Provided a probability density $\rho:\mathbb R_+\to\mathbb R_+$ (as nice as possible), consider the PDE

\begin{eqnarray} &&\partial_t p = \frac{\partial_{xx}p}{2\big(1+m(t)\big)^2} - \partial_x p,~ \forall t>0, ~ x>0;\quad p(0,x)=\rho(x),~ \forall x\ge 0;\quad p(t,0)=0,~ \forall t>0 \\ && m(t)=\int_0^{\infty}p(t,x)dx,~ \forall t\ge 0. \end{eqnarray}

Set

$$\Phi(z):=\int_{-\infty}^z\frac{1}{\sqrt{2\pi}} e^{-r^2/2}dr,\quad \forall z\in\mathbb R.$$

Define

$$ \color{blue}{p(s,y) := \int_0^{\infty}\rho(x)g(0,x,s,y)dx + \int_0^s \dot{m}(t)g(t,0,s,y)ds,}\quad\quad\quad\quad\quad\quad (\ast 1) $$

where

$$g(t,x,s,y):=\frac{1}{\sqrt{2\pi A_{t,s}}}\exp\left(-\frac{\big(y-x-(s-t)\big)^2}{2A_{t,s}}\right),~ \forall t\in [0,s),~ x,y\in\mathbb R,\quad \mbox{with } A_{t,s}:=\int_t^s\frac{du}{\big(1+m(u)\big)^2}$$

and $m$ satisfies

$$\color{blue}{m(s)=\int_0^{\infty}\Phi\left(\frac{x+s}{\sqrt{A_{0,s}}}\right)\rho(x)dx + \int_0^{s}\Phi\left(\frac{s-t}{\sqrt{A_{t,s}}}\right)\dot{m}(t)dt,~ \forall s\ge 0.}\quad\quad(\ast2)$$

Assuming the existence of $m$ s.t. $(\ast2)$ holds, can we verify $(p,m)$ is a solution to the above system?

Any answer, comments and references are highly appreciated.

PS : Let me describe in details where come from the above expressions in blue. First, integrating the PDE

$$\int_0^{\infty}\partial_t p(t,x)dx = \int_0^{\infty} \frac{\partial_{xx}p(t,x)}{2\big(1+m(t)\big)^2}dx - \int_0^{\infty} \partial_x p(t,x)dx$$

together with the boundary condition $p(t,0)=0$, one has

$$\dot{m}(t)=-\frac{\partial_{x}p(t,0)}{2\big(1+m(t)\big)^2},\quad \forall t>0.\quad\quad\quad\quad \quad (\star)$$

Next, using the fact

\begin{eqnarray} \partial_t g = -\frac{\partial_{xx}g}{2\big(1+m(t)\big)^2} - \partial_x g,~ \forall t\in [0,s), ~ x\in\mathbb R;\quad g(s,x)=\delta_y(x),~ \forall x\in\mathbb R, \end{eqnarray}

one may check

$$\partial_t(pg) + \partial_x(pg)- \frac{1}{2\big(1+m(t)\big)^2}\partial_x(\partial_x pg-\partial_x gp)=0,\quad \forall t\in (0,s),~ x>0.$$

Therefore, integrating this equality on $(0,s)\times (0,\infty)$ yields

$$\int_0^{\infty}\left(\int_0^s \partial_t(pg)dt\right)dx + \int_0^s \left( \int_0^{\infty}\partial_x(pg)dx\right)dt - \frac{1}{2\big(1+m(t)\big)^2}\int_0^s \left(\int_0^{\infty} \partial_x(\partial_x pg-\partial_x gp)dx\right)dt=0$$

and further by $p(0,\cdot)=\rho$ and $p(\cdot,0)=p(\cdot,\infty)=\partial_xp(\cdot,\infty)=0$

\begin{eqnarray} 0&=&\int_0^{\infty} \big(p(s,x)\delta_y(x)-\rho(x)g(0,x,s,y)\big)dx + \frac{1}{2\big(1+m(t)\big)^2}\int_0^s \partial_x p(t,0)g(t,0,s,y) dt \\ &=&p(s,y)- \int_0^{\infty} \rho(x)g(0,x,s,y)dx + \frac{1}{2\big(1+m(t)\big)^2}\int_0^s \partial_x p(t,0)g(t,0,s,y) dt \\ &=&p(s,y)- \int_0^{\infty} \rho(x)g(0,x,s,y)dx -\int_0^s \dot{m}(t)g(t,0,s,y) dt\quad \quad \mbox{ in view of } (\star), \\ \end{eqnarray}

which gives $(\ast 1)$. Integrating $(\ast 1)$ over $(0,\infty)$, one has

$$m(s)=\int_0^{\infty}\left(\int_0^{\infty}g(0,x,s,y)dy\right)\rho(x)dx -\int_0^s \left(\int_0^{\infty}g(t,0,s,y)dy\right)\dot{m}(t) dt,$$

which gives $(\ast 2)$.