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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?

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  • $\begingroup$ When you plug in what $f(t')$ is, shouldn't it be $\delta(t'-n\tau)?$ $\endgroup$ Aug 22, 2018 at 15:50
  • $\begingroup$ @AdrianKeister Sure should! I'll correct it thanks! $\endgroup$ Aug 22, 2018 at 15:51
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    $\begingroup$ Several typos: first, the prefactor of $x(0)$ in $x(t)$ is $e^{-at}$, not $e^{at}$; second, you forgot to replace $t$ by $k\tau$ in the second terms of $x_k$, thus missing the simplifications that follow. I believe that if you modify your try with respect to these two points, everything should run smoothly afterwards... $\endgroup$
    – Did
    Aug 22, 2018 at 15:57
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    $\begingroup$ I think your (2) is suspect. There has to be an additive relationship between $x_k$ and $x_{k-1},$ because the $x_{k-1}$ exponential sum doesn't go as far as the $x_k$ one does.I would write $$x_k=e^{\alpha\tau}e^{\alpha(k-1)\tau}x(0)+a\sum_{n=0}^{k-1}e^{-\alpha(t-n\tau)}+ae^{-\alpha(t-k\tau)}.$$ $\endgroup$ Aug 22, 2018 at 15:58
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    $\begingroup$ "Is that not the case?" No. As a way to memorize this, simply note that $y'=by$ is solved by $y(t)=y(0)e^{bt}$, for every $b$, thus, in the absence of $f$, your equation would read $x'=-\alpha x$ and be solved by $x(t)=x(0)e^{-\alpha t}$. $\endgroup$
    – Did
    Aug 22, 2018 at 16:02

1 Answer 1

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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.

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  • $\begingroup$ Great! Simple and ingenious. I was very close to it! Thank you very much for the help, I've greatly appreciated $\endgroup$ Aug 22, 2018 at 16:37
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    $\begingroup$ @DavideMorgante: Yes, I'd agree: you certainly did the heavy lifting here. $\endgroup$ Aug 22, 2018 at 16:37

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