I came across this problem while I was working through diode circuits in my textbook, doing problems where we have to graph the output voltage as a function of the input voltage or current. I approached the problem by looking at the end behavior of the function, when I_in is a really large positive or negative value. So when I_in is positive infinity, D2 is on and D1 is off and when I_in is negative infinity, D1 is on and D2 is off.
I thought that at the point where a diode turns on/off, the circuit is between two states and its behavior isn't really defined in a constant voltage model of a diode circuit. So I thought that the circuit has to be in both states and as a result, the two portions of the graph representing those two states must intersect at the point where the diode transitions. In other words, Vout must be a continuous function of I_in.
I found that when D1, D2 are off, Vout is 0V and when D1, D2 are on, Vout is the turn on voltage of D2. This meant in order to ensure continuity in between the two "edge states", D1 and D2 had to turn on and they never could have both been turned off (in general, I also think that a circuit with n diodes can only have n+1 "portions" in its graph).
With this, I was able to conclude that for -infinity < I_in < 2VD_on/R1, Vout = I_inR_1 - V_DON and for 2*VD_on/R1 < I_in < V_DON, Vout = V_DON.
However, to check my work, I simulated the circuit and found that I had the right equations for the graph but the transition between the two pieces happened at I_in = 0 so that at this point, V_out is both -V_DON and V_DON.
So now I'm really confused because this is the first circuit where my approach didn't work and I'm not sure how to get the actual answer if we don't have continuity of the graph.