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New numerical method for ordinary differential equations: Newton polynomial
Journal Pre-proof New numerical method for ordinary differential equations: Newton polynomial Abdon Atangana, Seda ˙I˘gret Araz
PII: DOI: Reference:
S0377-0427(19)30627-2 https://doi.org/10.1016/j.cam.2019.112622 CAM 112622
To appear in:
Journal of Computational and Applied Mathematics
Received date : 22 September 2019 Revised date : 28 October 2019 Please cite this article as: A. Atangana and S. ˙I˘gret Araz, New numerical method for ordinary differential equations: Newton polynomial, Journal of Computational and Applied Mathematics (2019), doi: https://doi.org/10.1016/j.cam.2019.112622. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
© 2019 Elsevier B.V. All rights reserved.
Journal Pre-proof
Abdon Atangana1 , Seda I·g¼ret Araz2 1
of
New numerical method for ordinary di¤erential equations : Newton polynomial [email protected], Institute for Groundwater Studies,
Faculty of Natural and Agricultural Sciences, University of the Free State, South Africa.
[email protected], Siirt University, Department of Mathematics Education, Faculty of Education, Turkey.
p ro
2
November 14, 2019 Abstract
1
Introduction
Pr e-
The Adams-Bashforth have been recognized to be a very e¢ cient numerical method to solve linear and nonlinear di¤erential equations, including those with non-integer orders. This method is based on the well-known Lagrange interpolation; however, it is well known that the Lagrange polynomial is less accurate than the Newton polynomial. In this paper, we introduced a new numerical scheme based on two steps Newton polynomial, we present some applications and illustrative examples for both ordinary di¤erential equations with classical and fractional derivative. Keywords: Newton polynomial, new numerical scheme, fractal calculus, fractional calculus.
Jo
urn
al
Numerical methods for ordinary di¤erential equations have been used in many …eld of science, technology and engineering due to their abilities to provide approximate solutions of nonlinear ordinary di¤erential equations arising in such …elds. Numerical methods are used where analytical methods failed. We shall recall that, these methods …nd their applications naturally in almost all …eld engineering and physical sciences, nevertheless, only within the 21 st century, their applications were extended in the life sciences, medicine, business, social sciences and researchers working within the …eld of arts have adopted them as elements of scienti…c computations. As humankind aim to control their environment via prediction,the development of computing power has revolutionized the utility of realistic mechanical and mathematical models in almost all …elds of science and engineering, thus subtle numerical analysis is needed to implement these detailed mathematical model depicting real world problems [[1]-[5]]. One can easily quote for instance, di¤erential equation arising in celestial mechanics, which of course depict the motion of planets, stars and galaxies; we can add stochastic di¤erential equations and the well-known Markovian chains, which are very essential in depicting living cells for medicine and biology [[6]-[9]]. We shall recall that, the …eld of numerical analysis predated the invention of computers by many decades. The concept of linear interpolation was introduced back more than 2000 years. We recognized that many outstanding mathematicians focused their attentions on numerical analysis, we can quote very important suggested algorithm like Newton’s method, Lagrange interpolation polynomial, Gaussian elimination or the Euler’s method [[10]-[14] ]. In the literature, there exist many numerical schemes for solving ordinary di¤erential equations specially nonlinear cases. One of the most used one is perhaps the well-known Adams-Bashforth method, which is based on the well-known Lagrange interpolation polynomial [[15],[16]]. Nevertheless, it was con…rmed in many published papers that, the Lagrange interpolation polynomial is less accurate than the Newton’s interpolation polynomial. Thus in this paper, we shall suggest new numerical scheme based on the Newton’s interpolation polynomial. Now, we present some de…nitions about the fractional calculus [[2]-[5]].
1
Journal Pre-proof
De…nition 1 Let f : R+ ! R and 2 (n 1; n) ; n 2 N. The left Caputo fractional derivative of order of the function f is given by the following equality; C 0 Dt
f (t) =
1 (n
)
Zt
n
(t
1
s)
(n)
f
(s) ds; t > 0:
0
ABC Dt a
f (t) =
2 [0; 1] then, the de…nition of the Atangana-Baleanu fractional
AB ( ) 1
Z
t
a
p ro
De…nition 3 Let f (t) 2 W21 (0; l) ; a < b; derivative in Caputo sense is given by;
of
De…nition 2 The fractional integral associate to the new fractional integral with power-law kernel is de…ned as: Z t 1 1 C f ( ) (t ) d : I (f (t)) = 0 t ( ) 0
d f ( )E d
where AB ( ) = 1
(t
1
+
( )
)
d
:
Pr e-
De…nition 4 The fractional integral associate to the new fractional integral with Mittag-Le- er kernel is de…ned as: Z t 1 1 ABC It (f (t)) = f (t) + f ( ) (t ) d : 0 AB ( ) AB ( ) ( ) 0 De…nition 5 Let f (t) 2 W21 (0; l) ; a < b; derivative is given by; CF a Dt
f (t) =
M( ) 1
2 [0; 1] then, the de…nition of the Caputo-Fabrizio fractional Z
a
t
d f ( ) exp d
1
(t
) d
al
where M ( ) is a normalization function such that M (0) = M (1) = 1:
urn
De…nition 6 The fractional integral associate to the new fractional integral with exponential decay kernel is de…ned as: Z t 1 FFE It (f (t)) = f (t) + f ( )d : 0 M( ) M( ) 0 De…nition 7 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having p86ower law type kernel is given by; Z t d 1 m 1 FFP D0;t; (f (t)) = (t s) f (s) ds (m ) dt 0 1< ;
m 2 N and
Jo
where m
f (t) df (s) = lim t!s dt t
f (s) : s
De…nition 8 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having power law type kernel is given by; FFP
J0;t; (f (t)) =
( )
Z
0
2
t
(t
s)
1
s
1
f (s) ds:
Journal Pre-proof
De…nition 9 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having exponentially decaying type kernel is given by; Z t M( ) d FFE exp (t s) f (s) ds D0;t; (f (t)) = (1 ) dt 0 1 > 0;
m 2 N and M (0) = M (1) = 1:
of
where
De…nition 10 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having exponentially decaying type kernel is given by; J0;t; (f (t)) =
Z
M( )
t 1
s
f (s) ds +
(1
) t 1 f (t) : M( )
p ro
FFE
0
De…nition 11 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having Mittag-Le- er type kernel is given by; FFM
J0;t;
(f (t)) =
AB ( ) ( )
Z
t
s
1
f (s) (t
0
s)
1
ds +
(1
) t 1 f (t) : AB ( )
where
> 0;
Pr e-
De…nition 12 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having exponentially decaying type kernel is given by; Z M( ) d t FFE D (f (t)) = (t s) f (s) ds exp 0 t 1 dt 0 1 m 2 N and M (0) = M (1) = 1:
2
urn
al
This study is organized as follows. In section 2, we suggest new numerical scheme for Cauchy problem with classical derivative. In sections 3, 4 and 5, we o¤er new numerical method for Cauchy problem with Caputo-Fabrizio, Atangana-Baleanu and Caputo fractional derivative, respectively. In sections 6, 7 and 8, we o¤er new numerical method for Cauchy problem with Caputo-Fabrizio, Atangana-Baleanu and Caputo fractal-fractional derivative, respectively. In section 9, we present numerical illustrations about the proposed method and the results are compared with Adams-Bashforth method. In section 10, we give important results about new numerical scheme.
New method with classical derivative
We consider the following Cauchy problem dy (t) = f (t; y (t)) dt
(1)
Jo
where the function f is non-linear. The aim is to provide a numerical scheme to solve the above equation. To do this, we convert the above into Z t y (t) y (0) = f ( ; y ( )) d (2) at the point tn+1 = (n + 1)
0
t; we have y (tn+1 )
y (0) =
Z
0
3
tn+1
f ( ; y ( )) d
(3)
Journal Pre-proof
at the point tn = n t; we have y (tn )
y (0) =
Z
tn
f ( ; y ( )) d :
(4)
0
Taking the di¤erence, we get y (tn+1 )
Z
y (tn ) =
tn+1
f ( ; y ( )) d :
(5)
tn
= f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn ))
f (tn
+
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn 2
2 ( t) (
tn
2) (
tn
1) :
p ro
Pn ( )
of
We can now consider the approximation of the function f (t; y (t)) as a polynomial more precisely the Newton polynomial tn
2)
(6)
Thus replacing such polynomial into the original equation, we get yn
= f (tn +
Z
2 ; y (tn 2 ))
tn+1
f (tn ; y n )
= f (tn +
tn+1
f tn
1; y
n 1
f tn
2; y
2f tn
1; y
n 1
+ f tn
2; y
f (tn ; y n )
f tn
2 )) t +
2f tn
1; y
n 1
f tn
t
1; y
n 1
+ f tn
2; y
n 2
2
2 ( t)
We can calculate the above integrals as follows; Z tn+1
(
urn
tn+1
2) d
(7)
2) (
tn
1) d
:
tn+1
(
tn
2) d
(8)
(
tn
2) (
tn
1) d
:
tn
=
5 2 ( t) 2
2) (
tn
1) d
=
23 3 ( t) : 6
tn
tn
tn
2) d
tn+1
tn
Z
tn
Z
n 2
tn
tn
(
2; y
(
al
Z
(
n 2
2 ( t)
2 ; y (tn
n 2
t
2
Integrating, we obtain yn
Z
tn
tn
y n+1
t+
Pr e-
y n+1
(9)
Replacing them above scheme, we have the following y n+1
= y n + f tn
2; y
+ f (tn ; y n )
Jo
and we get
y n+1 = y n +
3
5 f tn 12
2; y
