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Green's function for an impulsive force on a string

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img836.imageshack.us/img836/2479/stepvt.png [Broken]

    2. Relevant equations

    H'(t) = [itex]\delta[/itex](t)

    3. The attempt at a solution

    So far Ive taken the derivatives of G(x,t) with respect to xx and tt and gotten
    [itex]G_{xx}[/itex](x,t) = -[itex]\frac{θ^{2}}{c}[/itex] and
    [itex]G_{tt}[/itex](x,t) = [itex]θ^{2}[/itex]c

    which gives [itex]θ^{2}[/itex]c - [itex]c^{2}[/itex](-[itex]\frac{θ^{2}}{c}[/itex]) = [itex]\delta[/itex](x)[itex]\delta[/itex](t)

    = 2[itex]θ^{2}[/itex]c = [itex]\frac{dH(x)}{dx}[/itex][itex]\frac{dH(t)}{dt}[/itex]
    where H(x), H(t) are the heaviside step functions for x and t.

    I'm not sure how these are related or if I've gone about this in a completely wrong way. (Dirac functions are not my strong suit :frown: )
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 23, 2011 #2
    Is it possible that θ(ct-x) does not equal (θct - θx) which I assumed when I did the derivatives, but that theta is a function of ct -x ? In which case I'm really lost
     
  4. Oct 23, 2011 #3

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    Hi xago! :smile:

    No, θ(ct-x) does not equal (θct - θx).
    The first is zero for any x and t for which ct-x=0.
    The second is only zero if x=0 and t=0.

    I recommend working out the derivatives using product rule and chain rule.
    You can use that dθ/dx=δ(x) and dδ/dx=δ'(x).

    For the final expression you'll need some properties of δ in 2 dimensions to simplify it.
     
    Last edited: Oct 23, 2011
  5. Oct 23, 2011 #4
    I don't understand how dθ/dx=δ(x)
     
  6. Oct 23, 2011 #5

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    It's a property of the Heaviside step function.
    See for instance: http://en.wikipedia.org/wiki/Heaviside_step_function

    Edit: I meant what you already had as a relevant equation: H'(t) = δ(t).
     
  7. Oct 23, 2011 #6
    For [itex]G_{tt}[/itex](x,t) I'm getting [itex]\frac{1}{2}[/itex][itex]\frac{d^{2}}{dt^{2}}[/itex](θ)(ct-x)cθ(x+ct) + [itex]\frac{d}{dt}[/itex](θ)(ct-x)[itex]\frac{d}{dt}[/itex](θ)(x+ct)c + [itex]\frac{1}{2}[/itex]θ(ct-x)[itex]\frac{d^{2}}{dt^{2}}[/itex](θ)(x+ct)c

    For [itex]G_{xx}[/itex](x,t) I get pretty much the same thign except divied by c, [itex]\frac{1}{c}([/itex][itex]\frac{1}{2}[/itex][itex]\frac{d^{2}}{dx^{2}}[/itex](θ)(ct-x)cθ(x+ct) + [itex]\frac{d}{dx}[/itex](θ)(ct-x)[itex]\frac{d}{dx}[/itex](θ)(x+ct)c + [itex]\frac{1}{2}[/itex]θ(ct-x)[itex]\frac{d^{2}}{dx^{2}}[/itex](θ)(x+ct)c)

    When I plug them into [itex]G_{tt}[/itex](x,t) - [itex]c^{2}[/itex][itex]G_{xx}[/itex](x,t) I get 2c[itex]\frac{d}{dt}[/itex](θ)(ct-x)[itex]\frac{d}{dx}[/itex](θ)(x+ct)
     
  8. Oct 23, 2011 #7

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    Looks good!
    That's also what I have.

    Here's a few more properties of delta (from: http://en.wikipedia.org/wiki/Dirac_delta_function" [Broken]).

    [itex]\delta(x,y)=\delta(x)\delta(y)[/itex]

    [itex]\delta(ax)={\delta(x) \over |a|}[/itex]

    [itex]\delta(R(x,y))=\delta(x,y)[/itex] for any rotation or reflection R.
     
    Last edited by a moderator: May 5, 2017
  9. Oct 23, 2011 #8
    Well θ is a function of x and t right? So is take [itex]\frac{d}{dx}[/itex](θ)(x+ct) the same as δ(θ(x,t)) = δ(x,t) ?
    or [itex]\frac{d}{dx}[/itex](θ)(x+ct) = δ(x)?
     
  10. Oct 23, 2011 #9

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    Not quite.
    What you have is that θ'(y)=δ(y).
    Applying the chain rule, you get:
    [tex]{d \over dt}(\Theta(x+ct)) = \delta(x+ct) \cdot c[/tex]
    Apparently you have already applied the chain rule, so you should just replace d/dt(θ)(x+ct) by δ(x+ct).



    Edit: Note that δ(x+ct) is a one-dimensional dirac delta, while δ(x, ct) is a two-dimensional dirac delta.
     
  11. Oct 23, 2011 #10
    Right so what I'm trying to do get rid of the 2c term somehow, what I'm getting is

    2cδ(x+ct)δ(ct-x) which I need to make equal to δ(x)δ(t)
     
  12. Oct 23, 2011 #11

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    Yep.
     
  13. Oct 23, 2011 #12

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    Here's an alternative method to find it:

    Let f(u,v) be an unspecified function.
    Then:
    [tex]\iint f(ct+x, ct-x) \delta(ct+x) \delta(ct-x) dxdt = \iint f(u,v) \delta(u) \delta(v) |\det(J)| dudv[/tex]
    where J is the Jacobian matrix identified by the transformation from (u,v) to (x,t).

    Since this equation holds for any function f, it follows that:
    [tex]\delta(ct+x) \delta(ct-x) = \delta(x) \delta(t) |\det(J)|[/tex]
    which leaves only the Jacobian matrix to be identified.
     
    Last edited: Oct 23, 2011
  14. Oct 23, 2011 #13
    I understand it pretty good now, thanks for your help!
     
  15. Oct 23, 2011 #14

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    You're welcome! :smile:
     
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