Eigenvalue stability analysis of a finite difference method

For a normal amplification matrix $Q$, the bound $\rho (Q)\le 1$ with semisimple unit-modulus eigenvalues controls every power $Q^n$. For a nonnormal family, eigenvalues alone do not control transient growth or give a mesh-uniform bound; eigenvector conditioning, pseudospectra, or a direct energy estimate is then needed.

Stability of the partial differential equation requires a bound uniform over refining meshes and time steps on every fixed physical time interval. This is distinct from demanding an all-time bound for one fixed matrix. A fixed matrix is power bounded exactly when all eigenvalues are in the closed unit disk and its unit-modulus Jordan blocks have size one. A time-step family can nevertheless have harmless polynomial physical-time growth: $G_{\tau }=I+\tau N$ with $N^2=0$ gives $G_{\tau }^n=I+n\tau N$, uniformly bounded for $n\tau \le T$ despite its nontrivial unit-modulus Jordan block. Mesh dependence and the scaling of the step must therefore be kept in an eigenvalue argument.

Every eigenvalue of this nonnormal matrix is $-1$, but $e^{t{L}_h}=e^{-t}\left(\begin{array}{cc}1& t/h\\ 0& 1\end{array}\right)$. Applied to the second coordinate unit vector, its norm at $t=1$ is at least $1/(eh)$. Thus negative eigenvalues give no mesh-uniform semigroup bound by themselves. Normal matrices, uniformly conditioned eigenvector bases, or a direct dissipative energy estimate supply the missing control.

On a finite inflow grid, explicit upwinding has a triangular amplification matrix whose only eigenvalue is $1-\mu$. For $1<\mu <2$ this eigenvalue lies inside the unit disk, although interior high-frequency data are amplified by $|1-2\mu |>1$ before reaching the boundary. This shows why eigenvalue analysis without uniform normality can give the wrong stability range.