Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 4

Nonnegativity and unit mass give, by the triangle inequality for the Fourier transform, $|\hat{f}(\xi )|\le \int f=1=\hat{f}(0)$. The finite second moment implies a finite first absolute moment, so differentiating under the Fourier integral is justified andThe exponent convention has no $2\pi$, so no such factor occurs in this derivative.

If either vector vanishes, both arguments are zero. Otherwise choose a rotation matrix $R$ sending $x/|x|$ to $y/|y|$. The surface measure on a sphere is rotation invariant. Changing variables $\sigma =R\omega$ givesThus the two spherical integrals are equal. The property extends to complex continuous $\varphi$ by treating real and imaginary parts separately, as needed for the Fourier exponential.

The printed hint with unchanged $\sigma$ is not a valid change of variables: at fixed $\sigma$, the outgoing velocities forget the direction of $v-v_{\ast }$. A correct proof uses the angular exchange for elastic collisions.

Set $C=(v+v_{\ast })/2$ and $v-v_{\ast }=r\omega$, where $r>0$ and $\omega \in {\mathbb{S}}^2$. Then $dvd{v}_{\ast }=dC{r}^{2}drd\omega$, while $v^{\prime }=C+r\sigma /2$ and $v_{\ast }^{\prime }=C-r\sigma /2$. The gain integral becomesExchange the two independently integrated angular variables $\omega$ and $\sigma$. The measure is unchanged, and reverting to $v=C+r\omega /2$, $v_{\ast }=C-r\omega /2$ givesThe measure-zero set $r=0$ causes no difficulty. This proves the requested identity without invoking the false fixed-$\sigma$ Jacobian assertion.

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. 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.

Expansion of the two squares givesLet $D=d(f,g)$ be the Fourier distance of order two, with the supremum taken over $\xi \ne 0$. Both Fourier transforms have modulus at most one. Add and subtract $\hat{g}({\xi }^{+})\hat{f}({\xi }^{-})$, then use the triangle inequality:Dividing by $|\xi {|}^2$ proves the bound by $D$, and splitting the last expression into its two weighted terms gives the requested intermediate inequality. If one of ${\xi }^{\pm }$ vanishes, its unweighted difference is zero by equal mass; its weighted term is interpreted as zero, avoiding a $0/0$ quotient.

Equal mass and first moment also explain finiteness of this distance: subtract the constant and linear Taylor terms in the Fourier integral and use $|e^{-ia}-1+ia|\le a^2/2$. This bounds the difference by $|\xi {|}^2\int |v{|}^2(f+g)/2$. No direction-independent extension of the quotient at zero is required.

The reference to part (f) within this part is a printed self-reference; the needed product estimate is part (e). Since both masses are one, subtracting the Bobylev identities gives, for $\xi \ne 0$,The normalized angular average and part (e) giveFor completeness, the Duhamel principle for this scalar equation yields $d(t)\le e^{-t}d(0)+{\int }_0^t{e}^{-(t-s)}d(s)ds$. Apply the Gronwall inequality to $e^{t}d(t)$ to obtain the Fourier nonexpansion for Maxwell molecules, $\boxed{d(f_t,g_t)\le d(f_0,g_0)}$. This estimate is nonexpansion; by itself it does not prove strict decay or convergence to a specified equilibrium.