Given a system of n homogeneous linear ordinary differential equations $$\dot x = M(t)x(t)$$ where $M(t)$ is a Metzler matrix at any time $t$, prove that the fundamental matrix to this system of ODEs is non-negative(all entries are non-negative).
Edit: I know this holds if $A(t)$ is constant as the solution is a matrix exponential of the Metzler matrix, but I couldn't find reference the case when A(t) actually depends on t.
Let $\dot{\Phi}(t)=M(t)\Phi(t),\ \Phi(0)=I$, where $\Phi$ is the fundamental matrix function. Note that $\Phi(0)$ is nonnegative.
A Taylor expansion at $t=0$ gives $\Phi(\epsilon)=I+\epsilon M(t)+o(\epsilon)$, which shows that the matrix $\Phi(t)$ will be nonnegative on $[0,t_1]$ for some $t_1>0$.
Now let $t_2\ge t_1$ to be the first time for which (at least) one of the entries of $\Phi(t_2)$ is zero. Let $i,j$ be such that $\Phi_{ij}(t_2)=0$, then we have that
$$\dot{\Phi}_{ij}(t_2)=\sum_{k=1}^nm_{ik}(t_2)\Phi_{kj}(t_2)=m_{ii}(t_2)\Phi_{ij}(t_2)+\sum_{k\ne i}m_{ik}(t_2)\Phi_{kj}(t_2).$$
Since $\Phi_{ij}(t_2)=0$, then we have that $$\dot{\Phi}_{ij}(t_2)=\sum_{k\ne i}m_{ik}(t_2)\Phi_{kj}(t_2)$$
which is nonnegative since $M(t)$ is Metzler and $\Phi_{kj}(t_2)\ge0$ for all $k\ne i$. Therefore, when $\Phi_{ij}(t)$ reaches zero, its derivative is necessarily nonnegative at that point, which means that the value never becomes negative.