Minimizer of the Monge problem $\frac{1}{2}\int_A |x-T(x)|^2dx$ I'm studying for an exam and there I'm having trouble to solve this exercise, if you could help me it would be great.
Given $A =\{x\in \mathbb{R}^2:|x|<1\}$ and $B=\{x\in \mathbb{R}^2:1<|x|<\sqrt{2}\}$, consider the measures $\mu=\mathfrak{L}^2|_A$ and $\nu=\mathfrak{L}^2|_B$.
Compute the map $T:\nabla\varphi:\mathbb{R}^2\to \mathbb{R}^2$, with $\varphi:\mathbb{R}^2\to \mathbb{R}$ convex, solving Monge's minimum problem
$$ \min\bigg\{\frac{1}{2}\int_A|x-T(x)|^2dx:T\in \mathfrak{T}(\mu,\nu)\bigg\}.$$
Assume radial symmetry and regularity, $\varphi\in C^2(A)$.
In here $\tau\in \mathfrak{T}(\mu,\nu)$  if $\tau:\mathbb{R}^n\to \mathbb{R}^n$ is Borel and $\tau_\#\mu=\nu$, and
$ \tau_{\#}\mu(B)=\mu(\tau^{-1})B\; $ for $ B\subset\mathbb{R}^n$ Borel.
I know that a solution exists thanks to the Brenier Lemma:

Let $\mu,\nu$ be two Borel measures in $m\mathbb{R}^n$ with compact support, $\mu(\mathbb{R}^n)=\nu(\mathbb{R}^n)<\infty$ and $\mu<<\mathfrak{L}^n.$
Then there is a convex function $\varphi:\mathbb{
 R}^n\to \mathbb{R}$ such that $\tau=\nabla \varphi$ realizes the minimum
$$ \min \bigg\{
 \frac{1}{2}\int_{\mathbb{R}^n}|x-\tau(x)|^2d\mu|
 \tau_\#
 \mu=\nu 
 \bigg\}.$$

However I have problems in actually computing the minimizer, the course is about calculus of variations but until now I only solved problems using Euler-Lagrange equations and this doesn't seem the case to do so.
 A: The solutions to the Monge minimization problem can be constructed explicitely on $\mathbb{R}$, and since you assume radial symmetry, the problem you want to solve is $1$-dimensional.
For instance, how can you send optimally the measure with density $\mathbf{1}_{[0,1]} (x)$ to the measure with density $y \mathbf{1}_{[0,\sqrt{2}]} (y)$? Well, the minimizer $T$ is nondecreasing (from your point of view : it is the gradient of a convex function), and such that $T_* (\mathbf{1}_{[0,1]} (x) \text{d}x) = y \mathbf{1}_{[0,\sqrt{2}]} (y) \text{d}y$. Already, looking at the supports, we get $T(0) = 0$ and $T(1) = \sqrt{2}$. In addition, for all continuous and bounded functions $f$,
$$\int_0^1 f(x) \ \text{d}x =  \int_0^{\sqrt{2}} f(T^{-1} (y)) y \ \text{d}y  = \int_0^1 f(x) T(x) T' (x) \ \text{d}x,$$
the first equality being the constraint $T_* (\mathbf{1}_{[0,1]} (x) \text{d}x) = y \mathbf{1}_{[0,\sqrt{2}]} (y) \text{d}y$ and the second equality a change of variables $y = T(x)$. Since this holds for all $f$, we get, for (almost) all $x \in [0,1]$,
$$T(x) = \frac{1}{T' (x)}.$$
Solving this differential equation yields directly $T(x) = \sqrt{2x}$. Extend as you wish outside of $[0,1]$.
In your case, as I said, you can use the radial symmetry of the solution:
$$\int_A f(\mathbb{r}) \ \text{d}\mathbb{r} = \int_B f(T^{-1} (\mathbb{r}')) \ \text{d}\mathbb{r}' = \int_A f(\mathbb{r}) | \text{det} DT |(\mathbb{r}) \ \text{d}\mathbb{r},$$
whence $| \text{det} DT |(\mathbb{r}) = 1$ for (almost) all $\mathbb{r} \in A$. In addition, $T(\mathbb{0}) = \mathbb{0}$.The radial symmetry lets you compute $T$ from there.
You an also exploit the radial symmetry more directly. Writing $T(\mathbb{r}) = t(\mathbb{r}) \mathbb{u}_{\mathbb{r}}$, and restricting ourselves to radially symmetric functions,
$$\int_0^1 f(x) 2 \pi x \ \text{d}x = \int_A f(\mathbb{r}) \ \text{d}\mathbb{r} = \int_B f(T^{-1} (\mathbb{r}')) \ \text{d}\mathbb{r}' =  \int_1^{\sqrt{2}} f(t^{-1} (y)) 2 \pi y \ \text{d}y = \int_0^1 f(x) 2 \pi t(x) t'(x) \ \text{d}x,$$
and you can extract a differential equation to compute $t$.
