Example of a diffeomorphism $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that the image of an annulus is a bigger annulus.

Question

Let $R_1 =\{ (x,y) \in \mathbb R^{2} : 1 < x^{2}+y^{2} < 4 \}$ and $R_2=\{ (x,y) \in \mathbb R^{2} : 1< x^{2}+y^{2}<9 \}$. I need to find a diffeomorphism $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $f(R_1)=R_2$. I tried $f(x)=||x||^{\log_2 3 -1} \cdot x$, but its inverse isn't differentiable at $0$. How do you find such a function?

Answer 1

At first, I thought the most natural transformation is :

$$\tag{1} f: x \to \left(\dfrac83 \|x\|-\dfrac53\right) \dfrac{x}{\|x\|}$$

Why that ? Let $(r,\theta) \to (r',\theta)$ be the polar version of $f$.

What function can we choose for $r \to r'$ ?

A) One can choose an affine mapping $r'=ar+b$.

As we must have $\begin{cases}1=a1+b\\9=4a+b\end{cases}$, we find $a=\frac{8}{3} , b=-\frac{5}{3}$, thus, the correspondence is $r'= \dfrac83 r-\dfrac53$.

It remains to normalize the unit on the "r"-axis by taking $\dfrac{x}{\|x\|}$ and we get (1).

B) But, a better choice is the square mapping $r'=r^2$.

In this case, (1) is replaced by

$$\tag{2}f:x \to \|x\|^2\dfrac{x}{\|x\|}= \|x\|x$$

which is much simpler...

The cartesian version of (2) is :

$$f:(x,y) \to (x\sqrt{x^2+y^2},y\sqrt{x^2+y^2})$$