The coefficient matrix of $(u_y,v_y{)}^T$ is $A=\left(\begin{array}{cc}v& u\\ u^{-3}& v\end{array}\right)$. Its distinct eigenvalues are ${\lambda }_{\pm }=v\pm 1/u$ for every $u>0$, so the system is strictly hyperbolic. Define $D_{\pm }={\partial }_x+(v\pm 1/u){\partial }_y$. Thenthe derivative terms cancel, as do the sources $u-u$. Hence $R=v-1/u$ is constant on the plus-family characteristics.

The data give $R=0$, so throughout their domain of dependence $v=1/u$. The minus-family speed is then $v-1/u=0$: the other characteristics are the horizontal lines $y=a$. Substitution into the second PDE, or its minus-characteristic compatibility relation, gives $v_x=1/v$. Integrate at fixed $y$ using $v(0,y)=y$ and retain the positive branch:Indeed $u_x=-u^3$, $u_y=-y{u}^3$, $v_y=yu$, so $v{u}_y+u{v}_y=0$; similarly $u^{-3}{u}_y+v{v}_y=-y+y=0$.

For $x\ge 0$ the lower edge of the region determined solely by the given data is the plus characteristic from $(0,1)$. Write it $y=b(x)$, where the exact boundary characterization isThe determined region is $\boxed{x\ge 0,y\ge b(x)}$, with strict inequality for its interior. Both backward characteristic families there reach the supplied initial half-line. Below this boundary a backward plus characteristic would require initial data below $y=1$, which were not supplied; extending the displayed formulas there gives a possible continuation but is not forced by the data. An elementary expression for $b$ is unnecessary: its initial-value ODE defines it uniquely and the requested other family already has the explicit form $y=a$. The sketch integrates only this boundary for illustration.

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.