# Uniqueness of the solution to some degenerate SDE

Consider the one-dimensional stochastic differential equation:

$$dX_t = {\bf 1}_{\{X_t>0\}}\big(b(t,X_t)dt + a(t,X_t)dW_t\big),\quad \forall t>0,$$

or equivalently

$$dX_t = b(t,X_t)dt + a(t,X_t)dW_t,\quad \forall t\le \tau,\quad \mbox{with } \tau:=\inf\{t\ge 0: X_t\le 0\},$$

where $$(W_t)_{t\ge 0}$$ is a standard Brownian motion and $$\mathbb P(X_0>0)=1$$. Under which conditions on the coefficients $$b, a$$ does pathwise uniqueness or even uniqueness in law hold?

• @NawafBou-Rabee Thank you very kindly for the clarification. I'm interested in the uniqueness in law May 30 at 12:59
• @NawafBou-Rabee Indeed, both types of uniqueness are interesting to me (if this helps) May 30 at 13:05
• This is an SDE that is stopped/killed at zero, and both pathwise uniqueness and uniqueness in law should hold under mild/standard conditions on the coefficients … May 30 at 13:27
• @NawafBou-Rabee Thanks infinitely for the answer. I have taken a look at the book that you mention, while it seems that more details must refer to the books A First Course in Stochastic Processes and Markov Processes and Potential Theory. May I keep the chance to return towards you if I'm unable to find the references? Or could you please specify the conditions such that the uniqueness holds? May 30 at 13:39

The solution of the SDE in question is an example of a time-inhomogeneous diffusion process that is stopped/killed/terminated when it first hits the origin. For pathwise uniqueness until time $$t \wedge \tau$$, Theorem 3.7 in Chapter 5 of the following classic should do the trick.
Theorem (Pathwise Uniqueness). Let $$U \subset \mathbb{R}$$ be an open set, let $$T>0$$, and suppose that there exists a constant $$L$$ such that $$|b(t,x)-b(t,y)| \vee |a(t,x)-a(t,y)| \le L |x-y| \qquad 0 \le t \le T \quad x,y \in U \;.$$ Given two solutions $$X_t, Y_t$$ of the SDE, let $$\tau = \inf\{ t \ge 0 \mid X_t \notin U~~ \text{or}~~ Y_t \notin U \} \;.$$ Then $$P[X_0 = Y_0] = 1$$ implies that $$P[X_{t \wedge \tau} = Y_{t \wedge \tau}~~\text{for} ~~0 \le t \le T] = 1$$.
To be sure, choose the open set to be $$U = \{ x > 0 \}$$. The proof of this theorem is standard, and involves computing the mean-squared difference between $$X_{t \wedge \tau}$$ and $$Y_{t \wedge \tau}$$, applying Lipschitz continuity of the coefficients, and then invoking Grönwall's inequality.