I am trying to read through the paper titled IBV problems for linear PDEs with Variable Coefficients by P. A. Treharne and A. S. Fokas (link here).
This post is a follow up to my first that established the background and problem statement. Recall that the system we seek to solve is given in terms of the following two linear pdes.
$$ \begin{align*} L_0 \mu = \mu_{xxx} + (u(x) - (ik)^3) \mu &= q(x,t) \\ L_1 \mu = \mu_{t} + (ik)^3 \mu &= -q(x,t) \end{align*} $$
Question: Given the complexity of the authors' approach, I had trouble following their argument and tried to compute the main formula of Subsection 3.1 in a different way. If anyone could verify that my solution is correct, I would appreciate it immensely.
Attempt: Let's begin by noticing that $L_0$ has not time-dependence and $L_1$ has no spatial dependence. They commute. $L_0$ has a higher order derivative and variable coefficient, yielding the more difficult equation. We tackle it first by imagining that $t$ and $k$ are constant to apply variation of parameters in $x$. We don't have any reason to expect this solution will also obey $L_1 \mu = - q(x,t)$, so we'll name it $\mu_0 (x,t,k)$.
In my previous post, I discuss finding a family of solutions $\{\psi_i\}_i$ to the homgeneous equation $L_0 \psi_i = 0$. We take an ansatz that
$$\mu_0(x,t,k) = \sum_{i=1}^3 K_i(x,t,k) \psi_i(x,k)$$
Then, variation of parameters, really just boils down into Crammer's rule. (As noted in the paper, the Wronskian $W[\psi_1, \psi_2, \psi_3](k) = 3k^2 \Delta(k)$ has only spectral dependence.)
$$ \begin{align*} K_i (x,t,k) &= \int_0^x \frac{M_j(\xi, k) q(\xi, t) }{W[\psi_1, \psi_2, \psi_3](k)} d\xi \\ &= \int_0^x \frac{M_j(\xi, k) q(\xi, t) }{3k^2 \Delta(k)} d\xi \end{align*} $$
The $M$ matrices are just the cofactors of the Wronksian.
$$M_1(x,k) = \psi_2(x,k) \dot{\psi}_3 (x,k) - \dot{\psi}_2(x,k) \psi_3(x,k)$$ $$M_2(x,k) = \psi_3(x,k) \dot{\psi}_1 (x,k) - \dot{\psi}_3(x,k) \psi_1(x,k)$$ $$M_3(x,k) = \psi_1(x,k) \dot{\psi}_2 (x,k) - \dot{\psi}_1(x,k) \psi_2(x,k)$$
Okay, great. Now, I believe that we have solved $L_0 \mu_0(x,t,k) = q(x,t)$. We just need to find a correction term $h$ that will allow our solution to satisfy $L_1 \mu_p(x,t,k) = -q(x,t)$.
One last ansatz.
$$\mu_p(x,t,k) = \mu_0(x,t,k) + h(x,t,k)$$
where
$$h(x,t,k) = \sum_{i=1}^3 C_i(x,t,k) \psi_i(x,k)$$
Let's apply $L_1$ to the particular solution $\mu_p(x,t,k)$
$$ \begin{align*} L_1 \mu_p(x,t,k) &= \sum_{i=1}^3 [L_1 K_i(x,t,k) + L_1 C_i(x,t,k)] \psi_i(x,k) \\ &= 0 \\ \end{align*} $$
to notice $L_1 C_i(x,t,k) = - L_1 K_i(x,t,k)$
We use the same incompressibility condition mentioned in my first post
$$(e^{(ik)^3 t} q M)_t + e^{(ik)^3 t} (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$
and simplify
$$(ik)^3 qM + q_t M + (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$
I think we can just match coefficients and integrate.
$$ \begin{align*} (\partial_t + (ik)^3)C_i(x,t,k) &= -(\partial_t + (ik)^3)K_i(x,t,k) \\ &= -(\partial_t + (ik)^3) \int_0^x \frac{M_j(\xi, k) q(\xi, t) }{3k^2 \Delta(k)} d\xi \\ &= -(ik)^3 K_i(x,t,k) + \frac{1}{3k^2 \Delta(k)} \int_0^x (ik)^3 qM_j + [q(M_j)_{xx} - q_x (M_j)_x + q (M_j)_{xx}]_x d\xi \\ &= \frac{1}{3k^2 \Delta(k)} [q(M_j)_{xx} - q_x (M_j)_x + q (M_j)_{xx}] \end{align*} $$
Finally, $C_i(x,t,k)$ is just a first order ODE in time.
We introduce an integrating factor $$(e^{(ik)^3 t} C_i(x,t,k))_t = \frac{1}{3k^2 \Delta(k)} \int_0^t e^{(ik)^3 t} q(M_j)_{xx} - q_x (M_j)_x + q (M_j)_{xx} d\tau$$ $$C_i(x,t,k) = \frac{1}{3k^2 \Delta(k)} \int_0^t e^{(ik)^3 (t - \tau)} [q(M_j)_{xx} - q_x (M_j)_x + q (M_j)_{xx}] d\tau$$
$$ \begin{align*} \mu_p(x,t,k) &= \sum_{i=1}^3 [K_i(x,t,k) + C_i(x,t,k)] \psi_i(x,k) \\ &= \sum_{i=1}^3 \frac{\psi_i(x,k)}{3k^2 \Delta(k)} \left[\int_0^t e^{-ik^3 (\tau - t)} (q(M_j)_{xx} - q_x (M_j)_x + q (M_j)_{xx}) d\tau + \int_0^x M_j(\xi,k) q(\xi, t) d\xi\right] \end{align*} $$
The authors express the solution as
$$ \begin{align*} \mu_p(x,t,k) = \frac{1}{3k^2 \Delta(k)} \sum_{i=1}^3 \psi_i(x,k) \int^{(x,t)} e^{ik^3 (t - \tau)}[M_i(\xi, k) q(\xi, \tau) d \xi - (q_{xx} M_j - q_{x} (M_j)_{x} + q (M_j)_{xx} ) d \tau] \end{align*} $$
I am not quite sure how the authors' have used their contour integral notation. Are our solutions equivalent? If not, where did I go wrong in my reasoning. Thanks in advance.
