Past exam of the mathematics course of the University of Cambridge 2002 ia Paper 3 12A c Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 328 3 v Solution Created 2026-10-03 Updated 2026-10-06
A particularly simple constructive choice uses paired collocation points for a global relation. For , putFour suitable sets, one for each side, areThe braces in each set range over all . For large , compute to avoid cancellation. The corresponding pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.
The key advantage is that vanish on the vertical sides, while vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with , the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.
Scale each row by . Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled system is strictly diagonally dominant, with row margin . This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact , these are exactly the first sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.
Past exam of the mathematics course of the University of Cambridge 2022 ib Paper 3 14A a Solution Created 2026-09-24 Updated 2026-09-25
Apply Green second identity to and the free-space Green function on with a small ball about removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to becausehas unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity