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: $$ \begin{align} \label{1}\tag{1} & \int_{\mathbb R^d} \varphi_t(x)\,\rho_t(x)\,dx \,-\, \int_{\mathbb R^d} \varphi_0(x)\,\rho_0(x)\,d x \;=\\[5pt] &=\, \lim_{\epsilon\to0}\int_\epsilon^t\!\!\int_{\mathbb R^d} \Big(\partial_s\varphi_s(x) - \big(A\,\nabla\!\log\rho_s(x) \,+\, b(x)\big)\cdot\nabla\!\varphi_s(x)\,\Big)\,\rho_s(x)\;d x\,d s \end{align} $$ for all $t>0$, for all $\varphi\in C(\mathbb R^d\!\times\![0,\infty))\cap C^{2,1}(\mathbb R^d\!\times\!(0,\infty))$ such that $\textrm{supp}\,\varphi_t\subseteq K$ compact subset of $\mathbb R^d$ (the same $K$ for all $t$).
My question: Is the identity \eqref{1} still true when $\varphi$ is replaced by $\log\frac{\rho}{R}$ for a suitable $R\in C^\infty(\mathbb R^d)$, $R>0$, $\int R(x)\,dx=1$? Precisely: \begin{align} \label{1b}\tag{1'} & \int_{\mathbb R^d} \log\frac{\rho_t(x)}{R(x)}\,\rho_t(x)\,dx \,-\, \int_{\mathbb R^d} \log\frac{\rho_0(x)}{R(x)}\,\rho_0(x)\,d x \;=\\[5pt] &=\, -\int_0^t\!\!\int_{\mathbb R^d} \big(A\,\nabla\!\log\rho_s(x) \,+\, b(x)\big)\cdot\nabla\!\log\frac{\rho_s(x)}{R(x)}\,\rho_s(x)\;d x\,d s \quad ?\end{align} I know that for all $t\geq0$ : $$ \int_0^t\!\!\int_{\mathbb R^d} |b(x)|^2\,\rho_s(x) \,d x\,d s \ <\infty \;;\qquad\qquad \int_0^t\!\!\int_{\mathbb R^d} \frac{|\nabla\rho_s(x)|^2}{\rho_s(x)} \,d x\, ds\ <\infty \;;$$ and $$ \int_{\mathbb R^d} |\log\rho_t(x)|^2\,\rho_t(x)\,dx\,<\infty\;;\quad \int_{\mathbb R^d} |\log R(x)|^2\,\rho_t(x)\,dx\,<\infty\;;\quad \int_0^t\!\!\int_{\mathbb R^d} |\nabla\!\log R(x)|^2\,\rho_s(x) \,d x\,d s \ <\infty \;;$$ 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 $\epsilon\to0\,$.
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 $$ \Big| \int_{\mathbb R^d} \psi_N(x)\,\rho_t(x)\, dx \,-\, \int_{\mathbb R^d} \psi_N(x)\,\rho_0(x) \,d x \,\Big| \,+\, \|\rho_\epsilon-\rho_0\|_{L^1(\mathbb R^d)} $$ where the first quantity vanishes as $N\to\infty$ by dominated convergence (and does not depend on $\epsilon$), and we know the second quantity vanishes as $\epsilon\to0$. In particular we have proved that: $$ \tag{2'} \lim_{N\to\infty}\limsup_{\epsilon\to0}\Big|\int_\epsilon^t\!\!\int_{\mathbb R^d} \psi_N(x)\,\partial_s\rho_s(x)\,d x\,d s\Big| \,=\, 0 \;.$$
(*) $\mathbb H_0^{1,1}(B\times I)$ 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 $\int_I\int_B(|v_s(x)|+|\nabla v_s(x)|)\,d x\,d s<\infty$.
