# Long term analysis of splitting methods for charged-particle dynamics

Xicui Li  Bin Wang  School of Mathematics and Statistics, Xi’an Jiaotong
University, 710049 Xi’an, China
lixicui<EMAIL_ADDRESS>
###### Abstract

In this paper, we rigorously analyze the energy, momentum and magnetic moment
behaviours of two splitting methods for solving charged-particle dynamics. The
near-conservations of these invariants are given for the system under constant
magnetic field or quadratic electric potential. By the approach named as
backward error analysis, we derive the modified equations and modified
invariants of the splitting methods and based on which, the near-conservations
over long times are proved. Some numerical experiments are presented to
demonstrate these long time behaviours.

###### keywords:

Splitting method, charged particle dynamics, backward error analysis, near
conservations, long term analysis.

††journal:

Mathematics Subject Classification (2010): 65L05, 65P10, 78A35, 78M25

##  1 Introduction

In this paper, we formulate and analyze two splitting methods for solving the
charged-particle dynamics (CPD) which is described by

$\begin{split}&\ddot{x}=v\times B(x)+E(x),\quad\ x(0)=x_{0},\quad
v(0)=v_{0},\quad t\geq 0,\end{split}$ (1)

with the position $x(t)\in\mathbb{R}^{3}$ and the velocity
$v(t):=\dot{x}(t)\in\mathbb{R}^{3}$, where $B(x)=\nabla_{x}\times
A(x)\in\mathbb{R}^{3}$ denoted by $(B_{1}(x),B_{2}(x),B_{3}(x))^{\intercal}$
is a non-uniform magnetic field which is determined by the vector potential
$A(x)$, and $E(x)=-\nabla_{x}U(x)$ is a given electric field generated by some
scalar potential $U(x)$. Charged particle dynamics plays a fundamental role in
plasma physics ([2]) and there has been a lot of effective numerical methods
in recent years for solving this system. Boris algorithm [3] has been widely
used in the simulation of magnetized plasma due to its excellent properties
[29, 28]. Many other kinds of methods have also been developed and analyzed,
such as splitting methods [25, 21], exponential integrators [11],
asymptotically preserving methods [10, 9], filtered Boris algorithms [17],
uniformly accurate methods [7, 6, 8] and so on.

On the other hand, in the past over three decades, structure-preserving
algorithms have arisen in various fields of applied sciences. From the
scientific computing point of view, it is conventional to require numerical
methods to preserve the qualitative features of the true solution as much as
possible when applied to a differential equation. For the system of CPD (1),
we pay attention to three important invariants. With the Euclidean norm
$\left|\cdot\right|$, the total energy is denoted by ([4, 36])

$H(x,v)=\frac{1}{2}\left|v\right|^{2}+U(x).$ (2)

It is well known that the solution of (1) exactly conserves this energy. For
the scalar and vector potentials, if they are assumed to satisfy the
invariance properties

$U(e^{\tau S}x)=U(x),\quad e^{-\tau S}A(e^{\tau
S}x)=A(x),\quad\forall\tau\in\mathbb{R},$ (3)

the momentum

$M(x,v)=(v+A(x))^{\intercal}Sx$

is preserved by the solution of (1) ([12]), where $S$ is a skew-symmetric
matrix. Moreover, if we consider the system (1) under the constant magnetic
field, i.e. $B(x)\equiv B$ and $S$ satisfies $Sv=v\times B$, then
$A(x)=-\frac{1}{2}x\times B$ and the momentum becomes
$M(x,v)=v^{\intercal}(x\times B)-\frac{1}{2}\left|x\times B\right|^{2}.$ The
third invariant is the following magnetic moment

$I(x,v)=\frac{\left|v\times
B(x)\right|^{2}}{2\left|B(x)\right|^{3}}=\frac{\left|v_{\perp}\right|^{2}}{2\left|B(x)\right|},$

where $v_{\perp}:=\frac{v\times B(x)}{\left|B(x)\right|}$ is orthogonal to
$B(x)$. Based on the analysis of [1, 5, 27], it is known that this magnetic
moment is an _adiabatic invariant_.

From the point of long time scientific computation, it is of great interest to
investigate the long time behaviour of numerical methods. In order to get
numerical methods with near/exact conservation of qualitative features of the
CPD, various numerical methods have been formulated and analyzed. Broadly
speaking, up to the present, four categories of structure-preserving
algorithms for the CPD have been in the center of research: energy-preserving
methods [4, 23, 22, 30, 24, 34, 32] to preserve the energy, volume-preserving
schemes [19] to preserve the volume, symplectic algorithms [35, 31, 37, 20] to
preserve the symplecticity and numerical methods [16, 33, 12, 13, 15, 14] with
near conservation of qualitative features. Recently splitting methods have
been considered for solving CPD [21, 34, 24] and energy-preserving splitting
methods have been constructed. However, these publications focus on the
analysis of the accuracy and energy-preserving property of splitting methods.
It seems that the long-time analysis of classical splitting methods has not
been considered and the behaviour of energy-preserving splitting methods
concerning other structure-preserving aspects has not been investigated in the
literature, such as the long-time numerical conservation of momentum and
magnetic moment. _From the perspective of structure preserving algorithm, this
is far from enough_. Motivated by this point, this paper applies backward
error analysis for two splitting methods to gain insight into their long-term
energy, momentum and magnetic moment performance.

The rest of the paper is organized as follows. In section 2, we discribe two
different symmetric splitting methods and study their local errors. Then, in
section 3, the main results on the numeircal long time conservation are
presented and three numerical experiments are carried out to support the
theoretical results. The complete analysis of the results are rigorously given
in section 4. The lastsection is devoted to the conclusions of this paper.

##  2 Numerical Methods

In this section, we present two splitting methods and show their elementary
features. To formulate the methods, we split the equation (1) into two
subflows:

$\frac{d}{dt}\begin{pmatrix}x\\\ v\\\ \end{pmatrix}=\begin{pmatrix}0\\\
v\times B(x)\\\ \end{pmatrix},\qquad\frac{d}{dt}\begin{pmatrix}x\\\ v\\\
\end{pmatrix}=\begin{pmatrix}v\\\ E(x)\\\ \end{pmatrix}.$ (4)

Then for the first subflow, which is integrable, one can obtain its exact
solution:

$\Phi^{L}_{t}:\\\ \left(\begin{array}[]{c}\xi(t)\\\ \eta(t)\\\
\end{array}\right)=\left(\begin{array}[]{c}\xi(0)\\\
e^{t\tilde{B}(\xi(0))}\eta(0)\\\ \end{array}\right),\quad t\geq 0,$ (5)

where $\tilde{B}(x)=\begin{pmatrix}0&B_{3}(x)&-B_{2}(x)\\\
-B_{3}(x)&0&B_{1}(x)\\\ B_{2}(x)&-B_{1}(x)&0\\\ \end{pmatrix}$ is defined by
$v\times B(x)=\tilde{B}(x)v$. For the second subflow (a canonical Hamiltonian
system), we consider applying the average vector field (AVF) formula [26] to
get its numerical propagator:

$\Phi^{iL}_{t}:\ \ \begin{pmatrix}{\xi(t)}\\\ {\eta(t)}\\\
\end{pmatrix}=\begin{pmatrix}\xi(0)+t\eta(0)+\frac{t^{2}}{2}\int_{0}^{1}E\left(\rho\xi(0)+(1-\rho)\xi(t)\right)d\rho\\\
\eta(0)+t\int_{0}^{1}E\left(\rho\xi(0)+(1-\rho)\xi(t)\right)d\rho\\\
\end{pmatrix}.$ (6)

Based on this splitting, two methods are formulated as follows.

######  Algorithm 2.1 (Implicit splitting method)

We denote the numerical solution as $x^{n}\approx x(t_{n}),\,v^{n}\approx
v(t_{n})$ and choose $x^{0}=x_{0},\,v^{0}=v_{0}$. Taking a symmetric version
[18]:
$\Phi_{h}^{i}=\Phi_{\frac{h}{2}}^{L}\circ\Phi_{h}^{iL}\circ\Phi_{\frac{h}{2}}^{L},$
then we get its total formula for solving (1): for $n\geq 0$,

$\displaystyle x^{n+1}=$ $\displaystyle
x^{n}+he^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h^{2}}{2}\int_{0}^{1}E\left(\rho
x^{n}+(1-\rho)x^{n+1}\right)d\rho,$ (7) $\displaystyle v^{n+1}=$
$\displaystyle
e^{\frac{h}{2}\left({\tilde{B}(x^{n+1})+\tilde{B}(x^{n})}\right)}v^{n}+he^{\frac{h}{2}{\tilde{B}(x^{n+1})}}\int_{0}^{1}E\left(\rho
x^{n}+(1-\rho)x^{n+1}\right)d\rho.$

Obviously, it is an implicit and typical Strang splitting scheme, and we shall
use IMS-O2 to denote the method.

If we make some adjustments to (6), we can get the following explicit
algorithm.

######  Algorithm 2.2 (Explicit splitting method)

For the second subflow in (4), we linearize the nonlinear integrals in (6),
and then it is transformed to

$\Phi^{eL}_{t}:\\\ \begin{pmatrix}{\xi(t)}\\\ {\eta(t)}\\\
\end{pmatrix}=\begin{pmatrix}\xi(0)+t\eta(0)+\frac{t^{2}}{2}E(\xi(0))\\\
\eta(0)+\frac{t}{2}\left[E(\xi(0))+E(\xi(t))\right]\\\ \end{pmatrix}.$ (8)

In that way, the Strang splitting scheme
$\Phi_{h}^{e}:=\Phi_{\frac{h}{2}}^{L}\circ\Phi_{h}^{eL}\circ\Phi_{\frac{h}{2}}^{L}$
for solving (1) reads:

$\displaystyle x^{n+1}=$ $\displaystyle
x^{n}+he^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h^{2}}{2}E(x^{n}),$ (9)
$\displaystyle v^{n+1}=$ $\displaystyle
e^{\frac{h}{2}\left({\tilde{B}(x^{n+1})+\tilde{B}(x^{n})}\right)}v^{n}+\frac{h}{2}e^{\frac{h}{2}{\tilde{B}(x^{n+1})}}\left[E(x^{n})+E(x^{n+1})\right],$

and we shall call it as EXS-O2.

