Quadratic spline solution for boundary value
problem of fractional order
W. K. Zahra1 and S. M. Elkholy2
2
1 Department of Engineering Physics and Mathematics, Faculty of Engineering, Tanta Univ., Tanta, Egypt.
Department of Engineering Physics and Mathematics, Faculty of Engineering, Kafr El Sheikh Univ., Kafr El Sheikh,
Egypt.
Abstract
Fractional differential equations are widely applied in physics, chemistry as well as
engineering fields. Therefore, approximating the solution of differential equations of fractional
order is necessary. We consider the quadratic polynomial spline function based method to find
approximate solution for a class of boundary value problems of fractional order. We derive a
consistency relation which can be used for computing approximation to the solution for this
class of boundary value problems. Convergence analysis of the method is discussed. Four
numerical examples are included to illustrate the practical usefulness of the proposed method.
Keywords: Boundary value problems of fractional order ·
Riemann–Liouville fractional derivative · Caputo fractional derivative ·
Quadratic polynomial spline · Error bound
1. Introduction
In the last few years, considerable interest was paid to the fractional
calculus, because it becomes a fundamental tool to model many dynamic
systems. It was known that the classical calculus can provide a powerful tool for
describing many important dynamic processes in most applied sciences, but
recent studies proved that fractional calculus provides more accurate models
than classical ones. There are many applications of fractional derivative and
fractional integration in several complex systems such as physics, chemistry, fluid
mechanics, viscoelasticity, signal processing, mathematical biology, and
1
Corresponding Author :[email protected] , [email protected]
1
bioengineering and various applications in many branches of science and
engineering could be found [1-5,16,19-21,24-27,31,33-34]. Boundary value
problems of fractional order occur in the description of many physical processes
of stochastic transport and in the investigation of liquid filtration in a strongly
porous (fractal) medium [32]. Also, boundary value problems with integral
boundary conditions constitute a very interesting and important class of
problems. They include two, three, multipoint and nonlocal boundary value
problems as special cases. Integral boundary conditions appear in population
dynamic and Cellular systems [6-7]. They occur also in the mathematical model
which is developed for a micro-electro-mechanical system (MEMS) instrument
that has been designed primarily to measure the viscosity of fluids that are
encountered during oil well exploration [10].
Analysis and design of many systems require solution of fractional
differential equations (FDEs). Several methods have recently been proposed to
obtain the analytical solution of these equations. These methods include Laplace
and Fourier transforms, eigenvector expansion, method based on Laguerre
integral formula, direct solution based on Grunewald Letnikov approximation,
truncated Taylor series expansion and power series method [9,13-14,1718,21,23]. Also, several algorithms have been developed to solve FDEs
numerically such as fractional Adams–Moulton methods, explicit Adams
multistep methods, fractional difference method, decomposition method,
variational iteration method , least squares finite element solution, extrapolation
method and the Kansa method which is meshless, easy-to-use and has been used
to handle a broad range of partial differential equation models [6-8,1112,22,29,32]. The existence of at least one solution of fractional two-point
boundary value problems can be seen in [3, 25, 31-32,34].
We consider the numerical solution of the following fractional boundary
value problem (FBVP):
D -  y ¢¢(x ) + p (x ) y = g (x ) , 0 £  < 1,
x Î éëa, bùû,
(1.1)
subject to boundary conditions:
2
y(a) = y(b) = 0 ,
(1.2)
where the function g ( x ) and p (x ) are continuous on the interval [a, b ] and the
operator D  represents the Caputo fractional derivative. The analytical solution
of (1.1-1.2) cannot be obtained for arbitrary choices of p (x ) and g ( x ) . When
  0 , Eq. (1.1) is reduced to the classical second order boundary value problem.
The main objective of this work is to use quadratic polynomial spline
function to establish a new numerical method for the FBVP (1.1-1.2). This
approach has its own advantages. For example, once the solution has been
computed, the information needed for spline interpolation between mesh points
is available. This is important when the solution of the boundary value problem is
required at different locations in the interval [a, b ]. This approach has added
advantage that it not only provides continuous approximations to y ( x) , but also
for y ¢( x) at every point of the range of integration [2,30].
The paper is organized as follows: In section 2, we introduce some
definitions and theorem necessary to our work. Derivation of our method is
established in section 3. Convergence analysis of the new method is presented in
section 4. In section 5, numerical results are included to show the applications
and advantages of our method.
2. Preliminaries
In this section, definitions of fractional derivative and integral, used in our
work, will be presented. There are different definitions for fractional derivatives,
the most commonly used ones are the Riemann-Liouville and the Caputo
derivatives.
Let f ( x) be a function defined on (a, b) , then
Definition 1 [18]
The Riemann-Liouville fractional derivative:
3
R
1
dm
D f ( x) =
(m -  ) dx m

