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The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

Apply the exponential-symbol order criterion for a multistep method. The exact-solution residual of the linear multistep method is encoded by $E(z)=\rho (e^z)-z\sigma (e^z)$; vanishing through degree $p$ is equivalent to the order conditions for a linear multistep method through order $p$. Here $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)=1-\alpha$, and direct Taylor expansion givesFor example, for $n\ge 2$ the coefficient of $z^n$ can be calculated asFor $n=2,3$ this is zero, verifying the lower conditions explicitly. The constant and linear coefficients also vanish for every $\alpha$. Therefore the formal order is at least three for every $\alpha$.

The fourth-degree term vanishes exactly when $\alpha =-1$. For that value the fifth-degree coefficient is $-1/90\ne 0$, soAt $\alpha =1$ these are formal residual conditions only: the characteristic polynomial has a double root at one, so the method is not convergent. Canceling its common polynomial factor would change the recurrence and give a different order calculation. This distinction is relevant to part (b).

A fixed-step linear multistep method for $y^{\prime }=f(t,y)$ uses several previous solution and derivative values:The coefficients are fixed independently of the step size. It is explicit when ${\beta }_s=0$ and implicit otherwise. Its characteristic polynomials of a linear multistep method are $\rho (w)={\sum }_j{\alpha }_j{w}^j$ and $\sigma (w)={\sum }_j{\beta }_j{w}^j$. The contraction mapping theorem gives a uniquely solvable implicit step under a Lipschitz continuity bound $L$ whenever $k|{\beta }_s|L<|{\alpha }_s|$, by contraction of the new-value equation on the appropriate solution region. A suitable starting procedure must supply the first $s$ values; the method alone does not determine them.

Order is defined by inserting a smooth exact solution into the recurrence. If the residual is $O(k^{p+1})$ and its first nonzero coefficient occurs at that degree, the method has order $p$; dividing the residual by $k$ gives an $O(k^p)$ local truncation error. Taylor expansion gives the exact order conditions for a linear multistep methodwith the right side interpreted as zero at $q=0$, and failure at $q=p+1$ for exact order. Equivalently,The basic consistency of a numerical method conditions are $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$; for the ordinary nondegenerate consistent formulation this common value is nonzero. Exactness on constants and linear functions does not control propagation of numerical perturbations.

That propagation is the role of zero-stability. On $y^{\prime }=0$ the homogeneous recurrence is $\rho (E)y_n=0$. Its modes are $n^r{\xi }^n$, with powers of $n$ determined by root multiplicities. The theorem on the root condition for a multistep method states that zero-stability is equivalent to every root of $\rho$ having modulus at most one, with every unit-modulus root simple. Interior repeated roots are permitted, since geometric decay dominates their polynomial factors. An exterior root amplifies perturbations geometrically; a repeated unit root amplifies them polynomially and violates a step-independent perturbation bound.

The Dahlquist equivalence theorem states that a fixed-step consistent linear multistep method is convergent exactly when it is zero-stable. This concerns ordinary differential equations with a uniformly Lipschitz continuous right-hand side on the solution region over a fixed finite interval, convergent starting values, and the well-defined nearby branch of each small-step implicit solve. For order $p$ convergence of a numerical method, the solution must have the required smoothness and the starting errors must be $O(k^p)$.

The mechanism can be sketched explicitly. The root condition for a multistep method bounds the recurrence's discrete Green function. For an order-$p$ exact-solution defect $d_n=O(k^{p+1})$, the Lipschitz continuity error equation then gives an estimate of the formThe current-step term can be absorbed for small $k$. The discrete Gronwall inequality says that a bound $E_n\le A+Bk{\sum }_{j<n}{E}_j$ implies $E_n\le A(1+Bk{)}^n\le A{e}^{B{t}_n}$. There are $O(k^{-1})$ defects on a fixed interval, so their sum is $O(k^p)$ and this yields the claimed global order. Conversely, testing $y^{\prime }=0$ with vanishing starting perturbations in exterior-root or repeated-unit-root modes proves that convergence of a numerical method fails without the root condition for a multistep method.

For example, the two-step Adams-Bashforth method hasIt is explicit, order two and zero-stable, hence convergent with order-two starting values. By contrast, a recurrence with $\rho =(w-1{)}^2$ has solutions $y_n=y_0+n(y_1-y_0)$ on the zero differential equation. Starting differences $\sqrt{k}$ tend to zero yet become unbounded at a fixed positive time, since $n$ is of order $1/k$. A small local defect or formal high order cannot rescue this mode. Canceling a common factor may remove it only by changing the allowed starting relations, as the parameter endpoint in Question 2 illustrates.

Absolute stability addresses a different limit: apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=k\lambda$. The amplification roots satisfyAll roots must satisfy the disk condition, with unit roots simple; one cannot inspect only the root that tends to one as $z\to 0$. The degree and solvability of the recurrence must also be preserved. The resulting set of $z$ is the domain of absolute stability. A-stability means that it includes the whole closed left half-plane, so the numerical solutions of every Dahlquist test equation with $\mathrm{Re}\lambda \le 0$ are stable for every positive step size. It is a linear test-equation property, not a proof of arbitrary nonlinear contractivity or high accuracy at large steps.

