Research Journal of Applied Sciences, Engineering and Technology 7(22): 4696-4701, 2014
ISSN: 2040-7459; e-ISSN: 2040-7467
© Maxwell Scientific Organization, 2014
Submitted: October 25, 2013
Accepted: January 24, 2014
Published: June 10, 2014
Solution of the System of Fifth Order Boundary Value Problem using Quartic Spline
Ghazala Akram and Shahid S. Siddiqi
Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan
Abstract: The aim of the study is the solution of the system of fifth order boundary value problems associated with
obstacle, unilateral and contact problems using quartic spline. These problems arise in several branches of pure and
applied sciences and in engineering including transportation, equilibrium, optimization, mechanics, structural
analysis, fluid flow through porous media and image processing in the medical sciences. The results are compared
with the exact solution. Two examples are considered for the numerical illustration of the method developed and the
results are encouraging.
Keywords: Contact problems, obstacle problems, quartic spline, system of boundary value problems, variational
inequalities
INTRODUCTION
The contact, unilateral, obstacle and equilibrium
problems arising in different branches of pure and
applied sciences can be studied using the notion of
variational inequalities. During the past few years, this
has emerged as an interesting and important branch of
applied mathematics. The general variational
inequalities can be characterized by a system of
differential equations using the penalty function
technique, if the obstacle function is known. This
technique was used by Lewy and Stampacchia (1969)
to study the regularity of the solution of variational
inequalities. The main advantage of this technique is its
simple applicability in solving obstacle and unilateral
problems.
Al-Said (1996) developed the solution of system of
second order boundary value problems using quadratic
spline. Gao and Chi (2006) solved a system of thirdorder boundary value problems associated with thirdorder obstacle problems using the quartic B-splines and
the method is claimed to be of second order. Islam et al.
(2005) developed the solution of a system of thirdorder boundary value problems using non polynomial
spline and the method is claimed to be of second order
as well.
Siddiqi and Akram (2007a to c) solved the system
of fourth order boundary value problem using cubic
nonpolynomial spline, cubic spline and nonpolynomial
spline. Noor et al. (2011) solved fifth order obstacle
problem using variation of parameters method, but the
exact solution of the problem, given in the study is not
correct. Hence the method developed by Noor et al.
(2011) cannot be compared with the method developed
in this study.
In this study, the quartic spline function is used to
develop a technique for the solution of the following
system:
 f ( x),

y (5 ) ( x ) =  f ( x ) + y ( x ) g ( x ) + r ,
 f ( x),

a ≤ x ≤ c, 

c ≤ x ≤ d ,
d ≤ x ≤ b, 
(1)
Along with the boundary conditions:
y(a) = y(b) = α0 , y ( 1 )(a) = y( 1 )(b) =α1
y(c) = y(d) = α2 , y ( 1 )(c) = y( 1 )(d) =α3
(2)
y ( 2 )(a) = α4 , y ( 2 )(c) = y( 2 )(d) =α5
where r and α i , i = 0, 1, …, 5 are finite real constants
and the functions f (x) and g (x) are continuous on [a, b]
and [c, d], respectively. Such type of systems arises in
connection with contact, obstacle and unilateral
problems.
QUARTIC SPLINE METHOD
To develop the quartic spline approximation S to
the problem (1), the interval [a, b] is divided into k
equal subintervals (s.t k is divisible by 4 and k = 4n),
using the grid points x i = a + ih; i = 0, 1,…, k where
h = (b-a) /k.
Without loss of generality, the knots c and d may
be taken as:
c = (3a + b) / 4,
d = (a + 3b) / 4
The system (1), thus, leads to the following form:
yi
(5)
 fi ,
=
 f i + y ( xi ) g i + r ,
0 ≤ i ≤ n − 1 and 3n + 1 ≤ i ≤ 4n − 1, (3)

n ≤ i ≤ 3n,

Corresponding Author: Ghazala Akram, Department of Mathematics, University of the Punjab, Lahore 54590, Pakistan
4696
Res. J. Appl. Sci. Eng. Technol., 7(22): 4696-4701, 2014
f i = f ( x i ), g i = g ( x i ), i = 0,1,2,...,4n. The restriction S i of S to each subinterval [ x i , x i +1 ],
i = 0,1,2,..., k − 1, is defined as:
where,
S i ( x) = ai ( x − xi ) 4 + bi ( x − xi ) 3 + ci ( x − xi ) 2 + d i ( x − xi ) + ei
(4)
For,
S i ( xi ) = y i ,
S i( 2 ) ( xi ) = M i ,
S i(1) ( xi ) = mi ,

