Your formulas are correct, but I'm not sure where you are getting your numbers from.
Based on the 18V output voltage, the capacitor energy at the time of the break is 129.6mJ and you are adding 0.77mJ to it, so the capacitor voltage will change from 18V to 18.054V if all the energy of a pulse finds its way into the capacitor. Of course if more pulses come along that will add up quickly at 100kHz.
In general, \$\Delta V = -V +\sqrt{V^2 + I^2(\frac{L}{C})}\$
It may be more useful to focus on the inductor:
First assume it's fairly large and the switching frequency is fast.
Now the current through the inductor will remain fairly close to constant and always in the same direction.
With the switch ON, the current will gradually increase, from:
$$V =L \frac{dI}{dt}$$ where $$V = V_s-V_{out}$$
Because \$L\$ is large, \$\frac{dI}{dt}\$ is fairly small.
With the switch OFF, \$V\$ becomes \$0 - V_{out}\$ (negative) or more strictly, \$-0.6-V_{out}\$, so \$\frac{dI}{dt}\$ is negative, and the current gradually reduces.
And that's pretty much it. (for a buck operating in CCM, Continuous Current Mode)
Oh - the capacitor - it just helps Vout remain close to constant, despite the variation in incoming current.
(Switch OFF for too long and the curent eventually falls to \$0\$ : the maths for Discontinous Current Mode, DCM, are different.)
Best Answer
The average value of the output of a buck converter in discontinuous mode is still the same as continuous mode. However, the ripple voltage increases.
This happens on light loads when the switcher cannot produce pulses of low enough duty cycle and it transfers a little too much energy per switching cycle for the light load.
This means the average output voltage starts to rise too high and the control system shuts down the switcher for several cycles resulting in a slightly lower than normal output voltage. Until the load starts to take more power this situation persists.
Discontinuous mode also happens when the input voltage to the buck rises too high.
EDIT - more information:-
If the buck converter is operating at 100kHz and the inductor is 10uH. Let's also say that the supply is 12V and duty cycle is 50%. The "on" pulse of the mosfet will charge the inductor with current and this current is determined by input voltage, output voltage and inductance. Let's say the output voltage is 5V - this means the voltage across the inductor is 7V when the mosfet is "on".
V = L\$\dfrac{di}{dt} \therefore 7 = 10\times 10^{-6} \times\dfrac{di}{dt} \therefore\dfrac{di}{dt} =\space \$700,000
Because dt is 50% of 10us we can calculate I, which is \$700000\times 5\times 10^{-6}\$ = 3.5A.
The energy transferred per cycle is therefore \$\dfrac{L\times 3.5^2}{2}\$ = 57.8uJ.
This transfers 100k times per second \$\therefore\$ the power to the load is \$ 57.8\mu J \times 100000\$ = 5.78W.
If the load resistor is too high to take that power at 5V then the simple fact is that the output voltage will rise and the buck convertor will enter discontinuous mode unless the control loop reduces the duty cycle.
This duty cycle reduction will of course happen because it would be a poor buck converter that couldn't produce less than 50% duty but, it will have a minimum value and at this point, if the energy-load equation isn't in equilibrium the converter will enter discontinuous mode.