Example 12.2.5 (Nonlinear $1^{st}$ Order Example).label Consider
\begin{align*}\begin{cases}\pd_{1}u\pd_{2}u=u\\ u(0,x_{2})=(x_{2})^{2}\end{cases}\end{align*}
So $F(x_{1},x_{2},u,\xi_{1},\xi_{2})=\xi_{1}\xi_{2}-u$ and characteristic equations for this example is given by
\begin{align*}\begin{cases}\der{x_1}{t}=\xi_{1}&\der{x_2}{t}=\xi_{1}\\ \der{y}{t}=2\xi_{1}\xi_{2}\\ \der{\xi_1}{t}=\xi_{1}&\der{\xi_2}{t}=\xi_{2}\end{cases}\end{align*}
So
\begin{align*}\xi_{1}=\xi_{1}^{o}e^{t}&&\xi_{2}=\xi_{2}^{o}e^{t}\\ x_{1}=\xi_{2}^{o}e^{t}+x_{1}^{o}&&x_{2}=\xi_{1}^{o}e^{t}+x_{2}^{o}&&y=\xi_{1}^{o}\xi_{2}^{o}e^{2t}+u^{o}\end{align*}
Impose Cauchy data for characteristic strips
\begin{align*}x_{1}|_{t=0}&=0\iff\xi_{2}^{o}+x_{1}^{o}=0\implies x_{1}(t,s)=\xi_{2}^{o}(e^{t}-1)\\ x_{2}|_{t=0}&=s\iff\xi_{1}^{o}+x_{2}^{o}=s\implies x_{2}(t,s)=\xi_{1}^{o}(e^{t}-1)+s\\ u|_{t=0}&=s^{2}\iff \xi_{1}^{o}\xi_{2}^{o}+u^{o}=s^{2}\implies y(t,s)=\xi_{1}^{o}\xi_{2}^{o}(e^{2t}-1)+s^{2}\\ \xi_{2}(0,x_{2})&=\pder{u}{x_2}(0,x_{2})=2x_{2}\\ \implies \xi_{2}|_{t=0}&=2s\implies \xi_{2}(t,s)=2se^{t}\end{align*}
Use PDE to solve for the $\xi_{1}$
\begin{align*}\xi_{1}\xi_{2}-u=0\implies \xi_{1}(0,s)\xi_{2}(0,s)-u(0,s)=\xi_{1}|_{t=0}2s-s^{2}=0\implies \xi_{1}|_{t=0}=\frac{s}{2}\end{align*}
but here we divided by $s$ which is not allowed when $s=0$. Here we see $(g_{1}(s),g_{2}(s))=(0,s)$. This is where $\det\begin{pmatrix}0&2s\\1&\frac{s}{2}\end{pmatrix}=-2s$ which is $0$ when $s=0$.
Finally we carry out the inversion $(s,t)\mto x$. Rewrite the characteristic strips
\begin{align*}x_{1}(t,s)&=2s(e^{t}-1)\\ x_{2}(t,s)&=\frac{s}{2}(e^{t}-1)+s=\frac{s}{2}(e^{t}+1)\\ y(t,s)&=\frac{s}{2}2s(e^{2t}-1)+s^{2}=s^{2}e^{2t}\\ \xi_{1}(t,s)&=\frac{s}{2}e^{t}\\ \xi_{2}(t,s)&=2se^{t}\end{align*}
Inverting
\begin{align*}x_{1}-4x_{2}&=-4s\implies s=\frac{4x_{2}-x_{1}}{4}\\ x_{1}+4x_{2}&=4se^{t}\implies e^{t}=\frac{x_{1}+4x_{2}}{4x_{2}-x_{1}}\end{align*}
So
\begin{align*}u(x_{1},x_{2})&=y(s(x_{1},x_{2}),t(x_{1},x_{2}))\\&=\parens{s(x_1,x_2)}^{2}e^{2t(x_1,x_2)}\\&=\parens{\frac{4x_{2}-x_{1}}{4}}^{2} \parens{\frac{x_{1}+4x_{2}}{4x_{2}-x_{1}}}^{2}\\&=\frac{(x_{1}+4x_{2})^{2}}{16}\end{align*}
Check Cauchy data
\begin{align*}u(0,x_{2})=\frac{(4x_{2})^{2}}{16}=(x_{2})^{2}\end{align*}
Check PDE
\begin{align*}u_{1}=\frac{1}{8}(x_{1}+4x_{2})&&u_{2}=\frac{1}{8}4(x_{1}+4x_{2})\implies u_{1}u_{2}=u\end{align*}
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