Valeriy M. Imaykin

Moscow High School No 179

Moscow, Russian Federation

E-mail<EMAIL_ADDRESS>

ON STABILITY OF SOLITONS FOR A ROTATING

CHARGE AT REST IN THE MAXWELL FIELD

We consider the system of Maxwell equations and Lorentz torque equation which
describes a motion of charge in electromagnetic field. Under certain symmetry
conditions on charge distribution and on initial fields the charge remains at
rest but rotates. The system admits stationary soliton-type solutions. We
study the Lyapunov and the orbital stability of solitons exploiting the energy
conservation law the angular momentum argument for the system. The
bibliography: 20 refs.

##  1 Introduction

In this note we study some dynamical aspects of the Abraham model for a
classical charge with spin coupled to the Maxwell field. The model is known
since early 1900-ths, cf. [2, 3], and was the subject of a number of papers,
both at the physical and mathematical levels of rigor, see below the comment
on previous works and results.

First let us write down the equations of motion. The Maxwell field consists of
the electric field $E(x,t)$ and the magnetic field $B(x,t)$. The charge has
the center of mass $q$ with the velocity $\dot{q}$. We assume that the mass
distribution, $m\,\rho(x)$, and the charge distribution, $e\,\rho(x)$, are
proportional to each other. Here $m$ is the total mass, $e$ is the total
charge, and we use a system of units, where $m=1$ and $e=1$; $\rho(x)$ is a
smooth radially symmetric smearing function of compact support, explicitly,

$\rho\in C_{0}^{\infty}({\rm
I\kern-1.56693ptR}^{3}),\,\,\,\,\rho(x)=\rho_{rad}(|x|),\,\,\,\,\rho(x)=0,\,\,\,{\rm
for}\,\,\,|x|>R_{\rho}>0.$ (1.1)

The angular velocity of the charge is denoted by $\omega(t)\in{\rm
I\kern-1.56693ptR}^{3}$. In detail, $\omega(t)$ is the angular velocity “in
space” (in the terminology of [4]) of the charge. Namely, let us fix a
“center” point $O$ of the rigid body. Then the trajectory of each particular
point of the body is described by

$x(t)=q(t)+R(t)(x(0)-q(0)),$

where $q(t)$ is the position of $O$ at the time $t$, and $R(t)\in SO(3)$.
Respectively, the velocity reads

$\dot{x}(t)=\dot{q}(t)+\dot{R}(t)(x(0)-q(0))=\dot{q}(t)+\dot{R}(t)R^{-1}(t)(x(t)-q(t))=\dot{q}(t)+\omega(t)\wedge(x(t)-q(t)),$
(1.2)

where $\omega(t)\in{\rm I\kern-1.56693ptR}^{3}$ corresponds to the skew-
symmetric matrix $\dot{R}(t)R^{-1}(t)$ by the rule

$\dot{R}(t)R^{-1}(t)={\cal
J}\omega(t):=\left(\begin{array}[]{ccc}0&-\omega_{3}(t)&\omega_{2}(t)\\\
\omega_{3}(t)&0&-\omega_{1}(t)\\\
-\omega_{2}(t)&\omega_{1}(t)&0\end{array}\right).$ (1.3)

We assume that $x$ and $q$ refer to a certain Euclidean coordinate system in
${\rm I\kern-1.56693ptR}^{3}$, and the vector product $\wedge$ is defined in
this system by standard formulas. The identification (1.3) of a skew-symmetric
matrix and the corresponding angular velocity vector is true in any Euclidean
coordinate system of the same orientation as the initial one.

Then the Maxwell equations read, [3]

$\dot{E}(x,t)=\nabla\wedge
B(x,t)-(\dot{q}(t)+\omega(t)\wedge(x-q(t)))\rho(x-q(t)),\quad\dot{B}(x,t)=-\nabla\wedge
E(x,t),$ (1.4)

where the current has a contribution also from the internal rotation; together
with the constraints,

$\nabla\cdot E(x,t)=\rho(x-q(t)),\,\nabla\cdot B(x,t)=0.$ (1.5)

The back reaction of the field onto the charge is given through the Lorentz
force equation

$\ddot{q}(t)=\int\,[E(x,t)+(\dot{q}(t)+\omega(t)\wedge(x-q(t)))\wedge
B(x,t)]\rho(x-q(t))\,dx$ (1.6)

and the Lorentz torque equation

$I\,\dot{\omega}(t)=\int\,(x-q(t))\wedge[E(x,t)+(\dot{q}(t)+\omega(t)\wedge(x-q(t)))\wedge
B(x,t)]\rho(x-q(t))\,dx,$ (1.7)

with the moment of inertia

$I=\frac{2}{3}\int\,x^{2}\rho(x)d\,x.$ (1.8)

The important question is to obtain solutions having constant velocity and of
the form

$q(t)=q+vt,\,\omega(t)\equiv\omega,\,E(x,t)=E_{v,\omega}(x-vt),\,B(x,t)=B_{v,\omega}(x-vt).$
(1.9)

We will call them the “soliton solutions”, in brief, the “solitons”. If in
(1.4), (1.6), (1.14) we set $\omega=0$, by hand, then for every $v\in{\rm
I\kern-1.56693ptR}^{3}$ there is a unique solution of the form (1.9). However,
for the Abraham model including spin, the equation (1.14) can be satisfied
only if either $\omega\|v$ or $\omega\bot v$ [3]. This result is surprising at
first sight, but reflects the semirelativistic nature of the Abraham model.
The velocity singles out a direction, which is then taken by $\omega$.
Eventually one has to understand the domain of attraction of this soliton-like
solutions. In this paper we restrict ourselves however to a simpler situation,
where the charge remains at rest for all times, $q\equiv 0$. This can be
achieved by assuming the (anti-) symmetry conditions

$E(-x)=-E(x),\quad B(-x)=B(x)$ (1.10)

for the initial fields. Then this property would persist for all times:

$E(-x,t)=-E(x,t),\,\,\,B(-x,t)=B(x,t).$ (1.11)

The Lorentz force equation is automatically satisfied, the Maxwell equations
simplify to

$\dot{E}(x,t)=\nabla\wedge B(x,t)-(\omega(t)\wedge
x)\rho(x),\quad\dot{B}(x,t)=-\nabla\wedge E(x,t),$ (1.12)

with the constraints

$\nabla\cdot E(x,t)=\rho(x),\,\,\,\nabla\cdot B(x,t)=0,$ (1.13)

and the Lorentz torque equation simplifies to

$I\dot{\omega}(t)=\int x\wedge[E(x,t)+(\omega(t)\wedge x)\wedge
B(x,t)]\rho(x)\,dx.$ (1.14)

The system (1.11)–(1.14) is the subject of study of the present paper. We will
consider solutions of finite energy

$H(\omega,E,B)=\frac{I\omega^{2}}{2}+\frac{1}{2}\int\,\Big{(}|E(x)|^{2}+|B(x)|^{2}\Big{)}\,dx<\infty.$
(1.15)

The corresponding phase space will be equipped with a suitable topology below
and the existence and uniqueness of finite energy solutions will be briefly
explained. Note that the total momentum of the system

$P:=\int E(x,t)\wedge B(x,t)\,dx=0$ (1.16)

in view of the symmetry conditions (1.11).

