Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1 / a

The principal symbol of the equation isIn the standard notation $A{u}_{tt}+2B{u}_{tx}+C{u}_{xx}$, we have $A=tx$, $B=0$, and $C=-1$, soA second-order equation is a hyperbolic partial differential equation exactly where this discriminant is positive. Hence the hyperbolic set isthe union of the first and third open quadrants.

A characteristic hypersurface of the form $\varphi (t,x)=p(t)\pm q(x)$ must satisfyOn either connected component of $tx>0$, separation givesOne choice valid on both components isIndeed, $p^{\prime }(t)=|t{|}^{-1/2}$ and $q^{\prime }(x)=|x{|}^{1/2}$ away from the axes, so $tx{p}^{\prime 2}=q^{\prime 2}$. Thusare the two families of characteristics, and $p\pm q$ are characteristic coordinates.

The initial line is $t=t_0$, whose conormal is $dt$. Evaluating the principal symbol on it givesIt is therefore a non-characteristic hypersurface at $(t_0,x_0)$ exactly when $t_0{x}_0\ne 0$. At such a point the equation can be writtenwhose right-hand side is real analytic locally, and the prescribed Cauchy data are also real analytic. The Cauchy-Kovalevskaya theorem consequently gives a unique local analytic solution exactly at the points