This question is academic insofar as I am assuming a circuit with no inductance.
In a circuit having only passive capacitive elements, when driven with a square wave--is it possible to have a voltage greater than the drive voltage at any node? I am assuming linear components--e.g. no diodes or active elements--just a capacitor and resistor network.
From an energy perspective I think not. Overshoot from "ringing" for example, always (in my experience) is associated with inductances and a 2nd order circuit.

Fritz

 

I believe that's true.
Without inductive or active components, you can't get an output voltage greater than the input.
Of course, in the real world, you can't have a capacitor without some small wire inductance.

 

is it possible to have a voltage greater than the drive voltage at any node? I am assuming linear components--e.g. no diodes or active elements--just a capacitor and resistor network.

If I understand the question correctly, then yes, it is.

5 V square wave, 50/50 duty cycle, 0 ohm source impedance, 1 kHz. positive pulse width = 500 us
0.1 uF capacitor and 1.667K resistor in series; time constant = 166 us
Cap connected to square wave source, resistor tied to +5 V.

The pulse width is three time constants, so when the square wave is low for 500 us, the cap is charged to 95% of Vcc. When the squarewave goes high, the R-C node jumps up to 9.75 V and then starts decaying down to 5 V. As the C gets larger, the R-C node looks more and more like the input squarewave, now centered about 5.0 V (2.5 V to 7.5 V) by simple AC coupling.

ak