# Show that $dX_t=1_{X_t\not=0} dW_t$ does not have a pathwise unique solution.

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

The process

$$X_t := \xi+W_t$$

is also a solution of the SDE

$$dX_t = 1_{\{X_t \neq 0\}} dW_t.$$

Indeed: Obviously, $(X_t)_{t \geq 0}$ solves

$$dX_t = dW_t \qquad X_0 = \xi.$$

On the other hand,

$$\mathbb{E} \left[\left( \int_0^T 1_{\{X_s=0\}} \, dW_s \right)^2 \right] = \int_0^T \underbrace{\mathbb{E}(1_{\{X_s=0\}})}_{=\mathbb{P}(W_s = -\xi)=0} \, ds = 0$$

where we used that $\mathbb{P}(W_s = x) = 0$ for any $x \in \mathbb{R}$ and $s \geq 0$. Consequently, we find that $(X_t)_{t \geq 0}$ is a solution to the SDE

$$dX_t = 1_{\{X_t \neq 0\}} \, dW_t \qquad X_0 = \xi.$$

• Why the fact that $\mathbb E[(\int_0^T 1_{\{X_s=0\}}dW_s)^2]=0$ implies that $(X_t)$ solves $dX_t=1_{\{X_t\neq 0\}}dW_t$ ? May 1, 2021 at 9:36
• @joshua By definition of $X_t$, we have $dX_t = dW_t$. Now $$dX_t = dW_t = 1_{\{X_t \neq 0\}} \, dW_t + 1_{\{X_t=0\}} dW_t;$$ for the second one we have seen that it is zero (almost surely) since the expectation of its square is zero. Hence, $$dX_t = 1_{\{X_t \neq 0\}} dW_t.$$
– saz
May 2, 2021 at 16:32