# Asymptotics for $y'(t)=y(t-1)$

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?

• 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. – Lutz Lehmann Jan 23 at 14:49
• @LutzLehmann my mistake! It should be $(\lfloor t \rfloor +1)!$ indeed – Jacques Mardot Jan 23 at 14:59
• One solution is $y\propto e^{ct}$ with $ce^c=1$, i.e. $c=W(1)$. So you might want to study $z(t):=y(t)e^{-W(1)t}$. – J.G. Jan 23 at 16:18

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 $$t\to\infty$$.

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.