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Axisymmetric Stokes Flow

Let $ r$ , $ \theta $ , $ \varphi$ be standard spherical coordinates. Consider axisymmetric Stokes flow such that

$\displaystyle {\bf v}({\bf r}) = v_r(r,\theta)\,{\bf e}_r + v_\theta(r,\theta)\,{\bf e}_\theta.$ (10.84)

According to Equations (A.175) and (A.176), we can automatically satisfy the incompressibility constraint (10.80) by writing (see Section 7.4)

$\displaystyle {\bf v} =\nabla\varphi\times\nabla\psi,$ (10.85)

where $ \psi(r,\theta)$ is the Stokes stream function (i.e., $ {\bf v}\cdot\nabla\psi = 0$ ). It follows that

$\displaystyle v_r(r,\theta)$ $\displaystyle = - \frac{1}{r^{\,2}\,\sin\theta}\,\frac{\partial\psi}{\partial \theta},$ (10.86)
$\displaystyle v_\theta(r,\theta)$ $\displaystyle = \frac{1}{r\,\sin\theta}\,\frac{\partial\psi}{\partial r}.$ (10.87)

Moreover, according to Section C.4, $ \omega_r=\omega_\theta=0$ , and

$\displaystyle \omega_\varphi(r,\theta) = \frac{1}{r}\,\frac{\partial\,(r\,v_\th...
...r}\,\frac{\partial v_r}{\partial\theta} = \frac{{\cal L}(\psi)}{r\,\sin\theta},$ (10.88)

where (see Section 7.4)

$\displaystyle {\cal L} = \frac{\partial^{\,2}}{\partial r^{\,2}} + \frac{\sin\t...
...artial}{\partial\theta}\,\frac{1}{\sin\theta}\,\frac{\partial}{\partial\theta}.$ (10.89)

Hence, given that $ \vert\nabla\varphi\vert=1/(r\,\sin\theta)$ , we can write

$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle =\nabla\times{\bf v} = {\cal L}(\psi)\,\nabla\varphi.$ (10.90)

It follows from Equations (A.176) and (A.178) that

$\displaystyle \nabla\times$$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle = \nabla\varphi\times \nabla[-{\cal L}(\psi)].$ (10.91)

Thus, by analogy with Equations (10.85) and (10.90), and making use of Equations (A.173) and (A.177), we obtain

$\displaystyle \nabla\times (\nabla\times$$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle ) = -\nabla^{\,2}$$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle = - {\cal L}^{\,2}(\psi)\,\nabla\varphi.$ (10.92)

Equation (10.83) implies that

$\displaystyle {\cal L}^{\,2}(\psi) = 0,$ (10.93)

which is the governing equation for axisymmetric Stokes flow. In addition, Equations (10.82) and (10.91) yield

$\displaystyle \nabla P = \mu\,\nabla\varphi\times \nabla[{\cal L}(\psi)].$ (10.94)


next up previous
Next: Axisymmetric Stokes Flow Around Up: Incompressible Viscous Flow Previous: Stokes Flow
Richard Fitzpatrick 2016-03-31