Continuity equation: how to prove that these two notions of weak solution are equivalent?

Question

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 $$ \begin{align*} \int_0^1 \int_\Omega \partial_t \phi_t ( x) \, \mathrm d \mu_t (x) \, \mathrm d t& + \int_0^1 \int_\Omega \nabla \phi_t (x) \cdot v_t (x) \, \mathrm d \mu_t (x) \, \mathrm d t = 0,\\ &\forall \phi \in C^\infty_c ((0, 1) \times \Omega), \end{align*} \label{1}\tag{1} $$ where $\phi_t := \phi (t, \cdot)$.

• At page $185$ of Ambrosio's book Lectures on Optimal Transport, the author defines the weak solution of \eqref{a} as $$ \begin{align*} \frac{\mathrm d}{\mathrm d t} \int_\Omega \phi ( x) \, \mathrm d \mu_t (x) & + \int_\Omega \nabla \phi (x) \cdot v_t (x) \, \mathrm d \mu_t (x) = 0,\\ &\forall \phi \in C^\infty_c (\Omega). \end{align*}\label{2}\tag{2} $$

I would like to prove that the formulations (\ref{1}, \ref{2}) are equivalent. Could you please how to prove that (\ref{2}) implies (\ref{1})? Thank you so much for your help§!

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1. (\ref{1}) implies (\ref{2})

Fix $\phi \in C^\infty_c (\Omega)$ and $\varphi \in C^\infty_c ((0, 1))$. Then the map $\bar \phi :(t, x) \mapsto \varphi (t) \phi (x)$ belongs to $C^\infty_c ((0, 1) \times \Omega)$. By (\ref{1}) and integration by parts, $$ \begin{align*} \int_0^1 \int_\Omega \partial_t \bar \phi_t ( x) \, \mathrm d \mu_t (x) \, \mathrm d t& + \int_0^1 \int_\Omega \nabla \bar \phi_t (x) \cdot v_t (x) \, \mathrm d \mu_t (x) \, \mathrm d t &= 0\\ \implies \int_0^1 \varphi' (t) \int_\Omega \phi ( x) \, \mathrm d \mu_t (x) \, \mathrm d t& + \int_0^1 \varphi (t) \int_\Omega \nabla \phi (x) \cdot v_t (x) \, \mathrm d \mu_t (x) \, \mathrm d t &= 0\\ \implies \int_0^1 \varphi (t) \frac{\mathrm d}{\mathrm d t} \left ( \int_\Omega \phi ( x) \, \mathrm d \mu_t (x) \right ) \mathrm d t& + \int_0^1 \varphi (t) \left ( \int_\Omega \nabla \phi (x) \cdot v_t (x) \, \mathrm d \mu_t (x) \right ) \mathrm d t &= 0. \end{align*} $$

Because $\varphi \in C^\infty_c ((0, 1))$ is arbitrary, we get $$ \frac{\mathrm d}{\mathrm d t} \left ( \int_\Omega \phi ( x) \, \mathrm d \mu_t (x) \right ) + \int_\Omega \nabla \phi (x) \cdot v_t (x) \, \mathrm d \mu_t (x) = 0 $$ for a.e. $t \in (0, 1)$.

Answer 1

About your proof (1) $\Rightarrow$ (2): you show in the second line that $$ \int_0^1 \varphi'(t) \int_\Omega \phi(x) \mathrm{d}\mu_t(x) \mathrm{d}t = \int_0^1 \varphi(t) \int_\Omega \nabla\phi(x)\cdot v_t(x) \mathrm{d}\mu_t(x) \mathrm{d}t \quad \forall \varphi \in C^\infty_c(0,1). $$ This is already enough to claim that $t\mapsto\int_\Omega \phi(x) \mathrm{d}\mu_t(x)$ has a weak derivative! (And this derivative must equal $\int_\Omega \nabla\phi(x)\cdot v_t(x) \mathrm{d}\mu_t(x)$ for a.e. $t \in (0,1)$.)

