I've simulated a very simple circuit, as follows.
It has a step-up signal as input. And signal "n2" is interesting.
Have you tried a DC simulation and examined the voltage of node n2? I believe the steady-state voltage across n2 is 0 V - not 1V. The initial voltage of 1 V is due to displacement current and the charge is then re-distributed to set the voltage across the capacitor connected to ground to 0.
Put another way, if the rise time of the 2 V signal is very long (dv/dt is about 0), the node voltage at n1 will eventually rise to 2V, but the voltage at n2 will never change from 0.