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). For greater clarity, I split my question up into two posts. In the first post (this one), I ask how the authors' defined their contour integrals, and in the second post, I attempt to find an alternative method to derive the main formula of Section 3.1.

Background: The authors want to solve the equation

$$q_t + q_{xxx} + u(x) q= 0$$

on $(x, t) \in [0, \infty)^2$ with the boundary conditions

$$ \begin{cases} q(x,0) = q_0(x) &: 0 < x < \infty \\ q(0,t)= g_0(t) &: t >0 \end{cases} $$

The PDE can be reformulated as the compatibility condition of a system of linear operators, loosely referred to as the Lax pair of the system where $k \in \mathbb C$ is a spectral parameter.

$$ \begin{align*} L_0 \mu = \mu_{xxx} + (u(x) - (ik)^3) \mu &= q(x,t) \\ L_1 \mu = \mu_{t} + (ik)^3 \mu &= -q(x,t) \end{align*} $$

We can solve the original PDE if we find a function $\mu(x,t,k)$.

1. Find a class of homogeneous solutions $\{\psi_i(x,k)\}_i$ to the equation $L_0 \psi = 0$.
2. Determine a particular solution $\mu_p(x,t,k)$ to the Lax pair by a method I am calling "Multivariable Variation of Parameters."
3. Assemble the $\mu_p(x,t,k)$ into a sectionally analytic function $\mu(x,t,k)$ and form a Riemann-Hilbert problem.
4. Solve the Riemann-Hilbert problem.
5. Analyze the global relationship and eliminate the unknown boundary values using the Method of Fokas.

I am stuck on Step 2.

Paper's Solution:

I will paraphrase the author's solution to avoid copyright issues.

Solve the adjoint equation of the original PDE

$$- Q_t - Q_{xxx} + u(x) Q = 0$$

Write the adjoint equation and original PDE as the condition that a certain vector field is incompressible.

$$(Qq)_t + (Qq_{xx} - q_x Q_x + q Q_{xx})_{x} = 0$$

Ansatz: One solution is $Q(x,t) = e^{(ik)^3 t} M(x, k)$ where

$$M(x, k) = \begin{vmatrix} \psi_i(x,k) & \psi_j(x,k) \\ \dot{\psi}_i(x,k) & \dot{\psi}_j(x,k) \end{vmatrix}$$

where $\psi_i, \psi_j$ solve the homogeneous equation $L_0 \psi = 0$.

(Note: I am already unclear as to the motivation behind this ansatz.)

Substitute the ansatz into the above incompressibility condition.

$$(e^{(ik)^3 t} q M)_t + e^{(ik)^3 t} (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$

In the following,

$$M_1(x,k) = \psi_2(x,k) \dot{\psi}_3 (x,k) - \dot{\psi}_2(x,k) \psi_3(x,k)$$ $$M_2(x,k) = \psi_3(x,k) \dot{\psi}_1 (x,k) - \dot{\psi}_3(x,k) \psi_1(x,k)$$ $$M_3(x,k) = \psi_1(x,k) \dot{\psi}_2 (x,k) - \dot{\psi}_1(x,k) \psi_2(x,k)$$

Ansatz: The particular solution can be expressed via a formula akin to variation of parameters.

$$\mu_p (x,t,k) = \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^3 \psi_j(x,k) \nu_j(x,t,k)$$

where $\frac{1}{3k^2 \Delta(k)} $ is the Wronskian and $\{\nu_j\}_{j=1}^3$ obey

$$(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$$ $$(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$$

The following two substitutions hold

1. $$ \begin{align*} L_0 \mu_p(x,t,k) &= \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_{xx} \psi_j M_j + q_x (2 \psi_j \dot{M}_j + 3 \dot{\psi}_j M_j) + q \psi _j \ddot{M}_j + 3q(\dot{\psi}_j M_j)_x \\ &= q(x,t) \end{align*} $$

2. $$ \begin{align*} L_1 \mu_p(x,t,k) &= -\frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_xx \psi_j M_j - q_x \psi_j \dot{M}_j + q\psi_j \ddot{M}_j \\ &= -q(x,t) \end{align*} $$

because of the following identities

1. $\sum_{j=1}^3 \psi_j(x,k) M_j (x,k) = 0$
2. $\sum_{j=1}^3 \dot{\psi}_j(x,k) M_j(x,k) = 0$
3. $\sum_{j=1}^3 \psi_j(x,k) \dot{M}_j(x,k) = 0$
4. $\sum_{j=1}^3 \psi_j(x,k) \ddot{M}_j(x,k) = 3k^2 \Delta(k)$

(Note: I'm not sure how these identities were derived.)

