Let

$$Y_t:=1+\int_0^t b(s)ds + W_t,\quad\forall t\ge 0,$$

where $b:\mathbb R_+\to[1,2]$ is continuous and $(W_t)_{t\ge 0}$ is a standard Brownian motion. Denote $\tau:=\{t\ge 0: Y_t\le 0\}$ and $X_t:=Y_{t\wedge \tau}$. It is known from On the marginal distributions of an absorbed diffusion that the law of $X_t$, denoted by $\mu_t$, has the following decomposition:

$$\mu_t(dx) = \alpha(t)\delta_0(dx) + p_t(x)dx,\quad \forall t>0.$$

Can we show (under suitable conditions) $p_t(0+):=\lim_{x\to 0+}p_t(x)=0$ for every $t>0$?

PS : When $b$ is constant, e.g. $b\equiv 1$, we have

$$\int_x^{\infty}p_t(y)dy = \mathbb P[X_t>x] = \mathbb P[\inf_{0\le s\le t}Y_s>0, Y_t>x]= \mathbb P[\sup_{0\le s\le t}(-s+W_s)<1, -t+W_t<1-x],\quad \forall t,x>0.$$

Using the joint density of the drifted Brownian Motion and its running maximum, one has

$$\mathbb P[\sup_{0\le s\le t}(-s+W_s)<1, -t+B_t<1-x]=\int_0^1 dm \int_{-\infty}^{1-x}{\bf 1}_{\{y\le m\}} e^{-t/2-y}\frac{2(2m-y)}{\sqrt{2\pi t^3}}e^{-(2m-y)^2/2t}dy,$$

which yields by differentiating with respect to $x$

$$p_t(0+)=-\lim_{x\to 0+} \frac{\partial \mathbb P[X_t>x]}{\partial x}=0.$$

Can we extend to the general function $b$? The key is to show the existence of the joint density of $(Y_t, \inf_{0\le s\le t}Y_s)$ but I do not know how prove it.