Wild's sum for Boltzmann's equation

Question

Consider the spatially homogenous Boltzmann equation $$\partial_t f_t = Q^+(f_t,f_t) - f_t.$$ A semi-explicit representation formula for solutions of this Boltzmann equation can be written as (see for instance Villani's monograph) $$ f_t = \mathrm{e}^{-t}\sum_{n=1}^\infty \left(1-\mathrm{e}^{-t}\right)^{n-1} Q^+_n(f_0) \label{1}\tag{1} $$ where the $n$-fold nonlinear operator $Q^+_n$ is defined recursively by $$Q^+_1(f_0) = f_0,\quad Q^+_n(f_0) = \frac{1}{n-1}\,\sum_{k=1}^{n-1} Q^+\left(Q^+_k(f_0),Q^+_{n-k}(f_0)\right).$$ The author mentioned that one can easily check that \eqref{1} indeed solves the Boltzmann equation by plugging it into the integral formulation of the Boltzmann equation, which reads as $$f_t(v) = f_0(v)\,\mathrm{e}^{-t} + \int_0^t \mathrm{e}^{-(t-s)}\,Q^+(f_s,f_s)\,\mathrm{d}s.$$ However, I have no clue as to how one can verify that the Wild's sum representation \eqref{1} indeed solves the Boltzmann equation. Thus any help in filling in the missing details will be greatly appreciated.

Answer 1

Completely ignoring all convergence issues, this really is just following your nose.

Plugging in the representation you give, you want to check

$$ \sum_{n = 1}^\infty (1 - e^{-t})^{n-1} Q_n^+(f_0) \overset{?}{=} f_0 + \int_0^t e^{s} Q^+(f_s,f_s) ~ds $$

Using the representation you give again to replace $f_s$, you rewrite this as checking (using that $Q^+$ is bilinear)

$$ \sum_{n = 1}^\infty (1-e^{-t})^{n-1}Q_n^+(f_0) \overset{?}{=} \\Q_1^+(f_0) + \sum_{n = 1}^\infty \sum_{m = 1}^\infty \int_0^t e^{-s} (1 - e^{-s})^{n-1} (1-e^{-s})^{m-1} Q^+(Q_n^+(f_0), Q_m^+(f_0)) ~ds $$

The integral can now be explicitly evaluated, as $\frac{d}{ds} (1-e^{-s})^k = k (1 - e^{-s})^{k-1} e^{-s}$, so we reduce to $$ \sum_{n = 1}^\infty (1-e^{-t})^{n-1}Q_n^+(f_0) \overset{?}{=} Q_1^+(f_0) + \sum_{n,m = 1}^\infty \frac{1}{n+m-1} (1 - e^{-t})^{n+m-1} Q^+(Q_n^+(f_0), Q_m^+(f_0)) $$ Rewrite the sum in terms of $n$ and $k = n+m$, you get $$ \sum_{n = 1}^\infty (1-e^{-t})^{n-1}Q_n^+(f_0) \overset{?}{=} Q_1^+(f_0) + \sum_{k = 2}^\infty (1 - e^{-t})^{k-1}\underbrace{\frac{1}{k-1} \sum_{n = 1}^{k-1} Q^+(Q_n^+(f_0), Q_{k-n}m^+(f_0))}_{= Q_k^+(f_0)} $$ and so equality follows.

Convergence can be checked for small $t$: if the bilinear form satisfies the bound $|Q^+(f,g)| \leq C|f||g| $, then by induction you can show that $|Q_n^+(f_0)| \leq C^{n-1} |f_0|^n$. For $|t|\ll 1$ you have that $|(1-e^{-t})| \ll 1$ and hence the representation converges as it is dominated by a geometric series. And so for $t$ sufficiently small all of the interchanges of integrals and sums, and reindexing of summation, are valid in the formal computation above.