## 6.2. Wasserstein stability estimate

The study of Wasserstein stability estimates or Dobrušin-type estimates for measured-valued solutions to kinetic equations is a classical topic. Depending on the degree of regularity of the interaction kernel, an appropriate transportation distance has to be considered. In particular, the starting works by Dobrušin and Neunzert (see [24, 44]) show that the bounded-Lipschitz distance is appropriate for Lipschitz-continuous interaction kernels. This type of inequality has been generalized to some specific kernels with more limited regularity. In particular, the right transportation distance for gradient flows associated with $-\lambda$-convex is the quadratic Wasserstein distance $W_2$ (see [14]). Indeed, we do not necessarily need an underlying gradient structure, but only require that the interaction kernel is one-sided Lipschitz-continuous. This was proven in [49, Theorem 4.7] for the Kuramoto model with weakly singular weights, which in our case provides the following stability estimate for $W_2$:

$$W_2(f_t, \bar{f}_t) \le e^{(2K+\frac{1}{2})t} W_2(f_0, \bar{f}_0), \quad (6.4)$$

which holds for any two measured-valued solutions to (1.2). Notice that units are not correct in the above inequality, and this is again due to the fact that $W_2$ is not dimensionally correct in this problem (recall 3.2). Instead, we can replace $W_2$ with $SW_2$ (see Definition 3.2) to recover the following result.

**Lemma 6.3.** Consider $K > 0$ and let $f$ and $\bar{f}$ be weak measured-valued solutions to (1.2) with initial datum $f_0$ and $\bar{f}_0 \in \mathbb{P}_2(\mathbb{T} \times \mathbb{R})$. Then we have

$$SW_2(f_t, \bar{f}_t) \le e^{\frac{5}{2}Kt} SW_2(f_0, \bar{f}_0)$$

for every $t \ge 0$.

*Proof.* Consider an optimal transference plan $\gamma_0$ joining $f_0$ to $\bar{f}_0$, that is,

$$\gamma_0 \in \Pi(f_0, \bar{f}_0) := \{\gamma \in \mathbb{P}((\mathbb{T} \times \mathbb{R}) \times (\mathbb{T} \times \mathbb{R})) : (\pi_1)_{\#}\gamma = f_0 \text{ and } (\pi_2)_{\#}\gamma = \bar{f}_0\},$$

such that

$$SW_2(f_0, \bar{f}_0)^2 = \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} d_K((\theta_1, \omega_1), (\theta_2, \omega_2))^2 d((\theta_1, \omega_1), (\theta_2, \omega_2))\gamma_0.$$

Here $\pi_1$ and $\pi_2$ represent the projections

$$\pi_1((\theta, \omega), (\theta', \omega')) = (\theta, \omega),$$

$$\pi_2((\theta, \omega), (\theta', \omega')) = (\theta', \omega').$$

Let us consider the following competitor at time $t$ via push-forward, namely,

$$\gamma_t := (\mathbb{X}_{0,t} \otimes \bar{\mathbb{X}}_{0,t})_{\#}\gamma_0 \in \mathbb{P}((\mathbb{T} \times \mathbb{R}) \times (\mathbb{T} \times \mathbb{R})),$$
where $\mathbb{X}_{0,t}(\theta, \omega) = (\Theta_{0,t}(\theta, \omega), \omega)$ and $\bar{\mathbb{X}}_{0,t}(\theta, \omega) = (\bar{\Theta}_{0,t}(\theta, \omega), \omega)$ are the characteristic flows associated with the transport fields $v[f]$. Since $\gamma_t \in \Pi(f_t, \bar{f}_t)$,

$$ 
\begin{align*}
\frac{1}{2} \mathrm{SW}_2(f_t, \bar{f}_t)^2 &\le \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} \frac{1}{2} d_K((\theta_1, \omega_1), (\theta_2, \omega_2))^2 d((\theta_1, \omega_1), (\theta_2, \omega_2)) \gamma_t \\
&= \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} \frac{1}{2} d_K(\mathbb{X}_{0,t}(\theta_1, \omega_1), \bar{\mathbb{X}}_{0,t}(\theta_2, \omega_2))^2 d((\theta_1, \omega_1), (\theta_2, \omega_2)) \gamma_0 \\
&=: I(t).
\end{align*}
$$

