 Solution

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.