Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 11

The coordinate partial derivatives are $x_s=-x$, $y_s=-y$, $x_t=y$, and $y_t=-x$. Apply the multivariable chain rule to $v(s,t)=u(x(s,t),y(s,t))$:All partial derivatives of $u$ are evaluated at $(e^{-s}\sin t,e^{-s}\cos t)$. This is a logarithmic polar coordinate change; its Jacobian determinant is $x^2+y^2=e^{-2s}>0$, so it is locally invertible away from the origin. A full angular turn repeats the same point, as usual for polar coordinates.

Differentiate $v_t=y{u}_x-x{u}_y$ with respect to $t$, remembering that its coefficients also depend on $t$. The product rule and multivariable chain rule giveFor a twice continuously differentiable $u$, equality of mixed partial derivatives gives $u_{yx}=u_{xy}$. HenceThe first-order terms come from differentiating $x$ and $y$ themselves; treating $y{\partial }_x-x{\partial }_y$ as an operator with constant coefficients would miss them.

The last two terms in the preceding result equal $v_s$. ThereforeThe left-hand side of the original partial differential equation is consequently $v_{tt}-v_s$, givingIn these variables it has the form of a heat equation, with logarithmic radial coordinate $s$ playing the evolution variable and angular coordinate $t$ playing the spatial variable. If $u$ is single-valued around the whole punctured plane, then $v(s,t+2\pi )=v(s,t)$ as well; that periodicity is a global coordinate condition, not an extra local transformation term.