Consider an SDE $dX_t = a(t,X_t) dt + b(t,X_t)dB_t$, with initial conditions $X$ (or some distribution $\mu$ if thinking about weak solutions). I'm hoping to understand the difference between weak and strong solutions to SDEs better.
To the answer posted here, there is a comment by @Roberto Rastapopoulos that,
It may seem from this answer that a weak solution is in fact a strong solution for the right Brownian motion, but this is not the case. Indeed, the process $_$ in the weak solution need not be adapted to the filtration generated by the Brownian motion and initial condition. I think this is a key difference which deserves emphasis.
This is also mentioned in notes by Lalley (see page 1, def. 1).
However, I am curious of an example of a weak solution that is not adapted to the filtration generated by Brownian motion and the initial condition (Lalley calls this filtration the minimal filtration).
The only example I know of an SDE with a weak, but not strong, solution is the Tanaka equation. The above post describes the existence of its weak solution:
Example 3: The SDE $$dX_t = - \text{sgn}\,(X_t) \, dB_t, \qquad X_0 = 0 \tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be some Brownian motion and define $$W_t := -\int_0^t \text{sgn} \, (X_s) \, dX_s.$$ It follows from Lévy's characterization that $(W_t,\mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - \text{sgn} \, (X_t) \, dX_t$$ implies $$dX_t = - \text{sgn} \, (X_t) \, dW_t$$ this means that $(X_t)_{t \geq 0}$ is a weak solution to $(2)$.
But here, $X_t$ is (by first definition) adapted to $\mathcal{F} := (\mathcal{F}_t)_{t \geq 0}$. Moreover, it seems like the minimal filtration generated by $W_t$ is also this filtration. So the weak solution is also a strong solution with respect to this Brownian motion $(W_t,\mathcal{F}_t)$?
