I would like some help with the following problem. Thanks for any help in advance.
Determine the largest region $\Omega'$ in $\mathbb{R}^2$ spanned by $(u,v)$ such that the transformation $T$ that defines parabolic coordinates is one-to-one. Calculate the Jacobian $J$, find the set of zeros of $J$ in $\partial \Omega'$, and and investigate the transformation $T$ on $\partial \Omega'$ (by continuous extension). Is it one-to-one? Let $(u1, v1)\times(u2, v2) \subset \Omega'$ Sketch the region $\Omega=T(\Omega')$. Express in the parabolic coordinates the normal derivative $\frac{\partial f}{\partial n} = (n, \operatorname{grad} f)$ at the boundary of the region
Now, I have attempted part of the problem. We know that $x=\frac{v^2−u^2}{2}$, $y=uv$ from the definition of parabolic coordinates. In order for each point in the $xy$ plane correspond to a unique point in the $uv$ plane, we have to restrict the region spanned by $(u,v)$ to the region $(0,\infty)\times \mathbb{R}$. Calculating the Jacobian, we get that $J=u^2+v^2$, and we see that the set of zeros consists of the single point $(0,0)$. Additionally, since the boundary of the region is simply the $v$ axis in the $uv$ plane, and the transformation is not one-to-one on the boundary simply by inspection. I am now unsure how to sketch the region under the transformation and how to express in parabolic coordinates the normal derivative.