Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 336 (Perturbation Methods)

Write the Schläfli contour integral for Legendre polynomials asThe saddle points satisfyFor $\mu =\cos \theta$, they are $z_{\pm }=e^{\pm i\theta }$. Deforming the contour through both conjugate saddles and using the local Gaussian contributions from the method of steepest descent gives conjugate exponentials. Their sum is the Debye asymptotic for Legendre polynomialsBoth saddles are required for the real oscillatory answer.

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Put $s=|\xi |$. Since $\cosh \xi =\cosh s$, the saddle points are $e^{\pm s}$. The contour deformation relevant to the prescribed contour passes through the dominant saddle $z=e^s$. There,The quadratic saddle factor and the remaining amplitude give the Hyperbolic Debye asymptotic for Legendre polynomialsThe second saddle gives an exponentially subdominant contribution on this contour. At $\xi =0$ the saddles coalesce and this formula is nonuniform; the exact endpoint value is $P_n(1)=1$.

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Set $\delta =\sqrt{ϵ}$. The outer scale found below shows that the inner expansion requires the four asymptotic scalesSolving successively with $f_j(1)=0$ and matching the free homogeneous terms givesThis is valid for fixed $r=O(1)$.

To verify the differential-equation hierarchy, let $L[y]=y^{\prime \prime }+3y^{\prime }/(2r)$. The leading terms satisfyAt order $ϵ$,whose particular integral is $-\sqrt{r}+\log r+1$; the displayed homogeneous multiple of $1-r^{-1/2}$ is fixed by matching.

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Introduce the stretched coordinateThe equation becomesSeek the three-term outer expansionAt first order,so decay at infinity and matching giveAt the next order,Using the function $Y$ supplied in the question, the matched decaying solution isHere $Y$ is the stated particular solution normalized to decay at infinity.

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Use $\rho =ϵr={\delta }^{2}r$ in the supplied small-$\rho$ expansions. The outer approximation re-expanded in inner variables isThis is exactly the large-$r$ re-expansion of the inner result in part 1. In particular, the identityforces the switchback term $ϵ\log ϵ(1-r^{-1/2})$. The coefficient of the homogeneous $E_{3/2}$ contribution at outer order $ϵ$ is fixed to $-2\sqrt{\pi }$, while the remaining inner homogeneous constant is fixed to $\pi +C/4-1$. Thus the two matched asymptotic expansions agree term by term in $1\ll r\ll {ϵ}^{-1}$.

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Introduce the slow time $T=ϵt$ and writeAt leading order the relaxation equation givesThe last term is the freely decaying initial transient; it does not alter the long-time solvability condition.

At order $ϵ$, eliminating the resonant $e^{i\omega t}$ forcing is the solvability condition in the method of multiple scales. It givesWrite $A=R{e}^{i\phi }$ and let $R(0)=R_0$, $\phi (0)={\phi }_0$. ThenWiththeir explicit solutions are

The complete real leading approximation, uniform for $t=O({ϵ}^{-1})$, is thereforeThe constant $K$ is chosen from the initial value of $Z$ after subtracting the mean and second-harmonic pieces. If the initial transient is not required, set $K=0$.

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