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Green Function question.

  1. Mar 12, 2008 #1
    I need to find the green function of
    [tex](\frac{d^2}{dx^2}-k^2)\psi(x)=f(x)[/tex]
    s.t it equals zero when x approaches plus and minus ifinity.

    Now according to my lecturer I first need to solve the homogenoues equation, i.e
    its solution is: psi(x)=Ae^(kx)+Be^(-kx)
    and [tex]G(x,x')=\sum u_n^*(x)u_n(x)/(\lambda_n-\lambda)[/tex]
    so u_n(x)=Ae^(xk_n)+Be^(-xk_n)
    so as x approaches inifnity we get that A should be equal zero for green function to be zero, for x approaching minus infinity B should equal zero, so evetaully we get that G is identically zero, or am I missing something crucial, here?
    how do i find the nth-eigenvalue of the eigenfunction u_n?
     
  2. jcsd
  3. Mar 12, 2008 #2
    I know that [tex]\lambda=k^2[/tex]
    but don't know how to find the n-th eigenvalue.
     
  4. Mar 12, 2008 #3
    if you're domain is not finite, this is a not regular sturm-liouvile problem and you cannot find eigen-velues/vectors...
    and you did't specified wich are the boundary conditions also...
    search for sturm liouville regular problem...

    regards
    marco
     
    Last edited: Mar 12, 2008
  5. Mar 13, 2008 #4
    well I don't need to find here vectors but eigenfunctions.

    anyway this is the way I got the question, the condition that as x approaches plus minus infinity G(x,x') disappears is the boundary conditions on the green function, perhaps now it's better phrased.
     
  6. Mar 13, 2008 #5

    CompuChip

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    First off, the usual definition of a Green's function is that
    [tex](\partial_x^2 - m^2) G(x) = \delta(x)[/tex]
    (where [itex]\partial_x = \frac{d}{dx}[/itex]), sometimes people put factors like [itex]4\pi[/itex] on the right hand side, etc.
    This means that something special happens at zero, and the solutions on the negative and positive axes don't have to be the same. So if you have a Greens function which is a sum of exponentials, you should pick the right exponential for x < 0 and for x > 0 to make them vanish at infinity, and then properly sew them together around x = 0 (depending on the problem, you might want to make them continuous, continuously differentiable, or with a specified discontinuity).

    Then, once you have this, you can easily check that
    [tex]\psi(x) = \int G(x - y) f(y) \, \mathrm{d}y \equiv G * f[/tex]
    is a solution to the original equation (in fact, you should have done this at least once to see how this works out).

    A very efficient way to solve this is by using Fourier transforms. Have you learned about them yet?
     
    Last edited: Mar 13, 2008
  7. Mar 13, 2008 #6
    Yes I have learned about F.T, but let me go back to green's function.
    we were given one definition (inculding the one i gave in my first post), that green's function satisfy then next equation:
    [tex](L-\lambda)G(x,x')=\delta(x-x')[/tex]
    where L is some differential operator, i.e it could be d/dx or d^2/dx^2 etc.

    Now back to my question, or to be precise how to compute green's function with the two definitions i am given?
     
  8. Mar 13, 2008 #7
    p.s I am not given yet green's function I am just given the eigenfunctions which are the solutions of the homogeneous equation.
     
  9. Mar 14, 2008 #8
    Ok, you're problem, ill tell you another time, the decomposition you've made does not hold in general, it is true only for bounde operators, you dont have a set of discret eigenvalues...you can find G(p) via fourier tranform and regoularizing the pole you get.

    This operator has a continuum and residue spectrum.... you cannot find a discret spectrum.

    THe lesson to learn is that the very important thing is where is your L defined. So alaway s specify tha couple (L,DL).

    ciao
     
  10. Mar 14, 2008 #9

    HallsofIvy

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    The Green's function, G(x,t), for a linear differential operator, L(y) must
    1) Satisfy the equation L(G)= 0 for G as a function of x, for each t (except at x= t!)
    2) Satisfy the boundary conditions in x
    3) Be continuous even at x= t
    4) The derivative of G, with respect to x, must be continuous everywhere except at x= t and have [itex]lim_{x\rightarrow t^-}dG/dx- lim_{x\rightarrow t^+}dG/dx= 1[/itex].
    (#4 since the coefficient of d2y/dx2 is 1. If the coefficient were p(x), the righthand side would be p(t).)
    Any function that satisfies this particular equation and has a continuous derivative must be of the form you give: Cekt+ De-kt.

    Since dG/dx is continuous every where except at x= t, we can write G(x,y) as a piecewise function:
    G(x,t)= Aekx+ Be-kx if x< t, Cekx+ De-kx if x> t, where A, B, C, D may depend on t.

    Since we want G(x,t) to go to 0 as x goes to [itex]-\infty[/itex] (which requires that x< t) we must have B=0. Since we want G(x,t) to go to 0 as x goes to [itex]\infty[/itex](which requires that x> t) we must have C= 0.

    Since we want G(x,t) to be continuous, even at x= t, we must have Aekt= De-kt so D= Ae2kt.

    Differentiating, Gx(x,t)= kAekx if x< t and -kDe-kx= -kAe-k(x- 2t) if x> t.

