# Tau-Path Following Method for Solving the Riccati Equation with Fractional Order

Authors: Muhammed I. Syam; Hani I. Siyyam; Ibrahim Al-Subaihi
Journal: Journal of Computational Methods in Physics
 (2014)

---

## Abstract

A formulation for the fractional Legendre functions is constructed to find the solution of the fractional Riccati equation. The fractional derivative is described in the Caputo sense. The method is based on the Tau Legendre and path following methods. Theoretical and numerical results are presented. Analysis for the presented method is given.


---

## Body

## 1. Introduction

Recently, many papers on fractional boundary value problems have been studied extensively. Several forms of them have been proposed in standard models, and there has been significant interest in developing numerical schemes for their solutions. Several numerical techniques are used to solve such problems such as Laplace and Fourier transforms [1, 2], Adomian decomposition and variational iteration methods [3, 4], eigenvector expansion [5], differential transform and finite differences methods [6, 7], power series method [8], collocation method [9], and wavelet method [10, 11]. Many applications of fractional calculus on various branches of science such as engineering, physics, and economics can be found in [12, 13]. Considerable attention has been given to the theory of fractional ordinary differential equations and integral equations [14, 15]. Additionally, the existence of solutions of ordinary and fractional boundary value problems using monotone iterative sequences has been investigated by several authors [16–20].We consider the Riccati equation with fractional orders of the form(1)
a
(
x
)
D
α
y
(
x
)
+
b
(
x
)
y
(
x
)
+
c
(
x
)
y
2
(
x
)
=
g
(
x
)
,
x
∈
(
0,1
)
,
0
<
α
<
1
,
(2)
y
(
0
)
=
y
ini
,
where a
,
b
,
c, and g are continuous functions on [
0,1
] and  y
ini is a constant.Riccati equation with fractional order has been discussed by many researchers using different techniques such as collocation method based on Muntz polynomials [21], homotopy perturbation method [22], and series solution method [23].In this paper we study the Tau-path following method for solving the Riccati equation with fractional order. We organize this paper as follows. In Section2, we present basic definitions and results of fractional derivatives. We extend basic results to path following method for the fractional case. In Section 3, we introduce the fractional-order Legendre Tau method with path following method for solving the Riccati equation with fractional order. In Section 4, we present some numerical results to illustrate the efficiency of the presented method. Finally, we conclude with some comments in the last section.


