Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / B / 7 / Solution

Orient $\Gamma$ and label its left side plus and its right side minus. Assume the density is Hölder continuity near the point in question, or has sufficient regularity for the boundary limits below. The source does not explicitly state this regularity assumption. Parametrize locally by oriented arclength $\gamma (s)$, with $\gamma (0)=t$. Define the Cauchy principal value by removing equal arclengths on the two sides:Subtract $f(t)$ from the numerator near $t$. The Hölder continuity remainder is locally bounded in absolute value by a constant times $|s{|}^{\nu -1}$ for some $\nu >0$, so its integral has the same limit from the two sides. The constant part contributes half the residue, with sign determined by the side of the small semicircle. This proves the Sokhotski–Plemelj theoremIn particular $f=w_{+}-w_{-}$ and the Cauchy principal value term with coefficient $1/(\pi i)$ equals $w_{+}+w_{-}$. Substituting these two identities into the singular integral equation gives

Choose a canonical homogeneous solution $\tilde{w}$ whose boundary values do not vanish, and account for any permitted zeros or poles when forming a quotient. The assumptions $a\pm b\ne 0$ alone do not ensure that an arbitrary homogeneous solution has this property. Its jump relation isDividing the inhomogeneous relation by $(a+b){\tilde{w}}_{+}$ therefore givesThe Cauchy integral solves this additive jump. More generally every solution in the chosen analytic class haswhere $H$ has no jump across $\Gamma$ and obeys the applicable growth, endpoint and divisor conditions. Setting $H=0$ gives exactly the intermediate formula printed in the source.

Let $C\rho =(2\pi i{)}^{-1}\mathrm{PV}{\int }_{\Gamma }\rho (\zeta )/(\zeta -t)d\zeta$. Applying Sokhotski–Plemelj theorem to the displayed solution and subtracting boundary values yieldsUseThe normalized particular solution is consequentlyand the full solution isThus the printed expression is a particular solution; homogeneous terms need separate normalization. To see why “general” cannot hold without qualification, take the positively oriented unit circle, $a=(1+z)/2$, $b=(1-z)/2$, and $c=0$. Then $a+b=1$ and $a-b=z$ are nonzero on the contour. The analytic pair ${\tilde{w}}_{+}=1$, ${\tilde{w}}_{-}=1/z$ satisfies the homogeneous jump and decays on the exterior side. Its nonzero density $f=1-1/z$ has these very Cauchy boundary values and solves the original homogeneous equation, whereas the source formula with $c=0$ gives zero. This is homogeneous freedom in a scalar singular integral equation; endpoint and index conditions determine which jump-free additions are permitted.