The asymptotic expansion and numerical verification method for linear and nonlinear initial value problem

Applied Mathematics and Computation 180 (2006) 29–37 www.elsevier.com/locate/amc

The asymptotic expansion and numerical verification method for linear and nonlinear initial value problem I.H. Abdel-Halim Hassan Department of Mathematics, Faculty of Science, Zagazig University, Zagazig, Egypt

Abstract In this paper, the asymptotic expansion technique is applied to verify the order of accuracy of asymptotic expansion of linear and nonlinear initial value problems. Numerical solutions are obtained to illustrate the preciseness and effectiveness of verifying the validity of asymptotic expansions. Ó 2006 Published by Elsevier Inc. Keywords: Asymptotic expansion; Linear and nonlinear initial value problems; Regular perturbation method; Duffing’s equation; Runge–Kutta method

1. Introduction The perturbation method is iterative methods for obtaining approximate solution to problems involving a small parameter. The procedure of this method, expand the dependent variable in a power series depending on the small parameter e. One assumes the solution of the form 1 X yðtÞ ¼ en y n ðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ e3 y 3 ðtÞ þ    . ð1:1Þ n¼0

Substitute this series into the original equation, expand all of equations and equate the terms corresponding to different power of the small parameter en. One considers a part of solution by choosing a value of n, say, N. In this case, the order of approximation is O(eN+1). We can choose a test case and compare the asymptotic solution to either exact solution or the numerical solution, showing that the error is small. However, such a comparison may not be enough to verify the asymptotic of the expansion. Although the quantitative error may be small, it does not necessarily become small at the rate expected. Therefore, we need to further verify that the solution is indeed asymptotically accurate to the order to which it is constructed. Methods for constructing asymptotic expansions are discussed in a number of references (see [1,4–7]). The perturbation solution of

E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2006 Published by Elsevier Inc. doi:10.1016/j.amc.2005.11.146

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the linear initial value problems (LIVPs) or nonlinear initial value problems (NLIVPs)-involving a small parameter e is obtained by expanding the solution y(t) as the power series in e as the form 1 X yðtÞ ¼ en y n ðtÞ; ð1:2Þ n¼0

where y 0 ð0Þ ¼ 1;

y 00 ð0Þ ¼ 0;

n X

y i ð0Þ ¼ 0;

i¼0

n X

y 0i ð0Þ ¼ 0.

ð1:3Þ

i¼0

The first order asymptotic solution to the LIVPs or NLIVPs for N = 1, 2, 3, 4, 5 are N1 :

yðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ Oðe2 Þ;

N2 :

yðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ Oðe3 Þ;

N3 :

yðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ e3 y 3 ðtÞ þ Oðe4 Þ;

N4 :

2

yðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e y 2 ðtÞ þ e y 3 ðtÞ þ e y 4 ðtÞ þ Oðe Þ;

N5 :

yðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ e3 y 3 ðtÞ þ e4 y 4 ðtÞ þ e5 y 5 ðtÞ þ Oðe6 Þ.

3

4

ð1:4Þ 5

For general N the asymptotic solution is N X yðtÞ ¼ en y n ðtÞ þ OðeN þ1 Þ;

ð1:5Þ

n¼0

with error of order O(eN+1). A technique was first introduced in SIAM Review by Bosley [2]. In [3], Deeba and Xie have discussed the technique to verify the order of accuracy of the asymptotic expansion of Van der Pol’s equation. In the present paper, will be applied to verify the order of the accuracy of the asymptotic expansion and special values of the perturbation solution of the following two problems are considered: Problem 1. Consider the LIVP: d2 yðtÞ þ ð1  eðtÞÞyðtÞ ¼ 0; dt2

ð1:6Þ

with IC’s yð0Þ ¼ 1;

dyð0Þ ¼ 0. dt

ð1:7Þ

Problem 2. Consider the nonlinear oscillator problem, Duffing’s equation, NLIVP: d2 yðtÞ þ yðtÞ þ ey 3 ðtÞ ¼ 0; dt2

