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

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!

$$\newcommand{\id}{\operatorname{id}}$$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).
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.)
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').
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.