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For centered-space diffusion, a Fourier mode gives $z^2-(1-3a/2)z-a/2=0$, with $a=4\mu {\sin }^2(\theta /2)$. For $a>0$ the roots have opposite signs, the positive root lies in $(0,1)$, and the negative root lies in $[-1,0)$ exactly when $P(-1)=2-2a\ge 0$. At $a=0,1$ the unit-modulus roots are simple. Hence the root condition for a multistep method holds for every Fourier frequency exactly when $\mu \le 1/4$. Under fixed $\mu$, the unscaled one-step defect is $-\Delta t^2{u}_{xxxx}/(12\mu )+O(\Delta t^3)$; dividing by the step changes its order to $O(\Delta t)$.

For the symmetric two-step two-derivative formula, the closed-disk root condition for a multistep method fails off the imaginary axis. Positive real part makes the product of root moduli exceed one; negative real part fails the complex quadratic Schur criterion, as in the empty strict decay domain of the symmetric two-derivative formula proof. On $z=iy$, write $D=15-y^2-7iy=|D|e^{i\delta }$ and set $\xi =i{e}^{-i\delta }\eta$. The amplification polynomial becomes ${\eta }^2-16y\eta /|D|+1=0$. Its two roots are distinct and on the unit circle exactly when $|16y/|D||<2$, or $y^4-45y^2+225>0$. Thus the bounded-root set consists of the imaginary intervals $|y|<\sqrt{(45-15\sqrt{5})/2}$ and $|y|>\sqrt{(45+15\sqrt{5})/2}$, including $y=0$ and excluding the double-root endpoints. In particular, the set is unbounded. Unit-modulus oscillations are bounded but do not tend to zero.

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 a fixed-step multiderivative multistep method, the ordinary zero-stability root condition still controls propagation of the starting errors when all derivative evaluation maps are uniformly Lipschitz continuous on the relevant bounded region. A local defect $O(h^{p+1})$ then yields global error $O(h^p)$ over a fixed time interval, provided the starting errors are $O(h^p)$ and the implicit updates use the nearby solution branch.

For example, writing the numerical error equation as $\rho (E)e_n=h{F}_h(e_{n+s})+d_n$, with $F_h$ uniformly Lipschitz continuous, a bounded impulse response for the root condition for a multistep method givesAbsorbing the last current-step term for small $h$ and applying the discrete Gronwall inequality gives the stated order. The higher derivatives enter through $F_h$, whose bound stays uniform as $h\to 0$.

For a fixed-coefficient linear multistep method applied to a well-posed initial value problem with a continuous right-hand side uniformly Lipschitz continuous in the solution argument on the relevant region, consistency of a numerical method and zero-stability are equivalent to numerical convergence for arbitrary starting values whose errors tend to zero. The time interval is fixed while the step size tends to zero; an implicit numerical method must use its well-defined nearby solution branch for sufficiently small steps.

The quantitative statement is that a zero-stable method with consistency error $O(h^p)$ and starting errors $O(h^p)$ has global error $O(h^p)$, provided the exact solution is sufficiently smooth for that residual bound. A bounded response to step defects follows from the root condition for a multistep method, and the discrete Gronwall inequality controls the nonlinear error terms. Mere continuity of the right-hand side does not by itself supply these hypotheses or uniqueness of the exact solution.

For an ordinary differential equation $y^{\prime }=\Phi (t,y)$, this linear multistep method uses a centered derivative across the two-step interval $[t_{i-1},t_{i+1}]$, evaluated at its midpoint $t_i$. It requires two starting values, $y_0$ and $y_1$. For a sufficiently smooth exact solution, Taylor's theorem givesThe unscaled local residual is therefore $O(h^3)$ and the method has order two. Its first characteristic polynomial is $z^2-1$, with the simple roots $1,-1$, so it satisfies the root condition for a multistep method; the second root is a parasitic amplification root. This two-step method differs from the one-step midpoint method and the implicit midpoint rule.

For the special equation $G^{\prime }=g(t)$ and $G(a)=0$, an even number $n=2r$ of steps gives $y_n=2h{\sum }_{j=1}^{r}g(t_{2j-1})$, independently of the odd-index starting value. It is the composite midpoint quadrature rule on $r$ intervals of width $2h$. An odd final index depends on $y_1$, which must be specified separately.

