Weak differentiability and polar coordinates I have run into a problem/confusion regarding weak differentiability and polar coordinates for which I have found no useful references up to this point. Let $\Omega\subset\mathbb{R}^2$ be the unit disk with angle $\pi < \Theta < 2\pi$, that is
\begin{align}
\Omega = \{ (x,y)\in\mathbb{R}^2 : 0<x^2+y^2<1,\, 0<\mathrm{tan}^{-1}\left( \frac{x}{y}\right)<\Theta \}.
\end{align}
Now, define the rectangle $Q = (0,1)\times(0,\Theta)$. Given a function $\hat{u}:\Omega\to\mathbb{R}$, the change of variables from Cartesian coordinates $(x,y)$ into polar coordinates $(r,\theta)$ yields a new function $u:Q \to \mathbb{R}$ given by the composition $u = \hat{u}\circ F$, where $F:Q\to\mathbb{R}^2$ is given by $F(r,\theta) = (r\cos{\theta},r\sin{\theta})$.
Consider the weighted Lebesgue spaces on $Q$ defined by
\begin{align}
L_r^2(Q) &= \{ u : \int_0^1 \int_0^{\Theta}u^2r\mathrm{d}\theta\mathrm{d}r < \infty \}, \\
H^1_r(Q) &= \{u\in L^2(Q) : \int_0^1 \int_0^{\Theta}(\partial_r u)^2r\mathrm{d}\theta\mathrm{d}r + \int_0^1 \int_0^{\Theta}\frac{1}{r}(\partial_\theta u)^2\mathrm{d}\theta\mathrm{d}r < \infty \},
\end{align}
with the obvious norms.
Now, from a "formal" point of view I would state the following: if $\hat{u}\in H^1(\Omega)$, then by applying the chain rule and the change of variables formula we have that $u\in H^1_r(Q)$. However, all the theorems I have seen regarding weak differentiability and change of variables require the change of variables map to satisfy conditions which $F$ doesn't satisfy (e.g. Theorem 2.2.2. Ziemer1989, where we require $F$ to be bi-Lipschitz).
This leads to the following question:
Question 1. If $\hat{u}\in H^1(\Omega)$, do the weak derivatives $\partial_r u$ and $\partial_\theta u$ exist?
If the answer to the above question is "yes", and I denote by $\Phi$ the map $\hat{u} \mapsto u$, then we have $\Phi(H^1(\Omega))\subset H^1_r(Q)$. However, I am unable to prove whether this inclusion is proper or not. Intuitively, I would think both sets are equal, because if $u\in H^1_r(Q)$ then we must have that $\partial_{\theta}u(r,\theta)\to 0$ as $r\to 0$ for all $\theta\in(0,\Theta)$, indicating the function should be constant along the $r=0$ line in $Q$ and forcing $\Phi^{-1}u$ to behave properly around the origin. Hence my second question: 
Question 2. Do we have $\Phi(H^1(\Omega)) = H^1_r(Q)$?
Thank you very much!
 A: Let me try to answer your first question: if $\hat{u} \in H^1(\Omega)$, then $\hat{u} \in H^1(\Omega_{\rho})$, where the set $\Omega_{\rho} \subset \Omega$ is defined as $\Omega \cap B_{\rho}^c$. On $\Omega_{\rho}$, the map $F$ you defined is certainly bi-Lipschitz and therefore on $(\rho,1) \times (0,\Theta)$ the existence of $\partial_r u$ and $\partial_{\theta} u$ follows as in the standard result you quote. Recall also that if $\rho_1 < \rho_2$ then these partial derivatives must coincide on the smaller set, i.e., on $(\rho_2, 1) \times (0,\Theta)$. 
Notice also that by the change of variable formula, $$\int_{\rho}^1\int_0^{\Theta}u^2r = \int_{\Omega_{\rho}}\hat{u}^2 \le \int_{\Omega}\hat{u}^2,$$
and
$$\int_{\rho}^1\int_0^{\Theta} r(\partial_ru)^2 + r^{-1}(\partial_{\theta}u)^2 = \int_{\Omega_{\rho}}|\nabla \hat{u}| \le \int_{\Omega}|\nabla \hat{u}|.$$
Since the quantities on the RHS are uniform in $\rho$, by the monotone convergence theorem we also get that $u$, $\partial_r u$, and $\partial_{\theta}u$ are square integrable in the right weighted spaces.
