβ1
Γ(β)(αr + 1) n
N(αr+1)
2F11β, αr + 1, αr + 2; 1
xn
N1 .
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 D fC(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
M1(x)}. If we denote by fM(xτ) the best
approximation of f(xτ)Yα;τ
mon the interval n1
N1 ,n
N1i, we
get for the approximate solution fˆm(xτ) on the interval (0,1] the following
expression
kf(xτ)fˆm(xτ)kL2[0,1]supx(0,1] |DM α+αf(x)|
Γ(Mα +α+ 1)2M α + 2α+ 1 .(15)
Proof We define
f1(xτ) =
M
X
i=0
(xτ)
Γ( + 1) D f(0+).
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If x < τ ,f=fˆm. From Theorem 1, the generalized Taylor’s formula, we have
|f(xτ)f1(xτ)|≤ (xτ)+α
Γ(Mα +α+ 1) sup
xIα
n
|D+αf(xτ)|,
where Iα
n=n1
N1 ,n
N1i.
Since fM(xτ) is the best approximation of f(xτ) out of Yα;τ
mon the interval
Iα
nand f1(xτ)Yα;τ
m, we may write
kffˆmk2
L2[0,1] =
N
X
n=1
kffMk2
L2Iα
n
N
X
n=1
kff1k2
L2Iα
n
N
X
n=1 ZIα
n"(xτ)+α
Γ(Mα +α+ 1) sup
xIα
n
|D+αf(xτ)|#2
dx
Z1
τ"(xτ)+α
Γ(Mα +α+ 1) sup
x(0,1]
|D+αf(xτ)|#2
dx . (16)
Changing the variable xτx, Eq. (16) becomes
kffˆmk2
L2[0,1] Z1τ
0"x+α
Γ(Mα +α+ 1) sup
x(0,1]
|D+αf(x)|#2
dx
(1 τ)2+2α+1
Γ(Mα +α+ 1)2(2M α + 2α+ 1) sup
x(0,1]
|D+αf(x)|!2
.
Counting that 1 τ1, the theorem is proved by taking the square roots.
Corollary 1. We see from Eqs. (16) and (15) that if N , respectively
M , than kffˆmkL2[0,1]0.
Theorem 3. Considering the problem a). If his Lipshitz with the Lipshitz
ηthen
kDqfDqfˆmkL2[0,1]2ηsupx(0,1] |DM α+αf(x)|
Γ(Mα +α+ 1)2M α + 2α+ 1 .
Proof
kDqfDqfˆmkL2[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 2 completes the proof.
Corollary 2. We see from Corollary 1 that if N or M , than
kDqfDqfˆmkL2[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 D fC(0,1] with i= 0,1, ..., M ,
( ˆm=N(M+ 1)) and let Yα
m={βα
0(), βα
1(), . . . , βα
M1()}. If fM( )
is the best approximation of f()Yα
mon the interval n1
N1 ,n
N1i,
then the error bound of fˆm( ) on the interval (0,1] would satisfy the relation
kf()fˆm( )kL2[0,1] supx(0,1] |DM α+αf(x)|
Γ(Mα +α+ 1)2M α + 2α+ 1 .(17)
Proof We define
f1() =
M
X
i=0
()
Γ( + 1) D f(0+).
From Theorem 1, we have
|f()f1( )|≤ ( )M α+α
Γ(Mα +α+ 1) sup
xIα
n
|D+αf( )|,
where Iα
n=n1
N1 ,n
N1i.
