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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.

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.

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.

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.

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.

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.

Use the standard A-stability convention including the root condition at $z=h\lambda =0$. The amplification polynomial of a multistep method isZero-stability first restricts the possible parameters to $-1\le a<1$. For $-1<a<0$, as $z$ tends to negative infinity, one amplification root tends to $(3a-1)/(1+a)$, whose modulus exceeds one. For $a=-1$, the characteristic equation is ${\zeta }^2-2z\zeta -1=0$; large negative real $z$ likewise gives an exterior root. Thus $a\ge 0$ is necessary.

For sufficiency take $0\le a<1$. The leading coefficient cannot vanish in the closed left half-plane. On the unit circle $\zeta =e^{i\theta }$ the boundary-locus test for multistep A-stability uses $z=\rho (\zeta )/\sigma (\zeta )$ and giveswhere the denominator is nonzero. When $a=0$, the exceptional zero of $\sigma$ at $\zeta =-1$ is not a zero of $\rho$ and therefore cannot be an amplification root for a finite $z$. Near $z=0$ on the negative real axis, the root issuing from one is $1+z+O(z^2)$ and is strictly inside the disk; the other root is close to $a$ and is also inside. Roots vary continuously, cannot escape through infinity because the leading coefficient is nonzero, and cannot cross the unit circle anywhere in the open left half-plane by the displayed boundary formula. Thus every root remains inside there. Continuity gives the boundary case; for $a>0$ a unit root on the imaginary axis can occur only at $z=0$, where it is simple. For $a=0$, cancellation of the harmless zero root leaves the trapezoidal rule, whose amplification factor has modulus at most one.

ThereforeThe third-order member is convergent but not A-stable, consistent with the Second Dahlquist barrier. At $a=1$, $P_z=(\zeta -1)[(1-z)\zeta -1]$ has a permanent unit root and a double root at $z=0$: the unreduced method is not zero-stable and is not A-stable under the stated convention. Testing only the open half-plane while overlooking its zero-step behavior would give a weaker conclusion.

Apply the Dahlquist test equation and put $z=h\lambda$. Every recurrence mode must satisfyThe amplification polynomial of a multistep method, rather than just its root near one, determines absolute stability. Put $B=2a-1$ and use the Cayley transform between the half-plane and diskFor $\zeta \ne -1$, the characteristic equation becomesIf $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+B{w}^2=0$ would force $w=0$, a contradiction. Hence $\mathrm{Re}z<0$ forces $\mathrm{Re}w<0$, so all amplification roots have modulus less than one. On the imaginary axis the roots have modulus one and are simple: the transformed quadratic has discriminant $4(1-B{z}^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm 1$. Also $\zeta =-1$ cannot be a root when $B>0$ and $z\ne 0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.

If $a<1/2$, the root near $-1$ isFor small negative real $z$ it lies below $-1$, violating absolute stability. The endpoint $a=1/2$ deserves separate treatment:One root is always $-1$; the other is the trapezoidal rule multiplier $(1+z/2)/(1-z/2)$. They are distinct for every finite $z$ in the left half-plane. Consequently, with absolute stability understood as the bounded root condition for a multistep method,There is a convention at this reducible endpoint: if A-stability is defined to require every unreduced recurrence mode to decay for $\mathrm{Re}z<0$, the answer is $a>1/2$, since the $(-1{)}^n$ mode persists at $a=1/2$. Canceling the common factor $\zeta +1$ gives the A-stable trapezoidal rule, but cancellation removes an actual starting-error mode of the original two-step recurrence. The fourth-order member $a=1/3$ is outside either A-stability range.

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.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$, so the amplification polynomial of a multistep method becomesBoth roots of a polynomial, including the parasitic one, must satisfy the root condition for a multistep method.

For $\alpha =1/7$, multiply by seven and set $a=7-6z+2z^2$. The polynomial is $a{\xi }^2-8\xi +1$. The complex quadratic Schur criterion says that its roots of a polynomial are strictly in the unit disk ifHere is a direct verification on the entire left half-plane. Put $z=-r+iy$, $r\ge 0$, $t=y^2$, and $a_0=7+6r+2r^2$. Thenand expansion of the squared inequality givesEvery term is nonnegative, and at least one is positive unless $z=0$. Since $q-1>0$, taking square roots proves the strict complex quadratic Schur criterion for $z\ne 0$. At $z=0$, the roots of a polynomial are the simple values $1$ and $1/7$. The leading coefficient cannot vanish because $|a{|}^2\ge 49$. Thus the method at $\alpha =1/7$ is A-stable. Its order three is compatible with the Second Dahlquist barrier: that barrier concerns first-derivative linear multistep methods, whereas this method uses a second derivative.

For $\alpha =1/2$, choose $z=-4$, strictly in the left half-plane. The amplification polynomial of a multistep method reduces toOne root of a polynomial has modulus $(3+\sqrt{17})/4>1$. Thus the method at $\alpha =1/2$ is not A-stable.

Only $-1<\alpha <1$ needs consideration. On the Dahlquist test equation, the amplification polynomial of a multistep method is $P_z(w)=\rho (w)-z\sigma (w)$, where $z=h\lambda$. A-stability requires all amplification roots to have modulus at most one for every $\mathrm{Re}z\le 0$, with simple unit-modulus roots.

Choose the negative real value $z=-2/(1-\alpha )$. The leading coefficient of $P_z$ iswhereas $P_z(1)=-z\sigma (1)=4>0$. Hence $P_z(w)\to -\infty$ as real $w\to +\infty$. The intermediate value theorem supplies a real amplification root $w>1$, so this negative test value is unstable. ThereforeThis is a direct absolute stability obstruction, independent of the Second Dahlquist barrier, which by itself would not rule out the convergent second-order members.