Example 12.2.4 (Nonlinear $1^{st}$ Order PDE).label Finally we consider the Cauchy problem for a fully nonlinear $1^{th}$ order PDE.
\begin{align*}F(x_{1},...,x_{n},u,\pd_{1}u,...,\pd_{n}u)=0&&u|_{S}=\p\end{align*}
We assume that $F\in C^{2}$. We let $\xi_{i}\define \pd_{i}u$ and want $u$ such that image of the map
\begin{align*}x\mto (x,u(x),\xi(x))\end{align*}
lies in $F=0$. We note $x_{1},...,x_{n},u,\xi_{1},...,\xi_{n}$ is $2n+1$ dimensional and denoted $J^{1}(\R^{n},\R)$ the space of $1$-jets of maps $\R^{n}\to \R$ and $(x,u(x),\pd u(x))$ is a solution if and only if the graph is contained in $\curl{F=0}$. One can consider the PDE as a hypersurface in the space of $1$-jets.
By analogy with the quasilinear case Theorem 12.2.2, we write the Cauchy characteristic equation as follows
\begin{align*}\der{x_i}{t}=\pder{F}{\xi_i},1\leq i\leq n\end{align*}
to be augmented by ODEs for $y\geq \xi_{i}$
\begin{align*}\der{y}{t}=\sums{n}{i=1}\pder{y}{x_i}\der{x_i}{t}=\sums{n }{i=1}\pder{y}{x_i}\pder{F}{\xi_i}\end{align*}
thus we have
\begin{align*}\der{y}{t}=\sums{n }{i=1}\xi_{i}\pder{F}{\xi_i}\end{align*}
Furthermore
\begin{align*}\der{\xi_j}{t}&=\sums{n }{k=1}\pder{\xi_j}{x_k}\der{x_k}{t}=\sums{n }{k=1}\pder{x_j}{x_k}\pder{F}{\xi_k}\\&=\sums{n }{k=1}\pder{^2u}{x_j\pd x_k}\pder{F}{\xi_k}=\sums{n }{k=1}\pder{^2u}{x_k\pd x_j}\pder{F}{\xi_k}\\&=\sums{n }{k=1}\pder{\xi_k}{x_j}\pder{F}{\xi_k}\end{align*}
thus
\begin{align*}\pder{F}{x_j}+\pder{F}{u}\UBC{\pder{u}{x_j}}{\xi_j}+\UBC{\sums{n }{k=1}\pder{F}{\xi_k}\pder{x_k}{\xi_j}}{=-\der{\xi_j}{t}}=0\end{align*}
So the characteristic equations are
\begin{align*}\der{x_i}{t}&=\pder{F}{\xi_i}(x,y,\xi)\\ \der{y}{t}&=\sums{n }{i=1}\xi_{i}\pder{F}{\xi_i}(x,y,\xi)\\ \der{\xi_i}{t}&=\parens{-\pder{F}{x_i}-\xi_i\pder{F}{y}}(x,y,\xi)\end{align*}
We remark that $\pder{F}{u}=0$ then the characteristic equation describes Hamiltonian flow in $(x,\xi)$. The solutions of this ODE system is called characteristic strips. What about initial conditions for the characteristic system of ODEs?
\begin{align*}x(s,0)&=g(s)\\ y(s,0)&=\p(g(s))\end{align*}
To get conditions for $\xi$ we note that on $S$
\begin{align*}\pder{y}{s_j}=\pder{\p(g(s))}{s_j}=\sums{n }{k=1}\pder{y}{x_k}\pder{g_k}{s_j}=\sums{n }{k=1}\xi_{k}\pder{g_k}{s_j}&&F(g(s),\p(g(s)),\xi_{j}(g(s)))=0&&\end{align*}
for $1\leq j\leq n-1$. We need
\begin{align*}\det\begin{pmatrix}\pder{g_1}{s_1}&\cds&\pder{g_1}{s_{n-1}}&\pder{F}{\xi_1}\\ \vdots&&\vdots\\ \pder{g_n}{s_1}&\cds&\pder{g_n}{s_{n-1}}&\pder{F}{\xi_n}\end{pmatrix}\neq 0\end{align*}
so here the ambient space also include $\xi$ and thus more things can go wrong than just tangency. The first thing that can go wrong (for $\xi$) is related the $1$-jets the second (for $y$) comes from graph of function.
\begin{align*}(s_{1},...,s_{n-1},t)\bij (x_{1},...,x_{n})&&u(x)\define y(s(x),t(x))\end{align*}
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