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Linear equations for amplitude factors

  1. Oct 17, 2012 #1
    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?
     

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  3. Oct 18, 2012 #2

    ehild

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    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
  4. Oct 18, 2012 #3
    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
  5. Oct 18, 2012 #4

    ehild

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    You meant homogeneous linear system...

    ehild
     
  6. Oct 18, 2012 #5
    Thanks for pointing this out!
     
  7. Oct 18, 2012 #6
    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?
     
  8. Oct 18, 2012 #7
    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.
     
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