Maybe this is the right place for my question:
I have a system of coupled FBSDEs in 3 dimensions as follows (in cartesian coordinates):
$$ \mathrm{d}\vec{r}(t) = \vec{u}(\vec{r}(t))\mathrm{d}t + \sqrt{ \frac{\hbar}{\mu}}\,\mathrm{d}\vec{w}_\mathrm{f}(t)\\ \mathrm{d}\vec{u}(\vec{r}(t)) = \frac{1}{\mu}\frac{e^2_0}{r^2(t)}\hat{\vec{r}}(t)\mathrm{d}t + \frac{1}{\mu} \sigma \mathrm{J}_{\vec{u}}(\vec{r})\,\mathrm{d}\vec{w}_\mathrm{b}(t) $$
where $\vec{w}$ is a 3D Wiener process (f stands for forward, b for backward) and $\mathrm{J}_{\vec{u}}(\vec{r})$ denotes the Jacobi matrix of $\vec{u}$ with respect to $\vec{r}$.
I want to transform these SDEs to spherical coordinates $\rho = \sqrt{x^2_1+x^2_2+x^2_3},\ \vartheta = \arccos\left(\frac{x_3}{\rho}\right),\ \varphi = \mathrm{atan2}(x_2,x_1)$ with the use of Ito formula. For the forward equations I get (still not absolutely sure if this is correct, edit: latter equation corrected):
$$ \mathrm{d}\rho = \bigg(u^\rho(\rho,\vartheta,\varphi)+\frac{\sigma^2}{\rho}\bigg)\mathrm{d}t + \sigma\,\mathrm{d}w^\rho_\mathrm{f} \quad \text{with}\ u^\rho = \vec{u}\cdot \hat{\vec{\rho}} \\ \rho\,\mathrm{d}\vartheta = \bigg(u^\vartheta(\rho,\vartheta,\varphi) + \frac{\sigma^2}{2\rho}\cot \vartheta\bigg)\mathrm{d}t + \sigma\,\mathrm{d}w^\vartheta_\mathrm{f} \quad \text{with} \ u^\vartheta=\vec{u}\cdot\hat{\vec{\vartheta}}\\ \rho\sin\vartheta\,\mathrm{d}\varphi = \phantom{\bigg(}u^\varphi(\rho,\vartheta,\varphi)\mathrm{d}t + \sigma\,\mathrm{d}w^\varphi_\mathrm{f} \quad \text{with} \ u^\varphi= \vec{u}\cdot\hat{\vec{\varphi}} $$
If I want to use Ito formula again for $u^\rho (\rho,\vartheta,\varphi) = (u^1(\rho,\vartheta,\varphi), u^2(\rho,\vartheta,\varphi), u^3(\rho,\vartheta,\varphi))^T\cdot \hat{\vec{\rho}}(\vartheta,\varphi)$ for example, $u^\rho$ is a function of $u^1,u^2,u^3$ and $\rho,\vartheta,\varphi$. How do I have to use the Ito formula correctly? W.r.t. $u^1,u^2,u^3$ or $\rho,\vartheta,\varphi$ or both?
Thanks in advance.
