Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 14E (Further Complex Methods)

Set $\omega (k)=k^2-i\beta k$ and $E=e^{-ikx+\omega t}$. Differentiation gives $(Eu{)}_t=E(u_t+\omega u)$, whileThe two sides are equal precisely by the stated advection-diffusion equation. This supplies its divergence form in the $(x,t)$ plane.

Rapid decay allows two integrations by parts in the half-line Fourier transform. They giveSolving this scalar inhomogeneous equation by an integrating factor yieldsThe initial transform is holomorphic for $\mathrm{Im}k<0$ and continuous on the real axis under the given decay. The time integrals are entire functions of $\omega$ for fixed $t$ and smooth boundary data.

This identity is the half-line advection-diffusion global relation.

For $k=a+ib$, $\mathrm{Re}\omega =a^2-b^2+\beta b$. The upper contour is therefore the graphoriented from left to right. Between the real axis and $L$, $b\ge 0$ and $\mathrm{Re}\omega \ge 0$. Hence $e^{ikx-\omega (t-s)}$ is bounded by $e^{-bx}$ for $0\le s\le t$. The boundary transforms are entire in $k$; with the standard smooth, compatible boundary data, integration by parts in the time integral gives the required polynomial bounds. Closing the intervening sectors and using exponential decay for $x>0$ deforms their real-line integrals onto $L$. Small endpoint truncations, followed by dominated limits, handle the sector boundaries.

For $k^{\prime }=-k+i\beta$, $\omega (k^{\prime })=\omega (k)$ and $\mathrm{Im}{k}^{\prime }\le 0$ on $L$. Thus the already valid transformed equation at $k^{\prime }$ saysIt eliminates the unknown boundary derivative through

Write ${\hat{u}}_0(k)=\hat{u}(k,0)$ and retain the left-to-right orientation of $L$. Substituting the previous relation into the deformed inverse Fourier transform gives the data-only representationThe formally remaining integral is ${\int }_L{e}^{ikx}\hat{u}(-k+i\beta ,t)dk$. Above $L$, its reflected argument lies strictly in the lower half-plane, where the transform is analytic. At large radius, $L$ approaches the rays of angles $\pi /4$ and $3\pi /4$. Close it by the upper arc between these rays. Along that arc $\mathrm{Im}k$ is bounded below by a positive multiple of its radius, whereas the reflected transform is bounded by ${\int }_0^{\infty }|u(s,t)|ds$. Thus the arc integral is bounded by a constant times $R{e}^{-cRx}$ and tends to zero for $x>0$. There are no enclosed singularities, so Cauchy integral theorem provesThis is the needed closure argument; it does not assume the unknown transform vanishes. The representation holds for smooth decaying compatible data, and extends to weaker data by the usual limiting heat-kernel interpretation.