Full-support collision equilibrium has Maxwellian form

By [F2], $D_x\ge 0$ for almost every $x$. Since its spatial integral is zero, $D_x=0$ for almost every $x$. Fix any such $x$ and write $g(v)=\log (f(x,v)\mathrm{/}{f}_{\mathrm{r}\mathrm{e}\mathrm{f}})$, where $f_{\mathrm{r}\mathrm{e}\mathrm{f}}>0$ is a constant with the units of $f$. The integrand defining $D_x$ is continuous and nonnegative in the collision variables. If it were positive at any elastic quadruple, it would be bounded below by a positive constant on a relatively open neighborhood; full support would make that neighborhood have positive measure, contradicting $D_x=0$. Therefore $f(v)f(w)=f(v^{\mathrm{\prime }})f(w^{\mathrm{\prime }})$ for every elastic quadruple, equivalently $g(v)+g(w)=g(v^{\mathrm{\prime }})+g(w^{\mathrm{\prime }})$.

For arbitrary $a,z\in {\mathbb{R}}^3$ and any rotation $R\in SO(3)$, the incoming pair $(a+z,a-z)$ and outgoing pair $(a+Rz,a-Rz)$ have equal total momentum $2a$ and equal total squared speed $2\mathrm{\mid }a{\mathrm{\mid }}^2+2\mathrm{\mid }z{\mathrm{\mid }}^2$. The collision identity from step 1.1 gives $g(a+z)+g(a-z)=g(a+Rz)+g(a-Rz)\mathrm{.}$ Thus for fixed $a$ and radius $\mathrm{\mid }z\mathrm{\mid }$, the left side is independent of the direction of $z$.

Fix a unit vector $\xi$ and any other unit vector $\eta$. Choose a rotation taking $\xi$ to $\eta$, put $z=s\xi$ in step 2.1, and expand each side at $s=0$ using [F3]. The constant and linear terms cancel in each symmetric pair. Dividing the resulting equality by $s^2$ and taking $s\to 0$ gives ${\xi }^{\mathsf{T}}{D}^{2}g(a)\xi ={\eta }^{\mathsf{T}}{D}^{2}g(a)\eta$. Hence the quadratic form of the symmetric Hessian is constant on the unit sphere. Testing $e_i$ gives equal diagonal entries; testing $(e_i+e_j)\mathrm{/}\sqrt{2}$ gives zero off-diagonal entries. Therefore $D^{2}g(a)=\lambda (a)I$.

Since the mixed second derivatives vanish, ${\mathrm{\partial }}_{i}g(v)$ is constant along every coordinate segment in a direction $e_j$ with $j\mathrm{\ne }i$, by [F4]. Connecting points with the same $i$th coordinate by three coordinate segments shows ${\mathrm{\partial }}_{i}g(v)=G_i(v_i)$ for a differentiable one-variable function $G_i$. The Hessian identity in step 3.1 yields $G_i^{\mathrm{\prime }}(v_i)=G_j^{\mathrm{\prime }}(v_j)$ for every pair $i\mathrm{\ne }j$ and every independent choice of $v_i,v_j$. Fixing $v_j$ while varying $v_i$ shows $G_i^{\mathrm{\prime }}$ is constant; equality then makes the constants the same $\lambda$. Thus $G_i(s)=\lambda s+b_i$, and integrating the gradient along coordinate segments gives $g(v)=\alpha +b\cdot v+\frac{\lambda }{2}\mathrm{\mid }v{\mathrm{\mid }}^2$ for constants $\alpha \in \mathbb{R}$, $b\in {\mathbb{R}}^3$, and $\lambda \in \mathbb{R}$.

Because $f$ is integrable and positive on ${\mathbb{R}}^3$, $\lambda$ cannot be positive: the quadratic term then makes $e^{g(v)}$ nonintegrable at infinity. If $\lambda =0$, an affine exponential is either constant or grows on a half-space, and is again nonintegrable. Hence $\lambda <0$. Completing the square gives $g(v)={\alpha }^{\mathrm{\prime }}-\mathrm{\mid }v-U{\mathrm{\mid }}^2\mathrm{/}(2\theta )$ with $\theta =-1\mathrm{/}\lambda >0$ and $U=-b\mathrm{/}\lambda$. Exponentiating gives the asserted profile.

For the profile obtained in step 5.1, $A=B$ on every elastic quadruple because its logarithm is a constant plus a linear and a quadratic collision invariant. The Definition's weak identity for each compactly supported smooth test therefore gives $\int \phi Q_{\nu }(f(x,\cdot ))dv=0$; the collision integrand also vanishes, so $D_x=0$. Conversely, the same calculation applies to every profile of this form, with weak annihilation conditional on its membership in ${\mathcal{D}}_{\nu }$.