Noninverting schmitt trigger circuit LTP, UTP calculation
The voltage at the \$V^+\$ is actually, 

$$V^+=V_i\dfrac{R_2}{R_1+R_2}+V_o\dfrac{R_1}{R_1+R_2} $$

You can find that using superposition, for example. So if \$V_i\$ is indeed 0, you end up with what you've shown so far.

Now here is the thing, if doesn't take much to saturate the OP-Amp output to either +VSAT or -VSAT.  Under negative feedback, the OP-Amp is forced to work under the linear region, not the case under positive feedback. So if \$V^+ &gt; V^-\$ , the output is +VSAT, if \$V^+ &lt; V^-\$, the output is  -VSAT.

The trick here is that you have to assume an intial state for the output, either +VSAT or -VSAT. Say, it's +VSAT, then at the \$V^+\$ node, you'll have:

$$V^+=V_i\dfrac{R_2}{R_1+R_2}+V_{SAT}\dfrac{R_1}{R_1+R_2} $$

Since the \$V^-\$ node is fixed at ground, the OP-Amp output will remain at +VSAT so long as:

$$V_i\dfrac{R_2}{R_1+R_2}+V_{SAT}\dfrac{R_1}{R_1+R_2}&gt;0 $$

$$V_i\dfrac{R_2}{R_1+R_2}&gt;-V_{SAT}\dfrac{R_1}{R_1+R_2} $$
$$V_iR_2&gt;-V_{SAT}R_1$$
$$V_i&gt;-V_{SAT}\dfrac{R_1}{R_2}$$

So, as long as \$V_i\$ stays above that, the output will remain at +VSAT. If \$V_i\$ starts to decrease such that it doesn't meet the condition above, then it means that \$V^+ &lt;V^-\$ and the output will switch over to -VSAT. You can follow the same procedure as above to find what the threshold would be to get it back to +VSAT.