# Differential inequalities for a strictly diagonal dominant system of linear ODEs

Let $A$ be a real $d\times d$ matrix. The diagonal elements are strictly negative ($a_{ii}<0$) and the off-diagonal elements are non-negative ($a_{ij}\geq 0$ for $i\neq j$). $A$ is strictly column diagonally dominant ($\forall j, |a_{jj}|>\sum_{i\neq j}|a_{ij}|$).

Consider the system of differential equations given by ${\bf \dot x}(t)=A {\bf x}(t)$ and suppose that the set of inequalities $\dot y_i(t)\leq (A {\bf y}(t))_i$ with $i=1,\ldots,d$ holds for all $t$. Given the initial conditions ${\bf x}(0)={\bf y}(0)$ with $x_i(0)\geq 0 \,\forall i$, do we necessarily have $x_i(t)\geq y_i(t)\,\forall i$ at every $t>0$?

The point is that Metzler matrices, i.e., matrices with nonnegative off-diagonal entries are precisely the matrices for which $e^{At}$ has nonnegative entries for all $t\geq 0$.
The argument for the case that $y$ is continuously differentiable goes something like this. (A bit more care also yields the result with less regularity, see the reference). Assume that $y$ is about to overtake $x$ in some entry at time $t$ and $y(s) \leq x(s)$ for all $s \in [0,t]$. Thus, by continuity, $y_i(t) = x_i(t)$ and $y(t) \leq x(t)$ componentwise. Then $$\dot y_i(t) \leq (A y(t))_i = a_{ii} y_i(t) + \sum_{j\neq i} a_{ij} y_j(t) \leq a_{ii} x_i(t) + \sum_{j\neq i} a_{ij} x_j(t) = (A x(t))_i = \dot x_i(t)$$ so that $y_i(t+s) \not > x_i(t+s)$ for small $s>0$. Thus $y$ can never overtake $x$ in any component.
Note that for the inequality we use that $x_i(t) = y_i(t)$, as there is no information about the sign of $a_{ii}$ and the condition that the off-diagonal entries are nonnegative.