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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.

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  • $\begingroup$ $x$ and $y$ cannot cross into the negative area without the other variable getting unbounded. On the other hand, as long as $x,y>0$, the sum $x+y$ can grow at most exponentially in $t$. $\endgroup$ – fedja Nov 10 '18 at 2:11
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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.

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