 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/23f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

g = h e^{- Φ/2}, V = \frac{∣ Φ ^{'} ∣ ^{2}}{4} - \frac{Φ ^{''}}{2} .

\int_{R} ∣ h^{'} ∣^{2} e^{- Φ} = \int_{R} g^{'} + \frac{Φ ^{'}}{2} g^{2} = \int_{R} ∣ g^{'} ∣^{2} + \int_{R} V ∣ g ∣^{2} .

The boundary term vanishes because

h

, and hence

g

Since

V (x) \to ℓ > 0

, choose

R_{0}

so that

V \geq ℓ /2

when

∣ x ∣ \geq R_{0}

, and set

λ_{1} = \frac{ℓ}{2}, K_{1} = max {0, - min_{∣ x ∣ \leq R_{0}} V (x)} .

For every

R \geq R_{0}

, discard the nonnegative

∣ g^{'} ∣^{2}

term and the nonnegative contribution from

R_{0} \leq ∣ x ∣ \leq R

to obtain

\int_{R} ∣ h^{'} ∣^{2} e^{- Φ} \geq λ_{1} \int_{∣ x ∣ \geq R} ∣ g ∣^{2} - K_{1} \int_{∣ x ∣ \leq R} ∣ g ∣^{2} .

Because

∣ g ∣^{2} = ∣ h ∣^{2} e^{- Φ}

, this is exactly the required estimate.

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