x
ò (x -
t )m-  - 1 f (t )dt ,  > 0, m - 1 <  < m .
0
Definition 2 [18]
The left Riemann-Liouville fractional integral:
Da- +
f ( x) =
Da- 
x
1
f ( x) =
( x - t ) - 1 f (t )dt ,  > 0 ,
( ) ò
a
where  is the gamma function.
Definition 3 [18,26,29]
Right Riemann-Liouville fractional integral:
Db- -
b
1
f ( x) =
(t - x) - 1 f (t )dt ,  > 0
ò
( )
x
Definition 4 [4]
The Caputo fractional derivative:
D f ( x) =
x
1
( x - s)m-  - 1 f ( m) (s)ds ,  > 0, m - 1 <  < m .
ò
( m -  )
0
The relation between the Riemann–Liouville operator and Caputo operator is
given by:
D f ( x) = R D [ f ( x) -
m- 1
1
( x - a)k f (k ) (a)],  > 0, m - 1 <  < m
k= 0 k !
å
Theorem 1 (Leibniz' formula)[21](p.75)
Let f be continuous on [0,t ] and let g be analytical at a for all a Î [0,t ] .
Then, for  > 0 and 0<a £ t
D -  f (x ) g (x ) =
¥
å
(- 1)n C n D n f (x ) R D -  - n g (x ) ,
n= 0
4
R
D  f (x ) g (x ) =
¥
C n D n f (x ) R D  - n g (x ) .
å
n= 0
Where D n is the ordinary differential operator and C n 
(  1)
.
n !(  n  1)
Lemma 1 [12]
If f ( x) is continuous and  ,   0 , then the following relationships hold:
(1) R D (D-

f ( x)) =
R - 
D
f ( x)
(2) D-  D-  f ( x) = D-  D-  f ( x) = D- -  f ( x)
(3) D  x m 
(m  1)
x m
(m    1)
(4) D  exp(ax ) 
x
1
(x  t ) 1 exp(at ) dt

( ) 0
 x  exp(ax ) ( , ax ) ,
where,  ( , ax) 
1

x
x
t
( )
 1
exp(a t ) dt , is called incomplete gamma function.
0
Lemma 2 [26,29]
If f (x ) is continuous and 1   ,   0 , then the right Riemann-Liouville fractional
operators follow the following properties:
(1) Db- - Db- - f (x ) = Db- - Db- - f (x ) = Db- - -  f (x ) ,
(2) Db Db f (x )  f (x ) ,
(3) Db (b  x )k 
(4) D1--  x k =
(k  1)
(b  x )k  , k  N
(k    1)
k
(1- x )
(- 1)m (1+  )
[1+ k ! å
(1- x )m ], k Î N
(1+  )
(4
m
)!