n 2
t + f tn
2f tn
n 2
1; y
t
n 1
1; y
n 1
+ f tn
4 f tn 3
1; y
5 t 2
(10)
23 f (tn ; y n ) t: 12
(11)
f tn 2; y
n 1
n 2
t+
2; y
n 2
23 t 12
New method with Caputo-Fabrizio fractional derivative
In this section, we handle the following Cauchy problem with Caputo-Fabrizio fractional derivative CF 0 Dt
y (t) = f (t; y (t))
4
(12)
Journal Pre-proof
where the function f is non-linear. To present a numerical scheme for solution of our equation, we can reformulate the above equation as Z t 1 f ( ; y ( )) d : (13) f (t; y (t)) + y (t) y (0) = M( ) M( ) 0 t; we have 1 y (0) = f (tn ; y (tn )) + M( ) M( )
y (tn+1 )
Z
tn+1
0
at the point tn = n t; we have y (0) =
1 f (tn M( )
1 ; y (tn
1 )) +
If we take the di¤erence of this equations, we obtain y (tn )
=
y (tn+1 )
y (tn )
=
and
f ( ; y ( )) d :
(15)
tn
0
1 [f (tn ; y (tn )) f (tn 1 ; y (tn M( ) Z tn+1 f ( ; y ( )) d + M ( ) tn
Pr e-
y (tn+1 )
M( )
Z
(14)
p ro
y (tn )
f ( ; y ( )) d
of
At the point tn+1 = (n + 1)
1 [f (tn ; y (tn )) f (tn 1 ; y (tn M( ) n Z tj+1 X + f ( ; y ( )) d : M ( ) j=2 tj
1 ))]
(16)
1 ))]
(17)
Using the Newton polynomial, we can write the approximation of the function f (t; y (t)) as follows = f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn ))
+
f (tn
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn
tn
2)
(18)
2
al
Pn ( )
2 ( t)
(
tn
2) (
tn
1) :
yn
=
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) 8 9 f tj 2 ; y j 2 > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 f (tj ;y ) 2f (tj 1 ;y )+f (tj 2 ;y ) > d > M ( ) j=2 tj > > + 2 > > 2( t) > > ; : ( tj 2 ) ( tj 1 )
Jo
y n+1
urn
Thus putting this polynomial into the above equation, we write the following
and reorder as follows y n+1
yn
=
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) 8 f tj 2 ; y j 2 t > > > j 1 > f t ;y f (tj 2 ;y j 2 ) R tj+1 ( ) j 1 > n < ( tj 2 ) d + X t tj + f (tj ;y j ) 2f (tj 1 ;y j 1 )+f (tj 2 ;y j 2 ) M ( ) j=2 > + > > 2( t)2 > R tj+1 > : ( tj 2 ) ( tj 1 ) d tj
5
9 > > > > > = > > > > > ;
(19)
:
(20)
Journal Pre-proof
We can have the following calculations for the above integrals Z tj+1 ( tj 2 ) d = Z
5 2 ( t) 2
tj
tj+1
(
tj
2) (
tj
1) d
(21)
23 3 ( t) : 6
=
tj
=
yn + +
1 [f (tn ; y (tn )) M( ) n X f tj 2 ; y j
M( )
j=2
f (tn 2
+ f tj ; y
1 ; y (tn 1 ))]
t + f tj 1 ; y j 1 f tj 2 ; y j j 1 2f tj 1 ; y + f tj 2 ; y j 2
j
and we can rearrange as
4
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) n X 4 5 f tj 1 ; y j 1 t + f tj 2 ; y j + M ( ) j=2 3 12 yn +
=
2
t+
23 6
5 2
t
(22)
t
(23)
23 f tj ; y j 12
t :
Pr e-
y n+1
2
p ro
y n+1
of
If we replace them into above scheme, we obtain the following scheme
New method with Atangana-Baleanu fractional derivative
Now we deal with the following Cauchy problem with Atangana-Baleanu fractional derivative ABC Dt 0
y (t) = f (t; y (t)) :
(24)
al
In this section, we provide a numerical scheme to solve this equation. Applying Atangana-Baleanu integral, we convert the above equation into Z t 1 1 y (t) y (0) = f (t; y (t)) + f ( ; y ( )) (t ) d : (25) AB ( ) AB ( ) ( ) 0 At the point tn+1 = (n + 1)
Also we can write
1 f (tn ; y (tn )) + y (0) = AB ( ) AB ( ) ( )
urn
y (tn+1 )
t; we have
y (tn+1 ) = y (0) +
Z
tn+1
f ( ; y ( )) (tn+1
1
)
d :
(26)
0
n
X 1 f (tn ; y (tn )) + AB ( ) AB ( ) ( ) j=2
Z
tj+1
f ( ; y ( )) d :
(27)
tj
Jo
Here, for the approximation of the function f (t; y (t)) ; we use the Newton polynomial which is given by Pn ( )
= f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn )) (
+
f (tn
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn 2
2 ( t) tn
2) (
tn
1) :
6
tn
2)
(28)
Journal Pre-proof
Thus if we write this polynomial in equation (27), we have the following y n+1
= y0 +
1 f (tn ; y (tn )) AB ( )
(29)
and we can reorganize
Thus we have
1 f (tn ; y (tn )) AB ( ) 8 R tj+1 1 > f tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R > f (tj 1 ;y f (tj 2 ;y ) ) t n j+1 < X + tj ( tj 2 ) (tn+1 ) t + j 1 j 2 j R f t ;y 2f t ;y +f t ;y ( ) ( ) ( ) t j j 1 j 2 > AB ( ) ( ) j=2 > + tjj+1 > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d y0 +
y n+1
p ro
=
Pr e-
y n+1
1
)
d
of
8 9 f tj 2 ; y j 2 > > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 f (tj ;y ) 2f (tj 1 ;y )+f (tj 2 ;y ) > (tn+1 > AB ( ) ( ) j=2 tj > > + 2 > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
1 f (tn ; y (tn )) AB ( ) n X f tj 2 ; y j + AB ( ) ( ) j=2
= y0 +
+
n X f tj
1; y
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (tn+1
2
Z
t
tj+1
(tn+1
1
)
(30)
1
d
9 > > > > > = > > > > > ;
d
tj
j 1
f tj
2; y
j 2
(31)
t
)
:
1
d
(32)
tj
n X f tj ; y j
al
+
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (
tj
2f tj
1; y
j 1
+ f tj
2; y
j 2
2
2 ( t)
1 ) (tn+1
)
1
d :
urn
tj
When calculating the above integrals Z tj+1 (tn+1 ) tj+1
(
tj
tj+1
tj
(
d
=
)
1
d
=
)
1
d
=
( t)
[(n
j + 1)
(n
j) ]
tj
tj
2 ) (tn+1
Jo
Z
Z
1
tj
2) (
tj
1 ) (tn+1
+1
( t) ( + 1)
(n
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 )
(33)
+2
( t) ( + 1) ( + 2) 2 2 2 (n j) + (3 + 10) (n j) (n j + 1) 6 +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
and putting this equalities into above scheme, we can obtain the following scheme
7
3 7 7 5
Journal Pre-proof
y n+1
1 f (tn ; y (tn )) AB ( ) n X ( t) + f tj AB ( ) ( + 1) j=2
= y0 +
2; y
j 2
[(n
j + 1)
(n
j) ]
n
X ( t) f tj AB ( ) ( + 2) j=2
1; y
j 1
f tj
2; y
j 2
(34)
of
+
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) + f tj ; y j 2f tj 1 ; y j 1 + f tj 2AB ( ) ( + 3) j=2 3 2 2 2 (n j) + (3 + 10) (n j) 7 6 (n j + 1) +2 2 + 9 + 12 7: 6 2 5 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
2; y
j 2
New method with Caputo fractional derivative
Pr e-
5
p ro
(n
In this section, we concern with the following Cauchy problem C 0 Dt
y (t) = f (t; y (t))
(35)
where the derivative is Caputo fractional derivative. Here we aim to present a numerical scheme to solve the above equation. For this, …rstly we transform the above equation into Z t 1 1 y (t) y (0) = f ( ; y ( )) (t ) d : (36) ( ) 0
Also we have
al
t; we have the following Z tn+1 1 f ( ; y ( )) (tn+1 y (tn+1 ) y (0) = ( ) 0
urn
At the point tn+1 = (n + 1)
y (tn+1 ) = y (0) +
n
1 X ( ) j=2
Z
1
)
tj+1
f ( ; y ( )) (tn+1
)
d :
(37)
1
(38)
d :
tj
Jo
Replacing Newton polynomial into the above equation, we have 8 9 f tj 2 ; y j 2 > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > > n Z tj+1 > < = X + ( t ) 1 j 2 n+1 0 t y =y + (t j j 1 j 2 > + f (tj ;y ) 2f (tj 1 ;y 2 )+f (tj 2 ;y ) > > n+1 ( ) j=2 tj > > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
1
)
d :
(39)
Thus we can write the following
y n+1 = y 0 +
8 R tj+1 1 > f tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R tj+1 f (tj 1 ;y ) f (tj 2 ;y ) > n < + tj ( tj 2 ) (tn+1 ) 1 X t j 2 j j 1 R f t ;y 2f t ;y +f t ;y (j ) (j 1 ) (j 2 ) t > ( ) j=2 > + tjj+1 > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d 8
1
d
9 > > > > > = > > > > > ;
(40)
Journal Pre-proof
and rearrange such as Z
n
1 X f tj ( ) j=2
y0 +
=
n
1 X f tj + ( ) j=2
j 2; y
Z
tj+1
(tn+1
1; y
j 1
f tj
2; y
Z
j 2
t
1 X f tj ; y j ( ) j=2 tj
tj+1
(
2f tj
1; y
j 1
+ f tj
2
2 ( t)
2) (
tj
1 ) (tn+1
1
)
tj
+1
tj Z
2 ) (tn+1
1
)
tj+1
tj
(
tj
2
6 (n +6 4
2) (
tj
j + 1) (n
d =
j)
)
1
d
(41)
d :
j 21
( t) ( + 1)
j + 1)
(n
(n
j) ]
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) +2
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7: 2 5 2 (n j) + (5 + 10) (n j) +6 2 + 18 + 12
Pr e-
tj
(
2 ) (tn+1
p ro
For the above integrals, we can have as follows Z tj+1 ( t) 1 (tn+1 ) d = [(n tj+1
tj
2; y
tj
Z
d
tj
tj+1
(
1
)
tj
n
+
2
of
y n+1
1 ) (tn+1
)
1
d =
(42)
If we put them into above equality, we obtain the following scheme n
= y0 +
X ( t) f tj ( + 1) j=2
al
y n+1
2; y
j 2
[(n
j + 1)
f tj
2; y
(n
j) ]
n
+
X ( t) f tj ( + 2) j=2
1; y
j 1
j 2
6
Jo
urn
(n j + 1) (n j + 3 + 2 ) (n j + 1) (n j + 3 + 3 ) n X ( t) + f tj ; y j 2f tj 1 ; y j 1 + f tj 2 ( + 3) j=2 2 2 2 (n j) + (3 + 10) (n j) (n j + 1) 6 +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
(43) 2; y
j 2
3
7 7: 5
New method with Caputo-Fabrizio fractal-fractional derivative
In this section, we consider the following Cauchy problem with Caputo-Fabrizio fractal-fractional derivative FFE Dt ; 0
y (t) = f (t; y (t))
9
(44)
Journal Pre-proof
and if we integrate above equation, we obtain CF 0 Dt
y (t) =
1 t M( )
1
f (t; y (t)) +
M( )
Z
t 1
f ( ; y ( )) d :
0
Again integrating the above equation, we obtain the following equation Z t 1 y (t) y (0) = F ( ; y ( )) d : F (t; y (t)) + M( ) M( ) 0 f (t; y (t)). At the point tn+1 = (n + 1)
y (tn+1 )
y (0) =
1 F (tn ; y (tn )) + M( ) M( )
y (tn )
y (0) =
1 F (tn M( )
1 ; y (tn
0
Z
p ro
and at the point tn = n t; we have
t; we have Z tn+1 F ( ; y ( )) d
of
1
where F (t; y (t)) = t
1 )) +
M( )
y (tn )
=
Pr e-
y (tn+1 )
(46)
tn
F ( ; y ( )) d :
(47)
0
Taking the di¤erence these equations, we obtain the following 1 y (tn+1 ) y (tn ) = [F (tn ; y (tn )) F (tn 1 ; y (tn M( ) Z tn+1 + F ( ; y ( )) d : M ( ) tn and
(45)
1 [F (tn ; y (tn )) F (tn 1 ; y (tn M( ) n Z tj+1 X F ( ; y ( )) d : + M ( ) j=2 tj
1 ))]
(48)
1 ))]
(49)