In what follows, we will study the basic properties of these two methods. We
begin with their symmetry. A numerical method denoted by
$y^{n+1}=\Phi_{h}(y^{n})$ is called symmetric if exchanging
$y^{n}\leftrightarrow y^{n+1}$ and $h\leftrightarrow-h$ leaves the method
unaltered. It has been pointed out in [18] that symmetric methods have
excellent long time behavior and play a central role in the geometric
numerical integration of differential equations. The following theorem states
the symmetry of Algorithms 2.1 and 2.2.

######  Theorem 2.3 (Symmetry)

From the symmetric versions:
$\Phi_{h}^{i}=\Phi_{\frac{h}{2}}^{L}\circ\Phi_{h}^{iL}\circ\Phi_{\frac{h}{2}}^{L}$
and
$\Phi_{h}^{e}=\Phi_{\frac{h}{2}}^{L}\circ\Phi_{h}^{eL}\circ\Phi_{\frac{h}{2}}^{L}$,
it follows that Algorithms 2.1 and 2.2 are symmetric.

For these two algorithms, we next investigate and briefly show their local
errors.

######  Theorem 2.4 (Local errors)

For the two methods presented above, under the local assumptions
$x(t_{n})=x^{n},v(t_{n})=v^{n},$ the local errors are given as

$x^{n+1}-x(t_{n+1})=\mathcal{O}(h^{3}),\quad
v^{n+1}-v(t_{n+1})=\mathcal{O}(h^{3}),$

where the constants symbolized by $\mathcal{O}$ are independent of $n$ and
$h$.

Proof.  For simplicity of notations, we denote

$y:=\begin{pmatrix}x\\\ v\end{pmatrix},\quad f(y):=\begin{pmatrix}v\\\ v\times
B(x)+E(x)\end{pmatrix},\quad f_{1}(y):=\begin{pmatrix}0\\\ v\times
B(x)\end{pmatrix},\quad f_{2}(y)=\begin{pmatrix}v\\\ E(x)\end{pmatrix},$

and let $\Phi_{h}$ stand for the exact flow. Then the original system can be
rewritten as $\dot{y}=f(y)=f_{1}(y)+f_{2}(y)$.

By Taylor expansions, we reach

$\displaystyle\Phi_{h}(y)=\left[Id+hf+\frac{h^{2}}{2}f^{\prime}f\right](y)+\mathcal{O}(h^{3}),\quad\Phi_{h}^{L}(y)=\left[Id+hf_{1}+\frac{h^{2}}{2}f^{\prime}_{1}f_{1}\right](y)+\mathcal{O}(h^{3}),$
$\displaystyle\Phi_{h}^{iL}(y)=\left[Id+hf_{2}+\frac{h^{2}}{2}f^{\prime}_{2}f_{2}\right](y)+\mathcal{O}(h^{3}),\quad\Phi_{h}^{eL}(y)=\left[Id+hf_{2}+\frac{h^{2}}{2}f^{\prime}_{2}f_{2}\right](y)+\mathcal{O}(h^{3}).$

Note that $\Phi_{h}^{iL}(y)\neq\Phi_{h}^{eL}(y)$ due to their different
coefficients of the term $\mathcal{O}(h^{3})$. After some calculations, we get

$\displaystyle\Phi_{h}^{i}(y)$
$\displaystyle=\Phi_{\frac{h}{2}}^{L}\circ\Phi_{h}^{iL}\circ\Phi_{\frac{h}{2}}^{L}(y)=\left[Id+h(f_{1}+f_{2})+\frac{h^{2}}{2}(f^{\prime}_{1}f_{1}+f^{\prime}_{1}f_{2}+f^{\prime}_{2}f_{1}+f^{\prime}_{2}f_{2})\right](y)+\mathcal{O}(h^{3})$
$\displaystyle=\Phi_{h}(y)+\mathcal{O}(h^{3}).$

Even if EXS-O2 is the linearization of IMS-O2, the expansions of
$\Phi_{h}^{iL}$ and $\Phi_{h}^{eL}$ are similar. Hence, it can be verified
that this makes no difference to the error result, i.e.,
$\Phi_{h}^{e}(y)-\Phi_{h}(y)=\mathcal{O}(h^{3}).$ $\blacksquare$

##  3 Main results and numerical experiments

###  3.1 Main results

In this subsection, we assume that the scalar potential $U(x)$ and the
magnetic field $B(x)$, as functions of $x$, are arbitrarily differentiable. In
this paper, $B(x)\equiv B$ represents the constant magnetic field and
$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$ with a symmetric matrix $Q$
refers to the quadratic scalar potential. Meanwhile, it is assumed that there
exists a compact set (independent of $h$) such that the result $(x^{n},v^{n})$
produced by the considered method stays in this set. Then the energy, momentum
and magnetic moment behaviours of Algorithms 2.1 and 2.2 are given as follows
one by one.

######  Theorem 3.1 (Energy conservation)

The method IMS-O2 exactly preserves the energy (2) of CPD [24], i.e.

$IMS-O2:\qquad H(x^{n},v^{n})=H(x^{0},v^{0})\qquad for\qquad nh\leq T.$

We assume that $B(x)\equiv B$ or
$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$, then the energy (2) along
the numerical solution over long times is conserved as follows

$EXS-O2:\ \ {H(x^{n},v^{n})=H(x^{0},v^{0})+\mathcal{O}(h^{2})}\quad\text{for}\
\ nh\leq h^{2-N},$ (10)

with an arbitrarily large positive integer $N$.

Moreover, the method EXS-O2 has an exact modified energy conservation which is
stated as below. If the scalar potential $U(x)$ is quadratic, EXS-O2 (9)
exactly preserves the modified energy

$H_{h}(x,v)=\frac{1}{2}\left|v\right|^{2}+U(x)-\frac{h^{2}}{8}\left|\nabla
U(x)\right|^{2}$ (11)

at the discrete level, i.e.

$EXS-O2:\qquad H_{h}(x^{n+1},v^{n+1})=H_{h}(x^{n},v^{n})\qquad for\qquad
nh\leq T.$ (12)

######  Theorem 3.2 (Momentum conservation)

Under the conditions (3) and the following two assumptions:

  * 1.

$B(x)\equiv B$ and $Sv=v\times B$,

  * 2.

$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$ and $QS=SQ$,

the momentum $M(x,v)=(v+A(x))^{\intercal}Sx$ along the numerical solution over
long times is nearly preserved as

${M(x^{n},v^{n})=M(x^{0},v^{0})+\mathcal{O}(h^{2})}\quad\ \text{for}\quad\
nh\leq h^{2-N},$ (13)

where $N$ is an arbitrarily large positive integer.

######  Theorem 3.3 (Magnetic moment conservation)

Assume that the following two assumptions hold:

  * 1.

$B(x)\equiv B$ ,

  * 2.

$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$ and
$Q\widehat{B}=\widehat{B}Q$ with
$v\times\frac{B}{\left|B\right|}=\widehat{B}v$,

then the near conservation of magnetic moment $I(x,v)$ is

${I(x^{n},v^{n})=I(x^{0},v^{0})+\mathcal{O}(h^{2})}\quad\text{for}\quad nh\leq
h^{2-N},$ (14)

with an arbitrarily large positive integer $N$.

###  3.2 Numerical experiments

In this part, by means of MATLAB, we present three numerical experiments to
show the long time energy, momentum, and magnetic moment behaviour of the
above splitting methods. We choose Boris method for comparison, and solve all
the tests on $[0,10000]$ with the step size $h=0.01$ to show the long time
conservations. For IMS-O2, the fixed-point iteration is employed into implicit
iteration, where the error tolerance is $10^{-16}$ and the maximum number of
each iteration is 50. In addition, we make use of Gauss-Legendre to deal with
the nonlinear integral in (7). Denote the relative errors of energy, momentum
and magnetic moment respectively by

$e_{H}:=\frac{\left|H(x^{n},v^{n})-H(x^{0},v^{0})\right|}{\left|H(x^{0},v^{0})\right|},\
\
e_{M}:=\frac{\left|M(x^{n},v^{n})-M(x^{0},v^{0})\right|}{\left|M(x^{0},v^{0})\right|},\
\
e_{I}:=\frac{\left|I(x^{n},v^{n})-I(x^{0},v^{0})\right|}{\left|I(x^{0},v^{0})\right|}.$
(15)

For the momentum $M(x,v)=(v+A(x))^{\intercal}Sx$, throughout this subsection,
the skew-symmetric matrix is given as $S=\begin{pmatrix}0&1&0\\\ -1&0&0\\\
0&0&0\\\ \end{pmatrix}$ , i.e.
$M(x,v)=(v_{1}+A_{1}(x))x_{2}-(v_{2}+A_{2}(x))x_{1}.$

---

Figure 1: Problem 1. Evolution of the energy error $e_{H}$ as function of time
$t$.

---

Figure 2: Problem 1. Evolution of the energy error $e_{M}$ as function of time
$t$.

---

Figure 3: Problem 1. Evolution of the energy error $e_{I}$ as function of time
$t$.

Problem 1. (Quadratic scalar potential and constant magnetic field) We first
consider the charged particle dynamics (1) with the quadratic scalar potential
$U(x)=\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{100}$ and the constant magnetic
field
$B=-\nabla\times\frac{1}{2\varepsilon}(x_{2},-x_{1},0)^{\intercal}={\frac{1}{\varepsilon}}(0,0,1)^{\intercal}.$
Particularly when $\varepsilon=1$, we have
$\widehat{B}=\frac{\tilde{B}}{\left|B\right|}=\tilde{B}=S$ and $QS=SQ$. The
initial values are chosen as $x(0)=(0,1,0.1)^{\intercal}$ and
$v(0)=(0.09,0.05,0.20)^{\intercal}$. The errors in (15) are respectively
displayed in Figs. 1–3. In addition, let
$e_{H_{h}}:=\frac{\left|H_{h}(x^{n},v^{n})-H_{h}(x^{0},v^{0})\right|}{\left|H_{h}(x^{0},v^{0})\right|}$
with $H_{h}$ (11). The conservation of $e_{H_{h}}$ by EXS-O2 is displayed in
Fig. 4.

---

Figure 4: Problem 1. Evolution of the energy error $e_{H_{h}}$ for EXS-O2 and
$e_{H}$ for IMS-O2 as functions of time $t$ with
$\varepsilon=1,1/2^{3},1/2^{6}$.

