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

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

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 $(\ast 2)$.

Provided a probability density $\rho:\mathbb R_+\to\mathbb R_+$ (as nice as possible), consider the PDE $$ \begin{cases} \partial_t p = \dfrac{\partial_{xx}p}{2\big(1+m(t)\big)^2} - \partial_x p, & \forall t>0, ~ x>0;\\ m(t) =\displaystyle \int\limits_0^{\infty}p(t,x)dx,& \forall t\ge 0;\\ p(0,x)=\rho(x),& \forall x\ge 0;\\ p(t,0)=0,&\forall t>0. \end{cases} $$ Set $$ \Phi(z):=\int\limits_{-\infty}^z\frac{1}{\sqrt{2\pi}} e^{-r^2/2}dr,\quad \forall z\in\mathbb R. $$ Define $$ \color{blue}{p(s,y) := \int\limits_0^{\infty}\rho(x)g(0,x,s,y)dx + \int\limits_0^s \dot{m}(t)g(t,0,s,y)dt,} \label{1}\tag{$\ast$1} $$ where $$ \begin{split} g(t,x,s,y)&:=\frac{1}{\sqrt{2\pi A_{t,s}}}\exp\left(-\frac{\big(y-x-(s-t)\big)^2}{2A_{t,s}}\right)\\ \mbox{with } A_{t,s}&:=\int\limits_t^s\frac{du}{\big(1+m(u)\big)^2} \end{split}\quad \forall t\in [0,s),~ x,y\in\mathbb R, $$ and $m$ satisfies

$$\color{blue}{m(s)=\int\limits_0^{\infty}\Phi\left(\frac{x+s}{\sqrt{A_{0,s}}}\right)\rho(x)dx + \int\limits_0^{s}\Phi\left(\frac{s-t}{\sqrt{A_{t,s}}}\right)\dot{m}(t)dt,~ \forall s\ge 0.}\label{2}\tag{$\ast$2} $$

Assuming the existence of $m$ s.t. \eqref{2} holds, can we verify $(p,m)$ is a solution to the above system?

Any answer, comments and references are highly appreciated.

P.S. Let me describe in details where come from the above expressions in blue. First, integrating the PDE $$ \int\limits_0^{\infty}\partial_t p(t,x)dx = \int\limits_0^{\infty} \frac{\partial_{xx}p(t,x)}{2\big(1+m(t)\big)^2}dx - \int\limits_0^{\infty} \partial_x p(t,x)dx $$ together with the boundary condition $p(t,0)=0$, one has $$ \dot{m}(t)=-\frac{\partial_{x}p(t,0)}{2\big(1+m(t)\big)^2},\quad \forall t>0.\label{3}\tag{$\star$} $$ Next, using the fact $$ \begin{cases} \partial_t g = -\dfrac{\partial_{xx}g}{2\big(1+m(t)\big)^2} - \partial_x g, &\forall t\in [0,s), ~ x\in\mathbb R;\\ g(s,x)=\delta_y(x), &\forall x\in\mathbb R, \end{cases} $$ 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 \eqref{3}}, \\ \end{eqnarray}

which gives \eqref{1}. Integrating \eqref{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 \eqref{2}.

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 $(\ast 2)$.

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

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

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 $(\ast 2)$.