Problem getting my matrices in correct form

1. Dec 5, 2015

saybrook1

1. The problem statement, all variables and given/known data
Hello guys; I am currently dealing with a problem that I have faced before several times and I would like to know a consistent way on how to solve it. I think what I want to do is diagonalize a matrix but I'm not sure if that's exactly it. Basically I have two or three equations of motion and I want to put them in a good matrix form in order to begin finding solutions. In this particular problem I have three equations of motion and the book I'm using puts them in to a nice matrix form right away and I'm not sure how to get there. Here is a picture of what they've done:

I know how to create these matrices for 'm' and 'k' when they are more symmetric if you will but I don't know how to make it nice like this. This will help me so much on other problems I also have to deal with. For some stupid reason I've ran in to this wall multiple times. Thanks in advance.

2. Relevant equations
3 equations of motion put in to a nice diagonalized matrix

3. The attempt at a solution
I just don't understand how to get it into this nice 'diagonalized(?)' matrix form. If anyone can help out, this will greatly help me in the future.

Last edited by a moderator: Dec 5, 2015
2. Dec 5, 2015

Staff: Mentor

I'm not sure I understand what you want. The form you get there, which is not diagonal by the way, by tridiagonal, is due to the fact that there is only nearest neighbor coupling. If there was an additional spring between 1 and 3, the matrix would look different.

Note also that the result is due to the Ansatz that motion is harmonic, as given by the intermediate equation. This is what transforms the problem into an eigenproblem, with ω2 the eigenvalue.

3. Dec 5, 2015

saybrook1

Okay, thank you for your response. I guess I'm just not sure exactly how to use the Ansatz and arrive at this matrix form. It says "substitution gives"... Not sure what substitution they are making,

4. Dec 5, 2015

SteamKing

Staff Emeritus
It's just a matter of organizing and re-arranging the unknowns, x10, etc., after making the substitution of the trial solution into the equations in (1).

5. Dec 5, 2015

saybrook1

Okay, thanks, I'll give it a shot.

6. Dec 5, 2015

saybrook1

It's tough because after I substitute in the Ansatz, I still don't know where the $(k-\omega^2M)$ comes from or why it's only associated with the $\ddot{x}$ terms. Any hints as to how that works would be excellent. Also, if someone happens to know a situation where you need to diagonalize a matrix or simultaneous diagonalize two matrices and could point me in the direction of an example of such a situation that would be really great. Thanks

7. Dec 5, 2015

SteamKing

Staff Emeritus
Well, take the first equation in (1), namely $M\ddot{x_1} + k(x_1 - x_2) = 0$ and make the substitution $x_i = x_{i0} e^{i\omega t}$. Note that the i-subscript is not the same as i in the exponential function, which is the imaginary unit.

8. Dec 5, 2015

saybrook1

Right, I did that and I end up with $M(\ddot{x_{10}}e^{i\omega t})+k(x_{10}e^{i\omega t}-x_{20}e^{i\omega t}) = 0$. Not sure where to go after that.

9. Dec 6, 2015

saybrook1

Ended up figuring this out by creating a matrix for M and K and X all separately and then combining them. Still not exactly sure how to get there with the Ansatz but I appreciate the help you guys gave me. Thanks

10. Dec 6, 2015

Staff: Mentor

That's not correct. Remember that $\ddot{x}$ means $\frac{d^2x}{dt^2}$, so when you do the substitution:
\begin{align*} M \ddot{x_{1}} &= M \frac{d^2x}{dt^2} \\ &= M \frac{d^2}{dt^2} ( x_{10}e^{i\omega t} ) \\ &= -M \omega^2 x_{10} \mathbf{e^{i\omega t}} \end{align*}
($x_{10}$ is independent of time; the entire time dependence of $x_{1}$ is in the exponential)

Edit: missing exponential in bold added.

Last edited: Dec 7, 2015
11. Dec 7, 2015

saybrook1

That's great, thank you! why do we drop the exponential at the end though?

12. Dec 7, 2015

Ray Vickson

The general solution does not have quite as simple a form as you suggest: besides $x_k = x_{k0} \, e^{i \omega t}$ terms there are also terms of the form $x_1 = x_2 = x_3 = a + b t$. This is due in part to the fact that when written as a $6 \times 6$ linear first-order system we get a $6 \times 6$ matrix that is not quite diagonalizable: one of its Jordan blocks is non-diagonal. I can tell you more if you are interested.

13. Dec 7, 2015

Staff: Mentor

It is still there, I just forgot to add it. Eventually, you multiply everything by $e^{-i \omega t}$, so it goes away.

14. Dec 7, 2015

saybrook1

Gotchya okay, thanks a bunch for the help.