(1
+

+
m
)
m=1
5
Theorem 2[19]
Let f  C m 0,1 and   (m 1, m) , m  N and g  C 0,1 . Then for
x  0,1 :

D D  g ( x)  g ( x)

D D f ( x)  f ( x)  

m1 x k
f ( k ) (0)
k
!
k 0


lim D f ( x)  lim D f ( x)  0
x0
x0
n
 If i  (0,1], i  1, 2,..., n with    i are
i 1
such
that,
for
each
ik
k  1,2,..., m 1 , there exist i k  n with   j  k ,then the following
j 1
2
composition rule holds: D f ( x)  D ...D D1 f ( x) .
n

3. Quadratic spline solution for FBVPs
In order to develop the spline approximation for the fractional differential
equation (1.1) along with the boundary condition (1.2), we, firstly, use theorem 1
[19] to convert the FBVPs given by Eq.(1.1) into the following form:
y¢¢( x) + D p( x) y = D g ( x) 0 £  < 1,
x Î [a, b ] .
(3.1)
Now we introduce a finite set of grid points x i by dividing the interval [a, b ]
into n - equal parts.
xi = a + ih , x0 = a , xn = b, h =
b- a
n
, i = 0,1, 2,..., n .
(3.2)
Let y( x) be the exact solution of (2.1) and Si be an approximation to
yi  y( xi ) obtained by the spline function i ( x) passing through the points
( xi , Si ) and ( xi+ 1, Si+ 1 ).
3.1 Spline solution for FBVPs with left fractional integral operator:
6
Consider that each quadratic polynomial spline segment  i ( x) has the
form:
i ( x) = ai ( x - xi )2 + bi ( x - xi ) + ci , i = 0,1, 2,..., n - 1 ,
(3.3)
where ai , bi and ci are constants to be determined. To obtain the necessary
conditions for the coefficients introduced in (3.3), we do not only require that
i ( x) satisfies (3.1) at xi and xi 1 and that the boundary conditions (3.2) are
fulfilled, but also that the value of the slope be the same for the pair of segments
that join at each point ( xi , Si ) which means that  i ( x) satisfies the conditions:
(i) i (x ) Î C 1[a, b ]
(ii) S (x ) = i (x ) " x Î [x i , x i + 1] , i = 0,1, 2,..., n - 1
(3.4)
We express the three coefficients in Eq.(3.1) in terms of S i 1/ 2 , Di and
M i 1/ 2 [25],
i ( xi+ 1/ 2 ) = Si+ 1/ 2 ,
(1) ( xi ) = Di ,
(2)i ( xi+ 1/ 2 ) = Mi+ 1/ 2.
(3.5)
and we obtain:
ai = M i+ 1/ 2 / 2 ,
2
bi  Di and ci = Si+ 1/ 2 - h M i+ 1/ 2 - h Di .
8
2
(3.6)
Now from the continuity conditions that is the continuity of quadratic spline
S ( x) at the point (x i , S i
