$g'(t)-g'(t-T)=g(t)$$\text{(1)} \quad\quad\quad g'(t)-g'(t-T)=-g(t)$$
There is plenty of literature on
The asymptotics of solutions to
$g'(t)-Cg'(t-T)=Ag(t)+Bg(t-T)$
but everyone seems $\text{(2)} \quad\quad\quad g'(t)-Cg'(t-T)=Ag(t)+Bg(t-T),$$
including $C=1$, is governed by the characteristic function
$$H(\lambda)=\lambda(1-Ce^{-\lambda T})-A-Be^{-\lambda T}$$
via the following theorem
If $\alpha_0=\sup\{Re(\lambda)\,:\,H(\lambda)=0\}$ and $g(t)$ is a solution to be avoiding (2), then, for any statement on the case $C=1$, which \alpha>\alpha_0$, there is sometimes called "critical". I am aware a constant $K=K(\alpha)$ such that $$ \lvert g(t)\rvert\le Ke^{\alpha t}\sup_{-T\le s\le0}\lvert g_0(s)\rvert. $$
Because of the context in which equation (1) arises I expect $g(t)$ toconverge to $0$, so let's apply the theorem.
The roots of the characteristic function of (1) are solutions to equation
$\lambda(1-e^{-T\lambda})=1$ have some nasty asymptotic behavior, i.e$e^{\lambda T}=\frac{\lambda}{\lambda+1}.$Re \lambda_j\to0-$ as $|\lambda_j|\to\infty$, in fact that's
After taking the point in considering such equation.
1module of both sides we have
$$e^{Re(\lambda) Are there any techniques T}=\big\lvert\frac{\lambda}{\lambda+1}\big\rvert<1,$$
so $Re(\lambda)<0$. On the other hand, for approaching $\lvert\lambda\rvert\gg1$real part of $\lambda$ converges to $0-$, so $\alpha_0=0$ and the asymptotics lasttheorem is inconclusive in the matter of convergence of $g(t)$ to $0$so the "critical" case?
2question is:
How to establish rate of convergence (or divergence) Are there any papers dealing with eigenvalues that behave as describedof solutions toequation (1) and its (in)dependence on initial data?
The conjecture is that $g(t)\sim t^{\gamma}e^{-\beta t}$, where$\gamma$ and $\beta\ge0$ might depend on the differentiability classof initial data, but that's a wild guess.
