Consider the stochastic differential equation as follows:


$$X\_t=X\_0+t+\int\_0^t\frac{dW\_s}{1+m(s)},\quad \forall t\ge 0,~~~~~~~~~~~~~~~(\ast)$$


where $X\_0>0$ is square integrable and $m(t)=\mathbb P[\inf\_{0\le s\le t}X\_s>0]$ for all $t\ge 0$. Can we prove that $(\ast)$ admits at most one solution $(X,m)$ (assuming its existence)?


Any answer, comments or references are highly appreciated.



Uniqueness of the solution to some SDE
$\newcommand{\F}{\mathcal{F}}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$According to a comment by the OP, $X\_0$ and $(W\_t)$ are independent. So, without loss of generality (wlog), $X\_0$ is a real number $x\_0>0$.


Let $\F$ denote the set of all nonincreasing functions from $[0,\infty)$ to $[0,1]$. Define the (nonlinear) operator $F$ from $\F$ to $\F$ as follows: for each $f\in\F$ and each real $t\ge0$,
\begin{equation\*}
 F(f)(t):=P(\inf\_{s\in[0,t]}X^f\_s>0), \tag{1}
\end{equation\*}
where
\begin{equation\*}
 X^f\_t:=x\_0+t+\int\_0^t\frac{dW\_s}{1+f(s)}. \tag{2}
\end{equation\*}
We have to show that the equation
\begin{equation\*}
 F(f)=f \tag{3}
\end{equation\*}
has at most one solution $f$ in $\F$.


Before doing this, let us prove, for completeness, the existence of a solution of (3). Here we just detail the argument in the comment by user GJC20 on this [previous answer](https://mathoverflow.net/a/410689/36721).
Let $f\_0:=0$ and $f\_n:=F(f\_{n-1})$ for all natural $n$. Then $f\_1=F(f\_0)\ge0=f\_0$ and hence, by that previous answer, $f\_n\ge f\_{n-1}$ for all natural $n$. So, the uniformly bounded sequence $(f\_n)$ converges (as $n\to\infty$) pointwise to some $\bar f\in\F$, which implies $f\_{n+1}=F(f\_n)\to F(\bar f)$ pointwise. So, $F(\bar f)=\bar f$, that is, $\bar f$ is a solution of (3).


To prove that (3) has at most one solution in $\F$, suppose that functions $f\_1$ and $f\_2$ in $\F$ are solutions of (3). Let
\begin{equation\*}
 t\_0:=\sup\{t\in[0,\infty)\colon f\_1(s)=f\_2(s)\ \forall s\in[0,t)\}. \tag{4}
\end{equation\*}
Then wlog $t\_0<\infty$. Take any $h\in(0,1)$. To obtain a contradiction, suppose that  

\begin{equation\*}
 \ep:=\|f\_1-f\_2\|>0, \tag{5}
\end{equation\*}
where $\|g\|:=\sup\{|g(t)|\colon t\in[0,t\_0+h]\}$.


The crucial point is to use the same kind of time change as in the mentioned previous answer: The process
\begin{equation\*}
\text{$(X^{f\_i}\_t)$ equals the process $(x\_0+t+W\_{\tau\_i(t)})$ in distribution,} \tag{\*} 
\end{equation\*}
where
\begin{equation\*}
 \tau\_i(t):=\int\_0^t\frac{ds}{(1+f\_i(s))^2}; \tag{6}
\end{equation\*}
here and in what follows, $i\in\{1,2\}$.


Note that $\frac1{(1+z)^2}$ is $2$-Lipschitz in $z\ge0$. Therefore and in view of (6), (4), and (5),
\begin{equation\*}
 |\tau\_1(t)-\tau\_2(t)|\le\int\_{t\_0}^{\max(t\_0,t)} ds\,\Big|\frac1{(1+f\_1(s))^2}-\frac1{(1+f\_2(s))^2}\Big|
 \le 2h\ep \tag{7}
\end{equation\*}
for all $t\in[0,t\_0+h]$.


Since $\frac14\le\frac1{(1+f\_i)^2}\le1$, the functions $\tau\_i$ are Lipschitz-continuous and strictly increasing on $[0,\infty)$ from $\tau\_i(0)=0$ to $\tau\_i(\infty-)=\infty$, and the inverse functions $\tau\_i^{-1}$ are defined, strictly increasing, and $4$-Lipschitz on $[0,\infty)$. It follows that for all $u\in[0,\tau\_2(t\_0+h)]$
\begin{equation\*}
 |\tau\_1^{-1}(u)-\tau\_1^{-1}(u)|\le4\sup\_{t\in[0,t\_0+h]}|\tau\_1(t)-\tau\_2(t)|\le8h\ep, \tag{8}
\end{equation\*}
in view of (7).


