Solving $\psi_{xxx} + (u(x) - (ik)^3))\psi = 0$ for $x > 0$, $k \in \mathbb C$, and $u(x)$ smooth

Question

I was reading Initial-Boundary Value Problems for Linear PDEs with Variable Coefficients by P. Treharne and A. S. Fokas, when I came across the following ODE formulated as part of a Lax pair for a linear system $q_t + q_{xxx} + u(x)q = 0$.

The authors claim that for $x > 0$, $k \in \mathbb C$, and a smooth $u(x)$ that decays to $0$ as $x \to \infty$, the ODE. $$\psi_{xxx} + (u(x) - (ik)^3))\psi = 0$$ has a solution given in terms of the following inhomgenous Fredholm integral equation $$\psi(x,k) = f(x,k) + \frac{1}{3k^2} \int_0^x \left(\sum_{j=0}^{2}\alpha^j e^{i\alpha^jk(x - \xi)}u(\xi)\right) \psi(\xi,k)d\xi$$ where $$f(x,k) = e^{i\alpha^j k x}$$ for $0\leq j \leq 2$ and $\alpha = \exp(2\pi i /3)$.

I have two questions.

1. How does one derive this solution? It appears to be similar to solving a first order ODE by using an integrating factor.
2. Is the integral kernel separable in that it may be written as $\sum_{n=0}^N \alpha_j(\xi) \beta_j(k)$? I don't believe that the answer is affirmative due to the $k(x-\xi)$ term, but I am curious to know if there is a different means of separating it.

Note: Although this question is not research level, I originally posed it on Math Stack Exchange 4 days ago. No one answered it, so I decided to delete it and instead ask it here.

Answer 1

Treharne and Fokas claim that particular solutions of the ODE satisfy the integral equation, i.e. the integral equation that you quote is a reformulation of the ODE, not a solution of it.

There are numerous resources available that explain how this reformulation process works, e.g. https://math.stackexchange.com/questions/2901536/conversion-of-second-order-ode-into-integral-equation.

If you wish to dig deeper, several examples of the Fokas method are given in the paper by B. Deconinck, T. Trogdon and V. Vasan, The Method of Fokas for Solving Linear Partial Differential Equations, SIAM Review 56(1) (2014) pp. 159-186.