Let $$z(t)=\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}~,~ A=\begin{pmatrix}0 & -\beta \\ \alpha & -(\alpha + \beta)\end{pmatrix}$$ satisfies the following system of linear differential inequalities $$z'(t) \leq A\,z(t)$$ where $\alpha, \beta > 0$ are fixed constants, and $x(t), y(t) \in [0,\infty)$, I am wondering if there is a vectorized version of Gronwall lemma which allows me to deduce that $$z(t)\leq \mathrm{e}^{At}\,z(0).$$ It seems hard to find a literature which addresses precisely my question. Any help (including referring to precise literatures) will be greatly appreciated!




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Side note: by writing $u \leq v$ for $u,v \in \mathbb{R}^2\_+$ I mean that each component of $u$ is less than or equal to the corresponding component of $v$. Also (and unfortunately), the off-diagonal entries of $A$ are not all non-negative, which means that the discussion mentioned in [this post](https://mathoverflow.net/questions/239751/differential-inequalities-for-a-strictly-diagonal-dominant-system-of-linear-odes) may not be applicable...



Gronwall lemma for a $2$-dimensional system of linear differential inequalities
**Counter-example.** Let $\left(z(t)\right)\_{t\geq 0}$ be the solution to


$$\dot{z}(t)=A z(t)-\varepsilon \mathbf{e}\_2,$$


with $\mathbf{e}\_2:=\left[0 \,\,\,\,1\right]^{\top}$ and initial condition $z(0)$. Then, $\left(z(t)\right)\_{t\geq 0}$ obeys the inequality $\dot{z}(t)\leq Az(t)$ for all $t$.


Let $\left(x(t)\right)\_{t\geq 0}$ be the solution to $\dot{x}(t)=Ax(t)$ with initial condition $x(0)=z(0)$. Then, $x(t)=e^{At}z(0)$ for all $t\geq 0$.


Now, remark that,


$$ \dot{z}\_1(0)=\left[Az(0)\right]\_1=\left[Ax(0)\right]\_1=\dot{x}\_1(0)$$


and


$$ \ddot{z}\_1(0)=-\beta \dot{z}\_2(0)>-\beta \dot{x}\_2(0)=\ddot{x}\_1(0),$$


since $\dot{z}\_2(0)=\left[Az(0)\right]\_2-\varepsilon < \left[Az(0)\right]\_2 =\left[Ax(0)\right]\_2 = \dot{x}\_2(0).$
In other words, $w(t)\overset{\Delta}= z\_1(t)-x\_1(t)$ is analytic with $w(0)=0$, $\dot{w}(0)=0$ and $\ddot{w}(0)>0$. Therefore, $w(t)>0$ for all $t\in\left(0,\delta\right)$, for some $\delta>0$ small enough. That is,


$$z\_1(t)>\left[e^{At}z(0)\right]\_1$$


for all $t\in\left(0,\delta\right)$ which breaks Grönwall's.




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**For the record.** If the two off-diagonal entries of $A$ are negative (with arbitrary diagonal entries), then we can show that Grönwall's inequality holds w.r.t. the partial order $\leq\_a$ defined as $x\leq\_a y \Leftrightarrow (x\_1 \geq y\_1 \wedge x\_2 \leq y\_2)$
or to the partial order $\leq\_b$ defined as $x\leq\_b y \Leftrightarrow (x\_1 \leq y\_1 \wedge x\_2 \geq y\_2)$. In other words, for this regime, if $\dot{z}(t)\leq\_i Az(t)$ then, $z(t)\leq\_i e^{At} z(0)$ for $i\in \left\{a,b\right\}$.


This is done simply via comparing $\left(z(t)\right)$ and $\left(x(t)\right)$ (where $x(t)$ is the solution to the linear system with $x(0)=z(0)$) at the hitting moment $T$ for the possible cases: i) $z\_1(T)=x\_1(T)$ and $z\_2(T)<x\_2(T)$; ii) $z\_1(T)>x\_1(T)$ and $z\_2(T)=x\_2(T)$; or iii) $z\_1(T)=x\_1(T)$ and $z\_2(T)=x\_2(T)$. In the latter case you need to resort to the higher order derivatives as $\dot{z}\_1(T)=\dot{x}\_1(T)$ and $\dot{z}\_2(T)=\dot{x}\_{2}(T)$.


