We have the following equation with initial condition \begin{align*} u_y + uu_x &= 0\\ u(x,0)&=0 \end{align*} corresponding to the manifold $\Gamma$ in the $xyz$ space given by \begin{equation} x = s,\quad y = 0,\quad z = h (s) \end{equation} The characteristic differential equations \begin{equation*} \frac{dx}{dt}=z\quad \frac{dy}{dt}=1\quad \frac{dz}{dt}=0 \end{equation*} Integrating each of the expressions we have \begin{align*} \int dx&=\int z dt\\ x&=zt \end{align*} also \begin{align*} \int dy&=\int 1 dt\\ y&=t \end{align*} and \begin{align*} \int dz&=\int 0 dt\\ z&=0 \end{align*} combined with the initial condition for $t=0$ lead to the parametric representation \begin{equation*} x=s+zt,\quad y=t, \quad z=h(s) \end{equation*} now as \begin{equation*} s=x-zt\quad \text{and}\quad y=t \quad \text{and}\quad z=h(s) \end{equation*} For the solution $z=u(x,y)$ then yields the implicit equation \begin{equation*} u=h(x-uy) \end{equation*} for $u$ as a function of $x,y$.The characteristic (projection) $C_s$ in the $xy$-plane passing through the point $(s,0)$ \begin{align*} u(s,0)&=h(s-u\cdot(0))\\ &=h(s) \end{align*} then is the line \begin{equation*} x = s + h (s) y \end{equation*} along which $u$ has the constant value \begin{equation*} u(s,0)=h(s) \end{equation*} Now two characteristics $C_{s_1}$ and $C_{s_2}$ intersect at a point (x,y) with \begin{equation*} y=-\frac{s_2-s_1}{h(s_2)-h(s_1)} \end{equation*} it's going to be done $u_x$ \begin{equation*} u_x=\frac{h'(s)}{1+h'(s)y} \end{equation*} Hence for $h'(s) <0$ we find that $u_x$ becomes infinite at the positive time \begin{equation*} y=\frac{-1}{h'(s)} \end{equation*} I have two questions
I do not understand How do you get $u_x=\dfrac{h'(s)}{1+h'(s)y}$?
why if $ h '(s) <0 $ gives that $y=\dfrac{-1}{h'(s)}$?
I would be grateful for your help
