Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 14A (Methods)

By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.

The Fourier transform of a derivative givesHencefor $k\ne 0$. The assumed existence of the transforms excludes an arbitrary nonzero constant in $p$.

Using Euler's formula and the Inverse Fourier transforms of phase factors,andThese are distributional identities, with $\delta$ the Dirac delta function.

Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

Here the D'Alembert formula with initial velocity becomesWhen $t>1$ andthe integration interval contains the whole support $[-1,1]$. Since $g$ is odd,Thus the requested region containing the line $x=0$ isThis is the central zero region for odd compactly supported initial velocity. The initial disturbance splits into left-moving and right-moving waves; after they pass, their opposite signed impulses cancel and leave an expanding undisturbed region around the origin.