Applied Mathematics and Computation 206 (2008) 457–464
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Applied Mathematics and Computation
journal homepage: www.elsevier.com/locate/amc
Use of modified Bernstein polynomials to solve KdV–Burgers
equation numerically
Dambaru Bhatta
Department of Mathematics, The University of Texas-Pan American, Edinburg, TX 78541-2999, USA
a r t i c l e
i n f o
Keywords:
Bernstein polynomials
Galerkin’s method
Korteweg-de Vries Burgers equation
Runge–Kutta method
a b s t r a c t
Numerical solution of Korteweg-de Veries–Burgers (KdVB) equation is presented using
modified Bernstein polynomials (B-polynomials). Over the spatial domain, B-polynomials
are used to expand the desired solution requiring discretization with only the time variable. Galerkin method is used to determine the expansion coefficients to construct initial
trial functions. We use fourth-order Runge–Kutta method to solve the system of equations
for the time variable. The accuracy of the solutions is dependent on the size of the B-polynomial basis set. Numerical results obtained using this method are compared with existing
analytical results. Excellent agreement is found between the exact solution and approximate solution obtained by this method.
Published by Elsevier Inc.
1. Introduction
Many researchers have studied Korteweg-de Vries–Burgers (KdVB) equation because of its importance in various physical
phenomena. KdVB equation models the effects of dispersion, dissipation, and nonlinearity. Johnson [1] examined the travelling wave solution to the KdVB equation in phase plane by means of a perturbation method. Bona and Schonbeck [2] studied the existence and uniqueness of bounded travelling wave solution to KdVB equation. Jeffrey and Xu [3] and Jeffrey and
Mohamad [4] studied the progressive wave solution to the KdVB equation analytically and obtained right going and left
going waves. Demiray [5] presented a progressive wave solution using the hyperbolic tangent method. Helal and Mehanna
[6] described two methods to find the soliton solutions to the KdVB equation, one is a numerical method using finite difference and the other one is a semi-analytic method using Adomian decomposition.
In this paper, we present a numerical approach to solve KdVB equation which requires discretization of the time variable t
only. Modified B-polynomials are used as trial functions and Galerkin method is used with initial data to obtain a system of
equations in spatial variable x. The system of equations has been solved using a smaller time step in a fourth-order Runge–
Kutta scheme. In the following sections, we provide details of the numerical method to solve the nonlinear KdVB equation.
We also present our results and compare with the existing analytical results.
2. Mathematical formulation
We consider the following form of KdVB equation:
ou
ou
o2 u
o3 u
þ l1 u þ l2 2 þ l3 3 ¼ 0;
ot
ox
ox
ox
E-mail address: [email protected]
0096-3003/$ - see front matter Published by Elsevier Inc.
doi:10.1016/j.amc.2008.09.031
ð1Þ
458
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
where
l1 ; l2 , and l3 are constant coefficients with the initial condition:
uðx; 0Þ ¼ u0 ðxÞ:
ð2Þ
The second term of Eq. (1) describes nonlinearity, the third term corresponds to dissipation, and the last term represents
dispersion. As limiting cases, KdVB equation (1) reduces to KdV equation when l2 ! 0 and Burgers equation when
l3 ! 0. We seek a numerical solution to (1) using modified Bernstein polynomials.
2.1. B-polynomials
We use modified version of Bernstein polynomials [see Gelbaum [7]] as our basis function over a closed interval ½a; b.
Modified B-polynomials over ½a; b are extension of regular B-polynomials over ½0; 1: This provides greater flexibility on
the boundary.
The general form of the B-polynomials of nth degree is defined as
Bi;n ðxÞ ¼
n ðx aÞi ðb xÞni
ðb aÞn
i
ð3Þ
n
n!
¼ i!ðniÞ!