Our final goal is to derive some Grönwall-type inequality for $I$. Fix $(\theta_1, \omega_1)$, $(\theta_2, \omega_2) \in \mathbb{T} \times \mathbb{R}$ and define the following curves in $\mathbb{T}$:

$$ \Theta(t) := \Theta_{0,t}(\theta_1, \omega_1) \quad \text{and} \quad \bar{\Theta}(t) := \bar{\Theta}_{0,t}(\theta_2, \omega_2), $$

and the associated characteristic curves in $\mathbb{T} \times \mathbb{R}$,

$$ \mathbb{X}(t) := \mathbb{X}_{0,t}(\theta_1, \omega_1) = (\Theta(t), \omega_1), $$

$$ \bar{\mathbb{X}}(t) := \bar{\mathbb{X}}_{0,t}(\theta_2, \omega_2) = (\bar{\Theta}(t), \omega_2). $$

Set a minimizing geodesic $x_t: [0, 1] \to \mathbb{T} \times \mathbb{R}$ joining $\mathbb{X}(t)$ to $\bar{\mathbb{X}}(t)$ for every fixed $t > 0$. Notice that the function

$$ t \mapsto \frac{1}{2} d_K^2(\mathbb{X}(t), \bar{\mathbb{X}}(t)) $$

is Lipschitz continuous. Then we can take derivatives and show that

$$ \frac{d}{dt} \frac{1}{2} d_K^2(\mathbb{X}(t), \bar{\mathbb{X}}(t)) \le -\langle (v[f_t](\mathbb{X}(t)), 0), x'_t(0) \rangle - \langle (v[\bar{f}_t](\bar{\mathbb{X}}(t)), 0), -x'_t(1) \rangle \quad (6.5) $$

for almost every $t \ge 0$. Let us now consider $\theta(t) := \overline{\bar{\Theta}(t) - \Theta(t)}$, the representative of $\bar{\Theta}(t) - \Theta(t)$ modulo $2\pi$ that lies in $(-\pi, \pi]$. We find two different cases:

**Case 1:** $\theta(t) \in (-\pi, \pi)$. In this case, the only minimizing geodesic reads

$$ x_t(s) = (e^{i(\Theta(t)+s\theta(t))}, \omega_1 + s(\omega_2 - \omega_1)), \quad s \in [0, 1]. $$

Then (6.5) reads

$$ \frac{d}{dt} \frac{1}{2} d_K^2(\mathbb{X}(t), \bar{\mathbb{X}}(t)) \le (v[\bar{f}_t](\bar{\Theta}(t), \omega_2) - v[f_t](\Theta(t), \omega_1))\theta(t) $$

for almost every $t \ge 0$.

**Case 2:** $\theta(t) = \pi$. In this second case there are exactly two minimizing geodesics

$$ x_{t, \pm}(s) = (e^{i(\Theta(t) \pm \pi s)}, \omega_1 + s(\omega_2 - \omega_1)), \quad s \in [0, 1]. $$