ð1:8Þ

with IC’s yð0Þ ¼ 1;

dyð0Þ ¼ 0. dt

ð1:9Þ

The outline of this paper is as follows. In Section 2, we describe the order verification method. In Section 3, we discuss the asymptotic solution of the LIVP (1.6) and (1.7). In Section 4, we discuss the asymptotic solution of the NLIVP (1.8) and (1.9). In this paper, the job is carried out by using the MAPLE. 2. Description of order verification method As in [3], the technique of order verification enables us to verify that the order of the accuracy of the asymptotic expansion for LIVPs or NLIVPs with N terms is indeed N + 1. Expansion (1.5) implies that the error of the asymptotic expansion for LIVPs or NLIVPs is given by

I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

 Error ¼ EN ðt; eÞ ¼ y

   N X    n  ¼ y ðt; eÞ  y ðt; eÞ ðt; eÞ  e y ðtÞ  ¼ OðeN þ1 Þ ¼ KeN þ1 ; exact asymptotic n   exact n¼0

31

ð2:1Þ

where K is a constant. We can verify the order N + 1 by taking the logarithm of both sides of Eq. (2.1), thus logðErrorÞ ¼ log ðEN ðt; eÞÞ ¼ log K þ ðN þ 1Þ log e.

ð2:2Þ

From (2.2), we note that for a fixed value of t and for a series of small values of e the value of log(EN(t, e)) as a function of log(e) is linear with slop (N + 1). Therefore, we graph log(EN(t, e)) corresponding to log(e) for different values of e, these, points should be nearly a line and the linear equation, that interpolates these points using least-square fit should have slop (N + 1). 3. The asymptotic solution and the order verification for LIVP In this section, the technique of order verification is applied to the LIVP (1.6) and (1.7), We first find the perturbative solution of the LIVP (1.6) and (1.7) by expanding y(t) as a power series in e: 1 X yðtÞ ¼ en y n ðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ e3 y 3 ðtÞ þ    ; ð3:1Þ n¼0

where y 0 ¼ 1;

y 00 ð0Þ ¼ 0; . . . ; y n ð0Þ ¼ y 0n ð0Þ ¼ 0

for all n P 1.

ð3:2Þ

Substituting (3.1) into the differential Eq. (1.6) and equating the coefficients of like powers of e gives a sequence of linear differential equations for yi(t) (i = 0, 1, 2, 3, 4, 5): Oðe0 Þ :

y 000 ðtÞ þ y 0 ðtÞ ¼ 0; y 0 ð0Þ ¼ 1; y 00 ð0Þ ¼ 0;

ð3:3Þ

Oðe1 Þ :

y 001 ðtÞ þ y 1 ðtÞ ¼ ty 0 ðtÞ; y 1 ð0Þ ¼ 0; y 01 ð0Þ ¼ 0;

ð3:4Þ

Oðe2 Þ :

y 002 ðtÞ þ y 2 ðtÞ ¼ ty 1 ðtÞ; y 2 ð0Þ ¼ 0; y 02 ð0Þ ¼ 0;

ð3:5Þ

Oðe3 Þ :

y 003 ðtÞ þ y 3 ðtÞ ¼ ty 2 ðtÞ; y 3 ð0Þ ¼ 0; y 03 ð0Þ ¼ 0;

ð3:6Þ

Oðe Þ :

y 004 ðtÞ

¼ 0;

ð3:7Þ

Oðe5 Þ :

y 005 ðtÞ þ y 5 ðtÞ ¼ ty 4 ðtÞ; y 5 ð0Þ ¼ 0; y 05 ð0Þ ¼ 0.

ð3:8Þ

4

þ y 4 ðtÞ ¼ ty 3 ðtÞ; y 4 ð0Þ ¼ 0;

