Symmetric two-step two-derivative formula ($\alpha =7/15,\beta =1/15,\gamma =16/15$)

For the multiderivative multistep method $y_{n+1}-\alpha h{f}_{n+1}+\beta h^2{g}_{n+1}=\gamma h{f}_n+y_{n-1}+\alpha h{f}_{n-1}+\beta h^2{g}_{n-1}$, with $g=f^{\prime }f$, centered expansion gives defect coefficients $2-2\alpha -\gamma$, $1/3-\alpha +2\beta$, and $1/60-\alpha /12+\beta /3$ at powers $h,h^3,h^5$. Their simultaneous vanishing gives the displayed parameters. The next coefficient is $1/4725$ at $h^7$, so the formal order is six. Reflection symmetry removes every even defect power. The method satisfies fifth-order conditions but cannot have exact order five.

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.

On the Dahlquist test equation, write the amplification polynomial as $D{\xi }^2-16z\xi -A$, where $D=z^2-7z+15$ and $A=z^2+7z+15$. The identity $|A{|}^2-|D{|}^2=28\mathrm{Re}z(|z{|}^2+15)$ shows that both roots of a polynomial can lie inside the open unit disk only if $\mathrm{Re}z<0$. But the complex quadratic Schur criterion would then require $32|\mathrm{Re}z|(|z{|}^2+15)<28|\mathrm{Re}z|(|z{|}^2+15)$, which is impossible. Thus the strict linear stability domain is empty. This concerns robust damping of all modes, rather than convergence on fixed time intervals or the bounded-root stability set of the symmetric two-derivative formula.

On the Dahlquist test equation, the symmetric two-step two-derivative formula has amplification polynomial $F_z(w)=(z^2-7z+15)w^2-16zw-(z^2+7z+15)$. For every real $z<0$, its leading coefficient is positive and $F_z(-1)=2z<0$, whereas $F_z(w)\to +\infty$ as $w\to -\infty$. The intermediate value theorem therefore gives a real root below minus one. Thus the method is not A-stable, without any appeal to a multistep order barrier.