I'm very confused by this next step.

Rewrite the equations $(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$ and $(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$ as

$$d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dt + X_j (x,t,k) dx]$$

(I thought that the formula should be $d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dx - X_j (x,t,k) dt]$, so I must be missing how they selected their contour integral.)

Integrate to determine

$$\nu_j(x,t) = \int^{(x,t)} e^{(ik)^3 (t - \tau)} [q (\xi, \tau) M_j(\xi, k) d \tau + X_j(\xi, \tau, k) d \xi]$$

where $$X_j(x,t,k) = q_{xx}(x,t) M_j(x,k) - q_x(x,t) \dot{M}_j (x,k) + q(x,t) \ddot{M}_j(x,k)$$

so that

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \int^{(x,t)} e^{(ik)^3 (t - \tau)} [M_j(\xi, k) q(\xi, \tau) d\xi - X_j(\xi, \tau, k) d \tau] $$

(I don't know how the authors' jumped from the previous integral to the present one.)

Question: Can I write the result of integration as follows?

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \left[ \int_0^x M_j(\xi, k) q(\xi, t) d\xi - \int_0^t e^{(ik)^3 (t - \tau)} X_j(x, \tau, k) d \tau \right] $$

As for the sum identities (1 through 4) used to find the integrand, do they follow from Cofactor expansions? I can mentally verify that they hold, but I would be thankful if anyone had more motivation for them.

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). For greater clarity, I split my question up into two posts. In the first post (this one), I ask how the authors' defined their contour integrals, and in the second post, I attempt to find an alternative method to derive the main formula of Section 3.1.

Background: The authors want to solve the equation

$$q_t + q_{xxx} + u(x) q= 0$$

on $(x, t) \in [0, \infty)^2$ with the boundary conditions

$$ \begin{cases} q(x,0) = q_0(x) &: 0 < x < \infty \\ q(0,t)= g_0(t) &: t >0 \end{cases} $$

The PDE can be reformulated as the compatibility condition of a system of linear operators, loosely referred to as the Lax pair of the system where $k \in \mathbb C$ is a spectral parameter.

$$ \begin{align*} L_0 \mu = \mu_{xxx} + (u(x) - (ik)^3) \mu &= q(x,t) \\ L_1 \mu = \mu_{t} + (ik)^3 \mu &= -q(x,t) \end{align*} $$

We can solve the original PDE if we find a function $\mu(x,t,k)$.

1. Find a class of homogeneous solutions $\{\psi_i(x,k)\}_i$ to the equation $L_0 \psi = 0$.
2. Determine a particular solution $\mu_p(x,t,k)$ to the Lax pair by a method I am calling "Multivariable Variation of Parameters."
3. Assemble the $\mu_p(x,t,k)$ into a sectionally analytic function $\mu(x,t,k)$ and form a Riemann-Hilbert problem.
4. Solve the Riemann-Hilbert problem.
5. Analyze the global relationship and eliminate the unknown boundary values using the Method of Fokas.

I am stuck on Step 2.

Paper's Solution:

I will paraphrase the author's solution to avoid copyright issues.

Solve the adjoint equation of the original PDE

$$- Q_t - Q_{xxx} + u(x) Q = 0$$

Write the adjoint equation and original PDE as the condition that a certain vector field is incompressible.

$$(Qq)_t + (Qq_{xx} - q_x Q_x + q Q_{xx})_{x} = 0$$

Ansatz: One solution is $Q(x,t) = e^{(ik)^3 t} M(x, k)$ where

$$M(x, k) = \begin{vmatrix} \psi_i(x,k) & \psi_j(x,k) \\ \dot{\psi}_i(x,k) & \dot{\psi}_j(x,k) \end{vmatrix}$$

where $\psi_i, \psi_j$ solve the homogeneous equation $L_0 \psi = 0$.

