Find the limiting value of a Delayed (or Ordinary) Differential Equation

Question

General Question

Suppose we have a function $f(s)$ which satisfies an Ordinary Differential Equation (ODE) $f'(s) = A(s,f(s))$ which can not be solved explicitly. Is there some method to find $\lim_{s\rightarrow \infty} f(s)$? In particular I am interested in this question for Delayed Differential Equations (DDE). What I am looking for is some reference where such a method is outlined (I am also interested in this question with regard to Integro-Differential Equations).

Example Problem

Let $f(s)$ satisfy the DDE: \begin{align*} f'(s)&=a(f(s)^2-1) & s \in [0,1)\\ f'(s)&=a(f(s)^2-f(s-1)^2) & s \geq 1, \end{align*} $f(0) = a$. I know from numerical studies that in fact for $a \in (0,1)$ $\lim_{s\rightarrow\infty}f(s)=0$ for this case. However I do not know of any mathematical method to come to this result.

Answer 1

As I noted in my earlier comment, a solution of the DDE $$ f'(s) = a(f(s)^2 - f(s-1)^2) $$ need not converge to zero. However, one can ask if a solution converges to a constant. And indeed the answer is yes, at least for bounded solutions. Namely, in their 1984 paper Solutions of x′(t)=f(x(t),x(t−L)) have limits when f is an order relation, J. Kaplan, F. Sorg and J. Yorke proved that if a continuous function $g(x, y)$ that is locally Lipschitz wrt $x$ satisfies the conditions:

1. $g(x, y) > 0$ if and only if $g(y, x) < 0$,
2. $g(x, y) \ge 0$ and $g(y, z) \ge 0$ implies $g(x , z) \ge 0$,

then each bounded solution converges to a constant as $t \to \infty$. Now, the function $g(x, y) = a (x^2 - y^2)$ satisfies the above conditions.