Background
Consider the non homogeneous linear ODE in $\mathbb{R}^n$:
$$\dot{\xi}(t) = A(t)\xi(t) + \nu(t)\; \xi(\sigma_0) = x_0 \label{1}\tag{I}$$
With $A$ a matrix of size $n\times n$ whose entries are measurable and locally essentially bounded functions, and $\nu$ vector of $n$ locally integrable functions.
The solution of this problem is given by the variation of parameters formula:
$$ \xi(t) = \Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds $$
Where $\Phi(t,\sigma)$ is the fundamental solution of the homogeneous case of the system in $\eqref{1}$ i.e $\dot{\xi}(t) = A(t)\xi(t)$, $\xi(\sigma) = x_0$. More explicitly, $\Phi(t,\sigma_0)$ fulfills that $(\Phi(t,\sigma_0)x_0)\dot{} = A(t)\Phi(t,\sigma_0)x_0,$ with $\Phi(\sigma_0,\sigma_0)x_0 = x_0$.
My question/problem
So I was trying to verify the variation of parameters formula, that is, I was trying to see that indeed
$$\frac{d}{dt}\left(\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right) = A(t)\left(\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right) + \nu(t) \label{2}\tag{II}$$
And my problem is that I can't prove \eqref{2}.
Of course, what one should do is to simply develop, so here is my try:
- LHS
\begin{align} & \phantom{\; = \;}\dot{\xi}(t)=\frac{d}{dt}\left(\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right) = \frac{d}{dt}\left(\Phi(t,\sigma_0)x_0\right) + \frac{d}{dt}\left(\int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right)\\ & = A(t)\Phi(t,\sigma_0)x_0 + \Phi(t,t)\nu(t) \text{ (1}^{\text{st}}\text{ fundamental theorem of calculus)}\\ & = A(t)\Phi(t,\sigma_0)x_0 + \nu(t). \end{align}
- RHS
\begin{align} & \phantom{\; = \;}A(t)\xi(t) + \nu(t) = A(t)\left(\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right) + \nu(t)\\ & = A(t)\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t A(t)\Phi(t,s)\nu(s) \, ds + \nu(t). \end{align}
Here I don't know what to do. There are useful properties like $\Phi(t,s)^{-1} = \Phi(s,t)$, but I don't see how to use them.
I know there are a lot of questions about the variation of parameters formula:
How to derive the variation of parameters formula?,
Approach of Variation of parameters (Variation of Constants),
First Order ODE - Variation Of Parameters.
but as far as I've seen nobody has asked about this verification, it should be simple (so say all the books of theory of ODE that I've consulted) but I don't see how to proceed.
Any hints? what am I missing?
If this is a duped question my apologies.
EDIT (one day later).
Ok, after 24 hours, I think I manage to see how to do it. First, indeed in the RHS we have that:
\begin{align} & \phantom{\; = \;}A(t)\xi(t) + \nu(t) = A(t)\Phi(t,\sigma_0)x_0 + A(t)\int_{\sigma_0}^t\Phi(t,s)\nu(s) \, ds + \nu(t). \end{align}
The trick is in the LHS: using the fact that $\Phi(t,s) = \Phi(t,\sigma_0)\Phi(\sigma_0,s)$ for any $t,s,\sigma_0 \in \mathbb{R}$, we can see that:
\begin{align} \dot{\xi}(t) & =\frac{d}{dt}\left(\Phi(t,\sigma_0)x_0 + \int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right) = \frac{d}{dt}\left(\Phi(t,\sigma_0)x_0\right) + \frac{d}{dt}\left(\int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds\right)\\ & = A(t)\Phi(t,\sigma_0)x_0 + \frac{d}{dt}\left( \Phi(t,\sigma_0)\int_{\sigma_0}^t \Phi(\sigma_0,s)\nu(s) \, ds\right)\\ & = A(t)\Phi(t,\sigma_0)x_0 + \frac{d}{dt}\left( \Phi(t,\sigma_0)\right)\int_{\sigma_0}^t \Phi(\sigma_0,s)\nu(s) \, ds + \Phi(t,\sigma_0)\frac{d}{dt}\left(\int_{\sigma_0}^t \Phi(\sigma_0,s)\nu(s) \, ds\right) \; \text{(derivative product rule)}\\ & = A(t)\Phi(t,\sigma_0)x_0 + A(t)\Phi(t,\sigma_0)\int_{\sigma_0}^t \Phi(\sigma_0,s)\nu(s) \, ds + \Phi(t,\sigma_0)\Phi(\sigma_0,t)\nu(t)\\ & = A(t)\Phi(t,\sigma_0)x_0 + A(t)\int_{\sigma_0}^t \Phi(t,s)\nu(s) \, ds + \nu(t) \end{align}
Wich proves \eqref{2}.
So I suppose I just answered my own question.
Going to leave the question in case someone sees something weird in my procedure. And in case it helps someone else.