---

Figure 5: Problem 2. Evolution of the energy error $e_{H}$ as function of time
$t$.

Problem 2. (Constant magnetic field) As the second problem, we consider a
general scalar potential $U(x)=\frac{1}{100\sqrt{x_{1}^{2}+x_{2}^{2}}}.$ The
magnetic field and the initial values are the same as those in Problem 3.2.
The errors in (15) are respectively presented in Figs. 5$-$7.

Problem 3. (General case) In the third numerical experiment, we are interested
in a general case that the scalar potential is the same as it in Problem 3.2
and the magnetic field is

$B(x)=\nabla\times\frac{1}{3\varepsilon}(-x_{2}\sqrt{x_{1}^{2}+x_{2}^{2}},-x_{1}\sqrt{x_{1}^{2}+x_{2}^{2}},0)^{\intercal}={\frac{1}{\varepsilon}}(0,0,\sqrt{x_{1}^{2}+x_{2}^{2}})^{\intercal}.$

The initial values are the same as those in Problem 3.2. Then errors in (15)
are severally shown in Figs. 8$-$10. Besides, in order to show the accuracy of
the methods, we plot the global errors
$error:=\frac{\left|x^{n}-x(t_{n})\right|}{x(t_{n})}+\frac{\left|v^{n}-v(t_{n})\right|}{v(t_{n})}$
at $t_{end}=1$ in Fig 11, where the reference solution is obtained by using
“ode45” of MATLAB.

In accordance with these numerical results, we have the following
observations.

---

Figure 6: Problem 2. Evolution of the energy error $e_{M}$ as function of time
$t$.

---

Figure 7: Problem 2. Evolution of the energy error $e_{I}$ as function of time
$t$.

$\bullet$ Energy conservation. Clearly from Figs. 1, 5 and 8, we can see that
only IMS-O2 is energy-preserving, Boris and EXS-O2 show a long-term energy
behavior, and EXS-O2 behaves better than Boris. Meanwhile, Fig 4 demonstrates
that EXS-O2 exactly preserves the energy $H_{h}$.

$\bullet$ Momentum conservation. In the light of the results given by Figs. 2,
6 and 9, all the three methods are not momentum-preserving but they hold a
long time momentum conservation. Besides, the behaviour of our methods is
better than the Boris algorithm.

---

Figure 8: Problem 3. Evolution of the energy error $e_{H}$ as function of time
$t$.

---

Figure 9: Problem 3. Evolution of the energy error $e_{M}$ as function of time
$t$.

---

Figure 10: Problem 3. Evolution of the energy error $e_{I}$ as function of
time $t$.

---

Figure 11: Problem 3. The global errors _error_ with $t=1$ and
$h=\frac{1}{2^{k}}$ for $k=6,\dots,12$ under different $\varepsilon$.

$\bullet$ Magnetic Moment conservation. Observing Figs. 3, 7 and 10, all the
methods have similar magnetic moment conservations over long times.

$\bullet$ Accuracy. Fig. 11 confirms that our IMS-O2 and EXS-O2 both have
second-order accuracy and they perform better than the Boris algorithm.

The numerical results of Problems 1 and 2 support the theoretical conclusions
given in Section 3.1. It is noted that for the general system (Problem 3.2),
the methods also show a long time conservation in the energy, momentum, and
magnetic moment, which is more than expected.

##  4 Proofs of the main results

In this section, we rigorously prove the main results given in Section 3.1.
The technical tool so called the backward error analysis [18] which is
indispensible for the study of equations or numerical solutions over long
times will be used in the following proofs.

###  4.1 Backward error analysis

Following the analysis of backward error analysis, the main idea is to find a
modified differential equation whose solution $z(t)$ at $t=nh$ is equivalent
to the numerical result $x^{n}$ produced by the considered method. EXS-O2. On
the basis of the first equation in (9) and the symmetry of EXS-O2, we get

$x^{n-1}=x^{n}-he^{-\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h^{2}}{2}E(x^{n}).$

This result and the first one of (9) give

$\displaystyle
x^{n+1}-2x^{n}+x^{n-1}=h\left(e^{\frac{h}{2}\tilde{B}(x^{n})}-e^{-\frac{h}{2}\tilde{B}(x^{n})}\right)v^{n}+h^{2}E(x^{n}),$
$\displaystyle
x^{n+1}-x^{n-1}=h\left(e^{\frac{h}{2}\tilde{B}(x^{n})}+e^{-\frac{h}{2}\tilde{B}(x^{n})}\right)v^{n}.$

Elimination of $v^{n}$ leads to

$\frac{x^{n+1}-2x^{n}+x^{n-1}}{h^{2}}=\frac{2}{h}\frac{e^{\frac{h}{2}\tilde{B}(x^{n})}-e^{-\frac{h}{2}\tilde{B}(x^{n})}}{e^{\frac{h}{2}\tilde{B}(x^{n})}+e^{-\frac{h}{2}\tilde{B}(x^{n})}}\frac{x^{n+1}-x^{n-1}}{2h}+E(x^{n}).$

According to the special scheme of $\tilde{B}(x)$, it is obtained that

$\frac{2}{h}\frac{e^{\frac{h}{2}\tilde{B}(x^{n})}-e^{-\frac{h}{2}\tilde{B}(x^{n})}}{e^{\frac{h}{2}\tilde{B}(x^{n})}+e^{-\frac{h}{2}\tilde{B}(x^{n})}}=\frac{\tan\left(\frac{h}{2}\left|B(x^{n})\right|\right)}{\frac{h}{2}\left|B(x^{n})\right|}\tilde{B}(x^{n}),$

and such that

$\frac{x^{n+1}-2x^{n}+x^{n-1}}{h^{2}}=\frac{\tan\left(\frac{h}{2}\left|B(x^{n})\right|\right)}{\frac{h}{2}\left|B(x^{n})\right|}\frac{x^{n+1}-x^{n-1}}{2h}\times
B(x^{n})+E(x^{n}).$

For a fixed $t$, the function $z(t)$ has to satisfy

$\frac{z(t+h)-2z(t)+z(t-h)}{h^{2}}=\frac{\tan\left(\frac{h}{2}\left|B(z(t))\right|\right)}{\frac{h}{2}\left|B(z(t))\right|}\frac{z(t+h)-z(t-h)}{2h}\times
B(z(t))+E(z(t)).$

We let $z:=z(t)$ and expand the above functions in powers of $h$, so that

$\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots)\times
B(z)+E(z).$ (16)

IMS-O2. In an analogous way, the scheme (7) of IMS-O2 can be formulated as

$\displaystyle\frac{x^{n+1}-2x^{n}+x^{n-1}}{h^{2}}$ $\displaystyle=$
$\displaystyle\frac{2}{h}\frac{e^{\frac{h}{2}\tilde{B}(x^{n})}-e^{-\frac{h}{2}\tilde{B}(x^{n})}}{e^{\frac{h}{2}\tilde{B}(x^{n})}+e^{-\frac{h}{2}\tilde{B}(x^{n})}}\tilde{B}(x^{n})\left[\frac{x^{n+1}-x^{n-1}}{2h}-\frac{h}{4}\int_{0}^{1}\left[E\left(\rho
x^{n}+(1-\rho)x^{n+1}\right)-E\left(\rho
x^{n}+(1-\rho)x^{n-1}\right)\right]d\rho\right]$
$\displaystyle+\frac{1}{2}\int_{0}^{1}\left[E\left(\rho
x^{n}+(1-\rho)x^{n+1}\right)+E\left(\rho
x^{n}+(1-\rho)x^{n-1}\right)\right]d\rho$ $\displaystyle=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(x^{n})\right|\right)}{\frac{h}{2}\left|B(x^{n})\right|}\left[\frac{x^{n+1}-x^{n-1}}{2h}-\frac{h}{4}\int_{0}^{1}\left[E\left(x^{n}+\rho(x^{n+1}-x^{n})\right)-E\left(x^{n}+\rho(x^{n-1}-x^{n})\right)\right]d\rho\right]\times
B(x^{n})$
$\displaystyle+\frac{1}{2}\int_{0}^{1}\left[E\left(x^{n}+\rho(x^{n+1}-x^{n})\right)+E\left(x^{n}+\rho(x^{n-1}-x^{n})\right)\right]d\rho$
$\displaystyle=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(x^{n})\right|\right)}{\frac{h}{2}\left|B(x^{n})\right|}\left[\frac{x^{n+1}-x^{n-1}}{2h}-\left(\frac{h}{8}E^{\prime}(x^{n})(x^{n+1}-x^{n-1})+\cdots\right)\right]\times
B(x^{n})$
$\displaystyle+E(x^{n})+\frac{1}{4}E^{\prime}(x^{n})(x^{n+1}-2x^{n}+x^{n-1})+\cdots,$

where the following fact is used here

$\displaystyle\int_{0}^{1}{E}\left(\rho
x^{n}+(1-\rho)x^{n+1}\right)d\rho=\int_{0}^{1}{E}\left(\rho
x^{n+1}+(1-\rho)x^{n}\right)d\rho=\int_{0}^{1}{E}\left(x^{n}+\rho(x^{n+1}-x^{n})\right)d\rho$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\left[E(x^{n})+E^{\prime}(x^{n})\rho(x^{n+1}-x^{n})+\cdots\right]d\rho=E(x^{n})+\frac{1}{2}E^{\prime}(x^{n})(x^{n+1}-x^{n})+\cdots.$

Therefore, the function $z(t)$ satisfies

$\displaystyle\frac{z(t+h)-2z(t)+z(t-h)}{h^{2}}$ $\displaystyle=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(z(t))\right|\right)}{\frac{h}{2}\left|B(z(t))\right|}\left[\frac{z(t+h)-z(t-h)}{2h}-\left(\frac{h}{8}E^{\prime}(z(t))(z(t+h)-z(t-h))+\cdots\right)\right]\times
B(z(t))$
$\displaystyle+E(z(t))+\frac{1}{4}E^{\prime}(z(t))\left[z(t+h)-2z(t)+z(t-h)\right]+\cdots.$

Letting $z:=z(t)$ and expanding the above functions in powers of $h$, we
obtain

$\displaystyle\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}\left[\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)-\left(\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right)\right]\times
B(z)$ (17)
$\displaystyle+E(z)+\frac{h^{2}}{4}E^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots.$

We note that when $h=0$, the equations (16) and (17) are both identical to
(1). In fact, differentiating these two equations recursively and considering
$h=0$ shows that the third and higher derivatives depend on $(z,\dot{z})$.
Then we get a modified equation in even powers of $h$:

$\ddot{z}=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}\dot{z}\times
B(z)+E(z)+h^{2}F_{2}(z,\dot{z})+h^{4}F_{4}(z,\dot{z})+\cdots,$

where the coefficient functions depend on $(z,\dot{z})$. By taking inner
product on both sides of (16) and (17) with different expressions, we can
prove Theorem 3.1–3.3 in the following three subsections, respectively.