## 2. Preliminaries

In this section, we review the definition and some preliminary results of the Caputo fractional derivatives, as well as, the definition of the fractional-order Legendre functions and their properties.Definition 1.
The Riemann-Liouville fractional integral operatorI
α of order α
>
0 on the usual Lebesgue space L
1
[
0,1
] is given by
(3)
I
α
f
(
x
)
=
1
Γ
(
α
)
∫
0
x
f
(
t
)
(
x
-
t
)
1
-
α
d
t
,
I
0
f
(
x
)
=
f
(
x
)
,
where Γ
(
ζ
)
=
∫
0
∞
t
ζ
-
1
e
-
t
d
t is the Euler Gamma function, see [12, 24].For anyf
∈
L
1
[
0,1
]
,
α
,
β
⩾
0, and γ
>
-
1, the following properties hold:(1)
I
α exists for any x
∈
[
0,1
],
(2)
I
α
I
β
=
I
α
+
β,
(3)
I
α
I
β
=
I
β
I
α,
(4)
I
α
x
γ
=
(
Γ
(
γ
+
1
)
/
Γ
(
α
+
γ
+
1
)
)
x
α
+
γ.Definition 2.
The Caputo fractional derivative of orderα is defined by
(4)
D
α
f
(
x
)
=
I
n
-
α
D
n
f
(
x
)
=
1
Γ
(
n
-
α
)
∫
0
x
f
(
n
)
(
t
)
(
x
-
t
)
α
-
n
+
1
d
t
,
provided that the integral exists, where n
=
[
α
]
+
1 and [
α
] is the integer part of the positive real number α
,
x
>
0.The Caputo fractional derivative satisfies the following properties forf
∈
L
1
[
0,1
] and α
,
β
⩾
0:(1)
D
α
I
α
f
(
x
)
=
f
(
x
),
(2)
I
α
D
α
f
(
x
)
=
f
(
x
)
-
∑
k
=
0
n
-
1
‍
f
(
k
)
(
0
+
)
(
x
k
/
k
!
),
(3)
D
α
D
β
f
(
x
)
=
D
β
D
α
f
(
x
)
=
D
α
+
β
f
(
x
)
,
x
>
0,
(4)
D
α
c
=
0, where c is constant,
(5)
D
α
x
γ
=
{
0
,
γ
<
α
,
γ
∈
{
0,1
,
2
,
…
}
(
Γ
(
γ
+
1
)
/
Γ
(
γ
-
α
+
1
)
)
x
γ
-
α
,
otherwise
},
(6)
D
α
(
∑
k
=
0
m
‍
c
i
f
i
(
x
)
)
=
∑
k
=
0
m
‍
c
i
D
α
f
i
(
x
), where c
1
,
c
2
,
…
,
c
m are constants.The basic concept of this paper is the Legendre polynomials. For this reason, we study some of their properties.Definition 3.
The Legendre polynomials{
ℒ
k
(
x
)
:
k
=
0,1
,
2
,
…
} are the eigenfunctions of the Sturm-Liouville problem
(5)
(
(
1
-
x
2
)
ℒ
k
′
(
x
)
)
′
+
k
(
k
+
1
)
ℒ
k
(
x
)
=
0
,
x
∈
[
-
1,1
]
.Among the properties of the Legendre polynomials we list the following properties:(1)
∫
-
1
1
ℒ
i
(
x
)
ℒ
j
(
x
)
d
x
=
(
2
/
(
2
i
+
1
)
)
δ
i
j, where δ
i
j
=
{
0
,
i
≠
j
1
,
i
=
j
},
(2)
ℒ
i
+
1
(
x
)
=
(
(
2
i
+
1
)
/
(
i
+
1
)
)
x
ℒ
i
(
x
)
-
(
i
/
(
i
+
1
)
)
ℒ
i
-
1
(
x
), for i
⩾
1,
(3)
ℒ
i
(
±
1
)
=
(
±
1
)
i.In order to use these polynomials on the interval[
0,1
], we define the shifted Legendre polynomials by S
i
(
z
)
=
ℒ
i
(
2
z
-
1
). Using the change of variable x
=
2
z
-
1, S
i
(
z
) has the following properties:(1)