) , i1( xi )  i ( xi ) , and the continuity of the first
derivative, i1( xi )  i ( xi ) , at x  x i and using the values of the constants in
Eq.(3.6) yields to the following recurrence relation:
S i + 1/2 - 2S i - 1/2 + S i - 3/2 =
h2
[M
+ 6M i - 1/2 + M i - 3/2 ], i = 2,3,..., n - 1 ,
8 i + 1/2
where M i+ 1/ 2 = [- D p( x)S ( x) + D g ( x)]
x= xi+ 1/ 2
.
(3.7)
(3.8)
The relation (3.7) gives ( n - 2 ) algebraic equations in the n unknown S i 1 / 2 .
We need two more equations, one at each end of the range of the integration
7
interval. The two end conditions can be derived by Taylor series and the method
of undetermined coefficients. These two equations are, see [28]:
2S0 - 3S1/ 2 + S3/ 2 =
h2
[5M1/ 2 + M 3/ 2 ].
8
(3.9)
and
h2
Sn- 3/ 2 - 3Sn- 1/ 2 + 2Sn =
[M n- 3/ 2 + 5M n- 1/ 2 ].
8
(3.10)
Lemma 3 [28]
4
Let y  C [a, b ] then the local truncation errors ti , i  1, 2,..., n associated
with the scheme (3.7-3.10) are:
(1/ 64)h4 yi(4)  O(h5 ), i  1, n
ti  
4 (4)
5
(1/ 24)h yi  O(h ), i  2,3,..., n  1.
(3.11)
Proof
To obtain the local truncation errors t i , i  1, 2,..., n of Eqns.(3.7-3.10), we
first rewrite them in the form:
t1 = 2 y 0 - 3y 1/2 + y 3/2 -
t i = y i + 1/2 - 2 y i - 1/2 + y i - 3/2 -
h2
¢¢ + y 3/2
¢¢ ],
[5 y 1/2
8
(3.12)
h 2 ¢¢
[y
+ 6 y i¢¢- 1/2 + y i¢¢- 3/2 ], i = 2,3,..., n - 1,
8 i + 1/2
(3.13)
t n = y n - 3/2 - 3y n - 1/2 + 2 y n -
h2
[ y n¢¢- 3/2 + 5 y n¢¢- 1/2 ].
8
(3.14)
The terms y 1/2 and y1¢¢/ 2 , etc. in Eq.(3.12) are expanded around the point
x 0 = a using Taylor series and the expressions for t i , i
 1,
can be obtained. Also,
expressions for t i , i  2, 3,..., n  1 , in Eq.(3.13) can be obtained by expanding
these expressions around the point x i using Taylor series and the expressions for
t i , i  2, 3,..., n  1 , can be obtained. Finally expressions t i , i
n
are obtained as
expanding Eq.(3.14) around the point x n = b , then Eq.(3.11) is derived.
8
In order to determine the value of M i+ 1/ 2 , we use Eq.(3.8) and (3.4) and
using Theorem 1 [18],we get:
D S ( x) p( x)  D i ( x) p( x)
 i ( x) D p( x)   i ( x) D 1 p( x) 
 ( 1)
2
i ( x) D 2 p( x)
Substitute the spline function for the values of ai , bi and ci and x  xi 1/ 2 , we
get that:
D   i (x ) p (x )
x i 1/2