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.

For a consistent zero-stable real linear $k$-step method with constant coefficients and only first-derivative evaluations, the attainable order is at most $k+1$ for odd $k$ and $k+2$ for even $k$. For an explicit method the bound is $k$. These are order barriers under the root condition for a multistep method, not bounds on multiderivative methods or on arbitrary variable-step formulations. The Second Dahlquist barrier adds the stronger bound of order two for irreducible A-stable methods.

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.

Interpret the printed missing parenthesis as ending $f(y_{n+2})$ before the next summand. The characteristic polynomials of a linear multistep method are thenThey satisfy $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the consistency identities hold. The two nonprincipal roots are reciprocal. If $|\alpha |<1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$, distinct unit roots also distinct from the principal root one. If $|\alpha |>1$, one reciprocal root lies outside the unit circle. At $\alpha =1$ the root one is triple, and at $\alpha =-1$ the root minus one is double. The root condition for a multistep method therefore holds exactly when $-1<\alpha <1$. The Dahlquist equivalence theorem, with standard smoothness and consistent starting data, givesTo determine order rather than guess it from the number of steps, the exponential-symbol order criterion for a multistep method givesThus the formal order is three for $\alpha \ne -5$, and four for $\alpha =-5$. At the exceptional value the $z^5$ coefficient is $1/10$. That fourth-order formula is not zero-stable. Every convergent member consequently has global order three. At $\alpha =1$ the third-order Taylor cancellation is only a formal defect property of a degenerate, nonconvergent recurrence.

Here is a direct A-stability test for the convergent parameters. For a nonprincipal unit root $r=\alpha +i\sqrt{1-{\alpha }^2}$, the simple root of the amplification polynomial of a multistep method satisfiesReduction using $r^2-2\alpha r+1=0$ givesFor small negative real $z$, it follows thatSo even a small negative real test parameter has an amplifying parasitic mode. None of the convergent parameters is A-stable. Parameters outside this range already fail the necessary zero-step root condition, so there is no real $\alpha$ giving an A-stable original method.

The endpoint $\alpha =1$ is worth keeping separate from a misleading cancellation: there $\rho =(\zeta -1{)}^3$ and $\sigma =\zeta (\zeta -1{)}^2$. The original amplification polynomial retains the double root one for every $z$, giving unbounded polynomial parasitic modes for general starting values. Canceling $(\zeta -1{)}^2$ would yield a different, reduced Backward Euler method; it does not stabilize the original recurrence. This is the common-factor cancellation defect in a multistep recurrence. The family is summarized by the reciprocal-root three-step multistep family.

Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the characteristic polynomials of a linear multistep methodFor every admissible real $\alpha$, $\rho (1)=0$ and ${\rho }^{\prime }(1)=2\alpha =\sigma (1)$, giving consistency of a numerical method. Factorization givesThe second root is $r=(2-\alpha )/(2+\alpha )$. The root condition for a multistep method requires $|r|\le 1$ and forbids a repeated root at one. For real $\alpha \ne -2$,At $\alpha =0$ both roots equal one, violating zero-stability. For $\alpha >0$ the second root lies strictly inside the unit disk. The Dahlquist equivalence theorem therefore givesTo find the order rather than only consistency, expand the exact-solution residual through the generating function:Thus its unnormalized local truncation error begins with $(\alpha /3)h^3{y}^{(3)}$ when $\alpha >0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha =0$ has formal order three but is not convergent because of its double unit root:These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.

