Decay estimates for transport equations in massless case

Question

Considering the kinetic-transport equation $f(t,x,p)$ in the mass shell. It is well known that in the classic case (Vlasov non-relativistic)

\begin{align*} \int_{\mathbb{R}^3} f(t,x,v)dv &= \int_{\mathbb{R}^3} f^0(x-vt,v)dv \\ &\leq \sup_{w \in \mathbb{R}^3} \int_{\mathbb{R}^3}\ f^0(x-vt,w)dv, \end{align*} for $t>0$, we can apply the change of variable $y=x-vt$ and the appropriate scaling to get the decay estimates \begin{align*} \left|\int_{\mathbb{R}^3} f(t,x,v)dv \right| &\leq \frac{1}{t^3}\sup_{w \in \mathbb{R}^3 } \int_{\mathbb{R}^3} \left|f^0(y,w)\right|dy \\ &\leq \frac{C}{t^3}. \end{align*}

Otherwise, for the massive case (relativistic with $m>0$) applying the change of variable $y=x-\frac{v}{v^0}t$ we can obtain

\begin{align*} \int_{\mathbb{R}^3} | f(t,x,v)|dv &\leq \int_{\mathbb{R}^3}\ f^0(x-\frac{v}{v^0}t,v)dv. \\ & \leq \frac{C}{t^3} \int_{\mathbb{R}^3}\ \sup_{w \in \mathbb{R}^3}|f^0(y,w)|dy, \end{align*} where $C$ depends support in $v$.

For more details, see https://www.ljll.math.upmc.fr/smulevici/lgke.pdf.

My question is:

For massless case where $v^0 = |v|$ (with $v \in \mathbb{R}^3$ and $|v|=\sqrt(v_1^2+v_2^2+v_3^3)$ ) with the compact support assumption in $v$, the decay estimate falls off like $t^{-2}$, but the above changes of variables still gives me t^{-3}. \begin{align*} \int_{\mathbb{R}^3} f(t,x,v)dv &= \int_{\mathbb{R}^3} f^0(x-\frac{v}{|v|}t,v)dv \\ &\leq \ \ \ ? \end{align*}

Any help or suggestions on what to do? The idea is to continue using the characteristics method.

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EDIT:

The idea seems to be to pass the integral of $\mathbb{R}^3$ to an appropriate sphere (of radius $r$ and centered at $0$), i.e. using polar coordinates in the variable $v$.

1. Using $v= (v_1,v_2,v_3) = (r \sin \phi \cos \theta, r \sin \phi \sin \theta, r \cos \phi) = \omega$ , we have that $|v| = |r| = r$.

Thus, \begin{align*} \int_{\mathbb{R}^3} f(t,x,v)dv &= \int_{\mathbb{R}^3}\ f^0(x-\frac{v}{|v|}t,v)dv \\ & = \int_{\mathbb{S}^2}\ f^0(x-\frac{\omega}{r}t,\omega) r^2 \sin \phi d\omega \end{align*}

???? However, I can't see how to get the t^-2.

2. To write $\omega=\frac{v}{|v|}$, $r=|v|$ so that $dv=r^2 dr d\omega$ where $r\in(0,\infty)$ and $\omega \in S^2$

Thus, \begin{align*} \int_{v \in \mathbb{R}^3} f(t,x,v)dv &= \int_{v \in \mathbb{R}^3}\ f^0(x-\frac{v}{|v|}t,v)dv \\ & = \int_{\omega \in \mathbb{S}^2}\ f^0(x-\omega t,r\omega) r^2 dr d\omega \end{align*}

Now, should I change the variables again?

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Some questions about it:

1. Should take the sphere of radius $t$?, or
2. Does this change of variables make sense?

\begin{align*} \int_{\omega \in \mathbb{S}^2}\ f^0(x-\omega t,r\omega) r^2 dr d\omega & \leq C_r \sup_{r \geq 0} \int_{\omega \in \mathbb{S}^2}\ f^0(x-\omega t,\omega) d\omega \\ & \leq \frac{C_r}{t^{-2}} \sup_{r \geq 0} \int_{y \in \mathbb{S}^2}\ f^0(x-y, \frac{ry}{t}) dy \end{align*}

Answer 1

There are a few ways to do this, but one way is as follows. Using polar coordinates $r=|v|$, $v = r\omega$ we may write

$$\int f(t,x,v) dv = \int_0^\infty \int_{S^2} f^0(x-t\omega,r\omega) d\omega dr$$

Then, on the sphere (rather than the whole space) we make the change of variables $\omega\mapsto t\omega = \gamma$, and you can compute the Jacobian explicitly as $|J| = t^2$ (you can verify this by direct calculation using the usual coordinates on the sphere, but you don't even need an explicit choice of coordinates, because this map is a scaling of the identity). Then, $d\gamma = |J| d\omega$, so that

$$\int f(t,x,v) dv = t^{-2}\int_0^\infty \int_{S^2} f^0(x-\gamma,rt^{-1}\gamma) d\gamma dr \leq t^{-2}\sup_{x,v}\int_0^\infty\int_{S^2} f^0(x-\gamma,v)d\gamma dr$$