MATEC Web of Conferences 57, 05007 (2016)
DOI: 10.1051/ matecconf/20165705007
ICAET- 2016
The numerical simulation of convection delayed dominated diffusion
equation
P. Murali Mohan Kumar
1, a
, A.S.V. Ravi Kanth
1
Department of Mathematics, National Institute of Technology Kurukshetra, Haryana – 136119, India
1
Abstract. In this paper, we propose a fitted numerical method for solving convection delayed dominated diffusion
equation. A fitting factor is introduced and the model equation is discretized by cubic spline method. The error
analysis is analyzed for the consider problem. The numerical examples are solved using the present method and
compared the result with the exact solution.
1 Introduction
Consider the following convection delayed dominated
diffusion equation
y ''( x) a( x) y '( x) b( x) y( x ) 0 on [0,1] (1)
subject to the interval conditions
(2)
y( x) ( x) on x 0 , y(1) where 0 1 is a perturbation parameter and is a
small shifting parameter of order . It is also assumed
that a( x), b( x), ( x) are smooth functions and is a
constant. These convection diffusion delayed types with
dominated convection term problems play an important
role in the mathematical modelling of various practical
phenomena in the engineering and environmental
sciences, for examples include high Reynold’s number
flow in fluid dynamics, heat transport problems with
large Pecklet number, modelling the problems in
mathematical biology and semi-conductor devices etc.
It is challenging to develop efficient numerical methods
for solving convection diffusion with dominated
convection term due to the existence of boundary layers.
Standard discretization methods for solving such kind of
problems are unstable and fails to give accurate results
when perturbation parameter is small. Therefore, it is
challenging to develop suitable numerical methods to
these problems, whose accuracy does not depend on the
parameter . Lange and Miura[1] initiated the singular
perturbation analysis of boundary value problems for
differential difference equations with small shifts. The
numerical study of second order singularly perturbed
delay differential equations has been given in [2-3] and
references therein.
In this paper, we present an exponentially fitted method
on uniform mesh based on cubic spline method for the
convection delayed dominated diffusion equation. The
layout of the paper is organized as follows: Continuation
of the problem is presented in next section and follows
a
description of the method. In Section 3, the error analysis
of the method is discussed. Section 4 ends with the
Numerical results.
2 Continuous problem
An application of Taylor series in (1) yields
y ''( x) (a( x) b( x)) y '( x) b( x) y( x) 0,
xi 1 x xi 1
y( x) 0 , y(1) We assume that
a( x) b( x) M 0
(3)
throughout the
interval [0,1] , where M is positive constant, then the
problem (3) exhibits boundary layer at x 0 .
From the theory of singular perturbations [4],
y x y0 x y0 0 e
a 0 b 0 x
o (4)
where y0 x is the solution of the reduced problem given
by
a x b x y0 x b x y0 x 0 with y0 1 From Eq.(4) as h 0 , we obtain
lim y ih y0 0 0 y0 0 e
a 0 b 0 ih
h 0
Let h
, then
lim y ih y0 0 0 y0 0 e
h0
a 0 b 0 i
(5)
Now introducing an exponentially fitting factor to
the Eq.(3), we get
y x a x b x y x b x y x 0
(6)
with
y(0)=0 , y(1) (7)
Corresponding author:nitmurali@gmail.com
© The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative Commons Attribution
License 4.0 (http://creativecommons.org/licenses/by/4.0/).
MATEC Web of Conferences 57, 05007 (2016)
DOI: 10.1051/ matecconf/20165705007
ICAET- 2016
The fitting factor is to be determined in such a way
n Eq.(10) and using the following three
approximations for first order derivatives
y yi 1
yi i 1
2h
3 yi 1 4 yi yi 1
yi1 2h
yi 1 4 yi 3 yi 1
yi1 2h
We get the following difference scheme
Ei yi 1 Fi yi Gi yi 1 0,i 11 N 1
that the solution of Eq.(6) converges uniformly to the
solution Eq.(3) .
Lemma 1. Let u(x) be a smooth function satisfing the
and
then
.
Prof We can prove the above lemma by method of
contradiction.
