# Continuity equation $\partial_t \mu_t+\operatorname{div} (v_t \mu_t)=0$: are these two notions of weak solution equivalent?

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})

• Thank you @Daniele so much for your editorial service! Oct 11, 2023 at 19:01
• 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. Feb 2 at 1:45

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?