We will take into consideration the approximation of the function f (t; y (t)) as a polynomial more precisely the Newton polynomial = F (tn +
2 ; y (tn 2 ))
F (tn ; y (tn ))
+
F (tn
2F (tn
1 ; y (tn 1 ))
1 ; y (tn 2
al
Pn ( )
F (tn 2 ; y (tn 2 )) ( t 1 )) + F (tn 2 ; y (tn 2 ))
tn
2)
(50)
2 ( t)
(
tn
2) (
tn
1) :
Jo
urn
When replacing this polynomial into the above equation, we have 1 y n+1 y n = [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) 8 9 F tj 2 ; y j 2 > > > > > F (tj 1 ;yj 1 ) F (tj 2 ;yj 2 ) > > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 F (tj ;y ) 2F (tj 1 ;y )+F (tj 2 ;y ) > d > M ( ) j=2 tj > > + > > 2( t)2 > > : ; ( tj 2 ) ( tj 1 )
(51)
and we obtain
y n+1
yn
=
1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) 8 F tj 2 ; y j 2 t > > > j 1 > F t ;y F (tj 2 ;y j 2 ) R tj+1 ( ) j 1 > n < ( tj 2 ) d + X t tj + F (tj ;y j ) 2F (tj 1 ;y j 1 )+F (tj 2 ;y j 2 ) M ( ) j=2 > + > > 2( t)2 > R tj+1 > : ( tj 2 ) ( tj 1 ) d tj
10
9 > > > > > = > > > > > ;
:
(52)
Journal Pre-proof
We know that we have the following equalities Z tj+1 Z
(
tj
2) d
=
5 2 ( t) 2
2) (
tj
1) d
=
23 3 ( t) : 6
tj
tj+1
(
tj
tj
(53)
of
Replacing them above scheme, we write as follows 1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) n X F tj 2 ; y j 2 t + F tj 1 ; y j 1 F tj 2 ; y j + j j 1 + F tj ; y 2F tj 1 ; y + F tj 2 ; y j 2 M( )
y n+1
= yn +
and we get the following scheme
1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) n X 4 5 F tj 1 ; y j 1 t + F tj 2 ; y j + M ( ) j=2 3 12
= yn +
Thus we have y n+1
7
2
t+
Pr e-
y n+1
5 2
t
(54)
t
(55)
23 F tj ; y j 12
i (1 ) h 1 tn f (tn ; y n ) tn 11 f (tn 1 ; y (tn 1 )) M( ) ( n 1 4 5 j 1 X t + 12 tj 21 f tj 2 ; y j 3 tj 1 f tj 1 ; y + 1 23 M ( ) j=2 t + 12 tj f tj ; y j yn +
=
23 12
p ro
j=2
2
t :
(56) 2
)
t
:
New method with Atangana-Baleanu fractal-fractional deriva-
al
tive Let us consider the following Cauchy problem
FFM Dt ; 0
y (t) = f (t; y (t))
(57)
urn
where the derivative is Atangana-Baleanu fractal-fractional derivative. Integrating above equation, we obtain the following Z t 1 1 1 y (t) y (0) = t 1 f (t; y (t)) + f ( ; y ( )) (t ) d : (58) AB ( ) AB ( ) ( ) 0 1
If we take F (t; y (t)) = t
Jo
y (t)
f (t; y (t)) ; we reformulate the above equation as Z t 1 y (0) = F (t; y (t)) + F ( ; y ( )) (t AB ( ) AB ( ) ( ) 0
At the point tn+1 = (n + 1) y (tn+1 )
y (0) =
)
1
d :
(59)
t; we have
1 F (tn ; y (tn )) + AB ( ) AB ( ) ( )
Z
tn+1
F ( ; y ( )) (tn+1
)
1
d :
(60)
F ( ; y ( )) (tn+1
)
1
d :
(61)
0
Also we obtain
n
y (tn+1 ) = y (0) +
X 1 F (tn ; y (tn )) + AB ( ) AB ( ) ( ) j=2 11
Z
tj+1
tj
Journal Pre-proof
For the approximation of the function F (t; y (t)) ; we shall use the Newton polynomial which is given as follows = F (tn
2 ; y (tn 2 ))
F (tn ; y (tn ))
+
+
F (tn
2F (tn
1 ; y (tn 1 ))
1 ; y (tn 2
F (tn 2 ; y (tn 2 )) ( t 1 )) + F (tn 2 ; y (tn 2 ))
tn
2) (
tn
(62)
1) :
Here if we put this polynomial into above equation, we write such as = y0 +
2)
2 ( t)
(
y n+1
tn
of
Pn ( )
1 F (tn ; y (tn )) AB ( )
(63)
p ro
9 8 F tj 2 ; y j 2 > > > > > > j 1 j 2 > n Z tj+1 > = < + F (tj 1 ;y ) F (tj 2 ;y ) ( X t ) j 2 t + (tn+1 j j 1 j 2 F t ;y 2F t ;y +F t ;y ( ) ( ) ( ) j j 1 j 2 > > AB ( ) ( ) j=2 tj > > + > > 2( t)2 > > ; : ( tj 2 ) ( tj 1 )
If we order the above equality, we obtain y0 +
+
1 F (tn ; y (tn )) AB ( ) 8 R tj+1 > F tj 2 ; y j 2 (tn+1 > tj > > R > F (tj 1 ;y j 1 ) F (tj t > + tjj+1 n > < t X
AB ( ) ( )
and write the following
+
2 ;y
j
2
n X F tj
1; y
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (tn+1
2
Z
tj+1
(tn+1
)
d
)
1
( tj 2 ) (tn+1 ) d > R j j 1 )+F (tj tj+1 F (tj ;y ) 2F (tj 1 ;y > + tj > > 2( t)2 > : ( tj 2 ) ( tj 1 ) (tn+1 )
1 F (tn ; y (tn )) AB ( ) n X + F tj 2 ; y j AB ( ) ( ) j=2
= y0 +
1
)
j=2 > >
urn
y n+1
Pr e-
=
al
y n+1
1
2 ;y
j
1
2
d
Jo
1
d :
(64)
d
tj
j 1
F tj
2; y
j 2
t )
n X F tj ; y j
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (
9 > > > > > > > =
> > ) > > > > > ;
1
d
(65)
tj
+
)
tj
2F tj
1 ) (tn+1
tj
12
1; y
j 1 2
2 ( t) )
1
d :
+ F tj
2; y
j 2
Journal Pre-proof
If we calculate the above integrals Z tj+1 (tn+1
d =
( t)
[(n
j + 1)
(n
j) ]
tj
+1
tj+1
tj
tj Z
2 ) (tn+1
1
)
d =
( t) ( + 1)
+2
tj+1
(
tj
tj
2) (
2
6 (n 6 4
tj
j + 1) (n
j)
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7 2 5 2 (n j) + (5 + 10) (n j) 2 +6 + 18 + 12 1 ) (tn+1
1 F (tn ; y (tn )) AB ( ) n X ( t) + F tj AB ( ) ( + 1) j=2 y0 +
2; y
d =
j 2
[(n
j + 1)
Pr e-
=
1
)
and replace them into above scheme, we get the following y n+1
(n j + 1) (n j + 3 + 2 ) (n j + 1) (n j + 3 + 3 ) (66)
of
(
p ro
Z
1
)
(n
j) ]
n
+
X ( t) F tj AB ( ) ( + 2) j=2
j 1
F tj
2; y
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) F tj ; y j 2F tj 1 ; y j + 2AB ( ) ( + 3) j=2 2 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
j 2
(67) 1
1 t AB ( ) n
urn
=
y0 +
1
2; y
j 2
7 7: 5
Thus we have the following approximation y n+1
+ F tj 3
al
(n
1; y
(tn ; y (tn )) n
+
X ( t) t AB ( ) ( + 1) j=2 j
Xh ( t) t AB ( ) ( + 2) j=2 j
1 2f
tj
2; y
j 2
[(n
j + 1)
(n
j) ]
n
+
1 1f
tj
1; y
j 1
tj
1 2f
2; y
j 2
i
(n
Jo
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n h X ( t) t 1 f tj ; y j 2tj 11 f tj + 2AB ( ) ( + 3) j=2 j 22 2 2 (n j) + (3 + 10) (n j) (n j + 1) 66 +2 2 + 9 + 12 66 2 44 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
tj
13
(68) 1; y
j 1
33
77 77 : 55
+ tj
1 2f
tj
2; y
j 2
i
Journal Pre-proof
8
New method with Caputo fractal-fractional derivative
Let us handle the following Cauchy problem with Caputo fractal-fractional derivative FFP Dt ; 0
y (t) = f (t; y (t))
(69)
and if we integrate above equation, we get
1
Here we shall take as F (t; y (t)) = t y (t)
( )
Z
t 1
of
y (0) =
f ( ; y ( )) (t
f (t; y (t)) : Thus we have Z
1 ( )
y (0) =
1
)
0
d :
(70)
p ro
y (t)
t
F ( ; y ( )) (t
0
1
)
d
(71)
where the function F is non-linear. Here we will provide numerical scheme for solving the above equation. At the point tn+1 = (n + 1) t; we have 1 ( )
y (0) =
Also we obtain
Z
F ( ; y ( )) (tn+1
0
n
y (tn+1 ) = y (0) +
tn+1
1 X ( ) j=2
Z
1
)
Pr e-
y (tn+1 )
tj+1
F ( ; y ( )) (tn+1
)
d :
(72)
1
(73)
d :
tj
al
Thus replacing Newton polynomial into the original equation, we get the following 8 9 F tj 2 ; y j 2 > > > > > > j 1 j 2 > > Z F t ;y F t ;y ( ) ( ) j 1 j 2 n < = t j+1 X + ( t ) 1 j 2 n+1 0 t (tn+1 y =y + j j 1 j 2 > + F (tj ;y ) 2F (tj 1 ;y 2 )+F (tj 2 ;y ) > ( ) j=2 tj > > > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
or we write
urn
y n+1 = y 0 +
8 R tj+1 1 > F tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R > F t ;y F t ;y ( ) ( ) t n j 1 j 2 ( tj 2 ) (tn+1 ) 1 X < + tjj+1 t R tj+1 F (tj ;yj ) 2F (tj 1 ;yj 1 )+F (tj 2 ;yj 2 ) ( ) j=2 > > + tj > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d
1
)
1
9 > > > > > =
d
> > > > > ;
Thus we have the following equality
Z
n
=
y0 +
1 X F tj ( ) j=2
Jo
y n+1
n
1 X F tj + ( ) j=2
1; y
n
+
Z
j 2; y
1 X F tj ; y j ( ) j=2
2
tj+1
(tn+1
tj
2) (
1
(74)
:
(75)
d
tj
j 1
F tj
2; y
Z
j 2
t 2F tj
1; y
j 1
2 ( t) tj
tj+1
(
tj
tj
+ F tj
2
tj+1
(
)
d
1 ) (tn+1
tj
14
)
1
d :
2; y
j 2
2 ) (tn+1
)
1
d
(76)
Journal Pre-proof
We can calculate the above integrals as follows; Z tj+1 ( t) 1 (tn+1 ) d = [(n
(n
j) ]
+1
tj+1
tj
tj Z
2 ) (tn+1
1
)
d =
(n
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) +2
tj+1
(
tj
( t) ( + 1)
tj
2) (
2
6 (n 6 4
tj
j + 1) (n
j)
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7: 2 5 2 (n j) + (5 + 10) (n j) +6 2 + 18 + 12 1 ) (tn+1
1
)
d =
(77)
of
(
p ro
Z
j + 1)
tj
When replacing them into above equality, we have the following n
y n+1
X ( t) F tj ( + 1) j=2
y0 +
=
2; y
j 2
n
X ( t) F tj ( + 2) j=2
j 1
j + 1)
Pr e-
+
[(n
F tj
2; y
j) ]
j 2
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) F tj ; y j 2F tj 1 ; y j 1 + F tj 2 ; y j + 2 ( + 3) j=2 2 3 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) 7 +2 2 + 9 + 12 6 7: 2 4 5 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
Thus, we have
al
(n
1; y
(n
(78) 2
n
=
y0 +
( t) X t ( + 1) j=2 j
urn
y n+1
( t) X h t ( + 2) j=2 j
1 2f
tj
2; y
j 2
[(n
j + 1)
(n
n
+
1 1f
tj
1; y
j 1
tj
1 2f
tj
2; y
i
(n
Jo
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n ( t) X h 1 + t f tj ; y j 2tj 11 f tj 1 ; y j 2 ( + 3) j=2 j 2 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
j 2
j) ]
15
(79) 1
+ tj 3
7 7: 5
1 2f
tj
2; y
j 2
i
Journal Pre-proof
9
Numerical Illustrations and Simulation
Example 1. We consider the following d dt y (t)
= sin t:y (t) y (0) = exp ( 1) :
(80)
The exact solution is given as
of
y (t) = exp ( cos t) : We compare the exact solution with Adams-Bashforth and our proposed method . Here we consider the equivalent Adams-Bashforth 3 t f (tn ; y (tn )) 2
We consider the in…nite norm
t f (tn 2
1 ; y (tn 1 )) :
p ro
y n+1 = y n +
kf k1 = sup jf (t)j : t2 t
The numerical comparison are depicted in Figure 1.The error between the exact and numerical solution for this example t = 0:01; t = 0 to 2000:
kyex
But if
t = 0:001; we have t = 0 to 2000 kyex
while
=
5:7536
5
10
yprop k1
=
3:1197
yAdams k1 = 5:7536 yprop k1 = 4:2920
10
10
10
10
6
:
7
:
Jo
urn
al
kyex
yAdams k1
Pr e-
kyex
Figure 1. Numerical comparison with classical derivative.
