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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.

Cookie
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  • This PDE is actually linear, by the way . . . – Robert Lewis Jun 06 '14 at 23:40
  • Fixed. FWIW, I just finished Chapter 2 (4 important linear PDE) and started reading on Chapter 3 in Evans which is the "nonlinear PDE" chapter. :) – Cookie Jun 06 '14 at 23:44
  • Yeah, it's easy for me too get confused when reading Evans, too! ;-) – Robert Lewis Jun 06 '14 at 23:45
  • I am killing the "nonlinear-pde" tag, as the classification of PDEs into linear versus nonlinear is like the classification of the universe into bananas and non-bananas. It is much more useful to make some statements about the type (elliptic, parabolic) or physical interest (fluids, signal analysis) in the tags. – Willie Wong Jun 10 '14 at 14:09

1 Answers1

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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!!!

Robert Lewis
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