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Similar to an eigenvalue problem . . . help!

  1. Nov 28, 2009 #1
    Consider the following linear homogeneous ordinary differential equation system:
    (NB this system describes the movement of the natural response of a two degree of freedom structural system made up of two lumped masses connected by elastic rigidities) :

    [tex]
    \left( \begin{array}{cc}
    m_1 & 0 \\
    0 & m_2 \\
    \end{array} \right)

    \left( \begin{array}{cc}
    \ddot{u}_1 \\
    \ddot{u}_2 \\
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    (k_1 + k_2) & -k_2 \\
    -k_2 & k_2 \\
    \end{array} \right)

    \left( \begin{array}{cc}
    u_1 \\
    u_2 \\
    \end{array} \right)

    =
    \left( \begin{array}{cc}
    0 \\
    0 \\
    \end{array} \right)


    [/tex]

    which I shall compactly write as:

    [tex]
    [m] \vec{\ddot{u}} + [k] \vec{u}} = \vec{0}
    [/tex]

    Now, to solve, we assume a solution of the form:

    [tex]
    \vec{u}(t)=q_n(t) \vec{\phi _n}
    [/tex]

    where

    [tex]
    q_n(t) = A_n cos (\omega _n t) + B_n sin (\omega _n t)
    [/tex]

    and

    [tex]
    \vec{\phi _n}
    [/tex]

    is a constant vector.

    Then
    [tex]
    \vec{\ddot{u}}(t)=-\omega _n^2 q_n(t) \vec{\phi _n}
    [/tex]

    Substituting into the differential system,

    [tex]
    \left[-\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} \right] q_n(t) = \vec{0}
    [/tex]

    from which

    [tex]
    -\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} = \vec{0}
    [/tex]

    [tex]
    (-\omega _n^2 [m] + [k]) \vec{\phi _n} = \vec{0}
    [/tex]

    and for there to be a non trivial solution, we need:

    [tex]
    det(-\omega _n^2 [m] + [k]) = 0
    [/tex]

    from which we get two values of

    [tex]
    \omega _n^2
    [/tex]

    Now, my book (Dynamics of Structures by Chopra) says that the [tex] \omega _n^2 [/tex] are real and positive because [k] and [m] are real symmetric and positive definite.
    I don't see how this deduction is made!! I mean, I know that if a matrix [A] is a real symmetric matrix that is positive definite, then all its eigenvalues are real and positive (the proof is available in any standard text of linear algebra).
    But I just don't see how to prove the other statement! the [tex] \omega _n^2 [/tex] are not the eigenvalues of any matrix, are they? (even though it's a similar problem to an eigenvalue problem). Can someone help me see how that deduction is made?
     
  2. jcsd
  3. Nov 29, 2009 #2

    Mute

    User Avatar
    Homework Helper

    The matrix [itex][m][/itex] is invertible. Factor it out of

    [tex]-\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} = \vec{0}[/tex]

    to get

    [tex]-\omega _n^2 \vec{\phi _n} + [m]^{-1}[k] \vec{\phi _n} = \vec{0}[/tex]

    or

    [tex][m]^{-1}[k] \vec{\phi _n} = \omega _n^2 \vec{\phi _n},[/tex]

    which means the [itex]\omega_n^2[/itex] are eigenvalues of the matrix [m]^{-1}[k].
     
  4. Nov 30, 2009 #3
    Thank you Mute, I never thought of doing that!

    But I'm still not quite able to reach the desired conclusion...

    Okay so the [itex]
    \omega_n^2
    [/itex] are eigenvalues of the matrix [itex]
    [m]^{-1}[k]
    [/itex]

    And I've read that every positive definite matrix is invertible, and its inverse is also positive definite, so that means that if [m] is positive definite, then so is [itex]
    [m]^{-1}
    [/itex]

    So we have that both [itex]
    [m]^{-1}
    [/itex] and [itex]
    [k]
    [/itex] are positive definite, but in general that does not mean that [itex]
    [m]^{-1} [k]
    [/itex] is positive definite, does it?
    I've read that if two matrices [M] and [N] are positive definite, then their product is positive definite if [itex]
    [M] [N] = [N] [M]
    [/itex], but this is not the case with [itex]
    [m]^{-1}
    [/itex] and [k], their product is not commutative in general. So how can we see that [itex]
    [m]^{-1} [k]
    [/itex] is positive definite?

    Thanks for your help!
     
  5. Nov 30, 2009 #4
    Oh but wait! I just realised that even though [m] and [k] are real symmetric, [itex] [m]^{-1}[k] [/itex] is not even symmetric, so it wouldn't be of any use to prove that [itex] [m]^{-1}[k] [/itex] is positive definite, would it, because we don't know that the eigenvalues are real...

    [tex] [m]^{-1}[k] =

    \left( \begin{array}{cc}
    1/m_1 & 0 \\
    0 & 1/m_2 \\
    \end{array} \right)

    \left( \begin{array}{cc}
    (k_1 + k_2) & -k_2 \\
    -k_2 & k_2 \\
    \end{array} \right)

    =
    \left( \begin{array}{cc}
    (k_1 + k_2)/m_1 & -k_2/m_1 \\
    -k_2/m_2 & k_2/m_2 \\
    \end{array} \right)



    [/tex]

    So how do we see that the eigenvalues of [itex] [m]^{-1}[k] [/itex] are real and positive?
     
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