Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 2

Use the unit-period circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, with total measure one. The free characteristic flow map givesTranslations in $x$ preserve its periodic measure. The Tonelli theorem and the integral triangle inequality yieldThe Fubini's theorem therefore applies also to signed data, and the same translation givesThis holds for positive or negative time. The printed $L^1(\mathbb{R})$ in this subpart is a domain typo: the spatial variable is periodic, so the correct space is $L^1(\mathbb{T})$. For example, the smooth initial value $f_0(x,v)=e^{-v^2}$ gives the constant density $\sqrt{\pi }$, whose periodic extension is not integrable on the real line.

Fix the mixed Fourier transform conventionThe spatial derivative transforms to $2\pi ik\hat{f}$, and multiplication by $v$ transforms to $-(2\pi i{)}^{-1}{\partial }_{\xi }\hat{f}$. Hence the transformed free transport equation isIts characteristic equations for a transport equation give $\dot{\xi }=-k$, so the characteristic ending at $\xi$ at time $t$ began at $\xi +kt$. ConsequentlyThe 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.

The Fourier coefficient of the velocity-integrated density is its mixed transform at zero velocity frequency. ThusThe constant ${\rho }_{\infty }$ contributes only to the zero Fourier mode, where it equals ${\hat{f}}_0(0,0)$. ThereforeIt is important to remove the zero mode, which is conserved rather than mixed away.

Interpret the given velocity regularity in the stated Sobolev space sense. Repeated integration by parts, or the Fourier transform of a derivative in distributions, givesThe usual one-dimensional one-dimensional Sobolev representative or a smooth approximation justifies this identity without imposing extra decay of classical derivatives at specific boundary points. The transform of an integrable function has absolute value at most its $L^1$ norm. ThereforeSetting $\xi =kt$ and taking the supremum gives the required uniform weighted bound. The $k=0$ term on the left is zero; there is no division by that frequency in this argument.

Take the positive integer $n\ge 1$ and put $m=n+1\ge 2$. From the preceding estimate and the absent zero Fourier mode, for $t\ne 0$,Here $\zeta$ is the Riemann zeta function, and its displayed series is finite because $m>1$. One may replace the last derivative norm by the given full mixed Sobolev norm to obtain the requested constant depending only on $n$ and $f_0$.

An absolutely summable sequence of Fourier coefficients gives a uniformly and absolutely convergent Fourier series, with supremum bounded by the sum of their absolute values. Its sum agrees almost everywhere with $r_t$ by uniqueness of Fourier coefficients for integrable periodic functions. HenceThis is the uniform phase-mixing bound from velocity derivatives: uniform convergence for the continuous representative of the density, with rate $O(|t{|}^{-n-1})$. No uniform decay of the full phase-space distribution is asserted. The free-transport phase mixing acts by shifting nonzero spatial modes to large velocity frequency. If a convention allows $0\in \mathbb{N}$, that endpoint needs separate assumptions or an argument: the harmonic series in this proof would diverge.