01112012, 11:22 AM



Location: Wilkinsburg
1,661 posts, read 1,464,406 times
Reputation: 965


So I've been working on a problem at work that deals with the flow of fluids through annular passages. The following differential equation arose from a NavierStokes derivation:
U_theta represents fluid velocity in the azimuthal direction, r is the radius, and R distance to the outside of the channel.
I know I solved a lot of ODEs like this in college, but right now I can't figure out how to solve it. I tried doing a numerical solution, as I typically do when I can't figure out an analytical solution, but unfortunately I couldn't get it to converge.
Any math nuts out there that can offer advice?

01112012, 02:10 PM



Location: Not where you ever lived
11,003 posts, read 16,417,033 times
Reputation: 5518


If you don't get an answer I can ask my math whiz.

01112012, 05:20 PM



Location: Westwood, MA
1,840 posts, read 2,098,858 times
Reputation: 1802


Let me wolframalpha that for you
Give a man a fish, he'll eat for a day. Teach a man to fish, he'll eat for a lifetime. Teach a man about wolframalpha and he'll totally forget even the most basic of ODEs ;)
DSolve[r*u''[r] + u'[r] u[r]/r==0, u[r], r]  WolframAlpha
Matching the boundary conditions you'll find C[2] = i*C[1] and C[1] = const/R giving the final, super interesting solution
u = const * (R/r)
If you're interested, that's a CauchyEuler ODE, which has a fairly straightforward solution ( Cauchy
The trick is to see that each higher derivative has one lower power of r, so a polynomial would be a good trial solution. Substituting u = x^m (after simplifying) gives
m^2  1 = 0
Which has solutions m = +/ 1 or u = C[1]*r + C[2]/r
(clearly wolfram's solver uses a different method and comes out with a different combination)

01112012, 07:22 PM



Location: Westwood, MA
1,840 posts, read 2,098,858 times
Reputation: 1802


u = const * (r/R)
is the right way, I had reversed r and R before

01112012, 08:00 PM



Location: Wilkinsburg
1,661 posts, read 1,464,406 times
Reputation: 965


Quote:
Originally Posted by jayrandom
Give a man a fish, he'll eat for a day. Teach a man to fish, he'll eat for a lifetime. Teach a man about wolframalpha and he'll totally forget even the most basic of ODEs ;)
DSolve[r*u''[r] + u'[r] u[r]/r==0, u[r], r]  WolframAlpha
Matching the boundary conditions you'll find C[2] = i*C[1] and C[1] = const/R giving the final, super interesting solution
u = const * (R/r)
If you're interested, that's a CauchyEuler ODE, which has a fairly straightforward solution ( Cauchy
The trick is to see that each higher derivative has one lower power of r, so a polynomial would be a good trial solution. Substituting u = x^m (after simplifying) gives
m^2  1 = 0
Which has solutions m = +/ 1 or u = C[1]*r + C[2]/r
(clearly wolfram's solver uses a different method and comes out with a different combination)

Ah, thanks so much! Yeah, I recognized the form, but couldn't for the life of me think how to approach it. I psyched myself out, and started thinking that if anyone did reply they were going to tell me that the solution involved Bessel functions, at which point I was going to find someone else to do the work.

Please register to post and access all features of our very popular forum. It is free and quick. Over $68,000 in prizes has already been given out to active posters on our forum. Additional giveaways are planned.
Detailed information about all U.S. cities, counties, and zip codes on our site: Citydata.com.

