= Solution
Use the corrected <piecewise-deterministic Markov process> from part (b). For a smooth <test function> $f$, its time-dependent <infinitesimal generator> is
$$
(L_tf)(x)=a(t,x)f'(x)+\alpha(x)\left[\int_{\mathbb R}f(y)m_t(x,y)\,dy-f(x)\right].
$$
Indeed, over an interval of length $h$, deterministic motion contributes $ha(t,x)f'(x)$, while a jump has <probability> $h\alpha(x)+o(h)$ and changes the conditional mean of $f$ by $K_tf(x)-f(x)$. Terms involving both drift and a jump contribute only to the remainder. Thus $P_{t,t+h}f=f+hL_tf+o(h)$.
The <Dynkin formula>, or this infinitesimal expansion integrated against the current <probability distribution>, gives the weak evolution identity
$$
\frac d{dt}\int_{\mathbb R}f(x)p_t(x)\,dx
=\int_{\mathbb R}(L_tf)(x)p_t(x)\,dx.
$$
Take $f\in C_c^\infty(\mathbb R)$, so <integration by parts> has no boundary contribution. Assuming the displayed jump integrals are finite, <Fubini's theorem> gives
$$
\begin{aligned}
\int_{\mathbb R}a(t,x)f'(x)p_t(x)\,dx
&=-\int_{\mathbb R}f(x)\partial_x\bigl(a(t,x)p_t(x)\bigr)\,dx,\\
\int_{\mathbb R}\alpha(y)p_t(y)\int_{\mathbb R}f(x)m_t(y,x)\,dx\,dy
&=\int_{\mathbb R}f(x)\int_{\mathbb R}\alpha(y)m_t(y,x)p_t(y)\,dy\,dx.
\end{aligned}
$$
Consequently the <forward equation of a piecewise-deterministic Markov process> is
$$
\boxed{\begin{aligned}
\partial_tp_t(x)=(A_tp_t)(x)
&=-\partial_x\bigl(a(t,x)p_t(x)\bigr)-\alpha(x)p_t(x)\\
&\quad+\int_{\mathbb R}\alpha(y)m_t(y,x)p_t(y)\,dy.
\end{aligned}}
$$
Here $A_t=L_t^*$ is the formal <adjoint operator> acting on <probability density functions>. Its terms are transport, loss from $x$, and incoming mass from every $y$. In particular, the incoming rate is $\alpha(y)$, evaluated at the departure state. The operator denoted $A$ in the source is generally time-dependent. Because $\int m_t(y,x)\,dx=1$, the gain and loss integrals cancel; with vanishing transport flux at infinity, $\frac d{dt}\int p_t=0$.
\b[The deterministic initial state in part (b) does not actually yield an everywhere smooth full density.] For finite times the no-jump contribution is the positive measure
$$
\exp\left(-\int_0^t\alpha(\varphi(0,y_0,u))\,du\right)
\delta_{\varphi(0,y_0,t)},
$$
where $\delta_z$ is the <Dirac measure> at $z$. Smooth jump destination <probability density functions> cannot remove this positive atom. The density equation above is therefore conditional on the extra density hypothesis, for example with a suitable random initial <probability density function>. For the specified fixed initial state, write $\nu_t$ for the whole <probability measure> and use instead
$$
\boxed{\frac d{dt}\int f\,d\nu_t=\int L_tf\,d\nu_t,\qquad \nu_0=\delta_{y_0}.}
$$
This weak measure evolution includes both the atom and any continuous component. It becomes the displayed <partial differential equation> whenever $\nu_t(dx)=p_t(x)\,dx$ with the assumed regularity. It does not describe the literal hidden-reset construction without the correction in part (b).
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