Research Article
A Novel Method for Solving KdV Equation Based on Reproducing Kernel Hilbert Space Method
Table 3
The absolute error of Example
11 for initial condition at
,
.
| | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
| | 0.1 | 1.78 × 10−6 | 3.01 × 10−9 | 8.55 × 10−7 | 3.49 × 10−7 | 2.85 × 10−7 | 6.28 × 10−7 | | 0.2 | 6.38 × 10−7 | 6.98 × 10−7 | 6.52 × 10−7 | 4.51 × 10−7 | 8.33 × 10−6 | 2.42 × 10−7 | | 0.3 | 2.2 × 10−8 | 9.09 × 10−7 | 6.88 × 10−6 | 1.35 × 10−7 | 2.97 × 10−6 | 1.69 × 10−7 | | 0.4 | 1.70 × 10−7 | 1.03 × 10−7 | 5.38 × 10−7 | 1.20 × 10−6 | 3.98 × 10−7 | 1.68 × 10−7 | | 0.5 | 2.26 × 10−7 | 1.29 × 10−7 | 8.74 × 10−7 | 3.13 × 10−7 | 4.02 × 10−7 | 9.63 × 10−7 | | 0.6 | 8.94 × 10−7 | 7.83 × 10−7 | 4.34 × 10−7 | 9.79 × 10−7 | 2.77 × 10−7 | 1.45 × 10−8 |
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