. Here n! ¼ 1 2 3 n for n P 1 and 0! ¼ 1.
i
B-polys defined above form a complete basis over the interval ½a; b. There are ðn þ 1Þ polynomials with degree n. For convenience, we set Bi;n ðxÞ ¼ 0 if i < 0 or i > n. The details of these polynomials with their recurrence relations can also be found
in the work of Bhatti and Bracken [8]. It can easily be shown that each of the B-polys is positive and also the sum of all the BP
polys is unity for all real x 2 ½a; b, i.e., ni¼0 Bi;n ðxÞ ¼ 1. It is easy to show that any given polynomial of degree n can be expressed in terms of linear combination of the basis functions. Recurrence formula and derivatives of B-polys are given by
for i ¼ 0; 1; . . . ; n where the binomial coefficients are given by
bx
x
Bi;n1 ðxÞ þ
Bi1;n1 ðxÞ;
ba
ba
n
0
Bi;n ðxÞ ¼
½Bi1;n1 Bi;n1 ;
ba
nðn 1Þ
B00i;n ðxÞ ¼
½Bi2;n2 2Bi1;n2 þ Bi;n2 ;
ðb aÞ2
nðn 1Þðn 2Þ
½Bi3;n3 3Bi2;n3 þ 3Bi1;n3 Bi;n3 :
B000
i;n ðxÞ ¼
ðb aÞ3
Bi;n ðxÞ ¼
ð4Þ
ð5Þ
ð6Þ
ð7Þ
Bhatta and Bhatti [9] presented numerical results of the solution to the KdV equation using modified Bernstein polynomials. In the present work, an approximate solution in terms of linear combination of B-polys is assumed. The initial values
of the unknown coefficients, ai , are obtained with the help of Galerkin method by using the initial data. For time variable, the
system of equations is solved by using fourth-order Runge–Kutta method. In the following sections, the subscript n ¼ N of
Bi;N ðxÞ is dropped for simplicity.
2.2. Galerkin formulation
We assume an approximate solution of the following form:
ua ðx; tÞ ¼
N
X
ai ðtÞBi ðxÞ
ð8Þ
i¼1
for a 6 x 6 b.
Substitution of Eq. (8) into Eq. (1) produces a residual:
R¼
X
a_ i Bi þ l1
i
X
!
ai Bi
i
X
j
!
aj B0j
þ l2
X
i
!
ai B00i
þ l3
X
!
ai B000
i
:
ð9Þ
i
Here represents differentiation with respect to time t and 0 is used to denote differentiation with respect to x. Repeated evaluation of the inner product:
ðR; Bk Þ ¼ 0;
k ¼ 1; . . . ; N;
ð10Þ
produces a system of ordinary differential equations that can be written as
MA_ þ ðl1 D þ l2 E þ l3 FÞA ¼ 0;
where an element of A_ is a_ i and elements of M, D, E, and F are given by
ð11Þ
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
mki ¼ ðBi ; Bk Þ ¼
Z
459
b
Bi Bk dx;
!
X Z
X
aj Bi B0j ; Bk ¼
aj
ð12Þ
a
dki ¼
j
eki ¼ ðB00i ; Bk Þ ¼
fki ¼ ðB000
i ; Bk Þ ¼
Z
j
Bi B0j Bk dx;
ð13Þ
b
a
Z
a
b
B00i Bk dx;
ð14Þ
B000
i Bk dx:
ð15Þ
b
a
The dependence of dki on the summation of the unknown coefficients aj manifests the nonlinearity of the problem. The initial
values of the coefficients ai are obtained by applying the Galerkin method to the initial data:
ðua u0 ; Bk Þ ¼ 0;
ð16Þ
i.e.,
X
!
ai Bi ; Bk
¼ ðu0 ; Bk Þ;
i
X
ai ðBi ; Bk Þ ¼ ðu0 ; Bk Þ:
i
Here u0 ðxÞ ¼ uðx; 0Þ is the initial value of uðx; tÞ.
This yields a system of equations given by
MA ¼ G:
ð17Þ
Here the elements of M are given by (12) and the elements of G are given by
g k ¼ ðu0 ; Bk Þ ¼
Z
b
u0 Bk dx:
ð18Þ
a
3. Analytical solution
Let us take the initial value uðx; 0Þ as
uðx; 0Þ ¼
6l22
l2 x
1
l2 x
2
þ sech
:
1 þ tanh
10l3
2
10l3
25l1 l3
ð19Þ
The analytical solution of this problem with the above initial data has been obtained by various researchers as follows:
uðx; tÞ ¼
6l22
l2
6l22 t
1
l2
6l22 t
2
1 þ tanh
x
x
þ sech
:
25l3
2
25l3
25l1 l3
10l3
10l3
ð20Þ
This result was presented in various works by different authors such as in Eq. (2.11a) of Jeffrey and Xu [3], in Eq. (17a) of
Jeffrey and Mohamad [4], and in Eq. (18) by Demiray [5].
