Table 1 Numerical results of Example 4.3 for the initial points \(x_{0} = (-1,0)^{T}\) and \(x_{1} = (0,1)^{T}\) using the algorithm in Theorem 4.1
From: The zeros of monotone operators for the variational inclusion problem in Hilbert spaces
Type | \(\lambda _{n}\) for L = α = 1/2 | n | CPU (s) | \(x^{*}\) | \(\|x_{n+1}-x_{n}\|_{2}\) |
|---|---|---|---|---|---|
A1 | \(\frac{L}{10}\) | 272 | 0.078 | \((0.50000, 0.50000)^{T}\) | 9.92819 × 10−7 |
A2 | \(L-\frac{L}{10}\) | 44 | 0.016 | \((0.49999, 0.49999)^{T}\) | 9.80129 × 10−7 |
A3 | L | 45 | 0.015 | \((0.49999, 0.49999)^{T}\) | 9.90160 × 10−7 |
A4 | \(L+\frac{L}{10}\) | 47 | 0.031 | \((0.49999, 0.49999)^{T}\) | 9.30295 × 10−7 |
A5 | \(2\alpha -\frac{L}{10}\) | 105 | 0.031 | \((0.49999,0.49999 )^{T}\) | 9.41142 × 10−7 |
B1 | \(\frac{Ln}{n+1}\) | 45 | 0.015 | \((0.49999, 0.49999)^{T}\) | 9.60866 × 10−7 |
B2 | \(\frac{L(n+2)}{n+1}\) | 46 | 0.015 | \((0.49999, 0.49999)^{T}\) | 9.35467 × 10−7 |
C1 | \(L+\frac{(-1)^{n} L}{n+1}\) | 34 | 0.016 | \((0.49998, 0.49998)^{T}\) | 9.09566 × 10−7 |
C2 | \(L+\frac{(-1)^{n+1} L}{n+1}\) | 35 | 0.016 | \((0.49998, 0.49998)^{T}\) | 8.13627 × 10−7 |