Journal of Computational Methods in Physics

Volume 2014 (2014), Article ID 853198, 11 pages

http://dx.doi.org/10.1155/2014/853198

Research Article

A Single Sweep AGE Algorithm on a Variable Mesh Based on Off-Step Discretization for the Solution of Nonlinear Burgers’ Equation

R. K. Mohanty¹ and Jyoti Talwar²

¹Department of Applied Mathematics, Faculty of Mathematics & Computer Science, Akbar Bhawan, Chanakyapuri, New Delhi 110021, India

²Department of Mathematics, Faculty of Mathematical Sciences, University of Delhi, Delhi 110007, India

Received 4 October 2013; Accepted 1 December 2013; Published 16 January 2014

Academic Editor: Marta B. Rosales

Copyright © 2014 R. K. Mohanty and Jyoti Talwar. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We discuss a new single sweep alternating group explicit iteration method, along with a third-order numerical method based on off-step discretization on a variable mesh to solve the nonlinear ordinary differential equation subject to given natural boundary conditions. Using the proposed method, we have solved Burgers’ equation both in singular and nonsingular cases, which is the main attraction of our work. The convergence of the proposed method is discussed in detail. We compared the results of the proposed iteration method with the results of the corresponding double sweep alternating group explicit iteration methods to demonstrate computationally the efficiency of the proposed method.

1. Introduction

Consider the general nonlinear ordinary differential equation

subject to essential boundary conditions

where are finite constants.

We assume that for

(i) is continuous,(ii) and exist and are continuous,(iii) and for some positive constant .

These conditions ensure that the boundary value problem (1) and (2) possesses a unique solution (see Keller [1]).

With the advent of parallel computers, scientists are focusing on developing finite difference methods with the property of parallelism. Working on this, in the early 1980s, Evans [2, 3] introduced the Group Explicit methods for large linear system of equations. Further he discussed the Alternating Group Explicit (AGE) method to solve periodic parabolic equations in a coupled manner. Mohanty and Evans applied AGE method along with various high order methods [4, 5] for the solution of two-point boundary value problems. Later, Sukon and Evans [6] introduced a Two-parameter Alternating Group Explicit (TAGE) method for the two-point boundary value problem with a lower order accuracy scheme. In 2003 Mohanty et al. [7] discussed the application of TAGE method for nonlinear singular two point boundary value problems using a fourth-order difference scheme. In 1990, Evans introduced the Coupled Alternating Group Explicit method [8] and applied it to periodic parabolic equations. Many scientists are applying these parallel algorithms to solve ordinary and partial differential equations [9–11].

Recently, Mohanty [12] has proposed a high order variable mesh method for nonlinear two-point boundary value problem. Mohanty and Khosla [13, 14] also devised a new third-order accurate arithmetic average variable mesh method for the solution of the boundary value problem (1) and (2), using three grid points, which is applicable to both singular and nonsingular problems. No special technique is required to handle singular problems. They also discussed the application of two-parameter double sweep alternating group explicit methods. Here, we discuss a new single sweep group explicit iteration method along with a third-order accurate variable mesh method based on two extra off-step grid points for the solution of the boundary value problem (1) and (2).

2. Off-Step Discretization

Consider the solution interval with a nonuniform mesh such that . Let , , be the mesh size and let , , be the mesh ratio. Grid points are given by , . Let be the exact solution of at the grid point and approximated by . Let and .

The new third-order method is described as follows.

For , we first obtain the off-step discretization

where , , and .

Now, let

Then, at each interior grid point , the proposed differential equation (1) is discretized by

With the help of (3), from (5) it is easy to verify that , provided . However, for uniform mesh , the local truncation error associated with (5) becomes (see Chawla and Shivakumar [15]).

Note that the proposed method (5) is applicable to both singular and nonsingular problems. We do not require any special technique to handle singular problems (see Mohanty [13]). Further note that, the boundary values are given by and . If the differential equation (1) is linear we use the proposed single sweep AGE iterative method and in nonlinear case, we use the Newton-AGE iterative method to obtain the solution.

