Hello, I am trying to solve the functional equation below. Can anybody help? I don't expect a solution, but tips on how to do it, tricks or references to usefull books.

Thank you
Jan

The notation is following. Let $\displaystyle \pi>0,~\phi>0,~\delta
\in(0,1),~p\in(0,1)$, all of them parameters of the problem. Then
the problem is to find real-valued function
$\displaystyle q_{d}(x):\mathbb{R}\rightarrow\mathbb{R}$ which satisfies
$\displaystyle \frac{[q_{d}(x)-\pi-\phi\delta
p](1-p)}{\sqrt{D_{1}}}+\frac{[q_{d}(x)-\pi-\phi(1-\delta(1-p))]p}{\sqrt{D_{2}}}-\frac{q_{d}(x)-\pi-\phi
p}{\sqrt{D_{3}}}=0
$
where
$\displaystyle D_{1}=\frac{1}{1-\delta}[[q^{-1}_{d}(q_{a}(q_{d}(x)))-\pi-\phi(1-\delta(1-p))]^{2}-\delta[q_{a}(q_{d}(x))-\pi-\phi
p]^{2}]\\+\phi^2\delta(1-p)^{2}
$
$\displaystyle D_{2}=\frac{1}{1-\delta}[[q_{d}(x)-\pi-\phi(1-\delta(1-p))]^{2}-\delta[q_{d}(q_{d}(x))-\pi-\phi
p]^{2}]\\+\phi^2\delta(1-p)^{2}
$
$\displaystyle D_{3}=\frac{1}{1-\delta}[[x-\pi-\phi(1-\delta(1-p))]^{2}-\delta[q_{d}(x)-\pi-\phi
p]^{2}]\\+\phi^2\delta(1-p)^{2}$

where $\displaystyle q_{d}^{-1}(x)$ is inverse of $\displaystyle q_{d}(x)$ and

$\displaystyle q_{a}(x)=\pi+\phi p-\sqrt{\frac{(x-\pi-\phi\delta
p)^{2}}{\delta}+\phi^2p^{2}(1-\delta)}
$
is real-valued $\displaystyle \mathbb{R}\rightarrow\mathbb{R}$ function.

I already know
$\displaystyle q_{d}(\pi)=\pi
$and defining $\displaystyle q_{d}^{n}(x)=q_{d}(q_{d}^{n-1}(x))$ with
$\displaystyle q_{d}^{1}(x)=q_{d}(x)$.

$\displaystyle \lim_{n\rightarrow\infty}q_{d}^{n}(x)=\pi
$ that is successive applications of the $\displaystyle q_{d}$ function converge to
$\displaystyle \pi$.