Past exam of the mathematical methods II course of the University of Oxford / 2022 / 9 / b / ii / Solution

Let $F^{\prime }=f$ and $G^{\prime }=g$ on the relevant intervals. The chain rule givesThus the one-form is explicitly an exact differential. Fixing its potential to vanish at $(1,1)$ determines it uniquely on the connected domain:Continuity of $f,g$ on the integration ranges suffices. For arbitrary smooth data defined throughout $x>0$, that half-plane also supplies the simply connected domain in the previous argument.

For the specialization $f(s)=s^{-2}$, the potential based at $(1,1)$ is defined in the positive quadrant; it cannot be continued through $xy=0$ using this formula. Taking $a,b>0$ keeps both endpoints in that component. Their ratios $y/x$ are both $b/a$, so the $G$ term cancels andNo condition on $g$ beyond existence on the traversed ratios enters this difference.