∫
0
1
S
i
(
z
)
S
j
(
z
)
d
z
=
(
1
/
(
2
i
+
1
)
)
δ
i
j,
(2)
S
i
+
1
(
z
)
=
(
(
2
i
+
1
)
/
(
i
+
1
)
)
(
2
z
-
1
)
S
i
(
z
)
-
(
i
/
(
i
+
1
)
)
S
i
-
1
(
z
), for i
⩾
1,
(3)
S
i
(
0
)
=
(
-
1
)
i and S
i
(
1
)
=
1.The analytic closed form of the shifted Legendre polynomials of degreei is given by
(6)
S
i
(
z
)
=
∑
k
=
0
i
(
-
1
)
i
+
k
(
i
+
k
)
!
(
i
-
k
)
!
z
k
(
k
!
)
2
.One of the common and efficient methods for solving fractional differential equations of orderα
>
0is using the series expansion of the form ∑
k
=
0
i
‍
c
k
t
α
k. For this reason, we define the fractional-order Legendre function by F
i
α
(
t
)
=
S
i
(
t
α
). Using the properties of the shifted Legendre polynomials and the change of variable z
=
t
α, it is easy to show that(1)
(
(
t
-
t
1
+
α
)
F
i
α
′
(
t
)
)
′
+
i
(
i
+
1
)
α
2
t
α
-
1
F
i
α
(
t
)
=
0
,
t
∈
(
0,1
),
(2)
F
i
α
(
t
)
=
(
(
2
i
+
1
)
/
(
i
+
1
)
)
(
2
t
α
-
1
)
F
i
α
(
t
)
-
(
i
/
(
i
+
1
)
)
F
i
-
1
α
(
t
), for i
⩾
1,
(3)
F
0
α
(
t
)
=
1 and F
1
α
(
t
)
=
2
t
α
-
1,
(4)
F
i
α
(
0
)
=
(
-
1
)
i and F
i
α
(
1
)
=
1.In addition,{
F
i
α
(
t
)
:
i
=
0,1
,
2
,
…
} are orthogonal functions with respect to the weight function w
(
t
)
=
t
α
-
1 on (
0,1
) with
(7)
∫
0
1
F
i
α
(
t
)
F
j
α
(
t
)
w
(
t
)
d
t
=
1
(
2
i
+
1
)
α
δ
i
j
.The closed form ofF
i
α
(
t
) is given by
(8)
F
i
α
(
t
)
=
∑
k
=
0
i
(
-
1
)
i
+
k
(
i
+
k
)
!
(
i
-
k
)
!
t
α
k
(
k
!
)
2
.Using properties (4) and (5) of the Caputo fractional derivative, one can see that(9)
D
α
F
i
α
(
t
)
=
∑
k
=
1
i
(
-
1
)
i
+
k
(
i
+
k
)
!
(
i
-
k
)
!
(
k
!
)
2
×
Γ
(
k
α
+
1
)
Γ
(
(
k
-
1
)
α
+
1
)
t
(
k
-
1
)
α
.In the next theorem, we state one of the main properties of the fractional Legendre functions which will be used later in this paper.Theorem 4.
For any nonnegative integersp and q,
(10)
F
p
α
(
t
)
F
q
α
(
t
)
=
∑
k
=
0
[
(
p
+
q
)
/
2
]
A
2
k
F
p
+
q
-
2
k
α
(
t
)
,
where
(11)
A
2
k
=
(
(
(
p
+
q
-
k
)
!
)
2
(
2
p
+
2
q
-
4
k
+
1
)
(
2
k
)
!
(
2
p
-
2
k
)
!
×
(
2
q
-
2
k
)
!
2
p
+
q
-
k
(
(
p
+
q
-
k
)
!
)
2
)
×
(
(
(
p
-
q
)
!
)
2
(
(
q
-
k
)
!
)
2
2
p
+
q
-
k
(
2
p
+
2
q
-
2
k
+
1
)
(
k
!
)
2
×
(
(
p
-
q
)
!
)
2
(
(
q
-
k
)
!
)
2
(
2
p
+
2
q
-
2
k
)
!
)
-
1
.Proof.
For any nonnegative integersp and q,
(12)
ℒ
p
(
x
)
ℒ
q
(
x
)
=
∑
k
=
0
[
(
p
+
q
)
/
2
]
A
2
k
α
ℒ
p
+
q
-
2
k
(
x
)
,
where
(13)
A
2
k
=
(
(
(
p
+
q
-
k
)
!
)
2
(
2
p
+
2
q
-
4
k
+
1
)
(
2
k
)
!
(
2
p
-
2
k
