= Solution
Apply the spherical identity of part (b) with $x=v-v_*$, $y=\xi$ and $\phi(s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi|(v-v_*)\cdot\sigma/2}$. The full phase is then
$$
-i v\cdot\frac{\xi+|\xi|\sigma}{2}-i v_*\cdot\frac{\xi-|\xi|\sigma}{2}.
$$
The <Fubini theorem> factors the two velocity <integrals> into <Fourier transforms>. With $\xi^\pm=(\xi\pm|\xi|\sigma)/2$, the gain <integral> is exactly $\int_{\mathbb S^2}\widehat f(\xi^+)\widehat f(\xi^-)\,d\sigma$.
The loss <Fourier transform> is $\widehat f(\xi)\widehat f(0)$. Dividing the gain by the <sphere> area gives the <Bobylev identity> for the <Maxwell molecule collision operator>:
$$
\boxed{\partial_t\widehat f(\xi)=\frac1{|\mathbb S^2|}\int_{\mathbb S^2}\widehat f(\xi^+)\widehat f(\xi^-)\,d\sigma-\widehat f(\xi)\widehat f(0)}.
$$
The initial Fourier datum is $\widehat f_0$. The <surface measure on a sphere> here is two-dimensional surface area, not the restriction of ambient three-dimensional <Lebesgue measure>, which would give the <sphere> measure zero.
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