Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 31C (Asymptotic Methods)

The coefficient changes sign where $1-{\lambda }^2/x^2=0$, so the single turning point isPut $p(x)=({\lambda }^2/x^2-1{)}^{1/2}$ and $q(x)=(1-{\lambda }^2/x^2{)}^{1/2}$. The WKB approximation satisfying the boundary condition at $x=1$ iswhile decay at infinity selects

Near $x=\lambda$,Withthe leading equation is $y_{tt}-ty=0$. Decay for $t>0$ selects $y=C\mathrm{Ai}(t)$. The Airy turning-point connection formula gives on the oscillatory sideFor this to be proportional to the inner expression from part (a), the two phases must differ by an integer multiple of $\pi$. Thus the WKB quantization condition is

Write $\lambda =1+\delta$, where $\delta \downarrow 0$, and put $x=\lambda (1-u)$. The integration range is $0\le u\le 1-1/\lambda =\delta +O({\delta }^2)$, andConsequentlySubstitution into the quantization condition gives

The action integral can be evaluated exactly:As $\lambda \to \infty$,Its leading logarithmic term is $\lambda \log \lambda$. Hence the large eigenvalues satisfy, to leading order,