)
!
×
(
2
q
-
2
k
)
!
2
p
+
q
-
k
(
(
p
+
q
-
k
)
!
)
2
)
×
(
(
(
p
-
q
)
!
)
2
(
(
q
-
k
)
!
)
2
2
p
+
q
-
k
(
2
p
+
2
q
-
2
k
+
1
)
(
k
!
)
2
×
(
(
p
-
q
)
!
)
2
(
(
q
-
k
)
!
)
2
(
2
p
+
2
q
-
2
k
)
!
)
-
1
.
For the proof of this case, see [25]. Using the change of variables x
=
2
z
-
1 and z
=
t
α, we obtain the result of the theorem.Another important result which will facilitate applying the Tau method for fractional case is given in Theorem5.Theorem 5.
Letu
∈
C
[
0,1
] and u
′
(
t
) be a piecewise continuous function on [
0,1
]. Then, u
(
t
) can be written in the infinite expansion as u
(
t
)
=
∑
k
=
0
∞
u
k
F
k
α
(
t
), where
(14)
u
k
=
(
2
i
+
1
)
α
∫
0
1
u
(
t
)
F
k
α
(
t
)
w
(
t
)
d
t
,
and w
(
t
)
=
t
α.Proof.
Ifu
∈
C
1
[
-
1,1
] and u
′′
(
x
) is a piecewise continuous function on [
-
1,1
],  ∑
k
=
0
∞
v
k
ℒ
k
(
x
) converges uniformly to u
(
x
) on [
-
1,1
], see [26–28]. Let h
:
[
0,1
]
→
[
-
1,1
] by
(15)
h
(
t
)
=
2
t
α
-
1
.
Since h
(
t
) is a bijective continuous function, ∑
k
=
0
∞
u
k
F
k
α
(
t
) converges uniformly to u
(
t
) on [
0,1
]. The value of u
k follows from the orthogonality relation of {
F
i
α
(
t
)
:
i
=
0,1
,
2
,
…
} with respect to the weight function w
(
t
)
=
t
α
-
1 on [
0,1
]
Theorem6 gives the relation between the coefficient of the series solution of D
α
u
(
t
) and the coefficients of the series expansion of u
(
t
).Theorem 6.
Letu
∈
C
1
[
0,1
] and u
′′
(
t
) be a piecewise continuous function on [
0,1
]. Then, ∑
k
=
0
∞
u
k
(
α
)
F
k
α
(
t
) converges uniformly on [
0,1
] to D
α
u
(
t
), where
(16)
u
k
(
α
)
=
∑
j
=
k
+
1
∞
a
j
k
u
j
,
a
j
k
=
(
2
k
+
1
)
α
∫
0
1
D
α
F
j
α
(
t
)
F
k
α
(
t
)
w
(
t
)
d
t
f
o
r
k
=
0,1
,
2
,
…
,
j
=
k
+
1
,
k
+
2
,
…
.Proof.
LetS
n
(
t
)
=
∑
k
=
0
n
u
k
F
k
α
(
t
) for n
=
0,1
,
2
,
…. From Theorem 5, S
n
(
t
) converges uniformly to u
(
t
) on [
0,1
]
.Since u
∈
C
1
[
0,1
] and u
′′
(
t
) is a piecewise continuous function on [
0,1
],
(17)
d
d
t
(
Lim
n
→
∞
S
n
(
t
)
)
=
Lim
n
→
∞
(
d
d
t
S
n
(
t
)
)
and (
d
/
d
t
)
S
n
(
t
) converges uniformly to (
d
/
d
t
)
u
(
t
) on [
0,1
]. Thus, ∫
0
x
(
S
n
′
(
t
)
/
(
x
-
t
)
α
)
d
t converges uniformly to ∫
0
x
(
u
′
(
t
)
/
(
x
-
t
)
α
)
d
t on [
0,1
] which gives the result of the theorem. The value of a
j
k follows from the orthogonality relation of {
F
i
α
(
t
)
:
i
=
0,1
,
2
,
…
} with respect to the weight function w
(
t
)
=
t
α
-
1 on [
0,1
].


