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Green Function

  1. Feb 12, 2013 #1
    Solve...
    [tex]
    \frac{d^2}{dt^2}G(t,t') + \omega^2G(t,t') = \delta(t-t')
    [/tex]
    Solve (above) if G=0 and [itex]\frac{dG}{dt}=0[/itex] at t=0 to obtain:
    [tex]
    G(t,t')=\begin{cases}
    0 & 0<t<t' \\
    \frac{1}{\omega}\sin\omega(t-t') & 0<t'<t

    \end{cases}
    [/tex]


    I'm supposed to use Laplace Transforms to figure this out. (I'm going out of Boas Chapter 8, Section 12 problem 1) To be honest, I'm having a really difficult time getting my head around Green Functions so this is really pushing on me.

    The rest of the problem states, "Use laplace transforms to find the inverse transform".

    I don't even know where to begin.
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    Mute

    User Avatar
    Homework Helper

    Well, you know what a Laplace transform is, I'm assuming. Do you know the rules for evaluating the Laplace transforms of derivatives,

    $$\mathcal L\left[\frac{d^nf}{dt^n}\right],$$

    in terms of ##\mathcal L[f]##, for any function f(t)?

    What happens if you apply the Laplace transform to your differential equation? i.e., what is

    $$\mathcal L\left[ \frac{d^2}{dt^2}G(t,t') + \omega^2 G(t,t')\right]$$
    in terms of the Laplace transform of ##G(t,t')##? (transforming with respect to t, not t').

    Once you've found that, you know it must also be equal to the Laplace transform of the delta function, so you will then need to solve for ##\mathcal L[G]## and inverse transform to get G(t,t').
     
  4. Feb 12, 2013 #3
    Hmmm, I'm rereading the Laplace section carefully. I'll reaffirm that and then I'll try your hints. I'll come back with potential/probable issues. Thank you.
     
  5. Feb 13, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Personally, I have never liked "Laplace Transform methods"- they hide too much in the machinery. Here's how I would approach this problem:

    Write [itex]G_1(t, t')[/itex] for G(t, t') for 0<t< t' and [itex]G_2(t, t')[/itex] for G(t, t') for t'< t< 1.

    For 0< t< t', [itex]G_1(t, t')[/itex] must satisfy [itex]d^2/dt^2G_1+ \omega^2G_1= 0[/itex] so [itex]G_1(t, t')= A cos(\omega t)+ B sin(\omega t)[/itex]. Since the initial conditions are G(0, t')= 0, dG(0, t')/dt= 0, A= B= 0 so [itex]G_1(t, t')= 0 for all 0< t< t'.

    For t'< t< 1, [itex]G_2(t, t')[/itex] must satisfy the same differential equation so that [itex]G_2(t, t')= A cos(\omega t)+ B sin(\omega t)[/itex]. To find the "initial conditions" we use the facts that the Green's function is continuous at t= t' and has a unit jump discontinuity in the first derivative at t= t'.

    So [itex]G_2(t', t')= Acos(\omega t')+ B sin(\omega t')= 0[/itex] and [itex]dG(t',t')/dt= -\omega A sin(\omega t')+ \omega B cos(\omega t')= 0[/itex]. That gives two equations to solve for A and B in terms of t'.
     
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