Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 86 / 3 / c / Solution

First resolve the orientation sign in quadrant harmonic spectral inversion. The printed spectral definitions traverse the physical boundary clockwise, whereas both reconstruction rays run outward from zero. Integrating first in $k$ gives the kernel $i/(z-w)$. The resulting clockwise Cauchy integral is the negative of the desired holomorphic function. Thus, with exactly the spectral definitions used above, the consistent reconstruction isThe printed plus sign would reconstruct $-q_z$. For a concrete nonzero example, $q(z)=-2\mathrm{Re}[(1+i)/(z+1+i)]$ is real, harmonic, smooth on the closed quadrant and decays along both axes. With ${\beta }_1={\beta }_2=0$, it has $h_1(0)=h_2(0)=1$ and $q_z=(1+i)/(z+1+i{)}^2$. The printed orientation returns the negative of this derivative, so the issue persists even under the stated corner condition.

Let $a=\frac{1}{2}{e}^{-i{\beta }_1-i\gamma }$. In the expressions from the preceding part, the coefficient of $\Phi$ on the imaginary ray is $a$, and the coefficient on the real ray is $-a{e}^{4i\gamma }$. For each of the three allowed sums, $e^{4i\gamma }=1$. The unknown part is therefore proportional toIndeed, close the first spectral quadrant. For $z=x+iy$ with $x,y>0$, $|e^{ikz}|=e^{-(x\mathrm{Im}k+y\mathrm{Re}k)}$, so the large-arc contribution vanishes for bounded $\Phi$. The Cauchy integral theorem proves the equality and eliminates the single unknown function.

Substituting only the known pieces leavesThis already determines the answer from $h_1,h_2$ alone. Evaluating the elementary ray kernels gives the more direct rational-angle oblique derivative problem on a quadrant formulaNo denominator vanishes for an interior point of the first quadrant.

For explicit formulas in the three cases, defineCombining the paired fractions and using ${\beta }_2=\gamma -{\beta }_1$ yields

There is also a necessary corner compatibility for collinear oblique derivative data. When $\gamma =\pi /2$, the two prescribed derivative directions are opposites, so a continuous corner gradient requires $h_2(0)=-h_1(0)$. Together with the stipulated equality, this forces $h_1(0)=h_2(0)=0$ in that case. Existence was assumed, so admissible data must satisfy this additional consequence. Smoothness at the corner and decay at infinity also exclude the homogeneous corner ambiguity in a quadrant Laplace problem; no extra singular or growing term is permitted in the boxed reconstruction.