Inductor and Capacitor story

This was a fun problem. I solved numerically using two different circuits: one prior to switch closing (left of diagram) and one after switch closing (right of diagram). The differential equations are shown on each side. Prior to switch closing, the two inductors are lumped into one. The simulation is run for the first circuit until the time derivative of the inductor current becomes zero. At that point, currents $I_{L1}$ and $I_{L2}$ are initialized to the value of $I_{LS}$ , and the program switches over to the second circuit model and looks for the maximum switch current. This turns out to be $\frac{q_0}{\sqrt{2 L C}}$ .

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import math

dt = 10.0**(-5.0)

q0 = 10.0

C = 1.0
L = 1.0

C1 = 2.0*C
C2 = C

L1 = L
L2 = 2.0*L

LS = L1 + L2

###########################################

# Simulate up until switch closing
# While rate of change of inductor current is positive

t = 0.0
count = 0

VC1 = q0/C1
VC2 = 0.0
ILS = 0.0

VC1d = -ILS/C1
VC2d = -ILS/C2
ILSd = (VC1 + VC2)/LS

while ILSd >= 0.0:

    VC1 = VC1 + VC1d*dt
    VC2 = VC2 + VC2d*dt
    ILS = ILS + ILSd*dt

    VC1d = -ILS/C1
    VC2d = -ILS/C2
    ILSd = (VC1 + VC2)/LS

    t = t + dt
    count = count + 1

###########################################

# Simulate period after switch closing

tclose = t

IL1 = ILS
IL2 = ILS

VC1d = -IL1/C1
VC2d = -IL2/C2
IL1d = VC1/L1
IL2d = VC2/L2

Isw_max = 0.0

while t <= tclose + 20.0:

    VC1 = VC1 + VC1d*dt
    VC2 = VC2 + VC2d*dt
    IL1 = IL1 + IL1d*dt
    IL2 = IL2 + IL2d*dt

    Isw = IL1 - IL2

    VC1d = -IL1/C1
    VC2d = -IL2/C2
    IL1d = VC1/L1
    IL2d = VC2/L2

    t = t + dt
    count = count + 1

    if Isw > Isw_max:
        Isw_max = Isw

    #if count % 1000 == 0:
        #print t,IL1,IL2,Isw

###########################################

print dt
print t
print Isw_max

#>>> 
#1e-05
#22.2214599993
#7.07142124991
#>>> 

Initial circuit equation :

$3L\dfrac{d^2q_1}{dt^2}+\dfrac{3q}{2C}=0\implies \dfrac{d^2q_1}{dt^2}=-\omega^2q_1$ . ( $\omega =\frac{1}{\sqrt {2LC}}$ ).

Substituting initial conditions, we get

$q_1=q_0\cos (\omega t), I_{max}=\omega q_0$ .

Final circuit equation :

$L\dfrac{d^2q_2}{dt^2}=-\dfrac{q_2}{2C}\implies \dfrac{d^2q_2}{dt^2}=-\omega^2 q_2$ .

Again, substituting initial conditions we get

$q_2=q_0\sin (\omega t) \implies I_{final}=\omega q_0\cos (\omega t)$ .

Hence the maximum final current is

$\omega q_0=\dfrac{q_0}{\sqrt {2LC}}$ .

So, $α=1,β=2,α+β=1+2=\boxed 3$ .