Question about RC charge-discharge curve
Yes, it actually looks like this (doing a numerical integration of the differential equation):



At equilibrium, the peaks of the sawtooth are at \${1\over 1+e^{T\over 2RC}}V_S\$ and \${e^{T\over 2RC}\over 1+e^{T\over 2RC}}V_S\$, where \$T\$ is the period of the square wave (different from the \$T\$ in the question plot).  For this example, that's about \$0.12V_S\$ to \$0.88V_S\$.

In that case equilibrium is reached very fast.  For a higher frequency square wave, it can take a few cycles.  In this example, the period of the square wave is \$RC\over 2\$ instead of \$4RC\$: