The 3-dimensional Vlassov -Poisson equation I am studying at university is $$ \partial_t f (t,x,v) + v\cdot \nabla_x f (t,x,v) - \nabla_x \phi (t,x) \cdot\nabla_v f (t,x,v) =0,$$ where $$\Delta \phi = 4\pi\gamma\, \rho(t,x,v)\ \text{ and } \ \rho (t,x) = \int_{\mathbb{R}^3} f(t,x,v)\,\mathrm dv$$
I am trying to prove the momentum is conserved: $$q(t) = \int_{\mathbb{R}^6} v \,f(t,x,v)\,\mathrm dx \,\mathrm dv \Rightarrow q' (t)=0$$
I need some help to prove that This is my attempt, I first substitute the equation of $f$ into the expression: $$ q'(t) = \int \partial_t [v\,f(t,x,v)]\,\mathrm dx \,\mathrm dv \\ = \int v \left(-v\cdot \nabla_x f (t,x,v) + \nabla_x \phi (t,x) \cdot \nabla_v f (t,x,v)\right) \,\mathrm dx \,\mathrm dv $$ Secondly, I noticed that the first term is equal to zero after an integration by parts (passing the $x$-derivative to $|v|ˆ2$ which does not depende on $x$), therefore $$ q'(t)= \int v\,\nabla_x\phi(t,x)\cdot\nabla_v f(t,x,v)\,\mathrm dx \,\mathrm dv $$ What can I do next to show that $q' =0$? Thanks in advance!
Starting from your expression, integrating by parts and then using your definitions of $\rho$ and $\phi$ $$ q'(t)= \iint_{\mathbb R^6} v\,\nabla_x\phi(t,x)\cdot\nabla_v f(t,x,v)\,\mathrm dx \,\mathrm dv = - \iint_{\mathbb R^6} f(t,x,v)\,\nabla_x\phi(t,x)\,\mathrm dx \,\mathrm dv \\ = - \int_{\mathbb R^3} \rho(t,x)\,\nabla_x\phi(t,x)\,\mathrm dx = - 4\pi\gamma\int_{\mathbb R^3} \rho(t,x)\,\nabla_x (K*\rho)(t,x)\,\mathrm dx $$ where $K$ is the fundamental solution of the Poisson equation ($K(x) = \tfrac{c}{|x|}$). This can be written $$ q'(t) = 4\pi\gamma \iint_{\mathbb R^6} \rho(t,x)\,\rho(t,y)\,\nabla K(x-y)\,\mathrm d x\,\mathrm d y $$ Since $\nabla K(x-y)$ is an odd function, $q'(t) = 0$.