BobbyBear
- 162
- 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}<br /> m_1 & 0 \\<br /> 0 & m_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> \ddot{u}_1 \\<br /> \ddot{u}_2 \\<br /> \end{array} \right) <br /> +<br /> \left( \begin{array}{cc}<br /> (k_1 + k_2) & -k_2 \\<br /> -k_2 & k_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> u_1 \\<br /> u_2 \\<br /> \end{array} \right) <br /> <br /> =<br /> \left( \begin{array}{cc}<br /> 0 \\<br /> 0 \\<br /> \end{array} \right) <br /> <br /> [/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?
(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}<br /> m_1 & 0 \\<br /> 0 & m_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> \ddot{u}_1 \\<br /> \ddot{u}_2 \\<br /> \end{array} \right) <br /> +<br /> \left( \begin{array}{cc}<br /> (k_1 + k_2) & -k_2 \\<br /> -k_2 & k_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> u_1 \\<br /> u_2 \\<br /> \end{array} \right) <br /> <br /> =<br /> \left( \begin{array}{cc}<br /> 0 \\<br /> 0 \\<br /> \end{array} \right) <br /> <br /> [/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?