Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-1/38a/b/solution
Past exam of the mathematics course of the University of Cambridge 2005 ii Paper 1 38A b Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
For the Fourier mode , substitution givesThe denominator is never zero for real , and numerator and denominator are complex conjugates, so . Parseval's identity therefore makes each step norm-preserving on square-summable grid data. The method is von Neumann stable for every , with no upper Courant restriction. It is an implicit nondissipative scheme; unconditional stability does not remove phase error or guarantee accuracy at a large time step.
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