I am stuck on this problem:

Incompressible N-S

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

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

Incompressible,

$\displaystyle \vec{b} = 0$

$\displaystyle a < r < b$

Pressure is uniform with

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

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

$\displaystyle u=v=0$

$\displaystyle w=w(r)$

Show that

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

with general solution

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

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

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

__________________________________________________ _________

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

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

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

meaning

$\displaystyle w = {w}_{c} + {w}_{p} $

plugging w back into the differential equation I get

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

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 $\displaystyle B \log{r}$ term.