Then we restate (6.5) as

$$ \frac{d}{dt} \frac{1}{2} d_K^2(\mathbb{X}(t), \bar{\mathbb{X}}(t)) \le (v[\bar{f}_t](\bar{\Theta}(t), \omega_2) - v[f_t](\Theta(t), \omega_1))(\pm \pi) $$
for almost every $t \ge 0$. To sum up, we achieve the following estimate:

$$ \frac{d}{dt} \frac{1}{2} d_K^2(\mathbb{X}_{0,t}(\theta_1, \omega_1), \bar{\mathbb{X}}_{0,t}(\theta_2, \omega_2)) \\ \le (v[f_t](\Theta_{0,t}(\theta_1, \omega_1), \omega_1) - v[\bar{f}_t](\bar{\Theta}_{0,t}(\theta_2, \omega_2), \omega_2)) \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)} $$

for every $\theta_1, \theta_2 \in \mathbb{T}$, each $\omega_1, \omega_2 \in \mathbb{R}$, and almost every $t \ge 0$. Using the dominated convergence theorem, we show that $I$ is absolutely continuous and taking derivatives under the integral sign implies

$$ \frac{dI}{dt} \le \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} (v[f_t](\Theta_{0,t}(\theta_1, \omega_1), \omega_1) - v[\bar{f}_t](\bar{\Theta}_{0,t}(\theta_2, \omega_2), \omega_2)) \\ \times \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_1, \omega_2)} d((\theta_1, \omega_1), (\theta_2, \omega_2)) \gamma_0 \quad (6.6) $$

for almost every $t \ge 0$. Also, note that

$$ v[f_t](\theta, \omega) = \omega - K \int_{\mathbb{T} \times \mathbb{R}} \sin(\theta - \Theta_{0,t}(\theta'_1, \omega'_1)) d(\theta'_1, \omega'_1) f_0, $$

$$ v[\bar{f}_t](\theta, \omega) = \omega - K \int_{\mathbb{T} \times \mathbb{R}} \sin(\theta - \bar{\Theta}_{0,t}(\theta'_2, \omega'_2)) d(\theta'_2, \omega'_2) \bar{f}_0. $$

Since $(\pi_1)_{\#}\gamma_0 = f_0$ and $(\pi_2)_{\#}\gamma_0 = \bar{f}_0$,

$$ v[f_t](\theta, \omega) = \omega - K \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} \sin(\theta - \Theta_{0,t}(\theta'_1, \omega'_1)) d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2)) \gamma_0, \quad (6.7) $$

$$ v[\bar{f}_t](\theta, \omega) = \omega - K \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} \sin(\theta - \bar{\Theta}_{0,t}(\theta'_2, \omega'_2)) d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2)) \gamma_0. \quad (6.8) $$

Putting (6.7)–(6.8) into (6.6) amounts to

$$ \frac{dI}{dt} \le \int_{(\mathbb{T} \times \mathbb{R})^4} (\omega_1 - \omega_2) \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)} \quad (6.9) \\ \times d((\theta_1, \omega_1), (\theta_2, \omega_2)) \gamma_0 d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2)) \gamma_0 \\ - K \int_{(\mathbb{T} \times \mathbb{R})^4} (\sin(\Theta_{0,t}(\theta_1, \omega_1) - \Theta_{0,t}(\theta'_1, \omega'_1)) - \sin(\bar{\Theta}_{0,t}(\theta_2, \omega_2) - \bar{\Theta}_{0,t}(\theta'_2, \omega'_2))) \\ \times \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)} d((\theta_1, \omega_1), (\theta_2, \omega_2)) \gamma_0 d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2)) \gamma_0 $$

for almost every $t \ge 0$. By Young's inequality, it is clear that

$$ \begin{align*} (\omega_1 - \omega_2) \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)} &\le \frac{K}{2} \overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)}^2 + \frac{(\omega_1 - \omega_2)^2}{2K} \\ &= \frac{K}{2} d_K(\mathbb{X}_{0,t}(\theta_1, \omega_1), \bar{\mathbb{X}}_{0,t}(\theta_2, \omega_2))^2. \end{align*} $$
This, along with a clear symmetrization argument in the second term, implies