###  4.2 Proof of the energy conservation (Theorem 3.1)

Since it has been shown in [24] that IMS-O2 is an energy-preserving method,
here we just prove the exact conservation of modified energy and the long time
behaviour of energy for EXS-O2. To prove the result, we need to derive two
almost invariants which are close to the energy $H(x,v)$. The first one is
deduced from the following lemma and the second is obtained with the help of
Lemma 4.4.

###### Lemma 4.1

It is obtained a function

$H_{h}(x,v)=H(x,v)+h^{2}H_{2}(x,v)+h^{4}H_{4}(x,v)+\cdots$

such that

$\frac{d}{dt}H_{h}{(z,\dot{z})}=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}\dot{z}^{\intercal}\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)\times
B(z)+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16), where the
functions $H_{2j}(x,v)$ are independent of the step size $h$ and
$\mathcal{O}(h^{N})$ is the trunction term.

Proof.  Multiplying (16) with $\dot{z}^{\intercal}$ and using the fact that

$\dot{z}^{\intercal}z^{(2k)}=\frac{d}{dt}\left(\dot{z}^{\intercal}z^{(2k-1)}-\ddot{z}^{\intercal}z^{(2k-2)}+\cdots+\frac{(-1)^{k+1}}{2}\left(z^{(k)}\right)^{\intercal}z^{(k)}\right),$

we get

$\frac{d}{dt}\left(\frac{1}{2}\dot{z}^{\intercal}\dot{z}+U(z)+\frac{h^{2}}{12}\left(\dot{z}^{\intercal}\dddot{z}-\frac{1}{2}\ddot{z}^{\intercal}\ddot{z}\right)+\cdots\right)=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}\dot{z}^{\intercal}\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)\times
B(z)+\mathcal{O}(h^{N}).$

Then the result of this lemma is immediately obtained. $\blacksquare$

###### Corollary 4.2

(The first almost invariant close to energy) If the magnetic field is
constant, i.e., $B(x)\equiv B$, we have a function

$\tilde{H}_{h}(x,v)=H(x,v)+h^{2}\tilde{H}_{2}(x,v)+h^{4}\tilde{H}_{4}(x,v)+\cdots$

satisfing

$\frac{d}{dt}\tilde{H}_{h}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16), where the
functions $\tilde{H}_{2j}(x,v)$ are independent of $h$ and
$\mathcal{O}(h^{N})$ is the trunction term.

Proof.  Based on the results in Lemma 4.1 and the fact that

$\dot{z}^{\intercal}\left(z^{(2k+1)}\times
B\right)=\frac{d}{dt}\left(\dot{z}^{\intercal}\left(z^{(2k)}\times
B\right)-\ddot{z}^{\intercal}\left(z^{(2k-1)}\times
B\right)+\cdots+(-1)^{k+1}\left(z^{(k)}\right)^{\intercal}\left(z^{(k+1)}\times
B\right)\right),$

the first statement is obtained. Then, it is arrived that

$\frac{d}{dt}\left(\frac{1}{2}\dot{z}^{\intercal}\dot{z}+U(z)+\frac{h^{2}}{12}\left(\dot{z}^{\intercal}\dddot{z}-\frac{1}{2}\ddot{z}^{\intercal}\ddot{z}\right)-\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}\left(\frac{h^{2}}{3!}\dot{z}^{\intercal}(\ddot{z}\times
B)+\cdots\right)+\cdots\right)=\mathcal{O}(h^{N}),$

which completes the proof. $\blacksquare$

###### Remark 4.3

It is noted that based on this corollary, the method EXS-O2 will be shown to
have a long-time near-conservation of the energy for a constant magnetic field
$B$.

###### Lemma 4.4

There exists a function

$H_{h}(x,v)=H(x,v)+h^{2}H_{2}(x,v)+h^{4}H_{4}(x,v)+\cdots$

and it satisfies

$\frac{d}{dt}H_{h}{(z,\dot{z})}=\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)^{\intercal}E(z)+\mathcal{O}(h^{N})$

along solutions obtained by the modified differential equation (16), where the
functions $H_{2j}(x,v)$ (different from those in Lemma 4.1) don’t depend on
the step size $h$ and $\mathcal{O}(h^{N})$ is the trunction term.

Proof.  It is noted that if $l+m$ is odd, ${z^{(l)}}^{\intercal}z^{(m)}$ can
be written as a total differential. Then taking inner product on both sides of
(16) with $\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)$ and using the
same arguments of Lemma 4.1, one has

$\frac{d}{dt}\left(\frac{1}{2}\dot{z}^{\intercal}\dot{z}+U(z)+\frac{h^{2}}{12}\left(\dot{z}^{\intercal}\dddot{z}-\frac{1}{2}\ddot{z}^{\intercal}\ddot{z}\right)+\cdots\right)=\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)^{\intercal}E(z)+\mathcal{O}(h^{N}).$

The proof is complete. $\blacksquare$

###### Corollary 4.5

(The second almost invariant close to energy) If
$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$, we get a function

$\widehat{H}_{h}(x,v)=H(x,v)+h^{2}\widehat{H}_{2}(x,v)+h^{4}\widehat{H}_{4}(x,v)+\cdots$

such that

$\frac{d}{dt}\widehat{H}_{h}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16), where the
functions $H_{2j}(x,v)$ are independent of $h$ and $\mathcal{O}(h^{N})$ is the
trunction term.

Proof.  Since the expression ${z^{(2k+1)}}^{\intercal}\nabla
U(z)={z^{(2k+1)}}^{\intercal}(Qz+q)$ is a total differential:

$\left({z^{(2k+1)}}\right)^{\intercal}(Qz+q)=\frac{d}{dt}\left(\left({z^{(2k)}}\right)^{\intercal}(Qz+q)-\left({z^{(2k-1)}}\right)^{\intercal}Q\dot{z}+\cdots+\frac{(-1)^{k}}{2}\left({z^{(k)}}\right)^{\intercal}Qz^{(k)}\right),$

it is clear that

$\frac{d}{dt}\left(\frac{1}{2}\dot{z}^{\intercal}\dot{z}+U(z)+\frac{h^{2}}{12}\left(\dot{z}^{\intercal}\dddot{z}-\frac{1}{2}\ddot{z}^{\intercal}\ddot{z}\right)+\frac{h^{2}}{6}\left(\ddot{z}^{\intercal}(Qz+q)-\frac{1}{2}\dot{z}^{\intercal}Q\dot{z}\right)+\cdots\right)=\mathcal{O}(h^{N}).$

$\blacksquare$

###### Remark 4.6

This result confirms that the method EXS-O2 hold a long-term near-conservation
of the total energy as long as the scalar potential is quadratic.

Based on the above preparations, we are in the position to prove the results
(10) and (12) of Theorem 3.1.

Proof of (10). The result is firstly shown for a constant magnetic field $B$.
From Corollary 4.2, it follows that

$\displaystyle H(x^{n},v^{n})$
$\displaystyle=\tilde{H}_{h}(x^{n},v^{n})+\mathcal{O}(h^{2})=\tilde{H}_{h}(x^{0},v^{0})+\sum_{k=1}^{n}\left(\tilde{H}_{h}(x^{k},v^{k})-\tilde{H}_{h}(x^{k-1},v^{k-1})\right)+\mathcal{O}(h^{2})$
(18)
$\displaystyle=H(x^{0},v^{0})+\mathcal{O}(h^{2})+n\mathcal{O}(h^{N+1})+\mathcal{O}(h^{2})=H(x^{0},v^{0})+\mathcal{O}(h^{2}).$

The last equation is meaningful if and only if $nh^{N+1}\leq h^{2}$, i.e.
$nh\leq h^{2-N}$ and this gives the desired bound for the deviation of the
total energy along the numerical solution. Moreover, for the quadratic scalar
potential, Corollary 4.5 implies
$H(x^{n},v^{n})=\hat{H}_{h}(x^{n},v^{n})+\mathcal{O}(h^{2}).$ Using the same
derivation as stated above, it is easy to show that the result (10) also holds
for the quadratic scalar potential. $\blacksquare$ Proof of (12). The second
equation in (9) can be reformulated as

$e^{-\frac{h}{2}\tilde{B}(x^{n+1})}v^{n+1}=e^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h}{2}[E(x^{n})+E(x^{n+1})].$

This result and the fact that $\tilde{B}$ is skew-symmetric imply

$\displaystyle\frac{1}{2}({v^{n+1}})^{\intercal}v^{n+1}-\frac{1}{2}({v^{n}})^{\intercal}v^{n}$
$\displaystyle=\frac{1}{2}\left(e^{-\frac{h}{2}\tilde{B}(x^{n+1})}{v^{n+1}}\right)^{\intercal}\left(e^{-\frac{h}{2}\tilde{B}(x^{n+1})}v^{n+1}\right)-\frac{1}{2}\left(e^{\frac{h}{2}\tilde{B}(x^{n})}{v^{n}}\right)^{\intercal}\left(e^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}\right)$
$\displaystyle=\frac{1}{2}\left[E(x^{n})+E(x^{n+1})\right]^{\intercal}\left[he^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h^{2}}{4}\left[E(x^{n})+E(x^{n+1})\right]\right].$

Besides, the deviation of $U$ at $x^{n+1}$ and $x^{n}$ can be expressed as

$\displaystyle U(x^{n+1})-U(x^{n})$
$\displaystyle=\frac{1}{2}({x^{n+1}})^{\intercal}Qx^{n+1}+q^{\intercal}x^{n+1}-\left(\frac{1}{2}({x^{n}})^{\intercal}Qx^{n}+q^{\intercal}x^{n}\right)$
$\displaystyle=\frac{1}{2}({x^{n+1}}+x^{n})^{\intercal}Q(x^{n+1}-x^{n})+q^{\intercal}(x^{n+1}-x^{n})$
$\displaystyle=\left[\frac{1}{2}({x^{n+1}}+x^{n})^{\intercal}Q+q^{\intercal}\right]{\left(he^{\frac{h}{2}\tilde{B}(x^{n})}v^{n}+\frac{h^{2}}{2}E(x^{n})\right)}.$

Therefore

$\frac{1}{2}({v^{n+1}})^{\intercal}v^{n+1}-\frac{1}{2}({v^{n}})^{\intercal}v^{n}+U(x^{n+1})-U(x^{n})=\frac{h^{2}}{8}\left|\nabla
U(x^{n+1})\right|^{2}-\frac{h^{2}}{8}\left|\nabla U(x^{n})\right|^{2}$

and this further gives $H_{h}(x^{n+1},v^{n+1})=H_{h}(x^{n},v^{n}).$
$\blacksquare$

###  4.3 Proof of the momentum conservation (Theorem 3.2)

The proof is given by finding two almost invariants which are close to
momentum. To derive these invariants, we first present the following lemma.

