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For a fixed-step linear multistep method with a nonzero leading coefficient, applied to an ordinary differential equation with a locally Lipschitz continuous vector field, the Dahlquist equivalence theorem states thatHere convergence is uniform on each fixed finite time interval as the step size tends to zero, for every set of starting values tending to the corresponding exact values. Consistency of a numerical method means that the normalized local defect tends to zero; for the characteristic polynomials of a linear multistep method it gives $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability is equivalent to the root condition for a multistep method: every root of a polynomial $\xi$ of $\rho$ satisfies $|\xi |\le 1$, and roots with $|\xi |=1$ are simple. Order $p\ge 1$, together with zero-stability and starting errors $O(h^p)$, gives global error $O(h^p)$ under the usual smoothness assumptions.

To prove the necessity of the root condition for multistep convergence, use the test problem $y^{\prime }=0$, $y(0)=0$. Its discrete error satisfies $\rho (E)e_n=0$, where $E{e}_n=e_{n+1}$. Set $h=T/N$. If $|\xi |>1$, the exact recurrence solutionhas starting errors tending to zero at every fixed starting index, but $|e_N|=1$. Thus convergence fails.

If a unit-modulus root has multiplicity $m\ge 2$, the recurrence instead admitsAgain every fixed starting error tends to zero while $|e_N|=1$. These polynomial-times-exponential solutions follow from the repeated factor $(E-\xi {)}^m$. Complex modes can be interpreted as a real two-component system, or their real and imaginary parts can be used. Therefore convergence requires precisely the root condition for a multistep method. Repeated roots strictly inside the unit disk are allowed, since their polynomial factors are dominated by exponential decay.

For this Adams–Moulton method, the characteristic polynomials of a linear multistep method areExpansion at zero givesPart (a) therefore shows that the method has order exactly three.

It is consistent because its order is at least one. The roots of $\rho (w)=w(w-1)$ are $0$ and $1$; both lie in the closed unit disk, and the only unit-modulus root is simple. Thus the method is zero-stable. By the Dahlquist equivalence theorem, consistency and zero stability imply convergence. Hence the method is

The first characteristic polynomial of the linear multistep method comes only from its left side:Its roots are zero and one; both lie in the closed unit disc, and the only unit-modulus root is simple. Thus the method satisfies the root condition for a multistep method and is zero-stable. Part (b) showed consistency for every $\alpha ,\beta$. The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent exactly when it is zero-stable. Thereforeunder the usual Lipschitz hypotheses on $f$.

Every member of the family has order at least three, so it is consistent. By the Dahlquist equivalence theorem, convergence is therefore equivalent to zero-stability, which is characterized by the root condition for a multistep method.

The roots of $\rho$ are $1$ and $a$. They lie in the closed unit disk exactly when $|a|\le 1$. At $a=1$, however, the unit root $1$ is repeated, whereas at $a=-1$ the distinct unit roots $1$ and $-1$ are both simple. Consequently

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.