# 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.

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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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