The solitons for the system (1.11)-(1.14) have the form

$E(x,t)=E_{\omega}(x),\,\,\,B(x,t)=B_{\omega}(x),\,\,\,\omega(t)=\omega=const\in{\rm
I\kern-1.56693ptR}^{3}.$

The solitons satisfy the stationary equations

$E_{\omega}(-x)=-E_{\omega}(x),\,\,\,B_{\omega}(-x)=B_{\omega}(x),$ (1.17)
$\nabla\wedge B_{\omega}(x)-(\omega\wedge x)\rho(x)=0,\,\,\,\nabla\wedge
E_{\omega}(x)=0,$ (1.18) $\nabla\cdot E_{\omega}(x)=\rho(x),\,\,\,\nabla\cdot
B_{\omega}(x)=0,$ (1.19) $\int x\wedge[E_{\omega}(x)+(\omega\wedge x)\wedge
B_{\omega}(x)]\rho\,dx=0.$ (1.20)

To write down the exact formulas of solitons let us specify the version of
Fourier transform we use. The Fourier transform $F[f](k)=\hat{f}(k)$ of a
function $f(x)$ reads

$\hat{f}(k):=(2\pi)^{-3/2}\int e^{-ikx}f(x)\,dx.$ (1.21)

Then by the Fourier transform

$xf(x)\mapsto i\nabla_{k}\hat{f}(k),\,\,\,\nabla f(x)\mapsto ik\hat{f}(k).$
(1.22)

The Parseval equality holds,

$\int f(x)\overline{g(x)}\,dx=\int\hat{f}(k)\overline{\hat{g}(k)}\,dk.$ (1.23)

Note that the conditions (1.1) imply some special properties of the Fourier
transform $\hat{\rho}(k)$ of $\rho(x)$. First,

$\hat{\rho}(k)\,\,\,{\rm is\,\,\,a\,\,\,real-
valued\,\,\,spherically\,\,\,symmetric\,\,\,function:}\,\,\,\overline{\hat{\rho}(k)}=\hat{\rho}(k),\,\,\,\rho(k)=\rho_{r}(r),\,\,r:=|k|.$
(1.24)

Second, as for the spatial decay of $\hat{\rho}(k)$,

$\hat{\rho}(k)\,\,\,{\rm
is\,\,\,at\,\,\,least\,\,\,a\,\,\,fast\,\,\,decaying\,\,\,function\,\,\,of\,\,\,the\,\,\,Schwarz\,\,\,space}.$
(1.25)

Now, in Fourier space, the soliton fields are expressed [3, 5] by

$\hat{E}_{\omega}(k)=\frac{-ik\hat{\rho}(k)}{k^{2}},\,\,\,\hat{B}_{\omega}(k)=-\frac{k\wedge(\omega\wedge\nabla_{k}\hat{\rho}(k))}{k^{2}}.$
(1.26)

It follows from (1.24) to (1.26) that

$\hat{E}_{\omega}\in L^{2},\hat{B}_{\omega}\in L^{2}\,\,\,\,{\rm
in}\,\,\,k-{\rm space,\,\,\,hence}\,\,\,\,E_{\omega}\in L^{2},B_{\omega}\in
L^{2}\,\,\,\,{\rm in}\,\,\,x-{\rm space}.$ (1.27)

A stability of solitons of the system (1.12), (1.14) together with the
symmetry conditions (1.11) and the constraints (1.13) (= “spinning charge at
rest”) is the main result of our paper. It can be viewed as a partial result
on the way to establishing the soliton asymptotics and scattering behavior of
solutions to the system, see the following comment.

Let us comment on previous works. Note that, concerning the above system,
there is a number of formal analytical results, but very few mathematically
rigorous results on the qualitative behavior of the system’s solutions.

The system (1.4) to (1.7) is well known since Abraham’s works [2, 6]. The
direct derivation of the conservation laws from (1.7) is presented by
Kiessling in [7]. Soliton solutions to the system (1.4) to (1.7) were computed
first by Schwarzschild [8], see also the derivation in [3].

Some papers concern Lagrangian and Hamiltonian structure of the system. In [6,
Section 11] Abraham computed the Lagrangian as integral of
$-A_{0}\rho+\vec{A}\cdot\vec{j}$ for standing rotating spherically symmetric
electron subject to external fields obeying very special symmetry conditions.
In this case the Lagrangian depends only on one variable $\omega$, the angular
velocity. However, derivation of the torque equation (1.7) from the
variational Hamilton’s least action principle remained an open question. The
main goal of the Nodvik’s paper [9] is a variational derivation of the
Lorentz-covariant dynamics for the relativistic rotating charged particle in
the Maxwell field, and the proof of the corresponding conservation laws. The
system of Nodvick’s equations is overdetermined, since they do not include
rotational bare inertia. The situation was improved by Appel and Kiessling in
[10], where they develop the theory for the relativistic rotating particle
introducing a re-normalization limit. An invariant derivation of the non-
relativistic Abraham equation (1.7) from the Hamilton least action principle
relying on the Poincare equations on the Lie group SO(3) was provided in [11],
in [12] it was shown that the Kiessling conserved quantities are the Nöter
invariants of the system’s Lagrangian. Hamiltonian structure of the system was
clarified in [13].

The new interest for the rather old Abraham model is caused by the fact that a
broad class of models of this type display soliton-type asymptotics and
scattering behavior as it was discovered in recent years, see e.g. [14, 15,
16], where a charged particle moves in Maxwell or scalar field without
rotation. In particular, in [14] the orbital stability was established for
solitons of the Maxwell-Lorentz system for a moving but non-rotating particle.
The method is a thorough combination of the energy and the total momentum
conservation, for the system written in Hamiltonian form. Lagrangian and
Hamiltonian structure of the models play a significant role in these methods,
so it was of a considerable interest and importance, to include the Abraham
model with rotating charge into the class of Lagrangian and Hamiltonian
systems. Nevertheless, in view of rather complicated structure of the system,
the problem of establishing soliton-type asymptotics and scattering behavior
for the Abraham model with rotating charge remains open.

Some progress in this direction was made in the paper [5], where results on
soliton-type asymptotics in local energy seminorms and also on scattering of
solitons in global energy norms are obtained for the system (1.11) to (1.14).
The crucial assumption of the paper is that the norm of $\rho$ in $L^{2}$ is
sufficiently small that means a weak wave-particle interaction.