(Note: I am already unclear as to the motivation behind this ansatz.)

Substitute the ansatz into the above incompressibility condition.

$$(e^{(ik)^3 t} q M)_t + e^{(ik)^3 t} (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$

In the following,

$$M_1(x,k) = \psi_2(x,k) \dot{\psi}_3 (x,k) - \dot{\psi}_2(x,k) \psi_3(x,k)$$ $$M_2(x,k) = \psi_3(x,k) \dot{\psi}_1 (x,k) - \dot{\psi}_3(x,k) \psi_1(x,k)$$ $$M_3(x,k) = \psi_1(x,k) \dot{\psi}_2 (x,k) - \dot{\psi}_1(x,k) \psi_2(x,k)$$

Ansatz: The particular solution can be expressed via a formula akin to variation of parameters.

$$\mu_p (x,t,k) = \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^3 \psi_j(x,k) \nu_j(x,t,k)$$

where $\frac{1}{3k^2 \Delta(k)} $ is the Wronskian and $\{\nu_j\}_{j=1}^3$ obey

$$(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$$ $$(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$$

The following two substitutions hold

1. $$ \begin{align*} L_0 \mu_p(x,t,k) &= \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_{xx} \psi_j M_j + q_x (2 \psi_j \dot{M}_j + 3 \dot{\psi}_j M_j) + q \psi _j \ddot{M}_j + 3q(\dot{\psi}_j M_j)_x \\ &= q(x,t) \end{align*} $$

2. $$ \begin{align*} L_1 \mu_p(x,t,k) &= -\frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_xx \psi_j M_j - q_x \psi_j \dot{M}_j + q\psi_j \ddot{M}_j \\ &= -q(x,t) \end{align*} $$

because of the following identities

1. $\sum_{j=1}^3 \psi_j(x,k) M_j (x,k) = 0$
2. $\sum_{j=1}^3 \dot{\psi}_j(x,k) M_j(x,k) = 0$
3. $\sum_{j=1}^3 \psi_j(x,k) \dot{M}_j(x,k) = 0$
4. $\sum_{j=1}^3 \psi_j(x,k) \ddot{M}_j(x,k) = 3k^2 \Delta(k)$

(Note: I'm not sure how these identities were derived.)

I'm very confused by this next step.

Rewrite the equations $(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$ and $(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$ as

$$d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dt + X_j (x,t,k) dx]$$

(I thought that the formula should be $d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dx - X_j (x,t,k) dt]$, so I must be missing how they selected their contour integral.)

Integrate to determine

$$\nu_j(x,t) = \int^{(x,t)} e^{(ik)^3 (t - \tau)} [q (\xi, \tau) M_j(\xi, k) d \tau + X_j(\xi, \tau, k) d \xi]$$

where $$X_j(x,t,k) = q_{xx}(x,t) M_j(x,k) - q_x(x,t) \dot{M}_j (x,k) + q(x,t) \ddot{M}_j(x,k)$$

so that

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \int^{(x,t)} e^{(ik)^3 (t - \tau)} [M_j(\xi, k) q(\xi, \tau) d\xi - X_j(\xi, \tau, k) d \tau] $$

(I don't know how the authors' jumped from the previous integral to the present one.)

Question: Can I write the result of integration as follows?

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \left[ \int_0^x M_j(\xi, k) q(\xi, t) d\xi - \int_0^t e^{(ik)^3 (t - \tau)} X_j(x, \tau, k) d \tau \right] $$

As for the sum identities (1 through 4) used to find the integrand, do they follow from Cofactor expansions? I can mentally verify that they hold, but I would be thankful if anyone had more motivation for them.

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). For greater clarity, I split my question up into two posts. In the first post (this one), I ask how the authors' defined their contour integrals, and in the second post, I attempt to find an alternative method to derive the main formula of Section 3.1.

