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Green function question

  1. Feb 24, 2010 #1
    I was reading up on Classical Mechanics and the general method for solving for an undamped harmonic oscillator was given as

    [tex]\frac{d^{2}q}{dt^{2}} + \omega^{2}q = F(t)[/tex]

    was solved using the Green function, G, to the equation

    [tex]\frac{d^{2}G}{dt^{2}} + \omega^{2}q = \delta(t-t')[/tex]

    and then integrating via the normal procedure.

    Then the next case considered was the damped harmonic oscillator which had a damping term proportional to [tex]\frac{\omega}{Q}[/tex] times the velocity. The equation has the form

    [tex]\frac{d^{2}q}{dt^{2}} + \frac{\omega}{Q}\frac{dq}{dt} + \omega^{2}q = F(t)[/tex]

    Now, the author wants to find the particular solution to this equation and says that his Green function is of the form

    G = [tex] A e^{\frac{{\omega(t-t')}}{2Q} + i\omega(t-t')}[/tex] plus this term's complex conjugate. This is a solution to STEADY STATE ONLY but does not consider the transient part.

    BUT, he also claims that the above Green function is a solution the equation

    [tex]\frac{d^{2}G}{dt^{2}} + \frac{\omega}{Q}\frac{dG}{dt} + \omega^{2}G = \delta(t-t')[/tex]

    How can he do this? The Green function which will solve the above equation will also have a transient part apart from the steady state function which he has considered. G will be of the form above PLUS a dying out part. What happens to that?
     
    Last edited: Feb 24, 2010
  2. jcsd
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