 S i 1/2 D  p (x )  D  1 p (x ) 



 S i 1/2  D  1 p (x ) 
 x i 1/2
h
 x i 1/2
h
3 h  1
  (  1)  2

D p (x ) 
D p (x ) 
8
 2
x
 h
 M i 1/2 
 M i 1/2 
 8
i 1/2

D  1 p (x ) 
 x i 1/2
)3.15)
.
Then Eq.(3.8) can be written as:
(1  1 )M i 1/2  2 M i 1/2  g i 1/2  1S i 1/2  2S i 1/2 ,
i  1, 2,..., n  2 .
(3.16)
where,
1  [
 (  1)
1  [ D p( x) 
2

h
D 2 p( x) 
3 h  1
 h  1
, 2 
,
D p( x)]
D p( x)
8
8
x  xi1/ 2
x  xi1/ 2
D 1 p( x)]
x  xi1/ 2
, 2 

h
D 1 p( x)
x  xi1/ 2
and gi+ 1/ 2 = D g ( x)
x= xi+ 1/ 2
.
We need two more equations, one at each end of the range of the
integration interval. The two end conditions can be derived by computing the
values of the constants in Eq.(3.3) at each end with the help of Taylor series, we
have that:
2
D0  ( S1/ 2  S0 ) for x   x0 , x1  .
h
This leads to the following end conditions:
9
(1  3 ) M 1/2  ( g1/2  4S 0 )  3S1/2 .
(3.17)
Similarly, we can get the second end condition:
(1  3 ) M n 1/ 2  ( g n 1/ 2  4S n )  3S n 1/ 2 ,
where
3 
 (  1)
,  3  [ D p ( x ) 
D  2 p ( x )
2
(3.18)
x  xi 1/ 2
4 
2
h
D 1 p ( x)]
and
x  xi 1/ 2
2  1
D p (x )
, i  1, n .
x  x i 1/ 2
h
3.2 Spline solution for FBVPs with right fractional integral operator:
In this case, the fractional differential equation can be written as:
Db- - y ¢¢(x ) + f (x ) y = g (x ), 0 £  < 1,
x Î [a, b ] ,
(3.19)
along with the boundary condition (1.2).
In this case, each quadratic polynomial spline segment is:
i (x ) = ai (x i + 1 - x )2 + bi (x i + 1 - x ) + c i , " x Î [x i , x i + 1 ], i = 0,1, 2,..., n - 1.
(3.20)
Operating on Eq.(3.19) by the operator Db- , leads to:
y ¢¢(x ) + Db- f (x ) y = Db- g (x ), 0 £  < 1,
x Î [a, b ] .
As indicated in section 3.1, we express the three coefficients in Eq.(3.20) in terms
of S i 1/ 2 , Di and M i 1/ 2 [25],
i (x i + 1/2 ) = S i + 1/2 ,
i¢(x i + 1) = Di + 1,
i(2) (x i + 1/2 ) = M i + 1/2.
and we obtain:
ai = M i + 1/2 / 2 ,
2
bi  Di 1 and ci = S i + 1/2 - h M i + 1/2 + h Di + 1 .
Applying the continuity of quadratic spline S ( x) at the
8
2
point
( xi , Si ) ,  (i 1) (x i )  i (x i ) , and the continuity of the first derivative at the
10
point ( xi , Si ) , (i 1) (x i )  i (x i ) , yield the same recurrence relation (3.7-3.10)
obtained in section 3. Where:
M i + 1/2 = [- Db- p (x )S (x ) + Db- g (x )]
x = x i + 1/2
.
(3.21)
3.3 Spline solution for FBVPs with two sided fractional integral operator:
In this case, the two sided fractional differential equation have the following
form:
(qDa-  + (1- q )Db- - ) y ¢¢(x ) + f (x ) y = g (x ), 0 £  < 1,
x Î [a, b ], (3.22)
along with the boundary condition (1.2), where 0 £ q £ 1 .
In this case, we apply the algorithm developed in [35] to get the following
system:
2S 0 - 3S 1/2 + S 3/2 =
S i + 1/2 - 2S i - 1/2 + S i - 3/2 =
ù
h 2-  (1 +  ) é
(1 + 4 )
ê(5 - 4 )M 1/2 +
M 3/2 ú, i = 1,
3- 

