Can someone help verify the first part of my solution for this DDE?

Question

I'm trying to solve this delayed differential equation that I saw in a paper I'm studying $$y′(t)=ay(t)−ay(t−\tau)$$ with $a$ as a positive constant, $y(t−\tau)=0$ for $t<\tau$ and the initial condition as $y(0)=y_0$. (I posted about this a couple of days ago and was making a terrible mistake so I deleted that one. I'm doing it again now, but I compare my solution to the one that someone else got and it's slightly different, even though I think my working is correct mostly? I can't see their working so I'm not sure where/if I'm going wrong, so I just wanted to confirm if I'm making a mistake or not.)

I'll outline my working below:

On $[0,\tau]$ we have $$y′(t)−ay(t)=0$$ which gives us $$y(t)=y_0e^{at}$$ if we use an integrating factor of $e^{−at}$.

For $[\tau,2\tau]$: $$y′(t)−ay(t)=−ay_0e^{a(t-\tau)}$$

which gives me $$y(t) = y_0e^{at}[1-ae^{-a\tau}(t-\tau)]$$

Then, for $t \in [2\tau, 3\tau]$, I can just continue using the method of steps, but I get a super long expression that does not,,, particularly seem like it would be very generalizable, so I'm not sure if I can get a closed form solution or not (also why I think I might be making a mistake).

(Edit: Adding it here anyway)

For $t \in [2\tau, 3\tau]$:

$$y(t) = y_0 e^{at}\bigg[1-at+2a\tau + ae^{-a\tau}\bigg(a\bigg(\frac{t^2}{2} - 2\tau\bigg) + \tau - 2\tau^2\bigg) \bigg]$$

Could someone please confirm if I got the first bits that I've written up there correct though? I'd appreciate it a lot, as I do tend to make many many basic errors sometimes.

EDIT 2:

Got the answer by continuing to use the method of steps and changing how I wrote my solution for $t \in [2\tau, 3\tau]$ to $$y(t) = y_0\bigg[e^{at}-a(t-\tau)e^{a(t-\tau)} + \frac{a^2}{2}(t-2\tau)^2e^{a(t-2\tau)}\bigg]$$ instead of what it was before, which gave me a much nicer way to look at it.

Then I generalized this to pretty much the exact thing user DinosaurEgg wrote below except I realized it after looking at the answers fully once I was done lol. But yeah, so I get the sum

$$y(t) = y_0 \sum_{k=0}^{n} \frac{(-a)^k}{k!} (t-k\tau)^ke^{a(t-k\tau)}$$ which is again, the exact same as the answer below.

Thanks for all the help though! I do appreciate it! :)

Answer 1

Your formulas are correct, I think. But the argument does generalize.

Suppose $p_k(t)$ is a polynomial in $t$ such that $$y(t)=y_0p_k(t-k\tau)e^{a(t-k\tau)}$$ on $t\in[k\tau,(k+1)\tau]$.

Then \begin{align*} y_0p_k'(t-k\tau)e^{a(t-k\tau)}&=y_0p_k'(t-k\tau)e^{a(t-k\tau)}+y_0ap_k(t-k\tau)e^{a(t-k\tau)}-y_0ap_k(t-k\tau)e^{a(t-k\tau)} \\ &=y'(t)-ay(t) \\ &=-ay(t-\tau) \\ &=-y_0ap_{k-1}((t-\tau)-(k-1)\tau)e^{a(t-\tau-(k-1)\tau)} \\ &=-y_0ap_{k-1}(t-k\tau)e^{a(t-k\tau)} \end{align*} Canceling, we find that $p_k'=-ap_{k-1}$.

Moreover, to ensure continuity, we must have $$p_{k-1}(\tau)e^{a\tau}=y(k\tau)=p_k(0)$$ Thus $$p_k(t)=e^{a\tau}p_{k-1}(\tau)-a\int_0^t{p_{k-1}(s)\,ds}$$

Putting it all together, we obtain a simple recurrence for the $\{p_k\}_k$: \begin{align*} p_k(t)&=e^{a\tau}p_{k-1}(\tau)-a\int_0^t{p_{k-1}(s)\,ds} \\ p_0(t)&=1 \end{align*}

In particular, the recurrence isn't too bad, when expressed in coefficients: if $p_k(t)=\sum_j{c_{k,j}t^j}$, then $$c_{k,j}=\begin{cases} -\frac{a}{j}c_{k-1,j-1} & j>0 \\ e^{a\tau}\sum_l{c_{k-1,l}\tau^l} & j=0 \\ \end{cases}$$

Answer 2

I think what you are trying to get at can be expressed very concisely by applying a Laplace transform to the DDE above to obtain

$$Y(s)=\frac{y_0}{s-a+ae^{-s\tau}}$$

where we denote $Y(s)=\int_0^\infty y(t)e^{-st}dt$. To obtain the expansion you have above just expand the denominator around $\tau=\infty$ which can be achieved by expanding in powers of $e^{-s\tau}$. This yields

$$Y(s)=\frac{y_0}{s-a}\sum_{n=0}^{\infty}\left(-\frac{a}{s-a}\right)^ne^{-ns\tau}$$

Formally inverting the Laplace transform term by term now yields the desired result

$$y(t)=y_0\sum_{n=0}^{\infty}\frac{(-a)^n}{n!}\theta(t-n\tau)(t-n\tau)^ne^{a(t-n\tau)}$$

with $\theta(x)$ the Heaviside step function. Unpacking the formula we derived a little further, we see the solution is different for every interval $(m\tau,(m+1)\tau)~,~ m\in\mathbb{N}$ and is given by

$$y_m(t)=\sum_{n=0}^m\frac{(-a)^n}{n!}(t-n\tau)^ne^{a(t-n\tau)}~~,~~t\in \left(m\tau, (m+1)\tau\right)$$

and this is probably the closest one can get to a closed form expression. Note that continuity is automatically satisfied here, because

$$y_{m+1}(t)-y_m(t)\Big|_{t=(m+1)\tau}=\frac{(-a)^{m+1}}{(m+1)!}(t-(m+1)\tau)^{m+1}e^{a(t-(m+1)\tau)}\Big|_{t=(m+1)\tau}=0$$

With this you can easily verify that your solutions in $(0,2\tau)$ are correct.