Physics Asked by semmo on February 6, 2021
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?
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.
Answered by mike stone on February 6, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP