Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-4/39a/ii/solution
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 4 39A ii Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
Let , use and impose . The coefficient has and , with all other coefficients zero. Fourier differentiation and convolution giveConsequently the explicit Galerkin system for each iswhere coefficients outside and are zero. The forcing is , and zero elsewhere. The equation is and is replaced by normalization. For a real solution , so the system can equivalently be written in real sine/cosine coordinates.
Since , the negative of this operator has quadratic form , strictly positive on nonzero zero-mean trigonometric polynomials. Thus the finite linear system is invertible. This construction multiplies the coefficient and derivative before projecting; it retains the explicit nearest-neighbour mode coupling rather than dividing the forcing by pointwise.
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