
J. BIAZAR ET AL.
991
Table 3. Comparisons of of 10th-order HAM solu-
tions for different values of in Example 3.
t
0.2x 0.6x
1.0x 1.4x1.8x
1 [12] 1.5e–2 8.4e–2 6. 4e–3 7.49e–36.8e–3
1c 0.924 7.1e–7 4.3e–7 3.7e–7 2.0e–69.6e–6
=0.57 [12] 1.1e–2 1.2e–2 1.2e–2 1.2e–21.1e–2
5c =0.954 1.3e–4 1.3e–4 1.2e–4 1.2e–41.4e–4
0.634 [12] 1.3e–1 1.5e–1 1.5e–1 1.5e–11.3e–1
12c 0.655 1.0e–2 9.9e–3 9.4e–3 9.9e–31.1e–2
Here we will discuss following three cases of .
c
1) In the first case, we take as an example. For
and in (16), we get into a nonlinear
equation, which root is .
1c
0.924
1c10m
2) In second case, we take as an example.
Equation (16) with and has the real solu-
tion .
5c
105cm
=0.954
3) Finally, we take as an example. Equation
(16) with and , introduces
as a proper value for .
12c 10m
12c 0.655
Since the closed-form solution to the problem (22),
(23) is not available, the numerical solution numer is
calculated via the Runge-Kutta-Fehlberg 4-5 technique,
then, we compare the relative errors of the 10th order
u
HAM approximations at different points in the
appr
u
interval , using the formula
(0,1)
appr numer
numer
uu
tu
for different cases of , are reported in Table 3.
c
5. Conclusions
In this paper the solutions of a new way of finding the
control parameter in the homotopy analysis method is
proposed. It is shown for obtained values of such pa-
rameter, HAM approximation series leads to exact solu-
tion of problems or produces an approximate results
which are in a highly agreement with exact solution of
problems. All computations were done using Maple 13
with 15 digit floating point arithmetics (Digits: = 15).
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