Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 7 / 1

Use $(y,w)$ to label the initial point in phase space. The characteristic equations for a transport equation are $\dot{X}=a(V)$ and $\dot{V}=0$, with $X(s)=y$ and $V(s)=w$. Since the velocity remains fixed, the ordinary differential equations integrate directly:This characteristic flow map exists for all real times, even if $a$ grows rapidly at infinity: each trajectory samples just one fixed velocity. Its inverse is $S_{t,s}$, and its Jacobian matrix is the block triangular matrixConsequently its Jacobian determinant is one. The nonzero Jacobian determinant of $a$ is unnecessary for this construction.

The chain rule gives $d[f(t,X(t),V(t))]/dt=0$ along each characteristic curve. Pulling the initial datum back by the inverse characteristic flow map therefore givesDirect differentiation also gives $f_t=-a(v)\cdot {\nabla }_{x}f$, so this is a classical solution of the transport equation. Composition with a smooth function makes $f$ smooth jointly in $(t,x,v)$; nonnegative initial data remain nonnegative.

If $K=\mathrm{supp}{f}_{\mathrm{in}}$, then $\mathrm{supp}f(t)=S_{0,t}(K)$. A continuous image of a compact set is compact. Thus compact support is preserved at every finite time; there is no claim of one fixed compact support for all $t\ge 0$.

Choose $R_0$ with $K\subset \{(y,w):|y|\le R_0,|w|\le R_0\}$ and let $A_0={\max }_{|w|\le R_0}|a(w)|$, which is finite by continuity. Every point transported from $K$ satisfiesHence anycontains the support of a function in the required phase space ball throughout $[0,T]$. Equivalently, the image of the compact set $[0,T]\times K$ under the characteristic flow map is compact. This uniform support control puts the solution in ${\mathcal{C}}_T$ and justifies the following differentiations under the integral.

For two solutions with the same initial datum, let $h=f_1-f_2$. Their uniform compact support on $[0,T]$ puts $h$ and $h_t$ in one compact region of phase space. Smoothness there implies that $E(t)=\int |h(t,x,v){|}^{2}dxdv$ is finite and continuously differentiable, by differentiation under the integral sign. The transport equation and integration by parts in $x$ yieldThere is no boundary contribution because of compact support, and ${\mathrm{div}}_{x}a(v)=0$. Since $E(0)=0$, $h=0$ almost everywhere and then everywhere by continuity. The solution is unique in ${\mathcal{C}}_T$. The same calculation gives Lp conservation for incompressible transport at $p=2$.

For the free transport equation, the mixed Lebesgue norm dispersion estimate isIndeed, use the method of characteristics and then the change of variables $y=x-tv$:Taking the essential supremum over $x$ proves the result. This decay measures the spreading of a velocity average; the total phase space Lp norm is conserved rather than decaying. The datum on the right is the initial datum, not the solution at the same time.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

Fix the standard Jacobian matrix convention $(Da{)}_{ij}={\partial }_{v_j}{a}_i$. With column gradients, the commuting velocity derivative for kinetic transport isIf the symbol $J_a$ is instead defined with derivative indices as rows, the printed expression is this same formula. Under the standard convention it needs a transpose.

To prove the claim, put $D={\partial }_t+a(v)\cdot {\nabla }_x$. Differentiating $Df=0$ in $v_j$ givesThe terms cancel, so each component of $G$ satisfies the transport equation. Multiplication by the smooth coefficient $Da$ and differentiation preserve the same uniform compact support, hence $G\in {\mathcal{C}}_T$ componentwise. At $t=0$, $G(0)={\nabla }_v{f}_{\mathrm{in}}$. The L2 norm calculation in part (d), summed over components, givesOne can also differentiate the explicit solution: the $t(Da{)}^T{\nabla }_{x}f$ term cancels the derivative of $x-ta(v)$, leaving the initial velocity derivative transported along the characteristic curves.

The transpose matters. For the shear $a(v)=(v_1+v_2,v_2)$, $\det Da=1$ but $Da\ne (Da{)}^T$. The printed field satisfies $Dg=(Da-(Da{)}^T){\nabla }_{x}f$, which is nonzero for a generic compactly supported datum. The untransposed formula is false with the standard convention.