Numerical solution to some functional equation Let $z>0$ be fixed. Consider the function $p_a: \mathbb R^2_+\to\mathbb R_+$ given as
$$
p_a(t,x):=\frac{1}{\sqrt{2\pi N_a(t)}}\left[\exp\left(-\frac{(x-z)^2}{2N_a(t)}\right)-\exp\left(-\frac{(x+z)^2}{2N_a(t)}\right)\right],
$$
where $N_a:\mathbb R_+\to\mathbb R_+$ is defined by
$$N_a(t):=\int_0^t\frac{ds}{(1+a(s))^2}$$
and $a:\mathbb R_+\to [0,1]$ is some measurable function. I can show there exists a unique function $a^*$ (which is also decreasing) to the equation
$$a^*(t)=\int_0^\infty p_{a^*}(t,x)dx\equiv \text{Erf}\left(\frac{z}{\sqrt{2N_{a^*}(t)}}\right),\quad \forall t>0,$$
where $\text{Erf}$ is the Gauss error function (https://en.wikipedia.org/wiki/Error_function). Is there (efficient) numerical scheme to compute/approximate $a^*$?
 A: $\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}$The functional equation in question is
\begin{equation*}
    a=F(a) \tag{1}\label{1}
\end{equation*}
on $(0,\infty)$, where $a$ is in the closed convex set, say $A$, of all nonincreasing functions from $(0,\infty)$ to $[0,1]$ with norm $\|\cdot\|:=\|\cdot\|_\infty$ and
\begin{equation*}
    F(a)(t):=\erf\frac{z}{\sqrt{2N_a(t)}} 
\end{equation*}
for real $t>0$.
For any $a\in A$, any function $h$ from $[0,\infty)$ to $\R$ such that $a+uh\in A$ for all small enough $u>0$, and all such $u$, let $g_{a,h}(u):=F(a+uh)$. Then for all real $t>0$
\begin{equation*}
g'_{a,h}(0+):=\lim_{u\downarrow0}\frac{g_{a,h}(u)-g_{a,h}(0)}u \\ 
=\frac2{\sqrt\pi}\,\exp\Big(-\frac{z^2}{2N_a(t)}\Big)
\frac{-z}{2\sqrt2\,N_a(t)^{3/2}} \int_0^t\frac{-2ds\,h(s)}{(1+a(s))^3}
\end{equation*}
and $\big|\int_0^t\frac{-2ds\,h(s)}{(1+a(s))^3}\big|\le2N_a(t)\|h\|$,
so that, with $y:=\frac z{\sqrt{2\,N_a(t)}}>0$
\begin{equation*}
|g'_{a,h}(0+)|\le\frac2{\sqrt\pi}\,e^{-y^2}y\,\|h\|\le r\|h\|, 
\end{equation*}
where $r:=\sqrt{\frac2{\pi e}}\in(0,1)$.
So, the map $F$ is a contraction. So, there is a unique solution $a^*\in A$ of \eqref{1}, and the iterations $a_{n+1}=F(a_n)$ with any $a_0\in A$ converge to $a_*$ uniformly on $(0,\infty)$.
