Hamiltonian splitting

\[\mathcal{H}=\mathcal{H}_{p}+\mathcal{H}_{A}+\mathcal{H}_{E}+\mathcal{H}_{2}+\mathcal{H}_{3},\]

where

\[\begin{aligned} \mathcal{H}_{p} & = \frac{1}{2}\int p^2 f_0 \mathrm{d}{ x}\mathrm{d}{p},\\ \mathcal{H}_{A} &= \frac{1}{2}\int |{\mathbf A}_\perp|^2 f_0 \mathrm{d}{ x}\mathrm{d}{ p}+\frac{1}{2}\int \left|\frac{\partial {\mathbf A}_\perp}{\partial x}\right|^2 \mathrm{d}{x},\\ \mathcal{H}_{E} &= \frac{1}{2}\int |{\mathbf E}|^2 \mathrm{d}{ x}=\frac{1}{2}\int (E_x^2+|{\mathbf E_\perp}|^2 ) \mathrm{d}{ x}, \\ \mathcal{H}_{2} &= \int_{\Omega} \mathfrak{h} f_2 \frac{\partial A_z}{\partial x} \mathrm{d}x\mathrm{d}p,\\ \mathcal{H}_{3} &= \int_{\Omega} -\mathfrak{h} f_3 \frac{\partial A_y}{\partial x} \mathrm{d}x\mathrm{d}p. \end{aligned} \]

Subsystem for $\mathcal{H}_p$

The subsystem $\frac{\partial \mathcal{Z}}{\partial t} = \{ \mathcal{Z}, \mathcal{H}_p \}$ associated to $\mathcal{H}_{p} = \frac{1}{2}\int p^2 f_0 \mathrm{d}{ x}\mathrm{d}{p}$ is

\[\left\{ \begin{aligned} &\frac{\partial f_0}{\partial t} = \{f_0, \mathcal{H}_{p} \} = -p\frac{\partial f_0}{\partial x}, \\ &\frac{\partial \mathbf{f}}{\partial t} = \{\mathbf{f}, \mathcal{H}_{p} \}= -p\frac{\partial \mathbf{f}}{\partial x}, \\ %&\frac{\partial f_1}{\partial t} = \{f_1, \mathcal{H}_{p} \}= -p\frac{\partial f_1}{\partial x}, \\ %&\frac{\partial f_2}{\partial t} = \{f_2, \mathcal{H}_{p} \}= -p\frac{\partial f_2}{\partial x}, \\ %&\frac{\partial f_3}{\partial t} = \{f_3, \mathcal{H}_{p} \}= -p\frac{\partial f_3}{\partial x},\\ & \frac{\partial E_x}{\partial t} = \{ E_x, \mathcal{H}_{p} \} =- \int_{\mathbb{R}} p f_0\mathrm{d}{p},\\ & \frac{\partial {\mathbf E}_\perp}{\partial t} =\frac{\partial {\mathbf A}_\perp}{\partial t} =0. %& \frac{\partial {\mathbf E}_\perp}{\partial t} = \{ {\mathbf E}_\perp, \mathcal{H}_{p} \} = {\mathbf 0},\;\; \frac{\partial {\mathbf A}_\perp}{\partial t} = \{ {\mathbf A}_\perp, \mathcal{H}_{p} \} = {\mathbf 0}. \end{aligned} \right.\]

We denote the initial value as $(f_0^0(x,p),\mathbf{f}^0(x,p), E_x^0(x),{\mathbf E}_\perp^0(x),{\mathbf A}_\perp^0(x))$ at time $t=0$. The solution at time $t$ of this subsystem can be written explicitly,

\[\begin{aligned} &f_0(x,p,t)=f_0^0(x-pt,p), \;\; \mathbf{f}(x,p,t)=\mathbf{f}^0(x-pt,p),\\ %f_2(x,p,t)=f_2^0(x-pt,p), f_3(x,p,t)=f_3^0(x-pt,p), \\ &E_x(x,t)=E_x^0(x)-\int_0^t\int_{\mathbb{R}} pf_0(x,p,\tau) \mathrm{d}p\mathrm{d}\tau=E_x^0(x)-\int_0^t\int_{\mathbb{R}} pf_0^0(x-p\tau,p) \mathrm{d}p\mathrm{d}\tau, \\ &{\mathbf E}_\perp(x,t)={\mathbf E}_\perp^0(x), \;\; {\mathbf A}_\perp(x,t)={\mathbf A}_\perp^0(x). \end{aligned}\]

