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

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.