The stability convention matters. The Cambridge numerical analysis notes define a strict linear stability domain through decay $y_n\to 0$. For a multistep recurrence, robust decay requires every amplification root to lie strictly inside the unit disk. With that convention we can prove the stronger result that this method's strict linear stability domain is empty, and therefore bounded. Mere boundedness of oscillatory sequences gives a different answer, described below.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$. With $z=h\lambda$, the amplification polynomial of a multistep method isExclude $D=0$, where the intended two-step update cannot be solved. If both polynomial roots have modulus below one, their product requires $|A|<|D|$. Butso necessarily $\mathrm{Re}z<0$. The complex quadratic Schur criterion also requiresIts left-hand side equals $32|\mathrm{Re}z|(|z{|}^2+15)$, while for $\mathrm{Re}z<0$ its right-hand side is only $28|\mathrm{Re}z|(|z{|}^2+15)$. The inequality is impossible. ThusFor comparison, the bounded-root stability set of the symmetric two-derivative formula is unbounded. On $z=iy$, rotate the amplification polynomial into a real reciprocal quadratic. Its two roots are distinct and on the unit circle precisely whenequivalently $y^2<(45-15\sqrt{5})/2$ or $y^2>(45+15\sqrt{5})/2$. In particular, arbitrarily large imaginary parameters give bounded, undamped numerical oscillations. At the two equality thresholds, repeated unit-circle roots produce unbounded solutions. Hence the question's bounded-domain claim holds for the strict decay convention; it does not hold if the phrase is interpreted using the non-growing root condition for a multistep method.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

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.

Use a Fourier mode $u_m^n=z^n{e}^{im\theta }$. Put $a=4\mu {\sin }^2(\theta /2)\ge 0$. Its amplification polynomial of a multistep method isThe two roots are real, one nonnegative and one nonpositive. For $a>0$, evaluation at $1$ gives $a>0$ and at zero gives $-a/2<0$, so the positive root lies below one. The negative root is at least $-1$ exactly when the value at $-1$, namely $2-2a$, is nonnegative. Thus both roots have modulus at most one exactly for $0\le a\le 1$. At $a=0$ they are $1,0$, and at $a=1$ they are $1/2,-1$; all unit-modulus roots are simple, satisfying the multistep root condition for a multistep method. Since the largest available $a$ is $4\mu$, the Adams-Bashforth stability for centered diffusion condition isFor $\mu >1/4$, the highest-frequency mode has a root less than $-1$ and grows, proving necessity as well as sufficiency.

Along a smooth exact solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Expand the unscaled defect of the multiderivative multistep method about the central time $t_n$. The antisymmetric difference of the solution and second derivative, and the symmetric sum of first derivatives, giveAll derivatives are evaluated at $t_n$. Reflection symmetry removes every even power. The fifth-order conditions require the coefficients of $h,h^3,h^5$ to vanish. The last two equations give $\beta =1/15$ and $\alpha =7/15$, and the first then gives $\gamma =16/15$. ThusThese determine the symmetric two-step two-derivative formula. Substitution into the next coefficient givesThe requested fifth-order conditions actually produce formal order six. In the usual maximal-order terminology there is no member of this symmetric family with exact order five: cancelling the fifth-power defect also cancels the automatically zero sixth-power defect. Interpreting the printed request as order at least five gives precisely the boxed parameters.

At zero step size the recurrence polynomial is $w^2-1$, with simple roots $1$ and $-1$, so it satisfies the root condition for a multistep method and is zero-stable. With sufficiently smooth locally Lipschitz derivative maps, the nearby implicit solution branch and starting values of matching accuracy, the convergence of a zero-stable multiderivative method gives global order six. Formal order by itself does not imply decay of its additional numerical mode.

Consistency was established in part (a). By the Dahlquist equivalence theorem, convergence additionally requires the root condition for a multistep method. Since $\rho (w)=(w-1)(w-\alpha )$, its roots lie in the closed unit disk with each unit-modulus root simple exactly forThe endpoint $\alpha =-1$ is zero-stability although it has a simple parasitic root at minus one; $\alpha =1$ has a repeated root at one and is excluded.

For every parameter in this convergent range, the leading coefficient of $\sigma$ is $(5+\alpha )/12>0$, whileSince $\sigma (w)\to +\infty$ as $w\to -\infty$, it has a real root $w_{\ast }<-1$. Its other real root lies between minus one and one because $\sigma (1)=1-\alpha >0$, so the exterior root is simple.

On the Dahlquist test equation, the amplification polynomial of a multistep method is $P_z(w)=\rho (w)-z\sigma (w)$, with $z=h\lambda$. For $|z|\to \infty$, the equivalent polynomial $-P_z/z=\sigma -\rho /z$ tends coefficientwise to $\sigma$, and its leading coefficient stays nonzero. Choose a small closed disk about $w_{\ast }$ lying wholly outside the unit disk and containing no other root of $\sigma$. On its boundary $|\sigma |$ has a positive minimum, while $|\rho /z|$ is smaller for every sufficiently large $|z|$. Rouché's theorem then puts one root of $P_z$ inside this exterior disk for all such complex $z$.