3. Application to Singular Problems

3.1. Linear Singular Problems

We now discuss the application of the proposed numerical method (5) to the linear differential equation with variable coefficients

where represents a forcing function.

For and for and , the equation above represents singular equation in cylindrical and spherical symmetry, respectively.

Let us denote

Therefore applying the method (5) to the differential equation (6) and neglecting error term, we obtain a linear difference equation of the form

where

The linear difference scheme (8) has a local truncation error of and is free from the terms and therefore can very easily be solved for in the region .

3.2. Nonlinear Singular Problems

Now we consider the application to nonlinear differential equation (1). Neglecting the error term, we may rewrite the nonlinear difference equation (5) as

Let

We denote

The Jacobian may be represented as

4. Single Sweep AGE Method

4.1. Description of the Method

In this section we discuss the single sweep AGE iteration method. Using the boundary conditions , , the linear difference equation (8) may be written in the matrix form as

where

To implement the single sweep AGE iterative method, we split the coefficient matrix into two submatrices , where and satisfy the following conditions.

(i) and are nonsingular for suitable choice of and .(ii)For any vectors and and , , it is “convenient” to solve the system explicitly; that is, and for vectors and , respectively.

We will be concerned here with the situation where and are small block systems.

Now we discuss the case when is even (with , ).

Let

So that the system (14) can be rewritten as

Then a two-parameter AGE method for solving the afore mentioned system may be written as

where is an intermediate vector.

Eliminating and combining (18) and (19), we obtain the iterative method

or

where

The new iterative method (20) or (21) is called the two-parameter single sweep AGE iterative method and the matrix is called the iteration matrix.

Now we discuss the single sweep AGE algorithm, when is even.

For simplicity, we denote

and we define for .

By carrying out the necessary algebra in (20), we obtain the following algorithm.

For ,

For .

Let

and then

Finally, for ,

Similarly, we can write the single sweep AGE algorithm when is odd.

4.2. Convergence Analysis

The single sweep AGE iteration method is given by

or

where

The matrix is called the iteration matrix.

To prove the convergence of the method, we need to prove that , where denotes the spectral radius of .

Lemma 1. Let be sufficiently small. Then, the eigenvalues of and are all real.

Proof. Consider

Therefore, we have , for .

Let , be the eigenvalues of . Then ’s are the roots of the quadratic equation

Simplifying, we get

The discriminants of the quadratic equations are

Hence, the eigenvalues of are real.

In a similar manner, we can show that the eigenvalues of are real.

Now we give the sufficient condition for the convergence of the method.

Theorem 2. Let and , be the eigenvalues of and , respectively. If

then the iterative method is convergent for the system (14).

Proof. Let

Since the off-diagonal entries of are negative, therefore , . Therefore, the diagonal entries of are positive.

The iteration matrix is given by

Define

where and , and

It is easy to verify that and are symmetric. Therefore, the matrices and are also symmetric. Hence,

Also, is symmetric, and therefore

Therefore, we have

From (35) and (36), we have and for .

Hence,

Also, from (37), we have

Hence, we conclude that

Thus,

Similarly, we can prove that

Using (44), (48), and (49), we get .

Hence, the convergence of the method (20) follows.

4.3. Application of Single Sweep Newton-AGE Method

Now we discuss the single sweep Newton-AGE iterative method for the nonlinear difference equation (11). We follow the approaches given by Evans [8, 16].

Let us define

Then the Jacobian of can be written as the th-order tridiagonal matrix

Now with any initial vector , we define

where is the solution of the nonlinear system

For the single sweep Newton-AGE method, we consider the case when is even. We split the matrix as , where

and then we write single sweep Newton-AGE method as

where are relaxation parameters and and are nonsingular.

Since and consist of submatrices, they can be easily inverted as

with , , , and .

The matrices and can be evaluated in a manner suitable for parallel computing. In order for this method to converge, it is sufficient that the initial vector be close to the solution.

Similarly, we can write the single sweep Newton-AGE algorithm when is odd.

