# Implementable numerical scheme for the equation $a=\text{Erf}\big(z/\sqrt{2N_{a}}\big)$

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 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 $$\{0=t_0, the computer yields a output $$\{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.

$$\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}\newcommand{\de}{\delta}\newcommand\ep\epsilon$$In the previous answer, 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 $$0=t_0. For each $$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" $$0=t_0, assume without loss of generality that $$\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}

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}. $$\quad\Box$$

• Nice answer. Many thanks Feb 18 at 7:18