Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 19

Assume $c\ne 0$, so $(\xi ,\eta )=(x-ct,x+ct)$ is an invertible coordinate change. The chain rule givesFor a twice continuously differentiable solution the mixed partial derivatives commute. Squaring the two operators therefore givesThus the one-dimensional wave equation becomes

If $\psi =f(\xi )+g(\eta )$ with twice differentiable $f,g$, its mixed partial derivative vanishes because each summand depends on only one coordinate. Conversely ${\psi }_{\xi \eta }=0$ means ${\psi }_{\xi }$ is independent of $\eta$; integration in $\xi$ supplies $f(\xi )$ and an integration function $g(\eta )$. Hence, on a characteristic rectangle, the general solution isThese terms travel in opposite directions without changing profile, as in D'Alembert's formula.

Take the spatial interval $0<x<L$ with $L>0$, and write $v=|c|>0$. A separated nonzero solution satisfiesThe zero-endpoint Dirichlet boundary conditions rule out nonzero solutions for $\lambda \le 0$: linear or hyperbolic spatial factors cannot vanish at both ends. For $\lambda >0$, they give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$, $n\ge 1$. The temporal factors solve $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with the fixed-end standing wave frequencies ${\omega }_n=n\pi v/L$. Superposition givesThe constants are arbitrary subject to the convergence and regularity required of the solution. For initial displacement $f$ and initial velocity $g$, their Fourier sine series give $A_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$ and $B_n=(2/(L{\omega }_n)){\int }_0^{L}g(x)\sin (n\pi x/L)dx$. Smooth compatible data give classical solutions; weaker data have the corresponding energy interpretation. A negative printed $L$ describes the interval with endpoints reversed and gives the same modes after using its positive length.