y 04 ð0Þ

Solving Eqs. (3.3)–(3.8), the first six terms of this regular asymptotic solution as y 0 ðtÞ ¼ cos t; 1 1 1 y 1 ðtÞ ¼  t cos t  t2 sin t þ sin t; 4 4 4 7 2 1 4 7 5 y 2 ðtÞ ¼ t cos t  t cos t  t sin t þ t3 sin t; 32 32 32 48 35 3 35 7 5 35 2 7 1 6 35 t cos t þ t cos t  t cos t þ t sin t  t4 sin t þ t sin t  sin t; y 3 ðtÞ ¼  192 128 384 128 96 384 128 1365 2 1 8 119 6 1085 4 1 7 175 5 t cos t þ t cos t  t cos t þ t cos t  t sin t þ t sin t y 4 ðtÞ ¼  2048 6144 9216 6144 512 3072 1225 3 1365 t sin t þ t sin t;  3072 2048 15015 15015 109 8 15015 2 23 7 5005 3 t cos t þ sin t  t sin t  t sin t þ t cos t þ t cos t y 5 ðtÞ ¼  8192 8192 73728 8192 2304 4096 2275 4 1547 5 3661 6 1 10 11 9 t sin t  t cos t  t sin t  t sin t  t cos t. þ 4096 8192 73728 12280 73728

ð3:9Þ ð3:10Þ ð3:11Þ ð3:12Þ

ð3:13Þ

ð3:14Þ

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Therefore, the first order asymptotic solutions to the LIVP (1.6) and (1.7) for N = 1, 2, 3, 4 and 5 as same as in (1.4). Thus, we find the numerical solutions of the LIVP (1.6) and (1.7) at fixed point t = 5 for 140 different values of e starting from 0.005 and ending at 0.145 by step size 0.001. Next, we find the asymptotic solution y asymptotic ðt; eÞ ¼

N X

en y n ðtÞ;

ð3:15Þ

n¼0

for N = 1, 2, 3, 4 and 5 at the same 140 values of e with t also equal to 5. Table 1 lists the errors     N X   en y n ðtÞ EN ð5; eÞ ¼ y numerical    n¼0

ð3:16Þ

for N = 1, 2, 3, 4 and 5 at 15 points of e, instead of all 140. We then plot the logarithmic values of the errors at 15 points, namely, log E1(5, e), log E2(5, e), log E3(5, e), log E4(5, e) and log E5(5, e) as a function of log(e). The results are indicated in Figs. 1 and 2 for log E1(5, e) and log E5(5, e) as a function of log(e), respectively. In Table 1 The table of EN(5, e) for N = 1, 2, 3, 4, 5 for LIVP (1.6) and (1.7) E1(5, e)

E2(5, e)

E3(5, e)

E4(5, e)

E5(5, e)

0.0543764053 0.1654681427 0.2797190589 0.3971760163 0.5178836645 0.6418842654 0.7692175379 0.8999209650 1.034027115 1.171568449 1.312572121 1.457062278 1.605059340 1.756579788 1.911635931

0.0539907512 0.1619972546 0.2700777031 0.3782789589 0.4866456718 0.5952201035 0.7040419730 0.8131482632 0.9225730424 1.032347272 1.142498606 1.253051190 1.364025446 1.475437854 1.587300723

0.0539918102 0.1620258408 0.2702100877 0.3786422221 0.4874177385 0.5699297341 0.7063687640 0.8167226460 0.9277762849 1.039611477 1.152306712 1.265936973 1.380573515 1.496283655 1.613130537

0.0539918151 0.1620262348 0.2702130573 0.3786536296 0.3786536296 0.5366992959 0.7064044617 0.8169631733 0.9281731062 1.040230652 1.153230722 1.267266542 1.382429436 1.498808616 1.616490942

0.053918151 0.1620262253 0.2702129365 0.3786529797 0.3786529797 0.5966930075 0.7064901019 0.8169338047 0.9281181987 1.040134890 1.153072771 1.267017619 1.382051795 1.498253687 1.615697682

y

e 0.005 0.015 0.025 0.035 0.045 0.055 0.065 0.075 0.085 0.095 0.105 0.115 0.125 0.135 0.145

0.8 0.5 0.3 0.0 -0.3 -0.5 -0.8 -1.0 -1.3 -1.5 -1.8 -2.0 -2.3 -2.5 -2.8 -3.0

y = 2.656749259 + 1.060081327 x

-5.0

-4.5

-4.0

-3.5 x

-3.0

-2.5

-2.0

Fig. 1. The graph of log E1(5, e) versus log e for the linear initial value problem (1.6) and (1.7).

y

I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

0.8 0.5 0.3 0.0 -0.3 -0.5 -0.8 -1.0 -1.3 -1.5 -1.8 -2.0 -2.3 -2.5 -2.8

33

y = 2.416288601 + 1.008882209 x

-5.0

-4.5

-4.0

-3.5 x

-3.0

-2.5

-2.0

Fig. 2. The graph of log E5(5, e) versus log e for the linear initial value problem (1.6) and (1.7).

