Research Article
A Jacobi Collocation Method for Solving Nonlinear Burgers-Type Equations
Table 2
Absolute errors with
,
and various choices of
for Example
1.
| | | | | | | | | | | | |
| 0.0 | 0.1 | 0 | 1 | 0.001 | 0.001 | 0.001 | 12 | | −1 | 1 | | 0.1 | | | | | | | | | | | | 0.2 | | | | | | | | | | | | 0.3 | | | | | | | | | | | | 0.4 | | | | | | | | | | | | 0.5 | | | | | | | | | | | | 0.6 | | | | | | | | | | | | 0.7 | | | | | | | | | | | | 0.8 | | | | | | | | | | | | 0.9 | | | | | | | | | | | | 1 | | | | | | | | | | | |
| 0.0 | 0.2 | 0 | 1 | 0.001 | 0.001 | 0.001 | 12 | | −1 | 1 | | 0.1 | | | | | | | | | | | | 0.2 | | | | | | | | | | | | 0.3 | | | | | | | | | | | | 0.4 | | | | | | | | | | | | 0.5 | | | | | | | | | | | | 0.6 | | | | | | | | | | | | 0.7 | | | | | | | | | | | | 0.8 | | | | | | | | | | | | 0.9 | | | | | | | | | | | | 1 | | | | | | | | | | | |
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