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*}