The method of the present paper does not need this assumption. It exploits
energy conservation and angular momentum arguments.

Using the energy alone we prove, first, that the zero soliton with $\omega=0$
is Lyapunov stable and orbital stable. Second, we prove stability of an
arbitrary soliton with respect to a special class of perturbations of initial
data, namely, perturbations of uniformly compact support, see Theorem 2.3
below111These two results are published in [17]. We present them for
completeness, and also correct the computational error that was made in [17],
which did not affect the truth of its main result..

On the other hand, under the additional condition $\int\,\rho(x)\,dx=0$ the
solitons have some better spatial decay such that the angular momentum
(formula (2.29) below) is well-defined. We establish that a non-zero soliton
of finite angular momentum is unstable, see Theorem 2.5 below. Thus, generally
an arbitrary soliton is unstable.

The question of stability for an arbitrary non-zero soliton of infinite
angular momentum remains open.

##  2 Existence of dynamics and the main results

Consider the space $L=({\rm I\kern-1.56693ptR}^{3},L^{2}({\rm
I\kern-1.56693ptR}^{3};{\rm I\kern-1.56693ptR}^{3}),L^{2}({\rm
I\kern-1.56693ptR}^{3};{\rm I\kern-1.56693ptR}^{3}))$ with the norm

$\|(\omega,E,B)\|_{L}:=|\omega|+\|E\|_{L^{2}({\rm I\kern-1.09685ptR}^{3};{\rm
I\kern-1.09685ptR}^{3})}+\|B\|_{L^{2}({\rm I\kern-1.09685ptR}^{3};{\rm
I\kern-1.09685ptR}^{3})}.$

###### Definition 2.1

The phase space for the system (1.11)-(1.14) is the nonlinear submanifold
${\cal M}$ of states $(\omega,E,B)\in L$, where $E,B$ satisfy (1.11) and
(1.13). The topology of ${\cal M}$ is defined through the imbedding ${\cal
M}\subset L$.

Let us state a proposition on existence of dynamics for the system
(1.11)-(1.14) in the phase space ${\cal M}$.

###### Proposition 2.1

Let the assumptions (1.1) hold. Then

i) for any $(\omega_{0},E_{0},B_{0})\in{\cal M}$ the Cauchy problem for the
system (1.11)-(1.14) has a unique solution $(\omega(t),E(x,t),B(x,t))\in
C({\rm I\kern-1.56693ptR};{\cal M})$ with initial conditions
$\omega(0)=\omega_{0},E(x,0)=E_{0},B(x,0)=B_{0}$;

ii) for any $T\in{\rm I\kern-1.56693ptR}$ the map
$U_{T}:(\omega_{0},E_{0},B_{0})\mapsto(\omega(T),E(\cdot,T),B(\cdot,T))$ is
continuous in ${\cal M}$;

iii) the energy $H(t):=H(\omega(t),E(\cdot,t),B(\cdot,t))$ is conserved along
the solutions of the system:

$H(t)\equiv H(0),\,\,\,\,t\in{\rm I\kern-1.56693ptR}.$ (2.28)

The proof is similar to that of [18, Appendix], with the charge density
$\rho(x)$ and the current $j(x,t):=(\omega(t)\wedge x)\rho(x)$.

Now let us specify common definitions of Lyapunov stability, orbital
stability, and asymptotic stability for the particular case of soliton
solutions to the system (1.11)-(1.14).

###### Definition 2.2

1) A soliton solution $Y_{\omega}=(\omega,E_{\omega},B_{\omega})$ is called
Lyapunov stable if $\forall\varepsilon>0$ $\exists\delta>0$ such than for any
solution $Y(t)=(\omega(t),E_{(}t),B(t))$ from the condition
$\|Y(0)-Y_{\omega}(0)\|_{L}<\delta$ it follows that
$\|Y(t)-Y_{\omega}(t)\|_{L}<\varepsilon$ $\forall t\in{\rm
I\kern-1.56693ptR}$.

2) A soliton solution $Y_{\omega}=(\omega,E_{\omega},B_{\omega})$ is called
orbital stable if $\forall\varepsilon>0$ $\exists\delta>0$ such than for any
solution $Y(t)=(\omega(t),E_{(}t),B(t))$ with
$\|Y(0)-Y_{\omega}(0)\|_{L}<\delta$ the solution $Y(\cdot)$ remains in
$\varepsilon$-neighbourhood of $Y_{\omega}(\cdot)$ in the space $C({\rm
I\kern-1.56693ptR};L)$.

3) A soliton solution $Y_{\omega}=(\omega,E_{\omega},B_{\omega})$ is called
asymptotic stable if it is Lyapunov stable and $\|Y(t)-Y_{\omega}(t)\|_{L}\to
0$ as $t\to\infty$.

Note that in view of the energy conservation (2.28) one could not expect the
asymptotic stability of the solitons in the phase space ${\cal M}$ of finite
energy states. Our main result is then the Lyapunov and orbital stability. In
fact, for our case the properties of Lyapunov stability and orbital stability
of a soliton coincide, since solitons do not depend on $t$ and thus are
stationary solutions (fixed points of the dynamical system).

Further, the method of exploiting energy conservation arguments we use imposes
some restrictions on the class of possible perturbations. Namely, we assume
that the perturbations are of uniformly compact supports, see the exact
formulation below. The main result on stability is the following theorem.

###### Theorem 2.3

a) The zero soliton with $\omega=0$ is Lyapunov stable and orbital stable.

b) Let us fix a non-zero $\omega\in{\rm I\kern-1.56693ptR}^{3}$. Consider
solutions to the Cauchy problem for the system (1.11)-(1.14) with initial data
$\omega+\Omega_{0},\,\,E_{\omega}(x)+e_{0}(x),\,\,B_{\omega}(x)+b_{0}(x)$,
where $(\Omega_{0},e_{0},b_{0})\in L$, $e_{0}(x)$ is odd, $b_{0}(x)$ is even,
$\nabla\cdot e_{0}=0$, $\nabla\cdot b_{0}=0$.

The soliton $(\omega,E_{\omega}(x),B_{\omega}(x))$ is Lyapunov stable and as
well orbital stable with respect to perturbations of uniformly compact support

i.e. $\forall\varepsilon>0$ and $\forall R>0$ $\exists\delta>0$ such that for
any solution $(\omega(t),E(t),B(t))$ with the initial data
$\omega_{0}=\omega+\Omega_{0},E_{0}=E_{\omega}+e_{0},B_{0}=B_{\omega}+b_{0}$
from the condition $\|(\Omega_{0},e_{0},b_{0})\|_{L}<\delta$ it follows that
$\|(\omega(t)-\omega,E(t)-E_{\omega},B(t)-B_{\omega})\|_{L}<\varepsilon$
$\forall t\in{\rm I\kern-1.56693ptR}$ $\forall(\Omega_{0},e_{0}(x),b_{0}(x))$
such that ${\rm supp}\,e_{0}\subset\\{|x|\leq R\\}$ and ${\rm
supp}\,b_{0}\subset\\{|x|\leq R\\}$.

