MOTIVATION.


Let $f:[0,+\infty)\to \mathbb{R}$.
Solutions to $\partial\_tf(t) = -\lambda f(t)$, $f(0)\neq 0$ approach zero *exactly* as $e^{-\lambda t}$.
This property is preserved if we apply an exponentially decaying perturbation, i.e. we consider
$$\partial\_t f(t) = -\lambda f(t) +R(t) f(t)$$
where $|R(t)|< Ce^{-\delta t}$, $\delta >0$.
To see this we can use that $f(t) = f(0) e^{-\lambda t + \int\_{0}^tR(s) ds}$, and
$$e^{-\lambda t + \int\_{0}^tR(s) ds} \geq e^{-\lambda t + \int\_0^t Ce^{-\delta s}ds}
\geq e^{-\lambda t + C\_1}.$$
Therefore even in the perturbed case $f(t) = \Theta(e^{-\lambda t})$ for $t\to +\infty$.


QUESTION:



> 
> what happens in higher dimension? Suppose that $f:[0,+\infty) \to H$, $H$ Hilbert space, solves
> $$ \partial\_t f(t) = - Af + R(t)f(t)$$
> $$f(0) \neq 0 $$
> where $A, R(t) \in \mathcal{L}(H)$ and $||R(t)||< C e^{-\delta t}$ and $A$ is symmetric and has an (ortho)-eigenbasis $\{\psi\_\lambda\}\_{\lambda\in \sigma(A)}$, $\sigma(A) $ discrete, $0\not \in \sigma(A)$.
> Does still hold that $f\_\lambda(t) = \Theta(e^{-\lambda t})$ when not $0$?
> 
> 
> 


Here $f\_\lambda(t) = \langle f(t), \psi\_\lambda\rangle$ is the projection on the subspace generated by the eigenvector $\psi\_\lambda$.


Of course if $[A,R(t)]= 0$ then the result holds as we can reduce to the 1D case.



Decay rate for a small perturbation of a simple linear ODE
$\newcommand\R{\mathbb R}$The answer is no.


E.g., suppose that $H=\R^2$, with the standard basis $(e\_1,e\_2)$, $Ae\_1=3e\_1$, $Ae\_2=e\_2$, $R(t)e\_1=0$, $R(t)e\_2=e^{-t}e\_1$, and $f(0)=(0,1)$. So, $(e\_1,e\_2)$ is an ortho-eigenbasis of $A$, with the corresponding eigenvalues $3$ and $1$.


However, $f(t)\cdot e\_1=e^{-2t}-e^{-3t}\ne\Theta(e^{-3t})$.


