Solution

The initially irrotational inviscid flow remains irrotational, so

u = \nabla ϕ

; incompressibility gives

\nabla^{2} ϕ = 0

. At the plates, normal velocity matches their motion:

\frac{1}{r} ϕ_{θ} (r, α) = - Ω r, \frac{1}{r} ϕ_{θ} (r, - α) = Ω r .

With

ϕ = r^{2} f (θ)

f^{''} + 4 f = 0

. The symmetric solution is

ϕ = \frac{Ω r ^{2} c o s 2 θ}{2 s i n 2 α} .

Hence

u_{r} = \frac{Ω r c o s 2 θ}{s i n 2 α}, u_{θ} = - \frac{Ω r s i n 2 θ}{s i n 2 α} .

A streamfunction is

ψ = \frac{Ω r ^{2} s i n 2 θ}{2 s i n 2 α},

r^{2} sin 2 θ = constant

and fluid is expelled radially as the plates close. The outward flux through

r = R

\int_{- α}^{α} u_{r} R d θ = Ω R^{2} .

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