Comparison of homotopy analysis method and homotopy - Applied ...

and⎧⎨0, m ≤ 1,χ m =⎩1, m > 1.The solution of the mth-order deformation equation (24) for m ≥ 1 reads∫ tu m (x, t) = χ m u m−1 (x, t) + R m (u m−1 (x, t)) dτ + c 1 , (27)0where the constant of integration c 1 is determined by the initial condition (25).Using symbolic computation systems such as Maple or Mathematica, we recursivelyobtainu 0 (x, t) = e −x , (28)u 1 (x, t) = −e −x t, (29)u 2 (x, t) = e−x t(t + 2 − 2) , (30)2u 3 (x, t) = − e−x t ( 2 t 2 − 6 t + 6 2 t + 6 2 − 12 + 6 ) , (31)6u 4 (x, t) = e−x t ( 3 t 3 − 12 2 t 2 + 12 3 t 2 + 36 t24−72 2 t + 36 3 t − 24 + 72 − 72 2 + 24 3) , (32)u 5 (x, t) = − e−x t ( 4 t 4 − 20 3 t 3 + 20 4 t 3 + 120 2 t 2 − 240 3 t 2120+120 4 t 2 + 720 2 t − 240 t + 240 4 t − 720 3 t − 480 3−480 + 120 + 720 2 + 120 4) , (33)u 6 (x, t) = e−x t ( 5 t 5 − 30 4 t 4 + 30 5 t 4 + 300 5 t 3 + 300 3 t 3720−600 4 t 3 − 3600 4 t 2 + 1200 5 t 2 − 1200 2 t 2 + 3600 3 t 2−7200 2 t + 1800 t − 7200 4 t + 10800 3 t + 1800 5 t+7200 3 + 3600 − 720 − 7200 2 + 720 5 − 3600 4) . (34)When = −1, it is easily seen that the equations (30) up to (34) aboveare exactly the equations (3.17b) up to (3.17f) in [8], and the combinationof the equations (28) and (29) is exactly the equation (3.17a) in [8] (Ganjiet al made mistakes in the first two lines of (3.16) in [8], which makes thedifference). Furthermore, when = −1, the 6th-order approximationu(x, t) ≈ e−x720 (t6 + 66t 5 + 1470t 4 + 13320t 3 + 46440t 2 + 45360t + 720) (35)is exactly the same as the HPM solution (17). Therefore, the HPM solutionis indeed a special case of the HAM solution when = −1. This fact hasbeen pointed out by many researchers, such as Abbasbandy [2], Liao et al7