How is this interpolating curve well-defined in the minimizing movement scheme?

Question

Let $\Omega$ be a compact domain of $\mathbb R^d$. Let $\mathcal P (\Omega)$ be the space of probability measures on $\Omega$. For each $\tau >0$, let $(\varrho^\tau_{(k)})_{k \in \mathbb N} \subset \mathcal P (\Omega)$ be the sequence of minimizers in the minimizing movement scheme of the time step $\tau$.

The following is taken from section 8.3 in Santambrogio's Optimal transport for applied mathematicians:

Let us define two interpolations between the measures $\varrho_{(k)}^\tau$. With this time-discretized method, we have obtained, for each $\tau>0$, a sequence $(\varrho_{(k)}^\tau)_k$. We can use it to build at least two interesting curves in the space of measures:

• first we can define some piecewise constant curves, i.e., $\varrho_t^\tau:=\varrho_{(k+1)}^\tau$ for $t \in ] k \tau,(k+1) \tau]$; associated with this curve, we also define the velocities $\mathbf{v}_t^\tau=\mathbf{v}_{(k+1)}^\tau$ for $t \in] k \tau,(k+1) \tau]$, where $\mathbf{v}_{(k+1)}^\tau$ is defined as $\mathbf{v}_{(k+1)}^\tau=(\mathrm{id}-$ $\left.\mathrm{T}_{k+1}^\tau\right) / \tau$ where $\mathrm{T}_{k+1}^\tau$ is the optimal transport map from $\varrho_{(k+1)}^\tau$ to $\varrho_{(k)}^\tau$; we also define the momentum variable $E^\tau=\varrho^\tau \mathbf{v}^\tau$;
• then, we can also consider the densities $\tilde{\varrho}_t^\tau$ that interpolate the discrete values $(\varrho_{(k)}^\tau)_k$ along geodesics: $$ \tilde{\varrho}_t^\tau=\left(\frac{k \tau-t}{\tau} \mathbf{v}_{(k)}^\tau+\mathrm{id}\right)_{\sharp} \varrho_{(k)}^\tau, \quad \text {for} \quad t \in](k-1) \tau, k \tau[; \quad (8.11) $$ the velocities $\tilde{\mathbf{v}}_t^\tau$ are defined so that $\left(\tilde{\varrho}^\tau, \tilde{\mathbf{v}}^\tau\right)$ satisfy the continuity equation and $\left\|\tilde{\mathbf{v}}_t^\tau\right\|_{L^2\left(\tilde{\varrho}_t^\tau\right)}=\left|\left(\tilde{\varrho}^\tau\right)^{\prime}\right|(t)$. To do so, we take $$ \tilde{\mathbf{v}}_t^\tau=\mathbf{v}_t^\tau \circ\left((k \tau-t) \mathbf{v}_{(k)}^\tau+\mathrm{id}\right)^{-1}; $$ as before, we define a momentum variable: $\tilde{E}_\tau=\tilde{\varrho}^\tau \tilde{\mathbf{v}}^\tau$.

When $t^* := (k-1) \tau$ we have by (8.11) that $$ \tilde{\varrho}_{t^*} = \left(\mathbf{v}_{(k)}^\tau+\mathrm{id}\right)_{\sharp} \varrho_{(k)}^\tau. $$

I could not see how $\tilde{\varrho}_{t^*}^\tau = \varrho_{(k-1)}^\tau$. As such, I could not see how $(\tilde{\varrho}_t^\tau)_t$ ''interpolates'' $(\varrho^\tau_{(k)})_n$.

Could you elaborate on my confusion?

Thank you so much for your help!

Answer 1

$\newcommand{\id}{\operatorname{id}}$This is because there is a typo in (8.11). One could have guessed that (8.11) is wrong for homogeneity reasons: if $x\in \Omega$ has units [m], then $\id(x)=x$ has units [m]. But the veolcity $v_k^\tau$ has units [m.s-1] so $\frac{k\tau-t}{\tau}\mathbf{v}_{(k)}^\tau$ also has unit [m.s-1] and therefore (8.11) adds up [m] with [m.s-1].

The right formula should read $$ \tilde{\varrho}_t^\tau=\left(\id-(k\tau-t)\mathbf{v}_{(k)}^\tau\right)_{\sharp} \varrho_{(k)}^\tau, \quad \text {for} \quad t \in](k-1) \tau, k \tau[; \hspace{3cm} (8.11'). $$ (Note that this has now the correct physical units, time [s] $\times$ velocity [m.s-1] = displacement [m] for the $(k\tau-t)\mathbf{v}_{(k)}^\tau$ term).

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To recover (8.11'), think of it this way: by definition $$ \mathbf{v}_{(k)}^\tau(x)=\frac{x-T_k(x)}{\tau} $$ is the forward-in-time, constant velocity (effective terminal minus initial displacement $x-T_k(x)$ divided by travel time $\tau$) of a particle following a geodesic starting at $T_k(x)$ in the pointcloud $\varrho^\tau_{k-1}$ at time $t_{k-1}=(k-1)\tau$ and ending up at $x$ at time $t_k=k\tau$ in the pointcloud $\varrho^\tau_k$. (Here it is crucial that the trajectory is a geodesic, so that the velocity is constant.)

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Now, we want $\tilde{\varrho}_t^\tau$ to be the time-$\tau$ geodesic interpolation between $\varrho_{k-1}$ at time $t_{k-1}$ and $\varrho_k$ at time $t_k$. From standard optimal transport theory it suffices to perform the linear interpolation between transport maps. By definition the tansport map from $\varrho_k$ to $\varrho_{k-1}$ is $T_k$ (beware, $T_k$ is defined "backward in time" from $\varrho_k$ to $\varrho_{k-1}$, which is sometimes confusing) so one should simply take $$ T^t_k=\frac{t-t_{k-1}}{\tau}\id +\frac{t_k-t}{\tau}T_k \quad\text{and}\quad \tilde{\varrho}_t^\tau=\left(T^t_k\right)_\sharp \varrho^\tau_{(k)} \qquad t\in(t_{k-1},t_k). $$ This indeed interpolates linearly between $T^{t_{k-1}}_k(x)=T_k(x)$ at time $t_{k-1}$ and $T^{t_k}(x)=\id(x)=x$ at time $t_k$. Substituting for $\mathbf{v}_{(k)}^\tau=\frac{\id-T_k}{\tau}\Leftrightarrow T_k=\id-\tau \mathbf{v}_{(k)}^\tau$ readily gives $$ T^t_k=\id -(t_k-t)\mathbf{v}_{(k)}^\tau, $$ which is exactly (8.11').

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One could also have guessed it more directly: by construction $\mathbf{v}_{(k)}^\tau(x)$ is the forward-in-time velocity of the constant-speed geodesic starting from $T_{k-1}(x)$ at time $t_{k-1}$ and ending up at $x$ at time $t_k>t_{t_1}$. So, the intermediate position at time $t\in(t_{k-1},t_k)$ of this particle -- which by definition is the interpolating map $T^t_k(x)$ that is needed to define the geodesic interpolation -- is given by the rule-of-thumb "terminal position minus running time times forward velocity", i-e $T_k^t(x)=x-(t_k-t)\mathbf{v}_{(k)}^\tau(x)$. Here the minus sign is because we are running backward in time for a duration $t_k-t>0$ back from the terminal position $x$. It's a little mind game that one needs to play to figure out which objects are forward-in-time and which are backward-in-time, but once one gets used to this simply gymnastic things become much more clear (and easier!) so I'd strongly recommend thinking of it like that.