Say I have diffusion process $X(t) = (x(t),y(t))\in\mathbb{R}^2$, governed by $$ \begin{align} \mathrm{d} x(t) &= \sigma \mathrm{d}W_x(t)\\ \mathrm{d} y(t) &= \sigma \mathrm{d}W_y(t), \end{align} $$ where I'm just using the subscript notation to indicate that the Brownian increments are independent of each other.
It's straightforward to use Itô's lemma to transform $(x,y) \to (r,\theta)$, for which the evolution is $$ \begin{align}\label{sdes} \mathrm{d} r(t) &= \frac{\sigma^2}{2r}\mathrm{d}t + \sigma \mathrm{d}W_r(t)\\ \mathrm{d} \theta(t) &= \frac{\sigma}{r} \mathrm{d}W_\theta(t). \end{align} $$ From these, we can immediately write down the Fokker-Planck/forward Kolmogorov equation for $p(r,\theta,t)$, interpreting the SDEs in the Itô sense, $$ \partial_t p(r,\theta,t) = - \partial_r \left \{\frac{D}{r} p \right\} + D\partial_{rr} p + \frac{D}{r^2} \partial_{\theta\theta}p, $$ with $D:= \sigma^2/2$, which I've verified (in this book, pg 140).
However, it's clear by the diffusion coefficient remaining constant, the forward equation should just be the Laplacian $$ \partial_t \tilde{p} = D\Delta \tilde{p}, $$ which, when converted to polar coordinates yields $$ \partial_t \tilde{p} = D \Delta \tilde{p} = D \left\{ \frac{1}{r} \partial_r \tilde{p} + \partial_{rr} \tilde{p} + \frac{1}{r^2} \partial_{\theta\theta}\tilde{p} \right\}. \label{test} $$
My question: why do the PDEs for $p$ and $\tilde{p}$ differ?
I have a rough idea of where things might be going awry, which is basically that Itô and Fickian diffusion differ for non-constant diffusion $D(x)$. That is, $$\begin{align} \partial_t p &= \nabla \cdot \left \{ D(x) \nabla p\right\} \qquad \text{(Fick)}\\ \partial_t p &= \nabla^2\left\{ D(x) p\right\} \qquad \text{(Itô)}\\ \end{align} $$
Although we started with $D$ constant Cartesian, the $1/r$ term on the $\theta$ SDE might make the polar formulation more like the non-constant case, meaning these interpretations differ?
It's maybe worth pointing out that the SDEs for $r,\theta$ do produce the polar Laplacian for the backward Kolmogorov equation. However, the forward and backward operators are adjoints, and the Laplacian is self-adjoint under the appropriate inner product. This confuses me a bit. It doesn't seem true that the forward/backward operators are only adjoints in the $L^2$ inner product, so why this incompatibility?