Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 31D (Integrable Systems)

Let $C(z)=(2\pi i{)}^{-1}{\int }_L\varphi (\tau )/(\tau -z)d\tau$, with interior and exterior boundary values $C_{+},C_{-}$. The Sokhotski–Plemelj formula gives $\varphi =C_{+}-C_{-}$ and $C_{+}+C_{-}=(\pi i{)}^{-1}\mathrm{PV}{\int }_L\varphi (\tau )/(\tau -t)d\tau$. Write $K=(\pi i{)}^{-1}{\int }_L(\tau +1/\tau )\varphi (\tau )d\tau$ for the nonlocal scalar moment. Substitution into the integral equation yieldsThus the associated scalar Riemann-Hilbert problem isThe functions are analytic in their respective interior/exterior domains, and $K$ must ultimately equal the moment of the recovered jump; it is not an independently specified free constant.

The jump coefficient $G(t)=(t^2-1)/t$ has one enclosed pole at zero and no enclosed zeros, because $\pm 1$ are outside. Its winding index is therefore $\kappa =-1$. A homogeneous canonical pair isEach is analytic and nonvanishing in its required finite domain; the exterior growth $X_{-}\sim z$ encodes the negative index. Dividing the unknown by these factors gives an additive jumpThe Cauchy-integral solution for $D$ normally has an exterior $z^{-1}$ coefficient $-(2\pi i{)}^{-1}{\int }_{L}g$. Since $C_{-}=z{D}_{-}$ must decay, this coefficient must vanish. Thus the solvability condition is ${\int }_{L}g(t)dt=0$. Its only enclosed pole is zero, with residue $(1-K)/2$, soThese are the negative-index scalar Riemann-Hilbert moment conditions for index minus one. There is no arbitrary polynomial addition: an entire addition respecting $D_{-}=O(z^{-2})$ is zero by Liouville's theorem.

With $K=1$, the additive jump is $g(t)=1/[2(t^2-1)]$, analytic inside the contour. Its Cauchy integral has $D_{+}(z)=g(z)$ and $D_{-}(z)=0$. Therefore $C_{+}=(z^2-1)g=1/2$, $C_{-}=0$, andThe moment is indeed $(\pi i{)}^{-1}{\int }_L(\tau +1/\tau )(1/2)d\tau =1$, as required. The principal-value Cauchy sum is $1/2$, and the original left side becomes $[(t^2+t-1)-(t^2-t-1)]/2-1=t-1$. Thus the recovered function satisfies both the equation and the self-consistency condition, and the homogeneous negative-index argument proves uniqueness.