# Application of boundary conditions in determining the Green's function

• pondzo
G_{t<\tau}(t,\tau)f(\tau)d\tau$$It's the decision of which route to go down that I don't know how to make.In summary, when solving the given problem, there are two possible routes to take when applying the boundary conditions to the general solution for the Green's function. These routes have slightly different implications when performing the integration for the particular solution. The correct route to take may depend on the specific problem being solved, and intuition is needed to determine which route is appropriate. #### pondzo ## Homework Statement Find the Green's function G(t,\tau) that satisfies$$\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}$$## The Attempt at a Solution By solving the homogenous equation and letting it vary for ##t<\tau## and ##t>\tau## due to the discontinuity at ##t=\tau## in the first derivative of ##G(t,\tau)## I get:$$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}$$The way my teacher did it, he applied the boundary equations to the case where ##t<\tau## and deduced that ##A(\tau)=B(\tau)=0##. When I try solve this question, it only makes sense if I apply the boundary conditions to the case where ##t>\tau## since ##f(t)## is defined for ##-\infty<t<\infty## and the boundary conditions are given at ##t=0##. What I am trying to say, is I can never tell whether I am meant to apply the boundary conditions to the case where ##t<\tau## or ##t>\tau##, I always feel like I am guessing which one to choose. I want to develop some intuition for this part, the rest of the question I can do. I think this might have to do with the fact that we can choose the value for which ##\tau## satisfies the boundary conditions, but I am still not sure. Cheers. pondzo said: ## Homework Statement Find the Green's function G(t,\tau) that satisfies$$\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}$$## The Attempt at a Solution By solving the homogenous equation and letting it vary for ##t<\tau## and ##t>\tau## due to the discontinuity at ##t=\tau## in the first derivative of ##G(t,\tau)## I get:$$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 I am meant to apply the boundary conditions to the case where ##t<\tau## or ##t>\tau##, I always feel like I am 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 I am still not sure. Cheers. 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. Ray Vickson said: I think your homogeneous solutions are OK. Oops, brain fart...I deleted my earlier post. vela said: Oops, brain fart...I deleted my earlier post. No problem. Was questioning my sanity for a while then... Ray Vickson said: 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. 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$$. pondzo said: No problem. Was questioning my sanity for a while then... Hi Ray. I think I understand everything you say in this post, and It's not this that is the problem. Or maybe I am misinterpreting. I am having trouble knowing which of the following routes to go down when applying the boundary conditions to the general:$$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.

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
$$y_p(t) = \int_0^t F(t - \tau) f(\tau) \, d \tau .$$
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
$$y_p(t) = \begin{cases} 0 & \text{if} \; t < 0 \\ \int_0^t F(t - \tau) f(\tau) \, d \tau& \text{if} \; t > 0 \end{cases}$$
for any right-hand-side function of the form
$$\begin{cases} 0 & \text{if} \; t < 0 \\ f(t) & \text{if} \; t > 0 \end{cases}$$

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