Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 29C / iii / Solution

Start a characteristic curve at $(0,\eta )$ with $u=-\eta$. The Characteristic curve equations give $u=-\eta$, $y=\eta +s$, $x=-\eta s$, so $x={\eta }^2-\eta y$. Therefore the two branches areEach is $C^1$ on $D=\{y^2+4x>0\}$. The plus branch matches $u(0,y)=-y$ for $y>0$, whereas the minus branch matches it for $y<0$. Differentiating $u^2+yu-x=0$ gives $u_x=1/(2u+y)$ and $u_y=-u/(2u+y)$, verifying $u_y+u{u}_x=0$ wherever $2u+y\ne 0$.

On the initial line, the normal Characteristic curve speed is $u=-y$, nonzero away from the origin. Thus the local existence theorem gives precisely these separate local branches near positive and negative initial points. At $y^2+4x=0$ the branches merge and their derivatives become singular. There is no $C^1$ solution realizing the entire initial line in a neighborhood of the origin: the data force $u_y(0,0)=-1$, while the equation and $u(0,0)=0$ force $u_y(0,0)=0$. The failure at this Characteristic curve point is a genuine limitation of the printed global initial-data request.