= Solution
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 $s,\xi$ gives
$$
\boxed{u(x_1,x_2)=x_2^{5/2}g(x_1/\sqrt{x_2}),\qquad x_2>0.}
$$
Every 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.
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