16
Journal Pre-proof
Example 2. We consider the following simple equation y0 (t) =
2y (t) :
(81)
The exact solution is 2t
:
The comparisons are given in Figure 2 below. Also, the error for kyex
kyex
yprop k1
=
4:3875
10
=
2:6008
10
5
6
t = 0:001; we have the following error kyex
yAdams k1 = 6:2208 yprop k1 = 2:66
10
8
10
9
:
urn
al
Pr e-
kyex
:
p ro
For
yAdams k1
t = 0:01 is given as
of
y (t) = e
Figure 2. Numerical comparison with classical derivative.
Jo
Example 3. We consider the following non-homogeneous ordinary di¤erential equation y0 (t) = 2y + 3 y (0) = 1:
(82)
The exact solution is given as
5 2t 3 e : 2 2 We present the comparison in Figure 3 below. Thus for t = 0:0001, we have the following error y (t) =
kyex
yAdams k1 = 2:4853 17
10
8
Journal Pre-proof
while the proposed method has yprop k1 = 1:4415
10
11
:
Pr e-
p ro
of
kyex
Figure 3. Numerical comparison with classical derivative.
Example 4. We consider the following example
The exact solution is
al
y0 (t) =
y (t) =
cos (2t) y 2 (t) :
(83)
2 : 2 + sin 2t
urn
The following comparison can be obtained kyex
while
kyex
t = 0:01: For
yprop k1 = 3:1458
4
10 5
10
t = 0:0001; we have
Jo
for
yAdams k1 = 5:6138
kyex
yAdams k1 = 1:6267
kyex
yprop k1 = 1:6659
10
9
and the proposed method
18
10
12
:
Pr e-
p ro
of
Journal Pre-proof
Figure 4. Numerical comparison with classical derivative.
Example 5. We next consider the following example
y0 (t) = 5ty (t)
with initial condition
(84)
y (0) = 1:
al
The exact solution is given as y (t) = exp
5 2 t : 2
urn
We present the comparison in Figure 5. The comparison of error for kyex
yAdams k1 = 5:9306
while the error of the proposed method is given as yprop k1 = 0:0503:
Jo
kyex
19
t = 0:001; is given as
Pr e-
p ro
of
Journal Pre-proof
Figure 5. Numerical comparison with classical derivative.
Example 6. Now we consider the following chaotic problem =
(y (t)
= x (t) =
For simplicity, we shall take as f1 (x; y; z; t) = (y (t) y (t) :So, we can write the above as follows
urn
d x (t) dt d y (t) dt d z (t) dt
h)
y (t) + z (t)
(85)
y (t) :
al
d x (t) dt d y (t) dt d z (t) dt
h) ; f2 (x; y; z; t) = x (t) y (t)+z (t) and f3 (x; y; z; t) =
= f1 (x; y; z; t) = f2 (x; y; z; t)
(86)
= f3 (x; y; z; t) :
Jo
As we did before, we can have the following x (tn+1 )
y (tn+1 )
x (tn ) y (tn )
= =
Z
tn+1
tn Z tn+1
f1 (x; y; z; ) d f2 (x; y; z; ) d
tn tn+1
z (tn+1 )
z (tn )
=
Z
tn
20
f3 (x; y; z; ) d :
(87)
Journal Pre-proof
After putting Newton polynomial into above system and doing some manipulations, we get xn+1
= xn +
5 f1 t n 12
n 2 n 2 n 2 ;y ;z 2; x
23 f1 (tn ; xn ; y n ; z n ) t 12 5 = y n + f2 tn 2 ; xn 2 ; y n 12 23 + f2 (tn ; xn ; y n ; z n ) t 12 5 n = z + f3 tn 2 ; xn 2 ; y n 12 23 + f3 (tn ; xn ; y n ; z n ) t 12
t
4 f1 tn 3
n 1 n 1 n 1 ;y ;z 1; x
t
n 1 n 1 n 1 ;y ;z 1; x
t
+
; zn
2
t
4 f2 tn 3
2
; zn
2
t
4 f3 t n 3
of
z n+1
2
n 1 n 1 n 1 ;y ;z 1; x
p ro
y n+1
For the numerical solution of this problem, we take as h =
b sin
x(t) 2a
+d ;
= 10:82;
(88)
t
= 14:286; a =
urn
al
Pr e-
13; b = 0:11; c = 7; d = 2. Also we take the initial conditions as x (0) = 1; y (0) = 1; z (0) = 0: Numerical simulations are depicted in Figure 6,7,8 and 9.
Jo
Figure 6. Numerical simulation for chaotic problem with classical derivative in yz coordinates.
21
Pr e-
p ro
of
Journal Pre-proof
Jo
urn
al
Figure 7. Numerical simulation for chaotic problem with classical derivative in xz coordinates.
Figure 8. Numerical simulation for chaotic problem with classical derivative in xy coordinates. 22
Pr e-
p ro
of
Journal Pre-proof
Figure 9. Numerical simulation for chaotic problem with classical derivative in xyz coordinates.
Jo
urn
al
We shall take same problem with initial conditions x (0) = 11; y (0) = 114:286: Numerical simulations are depicted in Figure 10,11,12 and 13.
23
1; z (0) = 10: Also
=
Pr e-
p ro
of
Journal Pre-proof
Jo
urn
al
Figure 10. Numerical simulation for chaotic problem with classical derivative in yz coordinates.
Figure 11. Numerical simulation for chaotic problem with classical derivative in xz coordinates. 24
Pr e-
p ro
of
Journal Pre-proof
Jo
urn
al
Figure 12. Numerical simulation for chaotic problem with classical derivative in xy coordinates.
Figure 13. Numerical simulation for chaotic problem with classical derivative in xyz coordinates.
25
Journal Pre-proof
Example 7. For fractional Caputo case, we consider the following equation C 0 Dt
y (t) = t
(89)
with initial condition y (0) = 1: The exact solution of such equation is t ( + 1) : ( + + 1) We plot the above solution against the numerical counterpart to have the following solution. For the numerical solution of this problem, we shall take as = 0:87; = 0:9; h = 0:01: The error of the proposed method is given as kyex yprop k1 = 4:9855 10 4 :
p ro
of
y (t) =
al
Pr e-
Numerical simulation is depicted in Figure 14.
10
urn
Figure 14. Numerical simulation with Caputo derivative.
Conclusion
Jo
Modeling real world problems using mathematical models have been the focus of many researchers working in di¤erent …eld of sciences in the last years. The very reason for this attraction is the fact that such models are used for prediction, control and analysis. Noting that real world problems are complex, thus their counterparts mathematical models are also complex. For those depicting linear problems, their associated mathematical models can be solved analytically using some techniques like Green function, Laplace transform, method of separation of variable and others. Nevertheless those depicting nonlinear problems, their associated mathematical models cannot sometime be solved analytically, thus numerical techniques are requested. Adams-Bashforth, a powerful numerical technique to solve nonlinear equations has been used in many …elds. The method is based on the Lagrange polynomial, but it was reported in some published papers such polynomial has limitations. In this paper, we suggested the use of Newton polynomial that seems to be more accurate. Con‡ict of interest The authors declare that there is no con‡ict of interests regarding the publication of this paper. 26
Journal Pre-proof
References [1] Bonyah E., Chaos in a 5-D hyperchaotic system with four wings in the light of non-local and non-singular fractional derivatives, Chaos, Solitons & Fractals, 116, 2018, 316-331. [2] A. Atangana and D. Baleanu, New fractional derivative with non-local and non-singular kernel, Thermal Sci., 20 (2016), 757-763.
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[3] Owolabi K.M., Mathematical modelling and analysis of love dynamics: A fractional approach, Physica A: Statistical Mechanics and its Applications, 525, 2019, 849-865. [4] Igret Araz S., Numerical analysis of a new Volterra integro-di¤erential equation involving fractalfractional operators, Chaos, Solitons&Fractals, 2020, 130, (109396).
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[5] Tou…k M., Atangana A., New numerical approximation of fractional derivative with non-local and nonsingular kernel: Application to chaotic models, The European Physical Journal Plus, 132 (10), 2017, 444. [6] Ganji RM., Jafari H., A numerical approach for multi-variable order di¤erential equations using Jacobi polynomials, International Journal of Applied and Computational Mathematics, 2019, 5:34.
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[7] Zhou, Y., Manimaran, J., Shangerganesh, L., Debbouche A., Weakness and Mittag–Le- er Stability of Solutions for Time-Fractional Keller–Segel Models. International Journal of Nonlinear Sciences and Numerical Simulation, 19(7-8), 2018, 753-761. [8] Jafari H., Babaei A., Banihashemi S., A Novel Approach for Solving an Inverse Reaction–Di¤usion– Convection Problem, Journal of Optimization Theory and Applications, 183 (2), 2019, 688-704. [9] Peng L., Zhou Y., Debbouche A., Approximation techniques of optimal control problems for fractional dynamic systems in separable Hilbert spaces, Chaos, Solitons & Fractals, 118, 2019, 234-241. [10] Werner W., Polynomial interpolation: Lagrange versus Newton, Mathematics and Computation, 43 (1984), 205-217.
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[11] Fred T. Krogh, E¢ cient algorithms for polynomial interpolation and numerical di¤erentiation, Mathematics and Computation. 24 (1970), 185–190. [12] Yang Y., Gordon S.P., Visualizing and understanding the components of Lagrange and Newton Interpolation, Problems, Resources, and Issues in Mathematics Undergraduate Studies, 26 (1), 2015, 39-52.
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[13] Dimitrov D.K., Philipps G.M., A note on convergence of Newton interpolating polynomials, Journal of Computational and Applied Mathematics, 51, 1, 1994, 127-130. [14] Srivastava R.B., Shukla S., Numerical accuracies of Lagrange’s and Newton polynomial interpolation: Numerical accuracies of Interpolation formulas, LAP LAMBERT Academic Publishing, 2012. [15] Zhang, Tong & Jin, JiaoJiao & Jiang, Tao, The decoupled Crank–Nicolson/Adams–Bashforth scheme for the Boussinesq equations with nonsmooth initial data, Applied Mathematics and Computation, Elsevier, 337(C), 2018, 234-266.
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[16] Jain S., Numerical analysis for the fractional di¤usion and fractional Buckmaster equation by the twostep Laplace Adam-Bashforth method, The European Physical Journal Plus, 2018, 133:19.
27
Journal Pre-proof Highlights
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New numerical scheme based on Newton polynomial. New numerical scheme for ordinary fractional differential equations. New numerical scheme for ordinary fractal-fractional differential equations. Fractional and integral equations.
