1. Not finding help here? Sign up for a free 30min 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!

Application of boundary conditions in determining the Green's function

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the Green's function $G(t,\tau)$ that satisfies
    $$\frac{\text{d}^2G(t,\tau)}{\text{d}t^2}+\alpha\frac{\text{d}G(t,\tau)}{\text{d}t}=\delta(t-\tau)$$
    under the boundary conditions $$G(0,\tau)=0~~~\text{ and }~~~\frac{\text{d}G(t,\tau)}{\text{d}t}=0\big|_{t=0}$$
    Then, solve $$\frac{\text{d}^2x(t)}{\text{d}t^2}+\alpha\frac{\text{d}x(t)}{\text{d}t}=f(t)$$ for $$f(t)=\begin{cases}0&\text{if }~t<0\\Ae^{-\beta t}&\text{if }~t\geq 0\end{cases}$$

    3. The attempt at a solution

    By solving the homogenous equation and letting it vary for ##t<\tau## and ##t>\tau## due to the discontinuity at ##t=\tau## in the first derivative of ##G(t,\tau)## I get:

    $$G(t,\tau)=\begin{cases}A(\tau)e^{-\alpha t}+B(\tau)&\text{if }~t<\tau\\C(\tau)e^{-\alpha t}+D(\tau)&\text{if }~t> \tau\end{cases}$$

    The way my teacher did it, he applied the boundary equations to the case where ##t<\tau## and deduced that ##A(\tau)=B(\tau)=0##. When I try solve this question, it only makes sense if I apply the boundary conditions to the case where ##t>\tau## since ##f(t)## is defined for ##-\infty<t<\infty## and the boundary conditions are given at ##t=0##. What I am trying to say, is I can never tell whether im meant to apply the boundary conditions to the case where ##t<\tau## or ##t>\tau##, I always feel like im guessing which one to choose. I want to develop some intuition for this part, the rest of the question I can do. I think this might have to do with the fact that we can choose the value for which ##\tau## satisfies the boundary conditions, but im still not sure. Cheers.
     
  2. jcsd
  3. Apr 18, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Since ##f(t) = 0## for ##t < 0##, the particular solution ##y_p(t)=0## for ##t < 0## certainly satisfies the DE in the region ##\{ t < 0 \}##. For ##t > 0## you can write a particular solution ##y_p(t)## as ##y_p(t) = \int_0^{\infty} G(t, \tau) f(\tau) \, d\tau## because portion of the integration over ##\tau < 0## gives 0.

    If I were doing the problem I would take ##G(t,\tau) = F(t - \tau)##, with ##F(s) = 0## for ##s < 0##.

    I think your homogeneous solutions are OK.
     
  4. Apr 18, 2016 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Oops, brain fart...I deleted my earlier post.
     
  5. Apr 18, 2016 #4
    No problem. Was questioning my sanity for a while then...

    Hi Ray. I think I understand everything you say in this post, and It's not this that is the problem. Or maybe I am misinterpreting.

    I am having trouble knowing which of the following routes to go down when applying the boundary conditions to the general: $$G(t,\tau)=\begin{cases}A(\tau)e^{-\alpha t}+B(\tau)&\text{if }~t<\tau\\C(\tau)e^{-\alpha t}+D(\tau)&\text{if }~t> \tau\end{cases}$$

    1. Apply boundary conditions to ##G(t,\tau)## for ##t<\tau##:
    $$G(0,\tau)=0\implies A(\tau)+B(\tau)=0\implies A(\tau)=-B(\tau)$$
    $$\frac{\text{d}G(t,\tau)}{\text{d}t}\big|_{t=0}=0\implies-\alpha A(\tau)=0\implies A(\tau)=0~\&~ B(\tau)=0$$

    $$\text{or}$$

    2.Apply boundary conditions to ##G(t,\tau)## for ##t>\tau##:
    $$G(0,\tau)=0\implies C(\tau)+D(\tau)=0\implies C(\tau)=-D(\tau)$$
    $$\frac{\text{d}G(t,\tau)}{\text{d}t}\big|_{t=0}=0\implies -\alpha C(\tau)=0\implies C(\tau)=0~\&~ D(\tau)=0$$

