Solution (source code)

= Solution

Normalize the symmetric $N$-particle <phase-space distribution function> $D_N$ to one and write $z_i=(\mathbf x_i,\mathbf v_i)$, $dz_i=d^3x_i\,d^3v_i$. The number-normalized one- and <two-particle distribution functions> are
$$
f(z_1,t)=N\int D_N\,dz_2\cdots dz_N,\qquad f_2(z_1,z_2,t)=N(N-1)\int D_N\,dz_3\cdots dz_N.
$$
Thus $\int f\,dz=N$ and $\int f_2(z,z_\star)\,dz_\star=(N-1)f(z)$. These conventions absorb the otherwise explicit factor $N-1$ in the first <BBGKY hierarchy> equation.

Separate the <Hamiltonian> into <kinetic energy>, external <potential energy>, and <pair potential> terms. Their <Hamilton's equations> give
$$
\dot{\mathbf x}_i=\mathbf v_i,\qquad \dot{\mathbf v}_i=-\nabla U(\mathbf x_i)-\sum_{j\ne i}\nabla_{\mathbf x_i}\phi(|\mathbf x_i-\mathbf x_j|).
$$
The sign follows directly from the stated <Poisson bracket>: $\dot v_{i\alpha}=\{v_{i\alpha},H\}=-\partial_{x_{i\alpha}}H$. The <Hamiltonian Liouville equation> is consequently
$$
\partial_tD_N+\sum_i\left[\mathbf v_i\cdot\nabla_{\mathbf x_i}D_N-\left(\nabla U_i+\sum_{j\ne i}\nabla_{\mathbf x_i}\phi_{ij}\right)\cdot\nabla_{\mathbf v_i}D_N\right]=0.
$$
Integrate over $z_2,\ldots,z_N$ and multiply by $N$. Assume sufficient decay in <velocity space> and either decay or periodic <boundary conditions> in position, so that <integration by parts> removes all transport terms with $i\geq2$. The surviving interaction term is
$$
N\sum_{j=2}^N\int \nabla_{\mathbf x_1}\phi_{1j}\cdot\nabla_{\mathbf v_1}D_N\,dz_2\cdots dz_N
=\int\nabla_{\mathbf x_1}\phi_{12}\cdot\nabla_{\mathbf v_1}f_2\,dz_2,
$$
where indistinguishability makes all $N-1$ terms equal. Relabeling $(z_1,z_2)$ as $(z,z_\star)$ yields the exact result
$$
\boxed{\partial_tf+\mathbf v\cdot\nabla f-\nabla U\cdot\nabla_{\mathbf v}f=\int d^3x_\star\,d^3v_\star\,\nabla_{\mathbf x}\phi(|\mathbf x-\mathbf x_\star|)\cdot\nabla_{\mathbf v}f_2.}
$$
No assumption of <molecular chaos for the Boltzmann equation> has entered: the dependence on the <two-particle distribution function> is still exact.