 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-344/1/d/solution

For

F = \int [f (ϕ) + \frac{κ ( ϕ )}{2} ∣\nabla ϕ ∣^{2}] d r,

μ = f^{'} (ϕ) - \nabla \cdot [κ (ϕ) \nabla ϕ] + \frac{κ ^{'} ( ϕ )}{2} ∣\nabla ϕ ∣^{2} = f^{'} (ϕ) - κ (ϕ) \nabla^{2} ϕ - \frac{κ ^{'} ( ϕ )}{2} ∣\nabla ϕ ∣^{2} .

The constrained minimum again makes

μ

constant. Evaluating it in either uniform bulk phase gives

μ = f^{'} (\pm ϕ_{B}) = 0

, so

μ = 0

everywhere.

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