Solution (source code)

= Solution

Let $\Phi$ denote the common trace transform of the unique solution, or of the rotational average at an exceptional parameter. Because $q$ and $f$ are real and $\lambda$ is real, $\overline{\Phi(\bar k)}=\Phi(k)$ and $\overline{\Psi(\bar k)}=\Psi(k)$. Taking the Schwarz conjugate of the <global relation> means replacing $k$ by $\bar k$ and conjugating the whole equation. It yields the second of the <conjugate global relations for the modified Helmholtz equation>:
$$
E(ik)\Phi(k)+E(i\bar ak)\Phi(\bar ak)+E(iak)\Phi(ak)
=\frac i2[E(ik)\Psi(k)+E(i\bar ak)\Psi(\bar ak)+E(iak)\Psi(ak)].
$$
In this conjugation the rotated arguments interchange; after putting terms in the order $k,\bar ak,ak$, the coefficients have the displayed signs and arguments. Define
$$
\chi(k)=\frac l2\left(k+\frac\lambda k\right),\qquad e(k)=e^{\chi(k)}.
$$
The elementary cube-root identities give
$$
\begin{aligned}
E(iak)E(-ik)&=e(\bar ak),& E(iak)E(-i\bar ak)&=e(-k),\\
E(-iak)E(ik)&=e(-\bar ak),& E(-iak)E(i\bar ak)&=e(k).
\end{aligned}
$$
Multiply the first <global relation> by $E(iak)$ and the conjugate one by $E(-iak)$. Then the coefficient of $\Phi(ak)$ is one in both equations:
$$
\begin{aligned}
e(\bar ak)\Phi(k)+e(-k)\Phi(\bar ak)+\Phi(ak)&=A(k),\\
e(-\bar ak)\Phi(k)+e(k)\Phi(\bar ak)+\Phi(ak)&=B(k),
\end{aligned}
$$
where all the quantities on the right are known,
$$
\begin{aligned}
A(k)&=-\frac i2[e(\bar ak)\Psi(k)+e(-k)\Psi(\bar ak)+\Psi(ak)],\\
B(k)&=\frac i2[e(-\bar ak)\Psi(k)+e(k)\Psi(\bar ak)+\Psi(ak)].
\end{aligned}
$$
Subtracting eliminates $\Phi(ak)$. If
$$
C(k)=\cosh\chi(\bar ak)\Psi(k)+\cosh\chi(k)\Psi(\bar ak)+\Psi(ak),
$$
the resulting \b[solution for the requested transform] is
$$
\boxed{\Phi(\bar ak)=\frac{2\sinh\chi(\bar ak)\Phi(k)+iC(k)}{2\sinh\chi(k)}.}
$$
At a zero of the denominator this identity is interpreted in its undivided form. The transforms themselves remain finite for $k\ne0$; consequently the numerator must vanish. That cancellation, rather than evaluation of a singular quotient, determines the Fourier coefficients below.