Si $\omega < 1$, factor de subrelajación
Si $\omega > 1$, factor de sobrerelajación
Calcular $\omega_{0}$ de $\mathbf{A}$
\begin{equation*} \mathbf{A} = \begin{bmatrix} 10 & -1 & 2 & 0 \\ -1 & 11 & -1 & 3 \\ 2 & -1 & 10 & -1 \\ 0 & 3 & -1 & 8 \end{bmatrix} \end{equation*}La matriz diagonal es
\begin{equation*} \mathbf{D} = \begin{bmatrix} 10 & 0 & 0 & 0 \\ 0 & 11 & 0 & 0 \\ 0 & 0 & 10 & 0 \\ 0 & 0 & 0 & 8 \end{bmatrix} \end{equation*}La matriz triangular inferior es
\begin{equation*} \mathbf{L} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 3 & -1 & 0 \end{bmatrix} \end{equation*}La matriz triangular superior es
\begin{equation*} \mathbf{U} = \begin{bmatrix} 0 & -1 & 2 & 0 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}Reemplazando
\begin{equation*} \mathbf{T} = \begin{bmatrix} \frac{1}{10} & 0 & 0 & 0 \\ 0 & \frac{1}{11} & 0 & 0 \\ 0 & 0 & \frac{1}{10} & 0 \\ 0 & 0 & 0 & \frac{1}{8} \end{bmatrix} \begin{bmatrix} 0 & -1 & 2 & 0 \\ -1 & 0 & -1 & 3 \\ 2 & -1 & 0 & -1 \\ 0 & 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{10} & \frac{1}{5} & 0 \\ -\frac{1}{11} & 0 & -\frac{1}{11} & \frac{3}{11} \\ \frac{1}{5} & -\frac{1}{10} & 0 & -\frac{1}{10} \\ 0 & \frac{3}{8} & -\frac{1}{8} & 0 \end{bmatrix} \end{equation*}El radio espectral de $\mathbf{T}$ es el mayor en valor absoluto de los valores propios
\begin{equation*} \rho(\mathbf{T}) = 0.426437 \end{equation*}Reemplazando
\begin{equation*} \omega_{0} = \frac{2}{1 + \sqrt{1 - 0.426437^{2}}} = 1.05 \end{equation*}Si $w_{1} = \omega$
\begin{equation*} x_{i}^{(k)} = \omega x_{i}^{(k)} + w_{2} x_{i}^{(k-1)} \end{equation*}Si $x_{i}^{(k)} \approx x_{i}^{(k-1)}$
\begin{equation*} x_{i}^{(k)} = \omega x_{i}^{(k)} + w_{2} x_{i}^{(k)} \end{equation*}Despejando $w_{2}$
\begin{equation*} w_{2} = 1 - \omega \end{equation*}Reemplazando en \eqref{ecuacion3}
\begin{equation} x_{i}^{(k)} = \omega x_{i}^{(k)} + (1 - \omega) x_{i}^{(k-1)} \end{equation}