Decay rate for a small perturbation of a simple linear ODE

Question

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.

Answer 1

$\newcommand\R{\mathbb R}$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, $f(t)\cdot e_1=e^{-2t}-e^{-3t}\ne\Theta(e^{-3t})$.