For a unidirectional,

z

-independent velocity

u = u_{z} (r, θ) e_{z},

incompressibility holds automatically. The axial Stokes flow equation is

0 = - p_{z} + μ \nabla_{⊥}^{2} u_{z} .

No external pressure gradient is imposed, so

p_{z} = 0

and

\nabla_{⊥}^{2} u_{z} = 0 .

Thus the axial velocity is a harmonic function in the disk.

ii

Solution

The boundary data are

U sgn (sin θ)

sgn (sin θ) = \frac{4}{π} \sum_{n \geq 1 n odd} \frac{s i n ( n θ )}{n} .

Regular harmonic modes in a disk are

(r / R)^{n} sin (n θ)

and

(r / R)^{n} cos (n θ)

. Matching the odd boundary data gives

u_{z} (r, θ) = \frac{4 U}{π} n \geq 1 n odd \sum \frac{1}{n} (\frac{r}{R})^{n} sin (n θ) .

Every term is regular at

r = 0

, and the series approaches the prescribed values at every boundary point away from the two jump discontinuities.

For

∣ z ∣ < 1

\sum_{n \geq 1 n odd} \frac{z ^{n}}{n} = \frac{1}{2} lo g \frac{1 + z}{1 - z} .

z = (r / R) e^{i θ}

and take the imaginary part. Since

r < R

, the real part of the relevant denominator is positive, and the result is

u_{z} (r, θ) = \frac{2 U}{π} arctan (\frac{2 R r sin θ}{R ^{2} - r ^{2}}) .

r \to R^{-}

this tends to

U

for

0 < θ < π

and to

- U

for

- π < θ < 0

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