How can I solve $1-F(x)=F(F'(x))$, a particular functional differential equation?

Question

I'm trying to demonstrate a Probability result.

In order to do that, I have to solve the following equation, but I don't know how to do:

$$1-F(x)=F(F'(x))$$

What's the expression of the solution F?

Answer 1

First, follow dxiv's suggestion from chat to take $G(x)=F(x)-\frac{1}{2}$; this removes the constant term, so that $$-G(x)=G(G'(x))\tag{1}$$

Unfortunately, one cannot look for a power series solution to (1). The function composition is non-local, so one cannot justify it via e.g. Taylor's theorem. Worse, even if we accept purely formal solutions, "high-order" terms still have "lower-order" effects. (Formally, the approximating polynomials do not converge in the $x$-adic topology.)

On the other hand, an equivalent form of (1) is $$G^{-1}(-G(x))=G'(x)\tag{2}$$ This suggests changing variables: let $u=G(x)$ and $x=H(u)$.

Then $1=H'(u)G'(x)$ by the chain rule; substituting into (2) and rearranging, $$1=H(-u)H'(u)$$ This, too, is nonlocal…except around $u=0$. Taking a power series around $0$ as ansatz, one discovers that $$H(u)=\sum_{k=0}^{\infty}{\frac{a_0}{k!}\left(\frac{u}{a_0^2}\right)^k}=a_0e^{\frac{u}{a_0^2}}$$

Interestingly, there's also a sporadic singular solution: $$H(u)=\sqrt{\frac{2u}{i}}$$ where $i^2=-1$. When we invert back to $G$ and $F$, this solution regularizes to $$F(x)=\sqrt{i}x^2$$ I only went looking for it because I had seen it in an earlier draft of this solution. Proving uniqueness of smooth solutions to $H$ is pretty straightforward…but I have no idea if there are other sporadic singular solutions lurking that would give a reasonably well-behaved $F$.

In any case, I'm going to assume there aren't such hidden pitfalls. Then the main family of solutions gives $$G(x)=H^{-1}(x)=a_0^2\ln{\left(\frac{x}{a_0}\right)}$$ Thus $$F(x)=\frac{1}{2}+a_0^2\ln{\left(\frac{x}{a_0}\right)}$$ for some $a_0$.