1
$\begingroup$

In differential equations, Green's functions are only defined given boundary conditions. In fact, you need two of them for a second order differential equation.

In a lot of physics lecture notes, a solution to the Green's function is solved by going to the Fourier domain. For example, for the damped harmonic oscillator,

$$\left( \frac{d^2}{dt^2} + \gamma \frac{d}{dt} + \omega_0^2 \right) G(t,t') = \delta(t-t'),$$

which we can transform into the Fourier domain by asserting $G(t,t') = \int \frac{d \omega}{2 \pi} e^{-i \omega t} G(\omega,t')$, and using the property $\delta(t-t') = \int \frac{d \omega}{2 \pi} e^{-i \omega (t-t')}$, to become

$$ (-\omega^2 - i \gamma \omega + \omega_0^2) G(\omega, t') = e^{+i \omega t'}$$

And therefore

$$ G(\omega, t') = \frac{e^{+i \omega t'}}{-\omega^2 - i \gamma \omega + \omega_0^2},$$

and so

$$ G(t,t') = \int_{-\infty}^{\infty} \frac{d \omega}{2 \pi} e^{-i \omega(t-t')} \frac{1}{-\omega^2 - i \gamma \omega + \omega_0^2},$$

which we can solve using a contour integral for $t > t'$ and $t < t'$. The poles are both in the LHS, which results in causality, i.e. a factor of $\Theta (t)$, the Heaviside function.

However, boundary conditions were not once invoked in this derivation, yet alone two. What gives? Does one of the steps sneakily contain boundary conditions?

$\endgroup$
1
$\begingroup$

Let us assume that $\gamma>0$, so the motion is damped.

You already understand that a boundary condition is needed because the kick given to the system at $t'$ can either start the mass oscillating, (folowing which the friction will cause its motion to die away) or it can stop an already existing motion. In the latter case the motion was already dying away because of the damping --- but that means that the it must have had infinite amplitude at $t=-\infty$. This is where the Fourier transform imposes choice of BC's: the Fourier transform of a smooth function (and the GF is smooth) tends to zero at infinity, so the FT always outputs the bounded solution. If $\gamma$ were negative the motion is anti-damped and the FT will output the bounded anticausal solution by stopping stopping the exponentially growing motion rather than starting it.

$\endgroup$
6
  • $\begingroup$ But the Green's function is not smooth, it has a discontinuity at $t=0$ (due to the Heaviside function). Also, doesn't the Fourier transform allow for distributions (such as the dirac-delta function)? Is the important point the behaviour at infinity? $\endgroup$
    – semmo
    Jan 10 at 14:30
  • $\begingroup$ I should have said that the FT $(-\omega ^2 --i\gamma\omega +\omega_0^2)^{-1}$ of the Green function is smooth. $\endgroup$
    – mike stone
    Jan 10 at 14:33
  • $\begingroup$ I thought that FT of $(-\omega^2 - i \gamma \omega + \omega_0^2)^{-1}$ is not smooth, it's zero for $t<0$ and non-zero for $t>0$, which means that derivatives are discontinous, i.e. not smooth. $\endgroup$
    – semmo
    Jan 10 at 14:38
  • $\begingroup$ I mean that the FT of the discontinuous Green function is smooth. The rational function $(-\omega^2 -i\gamma \omega +\omega_0^2)^{-1}$ is smooth on the real axis, so its FT (the GF) must go to zero at infinity. $\endgroup$
    – mike stone
    Jan 10 at 14:41
  • $\begingroup$ Ah okay, that seems to make sense. So the two (required) boundary conditions would be that $G(-\infty,t') = G'(-\infty, t') = 0$, essentially? $\endgroup$
    – semmo
    Jan 10 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.