Table 2 The table lists the slope and linear equation for N = 1, 2, 3, 4, 5 for LIVP (1.6) and (1.7) N

Slope

Linear equation

1 2 3 4 5

1.060081327 1.060081327 1.008610426 1.008988829 1.008882209

y = 2.656749259 + 1.060081327x y = 2.395825566 + 1.060081327x y = 2.415194779 + 1.008610426x y = 2.416707894 + 1.008988829x y = 2.416288601 + 1.008882209x

the same way we plot log E2(5, e), log E3(5, e), and log E4(5, e) as a function of log(e) (see Table 2). The slope of the lines are approximated using a least-squares fit of the data. Table 2 lists the slope and linear equations are obtained for N = 1, 2, 3, 4 and 5. Note that, if we interchange the exact solution with the numerical solution we have the slight error in the slope. In this Section, we have shown a method for verifying the correctness of the order of approximation as applied to LIVPs. 4. The asymptotic solution and the order verification for NLIVP In this section, the technique of order verification is applied to the NLIVP (1.8) and (1.9). We first find the perturbative solution of the NLIVP (1.8) and (1.9) by expanding y(t) as a power series in e: yðtÞ ¼

1 X

en y n ðtÞ ¼ y 0 ðtÞ þ ey 1 ðtÞ þ e2 y 2 ðtÞ þ e3 y 3 ðtÞ þ    ;

ð4:1Þ

n¼0

where y 0 ¼ 1;

y 00 ð0Þ ¼ 0; . . . ; y n ð0Þ ¼ y 0n ð0Þ ¼ 0

for all n P 1.

ð4:2Þ

Substituting (4.1) into the differential Eq. (1.8) and equating the coefficients of like powers of e gives a sequence of linear differential equations for yi(t) (i = 0, 1, 2, 3, 4, 5):

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I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

Oðe0 Þ : y 000 ðtÞ þ y 0 ðtÞ ¼ 0; y 0 ð0Þ ¼ 1; y 00 ð0Þ ¼ 0;

ð4:3Þ

Oðe1 Þ : y 001 ðtÞ þ y 1 ðtÞ ¼ y 30 ðtÞ; y 1 ð0Þ ¼ 0; y 00 ð0Þ ¼ 0;

ð4:4Þ

Oðe2 Þ : y 002 ðtÞ þ y 2 ðtÞ ¼ 3y 0 ðtÞy 1 ðtÞ; y 2 ð0Þ ¼ 0; y 02 ð0Þ ¼ 0; 3

Oðe Þ :

y 003 ðtÞ

þ y 3 ðtÞ ¼

3y 0 ðtÞy 21 ðtÞ



3y 20 ðtÞy 2 ðtÞ;

y 3 ð0Þ ¼ 1;

ð4:5Þ y 03 ð0Þ

¼ 0;

ð4:6Þ

Oðe4 Þ : y 004 ðtÞ þ y 4 ðtÞ ¼ y 31 ðtÞ  6y 0 ðtÞy 1 ðtÞy 2 ðtÞ  3y 20 ðtÞy 3 ðtÞ; y 4 ð0Þ ¼ 0; y 04 ð0Þ ¼ 0;

ð4:7Þ

Oðe5 Þ : y 005 ðtÞ þ y 5 ðtÞ ¼ 3y 21 ðtÞy 2 ðtÞ  3y 0 ðtÞy 21 ðtÞ  6y 0 ðtÞy 1 ðtÞy 3 ðtÞ  3y 20 ðtÞy 4 ðtÞ; y 5 ð0Þ ¼ 1; y 05 ð0Þ ¼ 0: ð4:8Þ

Table 3 The table of EN(3, e) for N = 1, 2, 3, 4, 5 for LIVP (1.8) and (1.9) e

E1(3, e)

