Inverting the integral
We are interested in inverting $U(t)$ as defined above. Note that $U(t)$ is invertible if $u(t)$ is positive for all $t > t_1$. If we define
\[\begin{equation} U_J = \int_{t_1}^{\tilde{t}_J}u(\tau)\text{d}\tau = \sum_{j = 2}^{J} \int_{\tilde{t}_{j-1}}^{\tilde{t}_j} u(\tau)\text{d}\tau, \end{equation}\]
then solving $U(t) = V$ for $t$ where $V \in [U_{J-1}, U_J]$ amounts to solving
\[\begin{equation} \int_{\tilde{t}_{J-1}}^t u(\tau)\text{d}\tau = V - U_{J-1}. \end{equation}\]
For linear sections this yields a quadratic equation in $t$ with solution
\[\begin{equation} t = t_I + \left[-\frac{u_I}{\Delta u_{I+1}} + \text{sign}\left(\frac{u_{I+1}}{\Delta u_{I+1}}\right)\sqrt{\left(\frac{u_I}{\Delta u_{I+1}}\right)^2 +\lambda\left(\frac{u_I}{\Delta u_{I+1}} + \frac{\lambda}{4}\right) +2\frac{V - U_{J-1}}{\Delta t_{I+1}\Delta u_{I+1}}}\right]\Delta t_{I+1}. \end{equation}\]
For spline sections this leads to a quartic equation in $s$:
\[\begin{equation} \begin{aligned} 3(\Delta t_{I+1} - \Delta t_I)(\Delta u_{I+1} - \Delta u_I)s^4 + \\ 4\Delta t_I (\Delta u_{I+1} - \Delta u_I) s^3 + \\ 12(\Delta t_{I+1} - \Delta t_I)\left(\Delta u_I + \frac{u_{I - \frac{\lambda}{2}}}{\lambda}\right)s^2 + \\ 24 \Delta t_I \left(\Delta u_I + \frac{u_{I - \frac{\lambda}{2}}}{\lambda}\right) s + \\ - \frac{24}{\lambda^2}(V - U_{J-1}) = 0 \end{aligned} \end{equation}\]
This quartic equation can be solved with the quartic formula.