The circuit is shown below:

I recognize that this is a DC restorer circuit. When VI is high, the diode (assuming ideal diode model) will be a short, so we'll have a capacitor and resistor in series. When VI is low, the diode will be reverse biased, leaving us with an open circuit in which all of VI + VC will appear across Vo, assuming the capacitor is charged.

The solution for this problem is the following:

I'm thrown off by this solution for two reason:

1. When the input waveform starts at -10 V, we have an open circuit. So all of the -10 V input should appear across this open circuit since we have not charged the capacitor yet, but the solution states that -20 V will appear across Vo initially. Perhaps I'm supposed to assume the capacitor is already charged, but if I don't, would I be correct in thinking that -10 V would be at the output initially?

2. When VI goes to 10 V, we'll have a short circuit through which the capacitor is charged. But then I'm thrown off by the resistor: when does it come into play with regard to exponential decay (RC circuit)? Does no decay occur here? If so, why?

2. When VI goes to 10 V, we'll have a short circuit through which the capacitor is charged. But then I'm thrown off by the resistor: when does it come into play with regard to exponential decay (RC circuit)? Does no decay occur here? If so, why?