[proofplan] We argue by contradiction against global $W_2$-contractivity. For nearby initial data chosen from the two basins, contractivity would keep the two corresponding solutions at distance no larger than their initial $W_2$-distance for every time. Passing to the long-time limit using basin convergence and the triangle inequality for the metric $W_2$ forces $W_2(\bar\rho_0,\bar\rho_1)$ to be bounded by an arbitrarily small number, contradicting the assumed separation of the two stationary measures. The same basin convergence then shows that different initial measures can converge to different stationary states. [/proofplan]

[step:Assume a global $W_2$ contraction rule exists] Assume, toward a contradiction, that the flow is globally contractive in $W_2$ in the sense stated above. Thus there is a selection map $S$ assigning to each $\sigma\in\mathcal P_2(\mathbb R)$ an admissible curve \begin{align*} S(\sigma):[0,+\infty)\to\mathcal P_2(\mathbb R) \end{align*} with $S(\sigma)(0)=\sigma$ such that, for every $\sigma,\eta\in\mathcal P_2(\mathbb R)$ and every $t\ge 0$, \begin{align*} W_2(S(\sigma)(t),S(\eta)(t))\le W_2(\sigma,\eta). \end{align*} For each $k\in\mathbb N$, define the two selected admissible solutions \begin{align*} \alpha_k:[0,+\infty)\to\mathcal P_2(\mathbb R) \end{align*} by $\alpha_k(t):=S(\rho_{0,k})(t)$, and \begin{align*} \beta_k:[0,+\infty)\to\mathcal P_2(\mathbb R) \end{align*} by $\beta_k(t):=S(\rho_{1,k})(t)$. Since $\rho_{0,k}\in U_0$ and $\rho_{1,k}\in U_1$, the basin hypothesis gives \begin{align*} \lim_{t\to+\infty}W_2(\alpha_k(t),\bar\rho_0)=0 \end{align*} and \begin{align*} \lim_{t\to+\infty}W_2(\beta_k(t),\bar\rho_1)=0. \end{align*} The contraction assumption gives, for every $t\ge 0$, \begin{align*} W_2(\alpha_k(t),\beta_k(t))\le W_2(\rho_{0,k},\rho_{1,k}). \end{align*} [/step]

[step:Pass to the long-time limit using the triangle inequality]Fix $k\in\mathbb N$. Since $W_2$ is a metric on $\mathcal P_2(\mathbb R)$, the triangle inequality gives, for every $t\ge 0$, \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\bar\rho_0,\alpha_k(t))+W_2(\alpha_k(t),\beta_k(t))+W_2(\beta_k(t),\bar\rho_1). \end{align*} Using the contraction estimate on the middle term, we obtain \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\bar\rho_0,\alpha_k(t))+W_2(\rho_{0,k},\rho_{1,k})+W_2(\beta_k(t),\bar\rho_1). \end{align*} Letting $t\to+\infty$, using the ordinary limit law for a sum of real-valued functions, and using the two basin convergence limits yields \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\rho_{0,k},\rho_{1,k}). \end{align*}[/step]

[guided]Fix $k\in\mathbb N$. The goal is to compare the two limiting stationary measures $\bar\rho_0$ and $\bar\rho_1$ to the two time-dependent measures $\alpha_k(t)$ and $\beta_k(t)$, because the contraction assumption controls only the distance between the time-dependent solutions. The bridge is the triangle inequality for the metric $W_2$ on $\mathcal P_2(\mathbb R)$. Applying it first from $\bar\rho_0$ to $\alpha_k(t)$, then from $\alpha_k(t)$ to $\beta_k(t)$, and finally from $\beta_k(t)$ to $\bar\rho_1$, we get \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\bar\rho_0,\alpha_k(t))+W_2(\alpha_k(t),\beta_k(t))+W_2(\beta_k(t),\bar\rho_1). \end{align*} The contraction hypothesis applies to the selected solutions $\alpha_k$ and $\beta_k$, whose initial values are $\alpha_k(0)=\rho_{0,k}$ and $\beta_k(0)=\rho_{1,k}$. Hence, for every $t\ge 0$, \begin{align*} W_2(\alpha_k(t),\beta_k(t))\le W_2(\rho_{0,k},\rho_{1,k}). \end{align*} Substituting this estimate into the triangle inequality gives \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\bar\rho_0,\alpha_k(t))+W_2(\rho_{0,k},\rho_{1,k})+W_2(\beta_k(t),\bar\rho_1). \end{align*} Now we use exactly what the basin hypotheses provide. Since $\rho_{0,k}\in U_0$, the admissible solution $\alpha_k$ converges to $\bar\rho_0$ in $W_2$, so \begin{align*} \lim_{t\to+\infty}W_2(\bar\rho_0,\alpha_k(t))=0. \end{align*} Since $\rho_{1,k}\in U_1$, the admissible solution $\beta_k$ converges to $\bar\rho_1$ in $W_2$, so \begin{align*} \lim_{t\to+\infty}W_2(\beta_k(t),\bar\rho_1)=0. \end{align*} The term $W_2(\rho_{0,k},\rho_{1,k})$ is independent of $t$. By the ordinary limit law for sums of real-valued functions, passing to the limit $t\to+\infty$ therefore yields \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le W_2(\rho_{0,k},\rho_{1,k}). \end{align*}[/guided]

[step:Send the basin distance to zero to contradict separation] The previous step holds for every $k\in\mathbb N$. Taking the limit superior as $k\to\infty$ and using the hypothesis \begin{align*} \lim_{k\to\infty}W_2(\rho_{0,k},\rho_{1,k})=0 \end{align*} gives \begin{align*} W_2(\bar\rho_0,\bar\rho_1)\le 0. \end{align*} Since $W_2$ is nonnegative, this implies \begin{align*} W_2(\bar\rho_0,\bar\rho_1)=0, \end{align*} contradicting the assumption \begin{align*} W_2(\bar\rho_0,\bar\rho_1)>0. \end{align*} Thus no globally $W_2$-contractive selection of the flow exists. [/step]

[step:Identify the dependence of the long-time limit on the initial measure] Choose any $\sigma_0\in U_0$ and any $\sigma_1\in U_1$; such choices are possible because the sequences $(\rho_{0,k})_{k\in\mathbb N}\subset U_0$ and $(\rho_{1,k})_{k\in\mathbb N}\subset U_1$ show that both basins are nonempty. By the basin hypothesis in the statement, admissible solutions exist from $\sigma_0$ and $\sigma_1$. If $\eta_0:[0,+\infty)\to\mathcal P_2(\mathbb R)$ is an admissible solution with $\eta_0(0)=\sigma_0$, the basin hypothesis gives \begin{align*} \lim_{t\to+\infty}W_2(\eta_0(t),\bar\rho_0)=0. \end{align*} If $\eta_1:[0,+\infty)\to\mathcal P_2(\mathbb R)$ is an admissible solution with $\eta_1(0)=\sigma_1$, the basin hypothesis gives \begin{align*} \lim_{t\to+\infty}W_2(\eta_1(t),\bar\rho_1)=0. \end{align*} Because $W_2(\bar\rho_0,\bar\rho_1)>0$, the two limiting measures are distinct. Hence admissible solutions starting from data in different basins have different long-time limits. This proves that the long-time limit depends on the initial measure. [/step]