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Write the parameter along a characteristic curve as $r$. Charpit's equations areAt $r=0$, prescribe $x=x_0(s)$, $y=y_0(s)$, $u=u_0(s)$ and choose $p_0,q_0$ satisfying both $F(x_0,y_0,u_0,p_0,q_0)=0$ and the compatibility relation $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. Direct differentiation givesThus the initial constraint propagates. A chosen initial-gradient branch is locally unique whenThis is simultaneously the implicit-function determinant for the initial gradients and the projection map determinant for the surface generated by the method of characteristics. Uniqueness continues while that projection map is invertible, the flow remains regular, and distinct characteristic curves do not give competing values at the same point. Multiple initial-gradient branches, characteristic curve initial points or a caustic require separate analysis.

Here the initial coordinate is $s=x$, so $p_0=3s^2/4$ and the equation forces $q_0^3=-s^3$, hence $q_0=-s$. The projection map determinant on the initial line is $3s^2$, nonzero for $0<s\le 1$. Solving the characteristic curve ordinary differential equations giveswith $p=3s^2{e}^{-2r}/4$ and $q=-s{e}^{-r}$. These satisfy both the equation and all the noncharacteristic initial values.

The projection map determinant is $J=3s^2{e}^r(2e^{-2r}-1)$. Starting from $r=0$, the first caustic occurs at $r_c=\frac{1}{2}\log 2$. At the initial endpoint $s=0$ the non-characteristic condition fails, and its entire characteristic curve remains the single point $(0,0)$; it does not generate an open region of determination.

The maximal regular sheet connected to the initial line uses $r<r_c$. Its interior image is exactlyIndeed the inverse on this sheet isand the inequalities say $0<s<1$ and $s>x/\sqrt{2}$. The upper boundary is the caustic; the lower boundary is the characteristic curve from $s=1$. Endpoints of that latter boundary can be retained as one-sided data. For forward characteristic curves alone, also impose $r\ge 0$: this retains $y\ge 0$, with lower boundary zero for $x\le 1$ and the displayed endpoint characteristic curve for $1<x<\sqrt{2}$. The other square-root branch belongs to the folded sheet after the caustic and does not provide a unique regular continuation from the initial data.

The square-root reduction of a quadratic eikonal equation explains why two initial normal derivatives are possible. Set $F=p^2+q^2-4u$. Charpit's equations are $\dot{x}=2p$, $\dot{y}=2q$, $\dot{u}=2p^2+2q^2=8u$, $\dot{p}=4p$, $\dot{q}=4q$. Parametrize the initial semicircle by $(x_0,y_0)=(\cos s,\sin s)$, $-\pi /2<s<\pi /2$. Tangential compatibility gives $-p_0\sin s+q_0\cos s=0$, and $F=0$ gives $p_0^2+q_0^2=4$. Thus there are two normal-gradient choices, $p_0=2\epsilon \cos s$, $q_0=2\epsilon \sin s$, $\epsilon =\pm 1$.

Integration gives the requested parametric solutionsFor $\epsilon =+1$, $r=\sqrt{x^2+y^2}=e^{4t}$, so $u_{+}=r^2$ on the whole right half-plane. For $\epsilon =-1$, before focusing at the origin, $r=2-e^{4t}$, so $u_{-}=(2-r{)}^2$ on $x>0$, $0<r<2$. The origin is a caustic for this inward family. The circle $r=2$ is approached only as characteristic time tends to negative infinity, where $u$ and its gradient vanish. It cannot be crossed by this initial characteristic chart. Thus the initial data alone permit two branches; each is locally unique only after its normal-gradient sign is chosen.

On the added half-line write $x_0=a\le 0$, $y_0=1$, $u_0=1$. Tangential differentiation gives $p_0=0$ and $q_0=2\delta$, $\delta =\pm 1$. Its characteristics giveThe plus branch is $u=y^2$ on $x\le 0,y>0$; the minus branch is $u=(2-y{)}^2$ on $x\le 0,y<2$. On the positive $y$-axis, the semicircle traces are $y^2$ and $(2-y{)}^2$, respectively. Therefore only matching sign choices are continuous throughout their common positive-axis coverage. Mixed choices agree merely at the initial junction $y=1$.

For the matching-plus branch the data-covered open domain isFor the matching-minus branch it isIts formula is $(2-\sqrt{x^2+y^2}{)}^2$ in the right portion and $(2-y{)}^2$ in the left portion. The origin is excluded. The negative $y$-axis must also be a cut: the two formulas have different traces there, so including it would destroy continuity. Both matching solutions are $C^1$ across the included positive-axis seam; generally their second transverse derivatives differ, which is compatible with this first-order problem. Values beyond these characteristic charts can sometimes be continued, but are not selected by the initial data through a nonsingular characteristic graph.

For the general equation the Charpit's equations areAt parameter time zero set $(x,y,u)=(x_0,y_0,u_0)$ and choose $p_0,q_0$ to satisfy $F(x_0,y_0,u_0,p_0,q_0)=0$ and $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. The chain rule givesso $F$ remains zero. The non-characteristic condition is $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$, ensuring a local graph.

For the particular equation put $F=p^3+(y-1)q+u-2$ and label the initial curve by $y_0=1+s$, $0<s\le 1$, $x_0=1$, $u_0=2+2s^3$. Compatibility gives $q_0=6s^2$ and $p_0^3=-8s^3$, hence the unique real choice $p_0=-2s$. Integrating the characteristic equations givesThe projection determinant is $J=12s^2{e}^t(1-2e^{-2t})$. Initially $J=-12s^2\ne 0$; its first zero is $t=\frac{1}{2}\log 2$. At that fold,To describe the entire chart attached to the data before the fold, put $Y=y-1>0$ and $w=e^{-2t}>1/2$. Elimination gives $x-1=6Y^{2}w(1-w)$ andThe initial-endpoint constraint $s\le 1$ therefore gives the uniquely determined chartThe boundary $s=1$ is included as inherited endpoint data; open-domain statements use strict inequality there. The fold is excluded. For forward characteristic times only, $0\le t<\frac{1}{2}\log 2$, impose additionally $x\ge 1$. The source excludes $s=0$, so $(1,1)$ is not a non-characteristic initial point. The sheet beyond the fold does not extend this initial graph uniquely.

Use the initial-curve parameter $s$ and a characteristic parameter $t$. Charpit's equations areAt $t=0$ prescribe $x=x_0(s)$, $y=y_0(s)$ and $u=u_0(s)$. The two initial derivatives $p_0,q_0$ must satisfyThe second equation is the chain rule along the initial curve. For the usual local Cauchy construction also require the non-characteristic condition $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It ensures both local determination of an allowed branch of $p_0,q_0$ and local inversion of the projected characteristic coordinates.

Along a characteristic, the chain rule and the five equations giveSince $F=0$ initially, $\boxed{F\equiv 0}$ along each constructed curve. With the contact compatibility $du=pdx+qdy$ propagated by this system, an invertible characteristic chart reconstructs $p=u_x,q=u_y$ and hence a local classical solution.

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.

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.