## 3. Fractional-Order Legendre Tau-Path Following Method

In this section, we present a numerical method for solving problem (1)-(2). We use the fractional-order Legendre Tau method to discretize problem (1)-(2). The result is a nonlinear system. The initial guess that is used in the standard methods for solving the produced nonlinear system is one of the challenges. To overcome this problem, we use the path following method. Approximate the solution y
(
x
), a
(
x
), b
(
x
), c
(
x
), and g
(
x
) in terms of the fractional-order Legendre functions as follows:
(18)
Y
N
(
x
)
=
∑
k
=
0
N
+
1
y
k
F
k
α
(
x
)
,
a
N
(
x
)
=
∑
k
=
0
N
+
1
a
k
F
k
α
(
x
)
,
b
N
(
x
)
=
∑
k
=
0
N
+
1
b
k
F
k
α
(
x
)
,
c
N
(
x
)
=
∑
k
=
0
N
+
1
c
k
F
k
α
(
x
)
,
g
N
(
x
)
=
∑
k
=
0
N
+
1
g
k
F
k
α
(
x
)
.
Thus, D
α
y
(
x
) can be approximated by
(19)
D
α
Y
N
(
x
)
=
∑
k
=
0
N
y
k
(
α
)
F
k
α
(
x
)
,
where y
k
(
α
) is given by Theorem 6. For Y
N, the residual is given by
(20)
R
(
Y
N
)
=
a
N
(
x
)
D
α
Y
N
(
x
)
+
b
N
(
x
)
Y
N
(
x
)
+
c
N
(
x
)
Y
N
2
(
x
)
-
g
N
(
x
)
which can be written as
(21)
R
(
Y
N
)
=
(
∑
k
=
0
N
+
1
a
k
F
k
α
(
x
)
)
(
∑
k
=
0
N
y
k
(
α
)
F
k
α
(
x
)
)
+
(
∑
k
=
0
N
+
1
b
k
F
k
α
(
x
)
)
(
∑
k
=
0
N
+
1
y
k
F
k
α
(
x
)
)
+
(
∑
k
=
0
N
+
1
c
k
F
k
α
(
x
)
)
(
∑
k
=
0
N
+
1
y
k
F
k
α
(
x
)
)
2
-
∑
k
=
0
N
+
1
g
k
F
k
α
(
x
)
or
(22)
R
(
Y
N
)
=
∑
k
=
0
2
N
+
1
∑
j
=
0
k
a
j
y
k
-
j
(
α
)
F
j
α
(
x
)
F
k
-
j
α
(
x
)
+
∑
k
=
0
2
N
+
2
∑
j
=
0
k
b
j
y
k
-
j
F
j
α
(
x
)
F
k
-
j
α
(
x
)
+
∑
k
=
0
3
N
+
3
∑
i
=
0
k
∑
j
=
0
k
-
i
c
i
y
j
y
k
-
i
-
j
F
j
α
(
x
)
F
k
-
i
-
j
α
(
x
)
F
i
α
(
x
)
-
∑
k
=
0
N
+
1
g
k
F
k
α
(
x
)
,
where a
j
=
0
,
b
j
=
0
,
y
j
=
0
,
c
j
=
0 for j
≥
N
+
2, and y
j
(
α
)
=
0for j
≥
N
+
1. Using Theorem 4, we can rewrite the residual as
(23)
R
(
Y
N
)
=
∑
k
=
0
2
N
+
1
∑
j
=
0
k
∑
l
=
0
[
k
/
2
]
A
l
a
j
y
k
-
j
(
α
)
F
k
-
2
l
α
(
x
)
+
∑
k
=
0
2
N
+
2
∑
j
=
0
k
∑
l
=
0
[
k
/
2
]
A
l
b
j
y
k
-
j
F
k
-
2
l
α
(
x
)
+
∑
k
=
0
3
N
+
3
∑
i
=
0
k
∑
j
=
0
k
-
i
∑
l
=
0
[
(
k
-
i
)
/
2
]
∑
m
=
0
[
(
k
-
2
l
)
/
2
]
A
m
A
l
c
i
y
j
y
k
-
i
-
j
F
k
-
2
l
-
2
m
α
(
x
)
-
∑
k
=
0
N
+
1
g
k
F
k
α
(
x
)
.
Since we are interested in the first n terms only, we ignore higher order terms and we rewrite the residual as
(24)
R
(
Y
N
)
=
∑
k
=
0
N
d
k
F
k
α
(
x
)
.
