PDEs and Wasserstein Spaces

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

Some simple PDEs

Transport equations. Constant velocity

\[ \DeclareMathOperator{\diver}{div} \newcommand{\R}{{\mathbb R}} \newcommand{\Rd}{{\mathbb R^d}} \newcommand{\diff}{\,\mathrm{d}} \DeclareMathOperator*{\argmin}{argmin} \newcommand{\ee}{\varepsilon} \DeclareMathOperator{\Lip}{Lip} \]

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 (Evans, 1998:chap.3).

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

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

Transport of mass by characteristics

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

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