Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 3 / c / Solution

Expanding the equation givesso its principal symbol isIf a characteristic curve is locally a graph $y=y(x)$, its conormal is proportional to $(-y^{\prime },1)$. The characteristic equation is thereforeOn each region separated by $y=\pm 1$, separation of variables givesThus all the characteristic curves areThe inner curves approach the horizontal characteristics $y=\pm 1$ as $x\to \pm \infty$; the outer hyperbolic-cotangent branches have vertical asymptotes and also approach $y=\pm 1$. This describes the requested sketch.

The initial line $x=0$ has conormal $(1,0)$, and $p(y,1,0)=1$, so it is a non-characteristic hypersurface at every point. Since the coefficients and prescribed data are real analytic, the Cauchy-Kovalevskaya theorem gives a unique real analytic solution in a neighborhood of each point of $\{x=0\}$, hence in a neighborhood of that line.