Orthogonalize the residual with respect to the fractional-order Legendre function as follows
(25)
∫
0
1
R
(
Y
N
)
F
j
α
(
x
)
w
(
x
)
d
x
=
0
,
for
j
=
0
:
N
,
where w
(
x
)
=
x
α
-
1. Therefore, (25) leads to the elementwise equation
(26)
d
j
=
1
(
2
k
+
1
)
α
,
for
j
=
0
:
N
.
Therefore, we can write (26) as a system of N nonlinear equations in (
N
+
1
) unknowns as
(27)
Ω
(
Y
)
=
R
1
,
where Y
=
[
y
0
y
1
⋯
y
N
+
1
]
T
,
R
1
=
(
1
/
α
)
[
1
1
/
3
⋯
1
/
(
2
N
+
1
)
]
T, and T means the transpose of the matrix.From the initial condition (2), one can see that
(28)
y
ini
=
Y
N
(
0
)
=
∑
k
=
0
N
+
1
(
-
1
)
k
y
k
or
(29)
y
ini
=
Λ
Y
,
where Λ
=
[
1
-
1
⋯
(
-
1
)
N
+
1
]
T. From systems (27) and (29), we obtain a system of (
N
+
1
) nonlinear equations in (
N
+
1
) unknowns
(30)
Ψ
(
Y
)
=
R
,
where Ψ
:
ℜ
N
+
1
→
ℜ
N
+
1 by Ψ
(
Y
)
=
[
Ω
(
Y
)
Λ
] and R
=
[
R
1
y
ini
]
∈
ℜ
N
+
1.The standard methods for solving system (30), such as secant method and Newton method, need a good initial guess which is not available. To overcome this problem, we look for another method which does not depend on the initial guess. The promise technique is the path following method which is described as follows. From (23), we can see that the function Ωis the sum of three terms. Two of them are linear systems which are ∑
k
=
0
2
N
+
1
∑
j
=
0
k
∑
l
=
0
[
k
/
2
]
A
l
a
j
y
k
-
j
(
α
)
F
k
-
2
l
α
(
x
) and ∑
k
=
0
2
N
+
2
∑
j
=
0
k
∑
l
=
0
[
k
/
2
]
A
l
b
j
y
k
-
j
F
k
-
2
l
α
(
x
). The third term is nonlinear system which is
(31)
∑
k
=
0
3
N
+
3
∑
i
=
0
k
∑
j
=
0
k
-
i
∑
l
=
0
[
(
k
-
i
)
/
2
]
∑
m
=
0
[
(
k
-
2
l
)
/
2
]
A
m
A
l
c
i
y
j
y
k
-
i
-
j
F
k
-
2
l
-
2
m
α
(
x
)
.
Thus, we can rewrite Ω as
(32)
Ω
(
Y
)
=
A
1
Y
+
A
2
Y
+
Ω
1
(
Y
)
,
where A
1 and A
2 are two N
×
(
N
+
1
) matrices and Ω
1
:
ℜ
N
+
1
→
ℜ
N is a nonlinear function of Y
.Hence, system (30) can be written as
(33)
Ψ
(
Y
)
=
A
Y
+
Ψ
1
(
Y
)
=
R
,
where A
=
[
A
1
Λ
] is (
N
+
1
)
×
(
N
+
1
) matrix and Ψ
1
:
ℜ
N
+
1
→
ℜ
N
+
1 is a nonlinear function of Y given by Ψ
1
=
[
A
2
+
Ω
1
(
Y
)
0
1
×
(
N
+
1
)
]. Define the function H
:
ℜ
N
+
1
×
[
0,1
]
→
ℜ
N
+
1 by
(34)
H
(
Y
,
λ
)
=
λ
(
Ψ
(
Y
)
-
R
)
+
(
1
-
λ
)
A
Y
=
A
Y
+
λ
(
Ψ
1
(
Y
)
-
R
)
.
Thus, H satisfies the following properties(1)
H
∈
C
∞
(
ℜ
N
+
1
×
[
0,1
]
),
(2)
H
(
0
