 Solution

The characteristic equations for a first-order quasilinear equation are $\dot{x}=x$, $\dot{y}=y$, $\dot{u}=2u$, with characteristic parameter $s$. Through the prescribed point they giveFor a nonzero projected initial point, the characteristic projections are rays from the origin, not trajectories crossing through the origin to the opposite ray. If $x_0=y_0=0$, a classical solution graph requires $u_0=0$, because the original equation reduces there to $0=2u$.

Parameterize the initial line by $(x,y,u)=(a,1-a,a^3)$. Along its characteristic, $x+y=e^s>0$, $a=x/(x+y)$ and $u=a^3{e}^{2s}$. ThereforeDifferentiation verifies the equation, and substitution of $x+y=1$ verifies the initial data. Characteristic reachability for radial transport shows why the domain is this positive half-plane: each point there lies on precisely one characteristic from the data line. Rays in $x+y<0$ never meet that line, so their values are not determined by these data. Allowing a negative scaling factor would incorrectly join two different characteristic rays through a degenerate origin.