# Approximation of differential equation with convolution

I can't understand one approximation I've come across in paper on mathematical biology. I have equation:

$\frac{d\theta}{dt}=\omega+P\left(\frac{t}{\tau}\right)\Delta(\theta)$,

where $\omega$ is constant, $P\left(\frac{t}{\tau}\right)$ and $\Delta(\theta)$ are periodical with period 1.

If $P(\Phi)$ and $\Delta(\Phi)$ are smooth enough we can average this equation to:

$\frac{d\theta}{dt}=\omega+H\left(\frac{t}{\tau}-\theta\right)$,

where

$H(\Phi)=\int\limits_{0}^{1}P(s)\Delta(s-\Phi)ds$.

Do you have any idea what could be justification of this approximation? This comes from section 2 in article https://www.math.uh.edu/~zpkilpat/teaching/math4309/project/jmb91_ermentrout.pdf

• Has $P$ itself period $1$ or $τ$? What does "averaging" mean? Is it to consider $\bar θ(t)=\int_0^1θ(t+s)\,ds$, is there some weight function involved? – LutzL Nov 25 '17 at 10:26
• Inserting $Φ=\frac tτ−θ(t)$ you get $H(\frac tτ−θ(t))=\int_0^1 P(s)Δ(θ(t)+s−\frac tτ)ds$ which looks like a unit mismatch, mixing an angle with a unit-less fraction. – LutzL Nov 25 '17 at 10:50
• I guess averaging means just approximation. I think $P$ has period $1$. We can just simplify it to $P(t) := P\left(\frac{t}{\tau}\right)$. $\Delta(\theta)$ is so called phase response curve and it is some kind of weight function. – Marta Kowalska Nov 25 '17 at 11:01
• Yes, everything is $1$-periodic and $τ$ is close to $1$ so that the forcing is also nearly $1$-periodic. But also my attempts at reverse-engineering the integral to get an ensemble of situations that is averaged over make no sense to me. – LutzL Nov 25 '17 at 11:16
• I guess theorem from this paper math.colostate.edu/~shipman/47/volume2a2010/Zhang.pdf can be used here. – Marta Kowalska Nov 29 '17 at 20:08