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

  1. Nov 5, 2006 #1
    I've reduced a portion of the Navier Stokes to solve a flow problem, and am left with the following ODE:

    [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.
     
    Last edited: Nov 5, 2006
  2. jcsd
  3. Nov 6, 2006 #2

    dextercioby

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    HINT: Substitute [itex] \frac{\partial V_{z}}{\partial r} [/itex] by a new function, let's call it [itex] f(r,...) [/itex]. Then see what you get.

    Daniel.
     
  4. Nov 6, 2006 #3
    So, using the hint I get:

    uf''(r) + (u/r)f'(r) = 0

    Assuming an exponential function,
    [tex]f(r) = e^{nr}[/tex]
    [tex]f'(r) = ne^{nr}[/tex]
    [tex]f''(r) = n^2e^{nr}[/tex]

    Thus

    [tex]un^2e^{rn} + (u/r)ne^{rn} = 0[/tex]
    [tex]u(n^2+1/rn) = 0[/tex]
    [tex]un(n+1/r) = 0[/tex]
    [tex]n = 0, -1/r[/tex]

    So:
    [tex]Vz = Ae^{0} + Be^{-1/r} [/tex]

    Is that the correct methodology? Thanks again.
     
    Last edited: Nov 6, 2006
  5. Nov 7, 2006 #4

    dextercioby

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    I get the eq for "f"

    [tex] u\frac{df}{dr}+\frac{r}{u}f=0 [/tex]

    with the solution

    [tex] f(r)=Ce^{-\frac{r^{2}}{2u^{2}}} [/tex]

    and then finally

    [tex] V(r,\vartheta)=C\int Ce^{-\frac{r^{2}}{2u^{2}}} \ dr + g(\vartheta) [/tex]

    The integral brings in the erf function, while the angle dependence should be determined by boundary conditions.

    Daniel.
     
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