Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / B / 5

Orient the contour so that the plus side is on the left; for a positively oriented simple closed contour this is its interior. If $\zeta (s)$ is a smooth arclength parametrization with $\zeta (0)=t$, define the Cauchy principal value by deleting a symmetric interval $|s|<\eta$ and taking $\eta \downarrow 0$. This pairs the two singular sides at equal local arclength.

Near $t$, write $f(\zeta )=f(t)+[f(\zeta )-f(t)]$. The second term divided by $\zeta -t$ is regular because $f$ is analytic there. For the constant term, indent the contour on either side of $t$ by a small semicircle. Its limiting contribution to the Cauchy integral is $\pm f(t)/2$, while the remaining contour tends to its principal value. This proves the Sokhotski–Plemelj theorem:In particular $w_{+}-w_{-}=f$ and $w_{+}+w_{-}=g$ for the given integral equation.

Define $W=w$ inside and $W=-w$ outside. Its jump is $W_{+}-W_{-}=w_{+}+w_{-}=g$, and it vanishes at infinity. ThereforeIndeed, the difference between the two sides has no jump, extends analytically across the smooth contour, and vanishes at infinity; Liouville's theorem makes it zero. Applying the Sokhotski–Plemelj theorem again,Thus the normalized closed-contour singular integral operator is its own inverse.

For an admissible Laplace integral solution, differentiation under the integral sign and integration by parts giveChoose endpoints for which the boundary term vanishes. Setting the density in braces to zero givesTwo suitable contours are the rays ${\Gamma }_{+}:0\to +i\infty$ and ${\Gamma }_{-}:0\to -i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O({\zeta }^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.

To check independence rather than merely exhibit different paths, defineWith unit normalization, the two solutions satisfy $w_{\pm }(0)=e^{\pm i\pi /4}{J}_0$ and $w_{\pm }^{\prime }(0)=\pm i{e}^{\pm i\pi /4}{J}_1$. Their Wronskian at zero is $-2i{J}_0{J}_1\ne 0$, so they form an independent pair.

Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a null weight in a contour Laplace representation, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.