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Explicit Scheme, final stability analysis

The eigenvalues of \mathbf{A} are \lambda_i=1-\alpha\mu_i , with \mu_i being the eigenvalues of \hat{B} . To find \mu_i we note that the matrix elements of \hat{B} are \begin{equation*} b_{ij} = 2\delta_{ij}-\delta_{i+1j}-\delta_{i-1j}, \end{equation*} meaning that we have the following set of eigenequations for component i \begin{equation*} (\hat{B}\hat{x})_i = \mu_ix_i, \end{equation*} resulting in \begin{equation*} (\hat{B}\hat{x})_i=\sum_{j=1}^n\left(2\delta_{ij}-\delta_{i+1j}-\delta_{i-1j}\right)x_j = 2x_i-x_{i+1}-x_{i-1}=\mu_ix_i. \end{equation*}