# Contradicting solutions in the Fourier space.

Consider the following differential equations $$\begin{cases} \partial_t X(t) = -i \alpha X(t) + \beta Y(t), \\ \partial_t Y(t) = -i \alpha Y(t) - \beta X(t). \end{cases}\label{1}\tag{1}$$ with $$\alpha$$, $$\beta$$ some complex constants (independent of $$t$$).

Case-1 [Time domain]: The solution (with initial condition, say $$X(0) = 1, Y(0) = 0$$) reads $$\begin{cases} X(t) = \frac{1}{2}[e^{-ist} + e^{-idt}], \\ Y(t) = -\frac{1}{2} [e^{-ist} - e^{-id t}], \end{cases}\label{2}\tag{2}$$ where $$s = \alpha + \beta$$ and $$d = \alpha - \beta$$. Now, the Fourier transform $$\mathscr{F}[A(t)] = A(\omega) = \int_{-\infty}^{\infty} dt e^{i\omega t} A(t)$$ of \eqref{2} reads: $$\begin{cases} X(\omega) = \frac{1}{2}[ \delta (\omega - s) + \delta (\omega + s)], \\ Y(\omega) = -\frac{i}{2}[ \delta (\omega - s) - \delta (\omega + s)] \end{cases}\label{3}\tag{3}$$ Case-2 [Fourier domain]: Taking Fourier transform of \eqref{1}, we have $$\begin{cases} -i \omega X(\omega) = -i \alpha X(\omega) + \beta Y(\omega),\\ -i\omega Y(\omega) = -i\alpha Y(\omega) - \beta X(\omega). \end{cases}\label{4}\tag{4}$$ The solution of \eqref{4} (using Mathematica) turns out to be $$\begin{cases} X(\omega) = 0, \\ Y(\omega) = 0. \end{cases}\label{5}\tag{5}$$ Clearly solution \eqref{5} is not the same as solution \eqref{3}. Am I missing too many things here?

Edit: My question is simple. How to start with equation \eqref{4} and obtain equation \eqref{3}?

• Fourier transform is defined over the entire real line. Try using the Laplace Transform, which is defined for functions on the half line $[0,\infty)$
– Paul
Commented Apr 25 at 10:29
• Check if you are dividing by zero when $\omega=\pm s$ Commented Apr 25 at 13:22
• Yes, @user619894, thanks for pointing that out.
– Mike
Commented Apr 29 at 9:29

Equation \eqref{4} is actually $$\begin{bmatrix}-i\alpha & \beta \\-\beta & -i\alpha\end{bmatrix} \begin{bmatrix}X \\Y \end{bmatrix}=-i\omega\begin{bmatrix}X \\Y \end{bmatrix}$$ so besides the zero vector, there are 2 eigenvalues. This means that the solution to equation \eqref{4} is zero except when $$\omega$$ is an eigenvalue of the matrix equation. This will yield your equation \eqref{3}.

• And how does that explain $(3)$ and $(5)$?
– Mike
Commented Apr 25 at 20:27
• X,Y can be non zero, only when $i\omega$ is an eigenvalue Commented Apr 25 at 20:52
• @user619894, it would be nice if you elaborate a bit. Start with which of the $(3)$ and $(5)$ is correct and then correct the incorrect one showing that that two approaches yield consistent results. Commented Apr 26 at 7:11
• @User101, I edited the answer, is that clearer? Commented Apr 26 at 7:19
• @user619894, your answer seems going in the right direction, but it would help the reader if you could be more detailed. To be precise, show in a few steps as how to reach (3) from (4). Commented Apr 28 at 15:45

It is possible to disentangle/diagonalize the system of equations $$(4)$$ in the Fourier domain by defining $$Z_\pm = X \pm iY$$, hence $$i(\omega - s_\pm) Z_\pm(\omega) = 0$$, where $$s_\pm = \alpha \pm \beta$$. However, you cannot conclude $$Z_\pm(\omega) = 0$$ straightforwardly, because there is an implicit convention when taking the Fourier transform that we are working in a distributional space. Actually, the linear equation $$\omega f(\omega) = 0$$ is solved by $$f(\omega) \propto \delta(\omega)$$, because $$\omega\delta(\omega) \equiv 0$$. In other words, the multiplicative operator $$\omega$$ is not invertible anymore and its one-dimensional kernel is generated by the Dirac delta function.

In consequence, one has $$Z_\pm(\omega) = A_\pm \delta(\omega - s_\pm)$$, with $$A_\pm$$ some constants. How to find these constants ? They have to be determined from the initial conditions ! One has $$Z_\pm(t=0) = 1$$. From there, two strategies can be adopted. Either you apply the initial conditions after inverse Fourier transforming $$Z_\pm$$, i.e. $$Z_\pm(t) = A_\pm e^{-is_\pm t}$$, hence $$A_\pm = 1$$. Alternatively, you may take advantage of the relation $$f(t=0) = \int_\Bbb{R} F(\omega) \,\mathrm{d}\omega$$, which will lead to the same conclusion.

Finally, you switch back to the initial variables by writing $$\begin{cases} X = \,\frac{1}{2}(Z_+ + Z_-) \\ Y = \frac{1}{2i}(Z_+ + Z_-) \end{cases},$$ either after or before having inverted the Fourier transform, depending on the strategy you have followed, and you will obtain the same solutions as derived in the time domain only. In fact, the whole procedure could have been carried with $$X$$ and $$Y$$ instead of $$Z_\pm$$, but it permitted to avoid to consider manipulations over matrices, determinant, eigenvalues, etc.

• Thanks, @Abezhiko. If one denotes $M=\begin{pmatrix} -i \alpha & \beta \\ -\beta & -i \alpha \end{pmatrix}$, then in Fourier space, we have $M \cdot u(\omega)= -i \omega~ u(\omega)$, where $u(\omega)=(X(\omega)~~Y(\omega))^T$. In other words, equation $(4)$ reads $(M - i \omega \mathbb{1}) u(\omega) = 0.$ Could you now explain how can I obtain $(3)$ from here?
– Mike
Commented Apr 28 at 19:57
• Could you add a proof without diagonalising the system?
– Mike
Commented Apr 29 at 7:34
• @Mike I mean, how do you hope to solve the system, i.e. disentangle the variables, without diagonalization. In fact, $Z_\pm$ are the eigenvectors of $M$, associated to the eigenvalues $\omega-s_\pm$. You could avoid eventually its diagonalization by writing something like $\begin{pmatrix} X(\omega) \\ Y(\omega) \end{pmatrix} = \delta(\omega+iM) \begin{pmatrix} A \\ B \end{pmatrix}$, with $A$ and $B$ some constants, together with the integral representation of the Dirac delta, but it would be a somewhat awkward formal analog in Fourier space of the matrix exponential $e^{tM}$ in time domain. Commented May 1 at 13:50