E2(3, e)

E3(3, e)

E4(3, e)

E5(3, e)

0.039 0.051 0.063 0.075 0.087 0.099 0.111 0.123 0.135 0.147 0.159 0.171 0.183 0.195 0.207

0.9398767e3 0.16046736e2 0.24446912e2 0.34590163e2 0.46467105e2 0.60068065e2 0.56627601e2 0.92401819e2 0.111113700e1 0.131507721e1 0.153572528e1 0.177296426e1 0.202667305e1 0.229672690e1 0.258299694e1

0.47870e5 0.107570e4 0.203777e4 0.345621e4 0.542487e4 0.804048e4 0.1140275e3 0.1561462e3 0.2078239e3 0.201586e3 0.3442850e3 0.4313746e3 0.5326373e3 0.6993205e3 0.7827120e3

0.555e7 0.1761e6 0.4319e6 0.9105e6 0.17225e5 0.13537e5 0.49364e5 0.77105e5 0.115673e4 0.167780e4 0.236484e4 0.325241e4 0.437875e4 0.578602e4 0.752011e4

0.94e8 0.415e7 0.1190e6 0.2810e6 0.5831e6 0.109711e5 0.19174e5 0.31590e5 0.49620e5 0.74913e5 0.109376e4 0.1551944e4 0.214837e4 0.291051e4 0.386870e4

0.22e8 0.29e8 0.95e8 0.255e7 0.611e7 0.1320e6 0.2605e6 0.4803e6 0.8338e6 0.13804e6 0.21972e5 0.83783e5 0.50435e5 0.73375e5 0.104361e4

-3.3 -3.7

y = -0.5295663235 + 1.998836297 x

-4.0

-4.3

y

-4.7 -5.0

-5.3 -5.7 -6.0

-6.3 -6.7 -7.0

-3.0

-2.8

-2.5

-2.3 x

-2.0

-1.8

-1.5

Fig. 3. The graph of log E3(3, e) versus log e for the nonlinear initial value problem (1.8) and (1.9).

I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

35

Solving Eqs. (4.3)–(4.8), the first six terms of this regular asymptotic solution are: y 0 ðtÞ ¼ cos t; 3 1 1 y 1 ðtÞ ¼  t sin t þ cos3 t  cos t; 8 8 8 1 9 9 2 33 25 29 y 2 ðtÞ ¼ cos5 t  t cos2 t sin t  t cos t þ t sin t þ cos t  cos3 t; 64 64 128 256 256 256 15 81 177 55 189 2 1 t sin t cos4 t þ t cos2 t sin t  t sin t  cos5 t þ t cos t þ cos7 t y 3 ðtÞ ¼  512 512 2048 2048 2048 512 9 3 161 53 81 2 t sin t  cos t þ cos3 t  t cos3 t; þ 1024 2048 512 1024 465 17033 81 10431 21 t sin t cos4 t þ cos t  cos7 t  t cos2 t sin t  t sin t cos6 t y 4 ðtÞ ¼ 8192 262144 16384 65536 4096 17673 4203 2 24989 2297 225 2 t sin t þ t cos3 t  cos3 t þ cos5 t  t cos5 t þ 262144 32768 262144 65536 8192 441 3 1 13077 2 27 4 243 3 t sin t þ cos9 t  t cos t þ t cos t þ t sin t cos2 t;  32768 4096 131072 32768 8192 2187 4 27 20925 1 441 2 t cos3 t  t sin t cos8 t  t sin t cos4 t þ cos11 t  t cos7 t y 5 ðtÞ ¼ 262144 32768 262144 32768 65536 115217 2211 1125 3 20385 cos t þ cos4 t þ t sin t cos4 t þ t sin t cos2 t  2097152 262144 65536 131072 231 11709 107 46347 86211 t sin t cos6 t  t sin t  cos9 t þ cos3 t  cos5 t þ 16384 2097152 131072 524288 2097152 17037 2 16695 3 213165 2 459 4 81 t cos5 t þ t sin t þ t cos t  t cos t  t5 sin t þ 262144 1048576 2097152 65536 1310720 172719 2 7155 3 t cos3 t  t sin t cos2 t.  1048576 131072

