Apply a numerical method to the Dahlquist test equation $y' = \lambda y$ with $\lambda \in \mathbb{C}$, $\mathrm{Re}(\lambda) \leq 0$. The method produces $y_{n+1} = R(h\lambda) y_n$ where $R(z)$ is the stability function. The stability region is $\mathcal{S} = \{z \in \mathbb{C} : |R(z)| \leq 1\}$. A method is A-stable if $\mathcal{S} \supseteq \mathbb{C}^- = \{z : \mathrm{Re}(z) \leq 0\}$, meaning it is stable for all step sizes on all problems with decaying modes.

A system is stiff if it has widely separated time scales, i.e., eigenvalues of the Jacobian $\partial f/\partial y$ span several orders of magnitude with $\mathrm{Re}(\lambda) < 0$. A method is L-stable if it is A-stable and $R(z) \to 0$ as $z \to -\infty$. L-stability ensures that very stiff components ($|h\lambda| \gg 1$) are damped, not merely bounded. Backward Euler and BDF$k$ ($k \leq 2$) are L-stable; the trapezoidal rule is A-stable but not L-stable ($|R(z)| = 1$ on $i\mathbb{R}$).

A multistep method is zero-stable if the roots of its characteristic polynomial $\rho(\zeta) = \sum \alpha_j \zeta^{k-j}$ satisfy $|\zeta_j| \leq 1$, with roots on the unit circle being simple. The Dahlquist equivalence theorem states that a consistent multistep method is convergent if and only if it is zero-stable. Convergence then has order $p$: $\max_n |y(t_n) - y_n| = O(h^p)$ where $p$ is the order of the local truncation error.