\[prp:steady-stat-equat\] below, we identify the unique steady state $\mu^\star$ of as a suitable cut-off of $ \frac{1}{2} \psi_a'' \ast \omega $ — note that $ \frac{1}{2} \psi_a''\ast\omega=\omega$ if $q_a=1$. Theorem \[thm:q1\] provides convergence of transient solutions $\mu(t)$ towards $\mu^\star$ as $ t\rightarrow\infty $.

\[prp:steady-stat-equat\] Assume that either $q_a>1$, or that $q_a=1$ and $m\ge1$. Then, there are (unique, if $q_a>1$) real numbers $\underline x<x_0<\overline x$ such that $$\begin{aligned}
    \psi'_a\ast\omega(x_0)=0, \quad \psi'_a\ast\omega(\underline x)=-1,\quad \psi'_a\ast\omega(\overline x)=1.
  \end{aligned}$$ And there is a unique steady state in the sense of Definition \[dfn:steady\], which is given by $$\begin{aligned}
    \mu^\star = \frac12\big(\psi_a''\ast\omega\big)\, \mathbf{1}_{[\underline x,\overline x]}.
  \end{aligned}$$ If instead $q_a=1$ and $m<1$, then there is no steady state — but see Theorem \[thm:q1bis\].

See Figure \[fig:q1plot\] for an example of the resulting $ \mu^\star $, using the notation of Proposition \[prp:steady-stat-equat\].

We subdivide the proof in two cases depending on $q_a$.

*First, assume $q_a>1$.* Let $U:=\psi_a'\ast\omega$. Since $\psi_a'$ is strictly monotonically increasing, continuous, and tends to $ \pm \infty $ as $ x \rightarrow \pm \infty $, the same is true for $U$. Hence $x_0,\underline x,\overline x$ are well-defined. Formula is applicable, and the steady state equation is thus equivalent to $$2z - 1 = U(X^\star(z)), \quad 0\le 