$$ \frac{dI}{dt} \leq KI(t) - \frac{K}{2} \int_{(\mathbb{T} \times \mathbb{R})^4} (\sin(\Theta_{0,t}(\theta_1, \omega_1) - \Theta_{0,t}(\theta'_1, \omega'_1)) - \sin(\bar{\Theta}_{0,t}(\theta_2, \omega_2) - \bar{\Theta}_{0,t}(\theta'_2, \omega'_2))) \\ \times (\overline{\Theta_{0,t}(\theta_1, \omega_1) - \bar{\Theta}_{0,t}(\theta_2, \omega_2)} - \overline{\Theta_{0,t}(\theta'_1, \omega'_1) - \bar{\Theta}_{0,t}(\theta'_2, \omega'_2)}) \\ \times d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2))\gamma_0 d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2))\gamma_0 $$

for almost every $t \geq 0$. Now, using the Lipschitz property of the sine function we achieve the inequality

$$ \frac{dI}{dt} \leq (K + 4K)I \quad \text{for a.e. } t \geq 0. $$

Integrating the inequality and using that

$$ I(0) = \int_{\mathbb{T} \times \mathbb{R}} \int_{\mathbb{T} \times \mathbb{R}} \frac{1}{2} d_K((\theta_1, \omega_1), (\theta_2, \omega_2))^2 d((\theta'_1, \omega'_1), (\theta'_2, \omega'_2))\gamma_0 = \frac{1}{2} \mathrm{SW}_2(f_0, \bar{f}_0)^2, $$

yields the desired result.

### 6.3. Probability of mass concentration and diameter contraction

Now we are ready to begin the proof of Corollary 1.1. Let $L$ and $L_{1/2}$ be intervals of diameter 2/5 and 1/5 centered around the order parameter $\phi_\infty$ of $f_\infty$. Recall that by formula (1.22) we obtain

$$ R_{\infty} = \lim_{t \to \infty} R(t) \geq 3/5. $$

Looking at the structure of the stable equilibria $f_\infty$ in (1.18) (which corresponds to $g^- = 0$, that is, no antipodal mass), we observe that for any $(\theta, \omega)$ in supp $f_\infty$ we have the relation

$$ \theta = \phi_{\infty} + \arcsin\left(\frac{\omega}{KR_{\infty}}\right). $$

In particular,

$$ |\theta - \phi_{\infty}| \leq \arcsin\left(\frac{W}{KR_{\infty}}\right) \leq \arcsin\left(\frac{5}{3}\frac{W}{K}\right). $$

Then we can select $C$ in (1.20), so that we have

$$ \supp f_{\infty} \subset L_{\frac{1}{2}} \times [-W, W]. \qquad (6.10) $$

Notice that the choice of the diameter of $L$ is somehow arbitrary and is subordinated to the size of the universal constant $C$ in Theorem 1.1 (the smaller $C$, the smaller the diameter of $L$). For simplicity, we have set it to 2/5 but it can be generalized to sharper values. We divide the proof into the following steps:
Step a. We control the mass of $\mu_t^N$ and $f_t$ in $T\setminus L$, namely,

$$ \mu_t^N((T\setminus L) \times \mathbb{R}) \le 25 \mathrm{SW}_2(\mu_t^N, f_\infty)^2, \quad (6.11) $$

$$ \rho_t(T\setminus L) \le 25 \mathrm{SW}_2(f_t, f_\infty)^2, \quad (6.12) $$

for any $t > 0$.

Fix $t > 0$ and let $\gamma_t \in \mathbb{P}((T \times \mathbb{R}) \times (T \times \mathbb{R}))$ be an optimal transport plan between $\mu_t^N$ and $f_\infty$ for the scaled Wasserstein distance $\mathrm{SW}_2$. Then we have

