On the solution to $u'(x)=2u(2x+1)-2u(2x-1)$ and the Fabius function

Question

Looking for closed-form examples of Delay Differential Equations I found that $$u'(x)=2u(2x+1)-2u(2x-1)$$ could had as solution a modification of the Fabius function $F(x)$, which "is an example of an infinitely differentiable function that is nowhere analytic", so is not really a closed-form, but at least is a function which could be explored graphically as is shown in this answer.

But I also found some "incongruences" among different sources as what it is shown in this answer and on this paper, so I want to know if the following statements are true:

1. From what I think I have understood through other questions, I think that $u(x)=F(x+1)$ fulfills $u'(x)=2u(2x+1)-2u(2x-1)$ for $|x|<1$, Is this true?
2. Under the assumption point (1) is not mistaken, I believe that for extending the result to the whole real line, by itself the Fabius function $F(x)$ is not an alternative since it change values from over and below zero following an intrincated pattern (the Thue–Morse sequence), but since $u(x)$ allready work in the interval $[-1,\ 1]$ my intuition tells that $z(x)=|F(x+1)|$ should fulfill $z'(x)=2z(2x+1)-2z(2x-1)$ in the whole real line, Is this true?
3. Now under the assumption point (2) is true, my intuition tell that the relation should not be dependent of some location parameter, so for any real-valued constant $a$ the function $q(x)=z(x-a)$ also should fulfill $q'(x)=2q(2x+1)-2q(2x-1)$ in the whole real line, Is this true?

I tried to evaluate them but got confused with the aritmetic (is not that easy, there are a full papers related to it - as the allready mentioned one).

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Added later

So far using an approximation function that don't fulfill the differential equation, but has a similar behavior $$f(t)=\frac{1}{1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)}$$ I am trying on Desmos to extend the definition of point (1) which looks true, to fulfill the relation in the whole real line, but so far it looks like the hypothesis (2) and (3) are mistaken.

It is possible to extend definition (1) to the whole real line? How?

Answer 1

The link Recursive Integration over Piecewise Polynomials: Closed form? says:

$u(x)$ is a solution if it satisfies below equation,

For $-1 \leq x \leq 0$, $$u'(x) = 2u(2x+1)$$

For $0 \leq x \leq 1$, $$u'(x) = -2u(2x-1)$$

The paper https://arxiv.org/pdf/1702.06487.pdf says you can satisfy the above with the following definition:

For $-1 \leq x \leq 0$, $$u(x) = F(x+1)$$

For $0 \leq x \leq 1$, $$u(x) = F(1-x)$$

Where $F$ is the Fabius function satisfying $F'(x) = 2 F(x)$ as per definition given in https://en.wikipedia.org/wiki/Fabius_function.

So all the answers are consistent.

Regarding extending it to full real line, Not sure.

Answer 2

I found recently the following website that the Rvachëv function mentioned in the paper that fulfills in the $\mathbb{R}$ the relation: $$u'(x)=2u(2x+1)-2u(2x-1)$$ its indeed: $$u(x)=\begin{cases} F(x+1),\quad |x|<1\\0,\quad |x|\geq 1\end{cases}$$ with $F(x)$ the Fabius function... both the paper and the website signed by Juan Arias de Reyna, so I don't know why its definition is overcomplicated in the paper.

With this the hypothesis of point (2) and (3) are false, but it is interesting how they fail: if I create a train of everly spaced versions of $u(x)$ (using and approximation), the Right-Hand-Side of the DDE makes lobes that matches those of the Left-Hand-Side but also extra lobes were the function is separated by null values, and everything looks both sides are going to fit each others as the lobes in the train becomes near each other, but instead just in the point the lobes becomes joined instead of matching the DDE, the Right-Hand-Side kind of becomes the zero function (the are remaining lobes at the extremes of the finite train I use as approximated example). I don't really get what have happened, but is interesting to see it here in Desmos.