= Solution
For the general equation the <Charpit's equations> are
$$
\dot x=F_p,\quad\dot y=F_q,\quad\dot u=pF_p+qF_q,\quad\dot p=-F_x-pF_u,\quad\dot q=-F_y-qF_u.
$$
At 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 $p_0x_0'+q_0y_0'=u_0'$. The <chain rule> gives
$$
\frac{dF}{dt}=F_xF_p+F_yF_q+F_u(pF_p+qF_q)-F_p(F_x+pF_u)-F_q(F_y+qF_u)=0,
$$
so $F$ remains zero. The <non-characteristic> condition is $x_0'F_q-y_0'F_p\ne0$, ensuring a local graph.
For the particular equation put $F=p^3+(y-1)q+u-2$ and label the initial curve by $y_0=1+s$, $0<s\leq1$, $x_0=1$, $u_0=2+2s^3$. Compatibility gives $q_0=6s^2$ and $p_0^3=-8s^3$, hence the unique real choice $p_0=-2s$. Integrating the characteristic equations gives
$$
\boxed{\begin{aligned}
x&=1+6s^2(1-e^{-2t}),&y&=1+se^t,\\
p&=-2se^{-t},&q&=6s^2e^{-2t},\\
u&=2+8s^3e^{-3t}-6s^3e^{-t},&&0<s\leq1.
\end{aligned}}
$$
The projection determinant is $J=12s^2e^t(1-2e^{-2t})$. Initially $J=-12s^2\ne0$; its first zero is $t=\tfrac12\log2$. At that fold,
$$
x=1+3s^2,\qquad y=1+\sqrt2s,\qquad\boxed{x=1+\tfrac32(y-1)^2.}
$$
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 $x-1=6Y^2w(1-w)$ and
$$
w=\frac{1+\sqrt{1-2(x-1)/(3Y^2)}}2,\qquad s=Y\sqrt w.
$$
The initial-endpoint constraint $s\leq1$ therefore gives the uniquely determined chart
$$
\boxed{0<Y<\sqrt2,\qquad 7-\frac6{Y^2}\leq x<1+\frac32Y^2.}
$$
The 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, $0\leq t<\tfrac12\log2$, impose additionally $x\geq1$. 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.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2009-cubic-charpit-region.png]
{title=Initial segment, endpoint characteristic, and fold of the cubic-gradient solution}
Back to article page
