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In Chapter 3 of Evans' PDE Text, we're interested in solving nonlinear first-order PDEs of the form $$F(Du, u, x) = 0 \text{ in } U \\ u = g \text{ on } \Gamma$$

In the section where constructing local solutions via the method of characteristics is discussed, the first step is given to be "straighten out the boundary" by finding a smooth mapping $\Phi$ that straightens out $\partial U$ near some point $x^0 \in \partial U$ (some fixed point on the boundary).

My question is - why is this done? The idea is that this transformation gives you a PDE of the same form and now from the outset, given some point $x^0 \in \Gamma$ you can assume that $\Gamma$ is flat near $x^0$ (lying in the plane $x_n = 0$.

I suppose it has something to do with letting you take derivatives more easily (instead of on some general non-straightened domain) but I'm not seeing why this is helpful or how it's even used in subsequent calculations to construct solutions.

tastykakes
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    For the sake of being self-contained, would you mind providing the an example of "straightening out the boundary" or writing out what Evans has to say about it? – Michael L. Nov 26 '17 at 08:35
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    @MichaelLee - I gave it a shot, hope that's a bit better. – tastykakes Nov 26 '17 at 08:46

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The flatness of boundary is used throughout the section "Local solution". It begins by setting up the problem so that the boundary is $x_n=0$. Then the equation of characteristic curves becomes $\mathbf x(s) = \mathbf x(y_1,\dots, y_{n-1}, s)$ where $(y_1,\dots, y_{n-1}, 0)$ is the initial point of the curve. This structure of $\mathbf x$ is used to find its derivative matrix and conclude it's invertible.

Without the boundary being flat, how would we write the mapping $\mathbf x$? The initial point would be $(y_1, \dots, y_n)$ subject to some nonlinear constraint on $y_1,\dots, y_n$. Hence, $\mathbf x$ depends on $(n+1)$ variablves $y_1,\dots, y_n, s$ which are not independent. Not exactly what we need to apply the Inverse Mapping Theorem to $\mathbf x$.