 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/31e/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write the normal-form equation as

u^{''} = Q (x) u, Q (x) = \frac{1}{x ^{4}} - \frac{1}{4 x ^{2}} .

For sufficiently small positive

x

Q (x) > 0

. Moreover,

\frac{∣ Q ^{'} ( x ) ∣}{Q ( x ) ^{3/2}} = O (x) ⟶ 0,

and the corresponding higher derivative condition also tends to zero, so the amplitude varies slowly relative to the exponential phase and the Liouville-Green approximation applies.

The phase and amplitude have expansions

Q (x) Q (x)^{- 1/4} \int^{x} Q (s) d s = x^{- 2} (1 - \frac{x ^{2}}{8} - \frac{x ^{4}}{128} + O (x^{6})), = x (1 + \frac{x ^{2}}{16} + O (x^{4})), = - \frac{1}{x} - \frac{x}{8} - \frac{x ^{3}}{384} + O (x^{5}) .

Hence two independent Liouville-Green solutions satisfy

u_{\pm} (x) \sim Q (x)^{- 1/4} exp (\pm \int^{x} Q (s) d s) .

In particular, after choosing the signs according to growth and decay,

u_{grow} (x) \sim x e^{1/ x} (1 + \frac{x}{8} + \frac{9 x ^{2}}{128} + \dots), u_{decay} (x) \sim x e^{- 1/ x} (1 - \frac{x}{8} + \frac{9 x ^{2}}{128} + \dots) .

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