= Solution
The <method of characteristics> gives
$$
\frac{dx}{ds}=-y,\qquad
\frac{dy}{ds}=x,\qquad
\frac{du}{ds}=0.
$$
Along a characteristic,
$$
\frac d{ds}(x^2+y^2)=2x(-y)+2yx=0.
$$
Thus the characteristics are circles $x^2+y^2=C$, traversed counterclockwise, and $u$ is constant on each circle. Their intersections with the $x$-axis have $x^2=C$, where the boundary value is $f(C)$. Hence
$$
\boxed{u(x,y)=f(x^2+y^2)}.
$$
The characteristic sketch is the family of concentric circles centred at the origin.
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