Convergence of the solutions of a ODE system Consider this system of differential equations for $t\in[0,\infty)$:
$$ \frac{d}{dt}x(t) = a(t) + F(x(t), y(t)),$$
$$ \frac{d}{dt}y(t) = a(t) + G(x(t), y(t)),$$
with positive initial conditions: $y(0)>0, x(0) >0$.
Here $a(t)$ is a piecewise continuous function and $F, G, \frac{\partial}{\partial x} F, \frac{\partial}{\partial y} F, \frac{\partial}{\partial x} G, \frac{\partial}{\partial y} G$ are continuous on $\{(t,x,y)\,|\,  x>0, y>0\}$. So I guess there must a unique solution on some interval $I=[0,T]$.
Now assume that there exists a sequence of continuous functions $\{a_n(t)\}_{n=1}^{\infty}$ converging to $a(t)$ in $L^1[0,T]$, such that this system of differential equations:
$$ \frac{d}{dt}x_n(t) = a_n(t) + F(x_n(t), y_n(t)),$$
$$ \frac{d}{dt}y_n(t) = a_n(t) + G(x_n(t), y_n(t)),$$
with the same initial conditions $x_n(0)=x(0)>0$ and $y_n(0)=y(0)>0$, has positive continuous solutions $x_n(t)>0$ and $y_n(t)>0$ for $t \in I_n$ for some $I_n = [0, T_n]$. (Positivity of $x_n$ and $y_n$ is known which means that as long as there is a unique solution $x_n, y_n$ on some interval, they are strictly positive on that interval).
$\bf{1 -}$ Is it possible to prove that the solutions $x(t)$ and $y(t)$ of the original ODE system is nonnegative on some interval $0 \le t \le t'?, i.e,$
$$ x(t) \ge 0, \quad y(t) \ge 0,$$
for $t \in [0, t']$?
First I don't know how $T_n$ behaves maybe $T_n \rightarrow 0$ but maybe from the conditions of this problem we can find a wider range on which the solutions exist and then how to show the non-negativity of $x(t)$ and $y(t)?$ So these are my questions.
 A: Fact 1: therer exists $\tau>0$ such that, for all $n\geq n_0$ large enough, the ODE with $a_n(t)$ AND the limit ODE with $a(t)$ are uniquely solvable at least for times $t\in[0,\tau]$.
Indeed, writing $X=(x,y)$, the proof of existence is a fixed point for the map $X\mapsto \phi[X]$ in the space $C([0,\tau];\mathbb R^2)$ defined as
$$
\Phi_n[X](t):=X_0+\int_0^\tau \alpha_n(s) + f(X(s)) ds
$$
where for convenience $f=(F,G):\mathbb R^2\to\mathbb R^2$ and $\alpha_n(t)=a_n(t) (1,1)$.
For $n=\infty$ the map $\Phi$ is defined accordingly, simply taking $\alpha(t)=a(t)(1,1)$ instead of $\alpha_n$.
The classical proof of existence and uniqueness goes then as follows:
Pick first some large $R\geq 2|X_0|$, and let $M=M(R)=\sup \{|f(X)|,\,|X|\leq R\}$. For $\tau$ small enough, and because $\alpha_n\to \alpha$ in $L^1(0,T)$, it is easy to check the if $\tau$ is small enough so that
$$
|X_0|+ \int_0^\tau 2|\alpha_n(s)|ds + M\tau \leq R
$$
then $\Phi_n$ maps the ball $B_R$ in $X=C([0,\tau];\mathbb R^2)$ to itself. Similarly, the same choice of $\tau$ works for $\Phi$ instead of $\Phi_n$ (this is why I kept some room of maneuver and kept a factor $2$ for $\alpha_n$).
Moreover, if $L=L(R)$ is the local Lipschitz constant of $f$ on $\{|X|\leq R\}$, then choosing $\tau$ possibly smaller so that $L\tau<1$ guarantees that both $\Phi_n$ and $\Phi$ are contractions. Existence and uniqueness thus follow from Banach's contraction mapping theorem.

Fact 2: The solution $X_n\to X$ in $L^1(0,\tau)$. Indeed, from the proof of existence it is clear that $|X_n(t)|\leq R$ and $|X|(t)\leq R$ uniformly in $n\in\mathbb N$ and $t\in [0,\tau]$. In particular, $f$ is $L$-Lipschitz in this ball for some $L=L(R)$, hence
$$
|(X_n-X)'(t)| \leq |\alpha_n(t)-\alpha(t)| + |f(X_n(t))-f(X(t)|
\leq
|\alpha_n(t)-\alpha(t)| + L| X_n(t)-X(t)|.
$$
An immediate application of Grönwall's inequality gives
$$
|X_n(t)-X(t)|\leq \epsilon_n(t)+\int_0^t \epsilon_n(s) Le^{Ls}ds
\leq \epsilon_n(t)+ Le^{L\tau} \int_0^\tau\epsilon_n(s)ds
$$
with $\epsilon(t)=|\alpha_n(t)-\alpha(t)|$.
Integrating in time finally gives
$$
\|X_n-X\|_{L^1(0,\tau)}\leq \|\epsilon_n\|_{L^1(0,\tau)}(1+\tau Le^{\tau L})\to 0
$$
as $n\to\infty$.

Immediate consequence: since you know somehow that $x_n(t)\geq 0,y_n(t)\geq 0$ and because the $L^1$ convergence from step 2 implies in particular pointwise convergence, you can conclude that $x(t)\geq 0,y(t)\geq 0$ at least for times $t\in [0,\tau]$. In fact the whole argument works in the largest common interval of existence, i-e if you know that $X_n,X$ are all defined up to some fixed time $\bar T$ then you can get a priori bounds, fiddle a little bit with the constants $R,M,L$ above, and conclude that $X_n\to X$ in $L^1(0,\bar T)$. (The fact that $\tau$ was small enough was not really useful in step 2 above, just in step 1 for the contraction estimate)
