Page 99 of PDE Evans, 2nd edition, says
In summary, \begin{cases} \tag{17} \dot{\textbf{x}}(s)=\textbf{b}(\textbf{x}(s)) \\ \dot{z}(s) = -c(\textbf{x}(s))z(s)\end{cases}will comprise the characteristic equations for the linear, first-order PDE $$F(Du,u,x)=\textbf{b}(x)\cdot Du(x)+c(x)u(x)=0$$for all $x \in U \subset \mathbb{R}^n$.
Now, on page 100, the book proceeds to this:
Example 1. We demonstrate the utility of equations $(17)$ by explicitly solving the problem \begin{cases} \tag{18}x_1u_{x_2}-x_2u_{x_1}=u & \text{in }U \\ u=g &\text{on }\Gamma \end{cases} where $U$ is the quadrant $\{x_1 > 0,x_2 > 0\}$ and $\Gamma = \{x_1 > 0,x_2 > 0\} \subseteq \partial U$. The PDE in $(18)$ is of the form $F(\textbf{p}(s),z(s),\textbf{x}(s)) \equiv 0$, for $\color{blue}{\textbf{b}=(-x_2,x_1)}$ and $\color{#FF00FF}{c=-1}$.
My question is how were $(17)$ and $(18)$ used to find $\textbf{b}$ and $c$, as I highlighted above.
If equation (18) in the question is written as
$x_1u_{x_2}-x_2u_{x_1} - u = 0 \tag{1}$
and we know that
$Du = (u_{x_1}, u_{x_2}), \tag{2}$
then if we define $\mathbf b(x)$ via
$\mathbf b(x) = (-x_2, x_1), \tag{3}$
then
$\mathbf b(x) \cdot Du = -x_2 u_{x_1} + x_1 u_{x_2} = x_1 u_{x_2} - x_2 u_{x_1}, \tag{4}$
and if we again define $c$ via
$c(x) = -1, \tag{5}$
then again, re-writing (1),
$\mathbf b(x) \cdot Du + c(x) u = 0. \tag{6}$
These definitions were motivated by comparing the question's (17), which I take to be
$\mathbf{b}(x)\cdot Du(x)+c(x)u(x)=0, \tag{7}$
with (1), and just picking off coefficients of similar terms in $u$ (that is, of $u$, $u_{x_i}$). The equations for the characteristics, viz.,
$\dot{\textbf{x}}(s)=\textbf{b}(\textbf{x}(s)), \; \dot{z}(s) = -c(\textbf{x}(s))z(s), \tag{8}$
don't really have much to do with the selection of $\mathbf b$ and $c$ based on (1), (6). They come later.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!