Incompressible Navier-Stokes Problem

**gamerman315** · May 10th 2011, 03:38 PM

I am stuck on this problem:

Incompressible N-S

$\frac{d\vec{v}}{dt} + (\vec{v} \cdot \nabla)\vec{v} = - \nabla P + \vec{b} + \mu {\nabla }^{2} \vec{v}$

$\mathrm{div}{\vec{v}} = 0$

Incompressible,

$\vec{b} = 0$

$a < r < b$

Pressure is uniform with

$\frac{\mathrm{dP} }{\mathrm{dz} } = -m$

$\vec{v} = u \vec{{i}_{r}} + v \vec{{i}_{\theta}} + w \vec{{i}_{z}}$

$u=v=0$

$w=w(r)$

Show that

$\frac{\mathrm{{d}^{2}w}}{\mathrm{d{r}^{2}}} + \frac{1}{r}\frac{\mathrm{dw}}{\mathrm{dr}} = -\frac{m}{\mu}$

with general solution

$w = A + B \log{r} - \frac{1}{4}\frac{m}{\mu}{r}^{2}$

Choose A and B to satisfy the no-slip because at r = a, r = b. Derive

$w(r) = \frac{1}{4}\frac{m}{\mu}[({a}^{2} - {r}^{2}) + ({b}^{2} - {a}^{2})\frac{\log{\frac{r}{a}}}{\log{\frac{b}{a}} }}]$

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Since posting this I have gotten farther, solving the differential equation as a linear nonhomogenious equation I get

${w}_{c}={c}_{1} + \frac{{c}_{2}}{\exp}$

${w}_{p}={c}_{3} \cdot r$

meaning
$w = {w}_{c} + {w}_{p}$

plugging w back into the differential equation I get

${c}_{3} = -\frac{m \cdot r}{4 \cdot \mu}$

this equation does satisfy the differential equation but as far as matching the equation for w above I dont understand how they are getting the $B \log{r}$ term.

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