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The modified Helmholtz closed spectral one-form is exact in the simply connected quadrant. Writing its primitive as $\mu E$ gives the compatible first-order equationsTheir compatibility is precisely the PDE already proved in part (a). Set $H=E^{-1}=e^{ia(k)x-b(k)y}$. Construct three primitives: ${\mu }_0$ starts at the corner, ${\mu }_x$ starts at the infinite end of a horizontal line through the point, and ${\mu }_y$ starts at the infinite end of a vertical line. More explicitly,Path independence follows from closedness. The two infinite primitives exist in the lower and left spectral half-planes respectively. In their common third quadrant they coincide, because the global relation makes the difference of the two basepoint integrals zero.

Choose a sectionally analytic spectral function by using ${\mu }_0$ in quadrant I, ${\mu }_y$ in quadrant II and ${\mu }_x$ in quadrants III and IV. It has no jump across the negative axes. Let ${\ell }_x=(0,\infty )$ and ${\ell }_y=(0,i\infty )$, both oriented outward. With plus denoting the left side of an oriented ray, the two remaining jumps areFor example, on ${\ell }_x$ the jump is ${\mu }_0-{\mu }_x$, the primitive around the bottom boundary; on ${\ell }_y$ it is ${\mu }_y-{\mu }_0$, the left boundary traversed downward. This fixes both orientations and signs of the Riemann-Hilbert problem.

Integration by parts in the infinite primitive formulas, or in the finite primitive with its decaying basepoint contribution, givesFor instance the horizontal formula has leading value $-bq/a$, which tends to $-q$ at infinity and $q$ at zero; the vertical formula has leading value $-aq/b$ with the same limits. These expansions hold in closed subsectors, with the boundary limits obtained by oscillatory regularization. Along ${\ell }_x$, $H$ damps with $y>0$ at both spectral ends; along ${\ell }_y$, it damps with $x>0$. Thus the ray integrals converge in the interior and small and large spectral circles can be removed.

Solve the additive Riemann-Hilbert problem by a Cauchy integral formula, normalized at infinity:The Sokhotski–Plemelj theorem gives the stated jumps. The difference between this expression and the constructed primitive is analytic across both rays, bounded at zero and vanishes at infinity; the Liouville theorem makes that difference zero. Evaluate at zero, where $\mu (0)=q$. The difference between the two spectral normalizations is $2q$, hence the quadrant modified Helmholtz spectral reconstruction isThe factor $1/(4\pi i)$ is essential. A useful independent normalization check is to write $p=a(k)$ on the real ray, so $b=\sqrt{p^2+4\lambda }=B(p)$ and $dk/k=dp/B$. On the imaginary ray write $k=ir$, $p=r-\lambda /r$ and $a=iB(p)$. The formula becomesEach hat here is a half-line Fourier transform. This is exactly the spectral form of Green second identity with the decaying full-plane fundamental solution. The representation still contains both value and derivative traces; the next part eliminates the latter.