On the other hand, non-zero solitons of finite angular momentum, see below,
turn out to be instable. Perturbation providing instability have their
supports tending to infinity, so the result is consistent with Theorem 2.3.

###  2.1 Angular momentum

The angular momentum is defined by

$M(\omega,E,B):=I\omega+\int\,x\wedge(E(x)\wedge B(x))\,dx.$ (2.29)

Note that for $(E,B)\in(L^{2}({\rm I\kern-1.56693ptR}^{3};{\rm
I\kern-1.56693ptR}^{3}),L^{2}({\rm I\kern-1.56693ptR}^{3};{\rm
I\kern-1.56693ptR}^{3}))$ the angular momentum is generally not defined. It is
well defined for the fields $(E,B)$ with the finite weighted norms

$\int\,|x|\,|E(x,t)|^{2}dx,\,\,\,\,\int\,|x|\,|B(x,t)|^{2}dx.$ (2.30)

###  2.2 Faster spatial decay and angular momentum of solitons

From (1.26), by a straightforward computation we obtain that

$\partial_{j}\hat{E}_{\omega}\in L^{2},\,\,\,\,\partial_{j}\hat{B}_{\omega}\in
L^{2},\,\,\,\,j=1,2,3$

under the condition

$\hat{\rho}(0)=0.$ (2.31)

Hence,

$x_{j}E_{\omega}\in L^{2},\,\,\,\,x_{j}B_{\omega}\in
L^{2},\,\,\,\,j=1,2,3\,\,\,\,{\rm(in\,\,\,}x-{\rm space)}$ (2.32)

under the same condition (2.31) or the equivalent condition

$\int\,\rho(x)\,dx=0.$ (2.33)

As the result, we obtain the following

###### Proposition 2.2

Under the condition (2.31) or the equivalent condition (2.33) the weighted
norms of the soliton fields

$\int\,|x|\,|E_{\omega}(x)|^{2}dx,\,\,\,\,\int\,|x|\,|B_{\omega}(x)|^{2}dx$
(2.34)

are finite and the angular momentum of the soliton

$M_{\omega}:=I\omega+\int\,x\wedge(E_{\omega}(x)\wedge
B_{\omega}(x))\,dx<\infty$ (2.35)

is well-defined.

###### Remark 2.4

The angular momentum of a soliton can be computed exactly, see Appendix.

###  2.3 Instability of a soliton of finite angular momentum

###### Theorem 2.5

Let the condition (2.31), hold. Then a non-zero soliton (1.26) is not Lyapunov
stable and is not orbital stable.

Note that we need not construct a solution of finite angular momentum to the
system. The finiteness of angular momentum of the soliton and its
perturbations is the sufficient condition.

##  3 Proof of Theorem 2.3

###  3.1 Equations for perturbations and their solutions

Let us fix an arbitrary soliton $(\omega,E_{\omega},B_{\omega})$. To study its
stability we put

$\omega(t)=\omega+\Omega(t),\,\,E(x,t)=E_{\omega}(x)+e(x,t),\,\,B(x,t)=B_{\omega}(x)+b(x,t)$
(3.36)

with

$\nabla\cdot e=0,\,\,\nabla\cdot b=0,\,\,e(-x,t)=-e(x,t),\,\,b(-x,t)=b(x,t).$
(3.37)

Insert (3.36) into (1.12) and (1.14), take the stationary equations
(1.18)–(1.20) into account and obtain the following system for the
perturbations $e$, $b$, $\Omega$:

$\dot{e}=\nabla\wedge b-(\Omega\wedge x)\rho,\,\,\,\dot{b}=-\nabla\wedge e,$
(3.38) $I\dot{\Omega}=\int x\wedge[e+(\Omega\wedge x)\wedge
B_{\omega}+(\omega\wedge x)\wedge b+(\Omega\wedge x)\wedge b]\rho\,dx.$ (3.39)

The following remark is very important for our further analysis.

###### Remark 3.1

By the energy conservation (2.28), (1.27), and (3.36),

$(\Omega(t),e(x,t),b(x,t))\,\,\,{\rm is\,\,bounded\,\,in}\,\,\,L\,\,\,{\rm
uniformly\,\,in}\,\,\,t\in{\rm I\kern-1.56693ptR}.$ (3.40)

Let us study the system (3.37), (3.38), (3.39). First we find the exact
solutions of the Cauchy problem to the system with initial data
$(\Omega_{0},e_{0},b_{0})$.

Step 1. We find the solution to the system

$\dot{e}=\nabla\wedge b-j(x,t),\,\,\,\dot{b}=-\nabla\wedge e$ (3.41)

with zero charge density and the prescribed current $j(x,t)=(\Omega(t)\wedge
x)\rho(x)$. By [18, Appendix] the solution, in the Fourier space reads

$\hat{C}(k,t)=\hat{m}_{t}(k)\hat{C}_{0}(k)-\int_{0}^{t}\hat{m}_{t-s}(k)\hat{j}(k,s)\,ds,$
(3.42)

where $\hat{C}(k,t)=\hat{e}(k,t)+i\hat{b}(k,t)$,
$\hat{C}_{0}(k)=\hat{e}_{0}(k)+i\hat{b}_{0}(k)$;

$\hat{m}_{t}(k):=m\hat{K}_{t}(k)+\frac{d}{dt}\hat{K}_{t}(k),\,\,\,m:=k\wedge,\,\,\,\hat{K}_{t}(k):=\frac{\sin(|k|t)}{|k|}.$

To separate the fields we return to the $x$-representation. Let
$K_{t}(x)=F^{-1}[\hat{K}_{t}](x)$, then
$m_{t}(x)=F^{-1}[\hat{m}_{t}](x)=-i\nabla\wedge K_{t}(x)+\dot{K}_{t}(x)$. By
(3.42) we have

$e+ib=(-i\nabla\wedge
K_{t}(x)+\dot{K}_{t}(x))*(e_{0}+ib_{0})-\int_{0}^{t}(-i\nabla\wedge
K_{t-s}(x)+\dot{K}_{t-s}(x))*j(s)\,ds.$ (3.43)

Separating the real and the imaginary parts we obtain

$e=\dot{K}_{t}(x)*e_{0}+(\nabla\wedge
K_{t}(x))*b_{0}-\int_{0}^{t}\dot{K}_{t-s}(x)*j(s)\,ds,$ (3.44)
$b=-(\nabla\wedge
K_{t}(x))*e_{0}+\dot{K}_{t}(x)*b_{0}+\int_{0}^{t}\nabla\wedge
K_{t-s}(x)*j(s)\,ds.$ (3.45)

