In the following $\mu: [0,1] \to P_2(\mathbb{R}^d)$ is a curve in space of probability measures having finite second moment in $\mathbb{R}^d$ and $v:[0,1] \times \mathbb{R}^d \to \mathbb{R}^d$ is a vector field. The continuity equation is $$ \partial_t \mu_t + \nabla\cdot (\mu_t v_t) =0 \\ \mu(0)=\mu_0 $$ in the sense of distribution.
Suppose that the for every $T>0$ and $\psi: \mathbb{R}^+ \times\mathbb{R}^d \to \mathbb{R}$ the transport equation TE admits a solution $f$, where TE is $$ \frac{d}{dt} f+ v\cdot \nabla f = \psi \\ f(T,\cdot)=0 $$
Then prove that the continuity equation admits at most one absolutely continuous solution in the sense of distribution.
I do not know where to start, I wrote the continuity equation as the condition $ \forall \phi \in C_c^\infty (\mathbb{R}^d) $ $$ \frac{d}{dt} \int \phi \, d\mu_t = \int \nabla \phi\, \cdot v_t d\mu_t $$ and try to think of a smart choice of $\psi$ to get something interesting, but had no ideas. Any help?
I'm not really a PDE guy, so I'm deliberately neglecting regularity issues. Yet, I guess a rigorous proof will rely on an argument as follows:
Assume there is another solution to the CE, say $\mu'$. Then by linearity the difference $\nu=\mu-\mu'$ also satisfies CE, but with $\nu(0)=0$. Now let $f$ be a solution to TE. Then \begin{equation} \frac{d}{dt}\int f_{t}d\nu_{t}=\int(\dot{f_{t}}+\nabla f_{t}\cdot v_{t})d\nu_{t}=\int\psi d\nu_{t}. \end{equation} Integrate from $0$ to $T$ you observe that \begin{equation} 0=\int_{0}^{T}\int\psi d\nu_{t}. \end{equation} This being true for all $T>0$ and $\psi$ should imply $\nu\equiv 0$ which yields the claim.