Model-free ultra-local function approximation I've been reading a lot about model-free control and I came across the concept of the ultra-local model. There is a really intricate approach outlined here but I'm having an issue with one part outlined below.
Image of the equations I have the question about
I'm not sure how the authors went from $y+s\frac{dy}{ds} = -\frac{\phi}{s^2} + \alpha\frac{du}{ds}$ to the next equation for $\phi$. It is mentioned in the article that it has to do with $\int_{\tau_{i}}^{\tau_{f}} F(\tau)d\tau$ and said in a conference that it is the averaging integral, but I don't see how. The equations are outlined in section 3.4.1 in the article.
 A: I am considering the notation of the paper enter link description here which you mention in one the comments.
First, note that we have the following equality
$$\int_0^T\int_0^sz(\theta)d\theta ds=\int_0^T(T-s)z(s)ds,$$
which can be proven by integrating by parts.
The starting point is the following expression in the frequency domain
$$-\dfrac{\phi}{s^4}=\dfrac{1}{s^2}Y(s)+\dfrac{1}{s}\dfrac{d}{ds}Y(s)-\dfrac{\alpha}{s^2}\dfrac{d}{ds}U(s).$$
Let us consider first the terms in $Y(s)$ and convert them to the time-domain using the inverse Laplace transform. We have that
$$\dfrac{1}{s^2}Y(s)\rightarrow \int_0^T\int_0^\tau y(\theta)d\theta d\tau=\int_0^T(T-\tau)y(\tau)d\tau$$
and
$$\dfrac{1}{s}\dfrac{d}{ds}Y(s)\rightarrow -\int_0^T\tau y(\tau)d\tau.$$
If we sum the two, get the term
$$\int_0^T(T-2\tau)y(\tau)d\tau.$$
Similarly, the term in $U(s)$ becomes
$$\dfrac{\alpha}{s^2}\dfrac{d}{ds}U(s)\rightarrow -\alpha\int_0^T(T-\tau)\tau u(\tau)d\tau.$$
Therefore, the time-domain expression of the right-hand side is given by
$$\int_0^T\left[(T-2\tau)y(\tau)+\alpha(T-\tau)\tau u(\tau)\right]d\tau.$$
Now we need to consider the term in $\phi/s^4$ and we have that
$$\dfrac{1}{s^4}\rightarrow \dfrac{T^3}{3!}=\dfrac{T^3}{6}$$
which finally yields
$$-\dfrac{T^3}{6}\phi=\int_0^T\left[(T-2\tau)y(\tau)+\alpha(T-\tau)\tau u(\tau)\right]d\tau$$
or, equivalently,
$$\phi=-\dfrac{6}{T^3}\int_0^T\left[(T-2\tau)y(\tau)+\alpha(T-\tau)\tau u(\tau)\right]d\tau.$$
