Let $(\rho^{(i)}_t,{\bf v}^{(i)}_t)$ for $i = 1,2$ be two solutions of the continuity equation $$\partial \rho^{(i)}_t + \nabla\cdot \left({\bf v}^{(i)}_t \rho^{(i)}_t\right) = 0 \label{1}\tag{1}$$ on a compact domain $\Omega$ (with no-flux boundary conditions), and suppose that $(\rho^{(i)}_t,{\bf v}^{(i)}_t)$ are absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^d$ for every $t$ and that $\rho^{(i)}$ are absolutely continuous curves in $\mathbb{W}_p(\Omega)$. Then Corollary 5.25 in the monograph Optimal Transport for Applied Mathematicians stated that $$\frac{d}{dt}\left(\frac{1}{p} W^p_p\left(\rho^{(1)}_t,\rho^{(2)}_t\right)\right) = \int_{\Omega} \left(x-T_t(x)\right)\cdot \left({\bf v}^{(1)}_t(x) - {\bf v}^{(2)}_t(T_t(x))\right)\rho^{(1)}_t ~dx \label{2}\tag{2}$$ where $T_t$ is the optimal transport map from $\rho^{(1)}_t$ to $\rho^{(2)}_t$ for the cost $\frac{1}{p}|x-y|^p$.

On the other hand, in the paper Convergence to equilibrium in Wasserstein distance for Fokker-Planck equations the continuity equation \eqref{1} with the specific velocity field $${\bf v}_t = -(\nabla \log \rho_t + A) \label{3}\tag{3}$$ is considered. Here the vector field $A$ takes the form $A = \nabla V + F$ where $F$ satisfies $\nabla \cdot (\mathrm{e}^{-V} F) = 0$, so that the equilibrium solution/distribution is of the form $\nu = \mathrm{e}^{-V}$. Now Theorem 2.1 in the aforementioned paper claimed that $$\frac{d}{dt}\left(\frac{1}{2} W^2_2\left(\rho_t,\nu\right)\right) = -\int (x-\nabla \psi_t)\cdot \left(\nabla \log \rho_t + A\right) ~ d\rho_t\label{4}\tag{4}$$ where $\nabla \psi_t$ push forwards $\rho_t$ to $\nu$ for every $t \geq 0$.

Of course, we can take $T_t(x) = \nabla \psi_t(x)$ due to property of $W^2_2$. But I am failing to see the equivalence between \eqref{2} and \eqref{4} (when $p=2$). Specially, I think the formulation \eqref{2} contains the term $$\int (x-\nabla \psi_t)\cdot {\bf v}_t(T_t(x))~d\rho_t$$ while such term is missing in the formulation \eqref{4}. May I know what is going on here?


1 Answer 1


Without having read the sources in detail, I would notice the following:

In (4) the second measure is time independent. To satisfy (1) this then corresponds to $(\rho^{(2)}_t,v_t^{(2)}) = (\nu,0)$. But then since $v_t^{(2)}$ vanishes, so does the second term in (2).

All the other changes are already noticed in the question. One is just the specific definition of the vector field. The other is the fact that for absolutely continuous measures the unique solution $T_t$ can be expressed as a gradient $\nabla \psi_t$, where $\psi_t$ is a convex function. This is a well known theorem (due to Brenier, I believe) which is found in any textbook on optimal transport.

  • $\begingroup$ The theorem is due to Brenier when $p=2$, but the more general case was established by Gangbo and McCann. $\endgroup$
    – Gabe K
    Aug 18 at 14:25
  • $\begingroup$ Hello, I do not understand why $\rho^{(2)}_t = \nu$ (time-independent) implies ${\bf v}^{(2)}_t = 0$, it should be that $\nabla \cdot \left({\bf v}^{(2)}_t \rho^{(2)}_t \right) = \nabla \cdot \left({\bf v}^{(2)}_t \nu \right) = 0$ right? $\endgroup$
    – Fei Cao
    Aug 18 at 15:03
  • 1
    $\begingroup$ There may indeed be other velocity fields satisfying $\nabla\cdot v_t \nu=0$, the point is that (2) does NOT depend on which one you choose to "represent" the curve $\rho_t^{(2)}$. So in case of such a constant-in-time curve you just choose to use this particular representative $\endgroup$ Aug 19 at 8:55
  • $\begingroup$ @leomonsaingeon Ok thank you for this answer. I will wait to see if there are other answers before I award you the bounty. $\endgroup$
    – Fei Cao
    Aug 19 at 19:56
  • $\begingroup$ @leomonsaingeon I think I find the flaw in your argument. Basically, one has to stick with the definition of the velocity field ${\bf v}_t$ given in equation (3). If not, I can apply your argument to any $(\rho^{(2)}_t, {\bf v}^{(2)}_t) = (\rho, 0)$ for any probability measure $\rho$ (just by manually taking ${\bf v}^{(2)}_t = 0$), this means that identity (4) would be valid for any $\rho$ in place of the "true equilibrium“ $\nu$, which is absurd. $\endgroup$
    – Fei Cao
    Aug 20 at 15:38

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