###### Lemma 4.7

The following two functions

$\displaystyle
M_{h}^{e}(x,v)=M(x,v)+h^{2}M_{2}^{e}(x,v)+h^{4}M_{4}^{e}(x,v)+\cdots,$
$\displaystyle
M_{h}^{i}(x,v)=M(x,v)+h^{2}M_{2}^{i}(x,v)+h^{4}M_{4}^{i}(x,v)+\cdots,$

can be derived with the $h$-independent functions $M_{2j}^{e}(x,v)$ and
$M_{2j}^{i}(x,v)$, and they satisfy

$\frac{d}{dt}M_{h}^{e}{(z,\dot{z})}=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)\times
B(z)+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16) for EXS-O2 and

$\displaystyle\frac{d}{dt}M_{h}^{i}{(z,\dot{z})}$
$\displaystyle=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left[\left(\frac{h^{2}}{3!}\dddot{z}+{\frac{h^{4}}{5!}z^{(5)}}+\cdots\right)-\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{3!}\dddot{z}+\cdots\right)+\cdots\right]\times
B(y)$
$\displaystyle+\frac{h^{2}}{4}z^{\intercal}SE^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (17) for IMS-O2.

Proof.  We first multiply (16) and (17) with $\dot{z}^{\intercal}S$. Then
notice the fact that the expression $z^{\intercal}Sz^{(2k)}$ takes a form of
total derivative:

$z^{\intercal}Sz^{(2k)}=\frac{d}{dt}\left(z^{\intercal}Sz^{(2k-1)}-\dot{z}^{\intercal}Sz^{(2k-2)}+\cdots+(-1)^{k-1}\left({z^{(k-1)}}\right)^{\intercal}Sz^{(k)}\right).$

In addition, the invariance properties (3) show that $z^{\intercal}S\nabla
U(z)=0$ and $z^{\intercal}S(\dot{z}\times
B(z))=-\frac{d}{dt}(z^{\intercal}SA(z))$ [12]. Based on the above results, we
get

$\displaystyle\frac{d}{dt}\left(z^{\intercal}S\dot{z}+\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}SA(z)+\frac{h^{2}}{12}(z^{\intercal}S\dddot{z}-\dot{z}^{\intercal}S\ddot{z})+\cdots\right)$
$\displaystyle=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)\times
B(z)+\mathcal{O}(h^{N})$

and by expansion of $\frac{\tan x}{x}$, the above equation can be rewritten as

$\displaystyle\frac{d}{dt}\left(z^{\intercal}S\dot{z}+z^{\intercal}SA(z)+\frac{h^{2}}{3}\left(\frac{\left|B(z)\right|}{2}\right)^{2}z^{\intercal}SA(z)+\frac{h^{2}}{12}(z^{\intercal}S\dddot{z}-\dot{z}^{\intercal}S\ddot{z})+\cdots\right)$
$\displaystyle=\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left(\frac{h^{2}}{3!}\dddot{z}+\frac{h^{4}}{5!}z^{(5)}+\cdots\right)\times
B(z)+\mathcal{O}(h^{N})$

for EXS-O2.

Similarly, we can get

$\displaystyle\frac{d}{dt}\left(z^{\intercal}S\dot{z}+z^{\intercal}SA(z)+\frac{h^{2}}{3}\left(\frac{\left|B(z)\right|}{2}\right)^{2}z^{\intercal}SA(z)+\frac{h^{2}}{12}(z^{\intercal}S\dddot{z}-\dot{z}^{\intercal}S\ddot{z})+\cdots\right)$
$\displaystyle=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left[\left(\frac{h^{2}}{3!}\dddot{z}+{\frac{h^{4}}{5!}z^{(5)}}+\cdots\right)-\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{3!}\dddot{z}+\cdots\right)+\cdots\right]\times
B(z)$
$\displaystyle+\frac{h^{2}}{4}z^{\intercal}SE^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots+\mathcal{O}(h^{N})$

for IMS-O2. $\blacksquare$

The above results can be improved under some conditions, which is stated by
the following corollary.

###### Corollary 4.8

(The first almost invariant close to momentum) If $B(x)\equiv B$ and
$x^{\intercal}\nabla U(x)\times B=0$ for all $x$, it is obtained that the
functions

$\widehat{M}_{h}^{e}(x,v)=M(x,v)+h^{2}\widehat{M}_{2}^{e}(x,v)+h^{4}\widehat{M}_{4}^{e}(x,v)+\cdots,$
$\widehat{M}_{h}^{i}(x,v)=M(x,v)+h^{2}\widehat{M}_{2}^{i}(x,v)+h^{4}\widehat{M}_{4}^{i}(x,v)+\cdots,$

satisfy

$\frac{d}{dt}\widehat{M}_{h}^{e}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16) for EXS-O2 and

$\displaystyle\frac{d}{dt}\widehat{M}_{h}^{i}{(z,\dot{z})}=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B(z)\right|\right)}{\frac{h}{2}\left|B(z)\right|}z^{\intercal}S\left[-\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right]\times
B(z)$
$\displaystyle+\frac{h^{2}}{4}z^{\intercal}SE^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (17) for IMS-O2. Here
the functions $\widehat{M}_{2j}^{e}(x,v)$ and $\widehat{M}_{2j}^{i}(x,v)$ are
$h$-independent and $\mathcal{O}(h^{N})$ is the truncation term.

Proof.  Using the same way as that of Lemma 4.7 and noticing that
$z^{\intercal}S(z^{(2k+1)}\times B)$ is a total derivative (due to the
symmetry of $S^{2}$):

$z^{\intercal}S\left(z^{(2k+1)}\times
B\right)=z^{\intercal}S^{2}z^{(2k+1)}=\frac{d}{dt}\left(z^{\intercal}S^{2}z^{(2k)}-\dot{z}^{\intercal}S^{2}z^{(2k-1)}+\cdots+\frac{(-1)^{k}}{2}\left({z^{(k)}}\right)^{\intercal}S^{2}z^{(k)}\right),$

we get

$\displaystyle\frac{d}{dt}\left(z^{\intercal}S\dot{z}+z^{\intercal}SA(z)+\frac{h^{2}}{3}\left(\frac{\left|B\right|}{2}\right)^{2}z^{\intercal}SA(z)+\frac{h^{2}}{12}(z^{\intercal}S\dddot{z}-\dot{z}^{\intercal}S\ddot{z})-\frac{h^{2}}{6}\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}\left(z^{\intercal}S^{2}\ddot{z}{-\frac{1}{2}\dot{z}^{\intercal}S^{2}\dot{z}}\right)+\cdots\right)$
$\displaystyle=\mathcal{O}(h^{N})$

for EXS-O2 and

$\displaystyle\frac{d}{dt}\left(z^{\intercal}S\dot{z}+z^{\intercal}SA(z)+\frac{h^{2}}{3}\left(\frac{\left|B\right|}{2}\right)^{2}z^{\intercal}SA(z)+\frac{h^{2}}{12}(z^{\intercal}S\dddot{z}-\dot{z}^{\intercal}S\ddot{z})-\frac{h^{2}}{6}\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}\left(z^{\intercal}S^{2}\ddot{z}{-\frac{1}{2}\dot{z}^{\intercal}S^{2}\dot{z}}\right)+\cdots\right)$
(19) $\displaystyle=$
$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}z^{\intercal}S\left[-\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right]\times
B+\frac{h^{2}}{4}z^{\intercal}SE^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots+\mathcal{O}(h^{N})$

for IMS-O2. $\blacksquare$

###### Corollary 4.9

(The second almost invariant close to momentum) Suppose that the conditions in
Corollary 4.8 are satisfied and
$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$ with $QS=SQ$. Then the
function

$\tilde{M}_{h}^{i}(x,v)=M(x,v)+h^{2}\tilde{M}_{2}^{i}(x,v)+h^{4}\tilde{M}_{4}^{i}(x,v)+\cdots$

satisfies

$\frac{d}{dt}\tilde{M}_{h}^{i}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (17) for IMS-O2. Here
the functions $\tilde{M}_{2j}^{i}(x,v)$ don’t depend on the step size $h$ and
$\mathcal{O}(h^{N})$ stands for the truncation term.

