= Solution
A <local Maxwellian> can be written as $\mathcal M=\exp(a+b\cdot v+c|v|^2)$, with $c<0$ and coefficients depending on $(t,x)$. Its logarithm is a <linear combination> of <collision invariants>, so $\mathcal M'\mathcal M_*'=\mathcal M\mathcal M_*$ and $Q(\mathcal M,\mathcal M)=0$ at each spatial point.
Comparing powers of $v$ in $(\partial_t+v\cdot\nabla_x)\log\mathcal M=0$ gives
$$
\nabla_xc=0,\qquad\operatorname{sym}\nabla_xb+c_tI_3=0,\qquad b_t+\nabla_xa=0,\qquad a_t=0.
$$
Here $\operatorname{sym}\nabla b$ is the <symmetric part of a matrix> formed by the coefficients of the quadratic polynomial. For constants $A,\alpha,\beta>0$, choose $a=\log A-\alpha|x|^2$, $b=2\alpha tx$ and $c=-\beta-\alpha t^2$. All four coefficient equations hold. The resulting <expanding Gaussian solution of the Boltzmann equation> is
$$
\boxed{f(t,x,v)=A\exp\bigl(-\alpha|x-tv|^2-\beta|v|^2\bigr)}.
$$
Direct verification is particularly simple: $x-tv$ and $v$ are constant along free <characteristic curves>, so the transport <derivative> is zero. At fixed $(t,x)$ its logarithm is the displayed <Maxwellian> quadratic, so the collision term is also zero. Its spatial dependence is nonconstant, it is positive and rapidly decaying, and its Gaussian velocity <integral> is finite because $\beta+\alpha t^2>0$. It is not a compactly supported example, nor does this final part require one.
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