Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 3C (Vector Calculus)

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.