êë
ú
2
3
û
ù
h 2-  (1 +  ) éê
(7 - 5 )
ú, i = 2,3,..., n - 1 ,
M
+
M
+
M
i
+
1/2
i
1/2
i
3/2
ê
ú
23- 
3
êë
úû
and
S n - 3/2 - 3S n - 1/2 + 2S n =
ù
h 2-  (1 +  ) é(1 + 4 )
ê  M n - 3/2 + (5 - 4 )M n - 1/2 ú, i = n .
3- 
êë 3
ú
2
û
where M i = - fi si + gi , with fi  f ( xi ) and g i  g (x i ) .
For more details about this algorithm and its convergence analysis we can refer
to [35].
4. Convergence analysis of the method
11
Let Y  ( yi1/ 2 ), S  (Si1/ 2 ), C  (Ci ), T  (ti )
and
E  (ei1/ 2 )  Y  S
be
n-dimensional column vectors. Then with these notations we can write the
system given by (3.7-3.10) in matrix form as follows:
N S = h 2B M ,
(4.1)
Also, the system given by (3.16-3.18) can be written as follows:
K S V M G ,
(4.2)
where the matrices N , B , V , K and the vector G are given in Appndix1.
Rewrite Eq.(4.2) in the following form:
M V
1
G V
1
(4.3)
K S.
Substitute by Eq.(4.3) into Eq.(4.1), we get the standard matrix equations as
followed:
( N + h2 BV - 1 K ) S = h2 BV - 1 G ,
( N + h2 BV - 1 K ) Y = h2 BV - 1 G + T .
(4.4)
From which the error equation can be written as:
( N + h2 BV - 1 K ) E = T ,
(4.5)
where the vector T is:
æ- 1
T = h 4 ççç y 0(4)
è 64
- 1 (4)
y
24 1
...
- 1 (4)
y
24 n - 1
t
ö
- 1 (4) ÷
yn ÷
,
÷
ø
64
(4.6)
where t stands for the transpose of a matrix.
Writing the error Eq.(4.5) in the following form:
E = ( N + h2 BV - 1 K )- 1 T
E = (I + h2 N - 1BV - 1 K )-
1
N- 1 T
(4.7)
12
For simplicity, we will consider the case where p( x) is a constant function.
Then, Eq.(3.8) becomes:
M i+ 1/ 2 = [- p D S ( x) + D g ( x)]
In order to get a bound on E
x= xi+ 1/ 2
.
, we need the following lemma.
Lemma 4 [15,28,30]
If M is square matrix of order n and M  1 , then ( I  M )1 exists and
( I  M )1  1/(1  M ) .
Lemma 5
The infinite norm of V
- 1
V
£
1
satisfies the inequality
4(2 -  )
4(2 -  ) - p (3 +  )
provided that p (3 +  )
4(2 -  )
2- 
2- 
£ 1 and
,
(4.8)
 h / 2.
Proof
Rewrite the matrix V given by Eq.(4.5) in the form:
V =I+
p
4(2 -  )
2- 
V ,
where,
æ4 /(2 -  )
çç
çç
1
çç
çç
ç
V = çç
çç
çç
çç
çç
çç
è
2+ 
1
2+ 
1
2+ 
1
2+ 
ö
÷
÷
÷
÷
÷
÷
÷
÷
.
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
ø
4 /(2 -  )÷
Then,
13
V
- 1
= (I +
p
4(2 -  )
2- 
V )- 1 ,
(4.9)
with the help of Lemma 4, if
p
4(2 -  )
2- 
(4.10)
V <1
Then
V
- 1
1-
where
,
1
£
p
4(2 -  )
p
4(2 -  )
2- 
V =
2- 
(4.11)
V
p
4(2 -  )
2- 
(3 +  ).
(4.12)
Maximum value of Eq.(4.12) at   1 , in this case
p
4(2 -  )
2- 
V £ p ,
in order that Lemma 5 to be satisfied, the parameter p must satisfy the condition:
pmax  2 / h
Then
V
- 1
£
4(2 -  )
4(2 -  ) - p (3 +  )
2- 
.
(4.13)
The following Lemma gives a bound of E .
Lemma 6
The matrix ( N + h2 BV - 1 K ) given by (4.5) is nonsingular, provided that:
p 
(0.5(b  a )2  (5  )
4(2  )
2
)  1,
(4.14)
then
14
N 1 T
E 
1  h2 N 1 B V 1 K
 O( h 2 ) .
(4.15)
Proof
Using Lemma 5, we have:
N 1 T
E  max ei 
1  h2 N 1 B V 1 K
1i n
.
(4.16)
Provided that, h2 N 1 B V 1 K  1 .
It was shown that [12,25]
N 1 
h2
((b  a)2  h2 ) ,
8
(4.17)
p
(2   )
(4.18)
we also have that:
B  1 and
K 