,
0
)
=
0,
(3)
(
∂
H
/
∂
Y
)
(
0
,
0
)
=
A.It is worth to mention thatA is a nonsingular matrix since A
Y
=
R is the system produced by the following problem:
(35)
a
(
x
)
D
α
y
(
x
)
=
g
(
x
)
,
x
∈
(
0,1
)
,
0
<
α
<
1
,
y
(
0
)
=
y
ini
,
which has a unique solution. Hence, it follows from the implicit function theorem that there is a smooth curve γ
:
I
→
ℜ
N
+
1 for some open interval I containing zero such that γ
(
0
)
=
0
,
γ
′
(
0
)
≠
0
,
rank
⁡
(
H
′
(
γ
(
s
)
)
)
=
N
+
1, and
(36)
H
(
γ
(
s
)
)
=
0
,
for all s
∈
I. Consider the solution γ
(
s
)
=
(
Y
(
s
)
,
λ
(
s
)
) (parametrized for convenience with respect to arc length) such that γ
(
0
)
=
(
0
,
0
). The solution curve H
-
1
(
Y
(
0
)
,
λ
(
0
)
) should be either diffeomorphic to the circle or to the real line. Since the solution point (
0
,
0
) is unique for λ
=
0, it follows that γ cannot be closed, and hence, it is diffeomorphic to the real line. Since γ is a smooth curve, Y
(
s
) is bounded for λ
(
s
)
∈
[
0,1
]
.Moreover, the curve γ reaches the level λ
=
1after a finite arc length s
1. This means, γ
(
s
1
)
=
(
Y
-
,
1
), and hence, Y
- is the zero of Ψ
(
Y
)
-
R. Thus, we can take I
=
[
0,1
]. For more details about the proof, [29–31]. To apply path following method numerically, we differentiate (36) to get
(37)
H
′
(
γ
(
0
)
)
γ
′
(
0
)
=
0
,
which implies that γ
′
(
0
) is orthogonal to all rows of H
′
(
γ
(
0
)
)
.Thus, for all s
∈
[
0,1
],(1)
det
⁡
(
H
′
(
γ
(
s
)
)
γ
′
(
s
)
T
)
>
0,
(2)
∥
γ
′
(
s
)
∥
=
1, where ∥
·
∥ is the Euclidean norm,
(3)
H
′
(
γ
(
s
)
)
γ
′
(
s
)
=
0.We use the predictor-corrector method to numerically trace the curveγ. The predictor step is the Euler-predictor which is given by
(38)
X
=
Z
+
h
t
(
H
′
(
Z
)
)
,
where Z is a point along the curve γ, and h
>
0 is a fixed step size. The corrector step is the Gauss-Newton corrector which is given by
(39)
W
=
X
-
H
′
(
X
)
+
H
′
(
X
)
,
where H
′
(
X
)
+
=
(
H
′
(
X
)
)
T
(
H
′
(
X
)
H
′
(
X
)
T
)
-
1. For more details, see [31]. Thus, we will start from Z
0
=
(
0
,
0
) and then generate the sequence Z
1
,
Z
2
,
Z
3
,
… We stop our procedure at Z
k when the last component of Z
k
≤
1 and the last component of Z
k
+
1
>
1
.We can write Z
k as Z
k
=
(
Y
k
,
λ
k
). Thus, Y
k will be the approximate solution to the system (33).


