# Lotka Volterra existence of Caratheodory solution

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.

• $x$ 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 $t$. – fedja Nov 10 '18 at 2:11

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.