 Solution

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

For a planar equilibrium, multiply

μ = 0

ϕ^{'}

and use the chain rule. The result is

\frac{d}{d x} [f (ϕ) - \frac{κ ( ϕ )}{2} (ϕ^{'})^{2}] = 0.

Matching to either bulk phase fixes the constant:

f (ϕ) - \frac{κ ( ϕ )}{2} (ϕ^{'})^{2} = f (ϕ_{B}) .

Consequently

\frac{d x}{d ϕ} = \pm \frac{κ ( ϕ )}{2 [ f ( ϕ ) - f ( ϕ _{B} )]},

and the inverse profile is

x (ϕ) - x_{0} = \pm \int_{0}^{ϕ} \frac{κ ( ψ )}{2 [ f ( ψ ) - f ( ϕ _{B} )]} d ψ .

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