01112012, 11:22 AM



Location: Wilkinsburg
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Simple Differential Equation That For Some Reason Eludes Me
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
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If you don't get an answer I can ask my math whiz.

01112012, 05:20 PM



Location: Westwood, MA
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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
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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 1,154,832 times
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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.

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