I strive to prove that the following system of differential equations:


$$\begin{cases} x'=x-u(t)xy\\ y'= -y+u(t)xy \\ x(0)=x\_0>0\\ y(0)=y\_0>0 \end{cases}$$


has a unique Caratheodory solution on a given interval $[0,T]$, where $u:[0,T]\to [0,1]$ is a control, lets say measurable or continuous if necessary. I cannot apply Caratheodory existence theorem <https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_existence_theorem> since the function $f:[0,T]\times\mathbb{R}^2\to\mathbb{R}^2,\ f(t,x,y)=(x-u(t)xy,-y+u(t)xy)$ is not satisfying the Lipschitz-like condition, and the linearity-like growing in $(x,y)$. Someone told me that maybe it can be proven that the solution $(x,y)$ if it exists is bounded (via Lyapunov function or via first inegral), but I cannot find a solution yet.



Lotka Volterra existence of Caratheodory solution
Assuming that $u$ is Lebesgue integrable, $f$ **does** satisfy a Lipschitz-like condition, so we have (local) existence and uniqueness theorem. 


Whatever the controls, the sets $\{(0,0)\}$, $\{\, (x, 0): x > 0\,\}$ and $\{\, (0, y): y > 0\,\}$ are invariant. So, by uniqueness, any nonextendible solution starting in $\mathbb{R}\_{++} := \{\, (x , y): x > 0,\ y > 0 \,\}$ stays there (as fedja noticed in their comment).


Suppose to the contrary that for some initial conditions $(x\_0, y\_0) \in \mathbb{R}\_{++}$ the solution blows up at $\tau \in (0, T)$. Since $1 - u(t) y(t) \le 1$, by standard differential inequalities we have that $x(t) \le x\_0 e^{\tau}$ for all $t \in [0, \tau)$. Consequently, as $-1 + u(t) x(t) \le -1 + x\_0 e^{\tau}$, again by differential inequalities we have $y(t) \le y\_0 \exp{((-1 + x\_0 e^{\tau}) \tau)}$ for all $t \in [0, \tau)$, a contradiction.


