 Solution

ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2021/6/b/solution

Put

m = π / b

. With

r = 0

v (y) = \frac{F _{0}}{β} cos (m y) .

At an impermeable eastern wall,

u (L, y) = 0

. Continuity implies

u_{x} = - v_{y} = \frac{m F _{0}}{β} sin (m y), u (x, y) = \frac{m F _{0}}{β} (x - L) sin (m y) .

Equivalently,

ψ (x, y) = \frac{F _{0}}{β} (x - L) cos (m y)

satisfies the impermeability condition with the boundary streamfunction set to zero.

x = 0

, this solution has

u (0, y) = - m F_{0} L sin (m y) / β

, generally nonzero. A useful integral proof of the obstruction is

\frac{d}{d y} \int_{0}^{L} v d x = - [u (L, y) - u (0, y)] = 0

for two impermeable walls, whereas the proposed interior gives

\int_{0}^{L} v d x = L F_{0} cos (m y) / β

, which varies with

y

. Friction or another departure from the inviscid model is needed for a return flow that closes the circulation.

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