Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 67 2 b Solution Created 2026-10-03 Updated 2026-10-07
Let on a smooth interval, with . For the WKB approximation for a slowly varying oscillator, writeSubstitution into gives, at the first two orders,Thus and . The leading WKB approximation isThe amplitude is the transport correction accompanying the rapid wave phase. Real solutions are equivalent real sine and cosine combinations.
For validity, must be smooth, nonzero and slowly varying compared with the local wavelength. In physical derivatives, sufficient local checks are and . Indeed each displayed branch has relative residualSmooth positive bounded away from zero has these properties on fixed slow intervals. At a zero of the WKB approximation fails and a classical turning point requires a different local analysis, often an Airy turning-point connection formula. Rapid variations, coefficient singularities and excessively long accumulation of wave phase error also lie outside this leading approximation.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 4 30A a Solution Created 2026-09-24 Updated 2026-10-03
Between the classical turning points, , while boundedness selects exponentially decaying WKB approximation solutions in the forbidden regions and . The Airy turning-point connection formula at the left turning point matches the left decaying solution toin the allowed region. Matching the right decaying solution across givesFor these expressions to be proportional throughout , their phases must be compatible. Each simple turning point contributes a phase shift , giving the two-turning-point WKB quantization condition
Past exam of the mathematics course of the University of Cambridge 2022 ia Paper 4 9C c ii Solution Created 2026-09-24 Updated 2026-09-25
The equation has the one-dimensional effective potential per unit massThus the conserved specific energy isThe potential has its minimum at and tends to from below as . Therefore gives bounded oscillation between two turning points; gives marginal escape with asymptotic speed zero; and gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium .
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 3 30E a Solution Created 2026-09-24 Updated 2026-09-25
The endpoints are the two classical turning points, determined by . Since the potential is even, they are symmetric. For , they lie in the linear part of the potential and satisfy ; for , they lie in the outer part and satisfy . HenceThe two-turning-point WKB quantization condition uses
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 336 2 b Solution 2026-09-28
Set . The equation becomesSince is positive for every , there is no classical turning point. The leading WKB approximation for a slowly varying oscillator isThe condition selects a sine, and fixes its coefficient. Sincethe result isThe WKB validity measureis uniformly and tends to zero at both ends of . With no turning point, the approximation is therefore uniformly valid on the entire stated half-line.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 336 3 a Solution Created 2026-09-24 Updated 2026-09-25
With , the equation isThe leading WKB approximation for a slowly varying oscillator isIt requires smooth nonzero , , and distance from every classical turning point large compared with its turning-point scale.
Two-turning-point WKB quantization condition Created 2026-09-24 Updated 2026-09-24