Fluid mech problem

[Swordmaster Dragon]

This is a problem from my girlfriend's fluid mechanics problem set that both she and I have been having trouble with. I'm posting it here because while the setup is physical, the primary issue is mathematical. (Also, in the tradition of all good theoretical physics problems, the setup is absolutely ludicrous.)

Problem: A spherical hole of radius a suddenly forms in an incompressible fluid filling all space. Determine the time for the hole to be filled with fluid.

Subproblem: We think that the equation governing the boundary of the vacuum/hole/thingy R as a function of time t is
R*R'' + 3/2*(R')^2 = -P_0/rho
where P_0 is the pressure in the fluid at infinity (ambient pressure) and rho is the density. The RHS can also be written as -c^2, c being the speed of sound in the fluid. With this are the initial conditions
R'(0) = 0 (fluid initially at rest)
R(0) = a.
Note: It is certainly possible we messed up reasoning and/or algebra. If you're more interested in the physics, feel free to check these results. If you're just interested in the math, try to solve this equation or whatever corrected version someone else posts. I might post the derivation of the above equation after I work through it again.

The direct approach to this would be to solve the above ODE and invert it to find R(t1) = 0. In trying to do this I've found a 2-dimensional symmetry algebra and reduced it to a first-order equation, but I'm still struggling to solve that one.

However, if someone can find an indirect approach that can get the time t1 for the hole to close *without* actually solving for R, you will earn my everlasting respect and 100 internets. That would be totally slick.  

[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.  

[Swordmaster Dragon]

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?