= Solution
For a candidate curve $\rho$, define the <velocity field> by the <convolution>
$$
b_\rho(t,x)=K*\rho_t(x)=-\int J\nabla W(x-z)\,d\rho_t(z).
$$
It is well defined with a finite first <absolute moment>, because $|K(x-z)|\leq L(|x|+|z|)$. Equality of mixed <partial derivatives> gives
$$
\operatorname{div}K=\partial_xW_y-\partial_yW_x=0,\qquad\operatorname{div}b_\rho=0.
$$
The <Hessian matrix> bound justifies moving a spatial derivative through this <convolution>. For smooth densities, the <transport equation> consequently becomes the <continuity equation>
$$
\boxed{\partial_t\rho+\operatorname{div}(b_\rho\rho)=0.}
$$
This form defines the equation for singular <probability measures> as well. With initial <probability measure> $\rho_0$, a <weak solution> is a curve in $C([0,\infty),\mathcal P_1)$ such that every $\varphi\in C_c^1([0,\infty)\times\mathbb R^2)$ satisfies
$$
\boxed{\int_0^\infty\!\int\left[\partial_t\varphi(t,x)+b_\rho(t,x)\cdot\nabla\varphi(t,x)\right]d\rho_t(x)\,dt+\int\varphi(0,x)\,d\rho_0(x)=0.}
$$
Using smooth <test functions> gives the same definition by approximation on compact sets. Continuity in the <Wasserstein distance> bounds first <absolute moments> on compact time intervals, so the displayed integrals exist.
For the <empirical measure>, the <chain rule> and the particle equations give
$$
\begin{aligned}
\frac d{dt}\int\varphi(t,x)\,d\mu_t^N(x)&=\frac1N\sum_i\left[\partial_t\varphi(t,X^i)+\nabla\varphi(t,X^i)\cdot\dot X^i\right]\\
&=\int\left[\partial_t\varphi+b_{\mu^N}\cdot\nabla\varphi\right]d\mu_t^N.
\end{aligned}
$$
Integrating in time, with zero terminal contribution because $\varphi$ has compact <support>, gives precisely the weak identity and the correct initial <empirical measure>. Thus \b[$\mu^N$ is a global weak solution]. The self-interaction term vanishes since $K(0)=0$.
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