= Solution
Write the parameter along a <characteristic curve> as $r$. <Charpit's equations> are
$$
\boxed{\dot x=F_p,\quad\dot y=F_q,\quad\dot u=pF_p+qF_q,\quad
\dot p=-F_x-pF_u,\quad\dot q=-F_y-qF_u.}
$$
At $r=0$, 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 $p_0x_0'+q_0y_0'=u_0'$. Direct differentiation gives
$$
\frac{dF}{dr}=F_xF_p+F_yF_q+F_u(pF_p+qF_q)
-F_p(F_x+pF_u)-F_q(F_y+qF_u)=0.
$$
Thus the initial constraint propagates. A chosen initial-gradient branch is locally unique when
$$
\boxed{x_0'F_q-y_0'F_p\ne0.}
$$
This 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 $0<s\leq1$. Solving the <characteristic curve> <ordinary differential equations> gives
$$
\boxed{x=se^r,\quad y=\frac32s^2(1-e^{-2r}),\quad
u=1+s^3(e^{-3r}-\tfrac34e^{-r}),\quad0<s\leq1,}
$$
with $p=3s^2e^{-2r}/4$ and $q=-se^{-r}$. These satisfy both the equation and all the noncharacteristic initial values.
The <projection map> <determinant> is $J=3s^2e^r(2e^{-2r}-1)$. Starting from $r=0$, the first <caustic> occurs at $r_c=\tfrac12\log2$. 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 exactly
$$
\boxed{0<x<\sqrt2,\qquad\frac32(1-x^{-2})<y<\frac38x^2.}
$$
Indeed the inverse on this sheet is
$$
s^2=\frac{x^2+\sqrt{x^4-\tfrac83x^2y}}2,\qquad r=\log(x/s),
$$
and the inequalities say $0<s<1$ and $s>x/\sqrt2$. 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 $r\geq0$: this retains $y\geq0$, with lower boundary zero for $x\leq1$ and the displayed endpoint <characteristic curve> for $1<x<\sqrt2$. 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>.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2007-charpit-caustic.png]
{title=The regular characteristic sheet issued from the initial segment, its endpoint characteristic and the first caustic}
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