Solving a Simple ODE from the Navier-Stokes

[Peregrine]

I've reduced a portion of the Navier Stokes to solve a flow problem, and am left with the following ODE:

u (\frac {\partial^2 Vz} {\partial^2 r}) + \frac {r} {u}\frac {\partial Vz} {\partial r} = 0

I tried to solve this equation by assuming a power law solution with
Vz = Cr^n

Which yields

n(n-1)r^{n-2} + nr^{n-2} = 0,

Thus n^2 - n + n = 0

Which seems to indicate n^2 =0 => 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.

 

[dextercioby]

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

Daniel.

 

[Peregrine]

So, using the hint I get:

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

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

Thus

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

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

Is that the correct methodology? Thanks again.

 

[dextercioby]

I get the eq for "f"

u\frac{df}{dr}+\frac{r}{u}f=0

with the solution

f(r)=Ce^{-\frac{r^{2}}{2u^{2}}}

and then finally

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

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

Daniel.