## 4. Numerical Results

In this section, we implement the Tau-Path following method to the nonlinear fractional Riccati differential equations. Two examples of nonlinear fractional Riccati differential equations are solved to show the efficiency of the presented method.Example 7.
Consider the following initial value problem [32]:
(40)
D
α
y
(
x
)
-
2
y
(
x
)
+
y
2
(
x
)
=
1
,
x
∈
(
0,1
)
,
0
<
α
<
1
,
y
(
0
)
=
0
.
The matrices in (33) are
(41)
A
=
[
1
0
0
⋯
0
0
1
0
⋯
0
⋮
⋱
⋱
⋱
⋮
0
⋯
0
1
0
1
-
1
⋯
(
-
1
)
N
(
-
1
)
N
+
1
]
,
A
2
=
[
-
2
0
0
⋯
0
-
2
0
⋯
⋮
⋱
⋱
⋱
0
⋯
0
-
2
]
,
R
=
[
1
0
⋮
0
0
]
.
It is easy to see that|
det
⁡
(
A
)
|
=
1 which means A is nonsingular. Define the error by
(42)
e
(
x
,
α
)
=
D
α
Y
N
(
x
)
-
2
Y
N
(
x
)
+
Y
N
2
(
x
)
-
1
.
Then, the absolute maximum of the error for different values ofα are given in Table 1 for N
=
16.
The exact solution whenα
=
1 is y
(
x
)
=
1
+
2
tanh
(
2
x
+
(
1
/
2
)
log
⁡
(
(
2
-
1
)
/
(
2
+
1
)
)
). The graphs of the approximate solution whenα
=
0.9999 and the graph of the exact solution when α
=
1 are given in Figure 1.Table 1
α
Error
0.5
1
*
10
-
9
0.9
2
*
10
-
10
0.99
1.2
*
10
-
10
0.9999
1.7
*
10
-
12Figure 1
The graphs of the approximate solution whenα
=
0.9999 and the graph of the exact solution when α
=
1.Example 8.
Consider the following initial value problem [32]:
(43)
D
α
y
(
x
)
-
y
2
(
x
)
=
t
2
,
x
∈
(
0,1
)
,
0
<
α
<
1
,
y
(
0
)
=
1
.
The matrices in (33) are
(44)
A
=
[
1
0
0
⋯
0
0
1
0
⋯
0
⋮
⋱
⋱
⋱
⋮
0
⋯
0
1
0
1
-
1
⋯
(
-
1
)
N
(
-
1
)
N
+
1
]
,
A
2
=
[
0
0
0
⋯
0
0
0
⋯
⋮
⋱
⋱
⋱
0
⋯
0
0
]
.
It is easy to see that|
det
⁡
(
A
)
|
=
1 which means A is nonsingular. Define the error by
(45)
e
(
x
,
α
)
=
D
α
Y
N
(
x
)
-
Y
N
2
(
x
)
-
x
2
.
Then, the absolute maximum of the error for different values ofα is given in Table 2 for N
=
16.
The exact solution whenα
=
1 is y
(
x
)
=
(
x
(
J
-
3
/
4
(
x
2
/
2
)
Γ
(
1
/
4
)
+
J
3
/
4
(
x
2
/
2
)
Γ
(
3
/
4
)
)
)
/
(
J
1
/
4
(
x
2
/
2
)
Γ
(
1
/
4
)
-
2
J
-
1
/
4
(
x
2
/
2
)
Γ
(
3
/
4
)
) where J
ν
(
x
) is the Bessel function of first kind. The graphs of the approximate solution whenα
=
0.9999 and the exact solution when α
=
1 are given in Figure 2.Table 2
α
Error
0.5
2.7
*
10
-
10
0.9
8.6
*
10
-
1
1
0.99
5.4
*
10
-
1
1
0.9999
2.1
*
10
-
1
3Figure 2
The graphs of the approximate solution whenα
=
0.9999 and the exact solution when α
=
1.



---

*Source: 207916-2014-03-05.xml*