# Phase space density behavior in finite volume

I am trying to develop better intuition for the single-particle phase space density $$f(x,v,t)$$ and I am confused by the tendency of $$f(x,v,t)$$ to become wiggly in velocity space. Consider the simplest possible situation of non-interacting particles on the interval $$[0,L]$$ with periodic boundary conditions. If we consider the initial phase space density ($$|A|<1$$) $$f_0(x,v) = \frac{1+A\cos(k_1 x)}{L}\frac{e^{-\frac{v^2}{2\sigma^2}}}{\sqrt{2\pi \sigma^2}}$$ where $$k_n = (2\pi/L)n$$, then according to the non-interacting Vlasov equation $$\partial f/\partial t + v \partial f /\partial x = 0$$, the phase space density is given by $$f(x,v,t) = f_0(x-vt,v)$$. Marginalizing over velocity space one finds that the number density decays exponentially to a constant, $$\rho(x,t) = \frac{1+A\cos(k_1 x)e^{-\frac{1}{2}k_1^2\sigma^2 t^2}}{L} \longrightarrow \frac{1}{L}$$ whereas the conditional velocity density $$f(x,v,t)/\rho(x,t)$$ oscillates wildly as a function of $$v$$ as $$t\to\infty$$. How can one understand this behavior intuitively?

Below I include a plot of the conditional velocity density (conditioned on $$x=0$$) at time $$t=0$$ and $$t=4$$ and the Mathematica code used to generate it.

\[Sigma]=1;L=1;a=0.3;
k[n_]:=(2Pi/L)n;
f[x_,v_,t_]:=(1+a Cos[k[1](x-v t)])/L Exp[-v^2/(2\[Sigma]^2)]/Sqrt[2Pi \[Sigma]^2];
\[Rho][x_,t_]:=(1+a Cos[k[1]x]Exp[-(1/2)k[1]^2\[Sigma]^2t^2])/L
x=0;
Plot[{f[x,v,0]/\[Rho][x,0],f[x,v,4]/\[Rho][x,4]},{v,-3,3},AxesLabel->{Style["v",Italic],""}, LabelStyle -> {FontFamily -> "Times", FontSize -> 16,SingleLetterItalics -> True},PlotLegends->"Expressions"]

• It does not seem that your equation for $f(x,v,t)$ is consistent with the periodic boundary conditions. That might be the solution for the unbounded process.
– Javi
Commented Aug 24, 2023 at 8:36
• @Javi, it’s consistent. Note that the periodicity is only wrt $x$. Commented Aug 24, 2023 at 13:27
• I insist that there is a problem. The equation allows for having a non-zero density for positions outside the interval $[0,L]$. For example, when $x=2L$, $v$ is arbitrary, and $t=\frac{2L}{v}$, we have $f(2L,v,\frac{2L}{v}) = f_0(0,v) \neq 0$.
– Javi
Commented Aug 24, 2023 at 16:52
• For any fixed $t \in \mathbb{R}$, the function $f(x,v,t)$ is a normalized probability density on $[0, L] \times \mathbb{R}$, which satisfies the boundary condition $f(0, v, t)=f(L, v, t)$, the initial condition $f(x,v,0)=f_0(x,v)$ and the PDE. I don’t see what else you could want. Commented Aug 24, 2023 at 20:14