Let $z>0$ be fixed and $A$ be the set of non-increasing functions from $\mathbb R\_+$ to $[0,1]$ with norm $\|\cdot\|:=\|\cdot\|\_\infty$. Define by $F$ the operator on $A$ by


\begin{equation\*}
 F(a)(t):=\text{Erf}\left(\frac{z}{\sqrt{2N\_a(t)}} \right),\quad \forall t\ge 0, 
\end{equation\*}


where $\text{Erf}$ is the Gauss error function and $N\_a:\mathbb R\_+\to\mathbb R\_+$ is defined by


$$N\_a(t):=\int\_0^t\frac{ds}{(1+a(s))^2}.$$


Iosif has shown in [Numerical solution to some functional equation](https://mathoverflow.net/questions/440988/numerical-solution-to-some-functional-equation) that $F$ maps $A$ to $A$ and is a contraction map. I'm looking for a numerical approximation of the fixed point $a^\*$ of $F$. Let $a\_0\equiv 0$ and $a\_{n}:=F(a\_{n-1})$ for all $n\ge 1$. Then it follows that


$$\|a\_n-a^\*\|\le Cr^n,\quad \forall n\ge 1,$$


where $C>0, r\in (0,1)$ are some constants. How could we implement (code in computer) the iteration $a\_{n}:=F(a\_{n-1})$, and obtain a numerical approximation of $a\_N$ for any $N$?


PS : Here the numerical approximation of $a\_N$ means that: For any fixed $T>0$, any $\epsilon>0$ and any subset $\{0=t\_0<t\_1<\cdots<t\_M=T\}$, the computer yields a output $\{y\_0, y\_1,\ldots, y\_M\}$ s.t.


$$\max\_k |y\_k-a\_N(t\_k)|\le \epsilon.$$


Of course, any other numerical method (different from the above iteration) is highly appreciated.



Implementable numerical scheme for the equation $a=\text{Erf}\big(z/\sqrt{2N_{a}}\big)$
$\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}\newcommand{\de}{\delta}\newcommand\ep\epsilon
$In the [previous answer](https://mathoverflow.net/questions/440988/numerical-solution-to-some-functional-equation), it was shown that the operator $F$ on $A$ is $r$-Lipschitz for a certain universal constant $r\in(0,1)$ with respect to the norm $\|\cdot\|\_\infty$. The same proof holds for the corresponding operator $F\_T\colon A\_T\to A\_T$, where $A\_T$ is the set of all non-increasing functions from $(0,T]$ to $[0,1]$ with norm given by $\|a\|:=\sup\_{t\in(0,T]}|a(t)|$ for $a\in A\_T$ and
\begin{equation\*}
 F\_T(a)(t):=\text{Erf}\left(\frac{z}{\sqrt{2N\_a(t)}} \right) 
\end{equation\*}
for $a\in A\_T$ and $t\in(0,T]$; this follows because for any $t\in(0,T]$ the value of $N\_a(t)$ depends only on the values of $a$ on $(0,T]$.


The additional ingredient needed to answer the current question is the observation that $F\_T(a)(t)$ is $L$-Lipschitz in $t$ with the same $L$ for all $a\in A\_T$. Indeed, for any $a\in A\_T$ and $t\in(0,T]$, with $u:=z/\sqrt{N\_a(t)}$,
\begin{equation\*}
 F\_T(a)'(t):=\frac d{dt}\,F\_T(a)(t)
 =-\frac2{\sqrt\pi}\,e^{-u^2/2}u^3\,\frac1{2\sqrt2\,z^2}\frac1{(1+a(t))^2},
\end{equation\*}
so that
\begin{equation\*}
 |F\_T(a)'(t)|\le L:=c/{z^2},
\end{equation\*}
where $c:=(3/e)^{3/2}/\sqrt{2\pi}$.


As requested, take now any real $\ep>0$, any natural $K$, and any $t\_0,\dots,t\_K$ such that $0=t\_0<t\_1<\cdots<t\_K=T$. For each $n=0,1,\dots$, we want a computer to yield $y\_{n,0},\ldots,y\_{n,K}$ s.t.
\begin{equation\*}
 M\_n:=\max\_{k\in[K]}|a\_n(t\_k)-y\_{n,k}|\le\ep, \tag{$\*$}\label{1}
\end{equation\*}
where, as usual, $[K]:=\{1,\dots,K\}$.


Take any real $\de>0$. By refining the "partition" $0=t\_0<t\_1<\cdots<t\_K=T$, assume without loss of generality that
\begin{equation\*}
 \max\_{k\in[K]}|t\_k-t\_{k-1}|\le\de.
\end{equation\*}


Let (say) $a\_0:=1$ and $a\_n:=F\_T(a\_{n-1})$ for $n\ge1$.


For $k\in[K]$, let $y\_{0,k}:=0$ and for $n\ge1$ let (a computer compute)
\begin{equation\*}
 y\_{n,k}:=F\_T(Y\_{n-1})(t\_k),
\end{equation\*}
where
\begin{equation\*}
 Y\_n(t):=\sum\_{k\in[K]}y\_{n,k}\,1(t\_{k-1}<t\le t\_k). 
\end{equation\*}


Then, recalling that $F\_T$ is $r$-Lipschitz, for $n\ge1$ we have
\begin{equation\*}
\begin{aligned}
M\_n&=\max\_{k\in[K]}|F\_T(a\_{n-1})(t\_k)-F\_T(Y\_{n-1})(t\_k)| \\ 
&\le\|F\_T(a\_{n-1})-F\_T(Y\_{n-1})\| \\ 
&\le r\|a\_{n-1}-Y\_{n-1}\| \\ 
&=r\sup\_{t\in(0,T]}|a\_{n-1}(t)-Y\_{n-1}(t)| \\ 
&=r\max\_{k\in[K]}\sup\_{t\in(t\_{k-1},t\_k]}|a\_{n-1}(t)-y\_{n-1,k}|; 
\end{aligned}
\end{equation\*}
also, $|a\_{n-1}(t)-y\_{n-1,k}|\le|a\_{n-1}(t\_k)-y\_{n-1,k}|+L\de$, because ($F\_T(a)(t)$ is $L$-Lipschitz in $t$ and hence) $a\_{n-1}$ is $L$-Lipschitz.


It follows that for all $n\ge1$
\begin{equation\*}
 M\_n\le rM\_{n-1}+rL\de
\end{equation\*}
and hence (by induction on $n$, with $M\_0=0$)
\begin{equation\*}
 M\_n\le C\de,
\end{equation\*}
where $C:=rL/(1-r)$. Finally, choosing $\de=\ep/C$, we get \eqref{1}. $\quad\Box$


