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

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 $e^{-t}(\rho -1)$.