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