PII: DOI: Reference:
S0377-0427(19)30627-2 https://doi.org/10.1016/j.cam.2019.112622 CAM 112622
To appear in:
Journal of Computational and Applied Mathematics
Received date : 22 September 2019 Revised date : 28 October 2019 Please cite this article as: A. Atangana and S. ˙I˘gret Araz, New numerical method for ordinary differential equations: Newton polynomial, Journal of Computational and Applied Mathematics (2019), doi: https://doi.org/10.1016/j.cam.2019.112622. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
© 2019 Elsevier B.V. All rights reserved.
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Abdon Atangana1 , Seda I·g¼ret Araz2 1
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New numerical method for ordinary di¤erential equations : Newton polynomial [email protected], Institute for Groundwater Studies,
Faculty of Natural and Agricultural Sciences, University of the Free State, South Africa.
[email protected], Siirt University, Department of Mathematics Education, Faculty of Education, Turkey.
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2
November 14, 2019 Abstract
1
Introduction
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The Adams-Bashforth have been recognized to be a very e¢ cient numerical method to solve linear and nonlinear di¤erential equations, including those with non-integer orders. This method is based on the well-known Lagrange interpolation; however, it is well known that the Lagrange polynomial is less accurate than the Newton polynomial. In this paper, we introduced a new numerical scheme based on two steps Newton polynomial, we present some applications and illustrative examples for both ordinary di¤erential equations with classical and fractional derivative. Keywords: Newton polynomial, new numerical scheme, fractal calculus, fractional calculus.
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Numerical methods for ordinary di¤erential equations have been used in many …eld of science, technology and engineering due to their abilities to provide approximate solutions of nonlinear ordinary di¤erential equations arising in such …elds. Numerical methods are used where analytical methods failed. We shall recall that, these methods …nd their applications naturally in almost all …eld engineering and physical sciences, nevertheless, only within the 21 st century, their applications were extended in the life sciences, medicine, business, social sciences and researchers working within the …eld of arts have adopted them as elements of scienti…c computations. As humankind aim to control their environment via prediction,the development of computing power has revolutionized the utility of realistic mechanical and mathematical models in almost all …elds of science and engineering, thus subtle numerical analysis is needed to implement these detailed mathematical model depicting real world problems [[1]-[5]]. One can easily quote for instance, di¤erential equation arising in celestial mechanics, which of course depict the motion of planets, stars and galaxies; we can add stochastic di¤erential equations and the well-known Markovian chains, which are very essential in depicting living cells for medicine and biology [[6]-[9]]. We shall recall that, the …eld of numerical analysis predated the invention of computers by many decades. The concept of linear interpolation was introduced back more than 2000 years. We recognized that many outstanding mathematicians focused their attentions on numerical analysis, we can quote very important suggested algorithm like Newton’s method, Lagrange interpolation polynomial, Gaussian elimination or the Euler’s method [[10]-[14] ]. In the literature, there exist many numerical schemes for solving ordinary di¤erential equations specially nonlinear cases. One of the most used one is perhaps the well-known Adams-Bashforth method, which is based on the well-known Lagrange interpolation polynomial [[15],[16]]. Nevertheless, it was con…rmed in many published papers that, the Lagrange interpolation polynomial is less accurate than the Newton’s interpolation polynomial. Thus in this paper, we shall suggest new numerical scheme based on the Newton’s interpolation polynomial. Now, we present some de…nitions about the fractional calculus [[2]-[5]].
1
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De…nition 1 Let f : R+ ! R and 2 (n 1; n) ; n 2 N. The left Caputo fractional derivative of order of the function f is given by the following equality; C 0 Dt
f (t) =
1 (n
)
Zt
n
(t
1
s)
(n)
f
(s) ds; t > 0:
0
ABC Dt a
f (t) =
2 [0; 1] then, the de…nition of the Atangana-Baleanu fractional
AB ( ) 1
Z
t
a
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De…nition 3 Let f (t) 2 W21 (0; l) ; a < b; derivative in Caputo sense is given by;
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De…nition 2 The fractional integral associate to the new fractional integral with power-law kernel is de…ned as: Z t 1 1 C f ( ) (t ) d : I (f (t)) = 0 t ( ) 0
d f ( )E d
where AB ( ) = 1
(t
1
+
( )
)
d
:
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De…nition 4 The fractional integral associate to the new fractional integral with Mittag-Le- er kernel is de…ned as: Z t 1 1 ABC It (f (t)) = f (t) + f ( ) (t ) d : 0 AB ( ) AB ( ) ( ) 0 De…nition 5 Let f (t) 2 W21 (0; l) ; a < b; derivative is given by; CF a Dt
f (t) =
M( ) 1
2 [0; 1] then, the de…nition of the Caputo-Fabrizio fractional Z
a
t
d f ( ) exp d
1
(t
) d
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where M ( ) is a normalization function such that M (0) = M (1) = 1:
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De…nition 6 The fractional integral associate to the new fractional integral with exponential decay kernel is de…ned as: Z t 1 FFE It (f (t)) = f (t) + f ( )d : 0 M( ) M( ) 0 De…nition 7 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having p86ower law type kernel is given by; Z t d 1 m 1 FFP D0;t; (f (t)) = (t s) f (s) ds (m ) dt 0 1< ;
m 2 N and
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where m
f (t) df (s) = lim t!s dt t
f (s) : s
De…nition 8 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having power law type kernel is given by; FFP
J0;t; (f (t)) =
( )
Z
0
2
t
(t
s)
1
s
1
f (s) ds:
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De…nition 9 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having exponentially decaying type kernel is given by; Z t M( ) d FFE exp (t s) f (s) ds D0;t; (f (t)) = (1 ) dt 0 1 > 0;
m 2 N and M (0) = M (1) = 1:
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where
De…nition 10 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having exponentially decaying type kernel is given by; J0;t; (f (t)) =
Z
M( )
t 1
s
f (s) ds +
(1
) t 1 f (t) : M( )
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FFE
0
De…nition 11 Suppose that f (t) be continuous on an open interval (a; b), then the fractal-fractional integral of f (t) with order having Mittag-Le- er type kernel is given by; FFM
J0;t;
(f (t)) =
AB ( ) ( )
Z
t
s
1
f (s) (t
0
s)
1
ds +
(1
) t 1 f (t) : AB ( )
where
> 0;
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De…nition 12 Suppose that f (t) be continuous and fractal di¤ erentiable on an open interval (a; b) with order , then the fractal-fractional derivative of f (t) with order in the Riemann-Liouville sense having exponentially decaying type kernel is given by; Z M( ) d t FFE D (f (t)) = (t s) f (s) ds exp 0 t 1 dt 0 1 m 2 N and M (0) = M (1) = 1:
2
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This study is organized as follows. In section 2, we suggest new numerical scheme for Cauchy problem with classical derivative. In sections 3, 4 and 5, we o¤er new numerical method for Cauchy problem with Caputo-Fabrizio, Atangana-Baleanu and Caputo fractional derivative, respectively. In sections 6, 7 and 8, we o¤er new numerical method for Cauchy problem with Caputo-Fabrizio, Atangana-Baleanu and Caputo fractal-fractional derivative, respectively. In section 9, we present numerical illustrations about the proposed method and the results are compared with Adams-Bashforth method. In section 10, we give important results about new numerical scheme.
New method with classical derivative
We consider the following Cauchy problem dy (t) = f (t; y (t)) dt
(1)
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where the function f is non-linear. The aim is to provide a numerical scheme to solve the above equation. To do this, we convert the above into Z t y (t) y (0) = f ( ; y ( )) d (2) at the point tn+1 = (n + 1)
0
t; we have y (tn+1 )
y (0) =
Z
0
3
tn+1
f ( ; y ( )) d
(3)
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at the point tn = n t; we have y (tn )
y (0) =
Z
tn
f ( ; y ( )) d :
(4)
0
Taking the di¤erence, we get y (tn+1 )
Z
y (tn ) =
tn+1
f ( ; y ( )) d :
(5)
tn
= f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn ))
f (tn
+
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn 2
2 ( t) (
tn
2) (
tn
1) :
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Pn ( )
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We can now consider the approximation of the function f (t; y (t)) as a polynomial more precisely the Newton polynomial tn
2)
(6)
Thus replacing such polynomial into the original equation, we get yn
= f (tn +
Z
2 ; y (tn 2 ))
tn+1
f (tn ; y n )
= f (tn +
tn+1
f tn
1; y
n 1
f tn
2; y
2f tn
1; y
n 1
+ f tn
2; y
f (tn ; y n )
f tn
2 )) t +
2f tn
1; y
n 1
f tn
t
1; y
n 1
+ f tn
2; y
n 2
2
2 ( t)
We can calculate the above integrals as follows; Z tn+1
(