In particular, I think your last line is technically not correct a priori due to the weak derivative of $t\mapsto\int_\Omega \phi(x) \mathrm{d}\mu_t(x)$ not existing yet (but it does by the argument above). Correct me if I am wrong here.

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The proof (2) $\Rightarrow$ (1):

Let $\bar\phi \in C^\infty_c((0,1)\times\Omega)$.

Key observation: The following 'integration by parts' formula holds: $$ \int_0^1\frac{\mathrm{d}}{\mathrm{d}t} \left(\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_t(x)\right)_{|s=t} \mathrm{d}t = \int_0^1 \int_\Omega \partial_t\bar\phi_t(x) \mathrm{d}\mu_t(x) \mathrm{d}t. $$ Proof of observation: We will employ partial summation. Let $h > 0$ be small and $s\in(0,1-h)$, then, by the Fundamental Theorem of the Calculus,

\begin{eqnarray} \int_s^{s+h}\frac{\mathrm{d}}{\mathrm{d}t} \left(\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_t(x)\right)\mathrm{d}t& = \left[\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_t(x)\right]_{t=s}^{t=s+h}. \end{eqnarray}

Multiplying by $1/h$ and integrating over $s\in(0,1-h)$ yields \begin{eqnarray} &\frac{1}{h}& \int_0^{1-h} \int_s^{s+h} \frac{\mathrm{d}}{\mathrm{d}t} \left(\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_t(x)\right)\mathrm{d}t \mathrm{d}s \\ & = & \frac{1}{h} \int_0^{1-h} \left( \int_\Omega \bar\phi_s(x) \mathrm{d}\mu_{s+h}(x) - \int_\Omega \bar\phi_s(x) \mathrm{d}\mu_s(x) \right)\mathrm{d}s \\ & = & \frac{1}{h} \int_0^{h} \int_\Omega \bar\phi_s(x) \mathrm{d}\mu_{s+h}(x) + \frac{1}{h} \int_h^{1-h} \int_\Omega \bar\phi_s(x) \mathrm{d}\mu_{s+h}(x) \mathrm{d}s - \frac{1}{h} \int_0^{1-h}\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_s(x) \mathrm{d}s \end{eqnarray} We take $h>0$ small enough such that $\phi_s\equiv 0$ on $[0,2h)$ and $(1-2h,1]$. In particular, the first term vanishes. For the second term, we use the substitution $\tilde{s} = s+h$ and rewrite $s=\tilde{s}$. Then, we combine it with the last term to obtain \begin{eqnarray} & \frac{1}{h} & \int_0^{1-h} \int_s^{s+h} \frac{\mathrm{d}}{\mathrm{d}t} \left(\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_t(x)\right)\mathrm{d}t \mathrm{d}s \\ & = & \frac{1}{h} \int_{2h}^{1} \int_\Omega \bar\phi_{s-h}(x) \mathrm{d}\mu_{s}(x) \mathrm{d}s - \frac{1}{h} \int_0^{1-h}\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_s(x) \mathrm{d}s \\ & = & \frac{1}{h} \int_h^{1} \int_\Omega \bar\phi_{s-h}(x) \mathrm{d}\mu_{s}(x) \mathrm{d}s - \frac{1}{h} \int_h^{1}\int_\Omega \bar\phi_s(x) \mathrm{d}\mu_s(x) \mathrm{d}s \\ & = & \int_h^{1} \int_\Omega \frac{1}{h}\left(\bar\phi_{s-h}(x) - \bar\phi_s(x)\right) \mathrm{d}\mu_{s}(x) \mathrm{d}s. \end{eqnarray} Again, in the second step, we used that $\bar\phi_s$ vanishes for $s \in [0,2h)\cup(1-2h,1]$.

Now, all integrands in the integrals over $s$ are uniformly bounded w.r.t. $h$. In the limit $h\to 0$ we obtain the desired statement. Q.E.D.

It should be clear how to obtain your result; apply equation (2) for test function $\bar\phi_t$, $t\in(0,1)$ fixed, and integrate over $t\in(0,1)$.