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Math Help - Incompressible Navier-Stokes Problem

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    Incompressible Navier-Stokes Problem

    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}}  }}]

    __________________________________________________ _________

    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.
    Last edited by gamerman315; May 11th 2011 at 04:30 AM.
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