Let
be
such
that
and assume that
.
and
and
Now
Clearly
consider
(10)
Where
"
3"
Ei 1 ai 1 bi 1 "2 ai bi 2 ai 1 bi 1 "1hbi 1
2
2
2
Fi 2"1 ai 1 bi 1 2" 2 ai 1 bi 1 2" 2 hbi
3"
"
Gi 1 ai 1 bi 1 " 2 ai bi 2 ai 1 bi 1 "1hbi 1
2
2
Eq.(11) gives a system of N 1 equations with
N 1 unknowns. These N 1 equations together with the
Eq.(7) are sufficient to solve the system by using Thomas
algorithm.
Which is contradiction to our assumption. Hence
.
Lemma 2. Let u(x) is the solution of the boundary value
problem (3), then
Proof. Let
point
be two barrier functions defined by
Then, we have
and
as
.
Using this inquality in the above inequality, we get
2.2 Determination of fitting factor
Taking the limit as h 0 in Eq.(11), we obtain
2
yi 1 lim yi
a 0 b 0 "1 " 2 hlim
0
h 0
!
Therefore by the maximum principle[5], we obtain
which gives the required
estimate.
(11)
yi 1 0
a 0 b 0 "1 " 2 hlim
! 0
Substituting Eq.(5) in Eq.(12) and then simplifying, we
get the variable fitting factor as follows
2.1 Description of the method
The spline function
Let
S ( x, ) S ( x) satisfying in the interval [ xi , xi 1 ] and the
differential equation
x x
S '' ( x) S ( x) S '' ( xi ) S ( xi ) i 1
h !
i a xi b xi "1 " 2 a xi b xi coth 2
2
!
( 12)
x xi S ( xi 1 ) S ( xi 1 ) h !
''
(8)
Where S ( xi ) yi and 0 is termed as cubic spline in
compression. Following Aziz and Khan [6], we obtain the
tridiagonal system
h2 "1M i 1 2"2 M i "1M i 1 yi 1 2 yi yi 1
(9)
Where
" 1 1
"1 2 1 , " 2 2 " cot " 1 ,
sin " !
"
"
is a constant fitting factor for left end boundary layer.
3 Error analysis
Substituting M j a j b j y 'j b j y j , j i, i # 1 in
(10), we obtain
M i y xi , i 11 N 1.
Substituting
a 0 b 0 "1 " 2 a 0 b 0 coth 2
2
!
y x j b x j y x j , i
M j a x j b x j
j i,i # 1,
2
MATEC Web of Conferences 57, 05007 (2016)
DOI: 10.1051/ matecconf/20165705007
ICAET- 2016
yi 1 2 yi yi 1 h 2 (2)
ai ('1 )
2!
h2
ai 1 ai hai' ai(2) ('2 )
2!
2
h
bi 1 bi hbi' bi(2) ('3 )
2!
2
h
bi 1 bi hbi' bi(2) (' 4 )
2!
Where xi 1 '1 ,'2 ,'3 ,'4 xi 1
Using these expansions and Eq.(21), we have
( "1h 2bi 1 )ei 1 (2 2"2 h 2bi )ei
ai 1 ai hai' h 2 "1 (ai 1 bi 1 ) yi' 1 bi 1 yi 1 2"2 h 2 (ai bi ) yi' bi yi h 2 "1 (ai 1 bi 1 ) yi' 1 bi 1 yi 1 (13)
Putting exact solution in (14), we get
y ( xi 1 ) 2 y ( xi ) y ( xi 1 ) h 2 "1 (ai 1 bi 1 ) y ' ( xi 1 ) bi 1 y ( xi 1 ) 2"2 h 2 (ai bi ) y ' ( xi ) bi y ( xi ) h 2 "1 (ai 1 bi 1 ) y ' ( xi 1 ) bi 1 y ( xi 1 ) (21)
( "1h2bi 1 )ei 1 Ti (h)
Where
h4
h4
Ti (h) (ai bi ) y (3) ( xi ) (1 12"1 ) y (4) ( xi )
3
12
h6
"1 (ai' bi' ) y (4) ( xi ) O h6 6
(22)
Clearly, it can be seen that Ti (h) O h4 for the choice
(14)
Where
h4
h6
(1 12"1 ) y (4) ($ i ) (1 30"1 ) y (6) ($ i )
12
360
xi 1 $ i xi 1
(15)
for any choice of "1 and "2 whose sum is 1 .
