Green's function for a critically damped oscillator

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 9K views
shyta
Messages
56
Reaction score
0

Homework Statement


Consider critically damped harmonic oscillator, driven by a force F(t)
Find the green's function G(t,t') such that x(t) = ∫ dt' G(t,t')F(t') from 0 to T solves the equation of motion with x(0) =0 and x(T) =0

Homework Equations


x(t) = ∫ dt' G(t,t')F(t') from 0 to T

The Attempt at a Solution



Hi guys, I am completely new to green's function.. need a lot of help understanding the use, and how to use it >_<

I've been doing some readings and this is what i understand so far

x = [itex]x _{h}[/itex] + ∫G(t,t')f(t') dt'

i.e. G(t,t') will be the particular solution to the ode with F(t) = δ(t-t')

does this mean that I should let [itex]\ddot{x}[/itex] +2γ[itex]\dot{x}[/itex] + [itex]ω _{0}[/itex]² = δ(t-t') and solve this to get the green's function?
 
Physics news on Phys.org
shyta said:
does this mean that I should let [itex]\ddot{x}[/itex] +2γ[itex]\dot{x}[/itex] + [itex]ω _{0}[/itex]² = δ(t-t') and solve this to get the green's function?
Yes. Except that there's a small typo in your expression. It should be

[tex]\ddot{x} +2\gamma\dot{x} + \omega_{0}^2 x = \delta(t-t^\prime)[/tex]

A nice way to solve the above equation is to write the RHS as the inverse of a Fourier transform, meaning the RHS becomes a harmonic function of time.
 
Hello! thanks for you reply! :)

yeah you are right, i missed out the x there

not sure if i got it right but on the RHS, i got [itex]\frac{1}{2\pi}[/itex][itex]\frac{1}{i(t-t')}[/itex][itex]e^{i\omega(t-t')}[/itex]+C

is this right?

oh and (d²/dt² + 2γd/dt + ω0²) G(t,t') = δ(t-t')
so i should be solving as usual 2nd order ode to get my answer for homogeneous and inhomo am i doing the steps right?
 
shyta said:
Hello! thanks for you reply! :)

yeah you are right, i missed out the x there

not sure if i got it right but on the RHS, i got [itex]\frac{1}{2\pi}[/itex][itex]\frac{1}{i(t-t')}[/itex][itex]e^{i\omega(t-t')}[/itex]+C

is this right?

oh and (d²/dt² + 2γd/dt + ω0²) G(t,t') = δ(t-t')
so i should be solving as usual 2nd order ode to get my answer for homogeneous and inhomo am i doing the steps right?

Try taking the Fourier transform of both sides. To makes things simpler, recognize that the dependent of t and t' in G has this form: [itex]G(t, t^\prime) = G(t - t^\prime)[/itex]
 
im really bad at Fourier transform, so i didnt follow your advice..

but i did try considering the force being written as a Fourier transform
f(t) = 1/2[itex]\pi[/itex]∫F(w)[itex]e^{iwt}[/itex] dw
and the dirac delta
δ(t-t')= 1/2[itex]\pi[/itex]∫ [itex]e^{iwt}[/itex][itex]e^{-iwt'}[/itex] dw

so i went ahead to solve and i got that

G(t,t') = [itex]x_{h}[/itex] + 1/2[itex]\pi[/itex] ∫ ([itex]e^{iwt}[/itex][itex]e^{-iwt'}[/itex]) / ([itex]-\omega^{2}[/itex] + 2i[itex]\gamma\omega[/itex] + [itex]\omega_{0}^{2}[/itex]) dw



and I am stuck at this integration lol
 
shyta said:
im really bad at Fourier transform, so i didnt follow your advice..

but i did try considering the force being written as a Fourier transform
f(t) = 1/2[itex]\pi[/itex]∫F(w)[itex]e^{iwt}[/itex] dw
and the dirac delta
δ(t-t')= 1/2[itex]\pi[/itex]∫ [itex]e^{iwt}[/itex][itex]e^{-iwt'}[/itex] dw

so i went ahead to solve and i got that

G(t,t') = [itex]x_{h}[/itex] + 1/2[itex]\pi[/itex] ∫ ([itex]e^{iwt}[/itex][itex]e^{-iwt'}[/itex]) / ([itex]-\omega^{2}[/itex] + 2i[itex]\gamma\omega[/itex] + [itex]\omega_{0}^{2}[/itex]) dw
and I am stuck at this integration lol
The integration is a little tricky. You're going to need to use complex analysis, which I'm assuming you've at least met. Since we are dealing with a critically damped oscillator, we can set [itex]\gamma=1[/itex]. What we're going to do now is extend the integral into the complex plane. Let [itex]C\subset\mathbb{C}[/itex] be a semi-circle in the upper half plane, that is, [itex]C=\{\omega:|\omega|< a, \omega\geq0\}[/itex]. So, we have

[tex]I=\int_C \frac{e^{i\omega(t-t^\prime)}}{-\omega^2 + 2i\omega_0\omega + \omega_0^2}\;\text{d}\omega[/tex]

Can you do the next step, maybe using Cauchy's Residue Theorem?
 
Last edited:
Hmm nope I have never heard of the residue theorem. I tried wiki-ing it to see how it works, but it looks complicated. Is there any other way to do this?
 
shyta said:
Hmm nope I have never heard of the residue theorem. I tried wiki-ing it to see how it works, but it looks complicated. Is there any other way to do this?
I can't immediately see another straightforward method to evaluate the integral. Let me ask a few of the other homework helpers and get back to you.
 
Hootenanny said:
The integration is a little tricky. You're going to need to use complex analysis, which I'm assuming you've at least met. Since we are dealing with a critically damped oscillator, we can set [itex]\gamma=1[/itex].
The differential equation for the damped harmonic oscillator is usually
[tex]\ddot{x} + 2\zeta\omega_0 \dot{x} + \omega_0^2x = 0[/tex]There's a factor of [itex]\omega_0[/itex] in the damping term, and with this convention, a critically damped oscillator would correspond to [itex]\zeta=1[/itex], so I think you actually want [itex]\zeta=\omega_0[/itex] here.
 
vela said:
The differential equation for the damped harmonic oscillator is usually
[tex]\ddot{x} + 2\zeta\omega_0 \dot{x} + \omega_0^2x = 0[/tex]There's a factor of [itex]\omega_0[/itex] in the damping term, and with this convention, a critically damped oscillator would correspond to [itex]\zeta=1[/itex], so I think you actually want [itex]\zeta=\omega_0[/itex] here.
Good catch! I didn't notice that :redfaced:

I will correct my earlier posts.
 
Another way to find the Green's function is to solve the differential equation in the two regions 0≤t<t' and t'<t≤T. Apply the boundary conditions, and then match the two solutions at t=t'. The Green's function needs to be continuous at the boundary, but its derivative is not. By integrating the differential equation from t'-ε to t'+ε and letting ε→0, you can find the discontinuity in the derivative.