While reading about how a half wave rectified output is filtered using a capacitor, I came across the following waveforms:
corresponding to the circuit:
In the interval from 0 to 
That much is clear to me. However, I am not exactly sure about what happens in the interval from
... in the interval from \$ \frac {\pi} {2}\$ to \$ \pi \$. According to me, the input voltage is simply decreasing here, so the diode should remain forward biased, and hence conduct. But the above waveform shows otherwise.
The diode is only forward biased when the \$ V_{AC} \$ > \$ V_C \$. Once the capacitor charges to peak voltage and the AC falls faster than \$ V_C \$ then the diode is reverse biased.
Figure 1. OP's graph with diode current superimposed.
Figure 1 shows a typical current pulse in this situation. All the charge is passed in the short duration when \$ V_{AC} \$ > \$ V_C \$.
So, is the capacitor discharging in this interval?
Yes. \$ V_C \$ > \$ V_{AC} \$ so the diode is reverse biased.
If yes, then is it different from the discharge in the interval from \$ \pi \$ to \$ 2\pi \$?
No, it is a continuation of the same discharge because the diode is reverse biased in both cases.
Note: The original diagram might be a little clearer if the red line was 0.7 V below the blue waveform as would be the case with a real diode due to its forward voltage drop.
