International Journal of Scientific & Engineering Research, Volume 2, Issue 6, June-2011 1
ISSN 2229-5518
A semi-implicit finite-difference approach for two-dimensional coupled Burgers’ equations
Mohammad Tamsir, Vineet Kumar Srivastava
Index Terms —Burgers’ equations; fnite- difference; semi-implicit scheme; Reynolds number.
—————————— • ——————————
HE two-dimensional Burgers’ equation is a mathe- matical model which is widely used for various phys-
au + u au
at ax
+ v au
ay
1 a2u
(
Re ax2
a 2u
+
ay 2
) = 0, (1)
ical applications, such as modeling of gas dynamics and
av
av av
1 a2v
a2v
traffic flow, shock waves [1], investigating the shallow
water waves[2,3], in examining the chemical reaction- diffusion model of Brusselator[4] etc. It is also used for testing several numerical algorithms. The first attempt to solve Burgers’ equation analytically was given by Bate- man [5], who derived the steady solution for a simple one-dimensional Burgers’ equation, which was used by J.M. Burger in [6] to model turbulence. In the past several
years, numerical solution to one-dimensional Burgers’
+ u + v ( + ) = 0, (2)
at ax ay Re ax 2 ay 2
subject to the initial conditions:
u ( x , y , 0 ) = \jf 1 ( x , y ); (x , y ) ,
v ( x, y , 0 ) = \jf 2 ( x , y ); (x , y ) ,
and boundary conditions:
equation and system of multidimensional Burgers’ equa- tions have attracted a lot of attention from both scientists and engineers and which has resulted in various finite-
u ( x, y , t ) = � ( x, y , t ); (x, y ) a ,
v ( x, y , t ) = ( x, y , t ); (x, y ) a ,
t > 0,
t > 0,
difference, finite-element and boundary element me-
where
= {( x, y) : a x b, c y d } and a is
its boundary;
u( x, y, t ) and
v( x, y, t ) are the velocity
thods. Since in this paper the focus is numerical solution
components to be determined,\jf ,\jf , � and are
1 2
of the two-dimensional Burgers’ equations, a detailed
survey of the numerical schemes for solving the one- dimensional Burgers’ equation is not necessary. Interest- ed readers can refer to [7-13] for more details.
Consider two-dimensional coupled nonlinear viscous
Burgers’ equations:
————————————————
• Mohammad Tamsir, Faculty member, Department of Mathematics, The
ICFAI University, Dehradun, India, Email: tamsiriitm@gmail.com.
• Vineet Kumar Srivastava, Faculty member, Department of Mathematics,
The ICFAI University, Dehradun, India, E-mail: vineetsriiitm@gmail.com.
known functions and Re is the Reynolds number.
The analytic solution of eqns. (1) and (2) was proposed by
Fletcher using the Hopf-Cole transformation [14]. The
numerical solutions of this system of equations have been
solved by many researchers. Jain and Holla [15] devel-
oped two algorithms based on cubic spline method. Fletcher [16] has discussed the comparison of a number of different numerical approaches.Wubs and Goede [17] have applied an explicit–implicit method. Goyon [18] used several multilevel schemes with ADI. Recently A. R. Bahad1r [19] has applied a fully implicit method. Vineet etl.[20] have used Crank-Nicolson scheme for numerical solutions of two dimensional coupled Burgers’ eqations. The usual implicit schemes are obviously unconditionally stable with higher order truncation error
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International Journal of Scientific & Engineering Research Volume 2, Issue 6, June-2011 2
ISSN 2229-5518
O(( t )2 + ( x)2 + ( y)2 ) . However, they involve solv- ing a nonlinear algebraic system of equations which makes it inefficient in practice. In this paper, to resolve the above issue, the semi-implicit scheme proposed by Ozis [7] is used for solving two-dimensional Burgers’ eq-
uations which has a truncation error
O(( t ) + ( x)2 + ( y)2 ) . Three numerical experiments
boundary conditions for u( x, y, t ) and v( x, y, t ) are tak- en from the analytical solutions. The numerical computa- tions are performed using uniform grid, with a mesh
width x = y = 0.05 . From Tables 1-4, it is clear that
the results from the present study are in good agreement with the exact solution for different values of Reynolds number. Comparison of numerical and exact solutions for u
have been carried out and their results are presented to illustrate the efficiency of the proposed method.
and v for Re=100 at shown in Figs. 1-4.
