Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 328 / 1

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

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 $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ 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 $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; 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 $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, 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 $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.