# Solving a SDE with quadratic drift

I am wondering whether the following SDE can be solved explicitly?

$$d X_t = X_t^2 d t + X_t d B_t$$

where $B_t$ is a standard Brownian motion. If not, can we say some thing about the moments of the solution, i.e., $E(|X_t|^n)$?

Thank you very much for any hints!

Anand

• It will blow up for large time because $X_t^2$ will dominate when $X_t$ is big. – Anand Jul 12 '13 at 9:28
• Actually, the it does not satisfy sufficient conditions for existence and uniqueness when drift and diffusion terms has to be of at most linear growth (Lipschitz everywhere), so the solution of this SDE may not exist at all. – Ilya Jul 12 '13 at 12:04
• Yes, you are right. It will blow up at finite time. Can we say something for the moments for the blowup? – Anand Jul 12 '13 at 12:44

Solutions do exist locally. Globally they MAY blow up as you already know. The blowup will be dominated by the deterministic system. You did not write what is your initial condition - note that $0$ is a perfectly fine solution to your equation. If your initial condition is say $X_0=1$, then it may blow up, but with positive probability the solution may actually converge to $0$. This is because if the $X^2$ drift term were not there then the solution would be $X_0 e^{B_t-t/2}$ which converges to $0$, and if you add your drift, as long as $X_t$ is small then the drift is smaller than $+\delta X_t$ (e.g., as long as $X_t<\delta$). But the solution of the equation $dX_t=\delta X_t dt+X_t dB_t$ converges to $0$ as long as $\delta$ is small enough (in fact $\delta<1/2$ will work).
So with positive probability, starting from any $X_0$ you become smaller than say $1/4$ after a finite time, and then with positive probability you actually converge to $0$ (using comparison theorems for 1D SDE's as in Ikeda-Watanabe should be enough to prove this). On the other hand, with positive probability you blow up. So I am not sure what do you mean by moments of the blow up time''.
• Dear Professor Zeltouni, thank you very much for your explanation and clarification. By "moments", I mean $E(|X_t|^n)$. If $\tau=\inf\{t\ge 0: |X_t|=\infty\}$, can we say that $\tau$ is almost surely bounded away from zero? If so, we may ask $E(|X_t|^n)$ for $t<\tau$. I will have a careful look. – Anand Jul 12 '13 at 18:35
• $\tau$ is bounded away from zero but not by a deterministic amount. So $E(|X_t|^n)$ is simply infinity. You can ask about $E(|X_t|^n|\tau>t)$ but this is a completely different question and I don't know how to write an explicit formula. – ofer zeitouni Jul 13 '13 at 12:48
Actually your SDE may be solved explicitly. Look at the more general SDE, \begin{align} dX_{t} = (a X_{t}^{n} + b X_{t}) dt + c X_{t} dW_{t} \end{align} where $n > 1$ and $a,b,c \in \mathbb{R}$. It has a solution given by \begin{align} dX_{t} = \Theta_{t} \Bigl( X_{0}^{1-n} + a(1-n)\int_{0}^{t} \Theta^{n-1}_{s} ds \Bigr)^{\frac{1}{1-n}} \end{align} with \begin{align} \Theta_{t} = e^{ (b-\frac{1}{2}c^2) t + c W_{t}} \end{align}