Given $x=(x_1, \cdots , x_n)$ with $dx$ being the Lebesgue measure, suppose we do the change of variable $h(x) = w$, a vector field $v(x)$ is divergence free $$\text{div}(v(x)) = 0$$ then after the change of variable $\tilde v = dh(v)$, multiplied by the new density $\rho$ from $dx \mapsto \rho dw$, is still divergence free $$\text{div}(\rho\tilde v(w)) = 0.$$
I worked out an example:
- $v= (x,-y)$ is a divergence free vector field in the Cartesian coordinate.
- $h(x,y) = (\sqrt{x^2+y^2}, \arctan(y/x)) = (r,\theta)$ is the change of variable from Cartesian to polar.
- The density $dxdy \mapsto rdrd\theta$ is $r$.
- The new vector field in terms of $\partial_r$ and $\partial_\theta$: $$\tilde v = dh(v) = Jh(v) = (r(\cos^2\theta - \sin^2\theta), -2\sin\theta\cos\theta).$$
- $\text{div}(r\tilde v) = 2r(\cos^2\theta - \sin^2\theta) - 2r(\cos^2\theta - \sin^2\theta) = 0$.
Now I am confused because the divergence in polar coordinate of a vector field $v(r,\theta)= (v_1, v_2)$ in my calculation is given by $$ \text{div}(v) = \partial_r v_1 + \partial_\theta v_2$$ but there is also the formula of divergence in polar coordinate given by $$\text{div}(v)= \frac{1}{r}\partial_r (rv_1) + \frac{1}{r} \partial_\theta v_2.$$
I have a vague idea that in the second one, the directions of the vector field we are looking at are still in the Cartesian coordinate in terms of $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$. But this feels very unnatural to me, because with these $e_r$, $e_\theta$ instead of $\partial_r, \partial_\theta$, we do not have the formula $\tilde v = dh(v)$
Could you help me clarify this please.
For Completeness, here is a sketch of the proof for the statement at the top, it is based on Liouville's theorem:
The solution $\phi^t(x)$ of $$\begin{cases} \dot y = v(y) \\ y(0) = x \end{cases}$$ as a flow preserves measure $\rho dx$, $\int_D \rho dx = \int_{\phi^t(D)} \rho dx$ if and only if $\text{div}(\rho v) = 0$.
So now it suffices to show that the solution to our new IVP after the change of variable $h(\phi^t(h^{-1}(w)))$ preserves the new volume $Jh^{-1} dw$, i.e. for each $K$
$$\int_K Jh^{-1} dw = \int_{h(\phi^t(h^{-1}(K)))} Jh^{-1} dw$$
and this is clear once we rewrite $K = h(h^{-1}(K))$
$$\int_{h(h^{-1}(K))} Jh^{-1} dw = \int_{h(\phi^t(h^{-1}(K)))} Jh^{-1} dw$$
which reduces to
$$\int_{h^{-1}(K)} dw = \int_{\phi^t(h^{-1}(K))} dw$$
and this is true by our assumption $\text{div}(v) = 0$.