= Solution
{c}
Use the initial conditions $x(0,r)=1$, $y(0,r)=r$, $u(0,r)=g(r)$ in the <characteristic equations for a transport equation>. The resulting parametrized <solution surface> is
$$
\boxed{\begin{aligned}
x(s,r)&=e^s,\\
y(s,r)&=(r+\beta)e^{2s}-\beta e^s,\\
u(s,r)&=\min\{r,0\}+e^s-1+2s,
\end{aligned}\qquad s\in\mathbb R,\ -1\leq r\leq1.}
$$
The <Jacobian determinant> of $(s,r)\mapsto(x,y)$ is $e^{3s}>0$. In fact the inverse is explicit:
$$
s=\log x,\qquad r=q(x,y):=\frac{y+\beta x}{x^2}-\beta.
$$
Consequently the explicit <solution of a differential equation> on the region reached from the data is
$$
\boxed{u(x,y)=x-1+2\log x+\min\left\{\frac{y+\beta x}{x^2}-\beta,0\right\}.}
$$
To verify it away from $q=0$, set $D=x\partial_x+(2y+\beta x)\partial_y$. Direct <differentiation> gives $Dq=0$ and $D(x-1+2\log x)=x+2$. At $x=1$ we have $q=y$, giving precisely the prescribed <boundary condition>.
Across $q=0$ the formula remains <continuous> and <locally Lipschitz>, is a <smooth function> on each side, and solves the <partial differential equation> along every <characteristic curve>. It is also a <weak solution>, since its first <distributional derivatives> agree almost everywhere with its ordinary <derivatives>. It is not a <classical solution> throughout the region because of the inherited corner.
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