Let $\rho:\mathbb R^d\times[0,\infty)\to(0,\infty)$ such that $\int \rho_t(x)\,dx=1$ for all $t\geq0\,$, $\rho$ is Holder-continuous (in both variables) and $\rho_t\in W^{1,1}(\mathbb R^d)$ for a.e. $t>0$.

Let $A$ be a positive definite symmetric matrix and $b\in C^1(\mathbb R^d;\mathbb R^d)$. Suppose that $\rho$ solves the following weak version of a parabolic PDE:
$$
\begin{align} \label{1}\tag{1} & \int_{\mathbb R^d} \varphi_t(x)\,\rho_t(x)\,dx \,-\, \int_{\mathbb R^d} \varphi_0(x)\,\rho_0(x)\,d x \;=\\[5pt] &=\, -\,\lim_{\epsilon\to0}\int_\epsilon^t\!\!\int_{\mathbb R^d} \Big(\partial_s\varphi_s(x)\,\rho_s(x) \,+\, A\,\nabla\!\varphi_s(x)\cdot\nabla\!\rho_s(x) \,+\, b(x)\cdot\nabla\!\varphi_s(x)\,\rho_s(x)\Big)\,d x\,d s 
\end{align}
$$
for all $t>0$, for all $\varphi\in C(\mathbb R^d\!\times\![0,\infty))$ such that, for example, $\varphi_s(x)=\zeta(s)\,\psi(x)$ with $\zeta\in C^\infty(0,\infty)$ and $\psi\in C^\infty_c(\mathbb R^d)\,$.

**My question:** Is the identity \eqref{1} still true when $\varphi$ is replaced by $\log\rho$ ? Precisely:
\begin{align} \label{1b}\tag{1'} & \int_{\mathbb R^d} \log\rho_t(x)\,\rho_t(x)\,dx \,-\, \int_{\mathbb R^d} \log\rho_0(x)\,\rho_0(x)\,d x \;=\\[5pt] &=\, -\,\int_0^t\!\!\int_{\mathbb R^d} \Big( A\,\frac{\nabla\!\rho_s(x)}{\rho_s(x)}\cdot\nabla\!\rho_s(x) \,+\, b(x)\cdot\nabla\!\rho_s(x) \Big)\,d x\,d s \quad ?\end{align}
I know that:
$$ \int_0^t\!\!\int_{\mathbb R^d} |b(x)|^2\,\rho_s(x) \,d x\,d s \ <\infty \;;\qquad\qquad\int_0^t\!\!\int_{\mathbb R^d} \frac{|\nabla\rho_s(x)|^2}{\rho_s(x)} \,d x\, ds\ <\infty \;$$
so the integrals in \eqref{1b} are absolutely convergent.

Can we prove \eqref{1b} by some approximation argument?

I also know from theory that there exists $\partial_s\rho$ bounded in the *dual space of* $\mathbb H_0^{1,1}(B\times I)$ (*see note at the end) for every $B\times I$ compactly contained in $\mathbb R^d\!\times\!(0,\infty)$. If I'm not wrong this implies $\partial_s\rho\in L^\infty_\textrm{loc}(\mathbb R^d\times(0,\infty))\,$. Moreover I know that $\|\rho_\epsilon-\rho_0\|_{L^1(\mathbb R^d)}\to0\ $ as $\epsilon\to0\,$.
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Regarding the time-derivative term on the r.h.s. of \eqref{1}, it becomes $0$ in \eqref{1b} since I expect:
$$ \label{2}\tag{2} \lim_{\epsilon\to0}\int_\epsilon^t\!\!\int_{\mathbb R^d} \partial_s\rho_s(x)\,d x\,d s \;=\, 0 \;.$$
Observe that choosing a sequence $\psi_N\in C_c(\mathbb R^d)$ such that $\psi_N(x)=1$ for $|x|\leq N\,$, $0\leq\psi_N\leq1$, we have (using Fubini theorem since $\partial_s\rho\in L^1_\textrm{loc}(\mathbb R^d\!\times\!(0,\infty))$ ):
$$ \int_\epsilon^t\!\!\int_{\mathbb R^d} \psi_N(x)\,\partial_s\rho_s(x)\,d x\,d s \,=
\int_{\mathbb R^d} \psi_N(x) \!\int_\epsilon^t\!\!\partial_s\rho_s(x)\,d s\,d x \,=
\int_{\mathbb R^d} \psi_N(x)\big(\rho_t(x) - \rho_\epsilon(x)\big) \,d x $$
which in absolute value is bounded by
$$ \Big| \int_{\mathbb R^d} \psi_N(x)\,\rho_t(x)\, dx \,-\, \int_{\mathbb R^d} \psi_N(x)\,\rho_0(x) \,d x \,\Big| \,+\, \|\rho_\epsilon-\rho_0\|_{L^1(\mathbb R^d)} \,\xrightarrow[\substack{N\to\infty,\,\epsilon\to0}]{}\, 0 $$
by dominated convergence in $N$ and what we know about $\|\rho_\epsilon-\rho_0\|_{L^1(\mathbb R^d)}$.
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(*) $\mathbb H_0^{1,1}(B\times I)$ denotes the space of functions $v:B\times I\to\mathbb R$ such that $v_s\in W^{1,1}_0(B)$ for a.e. $s\in I$ and $\int_I\int_B(|v_s(x)|+|\nabla v_s(x)|)\,d x\,d s<\infty$.