$$
\begin{align*}
\mathrm{SW}_2(\mu_t^N, f_\infty)^2
&= \int_{(T \times \mathbb{R})^2} d_K((\theta, \omega), (\theta', \omega'))^2 d\gamma_t \\
&\ge \int_{((T \setminus L) \times \mathbb{R}) \times (L_{1/2} \times \mathbb{R})} d(\theta, \theta')^2 d\gamma_t \\
&\ge \frac{1}{25} \gamma_t(((T \setminus L) \times \mathbb{R}) \times (L_{1/2} \times \mathbb{R})) \\
&= \frac{1}{25} [\gamma_t(((T \setminus L) \times \mathbb{R}) \times (T \times \mathbb{R})) - \gamma_t(((T \setminus L) \times \mathbb{R}) \times ((T \setminus L_{1/2}) \times \mathbb{R}))] \\
&\ge \frac{1}{25} [\gamma_t(((T \setminus L) \times \mathbb{R}) \times (T \times \mathbb{R})) - \gamma_t((T \times \mathbb{R}) \times ((T \setminus L_{1/2}) \times \mathbb{R}))] \\
&= \frac{1}{25} [\mu_t^N((T \setminus L) \times \mathbb{R}) - f_\infty((T \setminus L_{1/2}) \times \mathbb{R})].
\end{align*}
$$

Thus, using the inclusion (6.10), we observe that the second term in the last line of the above inequality vanishes and we obtain (6.11). Similarly, using the above argument with $\mu_t^N$ replaced by $f_t$, we deduce (6.12).

Step b. We claim that we can select $T_0$ satisfying

$$ T_0 \lesssim \frac{1}{KR_0^2} \log \left( 1 + W^{1/2} \|f_0\|_2 + \frac{1}{R_0} \right), \quad (6.13) $$

and with the additional property that

$$ \mathrm{SW}_2(f_t, f_\infty) \le \frac{1}{\sqrt{500}} e^{-\frac{1}{40} K(t-T_0)} \quad (6.14) $$

for every $t$ in $[T_0, \infty)$.

To show this, take $Q_1$ large enough and $T_0$ verifying

$$ T_0 \le \frac{Q_1}{KR_0^2} \log \left( 1 + W^{1/2} \|f_0\|_2 + \frac{1}{R_0} \right), $$

so that we meet the constraints in Theorem 1.1. Then, using (3.24) and Proposition 3.2 we obtain

$$ \mathrm{SW}_2(f_t, f_\infty) \le Q_2 e^{-\frac{1}{40} K(t-T_0)} \quad (6.15) $$

for all $t$ in $[T_0, \infty)$ and some universal constant $Q_2$. Notice that by taking $Q_1$ large enough, we can make $Q_2$ arbitrarily small (e.g., $Q_2 = \frac{1}{\sqrt{500}}$). This concludes the proof of the claim.
Step c. We compute $N$ in $\mathbb{N}$ and $d_N > 0$ for each $N \ge N^*$ so that

$$ \mathbb{P}(\mathrm{SW}_2(\mu_t^N, f_t) \le \frac{1}{\sqrt{500}} e^{-\frac{1}{40} K(t-T_0)}) \ge 1 - C_1 e^{-C_2 N^{\frac{1}{2}}} \quad (6.16) $$

for any $t$ in $[T_0, T_0 + d_N]$ and any $N \ge N^*$.

First, for each $N$ in $\mathbb{N}$ let us set the scale

$$ \varepsilon_N := N^{-\frac{1}{8}}. \quad (6.17) $$

Now we define $N^*$ as

$$ N^* := \min\{N \in \mathbb{N} : \varepsilon_N e^{\frac{5K}{2}T_0} \le \frac{1}{\sqrt{500}}\}, \quad (6.18) $$

so that, by definition, we get the bound

$$ N^* \ge 500^4 e^{20K T_0}. $$

Fix any $N \ge N^*$. Notice that $N^*$ has been defined in (6.18) so that there exists $d_N > 0$ with the property

$$ \varepsilon_N e^{\frac{5K}{2}(T_0+d_N)} = \frac{1}{\sqrt{500}} e^{-\frac{1}{40}Kd_N}. \quad (6.19) $$