Return to the Fourier space and get

$\hat{e}=\frac{d}{dt}\hat{K}_{t}\hat{e}_{0}+im\hat{K}_{t}\hat{b}_{0}-\int_{0}^{t}\frac{d}{dt}\hat{K}|_{t-s}\hat{j}(s)\,ds,$
(3.46)
$\hat{b}=-im\hat{K}_{t}\hat{e}_{0}+\frac{d}{dt}\hat{K}_{t}\hat{b}_{0}+\int_{0}^{t}im\hat{K}_{t-s}\hat{j}(s)\,ds.$
(3.47)

For $\Omega$ we obtain the closed equation

$I\dot{\Omega}=\int x\wedge[e+(\Omega\wedge x)\wedge B_{\omega}+(\omega\wedge
x)\wedge b+(\Omega\wedge x)\wedge b]\rho\,dx,$ (3.48)

where $e,b$ are given by (3.44), (3.45). Then $\Omega(t)$ determined by (3.48)
and by an initial condition $\Omega(0)=\Omega_{0}$ with $e,b$ determined by
(3.44), (3.45) provide the solution to the Cauchy problem to the system (3.37)
– (3.39).

For $\omega=0$ the equation (3.48) reads

$I\dot{\Omega}=\int x\wedge[e+(\Omega\wedge x)\wedge b]\rho\,dx.$ (3.49)

The system (3.38), (3.49) is of the same type as the initial system (1.10) to
(1.14) and the corresponding energy

${\cal
H}(t):=\frac{I\Omega(t)^{2}}{2}+\frac{1}{2}\int\,\Big{(}|e(x,t)|^{2}+|b(x,t)|^{2}\Big{)}\,dx$

is conserved. This implies the statement a) of Theorem 2.3.

Further, we rewrite the equation (3.39) as

$I\dot{\Omega}=\int x\wedge[(\Omega\wedge x)\wedge
B_{\omega}+e+(\omega(t)\wedge x)\wedge b]\rho\,dx.$ (3.50)

In (3.50), we consider $\omega(t)$ as the known function, the first component
of the solution to the system (1.11) – (1.14). In particular, $\omega(t)$ is
uniformly bounded in $t\in{\rm I\kern-1.56693ptR}$ by the conservation of the
energy (2.28).

###  3.2 The structure of the equation (3.50)

We write the right-hand side of (3.50) as $T_{1}+T_{2}+T_{3}$ with

$T_{1}:=\int x\wedge[(\Omega\wedge x)\wedge
B_{\omega}]\rho\,dx,\,\,\,T_{2}:=\int(x\wedge e)\rho\,dx,\,\,\,T_{3}:=\int
x\wedge[(\omega(t)\wedge x)\wedge b]\rho\,dx.$

Step 1. For the first term we have

$T_{1}=\int(\Omega\wedge x)(x\cdot B_{\omega})\rho\,dx=K\wedge\Omega,$

where

$K:=-\int x(x\cdot B_{\omega})\rho\,dx$ (3.51)

is a constant vector in ${\rm I\kern-1.56693ptR}^{3}$. Let us compute $K$ in
detail. Since $\hat{\rho}$ is real-valued,

$K=-\int(x\cdot
B_{\omega})x\rho\,dx=-\int(i\nabla_{k}\cdot\hat{B}_{\omega})\overline{i\nabla_{k}\hat{\rho}}\,dk=-\int\nabla_{k}\cdot\hat{B}_{\omega}\cdot\overline{\nabla_{k}\hat{\rho}}\,dk=\int\hat{B}_{\omega}\Delta_{k}\hat{\rho}\,dk.$

Recall that

$\hat{B}_{\omega}=-\frac{k\wedge(\omega\wedge\nabla_{k}\hat{\rho})}{k^{2}}=-\frac{\omega(k\cdot\nabla_{k}\hat{\rho})-\nabla_{k}\hat{\rho}(k\omega)}{k^{2}}=\frac{k\tilde{\rho}(k\omega)}{k^{2}}-\omega\tilde{\rho}.$

Here we denote $r=|k|$, $\hat{\rho}=\rho_{r}(r)$, then
$\nabla_{k}\hat{\rho}=k\rho_{r}^{\prime}(r)/r$ and we put
$\tilde{\rho}(r):=\rho_{r}^{\prime}(r)/r$. Then for $K$ we obtain

$K=K_{1}-K_{2};\,\,\,\,K_{1}:=\int\frac{\tilde{\rho}}{k^{2}}(k\otimes
k)\omega\,\Delta_{k}\hat{\rho}\,dk;\,\,\,\,K_{2}:=\omega\int\,\tilde{\rho}\,\Delta_{k}\hat{\rho}\,dk.$

Further, by the skew-symmetry property of the non-diagonal terms of the matrix
$k\otimes k$ we have

$K_{1}=\int\frac{\tilde{\rho}}{k^{2}}\Delta_{k}\hat{\rho}\left(\begin{array}[]{c}k_{1}^{2}\,\omega_{1}\\\
k_{2}^{2}\,\omega_{2}\\\ k_{3}^{2}\,\omega_{3}\end{array}\right)dk.$

Each of the integrals

$\int\frac{\tilde{\rho}}{k^{2}}\Delta_{k}\hat{\rho}\,k_{j}^{2}\omega_{j}\,dk\,\,\,\,\,{\rm
equals}\,\,\,\,\,\frac{1}{3}\int\frac{\tilde{\rho}}{k^{2}}\Delta_{k}\hat{\rho}k^{2}\,\omega_{j}\,dk=\frac{1}{3}\int\tilde{\rho}\Delta_{k}\hat{\rho}\,\omega_{j}\,dk,$

hence,

$K_{1}=\frac{1}{3}{\omega}\int\tilde{\rho}\Delta_{k}\hat{\rho}\,dk.$

Finally

$K_{1}-K_{2}=-\frac{2}{3}{\omega}\int\tilde{\rho}\Delta_{k}\hat{\rho}\,dk.$

Thus, $K$ is proportional to $\omega$.