Next, we check that the solution propagates the Poisson equation. To do so, we assume that the Poisson equation holds initially, i.e.

\[\frac{\partial E_x^0}{\partial x}=\int_{\mathbb{R}} f_0^0\mathrm{d}{p}-1.\]

Then we have, by differentiating the expression of $E_x(t, x)$ with respect to $x$

\[\begin{aligned} \frac{\partial E_x(x,t)}{\partial x}&=\frac{\partial E_x^0}{\partial x}-\int_0^t\int_{\mathbb{R}} p\frac{\partial f_0^0(x-p\tau,p)}{\partial x} \mathrm{d}p\mathrm{d}\tau=\frac{\partial E_x^0}{\partial x}+\int_0^t\int_{\mathbb{R}} \frac{\partial f_0^0(x-p\tau,p)}{\partial \tau} \mathrm{d}p\mathrm{d}\tau\\ &=\frac{\partial E_x^0}{\partial x}+\int_{\mathbb{R}} f_0^0(x-pt,p)\mathrm{d}p-\int_{\mathbb{R}} f_0^0(x,p)\mathrm{d}p=\int_{\mathbb{R}} f_0(x,p,t)\mathrm{d}p-1, \end{aligned}\]

which proves that the Poisson equation is satisfied at time $t$.

Subsystem for $\mathcal{H}_A$

The subsystem $\frac{\partial \mathcal{Z}}{\partial t} = \{ \mathcal{Z}, \mathcal{H}_A \}$ associated to the sub-Hamiltonian $\mathcal{H}_{A} = \frac{1}{2}\int |{\mathbf A}_\perp|^2 f_0 \mathrm{d}{\mathbf x}\mathrm{d}{\mathbf p}+\frac{1}{2}\int |\frac{\partial {\mathbf A}_\perp}{\partial x}|^2 \mathrm{d}{\mathbf x}$ is

\[\left\{ \begin{aligned} &\frac{\partial f_0}{\partial t} = \{f_0, \mathcal{H}_{A} \} = {\mathbf A}_\perp \cdot \frac{\partial {\mathbf A}_\perp}{\partial x}\frac{\partial f_0}{\partial p}, \\ &\frac{\partial {\mathbf f}}{\partial t} = \{{\mathbf f}, \mathcal{H}_{A} \}= {\mathbf A}_\perp \cdot \frac{\partial {\mathbf A}_\perp}{\partial x} \frac{\partial {\mathbf f}}{\partial p}, \\ %&\frac{\partial f_2}{\partial t} = \{f_2, \mathcal{H}_{A} \}= {\mathbf A}_\perp \cdot \frac{\partial {\mathbf A}_\perp}{\partial x} \frac{\partial f_2}{\partial p}, \\ %&\frac{\partial f_3}{\partial t} = \{f_3, \mathcal{H}_{A} \}= {\mathbf A}_\perp \cdot \frac{\partial {\mathbf A}_\perp}{\partial x} \frac{\partial f_3}{\partial p},\\ %& \frac{\partial E_x}{\partial t} = \{ E_x, \mathcal{H}_{A} \} = 0,\\ & \frac{\partial {\mathbf E}_\perp}{\partial t} = \{ {\mathbf E}_\perp, \mathcal{H}_{A} \} = - \frac{\partial^2 {\mathbf A}_\perp}{\partial x^2} + {\mathbf A}_\perp \int_{\mathbb{R}} f_0 \mathrm{d}\mathrm{p},\\ & \frac{\partial E_x}{\partial t} =\frac{\partial {\mathbf A}_\perp}{\partial t} = 0. % \{ {\mathbf A}_\perp, \mathcal{H}_{A} \} = {\mathbf 0}. \end{aligned} \right.\]

