Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 29A / b / Solution

Characteristics satisfy $dx/ds=1$, $dy/ds=x$, $du/ds=0$. From $(\xi ,0)$ they are $x=\xi +s$, $y=\xi s+s^2/2$, $u=f(\xi )$, so $x^2-2y={\xi }^2$. The initial parameter is recovered on the two initial half-axes byThis is $C^1$ on the displayed open set, takes the prescribed data at $y=0,x\ne 0$, and substitution verifies $u_x+x{u}_y=0$. Each branch may extend separately beyond this chosen domain, but arbitrary data on both half-axes need not allow one common extension.

The initial line has normal $(0,1)$; the transport field is $(1,x)$, whose normal component is $x$. Thus it is non-characteristic exactly away from the origin. At the origin the equation itself would require $u_x(0,0)=f^{\prime }(0)=0$. For generic $f$ this fails, so a $C^1$ solution on a neighborhood of the entire initial line is not guaranteed. The parabola $y=x^2/2$ is the envelope where the recovery of $\xi$ degenerates. Characteristics can also meet both initial half-axes, imposing compatibility such as $f(\xi )=f(-\xi )$ on an attempted larger domain.