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 :)


1 Answer 1


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


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.

  • $\begingroup$ 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

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