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.


1 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.

  • $\begingroup$ 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 Cao
    Jul 6, 2023 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.