# Finding an explicit inverse on a neighborhood for a function $\mathbb R^2\rightarrow\mathbb R^2$

I have the continuously differentiable function $f(x,y)=(x^3y,xy)$. Its Jacobian is $$\begin{pmatrix}3x^2y&x^3\\y&x\end{pmatrix}.$$ At $(1,1)$, the Jacobian is non-zero, so by the inverse function theorem, there is an neighborhood of $(1,1)$ in which $f$ has a continuously differentiable inverse. I would like to find the inverse of $f$ in this neighborhood.

If I invert the Jacobian, I get $$\frac{1}{2x^3y}\begin{pmatrix}x&-x^3\\-y&3x^2y\end{pmatrix}.$$ However, I am not sure what function would has such a Jacobian.

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## 2 Answers

Just invert directly.

It is easy to see that if $x=0$ or $y=0$ then the function is not invertible.

If $(x,y) \neq 0$, then let $a=x^3y,b=xy$. Then dividing (or cubing and dividing) gives $x^2 = \frac{a}{b}$, $y^2 = \frac{b^3}{a}$.

Hence there are four possible solutions $x = \pm \sqrt{\frac{a}{b}}$, $y = \pm \sqrt{\frac{b^3}{a}}$. The neighborhood you are in will dictate which solution to use.

Note: I forgot to mention that $a$ and $b$ always have the same sign (since $x, x^3$ have the same sign), so the square root is well-defined.

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Let $(a,b) = f(x,y) = (x^3y,xy)$, since $(x,y)$ is near $(1,1)$, $a=x^3y\neq 0$ and $b=xy\neq 0$. Thus $$x = \sqrt{\frac{a}{b}}\ \text{and}\ \ y = \sqrt{\frac{b^3}{a}},$$where the positive roots are to be taken as is customary.

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