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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]
    with
    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

    dextercioby

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    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...

    Daniel.
     
  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

    dextercioby

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    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]

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

    dextercioby

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    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.

    Daniel.
     
  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

    dextercioby

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    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'.

    Daniel.
     
  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 ...
     
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