5. Results and Observations

We have applied the methods to the following three examples, whose exact solutions are known to us, and have compared the results with the corresponding double sweep AGE and Newton-AGE method [14]. For single sweep Newton-AGE method, we use the technique given by Evans [16]. The right hand side function and boundary conditions may be obtained using the exact solutions. Here, we have taken constant, . The value of the first mesh spacing on the left is given by

Therefore, given the value of and , we can calculate from the above relation and the remaining mesh points are determined by , . The initial vector is used in all iterative methods and the iterations were stopped when the absolute error tolerance was achieved.

Example 1 (linear singular problem). Consider

The exact solution is . The root mean square errors (RMSE), the maximum absolute errors (MAE), and number of iterations both for single and double sweep AGE methods are tabulated in Table 1 for , , , , , and , . The graph of Example 1 of exact solution versus numerical solution is plotted in Figure 1.

Table 1: Example 1.

TAGE
Single sweep AGE
MAE
RMS errors

Iter

Iter

10
0.441
0.44
38
0.47
0.34
20
0.4062 (−03)
0.2435 (−03)

20
0.286
0.286
59
0.33
0.25
33
0.3604 (−03)
0.1558 (−03)

30
0.25
0.247
77
0.32
0.23
44
0.3592 (−03)
0.1269 (−03)

40
0.218
0.218
75
0.24
0.20
47
0.3592 (−03)
0.1099 (−03)

50
0.203
0.203
86
0.26
0.21
54
0.3592 (−03)
0.9829 (−04)

60
0.202
0.199
96
0.26
0.20
62
0.3592 (−03)
0.8973 (−04)

70
0.194
0.194
100
0.26
0.20
65
0.3592 (−03)
0.8307 (−04)

80
0.149
0.149
79
0.25
0.17
62
0.3592 (−03)
0.7771 (−04)

10
0.529
0.524
39
0.51
0.40
23
0.2990 (−02)
0.2910 (−02)

20
0.298
0.297
78
0.34
0.26
44
0.9774 (−03)
0.9580 (−03)

30
0.213
0.213
113
0.29
0.20
63
0.6774 (−03)
0.6672 (−03)

40
0.169
0.168
146
0.16
0.15
81
0.5925 (−03)
0.5855 (−03)

50
0.142
0.141
178
0.14
0.13
101
0.5633 (−03)
0.5579 (−03)

60
0.124
0.124
180
0.14
0.12
113
0.5525 (−03)
0.5481 (−03)

70
0.112
0.112
201
0.14
0.12
118
0.5484 (−03)
0.5446 (−03)

80
0.103
0.103
221
0.13
0.10
128
0.5469 (−03)
0.5435 (−03)

10
0.57
0.37
47
0.46
0.34
26
0.4258 (−03)
0.3739 (−03)

20
0.31
0.23
115
0.25
0.239
75
0.3371 (−04)
0.2934 (−04)

30
0.20
0.19
204
0.203
0.183
128
0.7197 (−05)
0.6256 (−05)

40
0.17
0.15
309
0.19
0.17
225
0.2367 (−05)
0.2058 (−05)

50
0.14
0.14
445
0.1749
0.173
362
0.9928 (−06)
0.8634 (−06)

60
0.13
0.13
623
0.179
0.164
503
0.4864 (−06)
0.4232 (−06)

70
0.125
0.124
842
0.177
0.163
681
0.2654 (−06)
0.2310 (−06)

80
0.124
0.123
1117
0.176
0.163
888
0.1567 (−06)
0.1365 (−06)

10
0.59
0.42
40
0.45
0.39
24
0.5725 (−03)
0.5344 (−03)

20
0.32
0.25
86
0.34
0.25
45
0.4488 (−04)
0.4152 (−04)

30
0.20
0.19
131
0.21
0.18
73
0.9532 (−05)
0.8792 (−05)

40
0.16
0.14
180
0.16
0.14
99
0.3125 (−05)
0.2877 (−05)

50
0.12
0.12
228
0.11
0.10
126
0.1307 (−05)
0.1202 (−05)

60
0.1
0.1
274
0.10
0.09
151
0.6392 (−06)
0.5876 (−06)

70
0.08
0.08
332
0.09
0.08
176
0.3485 (−06)
0.1890 (−06)

80
0.07
0.07
382
0.08
0.07
202
0.2058 (−06)
0.1890 (−06)

Figure 1: Graph of exact solution and the numerical solution for alpha = 2.0, sigma = 1.1, and for Example 1.

