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I have this capacitance network in my circuit and I can't figure out the math behind why the final steady-state value of Vx=0.542v when SW2 closes to the other position. Coincidentally I found out this is Vx - 1.(c2 /(c1+c2+c3)) but I would like an intuitive approach. Please ignore the absence of resistance in the circuit to make it practical.
As P2000 suggests, if the capacitors are fully discharged prior to applying the 1V DC source then capacitors in series will have the same charge (Q=CV).
C1 and C2 are initially in parallel for an equivalent capacitance of 31fF. So, the voltage across the parallel combination of C1 and C2 will be V = Q/31fF. The voltage across C3 will be Vx = Q/31fF. From these 2 equations you can see that V = Vx in this state, which leads to Vx = 0.5V. You can reach this same conclusion by applying Laplace transform circuit theory where you would end up with $$Vx = \frac{1}{s}\frac{C3}{Ce+C3}$$ where Ce is the parallel combination of C1 and C2. This is clearly = 0.5 for initial case. The charge on C2 will be Q2 = C2V2 = 8f Coulombs. The charge on C3 will be Q3 = C3Vx = 15.5f Coulombs.
Next, however, the switch is thrown and C2 is placed in anti-parallel with C3. With the new parallel combination being in series with C1 and the source. The easiest way to find the resulting voltage is Laplace circuit theory. Using the following equivalent for the capacitors you will be left with a circuit with 4 sources - you can solve this with normal circuit methods(e.g. superposition). The 1V source becomes 1/s in frequency domain.
I solved by hand and i get Vx = C2/(C1+C2+C3).
All that said, the absence of resistance actually makes the problem impractical. Numerical solvers won't like it either because you need infinite current.