That root violates the stability root condition for a multistep method. Hence there is a finite radius $R_{\alpha }$ beyond which no point belongs to the absolute stability domain:This proves boundedness in every complex direction, rather than only excluding A-stability. It is an instance of exterior roots of the derivative polynomial bound multistep stability.

For this two-step family with a third-order member, use the exponential-symbol order criterion for a multistep method. Expanding the exact-solution residual symbol givesThe constant, linear and quadratic coefficients vanish for every parameter. ThusAt the exceptional value the fourth-degree coefficient is $-1/30$, so there is no further increase. This means an unscaled exact-solution residual of order $h^{p+1}$, or a residual divided by the step of order $h^p$.

The first characteristic polynomial factors as $\rho (w)=(w-1)(w-a)$. The root condition for a multistep method requires every root to have modulus at most one and each unit-modulus root to be simple. The root $a=-1$ is allowed because it is distinct from the simple root one; $a=1$ is not, since it makes one a double root. Hence zero-stability holds exactly forOn this interval, $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=1-a\ne 0$ establish consistency of a numerical method. The Dahlquist equivalence theorem, for a fixed-step consistent linear multistep method applied to an ordinary differential equation with a Lipschitz continuous right-hand side with convergent starting values and well-defined small-step updates, says that convergence of a numerical method is equivalent to zero-stability. Therefore the same interval is exactly the convergence of a numerical method range. With sufficiently accurate starting values the convergent members have the orders just computed.

At $a=1$ the Taylor expansion still gives a formal order-two residual, but ${\rho }^{\prime }(1)=\sigma (1)=0$ and the double root prevents a convergent method. Canceling the common factor $w-1$ produces the common-factor cancellation defect in a multistep recurrence and changes the starting relations and cannot repair convergence of a numerical method of the original recurrence. At $a=0$, cancellation of a zero root instead reveals the ordinary trapezoidal rule; it is consistent with the order-two result.

Apply the method to the Dahlquist test equation, put $z=h\lambda$, and retain every root of its amplification polynomial of a multistep method:An A-stable convergent method must satisfy the root condition for a multistep method for every $\mathrm{Re}z\le 0$.

First take $-1<a<0$. As negative real $z$ tends to minus infinity, one root tends to the nonzero root of $\sigma$,It follows by root continuity that some sufficiently negative finite $z$ is unstable. At $a=-1$, the recurrence has polynomial $w^2-2zw-1$; for any negative real $z$, its root $z-\sqrt{z^2+1}$ is below $-1$. Thus all negative convergent parameters are excluded.

For $0\le a<1$, use the boundary-locus test for multistep A-stability. A unit root $w=e^{i\theta }$ with $\sigma (w)\ne 0$ would require $z=\rho (w)/\sigma (w)$. Direct multiplication by the conjugate denominator givesThere is therefore no unit-root crossing in the open left half-plane. The leading coefficient of $P_z$ never vanishes there, since its only zero is the positive real value $2/(1+a)$. For a small negative real $z$, the simple root initially at one moves to $1+z+O(z^2)$, inside the disk, and the other root is close to $a$, also inside. Continuity of polynomial roots and connectedness of the open left half-plane then keep both roots strictly inside throughout that half-plane. Possible zeros of $\sigma$ cause no omitted crossing: for $a>0$ its other root lies strictly inside the unit disk, while at $a=0$ its unit-circle zero $w=-1$ has $\rho (-1)\ne 0$.

On the imaginary axis, if $a>0$ the displayed real part vanishes only at $w=1$, which corresponds to $z=0$ and is simple. Other imaginary-axis parameters retain strictly interior roots. If $a=0$, the two roots are zero and the stability function $(1+z/2)/(1-z/2)$, whose modulus is one on that axis; the unit root is simple. Thus the required boundary root condition for a multistep method holds as well.

Combining necessity and sufficiency givesIn particular, the third-order member $a=-1/5$ is convergent but not A-stable, in agreement with the Second Dahlquist barrier.