Example 2 (Burgers’ equation). Consider

The exact solution is given by , where . The RMSE, MAE, and the number of iterations for both single and double sweep AGE methods are tabulated in Table 2 for , , , , and , . The graph of Example 2 of exact solution versus numerical solution for , , is plotted in Figure 2.

Table 2: Example 2.

TAGE
Single sweep AGE
MAE
RMS errors

Iter

Iter

10
0.0578
0.0585
6
0.0568
0.0568
6
0.1067(−03)
0.6084(−04)

20
0.0368
0.0374
7
0.0281
0.0285
7
0.8511(−04)
0.3726(−04)

30
0.0318
0.0321
8
0.0320
0.0315
7
0.8442(−04)
0.3028(−04)

40
0.0282
0.0286
9
0.0170
0.0170
8
0.8440(−04)
0.2622(−04)

50
0.0236
0.0239
10
0.0160
0.0166
8
0.8440(−04)
0.2345(−04)

60
0.0161
0.0162
10
0.0161
0.0166
8
0.8440(−04)
0.2141(−04)

10
0.0152
0.0155
5
0.016
0.016
4
0.8156(−03)
0.3181(−03)

20
0.0093
0.0096
6
0.009
0.0089
5
0.3228(−04)
0.1508(−04)

30
0.0063
0.0068
7
0.006
0.0059
7
0.8624(−05)
0.4435(−05)

40
0.0044
0.0045
10
0.0032
0.0042
9
0.5159(−05)
0.2771(−05)

50
0.0036
0.0035
12
0.002
0.0031
12
0.4238(−05)
0.2232(−05)

60
0.0029
0.0029
10
0.0027
0.00278
15
0.3917(−05)
0.1968(−05)

10
0.006
0.014
5
0.005
0.014
4
0.3335(−02)
0.1197(−02)

20
0.012
0.012
5
0.008
0.0089
4
0.3326(−03)
0.9523(−04)

30
0.008
0.008
6
0.008
0.008
4
0.7487(−04)
0.2147(−04)

40
0.0061
0.0061
7
0.0051
0.0053
5
0.2502(−04)
0.7150(−05)

50
0.0049
0.0049
8
0.0058
0.0049
5
0.1055(−04)
0.3012(−05)

60
0.0041
0.0041
9
0.0043
0.0041
5
0.5186(−05)
0.1478(−05)

Figure 2: Graph of the exact solution and the numerical solution for , sigma = 1.0, and for Example 2.

Example 3 (Burgers’ equation in polar coordinates). Consider

The exact solution is given by . The RMSE, MAE, and the number of iterations for both single and double sweep AGE methods are tabulated in Table 3 for various values of , and . The graph of Example 3 of exact solution versus numerical solution for , , , is plotted in Figure 3.

Table 3: Example 3.

TAGE
Single sweep AGE
MAE
RMS errors

Iter

Iter

10
0.0368
0.0373
8
0.050
0.052
5
0.8036(−03)
0.2692(−03)

20
0.0329
0.0334
9
0.034
0.046
6
0.7645(−03)
0.1770(−03)

30
0.0297
0.0311
10
0.017
0.034
7
0.7605(−03)
0.1436(−03)

40
0.0283
0.0287
10
0.023
0.0242
8
0.7602(−03)
0.1243(−03)

50
0.0261
0.0269
10
0.025
0.027
8
0.7602(−03)
0.1112(−03)

60
0.0242
0.0251
10
0.0211
0.0218
9
0.7602(−03)
0.1015(−03)

10
0.0554
0.0562
7
0.0555
0.0559
5
0.6954(−03)
0.2261(−03)