4. Results and discussion
In this section, we present numerical results and compare those with analytical results. We apply the current approach to
nonlinear KdVB equation to compute solutions numerically and then compare these solutions with exact solutions at various
times. We use B-polynomials to form our trial functions for the approximate solution. First initial values of ai are obtained by
solving the system of equations given by MA ¼ G, then we use these initial values to solve the system of equations given by
(11) using a fourth-order Runge–Kutta technique. We solve Eq. (17) by using initial solution u0 ðxÞ as
u0 ðxÞ ¼
6l22
l2 x
1
l2 x
2
þ sech
:
1 þ tanh
10l3
2
10l3
25l1 l3
ð21Þ
Table 1
Values of the constants considered for computation
Case 1
Case 2
Case 3
l1
l2
l3
0.01
1.0
1.0
1.0
1.0
2.0
100.0
1.0
1.0
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D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
In order to demonstrate the dependency of the accuracy of the solution on the number of polynomials, we use different values of N, the number of the polynomials. We consider various values of l1 ; l2 , and l3 to show that the current numerical
procedure works in all these cases. The scenarios we consider are included in Table 1.
4.1. Results for case 1
For our computations of analytical and numerical work presented here, first we use l1 ¼ 0:01; l2 ¼ 1:0; l3 ¼ 100:0. The
time steps are kept smaller so that the errors associated with numerical solutions are negligible compared with the errors in
the approximation of Eq. (8). We choose time step as Dt ¼ 0:001.
We compute the error as the difference between the exact solution and the numerical solution. Varying N, we compute
the differences between the exact solution and the numerical solution. The errors are presented in Table 2 and Fig. 1 for
N ¼ 5, 7.
It can be seen from Fig. 1 that the error is distributed over the entire domain and that the error diminishes with increasing
order N. For this case, we need only seven polynomials (i.e. N ¼ 7) to obtain the numerical results correctly upto the order
1014 . Comparisons of the results obtained by the current numerical procedure and the exact solution with
l1 ¼ 0:01; l2 ¼ 1:0; l3 ¼ 100 for different times using N ¼ 7. These results are presented in Tables 3–5 for times
t ¼ 0:001, 0.08, 1.05, respectively. We see an excellent agreement of the numerical results with the analytical results.
Table 2
Errors in the numerical solutions for various N for case 1
x
N¼5
N¼7
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
4.25826041094978E13
4.38538094726936E15
1.533773108519653E13
1.413313910347824E13
5.639932965095795E14
4.057865155004947E14
1.121325254871408E13
1.406652572200073E13
1.233457780358549E13
6.994405055138486E14
3.219646771412954E15
7.555067682574190E14
1.267319582609616E13
1.406097460687760E13
1.083577672034152E13
3.302913498259841E14
6.727951529228449E14
1.552091788425968E13
1.713629238508929E13
2.947642130379790E14
3.859135233597044E13
2.997602166487922E15
1.110223024625156E16
6.106226635438361E16
2.220446049250313E16
4.996003610813204E16
8.326672684688674E16
7.216449660063518E16
4.996003610813204E16
1.665334536937734E16
4.996003610813204E16
7.216449660063518E16
6.661338147750939E16
2.220446049250313E16
1.665334536937734E16
6.106226635438361E16
6.661338147750939E16
4.440892098500626E16
1.665334536937734E16
6.661338147750939E16
7.216449660063518E16
1.332267629550187E15
N=7
N=5
-11
3.3*10
-11
Error
2.2*10
-11
1.1*10
0
-11
-1.1*10
-11
-2.2*10
-18
-12
-6
0
6
12
18
x ----->
Fig. 1. Errors in the numerical solutions with N ¼ 5, 7 for case 1.
