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An explicit linear multistep method that integrates a polynomial extrapolation of the past derivative values. The displayed two-step member is second order in time: the linear interpolant through $f_{n-1},f_n$, integrated over the next step, gives coefficients $3/2,-1/2$. For a spatially discretized partial differential equation, the spatial error must also be included in the total local defect.

An Adams–Moulton method is an implicit linear multistep method obtained by integrating an interpolation polynomial for the vector field that includes the new time level.

Applying a linear multistep method to the Dahlquist test equation with $z=h\lambda$ produces a constant-coefficient recurrence whose modes satisfy $P_z(w)=0$. Absolute stability requires the resulting roots of a polynomial to lie in the unit disk, with unit-modulus roots simple. Thus multistep stability depends on every recurrence mode, rather than a single one-step stability function.

A unit-modulus amplification root of a linear multistep method puts its test-equation parameter on this boundary locus, except when numerator and denominator both vanish. If the locus avoids the open left half-plane, the leading coefficient never vanishes there and the roots are initially inside the disk at one point there, continuity keeps them inside throughout that connected region. Checking zero-stability, imaginary-axis roots and common polynomial factors is still necessary. A boundary plot alone is not a proof of A-stability.

For a real consistent zero-stable linear multistep method, transform its first characteristic polynomial by $z=(1+x)/(1-x)$ to $(1-x{)}^s\rho (z)=xP(x)$. The root condition for a multistep method puts all finite roots of $P$ in the closed left half-plane; real linear factors and conjugate quadratic factors consequently give nonnegative coefficients, after choosing $P(0)>0$. The order criterion requires a polynomial of degree at most $s$ to agree through degree $p-1$ with $P(x)x/(2\mathrm{arctanh}x)$. The latter quotient has strictly negative nonconstant even coefficients. Its first even coefficient beyond degree $s$ includes a strictly negative multiple of $P(0)$ and cannot vanish. This gives the stated bound.

For an initial value problem, convergence means that the largest global error on each fixed finite time interval tends to zero as the step size tends to zero and the starting errors vanish. For linear multistep methods, the Dahlquist equivalence theorem separates this requirement into consistency of a numerical method and zero-stability.

For an explicit linear multistep method, the highest-degree coefficient of $\rho (w)-z\sigma (w)$ is fixed while at least one lower coefficient grows unboundedly with negative real $z$. If all roots stayed in the unit disk, the elementary symmetric polynomials of those roots would keep every monic coefficient bounded. This contradiction proves that no nontrivial consistent explicit method is A-stable.

The Taylor expansion of the exact-solution residual for a linear multistep method has the same coefficient sequence asVanishing through degree $p$ is therefore equivalent to order at least $p$. Exact order $p$ additionally requires a nonzero coefficient at degree $p+1$. This symbol conveniently collects all polynomial consistency and order conditions.

If the derivative characteristic polynomial $\sigma$ of a linear multistep method has a simple root outside the unit disk and keeps its degree in the limit, then its absolute stability region is bounded. Otherwise a stable sequence with $|z|\to \infty$ would give amplification polynomials $\sigma -\rho /z$ tending coefficientwise to $\sigma$. A root near the exterior root would persist by continuity, contradicting the root condition for a multistep method. This argument excludes stability at every sufficiently large complex $z$, not only along the negative real axis.

An implicit time-stepping method determines a new solution value through an equation involving that same value. A small-step contraction mapping argument often establishes local solvability when the vector field is Lipschitz continuous. Examples include the Backward Euler method, implicit Runge-Kutta methods and implicit linear multistep methods. Solvability of the update is a separate issue from stability of a numerical method.

A multiderivative multistep method uses higher time derivatives as well as values and first derivatives from several time levels. For an autonomous ordinary differential equation $y^{\prime }=f(y)$, the second time derivative is $y^{\prime \prime }=f^{\prime }(y)f(y)$ by the chain rule. Its local truncation error must include the higher-derivative terms. Consequently the usual second-order barrier for A-stable linear multistep methods, which only use first derivatives, does not apply.

For a linear multistep method, convergence for every set of consistent starting values requires the root condition for a multistep method. Apply the recurrence to $y^{\prime }=0$ over $N$ steps. A root $\xi$ with $|\xi |>1$ gives the error mode $e_n=|\xi {|}^{-N}{\xi }^n$: its fixed-index starting errors vanish as $N\to \infty$, but $|e_N|=1$. A unit-modulus root of multiplicity $m\ge 2$ instead gives $e_n=N^{-(m-1)}{n}^{m-1}{\xi }^n$, with the same failure. Repeated roots inside the unit disk are harmless because their polynomial growth is dominated by exponential decay.

Using Newton-Cotes closed quadrature for the integrated derivative yields the displayed linear multistep method. Its first characteristic polynomial is $\rho (z)=z^s-1$, whose unit-circle roots are all simple, proving zero-stability. Quadrature exactness gives order $s+1$ for odd $s$ and at least $s+2$ for even $s$. The first Dahlquist barrier gives equality, while the Dahlquist equivalence theorem gives convergence with consistent starts. These are finite-time convergence statements, not claims of A-stability.