Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 4 / ii

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- Solution ii

Solution

At fixed time write the azimuthal shift as

δ = 2 A L_{x} t

L_{y}

. Integrating a radial derivative over the box gives a boundary difference proportional to

\int_{0}^{L_{z}} \int_{0}^{L_{y}} [Q (L_{x}, y, z) - Q (0, y, z)] d y d z = \int_{0}^{L_{z}} \int_{0}^{L_{y}} [Q (0, y + δ, z) - Q (0, y, z)] d y d z .

The two integrals agree by translation invariance of an integral over a full azimuthal period. Hence

⟨ \partial_{x} Q ⟩ = 0

. The ordinary periodic boundary conditions make the

y

and

z

boundary differences zero as well. Finally,

\int_{0}^{L_{y}} x \partial_{y} Q d y = x [Q (x, L_{y}, z) - Q (x, 0, z)] = 0

for every

x, z

. Thus the volume averages in a shearing box satisfy

⟨ \partial_{x} Q ⟩ = ⟨ \partial_{y} Q ⟩ = ⟨ x \partial_{y} Q ⟩ = ⟨ \partial_{z} Q ⟩ = 0 .

For smooth shearing-periodic extensions, spatial derivatives and products inherit the same shearing-periodic boundary condition. Their divergence therefore has zero volume average, and integration by parts is available without a residual boundary flux. Time differentiation of an average over this fixed physical box commutes with integration, even though the radial identification shifts with time.

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