Background: The authors want to solve the equation

$$q_t + q_{xxx} + u(x) q= 0$$

on $(x, t) \in [0, \infty)^2$ with the boundary conditions

$$ \begin{cases} q(x,0) = q_0(x) &: 0 < x < \infty \\ q(0,t)= g_0(t) &: t >0 \end{cases} $$

The PDE can be reformulated as the compatibility condition of a system of linear operators, loosely referred to as the Lax pair of the system where $k \in \mathbb C$ is a spectral parameter.

$$ \begin{align*} L_0 \mu = \mu_{xxx} + (u(x) - (ik)^3) \mu &= q(x,t) \\ L_1 \mu = \mu_{t} + (ik)^3 \mu &= -q(x,t) \end{align*} $$

We can solve the original PDE if we find a function $\mu(x,t,k)$.

1. Find a class of homogeneous solutions $\{\psi_i(x,k)\}_i$ to the equation $L_0 \psi = 0$.
2. Determine a particular solution $\mu_p(x,t,k)$ to the Lax pair by a method I am calling "Multivariable Variation of Parameters."
3. Assemble the $\mu_p(x,t,k)$ into a sectionally analytic function $\mu(x,t,k)$ and form a Riemann-Hilbert problem.
4. Solve the Riemann-Hilbert problem.
5. Analyze the global relationship and eliminate the unknown boundary values using the Method of Fokas.

I am stuck on Step 2.

Paper's Solution:

I will paraphrase the author's solution to avoid copyright issues.

Solve the adjoint equation of the original PDE

$$- Q_t - Q_{xxx} + u(x) Q = 0$$

Write the adjoint equation and original PDE as the condition that a certain vector field is incompressible.

$$(Qq)_t + (Qq_{xx} - q_x Q_x + q Q_{xx})_{x} = 0$$

Ansatz: One solution is $Q(x,t) = e^{(ik)^3 t} M(x, k)$ where

$$M(x, k) = \begin{vmatrix} \psi_i(x,k) & \psi_j(x,k) \\ \dot{\psi}_i(x,k) & \dot{\psi}_j(x,k) \end{vmatrix}$$

where $\psi_i, \psi_j$ solve the homogeneous equation $L_0 \psi = 0$.

(Note: I am already unclear as to the motivation behind this ansatz.)

Substitute the ansatz into the above incompressibility condition.

$$(e^{(ik)^3 t} q M)_t + e^{(ik)^3 t} (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$

In the following,

$$M_1(x,k) = \psi_2(x,k) \dot{\psi}_3 (x,k) - \dot{\psi}_2(x,k) \psi_3(x,k)$$ $$M_2(x,k) = \psi_3(x,k) \dot{\psi}_1 (x,k) - \dot{\psi}_3(x,k) \psi_1(x,k)$$ $$M_3(x,k) = \psi_1(x,k) \dot{\psi}_2 (x,k) - \dot{\psi}_1(x,k) \psi_2(x,k)$$

Ansatz: The particular solution can be expressed via a formula akin to variation of parameters.

$$\mu_p (x,t,k) = \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^3 \psi_j(x,k) \nu_j(x,t,k)$$

where $\frac{1}{3k^2 \Delta(k)} $ is the Wronskian and $\{\nu_j\}_{j=1}^3$ obey

$$(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$$ $$(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$$

The following two substitutions hold

1. $$ \begin{align*} L_0 \mu_p(x,t,k) &= \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_{xx} \psi_j M_j + q_x (2 \psi_j \dot{M}_j + 3 \dot{\psi}_j M_j) + q \psi _j \ddot{M}_j + 3q(\dot{\psi}_j M_j)_x \\ &= q(x,t) \end{align*} $$

2. $$ \begin{align*} L_1 \mu_p(x,t,k) &= -\frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_xx \psi_j M_j - q_x \psi_j \dot{M}_j + q\psi_j \ddot{M}_j \\ &= -q(x,t) \end{align*} $$

because of the following identities

1. $\sum_{j=1}^3 \psi_j(x,k) M_j (x,k) = 0$
2. $\sum_{j=1}^3 \dot{\psi}_j(x,k) M_j(x,k) = 0$
3. $\sum_{j=1}^3 \psi_j(x,k) \dot{M}_j(x,k) = 0$
4. $\sum_{j=1}^3 \psi_j(x,k) \ddot{M}_j(x,k) = 3k^2 \Delta(k)$

(Note: I'm not sure how these identities were derived.)

