Consider the following Stochastic Differential Equation:
$$dX_t = \sigma(X_t) \, dB_t \tag{1}$$
Where
$$\sigma(x)= \begin{cases}1 & x \geq 0\\ -1 & x < 0 \end{cases} $$
This exemple shows that one can fail to have pathwise uniqueness of solutions to an $SDE$ while having unicity in law.
If a pair $(X,B)$ is a solution of $(1)$ then $(-X,B)$ is also a solution of $(1)$.
I am having trouble understanding how such a solution might behave.
For instance if we start positive then the behaviour of the solution is the same as the one of the brownian motion. But what happens when you reach $0$ as $\sigma$ is discontinuous at $0$ one might guess that our trajectory might "choose" wich side to go. This is reasonable since the solution $X$ is a Brownian motion and also $-X$ is a solution.
the trouble is, How might the path continue after it reaches $0$?
The pictures might clarify my point.
It suffices to consider $X_t = \xi + B(t)$ where $B_t$ is a brownian motion and $\xi$ is an initial condition. In this setting define $\tilde{B}_t = \int_0^t \sigma(X_t) \, dB_t$, note that $\tilde{B}_t $ is a Brownian motion and that
$$X_t = \xi + \int_0^t \sigma(X_t)\, d\tilde{B}_t $$
therefore $(X, \tilde{B})$ is a solution to eq $(1)$