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tn+1
2) d
(7)
2) (
tn
1) d
:
tn+1
(
tn
2) d
(8)
(
tn
2) (
tn
1) d
:
tn
=
5 2 ( t) 2
2) (
tn
1) d
=
23 3 ( t) : 6
tn
tn
tn
2) d
tn+1
tn
Z
tn
Z
n 2
tn
tn
(
2; y
(
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Z
(
n 2
2 ( t)
2 ; y (tn
n 2
t
2
Integrating, we obtain yn
Z
tn
tn
y n+1
t+
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y n+1
(9)
Replacing them above scheme, we have the following y n+1
= y n + f tn
2; y
+ f (tn ; y n )
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and we get
y n+1 = y n +
3
5 f tn 12
2; y
n 2
t + f tn
2f tn
n 2
1; y
t
n 1
1; y
n 1
+ f tn
4 f tn 3
1; y
5 t 2
(10)
23 f (tn ; y n ) t: 12
(11)
f tn 2; y
n 1
n 2
t+
2; y
n 2
23 t 12
New method with Caputo-Fabrizio fractional derivative
In this section, we handle the following Cauchy problem with Caputo-Fabrizio fractional derivative CF 0 Dt
y (t) = f (t; y (t))
4
(12)
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where the function f is non-linear. To present a numerical scheme for solution of our equation, we can reformulate the above equation as Z t 1 f ( ; y ( )) d : (13) f (t; y (t)) + y (t) y (0) = M( ) M( ) 0 t; we have 1 y (0) = f (tn ; y (tn )) + M( ) M( )
y (tn+1 )
Z
tn+1
0
at the point tn = n t; we have y (0) =
1 f (tn M( )
1 ; y (tn
1 )) +
If we take the di¤erence of this equations, we obtain y (tn )
=
y (tn+1 )
y (tn )
=
and
f ( ; y ( )) d :
(15)
tn
0
1 [f (tn ; y (tn )) f (tn 1 ; y (tn M( ) Z tn+1 f ( ; y ( )) d + M ( ) tn
Pr e-
y (tn+1 )
M( )
Z
(14)
p ro
y (tn )
f ( ; y ( )) d
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At the point tn+1 = (n + 1)
1 [f (tn ; y (tn )) f (tn 1 ; y (tn M( ) n Z tj+1 X + f ( ; y ( )) d : M ( ) j=2 tj
1 ))]
(16)
1 ))]
(17)
Using the Newton polynomial, we can write the approximation of the function f (t; y (t)) as follows = f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn ))
+
f (tn
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn
tn
2)
(18)
2
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Pn ( )
2 ( t)
(
tn
2) (
tn
1) :
yn
=
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) 8 9 f tj 2 ; y j 2 > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 f (tj ;y ) 2f (tj 1 ;y )+f (tj 2 ;y ) > d > M ( ) j=2 tj > > + 2 > > 2( t) > > ; : ( tj 2 ) ( tj 1 )
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y n+1
urn
Thus putting this polynomial into the above equation, we write the following
and reorder as follows y n+1
yn
=
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) 8 f tj 2 ; y j 2 t > > > j 1 > f t ;y f (tj 2 ;y j 2 ) R tj+1 ( ) j 1 > n < ( tj 2 ) d + X t tj + f (tj ;y j ) 2f (tj 1 ;y j 1 )+f (tj 2 ;y j 2 ) M ( ) j=2 > + > > 2( t)2 > R tj+1 > : ( tj 2 ) ( tj 1 ) d tj
5
9 > > > > > = > > > > > ;
(19)
:
(20)
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We can have the following calculations for the above integrals Z tj+1 ( tj 2 ) d = Z
5 2 ( t) 2
tj
tj+1
(
tj
2) (
tj
1) d
(21)
23 3 ( t) : 6
=
tj
=
yn + +
1 [f (tn ; y (tn )) M( ) n X f tj 2 ; y j
M( )
j=2
f (tn 2
+ f tj ; y
1 ; y (tn 1 ))]
t + f tj 1 ; y j 1 f tj 2 ; y j j 1 2f tj 1 ; y + f tj 2 ; y j 2
j
and we can rearrange as
4
1 [f (tn ; y (tn )) f (tn 1 ; y (tn 1 ))] M( ) n X 4 5 f tj 1 ; y j 1 t + f tj 2 ; y j + M ( ) j=2 3 12 yn +
=
2
t+
23 6
5 2
t
(22)
t
(23)
23 f tj ; y j 12
t :
Pr e-
y n+1
2
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If we replace them into above scheme, we obtain the following scheme
New method with Atangana-Baleanu fractional derivative
Now we deal with the following Cauchy problem with Atangana-Baleanu fractional derivative ABC Dt 0
y (t) = f (t; y (t)) :
(24)
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In this section, we provide a numerical scheme to solve this equation. Applying Atangana-Baleanu integral, we convert the above equation into Z t 1 1 y (t) y (0) = f (t; y (t)) + f ( ; y ( )) (t ) d : (25) AB ( ) AB ( ) ( ) 0 At the point tn+1 = (n + 1)
Also we can write
1 f (tn ; y (tn )) + y (0) = AB ( ) AB ( ) ( )
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y (tn+1 )
t; we have
y (tn+1 ) = y (0) +
Z
tn+1
f ( ; y ( )) (tn+1
1
)
d :
(26)
0
n
X 1 f (tn ; y (tn )) + AB ( ) AB ( ) ( ) j=2
Z
tj+1
f ( ; y ( )) d :
(27)
tj
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Here, for the approximation of the function f (t; y (t)) ; we use the Newton polynomial which is given by Pn ( )
= f (tn +
2 ; y (tn 2 ))
f (tn ; y (tn )) (
+
f (tn
2f (tn
1 ; y (tn 1 ))
f (tn 2 ; y (tn 2 )) ( t 1 )) + f (tn 2 ; y (tn 2 ))
1 ; y (tn 2
2 ( t) tn
2) (
tn
1) :
6
tn
2)
(28)
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Thus if we write this polynomial in equation (27), we have the following y n+1
= y0 +
1 f (tn ; y (tn )) AB ( )
(29)
and we can reorganize
Thus we have
1 f (tn ; y (tn )) AB ( ) 8 R tj+1 1 > f tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R > f (tj 1 ;y f (tj 2 ;y ) ) t n j+1 < X + tj ( tj 2 ) (tn+1 ) t + j 1 j 2 j R f t ;y 2f t ;y +f t ;y ( ) ( ) ( ) t j j 1 j 2 > AB ( ) ( ) j=2 > + tjj+1 > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d y0 +
y n+1
p ro
=
Pr e-
y n+1
1
)
d
of
8 9 f tj 2 ; y j 2 > > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 f (tj ;y ) 2f (tj 1 ;y )+f (tj 2 ;y ) > (tn+1 > AB ( ) ( ) j=2 tj > > + 2 > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
1 f (tn ; y (tn )) AB ( ) n X f tj 2 ; y j + AB ( ) ( ) j=2
= y0 +
+
n X f tj
1; y
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (tn+1
2
Z
t
tj+1
(tn+1
1
)
(30)
1
d
9 > > > > > = > > > > > ;
d
tj
j 1
f tj
2; y
j 2
(31)
t
)
:
1
d
(32)
tj
n X f tj ; y j
al
+
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (
tj
2f tj
1; y
j 1
+ f tj
2; y
j 2
2
2 ( t)
1 ) (tn+1
)
1
d :
urn
tj
When calculating the above integrals Z tj+1 (tn+1 ) tj+1
(
tj
tj+1
tj
(
d
=
)
1
d
=
)
1
d
=
( t)
[(n
j + 1)
(n
j) ]
tj
tj
2 ) (tn+1
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Z
Z
1
tj
2) (
tj
1 ) (tn+1
+1
( t) ( + 1)
(n
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 )
(33)
+2
( t) ( + 1) ( + 2) 2 2 2 (n j) + (3 + 10) (n j) (n j + 1) 6 +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
and putting this equalities into above scheme, we can obtain the following scheme
7
3 7 7 5
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y n+1
1 f (tn ; y (tn )) AB ( ) n X ( t) + f tj AB ( ) ( + 1) j=2
= y0 +
2; y
j 2
[(n
j + 1)
(n
j) ]
n
X ( t) f tj AB ( ) ( + 2) j=2
1; y
j 1
f tj
2; y
j 2
(34)
of
+
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) + f tj ; y j 2f tj 1 ; y j 1 + f tj 2AB ( ) ( + 3) j=2 3 2 2 2 (n j) + (3 + 10) (n j) 7 6 (n j + 1) +2 2 + 9 + 12 7: 6 2 5 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
2; y
j 2
New method with Caputo fractional derivative
Pr e-
5
p ro
(n
In this section, we concern with the following Cauchy problem C 0 Dt
y (t) = f (t; y (t))
(35)
where the derivative is Caputo fractional derivative. Here we aim to present a numerical scheme to solve the above equation. For this, …rstly we transform the above equation into Z t 1 1 y (t) y (0) = f ( ; y ( )) (t ) d : (36) ( ) 0
Also we have
al
t; we have the following Z tn+1 1 f ( ; y ( )) (tn+1 y (tn+1 ) y (0) = ( ) 0
urn
At the point tn+1 = (n + 1)
y (tn+1 ) = y (0) +
n
1 X ( ) j=2
Z
1
)
tj+1
f ( ; y ( )) (tn+1
)
d :
(37)
1
(38)
d :
tj
Jo
Replacing Newton polynomial into the above equation, we have 8 9 f tj 2 ; y j 2 > > > > > f (tj 1 ;yj 1 ) f (tj 2 ;yj 2 ) > > n Z tj+1 > < = X + ( t ) 1 j 2 n+1 0 t y =y + (t j j 1 j 2 > + f (tj ;y ) 2f (tj 1 ;y 2 )+f (tj 2 ;y ) > > n+1 ( ) j=2 tj > > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
1
)
d :
(39)
Thus we can write the following
y n+1 = y 0 +
8 R tj+1 1 > f tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R tj+1 f (tj 1 ;y ) f (tj 2 ;y ) > n < + tj ( tj 2 ) (tn+1 ) 1 X t j 2 j j 1 R f t ;y 2f t ;y +f t ;y (j ) (j 1 ) (j 2 ) t > ( ) j=2 > + tjj+1 > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d 8
1
d
9 > > > > > = > > > > > ;
(40)
Journal Pre-proof
and rearrange such as Z
n
1 X f tj ( ) j=2
y0 +
=
n
1 X f tj + ( ) j=2
j 2; y
Z
tj+1
(tn+1
1; y
j 1
f tj
2; y
Z
j 2
t
1 X f tj ; y j ( ) j=2 tj
tj+1
(
2f tj
1; y
j 1
+ f tj
2
2 ( t)
2) (
tj
1 ) (tn+1
1
)
tj
+1
tj Z
2 ) (tn+1
1
)
tj+1
tj
(
tj
2
6 (n +6 4
2) (
tj
j + 1) (n
d =
j)
)
1
d
(41)
d :
j 21
( t) ( + 1)
j + 1)
(n
(n
j) ]
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) +2
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7: 2 5 2 (n j) + (5 + 10) (n j) +6 2 + 18 + 12
Pr e-
tj
(
2 ) (tn+1
p ro
For the above integrals, we can have as follows Z tj+1 ( t) 1 (tn+1 ) d = [(n tj+1
tj
2; y
tj
Z
d
tj
tj+1
(
1
)
tj
n
+
2
of
y n+1
1 ) (tn+1
)
1
d =
(42)
If we put them into above equality, we obtain the following scheme n
= y0 +
X ( t) f tj ( + 1) j=2
al
y n+1
2; y
j 2
[(n
j + 1)
f tj
2; y
(n
j) ]
n
+
X ( t) f tj ( + 2) j=2
1; y
j 1
j 2
6
Jo
urn
(n j + 1) (n j + 3 + 2 ) (n j + 1) (n j + 3 + 3 ) n X ( t) + f tj ; y j 2f tj 1 ; y j 1 + f tj 2 ( + 3) j=2 2 2 2 (n j) + (3 + 10) (n j) (n j + 1) 6 +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
(43) 2; y
j 2
3
7 7: 5
New method with Caputo-Fabrizio fractal-fractional derivative
In this section, we consider the following Cauchy problem with Caputo-Fabrizio fractal-fractional derivative FFE Dt ; 0
y (t) = f (t; y (t))
9
(44)
Journal Pre-proof
and if we integrate above equation, we obtain CF 0 Dt
y (t) =
1 t M( )
1
f (t; y (t)) +
M( )
Z
t 1
f ( ; y ( )) d :
0
Again integrating the above equation, we obtain the following equation Z t 1 y (t) y (0) = F ( ; y ( )) d : F (t; y (t)) + M( ) M( ) 0 f (t; y (t)). At the point tn+1 = (n + 1)
y (tn+1 )
y (0) =
1 F (tn ; y (tn )) + M( ) M( )