461
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
Table 3
Comparison of numerical and exact solutions of KdVB equation at t ¼ 0:001 for case 1
x
Numerical
Exact
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0.3575880802150530
0.3578303382618294
0.3580723607060486
0.3583141470515330
0.3585556968039936
0.3587970094710335
0.3590380845621520
0.3592789215887491
0.3595195200641290
0.3597598795035044
0.3599999994239998
0.3602398793446566
0.3604795187864354
0.3607189172722210
0.3609580743268260
0.3611969894769939
0.3614356622514037
0.3616740921806729
0.3619122787973615
0.3621502216359760
0.3623879202329717
0.3575880802150531
0.3578303382618296
0.3580723607060488
0.3583141470515334
0.3585556968039939
0.3587970094710337
0.3590380845621522
0.3592789215887493
0.3595195200641293
0.3597598795035045
0.3599999994240000
0.3602398793446566
0.3604795187864352
0.3607189172722209
0.3609580743268257
0.3611969894769936
0.3614356622514032
0.3616740921806724
0.3619122787973610
0.3621502216359755
0.3623879202329713
Table 4
Comparison of numerical and exact solutions of KdVB equation at t ¼ 0:8 for case 1
x
Numerical
Exact
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0.3575876154350110
0.3578298739331014
0.3580718968295877
0.3583136836282889
0.3585552338349122
0.3587965469570571
0.3590376225042194
0.3592784599877957
0.3595190589210860
0.3597594188192997
0.3599995391995578
0.3602394195808976
0.3604790594842763
0.3607184584325748
0.3609576159506022
0.3611965315650981
0.3614352048047378
0.3616736352001351
0.3619118222838463
0.3621497655903736
0.3623874646561693
0.3575876154350127
0.3578298739331028
0.3580718968295888
0.3583136836282899
0.3585552338349131
0.3587965469570579
0.3590376225042202
0.3592784599877962
0.3595190589210864
0.3597594188192998
0.3599995391995576
0.3602394195808969
0.3604790594842751
0.3607184584325733
0.3609576159506000
0.3611965315650955
0.3614352048047347
0.3616736352001316
0.3619118222838426
0.3621497655903700
0.3623874646561660
4.2. Results for case 2
As our second example, we use l1 ¼ 1:0; l2 ¼ 1:0; l3 ¼ 1:0. The errors are presented in Fig. 2 for N ¼ 15, 17, 21.
Observing the dependence of the error on the number of polynomials used in Fig. 2, we choose the number of B-polynomials used in Galerkin method as 21 ðN ¼ 21Þ for our numerical computations in this case. Comparisons of the results obtained by the current numerical procedure and the exact solution with l1 ¼ 1:0; l2 ¼ 1:0; l3 ¼ 1:0 are presented in Table 5
for time t ¼ 0:001 and 0.08.
We see that the numerical results are in agreement with the analytical results.
4.3. Results for case 3
As our third and final example, we use l1 ¼ 1:0; l2 ¼ 2:0; l3 ¼ 1:0. The errors are presented in Fig. 3 for N ¼ 11, 15, 21.
Observing the dependence of the error on the number of polynomials used in Fig. 3, we choose the number of B-polynomials used in Galerkin method as 21 for our numerical computations in this case. Comparisons of the results obtained by the
462
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
Table 5
Comparison of numerical and exact solutions of KdVB equation at t ¼ 1:05 for case 1
x
Numerical
Exact
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0.3575874700092876
0.3578297286485898
0.3580717516865861
0.3583135386270945
0.3585550889758210
0.3587964022403640
0.3590374779302182
0.3592783155567788
0.3595189146333450
0.3597592746751249
0.3599993951992382
0.3602392757247214
0.3604789157725303
0.3607183148655449
0.3609574725285726
0.3611963882883524
0.3614350616735582
0.3616734922148025
0.3619116794446406
0.3621496228975736
0.3623873221100524
0.3575874700092906
0.3578297286485922
0.3580717516865882
0.3583135386270964
0.3585550889758227
0.3587964022403655
0.3590374779302194
0.3592783155567796
0.3595189146333456
0.3597592746751250
0.3599993951992379
0.3602392757247204
0.3604789157725286
0.3607183148655425
0.3609574725285696
0.3611963882883486
0.3614350616735537
0.3616734922147975
0.3619116794446353
0.3621496228975682
0.3623873221100471
-5
3.6*10
N=15
N=17
N=21
-5
2.4*10
-5
Error
1.2*10
0
-5
-1.2*10
-5
-2.4*10
-18
-12
-6
0
6
12
18
x ----->
Fig. 2. Errors in the numerical solutions with N ¼ 15, 17, 21 for case 2.