Proof.  Based on the assumptions, the right-hand side of (19) can be
simplified as

$\displaystyle\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}z^{\intercal}S\left[-\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right]\times
B+\frac{h^{2}}{4}z^{\intercal}SE^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots+\mathcal{O}(h^{N})$
$\displaystyle=\frac{\tan\left(\frac{h}{2}\left|B\right|\right)}{\frac{h}{2}\left|B\right|}z^{\intercal}S\left[\frac{h^{2}}{4}Q\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)\right]\times
B-\frac{h^{2}}{4}z^{\intercal}SQ\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+{\mathcal{O}(h^{N}).}$

According to the properties of $S$ and $Q$, it is obtained that $SQ$ is skew-
symmetric and $S^{2}Q$ is symmetric. Then, we get

$z^{\intercal}S\left(\left(Qz^{(2k+1)}\right)\times
B\right)=z^{\intercal}S^{2}Qz^{(2k+1)}=\frac{d}{dt}\left(z^{\intercal}S^{2}Qz^{(2k)}-\dot{z}^{\intercal}S^{2}Qz^{(2k-1)}+\cdots+\frac{(-1)^{k}}{2}\left({z^{(k)}}\right)^{\intercal}S^{2}Qz^{(k)}\right)$

and

$z^{\intercal}SQz^{(2k)}=\frac{d}{dt}\left(z^{\intercal}SQz^{(2k-1)}-\dot{z}^{\intercal}SQz^{(2k-2)}+\cdots+(-1)^{k-1}\left({z^{(k-1)}}\right)^{\intercal}SQz^{(k)}\right).$

Therefore, the proof is complete. $\blacksquare$

Proof of (13). With the results stated above, the method EXS-O2 conserves the
momentum with the accuracy

$\displaystyle M(x^{n},v^{n})$
$\displaystyle=\tilde{M}_{h}^{e}(x^{n},v^{n})+\mathcal{O}(h^{2})=\tilde{M}_{h}^{e}(x^{0},v^{0})+\sum_{k=1}^{n}\left(\tilde{M}_{h}^{e}(x^{k},v^{k})-\tilde{M}_{h}^{e}(x^{k-1},v^{k-1})\right)+\mathcal{O}(h^{2})$
(20)
$\displaystyle=M(x^{0},v^{0})+\mathcal{O}(h^{2})+n\mathcal{O}(h^{N+1})+\mathcal{O}(h^{2})=M(x^{0},v^{0})+\mathcal{O}(h^{2}),$

as long as $nh\leq h^{2-N}$. For IMS-O2, the same discussion applies to
$M_{h}^{i}(x,v)$ and then (13) can be proved.

$\blacksquare$

###  4.4 Proof of the magnetic moment conservation (Theorem 3.3)

In this section, we just considered magnetic moment in constant magnetic field
$B$, that is

$I(x,v)=\frac{\left|v\times
B\right|^{2}}{2\left|B\right|^{3}}=\frac{\left|v\times
b\right|^{2}}{2\left|B\right|}$

with $b=\frac{B}{\left|B\right|}$ and a skew-symmetric matrix
$\widehat{B}=\frac{\tilde{B}}{\left|B\right|}$ satisfying $v\times
b=\widehat{B}v$.

Specifically, the modified equations (16) and (17) are respectively equivalent
to

$\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)\times
b=\frac{h}{2\tan\left(\frac{h}{2}\left|B\right|\right)}\left[\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)-E(z)\right]$
(21)

and

$\displaystyle\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)\times b=$
$\displaystyle\frac{h}{2\tan\left(\frac{h}{2}\left|B\right|\right)}\left[\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)-\left(E(z)+\frac{h^{2}}{4}E^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots\right)\right]$
(22)
$\displaystyle+\left(\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right)\times
b.$

Based on these results, we can derive two almost invariants close to magnetic
moment, which will complete the proof of Theorem 3.3. To this end, we first
prove the following result.

###### Lemma 4.10

If $B(x)\equiv B$, one gets that the functions

$I_{h}^{e}(x,v)=I(x,v)+h^{2}I_{2}^{e}(x,v)+h^{4}I_{4}^{e}(x,v)+\cdots,$
$I_{h}^{i}(x,v)=I(x,v)+h^{2}I_{2}^{i}(x,v)+h^{4}I_{4}^{i}(x,v)+\cdots,$

satisfy

$\frac{d}{dt}I_{h}^{e}{(z,\dot{z})}=-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}(\ddot{z}\times
b)^{\intercal}E(z)+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16) for EXS-O2 and

$\displaystyle\frac{d}{dt}I_{h}^{i}{(z,\dot{z})}=$
$\displaystyle-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}(\ddot{z}\times
b)^{\intercal}\left[E(z)+\frac{h^{2}}{4}E^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots\right]$
$\displaystyle+\frac{1}{\left|B\right|}(\ddot{z}\times
b)^{\intercal}\left[\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right]\times
b+\mathcal{O}(h^{N})$

along solutions of the modified differential equation (17) for IMS-O2. Here
the functions $I_{2j}^{e}(x,v)$ and $I_{2j}^{i}(x,v)$ are $h$-independent and
$\mathcal{O}(h^{N})$ is the truncation term.

Proof.  Multiply (21) and (22) with $\frac{1}{\left|B\right|}(\ddot{z}\times
b)^{\intercal}$. It is clear that

$\displaystyle\frac{1}{\left|B\right|}(\ddot{z}\times
b)^{\intercal}\left(z^{(2k+1)}\times b\right)$
$\displaystyle=\begin{cases}\frac{d}{dt}I(z,\dot{z}),&\text{$k=0$}\\\
\frac{d}{dt}\frac{1}{\left|B\right|}\left((\ddot{z}\times
b)^{\intercal}\left(z^{(2k)}\times b\right)-\left(\dddot{z}\times
b\right)^{\intercal}\left(z^{(2k-1)}\times
b\right)+\cdots+\frac{(-1)^{k+1}}{2}\left(z^{(k+1)}\times
b\right)^{\intercal}\left(z^{(k+1)}\times
b\right)\right),&\text{$k\in\mathbb{N}^{*}$}\end{cases}$

and

$\displaystyle\frac{1}{\left|B\right|}(\ddot{z}\times b)^{\intercal}z^{(2k)}$
$\displaystyle=\begin{cases}0,&\text{$k=1$}\\\
\frac{d}{dt}\frac{1}{\left|B\right|}\left((\ddot{z}\times
b)^{\intercal}z^{(2k-1)}-\left(\dddot{z}\times
b\right)^{\intercal}z^{(2k-2)}+\cdots+(-1)^{k}\left(z^{(k)}\times
b\right)^{\intercal}z^{(k+1)}\right),&\text{$k\in\mathbb{N}^{*}-\\{1\\}$}\end{cases}$

These immediately demonstrate that

$\displaystyle\frac{d}{dt}\left[I(z,\dot{z})+\frac{h^{2}}{12\left|B\right|}(\ddot{z}\times
b)^{\intercal}(\ddot{z}\times
b)-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}\left(\frac{h^{2}}{12}(\ddot{z}\times
b)^{\intercal}\dddot{z}+\cdots\right)\right]$
$\displaystyle=-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}(\ddot{z}\times
b)^{\intercal}E(z)+\mathcal{O}(h^{N})$

for EXS-O2.

On the other hand and with the same arguments, we get

$\displaystyle\frac{d}{dt}\left[I(z,\dot{z})+\frac{h^{2}}{12\left|B\right|}(\ddot{z}\times
b)^{\intercal}(\ddot{z}\times
b)-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}\left(\frac{h^{2}}{12}(\ddot{z}\times
b)^{\intercal}\dddot{z}+\cdots\right)\right]$ (23) $\displaystyle=$
$\displaystyle-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}(\ddot{z}\times
b)^{\intercal}\left[E(z)+\frac{h^{2}}{4}E^{\prime}(z)\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)+\cdots\right]$
$\displaystyle+\frac{1}{\left|B\right|}(\ddot{z}\times
b)^{\intercal}\left[\frac{h^{2}}{4}E^{\prime}(z)\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)+\cdots\right]\times
b+\mathcal{O}(h^{N})$

for IMS-O2. $\blacksquare$

###### Corollary 4.11

(The almost invariants close to magnetic moment) If $B(x)\equiv B$ and
$U(x)=\frac{1}{2}x^{\intercal}Qx+q^{\intercal}x$ with
$Q\widehat{B}=\widehat{B}Q$, one obtains that the functions

$\widehat{I}_{h}^{e}(x,v)=I(x,v)+h^{2}\widehat{I}_{2}^{e}(x,v)+h^{4}\widehat{I}_{4}^{e}(x,v)+\cdots,$
$\widehat{I}_{h}^{i}(x,v)=I(x,v)+h^{2}\widehat{I}_{2}^{i}(x,v)+h^{4}\widehat{I}_{4}^{i}(x,v)+\cdots,$

satisfy

$\frac{d}{dt}\widehat{I}_{h}^{e}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16) for EXS-O2, and

$\frac{d}{dt}\widehat{I}_{h}^{i}{(z,\dot{z})}=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (17) for IMS-O2. Here
the functions $\widehat{I}_{2j}^{e}(x,v)$ and $\widehat{I}_{2j}^{i}(x,v)$ are
$h$-independent and $\mathcal{O}(h^{N})$ refers to the truncation term.

Proof.  On the basis of the conditions above, $-(\ddot{z}\times
b)^{\intercal}E(z)=(\ddot{z}\times
b)^{\intercal}(Qz+q)=\frac{d}{dt}\left((\dot{z}\times
b)^{\intercal}(Qz+q)\right).$ The hand-side of (23) can be written as

$\displaystyle\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}(\ddot{z}\times
b)^{\intercal}\left[(Qz+q)+\frac{h^{2}}{4}Q\left(\ddot{z}+\frac{h^{2}}{12}\ddddot{z}+\cdots\right)\right]$
$\displaystyle-\frac{1}{\left|B\right|}(\ddot{z}\times
b)^{\intercal}\left[\frac{h^{2}}{4}Q\left(\dot{z}+\frac{h^{2}}{6}\dddot{z}+\cdots\right)\right]\times
b+\mathcal{O}(h^{N}).$

Since $\widehat{B}Q$ is skew-symmetric and $\widehat{B}^{2}Q$ is symmetric, we
have

$\displaystyle(\ddot{z}\times b)^{\intercal}Qz^{(2k)}$
$\displaystyle=-{\ddot{z}}^{\intercal}\widehat{B}Qz^{(2k)}=\begin{cases}0,&\text{$k=1$}\\\
-\frac{d}{dt}\left({\ddot{z}}^{\intercal}\widehat{B}Qz^{(2k-1)}-{\dddot{z}}^{\intercal}\widehat{B}Qz^{(2k-1)}+\cdots+(-1)^{k}\left({z^{(k)}}\right)^{\intercal}\widehat{B}Qz^{(k+1)}\right),&\text{$k\in\mathbb{N}^{*}-\\{1\\}$}\end{cases}$

and

$\displaystyle(\ddot{z}\times b)^{\intercal}\left(Qz^{(2k+1)}\right)\times b$
$\displaystyle=-\ddot{z}^{\intercal}\widehat{B}^{2}Qz^{(2k+1)}=\begin{cases}-\frac{1}{2}\dot{z}^{\intercal}\widehat{B}^{2}Q\dot{z},&\text{$k=0$}\\\
-\frac{d}{dt}\left(\ddot{z}^{\intercal}\widehat{B}^{2}Qz^{(2k)}-\dddot{z}^{\intercal}\widehat{B}^{2}Qz^{(2k-1)}+\cdots+\frac{(-1)^{(k+1)}}{2}\left({z^{(k+1)}}\right)^{\intercal}\widehat{B}^{2}Qz^{(k+1)}\right),&\text{$k\in\mathbb{N}^{*}.$}\end{cases}$

