2
$\begingroup$

The question is broad, I will specify an example to elaborate what I'm asking.

Suppose I have two different LC circuits inductively coupled (or capacitively, but the question I have will be relevant to both so for simplicity lets consider inductive coupling) as below.

Non-identical coupled LC circuits

Starting from Kirchoff's Voltage and Current Laws, and using the constitutive relations one can arrive at the equations of motion which look something like $$ \boldsymbol{A}\ddot{\vec{v}}= \boldsymbol{B}\vec{v}$$ here $\vec{v}=(v_1 \, v_5)^{T}$ where $v_1$ is the voltage across $C_1$ and $v_5$ is the voltage across $C_5$. One can easily check that this circuit has 2 degrees of freedom. Now I left $\boldsymbol{A},\boldsymbol{B}$ without explicitly stating them because the question is more general than this particular equation. Now I'm trying to find the normal modes of this circuit. But it occurs to me that $\boldsymbol{A},\boldsymbol{B}$ are guaranteed to be of rank $2$ and diagonalizable so we can find the eigenvectors for this generalized eigenvalue problem and be guaranteed they form a basis for $\mathbb{R}^2$. So we can decouple the equations of motion into 2 independent second order ODEs.

The problem is that normal modes are defined to be the orthogonal, that is the resulting eigenvectors are orthogonal and hence it is a different decoupling. It specifically requires $\boldsymbol{A},\boldsymbol{B}$ to be symmetric such that the resulting eigenvectors form an orthogonal basis.

Check this discussion on transforming second order linear systems of ODEs of same form as this.

So this leads me to think that if the matrices are not symmetric, the eigenvectors will not be orthogonal therefore no normal modes. But a decoupling is possible it seems. What is this decoupling and is this true - non-identical coupled oscillators don't have normal modes?

One note is that the lack of symmetry in the oscillators could justify not being able to exploit symmetric and anti symmetric normal modes (for a basic example like this) so perhaps not too surprising, but I was under the impression all coupled oscillators have normal modes.

Any help is appreciated!

$\endgroup$

2 Answers 2

1
$\begingroup$

The math in the link or post is not very clear. When done carefully, the prescription to find the modes becomes intuitive: it is equation $(*)$ below.

Start by writing $\boldsymbol{A}\ddot{\vec{v}}= \boldsymbol{B}\vec{v}$ as

$$ -\boldsymbol{A}\omega^2 \vec{v}_0= \boldsymbol{B}\vec{v}_0 \quad \Rightarrow \quad (\boldsymbol{A}\omega^2+\boldsymbol{B}) \vec{v}_0= 0, $$

thanks to the usual substitution $\vec{v}(t) = \vec{v}_0 e^{-i \omega t}$.

Now, call $\boldsymbol{C}(\omega) = \boldsymbol{A}\omega^2+\boldsymbol{B}$. The above system only has the trivial solution $ \vec{v}_0=0$ if $\boldsymbol{C}(\omega)$ has maximum rank (in this case 2). Therefore, to have non-trivial oscillatory solutions you have to require that $\omega$ takes some special value so that a solution with a non-zero amplitude $\vec{v}_0$ can exist. This amounts to require that the determinant of $ \boldsymbol{C}(\omega)$ is zero.

To summarize, you can find the frequencies of the natural modes of the system by solving the equation

$$ det[\boldsymbol{C}(\omega)] = 0 \quad \quad (*) $$

that is a polynomial equation in $\omega$. The solutions are the frequencies of the mode: for you 2D case you will find two (complex) solutions. As usual, the real part tells you the frequency of the oscillation, the imaginary part tells you how fast the mode decays or grows (depending on its sign).

$\endgroup$
3
  • $\begingroup$ That is what I'm doing, and if you click the cited link above - I further explain in there - that you obtain "modes" but clearly they are not "normal modes" since those need to be orthogonal. So what you described is what I've done and is generally valid as long as the matrix you mention has rank $2$ AND additionally it must be diagonalizable, otherwise the matrix's eigenvectors do not form a basis for $\mathbb{R}^2$. $\endgroup$ Commented Sep 4, 2021 at 18:52
  • 3
    $\begingroup$ @LostInEuclids5thPostulate, sorry but what you think it's clear from your exposition it was not clear to me (same for the link). I do not see the word "determinant", or a clear explanation of what you are already doing. $\endgroup$
    – Quillo
    Commented Sep 6, 2021 at 8:57
  • 1
    $\begingroup$ yes forgive me perhaps I should've explicitly stated determinant and generalized eigenvalue. But actually your comment brought up an interesting part helping me resolve a confusion. I will write a detailed answer to my own question when I can. Thanks for your input. $\endgroup$ Commented Sep 6, 2021 at 17:24
0
$\begingroup$

It's not clear to me that the matrices $A$ and $B$ are actually non-symmetric in the circuit case you've outlined. In general, we can define notions of "kinetic energy" and "potential energy" for an oscillating system as the part of the energy quadratic in the velocities and the part of the energy quadratic in the positions, respectively. Use of Lagrangian mechanics then guarantees that the resulting $A$ and $B$ matrices for the system are symmetric. So as long as you have notions of "kinetic energy" and "potential energy" for the system, you should be able to find a system of equations describing the system such that $A$ and $B$ are symmetric.

It's entirely possible that you've written down a system of equations that is equivalent to this "canonical" form but involve a non-symmetric $A$. Maybe your equations are just various linear combinations of the "canonical" equations, for example. If they are, it should be possible to invert this transformation. Some tips on how to do this can be found on this Math.SE Q&A, which has links explaining the process and an example of how this can be done.

But let's suppose you have a truly weird system without a proper notion of energy, and it is truly the case that the system is not "symmetrizable" in this way. As an example, take the system $$ \ddot{x}_1 + x_1 + x_2 = 0 \qquad \ddot{x}_2 + x_2= 0 $$ In that case we have $A = I$ and $$ B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}. $$ If we look for solutions of the form $\mathbf{x}(t) = e^{i \omega t} \mathbf{x}$, then we find that we must have $$ \det( - \omega^2 I + B) = (1 - \omega^2)^2 = 0 $$ as our characteristic equation. This equation has a double root at $\omega^2 = 1$. But if we try to find the vectors which correspond to this root, we find that they must satisfy $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \mathbf{x} = 0 $$ and this matrix only has a one-dimensional null space ($x_2 = 0$), rather than the 2-D nullspace we expect. This is the distinction between algebraic and geometric multiplicities of eigenvalues and eigenvectors, which never occurs for symmetric matrices but which can happen in more general cases.

But all this means is that there are solutions to this system that are not of the form $\mathbf{x}(t) = e^{i \omega t} \mathbf{x}$. In fact, we can see that we have $x_2(t) = C \sin (t - t_0)$, and then the equation for $x_1$ becomes $$ \ddot{x}_1 + x_1 = - C \sin (t - t_0) $$ which is the equation for an undamped driven oscillator. I believe the solution to this equation will include a term of the form $D t \sin t$, which is not of the form we postulated when we started solving the problem. In other words, it's not that non-symmetric matrices cause the physics to break, it's just that we need to look for more general solutions than we thought we did.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.