How to solve this DDE: \begin{align} \frac{1}{N_t} \frac{dN_t}{dt}=r\left(1-\frac{N_{t-\tau}}{K}\right) \end{align} where $N_0,K,r,\tau$ are constant?

This differential equation is based on logistic model, which adds a delay $\tau$ to its equation, namely delayed logistic model.

  • $\begingroup$ This equation has infinitely many solutions (depending n an arbitrary function). You should specify further conditions to single out one of them. Specifying $N_0$ is not enough. $\endgroup$ Jan 31, 2022 at 13:51

1 Answer 1


To simplify, denote $y(t)=\log N_t$. Then we have a simple differential-difference equation $$y'(t)=f(y(t-\tau)),\quad f(y)=r(1-y/K).$$ Let $\phi$ be an arbitrary continuous function on $[0,\tau]$ which has one sided derivative at $\tau$ satisfying $\phi'(\tau-)=\phi(0)$. Then the formulas $Y(t)=\phi(t), 0\leq t\leq \tau$, and $$y(t)=y(n\tau)+\int_{n\tau}^t f(y(t-\tau))dt, \quad n\tau\leq t\leq (n+1)\tau,\quad n\geq 1$$ defines a continuous solution which is everywhere continuous and continuously differentiable for $t\geq \tau$. Then $N_t=\exp(y(t))$ defines a solution of your problem.

  • $\begingroup$ You mean you define $y(t)=\phi(t)$ on $[0,\tau]$? $\endgroup$ Jan 31, 2022 at 17:57
  • $\begingroup$ @მამუკა ჯიბლაძე: Yes, sure. $\endgroup$ Jan 31, 2022 at 22:28
  • $\begingroup$ Then you need $\phi$ to be everywhere differentiable too, not just one-sidedly at $\tau$, right? $\endgroup$ Feb 1, 2022 at 5:15
  • $\begingroup$ @მამუკა ჯიბლაძე: On $[0,\tau]$ it is as good as $\phi$, but for $t>\tau$ its smoothness improves: if $\phi$ is only continuous, $f$ is continuously differentiable for $t>\tau$, twice continuously differentiable for $t>2\tau$ and so on. $\endgroup$ Feb 1, 2022 at 12:25
  • $\begingroup$ Oh I see, thanks $\endgroup$ Feb 1, 2022 at 13:38

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