left edge of $\mu(t)$’s support remains at constant distance to $\omega$’s support.

\[thm:q1bis\] Assume that $q_a=q_r=1$ and $m<1$. Let $\mu:[0,\infty)\to{\mathcal{P}}({\mathbb{R}})$ be a transient solution in the sense of Definition \[dfn:transient\]. Then $\mu$ converges to $\omega$ vaguely as $t\to\infty$.

First, we are going to prove that $X(t;z)\to-\infty$ as $t\to\infty$ for all $z<(1-m)/2$. Indeed, if $z<(1-m)/2$, then in combination with $m -2 G(X(t,z)) \leq m$ implies that $$\begin{aligned}
    \partial_t X(t,z) = 2z - 1 + m - 2 G(X(t,z)) \leq 2z - 1 + m <0,
  \end{aligned}$$ hence $X(t,z)\le X_0(z)+(2z-1+m)t\to-\infty$. An analogous argument proves that $X(t;z)\to+\infty$ for all $z>(1+m)/2$.

Next, observe that the stationary equation is satisfied by $\widetilde{X}^\star(z) := Y(z - \frac{1-m}{2})$ for all $ z \in \left[ (1-m)/2, (1+m)/2 \right] $, i.e., for those $z$ we have $2z-1=G(\widetilde X^\star(z))$. By applying now the same arguments as in the proof of Theorem \[thm:q1\] — restricting $z$ to the interval $\left[ (1-m)/2, (1+m)/2 \right]$ — one shows that $X(t,z)\to\widetilde X^\star(z)$ monotonically for all feasible $z$. We conclude that $ \mu(t) $ converges vaguely to $ \mu^\star=\omega $: For given $ f \in C^0_c(\mathbb{R}) $, we have $$\begin{aligned}
    \int_{\mathbb{R}} f(x) {\mathop{}\!\mathrm{d}}\mu(t,x) = {} & \int_{0}^{1} f(X(t,z)) {\mathop{}\!\mathrm{d}}z\\
    = {} & \int_{(1-m )/2}^{1-(1-m)/2} f(X(t,z)) 