I'm trying to calculate divergence for parabolic coordinates with a goal to check if i understand all notions of vector analysis correctly.
Definition of parabolic coordinates:
$$x = uv \\ y = \frac{1}{2}(u^2 - v^2)$$
Firstly, I calculate $du$ and $dv$: $$ dx = v du + udv \\ dy = udu-vdv $$
$$ dv = \frac{u}{u^2+v^2}dx - \frac{v}{u^2+v^2}dy\\ dv = \frac{v}{u^2+v^2}dx + \frac{u}{u^2+v^2}dy $$ Then I calculate $e_u$ and $e_v$ - basis vectors for tangent plane in new variables: $$ x_u = v \\ y_u = u \\ e_u = \frac{(x_u,y_u)}{||(x_u,y_u)||} = (\frac{v}{\sqrt{u^2+v^2}}, \frac{u}{\sqrt{u^2+v^2}}) $$
$$ x_v = u \\ y_v = -v \\ e_v = \frac{(x_v,y_v)}{||(x_v,y_v)||} = (\frac{u}{\sqrt{u^2+v^2}}, -\frac{v}{\sqrt{u^2+v^2}}) $$
Calculating $du(e_u)$ and $dv(e_v)$: $$ du(e_u) = \frac{u^2+v^2}{\sqrt{u^2+v^2}(u^2+v^2)} = \frac{1}{\sqrt{u^2+v^2}} (=dv(e_v)) $$ Noting that $du(e_v) = dv(e_u) = 0$, let $(F, H)$ be a vector field.
Now I need to calculate area form: $$dx\wedge dy = (u^2 + v^2)du\wedge dv$$ And take interiour product with my field: $$(u^2 + v^2)((du)(F,H) dv - dv(F,H) du) \\ = (u^2 + v^2)(\frac{F}{\sqrt{u^2+v^2}}dv - \frac{H}{\sqrt{u^2+v^2}}du) \\ = \sqrt{u^2+v^2}(Fdv-Hdu)$$ Then I apply differential on top of it to get my divergence as coefficient: $$ d(\sqrt{u^2+v^2}(Fdv-Hdu)) \\ = (\frac{u}{\sqrt{u^2+v^2}}du + \frac{v}{\sqrt{u^2+v^2}}dv)\wedge(Fdv-Hdu) + \sqrt{u^2+v^2}(F_u du\wedge dv - H_v dv \wedge du) \\ = \frac{1}{\sqrt{u^2+v^2}}(uFdu \wedge dv - vH dv \wedge du) + \sqrt{u^2+v^2}(F_u + H_v) du \wedge dv \\ = (\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)) du \wedge dv $$ So divergence of $(F, H)$ should be of form $$ \frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v) $$ But there I can see that another $\frac{1}{u^2 + v^2}$ was lost somewhere. But I can't find where it was lost, so I'm asking for help to find it.
Thanks for clarifying. OK, so the vector field is dual to the $1$-form $\omega = F\omega_1 + H\omega_2$, where $\omega_1 = \sqrt{u^2+v^2}\,du$ and $\omega_2=\sqrt{u^2+v^2}\,dv$ give an orthonormal coframe. [Note that the metric $dx^2 + dy^2 = \omega_1^2 + \omega_2^2$.]
Of course we have $\star\omega_1 = \omega_2$ and $\star\omega_2=-\omega_1$. So \begin{align*} d(\star\omega) &= d(F\omega_2) - d(H\omega_1) = d\big(F\sqrt{u^2+v^2}dv\big) - d\big(H\sqrt{u^2+v^2}du\big)\\ &= \left(\sqrt{u^2+v^2}(F_u+H_v) + F\frac u{\sqrt{u^2+v^2}}+H\frac v{\sqrt{u^2+v^2}}\right)du\wedge dv \\ &= \left(\frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}\right)\omega_1\wedge\omega_2. \end{align*} Finally, using $\star(\omega_1\wedge\omega_2) = 1$, we have $$\text{div} (F,H) = \frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}.$$