Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 328 1 i Solution Created 2026-10-03 Updated 2026-10-06
Put and apply the Dirichlet gauge transform for constant drift . Taking derivatives directly removes the advection term and gives the damped heat equation . Thus , while the two Dirichlet boundary conditions become and .
The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for by the backward heat kernel and use integration by parts in . Since the kernel vanishes at , the surviving boundary expression is . Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires and ; otherwise the same formula describes the solution away from those corners.
Here is a second integral representation, useful when the integral transforms are explicit. Extend by zero after ; the values of the extension cannot affect a solution at . Write their Laplace transforms as and set using the principal square root. The transformed Dirichlet Green function for isIt vanishes at both endpoints and its first derivative in jumps by , which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in . Their only spatial resolvent operator poles are . This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.