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Let $y : \mathbb{R}\to\mathbb{R}$ be a differentiable function satisfying $y'(t)=y(t-1)$ for all $t$.

Is it possible to give an asymptotics for $y(t)$ as $t\to \infty$?

It is clear to me that there should be an asymptotics of the form $y(t) \sim C\frac{t^{\lfloor t \rfloor +1}}{(\lfloor t\rfloor +1)! }$, inducting on intervals of lenghts 1. How is it possible to prove it?

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  • $\begingroup$ Should it not be $([t+1]+1)!$ in the denominator? Then using Stirling cancels the numerator so that $\sim C\sqrt{t}e^{t}$ remains. $\endgroup$ – Lutz Lehmann Jan 23 at 14:49
  • $\begingroup$ @LutzLehmann my mistake! It should be $(\lfloor t \rfloor +1)!$ indeed $\endgroup$ – Jacques Mardot Jan 23 at 14:59
  • $\begingroup$ One solution is $y\propto e^{ct}$ with $ce^c=1$, i.e. $c=W(1)$. So you might want to study $z(t):=y(t)e^{-W(1)t}$. $\endgroup$ – J.G. Jan 23 at 16:18
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This is a delayed DE. Let $y(t)=ce^{\lambda t}$ and then the characteristic equation is $$ \lambda=e^{-\lambda} $$ which has a unique solution $\lambda>0$. Therefore $y(t)\to\infty$ as $t\to\infty$.

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With the help of the Laplace transform

$$ Y(s) = \frac{\int_{-1}^0 e^{-s t} y(t) \, dt+e^s y(0)}{e^s s-1} $$

now considering $y(t)=0,\ \ -1\le t\le 0$ and $y(0)=1$

$$ Y(s) = \frac{1}{s-e^{-s}} $$

now taking the denominator $d(s) = s-e^{-s}$ and making $s=x + iy$ we have

$$ d(s) =x - e^{-x} \cos (y)+i \left(e^{-x} \sin (y)+y\right) $$

so the denominator has infinite zeros but we pick one of them which is located at $x = W(1), y = 0$ where $W(\cdot)$ indicates the Lambert function. Here $W(1) = 0.567143$ so $y(t)$ is unstable because $Y(s)$ has at least one pole in the right complex plane hence it grows exponentially.

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