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Let $b\in \mathbb R$ and $\sigma>0$ be given. For a fixed probability distribution $\mu_0$ on $\mathbb R$ s.t.

$$\int_{(0,\infty)}\mu_0(dx)=1,$$

consider the mean field SDE :

$$dX_t = \mathbf{1}_{\{X_t>0\}} \left[ bdt + \frac{\sigma}{1 + m_t {\bf 1}_{\{b>0\}}} dW_t \right],\quad \mbox{for all } t\ge 0,~~~~~~~~~~~~~~~~(\ast)$$

where $X_0\sim \mu_0$ is independent of the Brownian motion $(W_t)_{t\ge 0}$ and

$$m_t:=\int_{(0,\infty)}\mu_t(dx),\quad \mbox{for all } t\ge 0.$$

How can we show the existence and uniqueness of the (weak) solution to $(\ast)$?

Any answers, remarks or references are highly appreciated!

REMARK :

The case for $b\le 0$ is trivial. Indeed, $(\ast)$ reduces to $dX_t = \mathbf{1}_{\{X_t>0\}} \big[ bdt + \sigma dW_t \big]$ and the solution is given as $X_t=Y_{t\wedge \tau}$, where $Y_t:=X_0+bt+\sigma W_t$ and $\tau:=\inf\{t\ge 0: Y_t\le 0\}$. For the case $b>0$, $(\ast)$ turns to be

$$dX_t = \mathbf{1}_{\{X_t>0\}} \left[ bdt + \frac{\sigma}{1 + m_t} dW_t \right].$$

I do not find any literature on the existence of its solution.

REMARK 2 :

A heuristic argument is as follows : Let $\ell:\mathbb R_+\to [0,1]$ be some "nice" function. Consider the process

$$Y^{\ell}_t: = X_0+ bt+\int_0^t\frac{\sigma}{1+\ell(s)}dW_s.$$

Then we have $(Y^{\ell}_{t\wedge \tau^{\ell}})_{t\ge 0}$ is a solution to $(\ast)$ if $\mathbb P[\tau^{\ell}>t]=m_t$ for all $t\ge 0$, where $\tau^{\ell}:=\inf\{t\ge 0: Y^{\ell}_t\le 0\}$. Thus it remains to calculate the probability $\mathbb P[\tau^{\ell}>t]$ in terms of $\ell$. Is there any reference for this computation?

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This is a rough idea. Fix an arbitrary $T>0$. For $0\le t\le T$ and $x\ge 0$, define $u(t,x;T):=\mathbb P[\tau^{\ell}>T|Y^{\ell}_t=x]$. One has by definition $u(T,x;T)={\bf 1}_{\{x>0\}}$ for all $x\ge 0$ and $u(t,x;T)=0$ for all $0\le t\le T$.

For $0\le t<T$ and $x>0$, it follow from the tower property that

$$u(t,x;T)=\mathbb E\big[\mathbb E[{\bf 1}_{\{\tau^{\ell}>T\}}|Y^{\ell}_{s+h}]|Y^{\ell}_t=x\big]=\mathbb E\big[u(t+h,Y^{\ell}_{t+h};T)|Y^{\ell}_t=x\big], \quad \mbox{for all } 0<h<T-t.$$

Admitting the regularity of $u$, one has by Ito's formula

$$\frac{1}{h}\mathbb E\left[\int_{t}^{t+h}\left(\partial_tu(s,Y^{\ell}_s;T) + b\partial_xu(s,Y^{\ell}_s;T) +\frac{\sigma^2}{2(1+\ell(s))^2}\partial_{xx}^2 u(s,Y^{\ell}_s;T)\right)ds~\Big|Y^{\ell}_t=x\right]~~=~~0.$$

Letting $h\to 0$, it yields

$$\partial_tu(t,x;T) + b\partial_xu(t,x;T) +\frac{\sigma^2}{2(1+\ell(t))^2}\partial_{xx}^2 u(t,x;T)=0.$$

Assume $X_0=\bar x>0$. Then one has coupled equations:

\begin{equation} \partial_tu(t,x;T) + b\partial_xu(t,x;T) +\frac{\sigma^2}{2(1+\ell(t))^2}\partial_{xx}^2 u(t,x;T)=0,~ \forall 0\le t<T, x>0 \\ u(T,x;T)={\bf 1}_{\{x>0\}},~ \forall x\ge 0 \\ u(t,0;T) = 0,~ \forall 0\le t\le T \\ u(0,\bar x,;T) = \ell(T),~ \forall T\ge 0. \end{equation}

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