N=11
N=15
N=21
-2
4.4*10
Error
-2
2.2*10
0
-2
-2.2*10
-2
-4.4*10
-18
-12
-6
0
6
12
18
x ----->
Fig. 3. Errors in the numerical solutions with N ¼ 11, 15, 21 for case 3.
current numerical procedure and the exact solution with l1 ¼ 1:0; l2 ¼ 2:0; l3 ¼ 1:0 are illustrated in Fig. 4 for time
t ¼ 0:1. Numerical results are comparable with the analytical results (see Table 6).
463
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
Solution --->
2
Numerical
Exact
1.5
1
0.5
0
-10
-5
5
0
10
x ----->
Fig. 4. Comparison of numerical and exact solutions for case 3 at t ¼ 0:1.
Table 6
Comparison of numerical and exact solutions of KdVB equation for case 2
t ¼ 0:001
t ¼ 0:08
x
Numerical
Exact
x
Numerical
Exact
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
0.1711146167
0.1835763299
0.1964911402
0.2098062180
0.2234589883
0.2373776517
0.2514820978
0.2656852179
0.2798946003
0.2940145643
0.3079484571
0.3216011147
0.3348813666
0.3477044549
0.3599942360
0.3716850466
0.3827231371
0.3930676049
0.4026907952
0.4115781763
0.4197277288
0.4271489190
0.4338613461
0.4398931679
0.4452794071
0.4500602374
0.4542793331
0.4579823481
0.4612155721
0.1711146186
0.1835763279
0.1964911356
0.2098062138
0.2234589869
0.2373776540
0.2514821023
0.2656852220
0.2798946017
0.2940145623
0.3079484529
0.3216011106
0.3348813650
0.3477044566
0.3599942399
0.3716850506
0.3827231391
0.3930676038
0.4026907917
0.4115781724
0.4197277267
0.4271489198
0.4338613493
0.4398931716
0.4452794091
0.4500602365
0.4542793300
0.4579823449
0.4612155708
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.50
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
0.1706515921
0.1830952173
0.1959937467
0.2092947262
0.2229359437
0.2368459360
0.2509448867
0.2651459237
0.2793568008
0.2934819203
0.3074246220
0.3210896394
0.3343856060
0.3472274791
0.3595387521
0.3712533352
0.3823170067
0.3926883657
0.4023392538
0.4112546498
0.4194320764
0.4268805875
0.4336194260
0.4396764558
0.4450864705
0.4498894789
0.4541290510
0.4578507913
0.4611009876
0.1706515926
0.1830952134
0.1959937407
0.2092947218
0.2229359433
0.2368459401
0.2509448933
0.2651459294
0.2793568032
0.2934819190
0.3074246184
0.3210896364
0.3343856060
0.3472274828
0.3595387582
0.3712533411
0.3823170100
0.3926883652
0.4023392504
0.4112546459
0.4194320746
0.4268805890
0.4336194301
0.4396764601
0.4450864726
0.4498894778
0.4541290475
0.4578507879
0.4611009867
t=0
t=3
t=6
Solution --->
2
1.5
1
0.5
0
-10
-5
0
5
10
x ----->
Fig. 5. Solutions at various times for case 3.
15
464
D. Bhatta / Applied Mathematics and Computation 206 (2008) 457–464
Solutions for this case at various times are depicted in Fig. 5. The solutions travel to the right as seen in Fig. 5.
The numerical solutions obtained by using the procedure presented in this work show excellent agreement with exact
solutions. Throughout the calculation, the time steps are kept sufficiently smaller that the errors associated with the numerical integration are negligible. During the calculation, it is noticed that with larger time steps, the solution may not converge.
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