Linear multistep method

Remarks

The characteristic polynomials of the method are $\rho(\zeta) = \sum_i \alpha_i \zeta^i$ and $\sigma(\zeta) = \sum_i \beta_i \zeta^i$. The method is consistent if $\rho(1) = 0$ and $\rho'(1) = \sigma(1)$, and zero-stable if every root of $\rho$ satisfies $\abs{\zeta} \le 1$ and the roots with $\abs{\zeta} = 1$ are simple. Dahlquist's equivalence theorem says that a consistent method converges for all smooth problems with consistent starting values if and only if it is zero-stable. Example: the two-step Adams–Bashforth method $y_{j+2} - y_{j+1} = h \left( \frac32 f_{j+1} - \frac12 f_j \right)$ has $\rho(\zeta) = \zeta^2 - \zeta$ and $\sigma(\zeta) = \frac32 \zeta - \frac12$, so $\rho(1) = 0$ and $\rho'(1) = 1 = \sigma(1)$; it is explicit, consistent and zero-stable. The Euler method is the case $k = 1$, $\alpha_0 = -1$, $\beta_0 = 1$, $\beta_1 = 0$. Non-example: the classical fourth-order Runge–Kutta method is not a linear multistep method, since its update uses values of $f$ at intermediate points that are not of the form $f_{j+i}$. The explicit method $y_{j+2} + 4 y_{j+1} - 5 y_j = h (4 f_{j+1} + 2 f_j)$ is consistent of order $3$, but $\rho(\zeta) = (\zeta - 1)(\zeta + 5)$ has the root $-5$, so it is not zero-stable and its errors grow without bound.