# Diverging solution to a SDE

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 :)

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.

• 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. Mar 25, 2020 at 0:19
• You are absolutely right! I forgot to add the end of the argument, I'll edit in a bit. Mar 25, 2020 at 12:28
• 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. Mar 25, 2020 at 14:34
• 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 :) Mar 25, 2020 at 16:06