exponents $1\leq q_r\le q_a \leq 2$ in detail.

The case $q_r = q_a = 2$ {#sec:case-q-2}
------------------------

The most special case that we consider is $q_a = q_r = 2$. As $\psi_r'(x) = 2x$, equation simplifies significantly: $$\begin{aligned}
  \label{eq:233}
  \begin{split}
    \partial_t X(t,z) &= - \int_0^m 2(X(t,z) - Y(\zeta)) \, \mathrm{d} \zeta + \int_0^1 2(X(t,z) - X(t,\zeta)) \, \mathrm{d} \zeta \\
    &= -2(m-1)X(t,z) + 2\int_0^m Y(\zeta) \, \mathrm{d} \zeta - 2\int_0^1 X(t,\zeta)\, \mathrm{d} \zeta .
          \end{split}\end{aligned}$$ This integro-differential equation can be easily solved as follows. First, integrate [w.r.t.ifnextchar,]{}$z\in(0,1)$ on both sides to obtain $$\begin{aligned}
  \frac{\mathrm{d}}{\mathrm{d} t} \bigg(\int_0^1 X(t,z){\, \mathrm{d}}z - \frac1m \int_0^m Y(z){\, \mathrm{d}}z\bigg)
  = - 2m \bigg(\int_0^1 X(t,\zeta){\, \mathrm{d}}\zeta - \frac1m \int_0^m Y(\zeta){\, \mathrm{d}}\zeta\bigg).\end{aligned}$$ Therefore, $$\begin{aligned}
  \label{eq:666}
  \int_0^1 X(t,\zeta){\, \mathrm{d}}\zeta - \frac1m \int_0^m Y(\zeta){\, \mathrm{d}}\zeta = e^{-2mt}\bigg(\int_0^1 X(0,\zeta){\, \mathrm{d}}\zeta - \frac1m \int_0^m Y(\zeta){\, \mathrm{d}}\zeta \bigg).\end{aligned}$$ Since $\int_0^1X(t,z){\, \mathrm{d}}z$ and $\frac1m\int_0^mY(z){\, \mathrm{d}}z$ equal to the center of mass of $\mu(t)$ and of $\omega$, respectively, it follows that the center of mass of $\mu(t)$ converges to that of $\omega$ with the exponential rate $\exp(-2mt)$. Now insert into and solve for $X(t,z)$: $$\begin{aligned}
  X(t,z) = e^{-2(m-1)t}\bigg(X(0,z) - \big(1-e^{-2t}\big) \int_0^1X(0,\zeta) {\, \mathrm{d}}\zeta \bigg)  + \frac{1-e^{-2mt}}{m}\int_0^mY(\zeta){\, \mathrm{d}}\zeta .\end{aligned}$$ It is easily seen that $X(t,z)$ converges 