Question

Let F : R² → R² be defined by
F(x,y)= (x²-y²,2xy)

(a) Using the inverse function theorem, determine the points in R²
at which F has a local inverse
(b) Determined whether F has an inverse defined on all of R²
2. Is such an inverse of F unique?

Answer 1

(a) We remind the Inverse function theorem:

Theorem.

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be continuously differentiable on some open set containing a, and suppose that $det Jf(a)\neq0$ , where $Jf(a)$ denotes a Jacobi matrix at the point $a.$ Then, there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\rightarrow W$ has a continuous inverse $f^{-1}:W\rightarrow V$ , which is differentiable for all $y\in W$ .

1.(a). We compute $JF(x,y)=\begin{pmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} \end{pmatrix}=\begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}$ .

$det JF(x,y)=4x^2+4y^2$ . The latter is greater than 0 for all $x,y$ except of the point (0,0). Thus, the inverse function exists for all points (any neighborhood of (0,0) contains points, where function has a continuous inverse)

(b). $F(x,y)$ has an inverse for all points

2. The inverse is not unique. E.g, we can take F(-2,-1)=F(2,1)=(3,4). More generally, F(-x,-y)=F(x,y)