1
$\begingroup$

I'm considering the SDE, with $B$ the brownian motion and $\beta$ a scalar (it can be negative) $$ X_t = x_0 + \int_0^t (\beta + X_s^2) ds + B_t $$

and I would like to show that $X_t$ almost surely diverges to infinity in finite time.

I really don't know how to do it since usually I have to control $X_t$, does anyone have a tip ?

Thanks :)

$\endgroup$

1 Answer 1

1
$\begingroup$

Let us begin by showing that $X$ diverges to $+\infty$, possibly as time goes to infinity. Set $$f : x\mapsto \int_0^x\exp\left(-2\beta y-\frac 23y^3\right)\mathrm dy. $$ The point of $f$ is that $f(X)$ is a local martingale, possibly up to the explosion time $\tau$ of $X$ ($f$ is solution to $(\beta+x^2)\partial_xf+\frac12\Delta f=0$, in fact all such solutions can be written as $af+b$).

Notice that $f$ is increasing. Moreover, it is bounded above, since the exponential term goes to zero fast enough. Since the explosion of $X$ can only occur if $X$ diverges to $+\infty$ ($X_t$ is at least $x_0+\beta t+B_t$), $f(X)$ is in fact a local martingale for all times, setting by convention $f(X_t)=\sup f$ for all $t\geq\tau$.

As a local martingale bounded above, $f(X)$ converges almost surely, and given that $f$ is increasing and isn't bounded below, $X$ converges in $\mathbb R\cup\{+\infty\}$. We need to show that $\lim X_t$ cannot be finite with positive probability.

Notice that $$ B_{t+1}-B_t = (X_{t+1}-X_t) - \int_t^{t+1}(\beta+X_s^2)\mathrm ds. $$ In particular, if $X$ converges to a finite limit $\ell\in\mathbb R$, then $B_{t+1}-B_t$ converges to $ -\beta-\ell^2$. Hence the event that $X$ converges to a finite limit is included in the event that $B_{t+1}-B_t$ converges, which obviously has measure zero.

So $X$ diverges to $+\infty$. I will now try to use deterministic arguments. Let us assume for now the following

Fact.

Let $Y$ be a process such that $$ Y_t \geq -C + \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds $$ for some constants $C,\alpha>0$, and all $t$ possibly up to some explosion time $\tau$ (explosion means “leaves all compact subsets”).

Then either $\liminf_{t\to\tau} Y_t\leq\alpha$ or $Y$ diverges to $+\infty$ in finite time.

Note that for any $\alpha>0$ such that $\alpha^2>-\beta$, almost surely there exists a (random) constant $C>0$ such that $B_t\geq -C-(\alpha^2+\beta)t$ for all $t>0$. In particular, $$ X_t \geq x_0 - C + \int_0^t\left(X_s^2 - \alpha^2\right)\mathrm ds. $$ Since $X$ diverges to $+\infty$, obviously its limit inferior is not bounded above, so the fact implies that $X$ must undergo explosion in finite time.

Now onto the proof of the fact. Suppose that the limit inferior of $Y$ is larger than $\alpha$. Then for all $t$ large enough (which here means close enough to $\tau$), $Y_t^2>\alpha^2+\varepsilon$ for some $\varepsilon>0$. According to the inequality, $Y$ must then diverge, possibly in infinite time.

Setting $$ I_t = \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds, $$ we see that $$ I'_t = Y_t^2 - \alpha^2 \geq \frac12(Y_t+C)^2 \geq \frac12I_t^2 $$ for all $t$ large enough. In particular (note that $I_t>0$ for $t$ large enough), $$ \frac{\mathrm d}{\mathrm dt}\left(-\frac1{I_t}\right) = \frac{I'_t}{I_t^2} \geq \frac12 $$ for all $t$ large enough, hence $$ Y_t\geq -C + I_t\geq-C+\frac 2{T-t} $$ for some $T>0$ and all $t$ large enough, so that $Y$ explodes in finite time.

$\endgroup$
4
  • $\begingroup$ Great, thanks for the trick ! But if I'm not missing something, I think that we have just showed that $X_t$ diverges almost surely to infinity. We do not have showed that it's in finite time. But I should be able to conclude by myself though. $\endgroup$
    – Fulgrim
    Mar 25, 2020 at 0:19
  • $\begingroup$ You are absolutely right! I forgot to add the end of the argument, I'll edit in a bit. $\endgroup$
    – Pierre PC
    Mar 25, 2020 at 12:28
  • $\begingroup$ It should now be complete. I feel like the proofs of the fact are somewhat inefficient, but I cannot find something more convincing. Tell me which one you prefer and I will delete the other. $\endgroup$
    – Pierre PC
    Mar 25, 2020 at 14:34
  • $\begingroup$ Yes thank you very much ! I achieved the proof by myself yesterday, and I used the comparison with the deterministic situation as well. I have previously solved the deterministic equation in all possible case so the fact is implied by this previous work. Your second proof is quite clear though :) $\endgroup$
    – Fulgrim
    Mar 25, 2020 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.