0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.