t = 0.5
with
t = 0.001
are
3.2 Problem 2.
Here the computational domain is taken as
The computational domain is discretized with uniform
= {( x, y) : 0
x 0.5, 0
y 0.5} and Burgers’
grid. Denote the discrete approximation of u( x, y, t ) and
n
equations (1) and (2) are taken with the initial conditions:
v( x, y, t ) at the grid point (i x, j y, n t) by
n
ui , j and
u( x, y, 0) = sin(n x) + cos(n y)
vi , j respectively (i = 0,1, 2......, nx ; j = 0,1, 2....., ny ;
v( x, y, 0) = x + y
n = 0,1, 2......), where
x = 1/ nx
is the grid size in x-
and boundary conditions:
direction,
y = 1/ ny
is the grid size in y-direction, and
t represents the increment in time.
u(0, y,t) = cos(n y),
u(0.5, y, t) =1+cos(n y)
Semi-implicit finite-difference approximation to (1) and
(2) are given by:
v(0, y,t) = y,
v(0.5, y, t) = 0.5+ y
0 y
0.5, t � 0,
u n+1 un
u n+1
un+1
u n+1
un+1
u(x,0,t) =1+sin(nx),
u(x,0.5,t) =sin(nx)
i , j i, j +
un (
i +1, j i 1, j ) + vn (
i , j +1
i, j 1 )
0 x
0.5, t � 0,
t i, j
2 x
i, j
2 y
v(x,0,t) = x,
v(x,0.5, t) = x +0.5
1 u n+1
2u n+1 + un+1
un+1
2u n+1 + u n+1
[( i +1, j i, j i 1, j ) + (
i, j +1
i, j i , j 1 )] = 0
Re ( x)2
( y)2
The numerical computations are performed using
20 x 20
grids and
t = 0.0001 . The steady state solu-
vn +1 vn
v n+1
vn+1
vn+1
vn +1
tions for Re = 50 and Re = 500 are obtained at
i, j i , j +
u n (
i +1, j i 1, j ) + vn
( i , j +1
i, j 1 )
t i , j
2 x
i , j
2 y
t = 0.625 . Perspective views of u and v for Re = 50 at
1 vn +1
2vn +1 + vn +1
vn +1
2vn+1 + vn+1
t = 0.0001 are given in Figs. 5 and 6 respectively. The
results given in Tables 5-8 at some typical mesh points
[( i +1, j i, j i 1, j ) + (
i, j +1
i , j i , j 1 )] = 0
( x, y)
Re ( x)2
( y)2
demonstrate that the proposed scheme achieves
similar results given by [15, 19].
The above linear system of equations is solved by direct method.
3.3 Problem 3.
In this problem the computational domain is
= {( x, y) : 0
x 1, 0
y 1} and Burgers’ equa-
3.1 Problem 1
The exact solutions of Burgers’ equations (1) and (2) can
tions (1) and (2) are taken with the initial conditions:
be generated by using the Hopf–Cole transformation [3]
which is:
u( x, y, 0) =
4n cos(2n x) sin(n y)
Re(2 + sin(2n x) sin(n y))
, (x,y)
u( x, y, t ) = 3 1 ,
4 4[1 + exp( ( 4x + 4 y t) Re/ 32 )]
v( x, y, 0) =
2n sin(2n x) cos(n y)
, (x,y)
v( x, y, t) = 3 + 1 ,
Re(2 + sin(2n x) sin(n y))
4 4[1 + exp( ( 4 x + 4 y t ) Re/ 32 )]
Here the computational domain is taken as a square do-
with boundary conditions:
main
= {( x, y) : 0
x 1, 0
y 1} . The initial and
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International Journal of Scientific & Engineering Research Volume 2, Issue 6, June-2011 3
ISSN 2229-5518
2n e
5n 2 t
Re
sin(n y)
Table 2.
