How to solve using markov chain

Question

Two cars $X$ and $Y$ are in a race. The position of car $X$ as a function of time is described by $x(t)$. The position of car $Y$ is described by $y(t)$. Initially $x(t),y(t)>0$. Whenever $x(t)=y(t)$,we have $x'(t)≤y'(t)$. Further we know that $y'(t) < 0$ for all times $t ≥ 0$. Prove/ Disprove: Car $X$ either crosses the position $0$ in finite time, or $\lim_{t→∞} x(t) = 0$. Assume that $x(t)$ and $y(t)$ are infinitely differentiable.

Answer 1

Perhaps I am misunderstanding the problem, but consider the counterexample to disprove the statement.

$$y(t) = 2-\exp(t)$$ $$ x(t) = 3 + \exp(t).$$

Then we have of course that $$y(0),x(0)>0.$$ Since $$x(t)\neq y(t), \quad t\in \mathbb{R},$$ we don't have to worry about the condition ordering the derivatives. And $$y'(t) = -\exp(t) < 0, \qquad \forall t\geq 0.$$

Clearly though, $x(t) > 0, \; \forall t$ and so it never crosses the $0$, additionally, $$\lim_{t\to \infty}x(t) = \infty \neq 0.$$

Are you missing other conditions in the problem statement? Otherwise, it seems to be false.

Also, everything is deterministic, so it is still unclear why you are trying to use Markov chains with this problem.