I have a question about the following paper:
https://web.stanford.edu/class/math220a/handouts/firstorder.pdf
I have two questions about two points that I thought I understood, but I'm not sure now.
- "Now let $S$ be the integral of surface formed from a union of these characteristic curves. In doing so, we see that $z(x,t)$ is constant along the lines $x - at = x_{0}$"
The characteristic curves are $x(s) = as + c_{1},y(s) = s + c_{2}, z(s) = c_{3}$. Obviously, $z$ is a constant function, but why de construction of $S$ implies $z(x,t)$ constant along the $x - at = x_{0}$? At first, went unnoticed because $z$ is constant, but is not the same thing.
- Why we can take $u(x,t) = z(x,t)$?
The first time he mentions $S$, he writes $S = \{(x,y,u(x,y)\}$, so it's simply because $(x,y,z) = (x,y,u(x,y))$?
For the last question: we can interpret $u$ geometrically, as the third coordinate of a surface $z$ (it is interpreted too $t$ geometrically, maybe explaining why you alternate $t$ and $y$ in your questions). Probably you think of $u$ as $z$ without being aware :)
For the first question. $z(s)=c_3$ is constant as function of $s$, the parameter driving the characteristic curve. As $s$ varies, $x$ and $t$ do so too, but not $z$, so is, $z$ is constant along the characteristics. Now, we can see what is the graph of one of the characteristic curves by eliminating $s$. From your equations (fixing the mistake that takes $y$ as $t$):
$s=\dfrac{x-c_1}{a}$ and $s=t-c_2$, leading to:
$\dfrac{x-c_1}{a}=t-c_2$ and $x-at=c_1-ac_2$. But as the RHS are all constants, we can absorb them in a single one $x_0=c_1-ac_2\implies x-at=x_0$, as expected. Summarizing: this is the graph for a characteristic curve, but $u$ is constant along the characteristics, so, $u$ is constant for points into that graph.