= Solution
Apply the <Dahlquist test equation> and put $z=h\lambda$. Every recurrence mode must satisfy
$$
P_z(\zeta)=\zeta^2-1-z\{a\zeta^2+2(1-a)\zeta+a\}=0.
$$
The <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 disk>
$$
\zeta=\frac{1+w}{1-w},\qquad |\zeta|\leq1\ \Longleftrightarrow\ \operatorname{Re}w\leq0.
$$
For $\zeta\ne-1$, the characteristic equation becomes
$$
2w=z(1+Bw^2),\qquad
\operatorname{Re}z=\frac{2\operatorname{Re}w(1+B|w|^2)}{|1+Bw^2|^2}.
$$
If $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+Bw^2=0$ would force $w=0$, a contradiction. Hence $\operatorname{Re}z<0$ forces $\operatorname{Re}w<0$, 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 $4(1-Bz^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm1$. Also $\zeta=-1$ cannot be a root when $B>0$ and $z\ne0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.
If $a<1/2$, the root near $-1$ is
$$
\zeta_-(z)=-1+(1-2a)z+O(z^2).
$$
For small negative real $z$ it lies below $-1$, violating <absolute stability>. The endpoint $a=1/2$ deserves separate treatment:
$$
P_z(\zeta)=(\zeta+1)\left\{\zeta-1-\frac z2(\zeta+1)\right\}.
$$
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>,
$$
\boxed{\text{A-stability holds exactly for }a\geq\tfrac12.}
$$
There is a convention at this reducible endpoint: if <A-stability> is defined to require every unreduced recurrence mode to decay for $\operatorname{Re}z<0$, the answer is \b[$a>1/2$], 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.
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