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

## 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 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.

Ray Vickson
Homework Helper
Dearly Missed

## 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 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.

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.

vela
Staff Emeritus
Homework Helper
I think your homogeneous solutions are OK.
Oops, brain fart...I deleted my earlier post.

Oops, brain fart...I deleted my earlier post.

No problem. Was questioning my sanity for a while then...

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: $$y_p(t)=\int_t^{\infty}G_{t<\tau}(t,\tau)f(\tau)d\tau$$.

Ray Vickson
Homework Helper
Dearly Missed
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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