Since fM() is the best approximation of f( ) out of Yα
mon the interval Iα
n
and f1()Yα
m, we may write
kffˆmk2
L2[0,1] =
N
X
n=1
kffMk2
L2Iα
n
N
X
n=1
kff1k2
L2Iα
n
N
X
n=1 ZIα
n"()M α+α
Γ(Mα +α+ 1) sup
xIα
n
|D+αf( )|#2
dx
Z1
0"()M α+α
Γ(Mα +α+ 1) sup
x(0,1]
|D+αf( )|#2
dx . (18)
Changing the variable x, Eq. (18) becomes
kffˆmk2
L2[0,1] Zτ
0"x+α
Γ(Mα +α+ 1) sup
x(0,1]
|D+αf(x)|#2
dx
τ2+2α
Γ(Mα +α+ 1)2(2M α + 2α+ 1) sup
x(0,1]
|D+αf(x)|!2
.
Counting that τ1, the theorem is proven by taking the square roots.
Corollary 3. We see from Eqs. (18) and (17) that if N , respectively
M , than kffˆmkL2[0,1]0.
Theorem 5. If his Lipshitz with the Lipshitz ηthen
kDqfDqfˆmkL2[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
kDqfDqfˆmkL2[0,1] =kh(x, IqDqf(x), I qDqf())
h(x, IqDqfˆm(x), I qDqfˆm())kL2[0,1]
ηkIqDqf(x)IqDqfˆm(x)kL2[0,1]
+ηkIqDqf()IqDqfˆm( )kL2[0,1]
=ηkf(x)fˆm(x)kL2[0,1] +ηkf()fˆm( )kL2[0,1] .
The Theorem 4 completes the proof.
Corollary 4. We see from Corollary 3 that if N or M , than
kDqfDqfˆmkL2[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(xτ)f(x) + 2
Γ(3q)t2q1
Γ(2q)t1q+ 2τx τ2τ ,
f(x)=0, x 0,(19)
where 0 < q 1. The exact solution when q= 1 is f(x) = x2x.
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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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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(xτ) as
f(xτ) = 0,0< x < τ ,
CTBα(xτ, q ) = CTMα
τBα(x, q), τ x1.(22)
Introducing Eqs. (20), (21) and (22) into (19), we can write
CTBα(x) = CTBα(x, q) + 2
Γ(3q)t2q1
Γ(2q)t1q+ 2τx τ2τ ,
0< x < τ ,
CTBα(x) = CTMα
τBα(x, q)CTBα(x, q) + 2
Γ(3q)t2q1
Γ(2q)t1q
+2τx τ2τ , τ x1.
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) = x2x.
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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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(3q)t2q+1
Γ(2q)t1q2τx +τ2+τ|,
0xτ ,
|Dqfˆm(x)fˆm(xτ) + fˆm(x)2
Γ(3q)t2q+1
Γ(2q)t1q
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(x0.3) + e(t+0.3) ,0<x<1,2< q < 3,
f(0) = 1 , , f0(0) = 1, f 00(0) = 1 ,
f(x) = ex, x < 0.
(23)
The exact solution when q= 3 is f(x) = ex.
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)+1x+1
2x2.(25)
We may write f(x0.3) as
f(x0.3) = e(t+0.3) ,0x < 0.3,
CTMα
0.3Bα(x, q)+1.345 1.3x+ 0.5x2,0.3x1.(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 + x1
2x2,0<x<0.3,
CTBα(x) = CTBα(x, q)CTMα
0.3Bα(x, q) + e(t+0.3)
2.345 + 2.3xx2,0.3x1.
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)|,0x < 0.3,
|Dqfˆm(x) + fˆm(x) + fˆm(x0.3) e(t+0.3) |,0.3x1.
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)|0x1},
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
221.08(5) 1.19(8) 2.30(8) 1.05(8) 5.37(10)
233.81(5) 8.50(8) 2.05(8) 5.79(9) 5.33(10)
241.26(5) 3.32(7) 3.43(8) 2.00(8) 8.03(10)
254.09(5) 1.52(7) 3.29(8) 3.70(9) 8.71(10)
261.20(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
2eτx ,0<x<1,0< q 1,
f(0) = 1 ,(31)
with τ= 0.2. The exact solution, when q= 1 is f(x) = ex.
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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
21τ
2eτx .(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
2eτx |,
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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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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