Step 2. For $T_{2}$ we obtain

$T_{2}=-\int(e\wedge x)\rho\,dx=-\int(\hat{e}\wedge
i\nabla_{k})\hat{\rho}\,dk.$

Recall that $\nabla_{k}\hat{\rho}=\tilde{\rho}(r)k$, then

$T_{2}=i\int(k\wedge\hat{e})\tilde{\rho}\,dk=i\int
dk\,\tilde{\rho}\left[k\wedge\left(\frac{d}{dt}\hat{K}_{t}\hat{e}_{0}+im\hat{K}_{t}\hat{b}_{0}-\int_{0}^{t}\frac{d}{dt}\hat{K}|_{t-s}\hat{j}(s)\,ds\right)\right].$

Put

$T_{21}(t):=i\int
dk\,\tilde{\rho}\left[k\wedge\left(\frac{d}{dt}\hat{K}_{t}\hat{e}_{0}+im\hat{K}_{t}\hat{b}_{0}\right)\right]=i\int
dk\,\tilde{\rho}\left(\cos|k|t(k\wedge\hat{e}_{0})+i\hat{K}_{t}(k(k\cdot\hat{b}_{0})-k^{2}\hat{b}_{0}\right)=$

$\int
dk\,\tilde{\rho}\left(i\cos|k|t(k\wedge\hat{e}_{0})+\frac{\sin|k|t}{|k|}k^{2}\hat{b}_{0}\right)=\int
dk\,\tilde{\rho}\left(i\cos|k|t(k\wedge\hat{e}_{0})+|k|\sin|k|t\,\hat{b}_{0}\right),$
(3.52)

since $k\cdot\hat{b}_{0}=0$.

Further, $\hat{j}(k,s)=\Omega(s)\wedge i\nabla_{k}\hat{\rho}=i\Omega(s)\wedge
k\tilde{\rho}$. Then

$T_{22}(t):=-i\int
dk\,\tilde{\rho}\left[k\wedge\int_{0}^{t}\frac{d}{dt}\hat{K}|_{t-s}\hat{j}(s)\,ds\right]=-i\int_{0}^{t}\,ds\int
dk\,\tilde{\rho}^{2}\cos|k|(t-s)[k\wedge(i\Omega(s)\wedge k)]=$
$\int_{0}^{t}\,ds\int
dk\,\tilde{\rho}^{2}\cos|k|(t-s)[\Omega(s)k^{2}-k(k\cdot\Omega(s))]=\int_{0}^{t}\,ds\,M_{2}(t-s)\Omega(s),$

where

$M_{2}(t-s):=\int dk\,\tilde{\rho}^{2}\cos|k|(t-s)[k^{2}E-k\otimes k].$ (3.53)

The matrix $k^{2}E-k\otimes k$ reads

$\left(\begin{array}[]{ccc}k_{2}^{2}+k_{3}^{2}&-k_{1}k_{2}&-k_{1}k_{3}\\\
-k_{2}k_{1}&k_{1}^{2}+k_{3}^{2}&-k_{2}k_{3}\\\
-k_{3}k_{1}&-k_{3}k_{2}&k_{1}^{2}+k_{2}^{2}\end{array}\right).$

Thus, every non-diagonal element in (3.53) is zero, since the integrand
function is odd w.r.t. corresponding variable. (Note that convergence of each
integral above and below is provided by sufficient decay of $\hat{\rho}$ and
its derivatives in $k$ due to (1.1).) Further,

$\int dk\,\tilde{\rho}^{2}\cos|k|(t-s)k_{j}^{2}=\frac{1}{3}\int
dk\,\tilde{\rho}^{2}\cos|k|(t-s)k^{2},\,\,\,j=1,2,3$

by the change of variables $k_{i}\mapsto k_{j}$ and we obtain

$M_{2}(t-s):=\frac{2}{3}\int dk\,\tilde{\rho}^{2}\cos|k|(t-s)k^{2}E.$ (3.54)

Finally, we change once more the order of integration and obtain

$T_{22}(t)=\frac{2}{3}\int
dk\,\tilde{\rho}\,^{2}k^{2}\int_{0}^{t}\,ds\cos|k|(t-s)\Omega(s).$ (3.55)

Step 3. $T_{3}=\int\,x\wedge[(\omega(t)\wedge x)\wedge
b]\rho\,dx=\omega(t)\wedge\int\,(x\cdot b)\,x\rho\,dx.$ Further,

$\int\,(x\cdot
b)\,x\rho\,dx=\int\,(i\nabla_{k}\cdot\hat{b})(-i\nabla_{k}\hat{\rho})\,dk=\int\,dk(\partial_{1}\hat{b}_{1}+\partial_{2}\hat{b}_{2}+\partial_{3}\hat{b}_{3})k\tilde{\rho}=$
$-\int\,dk(\hat{b}_{1}\partial_{1}(k\tilde{\rho})+\hat{b}_{2}\partial_{2}(k\tilde{\rho})+\hat{b}_{3}\partial_{3}(k\tilde{\rho})).$
(3.56)

We have

$\partial_{j}(k\tilde{\rho})=\delta_{jm}\tilde{\rho}+\partial_{j}\tilde{\rho}\,k,\,\,\,j,m=1,2,3.$

Then (3.56) reads

$-\int\,dk(\tilde{\rho}\hat{b}+(\hat{b}\cdot\nabla_{k}\tilde{\rho})=-\int\,dk(\tilde{\rho}\hat{b}),$

since $\hat{b}\cdot\nabla_{k}\tilde{\rho}=0$ because $\nabla_{k}\tilde{\rho}$
is proportional to $k$. Introduce (3.47) and obtain

$-\int\,dk(\tilde{\rho}\hat{b})=\int\,dk\left[im\hat{K}_{t}\hat{e}_{0}-\frac{d}{dt}\hat{K}_{t}\hat{b}_{0}-\int_{0}^{t}\,ds\,im\hat{K}_{(t-s)}\hat{j}(k,s)\right]\tilde{\rho}.$

Put

$T_{31}(t):=\omega(t)\wedge\int\,dk\,\left[im\hat{K}_{t}\hat{e}_{0}-\frac{d}{dt}\hat{K}_{t}\hat{b}_{0}\right]\tilde{\rho}=\omega(t)\wedge\int\,dk\,\left[i\frac{\sin|k|t}{|k|}k\wedge\hat{e}_{0}-\cos|k|t\,\hat{b}_{0}\right]\tilde{\rho}.$
(3.57)

