Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 31E / a / Solution

Suppress $(x,t)$, put $R(k)=b(k)/a(k)$ and $C=c{e}^{-2px+8p^{3}t}N(ip)$. Removing the pole gives the additive jumpThe upper analytic function is $\mathcal{M}-1$ and the lower analytic function is $N(-k)-1$. The Cauchy integral formula and the Sokhotski–Plemelj formula solve this jump as the upper and lower boundary values of $(2\pi i{)}^{-1}{\int }_{\mathbb{R}}G(s)/(s-z)ds$. At the lower point $z=-k$, with $\Im k>0$, this gives $N(k)-1$.

The rational part can be evaluated by closing the contour upward: ${\int }_{\mathbb{R}}[(s-ip)(s+k){]}^{-1}ds=2\pi i/(k+ip)$. Thus the required linear integral equation isIt is supplemented by its value at $k=ip$, and follows under the usual decay or principal-value hypotheses of the stated Riemann-Hilbert problem.