Substitute Eqns. (4.13), (4.17-4.18) into Eq. (4.16) and using Eq.(4.6), we obtain
that:
E 
M 4(1  (3  )
24[ 1  (  (3  )
2
where,   0.5(b  a )2  2
2
, =
2
2
)
 O (h 2 ),
)]
(4.19)
p -
and M 4 = max y(4) ( x) .
4(2 -  )
a£ x£ b
Theorem 3
Let y( x) be the exact solution of the continuous boundary value problem
(1.1) with the boundary condition (1.2) and let yi+ 1/ 2 , i = 0,1, 2,..., n - 1 , satisfy the
discrete boundary value problem (4.4). Furthermore, if ei+ 1/ 2 = yi+ 1/ 2 - Si+ 1/ 2 ,
15
then E  O (h2 ) , which is given by Eq. (4.19), neglecting all errors due to round
off.
5. Numerical examples
We now consider some numerical examples illustrating the solution using
quadratic spline methods. All calculations are implemented with MATLAB 7.
Example 1: Consider the boundary value problem:
D  y (x )  y (x ) 
720
40320 6
x 4 
x
 (1  x 2 )x 6 ,
(5  )
(7  )
y (0)  y (1)  0 .
(5.1)
The analytical solution of Eq.(5.1) is
y (x )  x 6 (1  x 2 ) .
(5.2)
The numerical solutions for various values of  are represented in Fig (1) and
Table (1).
0.12
exact
α=0
α=0.2
α=0.4
α=0.6
α=0.8
0.1
0.08
0.06
0.04
0.02
0
0
0.2
0.4
0.6
0.8
1
-0.02
16
Fig (1)
Table (1): Observed maximum errors for example 1 and order of convergence
(O. C.).
=0
=0.2
=0.4
h
Error
O. C.
Error
O. C.
1.06E-01
Error
O. C.
1/8
9.29E-02
1.43E-01
1/16
2.57E-02
1.85
2. 91E-02
1.87
4.11E-02
1.80
1/32
7.15E-03
1.85
8.05E-03
1.85
1.10E-02
1.90
1/64
1.85E-03
1.95
2.21E-03
1.87
3.06E-03
1.85
Example 2: Consider the boundary value problem
D  y (x )  y (x )  [( x 2  (3  2)x   2 )x   (x , )  ( x  x 2 )]exp( x ),
y (0)  y (1)  0.
(5.3)
The analytical solution of Eq.(5.3) is:
y( x) = x(1- x)e x
(5.4)
The numerical solutions for various values of  are represented in Fig (2) and
0.5
Table (2).
exact
α=0
α=0.2
α=0.4
α=0.6
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.2
0.4
0.6
0.8
1
17
Fig (2)
Table (2): Observed maximum errors for example 2 and order of convergence
( O. C.).
=0
=0.2
=0.4
h
Error
O. C.
Error
O. C.
4.281E-03
Error
O. C.
1/8
3.735 E-03
6.875E-03
1/16
9.674 E-04
1.95
1.160E-03
1.88
1.940E-03
1.82
1/32
2.470 E-04
1.97
3.130E-04
1.89
5.310E-04
1.87
1/64
6.250E-05
1.98
8.460E-05
1.89
1.480E-04
1.84
For comparison, we plotted example 2 with various values of  and the
results in [35], as shown in Figs.(3-4).
0.5
exact
New results
method 1[30]
Method 2[30]
Method 3[30]
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
18
0
0
0.2
0.4
0.6
0.8
1
Fig (3) A comparison between the new results and the results cited in [30] for
  0.2 .
exact
New results
method 1[30]
Method 2[30]
Method 3[30]
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
Fig (4) A comparison between the new results and the results cited in [30] for
  0.4 .
Figs (3-4) show that as increases, our new method is more accurate than
methods in [35] for all values of . Moreover in the limit, as approaches zero,
19
the scheme provides solution for the integer order system. Results also suggest
that the scheme is numerically stable compared with [35].
Example 3: Consider the boundary value problem:
D  y (x )  x y (x ) 
120
(5  ) 22
x 3 
x
 x 6  x 5 ,
(4  )
(3  2)