Indeed, by dividing (6.19) over (6.18), we can quantify $d_N$ in terms of $N^*$ as

$$ \frac{\varepsilon_N}{\varepsilon_{N^*}} e^{\frac{5K}{2}d_N} \ge e^{-\frac{1}{40}Kd_N}. $$

Consequently, we have

$$ d_N \ge \frac{5}{101K} \log \frac{N}{N^*}. $$

By construction, letting $\varepsilon = \varepsilon_N$ in the concentration inequality (6.3) of Lemma 6.2, we obtain the following quantification:

$$ \mathbb{P}(\mathrm{SW}_2(\mu_0^N, f_0) \ge \varepsilon_N) \le C_1 e^{-C_2 N^{\frac{1}{2}}} \quad (6.20) $$

for every $N \in \mathbb{N}$. Thus, by monotonicity of the exponential function, we conclude that for any $t \in [T_0, T_0 + d_N]$ we have

$$ \begin{align*} C_1 e^{-C_2 N^{\frac{1}{2}}} &\ge \mathbb{P}(\mathrm{SW}_2(\mu_0^N, f_0) \ge \varepsilon_N) \\ &\ge \mathbb{P}(\mathrm{SW}_2(\mu_t^N, f_t) \ge \varepsilon_N e^{\frac{5K}{2}t}) \\ &\ge \mathbb{P}(\mathrm{SW}_2(\mu_t^N, f_t) \ge \varepsilon_N e^{\frac{5K}{2}(T_0+d_N)}) \\ &= \mathbb{P}(\mathrm{SW}_2(\mu_t^N, f_t) \ge \frac{1}{\sqrt{500}} e^{-\frac{1}{40}Kd_N}) \\ &\ge \mathbb{P}(\mathrm{SW}_2(\mu_t^N, f_t) \ge \frac{1}{\sqrt{500}} e^{-\frac{1}{40}K(t-T_0)}), \end{align*} $$

where in the first inequality we have used the concentration inequality (6.20), in the second one we have used the stability estimate in Lemma 6.3, and the remainder follow from our choice of $d_N$ in (6.19) and $t$ in $[T_0, T_0 + d_N]$. That ends the proof of (6.16).
Step d. We quantify the probability of mass concentration of $\mu_t^N$ in the interval $L$, namely,

$$ \mathbb{P}\left(\mu_t^N(L \times \mathbb{R}) \geq 1 - \frac{1}{5}e^{-\frac{1}{20}K(t-T_0)}\right) \geq 1 - C_1 e^{-C_2 N^{\frac{1}{2}}} \quad (6.21) $$

for every $t$ in $[T_0, T_0 + d_N)$ and any $N \geq N^*$.

Now, by (6.11), (6.14), and the triangular inequality we have

$$ \begin{aligned} \mu_t^N((T \setminus L) \times \mathbb{R}) &\leq 25 \mathrm{SW}_2(\mu_t^N, f_\infty)^2 \\ &\leq 50[\mathrm{SW}_2(\mu_t^N, f_t)^2 + \mathrm{SW}_2(f_t, f_\infty)^2] \\ &\leq 50[\mathrm{SW}_2(\mu_t^N, f_t)^2 + \frac{1}{500}e^{-\frac{1}{20}K(t-T_0)}] \end{aligned} $$

for every $t$ in $[T_0, T_0 + d_N)$. Hence, we obtain

$$ \mu_t^N(L \times \mathbb{R}) \geq 1 - \frac{1}{10}e^{-\frac{3}{10}K(t-T_0)} - 50 \mathrm{SW}_2(\mu_t^N, f_t)^2 $$

for each $t$ in $[T_0, T_0 + d_N]$. This, along with (6.16), concludes the proof of (6.21).

Step e. We quantify the probability of mass concentration and diameter contraction along the time interval $[s, \infty)$ for any $s$ in $[T_0, T_0 + d_N]$.

