A problem in single variable calculus

Question

Given a function $a(t)$, and the condition $$\dot{a}^2 \geq 1 - \frac{a^2}{L^2} $$ and $$a(t) \leq L $$ I need to prove that if there exists a $t_0$ such that $\dot{a}(t_0)<0$ and $a(t_0)>0$, $a(t)$ is eventually zero that is there exists a $\tilde{t}$ such that $a(\tilde{t})=0$. Now from intuition as the lower bound gets bigger, it seems $\dot{a}$ gets more negative. Now when trying to put these in equations the first step I am attempting is that for any $t_1>t_0$, I am trying to show $\dot{a}(t_1)<\dot{a}(t_0)$. But that doesn't seem to follow from the given inequality that I am allowed to use. I thought if I took $t_1=t_0 +dt$ I would get some information about $\ddot{a}$. Any leads on how to go about this?

Answer 1

Either I am missing something, or this is not necessarily true.

Let $a(t) = t^2 + 1$, $L = 1$.

$$ \dot{a}^2 + \frac{a^2}{L^2} = (2t)^2 + (t^2 + 1)^2 \geq 1 $$

At $t_0 < 0$:

$$ \begin{cases} \dot{a}(t_0) = 2t_0 < 0 \\ a(t_0) = t_0^2 + 1 > 0 \end{cases} $$

But $a(t)$ is always greater than zero.

UPDATE

With the new condition, $a(t) \leq L$, it's actually provable, assuming $a$ is differentiable on $\mathbb{R}$.

Case $\exists t' : a(t') < 0$

Since $a$ is differentiable, it's continuous. $a(t_0) > 0$, $a(t') < 0$. The intermediate value theorem applied to $t_0$ and $t'$ implies $\exists \tilde{t} : a(\tilde{t}) = 0$.

Case $\forall t : a(t) \geq 0$

Note that the differential condition as written implies $L \neq 0$.

$$a \leq L \implies L > 0 \land \frac{a}{L} \leq 1$$

Now let's have a look at the first inequality again.

$$\dot{a}^2 \geq 1 - \frac{a^2}{L^2} \iff \dot{a}^2 + \frac{a^2}{L^2} \geq 1 \implies \left ( \dot{a}(t) = 0 \implies a(t) = L \right )$$

We know that $\exists t_0 : \dot{a}(t_0) < 0$. If $\exists t_1 : \dot{a}(t_1) = 0$ then $t_1 < t_0$ — otherwise select $\alpha = \operatorname{argmin}(\{a(t) : t_0 \leq t \leq t_1\})$, by the extreme value theorem $\dot{a}(\alpha) = 0 \implies a(\alpha) = L \implies \forall t \in [t_0, t_1] \, a(t) = L \implies \dot{a}(t_0) \not< 0$, contradiction. Note that due to Darboux's theorem $\dot{a}$ cannot become positive without going through $0$.

Unboundedness

So $\forall t \geq t_0 \, \dot{a}(t) < 0$, the last problem we could have is $a$ being limited below. Let $M > 0$ be the lower bound. By Weierstrass's theorem $lim_{t \to \infty} a(t) = M$, or $\forall \epsilon > 0 \, \exists \delta \, \forall t > \delta \, |a(t) - M| < \epsilon$.

$$ \dot{a} \leq -\sqrt{1 - \frac{a^2}{L^2}} \leq -\sqrt{1 - \frac{(M + \epsilon)^2}{L^2}} = -C $$

By the mean value theorem $a(t + 1) - a(t) = \dot{a}(\gamma) \leq -C \implies a(t) - C \geq a(t + 1)$, therefore $a(t + \left\lceil \frac{L}{C} \right\rceil + \ldots) < 0$, which brings us to the first case again.

Conclusion

To give an example, consider $a(t) = \sin(t)$ and $L = 1$.

$$ \begin{cases} (cos(t))^2 + \frac{(sin(t))^2}{1^2} \geq 1 \\ sin(t) \leq 1 \end{cases} $$

At $t_0$ in the second quarter $\sin(t_0) > 0$ and $\cos(t_0) < 0$, also $\sin(0) = 0$.