Neutral derivative feedback amplifies outside the stable branch

Differentiating the displayed polynomial gives ${\varphi }^{\mathrm{\prime }}=q$, $q(-\tau )=1$, and $q(0)=a$. With $f=0$, the compatibility equation [F1] reads $a=a\cdot 1+0$, so it holds. The zero right-hand side is continuous and Lipschitz in the two state variables. Therefore [F2] supplies the unique solution on every finite horizon $K\tau$.

On the first segment, the equation gives $y^{\mathrm{\prime }}(s)=a{\varphi }^{\mathrm{\prime }}(s-\tau )=aq(s-\tau )$. If on segment $k-1$ the derivative is $a^{k}q(s-\tau )$, then on segment $k$ the same equation gives $y^{\mathrm{\prime }}(k\tau +s)=a{y}^{\mathrm{\prime }}((k-1)\tau +s)=a^{k+1}q(s-\tau )$. Compatibility makes the left and right endpoint values agree at every breakpoint, so induction proves the formula for $0\le k<K$. Since $q(-\tau )=1$, each segment derivative has supremum at least $\mathrm{\mid }a{\mathrm{\mid }}^{k+1}$; for $\mathrm{\mid }a\mathrm{\mid }>1$ this is geometric amplification.

For $a=1$, on the $K$-segment space from [F1], let $r_k(s)=1$ and $u_k(s)=k+1$ for $0\le k<K$. The defined shift has $(S_{K}u{)}_0=0$ and $(S_{K}u{)}_k=u_{k-1}$ for $k\ge 1$, so $(I-S_K)u=r$. Also $S_K^K=0$, hence $(I-S_K{)}^{-1}=I+S_K+\cdots +S_K^{K-1}$ by multiplying the finite sum. Since $\mathrm{\parallel }r{\mathrm{\parallel }}_{{\mathcal{X}}_K}=1$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{{\mathcal{X}}_K}=K$, the inverse norm is at least $K$. Thus no horizon-independent bound follows at $a=1$, consistent with the strict hypothesis $\mathrm{\mid }a\mathrm{\mid }<1$ in [F3].