Forward Euler method has order one and factor $1+z$, so its stability region is the disk $|1+z|\le 1$. Backward Euler method has factor $(1-z{)}^{-1}$ and is A-stable. The trapezoidal rule is order two with factor $(1+z/2)/(1-z/2)$; its modulus is at most one throughout the left half-plane. These one-step methods are included as the simplest multistep examples. For the two-step Adams-Bashforth method, the negative-real stability interval is $-1\le z\le 0$: its polynomial is $w^2-(1+3z/2)w+z/2$, and the root reaches $-1$ at $z=-1$. It therefore imposes a step restriction when applied to strongly decaying modes.

The genuinely two-step BDF2 method isIt has order two, roots $1,1/3$ at zero step and is A-stable. For its normalized polynomials $\rho =3w^2/2-2w+1/2$, $\sigma =w^2$, the unit-circle boundary locus hasThe leading coefficient has no zero in the left half-plane, and the roots at a small negative $z$ are both inside the disk. Root continuity then proves open-half-plane stability; on the imaginary boundary the only possible unit root is the simple root one at $z=0$. This gives a direct proof of the stated A-stability rather than relying on a boundary plot.

Two sharp order barriers explain the tradeoff. The first Dahlquist barrier states that a consistent zero-stable real linear $s$-step method using only first derivatives has order at most $s+1$ for odd $s$ and $s+2$ for even $s$; an explicit method has order at most $s$. The Second Dahlquist barrier states that an irreducible A-stable real linear multistep method has order at most two. These statements concern fixed constant coefficients and the ordinary first-derivative multistep class, not multiderivative methods or arbitrary variable-step formulations. Thus higher order alone is not a route to unconditional stiff stability. Question 2 provides a concrete illustration: its third-order member is zero-stable but not A-stable.

No nontrivial consistent explicit multistep method is A-stable. Its amplification polynomial of a multistep method has a fixed highest coefficient, while at least one lower coefficient grows without bound as real $z\to -\infty$. Were all roots uniformly in the unit disk, their elementary symmetric polynomials would keep all the monic coefficients bounded, a contradiction. This is the coefficient argument behind explicit multistep methods cannot be A-stable.

In practice, selecting a method therefore involves accuracy, startup, the presence of a stiff differential equation, implicit-solve cost and step-size control. Zero-stability handles propagation as the step tends to zero; consistency of a numerical method supplies a vanishing defect; A-stability handles the numerical solutions of all decaying Dahlquist test equations at finite steps. None substitutes for the others, and variable steps or PDE discretizations require additional uniform stability arguments appropriate to their setting.

Consider an initial value problem $y^{\prime }=f(t,y)$ on $[0,T]$, with a Lipschitz continuous vector field in the solution variable and enough smoothness for the stated error expansions. Convergence of a numerical method meanswith consistent starting values. An order of a numerical method $p$ quantifies this approximation: with sufficiently accurate starting data and the required stability, the global error is $O(h^p)$. Stability of a numerical method controls the amplification and accumulation of perturbations; it is the link between a small local defect and small global error.

For a one-step formula $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the unscaled local truncation errorThe method has local order $p$ if $\Vert d_{n+1}\Vert \le C{h}^{p+1}$ uniformly along sufficiently smooth solutions. If the defect is divided by $h$, its order is instead $O(h^p)$; the convention must be specified. Assume also the perturbation estimateThe global error $e_n=Y_n-y(t_n)$ then satisfiesIterating, or applying the discrete Gronwall inequality, givesThere are $O(1/h)$ steps on a fixed interval, so stable propagation turns $O(h^{p+1})$ local defects into $O(h^p)$ accumulated error. Starting error $O(h^p)$ is needed to retain order $p$. This explains why matching an exact Taylor expansion is insufficient unless errors are controlled during repeated steps.

For a fixed-coefficient linear multistep method, writeThe consistency of a numerical method conditions are $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. The higher order conditions for a linear multistep method arewith right side zero for $q=0$; exact order $p$ means first failure at $q=p+1$. They come from inserting the exact solution and comparing its Taylor expansion.

Errors at $h=0$ follow the homogeneous recurrence determined by $\rho$. Zero-stability is equivalent to the root condition for a multistep method: all roots of a polynomial have modulus at most one, and every unit-modulus root is simple. Roots outside the disk give exponential growth in the number of steps; repeated unit roots give polynomial growth in that number. Repeated interior roots are harmless because their polynomial factors are dominated by decay. Under the usual Lipschitz continuity and initialization assumptions, the Dahlquist equivalence theorem givesFor a zero-stable order-$p$ linear multistep method, local defects and starting errors of the stated orders yield global error $O(h^p)$ by a recurrence bound and the discrete Gronwall inequality. Every required starting value matters.

