3B (Vector Calculus)

[\nabla \times (ψ A)]_{i} = ϵ_{ijk} \partial_{j} (ψ A_{k}) = ψ ϵ_{ijk} \partial_{j} A_{k} + ϵ_{ijk} (\partial_{j} ψ) A_{k} .

This is

\nabla \times (ψ A) = ψ \nabla \times A + \nabla ψ \times A .

Similarly,

\nabla \cdot (A \times B) = \partial_{i} (ϵ_{ijk} A_{j} B_{k}) = ϵ_{ijk} (\partial_{i} A_{j}) B_{k} + ϵ_{ijk} A_{j} (\partial_{i} B_{k}) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B),

after cyclically relabelling the first term and swapping two indices in the second. Hence

\nabla \cdot (A \times B) = B \cdot \nabla \times A - A \cdot \nabla \times B .

b

The data are rotationally symmetric, so seek a radial solution. The radial Laplacian in two dimensions gives

\frac{1}{r} (r u^{'})^{'} = 16 r^{2} .

Twice integrating,

u (r) = r^{4} + C lo g r + D .

Regularity at the center forces

C = 0

, and

u (1) = 1

gives

D = 0

. Thus

u (r) = r^{4} .

On the annulus the logarithmic term is allowed, so

u (r) = r^{4} + C lo g r + D .

The condition

u (1) = 1

gives

D = 0

, while

u (2) = 1

gives

16 + C lo g 2 = 1.

Therefore

u (r) = r^{4} - \frac{15 lo g r}{lo g 2} .

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