 Solution

Yes: this is an A-stable third-order method. Its second derivative is essential; the Second Dahlquist barrier applies to ordinary first-derivative linear multistep methods, not this multiderivative multistep method.

Apply the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification polynomial isWe verify the root condition for a multistep method throughout $\mathrm{Re}z\le 0$. For a quadratic $a{\zeta }^2+b\zeta +c$ with $|a|>|c|$, the complex quadratic Schur criterion isfor both roots to be strictly inside the unit disk. Here is a brief justification of the criterion rather than just a root plot. Set $P^{\#}(\zeta )={\zeta }^2\overline{P(1/\overline{\zeta })}$. The polynomial $\overline{a}P-c{P}^{\#}$ is $\zeta (D\zeta +E)$, where $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Its two roots are inside precisely when $|E|<D$. On the unit circle $|P^{\#}|=|P|$, and $|c|<|a|$, so the Rouche theorem equates its interior root count with that of $P$. A boundary zero of $P$ would also be a boundary zero of the transformed polynomial, which the strict inequality excludes. This proves the strict criterion; the non-strict form follows by continuity, with any unit root simple because $|c/a|<1$.

For the present polynomial it is enough to show $8|Q-1|<|Q{|}^2-1$, except at $z=0$. Write $z=-r+iy$, $r\ge 0$, and put $k=r^2+3r$, $s=y^2$. ThenDirect expansion gives the nonnegative factorizationEvery term is nonnegative, and the sum is strictly positive unless $r=y=0$. Also $|Q{|}^2=X+1>1$, so the criterion applies. For every nonzero $z$ in the closed left half-plane both amplification roots are strictly inside the unit disk. At $z=0$ they are $1$ and $1/7$, and the unit root is simple. ThereforeThis proof establishes stability of both the principal and parasitic modes, including on the imaginary axis.