I am trying to find optimal transport in 2D from uniform distribution $\mu$ on unit disk $X = \{x \hspace{0.4em} | \hspace{0.4em} ||x||\leq 1\}$ to uniform distribution $\nu$ on annulus $Y = \{y \hspace{0.4em}|\hspace{0.4em} 2\leq||y||\leq3\}$ with distance cost $c(x,y) = ||x-y||$.
I guess transport map should send each point with radius $r$ on disk to point on annulus with radius $r+2$ and preserve the angle. So optimal transport map and potential from dual should be
$T(x,y) = (x + \frac{2x}{\sqrt{x^2+y^2}}, y + \frac{2y}{\sqrt{x^2+y^2}})$
$\phi(x,y) = 0.5x^2+0.5y^2 + 2\sqrt{x^2+y^2}$
However, these map and potential do not take into account that uniform probability of each point on $\mu$ is $\frac{1}{\pi}$ and uniform probability of each point on $\nu$ is $\frac{1}{5\pi}$.
So, any hints how to fix it? And how can I show calculations rigorously?
Thanks.
The annulus has a larger area than the disk, so the density will be lower. But since we want a uniform density it must be lower by the same factor everywhere in the annulus.
Let the points at radius $r$ on the disk be mapped to points at radius $r'$ on the annulus. If we assume the mapping from $r$ to $r'$ is a continuous increasing function, the CDF of $r$ (choosing a random point in the disk) must equal the CDF of $r'$ (choosing a random point in the annulus). That is,
$$\frac{\pi r^2}{\pi} = \frac{\pi (r'^2 - 4)}{5\pi}.$$
From this we derive $$ r' = \sqrt{5r^2 + 4}.$$
Proving rigorously that the points should be mapped radially outward (preserving the angle) from concentric circles to concentric circles is another matter altogether. But if we assume that at least this much is true, the total cost of the transport is $\lvert r' - r\rvert$ integrated over all pairs of points, which works out to the integral of $r'$ on the annulus (weighted by its density) minus the integral of $r$ on the disk (weighted by its density). That is, the cost is independent of the particular function from $r$ to $r'$, whether it is increasing, decreasing, or has discontinuities. So the continuous increasing function is as good as any other.