The question is based on the McKean-Vlasov problem formulated in the paper "A McKean-Vlasov equation with positive feedback and blow-ups", namely:
\begin{equation} \begin{cases} X_t = X_0 + B_t -\alpha L_t \\ \tau = \inf\{t\geq 0 : X_t\leq 0\} \\ L_t = \mathbb{P}(\tau\leq t) \end{cases} \end{equation}
which is in this case supposed to model the distance-to-default of a bank for constant $\alpha$. $\tau$ then is the stopping time for when the distance-to-default hits 0.
Now I'd like to introduce you to the actual problem:
Taking $\nu_t$ with $\nu_t(\phi):=\mathbb{E}[\phi(X_t)\mathbb{1}_{t<\tau}]$ for $\phi \in C^2$ and $\phi(0)=0$, $t\geq 0$. An application of Itô's formula is supposed to give
\begin{equation}
\begin{cases}
\nu_t(\phi)=\nu_0(\phi)+\frac{1}{2}\int_0^t\nu_s(\phi'')ds-\alpha\int_0^t\nu_s(\phi')dL_s\\
L_s = 1-\int_0^\infty\nu_s(dx)
\end{cases}
\end{equation}
I have two questions:
1) Since the dynamics of $X_t$ should be like "$dX_t=dB_t-\alpha dL_t$". I wonder where the $B_t$ dynamics went in the second part.
2) The authors of the paper proved that the density $V_t$ of $\nu_t$ exists and say that by integrating by parts in the above formula for $\nu_t$ one gets: \begin{equation} \partial_t V_t(x) = \frac{1}{2}\partial_{xx}V_t(x)+\alpha L'_t\partial_xV_t(x),\; L'_t=\frac{1}{2}\partial_xV_t(0) \end{equation} How does one integrate this by parts (I guess in the Itô kind of definition) to get the above equation?
Especially the second question is a concern of mine as I haven't dealt with such problems in a long time and although seeming kind of obvious I lack intuition and try to get it back solving and looking at solutions, respectively.
Thanks in advance for answers and I apologize if any insufficient specifications have been made.
So I've got an answer to the first question:
Applying Itô's formula to $\phi(X_t)$ one gets: \begin{align*} d\phi(X_t)\textbf{1}_{t<\tau}&=\frac{1}{2}\partial_{xx}\phi(X_t)\textbf{1}_{t<\tau}d\langle X\rangle_t +\partial_x\phi(X_t)\textbf{1}_{t<\tau}dX_t\\ &= \frac{1}{2}\partial_{xx}\phi(X_t)\textbf{1}_{t<\tau}d t + \partial_x\phi(X_t)\textbf{1}_{t<\tau}dB_t-\partial_x\phi(X_t)\textbf{1}_{t<\tau}\alpha dL_t \end{align*}
Obtaining $\nu_t(\phi)$ by taking expectations, the stochastic integral w.r.t. $B_t$ vanishes as one can interchange the expectation and integrals (under certain circumstances which are fullfilled here) like: \begin{align*} \mathbb{E}[\int_0^t\partial_x\phi(X_s)\textbf{1}_{t<\tau}\alpha dL_s]=\int_0^t\mathbb{E}[\partial_x\phi(X_s)\textbf{1}]\alpha dL_s \end{align*} and the expected value of a stochastic integral w.r.t. Brownian motions is 0.