The <principal symbol> of the equation is
$$
P(t,x;\tau,\xi)=tx\tau^2-\xi^2.
$$
In the standard notation $Au_{tt}+2Bu_{tx}+Cu_{xx}$, we have $A=tx$, $B=0$, and $C=-1$, so
$$
B^2-AC=tx.
$$
A second-order equation is a <hyperbolic partial differential equation> exactly where this discriminant is positive. Hence the hyperbolic set is
$$
\boxed{\{(t,x):tx>0\}},
$$
the union of the first and third open quadrants.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-105/1/a/i/solution