Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 30E / a

The characteristic equations for a transport equation are ${\dot{x}}_1=x_1$, ${\dot{x}}_2=2x_2$, $\dot{u}=5u$. Starting from $(\xi ,1)$ at parameter zero gives $(x_1,x_2)=(\xi e^s,e^{2s})$, $u=e^{5s}g(\xi )$. Eliminating $s,\xi$ givesEvery point of this upper half-plane lies on exactly one characteristic from the initial line, so it is the maximal characteristic domain determined by the data. Neither the line $x_2=0$ nor the lower half-plane is reached in finite characteristic time; extra data would be needed there.

For $g\in C^1$ this is a classical solution, directly verified by differentiation. The printed assumption is only continuity, which in general does not give a classical $C^1$ solution. Under that assumption the same formula is the unique continuous characteristic solution and solves the transport equation distributionally; smoothing $g$ locally and passing to limits verifies the latter interpretation. A classical interpretation needs the additional regularity just stated.