Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 30A (Partial Differential Equations)

Use $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$. For a Schwartz function, Fourier inversion is $f(x)=(2\pi {)}^{-1}\int e^{ix\xi }\hat{f}(\xi )d\xi$. For a tempered distribution define $⟨\hat{T},\varphi ⟩=⟨T,\hat{\varphi }⟩$ for Schwartz functions; this agrees with the function transform in the stated convention. The half-indicator kernel has transformwith value $t$ at zero. Transforming the wave equation gives ${\hat{u}}_{tt}+{\xi }^2\hat{u}=0$, zero initial value and initial derivative $\hat{g}$. Hence $\hat{u}=\hat{g}\sin (t\xi )/\xi$, and convolution yieldsFinite propagation speed means that the solution at $(t,x)$ depends only on initial data in this interval, so disturbances cannot travel faster than one. The formula also makes sense for any smooth $g$, regardless of its growth at infinity, because the integration interval is compact. Direct differentiation verifies the equation and both initial conditions. Equivalently, locally replace $g$ by a compactly supported smooth function agreeing throughout the relevant domain of dependence.

A smooth $2\pi$-periodic function has coefficients $g_k=(2\pi {)}^{-1}{\int }_0^{2\pi }g(x)e^{-ikx}dx$ decaying faster than any power, by repeated integration by parts. Thus its Fourier series and every differentiated series converge absolutely and uniformly to the corresponding function and derivatives. Solving each spatial Fourier mode givesThe zero-frequency term is linear in time, not obtained by division by zero. Uniform convergence on compact time intervals permits termwise differentiation and verifies the initial conditions and the wave equation.

Insert the uniformly convergent Fourier series of $g$ into the interval formula and integrate termwise. The constant mode contributes $t{g}_0$, while for $k\ne 0$,The sum is exactly the expression from part (b), proving agreement of the transform and Fourier-series formulas.