Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 / 4 / b

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

The all-time conclusion in the original PDF is false. The expanding-flow transformation of a wave equation makes both the correct forward statement and a counterexample transparent. PutFor $D={\partial }_t+\rho {\partial }_{\rho }$, the expanded operator from part (i) is $D^2+D-{\partial }_{\rho }^2$. In the new variables $D=-S{\partial }_S$ and ${\partial }_{\rho }=S{\partial }_X$, henceThus it is the ordinary wave equation on the half-plane $S>0$. By the D'Alembert formula, any $C^2$ solution has the formAt $t=0$, differentiating the zero displacement with respect to $\rho$ and using the zero velocity gives, for $-1<\rho <1$,Subtracting $(\rho -1)$ times the first equation from the second gives $2F^{\prime }(\rho +1)=0$, and then $G^{\prime }(\rho -1)=0$. Thus $F=c$ on $(0,2)$ and $G=-c$ on $(-2,0)$. If $t\ge 0$ and $-1<\rho <1$, both arguments remain in those intervals. Consequently the valid forward result isFor negative time the arguments need not stay in the initial intervals. Take $F(s)=(s-2{)}_{+}^3$ and $G=0$, where $q_{+}=\max \{q,0\}$. Thenis a global $C^2$ solution, has the prescribed zero initial displacement and velocity on $(-1,1)$, but $u(0,-\log 3)=1$. This directly disproves the assertion for $t\in \mathbb{R}$.

The connection with part (i) is directional finite propagation speed. At a future point in the strip, the two characteristics traced back to $t=0$ meet it at ${\rho }_0=e^{-t}(\rho +1)-1$ and ${\rho }_0=e^{-t}(\rho -1)+1$, both in $(-1,1)$. For negative time these intersections can lie outside that interval. The lines $\rho =\pm 1$ are characteristic barriers for this forward domain of dependence, not barriers in both time directions.