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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 $f$.

Insert the Fourier modeinto the leapfrog advection scheme. This givesand hence the amplification polynomialIts roots areIf $0<\mu <1$, the square root is real for every $\theta$, andIf $\mu >1$, choosing $|\sin \theta |>1/\mu$ makes the roots non-real multiples of $i$ with reciprocal moduli, one of which exceeds one, so the method is unstable.

The endpoint $\mu =1$ requires care. At $\theta =\pi /2$ the polynomial isso a generic mode has the form $(c_1+c_{2}n)i^n$ and grows linearly. It satisfies the weaker test $|{\zeta }_{\pm }|\le 1$ often quoted in elementary Fourier analysis, but it violates the root condition for a multistep method because the unit-modulus root is repeated. Therefore the rigorous uniform stability range for arbitrary two-level initial data isIf “stable” is being used only for the non-amplification condition on root moduli, the conventionally quoted range is $0<\mu \le 1$, with $\mu =1$ a defective marginal endpoint rather than a uniformly stable case.

Along an exact autonomous-ODE solution, $f(y)=y^{\prime }$ andMove every term to the left and substitute the Taylor expansions about $t_n$. The coefficients of $h^j{y}^{(j)}(t_n)$ vanish for $0\le j\le 4$, while the first nonzero coefficient isThus the local defect is $O(h^5)$. At $h=0$ the first characteristic polynomial iswhich satisfies the root condition for a multistep method. The method is therefore zero-stable and has

The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable. HereThe consistency conditions hold for every $a$. The two quadratic roots have product one. They are distinct and on the unit circle exactly when $0<a<2$; at $a=2$ the root $-1$ is repeated, while outside this interval one root has modulus greater than one. The root condition for a multistep method therefore givesThe polynomial exactness conditions hold through degree two, while the degree-three residual is $6-a$. Every convergent member thus has(The exceptional value $a=6$ has formal order four but is not zero-stable.)

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.

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.

At $h=0$ the first characteristic polynomial isIts roots satisfy the root condition for a multistep method: the only unit-modulus root is the simple root $1$, and the other root has modulus $1/5$. The method is consistent because part a gives positive order. The Dahlquist equivalence theorem, in the form allowed by the question for this derivative-augmented method, therefore proves convergence.

The polynomial $\rho (w)=(w-1)(w^2+w+a)$ satisfies the root condition for a multistep method exactly for $0\le a\le 1$. For $a<0$ a root is below $-1$; for $a>1$ the quadratic root product forces a root outside the unit disc. The endpoints have simple unit-modulus roots, and the repeated root at $a=1/4$ is strictly inside. A consistent method with this polynomial is convergent throughout the closed interval.