We denote by $(f_0^0(x,p),{\mathbf f}^0(x,p), E_x^0(x),{\mathbf E}_\perp^0(x),{\mathbf A}_\perp^0(x))$ the initial value at time $t=0$. The exact solution at time t is,

\[\begin{aligned} &f_0(x,p,t)=f_0^0 \left( x,p+t{\mathbf A}_\perp^0(x) \cdot \frac{\partial {\mathbf A}_\perp^0(x)}{\partial x} \right) ,\\ & {\mathbf f}(x,p,t)={\mathbf f}^0 \left( x,p+t{\mathbf A}_\perp^0(x) \cdot \frac{\partial {\mathbf A}_\perp^0(x)}{\partial x} \right), \\ &{\mathbf E}_\perp(x,t)={\mathbf E}_\perp^0(x)-t\frac{\partial^2 {\mathbf A}_\perp^0(x)}{\partial x^2} +t{\mathbf A}_\perp^0(x)\int_{\mathbb{R}} f_0^0(x,p) \mathrm{d}p, \\ &E_x(x,t)=E_x^0(x), \;\; {\mathbf A}_\perp(x,t)={\mathbf A}_\perp^0(x), \\ \end{aligned}\]

Subsystem for $\mathcal{H}_E$

The subsystem $\frac{\partial \mathcal{Z}}{\partial t} = \{ \mathcal{Z}, \mathcal{H}_E \}$ associated to the sub-Hamiltonian $\mathcal{H}_{E} = \frac{1}{2}\int |{\mathbf E}|^2 \mathrm{d}{\mathbf x}$ is

\[\left\{ \begin{aligned} &\frac{\partial f_0}{\partial t} = \{f_0, \mathcal{H}_{E} \} = -E_x \frac{\partial f_0}{\partial p}, \\ &\frac{\partial {\mathbf f}}{\partial t} = \{{\mathbf f}, \mathcal{H}_{E} \}= -E_x \frac{\partial {\mathbf f}}{\partial p}, \\ %&\frac{\partial f_2}{\partial t} = \{f_2, \mathcal{H}_{E} \}= -E_x \frac{\partial f_2}{\partial p}, \\ %&\frac{\partial f_3}{\partial t} = \{f_3, \mathcal{H}_{E} \}=-E_x \frac{\partial f_3}{\partial p},\\ %& \frac{\partial E_x}{\partial t} = \{ E_x, \mathcal{H}_{E} \} = 0,\\ & \frac{\partial {\mathbf E}_\perp}{\partial t} = \{ {\mathbf E}_\perp, \mathcal{H}_{E} \} = {\mathbf 0},\\ & \frac{\partial E_x}{\partial t} = \frac{\partial {\mathbf A}_\perp}{\partial t} = 0. % \{ {\mathbf A}_\perp, \mathcal{H}_{E} \} = -{\mathbf E}_\perp. \end{aligned} \right.\]

With the initial value$(f_0^0(x,p),{\mathbf f}^0(x,p),E_x^0(x),{\mathbf E}_\perp^0(x),{\mathbf A}_\perp^0(x))$ at time $t=0$, the solution at time t is as follows,

\[\begin{aligned} & f_0(x,p,t)=f_0^0 ( x,p-tE_x^0(x) ), \\ & {\mathbf f}(x,p,t)={\mathbf f}^0 ( x,p-tE_x^0(x) ), \\ & E_x(x,t)=E_x^0(x), \\ & {\mathbf E}_\perp(x,t)={\mathbf E}_\perp^0(x), \\ & {\mathbf A}_\perp(x,t)={\mathbf A}_\perp^0(x) -t{\mathbf E}_\perp^0(x). \end{aligned}\]

