# Backward stochastic differential equation

Let $$W_t$$ be a standard Brownian motion. Let $$T$$ be the terminal date, $$X_T=x$$, and $$dX_t=f_tdt+B_tdW_t$$ where $$f_t$$ and $$B_t$$ (yet to be determined) have to be adapted to the filtration generated by $$W$$.

Assume $$x$$ is a constant. One possible solution is that $$f_t=B_t=0$$ so that $$X_t=x, \forall t$$. Is it possible to have other solutions where $$f$$ or $$B$$ are not always 0?.

The are surely many ways to do this. One classical example of this kind of process is the Brownian Bridge from $$0$$ to $$x$$, given by the SDE
$$dX_t = \frac{x - X_t}{1-t}dt + dW_t.$$
This is solved by $$X_t = tx + (1-t)\int_0^t{\frac{dW_s}{1-s}}$$. As $$t$$ approaches $$1$$, $$X_t$$ approaches $$x$$ almost surely.
(Cross-listed on Math Stackexchange) So I think I figured out an answer to my question. Here is an example where $$f$$ and $$B$$ are not necessarily zero.
Let $$Y_t$$ be some Ito process with $$dY_t=\tilde f_tdt+\tilde B_t dW_t$$ for some adapted processes $$\tilde f, \tilde B$$. Define $$X_t=(1-e^{T-t})Y_t+x$$. We have that $$X_T=x$$ and \begin{align} dX_t=&e^{T-t}Y_tdt+(1-e^{T-t})dY_t\\[8pt] =&[e^{T-t}Y_t+(1-e^{T-t})\tilde f_t]dt+(1-e^{T-t})\tilde B_tdW_t\\[8pt] =&f_tdt+B_t dW_t \end{align} where $$f_t=e^{T-t}Y_t+(1-e^{T-t})\tilde f_t$$ and $$B_t=(1-e^{T-t})\tilde B_t$$. As $$Y, \tilde f, \tilde B$$ are adapted processes, I am pretty sure that $$X$$ is a well-defined BSDE.