A concrete failure of “consistency implies convergence” isHere $\rho =(\zeta -1)(\zeta -2)$ and $\sigma =-1$, so $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=-1$: the method is consistent, of order one. But its root $2$ violates zero-stability. For $y^{\prime }=0$ with exact solution zero and starting values $Y_0=0$, $Y_1=h^2$, the computed solution is $Y_n=h^2(2^n-1)$. These starting errors vanish, yet at $n\approx T/h$ the error diverges. Local accuracy cannot compensate for an unstable recurrence.

A different meaning of stability concerns a fixed step on decaying equations. The Dahlquist test equation $y^{\prime }=\lambda y$ introduces $z=h\lambda$. A one-step method gives $Y_{n+1}=R(z)Y_n$ and is absolutely stable when $|R(z)|\le 1$. For a linear multistep method, the corresponding amplification polynomial of a multistep method is $\rho (\zeta )-z\sigma (\zeta )$, and every root must satisfy the bounded root condition, not just the root approximating $e^z$. Strict asymptotic decay requires the roots to have modulus less than one for $\mathrm{Re}z<0$. This distinction matters at reducible methods with an undamped parasitic mode, as in question 1(c).

The Forward Euler method has order one and $R(z)=1+z$, so its linear stability domain is $|1+z|\le 1$. For $y^{\prime }=-\kappa y$, with $\kappa >0$, it requires $0\le h\kappa \le 2$. It converges as $h\to 0$, yet an excessively large step can produce growing oscillations on an exactly decaying problem. The Backward Euler method has the same order, but $R(z)=(1-z{)}^{-1}$ makes it A-stable and L-stable. For a stiff differential equation, fast decay can impose a severe explicit time-step restriction even when accuracy of the slow dynamics would permit a much larger step.

The trapezoidal rule has order two and is A-stable, but is not L-stable, since its stability function tends to $-1$ along the negative real axis. The Second Dahlquist barrier limits an irreducible A-stable linear multistep method to order at most two. This does not limit implicit Runge-Kutta methods: question 4's Lobatto IIIA method is fourth-order and A-stable. Conversely, question 1's fourth-order member is convergent but not A-stable. These examples show that order of a numerical method, zero-stability and A-stability answer different questions.

For nonlinear dissipative problems, B-stability controls distances between numerical solutions; algebraic stability of a Runge-Kutta method is a sufficient coefficient criterion. Question 4 shows that A-stability alone does not imply that coefficient criterion. Variable time steps require additional analysis of step ratios and error propagation, and stiff problems may require uniform-in-stiffness error estimates beyond the fixed-problem convergence result. The useful chain is local accuracy plus the appropriate perturbation bound, giving global accuracy; absolute and nonlinear stability then determine which practical time steps preserve the intended dynamics.

With ${\rho }_0=a$, ${\rho }_1=\theta$, ${\rho }_2=1$, the constant order condition gives $a=-1-\theta$. The next three order conditions for a linear multistep method giveSolving this linear system yields the unique coefficients of order at least three:The characteristic polynomial controlling zero-stability factors asThe other root is $-1-\theta$. Requiring its modulus to be at most one gives $-2\le \theta \le 0$, but $\theta =-2$ produces a repeated root at one and violates the root condition for a multistep method. At $\theta =0$, the roots $1,-1$ are both simple, so that endpoint is allowed. By the Dahlquist equivalence theorem, the method converges precisely forThe fourth defect coefficient is $C_4=\theta$, so the method has order exactly three for nonzero $\theta$. At $\theta =0$ it is the two-step Simpson method, of order four; thus the extra-order endpoint has correctly remained in the answer.

Factor the first characteristic polynomial asThe other two roots of a polynomial have product one. If $-1<\alpha <1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$: distinct unit-modulus roots, both different from one. At $\alpha =-1$, the root $-1$ is double; at $\alpha =1$, the root one is triple. Outside $[-1,1]$, one of the reciprocal real roots has modulus greater than one. Thus the root condition for a multistep method holds exactly for $-1<\alpha <1$.

For the order conditions for a linear multistep method, expand the exponential defect:The constant, linear and quadratic coefficients vanish for every $\alpha$. The cubic coefficient vanishes only at $\alpha =11/5$, where the quartic coefficient is $-3/5\ne 0$. Consequently the formal order isFor all parameters satisfying the root condition for a multistep method, ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )\ne 0$, so this formal calculation gives ordinary second-order consistency of a numerical method. The Dahlquist equivalence theorem therefore yieldsThe exceptional third-order formula is not zero-stable. At $\alpha =1$, $\sigma \equiv 0$ and $\rho =(w-1{)}^3$: the recurrence contains no vector-field evaluations. It satisfies the displayed Taylor identities only formally and is a degenerate, nonconvergent formula, not a usable second-order ODE solver. This distinction avoids interpreting formal cancellation as convergence.

Substitute the exact solution and expand every value about $t_n$. The coefficient of $h^q{y}^{(q)}(t_n)$ in the defect is zero for $q=0,\dots ,p$ exactly whenThese are the order conditions for a linear multistep method, specialized to the two nonzero derivative coefficients. The first failed identity determines the leading local truncation error.

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.