Solution
= Solution
The substitution gives $x^2-y^2=u^2\cos2v$. For this function $f_{uu}=2\cos2v$, $f_u/u=2\cos2v$, and $f_{vv}/u^2=-4\cos2v$, whose sum is zero. Thus it satisfies the polar form of <Laplace's equation> for $u>0$. The apparent coordinate singularity at $u=0$ does not make the original Cartesian <polynomial> singular; it is a smooth <harmonic function> through the origin.