    Taking the limits as x goes to t, from above and below, kAekt+ kAekt= 2kAekt= 1 so A= (1/2k)e-kt.

    Then D= Ae2kt= (1/2k)ekt.

    B= C= 0 so you can now write down the Green's function:

    G(x,t)= (1/2k)ek(x+t) if x< t, (1/2)e-k(x+t) if x[itex]\ge[/itex] t.
     
    Last edited: Mar 15, 2008
  11. Mar 14, 2008 #10
    this is why I hate theoretical methods for physicists, none of the four conditions given by halls were introduced to us, I went there blindfolded.
    At least someone can reassure me that this info given by halls is in mary boas book?
    I have it, but I thought first to try my notes from class.
     
  12. Mar 14, 2008 #11

    CompuChip

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    Since you know about Fourier transforms, here is the way I usually try to tackle these problems first (maybe it'll help you some day :smile:):
    For the homogeneous solution, you have an equation like
    [tex](\frac{d^2}{dx^2} - \lambda^2) G(x) = -\frac{1}{2\pi} \delta(x)[/tex]
    (in 2D, you would take [itex]2\pi\delta[/itex], in 3D, [itex]4\pi\delta[/itex], etc. -- this is somewhat arbitrary and usually chosen to make the results look good, otherwise they are also valid but you keep dragging those factors around).

    Perform a Fourier transform
    [tex]G(x) = \int G_k e^{i k x} \, dk[/tex]
    and
    [tex]\delta(x) = \int 1 e^{i k x} \, dk.[/tex]
    (Again, note the factors are somewhat arbitrary, I chose them to match Ivy's result, sometimes people put [itex]\sqrt{2\pi}[/itex] or [itex]2\pi[/itex] here to make the Fourier integrals look better).
    The left hand side then reads
    [tex](\partial_x^2 - \lambda^2) \int G_k e^{i k x} \, dk = \int (-k^2 - \lambda^2) G_k e^{i k x} \, dk = - \frac{1}{2\pi} \int e^{i k x} \, dk.[/tex]
    Because the exponentials are a complete (but not over-complete) basis, we can equate their coefficients, to get
    [tex](-k^2 - \lambda^2) G_k = -\frac{1}{2\pi} \quad\implies\quad G_k = \frac{1}{2\pi} \frac{1}{k^2 + m^2}.[/tex]
    Transforming back,
    [tex]G(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{i k x}}{k^2 + m^2} \, dk[/tex]
    If you now assume that x is real, you will find that the integral converges and for [itex]\lambda > 0[/itex] it is equal to
    [tex]G(x) = \frac{e^{- \lambda |x|}}{2 \lambda}[/tex]

    Of course, I have chosen the factors to exactly get this solution, otherwise, you would have found a multiple of this (for example, if you have naively taken just the delta function, you would have gotten [itex]-\pi[/itex] times this solution).
     
  13. Mar 15, 2008 #12
    what i am trying to say is that you have eigenfunctions (if you prefer) that are so only in a generalized sense, as the plane waves.... so you get a continuum spectrum.....

    you're lambdas belongs to R...

    ciao
     
  14. Mar 15, 2008 #13

    HallsofIvy

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    As CompuChip said, the "Green's function" for a linear differential operator, L(f(x)), is the function, G(x,t), thought of as a function of x with parameter t, satisfying L(G(x,t))= [itex]\delta(x-t)[/itex] (I have dropped the arbitrary constant but notice this is [itex]\delta(x-t)[/itex] not [itex]\delta(x)[/itex]. [itex]\delta(x-t)[/itex] is sometimes called the "influence function". It is like a "point" force- imagine plucking a string at a single point. And "force" along the string can be thought of as a sum (integral) of such things.) with certain boundary conditions.

    As long as [itex]x\ne t[/itex] that is just L(G)= 0. Certainly we must have
    1) L(G)= 0 for G as a function of x, for each t (except at x= t!)
    You might notice that I added the last in edit.

    Of course, the boundary conditions here are the same as for the problem itself:
    2) Satisfy the boundary conditions in x

    In order to be able to satisfy that differential equation the function certainly must
    3) Be continuous even at x= t

    And, by the definition of "delta function",
    4) The derivative of G, with respect to x, must be continuous everywhere except at x= t and have [itex]lim_{x\rightarrow t^-}dG/dx- lim_{x\rightarrow t^+}dG/dx= 1[/itex].
     
  15. Mar 15, 2008 #14

    HallsofIvy

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    It could be worse. I once taught "Calculus for Economics and Business Administration" using a textbook chosen by the Business Administration Department. On one page they gave the "4 rules of limits":
    1) If lim f= F and lim g= G, then lim(f+ g)= F+ G.
    2) If lim f= F and lim g= g, then lim(f- g)= F- G.
    3) If lim f= F and lim g= g, then lim(fg)= FG.
    4) If lim f= F and lim g= g, then lim(f/g)= F/G provided G is not 0

    On the very next page, they introduced the derivative, not bothering to mention that the limit defining the derivative always has the denominator going to 0 so those "rules of limits" don't help you!
     
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