Conservation of Kinetic Energy in Vlasov-Poisson System

450 Views Asked by At

I'm studying the very basics of kinetic theory in Vlasov Poisson Systems, and the first equation I'm studying is the free transport equation, i.e.: $$\frac{\partial f}{\partial t}+v\cdot\nabla_{x}f=0$$$$f(0,x,v)=f^{0}(x,v)$$

where $\,f(t,x,v)\geq 0\,$ is the distribution function of particles. One of its properties is that we can define a conservated quantity called the kinetic energy: $$E_{kin}(t)=\frac{1}{2}\int_{\mathbb{R}^{3}_{x}}\int_{\mathbb{R}^{3}_{v}}|v|^{2}f(t,x,v)\,dx\,dv.$$
Now, I know that this is a conservated quantity (i.e. $\frac{d}{dt}E_{kin}=0$) but I want to prove it, at least formally. I started multiplying the free transport equation by $\frac{|v|^{2}}{2}$ and then integrating everything with respect of x and v, and perhaps this is a dumb step but I can´t get rid of the $\int_{R^{3}_{x}}\int_{R^{3}_{v}}\frac{|v|^{2}}{2}v\cdot\nabla_{x}f\,dx\,dv$ term, how can I see that this term is zero? Any help is welcome, regards.

1

There are 1 best solutions below

0
On BEST ANSWER

Usual steps: (1) differentiate energy with respect to time; (2) use the PDE. $$\frac{1}{2}\int_{\mathbb{R}^{3}_{x}}\int_{\mathbb{R}^{3}_{v}}|v|^{2}\frac{\partial f}{\partial t}(t,x,v)\,dx\,dv = - \frac{1}{2}\int_{\mathbb{R}^{3}_{x}}\int_{\mathbb{R}^{3}_{v}}|v|^{2}v\cdot \nabla_x f(t,x,v)\,dx\,dv $$ Often there is step (3): integrate by parts. It's not really needed here, though. For each fixed $v$, we have $$\int_{\mathbb{R}^{3}_{v}} v\cdot \nabla_x f(t,x,v)\,dx = 0$$ because this is the directional derivative of $f$ in direction $v$, integrated over the entire space. We must assume, of course, that $f$ vanishes at infinity.

Solutions with mass going off into infinity or coming from infinity don't have to obey conservation laws since the system isn't really closed.