I am trying to learn how differential amplifiers work, and I would like to be able to visualize how the voltages and currents stabilize if the input voltages are changed. Specifically, I'm caught up on the circular dependency between the currents going through either transistor Q1 or Q2, and the source terminal voltage which is connected for them.

I do understand that what's happening with the shared source voltage is it'sits changing to be in balance with the currents coming through the transistors (which is capped by the source current below). But I'm used to thinking in terms of one transistor, where I deduce the current from knowing the terminal voltages. In this case, it seems to be the source terminal voltage follows from the currents, but the currents also still follow from the voltage \$v_{GS}\$ which is a function of \$v_{IN}\$ (fixed) and \$v_{S}\$ (not fixed).

I tried solving for this source terminal voltage to see if this would shed any insight but the result was messy:

$$I_S =I_1 + I_2$$

$$i_{D1} = \frac{K}{2}(v_{GS1}-V_T)^2$$

which is equal to:

$$\frac{K}{2}(v_{GS1}-V_T)^2 = \frac{K}{2}((v_{IN1}-V_T)-v_{S})^2$$

and so the same for the other transistor

$$i_{D2} = \frac{K}{2}((v_{IN2}-V_T)-v_{S})^2$$

combining these and solving for the quadratic equation, I got:

$$ v_{S} = \frac{2(v_{IN1} + v_{IN2} - 2V_T) \pm \sqrt{(2(v_{IN1} + v_{IN2})^2 - 8(\frac{2 I_S}{K}-(v_{IN1}-V_T)^2-(v_{IN2}-V_T)^2))}}{4} $$

This didn't really help me get a much better sense of what's happening.

What I'd like

is to be able to roughly imagine how the electric fields propagate from the disturbance (change in either voltage input), in order to understand how the circuit arrives at a new equilibrium.