    As I understand it, going down either of these routes will have slightly different implications when we perform the integration for the particular solution: $$y_p(t) = \int_0^{\infty} G(t, \tau) f(\tau) \, d\tau=\int_0^tG_{t>\tau}(t,\tau)f(\tau)d\tau+\int_t^{\infty}G_{t<\tau}(t,\tau)f(\tau)d\tau$$


    The implications being that if I go down route (1) then ##y_p(t)## is reduced to: $$y_p(t)=\int_0^tG_{t>\tau}(t,\tau)f(\tau)d\tau$$ and if I go down route (2) then ##y_p(t)## is reduced to: $$y_p(t)=\int_t^{\infty}G_{t<\tau}(t,\tau)f(\tau)d\tau$$.
     
  6. Apr 18, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you choose the form ##G(t,\tau) = F(t - \tau)## you would have ##y_p(t) = \int_0^{\infty} F(t - \tau) f(\tau) \, d\tau##. For fixed ##t##, the argument ##t - \tau \to -\infty## as ##\tau \to +\infty##, so for some functions ##f(\tau)## at least, we would encounter a divergent integral if ##\alpha >0##. (For example, rather than using your ##f(\tau) = A e^{-\beta \tau}##, if we were to use ##f = ## trigonometric function or a polynomial function, then we would encounter divergence.) Why? Well, in this case the two roots of the characteristic equation are ##r = \alpha>0## and ##r = 0##, so the general form of ##F(s)## for ##s < 0## would either be identically 0 or something of the form ##a + b e^{-\alpha s}##. If ##\alpha > 0## the exponential term ##e^{-\alpha s} \to \infty## when ##s \to -\infty##, so we would need ##b = 0##. Even if we have only the form ##F(s) = a \neq 0 ## for ##s < 0## we could still (for SOME reasonable ##f(\tau)##) be encountering a divergent integral. The only way to be able to handle a good variety of functions ##f(\tau)## is to have ##a = 0## also; that is, we should have ##F(s) = 0## for ##s < 0##.

    Note that this is not arbitrary: besides boundary conditions at ##\tau = t## we also need boundary conditions at ##\mathbf{\pm \infty}##. In particular, we want ##G \to 0## reasonable quickly as its arguments go to ##+\infty## or ##-\infty##.

    So, in this problem, we are more-or-less forced to use ##G(t,\tau) = F(t-\tau)## with ##F(s) = 0## for ##s < 0##. That will give us the particular solution
    [tex] y_p(t) = \int_0^t F(t - \tau) f(\tau) \, d \tau .[/tex]
    The other form, ##\int_t^{\infty} F(t - \tau) f(\tau) \, d \tau## is a non-starter in this particular problem for most interesting functions ##\mathbf{f(\tau)}##.

    Note that for your particular ##f(\tau) = A e^{-\beta \tau}, \: \beta > 0## you could take ##F(s) = a = \text{const.} ## for ##s < 0##, because the integral ##\int_t^{\infty} e^{-\beta \tau} a \,d \tau## would converge. You could also include a term ##b e^{-\alpha s}## in ##F(s)## for ##s < 0##, provided that ##\beta > \alpha##. However, this seems pointless; why not avoid possible problems altogether, by just putting ##F(s) = 0## for ##s < 0##? You could apply the formula
    [tex] y_p(t) = \begin{cases} 0 & \text{if} \; t < 0 \\
    \int_0^t F(t - \tau) f(\tau) \, d \tau& \text{if} \; t > 0
    \end{cases}
    [/tex]
    for any right-hand-side function of the form
    [tex] \begin{cases} 0 & \text{if} \; t < 0 \\
    f(t) & \text{if} \; t > 0
    \end{cases}
    [/tex]
     
    Last edited: Apr 18, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Application of boundary conditions in determining the Green's function
Loading...