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

AbstractIn this paper, a semi-implicit finite-difference method is used to find the numerical solution of two-dimensional Coupled Burgers’ equation. The proposed scheme forms a system of linear algebraic difference equations to be solved at each time-step. The linear system is solved by direct method. Numerical results are compared with those of exact solutions and other available results. The present method per- forms well. The proposed scheme can be extended for solving non-linear problems arising in mechanics and other areas of engineering and science.

Index Terms Burgers’ equations; fnite- difference; semi-implicit scheme; Reynolds number.

—————————— ——————————

1 INTRODUCTION

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

IJSER © 2011 http://www.ijser.org

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

2 THE SOLUTION PROCEDURE

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

3 NUMERICAL EXAMPLES AND DISCUSSION

= {( 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

IJSER © 2011 http://www.ijser.org

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

IJSER © 2011 http://www.ijser.org

International Journal of Scientific & Engineering Research Volume 2, Issue 6, June-2011 4

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 .

IJSER © 2011 http://www.ijser.org

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 .

IJSER © 2011 http://www.ijser.org

International Journal of Scientific & Engineering Research Volume 2, Issue 6, June-2011 6

ISSN 2229-5518

Fig.10. The exact value of v at 20 x 20 grids for

difference methods,” Journal of Computational and Applied

Mathematics 1999; 103:251-261.

[12] Kutluay S, Rsen A,”A linearized numerical scheme for Burgers- like equations,” Applied Mathematics and Computation 2004;

156:295-305.

[13] Wenyuan Liao,” An implicit fourth-order compact finite differ- ence scheme for one-dimensional Burgers’ equation,” Applied Mathematics and Computation 2008; 206:755-764.

[14] C.A.J. Fletcher, “Generating exact solutions of the two- dimensional Burgers’ equation”, Int. J. Numer. Meth. Fluids 3 (1983) 213–216.

[15] P.C. Jain, D.N. Holla, “Numerical solution of coupled Burgers_

equations, “Int. J. Numer. Meth.Eng. 12 (1978) 213–222.

[16] C.A.J. Fletcher,” A comparison of finite element and finite dif- ference of the one- and two-dimensional Burgers’ equations,” J. Comput. Phys., Vol. 51, (1983), 159-188.

Re = 1000 and at time level t = 1.0 with

4 CONCLUSION

t = 0.001 .

[17] F.W. Wubs, E.D. de Goede, “An explicit–implicit method for a class of time-dependent partial differential equations, “Appl. Numer. Math. 9 (1992) 157–181.

[18] O. Goyon,” Multilevel schemes for solving unsteady equa-

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.

REFERENCES

[1] Cole JD, “On a quasilinear parabolic equations occurring in aerodynamics,” Quart Appl Math 1951; 9:225-36.

[2] J.D. Logan, ”An introduction to nonlinear partial differential equations, “ Wily-Interscience, New York, 1994.

[3] L. Debtnath, ”Nonlinear partial differential equations for scien- tist and engineers,” Birkhauser, Boston, 1997.

[4] G. Adomian, ”The diffusion-Brusselator equation,” Comput.

Math. Appl. 29(1995) 1-3.

[5] Bateman H., “some recent researches on the motion of fluids,” Monthly Weather Review 1915; 43:163-170

[6] Burger JM, “A Mathematical Model Illustrating the Theory of

Turbulence, ”Advances in Applied mathematics 1950; 3:201-230 [7] T.Ozis, Y.Aslan, “The semi-approximate approach for solving Burgers’ equation with high Reynolds number,” Appl. Math.

Comput. 163 (2005) 131-145.

[8] Hon YC, Mao XZ,”An efficient numerical scheme for Burgers- like equations,”Applied Mathematics and Computation 1998;

95:37-50.

[9] Aksan EN, Ozdes A, ”A numerical solution of Burgers’ equa- tion,”Applied Mathematics and Computation 2004; 156:395-402.

[10] Mickens R, ”Exact solutions to difference equation models of Burgers’ equation,”Numerical Methods for Partial Differential Equations 1986; 2(2):123-129.

[11] Kutluay S, Bahadir AR, Ozdes A,”Numerical solution of one-

dimensional Burgers’ equation: explicit and exact-explicit finite

tions,” Int. J. Numer. Meth. Fluids 22(1996) 937–959.

[19] Bahadir AR. “A fully implicit finite-difference scheme for two- dimensional Burgers’ equation,” Appl Math Comput 2003;

137:131-7.

[20] Vineet Kumar Srivastava, Mohammad Tamsir, Utkarsh Bhard- waj, YVSS Sanyasiraju, Crank-Nicolson scheme for numerical solutions of two dimensional coupled Burgers’ equations,” In- ternational Journal of Scientific & Engineering Research Vo- lume 2, Issue 5, May-2011.

IJSER © 2011 http://www.ijser.org