Taking moments of the Vlasov equation

Question

Given the following term:

$\nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]$

where $\textbf{E}$ and $\textbf{B}$ are the electric and magnetic fields respectively experienced by a particle travelling with velocity $\textbf{v}$ and $f_{s}$ is the distribution function of particle velocities (assumed to be Maxwellian). I want to know how simply multiplying this expression by some arbitrary polynomial function, $X(\textbf{v})$ and integrating over velocity will give:

$- \frac{e}{m_{s}} \int (\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s} d^{3}v$

According to T.J.M. Boyd & J.J.Sanderson in the book Physics of Plasmas one needs to use integration by parts and use the limit:

$\lim_{|\textbf{v}|\to\infty} (Xf_{s})=0 $

I am struggling to see how this follows and would appreciate a detailed explanation with as few a step skipped as reasonable.

Answer 1

I am assuming $\nabla_{v}(Xf_{s})$...?

Integration by parts:

$$ - \frac{e}{m_{s}} \int \underbrace{(\textbf{E} + \textbf{v} \times \textbf{B})}_{f} \cdot \underbrace{\nabla_{v}Xf_{s}}_{g'} d^{3}v =$$

$$ \propto \underbrace{(\textbf{E} + \textbf{v} \times \textbf{B})}_{f} \cdot \underbrace{Xf_{s}}_{g} \Biggr|_{\text{limits}} - \int \underbrace{\nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]}_{f'}\cdot \underbrace{Xf_s}_g.$$

The limits will be $v$ from $0$ to $\infty$, where $v = |\mathbf{v}|$ and its angular part is just a constant.

I know that $\lim_{|\textbf{v}|\to\infty} (Xf_{s})=0$ and I want:

$$ Xf_{s} \Biggr|_{\text{limits}} = (Xf_s)\Biggr|_{\infty} - (Xf_s)\Biggr|_{0} = 0 - 0,$$

because $f_s \propto v^2 =0 $ at $0$.

So:

$$ - \frac{e}{m_{s}} \int (\textbf{E} + \textbf{v} \times \textbf{B}) \cdot \nabla_{v}Xf_{s}d^{3}v = - \int \nabla_{v} \cdot \left[ \frac{e}{m_{s}} (\textbf{E} + \textbf{v} \times \textbf{B})f_{s} \right]\cdot Xf_s.$$