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For a first-order nonlinear partial differential equation $F(x,y,u,p,q)=0$, with $p=u_x,q=u_y$, Charpit's equations extend the method of characteristics to the five-dimensional space of position, value and gradient:The chain rule gives $dF/dt=0$ by direct cancellation, so a trajectory starting on $F=0$ remains there. Initial data $u_0(s)$ on $(x_0(s),y_0(s))$ require both $F(x_0,y_0,u_0,p_0,q_0)=0$ and $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. The non-characteristic condition is $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It makes the projection onto $(x,y)$ locally invertible. Compatible initial contact data then propagate as $du=pdx+qdy$, reconstructing a classical solution. Multiple permitted initial normal derivatives can give different local branches; loss of the projection Jacobian determinant marks a characteristic caustic.

Set $F=xp+q^2/4+u-1$. Parametrize the data by $x_0=s$, $y_0=0$, $u_0=1-s^2/3$. Tangential differentiation gives $p_0=-2s/3$; substitution into $F=0$ then gives $q_0^2=4s^2$. Thus the data allow two normal-derivative branches, indexed by $\epsilon =\pm 1$, with $q_0=2\epsilon s$. For $s>0$ each branch is non-characteristic; the endpoint $s=0$ is degenerate.

The Charpit's equations reduce to $\dot{x}=x$, $\dot{y}=q/2$, $\dot{p}=-2p$, $\dot{q}=-q$ and $\dot{u}=xp+q^2/2$. Integrating with these initial values givesDirect substitution checks $xp+q^2/4+u-1=0$, and $t=0$ recovers the initial data. The projection onto the $(x,y)$ plane has Jacobian determinantIt first vanishes at $t=\log 2$. These points form the characteristic caustic $y=\epsilon x/4$, $0<x\le 2$; a single classical solution cannot in general be continued through the fold by treating all projected characteristic curves as a graph.

For a precise description put $r=\epsilon y$. On the chart containing the initial curve and lying before the fold, $t<\log 2$, elimination givesThe plus square-root is forced by $s/x>1/2$, rather than a free second choice. The conditions $0<s\le 1$ give the complete unfolded chartInterior points of the domain have strict inequality at the boundary generated by $s=1$ as well. The other boundary is the characteristic caustic; the original endpoint $s=0$ contributes no open chart. On the portion obtained by following characteristics forward from the data, $0\le t<\log 2$, the region is more specificallyThe $s=1$ boundary is $y=\epsilon (1-1/x)$ and meets the characteristic caustic at $(2,\epsilon /2)$. Negative characteristic times give the continuation across the initial segment into the opposite half-plane. Each sign branch is a distinct local solution; the initial values alone do not select one of them.

The solid shading follows $t\ge 0$; the lighter shading shows the same smooth chart at $t<0$. Reflected panels correspond to the two allowed signs of the initial normal derivative. Beyond the dashed characteristic caustic the characteristic parametrization no longer gives this smooth graph.

Take $F(p,q)=p^2+pq-1$. Parametrize the initial line by $x_0=s>0$, $y_0=0$, $u_0=2s^2$. Initial compatibility gives $p=4s$, and $F=0$ then gives $q=1/(4s)-4s$. The characteristic formula in part (a) yieldsSince $y>0$, take $\tau >0$. To select the branch connected smoothly to the initial data, stop each characteristic before its coordinate map folds. Its Jacobian determinant isso this branch has $0<\tau <2s^2$. Eliminating $\tau =y/(4s)$ givesFor fixed $y>0$, the right side defining $x-y$ is strictly increasing on that selected branch. At its lower endpoint it equals $3y^{1/3}/4$, and it tends to infinity as $s\to \infty$. Thus it gives exactly one $s$ for every point inSubstituting that unique $s$ into the formula for $u$ gives the required classical solution. As $y\downarrow 0$ at fixed $x>0$, $s\to x$, and $u\to 2x^2$ as required.

Without the pre-fold branch restriction, the algebraic equation also has a smaller positive root. That root tends to zero as $y\downarrow 0$, and its $u$ tends to zero rather than $2x^2$, so it does not solve the prescribed initial-boundary problem. The fold $J=0$ is the characteristic caustic; uniqueness here is for the classical branch continued from the given initial data before that fold.

On an open set where $u>0$, put $v=\sqrt{u}$. The chain rule gives $\nabla u=2v\nabla v$, so the quadratic eikonal equation is equivalent to $|\nabla v|=1$. For constant positive data $v=v_0$ on a smooth initial curve, the tangential derivative vanishes and the two allowed normal derivatives are ${\partial }_{n}v=\pm 1$. Locally the two solutions are $v=v_0\pm d$, where $d$ is oriented distance along the normal rays; hence $u=(v_0\pm d{)}^2$ wherever the chosen $v$ stays positive and the normal-coordinate map remains invertible.

There are two separate obstructions to using this initial-data chart. A characteristic caustic occurs when normal rays focus and the projection Jacobian determinant vanishes. Alternatively $v$ can reach zero while the geometric rays remain separated: the square-root transformation then ceases to apply and the original Charpit's equations have a degenerate zero-gradient state. A continuation of $u$ beyond either obstruction needs separate justification; squaring an algebraic expression alone does not preserve a prescribed positive square-root branch.