THE SOLUTION OF COUPLED MODIFIED KDV SYSTEM BY THE ...

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M. GHOREISHI, A.I.B.MD. ISMAIL, A. RASHID: COUPLED MODIFIED KDV SYSTEM ... 131 and R 2m (⃗v m−1 ) = (v m−1 ) t + (v m−1 ) xxx +3 ( m−1 ) ( ∑ m−1 ) ∑ v i (v m−1−i ) x + 3 (u i ) x (v m−1−i ) x − i=0 i=0 ∑ m−1 −3 i=0 ( m−1−i ) ∑ u i u k (v m−1−i−k ) x − 3λ(v m−1 ) x . k=0 (40) We start with the initial conditions (35) and (36). By means of the (21) and (22), we can easily obtain directly the other component of the HAM solution in the series form of (12) and (13). Thus, we have u 1 (x, t) = t[3λk 2 sec h 2 (kx) + 3k 4 sec h 2 (kx) tanh 2 (kx) + 3k 2 (2k 2 sec h 4 (kx) − 4k 2 sec h 2 (kx)× × tanh 2 (kx) − k 2 (−2k3 sec h 4 (kx) + 4k 3 sec h 2 (kx) tanh 2 (kx)) − 3(−4k 2 sec h 2 (kx)× × tanh 2 (kx) + k 2 sec h 2 (kx)( 1 2 (λ + 4k2 ) − 2k 2 tanh 2 (kx))))]. v 1 (x, t) = t[12λk 3 sec h 2 (kx) tanh(kx) − 12k 5 sec h 4 (kx) tanh(kx) + 12k 5 sec h 2 (kx) tanh 3 (kx)− −12 sec h 2 (kx) tanh(kx)( 1 2 (4k2 + λ) − 2k 2 tanh 2 (kx)) − 2k 2 (−16k 3 sec h 4 (kx)× × tanh(kx) + 8k 3 sec h 2 (kx) tanh 3 (kx))] and so on. Thus u(x, t) and v(x, t) can be written as follows u(x, t) = v(x, t) = ∞∑ u n (x, t) = u 0 (x, t) + u 1 (x, t) + u 2 (x, t) + . . . , n=0 ∞∑ v n (x, t) = v 0 (x, t) + v 1 (x, t) + v 2 (x, t) + . . . . n=0 Table 4 and 5 show the absolute error between solution obtained using HAM with four terms and the exact solution for u(x, t) and v(x, t), respectively, at = −1, k = 0.1 and λ = 0.1. The errors are very small in these tables. The results show that the homotopy analysis method is very useful to obtain explicit approximate analytical solutions. A very good approximation with the actual solution of the equations was achieved by using four terms only. We have ascertained that the overall errors can be made smaller by adding new terms of the HAM series (12) and (13). Table 4: Absolute error E 4 for variables x and t at = −1, for u(x, t). x/t 0.1 0.2 0.3 0.4 -100 1.24447×10 −19 1.13618×10 −17 5.58676×10 −18 1.01850×10 −17 -40 6.46940×10 −13 2.59019×10 −12 5.83872×10 −12 1.04084×10 −11 -10 4.83626×10 −8 1.93467×10 −7 4.35339×10 −7 7.74002×10 −7 10 4.83541×10 −8 1.93399×10 −7 4.35109×10 −7 7.73457×10 −7 40 6.47442×10 −13 2.59443×10 −12 5.85271×10 −12 1.04417×10 −11 100 2.51949×10 −18 6.81357×10 −18 6.13444×10 −19 7.00450×10 −18
132 TWMS JOUR. PURE APPL. MATH., V.3, N.1, 2012 Absolute error u forħ⩵1, k⩵0.1,Λ⩵0.1 at x⩵100 Absolute error u forħ⩵1, k⩵0.1,Λ⩵0.1 at x⩵100 1.410 17 t 4.10 12 t 1.210 17 1.10 17 3.10 12 E4 8.10 18 6.10 18 E4 2.10 12 4.10 18 1.10 12 2.10 18 0 0.0 0.1 0.2 0.3 0.4 0 0.0 0.1 0.2 0.3 0.4 Figure 7. The absolute error E 4 for = −1, k = 0.1, λ = 0.1 at x = −100. Absolute error u forħ⩵1, k⩵0.1,Λ⩵0.1 at x⩵0 Absolute error u forħ⩵1, k⩵0.1,Λ⩵0.1 at x⩵0 2.10 9 t 1.510 9 E4 1.10 9 E4 1.510 7 t 1.10 7 5.10 10 5.10 8 0 0.0 0.1 0.2 0.3 0.4 0 0.0 0.1 0.2 0.3 0.4 Figure 8. The absolute error E 4 for = −1, k = 0.1, λ = 0.1 at x = 0. 0.05 ħcurve for u t 0,0 0.0010 ħcurve for v t 0.5,0 0.04 0.0008 0.03 0.0006 u t 0,0 0.02 0.01 v t 0.5,0 0.0004 0.0002 0.00 0.0000 0.01 0.0002 2.0 1.5 1.0 0.5 0.0 0.5 1.0 ħ 2.0 1.5 1.0 0.5 0.0 0.5 1.0 ħ Figure 9. The -curve for u t (0, 0) and v t (0.5, t) given by the 4th-order HAM approximation solution when H 1(x, t) = H 2(x, t) = 1.
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