Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 5

Use the Fourier transform convention $\tilde{h}(k)={\int }_{\mathbb{R}}{e}^{-ikx}h(x)dx$. Split at zero and integrate the decaying exponentials:The transform is real and even because $h(x)=e^{-|x|/L}$ is real and even.

Put $a=1/L\in (0,1)$. The Fourier transform of a derivative turns the differential operator into $k^4+5k^2+4=(k^2+1)(k^2+4)$, soUse the partial fraction decompositionThe inverse-transform pair ${\mathcal{F}}^{-1}[(k^2+b^2{)}^{-1}]=e^{-b|x|}/(2b)$ is the one-dimensional modified Helmholtz Green function. HenceIn terms of the original length scale,This is the convolution of $h$ with the fourth-order Green kernel with two decay scales, here $K(x)=e^{-|x|}/6-e^{-2|x|}/12$. It decays at both infinities. Although the individual exponential pieces have cusps, their derivative jumps through order three cancel, so the result has no unintended Dirac delta distribution at zero. The decaying solution is unique: the homogeneous solutions $e^{\pm x},e^{\pm 2x}$ have no nonzero globally decaying combination.

At the origin, combine the rational coefficients before taking the limit:Equivalently $y(0)=L(3L+1)/[6(L+1)(2L+1)]$. The apparent poles at $L=1$ cancel, givingThe entire limiting profile is also finite:which independently gives the same central value.