For the sake of formal unity, we formulate
$-{z^{(l)}}^{\intercal}\widehat{B}^{2}Qz^{(m)}$ as $\left(z^{(l)}\times
b\right)^{\intercal}\left(Qz^{(m)}\right)\times b$ for any integers $l$ and
$m$. Hence,

$\displaystyle\frac{d}{dt}\left[I(z,\dot{z})+\frac{h^{2}}{12\left|B\right|}(\ddot{z}\times
b)^{\intercal}(\ddot{z}\times
b)-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}\left(\left(\frac{h^{2}}{12}(\ddot{z}\times
b)^{\intercal}\dddot{z}+\cdots\right)+(\dot{z}\times
b)^{\intercal}(Qz+q)\right)\right]=\mathcal{O}(h^{N})$

along solutions of the modified differential equation (16) for EXS-O2, and

$\displaystyle\frac{d}{dt}\left[I(z,\dot{z})+\frac{h^{2}}{12\left|B\right|}(\ddot{z}\times
b)^{\intercal}(\ddot{z}\times
b)-\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}\left(\frac{h^{2}}{12}(\ddot{z}\times
b)^{\intercal}\dddot{z}+\cdots\right)\right]$
$\displaystyle-\frac{d}{dt}\left[\frac{h}{2\left|B\right|\tan\left(\frac{h}{2}\left|B\right|\right)}\left((\ddot{z}\times
b)^{\intercal}(Qz+q)+\frac{h^{2}}{4}\left(\frac{h^{2}}{12}(\ddot{z}\times
b)^{\intercal}Q\dddot{z}+\cdots\right)\right)\right]$
$\displaystyle+\frac{d}{dt}\left[\frac{h^{2}}{4\left|B\right|}\left(\frac{1}{2}{(\dot{z}\times
b)}^{\intercal}(Q\dot{z})\times b+\cdots\right)\right]=\mathcal{O}(h^{N}).$

$\blacksquare$

Proof of (14).  The proof of (14) is the same as that of (10) and (13), and we
omit it for brevity. $\blacksquare$

##  5 Conclusion

In this paper, we presented two splitting algorithms and studied their long-
term behaviour for solving charged-particle dynamics. Using backward error
analysis, it was shown that these two algorithms have good conservations of
energy, momentum and magnetic moment in some special cases. Furthermore, one
algorithm was proved to conserve a modified energy exactly. All the results
were illustrated by some numerical tests.

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# On groups in which subnormal subgroups of infinite rank are commensurable
with some normal subgroup

Ulderico Dardano and Fausto De Mari

###### Abstract.

We study soluble groups $G$ in which each subnormal subgroup $H$ with infinite
rank is commensurable with a normal subgroup, i.e. there exists a normal
subgroup $N$ such that $H\cap N$ has finite index in both $H$ and $N$. We show
that if such a $G$ is periodic, then all subnormal subgroups are commensurable
with a normal subgroup, provided either the Hirsch-Plotkin radical of $G$ has
infinite rank or $G$ is nilpotent-by-abelian (and has infinite rank).

2020 Mathematics Subject Classification: Primary: 20F16, Secondary: 20E07,
20E15.
Keywords: transitivity, core-finite, normal-by-finite, close to normal

##  1\. Introduction and statement of results

A group $G$ is said to be a $T$-group if normality in $G$ is a transitive
relation, i.e. if all subnormal subgroups are normal. The structure of soluble
$T$-groups was well described in the 1960s by Gaschütz, Zacher and Robinson
(see [14]). Then, taking these results as a model, several authors have
studied soluble groups in which subnormal subgroups have some embedding
property which “approximates” normality. In particular, Casolo [2] considered
${T}_{*}$-groups, that is groups in which any subnormal subgroup $H$ has the
property $nn$ (nearly normal), i.e. the index $|H^{G}:H|$ is finite. Then
Franciosi, de Giovanni and Newell [11] considered ${T}^{*}$-groups, that is
groups in which any subnormal subgroup $H$ has the property $cf$ (core-finite,
normal-by-finite), i.e. the index $|H:H_{G}|$ is finite. Here, as usual,
$H^{G}$ (resp. $H_{G}$) denotes the smallest (resp. largest) normal subgroup
of $G$ containing (resp. contained in) $H$

Recently in [4], in order to put those results in a common framework, we
considered $T[*]$-groups, that is groups in which each subnormal subgroup $H$
is $cn$, i.e. commensurable with a normal subgroup of $G$. Recall that two
subgroups $H$ and $K$ are called commensurable if $H\cap K$ has finite index
in both $H$ and $K$, hence both $nn$ and $cf$ imply $cn$. Clearly all the
above results rely on corresponding previous results on groups in which all
subgroups are $nn$, $cf$, $cn$ resp. (see [13, 1, 3] resp.). A similar
approach was adopted in [7] where finitely generated groups in which subnormal
subgroups are inert have been considered, where the term inert refers to a
different generalization of both $nn$, $cf$ (namely, an inert subgroup is a
subgroup which is commensurable with each of its conjugates).

In the last decade, several authors have studied the influence on a soluble
group of the behavior of its subgroups of infinite rank (see for instance [5,
10] or the bibliography in [9]). Recall that a group $G$ is said to have
finite rank $r$ if every finitely generated subgroup of $G$ can be generated
by at most $r$ elements, and $r$ is the least positive integer with such
property and infinite rank is there is no such $r$. For example, in [8] it was
proved that if $G$ is a periodic soluble group of infinite rank in which every
subnormal subgroup of infinite rank is normal, then $G$ is a $T$-group indeed.
Then in [9], authors have considered groups of infinite rank with properties
$T_{+}$ ($T^{+}$, resp.), that is groups in which the condition of being $nn$
(resp. $cf$) is imposed only to subnormal subgroups with infinite rank. In
fact, it has been shown that a periodic soluble group of infinite rank $G$
with property $T_{+}$ ($T^{+}$, resp.) has the full $T_{*}$ ($T^{*}$, resp.)
property, provided one of the following holds:
(A) the Hirsch-Plotkin radical of $G$ has infinite rank,
(B) the commutator subgroup $G^{\prime}$ is nilpotent.

In this paper we show that a similar statement is true also for the property
$cn$. Moreover, by a corollary, we give some further information about the
property $cf$ as well. Let us call $T[+]$-group a group in which each
subnormal subgroup of infinite rank is a $cn$-subgroup.

Theorem A Let $G$ be a periodic soluble $T[+]$-group whose Hirsch-Plotkin
radical has infinite rank. Then $G$ is a $T[\ast]$-group.

Corollary Let $G$ be a periodic soluble $T[+]$-group (resp. $T_{+}$-group) of
infinite rank such that $\pi(G^{\prime})$ is finite. Then all subgroups of $G$
are $cn$ (resp. $cf$).

Theorem B Let $G$ be a periodic $T[+]$-group of infinite rank with nilpotent
commutator subgroup. Then $G$ is a $T[\ast]$-group.

Note that if $G=A\rtimes B$ is the holomorph group of the additive group $A$
of the rational numbers by the multiplicative group $B$ of positive rationals
(acting by usual multiplication), then, as noticed in [8], the only subnormal
non-normal subgroups of $G$ are those contained in $A$ (which has rank $1$) so
that $G$ is $T[+]$. However all proper non-trivial subgroups of $A$ are not
$cn$, since if they were $cn$ then they were $cf$ (see [6], Proposition 1)
contradicting the fact that $A$ is minimal normal in $G$.

Our notation and terminology is standard and can be found in [15, 16]

##  2\. Proofs

By a standard argument one checks easily that if $H_{1}$ and $H_{2}$ are
$cn$\- (resp. $cf$-) subgroups of $G$, then $H_{1}\cap H_{2}$ is likewise $cn$
(resp. $cf$). The same holds for $H_{1}H_{2}$, provided this set is a
subgroup.

######  Lemma 1.

Let $G$ be a $T[+]$-group and let $A$ be a subnormal subgroup of $G$. If $A$
is the direct product of infinitely many non-trivial cyclic subgroups, then
any subgroup of $A$ is a $cn$-subgroup of $G$.

###### Proof.

Let $X$ be any subgroup of $A$, then $X$ is a subnormal subgroup of $G$ and
$X$ is likewise a direct product of cyclic groups (see [16], 4.3.16). In order
to prove that $X$ is a $cn$-subgroup of $G$ we may assume that $X$ has finite
rank. Then there exist subgroups $A_{1}$, $A_{2}$, $A_{3}$ of $A$ with
infinite rank such that $X\leq A_{3}$ and $A=A_{1}\times A_{2}\times A_{3}$.
Thus $XA_{1}$ and $XA_{2}$ are subnormal subgroups of infinite rank, so that
they are both $cn$-subgroups of $G$. Therefore $X=XA_{1}\cap XA_{2}$ is
likewise $cn$ in $G$. ∎

######  Lemma 2.

Let $G$ be a periodic $T[+]$-group. If $G$ contains an abelian subnormal
subgroup of infinite rank $A$, then $G$ is a $T[\ast]$-group.

