Let $(W\_t)\_{0\leq t\leq 1}$ be a standard Wiener process on $[0,1]$, and let $\mathcal{F}\_t$ be the natural filtration. Consider a BSDE
$$
dX\_t=f(t,X\_t)dt+\sigma(t,X\_t) dW\_t
$$
with terminal condition $X\_1=x$, where $f(t,\cdot)$ and $\sigma(t, \cdot)$ are $\mathcal{F}\_t$-adapted square integrable processes.


My question: is it possible for the BSDE to be well defined if $\sigma(t,X)=0$ for all $t\in [0,1]$ and all $X$? Also, it seems unlikely to me that I can treat such a case as an ODE since reversing time would screw with the progressive measurability of $f$. Am I wrong?



BSDE without volatility
I am not sure if I understand your question correctly. A typical Brownian BSDE has the form


$$dY\_t = f(\omega, t, Y\_t, Z\_t)dt - Z\_t dW\_t$$


with terminal condition


$$Y\_T = \xi \in \mathcal{F}^{W}\_T$$


where $Y$ and $Z$ are two parts of the solution and required to be adapted to the Brownian filtration. If your question boils down to if there is a BSDE with deterministic terminal condition $\xi = x$ and (second part of the) solution process constant zero, the answer is yes. E.g., take


$$f(\omega, t, Y\_t, Z\_t) = W\_{\frac{T}{2}}1\_{(\frac{T}{2},\frac{3T}{4}]} - W\_{\frac{T}{2}}1\_{(\frac{3T}{4},T]}$$


and $x=0$.


