Let $u,v:[0,\infty)\to[0,\infty)$ satsfying the following system of differential inequalities: $$ u'(t)\leq a_1\,u(t) + a_2\,v(t) + a_0 \\[4pt] v'(t)\leq b_1\,u(t) + b_2\,v(t) + b_0 $$ for suitable coefficients $a_0,a_1,a_2,b_0,b_1,b_2\in\mathbb R\,.$ In particular I have $a_1,b_2<0\,$ and $\,0<a_2<|a_1|\,$, $0<b_1<|b_2|\,$.
Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type $$ u(t) \leq F(t) \\ v(t) \leq G(t)$$ where the functions $F,G:[0,\infty)\to[0,\infty)$ depend on $u,v$ only through their initial values $u(0),v(0)$?
I remind that for a single differential inequality $$ u'(t) \leq a_1\,u(t) + a_0 $$ the Gronwall lemma guarantees that $$u(t)\leq\, u(0)\,e^{a_1 t} + \frac{a_0}{a_1}\, (e^{a_1t}-1) $$ and it can be proven for example by bounding the derivative of $U(t)\equiv u(t) e^{-a_1t}$ and integrating on the inverval $[0,t]$. Notice also that the bound is the solution of the differential equation $y'(t)=a_1\,y(t)+a_0\,$, $y(0)=u(0)\,$.
Here is research that contains a generalization of the problem at hand, based on properties of the resolvent operator (which is nothing but iterating the inequality an arbitrary amount of times and studying the resulting object). The following reduction suffices to treat the present case:
If $x\in\mathbb{R}^n$, $A\in \mathbb{M}_{n\times n},$ and $a$ is a vector function of one variable with $$X(t)\leq a(t)+\int_{t_0}^tAX(s)ds$$
then it is true that
$$X(t)\leq a(t)+\int_{t_0}^tV(t,s)Aa(s)ds$$
where the kernel is an $n\times n$ matrix that satisfies the relation
$$V(t,s)=I+\int_s^tAV(y,s)dy$$
Here the inequalities are taken component-wise. Basically what this theorem really says is that, indeed, a system of linear inequalities is upper bounded by the solution of the corresponding linear system of equalities. One can verify doing some trivial matrix algebra that whenever $A$ is constant, the kernel has the form
$$V(t,s)=e^{A(t-s)}$$
Finally, we turn to our specific problem. To transform it into the form required by the theorem, set $$X(t)=\begin{pmatrix}u(t)\\v(t)\end{pmatrix}~,~ A=\begin{pmatrix}a_1&a_2\\b_1&b_2\end{pmatrix}~,~ c=\begin{pmatrix}a_0\\b_0\end{pmatrix}$$
and integrate all inequalities once to obtain
$$X(t)\leq X_0+ct+\int_{0}^t A X(s)ds$$
for which the extended Gronwall's lemma implies that
$$X(t)\leq X_0+ct+\int_{0}^t e^{A(t-s)}A (X_0+cs)ds$$
which, noting that $A$ as defined above is invertible, with the help of the following integrals
$$\int_{0}^te^{-As}ds=A^{-1}(I-e^{-At})$$ $$\int_{0}^t s e^{-As}ds=A^{-2}(I-(I+At)e^{-At})$$
can be written in the simple form
$$X(t)\leq X_0e^{At}+A^{-1}(e^{At}-1)c$$
Calculating matrix exponentials for general coefficients requires diagonalization of the matrix $A$, which is tractable since the matrix is only $2\times 2$.