I'm very confused by this next step.

Rewrite the equations $(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$ and $(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$ as

$$d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dt + X_j (x,t,k) dx]$$

(I thought that the formula should be $d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dx - X_j (x,t,k) dt]$, so I must be missing how they selected their contour integral.)

Integrate to determine

$$\nu_j(x,t) = \int^{(x,t)} e^{(ik)^3 (t - \tau)} [q (\xi, \tau) M_j(\xi, k) d \tau + X_j(\xi, \tau, k) d \xi]$$

so that

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \int^{(x,t)} e^{(ik)^3 (t - \tau)} [M_j(\xi, k) q(\xi, \tau) d\xi - X_j(\xi, \tau, k) d \tau] $$

(I don't know how the authors' jumped from the previous integral to the present one.)

Question: Can I write the result of integration as follows?

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \left[ \int_0^x M_j(\xi, k) q(\xi, t) d\xi - \int_0^t e^{(ik)^3 (t - \tau)} X_j(x, \tau, k) d \tau \right] $$

As for the sum identities (1 through 4) used to find the integrand, do they follow from Cofactor expansions? I can mentally verify that they hold, but I would be thankful if anyone had more motivation for them.

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). For greater clarity, I split my question up into two posts. In the first post (this one), I ask how the authors' defined their contour integrals, and in the second post, I attempt to find an alternative method to derive the main formula of Section 3.1.

Background: The authors want to solve the equation

$$q_t + q_{xxx} + u(x) q= 0$$

on $(x, t) \in [0, \infty)^2$ with the boundary conditions

$$ \begin{cases} q(x,0) = q_0(x) &: 0 < x < \infty \\ q(0,t)= g_0(t) &: t >0 \end{cases} $$

The PDE can be reformulated as the compatibility condition of a system of linear operators, loosely referred to as the Lax pair of the system where $k \in \mathbb C$ is a spectral parameter.

$$ \begin{align*} L_0 \mu = \mu_{xxx} + (u(x) - (ik)^3) \mu &= q(x,t) \\ L_1 \mu = \mu_{t} + (ik)^3 \mu &= -q(x,t) \end{align*} $$

We can solve the original PDE if we find a function $\mu(x,t,k)$.

1. Find a class of homogeneous solutions $\{\psi_i(x,k)\}_i$ to the equation $L_0 \psi = 0$.
2. Determine a particular solution $\mu_p(x,t,k)$ to the Lax pair by a method I am calling "Multivariable Variation of Parameters."
3. Assemble the $\mu_p(x,t,k)$ into a sectionally analytic function $\mu(x,t,k)$ and form a Riemann-Hilbert problem.
4. Solve the Riemann-Hilbert problem.
5. Analyze the global relationship and eliminate the unknown boundary values using the Method of Fokas.

I am stuck on Step 2.

Paper's Solution:

I will paraphrase the author's solution to avoid copyright issues.

Solve the adjoint equation of the original PDE

$$- Q_t - Q_{xxx} + u(x) Q = 0$$

Write the adjoint equation and original PDE as the condition that a certain vector field is incompressible.

$$(Qq)_t + (Qq_{xx} - q_x Q_x + q Q_{xx})_{x} = 0$$

Ansatz: One solution is $Q(x,t) = e^{(ik)^3 t} M(x, k)$ where

$$M(x, k) = \begin{vmatrix} \psi_i(x,k) & \psi_j(x,k) \\ \dot{\psi}_i(x,k) & \dot{\psi}_j(x,k) \end{vmatrix}$$

where $\psi_i, \psi_j$ solve the homogeneous equation $L_0 \psi = 0$.

(Note: I am already unclear as to the motivation behind this ansatz.)

