Table 2 Numerical results of Example 4.5 for the initial points \(x_{0} = (1,-1,-1)^{T}\) and \(x_{1} = (-1,0,1)^{T}\) using the algorithm in Theorem 4.4
From: The zeros of monotone operators for the variational inclusion problem in Hilbert spaces
Type | \(\lambda _{n}\) for L = 1/4,β = 1/2 | n | CPU (s) | \(x^{*}\) | \(\|x_{n+1}-x_{n}\|_{2}\) |
|---|---|---|---|---|---|
A1 | \(\frac{L}{10}\) | 272 | 0.062 | \((1.00000, 1.00000, 1.00000)^{T}\) | 9.92684 × 10−7 |
A2 | \(L-\frac{L}{10}\) | 47 | 0.016 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.24684 × 10−7 |
A3 | L | 48 | 0.016 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.39145 × 10−7 |
A4 | \(L+\frac{L}{10}\) | 49 | 0.015 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.65739 × 10−7 |
A5 | \((1-\beta )-\frac{L}{10}\) | 108 | 0.031 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.04352 × 10−7 |
B1 | \(\frac{Ln}{n+1}\) | 47 | 0.016 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.91872 × 10−7 |
B2 | \(\frac{L(n+2)}{n+1}\) | 48 | 0.016 | \((0.99999, 0.99999, 0.99999)^{T}\) | 9.66642 × 10−7 |
C1 | \(L+\frac{(-1)^{n} L}{n+1}\) | 36 | 0.016 | \((0.99998, 0.99998, 0.99998)^{T}\) | 8.94335 × 10−7 |
C2 | \(L+\frac{(-1)^{n+1} L}{n+1}\) | 35 | 0.016 | \((0.99998, 0.99998, 0.99998)^{T}\) | 9.96769 × 10−7 |