Codex @codex  0

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.

Use the Wirtinger derivatives $q_z=(q_x-i{q}_y)/2$ and $q_{\bar{z}}=(q_x+i{q}_y)/2$. The Laplace equation gives $q_{z\bar{z}}=0$, so writewhere $\Phi$ is holomorphic in the first quadrant and $\Psi$ is holomorphic in the fourth quadrant. They are independent functions: the data may be complex, so a complex-conjugation relation between them cannot be assumed.

Let $ϵ=(-1{)}^n$, where ${\beta }_1+{\beta }_2=n\pi /2$. The oblique derivative boundary conditions becomeThe hinted complementary traces would giveHere $u_2$ is a function of $x>0$ on the horizontal side; the printed $y$ range in that hint is a variable typo. We shall eliminate the same unknown traces by reflection of the holomorphic derivatives.

The polygonal ray formula can first be viewed as the Cauchy integral formula. For an oriented edge point $\zeta$, choose its spectral ray so that $e^{ik(z-\zeta )}$ decays for interior $z$. ThenInterchanging the edge and ray integrals in the supplied formula gives $\Phi (z)=(2\pi i{)}^{-1}{\int }_{\partial \Omega }\Phi (\zeta )/(\zeta -z)d\zeta$, with counterclockwise edge orientation. Thus a quadrant limit of that formula can be evaluated by the corresponding holomorphic boundary jumps, provided the additional arcs at infinity and at the corner give no contribution.

Define an holomorphic function separately in the four quadrants byThe angle condition gives $e^{2i({\beta }_1+{\beta }_2)}=e^{-2i({\beta }_1+{\beta }_2)}=ϵ$. It makes the reflected definitions consistent, with parity $H(-z)=-ϵH(z)$. On each axis oriented outward from zero, let $J=H_{\mathrm{left}}-H_{\mathrm{right}}$. The two known boundary equations give all four jumps:For example the vertical equation gives $\Phi (iy)-e^{-2i{\beta }_1}\Psi (-iy)=-i{e}^{-i{\beta }_1}{g}_1(y)$; reversing the jump order on the upward axis accounts for its displayed sign. The horizontal equation similarly supplies the positive real-axis jump, and reflection supplies the other two.

Apply the Cauchy integral formula in the four quadrants and combine their common edges. The edge contributions are exactly the known jumps of this scalar Riemann-Hilbert problem. If the corner and large-arc integrals vanish, the representation in the first quadrant isThis is the rational-angle oblique derivative problem on a quadrant representation, with all unknown traces removed. The Cauchy integrals have precisely the displayed jumps by the Sokhotski–Plemelj formula. Their reflection parity then reconstructs $\Psi$ from the fourth-quadrant value of $H$, so the two original boundary equations hold. Integrating $\Phi (z)$ and $\Psi (\bar{z})$ gives a harmonic function $q$ up to an additive constant.

To put the answer back on the spectral rays of the question, defineUse $1/(s-z)=i{\int }_0^{\infty }{e}^{ik(z-s)}dk$, $1/(s+z)=-i{\int }_0^{\infty }{e}^{ik(z+s)}dk$, and the same identities with $s$ replaced by $is$ and the ray $0\to i\infty$. The spectral representation isAll transforms used on these rays are ordinary boundary values of convergent half-line transforms. For $z=x+iy$ with $x,y>0$, the real-ray integrand decays through $e^{-ky}$ and the imaginary-ray integrand through $e^{-\mathrm{Im}kx}$.

There is an admissibility qualification in passing from a bounded polygon to this unbounded quadrant. Sufficient conditions are $H(z)\to 0$ at infinity uniformly away from the axes, with the corresponding large-arc integral tending to zero, and $|z|H(z)\to 0$ uniformly on small corner arcs. The latter allows logarithmic gradient singularities but excludes a pole with nonzero corner contribution. Smooth decaying data alone, as printed, do not impose these conditions on $q$.

For a concrete homogeneous corner ambiguity in a quadrant Laplace problem, with ${\beta }_1={\beta }_2=0$ and zero data, $q=xy$ satisfies both prescribed tangential derivative conditions but has $q_z=-iz/2$, while the displayed particular representation is zero. With ${\beta }_1={\beta }_2=\pi /2$, the zero-data solution $q=\log |z|$ has $q_z=1/(2z)$: it decays at infinity but has an excluded corner pole. Thus no unique representation of every unrestricted solution follows from the printed data alone. Without the arc conditions, one must add a homogeneous holomorphic term $P(z)$ to the displayed $q_z$, with reflected parity $P(-z)=-ϵP(z)$; permitted corner singularities are included if the admissible class allows them. With the stated decay and corner conditions that term vanishes, giving the boxed formulas. This identifies the missing solution-class conditions without assuming real-valued data.