# Linear equations for amplitude factors

1. Oct 17, 2012

### aaaa202

My book discusses the oscillations of systems for small pertubations around an equilibrium point. By considering taylors expansions of the potential and kinetic energy they are led to a set of equations for the amplitude factors. I have attached the crucial lines as a picture.
My question is to how they solve the system of equations. They say that we have n solutions so any nontrivial solution requires the determinant of the system of equations to be zero. I think I am misunderstanding something here, because how does that solve the system of equations? Everyone knows that a system has no solutions if the determinant is zero. So what information does requiring the determinant to be zero give?

#### Attached Files:

• ###### oscillations.png
File size:
62 KB
Views:
52
2. Oct 18, 2012

### ehild

An inhomogeneous system of equations has no solution if the determinant is zero, but a homogeneous one has (nonzero) solution only when the
determinant is zero.

See the following equations:

A
x+y=0
2x+2y=0

B

x+y=0
x+2y=0

which one has nonzero solution?

In case of small oscillation, you have an unknown parameter ω and you get the possible values of ω from the condition that the determinant is zero.

ehild

Last edited: Oct 18, 2012
3. Oct 18, 2012

### voko

A homogeneous linear system always has a trivial - all zeros - solution. This corresponds to the stationary solution of the original system of ODEs. But they are are looking for (non-stationary) perturbations in the vicinity of the stationary solution. So the linear system must have a non-trivial solution in that case. A non-trivial solution exists only if the determinant of the system if zero.

Last edited: Oct 18, 2012
4. Oct 18, 2012

### ehild

You meant homogeneous linear system...

ehild

5. Oct 18, 2012

### voko

Thanks for pointing this out!

6. Oct 18, 2012

### aaaa202

oh okay. I just always learnt that a matrix is invertible if determinant≠0. So there exists one solution. What is the basic difference between that case for inhomogenous systems and this one, in which you somehow find solutions when the matrix is non-invertible?

7. Oct 18, 2012

### voko

The basic difference is that (when det A = 0) the inhomogeneous system may have no solutions or may have infinitely many. The homogeneous system always has at least one solution (all zeros), and (when det A = 0) it has infinitely many non-zero solutions. You may want to read up on the theory of linear systems.