From S.D.E to Fokker-Plank-Smoluchowski equation Let me use as reference these lecture notes : http://wwwf.imperial.ac.uk/~pavl/lecture_notesM4A42.pdf  In section 5.4 (page 59) here the author argues how the sample paths for a diffusion (from which they derived the FPS equation) follows a standard S.D.E

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*But its not clear to me as to if one can start from a standard S.D.E (as say given in their page 60) one can derive the FPS equation.


*As a specific example : If one sees equation 6.33 (on page 75) this is exactly what they seem to be alluding to be possible : that one can start from the S.D.E ``$dX_t = -\nabla V (X_t)dt + \sqrt{2D}dW_t$" and get the FPS for it as, $\frac{\partial p}{\partial t} = \nabla \cdot (\nabla (V\cdot p)) + D \nabla^2 p$. Can someone kindly refer me to a derivation of this ? (or write it in if that is not too long!)
 A: The function $p(t,x)$ in the Fokker-Planck equation is the probability density at time $t$ of the process $X_t$: $P(X_t \in A) = \int_A p(t, x)dx.$ In general, the evolution equation for the density of a Feller process (i.e. a "nice" Markov process) is
$$\frac{\partial}{\partial t}p(t,x) = \mathcal{L}^* p(t,x),\tag{1}$$
where $\mathcal{L}^*$ is the adjoint of the infinitesimal generator (equation 4.23 on p. 49 of the lecture notes you linked).
(Aside: To make (1) plausible, I find it helpful to remember that $\pi$ is an invariant distribution for a Markov chain if $P^T \pi = \pi$, where $\pi$ is a column vector and $P$ is a transition matrix, or $(P^T - I)\pi = 0$. The analogy to continuous time is $\mathcal{L}^* = P^T - I$, and $\partial_t p(t,x) = 0$. See also (4.21) in your lecture notes and the paragraph that precedes it.)
Here are the key calculations. The infinitesimal generator $\mathcal{L}$ of a Feller process $X_t$ is defined by
$$\mathcal{L}f(x) = \lim_{t\rightarrow 0}\frac{E_x[f(X_t)] - f(x)}{t}$$
for a suitable class of functions $f$. To compute the infinitesimal generator of your diffusion, you can apply Itô's lemma and compute $E_x[f(X_t)]$ directly. For heuristics, you can Taylor expand (in one dimension for simplicity)
\begin{align*}
E_x[f(X_t)] &\approx E_x[f(x) + f'(x) (X_t - X_0) + (1/2) f''(x) (X_t - X_0)^2] \\
&= f(x) + f'(x)(-V'(x)t)+(1/2)f''(x)(2Dt),
\end{align*}
because for your diffusion the "mean" displacement for time $t$ starting at $x$ is $-V'(x)t$ and the "variance" is $(\sqrt{2Dt})^2 = 2Dt$.
The adjoint operator $\mathcal{L}^*$ is defined through the $L^2$ inner product: for any "nice" functions $f,g$, $\langle f, \mathcal{L}^* g\rangle = \langle \mathcal{L} f, g\rangle$. This is typically computed using integration by parts. This calculation is done in one dimension for $V(x)=x^2/2$ on p. 52, Example 4.7.3. of your lecture notes.
