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Matrix representation

  • Thread starter Anro
  • Start date
  • #1
10
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Hello everyone,
This is a normal modes problem that I’m working on, where the details are a bit tedious, but what I need to do is to write the following system:
mx`` = –2kx + ky + c
my`` = kx – ky + c
In the following form:
| m 0 | |x``| = |–2k k| |x|
| 0 m | |y``| = |k –k| |y|

The above are matrices with one “=” sign instead of two.
My problem is how to fit “c” into this system of matrices, where c is a constant; any help would be appreciated.
 

Answers and Replies

  • #2
525
6
First, here's a nice tex picture of your equation (quote me to see how you make these things):

[tex]
m\frac{d^2}{d t^2}\begin{pmatrix}x \\ y \end{pmatrix} = k\begin{pmatrix} -2 & 1 \\ 1 & 2\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} [/tex]

However, the problem with what you want to do is that your differential equation is not homogeneous in x and y (because of the constant c). Normally, for a homogeneous differential equation, the set of solutions form a vector space - which is why you can switch to this matrix notation in the first place. But for non-homogeneous systems this is no longer true.

Watch you can do is set c=0, which makes the system homogeneous. Next you find the solution to these equations. This gives you a set of independent solutions. Then you find one solution to your nonhomogeneous equation. The general solution is then this solution + a linear combination of the independent solutions. But maybe I'm running ahead of things here..
 
  • #3
10
0
Thanks for your reply xepma; I understand your idea of solving this system using the method of undetermined coefficients to find complementary and particular solutions, and then you add them up to get the general solution. But this idea will give me a messy solution, and what I need to do is to find a way of writing “c” in terms of x or y so that the matrix on the right contains only k’s; as my next step in the solution will be calculating a determinant [det(k – ω^2m)], and then solve for ω; that’s why I can’t afford to put messy values in it. Thanks again for your help.

By the way, please accept my apologies for my tex ignorance; hopefully I’ll get to it sometime.:smile:
 
  • #4
525
6
Well, first thing that comes to mind is that you can make the substitution:
[tex] x = u + A c[/tex]
[tex] y = v + B c[/tex]

where A and B are constants yet to be determined. These are determined by demanding that the contributions due to c vanishes (i.e. such that the differential equation in terms of u and v are homogeneous).

Answer:
A = 2/k
B = 3/k

Then your differential equation in terms of u and v just becomes the same as for x and y with c = 0.
 
Last edited:
  • #5
10
0
Now that’s more like it, and we don’t have to worry about any messy terms when solving the determinant for ω; thank you very much for your help xepma.
 

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