= Solution
{c}
The <characteristic equations for a transport equation> are $\dot x=1$, $\dot y=2$, $\dot z=-3z$. Label an initial point by $s\in[0,1]$ and choose <characteristic> time zero on the data segment. Integration gives
$$
x=s+t,\qquad y=2t,\qquad z=e^s e^{-3t}.
$$
Thus $s=x-y/2$ and $t=y/2$, giving
$$
\boxed{z(x,y)=e^{x-2y},\qquad0\leq x-y/2\leq1.}
$$
This is the full infinite strip between the parallel <characteristic> lines $y=2x$ and $y=2x-2$. Both positive and negative <characteristic> times are allowed because the problem specifies no forward-time restriction. Each <characteristic> in this strip meets the data segment exactly once. Outside it, the general solution $z=e^{-3y/2}F(x-y/2)$ contains unrestricted values of $F$, so the data do not determine a unique solution there.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2009-characteristic-strip.png]
{title=The data segment determines the solution throughout the strip swept out by its characteristics}
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