Subsystem for $\mathcal{H}_2$

The subsystem $\frac{\partial \mathcal{Z}}{\partial t} = \{ \mathcal{Z}, \mathcal{H}_2 \}$ associated to the sub-Hamiltonian $\mathcal{H}_{2} = \int_{\Omega} \mathfrak{h} f_2 \frac{\partial A_z}{\partial x} \mathrm{d}x\mathrm{d}p$ is

\[\left\{ \begin{aligned} &\frac{\partial f_0}{\partial t} = \{f_0, \mathcal{H}_{2} \} = \mathfrak{h}\frac{\partial^2 A_z}{\partial x^2}\frac{\partial f_2}{\partial p}, \\ &\frac{\partial f_1}{\partial t} = \{f_1, \mathcal{H}_{2} \}= \frac{\partial A_z }{\partial x} f_3, \\ &\frac{\partial f_2}{\partial t} = \{f_2, \mathcal{H}_{2} \}= {{\frac{\mathfrak{h}}{3}}} \frac{\partial^2 A_z}{\partial x^2}\frac{\partial f_0}{\partial p}, \\ &\frac{\partial f_3}{\partial t} = \{f_3, \mathcal{H}_{2} \}= -\frac{\partial A_z }{\partial x} f_1,\\ & \frac{\partial { E}_z}{\partial t} = \{ {E}_z, \mathcal{H}_{2} \} =-\mathfrak{h}{ \int_{\mathbb{R}} \frac{\partial f_2}{\partial x}\mathrm{d}{p}},\\ & \frac{\partial E_x}{\partial t} =\frac{\partial { E}_y}{\partial t} =\frac{\partial {\mathbf A}_\perp}{\partial t} =0. \end{aligned} \right.\]

In this subsystem, we observe some coupling between the distribution functions. To write down the exact solution, we reformulate the equations on $(f_0, {\mathbf f})$ as, using $A_z(x, t)=A_z^0(x)$

\[ \begin{aligned} & \partial_t \begin{pmatrix} f_1 \\ f_3 \end{pmatrix}-\frac{\partial A_z^0 }{\partial x} J \begin{pmatrix} f_1 \\ f_3 \end{pmatrix} =0, \\ & \partial_t \begin{pmatrix} f_0 \\ f_2 \end{pmatrix}-\mathfrak{h}\frac{\partial^2 A_z^0}{\partial x^2} \begin{pmatrix} 0 & 1 \\ \frac{1}{3}& 0 \end{pmatrix} \partial_p \begin{pmatrix} f_0 \\ f_2 \end{pmatrix} =0, \\ \end{aligned}\]

where $J$ denotes the symplectic matrix

\[J=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\]

With the initial value $(f_0^0(x,p),{\mathbf f}^0(x,p),E_x^0(x),{\mathbf E}_\perp^0(x),{\mathbf A}_\perp^0(x))$ at time $t=0$, the exact solution for the first system is

\[\begin{pmatrix} f_1 \\ f_3 \end{pmatrix}(x,p,t)=\exp{\left(\frac{\partial A_z^0(x) }{\partial x} J t\right)}\begin{pmatrix} f_1^0(x,p) \\ f_3^0(x,p) \end{pmatrix},\ \text{with}\ \exp{(Js)}=\begin{pmatrix} \cos(s) & \sin(s) \\ -\sin(s) & \cos(s) \end{pmatrix}.\]

Let us now focus on the second system

\[\partial_t \begin{pmatrix} f_0 \\ f_2 \end{pmatrix}-\mathfrak{h}\frac{\partial^2 A_z^0}{\partial x^2} \begin{pmatrix} 0 & 1 \\ \frac{1}{3}& 0 \end{pmatrix} \partial_p \begin{pmatrix} f_0 \\ f_2 \end{pmatrix} =0. \]

By the eigen-decomposition

\[\begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{1}{2}& -\frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ \frac{1}{3}& 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ \frac{1}{\sqrt{3}}& \frac{-1}{\sqrt{3}} \end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{3}}& 0 \\ 0& -\frac{1}{\sqrt{3}} \end{pmatrix},\]

then, one can diagonalize the transport equation to get

\[\partial_t \begin{pmatrix} \frac{1}{2}f_0+\frac{\sqrt{3}}{2}f_2 \\ \frac{1}{2}f_0-\frac{\sqrt{3}}{2}f_2 \end{pmatrix}- \frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2} \begin{pmatrix} 1 & 0 \\ 0& -1 \end{pmatrix} \partial_p \begin{pmatrix} \frac{1}{2}f_0+\frac{\sqrt{3}}{2}f_2 \\ \frac{1}{2}f_0-\frac{\sqrt{3}}{2}f_2 \end{pmatrix} =0. \]

