Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 / 1 / b

The principal symbol ofis $p(x,t;\xi ,\tau )=(1-x^2){\xi }^2$. For a regular curve $s\mapsto (x(s),t(s))$, the characteristic curve equation isAway from $x=\pm 1$ this gives $t=\text{constant}$. The two degenerate lines $x=1$ and $x=-1$ are also characteristic. These are the characteristic curves, apart from reparametrization and pieces joined at the degenerate lines.

Along $x=\sin s$ and $t=s^2$, the characteristic expression isIt vanishes exactly when

The initial line $x=0$ has conormal $dx$, and the principal symbol on this conormal is $1$. It is therefore a non-characteristic hypersurface at every $(0,t_0)$. The equation's coefficients and the prescribed Cauchy data $u(0,t)=\cos t$ and $u_x(0,t)=0$ are real analytic functions. The Cauchy-Kovalevskaya theorem consequently gives a unique analytic solution in a neighbourhood of $(0,t_0)$ for every