most countable number of values $\tilde z\in(0,1)$ for which $X^\star(\tilde z)$ is the end point of one such interval. For all other (i.e., almost all) values $z\in(0,1)$, $U$ is strictly monotone at $X^\star(z)$. For each of those $z$, we conclude $y_\infty=y^\star$ as well. In summary, for (almost, if $q_a=1$) all $z\in(0,1)$, the values $X(t,z)$ converge to their respective limit $X^\star(z)$ monotonically as $t\to\infty$. The monotone convergence theorem now implies that $$\begin{aligned}
    W_2(\mu(t),\mu^\star)^2 = \int_0^1 |X(t,\zeta)-X^\star(\zeta)|^2{\, \mathrm{d}}\zeta 
  \end{aligned}$$ tends to zero monotonically.

If $ q_r = q_a = m = 1 $, then $ X^\star = Y $. Above, we have proved that $ \mu(t) \rightarrow \omega $ in $ \mathcal{P}_2(\mathbb{R}) $. It is easily seen from that the support of $\mu(t)$ remains bounded uniformly in time. Consequently, convergence holds in *any* Wasserstein distance $W_p$ with finite $p$ as well. However, we cannot in general expect convergence in $W_\infty$. Indeed, assume that $ \underline X:=\inf {\operatorname{supp}}\mu_0 < \inf {\operatorname{supp}}\omega $. Then the left edge $X(t,0)$ of $\mu(t)$’s support satisfies $$\partial_t X(t,0) = 2G(X(t,0)), \quad X(0,0) = \underline X.$$ Since $G(\underline X)=0$, this initial value problem has the (unique, since $G$ is Lipschitz continuous) solution $X(t,0)=\underline X$ for all $t\ge0$. Hence the 