Given the SDE : $$dX_t=1_{X_t\not=0} dW_t \qquad \text{with} \quad X_{0}=\xi $$
how can I construct two obvious strong solutions to prove that SDE has non pathwise uniquenss
Indeed
Consider the stopping time $$ \sigma = inf \left\{ t ≥ 0: \xi +Wt \preceq 0 \right\} $$ and set
$$X_t= \xi +W_{t \wedge \sigma} $$
This process X is a strong solution of SDE $dX_t=1_{X_t\not=0} dW_t$ , $X_{0}=\xi $
Indeed, it is $ \mathcal f_t^{(\xi , \mathcal W)} $ adapted, $ X_{0}=\xi $
we have $$X_t-X_0=\int_0^t 1_{({s \prec \sigma })}dW_s=\int_0^t1_{({\xi +W_{s\wedge \sigma}>0})} dX_t=\int_0^t 1_{{X_t\not=0}}dW_s $$ which means that $$dX_t=1_{X_t\not=0}dW_s,\qquad X_0=\xi $$ so $X_t $ is solution of our SDE
my question how can i construct just two obvious strong solutions to prove that SDE has non pathwise uniquenss
i'll be grateful for any help
best regards, Educ