Stokes’ 1st

Stokes' first problem, also erroneously called Rayleigh flow:

$$\hbox{\epsffile{figures/7-7.ps}}$$

Continuity:

$$\mbox{div}\left(\vec v\right) = 0 = \frac{\partial u}{\par... ...ac{\partial v}{\partial y} \quad\Longrightarrow\quad u=u(y,t)$$

$y$-momentum:

$$\rho \rlap{\kern 3pt\smash{\Bigg/}}\frac{Dv}{Dt} = - \rho g... ...rrow\quad p = -\rho g y + P(\rlap{\kern 0pt\smash{\bigg/}}x,t)$$

$x$-momentum:

$$\rho \frac{\partial u}{\partial t} + \rho u \rlap{\kern 3p... ...u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)$$

The $x$-momentum equation becomes:

$$u_t = \nu u_{yy}$$

where $\nu=\mu/\rho$ is the kinematic viscosity.

$$\hbox{\epsffile{figures/7-7b.ps}}$$

Exercise:

How would you normally find $u$?

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A simpler way to solve is to guess that the solution is similar: after rescaling $u$ and $y$, all velocity profiles look the same.

Original profiles:

$$\hbox{\epsffile{figures/7-7c.ps}}$$

Supposed shape after scaling $u$ with $V_0$, and $y$ with a characteristic boundary layer thickness $\delta$ that increases with time:

$$\hbox{\epsffile{figures/7-7d.ps}}$$

Mathematical form of the similarity assumption:

$$\frac{u}{V_0} = f\left(\frac{y}{\delta(t)},\rlap{\kern -3pt\smash{\bigg/}}t\right)$$

The proof is in the pudding; if it satisfies the P.D.E., I.C., and B.C., it is OK.

$$u_t = \nu u_{yy} \quad\Longrightarrow\quad -V_0 f' \frac{y}{\delta^2} \delta_t = \nu V_0 f'' \frac{1}{\delta^2}$$

Put $\eta=y/\delta$:

$$-V_0 f' \eta \frac{\delta_t}{\delta} = \nu V_0 f'' \frac{1}{\delta^2}$$

Separate into terms depending only on $\eta$ and terms depending only on $t$:

$$- \frac{f' \eta}{f''} = \frac{\nu}{\delta \delta_t} = \mbox{constant} = \frac12$$

It does not make a difference what you take the constant; this merely changes the value of $\delta$, not the physical solution.

Solving the O.D.E.s for $\delta$ and $f$, we solve the P.D.E. For the boundary layer thickness $\delta \delta_t = 2 \nu$ so

$$\delta = \sqrt{4\nu t}$$

For the velocity profile $f''=-2\eta f'$ hence

$$f = \mbox{erfc}(\eta)$$

where erfc is the complementary error function defined as

$$\fbox{$\displaystyle \mbox{erfc}(x) \equiv \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-\xi^2}\d \xi$}$$

Exercise:

Derive the expressions for $\delta$ and $f$.

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Total:

$$\fbox{$\displaystyle u = V_0 \mbox{ erfc}\left(\frac{y}{\delta}\right) \qquad \delta = \sqrt{4\nu t} $}$$

You should now be able to do 7.14, 16, 17