Find function $f(x)$ that satisfying differential relation

Question

Suppose the functions $F(x)$ and $G(x)$ satisfying $$F(x)=f(x)-\frac{1}{f(x)}$$ $$G(x)=f(x)+\frac{1}{f(x)}$$ such that $F'(x)=(G\circ G)(x)$, with initial condition $f(\frac{\pi}{4})=1$ is given. Find $f(x)$.

I have attempted $F(x)+G(x)=2f(x)$, and try to relate functions $F(x)$ and $G(x)$, but stuck in the composite function $(G\circ G)(x)$. Taking integration for $F'(x)$ and $(G\circ G)(x)$ on both sides with respect to $x$ or $F(x)$ do not help that much. Any clue?

Answer 1

Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that \begin{align*} F'&=f'-\frac{-f'}{f^2}=\frac{f'(1+f^2)}{f^2}, \; \text{and} \\ G^2&=f^2+2+\frac{1}{f^2}. \end{align*} Then $$\frac{f'(1+f^2)}{f^2}=f^2+2+\frac{1}{f^2},$$ or $$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$ and therefore $$f'=1+f^2,$$ which is separable.

Answer 2

Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(G\circ G)(x)$.

Note that $$(G\circ G)(x)=F'(x)=\left(1+\frac{1}{\big(f(x)\big)^2}\right)\,f'(x)=G(x)\,\frac{f'(x)}{f(x)}\,.$$ We then have $$\frac{f'(x)}{f(x)}=\frac{(G\circ G)(x)}{G(x)}\,.$$ Similarly, observe that $$G'(x)=\left(1-\frac{1}{\big(f(x)\big)^2}\right)\,f'(x)=F(x)\,\frac{f'(x)}{f(x)}=F(x)\,\left(\frac{(G\circ G)(x)}{G(x)}\right)\,.$$ Therefore, $$F(x)=\frac{G(x)\,G'(x)}{(G\circ G)(x)}\,.$$ Taking derivative, we get $$(G\circ G)(x)=F'(x)=\frac{\big(G'(x)\big)^2+G(x)\,G''(x)}{(G\circ G)(x)}-\frac{G(x)\,G'(x)}{\big((G\circ G)(x)\big)^2}\,G'\big(G(x)\big)\,G'(x)\,.$$ Consequently, $$(G\circ G)(x)\,\big(G'(x)\big)^2+(G\circ G)(x)\,G(x)\,G''(x)=G(x)\,\big(G'(x)\big)^2\,G'\big(G(x)\big)+\big((G\circ G)(x)\big)^3\,,$$ with $G\left(\dfrac{\pi}{4}\right)=2$ and $G'\left(\dfrac{\pi}{4}\right)=0$. Well, this looks hopeless.