# A radius for which the inverse of a function is well defined (Inverse Function Theorem)

Let $f(x,y)=(\frac{1}{2} x^2 + x(y-1)^3, xy-x)$
(1) Find a formula for $f^{-1}$ in a small ball $B(b,r)$ where $b=(\frac{1}{2}, 0)$.
(2) Give an example of a radius $r>0$ for which the inverse function is well defined.

I showed that the inverse was well defined in some neighbourhood of b in a previous question using the Inverse Function Theorem.

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Sorry, I may be missing something, but how can you invert a function of two variables? Your function is from $R^2$ to $R$, so it cannot be a simple function, so it can't be (fully) inverted. The inverse function would have to be from $R$ to $R^2$, right? – Luboš Motl Jun 9 '11 at 16:28
@Lubos: The function is from $R^2$ to $R^2$. – Did Jun 9 '11 at 16:33
I see, thanks, it's a comma. ;-) – Luboš Motl Jun 9 '11 at 16:36

The advantage of not being Mathematica is that I can use simpler symbols.

Suppose that our function maps $(x,y)$ to $(s,t)$. We want to recover $x$ and $y$ given $s$ and $t$. It turns out that we don't really need Mathematica.

We know that $$\frac{1}{2}x^2+x(y-1)^3=s \qquad\text{and}\qquad xy-x=t.$$ The second equation looks more tractable. Trusting that $x=0$ is not going to give us a problem, we rewrite the equation as $$y-1 =\frac{t}{x}.$$ Substituting in the first equation, we see that perhaps we are doing something right, since we get $$\frac{1}{2}x^2 + \frac{t^3}{x^2}=s.$$ Solve for $x$. The rest will be straightforward, since if we have $x$ in terms of $s$ and $t$, then from $$y=1 +\frac{t}{x},$$ we also have $y$ in terms of $s$ and $t$.

Rewrite the equation for $x$ as $$x^4 -2sx^2 +2t^3=0.$$ This is a quadratic in $x^2$, readily solved for $x^2$, and hence for $x$. We get $$x=\pm\sqrt{s\pm \sqrt{s^2-2t^3}}.$$

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I will cheat. The command in Mathematica is:

Solve[{f1, f2} == {x^2/2 + x (y - 1)^3, x y - x}, {x, y}]


The result is:

{x -> (1/(2 f2))((2 Sqrt[2] f1 Sqrt[-f2 (-f1 + Sqrt[f1^2 - 2 f2^3])])/f2 - (Sqrt[2] (-f2 (-f1 + Sqrt[f1^2 - 2 f2^3]))^(3/2))/f2^2),y -> (2 f2 + Sqrt[2] Sqrt[f1 f2 ->f2 Sqrt[f1^2 - 2 f2^3]])/(2 f2)}

and similar solutions with different signs. You have to choose the right signs. They're just playing "I Saw the Sign" on radio. ;-) All of the four solutions returned by Mathematica look pretty messy but one of them is the right one.

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Thanks @Luboš. I was hoping that there was an easier way to do this though. The question is from a PDE exam paper and is worth only 5% of the exam. Perhaps there is a result related to the Inverse Function Theorem that I could simply apply here? – Faust Jun 9 '11 at 16:50
Maybe they just want an approximate formula for the inverse? I can't imagine how the formulae could simplify. Or maybe you're allowed to use some symbols inside? Something that makes the solution "non-explicit"? – Luboš Motl Jun 9 '11 at 16:57
The four derivatives $\partial f_{1,2} / \partial (x,y)$ are all nonzero at $(1/2,0)$ and the matrix is non-singular over there, so it's clear that in a vicinity of the point, the inverse exists. – Luboš Motl Jun 9 '11 at 16:59
Well, the Jacobian of the inverse of f is the inverse of the Jacobian of f, so the Jacobian of the inverse is the identity matrix $I_{2}$. Can we deduce something from this about $f^{-1}$? – Faust Jun 9 '11 at 17:11