The numerical results for v in comparison with the exact solution
u(0, y, t ) =
Re
5n 2 t
, t � 0 ;
at t = 0.01 and t = 1.0 with
(x, y) t=0.01
t = 0.0001 ,and Re = 10 . t=1.0
2n e
Re sin(n y)
Numerical Exact
Numerical Exact
u( x, 0, t ) = 0, t � 0 ;
v(0, y, t ) = 0, t � 0 ;
Re
u( x,1, t ) = 0, t � 0
v(1, y, t ) = 0, t � 0
(0.1, 0.1) (0.5, 0.1) (0.9, 0.1) (0.3, 0.3)
0.875195 0.875195
0.905798 0.905798
0.932918 0.932918
0.875195 0.875195
0.894374 0.894374
0.923160 0.923160
0.946983 0.946983
0.894373 0.894374
v( x, 0, t) =
n e
5n 2t
5n 2t
Re sin(2n x)
Re
, t � 0 ;
(0.7, 0.3)
(0.1, 0.5) (0.5, 0.5) (0.9, 0.5) (0.3, 0.7)
0.905798 0.905798
0.844569 0.844569
0.875195 0.875195
0.905798 0.905798
0.844569 0.844569
0.923160 0.923160
0.863315 0.863315
0.894372 0.894374
0.923160 0.923160
0.863313 0.863315
v( x,1, t) = n e
Re sin(2n x) Re
, t � 0
(0.7, 0.7)
(0.1, 0.9) (0.5, 0.9) (0.9, 0.9)
0.875195 0.875195
0.817389 0.817389
0.844569 0.844569
0.875195 0.875195
0.894371 0.894374
0.833647 0.833647
0.863313 0.863315
0.894372 0.894374
for which the exact solutions are:
5n 2 t
4n e
Re cos(2n x) sin(n y)
Table 3
Re(2 + e
5n 2 t
Re
sin(2n x) sin(n y))
at t = 0.01 and t = 1.0 with
(x, y) t=0.01
t = 0.0001 ,and Re = 100 . t=1.0
v( x, y, t) =
2n e
5n 2 t
Re
sin(2n x) cos(n y)
(0.1, 0.1)
Numerical Exact
0.623106 0.623047
Numerical Exact
0.510307 0.510522
Re(2 + e
5n 2 t
Re
sin(2n x) sin(n y))
(0.5, 0.1) (0.9, 0.1) (0.3, 0.3)
0.501617 0.501622
0.500011 0.500011
0.623106 0.623040
0.500072 0.500074
0.500000 0.500000
0.509824 0.510522
The computed solutions for u and v are ploted in Figs 7 and 9 respestively while the analytical solutions for u and v are shown in Figs 8 and 10 at 20 x 20 grids and
(0.7, 0.3) (0.1, 0.5) (0.5, 0.5) (0.9, 0.5)
0.501617 0.501622
0.748272 0.748274
0.623106 0.623047
0.501617 0.501622
0.500067 0.500074
0.716947 0.716759
0.509499 0.510522
0.500063 0.500074
at time level
t = 1.0
with
t = 0.001 for Re = 1000 .
(0.3, 0.7)
0.748272 0.748274
0.717266 0.716759
From these figures it is obvious that numerical solutions | (0.7, 0.7) | 0.623106 | 0.623047 | 0.509314 | 0.510522 |
are in excellent agreement with the corresponding analyt- | (0.1, 0.9) | 0.749988 | 0.749988 | 0.749738 | 0.749742 |
ical solutions. | (0.5, 0.9) | 0.748272 | 0.748274 | 0.717530 | 0.716759 |
(0.9, 0.9) | 0.623106 | 0.623047 | 0.509172 | 0.510522 |
Table 1.
The numerical results for u in comparison with the exact solution
Table 4
The numerical results for v in comparison with the exact solution
at t = 0.01 and t = 1 with
t = 0.0001 ,and Re = 10 .
at t = 0.01 and t = 1.0 with
t = 0.0001 ,and Re = 100 .
