Table 2: Comparison of the first eigenvalue and solutions of Example 3.1 using Taylor’s decomposition method, exact values, and Table 4 from [9 ], when
.
HWSM FDM TDM
Exact Errors of TDM Errors of TDM Errors of TDM (
) (
) (
) (
) (
) (
) (
) (
) (
) (
) 0 0 0 0 0 0 0 0 0.0625 0.27521 0.278599 0.275999 0.275899
— — 0.125 0.54181 0.541196 0.541196 0.541196
— 0.1875 0.78549 0.785695 0.785695 0.785695
— — 0.25 1.00482 1 1 1
0.3125 1.17851 1.17588 1.17588 1.17588
— — 0.375 1.31285 1.30656 1.30656 1.30656
— 0.4375 1.38376 1.38704 1.38704 1.38704
— — 0.5 1.41103 1.41421 1.41421 1.41421
0 0.5625 1.38376 1.38704 1.38704 1.38704
— — 0.625 1.31285 1.30656 1.30656 1.30656
— 0.6875 1.17851 1.17588 1.17588 1.17588
— — 0.75 1.00482 1 1 1
2.06E-11 0.8125 0.78549 0.785695 0.785695 0.785695
— — 0.875 0.54181 0.541196 0.541196 0.541196
— 0.9375 0.27521 0.275899 0.275999 0.275899
— — 1 0 0 0 0 0 0 0
(HWSM), 10.8379 (FDM), 10.8696 (TDM), 10.8696 (Exact).