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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.

Every convergent member is also A-stable. For $0<\alpha <2$, put $\zeta =e^{i\theta }$ and calculate its boundary-locus test for multistep A-stability:The denominator never vanishes on the unit circle in this parameter interval. It follows thatTherefore no amplification root can cross the unit circle as $z=h\lambda$ moves in the open left half-plane. For small negative real $z$, the root near one moves to $1+z+O(z^2)$, inside the circle; the other begins at $\alpha -1$, also strictly inside. The leading coefficient $1-(1-\alpha /4)z$ never vanishes in the left half-plane. Continuity of polynomial roots and connectedness of that half-plane now keep both roots inside everywhere. On its imaginary boundary the same locus has zero real part only at $\zeta =1$, corresponding to $z=0$; that root is simple because $2-\alpha >0$.

At $\alpha =0$, the two interlaced subsequences follow the trapezoidal rule with step $2h$:The right side has modulus at most one for $\mathrm{Re}z\le 0$. Its unit-circle roots are simple; the double zero at $z=-1$ lies strictly inside and is harmless. Thus the endpoint is also A-stable. Combining this with part 1, the requested range is $0\le \alpha <2$, including the BDF2 method at $\alpha =4/3$.

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.

This real-parameter linear multistep method has formal order two except at the degenerate value $\alpha =2$, where its local symbol vanishes to fourth degree. The zero-stability roots are $1$ and $\alpha -1$, so it is convergent exactly for $0\le \alpha <2$. Every convergent member is A-stable: the boundary-locus test for multistep A-stability gives nonnegative real part $\alpha (2-\alpha )(1-\cos \theta {)}^2/(2|\sigma (e^{i\theta }){|}^2)$, with no unit-circle denominator zero for $0<\alpha <2$. At $\alpha =0$ the two interlaced sequences obey a trapezoidal step of length $2h$. At $\alpha =4/3$ it is the BDF2 method. The endpoint two fails the root condition for a multistep method even though canceling a common polynomial factor resembles the trapezoidal rule.