###### Proof.

By hypothesis there exists a normal subgroup $N$ of $G$ which is commensurable
with $A$. Then $A\cap N$ has finite index in $AN$ and hence $N$ is an abelian-
by-finite group of infinite rank. In particular, $N$ contains a characteristic
subgroup $N_{\ast}$ of finite index which is an abelian group of infinite
rank; hence replacing $A$ by $N_{\ast}$ it can be supposed that $A$ is a
normal subgroup. Since $G$ is periodic and $A$ has infinite rank, it follows
that the socle $S$ of $A$ is a normal subgroup of $G$ which is the direct
product of infinitely many non-trivial cyclic subgroups. Application of Lemma
1 yields that all subgroups of $S$ are $cn$-subgroups of $G$ and hence by
Lemma 2.8 of [3], there exist $G$-invariant subgroups $S_{0}\leq S_{1}$ of $S$
such that $S_{0}$ and $S/S_{1}$ are finite and all subgroups of $S$ lying
between $S_{0}$ ad $S_{1}$ are normal in $G$

Let $X$ be any subnormal subgroup of finite rank of $G$. Then $X\cap S_{1}$ is
finite, hence $S_{2}=S_{0}(X\cap S_{1})$ is likewise finite. Since $S_{2}X$ is
commensurable with $X$, we may assume $S_{2}=\\{1\\}$. Clearly there exist
subgroups $S_{3}$ and $S_{4}$ with infinite rank such that $S_{1}=S_{3}\times
S_{4}$. Since both $S_{3}$ and $S_{4}$ are normal subgroups of $G$, we have
that both $XS_{3}$ and $XS_{4}$ are subnormal subgroups of infinite rank of
$G$ and hence they are both $cn$. Thus $X=XS_{3}\cap XS_{4}$ is likewise a
$cn$-subgroup of $G$. ∎

Recall that any primary locally nilpotent group of finite rank is a Chernikov
group (see [15] Part 2, p.38).

######  Lemma 3.

Let $G$ a $T[+]$-group of infinite rank. If $G$ is a Baer $p$-group, then $G$
is a nilpotent $T[*]$-group.

###### Proof.

Let $X$ be any subnormal subgroup of $G$ with finite rank. Then $X$ is a
Chernikov group (see [15] Part 2, p.389) and $X$ contains an abelian divisible
normal subgroup $J$ of finite index. Hence $J$ is subnormal in $G$, and so
$J^{G}$ is abelian and divisible (see [15] Part 1, Lemma 4.46). If $A$ is any
abelian subnormal subgroup of $G$, the subgroup $J^{G}A$ is nilpotent and
$[J,A]=\\{1\\}$ (see [15] Part 1, Lemma 3.13). Since $G$ is generated by its
subnormal abelian subgroups, it follows that $J\leq Z(G)$ and so $X/X_{G}$ is
finite. This proves that $G$ is a $T[\ast]$-group, hence nilpotent (see [4],
Proposition 20). ∎

The following lemma is probably well-known but we are not able to find it in
the literature, hence we write also the proof.

######  Lemma 4.

Let $G$ be a periodic finite-by-abelian group of finite rank. Then $G/Z(G)$ is
finite.

###### Proof.

Clearly $C=C_{G}(G^{\prime})$ is a normal subgroup of finite index of $G$
which is nilpotent and has finite rank; in particular, any primary component
of $C$ is a Chernikov group. Let $\pi=\pi(G^{\prime})$ be the set of all
primes $p$ such that $G^{\prime}$ contains some element of order $p$. Then
$\pi$ is finite and so the subgroup $C_{\pi}$ is a Chernikov group; hence
$C_{\pi}Z(G)/Z(G)$ is finite (see [15] Part 1, Lemma 4.3.1). On the other hand
$C_{\pi^{\prime}}$ is abelian, and so it follows that $C/Z(C)$ is finite. Thus
$G$ is both abelian-by-finite and finite-by-abelian and hence $G/Z(G)$ is
finite. ∎

Proof of Theorem A. Assume, for a contradiction, that the statement is false
and let $X$ be a subnormal subgroup $G$ which is not a $cn$-subgroup; in
particular, $X$ has finite rank. Among all counterexamples choose $G$ in such
a way that $X$ has the smallest possible derived length. Then the derived
subgroup $Y=X^{\prime}$ of $X$ is a $cn$-subgroup by the minimal choice on the
derived length of $X$; on the other hand, $Y$ has finite rank and so $Y/Y_{G}$
is finite (see [6], Proposition 1). Then $X/Y_{G}$ is a finite-by-abelian
group of finite rank, and hence its centre $Z/Y_{G}=Z(X/Y_{G})$ has finite
index in $X/Y_{G}$ by Lemma 4. Thus $Z$ is a subnormal subgroup of $G$ which
has finite index in $X$, so that the index $|Z:Z_{G}|$ is infinite and hence
$Z$ cannot be a $cn$-subgroup of $G$ (see [6], Proposition 1). Since $Y_{G}$
has finite rank, the Hirsch-Plotkin radical of $G/Y_{G}$ has infinite rank and
so $G/Y_{G}$ is also a counterexample; thus replacing $G$ with $G/Y_{G}$ and
$X$ with $Z/Y_{G}$ it can be supposed that $X$ is abelian. Hence $X$ is
contained in the Hirsh-Plotkin radical $H$ of $G$.

Let $P$ any primary component of $H$, and suppose that $P$ has infinite rank.
If $F$ is the Fitting subgroup of $P$, then $F$ is nilpotent by Lemma 3. Let
$A$ be a maximal abelian normal subgroup of $F$, then $A=C_{F}(A)$ (see [15]
Part 1, Lemma 2.19.1) and so $A$ has infinite rank (see [15] Part 1, Theorem
3.29). Hence $G$ is a $T[\ast]$-group by Lemma 2. This contradiction proves
that each primary component of $H$ has finite rank. In particular, as $H$ has
infinite rank, there exist $H_{1}$ and $H_{2}$ subgroups of infinite rank such
that $H=H_{1}\times H_{2}$ and $\pi(H_{1})\cap\pi(H_{2})=\emptyset$. By the
same reason, for $i\in\\{1,2\\}$, two subgroups of infinite rank $H_{i,1}$ and
$H_{i,2}$ can be found such that $H_{i}=H_{i,1}\times H_{i,2}$ and
$\pi(H_{i,1})\cap\pi(H_{i,2})=~{}\emptyset$. If $i,j\in\\{1,2\\}$ and $i\neq
j$, considered $\pi_{i}=\pi(H_{i})$ and denoted by $X_{i}$ the
$\pi_{i}$-component of $X$, the subgroups $X_{i}H_{j,1}$ and $X_{i}H_{j,2}$
are subnormal subgroups of infinite rank of $G$, so that they are both
$cn$-subgroups of $G$ and hence $X_{i}=X_{i}H_{j,1}\cap X_{i}H_{j,2}$ is
likewise a $cn$-subgroup of $G$. Therefore $X=X_{1}X_{2}$ is a $cn$-subgroup
of $G$ and this final contradiction concludes the proof. ∎

Proof of Corollary. One may refine the derived series of $G^{\prime}$ to a
series $G_{1}=G^{\prime}\geq\cdots\geq G_{n}=\\{1\\}$ whose factors
$A_{i}=G_{i}/G_{i+1}$ are $p$-groups (for possibly different primes). Let
$C_{i}=C_{G}(A_{i})$ for each $i$. If $A_{i}$ has finite rank, then $A_{i}$ is
a Chernikov group and the same holds for $G/C_{i}$ as a periodic group of
automorphisms of a Chernikov group (see [15] Part 1, Theorem 3.29). If $A_{i}$
has infinite rank, then Lemma 2 yields that each subgroup of $A_{i}$ is a
$cn$-subgroup (resp $cf$-) of $G$. Hence, according to Proposition 14 in [4],
$G/C_{i}$ is finite as a periodic group of power automorphisms of $p$-groups
(see [14], Lemma 4.1.2). Thus if $C$ is the intersection of all $C_{i}$’s,
then $G/C$ is a Chernikov group and therefore has finite rank. It follows that
$C$ has infinite rank. Om the other hand, $C$ is nilpotent by a well-known
fact (see [12]). Then by Theorem $A$, the group $G$ has property $T[*]$ (resp.
$T_{*}$). Further, by Theorem 15 in [4], $G$ all subgroups are $cn$ (resp.
$cf$). ∎

Proof of Theorem B. Assume that the statement is false. As in the first part
of proof of Theorem A, there exists a counterexample $G$ containing an abelian
subnormal subgroup $X$ that is not a $cn$-subgroup; in particular, $X$ has
finite rank and the index $|X:X_{G}|$ is infinite. Then $L=XG^{\prime}$ is a
nilpotent normal subgroup and hence has finite rank by Theorem A. Let
$p\in\pi(X)$, then $L/L_{p^{\prime}}$ is a nilpotent $p$-group of finite rank
and hence it is a Chernikov group; thus $G/C_{G}(L/L_{p^{\prime}})$ is finite
(see [15] Part 1, Corollary p.85) and hence
$C_{G/L_{p^{\prime}}}(L/L_{p^{\prime}})$ is a nilpotent normal subgroup of
infinite rank of $G/L_{p^{\prime}}$. Thus Theorem A yields that
$G/L_{p^{\prime}}$ is a $T[\ast]$-group. Therefore $X_{p}L_{p^{\prime}}$ is a
$cn$-subgroup of $G$, and hence it is even $cf$ because it has finite rank
(see [6], Proposition 1). The $p$-component of the core
$(X_{p}L_{p^{\prime}})_{G}$ of $X_{p}L_{p^{\prime}}$in $G$ is $G$-invariant,
it coincides with the subgroup $X_{p}\cap(X_{p}L_{p^{\prime}})_{G}$ and so has
finite index in $X_{p}$, therefore $X_{p}$ is $cf$. In particular, the set
$\pi$ of all primes $p$ in $\pi(X)$ such that $X_{p}$ is not normal in $G$ is
infinite. Replacing $G$ by $G/L_{\pi^{\prime}}$ it can be supposed that
$\pi=\pi(L)$. Then there exists an infinite subset $\pi_{0}$ of $\pi$ such
that $G/L_{\pi_{0}^{\prime}}$ contains a nilpotent normal subgroup of infinite
rank (see [9], Corollary 11); hence $G/L_{\pi_{0}^{\prime}}$ is a
$T[\ast]$-group by Theorem A. Therefore $X_{\pi_{0}}L_{\pi_{0}^{\prime}}$ is a
$cn$-subgroup of $G$ and so even a $cf$-subgroup (see [6], Proposition 1);
hence $X_{\pi_{0}}$ is $cf$ and this is a contradiction because $X_{p}$ is not
normal in $G$ for each $p\in\pi_{0}$. ∎

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Ulderico Dardano, Fausto De Mari
Dipartimento di Matematica e Applicazioni “R.Caccioppoli”,
Università di Napoli “Federico II”,
Via Cintia - Monte S. Angelo, I-80126 Napoli, Italy
email<EMAIL_ADDRESS><EMAIL_ADDRESS>