= Solution
Fix the mixed <Fourier transform> convention
$$
\widehat f(t,k,\xi)=\int_0^1\int_{\mathbb R}
f(t,x,v)e^{-2\pi i(kx+\xi v)}\,dv\,dx,
\qquad k\in\mathbb Z,\quad\xi\in\mathbb R.
$$
The spatial derivative transforms to $2\pi ik\widehat f$, and multiplication by $v$ transforms to $-(2\pi i)^{-1}\partial_\xi\widehat f$. Hence the transformed <free transport equation> is
$$
\boxed{\partial_t\widehat f-k\partial_\xi\widehat f=0.}
$$
Its <characteristic equations for a transport equation> give $\dot\xi=-k$, so the characteristic ending at $\xi$ at time $t$ began at $\xi+kt$. Consequently
$$
\boxed{\widehat f(t,k,\xi)=\widehat f_0(k,\xi+kt).}
$$
The same sign follows directly by substituting $x=y+tv$ in the <Fourier transform> of $f_0(x-tv,v)$. No first velocity moment is assumed, so the differential equation may be understood in the sense of <tempered distributions>; the explicit transform formula is valid pointwise because $f_0$ is integrable.
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