2
Subtracting Eq.(14) from Eq.(15) and substituting
e j y( x j ) y j , j i, i # 1 , we get
Tio (h) ( "1h 2bi 1 )ei 1 (2 2"2 h 2bi )ei ( "1h 2bi 1 )ei 1 h 2 "1 (ai 1 bi 1 )ei' 1 2"2 (ai bi )ei' "1 (ai 1 bi 1 )ei' 1 Tio (h)
h 2 (3)
h2
h 4 (5) i
y ( xi ) y (4) ( xi ) y ($1 )
3
12
30
y ' ( xi 1 ) yi' 1
2
4
h (3)
h (4)
h (5) i
y ( xi ) y ( xi ) y ($ 2 )
3
12
30
ei' y ' ( xi ) yi'
4
h (3)
h (5) i
y ( xi ) y ($3 )
6
120
and
matrices
E e1 , e2 ,
and
, eN 1 t
are
, S N 11 of A are
S1 1 2"2 h b1 "1h b2
2
2
Si h 2 "1bi 1 2"2 bi "1bi 1 h 2 Bi , i 2(1) N 2
(18)
S N 1 N 1 "1h 2bN 2 2"2 h 2bN 1
=
2
, bN 1 Clearly, the row sums S1 , S2 ,
(17)
=
2
tridiagonal
t
( N 1) component vectors. Eq.(21) can be written in
matrix vector form as
where A ( J h2 DQ)
(23)
AE Ti (h) ,
we have
ei' 1 y ' ( xi 1 ) yi' 1
= ( N 1) ( ( N 1)
Q b1 , b2 ,
(16)
ei' 1
of "1 and "2 whose sum is 1 .
2
Let J trid % 2 & , D trid %"1 2"2 "1 & are
For sufficiently small h , the matrix A is irreducible and
monotone which implies A1 exists. Hence from
Eq.(24), we have E A1Ti (h) .
From the theory of matrices we have,
(19)
xi 1 $1i , $2i , $3i xi 1
Using Eqs.(18)-(20) in Eq.(17), we get
( "1h 2bi 1 )ei 1 (2 2"2 h 2bi )ei ( "1h 2bi 1 )ei 1 N 1
)p
i 1
k ,i
Si 1 , k 1(1) N 1
Where pk ,i is the (k , i)th element of the matrix A1 .
h4
% "1 (ai 1 bi 1 ) 2"2 (ai bi ) "1 (ai 1 bi 1 ) & y (3) ( xi )
3
h5
"1 % (ai 1 bi 1 ) (ai 1 bi 1 ) & y (4) ( xi )
12
h6
"1 (ai 1 bi 1 ) y (5) ($1 ) (ai 1 bi 1 ) y (5) ($ 2 ) 30
h6
"2 (ai bi ) y (5) ($3 ) Ti 0 (h)
60
(20)
Therefore
N 1
)p
i 1
k ,i
1
1
1
min Si h 2 Bi h 2 Bi
1 i N 1
N 1
E ) pk ,iTi (h), k 1(1) N 1
i 1
Therefore
E Kh 2
Bi
Where K is constant independent of h .
E O h
Let
3
2
(24)
Therefore
for the choice of parameters "1 and "2
MATEC Web of Conferences 57, 05007 (2016)
DOI: 10.1051/ matecconf/20165705007
ICAET- 2016
Table 2. Maximum absolute errors for the Example 1.
whose sum is 1 . From the Eq.(25), it is observed that
2
the proposed method is uniform convergent since the
error is of order of the form E K * h2 , where K * is
"1 112 , "2 512
*N
independent of perturbation parameter .
1
2
22
23
24
28
212
216
232
4 Numerical Results
To demonstrate the applicability of the method, we
consider two convective diffusion with dominated
convection term exhibiting boundary layer at left end of
the interval and one right end of the interval. The exact
solution of the boundary value problem (1) with constant
coefficients is given by
y x em2
e
m1
e
m2
em1x em1 e
m1
e
m2
m1 m2 a b 2 4b
2
a b 256
512
1024
3.90E-06
9.76E-07
2.44E-07
6.10E-08
7.65E-06
1.91E-06
4.78E-07
1.19E-07
1.52E-05
3.78E-06
9.46E-07
2.36E-07
3.04E-05
7.55E-06
1.88E-06
4.70E-07
1.33E-04
6.55E-05
2.77E-05
8.29E-06
1.33E-04
6.65E-05
3.32E-05
1.66E-05
1.33E-04
6.65E-05
3.32E-05
1.66E-05
1.33E-04
6.65E-05
3.32E-05
1.66E-05
Example 2.
