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.