ð4:9Þ ð4:10Þ ð4:11Þ

ð4:12Þ

ð4:13Þ

ð4:14Þ

y

Therefore, the first order asymptotic solutions to the NLIVP (1.8) and (1.9) for N = 1, 2, 3, 4 and 5 as same as in (1.4). Thus, we find the numerical solutions of the NLIVP (1.8) and (1.9) at fixed point t = 3 for 60 different values of e starting from 0.039 and ending at 0.207 by step size 0.003. Next, we find the asymptotic solution

-9.0 -9.5 -10.0 -10.5 -11.0 -11.5 -12.0 -12.5 -13.0 -13.5 -14.0 -14.5 -15.0 -15.5 -16.0 -16.5

y = -2.706776026 + 4.345232604 x

-3.0

-2.8

-2.5

-2.3 x

-2.0

-1.8

-1.5

Fig. 4. The graph of log E3(3, e) versus log e for the nonlinear initial value problem (1.8) and (1.9).

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I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

-11

y = -3.110128035 + 5.443381101 x

-12

-13 -14

y

-15 -16

-17 -18

-19 -20 -3.0

-2.8

-2.5

-2.3 x

-2.0

-1.8

-1.5

Fig. 5. The graph of log E5(3, e) versus log e for the nonlinear initial value problem (1.8) and (1.9).

Table 4 The table lists the slope and linear equation for N = 1, 2, 3, 4, 5 for NLIVP (1.8) and (1.9) N

Slope

Linear equation

1 2 3 4 5

1.998836297 3.053973060 4.345232604 4.936738091 5.443381101

y = 0.5295663235 + 1.998836297x y = 2.356910270 + 3.053973060x y = 2.706776026 + 4.345232604x y = 2.340274561 + 4.936738091x y = 3.110128035 + 5.443381101x

y asymptotic ðt; eÞ ¼

N X

en y n ðtÞ;

ð4:15Þ

n¼0

for N = 1, 2, 3, 4 and 5 at the same 60 values of e with t also equal to 3. Table 3 lists the errors     N X   en y n ðtÞ EN ð3; eÞ ¼ y numerical    n¼0

ð4:16Þ

for N = 1, 2, 3, 4 and 5 at 15 points of e, instead of all 140. We then plot the logarithmic values of the errors at 15 points, namely, log E1(3, e), log E2(3, e), log E3(3, e), log E4(3, e) and log E5(3, e) as a function of log(e). The results are indicated in Figs. 3–5 for log E1(3, e), log E3(3, e) and log E5(3, e) as a function of log(e), respectively. In the the same way we plot log E2(3, e) and, log E4(3, e) as a function of log(e) (see Table 4). The slope of the lines are approximated using a least-squares fit of the data. Table 4 lists the slope and linear equations are obtained for N = 1, 2, 3, 4 and 5. Note that, if we interchange the exact solution with the numerical solution we have the slight error in the slope. From values in Table 3, we have shown the error is small and a method for verifying the correctness of the order of approximation as applied to NLIVPs. References [1] C.M. Bender, S.A. Orszag, Advanced Mathematics Methods for Scientists and Engineers, McGraw-Hill Inc., New York, 1978. [2] D.L. Bosley, A technique for numerical verification of asymptotic expansions, SIAM Review 38 (1996) 128–235.

I.H. Abdel-Halim Hassan / Applied Mathematics and Computation 180 (2006) 29–37

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[3] E. Deeba, S. Xie, The asymptotic expansion and numerical verification of Van der Pol’s equation, Journal of Computational Analysis and Applications 3 (2001) 165–171. [4] R. Haberman, Elementary Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, third ed., Prentice-Hall, Inc., Englewood Cliffs, NJ, 1998. [5] J. Kevorkian, J.D. Cole, Perturbation Methods in Applied Mathematics, Springer-Verlag, New York, 1981. [6] A.H. Nayfeh, Problems in Perturbation, John Wiley & Sons, Inc., New York, 1993. [7] D. Zwillinger, Handbook of Differential Equations, Academic Press, Inc., New York, 1992.