B Particle System has an Invariant Measure with Mean Zero

Consider the evolution described by , with \(\varphi\equiv 1\), that is the space homogeneous particle system with uniform interaction: \[\begin{align} \mathrm{d} {v}_t^{i,N} & = - v_t^{i,N} \mathrm{d} t + G\left(\frac{ \sum_{j=1}^N v_t^{j,N}}{N}\right) \mathrm{d} t + \sqrt{2\sigma} \mathrm{d} W_t^{i}, \quad i = 1,\dots N. \tag{B.1} \end{align}\] Notice that this is equivalent to the globally scaled particle model (2.12)-(2.13) when \(\varphi\equiv1\). We will show that this equation has an invariant measure with mean zero. We believe this to be unique and will prove as such in a future work.

Let \(G\) be smooth and define \(\bar{v}_t = \sum_{i=1}^N v_t^{i,N}\). Then, by summing over \(i\) in (B.1), \(\bar{v}_t\) solves \[\begin{equation} \tag{B.2} \mathrm{d}\bar{v}_t = \left[-\bar{v}_t+G(\bar{v}_t)\right]\mathrm{d}t + \sqrt{2\sigma}\mathrm{d}W_t, \quad \bar{v}_t \in \mathbb{R}. \end{equation}\] Let \(V(x) = \frac{x^2}{2} + \tilde{V}(x)\), where \(\tilde{V}(x)\) is such that \(G(x) = -\tilde{V}'(x)\) so that \[ -V'(\bar{v}) = -\bar{v} + G(\bar{v}). \] Equation (B.2) can be rewritten as \[\begin{equation} \tag{B.3} \mathrm{d}\bar{v}_t = -V'(\bar{v})\mathrm{d}t + \sqrt{2\sigma}\mathrm{d}W_t, \quad \bar{v}_t \in \mathbb{R}. \end{equation}\] Now $({v}) = - _{-}^{{v}} G(u) u $ by definition, so that if \(G\) is bounded then \(\tilde{V}(x)\) grows at most linearly. Hence, \(V(\bar{v}) \to +\infty\) as \(|\bar{v}|\to \infty\) so that \(\mathrm{e}^{-V(\bar{v})}\) is integrable. Thus (B.3) admits an invariant measure, which is \[ \rho(\bar{v}) = \frac{1}{Z}\mathrm{e}^{-V(\bar{v})}, \] where \(Z\) is a normalising constant that we neglect from now on. Note that \(\tilde{V}(\bar{v})\) is an even function as \(G(\bar{v})\) is odd. This implies that \[ V(\bar{v})=\frac{\bar{v}^2}{2}+ \tilde{V}(\bar{v}) \] is even, and so \[ \int_{\mathbb{R}} \bar{v}\mathrm{e}^{-V(\bar{v})} \mathrm{d}\bar{v} = 0. \] That is, as \(t\to \infty\), \(\mathbb{E}[\bar{v}_t] \to 0\).