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For $s=1$, imposing order two givesThe method is therefore the trapezoidal ruleIts first characteristic polynomial is $\rho (\zeta )=\zeta -1$, so it satisfies the root condition for a multistep method.

For $s=2$, the four conditions through order three giveand henceHereso the root condition for a multistep method again holds. Both methods are consistent and zero-stable, and the Dahlquist equivalence theorem therefore proves that both are convergent.

The characteristic polynomials of a linear multistep method areandFor the order conditions for a linear multistep method, put $\alpha =(a,-1-a,1)$ and $\beta =(-(1+5a)/12,,8(1-a)/12,,(5+a)/12)$. The defectsvanish for $q=0,1,2,3$, whileThereforeIndeed, at $a=-1$ one has $C_4=0$ but $C_5=-4/3\ne 0$.

A linear multistep method for $y^{\prime }=f(t,y)$ has the formIts characteristic polynomials of a linear multistep method are $\rho (\zeta )={\sum }_j{\alpha }_j{\zeta }^j$ and $\sigma (\zeta )={\sum }_j{\beta }_j{\zeta }^j$. Substitution of the exact solution, or equivalently expansion of $\rho (e^z)-z\sigma (e^z)$ at $z=0$, gives the order conditions for a linear multistep method. The method has order $p$ whenwith a nonzero coefficient at the next power. Its one-step defect is then $O(h^{p+1})$, while its global error is $O(h^p)$ when stability prevents accumulation from being amplified.

Order alone does not ensure convergence because the recurrence may contain growing parasitic modes. Zero-stability is the root condition for a multistep method: every root of $\rho$ lies in $|\zeta |\le 1$, and every root on the unit circle is simple. Consistency is the first-order conditionThe Dahlquist equivalence theorem states that a consistent linear multistep method is convergent if and only if it is zero-stable. For example, the leapfrog method $y_{n+2}-y_n=2h{f}_{n+1}$ has $\rho (\zeta )={\zeta }^2-1$; its roots $\pm 1$ are simple, so this second-order method is convergent, although its parasitic $-1$ mode can oscillate.

To study stiff decay, apply the method to $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification factors are the roots ofThe linear stability domain consists of the $z$ for which these roots satisfy the strict interior condition appropriate away from the boundary, with simple unit roots where boundary stability is admitted. The method is A-stable when this domain contains the left half-plane, so every mode with $\mathrm{Re}\lambda <0$ remains bounded for every positive step size. The Second Dahlquist barrier says that an A-stable multistep method has order at most two.

The main families illustrate the tradeoffs. The explicit Adams-Bashforth methods interpolate past derivative values and are inexpensive, but their stability domains are bounded. The implicit Adams-Moulton methods include the new derivative value; the one-step second-order member is the trapezoidal rule, which is A-stable. The backward differentiation formulas interpolate past solution values and differentiate the interpolant; the first two are A-stable, while higher orders retain useful sectors of the left half-plane. Implicit methods require a nonlinear solve at each step, commonly initialized by an explicit predictor to form a predictor-corrector pair. Thus order controls consistency error, the root condition controls convergence as $h\to 0$, and the amplification polynomial controls whether a convergent method remains useful on stiff equations.