= Solution
Suppress $(x,t)$, put $R(k)=b(k)/a(k)$ and $C=c e^{-2px+8p^3t}N(ip)$. Removing the pole gives the additive jump
$$
\mathcal M(k)-N(-k)=R(k)e^{2ikx+8ik^3t}N(k)-\frac C{k-ip}\equiv G(k),\qquad k\in\mathbb R.
$$
The 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 is
$$
\boxed{N(k)=1-\frac{c e^{-2px+8p^3t}N(ip)}{k+ip}+\frac1{2\pi i}\int_\mathbb R\frac{R(s)e^{2isx+8is^3t}N(s)}{s+k}\,ds.}
$$
It is supplemented by its value at $k=ip$, and follows under the usual decay or principal-value hypotheses of the stated <Riemann-Hilbert problem>.
Back to article page
