# 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 $$a and $$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.

• 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. – William Elliot Dec 31 '18 at 14:12
• 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) – HolyMonk Jan 2 at 0:16