 Advances in Pure Mathematics, 2013, 3, 159-163 http://dx.doi.org/10.4236/apm.2013.31A022 Published Online January 2013 (http://www.scirp.org/journal/apm) Numerical Solutions of the Regularized Long-Wave (RLW) Equation Using New Modification of Laplace-Decomposition Method Nawal A. Al-Zaid, Huda O. Bakodah, Fathiah A. Hendi Department of Mathematics, Science Faculty for Girls, King Abdulaziz University, Jeddah, KSA Email: nalzaid@kau.edu.sa, hbakodah@kau.edu.sa, falhendi@kau.edu.sa Received July 28, 2012; revised September 30, 2012; accepted October 8, 2012 ABSTRACT In this paper the new modification of Laplace Adomian decomposition method (ADM) to obtain numerical solution of the regularized long-wave (RLW) equation is presented. The performance of the method is illustrated by solving two test examples of the problem. To see the accuracy of the method, L2 and L∞ error norms are calculated. Keywords: Adomian Decomposition Method; Regularized Long-Wave (RLW); Laplace Transform 1. Introduction The regularized long wave (RLW) equation which can be shown in the form 0txx xxtuu uuu  (1) where ,  are positive parameters, is an important nonlinear wave equation. This equation plays a major role in the study of nonlinear dispersive waves. The RLW equation particularly describes the behavior of the undular bore [1-3], it has also been derived from the study of water waves and ion acoustic plasma waves. The RLW equation has been solved analytically only for restricted set of boundary and initial conditions. There- fore, the numerical solution of this equation has been the subject of many papers [4-7]. Recently a great deal of interest has been focused on application of Adomian de- composition method (ADM) to solve a wide variety of nonlinear problems [8,9]. In this paper, we will apply the new modification of Laplace ADM to the RLW Equation (1). The soliton solution of RLW equation has the form 20sech px vtx, 3uxt c (2) where is an arbitrary constant and p1221cc1vc 11p, , 0 ,0uxf x, and c is a constant . In this work, a new modification of Laplace ADM is used to solve the RLW equation with the initial condition (3) x is a localized disturbance inside the con-sidered interval. where 2. Description of Method We begin by consider Equation (1) in an operator form 0txLuuRuNuf (4) where tLt is a linear operator and R its remainder of the linear operator. The nonlinear term is represented by Nu txLuuR uNu  0,,nnuxtu xt0. Thus we get (5) We represent solution as an infinite series given be-low, (6) The nonlinear term Nu can be decomposed into infi-nite series of polynomial given by: xnnNu uuA,, ,nuu u000101102021120xxxxxxAuuAuuuu (7) where An are Adomian polynomials  of 01 and it can be calculated by formula given below: uuuu uu Aand so on. The rest of the polynomials can be constructed Copyright © 2013 SciRes. APM N. A. AL-ZAID ET AL. 160 in a similar manner. By applying the Laplace transform to both sides of Equation (5), we obtain txLu u  Ru Nu (8) Thus 11,,0xuxt uxuR u Nuss (9) In the new modification of ADM , Wazwaz re-placed the initial condition ,0ux0,0nnu x by a series of infi-nite components i.e., ,0ux (10) and the new recursive relationship can be expressed in the form   0 0,010,11,,0,1,,01nnxUxsu xsUxs uxsuRu,0Nuns 0,nnU xs (11) where ,,uxt Uxs (12) Now, by applying inverse Laplace transformation we get: ,,,0xsxsnmu0,mmiiuuxtu e100111,,nnuxt UuxtU (13) Using (13) the series solution follows immediately. 3. Numerical Examples and Results In this section, the new modification of Laplace ADM will be demonstrated on illustrative examples and we compare the approximate solution obtained for our RLW equation with known exact solutions. We define to be m-term approximate solution, i.e. e the exact solution and m the absolute error between the exact solution and the approximate solution memeuu In order to show how good the numerical solutions are in comparison with the exact ones, we will use the L2 and L∞ error norms defined by 12222 ,,1mem eimiiLuuxu u  ,,maxem eimiiLuu uu Example (1) We consider Equation (1) with the initial condition 20,03 sechuxcpx x The exact solution of this problem is given by Equation (2). This solution corresponds to the motion of a single solitary wave with amplitude 3c and width p, initially centered at 0x, where 1vc1 is the wave velocity. We use the new modification of Laplace ADM to solve this equation, all computations are done for the parameters 00x, 1 and . 