Solving system of ODEs using Fourier transforms

Question

I have a system of ODEs I'd like to solve. It seems in the literature that the best way of getting a complete solution is with Fourier transforms (and that under some conditions perturbative solutions do not work).
\begin{align} \frac{d P_{11}(t)}{dt} &= V(t) P_{21}- V(t)^* P_{12}\\ \frac{d P_{12}(t)}{dt} &= V(t)(P_{22}- P_{11})-c P_{12}\\ \frac{d P_{21}(t)}{dt} &= V^*(t)(P_{22}- P_{11})-c P_{21}\\ \frac{d P_{22}(t)}{dt} &= V^*(t) P_{12}- V(t) P_{21} \end{align} where ${P_{11},P_{12},P_{21},P_{22}}$ are functions of t that I'm solving for; c is a constant, V* is the complex conjugate of V; and $V = E_1 e^{-i \omega_1 t} + E_2 e^{-i \omega_2 t}$, with $E_1$ and $E_2$ being constants.

My plan is to Fourier transform this set of ODEs, solve the resultant linear algebra, and then inverse Fourier transform back to get my solution.

So first I apply some standard rules to Fourier transforms (convolution, $F(P(t) \cdot e^{i\omega_1 t}) = F(P)(\omega-\omega_1)$, and the rules for Fourier transformed derivatives $F(\dot{\rho}) = i \omega F(\rho)$).

But in doing so, I see that my system of equations (after transforming) has the form:

$$ \begin{equation} i \omega \hat{\rho}_{11} + \hat{\rho}_{21}(\omega-\omega_1)-\hat{\rho}_{12}(\omega - \omega_2) = 0 \\ -(\hat{\rho}_{11}(\omega-\omega_1) -\hat{\rho}_{11}(\omega-\omega_2) + (i\omega + c)\hat{\rho}_{12}+ \hat{\rho}_{22}(\omega-\omega_1)+\hat{\rho}_{22}(\omega-\omega_2)) = 0 \end{equation} \\ ... $$ My four equations now have terms like $\hat{\rho}_{21}(\omega)$ and $\hat{\rho}_{21}(\omega-\omega_1)$. In the basic methods of Fourier transforming to solve different equations, we solve algebraically for the fouriered functions. But in this case it's not obvious how to solve this system algebraically. I've thought about manually solving this system by hand using substitution (of both the input arguments and the functions themselves), but haven't had any luck.

Any ideas on how I can move forward to solve the system?

Edit: The fourier transform of a derivative is known to be ${\widehat {f'\;}}(\xi )=2\pi i\xi {\hat {f}}(\xi )$. This is found by performing integration by parts and identifying that the function at the boundaries is zero. In my case, I'm not completely sure if $ \lim_{x \to \infty} \hat{\rho}_{12}(\omega)$ (but, as discussed in the comments, I think that it's likely that there's some "energy" constraints limiting the maximum amount of frequencies allowed) And so, for now, I have omitted any potential contribution due to the boundaries.

Answer 1

Personally, I have never been a big fan of "Fourier transform" methods- or transform methods in general. Since these are linear equations with constant coefficients, I would write them in matrix form: $\frac{d}{dt}\begin{pmatrix}P_{11} \\ P_{12} \\ P_{21} \\ P_{22}\end{pmatrix}= \begin{pmatrix}0 & -V^* & V & 0 \\ -V & -c & 0 & V \\ -V^* & 0 & -c & V^* \\ 0 & V^* & V & 0 \end{pmatrix}\begin{pmatrix}P_{11} \\ P_{12} \\ P_{21} \\ P_{22}\end{pmatrix}$.

Now determine the eigenvalues and eigenvectors of that matrix.

Answer 2

A partial answer, but hopefully somewhat useful:

First off, it would be instructive to know how you obtained this system, but in the absense of this context, we can still do a few things.

Notice that $\frac{\text{d}}{\text{d} t}\left(P_{11}+P_{22}\right) = 0$; hence, the system reduces to \begin{align} \frac{\text{d} x}{\text{d} t} &= 2 V\,z - 2 \,V^*y\\ \frac{\text{d} y}{\text{d} t} &= V\,x - c\,y\\ \frac{\text{d} z}{\text{d} t} &= V^*x - c\,z\\ \end{align} with $x = P_{11}-P_{22}$, $y=P_{12}$ and $z=P_{21}$. Now, when the frequencies $\omega_1$ and $\omega_2$ are commensurable, the matrix $$ A(t) = \begin{pmatrix} 0 & -2 V^* & 2 V \\ V & -c & 0 \\ V^* & 0 & -c\end{pmatrix} $$ is periodic. In that case, you can try to attack the above system using Floquet theory, see e.g. here.

Alternatively, you can rewrite $y = e^{-c t} Y$, $z = e^{-c t} Z$ and $V = e^{- c t} W$ to obtain the even simpler system \begin{align} \frac{\text{d} x}{\text{d} t} &= 2 W\,Z - 2 W^*Y\\ \frac{\text{d} Y}{\text{d} t} &= W\,x \\ \frac{\text{d} Z}{\text{d} t} &= W^*x. \end{align} Unfortunately, $W$ is not periodic anymore; however, you might be able to make some headway with this formulation.