Green's function for a critically damped oscillator

In summary: I'm not sure why that would make a difference.In summary, the author is looking for help understanding green's function and how to use it. They explain that the equation of motion is x(t) = ∫ dt' G(t,t')F(t') from 0 to t, where G(t,t') is the particular solution to the ode with F(t) = δ(t-t'). They explore using the Fourier transform to solve for G(t,t'), and find that G(t,t') = x_{h} + 1/2\pi ∫ (e^{iwt}e^{-iwt'}). However,
  • #1
shyta
56
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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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]
 
  • #5
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
 
  • #6
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?
 
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  • #7
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?
 
  • #8
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.
 
  • #9
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.
 
  • #10
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.
 
  • #11
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.
 

1. What is a Green's function for a critically damped oscillator?

A Green's function for a critically damped oscillator is a mathematical tool used to solve differential equations that describe the motion of a critically damped oscillator. It represents the response of the system to an impulse input at a specific point in time.

2. How is a Green's function for a critically damped oscillator calculated?

A Green's function for a critically damped oscillator can be calculated by solving the differential equation for the critically damped oscillator and using the initial conditions to determine the coefficients. Another method is to use the Laplace transform to convert the differential equation into an algebraic equation, which can be solved for the Green's function.

3. What are the applications of a Green's function for a critically damped oscillator?

A Green's function for a critically damped oscillator can be used to solve a variety of physical problems, such as analyzing the behavior of mechanical systems, electrical circuits, and acoustic systems. It is also useful in signal processing and control systems.

4. What is the significance of a critically damped oscillator in relation to Green's function?

A critically damped oscillator is a type of damped oscillator that experiences the fastest decay to equilibrium without any oscillations. It is important in relation to Green's function because it has a unique solution that can be represented by a Green's function.

5. Are there any limitations of using a Green's function for a critically damped oscillator?

One limitation of using a Green's function for a critically damped oscillator is that it assumes the system is linear and time-invariant, meaning that the response to the input is independent of the time at which it is applied. Additionally, it may not be applicable to more complex systems that cannot be described by a simple differential equation.

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