Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 1 / g

Table of contents
- Solution g

Solution

For the quartic free energy,

f (ϕ) - f (ϕ_{B}) = \frac{b}{4} (ϕ^{2} - ϕ_{B}^{2})^{2} .

With constant

κ

, part e gives, between the two bulk values,

\frac{d x}{d ϕ} = \pm \frac{2 κ / b}{ϕ _{B}^{2} - ϕ ^{2}} .

Integration gives

x - x_{0} = \pm \frac{2 κ / b}{ϕ _{B}} artanh \frac{ϕ}{ϕ _{B}} = \pm ξ_{0} artanh \frac{ϕ}{ϕ _{B}},

whose inversion is exactly the hyperbolic-tangent profile in part c.

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