Limit of $y(t) = \frac{1}{1 - e^{- \gamma T}} \int_0^T e^{\gamma (u - T)} f(x(u + t)) du$

Question

Consider the function

\begin{align} y(t) = \frac{1}{1 - e^{- \gamma T}} \int_0^T e^{\gamma (u - T)} f(x(u + t, T)) du \end{align}

with $f: \Bbb R \to [0, 1], f(x) = x^n/(\kappa^n + x^n)$, $x: \Bbb R \to [0, 1], x(t, T) = (1 + sin(2 \pi t / T))/2$, $n > 0$, $\kappa > 0$, $\gamma > 0$ and $T > 0$.

From numerical results we know that

$\lim\limits_{T \rightarrow \infty}{y(t)} = f(x(t + t_0))$

with $t_0 \in \Bbb R$.

My question:

Does anyone know how to prove this? I think it may be easy to show from the above expression, but don't have an idea so far.

Answer 1

Some thoughts on the problem. Not an answer.

Let's consider a more general question first:

$$F(t) = \lim_{T \to \infty} \frac{1}{e^{\gamma T} - 1} \int_0^T e^{\gamma u} f \left( \frac{u + t}{T} \right) du$$

Where we have $ \gamma>0$ and $0 < f(x) \leq 1$ for $ x \in \mathbb{R}$ as is the OP's case.

Then clearly, it follows from the conditions on $f$ that:

$$0< F(t) \leq \frac{1}{\gamma}$$

For the general case, let's try transforming the integral:

$$\int_0^T e^{\gamma u} f \left( \frac{u + t}{T} \right) du=T \int_0^1 e^{\gamma T v} f \left(v+ \frac{t}{T} \right) dv$$

Then we can write:

$$F(t) = \lim_{T \to \infty} \frac{T}{e^{\gamma T} - 1} \int_0^1 e^{\gamma T v} f \left(v+ \frac{t}{T} \right) dv$$

Or:

$$F(t) = \frac{1}{\gamma} \lim_{T \to \infty} \frac{\int_0^1 e^{\gamma T v} f \left(v+ \frac{t}{T} \right) dv}{\int_0^1 e^{\gamma T v} dv} $$

Let's expand our function as a Taylor series around $t/T=0$:

$$f \left(v+ \frac{t}{T} \right)=f(v)+f'(v) \frac{t}{T}+\frac{f''(v)}{2} \frac{t^2}{T^2}+ O \left(\frac{t^3}{T^3} \right)$$

For nice functions, which don't have derivatives blowing up, we can assume that all the terms except the first one should vanish in the limit, which means that we are looking for:

$$F[f] = \frac{1}{\gamma} \lim_{T \to \infty} \frac{\int_0^1 e^{\gamma T v} f (v) dv}{\int_0^1 e^{\gamma T v} dv} $$

It is just a functional, a number that doesn't depend on $t$. That's what I assume should happen in the OP's case, which contradicts their own conclusions.

I'm pretty sure that we have:

$$F[f] = \frac{f(0)}{\gamma} $$

Numerical experiments with some trial functions support this.

Using the notation from the OP, I think we have:

$$f(x(0))= \frac{1}{2^n \kappa^n+1 }$$