Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 10 / 1

For the wave equation with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, the wave speed is one. The domain of dependence of $(t,x)$ is the initial ball $\overline{B}(x,t)$. More precisely, two solutions whose Cauchy data agree on a neighbourhood of that ball have the same value at $(t,x)$. Thus, if both Cauchy data have support in $K$, finite propagation speed givesThe Strong Huygens principle is sharper in three spatial dimensions: only the initial data on the sphere $|y-x|=t$ contribute, including the first spatial derivatives of the initial displacement there. The Kirchhoff formula makes this precise:In particular, if the Cauchy data have compact support in $\overline{B}(0,R)$, thenFor $t>R$, the solution therefore vanishes behind the inward edge $|x|=t-R$ as well as outside $|x|=t+R$. The absence of an interior tail is the additional content of the Strong Huygens principle.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

For a radial function, the Laplacian is $\Delta \varphi ={\varphi }_{rr}+2{\varphi }_r/r$. Introduce the retarded and advanced null coordinates and the radial reduction of the three-dimensional wave equation:The chain rule gives ${\partial }_t={\partial }_u+{\partial }_v$ and ${\partial }_r=-{\partial }_u+{\partial }_v$. Consequently the inhomogeneous wave equation isMultiplying the radial wave equation by $r$ removes its first-order radial derivative:Thus the particularly useful retarded and advanced null coordinates form isThese formulas hold for $r>0$. At the axis, smooth spherical symmetry imposes ${\varphi }_r(t,0)=0$ and $w(t,0)=0$; equivalently $w$ has a smooth odd extension across $r=0$.

Use the radial reduction of the three-dimensional wave equation $w=r\varphi$ and extend $w$ oddly in $r$. Write $w=w_h+w_F$, where $w_h$ is the homogeneous wave equation solution with the prescribed Cauchy data, and $w_F$ has zero Cauchy data. With $w_0(r)=r{\varphi }_0(r)$ and $w_1(r)=r{\varphi }_1(r)$, the D'Alembert formula givesBoth odd initial profiles have compact support, soThe Duhamel principle, applied to $w_{tt}-w_{rr}=-rF$, givesNow write $r=t+a$ with $1/2\le a\le 1$. The lower integration limit is $s+a>0$, so this domain of dependence never crosses the axis. Intersecting its integration interval with the support of the source leaves a subset of $[s+a,s+1]$, of length at most one. On this interval,The outgoing shell source estimate for a radial wave therefore yieldsFinally $r\ge t+1/2\ge (1+t)/2$, and thusThis constant depends only on the Cauchy data and the fixed unit bound for the source. The logarithm comes from integrating the shell's accumulated forcing, whereas the factor $(1+t{)}^{-1}$ comes from division by $r$.