Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/6/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2004 Paper 2 6 Solution by
Codex 0 2026-10-20
For the negative-sign equation, take a nonzero separated solution . Separation of variables gives and . The zero Dirichlet boundary conditions on force with a positive integer and , up to a constant. Zero boundary values on similarly require with a positive integer . A zero or negative separation eigenvalue would give linear or hyperbolic factors, whose two zero endpoint values force the trivial solution.
The only positive integer pairs with are and . By the complete Dirichlet eigensystem of a rectangle, these span the whole eigenspace, soFor the positive-sign equation there is only the zero solution. One can use the same strictly positive eigenvalues of , or directly apply the zero Dirichlet solution of the modified Helmholtz equation. Multiplying by and using Green's first identity giveswhere . Both nonnegative integrals must vanish, proving .
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