 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-154/2/2/3/solution

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

I_{χ} (t) = \int χ ∣ u ∣^{2}, V_{χ} (t) = Im \int \nabla χ \cdot \nabla u \overline{u} .

Substituting

u_{t} = i (Δ u + ∣ u ∣^{p - 1} u)

and integrating by parts gives the localized virial identity

\frac{1}{2} I_{χ}^{'} (t) = V_{χ} (t) .

Differentiating once more, integrating the Laplacian terms twice, and observing that the gauge-invariant nonlinearity contributes only through

Δ χ

, gives

\frac{1}{2} V_{χ}^{'} (t) = \int Hess χ (\nabla u, \nabla \overline{u}) - \frac{1}{4} \int Δ^{2} χ ∣ u ∣^{2} - (\frac{1}{2} - \frac{1}{p + 1}) \int Δ χ ∣ u ∣^{p + 1} .

These are the required formulas, with

χ^{''} ∣\nabla u ∣^{2}

interpreted as the Hessian quadratic form for a radial weight.

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