1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Green fuktion on the wave operator in 1 dimension

  1. Jan 10, 2005 #1
    I have a Homework here i'm trying to solve like the last 3-4 hours but somehow i'm stuck so I ask you guys for help:

    They gave me the Green Function like this:
    http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img50.png [Broken]
    with http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img53.png [Broken] for y >= 0 and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img55.png [Broken] for y<0
    Now i'm supposed to solve this:
    http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img61.png [Broken]
    http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img36.png [Broken] and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo5/img39.png [Broken] and http://idefix.physik.uni-freiburg.de/~aufgabe/THEO204/theo9/img62.png [Broken]

    Would be so nice if someone could help me! thx in advance!
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Jan 10, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    So you're basically saying to solve
    [tex] box_{1}[I_{G_{1}}(\phi)] [/tex]

    That means [itex] box_{1} \int G_{1}(x,t)\phi (x) dx [/itex]
    with he "box"/d'alembertian taken wrt to "y" and "t'"("t" prime)
    Then u can insert the d'alembertian under the integral and use the differential eq.verified by the propagator.

    It think the integration will be immediate...

  4. Jan 10, 2005 #3
    thx for your quick reply! But I don't quite following that one, could you explain it to me with more details? and phi is also a function of t ;)
    I Expect to get phi(0,0) at the end. I tried some more and got something similar except a minus in front of it but i really don't think that it was mathematically correct what I did there over those 2 pages x_X have to get this one done by tomorrow morning so i would appreciate any kind of help! thx!
    What i basically did until now is that I made my integration invervals smaller:
    for all t < 0 Teta is 0 anyway so my integral over dt goes from 0 to infinite, now with my t >0 I started to change the interval for dx by only integrating over the support of Teta that means my Interval changed to [-ct , ct]
    Now my first problem was that my Integration Interval is a function of t. I did get one differentation to t inside the Integral over dx and got 3 terms and well ... here I'm not sure anymore if that was right x_X
    This is the last thing i have where i'm quite sure that it could be right:

    [itex] ((1/c^2) \partial_{t}^2 - \partial_{x}^2) \int_{0}^{\infty} dt \int_{-ct}^{ct} dx ( (c/2)\phi (x,t) ) [/itex]
    Last edited: Jan 10, 2005
  5. Jan 10, 2005 #4
    anyone? still can't solve it x_X
  6. Jan 10, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    Okay,first question:I find kind of awkward the shape of the Green's function.
    It should always look like this:
    [tex] G_{1}(x,x',t,t') [/tex] and should verify identically the differential equation
    [tex] \hat{O}_{x,t} G(x,x',t,t')=\delta(x-x')\delta(t-t') [/tex](1)

    U can make the substitution
    [tex] x-x' \rightarrow R [/tex](2)
    [tex] t-t' \rightarrow \tau [/tex] (3)

    and the new propagator would be
    [tex] G_{1}(R,\tau}) [/tex],which would check the LPDE

    [tex] \hat{O}_{R,\tau} G_{1}(R,\tau)=\delta(R)\delta(\tau) [/tex](4)

    If u say that your propagator is
    [tex] G_{1}(R,\tau)=\theta(R-c\tau) [/tex](5)
    and u want to compute
    [tex] I_{G_{1}}=\int [O_{R,\tau}G_{1}(R,\tau)]\phi(R,\tau)=\int \delta(R)\delta(\tau)\phi(R,\tau) =\phi(0,0) [/tex]

    Last edited: Jan 10, 2005
  7. Jan 10, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    BTW,your problem is set up wrongly.
    If u were to do exactly what was indicated,u'd end up with a dalembertian applied to a number (the value of the integral) which would be identically zero.

  8. Jan 10, 2005 #7
    yes, it should look like G(t,t',x,x') but t' and x' are set 0 we kinda did it in the couse that we subsitute it.
    [tex] G_{1}(R,\tau)=\theta(R-c\tau) [/tex]
    Is practically the same
    and this:
    [tex] I_{G_{1}}=\int [O_{R,\tau}G_{1}(R,\tau)]\phi(R,\tau)=\int \delta(R)\delta(\tau)\phi(R,\tau) =\phi(0,0) [/tex]
    is exactely what I'm supposed to show
    Yes the notation looks like ****, but the "box" operator is kinda a differentiation of a distribution here:
    [tex] \partial I_{G_{1}}( \phi(t,x)) = I_{\partial G_{1}}( \phi(t,x)) = - I_{G_{1}}( \partial \phi(t,x)) [/tex]
    could you just help me how you got the last line? or how to solve this one (hope it's right this time):
    [itex] \int_{0}^{\infty} dt \int_{-ct}^{ct} dx ( ((1/2c) \partial_{t}^2 - (c/2)\partial_{x}^2)\phi (x,t)) [/itex]
    Last edited: Jan 10, 2005
  9. Jan 10, 2005 #8


    User Avatar
    Science Advisor
    Homework Helper

    Well,i just i've given a proof.Remember that the Green function/propagator is a solution of this eq.
    [tex] \hat{O}G(R,\tau)=\delta(R)\delta\tau [/tex]
    which means
    [tex] \hat{O}\theta(R-c\tau)=\delta(R)\delta\tau [/tex]
    ,which is just i used to get the integral involving the product of delta functionals and the 'phi'.

  10. Jan 10, 2005 #9
    I know, that this is the solution but in this homework they say I have to show explicit, that this given G actually is a Green function to that Operator by solving that equasion. So I'm not allowed to use that equasion above with the delta distribution because that is exactely what i have to proof ^^
    I Just cleared up that integral and finally got a [tex] \phi(0,0) [/tex] term. still there are some ugly integrals left which i hope will result to 0 somehow ...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook