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Consider a mapping $f:\mathbb{R}^2\to \mathbb{R}^2$ given by $$ f(x,y) = \left(-x+\sqrt{x^2+y^2},-x-\sqrt{x^2+y^2}\right). $$ I want to find an explicit inverse mapping, on the neighborhoods for which $f$ is locally bijective.

A computation shows that the derivative of $f$ is given by $$ Df(x,y) = \frac{1}{\sqrt{x^2+y^2}}\begin{bmatrix} -\sqrt{x^2+y^2}+x & y \\ -\sqrt{x^2+y^2}+x & -y \end{bmatrix} $$ so the partial derivative vectors $\partial_1f(x,y)$ and $\partial_2 f(x,y)$ are orthogonal. Also, the derivative has determinant $$ \det Df(x,y) = \frac{2y}{\sqrt{x^2+y^2}} $$ which is nonzero if and only if $y\neq0$. Hence the inverse function theorem tells us that for each $(x,y)$ with $y\neq0$, the map $f$ is a bijection on some neighborhood of $(x,y)$. However, I am asked to explicitly find an inverse, which I cannot seem to do. I've bashed the problem with algebra for a while, and nothing nice comes out. My professor calls the system of functions $f_1,f_2:\mathbb{R}^2\to \mathbb{R}$ a "parabolic coordinate system". The graphs of the level sets of $f_1$ and $f_2$ make this clear. Can you help me explicitly calculate what the inverse mapping must be, on a neighborhood $U$ where $f|_U$ is a bijection?

(Also, the wikepedia entry and every other page on "parabolic coordinates" uses a very different set of functions to define what they are labeling parabolic coordinates.)

The algebra I've tried: Set $$ -x+\sqrt{x^2+y^2} = u \text{ and } -x-\sqrt{x^2+y^2}=v$$ like in high-school algebra; we want to express $x$ and $y$ in terms of $u,v$. Adding the first equation to the second yields $$ -2x = u + v $$ which implies $ x= -(u+v)/2 $. Now subtracting the second equation from the first gives $$ 2\sqrt{x^2+y^2} = u-v $$ so we square both sides and obtain $$ 4(x^2+y^2) = u^2-2uv+v^2. $$ Now using the formula we derived for $x$ gives $$ 4(\left(-\frac{u+v}{2}\right)^2+y^2) = u^2-2uv+v^2 $$ which simplifies to $$ 4y^2 + u^2 + 2uv + v^2 = u^2-2uv+v^2 $$ and again to $ 4y^2 = -4uv $ which means $y=\pm\sqrt{-uv}$. Is this correct? This attempt, I got further than any other.

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  • $\begingroup$ Finding a formula for an inverse is just algebra that doesn't depend on knowledge of derivatives or the inverse function theorem. And then once you have a formula, you can think about what neighborhoods it works for. Since you said "bashed the problem with algebra for a while", can you focus your question on that and show your work? $\endgroup$
    – Mark S.
    Commented Oct 29, 2020 at 12:45
  • $\begingroup$ Yes, solving the algebraic equations $f_1(x,y)=u,f_2(x,y)=v$ for $x,y$ in terms of $u,v$ is what I'm stuck on. All I've been able to show is that such an inverse function must exist locally via the inverse function theorem $\endgroup$
    – Milk
    Commented Oct 29, 2020 at 12:46
  • $\begingroup$ Again, can you show your work for that algebra? Then someone can give a hint or point out a mistake. $\endgroup$
    – Mark S.
    Commented Oct 29, 2020 at 12:48
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    $\begingroup$ Hmm, considering "algebraic bashing": is this really such a big deal? $x = - \frac12 (u+v)$ and $y = \pm \sqrt{- u \cdot v}$ is a candidate... $\endgroup$
    – Andreas
    Commented Oct 29, 2020 at 12:55
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    $\begingroup$ You still want to check local bijectivity. $\endgroup$
    – Andreas
    Commented Oct 29, 2020 at 13:03

1 Answer 1

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$$ \begin{cases}u+v=-2x,\\uv=-y^2,\end{cases}$$

so

$$ \begin{cases}x=-\dfrac{u+v}2,\\y=\pm\sqrt{-uv}.\end{cases}$$

Check:

$$ \begin{cases}-x+\sqrt{x^2+y^2}=\dfrac{u+v}2+\left|\dfrac{u-v}2\right|=u,\\-x-\sqrt{x^2+y^2}=\dfrac{u+v}2-\left|\dfrac{u-v}2\right|=v.\end{cases}$$

The last equalities are correct because $u\ge v$. The sign of $y$ is indeterminate. This is no surprise, as $f$ is even in $y$.

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