Consider the spatially homogenous Boltzmann equation $$\partial_t f_t = Q^+(f_t,f_t)  f_t.$$ A semiexplicit representation formula for solutions of this Boltzmann equation can be written as (see for instance Villani's monograph) $$ f_t = \mathrm{e}^{t}\sum_{n=1}^\infty \left(1\mathrm{e}^{t}\right)^{n1} Q^+_n(f_0) \label{1}\tag{1} $$ where the $n$fold nonlinear operator $Q^+_n$ is defined recursively by $$Q^+_1(f_0) = f_0,\quad Q^+_n(f_0) = \frac{1}{n1}\,\sum_{k=1}^{n1} Q^+\left(Q^+_k(f_0),Q^+_{nk}(f_0)\right).$$ 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 $$f_t(v) = f_0(v)\,\mathrm{e}^{t} + \int_0^t \mathrm{e}^{(ts)}\,Q^+(f_s,f_s)\,\mathrm{d}s.$$ 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.
1 Answer
Completely ignoring all convergence issues, this really is just following your nose.
Plugging in the representation you give, you want to check
$$ \sum_{n = 1}^\infty (1  e^{t})^{n1} Q_n^+(f_0) \overset{?}{=} f_0 + \int_0^t e^{s} Q^+(f_s,f_s) ~ds $$
Using the representation you give again to replace $f_s$, you rewrite this as checking (using that $Q^+$ is bilinear)
$$ \sum_{n = 1}^\infty (1e^{t})^{n1}Q_n^+(f_0) \overset{?}{=} \\Q_1^+(f_0) + \sum_{n = 1}^\infty \sum_{m = 1}^\infty \int_0^t e^{s} (1  e^{s})^{n1} (1e^{s})^{m1} Q^+(Q_n^+(f_0), Q_m^+(f_0)) ~ds $$
The integral can now be explicitly evaluated, as $\frac{d}{ds} (1e^{s})^k = k (1  e^{s})^{k1} e^{s}$, so we reduce to $$ \sum_{n = 1}^\infty (1e^{t})^{n1}Q_n^+(f_0) \overset{?}{=} Q_1^+(f_0) + \sum_{n,m = 1}^\infty \frac{1}{n+m1} (1  e^{t})^{n+m1} Q^+(Q_n^+(f_0), Q_m^+(f_0)) $$ Rewrite the sum in terms of $n$ and $k = n+m$, you get $$ \sum_{n = 1}^\infty (1e^{t})^{n1}Q_n^+(f_0) \overset{?}{=} Q_1^+(f_0) + \sum_{k = 2}^\infty (1  e^{t})^{k1}\underbrace{\frac{1}{k1} \sum_{n = 1}^{k1} Q^+(Q_n^+(f_0), Q_{kn}m^+(f_0))}_{= Q_k^+(f_0)} $$ and so equality follows.
Convergence can be checked for small $t$: if the bilinear form satisfies the bound $Q^+(f,g) \leq Cfg $, then by induction you can show that $Q_n^+(f_0) \leq C^{n1} f_0^n$. For $t\ll 1$ you have that $(1e^{t}) \ll 1$ 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.

$\begingroup$ Thank you very much. I actually happen to figure it out by myself, after looking at the original paper written by Wild. But thank you very much! $\endgroup$– Fei CaoJul 6, 2023 at 16:16