A Product Measure is a Solution to Kinetic PDE

A.1 Product Distribution is a Solution

Consider the stationary problem, \[\begin{equation} \tag{A.1} - v\partial_x f_t(x,v) + \partial_v v f_t(x,v)- \partial_v \left[ G\left(M(t,x)\right)f_t(x,v))\right] + \sigma \partial_{vv} f_t(x,v) = 0. \end{equation}\]

Here we will show that a product distribution is a solution of the stationary problem (A.1) if and only if it is uniform in the space variable and Gaussian in the velocity variable. Moreover, the Gaussian velocity can only have variance \(\sigma\) and mean \(u\) where \(u\) solves \(G(u)=u\). Let \(f(x,v) = g(x)h(v)\), for some functions \(g,h\). Note the lack of dependence on \(t\) as we seek stationary solutions. Then,

\[\begin{align*} \partial_x g(x)h(v) &= h(v)\partial_xg(x)\\ \partial_v g(x)h(v) &= g(x)\partial_v h(v)\\ \partial_{vv} g(x)h(v) &= g(x)\partial_{vv}h(v) \end{align*}\]

The averaging function \(M(t,x)\) can also be simplified. \[ \begin{aligned}[t] M(x) &= \frac{\int_\mathbb{T}\mathrm{d} y\int_\mathbb{R}\mathrm{d} w \,g(y)h(w)\varphi(x-y) w}{\int_\mathbb{T}\mathrm{d} y\int_{\mathbb{R}} \mathrm{d} w \,g(y)h(w)\varphi(x-y)}\\ &= \frac{\left( \int_\mathbb{R}\mathrm{d} w \,w h(w)\right) \int_\mathbb{T}\mathrm{d} y\,g(y)\varphi(x-y)}{\left( \int_{\mathbb{R}}\mathrm{d} w \,h(w)\right) \int_{\mathbb{T}}\mathrm{d} y\,g(y)\varphi(x-y)} && \text{by independence}\\ &= \int_{\mathbb{R}} \mathrm{d} w \,w h(w) && \text{}\\ &= \bar{v} \end{aligned} \]

Note here that the average velocity depends on the distribution. Substituting this all in to (A.1) gives, \[\begin{equation} - vh(v)\partial_x g(x) + g(x)\partial_v\left[(v-G(\bar{v}))h(v) + \sigma \partial_v h(v)\right] = 0. \tag{A.2} \end{equation}\] Then, integrating in space, \[\begin{align*} &\partial_v\left[(v-G(\bar{v}))h(v) + \sigma \partial_v h(v)\right] = 0\\ \implies & \left[(v-G(\bar{v}))h(v) + \sigma \partial_v h(v)\right] = \mathrm{const.} \end{align*}\] where the first addend is zero by periodicity of the torus. So, \[ \partial_v h - \left(\frac{G(\bar{v})-v}{\sigma}\right)h + \mathrm{const.} = 0. \] This can be solved using an integrating factor to give \[ h(v) = C\mathrm{e}^{\frac{2G(\bar{v})v-v^2}{2\sigma}}. \] Applying the normalising constraint gives \[ h(v) = \frac{1}{\sqrt{2\pi\sigma}}\mathrm{e}^{\frac{-(v-G(\bar{v}))^2}{2\sigma}}. \] If we now multiply by \(v\) and integrate with respect to velocity we obtain \[ \bar{v} = \int v h(v)\mathrm{d}v = \int v \frac{1}{\sqrt{2\pi\sigma}}\mathrm{e}^{\frac{-(v-G(\bar{v}))^2}{2\sigma}}\mathrm{d}v = G(\bar{v}). \]

So the velocity distribution at stationarity is Gaussian with mean \(G(\langle w \rangle )\) and variance \(\sigma^2\). If we take the smooth herding function \(G(u) = \frac{\mathrm{atanh}(u)}{\mathrm{atanh}(1)}\), this corresponds to a Gaussian with mean \(-1,0\) or \(+1\). Returning to Equation (A.2), we see that the first addend must be identically zero for the equation to hold. In particular, this implies \(\partial_{x} g(x) = 0\), due to the positivity of \(h(v)\) (as it is a Gaussian). We thus deduce that \(g\) must be the uniform measure on \(\mathbb{T}\). Hence, \(f(x,v)=g(x)h(v)\) is a stationary solution of the PDE (A.1).

A.2 Convergence Depends on Initial Average Velocity

If we further assume space homogeneity of the PDE, that is, \(f_t(x,v) = h_t(v)\), we can infer exactly how the average velocity controls the dynamics. If we multiply the space homogeneous PDE by \(v\) and integrate with respect to velocity we obtain \[ \int v \partial_t h_t(v)\mathrm{d} v = -\int v \partial_v G(M(t)) h_t(v)\mathrm{d} v + \int v \partial_v vh_t(v)\mathrm{d} v + \sigma \int v \partial_vv h_t(v)\mathrm{d} v \] Integrating by parts then gives an autonomous first order ODE. \[ \partial_t M(t) = G(M(t) ) - M(t) \] As this equation is autonomous, the average velocity must be monotone. There are stationary points whenever the average velocity solves $G(M(t)) =M(t) $, as expected from the product solution found previously. For example, if we take \(G(u) = \frac{\mathrm{atanh}(u)}{\mathrm{atanh}(1)}\), then the average velocity converges to $ -1, 0 $ or \(+1\), depending on the sign of the initial average velocity. If \(M(0) > 0\), then \(M(t) \to +1\). Similarly, the stationary point at \(-1\) is stable and the system will converge there for any configuration such that \(M(0) < 0\). Finally, \(0\) is an unstable equilibrium, only attainable when \(M(0) = 0\). See Figure for a phase plane diagram of the system when \(G\) is the inverse hyperbolic tangent.

Figure A.1: Phase Plane Diagram of the Homogeneous PDE]{Dynamics of the space homogeneous ODE.

In a similar manner, we can also obtain equations for other moments. Denoting the \(n^{\text{th}}\) moment at time \(t\) by \(M_n(t)\), we obtain: \[\begin{align*} \dot{M}_0(t) &= 0\\ \dot{M}_1(t) &= G(M_1(t))-M_1(t) && \text{as above}\\ \dot{M}_2(t) &= 2\left[G(M_1)M_1-M_2+\sigma \right]. \end{align*}\] The first of these equations shows that the system conserves mass, an important property we must preserve when simulating numerically. The second is a restatement of the result above, and gives convergence of the average velocity. The final equation gives the convergence of the variance.