Thus, we can solve the transport equation

\[\Big(\frac{1}{2}f_0\pm\frac{\sqrt{3}}{2}f_2\Big)(x,p,t)=\Big(\frac{1}{2}f_0^0 \pm\frac{\sqrt{3}}{2}f_2^0\Big)( x,p\pm t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2}(x)),\]

and compute the exact solution at time $t$ as follows,

\[\begin{aligned} &f_1(x,p,t)=\cos(t \frac{\partial A_z^0(x) }{\partial x} )f_1^0 ( x,p)+\sin(t \frac{\partial A_z^0(x) }{\partial x} )f_3^0 ( x,p), \\ &f_3(x,p,t)=-\sin(t \frac{\partial A_z^0(x) }{\partial x} )f_1^0 ( x,p)+\cos(t \frac{\partial A_z^0(x) }{\partial x} )f_3^0 ( x,p) \\ &f_0(x,p,t)=\Big(\frac{1}{2}f_0^0 +\frac{\sqrt{3}}{2}f_2^0\Big)( x,p+t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2}(x) )+\Big(\frac{1}{2}f_0^0 -\frac{\sqrt{3}}{2}f_2^0\Big)( x,p-t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2}(x) ),\\ &f_2(x,p,t)=\frac{1}{\sqrt{3}}\Big(\frac{1}{2}f_0^0 +\frac{\sqrt{3}}{2}f_2^0\Big)( x,p+t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2}(x) )-\frac{1}{\sqrt{3}}\Big(\frac{1}{2}f_0^0 -\frac{\sqrt{3}}{2}f_2^0\Big)( x,p-t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_z^0}{\partial x^2}(x) ),\\ &{\mathbf A}_\perp(x,t)={\mathbf A}_\perp^0(x), E_x(x,t)=E_x^0(x), E_y(x,t)=E_y^0(x),\\ &E_z(x,t)=E_z^0(x)-t\mathfrak{h}\int_{\mathbb{R}} \frac{\partial f_2^0}{\partial x}\mathrm{d}{p}. \end{aligned}\]

Subsystem for $\mathcal{H}_3$

The subsystem $\frac{\partial \mathcal{Z}}{\partial t} = \{ \mathcal{Z}, \mathcal{H}_3 \}$ associated to the sub-Hamiltonian $\mathcal{H}_{3} = -\int_{\Omega} \mathfrak{h} f_3 \frac{\partial A_y}{\partial x} \mathrm{d}x\mathrm{d}p$ is

\[\left\{ \begin{aligned} &\frac{\partial f_0}{\partial t} = \{f_0, \mathcal{H}_{3} \} = -\mathfrak{h}\frac{\partial^2 A_y}{\partial x^2} \frac{\partial f_3}{\partial p}, \\ &\frac{\partial f_1}{\partial t} = \{f_1, \mathcal{H}_{3} \}= \frac{\partial A_y }{\partial x} f_2, \\ &\frac{\partial f_2}{\partial t} = \{f_2, \mathcal{H}_{3} \}= - \frac{\partial A_y }{\partial x} f_1,\\ &\frac{\partial f_3}{\partial t} =- \{f_3, \mathcal{H}_{3} \}= -{{\frac{\mathfrak{h}}{3}}} \frac{\partial^2 A_y}{\partial x^2} \frac{\partial f_0}{\partial p}, \\ & \frac{\partial { E}_y}{\partial t} = \{ {E}_y, \mathcal{H}_{3} \} = {\mathfrak{h} \int_{\mathbb{R}} \frac{\partial f_3}{\partial x}\mathrm{d}{p}},\\ & \frac{\partial E_x}{\partial t} = \frac{\partial { E}_z}{\partial t} = \frac{\partial {\mathbf A}_\perp}{\partial t} =0. \end{aligned} \right.\]

