Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 68 2 Solution Created 2026-10-03 Updated 2026-10-14
Let and suppose stays positive and sufficiently smooth in the region considered. Write in . The leading and first transport equations of the WKB approximation areWith , they give and . Thus the leading WKB approximation for a slowly varying oscillator isThe amplitude is fixed by the transport equation, not just by replacing the constant wave number in a sinusoid. The local validity condition is that variation of over a wavelength is small, for example .
If , the same calculation gives the evanescent formAt a classical turning point where vanishes, these amplitudes diverge and the separated WKB descriptions cease to be uniform. For a simple zero, an Airy turning-point connection formula replaces them in an interval in ; it connects oscillation to exponential behavior with the usual quarter-phase shift. A higher-order zero needs a different local canonical problem, so the Airy statement assumes a simple turning point.
For a smooth positive weight on the closed interval, let . The local wave number in the Sturm-Liouville problem is , and the condition at zero selects a sine. Imposing the condition at gives the WKB quantization conditionFor weighted normalization , rapidly oscillating averaging gives . There is no turning-point phase correction here because stays positive at both endpoints with Dirichlet boundary conditions.
In the specified example , so and . To solve exactly, set and . Direct differentiation reduces the Euler-Cauchy equation to on , with zero endpoint values. Hence the inverse-square weighted Dirichlet spectrum isValues give no nonzero solution satisfying both endpoint conditions. The WKB eigenfunction shape after leading quantization is actually identical to this exact shape; the leading eigenvalue omits .
Using the exact eigenvalue as the reference, its relative error isThus is sufficient and necessary among positive integers. Referencing the approximate value instead gives , and the same integer answer.