VorpalNeko
Your differential equation is correct for an inviscid fluid of negligible surface tension. Starting with
[1] RR" + (3/2)R'² = -P/ρ
and applying the integration factor R²R' gives
[2] (d/dt)[ R³R'² ] = -2(P/ρ)R²R',
integrating which results in
[3] R'² = (2/3)(P/ρ)(a³/R³-1).
This is probably the first-order equation you're talking about. Substituting u = R³/a³ gets you Int_0^1[ u^{-1/6} (1-u)^{-1/2} du ] = Β(1/2,5/6). The end result is
[4] T = Β(1/2,5/6) (a/3) sqrt[ (3/2)(ρ/P) ],
where Β(x,y) is the beta function. If you pull all the constants together, you get about 0.9147.
I wonder whether including Laplace pressure 2σ/R still gives a nice solution.
Heh, I had come to the same answer in a *much* more roundabout manner. I also hadn't thought about looking up possible representations for that integral. Since this is a problem from the second week of an intro fluids class, you're right to assume inviscid fluid with no surface tension
xd If anyone still wants to work on the problem, I think we'd be justified in adding a Laplace pressure. We could also try to find the velocity field in the rest of the fluid before the hole closes or track the pressure shockwave afterwards.
I think it's interesting that the final coefficient of 0.9 < 1 implies that at some point the boundary of the hole starts traveling faster than the speed of sound. Maybe we can try to figure out exactly when?
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