Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2006 / Differential equations / 8

With parameter $t$ along a characteristic curve, the equations areTheir solutions are $y=t$, $x=s+t^2/2$ and $u=s{e}^t$. Thus the projected curve satisfies $y^2=2(x-s)$, as required. Eliminating the parameters givesIndeed $u_x=e^y$ and $u_y=e^y(x-y^2/2-y)$, so $y{u}_x+u_y=u$, and the initial values hold.

The parameter map $(s,t)\mapsto (s+t^2/2,t)$ has Jacobian one and is globally invertible. Every point whose characteristic curve meets the initial segment therefore lies in the stripThe uniquely determined open region is its interior, with the boundary values determined on the two endpoint characteristic curves. Both signs of $y$ are included. Outside the strip, a general solution is $u=e^{y}F(x-y^2/2)$ with $F(s)=s$ only required for $1\le s\le 2$. Smooth perturbations of $F$ supported outside that interval produce different solutions with the same prescribed data, proving the stated region is exact.