0.1cWe consider , as in , so the initial condition ,0ux can expressed as a series of infinite components i.e. 241268 10,00.3 0.006818180.0001033061.330451.5754 1.77355uxx xxxxOx    Using recursive relation (11) yield the components 11000.3,, 0.3uxt Uxss  11 21120.00681818,,0.00681818uxt Uxsxsx 1224,,0.0177273 0.000103306uxtUxstx x 1332366,,0.0115227 0.0006301651.3304510uxt Uxsttxx  And so on, in this manner the rest of components of the decomposition series were obtained. The results are given in Table 1. The error norms for (c = 0.1) are re-corded in Table 2 for different value of m. Also in Figure 1 we show the exact solution and nu-merical solution with new modification of Laplace ADM for t = 0.1 and t = 0.5. Figure 2 shows the exact solution and numerical solution with new modification for t = 0.5 at the interval 5 ≤ x ≤ 5. Example (2) In the second test problem , a smaller solitary wave of amplitude 0.109 (c = 0.05), has been modeled. The initial condition ,0ux can expressed as a se-ies of infinite components i.e. r Copyright © 2013 SciRes. APM N. A. AL-ZAID ET AL. Copyright © 2013 SciRes. APM 161 Table 1. Absolute errors for Example (1) with c = 0.1 and m = 10. x/t 0.01 0.02 0.03 0.04 0.05 0.1 4.60152 × 10−7 4.29123 × 10−7 9.14459 × 10−8 1.09975 × 10−6 2.59382 × 10−6 0.2 1.15346 × 10−6 1.8112 × 10−6 1.97626 × 10−6 1.65183 × 10−6 8.41272 × 10−7 0.3 1.82435 × 10−6 3.1446 × 10−6 3.96518 × 10−6 4.29071 × 10−6 4.12595 × 10−6 0.4 2.4624 × 10−6 4.40862 × 10−6 5.84456 × 10−6 6.77623 × 10−6 7.20981 × 10−6 0.5 3.05691 × 10−6 5.58214 × 10−6 7.58301 × 10−6 9.06702 × 10−6 1.00418 × 10−5 Table 2. L2 and L∞ errors for Example (1) with m = 4, 6 and 10. n 4 6 10 x L2 L∞ L2 L∞ L2 L∞ 0.1 7.08453 × 10−7 5.4329 × 10−6 2.30062 × 10−7 2.44677 × 10−7 1.88979 × 10−6 2.59382 × 10−6 0.2 1.71987 × 10−6 1.8958 × 10−5 9.54522 × 10−7 5.43345 × 10−6 9.42236 × 10−7 8.41272 × 10−7 0.3 2.7237 × 10−6 3.2340 × 10−5 1.70526 × 10−6 1.05194 × 10−5 9.18072 × 10−7 4.12595 × 10−6 0.4 3.71444 × 10−6 4.5525 × 10−5 2.43975 × 10−6 1.54669 × 10−5 1.73297 × 10−6 7.20981 × 10−6 0.5 4.6877 × 10−6 5.8456 × 10−5 3.15018 × 10−6 2.02399 × 10−5 2.61274 × 10−6 1.00418 × 10−5 (a) (b) Figure 1. The exact solution and numerical solution with new modification of Laplace ADM for Example (1), for t = 0.1 and (b) for t = 0.5. (a) (b) Figure 2. (a) The exact solution for t = 0.5 and −5 ≤ x ≤ 5; (b) Numerical solution with new modification for t = 0.5 and −5 ≤ x 5. ≤ N. A. AL-ZAID ET AL. 162 210 8,00.15 0.001785710.0000141723 9.565.93001 10ux x8461207 10x1332386,,0.00236161 0.00007157039.560710uxt Uxsttxx xxxO Using recursive relation (11) yield the components 11uxt 00,,0.15 0.15Uxss 111120.00,,0.00178571uxtUxsx 2178571 xs4000141723x 122,,0.00410714 0.0uxtUxstxAnd so on, in this manner the rest of components of the decomposition series were obtained. The results are given in Table 3 and the error norms for (c = 0.05) are recorded in Table 4 for different value of m.Also in Figure 3 we show the exact solution and nu-merical solution with new modification of Laplace ADM for t = 0.1 and t = 0.5. Figure 4 show the exact solution and numerical solution with new modification for t = 0.5 at the interval 10 ≤ x ≤10. 4. Conclusion In this paper, we use the new modification of Laplace Table 3. Absolute errors for Example (2) with c = 0.05, k = 0.109 and m = 10. x/t 0.01 0.02 0.03 0.04 0.05 0.1 5.68488 × 10−9 2.48779 × 10−9 9.54769 × 10−9 3.03715 × 10−8 5.99272 × 10−8 0.2 1.56382 × 10−8 2.23134 × 10−8 2.01032 × 10−8 9.09157 × 10−9 1.06314 × 10−8 0.3 2.52853 × 10−8 4.14523 × 10−8 4.86129 × 10−8 4.6885 × 10−8 3.63928 × 10−8 0.4 3.44902 × 10−8 5.96351 × 10−8 7.55814 × 10−8 8.2482 × 10−8 8.04953 × 10−8 0.5 4.31131 × 10−8 7.65863 × 10−8 1.00602 × 10−7 1.15349 × 10−7 1.21020 × 10−7 Table 4. L2 and L∞ errors for Example (2) with m = 4, 6 and 10. n 4 6 10 x L2 L∞ L2 L∞ L2 L∞ 0.1 4.34078 × 10−6 1.3424 × 10−6 1.37871 × 10−6 4.32171 × 10−7 4.61577 × 10−7 1.48914 × 10−7 0.2 4.97100 × 10−6 1.70414 × 10−6 1.59318 × 10−6 5.56933 × 10−7 5.4413 × 10−7 1.98079 × 10−7 0.3 5.60954 × 10−6 2.0710 × 10−6 1.80913 × 10−6 6.83084 × 10−7 6.26085 × 10−7 2.47459 × 10−7 0.4 6.26191 × 10−6 2.4466 × 10−6 2.0284 × 10−6 8.11842 × 10−7 7.08138 × 107 2.97527 × 10−7 0.5 6.93387 × 10−6 2.8344 × 10−6 2.25293 × 10−6 9.44397 × 10−7 7.91008 × 10−7 3.48733 × 10−7 (a) (b) Figure 3. The exact solution and numerical solution with new modification of Laplace ADM for Example (1), (a) for t = 0.1 and (b) for t = 0.5. Copyright © 2013 SciRes. APM N. A. AL-ZAID ET AL. 163 (a) (b) Figure 4. (a) The exact solution for t = 0.5 and −10 ≤ x ≤ 10; (b) Numerical solution with new modification for t = 0.5 and −10 ≤ x ≤ 10. ADM to solve the RLW equation. 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