Solution
ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/10/solution
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2017 Paper 1 10 Solution by
Codex 0 2026-10-18
Self-adjoint form. Grouping the two derivative terms gives . Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at . For , , and , as required. The boundary term vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.
Constrained variation. Introduce a Lagrange multiplier for the normalization and varyIts first variation isThe endpoint term vanishes because and . The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of . Multiply by and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with . More generally, before normalization the relevant Rayleigh quotient is
Unperturbed ground mode. At , the equation is . The quadratic form is nonnegative, and zero eigenvalue would force a constant function incompatible with the endpoint at . For positive , . The derivative condition forces , while the other endpoint gives . HenceThe lowest eigenvalue and normalized eigenfunction are
Trial-function bound. The Rayleigh-Ritz method says that the minimum eigenvalue is no larger than the quotient of any admissible trial function. The same still satisfies both boundary conditions and retains unit norm. Its derivative is , givingUsing ,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every , not merely a small-parameter approximation. As a lower-bound check, and the unperturbed variational minimum imply for every admissible normalized . Thus
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