Let $X_t = x + bt + \sigma W_t$ be an arithmetic Brownian motion, where $x$ is a random variable independent to $W$, and $\sigma>0$. Suppose the initial distribution is given by $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ can be solved from the following Fokker-Planck equation:
$$\partial_t m = G^* m, \quad m(0, y) = m_0(y),$$
where $$G^* f(t, y) = (- b \partial_y f + \frac 1 2 \sigma^2 \partial_{yy} f)(t, y).$$

Now, Let $T = \inf\{t>0, X_t \notin (0, 1)\}$ be the first exit time of $X$ from $(0, 1)$, and $Y_t$ is the process $X_t$ truncated by $T$, i.e. 
$$Y_t = X_{t \wedge T}.$$

[Q.] Is the density $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ can be solved by the following equation?
$$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, 0) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

Thanks.