# How to solve this delay differential equation?

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.

• 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. Jan 31, 2022 at 13:51

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.
• You mean you define $y(t)=\phi(t)$ on $[0,\tau]$? Jan 31, 2022 at 17:57
• Then you need $\phi$ to be everywhere differentiable too, not just one-sidedly at $\tau$, right? Feb 1, 2022 at 5:15
• @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. Feb 1, 2022 at 12:25