y (0)  y (1)  0.
(5.5)
The analytical solution of Eq.(5.5) is
y (x )  x 5 ( x  x  ) .
(5.6)
The numerical solutions for   0.2 and   0.6 are represented in Fig (5), Fig
(6) and
Table(3).
0
-0.01
0
0.2
0.4
0.6
0.4
0.6
0.8
1
-0.02
-0.03
-0.04
-0.05
-0.06
-0.07
exact
-0.08
Approx
.
-0.09
0
0
0.2
0.8
1
-0.005
-0.01
-0.015
Fig (5)
-0.02
exact
-0.025
Approx.
20
-0.03
-0.035
Fig(6)
Table (3): Observed maximum errors for example 3and order of convergence
( O. C.).
=0
=0.2
=0.4
h
Error
O. C.
Error
O. C.
8.41E-3
Error
O. C.
1/8
8.382E-3
1.14E-2
1/16
2.155E-3
1.96
2.27E-3
1.89
3.12E-3
1.87
1/32
5.48E-4
1.98
6.11E-4
1.89
8.54E-4
1.87
1/64
1.38E-4
1.99
1.65E-4
1.89
2.35E-4
1.86
21
Example 4: Consider the boundary value problem:
D1--  y ¢¢(x ) + y (x ) = g (x ),
y (0)  y (1)  0.
(5.7)
Where:
g (x ) = x 6 - x 4 +
4
2
(1- x )
(- 1) m (1 +  )
(- 1) m (1 +  )
[18 + 720 å
(1- x ) m - 24 å
(1- x ) m ].
(1 +  )
m = 1 (4 - m )!(1 +  + m )
m = 1 (2 - m )!(1 +  + m )
The analytical solution of Eq.(5.7) is
y (x )  x 4 (x 2  1).
(5.8)
The numerical solutions for various values of  are represented in Fig (7) and
Table (4).
0
0
0.2
0.4
0.6
0.8
1
-0.02
-0.04
-0.06
-0.08
-0.1
exact
-0.12
α=0
α=0.2
-0.14
α=0.4
α=0.6
-0.16
Fig(7)
22
Table (4): Observed maximum errors for example 4and order of convergence
( O. C.).
=0
=0.2
=0.4
h
Error
O. C.
Error
O. C.
3.175E-2
Error
O. C.
1/8
2.774E-2
5.28E-2
1/16
7.33E-3
1.92
8.78E-3
1.85
1.47E-2
1.84
1/32
1.89E-3
1.96
2.393E-3
1.88
4.16E-3
1.82
1/64
4.81E-4
1.97
6.51E-4
1.89
1.16E-3
1.84
Conclusion:
New method for solving fractional boundary value problem is presented
using quadratic spline method. This method is shown to be a second ordered
convergent method. Convergence analysis for this method is presented.
Numerical comparisons between the solution using this new method and the
methods introduced in [30] are presented. The obtained numerical results show
that the proposed method maintain a remarkable high accuracy which make it
encouraging for dealing with the solution of two-point boundary value problem
of fractional order.
Appendix 1
The matrices N , B , V , K and the vector G are given in Eq.(4.7) are:
23
æ- 3 1
çç
çç 1 - 2 1
çç
çç
1 - 2
çç
N =ç
çç
çç
çç
çç
çç
çè
æv 0
ç
ç
ç
v1
ç
ç
ç
ç
ç
ç
V =ç
ç
ç
ç
ç
ç
ç
ç
ç
ç
ç
ç
è
1
1
æ5
ö
÷
ç
÷
ç
÷
ç
1
÷
ç
÷
ç
÷
ç
÷
ç
÷
,
÷
÷
1ç
ç
÷
ç
B =
÷
ç
÷
ç
÷
8
ç
÷
ç
÷
- 2 1
÷
ç
÷
ç
÷
ç
1 - 2 1÷
ç
÷
ç
÷
ç
ç
÷
è
1 - 3÷
ø
÷
v2
v1
v2
v1 v 2
v1
v2
æk 0
çç
ö
÷
÷
ççk 1
÷
÷
çç
÷
÷
çç
÷
÷
,
÷
ç
÷
÷
K =ç
çç
÷
÷
÷
çç
÷
÷
çç
÷
÷
çç
÷
÷
÷
çç
÷
çè
÷
v 0÷
ø
1
6
1
1
6
1
1
6
1
1
6
1
k2
k1
k2
k1
k2
k1
k2
ö
÷
÷
÷
÷
÷
÷
÷
÷
,
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
1÷
÷
÷
÷
5ø
ö
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
k 0ø
and
ìï
ïï g + p S0 -  ,
ïï 1/ 2 (2 -  )
ï
G = ïí gi+ 1/ 2 ,
ïï
ïï
p Sn - 
,
ïï g n- 1/ 2 +
(2 -  )
ïî
i= 1
i = 2,3,..., n - 1
i = n.
Where:
v 0  1
p
(3   )
2
, v1 
p
4(2   )
2
, v 2  1
and k 1  k 2 
p (2   )
4(2   )
p
2(2   )

2
, k0 
p
(2   )

.
Acknowledgements The authors are grateful to the referees for their suggestions
and valuable comments.
24
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