 i = 0,1,...,k
( 3)
S i ( xi ) = Ti , 

(5)
And assuming y (x) to be the exact solution of the system (1) and y i be an approximation to y (x i ), obtained by
the spline S (x i ).
After, applying the first and second derivative continuities at the knots, i.e., S i(−µ1) ( xi ) = S i( µ ) ( xi ) for μ = 1, 2, the
following consistency relation is derived which is necessary to find the solution of the problem (1):
Ti−2 + 11Ti−1 + 11Ti + Ti+1 =
24
[− yi−2 + 3 yi−1 − 3 yi + yi+1 ]; i = 2,3,..., k − 1
h3
(6)
To make the system (6) consistent with the BVP (1), the finite difference formula of O (h4) is used, which leads
to the following relation:
− Ti + 2 + 16Ti +1 − 30Ti + 16Ti −1 − Ti −2
= 12 yi( 5) + O(h 4 ); i = 2,3,...,k − 2
h2
(7)
From Eq. (6) and (7), it can be written as:
28Ti −1 − 8Ti + 28Ti +1 =
24
[− yi − 2 + 2 yi −1 − 2 yi +1 + yi + 2 ] + 12h 2 yi( 5) + O(h 6 ); i = 2,3,..., k − 2
h3
(8)
Moreover, following Lucas (1974) for quartic spline, it can be written as:
Ti −1 − 2Ti + Ti +1 = h 2 yi( 5) + O(h 7 ); i = 2,3,...,k − 1
(9)
Equation (8) and (9), leads the following relation:
Ti =
1
1
[− yi −2 + 2 yi −1 − 2 yi +1 + yi +2 ] − h 2 yi(5) + O(h 6 ); i = 2,3,...,k − 2
2h 3
3
(10)
Substituting Ti s; (i = 2, 3,..., k − 2), in Eq. (9) gives the following relation which is used for the quartic spline
solution of the system (3):
(
)
3 yi −3 − 12 yi −2 + 15 yi −1 − 15 yi +1 + 12 yi + 2 − 3 yi +3 = −h 5 2 yi(−51) + 2 yi(5) + 2 yi(+51) ,
3 ≤ i ≤ n − 3, n + 3 ≤ i ≤ 3n − 3,
3n + 3 ≤ i ≤ 4n − 3
(11)
The above system does not provide the complete solution of the system (1). Hence, the required equations are
obtained as:
Ti−1
5

 140
yi−1 + 57 yi − 12 yi+1 + yi+2 − 38 hy i(−11) − 12 h 2 yi(−21) 
−

3
3
1
, i = 1, n + 1, 3n + 1
= 3
h  5  5 ( 5) 227 ( 5) 67 ( 5) 

yi−1 −
yi −
yi+1 
+ h −


420
840
 168

4697
(12)
Res. J. Appl. Sci. Eng. Technol., 7(22): 4696-4701, 2014
38859
10299
721
4239 (1) 1467 2 ( 2 ) 
10933
yi −
yi+1 +
yi + 2 −
yi + 3 +
hy i−1 +
h yi−1 

6070
3035
1214
3035
3035
1 3035
, i = 1, n + 1, 3n + 1
Ti = 3 
31673
h 

 125207 (5) 319169 (5)
yi +
yi+1 +
yi(+52) 
+ h5 

1274700
1274700

1274700
4073
852
679
163
224 (1)

 4229
yi +1 +
yi −
yi −1 +
yi − 2 −
yi −3 +
hy i +2
−

879
293
879
1758
293
1  1758
, i = n − 2, 3n − 2, 4n − 2
Ti +1 = 3 
943 ( 5) 
h 
 5201 ( 5) 5189 ( 5)
yi +1 +
yi −
yi −1 
+ h 5 −

52740
21096

 105480
Ti+2
447
204
49
9
168 (1) 
 23
yi+2 +
yi+1 −
yi +
yi−1 −
yi−2 +
hy i+2 
−
25
25
25
50
25
1  2
, i = n − 2, 3n − 2, 4n − 2
= 3
h 
 233 ( 5) 119 ( 5) 41 ( 5) 
yi+2 +
yi+1 −
yi 
+ h5 

300
600
 3000

(13)
(14)
(15)
APPLICATIONS
To illustrate the implementation of the method developed, the following fifth order obstacle boundary value
problem can be considered as:




( 5)
y ( x) − f ( x) ( y ( x) − ψ ( x) ) = 0,

y (−1) = y (1) = y (1) (−1) = y (1) (1) = y ( 2 ) (−1) = 0, o n Ω = [−1, 1],

− y ( 5) ( x) ≥ f ( x),
y ( x) ≥ ψ ( x),
(
)
(16)
where, f is a given force acting on string and ψ (x) is the elastic obstacle. The problem (16) arise in several branches
of pure and applied sciences including transportation, equilibrium, optimization, mechanics, structural analysis, fluid
flow through porous media and image processing in the medical sciences. Using the ideas and technique of Lewy
and Stampacchia (1969), the obstacle problem (16) can be characterized by the following system of variational
inequality problem:
− y (5) + µ ( y −ψ ) (y −ψ ) = f ( x) ,
−1 < x < 1
(17)
y (−1) = y (1) = 0, y (1) (−1) = y (1) (1) = y ( 2) (−1) = ε
(18)
where,
ε
= A small positive constant
ψ
= The obstacle function
μ (t)
= The penalty function defined by:
1,
0,
µ (t ) = 
t ≥ 0, 

t < 0.
(19)
Since the obstacle function ψ is known, it is possible to find the exact solution of the problem in the interval
− 1 / 2 ≤ x ≤ 1 / 2.
Assuming that the obstacle function ψ is defined by:
− 1,
ψ ( x) = 
1,
− 1 ≤ x ≤ −1 / 2, 1 / 2 ≤ x ≤ 1,
− 1 / 2 ≤ x ≤ 1 / 2.
(20)
From Eq. (17)-(20), the following system of equations can be obtained as:
4698
Res. J. Appl. Sci. Eng. Technol., 7(22): 4696-4701, 2014
− 1 ≤ x ≤ −1 / 2, 1 / 2 ≤ x ≤ 1,
− 1 / 2 ≤ x ≤ 1 / 2,
f,
y (5) = 
− 1 + y + f ,
(21)
With the boundary conditions:
y (−1) = y (−1 / 2) = y (1 / 2) = y (1) = 0
(22)
y (1) (−1) = y (1) (−1 / 2) = y (1) (1 / 2) = y (1) (1) = y ( 2) (−1) = y ( 2) (−1 / 2) = y ( 2) (1 / 2) = ε
(23)
And the conditions of continuity for y, y(1) and y(2) at x = -1/2 and 1/2.
If μ (t) and ψ (x) are taken as:
4,
0,
µ (t ) = 
− 1 / 4,
ψ ( x) = 
1 / 4,
t ≥ 0,
t < 0,
(24)
− 1 ≤ x ≤ −1 / 2, 1 / 2 ≤ x ≤ 1,
− 1 / 2 ≤ x ≤ 1 / 2,
(25)
Then the following system of equations can be obtained as:
f,
y (5) = 
1 − 4 y + f ,
− 1 ≤ x ≤ −1 / 2, 1 / 2 ≤ x ≤ 1,
− 1 / 2 ≤ x ≤ 1 / 2.
(26)
It is to be mentioned that the system of equations associated with the obstacle problem (16) is a special case of
the system of fifth order boundary value problem (1):
Numerical examples:
Example 1: The following problem is considered, as:
1,
y (5) ( x) = 
2 − 4 y ( x),
− 1 ≤ x ≤ −1 / 2, 1 / 2 ≤ x ≤ 1,
− 1 / 2 ≤ x ≤ 1 / 2,
(27)
Together with the boundary conditions at x = -1, x = 1 and the conditions of continuity for y, y(1) and y(2) at x = 1/2 and x = 1/2. The analytic solution of the system (27) is:
 1 / 480(1 + 7 x + 19x 2 + 25x 3 + 16x 4 + 4 x 5 ) ,

 − 1 ≤ x ≤ −1 / 2,
 y ( −1) = y ( −1 / 2) = 0,

(1)
(1)
( 2)
 y ( −1) = y ( −1 / 2) = y ( −1) = 0,


2/5
1 / 2 − β1 exp(−2 x ) − β 2 exp(α 1 x ) cos(α 2 x )
− β exp(α x ) cos(α x ) + β exp(α x ) sin(α x ) + β exp(α x ) sin(α x ),
3
3
4
4
3
4
5
1
2

− 1 / 2 ≤ x ≤ 1 / 2,
y ( x) = 
 y ( −1 / 2) = y (1 / 2) = 0,
 y (1) ( −1 / 2) = y (1) (1 / 2) = y ( 2 ) ( −1 / 2) = 0,