Further, $\,\,\,\,T_{32}(t):=$

$\omega(t)\wedge\int\,dk\left[-\int_{0}^{t}\,ds\,im\hat{K}_{(t-s)}\hat{j}(k,s)\right]\tilde{\rho}=\omega(t)\wedge\int_{0}^{t}\,ds\,\int\,dk\tilde{\rho}^{2}\left[\hat{K}_{(t-s)}k\wedge(\Omega(s)\wedge
k)\right]=$
$\omega(t)\wedge\int_{0}^{t}\,ds\,\int\,dk\,\tilde{\rho}^{2}\hat{K}_{(t-s)}(\Omega(s)k^{2}-k(\Omega(s)\cdot
k))=\omega(t)\wedge\frac{2}{3}\int\,dk\,\tilde{\rho}^{2}k^{2}\int_{0}^{t}\,ds\,\frac{\sin|k|(t-s)}{|k|}\Omega(s),$
(3.58)

similarly to the case of $T_{22}$. Now, the equation (3.50) reads

$I\dot{\Omega}=K\wedge\Omega+T_{21}(t)+T_{31}(t)+\frac{2}{3}\int\,dk\,\tilde{\rho}^{2}k^{2}\int_{0}^{t}\,ds\,(\cos|k|(t-s))\Omega(s)+$
$\omega(t)\wedge\frac{2}{3}\int\,dk\,\tilde{\rho}^{2}k^{2}\int_{0}^{t}\,ds\,(\frac{\sin|k|(t-s)}{|k|})\Omega(s),$
(3.59)

where $K,T_{21},T_{31}$ are given by (3.51), (3.52), (3.57) respectively.
Finally, we rewrite (3.50) as

$\dot{\Omega}=M\Omega+T(t),$ (3.60)

where $M$ is the skew-adjoint matrix of $m:=(K/I)\wedge$ and

$T(t):=(1/I)(T_{21}(t)+T_{31}(t)+T_{22}(t)+T_{32}(t)).$ (3.61)

Note that, in detail,

$\frac{K}{I}=\frac{-\frac{2}{3}{\omega}\int\tilde{\rho}\Delta_{k}\hat{\rho}\,dk}{\frac{2}{3}\int
x^{2}\rho(x)\,dx}=\frac{-\frac{2}{3}{\omega}\int\tilde{\rho}\Delta_{k}\hat{\rho}\,dk}{-\frac{2}{3}\int\Delta_{k}\hat{\rho}(k)\,dk}=\alpha(\rho)\omega,$

where

$\alpha(\rho):=\frac{\int\tilde{\rho}\Delta_{k}\hat{\rho}\,dk}{\int\Delta_{k}\hat{\rho}(k)\,dk}.$

The equation (3.60) is an integro-differential equation of the type which is
considered in [21] in the case of real-valued functions; the problems of
existence and stability of solutions are studied. The extension of results of
[21] to the case of vector-valued functions is evident.

From [21, Lemma 1.2, formula (1.4)] it follows that for any finite $T\in{\rm
I\kern-1.56693ptR}$ the solution $\Omega(\cdot)$ depends continuously in
$C(0,T;{\rm I\kern-1.56693ptR}^{3})$ on initial datum $\Omega_{0}$:

$\Omega(\cdot)\to 0\,\,\,{\rm C(0,T;{\rm I\kern-1.56693ptR}^{3})}\,\,\,{\rm
as}\,\,\,\Omega_{0}\to 0\,\,\,{\rm in}\,\,\,{\rm
I\kern-1.56693ptR}^{3}\,\,\,{\rm in\,\,\,particular,}\,\,\,\Omega(T)\to 0.$
(3.62)

By (3.62), (3.46), (3.47)

$(\Omega(T),e(\cdot,T),b(\cdot,T))\to(0,0,0)\,\,\,{\rm in}\,\,\,L\,\,\,{\rm
as}\,\,\,(\Omega_{0},e_{0},b_{0})\to(0,0,0)\,\,\,{\rm in}\,\,\,L.$ (3.63)

To shorten the notations let us rewrite (3.44), (3.45) as

$\left(\begin{array}[]{c}e(x,t)\\\
b(x,t)\end{array}\right)=U(t)\left(\begin{array}[]{c}e_{0}(x)\\\
b_{0}(x)\end{array}\right)-\int_{0}^{t}U(t-s)\left(\begin{array}[]{c}j(x,s)\\\
0\end{array}\right)\,ds,$ (3.64)

where $U(t)$ is the group of the free Maxwell equation. Note that the group is
isometric in the space $[L^{2}({\rm I\kern-1.56693ptR}^{3};{\rm
I\kern-1.56693ptR}^{3})]^{2}$ by the corresponding energy conservation law for
free Maxwell equations, [16, 18, 19, 20]. Put

$\left(\begin{array}[]{c}e_{(0)}(x,t)\\\
b_{(0)}(x,t)\end{array}\right):=U(t)\left(\begin{array}[]{c}e_{0}(x)\\\
b_{0}(x)\end{array}\right).$ (3.65)

Then, in $x$-representation,

$T_{21}(t)=\int(x\wedge
e_{(0)}(x,t))\rho(x)\,dx,\,\,\,\,T_{31}(t)=\omega(t)\wedge\int(x\cdot
b_{(0)}(x,t))x\rho(x)\,dx.$ (3.66)

###  3.3 The special case of zero initial fields perturbation

Let us consider the special case of initial data $e_{0}(x)=0$, $b_{0}(x)=0$,
and $\Omega_{0}$ is arbitrary. In this case one has $e_{(0)}(x,t)=0$,
$b_{(0)}(x,t)=0$ and hence, $T_{21}(t)=0$ and $T_{31}(t)=0$ by (3.66). Then
the equation (3.60) becomes a linear homogeneous integro-differential equation
and its solution reads

$\Omega(t)=A(t)\Omega_{0},$ (3.67)

where $A(t)$ is a $3\times 3$-matrix and $\|A(t)\|$ is bounded uniformly in
$t\in{\rm I\kern-1.56693ptR}$ by (3.40).

Recall that

$j(x,s)=(\Omega(s)\wedge x)\rho(x).$ (3.68)

Then by (3.64), (3.68), and (3.67),

$\left(\begin{array}[]{c}e(x,t)\\\
b(x,t)\end{array}\right)=W(x,t)\Omega_{0}\,,$ (3.69)

where $W(x,t)$ is a $6\times 3$ matrix. The components $w_{ij}(t,x)$,
$i=1,...,6$, $j=1,...,3$ of the matrix are functions bounded in $L^{2}$
uniformly in $t\geq 0$ due to (3.40). Then we obtain the following preliminary
result on stability:

###### Proposition 3.1

For the system (1.12)-(1.14), in the phase space ${\cal M}$, the soliton
$(\omega,E_{\omega},B_{\omega})$ is Lyapunov stable (and as well orbital
stable) with respect to perturbations of type $(\Omega_{0},e_{0}=0,b_{0}=0)$.

###  3.4 Completing the proof of Theorem 2.3

Consider initial perturbations $(e_{0},b_{0},\Omega_{0})$ such that

${\rm supp}\,e_{0}\subset\\{|x|\leq R\\},\,\,\,{\rm
supp}\,b_{0}\subset\\{|x|\leq R\\};$

with a fixed $R>0$.

By the strong Huygens principle for the group of the free Maxwell equations
[16, 18, 19, 20] the supports of $e_{(0)}(x,t)$ and of $b_{(0)}(x,t)$ are
subsets of the region $\\{|x|>t-R\\}$.