(x, y) t=0.01 t=1.0 Numerical Exact Numerical Exact | (x, y) t=0.01 t=1.0 Numerical Exact Numerical Exact | |
(0.1, 0.1) 0.624805 0.624805 0.605626 0.605626 (0.5, 0.1) 0.594202 0.594202 0.576840 0.576840 (0.9, 0.1) 0.567082 0.567082 0.553017 0.553017 (0.3, 0.3) 0.624805 0.624805 0.605627 0.605626 (0.7, 0.3) 0.594202 0.594202 0.576840 0.576840 (0.1, 0.5) 0.655431 0.655431 0.636685 0.636685 (0.5, 0.5) 0.624805 0.624805 0.605628 0.605626 (0.9, 0.5) 0.594202 0.594202 0.576840 0.576840 (0.3, 0.7) 0.655431 0.655431 0.636687 0.636685 (0.7, 0.7) 0.624805 0.624805 0.605629 0.605626 (0.1, 0.9) 0.682611 0.682611 0.666353 0.666353 (0.5, 0.9) 0.655431 0.655431 0.636687 0.636685 (0.9, 0.9) 0.624805 0.624805 0.605628 0.605626 | (0.1, 0.1) 0.876894 0.876953 0.989693 0.989478 (0.5, 0.1) 0.998383 0.998378 0.999928 0.999926 (0.9, 0.1) 0.999989 0.999989 1.00000 0 1.000000 (0.3, 0.3) 0.876894 0.876953 0.990176 0.989478 (0.7, 0.3) 0.998383 0.998378 0.999933 0.999926 (0.1, 0.5) 0.751728 0.751726 0.783053 0.783241 (0.5, 0.5) 0.876894 0.876953 0.990501 0.989478 (0.9, 0.5) 0.998383 0.998378 0.999937 0.999926 (0.3, 0.7) 0.751728 0.751726 0.782734 0.783241 (0.7, 0.7) 0.876894 0.876953 0.990686 0.989478 (0.1, 0.9) 0.750012 0.750012 0.750262 0.750258 (0.5, 0.9) 0.751728 0.751726 0.782470 0.783241 (0.9, 0.9) 0.876894 0.876953 0.990828 0.989478 |
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ISSN 2229-5518
Table 5.
Comparison of computed values of u for Re = 50 at
t = 0.625 .
(x, y) | Present work | A.R.Bahadir | Jain and Holla |
(0.1, 0.1) | 0.97146 | 0.96688 | 0.97258 |
(0.3, 0.1) | 1.15280 | 1.14827 | 1.16214 |
(0.2, 0.2) | 0.86308 | 0.85911 | 0.86281 |
(0.4, 0.2) | 0.97984 | 0.97637 | 0.96483 |
(0.1, 0.3) | 0.66316 | 0.66019 | 0.66318 |
(0.3, 0.3) | 0.77232 | 0.76932 | 0.77030 |
(0.2, 0.4) | 0.58181 | 0.57966 | 0.58070 |
(0.4, 0.4) | 0.75860 | 0.75678 | 0.74435 |
Table 6.
Comparison of computed values of v for Re = 50 at
t = 0.625 .
Fig.1. The numerical value of u for Re = 100 at time level
(x, y) Present work A.R.Bahadir Jain and Holla
t = 0.5 with
t = 0.0001 .
(0.1, 0.1) (0.3, 0.1) (0.2, 0.2) (0.4, 0.2) (0.1, 0.3) (0.3, 0.3) (0.2, 0.4) (0.4, 0.4)
0.09869
0.14158
0.16754
0.17110
0.26378
0.22655
0.32851
0.32501
0.09824
0.14112
0.16681
0.17065
0.26261
0.22576
0.32745
0.32441
0.09773
0.14039
0.16660
0.17397
0.26294
0.22463
0.32402
0.31822
Table 7.
Comparison of computed values of u for Re = 500 at
t = 0.625 .
(x, y) Present work A.R.Bahadir Jain and Holla
Fig.2. The exact value of u for Re = 100 at time level
t = 0.5 with
t = 0.0001 .
Table 8.
Comparison of computed values of v for Re = 500 at
t = 0.625 .
(x, y) Present work A.R.Bahadir Jain and Holla
Fig.3. The numerical value of v for Re = 100 at time level
t = 0.5 with
t = 0.0001 .
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International Journal of Scientific & Engineering Research Volume 2, Issue 6, June-2011 5
ISSN 2229-5518
Fig.4. The exact value of v for Re = 100 at time level
Fig.7. The numerical value of u at 20 x 20 grids for
t = 0.5 with
t = 0.0001 .
Re = 1000 and at time level t = 1.0 with
t = 0.001 .
Fig.5. The computed value of u for Re = 50
at time level
Fig.8. The exact value of u at 20 x 20 grids for
t = 0.625 .
Re = 1000 and at time level t = 1.0 with
t = 0.001 .
Fig.6. The computed value of v for Re = 50
at time level
Fig.9. The numerical value of v at 20 x 20 grids for
t = 0.625 .
Re = 1000 and at time level t = 1.0 with
t = 0.001 .
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Fig.10. The exact value of v at 20 x 20 grids for
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A semi-implicit finite-difference method based on Ozis [7] has been presented for solving two-dimensional coupled nonlinear viscous Burgers’ equations. The efficiency and numerical accuracy of the present scheme are validated through three numerical examples. Numerical results are compared well with those from the exact solutions and previous available results.
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