By (\*), for all real $t\ge0$,
\begin{equation\*}
 F(f\_1)(t)-F(f\_2)(t)=D\_1(t)+D\_2(t), \tag{9}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
 D\_1(t)&:=P(\inf\_{u\in[0,\tau\_1(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\ 
 &-P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0)
\end{aligned}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
 D\_2(t)&:=P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\ 
 &-P(\inf\_{u\in[0,\tau\_2(t)]}(x\_0+\tau\_2^{-1}(u)+W\_u)>0). 
\end{aligned}
\end{equation\*}


Using e.g. (as an overkill) [Lemma 8, p. 407](https://epubs.siam.org/doi/10.1137/1115047) (or its Russian original, [Lemma 8, p. 423](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tvp&paperid=1854&option_lang=eng)) and (8), we get
\begin{equation\*}
 \|D\_2\|\le C\Big(1+\frac1{\sqrt{t\_0+h}}\Big)h\ep\le2C\sqrt h\,\ep, \tag{10}
\end{equation\*}
where $C$ is some universal positive real constant.


Let $\tau\_{\min}(t):=\min(\tau\_1(t),\tau\_2(t))$ and $\tau\_{\max}(t):=\max(\tau\_1(t),\tau\_2(t))$.
Take any $t\in[0,t\_0+h]$ and then let $u\_1:=\tau\_{\min}(t)$, $\de:=\tau\_{\max}(t)-\tau\_{\min}(t)$, $x\_1:=x\_0+\tau\_1^{-1}(u\_1)$, and $G:=1-\Phi$, where $\Phi$ is the standard normal cdf. Let $G\big(\frac{x\_1}{\sqrt{u\_1}}\big):=0$ if $u\_1=0$.
Then
\begin{equation\*}
\begin{aligned}
 &|D\_1(t)| \\
 &=P(\inf\_{u\in[0,\tau\_{\min}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\ 
 &-P(\inf\_{u\in[0,\tau\_{\max}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0) \\ 
 &=P(\inf\_{u\in[0,\tau\_{\min}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)>0, \\ 
 &\inf\_{u\in(\tau\_{\min}(t),\tau\_{\max}(t)]}(x\_0+\tau\_1^{-1}(u)+W\_u)\le0) \\ 
 &\le P(\inf\_{u\in[0,u\_1]}W\_u>-x\_1,\inf\_{u\in(u\_1,u\_1+\de]}W\_u\le-x\_1) \\ 
 &=P(\inf\_{u\in[0,u\_1+\de]}W\_u\le-x\_1)
 -P(\inf\_{u\in[0,u\_1]}W\_u\le-x\_1) \\ 
 &=2G\Big(\frac{x\_1}{\sqrt{u\_1+\de}}\Big)-2G\Big(\frac{x\_1}{\sqrt{u\_1}}\Big) \\ 
 &\le\frac\de{x\_1^2}\le\frac\de{x\_0^2}\le\frac{2h\ep}{x\_0^2}. 
\end{aligned}
\tag{11} 
\end{equation\*}
The first inequality in (11) follows because the function $\tau\_1^{-1}$ is increasing and $x\_0+\tau\_1^{-1}(u\_1)=x\_1$.
The fourth, last equality in (11) follows by the reflection principle.  

The second inequality there follows because $\frac{\partial}{\partial u}G\big(\frac x{\sqrt u}\big)\le\frac1{2x^2}$ all real $x>0$ and $u>0$.
The last inequality in (11) follows by (7), since $\de=\tau\_{\max}(t)-\tau\_{\min}(t)$.


Collecting (5), (9), (11), and (10), for any real $h\in(0,1)$ such that $\frac{2h}{x\_0^2}+2C\sqrt h<1$, we have
\begin{equation\*}
 \ep=\|f\_1-f\_2\|=\|F(f\_1)-F(f\_2)\|\le\|D\_1\|+\|D\_2\|\le\Big(\frac{2h}{x\_0^2}+2C\sqrt h\Big)\ep<\ep,
\end{equation\*}
which is the mentioned desired contradiction with (5).


So, $\ep=0$, which means that $f\_1(s)=f\_2(s)\ \forall s\in[0,t\_0+h)$, which contradicts (4). This final contradiction shows that $t\_0=\infty$ in (4), and we are done.