y (tn )
y (0) =
1 F (tn M( )
1 ; y (tn
0
Z
p ro
and at the point tn = n t; we have
t; we have Z tn+1 F ( ; y ( )) d
of
1
where F (t; y (t)) = t
1 )) +
M( )
y (tn )
=
Pr e-
y (tn+1 )
(46)
tn
F ( ; y ( )) d :
(47)
0
Taking the di¤erence these equations, we obtain the following 1 y (tn+1 ) y (tn ) = [F (tn ; y (tn )) F (tn 1 ; y (tn M( ) Z tn+1 + F ( ; y ( )) d : M ( ) tn and
(45)
1 [F (tn ; y (tn )) F (tn 1 ; y (tn M( ) n Z tj+1 X F ( ; y ( )) d : + M ( ) j=2 tj
1 ))]
(48)
1 ))]
(49)
We will take into consideration the approximation of the function f (t; y (t)) as a polynomial more precisely the Newton polynomial = F (tn +
2 ; y (tn 2 ))
F (tn ; y (tn ))
+
F (tn
2F (tn
1 ; y (tn 1 ))
1 ; y (tn 2
al
Pn ( )
F (tn 2 ; y (tn 2 )) ( t 1 )) + F (tn 2 ; y (tn 2 ))
tn
2)
(50)
2 ( t)
(
tn
2) (
tn
1) :
Jo
urn
When replacing this polynomial into the above equation, we have 1 y n+1 y n = [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) 8 9 F tj 2 ; y j 2 > > > > > F (tj 1 ;yj 1 ) F (tj 2 ;yj 2 ) > > n Z tj+1 > < + = X ( t ) j 2 t + j j 1 j 2 F (tj ;y ) 2F (tj 1 ;y )+F (tj 2 ;y ) > d > M ( ) j=2 tj > > + > > 2( t)2 > > : ; ( tj 2 ) ( tj 1 )
(51)
and we obtain
y n+1
yn
=
1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) 8 F tj 2 ; y j 2 t > > > j 1 > F t ;y F (tj 2 ;y j 2 ) R tj+1 ( ) j 1 > n < ( tj 2 ) d + X t tj + F (tj ;y j ) 2F (tj 1 ;y j 1 )+F (tj 2 ;y j 2 ) M ( ) j=2 > + > > 2( t)2 > R tj+1 > : ( tj 2 ) ( tj 1 ) d tj
10
9 > > > > > = > > > > > ;
:
(52)
Journal Pre-proof
We know that we have the following equalities Z tj+1 Z
(
tj
2) d
=
5 2 ( t) 2
2) (
tj
1) d
=
23 3 ( t) : 6
tj
tj+1
(
tj
tj
(53)
of
Replacing them above scheme, we write as follows 1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) n X F tj 2 ; y j 2 t + F tj 1 ; y j 1 F tj 2 ; y j + j j 1 + F tj ; y 2F tj 1 ; y + F tj 2 ; y j 2 M( )
y n+1
= yn +
and we get the following scheme
1 [F (tn ; y (tn )) F (tn 1 ; y (tn 1 ))] M( ) n X 4 5 F tj 1 ; y j 1 t + F tj 2 ; y j + M ( ) j=2 3 12
= yn +
Thus we have y n+1
7
2
t+
Pr e-
y n+1
5 2
t
(54)
t
(55)
23 F tj ; y j 12
i (1 ) h 1 tn f (tn ; y n ) tn 11 f (tn 1 ; y (tn 1 )) M( ) ( n 1 4 5 j 1 X t + 12 tj 21 f tj 2 ; y j 3 tj 1 f tj 1 ; y + 1 23 M ( ) j=2 t + 12 tj f tj ; y j yn +
=
23 12
p ro
j=2
2
t :
(56) 2
)
t
:
New method with Atangana-Baleanu fractal-fractional deriva-
al
tive Let us consider the following Cauchy problem
FFM Dt ; 0
y (t) = f (t; y (t))
(57)
urn
where the derivative is Atangana-Baleanu fractal-fractional derivative. Integrating above equation, we obtain the following Z t 1 1 1 y (t) y (0) = t 1 f (t; y (t)) + f ( ; y ( )) (t ) d : (58) AB ( ) AB ( ) ( ) 0 1
If we take F (t; y (t)) = t
Jo
y (t)
f (t; y (t)) ; we reformulate the above equation as Z t 1 y (0) = F (t; y (t)) + F ( ; y ( )) (t AB ( ) AB ( ) ( ) 0
At the point tn+1 = (n + 1) y (tn+1 )
y (0) =
)
1
d :
(59)
t; we have
1 F (tn ; y (tn )) + AB ( ) AB ( ) ( )
Z
tn+1
F ( ; y ( )) (tn+1
)
1
d :
(60)
F ( ; y ( )) (tn+1
)
1
d :
(61)
0
Also we obtain
n
y (tn+1 ) = y (0) +
X 1 F (tn ; y (tn )) + AB ( ) AB ( ) ( ) j=2 11
Z
tj+1
tj
Journal Pre-proof
For the approximation of the function F (t; y (t)) ; we shall use the Newton polynomial which is given as follows = F (tn
2 ; y (tn 2 ))
F (tn ; y (tn ))
+
+
F (tn
2F (tn
1 ; y (tn 1 ))
1 ; y (tn 2
F (tn 2 ; y (tn 2 )) ( t 1 )) + F (tn 2 ; y (tn 2 ))
tn
2) (
tn
(62)
1) :
Here if we put this polynomial into above equation, we write such as = y0 +
2)
2 ( t)
(
y n+1
tn
of
Pn ( )
1 F (tn ; y (tn )) AB ( )
(63)
p ro
9 8 F tj 2 ; y j 2 > > > > > > j 1 j 2 > n Z tj+1 > = < + F (tj 1 ;y ) F (tj 2 ;y ) ( X t ) j 2 t + (tn+1 j j 1 j 2 F t ;y 2F t ;y +F t ;y ( ) ( ) ( ) j j 1 j 2 > > AB ( ) ( ) j=2 tj > > + > > 2( t)2 > > ; : ( tj 2 ) ( tj 1 )
If we order the above equality, we obtain y0 +
+
1 F (tn ; y (tn )) AB ( ) 8 R tj+1 > F tj 2 ; y j 2 (tn+1 > tj > > R > F (tj 1 ;y j 1 ) F (tj t > + tjj+1 n > < t X
AB ( ) ( )
and write the following
+
2 ;y
j
2
n X F tj
1; y
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (tn+1
2
Z
tj+1
(tn+1
)
d
)
1
( tj 2 ) (tn+1 ) d > R j j 1 )+F (tj tj+1 F (tj ;y ) 2F (tj 1 ;y > + tj > > 2( t)2 > : ( tj 2 ) ( tj 1 ) (tn+1 )
1 F (tn ; y (tn )) AB ( ) n X + F tj 2 ; y j AB ( ) ( ) j=2
= y0 +
1
)
j=2 > >
urn
y n+1
Pr e-
=
al
y n+1
1
2 ;y
j
1
2
d
Jo
1
d :
(64)
d
tj
j 1
F tj
2; y
j 2
t )
n X F tj ; y j
AB ( ) ( ) j=2 Z tj+1 ( tj 2 ) (
9 > > > > > > > =
> > ) > > > > > ;
1
d
(65)
tj
+
)
tj
2F tj
1 ) (tn+1
tj
12
1; y
j 1 2
2 ( t) )
1
d :
+ F tj
2; y
j 2
Journal Pre-proof
If we calculate the above integrals Z tj+1 (tn+1
d =
( t)
[(n
j + 1)
(n
j) ]
tj
+1
tj+1
tj
tj Z
2 ) (tn+1
1
)
d =
( t) ( + 1)
+2
tj+1
(
tj
tj
2) (
2
6 (n 6 4
tj
j + 1) (n
j)
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7 2 5 2 (n j) + (5 + 10) (n j) 2 +6 + 18 + 12 1 ) (tn+1
1 F (tn ; y (tn )) AB ( ) n X ( t) + F tj AB ( ) ( + 1) j=2 y0 +
2; y
d =
j 2
[(n
j + 1)
Pr e-
=
1
)
and replace them into above scheme, we get the following y n+1
(n j + 1) (n j + 3 + 2 ) (n j + 1) (n j + 3 + 3 ) (66)
of
(
p ro
Z
1
)
(n
j) ]
n
+
X ( t) F tj AB ( ) ( + 2) j=2
j 1
F tj
2; y
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) F tj ; y j 2F tj 1 ; y j + 2AB ( ) ( + 3) j=2 2 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
j 2
(67) 1
1 t AB ( ) n
urn
=
y0 +
1
2; y
j 2
7 7: 5
Thus we have the following approximation y n+1
+ F tj 3
al
(n
1; y
(tn ; y (tn )) n
+
X ( t) t AB ( ) ( + 1) j=2 j
Xh ( t) t AB ( ) ( + 2) j=2 j
1 2f
tj
2; y
j 2
[(n
j + 1)
(n
j) ]
n
+
1 1f
tj
1; y
j 1
tj
1 2f
2; y
j 2
i
(n
Jo
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n h X ( t) t 1 f tj ; y j 2tj 11 f tj + 2AB ( ) ( + 3) j=2 j 22 2 2 (n j) + (3 + 10) (n j) (n j + 1) 66 +2 2 + 9 + 12 66 2 44 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
tj
13
(68) 1; y
j 1
33
77 77 : 55
+ tj
1 2f
tj
2; y
j 2
i
Journal Pre-proof
8
New method with Caputo fractal-fractional derivative
Let us handle the following Cauchy problem with Caputo fractal-fractional derivative FFP Dt ; 0
y (t) = f (t; y (t))
(69)
and if we integrate above equation, we get
1
Here we shall take as F (t; y (t)) = t y (t)
( )
Z
t 1
of
y (0) =
f ( ; y ( )) (t
f (t; y (t)) : Thus we have Z
1 ( )
y (0) =
1
)
0
d :
(70)
p ro
y (t)
t
F ( ; y ( )) (t
0
1
)
d
(71)
where the function F is non-linear. Here we will provide numerical scheme for solving the above equation. At the point tn+1 = (n + 1) t; we have 1 ( )
y (0) =
Also we obtain
Z
F ( ; y ( )) (tn+1
0
n
y (tn+1 ) = y (0) +
tn+1
1 X ( ) j=2
Z
1
)
Pr e-
y (tn+1 )
tj+1
F ( ; y ( )) (tn+1
)
d :
(72)
1
(73)
d :
tj
al
Thus replacing Newton polynomial into the original equation, we get the following 8 9 F tj 2 ; y j 2 > > > > > > j 1 j 2 > > Z F t ;y F t ;y ( ) ( ) j 1 j 2 n < = t j+1 X + ( t ) 1 j 2 n+1 0 t (tn+1 y =y + j j 1 j 2 > + F (tj ;y ) 2F (tj 1 ;y 2 )+F (tj 2 ;y ) > ( ) j=2 tj > > > > 2( t) > > : ; ( tj 2 ) ( tj 1 )
or we write
urn
y n+1 = y 0 +
8 R tj+1 1 > F tj 2 ; y j 2 (tn+1 ) d > tj > > j 1 j 2 R > F t ;y F t ;y ( ) ( ) t n j 1 j 2 ( tj 2 ) (tn+1 ) 1 X < + tjj+1 t R tj+1 F (tj ;yj ) 2F (tj 1 ;yj 1 )+F (tj 2 ;yj 2 ) ( ) j=2 > > + tj > > 2( t)2 > : 1 ( tj 2 ) ( tj 1 ) (tn+1 ) d
1
)
1
9 > > > > > =
d
> > > > > ;
Thus we have the following equality
Z
n
=
y0 +
1 X F tj ( ) j=2
Jo
y n+1
n
1 X F tj + ( ) j=2
1; y
n
+
Z
j 2; y
1 X F tj ; y j ( ) j=2
2
tj+1
(tn+1
tj
2) (
1
(74)
:
(75)
d
tj
j 1
F tj
2; y
Z
j 2
t 2F tj
1; y
j 1
2 ( t) tj
tj+1
(
tj
tj
+ F tj
2
tj+1
(
)
d
1 ) (tn+1
tj
14
)
1
d :
2; y
j 2
2 ) (tn+1
)
1
d
(76)
Journal Pre-proof
We can calculate the above integrals as follows; Z tj+1 ( t) 1 (tn+1 ) d = [(n
(n
j) ]
+1
tj+1
tj
tj Z
2 ) (tn+1
1
)
d =
(n
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) +2
tj+1
(
tj
( t) ( + 1)
tj
2) (
2
6 (n 6 4
tj
j + 1) (n
j)
( t) ( + 1) ( + 2) 3 2 2 (n j) + (3 + 10) (n j) 7 +2 2 + 9 + 12 7: 2 5 2 (n j) + (5 + 10) (n j) +6 2 + 18 + 12 1 ) (tn+1
1
)
d =
(77)
of
(
p ro
Z
j + 1)
tj
When replacing them into above equality, we have the following n
y n+1
X ( t) F tj ( + 1) j=2
y0 +
=
2; y
j 2
n
X ( t) F tj ( + 2) j=2
j 1
j + 1)
Pr e-
+
[(n
F tj
2; y
j) ]
j 2
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n X ( t) F tj ; y j 2F tj 1 ; y j 1 + F tj 2 ; y j + 2 ( + 3) j=2 2 3 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) 7 +2 2 + 9 + 12 6 7: 2 4 5 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
Thus, we have
al
(n
1; y
(n
(78) 2
n
=
y0 +
( t) X t ( + 1) j=2 j
urn
y n+1
( t) X h t ( + 2) j=2 j
1 2f
tj
2; y
j 2
[(n
j + 1)
(n
n
+
1 1f
tj
1; y
j 1
tj
1 2f
tj
2; y
i
(n
Jo
j + 1) (n j + 3 + 2 ) (n j) (n j + 3 + 3 ) n ( t) X h 1 + t f tj ; y j 2tj 11 f tj 1 ; y j 2 ( + 3) j=2 j 2 2 2 (n j) + (3 + 10) (n j) 6 (n j + 1) +2 2 + 9 + 12 6 2 4 2 (n j) + (5 + 10) (n j) (n j) +6 2 + 18 + 12
j 2
j) ]
15
(79) 1
+ tj 3
7 7: 5
1 2f
tj
2; y
j 2
i
Journal Pre-proof
9
Numerical Illustrations and Simulation
Example 1. We consider the following d dt y (t)
= sin t:y (t) y (0) = exp ( 1) :
(80)
The exact solution is given as
of
y (t) = exp ( cos t) : We compare the exact solution with Adams-Bashforth and our proposed method . Here we consider the equivalent Adams-Bashforth 3 t f (tn ; y (tn )) 2
We consider the in…nite norm
t f (tn 2
1 ; y (tn 1 )) :
p ro
y n+1 = y n +
kf k1 = sup jf (t)j : t2 t
The numerical comparison are depicted in Figure 1.The error between the exact and numerical solution for this example t = 0:01; t = 0 to 2000:
kyex
But if
t = 0:001; we have t = 0 to 2000 kyex
while
=
5:7536
5
10
yprop k1
=
3:1197
yAdams k1 = 5:7536 yprop k1 = 4:2920
10
10
10
10
6
:
7
:
Jo
urn
al
kyex
yAdams k1
Pr e-
kyex
Figure 1. Numerical comparison with classical derivative.
