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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 $a^*$?

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  • $\begingroup$ your defining equation for $a^\ast(t)$ could be written more compactly as $$a^\ast(t)=\text{Erf}\,\left(\frac{z}{\sqrt{2N_{a^\ast}(t)}}\right).$$ $\endgroup$
    – Carlo Beenakker
    Commented Feb 16, 2023 at 12:47
  • $\begingroup$ @CarloBeenakker Thanks for the comment which definitively simplifies the equation $\endgroup$
    – Fawen90
    Commented Feb 16, 2023 at 12:52
  • $\begingroup$ your equation for $N_a(t)$ contains $a$ in the integration limit and in the integrand; how should I understand this? are these different objects? $\endgroup$
    – Carlo Beenakker
    Commented Feb 16, 2023 at 12:58
  • $\begingroup$ @CarloBeenakker Many thx for pointing out this typo. It is indeed $0$ instead of $a$ $\endgroup$
    – Fawen90
    Commented Feb 16, 2023 at 13:00
  • $\begingroup$ How is the function $a$ given? There are "some measurable function" which are awful for numeric approximation. And what precision do you need? What means "efficiency" to you? $\endgroup$
    – g g
    Commented Feb 16, 2023 at 13:05

1 Answer 1

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$\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}$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 $(0,\infty)$.

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  • $\begingroup$ Many thx Iosif for this strengthen proof that ensures the wellposedness of iteration on any interval $[0,T]$. Now fix any horizon $T>0$, and let us focus on the numerical solution on $[0,T]$. Taking $a_0\equiv 0$, set $a_{n+1}:=F(a_n)$. From the computational viewpoint, we pick, for each $n\ge 1$, a division of the interval $\{0=t^n_0<t^n_1<\cdots<t^n_{N_n}=T\}$ and suppose we obtain the approximating values of $a_n$ on the points $\{t^n_0,t^n_1,\ldots, t^n_{N_n}\}$. According to which criterion we should choose $\{0=t^{n+1}_0<t^{n+1}_1<\cdots<t^{n+1}_{N_{n+1}}=T\}$ (for $n+1$)? $\endgroup$
    – Fawen90
    Commented Feb 16, 2023 at 16:22
  • $\begingroup$ More precisely, computer can stock at each iteration a finite number of values for $a_n$ (w.r.t. some grid $\{0=t^n_0<t^n_1<\cdots<t^n_{N_n}=T\}$), and how can we use this grid for the next step? Here we should also deal with the numerical integral to approximate $N_a$ $\endgroup$
    – Fawen90
    Commented Feb 16, 2023 at 16:24
  • $\begingroup$ To make the iteration implementable, an error analysis for $\tilde a_n-a_n$ is needed, where $\tilde a_n$ stands for the numerical approximation (that can be stocked in computer) $\endgroup$
    – Fawen90
    Commented Feb 16, 2023 at 16:26
  • $\begingroup$ @Fawen90 : I think the questions about the $t^n_j$'s are important, but they should be posted separately. $\endgroup$
    – Iosif Pinelis
    Commented Feb 16, 2023 at 17:16
  • $\begingroup$ Thanks for the answer. I will post it separately $\endgroup$
    – Fawen90
    Commented Feb 17, 2023 at 9:27

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