Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 30A / a

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.