Consider the following variable coefficient of left end
layer singularly perturbed problem
1
x
y x 1 y x y x 0 on %0,1&
2!
2
subject to the interval conditions
y x 0 on x 0, y 1 1
em2 x
where
a b 128
A valid solution for the above problem is given by
a b 2 4b
2
x2 x 4 !
e
1
2 x
Figure 2. indicates the numerical solution for different and values, it is observed that the boundary layer
behavior is not only on perturbation parameter and also
depends on the delay parameter.
Example 1.
Consider the following left end layer singularly perturbed
of convection delayed dominated diffusion
y x 5 y x y x 0 on %0,1&
u x subject to the interval conditions
y x 1 on x 0, y 1 0
4.1 Determination of fitting factor for right end
problem
In Table 1, represents the comparison between the
present method and the method in [7] with fixed and
. It is observed that the present method gives more
accurate results than the method in [7]. Table 2 indicates
the maximum absolute error for different N and , it is
observed from the results, the method is uniform
convergent. Figure 1 displays the numerical solution for
different values, the solution of the problem exhibits
the boundary layer behavior to the left side while the
perturbation parameter tends to zero.
2
Table 1. Numerical solution for Example 1. with 2 and
We assume that a( x) b( x) M 0 , M is a constant
then, the Eq.(3) exhibits boundary layer at x 1 .
y x y0 x y0 1 e
a1 b1 ( 1 x )
o (25)
As h 0 , Eq.(26) becomes
lim y ih y0 x y0 1 e
1
a 1 b1 i !
h 0
=0.1, "1 112 , "2 512
x
Exact
Proposed
Method in [7]
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.14379381821116
0.02067665891105
0.00297317208765
0.00042752012899
0.00006147117862
0.00000883567395
0.00000126708504
0.00000017883874
0.00000002242424
0.14380334377735
0.02067939843477
0.00297376299483
0.00042763342218
0.00006149154060
0.00000883918533
0.00000126767192
0.00000017893301
0.00000002243736
0.14085841827929
0.01984109173050
0.00279478220875
0.00039366596699
0.00005544852472
0.00000780774988
0.00000109714556
0.00000015190043
0.00000001875470
Introducing ( ) in Eq.(3) and by applying the same
procedure as in Section(2) and simplifying, we get
i a xi b xi "1 " 2 a xi b xi coth 2
2
!
a 1 b 1 "1 " 2 a 1 b 1 coth 2
2
!
is a constant fitting factor for right end boundary value
problem.
4
MATEC Web of Conferences 57, 05007 (2016)
DOI: 10.1051/ matecconf/20165705007
ICAET- 2016
Example 3.
Finally we consider the right end layer singularly
perturbed of convection diffusion problem
y x 5 y x 2 y x 0 on %0,1&
subject to the interval conditions
y x 0 on x 0, y 1 2
Table 3 gives the comparison between present method
and the method in [7] for fixed and . It is observed
that the method yields results better than the existing
method in [7]. Figure 2 indicates the numerical solution
for different and values.
The perturbation
parameter tends to zero the solution of the problem
exhibits boundary layer behavior to the right side of the
interval.
Figure 2. Numerical solution of the Example 2
Acknowledgements
Authors would like to thank National Board for Higher
Mathematics (NBHM), Government of India for
providing financial support under the grant number
2/48(12)/2013/NBHM(R.P.)/R&D II/1084.
Table 3. Numerical solution for Example 2. with
22 and
=0.1 "1 112 , "2 512
x
Exact
Proposed
Method in [7]
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.00000001824795
0.00000015942757
0.00000124681219
0.00000961694147
0.00007404065246
0.00056989545628
0.00438637260011
0.03376088858255
0.25984953872717
0.00000001827085
0.00000015960776
0.00000124804968
0.00000962512936
0.00007409319019
0.00057021895408
0.00438823990526
0.03377046939265
0.25988640673333
0.00000001297418
0.00000011682562
0.00000094810121
0.00000760202085
0.00006086311380
0.00048718987663
0.00389970919703
0.03121511401660
0.24986042277630
Figure3. Numerical solution of the Example 3 for
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0.1
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