A mean field SDE with hitting time 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?
 A: 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}
