for now that equality holds in the inequalities . Then we can, at least formally, compute the derivatives of these identities. From $F(t,X(t,z)) = z$, we get by differentiating with respect to time and space, respectively: $$\begin{gathered}
  \partial_t F(t, X(t,z)) + \partial_x F (t,X(t,z)) \cdot \partial_t X(t,z) = 0, \label{eq:222}\\
  \partial_x F(t, X(t,z)) \cdot \partial_z X(t,z) = 1. \label{eq:223}\end{gathered}$$ From , we obtain $$\partial_t X = \left( -(\partial_x F)^{-1} \cdot \partial_t F\right) \circ X.$$ Now we can integrate w.r.t. $x\in{\mathbb{R}}$ to derive an equation for $\partial_t F$, namely $$\partial_tF = (\psi_a' \ast \omega - \psi_r' \ast \mu)\mu,$$ where at the moment we interpret $\mu$ as a density. Using $\partial_x F = \mu$ and combining and , we see that $$\partial_t X = - (\psi_a' \ast \omega - \psi_r' \ast \mu) \circ X .
  $$ In the case where $1<q_a\le2$ and $1<q_r\le2$, the functions $\psi_a'$ and $\psi_b'$ are continuous with sublinear growth, so we can use the representation . This yields the formulation which we want to work with: $$\label{maineqM}
  \partial_t X(t,z) = - \int_0^m \psi_a'(X(t,z) - Y(\zeta)) \, \mathrm{d} \zeta + \int_0^1 \psi_r'(X(t,z) - X(t,\zeta)) \, \mathrm{d} \zeta.$$ If instead $q_r=1$, then $\psi_r' = \operatorname{sgn}$ is not continuous and hence 