The Schmitt Trigger is a comparator circuit with hysteresis.

We need to find out what the trigger point of the circuit is given different values of $v_{OUT}$. The output will switch when $v_+$ goes above or below ground. Remember the transfer function of an Op-Amp is $A(v_{+}-v_{-})$ where A is a very large number (> 1e3). A small difference between the inputs will create a large change in the output. The output can't grow forever. The output can't go above or below the supply voltages, so it stops there.

Let's build an expression for v_{+}.

The current flowing from $v_{out}$ to $v_{in}$ is given by Ohm's Law:

\[i=\frac{v_{OUT}-v_{IN}}{R_1+R_2}\]

We can use Ohm's law to find $v_{+}$ too.

\[v_{+} = v_{IN} + \frac{(v_{OUT}-v_{IN})R_1}{R_1+R_2} \]

Rearranging the terms…

\[v_{+} = \frac{v_{OUT}R_1}{R_1+R_2} + \frac{v_{IN}R_2}{R_1+R_2} \]

The Schmitt Trigger switches when $v_{+}$ cross zero.

\[0 = \frac{v_{OUT}R_1}{R_1+R_2} + \frac{v_{IN}R_2}{R_1+R_2} \]

Finally we get

\[-v_{OUT}\frac{R_1}{R_2}=v_{in}\]