We are now ready to finish the proof of Corollary 1.1. Let us consider $N \geq N^*$, $s$ in $[T_0, T_0 + d_N]$, and any realization of the random empirical measure $\mu^N$ (recall Definition 6.1) so that the condition within (6.21) holds. Hence, by construction, we obtain that at such a realization,

$$ p := \inf_{\theta, \theta' \in L} \cos(\theta - \theta') \geq \frac{4}{5} \quad \text{and} \quad m := \mu_s^N(L \times \mathbb{R}) \geq 1 - \frac{1}{5}e^{-\frac{1}{20}K(s-T_0)} \geq \frac{4}{5}. $$

Then we obtain the relation

$$ mp - (1 - m) = \frac{4}{5} \cdot \frac{4}{5} - \left(1 - \frac{4}{5}\right) = \frac{11}{25}. $$

In particular, take $\sigma := 2/5$ and notice that the above relations along with assumption (1.20) in Theorem 1.1 guarantee condition (4.5) within the hypotheses of Lemma 4.2. Notice that the result also holds true for the particle system. Consequently, it asserts that for such a realization of $\mu^N$ we can consider a time-dependent interval $L_s^N(t)$ with $t \geq s$ so that $L_s^N(s) = L$ and

$$ \begin{aligned} \mu_t^N(L_s^N(t) \times \mathbb{R}) &\geq 1 - \frac{1}{5}e^{-\frac{1}{20}K(s-T_0)}, \\ 1 - \inf_{\theta, \theta' \in L_s^N(t)} \cos(\theta - \theta') &\leq \max\left\{\frac{1}{5}e^{-\frac{K}{10}(t-s)}, 25\frac{W^2}{K^2}\right\}, \end{aligned} \quad (6.22) $$

for any $t \geq s$. Indeed, we have $L_s^N(t) = \pi_\theta(\mathbb{X}_{s,t}^N(L \times [-W, W]))$, where $\mathbb{X}_{s,t}^N$ represents the flow of the particle system, that is, the flow of $v[\mu^N]$. Our final goal is to simplify the
last condition in (6.22). To this end, let us consider $D_s^N(t) := \text{diam}(L_s^N(t))$ and notice
that the inequality implies

$$2 \frac{(D_s^N(t))^2}{5} \le 1 - \cos(D_s^N(t)) \le \max\left\{\frac{1}{5}e^{-\frac{K}{10}(t-s)}, 25 \frac{W^2}{K^2}\right\} \quad (6.23)$$

for any $t \ge s$. In particular, we obtain (D). Thus, Corollary 1.1 follows.

**Acknowledgments.** Part of this research was done while the second author was visiting CSCAMM at the University of Maryland, USA. We are grateful for the hospitality of CSCAMM, Prof. P.-E. Jabin, and Prof. E. Tadmor. We also thank the anonymous referee for useful comments and for pointing us to reference [34].

**Funding.** The work of J. Morales was supported by the NSF grants DMS16-13911, RNMS11-07444 (KI-Net) and ONR grant N00014-1812465. The work of D. Poyato was partially supported by the MECD (Spain) research grant FPU14/06304, the MINECO-Feder (Spain) research grant number RTI2018-098850-B-I00, the Junta de Andalucia (Spain) projects PY18-RT-2422 and A-FQM-311-UGR18, and the European Research Council (ERC) grant agreement no. 639638.

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Received 19 June 2020; revised 19 December 2021; accepted 20 December 2021.

**Javier Morales**

Center for Scientific Computation and Mathematical Modeling (CSCAMM), University of Maryland, College Park, MD 20742-3289, USA; javierm1@cscamm.umd.edu

**David Poyato**

Institut Camille Jordan (ICJ), UMR 5208 CNRS & Université Claude Bernard Lyon 1, 69100 Villeurbanne, France; and Research Unit “Modeling Nature” (MNat), Universidad de Granada, Granada, 18071, Spain; poyato@math.univ-lyon1.fr