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Let $\Omega$ be an open connected convex subset of $\mathbb R^d$. Let $\mathcal P (\Omega)$ be the space of Borel probability measures on $\Omega$. Let $C_0 (\Omega)$ be the space of real-valued continuous functions on $\Omega$ that vanish at infinity. We endow $\mathcal P (\Omega)$ with the topology of weak$^*$ convergence, i.e., $\mu_n \to \mu$ if and only if $\int \varphi \, \mathrm d \mu_n \to \int \varphi \, \mathrm d \mu$ for every $\varphi \in C_0 (\Omega)$.

Let $\mu : [0, T] \to \mathcal P (\Omega), t \mapsto \mu_t$ be continuous. We fix a Borel vector field $v:[0, 1] \times \Omega \to \mathbb R^d$ such that $v_t := v(t, \cdot) \in L^1 (\Omega, \mu_t, \mathbb R^d)$ for all $t \in [0, 1]$.

We consider the continuity equation $$ \partial_t \mu_t+\operatorname{div} (v_t \mu_t)=0.\label{a}\tag{$\ast$} $$

  • At page 123 of Santambrogio's book Optimal Transport for Applied Mathematicians, the author defines the weak solution of \eqref{a} as $$ \begin{align} \int_0^1 \int_\Omega \partial_t \phi_t ( x) \, \mathrm d \mu_t (x) & + \int_0^1 \int_\Omega \nabla \phi_t (x) \cdot v_t (x) \, \mathrm d \mu_t (x) \, \mathrm d t = 0,\\ &\forall \phi \in C^\infty_c ((0, 1) \times \Omega). \end{align} \label{1}\tag{1} $$

  • At page 3 of this note, the author defines the weak solution of \eqref{a} as $$ \begin{align} \frac{\mathrm d}{\mathrm d t} \int_\Omega \phi_t ( x) \, \mathrm d \mu_t (x) & + \int_\Omega \nabla \phi_t (x) \cdot v_t (x) \, \mathrm d \mu_t (x) = 0,\\ & \forall \phi \in C^\infty_c ((0, 1) \times \Omega). \end{align}\label{2}\tag{2} $$

  • At page $2$ of this note, the author defines the weak solution of \eqref{a} as $$ \begin{align} \frac{\mathrm d}{\mathrm d t} \int_\Omega \phi ( x) \, \mathrm d \mu_t (x) & + \int_\Omega \nabla \phi (x) \cdot v_t (x) \, \mathrm d \mu_t (x) = 0,\\ &\forall \phi \in C^\infty_c (\Omega). \end{align}\label{3}\tag{3} $$

The formulation \eqref{3} is somehow different from (\ref{1}, \ref{2}) because the test functions depend only on $x$.

Are the formulations (\ref{2}, \ref{3}) equivalent?

Thank you so much for your elaboration!


Update I have added below the screenshot of (\ref{2})

enter image description here

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    $\begingroup$ Thank you @Daniele so much for your editorial service! $\endgroup$
    – Analyst
    Oct 11, 2023 at 19:01
  • $\begingroup$ I don't have the reputation to comment but it looks like you introduced a sign error when transcribing, both in (2) and (3). In the sources the second term is on the RHS with a + sign. $\endgroup$ Feb 2 at 1:45

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This is more a long comment than an answer, since there's a peculiarity of definition \eqref{2} that I don't understand and makes me think the two definition aren't equivalent. To start, let's consider the part containing explicitly the time derivative in definition \eqref{2}: we have $$ \begin{split} \frac{\mathrm d}{\mathrm d t} \int_\Omega \phi_t ( x) \, \mathrm d \mu_t (x) &= \lim_{h\to 0} \frac{1}{h} \int_\Omega \big[\phi_{t+h} ( x)\, \mathrm d \mu_{t+h} (x) - \phi_t ( x) \, \mathrm d \mu_t (x)\big]\\ &= \lim_{h\to 0} \frac{1}{h} \int_\Omega \big[\phi_{t+h} ( x)\, \mathrm d \mu_{t+h} (x) -\phi_{t+h} ( x)\,\mathrm d\mu_t (x) \\ &\qquad\qquad\quad\quad +\phi_{t+h} ( x)\, \mathrm d\mu_t (x) - \phi_t ( x) \, \mathrm d \mu_t (x)\big]\\ &=\lim_{h\to 0} \frac{1}{h} \int_\Omega \phi_{t+h} ( x)\big[\mathrm d \mu_{t+h} (x) -\mathrm d\mu_t (x)\big] \\ &\qquad\qquad + \lim_{h\to 0} \frac{1}{h} \int_\Omega\Big[\phi_{t+h} ( x) - \phi_t ( x)\big] \mathrm d \mu_t (x)\\ &=\left.\frac{\mathrm d}{\mathrm d t} \int_\Omega \phi_{s} ( x)\,\mathrm d \mu_{t} (x) \right|_{s=t} + \left. \int_\Omega \frac{\mathrm d}{\mathrm d s}\phi_{s} ( x) \, \mathrm d \mu_t (x)\right|_{s=t}\\ &=\left.\frac{\mathrm d}{\mathrm d t} \int_\Omega \phi_{s} ( x)\,\mathrm d \mu_{t} (x) \right|_{s=t} + \int_\Omega \Big(\frac{\mathrm d}{\mathrm d t}\phi_{t} ( x) \Big) \, \mathrm d \mu_t (x) \end{split} $$ Loosely speaking, I got this identity by "freezing" the time dependence of the test function in the first term of the equation: this implies that if $\mu_t(x)$ is a solution of \eqref{a} according to definition \eqref{3}, then it is not a solution according to definition \eqref{2}, unless we have $$ \int_\Omega \Big(\frac{\mathrm d}{\mathrm d t}\phi_{t} ( x) \Big) \, \mathrm d \mu_t (x) \equiv 0 \quad\forall\phi \in C^\infty_c ((0, 1) \times \Omega)\label{4}\tag{vc} $$ But this leads straight to the reason for my doubt: the vanishing condition \eqref{4} cannot be satisfied unless $\mu_t(x)\equiv0$ identically!
This can be seen by constructing a class of test functions $\{_{\tiny M,N}\phi_t(x)\}_{MN\in\Bbb N}$ by using any smooth partition of unity of the domain $\Omega$, say $\{\psi_k(x)\}_{k\in \Bbb N}$ and any smooth partition of unity of the closed interval $[0,1]$ say $\{\chi_n(t)\}_{n\in \Bbb N}$. Formally speaking, $$ {_{\tiny M,N}\phi}_t(x) = t\sum_{n=1}^N\chi_n(t)\sum_{k=1}^M\psi_k(x) \quad M, N\in\Bbb N. $$ Am I wrong?

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    $\begingroup$ I have added a screenshot of (\ref{2}). The author of (2) possibly meant (\ref{3}) but had a typo in the intended space of test functions. $\endgroup$
    – Analyst
    Oct 12, 2023 at 7:09
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    $\begingroup$ @Analyst I agree with you: it should be a typo. $\endgroup$ Oct 12, 2023 at 7:13

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