# PDEs and Wasserstein Spaces

Seminario de Análisis Matemático y Matemática Aplicada UCM
Nov. 3, 2022

Author

David Gómez-Castro

# Some simple PDEs

## Transport equations. Constant velocity

One of the easiest PDEs to solve is, $$t,x \in \mathbb R$$ $\frac{\partial \rho_t}{\partial t} (x) + a \frac{\partial \rho_t}{\partial x} (x) = 0$

It can be solved by characteristics $$\rho_t (X_t(y)) = \rho_0(y)$$. Plugging this in

$0 = \frac{\partial }{\partial t} \Big( \rho_t \circ X_t \Big) = \frac{\partial \rho_t}{\partial t} (X_t) + \frac{\partial \rho_t}{\partial x} (X_t) \frac{\partial X_t}{\partial t}$

It suffices that $$\frac{\partial X_t}{\partial t} = a$$. And $$X_0(y) = y$$ to meet the initial datum.

Eventually $$X_t(y) = y + at$$.

$\rho_t(x) = \rho_0 (x - at)$

## Transport equation. Non-divergence form

More generally, if $$x \in \Rd$$ $\frac{\partial \rho}{\partial t} + v_t(x) \cdot \nabla \rho = 0$ Still admits solutions by characteristics $$\rho_t \circ X_t = \rho_0$$.

If $$v_t$$ is Lipschitz in $$x$$, the field of characteristics is the unique solution of

$\begin{cases} \dfrac{\partial X_t}{\partial t} = v_t(X_t) & t > 0, \\ X_0(y)_ = y. \end{cases}$

The map is $$X_t: \Rd \to \Rd$$ is bijective, since we solve “backwards” in time

$\begin{cases} \dfrac{\partial Y_s}{\partial s} = - v_{t-s}(Y_s) & t > 0, \\ Y_0(x) = x. \end{cases}$

Clearly $$X_t(Y_t(x)) = x$$. So $$\rho_t = \rho_0 \circ Y_t$$.

## Transport equation. Divergence form

$\frac{\partial \rho}{\partial t} + \diver( \rho v_t(x) ) = 0$

We can no longer solve by normal characteristics.

But we can use generalised characteristics .

We write $\frac{\partial \rho}{\partial t} + \nabla \rho \cdot v_t + \rho \diver v_t = 0$

In this case it suffices $$\rho_t \circ X_t = A_t \rho_0$$ with $$A_t \in \R$$.

$\begin{cases} \dfrac{\partial X_t}{\partial t} = v_t(X_t) & t > 0, \\ X(0,y) = y. \end{cases} \qquad \qquad \begin{cases} \dfrac{\partial A_t}{\partial t} = -\diver v_t(X_t) & t > 0, \\ A_0(y) = 1. \end{cases}$

Eventually, we can write

$\rho_t(X_t(y)) = \rho_0(y) e^{-\int_0^t \diver v_s(X_s(y)) \diff s}$

## Conservation

Let us consider $\frac{\partial \rho}{\partial t} + \diver( \rho v_t(x) ) = 0$

When $$d = 1$$, it is easy to see from the explicit solution that $\int_{X_t(a)}^{X_t(b)} \rho_t(x) \diff x = \int_a^b \rho_0(y) \diff y.$

For the $$d > 1$$, it is easier to compute for any solution that for $$A \subset \Rd$$ smooth $\frac{\diff}{\diff t} \int_{X_t(A)} \rho_t(x) \diff x = 0$

so

$\int_{A} \rho_t(x) \diff x = \int_{X_t^{-1}(A)} \rho_0(y) \diff y.$

## Push-forward

Let $$X, Y$$ be measure spaces,

$$T: X \to Y$$ be a measurable map,

$$\mu \in \mathcal M(X)$$ be a measure

The push-forward is the measure $$T_\# \mu = \nu \in \mathcal M(Y)$$ such that

$\nu (B) = \mu (T^{-1} (B)), \qquad \forall B \subset Y \text{ measurable.}$