 1 / 960( −1 + 8 x − 25x 2 + 38x 3 − 28x 4 + 8 x 5 ),

 1 / 2 ≤ x ≤ 1,
 y (1 / 2) = y (1) = 0,

 y (1) (1 / 2) = y (1) (1) = y ( 2 ) (1 / 2) = 0,


4699
(28)
Res. J. Appl. Sci. Eng. Technol., 7(22): 4696-4701, 2014
where,
β1
β2
β3
β4
β5
= 0.083472402636811683486236902194475814000555110022391306503
= 0.197226655875097793394145018248435611353617476770476389696
= 0.218780476045802273706722229445094829786402746435753855139
= 0.005598628986902192748619920098904556943240511107281886036
= 0.031624084344408001357285362723625687421220160348823034071
 1
 1
5− 5
5 
5+ 5
5 
α 1 =  3 / 5 − 3 / 5 , α 2 =
, α 3 =  3 / 5 + 3 / 5 , α 4 =
1 / 10
221 / 10
22 
22
22 
 22
 22
The observed maximum errors (in absolute values) are summarized in Table 1. It is confirmed from the Table 1
that if h is reduced by factor 1/2, then ||E|| is reduced by a factor 1/4, which indicates that the present method gives
second order results.
Example 2: The following problem is considered, as:
2,

y ( 5) ( x) = 
y ( x) ,
−
1
+

− 1 ≤ x ≤ −1 / 2,
− 1 / 2 ≤ x ≤ 1 / 2,
1 / 2 ≤ x ≤ 1,
(29)
Together with the boundary conditions at x = -1, x = 1 and the conditions of continuity for y, y(1) and y(2) at x = 1/2 and x = 1/2.
The analytic solution of the system (29) is:
 1 / 240(1 + 7 x + 19x 2 + 25x 3 + 16x 4 + 4 x 5 ),

 − 1 ≤ x ≤ −1 / 2,
 y ( −1) = y ( −1 / 2) = 0,
 (1)
(1)
( 2)
 y ( −1) = y ( −1 / 2) = y ( −1) = 0,


1 − β1 exp(α 1 x ) cos(α 2 x ) − β 2 exp(α 3 x ) cos(α 4 x )
− β 3 exp(x ) + β 4 exp(α 3 x ) sin(α 4 x ) + β 5 exp(α 1 x ) sin(α 2 x ),

− 1 / 2 ≤ x ≤ 1 / 2,
y ( x) = 
 y ( −1 / 2) = y (1 / 2) = 0,
 y (1) ( −1 / 2) = y (1) (1 / 2) = y ( 2 ) ( −1 / 2) = 0,


 1 / 480( −1 + 8 x − 25x 2 + 38x 3 − 28x 4 + 8 x 5 ),

 1 / 2 ≤ x ≤ 1,
 y (1 / 2) = y (1) = 0,

 y (1) (1 / 2) = y (1) (1) = y ( 2 ) (1 / 2) = 0,


where,
β1 = 0.421971253377808180784844197586481579304275360089514777474
β 2 = 0.363965259605979866780921476737942599547341650816790531421
β 3 = 0.214323949709503239977271742174791615518760110013465674423
β 4 = 0.031318976821447235311388883995038835884137184947278662482
β 5 = 0.032949018430124054163299461434266313081077383697306730612
 −1
 −1
5
1 1
5
1 1
, α 2 =
, α 4 =
+
−
α 1 = 
(5 + 5 ) , α 3 = 
(5 − 5 )


4 
2 2
4 
2 2
 4
 4
4700
(30)
Res. J. Appl. Sci. Eng. Technol., 7(22): 4696-4701, 2014
Table 1: Maximum absolute errors for problem (27)
h
1/14
1/28
1/56
1/112
1/224
|y (x i ) yi|
2.10×10-4
2.93×10-5
3.85×10-6
4.93×10-7
6.23×10-8
Table 2: Maximum absolute errors for problem (29)
The method
presented
4.39×10-4
6.12×10-5
8.05×10-6
1.03×10-6
1.30×10-7
h
1/14
1/28
1/56
1/112
1/224
To analyze the usefulness of the method, the results
are summarized in Table 2 and it is evident from the
Table 2 that ||E|| is reduced by a factor 1/4, if h is
reduced by factor 1/2, which shows that the method is
of second order.
CONCLUSION
Quartic spline method is developed for the
approximate solutions of system of fith order BVPs.
The obtained numerical results are compared with the
exact solution. Noor et al. (2011) developed variation
of parameters method for the solution of fifth order
obstacle problem but the exact solution of the problem
given in the study is not correct. Hence the method
developed by Noor et al. (2011) cannot be compared
with the quartic spline method. We showed that the
quartic spline method is a powerful mathematical tool
for the solution of system of fifth order BVPs.
Numerical examples also illustrate the accuracy of the
method.
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Islam, S., M.A. Khan, I.A. Tirmizi and E.H. Twizell,
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