I've reduced a portion of the Navier Stokes to solve a flow problem, and am left with the following ODE:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]u (\frac {\partial^2 Vz} {\partial^2 r}) + \frac {r} {u}\frac {\partial Vz} {\partial r} = 0 [/tex]

I tried to solve this equation by assuming a power law solution with

[tex]Vz = Cr^n [/tex]

Which yields

[tex] n(n-1)r^{n-2} + nr^{n-2} = 0 [/tex],

Thus n^2 - n + n = 0

Which seems to indicate [tex] n^2 =0 [/tex] => n = 0

So Vz = C. But, I don't think this is what physically happens, Iam expecting a radial profile and not a flat profile, so I'm looking for where I took a wrong turn. Any ideas? Thanks.

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# Homework Help: Solving a Simple ODE from the Navier-Stokes

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