Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-69/2/ii/solution
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 69 2 ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
Work first with in the Schwartz space, so that all Fourier manipulations and spectral contour integrals are justified; weaker classes follow by the usual density or distribution arguments. DefineThe phase is purely imaginary. Since , setting reduces the spectral equation to . The whole-plane Cauchy-Pompeiu formula therefore constructs the solution decaying spatially at infinity:The freedom to add times an entire function is removed by this decay condition. For every fixed , the integral is a spatial Cauchy-Green operator applied to a modulated source.
Now differentiate in the conjugate spectral parameter. The two exponential derivatives produce , canceling the Cauchy denominator, soThis is the spectral dbar equation: its right-hand side is the forward transform of multiplied by a known plane wave. Apply the whole-plane Cauchy-Pompeiu formula again, now in :The spatial spectral equation also gives as . One way to justify this is to integrate by parts in the first Cauchy integral: , where the modulated Cauchy integral tends to zero by the Riemann-Lebesgue lemma. Comparing the coefficient of the spectral contour integral therefore givesThis derives the transform pair from two uses of the Cauchy-Pompeiu formula, not from an assumed inversion formula.
To identify the usual normalization, write and . Then . Set , , so . The result is exactly the two-dimensional Fourier transform pairThe factor four in the real-frequency change of variables is essential.
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