How to linearize time-variant ordinary differential equation (ODE), in particular, if there is a convolution term?

Question

Letting $(\ * \ )$ denote convolution, I have an ODE of the form \begin{align} \dot{r} &= -7.4r -1.6f - 8.8 (f(t)*e^{-t}) - 10.4(f(t)*(te^{-t})) \\ \dot{f} &= 0.25r \end{align} with initial conditions $r(0) = f(0) = 0.$

I would like to linearize about $[r=0, f=0]^t$. I realized that in general I can linearize when I have a time invariant ODE; e.g., letting $x(t) = [r,f]^t$ I can linearize $\dot{x} = g(x)$.

I realized I do not know how to linearize a general equation of the form $\dot{x} = g(x,t)$ (time varying). How do we linearize time-variant ODEs?

Aside from the general question, there is a more specific question--how do we linearize the convolution terms? It seems I need to compute $\frac{\partial }{\partial f}$ of the operator $f\mapsto (f(t)*e^{-t})$. This implies we are considering the derivative of an operator defined on a function space---what are the function spaces and norms/topologies in play here?

Answer 1

The hard-wired convolutions in the first rhs "hide" the actual linear, autonomous (rhs does not explicitely depend on $t$, which is actually mute in your notation $f(t)*e^{-t}$) 4th order equivalent system: $$ \begin{align*} r'&=-7.4r-1.6f-8.8q-10.4s\\ f'&=0.25r\\ q'&=-q+f\\ s'&=-s+q \end{align*} $$ with $q(0)=s(0)=0$. Hence as already said, no linearization is needed.

Answer 2

The set of odes is linear. You can solve it using the Laplace Transform.

\begin{align} s\hat{r} &= -7.4\hat r -1.6\hat f - 8.8 \hat f\frac{1}{s+1}- 10.4\hat f\frac{1}{(s+1)^2} + r_0 \\ s\hat {f} &= 0.25\hat r+ f_0 \end{align}

thus obtaining

$$ \cases{ \hat r = \frac{r_0 s (s+1)^2-f_0 ((1.6 s+12.) s+20.8)}{s^4+9.4 s^3+16.2 s^2+10.4 s+5.2}\\ \hat f=\frac{s^2 (f_0 (1. s+7.4)+0.25 r_0)}{s^4+9.4 s^3+16.2 s^2+10.4 s+5.2} } $$

NOTE

If $r_0 = f_0 = 0$ then the solution is the trivial solution: $\hat f = \hat r = 0$