= Solution
Since $q$ is <harmonic>, $q_z$ is <holomorphic> and the <one-form> $e^{-ikz}q_z\,dz$ is closed. Closing the physical quadrant at infinity gives the <global relation>
$$
\widehat q_1(k)+\widehat q_2(k)=0\qquad
(\operatorname{Re}k<0,\ \operatorname{Im}k<0).
$$
In this third spectral quadrant the exponential decays on both boundary axes. Set $\gamma=\beta_1+\beta_2$. The relation becomes
$$
H_1(k)-iJ_1(k)=e^{i\gamma}[H_2(k)-iJ_2(k)].
$$
All four boundary <functions> are real. Consequently their transforms obey
$$
\overline{H_1(\bar k)}=H_1(k),\qquad
\overline{H_2(-\bar k)}=H_2(k),
$$
and the same identities for $J_1,J_2$. <Complex conjugation> of the <global relation> therefore gives its upper-left consequence
$$
H_1(k)+iJ_1(k)=e^{-i\gamma}[H_2(-k)+iJ_2(-k)]
\qquad(\operatorname{Re}k<0,\ \operatorname{Im}k>0).
$$
Introduce just one unknown <function>,
$$
\boxed{\Phi(k)=H_2(-k)-iJ_2(-k)
=\int_0^\infty e^{ikx}[h_2(x)-ij_2(x)]\,dx,\qquad\operatorname{Im}k>0.}
$$
It is <holomorphic> in the <upper half-plane>. Under the transform-admissible decay assumptions, $|\Phi(k)|\leq\int_0^\infty(|h_2|+|j_2|)\,dx$, so it is bounded there.
For $k$ on the positive real ray, use both consequences of the <global relation> at $-k$. Adding them removes $J_1(-k)$ and gives
$$
H_2(k)+iJ_2(k)=2e^{i\gamma}H_1(-k)-e^{2i\gamma}\Phi(k).
$$
For $k$ on the positive imaginary ray, use the upper-left relation and $H_2(-k)+iJ_2(-k)=2H_2(-k)-\Phi(k)$. The resulting <single-function spectral elimination for a rational-angle quadrant> is
$$
\boxed{\begin{aligned}
\widehat q_2(k)&=-e^{i\beta_2}[H_2(k)-e^{i\gamma}H_1(-k)]
-\frac{e^{i\beta_2+2i\gamma}}2\Phi(k),\qquad k>0,\\
\widehat q_1(k)&=e^{-i\beta_1}[H_1(k)-e^{-i\gamma}H_2(-k)]
+\frac{e^{-i\beta_1-i\gamma}}2\Phi(k),\qquad k\in i\mathbb R_+.
\end{aligned}}
$$
Boundary limits from the appropriate transform domains are understood. Every remaining unknown boundary <derivative> is contained in the one bounded upper-half-plane <function> $\Phi$.
Back to article page
