This is the exercise
i). Find a transport map between $d\mu_d=\frac{1}{\pi}\mathbb{I}_{B_1(0)}dxdy$ and $d\mu_c=\frac{1}{8\pi}\left(4-|x|^2\right)_+dxdy$.
ii) Compute $C^K_{\rho}(\mu_d,\mu_c)$ where $\rho(x,y)=|x-y|^2$.
My attempt
Both of these are radial measures. Intuitively, rotating is “wasted” work, so we should look for a map that just moves things out radially: $T(x)=\left(\frac{x}{|x|}\right)\phi(x)$ for increasing $\phi$. Begin by checking on thin annuli ($\epsilon\rightarrow 0$) of the form $A_{\phi(r),\phi(r+\epsilon)}$, where
$$ A_{r,R}=\{\bar{x}:r\le \|\bar{x}\|\le R\}. $$
This is what I was told to do
Get a differential equation for $\phi$ and $\phi'$.
Could anyone please help me here?
The trick here is to exploit rotational symmetry and transform the problem from 2D to 1D.
\begin{align*} d\mu_d &= \frac{1}{\pi} \mathbb{I}_{B_1(0)} dxdy = \frac{1}{\pi} 2\pi rdr = 2r dr\\ d\mu_c &= \frac{1}{8\pi} (4-|x|^2)_+ dx dy = \frac{1}{4} (4r-r^3) r dr \end{align*}
Here we have two 1D measure and we can compute the transport map, by first computing the CDFs \begin{align*} D(t) &= \int_{-\infty}^t 2r dr = t^2, \ D^{-1}(t) = \sqrt{t}\\ C(t) &= \int_{-\infty}^t \frac{1}{4} (4r-r^3) dr =\frac{t^2}{2} - \frac{t^4}{16}, \ C^{-1}(t) = 2 \sqrt{1-\sqrt{1-t}} \end{align*} so the transport map is $$ t(r) = C^{-1}(D(r)) = 2 \sqrt{1-\sqrt{1-r^2}} $$ with $t(r)$ we move $\mu_d$ to $\mu_c$ (in polar coordinate).
For the second part \begin{align*} C_\rho^K(\mu_d, \mu_c) &= \int_0^1 |D^{-1}(t) - C^{-1}(t)|^2 dt\\ & = \int_0^1 \left| \sqrt{t}-2 \sqrt{1-\sqrt{1-t}}\right|^2 dt\\ & = \frac{-319 + 256 \sqrt{2}}{210} \end{align*}