# Explicitly solving for inverse function

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.

• 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? Commented Oct 29, 2020 at 12:45
• 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
– Milk
Commented Oct 29, 2020 at 12:46
• Again, can you show your work for that algebra? Then someone can give a hint or point out a mistake. Commented Oct 29, 2020 at 12:48
• 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... Commented Oct 29, 2020 at 12:55
• You still want to check local bijectivity. Commented Oct 29, 2020 at 13:03

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

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 $$y$$.