Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 3

Let $E=L^1({\mathbb{R}}_x^d\times {\mathbb{R}}_v^d)$, and denote the free-transport semigroup by $(U_{t}g)(x,v)=g(x-tv,v)$. For fixed $v$, the spatial translation has unit Jacobian determinant, so the Tonelli theorem givesThus the given free term is an isometry on $E$. The operators form a strongly continuous semigroup: continuity first holds for smooth compactly supported functions by dominated convergence, and density plus the isometry extends it to every $L^1$ function.

The Fubini's theorem and integral triangle inequality giveDefine the normalized velocity-reset collision operator using the normalized velocity-reset projection $Pg(x,v)=\rho (g)(x)M(v)$ and $B=P-I$. Since $M\ge 0$ and $\int M=1$,Also $P^2=P$; the collision gain replaces the velocity distribution by $M$ while preserving the spatial mass. In Bochner integral notation the printed, undamped source operator isUsing the transport isometry and the integral triangle inequality provesThese estimates hold for measurable, locally time-bounded $E$-valued functions. If the displayed supremum is infinite, the numerical bound is interpreted in the extended sense; the construction below works in a space where it is finite.

Fix $T$ and set $A_T={\sup }_{0\le s\le T}\Vert f(s){\Vert }_1$. The preceding integral estimate gives the result for one iterate. If for some $n\ge 0$ the bound $\Vert {\tau }^{n}f(s){\Vert }_1\le A_T{2}^n{s}^n/n!$ holds for $s\le T$, thenTaking $T=t$ yields exactlyThe case $n=0$ uses the identity operator. This factorial bound for a Volterra iterate comes from time ordering, so no commutation between free transport and the collision projection is assumed.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

For $0\le s\le t$, sum the same absolutely convergent Neumann series estimate:ConsequentlyThis coarse bound is sufficient for the requested locally uniform control and the uniqueness argument. It is not claimed to be the sharp dissipative estimate for the collision model.

If $f$ and $\tilde{f}$ are two weak solutions with the same initial value and the required local time bound, their difference $w$ satisfies $w=\tau w$. By linearity this implies $w={\tau }^{n}w$ for every $n$. On $[0,T]$ put $A_T={\sup }_{s\le T}\Vert w(s){\Vert }_1<\infty$. The factorial bound for a Volterra iterate now givesFor fixed $T$, the factor tends to zero; its successive-term ratio is $2T/(n+1)$. Therefore $w(t)=0$ as an $L^1$ element at every time on this interval. Since $T$ is arbitrary, the locally time-bounded $L^1$ weak solution is unique globally. This proof uses precisely the additional condition requested, rather than assuming arbitrary pointwise-in-time integrability alone supplies a finite uniform bound.