Substitute the ansatz into the above incompressibility condition.

$$(e^{(ik)^3 t} q M)_t + e^{(ik)^3 t} (q_{xx} M - q_x M_x + q M_{xx})_x = 0$$

In the following,

$$M_1(x,k) = \psi_2(x,k) \dot{\psi}_3 (x,k) - \dot{\psi}_2(x,k) \psi_3(x,k)$$ $$M_2(x,k) = \psi_3(x,k) \dot{\psi}_1 (x,k) - \dot{\psi}_3(x,k) \psi_1(x,k)$$ $$M_3(x,k) = \psi_1(x,k) \dot{\psi}_2 (x,k) - \dot{\psi}_1(x,k) \psi_2(x,k)$$

Ansatz: The particular solution can be expressed via a formula akin to variation of parameters.

$$\mu_p (x,t,k) = \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^3 \psi_j(x,k) \nu_j(x,t,k)$$

where $\frac{1}{3k^2 \Delta(k)} $ is the Wronskian and $\{\nu_j\}_{j=1}^3$ obey

$$(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$$ $$(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$$

The following two substitutions hold

1. $$ \begin{align*} L_0 \mu_p(x,t,k) &= \frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_{xx} \psi_j M_j + q_x (2 \psi_j \dot{M}_j + 3 \dot{\psi}_j M_j) + q \psi _j \ddot{M}_j + 3q(\dot{\psi}_j M_j)_x \\ &= q(x,t) \end{align*} $$

2. $$ \begin{align*} L_1 \mu_p(x,t,k) &= -\frac{1}{3k^2 \Delta(k)} \sum_{j=1}^{3} q_xx \psi_j M_j - q_x \psi_j \dot{M}_j + q\psi_j \ddot{M}_j \\ &= -q(x,t) \end{align*} $$

because of the following identities

1. $\sum_{j=1}^3 \psi_j(x,k) M_j (x,k) = 0$
2. $\sum_{j=1}^3 \dot{\psi}_j(x,k) M_j(x,k) = 0$
3. $\sum_{j=1}^3 \psi_j(x,k) \dot{M}_j(x,k) = 0$
4. $\sum_{j=1}^3 \psi_j(x,k) \ddot{M}_j(x,k) = 3k^2 \Delta(k)$

(Note: I'm not sure how these identities were derived.)

I'm very confused by this next step.

Rewrite the equations $(e^{(ik)^3 t} \nu_j)_x = e^{(ik)^3 t} q M_j$ and $(e^{(ik)^3 t} \nu_j)_t = -e^{(ik)^3 t} X_j$ as

$$d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dt + X_j (x,t,k) dx]$$

(I thought that the formula should be $d[e^{(ik)^3 t} \nu_j] = e^{(ik)^3 t} [q(x,t) M_j(x,k) dx - X_j (x,t,k) dt]$, so I must be missing how they selected their contour integral.)

Integrate to determine

$$\nu_j(x,t) = \int^{(x,t)} e^{(ik)^3 (t - \tau)} [q (\xi, \tau) M_j(\xi, k) d \tau + X_j(\xi, \tau, k) d \xi]$$

where $$X_j(x,t,k) = q_{xx}(x,t) M_j(x,k) - q_x(x,t) \dot{M}_j (x,k) + q(x,t) \ddot{M}_j(x,k)$$

so that

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \int^{(x,t)} e^{(ik)^3 (t - \tau)} [M_j(\xi, k) q(\xi, \tau) d\xi - X_j(\xi, \tau, k) d \tau] $$

(I don't know how the authors' jumped from the previous integral to the present one.)

Question: Can I write the result of integration as follows?

$$ \mu_p(x,t,k) = \sum_{j=1}^3 \frac{\psi_j(x,k)}{3k^2 \Delta(k)} \left[ \int_0^x M_j(\xi, k) q(\xi, t) d\xi - \int_0^t e^{(ik)^3 (t - \tau)} X_j(x, \tau, k) d \tau \right] $$

As for the sum identities (1 through 4) used to find the integrand, do they follow from Cofactor expansions? I can mentally verify that they hold, but I would be thankful if anyone had more motivation for them.