I got stuck in my calculations trying to solve the following problem:
Given the ODE $$\dot{x} = -\alpha x + a\sum_{n=0}^\infty\delta(t-n\tau)$$ where $\alpha \gt 0$, define $$x_k = x(k\tau +0 )$$ and find $x_k$ as a function of $x_{k-1}$. What is the value of the following limit $$\lim_{k\rightarrow \infty}x_k?$$
My attempt
First of all we define the following function
$$f(t) = a\sum_{n=0}^\infty\delta(t-n\tau)$$ then the ODE becomes pretty much standard $$\dot{x}(t) = -\alpha x(t) + f(t)$$
for which the general solution is
$$x(t) = e^{-\alpha t} x(0)+\int_0^t e^{-\alpha(t-t')}f(t')\,dt'\\ x(t) = e^{-\alpha t} x(0)+\int_0^t e^{-\alpha(t-t')}a\sum_{n=0}^\infty\delta(t'-n\tau)\,dt'.$$
Now we interchange the sum with the integral (being a physicist I impudently change them without checking uniform convergence!) and get
$$x(t)= e^{-\alpha t} x(0)+\sum_{n=0}^\infty a\int_0^t e^{-\alpha(t-t')}\delta(t'-n\tau)\,dt'$$
By the definition of $x_k$
$$\begin{align}x_k = x(k\tau+0) &= e^{-\alpha k\tau} x(0)+\sum_{n=0}^\infty a\int_0^{k\tau} e^{-\alpha(k\tau-t')}\delta(t'-n\tau)\,dt'\\&=\underbrace{e^{-\alpha k\tau}x(0)}_{\text{first term}}+\underbrace{\sum_{n=0}^kae^{-\alpha(k\tau-n\tau)}}_{\text{second term}}\end{align}$$
We can easily calculate $x_{k-1}$ from the value of $x_k$
$$ x_{k-1} = \underbrace{e^{-\alpha(k-1)\tau}x(0)}_{\text{first term}}+\underbrace{a\sum_{n=0}^{k-1}e^{-\alpha(k\tau-n\tau)}}_{\text{second term}}.$$
It is clear that is we want to make a relation between $x_k$ and $x_{k-1}$, the first terms of both can be written as $$e^{-\alpha k\tau}x(0) = e^{-\alpha\tau}e^{-\alpha(k-1)\tau}x(0)$$ so I'm bound to say that, at list for the first terms $$x_k = e^{-\alpha\tau}x_{k-1}\tag2$$ The real problem arises when we try to make adjustments to $(2)$ to make the second terms equal, mainly from definition $(2)$ we get
$$x_k = e^{-\alpha\tau}e^{-\alpha(k-1)\tau}x(0) + \color{red}{e^{-\alpha\tau}a\sum_{n=0}^{k-1}e^{-\alpha(k-n)\tau}}$$
My questions now are
Question 1: How can I adjust that second term to get $x_k$ as a function of $x_{k-1}$?
Question 2: Is there an easier way to solve this problem?
So let's start from $$x_k(\tau) = x(k\tau)= e^{-\alpha k\tau}x(0)+a\sum_{n=0}^ke^{-\alpha\tau(k-n)}.$$ Then $$x_{k-1}(\tau)=e^{-\alpha (k-1)\tau}x(0)+a\sum_{n=0}^{k-1}e^{-\alpha\tau((k-1)-n)}.$$ In comparing the two, we want to get $x_k$ to look like $x_{k-1}.$ So we write \begin{align*}x_k(\tau) &= e^{-\alpha\tau}e^{-\alpha (k-1)\tau}x(0)+a\sum_{n=0}^{k-1}e^{-\alpha\tau(k-n)}+a \\ &=e^{-\alpha\tau}e^{-\alpha (k-1)\tau}x(0)+ae^{-\alpha\tau}\sum_{n=0}^{k-1}e^{-\alpha\tau((k-1)-n)}+a \\ &=e^{-\alpha\tau}x_{k-1}(\tau)+a.\end{align*}
To compute the limit, I would figure exactly what $x_{0}(\tau)$ is (looks like $x_0(\tau)=x(0)+a$), then write $x_{k}(\tau)$ in terms of that. That's something you could take the limit of.