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 $a^\*$?



Numerical solution to some functional equation
$\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}$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 $(0,\infty)$.


