method of characteristics: finding ode; evans i have a question regarding the method of characteristics for first order nonlinear pde, as done by evans (§3.2.3.b)
a fact that i cannot illuminate myself turns up discussing compatibility conditions, towards local existence. in order to construct an admissible triple, near a starting point on the boundary, we need to specify three initial values to obtain a solution by the characteristic ODE.
so for the variables $\mathbf p, z, \mathbf x$ which are the unknown in the characteristic ODE we specify the initial conditions: $z(0) := g(\mathbf x(0))$ and $p^i(0) = g_{x_i}(\mathbf x (0))$ for $1 \leq i \leq n-1$.
now in the normal direction $e_n$ it wouldn't make sense to se $p^n(0) = g_{x_n}(\mathbf x(0))$, rather he defines the $n$-th component of $\mathbf p(0)$ implicitly by $F(\mathbf p(0), z(0), \mathbf x(0)) = 0$, which means that the PDE is solved on the boundary point $\mathbf x(0)$.


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*how does this make sense? usually one doesnt require the solution to satisfy the PDE on boundary points.

 A: You are trying to solve the PDE $F(Du,u,x)=0$ in $U$, with $u=g$ on $\Gamma\subset\partial U$. If the solution you expect to find is a function $u\in C(\overline{U})\cap C^1(U)$ (for example this happens if $g$ is only continuous), then you are right, you do not expect the solution to satisfy the PDE on $\Gamma$. However, if $g$ is $C^1$ and $F$ is a continuous function defined also for $x\in \partial U$, then you expect/hope to find a solution $u\in C^1(\overline{U})$. In this case, if the PDE is satisfied in $U$ and you consider a point $x_0\in \partial U$, then by continuity,
$F(Du(x_0),u(x_0),x_0)=0.$
Now since $u=g$, if, say $\Gamma=\partial U$ is a smooth manifold, then at every point $x_0\in \partial U$ you have a tangent space and for every tangent vector $t$ to $\partial U$ at $x_0$, $\frac{\partial u}{\partial t}(x_0)=\frac{\partial g}{\partial t}(x_0)$. Since the tangent space at $x_0$ has dimension $n-1$, you have that $\frac{\partial u}{\partial t_i}(x_0)$ are determined for $i=1,\ldots, n-1$, where $t_i$ are linearly independent vectors. So only the derivative of $u$ in the normal direction is missing. To find it, you have to use the equation $F(Du(x_0),u(x_0),x_0)=0.$
We use the same trick to study regularity for elliptic equations, say $\Delta u=f$ in $U$ with Dirichlet boundary conditions $u=g$ on $\partial U$. 
