Table 3 Numerical results of Example 4.8 for the initial points \(x_{0} = (1,-2,-1)^{T}\) and \(x_{1} = (-2,-1,2)^{T}\) using the algorithm in Theorem 4.6
From: The zeros of monotone operators for the variational inclusion problem in Hilbert spaces
Type | \(\lambda _{n}\) for L = 1 | n | CPU (s) | \(x^{*}\) | \(\|x_{n+1}-x_{n}\|_{2}\) |
|---|---|---|---|---|---|
A1 | \(\frac{L}{10}\) | 272 | 0.109 | \((0.99999,-2.48737 \times 10^{-2335},2.00000)^{T}\) | 9.92500 × 10−7 |
A2 | \(L-\frac{L}{10}\) | 44 | 0.047 | \((0.99999,-1.47065 \times 10^{-351},1.99999)^{T}\) | 9.79481 × 10−7 |
A3 | L | 45 | 0.032 | \((0.99999, +6.39411 \times 10^{-360},1.99999)^{T}\) | 9.89477 × 10−7 |
A4 | \(L+\frac{L}{10}\) | 47 | 0.031 | \((0.99999, +1.13371 \times 10^{-376},1.99999)^{T}\) | 9.29597 × 10−7 |
A5 | \(2L-\frac{L}{10}\) | 123 | 0.063 | \((0.99999, +7.06031 \times 10^{-1032},1.99999)^{T}\) | 8.86229 × 10−7 |
B1 | \(\frac{Ln}{n+1}\) | 45 | 0.016 | \((0.99999, +2.30611 \times 10^{-353},1.99999)^{T}\) | 9.60213 × 10−7 |
B2 | \(\frac{L(n+2)}{n+1}\) | 46 | 0.046 | \((0.99999, -2.72090 \times 10^{-368},1.99999)^{T}\) | 9.34783 × 10−7 |
C1 | \(L+\frac{(-1)^{n} L}{n+1}\) | 34 | 0.015 | \((0.99998, +2.17358 \times 10^{-255},1.99998)^{T}\) | 9.10791 × 10−7 |
C2 | \(L+\frac{(-1)^{n+1}L}{n+1}\) | 35 | 0.016 | \((0.99998, +9.23665 \times 10^{-277},1.99998)^{T}\) | 8.14783 × 10−7 |