20
0.0457
0.0464
9
0.035
0.037
7
0.4545(−03)
0.1034(−04)

30
0.0403
0.0412
10
0.027
0.029
8
0.4245(−03)
0.7896(−04)

40
0.0381
0.0393
10
0.025
0.030
9
0.4198(−03)
0.6765(−04)

50
0.0359
0.0364
10
0.03
0.033
9
0.4191(−03)
0.6040(−04)

60
0.0326
0.0335
10
0.022
0.029
10
0.4190(−03)
0.5512(−04)

10
0.0204
0.0211
8
0.016
0.020
7
0.8035(−03)
0.3961(−03)

20
0.0183
0.0189
8
0.011
0.014
6
0.5906(−03)
0.1492(−03)

30
0.0079
0.0084
9
0.007
0.0072
7
0.5252(−03)
0.1101(−03)

40
0.0056
0.0067
11
0.0051
0.0052
9
0.5044(−03)
0.9227(−04)

50
0.0036
0.0041
15
0.0043
0.0043
12
0.4973(−03)
0.8159(−04)

60
0.0034
0.0035
17
0.0032
0.0034
16
0.4949(−03)
0.7419(−04)

10
0.0198
0.0211
8
0.016
0.017
7
0.4542(−03)
0.3755(−03)

20
0.0172
0.0181
8
0.011
0.018
6
0.6976(−04)
0.4929(−04)

30
0.0069
0.0072
9
0.007
0.008
7
0.4018(−04)
0.2422(−04)

40
0.0064
0.0064
10
0.0065
0.0057
8
0.3333(−04)
0.1789(−04)

50
0.0045
0.0043
13
0.0036
0.0041
12
0.3124(−04)
0.1525(−04)

60
0.0035
0.0034
17
0.0043
0.0033
15
0.3054(−04)
0.1372(−04)

10
0.0174
0.0177
10
0.01
0.027
8
0.1217(−02)
0.5654(−03)

20
0.009
0.011
12
0.0113
0.0115
9
0.1044(−03)
0.4913(−04)

30
0.0063
0.0065
14
0.0072
0.0077
10
0.4547(−04)
0.1367(−04)

40
0.0063
0.0069
16
0.0054
0.0056
11
0.2617(−04)
0.6119(−05)

50
0.0042
0.0046
18
0.0044
0.0045
12
0.1697(−04)
0.3416(−05)

60
0.0034
0.0039
21
0.0032
0.0041
13
0.1188(−04)
0.2152(−05)

10
0.017
0.025
9
0.015
0.026
7
0.1234(−02)
0.5767(−03)

20
0.0101
0.011
11
0.0118
0.0118
9
0.1058(−03)
0.4505(−04)

30
0.0071
0.0083
13
0.0075
0.0075
10
0.2644(−04)
0.1025(−04)

40
0.0054
0.0063
15
0.0049
0.0058
11
0.1502(−04)
0.3813(−05)

50
0.0043
0.0049
17
0.0038
0.0049
12
0.9686(−05)
0.1870(−05)

60
0.0035
0.004
20
0.0042
0.0044
11
0.6762(−05)
0.1084(−05)

Figure 3: Graph of the exact solution and the numerical solution for alpha = 2.0, , sigma = 0.9, and for Example 3.

6. Discussion and Conclusion

We have discussed a new single sweep AGE iterative method and three-point off-step method of accuracy on a variable mesh for the solution of nonlinear two point boundary value problems. To demonstrate the efficiency of the method, three examples including two nonlinear and singular cases are presented. The results obtained are compared with the corresponding double sweep AGE method and show superiority over the latter. The double sweep AGE method requires two sweeps to solve a problem, whereas the single sweep AGE method requires only one sweep to solve the problem. Experimentally, as compared to the double sweep method the corresponding single sweep method requires much less number of iterations as it uses less intermediate variables. The method can be extended to solve multidimensional problems and is suitable for use on parallel computers.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was supported by the Council of Scientific and Industrial Research under research Grant no. 09/045(0836)2009-EMR-I.

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