3
$\begingroup$

Short Description of the General Question

Suppose we have some Delayed Differential Equation (DDE) which depends on a parameter $a$, $x_a'(t)=f(a,x_a(t),x_a(t-s))$ for some fixed $a$ and $s$. I would like to prove that $x_a(t)$ is increasing in $a$. I.e. if we have $a<b$ and $x_a'(t) = f(a,x_a(t),x_a(t-s))$ and $x_b'(t)=f(b,x_b(t),x_b(t-s))$ then $x_a(t) \leq x_b(t)$ for all $t$.

Specific Scenario

Consider the following Delayed Differential Equation: \begin{align*} x_a(0) &= a\\ x_a'(t) &= - a (1 - x_a(t)^2) & t \leq 1\\ x_a'(t) &= -a(x_a(t-1)^2 - x_a(t)^2) & t > 1. \end{align*} I have found numerically that for all $a \in (0,1)$ we have: $$ a \leq b \Rightarrow x_a(t) \leq x_b(t), $$ but I am unable to prove this statement. I can solve the ODE in $[0,1]$ exactly, thus on this interval it is easily verified that $x_a(t) \leq x_b(t)$. I then use that solution to find a solution on $[1,2]$ and so on. But as $t$ grows large we can't find an exact solution anymore.

$\endgroup$
  • $\begingroup$ In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set. $\endgroup$ – William Elliot Dec 31 '18 at 14:12
  • $\begingroup$ Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example) $\endgroup$ – HolyMonk Jan 2 at 0:16

Your Answer

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

Browse other questions tagged or ask your own question.