This subsystem is very similar to the $\mathcal{H}_2$ one, hence, as previously, we reformulate the equations on the distribution functions as

\[\begin{aligned} & \partial_t \begin{pmatrix} f_1 \\ f_2 \end{pmatrix} -\frac{\partial A_y^0 }{\partial x} J %\begin{pmatrix} % 0 & -1 \\ % 1 & 0 %\end{pmatrix} \begin{pmatrix} f_1 \\ f_2 \end{pmatrix} =0, \\ & \partial_t \begin{pmatrix} f_0 \\ f_3 \end{pmatrix}+\mathfrak{h}\frac{\partial^2 A_y^0}{\partial x^2} \begin{pmatrix} 0 & 1 \\ \frac{1}{3}& 0 \end{pmatrix} \partial_p \begin{pmatrix} f_0 \\ f_3 \end{pmatrix} =0, \\ %& \frac{\partial E_x}{\partial t}= 0,\ %\frac{\partial { E}_y}{\partial t} = {\mathfrak{h} \int_{\mathbb{R}} \frac{\partial f_3}{\partial x}\mathrm{d}{p}},\\ %& \frac{\partial { E}_z}{\partial t} = \{ {E}_z, H_{3} \} = { 0},\ %\frac{\partial {\mathbf A}_\perp}{\partial t} = \{ {\mathbf A}_\perp, H_{3} \} = 0, \end{aligned}\]

with initial value

$(f_0^0(x,p),{\mathbf f}^0(x,p),E_x^0(x),{\mathbf E}_\perp^0(x),{\mathbf A}_\perp^0(x))$ at time $t=0$. We derive similar formula with $\mathcal{H}_2$ for the exact solution at time $t$

\[\begin{aligned} &f_1(x,p,t)=\cos(t \frac{\partial A_y^0(x) }{\partial x} )f_1^0 ( x,p)+\sin(t \frac{\partial A_y^0(x) }{\partial x} )f_2^0 ( x,p), \\ &f_2(x,p,t)=-\sin(t \frac{\partial A_y^0(x) }{\partial x} )f_1^0 ( x,p)+\cos(t \frac{\partial A_y^0(x) }{\partial x} )f_2^0 ( x,p), \\ &f_0(x,p,t)=\Big(\frac{1}{2}f_0^0 +\frac{\sqrt{3}}{2}f_3^0\Big)( x,p-t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_y^0}{\partial x^2}(x) )+\Big(\frac{1}{2}f_0^0 -\frac{\sqrt{3}}{2}f_3^0\Big)( x,p+t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_y^0}{\partial x^2}(x) ),\\ &f_3(x,p,t)=\frac{1}{\sqrt{3}}\Big(\frac{1}{2}f_0^0 +\frac{\sqrt{3}}{2}f_3^0\Big)( x,p-t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_y^0}{\partial x^2}(x) )-\frac{1}{\sqrt{3}}\Big(\frac{1}{2}f_0^0 -\frac{\sqrt{3}}{2}f_3^0\Big)( x,p+t\frac{\mathfrak{h}}{\sqrt{3}}\frac{\partial^2 A_y^0}{\partial x^2}(x) ), \end{aligned} \]

\[\begin{aligned} & E_y(x,t)=E_y^0(x)+t\mathfrak{h}\int_{\mathbb{R}} \frac{\partial f_3^0}{\partial x}\mathrm{d}{p}, \\ & {\mathbf A}_\perp(x,t)={\mathbf A}_\perp^0(x), \\ & E_x(x,t)=E_x^0(x), \\ & E_z(x,t)=E_z^0(x). \end{aligned} \]

To compute the solution $E_y(x,t)$, we use the fact that

\[\int_{\mathbb{R}} f_3(x,p,t) \mathrm{d}p =\int_{\mathbb{R}} f_3^0(x,p)\mathrm{d} p.\]