in the presence of a given, fixed background $\phi \in C_c^1(\bfR)$ that is neutralizing in the sense that $$\int \rho(0,y) \ dy = 0.$$ The electric and magnetic fields are given by $E(t,x) = \langle E_1(t,x),E_2(t,x) \rangle$ and $B(t,x)$, respectively. Finally, the matrix $D = v_0^{-1} ( \mathbb{I} + v \otimes v) \in \bfR^{2 \times 2}$ is the relativistic diffusion operator and possesses some desirable properties, as discovered for its three-dimensional variant in [@CalogeroRVMFP]. We note, however, that the operator $\nabla_v \cdot (D \nabla_v f)$ is not uniformly elliptic and provides less dissipation than the Laplacian $\Delta_v f$. Namely, for any $u \in \bfR^2$, $D$ satisfies $$\label{D1}
v_0^{-1} \vert u \vert^2  \leq \vert u \cdot Du \vert \leq v_0 \vert u \vert^2.$$ For initial data we take a nonnegative particle density $f^0$ with compact $x$-support and bounded moments $v_0^\b {\partial}_x^k f^0 \in L^2(\bfR^3)$, along with fields $E^0_2, B^0 \in H^2(\bfR)$. Additionally, we specify particular data for $E_1$, namely $$E_1(0,x) = \int_{-\infty}^x \left ( \int  f^0(y,w) \ dw - \phi(y) \right ) \ dy.$$ In fact, this particular choice of data for $E_1$ is the only one which leads to a solution possessing finite energy (see Lemma \[L2\] and [@GlaSch90]). The 