Global inverse of $G(x,y)=(\ln (xy),\frac{1}{x^2+y^2})$

Question

Let $G(x,y)=(\ln (xy),\frac{1}{x^2+y^2})$, with Dom$(G)=\{(x,y)\in\mathbb{R^2:0<y<x}\}$. I am asked to find the global inverse of $G$.

I have tried to proceed in the following way:

$$u=\ln(xy)=\ln(x)+\ln(y),\quad x=\frac{e^u}{y}$$

$$v=\frac{1}{x^2+y^2}=\frac{1}{\frac{e^{2u}}{y^2}+y^2}=\frac{y^2}{e^{2u}+y^4},\ y=+\sqrt{\frac{1\pm\sqrt{1-4v^2e^{2u}}}{2v}}=\frac{\sqrt{1\pm\sqrt{1-4v^2e^{2u}}}}{\sqrt {2v}}$$

But I'm not sure this is right, as my textbook says correct solution is $$G^{-1}(u,v)=(\frac{\sqrt{1+2e^uv}+\sqrt{1-2e^uv}}{2\sqrt v}, \frac{\sqrt{1+2e^uv}-\sqrt{1-2e^uv}}{2\sqrt v})$$

Could you help me? Thanks in advance!

Answer 1

One place you go wrong is the claimed equality

$$\sqrt{\frac{1 + \sqrt{1 - v^2e^{2u}}}{2v}} = \frac{\sqrt{2} + \sqrt{2}\sqrt{1 - v^2e^{2u}}}{2\sqrt{v}}$$

since $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$ in general, and also $\sqrt{\sqrt{a}} = a^{1/4}$...

Here's a way of getting the book's answer:

Let $u = \ln(xy)$, so that $xy = e^u$

Let $v = 1/(x^2+y^2)$, so that $v^{-1} = x^2 + y^2$.

Then $(x+y)^2 = v^{-1} + 2e^u$ and $(x-y)^2 = v^{-1} - 2e^u$. Since the domain is $0 < y < x$, we can take square roots of both expressions to get

$x+y = \sqrt{v^{-1} + 2e^u}$ and $x-y = \sqrt{v^{-1} - 2e^u}$.

Now $x = (x+y)/2 + (x-y)/2$ and $y=(x+y)/2 - (x-y)/2$, which simplifies to the book's solution.

Answer 2

You seem to be missing the $4$ from the quadratic solution formula!

If the equation is $ax^2+bx+c=0$ the solutions are $$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Applying this to the biquadratic formula $vy^4-y^2+e^{2u}v=0$, which you seems to have done, leads to

$$(y^2)_{1,2}= \frac{1\pm\sqrt{1-{\mathbf 4}v^2u^{2u}}}{2v}$$

which is the same as you got, except for the marked $4$, and the choice of the $\pm$ sign. It turns out that because $y$ is the samller of the solutions ($x$ is the other), you need to use the "-" sign. That is then the same as the solution for $y$ given in the textbook, though it is non-trivial to see that

$$\sqrt{1-\sqrt{1-4v^2u^{2u}}}=\frac{\sqrt{1+2ve^u}-\sqrt{1-2ve^u}}{\sqrt2}$$

Once you know it, you can easily verify it by squaring the right side.

Jane's answer shows you a much easier way to get to the solution, though!