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Discretization formula for a system of two differential equations. "Solution to one of these is the initial condition of the other". In which sense?

Computational Science Asked on August 29, 2021

Consider the following stochastic differential equation
begin{equation}
dy=left(A-left(A+Bright)yright)dt+Csqrt{yleft(1-yright)}dWtag{1}
end{equation}

where $A$, $B$ and $C$ are parameters and $dW$ is a Wiener increment.
Equation $(1)$ will be our point of reference in what follows.


Now, first let us consider a "method" for equation $left(1 right)$ which can be described by the following one-step discretization scheme:
begin{equation}
y_{n+1}=y_n+left(A-left(A+Bright)y_nright)Delta t +Csqrt{y_nleft(1-y_nright)}Delta W_n + Dleft(y_nright)left(y_n-y_{n+1}right)tag{2}
end{equation}

where $Delta t$ is the length of the time discretization interval, $Delta W_n$ is a Wiener increment and $D(y_n)$ is the system of control functions and takes the form
$$
D(y_n)=d^0(y_n)Delta t + d^1left(y_nright)|Delta W_n|
$$

with
$$
d^1(y)=
begin{cases}
Csqrt{frac{1-varepsilon}{varepsilon}}hspace{0.5cm}text{if }y<varepsilon
Csqrt{frac{1-y}{y}}hspace{0.5cm}text{if }varepsilonle y<frac{1}{2}
Csqrt{frac{y}{1-y}}hspace{0.5cm}text{if }frac{1}{2}le yle 1-varepsilon
Csqrt{frac{1-varepsilon}{varepsilon}}hspace{0.5cm}text{if }y>1-varepsilon
end{cases}
$$

At this point, let us consider a "method" which decomposes $left(1right)$ into two equations. Specifically, the first equation is a stochastic one, that consists of the diffusion term of $left(1right)$ only (see eqtn $left(3right)$), while the second one is an ordinary differential equation (see eqtn $left(4right)$) that consists of the drift part of $left(1right)$. We have:

$begin{equation}
dy_1=Csqrt{y_1left(1-y_1right)}dWtag{3}
end{equation}$

$begin{equation}
dy_2=left(A-left(A+Bright)y_2right)dttag{4}
end{equation}$

This last method approximates the solution to $left(3right)$ at each time step using $left(2right)$ (and numerical solution to $left(3right)$ is used as the initial condition in $left(4right)$), while $left(4right)$ can be solved using the Euler method. Thus, such a method can be described by the following one step discretization formula:
$$
y_{n+1}=y_n+left(A-left(A+Bright)y_nright)Delta t + dfrac{Csqrt{y_nleft(1-y_nright)}Delta W_n}{1+d^1left(y_nright)|Delta W_n|}left(1-left(A+Bright)Delta tright)tag{5}
$$

My doubts:

  1. I cannot understand in which way the last method approximates solution to $left(3right)$ at each time step using $left(2right)$. Could you please explicit such an approximation? How is it obtained by means of $left(2right)$?
  2. In which sense numerical solution to $left(3right)$ is used as the initial condition in $left(4right)$? Which is such an initial condition?
  3. Could you please explicit the way in which solution to $left(3right)$ and solution to $left(4right)$ are combined so as to obtain discretization formula $left(5right)$?

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