How to decouple this system of 2nd order partial differential equations?

Question

Solving a problem I found this system of equations:

$$ (\partial _t ^2 + \partial _x ^2 + \partial_y ^2 + \partial_z ^2)a_1(\vec{x},t) - 4gB(x\partial_y - y\partial_x)a_{\color{Red}2}(\vec{x},t) - 4g^2B^2(x^2 + y^2)a_1 - \lambda a_1 = 0 $$

$$ (\partial _t ^2 + \partial _x ^2 + \partial_y ^2 + \partial_z ^2)a_2(\vec{x},t) + 4gB(x\partial_y - y\partial_x)a_{\color{Red}1}(\vec{x},t) - 4g^2B^2(x^2 + y^2)a_2 - \lambda a_2 = 0 $$

taking into account that $g,B,\lambda=$ constants,

Any ideas to separate these equations in terms of $a_1$ and $a_2$?

Thanks a lot!

Answer 1

I found a way to decouple the equation, but I don't know if it is the only way or the most efficient way. Anyway, at least it helps the purpose.

The process for decoupling the equations consists of considering eigenfunctions of the form $ a^\pm = a^1 \pm ia^2$, where $i$ is the complex imaginary number. This is achieved after multiplying the second of the equations shown by the factor $ i $. Then add the first and second, then subtract the first and second (remember that this second is multiplied by $ i $).

Finally you will have: \begin{split} (\partial^2 _0 - \vec{\nabla}^2)a^+ &+ 4igB(x\partial_y - y\partial_x)a^+ -4g^{2}B^{2}(x^2 + y^2)a^+ -\lambda\; a^+ (\vec{x},t)=0 \\ (\partial^2 _0 - \vec{\nabla}^2)a^{-} &- 4igB(x\partial_y - y\partial_x)a^{-} -4g^{2}B^{2}(x^2 + y^2)a^{-} -\lambda\; a^{-}(\vec{x},t)=0 \\ \end{split}

I hope this can be of use.

Answer 2

Here's a linear algebra approach. I'll rearrange this slightly to make the next steps a bit clearer. I'm going to use $\Delta$ to represent $\partial_t^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$. If it were $-\Delta_t^2$, I'd use the standard $\square$ d'Alembert notation instead. I'll also write $x\partial_y - y\partial_x$ as $iL_z$, taking cues from quantum mechanics, where $L$ is the angular momentum operator. (I suspect this is a relativistic quantum particle in an external magnetic field anyway..)

\begin{align} (\Delta - 4g^2 B^2(x^2+y^2) - \lambda) a_1 - 4igB L_z a_2 &= 0 \\ (\Delta - 4g^2 B^2(x^2+y^2) - \lambda) a_2 + 4igB L_z a_1 &= 0 \end{align}

Writing this as a matrix equation, this would be

$$ \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) & -4igB L_z \\ 4igB L_z & \Delta - 4g^2 B^2(x^2+y^2) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$

This matrix is of the form

$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}. $$

A quick check shows that this matrix is diagonalized by

$$ P = \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}, \qquad P^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix} $$

so we can turn the above matrix equation into..

$$ \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix} \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) & -4igB L_z \\ 4igB L_z & \Delta - 4g^2 B^2(x^2+y^2) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$

simply by multiplying on the left by $P$. Multiplying by $P^{-1}P$ on the left of $\begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$, we get

$$ \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) + 4gB L_z & 0 \\ 0 & \Delta - 4g^2 B^2 (x^2+y^2) - 4gB L_z\end{pmatrix} \begin{pmatrix} ia_1 + a_2 \\ -ia_1 + a_2\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$

This matches with what you have. The strength of this approach is it gives you a prescription to go by. Decoupling a system of linear ODEs/PDEs is tantamount to diagonalizing the matrix-of-operators system.