I have to deal with the following PDE which describes the evolution of the generating function $\mathcal{Z}(h,t)$ for a population of cells dividing with rate $\gamma$: $$ \partial_t\mathcal{Z}(h,t)=\gamma(h^2-h)\partial_h\mathcal{Z}(h,t) $$ where the initial condition is given by $\mathcal{Z}(h,t=0)=h$.\
Bellman and Harris showed also that the following integral equation holds for such generating function: $$ \mathcal{Z}(h,t)=e^{-\gamma t}h+\int_0^t\mathcal{Z}^2(h,t-x)e^{-\gamma x}dx $$ I need to use the Characteristic curvers method to convert the PDE into the previous Integral non linear Volterra equation\
I know that such PDE can be solved exactly but it is not my goal
The characteristic curves system is given by: $$ \dot{t}(s,\tau)=1\\ \dot{h}(s,\tau)=-\gamma (h^2-h)\\ \dot{\mathcal{Z}}=0 $$ The second ODE can be cast as: $$ \dot{h}(s,\tau)=-\gamma (h^2-h)\\ \frac{d}{ds}\big(e^{-\gamma s}h(s,\tau)\big)=-\gamma h^2(s,\tau)e^{-\gamma s}\\ e^{-\gamma s}h(s,\tau)-h(0,\tau)=-\int_0^s\gamma h^2(u,\tau)e^{-\gamma u}du\\ h(0,\tau)=e^{-\gamma s}h(s,\tau)+\int_0^s\gamma h^2(u,\tau)e^{-\gamma u}du\\ $$ Because of the fact that the generating functional is constant along the Characteristic curves: $$ \dot{\mathcal{Z}}=0\rightarrow \mathcal{Z}(h(s,\tau),t(s,\tau))=\mathcal{Z}(h(0,\tau),t(0,\tau))=h(0,\tau) $$ and also: $$ t(s,\tau)=s\rightarrow t(0,\tau)=0 $$ Then: $$ \mathcal{Z}(h(s,\tau),t(s,\tau))=e^{-\gamma s}h(s,\tau)+\int_0^s\gamma h^2(u,\tau)e^{-\gamma u}du\\ \mathcal{Z}(h,t)=e^{-\gamma t}h+\int_0^t\gamma h^2(u,\tau)e^{-\gamma u}du\\ $$ Furthermore, comparing the two equations: $$ \mathcal{Z}(h,t)=e^{-\gamma t}h+\int_0^t\gamma \color{red}{h^2(u,\tau)}e^{-\gamma u}du\\ \mathcal{Z}(h,t)=e^{-\gamma t}h+\int_0^t\gamma \color{red}{\mathcal{Z}^2(h,t-u)}e^{-\gamma u}du $$ My question is: why I can say that $h(s,t)=\mathcal{Z}(h,t-s)$???\
**I know that we can prove such relation solving the PDE exactly but I need to understand why such relation holds without using the exact solution **