By not much more than I described in the first post I made.
1. You pick an arbitrary starting state of the output. The device operates in open loop, so it'll be one rail or the other. The amplifier is not operating in a linear mode. Suppose you start with the output at the low rail, say -10.0V.
2. Identify a starting condition which makes the output state valid. If Vo=(-10), then Vp the non-inverting terminal must be at a lower potential than Vn, the inverting terminal. If Vn is grounded, then Vp < 0. Suppose Vp=-1.0V
3. The opamp is then trying to use its massive open-loop gain to take (-1.0) - 0 = -1 and multiply it up by 100,000V/V, attempting to make the output -100,000V. Your supplies are limited so it stops at, say, -10V.
4. Vo is then ~= 100,000 * (Vp - Vn)
when Vp is slightly more positive than Vn, which is to say Vp > 0, the output slams to the positive rail, and then outputs +10V.
5. A schmitt trigger is a circuit design which leverages the "well defined output states" of the comparator to force hysteresis.
6. Go back to the initial state where Vo = -10V again. If you wanted the state to only change once the applied voltage were +0.5V instead of 0V, you then need to use a resistor from the input to the Vp terminal, and a resistor from Vp to vout.
7. Since the change happens roughtly around Vp=0, you pick a starting current, then find an appropriate hysteresis resistor: Rh = (0-(-10))/current, suppose I choose 1mA, then Rh=10kOhm.
8. So now, once I apply 0.5V, I need to have 1mA flow from the input to this targeted value of vp~=0: (0.5-0)/0.001 = 499 Ohms. So I install a 499 ohm resistor from the input to the Vp node and call this Ri.
9. When Vin = 0.5V, Vp crosses the zero and the huge amount of gain causes an output change.
10. Now the output is +10V. The applied input now sinks (10-0.5)/10.5kOhm = 904uA and Vp is then 0.5 + 904u*499 = 0.951V.
11. Your input begins going negative. What applied input will change the output? Whatever input will cause Vp to cross 0V.
12. Going the other way, 1mA will flow through Rh when Vp=0, the input will be sinking this 1mA. If the input is sinking, then Vin = (0 - 499 *0.001) ~= -0.5V.
13. If the input goes from 0.5 to 0V, and never reaches -0.5V, then the output will never switch. At 0V, Vp=0.475V, which is still above 0V.
A schmitt trigger is a non-linear circuit, so you have to do non-linear analysis.