Given is an oscillating function $y(t)$ of period $T$ with $y(t) \, \ge \, 0 \, \forall \, t \in \Bbb R$.
Consider the differential equation
$\frac{\partial g}{\partial t}(t) = a f(y(t)) - \gamma g(t)$.
Hereby, $a > 0$ is a constant and $f(y)$ is a continuous, strictly monotonously increasing function with $f(0) \ge 0$.
My question:
The differential equation has oscillating solutions $g(t)$ of the same period $T$ as the input function $y(t)$. Numerical analysis shows that the oscillating solutions $g(t)$ have a phase shift relative to the oscillations of $y(t)$. I want to determine this phase shift.
What I have so far:
The solutions of the differential equation can be written as
$g(t) = C_1 e^{-\gamma t} + e^{-\gamma t} \int_1^t e^{t'} f(y(t')) dt'$.
We choose $C_1 = 0$ in the following.
From here, I'm not sure how to continue. A possible ansatz could be to determine the minima of $g(t)$ and compare them to the minima of $y(t)$.
A necessary condition would be
\begin{align} \frac{\partial g}{\partial t} = 0 \\ \Leftrightarrow a f(y(t)) - \gamma g(t) = 0 \\ \Leftrightarrow a f(y(t)) = \gamma e^{-\gamma t} \int_1^t e^{t'} f(y(t')) dt'. \end{align}
There is a minimum at $t$ if it holds
\begin{align} \frac{\partial^2 g}{\partial t^2} > 0 \\ \Leftrightarrow a \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} (t) - \gamma \frac{\partial g}{\partial t}(t) > 0 \\ \Leftrightarrow a \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} (t) > 0 \end{align}
Does someone have another idea to approach this question?
You want first a periodic solution. Let $T$ be the period of $y$, then you want $$ g(t)=g(t+T)=g(t)e^{-γT}+\int_0^Taf(y(t+s))e^{-γ(T-s)}ds\implies g(t)=\frac{\int_0^Taf(y(t+s))e^{-γ(T-s)}ds}{1-e^{-γT}}. $$ The weight factor $e^{-γ(T-s)}$ is largest for $s$ close to $T$, that is, values of $f\circ y$ shortly before $t$ will dominate, which can be interpreted as a small, unspecified phase shift of about $O(γ^{-1})$. I do not see whether one can say any more from the exact solution.
Using Taylor arguments, for $γ$ large enough $γg(t)+g'(t)=γg(t+γ^{-1})+O(γ^{-1})$, so that in that approximation $$g(t)=\frac{a}{γ}f(y(t-γ^{-1}))+O(γ^{-2}).$$ However, the difference from the phase shift is now as large as the uncertainty from the Taylor remainder term, thus still rather unspecific.