16
Journal Pre-proof
Example 2. We consider the following simple equation y0 (t) =
2y (t) :
(81)
The exact solution is 2t
:
The comparisons are given in Figure 2 below. Also, the error for kyex
kyex
yprop k1
=
4:3875
10
=
2:6008
10
5
6
t = 0:001; we have the following error kyex
yAdams k1 = 6:2208 yprop k1 = 2:66
10
8
10
9
:
urn
al
Pr e-
kyex
:
p ro
For
yAdams k1
t = 0:01 is given as
of
y (t) = e
Figure 2. Numerical comparison with classical derivative.
Jo
Example 3. We consider the following non-homogeneous ordinary di¤erential equation y0 (t) = 2y + 3 y (0) = 1:
(82)
The exact solution is given as
5 2t 3 e : 2 2 We present the comparison in Figure 3 below. Thus for t = 0:0001, we have the following error y (t) =
kyex
yAdams k1 = 2:4853 17
10
8
Journal Pre-proof
while the proposed method has yprop k1 = 1:4415
10
11
:
Pr e-
p ro
of
kyex
Figure 3. Numerical comparison with classical derivative.
Example 4. We consider the following example
The exact solution is
al
y0 (t) =
y (t) =
cos (2t) y 2 (t) :
(83)
2 : 2 + sin 2t
urn
The following comparison can be obtained kyex
while
kyex
t = 0:01: For
yprop k1 = 3:1458
4
10 5
10
t = 0:0001; we have
Jo
for
yAdams k1 = 5:6138
kyex
yAdams k1 = 1:6267
kyex
yprop k1 = 1:6659
10
9
and the proposed method
18
10
12
:
Pr e-
p ro
of
Journal Pre-proof
Figure 4. Numerical comparison with classical derivative.
Example 5. We next consider the following example
y0 (t) = 5ty (t)
with initial condition
(84)
y (0) = 1:
al
The exact solution is given as y (t) = exp
5 2 t : 2
urn
We present the comparison in Figure 5. The comparison of error for kyex
yAdams k1 = 5:9306
while the error of the proposed method is given as yprop k1 = 0:0503:
Jo
kyex
19
t = 0:001; is given as
Pr e-
p ro
of
Journal Pre-proof
Figure 5. Numerical comparison with classical derivative.
Example 6. Now we consider the following chaotic problem =
(y (t)
= x (t) =
For simplicity, we shall take as f1 (x; y; z; t) = (y (t) y (t) :So, we can write the above as follows
urn
d x (t) dt d y (t) dt d z (t) dt
h)
y (t) + z (t)
(85)
y (t) :
al
d x (t) dt d y (t) dt d z (t) dt
h) ; f2 (x; y; z; t) = x (t) y (t)+z (t) and f3 (x; y; z; t) =
= f1 (x; y; z; t) = f2 (x; y; z; t)
(86)
= f3 (x; y; z; t) :
Jo
As we did before, we can have the following x (tn+1 )
y (tn+1 )
x (tn ) y (tn )
= =
Z
tn+1
tn Z tn+1
f1 (x; y; z; ) d f2 (x; y; z; ) d
tn tn+1
z (tn+1 )
z (tn )
=
Z
tn
20
f3 (x; y; z; ) d :
(87)
Journal Pre-proof
After putting Newton polynomial into above system and doing some manipulations, we get xn+1
= xn +
5 f1 t n 12
n 2 n 2 n 2 ;y ;z 2; x
23 f1 (tn ; xn ; y n ; z n ) t 12 5 = y n + f2 tn 2 ; xn 2 ; y n 12 23 + f2 (tn ; xn ; y n ; z n ) t 12 5 n = z + f3 tn 2 ; xn 2 ; y n 12 23 + f3 (tn ; xn ; y n ; z n ) t 12
t
4 f1 tn 3
n 1 n 1 n 1 ;y ;z 1; x
t
n 1 n 1 n 1 ;y ;z 1; x
t
+
; zn
2
t
4 f2 tn 3
2
; zn
2
t
4 f3 t n 3
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z n+1
2
n 1 n 1 n 1 ;y ;z 1; x
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y n+1
For the numerical solution of this problem, we take as h =
b sin
x(t) 2a
+d ;
= 10:82;
(88)
t
= 14:286; a =
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13; b = 0:11; c = 7; d = 2. Also we take the initial conditions as x (0) = 1; y (0) = 1; z (0) = 0: Numerical simulations are depicted in Figure 6,7,8 and 9.
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Figure 6. Numerical simulation for chaotic problem with classical derivative in yz coordinates.
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Figure 7. Numerical simulation for chaotic problem with classical derivative in xz coordinates.
Figure 8. Numerical simulation for chaotic problem with classical derivative in xy coordinates. 22
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Figure 9. Numerical simulation for chaotic problem with classical derivative in xyz coordinates.
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We shall take same problem with initial conditions x (0) = 11; y (0) = 114:286: Numerical simulations are depicted in Figure 10,11,12 and 13.
23
1; z (0) = 10: Also
=
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Figure 10. Numerical simulation for chaotic problem with classical derivative in yz coordinates.
Figure 11. Numerical simulation for chaotic problem with classical derivative in xz coordinates. 24
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Figure 12. Numerical simulation for chaotic problem with classical derivative in xy coordinates.
Figure 13. Numerical simulation for chaotic problem with classical derivative in xyz coordinates.
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Example 7. For fractional Caputo case, we consider the following equation C 0 Dt
y (t) = t
(89)
with initial condition y (0) = 1: The exact solution of such equation is t ( + 1) : ( + + 1) We plot the above solution against the numerical counterpart to have the following solution. For the numerical solution of this problem, we shall take as = 0:87; = 0:9; h = 0:01: The error of the proposed method is given as kyex yprop k1 = 4:9855 10 4 :
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y (t) =
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Numerical simulation is depicted in Figure 14.
10
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Figure 14. Numerical simulation with Caputo derivative.
Conclusion
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Modeling real world problems using mathematical models have been the focus of many researchers working in di¤erent …eld of sciences in the last years. The very reason for this attraction is the fact that such models are used for prediction, control and analysis. Noting that real world problems are complex, thus their counterparts mathematical models are also complex. For those depicting linear problems, their associated mathematical models can be solved analytically using some techniques like Green function, Laplace transform, method of separation of variable and others. Nevertheless those depicting nonlinear problems, their associated mathematical models cannot sometime be solved analytically, thus numerical techniques are requested. Adams-Bashforth, a powerful numerical technique to solve nonlinear equations has been used in many …elds. The method is based on the Lagrange polynomial, but it was reported in some published papers such polynomial has limitations. In this paper, we suggested the use of Newton polynomial that seems to be more accurate. Con‡ict of interest The authors declare that there is no con‡ict of interests regarding the publication of this paper. 26
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References [1] Bonyah E., Chaos in a 5-D hyperchaotic system with four wings in the light of non-local and non-singular fractional derivatives, Chaos, Solitons & Fractals, 116, 2018, 316-331. [2] A. Atangana and D. Baleanu, New fractional derivative with non-local and non-singular kernel, Thermal Sci., 20 (2016), 757-763.
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[3] Owolabi K.M., Mathematical modelling and analysis of love dynamics: A fractional approach, Physica A: Statistical Mechanics and its Applications, 525, 2019, 849-865. [4] Igret Araz S., Numerical analysis of a new Volterra integro-di¤erential equation involving fractalfractional operators, Chaos, Solitons&Fractals, 2020, 130, (109396).
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[5] Tou…k M., Atangana A., New numerical approximation of fractional derivative with non-local and nonsingular kernel: Application to chaotic models, The European Physical Journal Plus, 132 (10), 2017, 444. [6] Ganji RM., Jafari H., A numerical approach for multi-variable order di¤erential equations using Jacobi polynomials, International Journal of Applied and Computational Mathematics, 2019, 5:34.
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[7] Zhou, Y., Manimaran, J., Shangerganesh, L., Debbouche A., Weakness and Mittag–Le- er Stability of Solutions for Time-Fractional Keller–Segel Models. International Journal of Nonlinear Sciences and Numerical Simulation, 19(7-8), 2018, 753-761. [8] Jafari H., Babaei A., Banihashemi S., A Novel Approach for Solving an Inverse Reaction–Di¤usion– Convection Problem, Journal of Optimization Theory and Applications, 183 (2), 2019, 688-704. [9] Peng L., Zhou Y., Debbouche A., Approximation techniques of optimal control problems for fractional dynamic systems in separable Hilbert spaces, Chaos, Solitons & Fractals, 118, 2019, 234-241. [10] Werner W., Polynomial interpolation: Lagrange versus Newton, Mathematics and Computation, 43 (1984), 205-217.
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[11] Fred T. Krogh, E¢ cient algorithms for polynomial interpolation and numerical di¤erentiation, Mathematics and Computation. 24 (1970), 185–190. [12] Yang Y., Gordon S.P., Visualizing and understanding the components of Lagrange and Newton Interpolation, Problems, Resources, and Issues in Mathematics Undergraduate Studies, 26 (1), 2015, 39-52.
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[13] Dimitrov D.K., Philipps G.M., A note on convergence of Newton interpolating polynomials, Journal of Computational and Applied Mathematics, 51, 1, 1994, 127-130. [14] Srivastava R.B., Shukla S., Numerical accuracies of Lagrange’s and Newton polynomial interpolation: Numerical accuracies of Interpolation formulas, LAP LAMBERT Academic Publishing, 2012. [15] Zhang, Tong & Jin, JiaoJiao & Jiang, Tao, The decoupled Crank–Nicolson/Adams–Bashforth scheme for the Boussinesq equations with nonsmooth initial data, Applied Mathematics and Computation, Elsevier, 337(C), 2018, 234-266.
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[16] Jain S., Numerical analysis for the fractional di¤usion and fractional Buckmaster equation by the twostep Laplace Adam-Bashforth method, The European Physical Journal Plus, 2018, 133:19.
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New numerical scheme based on Newton polynomial. New numerical scheme for ordinary fractional differential equations. New numerical scheme for ordinary fractal-fractional differential equations. Fractional and integral equations.