Then, since $\rho$ is compact supported, there is a
$\overline{T}=\overline{T}(R,R_{\rho})$ such that $T_{21}(t)=0$ and
$T_{31}(t)=0$ for $t\geq\overline{T}$ by (3.66). Then for $t\geq\overline{T}$
the equation (3.60) reads

$\dot{\Omega}=M\Omega+\frac{1}{I}(T_{21}(t)+T_{31}(t))\,\,\,{\rm
with\,\,the\,\,initial\,\,condition}\,\,\,\overline{\Omega}:=\Omega(\overline{T}),$
(3.70)

where $\Omega(t)$ is the solution to (3.60) for $0\leq t\leq\overline{T}$. The
equation (3.70) is a linear homogeneous integro-differential equation w.r.t.
$\Omega$.

By (3.63)

$\overline{\Omega}\to 0\,\,\,{\rm in}\,\,\,{\rm
I\kern-1.56693ptR}^{3}\,\,\,{\rm as}\,\,\,(\Omega_{0},e_{0},b_{0})\to
0\,\,\,{\rm in}\,\,\,L.$ (3.71)

Further, for $t\geq\overline{T}$ the solution in (3.64) reads

$\left(\begin{array}[]{c}e(x,t)\\\
b(x,t)\end{array}\right)=U(t)\left(\begin{array}[]{c}e_{0}(x)\\\
b_{0}(x)\end{array}\right)-\int_{0}^{\overline{T}}U(t-s)\left(\begin{array}[]{c}j(x,s)\\\
0\end{array}\right)\,ds-\int_{\overline{T}}^{t}U(t-s)\left(\begin{array}[]{c}j(x,s)\\\
0\end{array}\right)\,ds.$ (3.72)

i) In the right-hand side of (3.72), for the first term $(e_{(0)},b_{(0)})$ we
have

$\|e_{(0)}(\cdot,t)\|_{L^{2}}^{2}+\|b_{(0)}(\cdot,t)\|_{L^{2}}^{2}=\|e_{0}\|_{L^{2}}^{2}+\|b_{0}\|_{L^{2}}^{2},$
(3.73)

since the group $U(t)$ is unitary.

ii) For the second term let us observe that $s\in[0;\overline{T}]$ with a
fixed $\overline{T}$, the group $U(t-s)$ is unitary, and
$j(x,s)=(\Omega(s)\wedge x)\rho(x)$. Then, by continuous dependence, see
(3.62), (3.63), this term is bounded in $[L^{2}({\rm
I\kern-1.56693ptR}^{3};{\rm I\kern-1.56693ptR}^{3})]^{2}$ uniformly in
$t\in{\rm I\kern-1.56693ptR}$ and tends to zero as
$(\Omega_{0},e_{0},b_{0})\to 0$.

iii) For the third term it follows from i), ii), and (3.40) that it is also
bounded in $[L^{2}({\rm I\kern-1.56693ptR}^{3};{\rm
I\kern-1.56693ptR}^{3})]^{2}$ uniformly in $t\in{\rm I\kern-1.56693ptR}$.
Further, the current reads

$j(x,s)=(\Omega(s)\overline{\Omega}\wedge x)\rho,\,\,\,s\geq\overline{T},$

and after integrating the term becomes $\overline{W(x,t)}\,\overline{\Omega}$,
where the components

$\overline{w}_{ij}(x,t)\,\,\,{\rm
of\,\,the\,\,matrix}\,\,\,\overline{W}(x,t)\,\,\,{\rm
are\,\,bounded\,\,in}\,\,\,L^{2}\,\,\,{\rm uniformly\,\,in}\,\,\,t,$ (3.74)

because of the uniform boundness (3.40). By (3.71) this term tends to zero in
$L$ as $(\Omega_{0},e_{0},b_{0})\to(0,0,0)$ in $L$.

iv) Finally, for $\Omega(t)$ itself we have $\Omega(t)\to 0$ in ${\rm
I\kern-1.56693ptR}^{3}$ as $(\Omega_{0},e_{0},b_{0})\to(0,0,0)$ in $L$ by
(3.62) for $t\leq\overline{T}$ and by (3.67) for $t\geq\overline{T}$.

Then the conclusion of Theorem 2.3 follows from i) to iv). The proof is
complete.

##  4 Proof of Theorem 2.5

###  4.1 Energy variation on the surface of constant angular momentum

Let us fix an $\omega\neq 0$ and consider the soliton
$(\omega,E_{\omega},B_{\omega})$ with the angular momentum $M=M_{\omega}$.
Consider the surface of constant angular momentum

$\\{(\Omega,E,B):M(\Omega,E,B)=M=M_{\omega}\\}.$

where the fields $(E,B)$ obey the constraints (1.5). Let us make variations of
the energy $H(\Omega,E,B,)$ on this surface. For this purpose we express
$\Omega$ in $M$,

$\Omega=\frac{1}{I}(M-\int x\wedge(E\wedge B)dx)$

and make variations of

$H_{M}(E,B)=\frac{1}{2I}\left(M-\int x\wedge(E\wedge
B)dx\right)^{2}+\frac{1}{2}\int(|E(x)|^{2}+|B(x)|^{2})dx$

in $E$ and $B$.

Let $e$, $b$ be variations in $E,B$ of finite weighted norms, such that
$H_{M}(E+e,B+b)$ is finite. Since $\nabla\cdot E=\rho$ and
$\nabla\cdot(E+e)=\rho$, we have

$\nabla\cdot e=0.$ (4.75)

Similarly

$\nabla\cdot b=0.$ (4.76)

Let us make variation in $E$,

$\frac{d}{dt}\Big{|}_{t=0}H_{M}(E+te,B).$

We have

$\frac{d}{dt}\Big{|}_{t=0}\left[\frac{1}{2I}\left(M-\int x\wedge((E+te)\wedge
B)dx\right)^{2}+\frac{1}{2}\int(|E+te|^{2}+|B|^{2})dx\right]=$
$\int(E+te)\cdot e\,dx\Big{|}_{t=0}-\frac{1}{I}\left(M-\int
x\wedge((E+te)\wedge B)dx\right)\Big{|}_{t=0}\cdot\int x\wedge(e\wedge B)dx=$
$\int E\cdot e\,dx-\frac{1}{I}\left(M-\int x\wedge(E\wedge
B)dx\right)\cdot\int x\wedge(e\wedge B)dx.$

Similarly, variation in $b$ results in

$\int B\cdot b\,dx-\frac{1}{I}\left(M-\int x\wedge(E\wedge
B)dx\right)\cdot\int x\wedge(E\wedge b)dx.$

###### Lemma 4.1

The soliton fields $(E_{\omega},B_{\omega})$ satisfy the stationary
variational equations

$\int E\cdot e\,dx-\frac{1}{I}\left(M-\int x\wedge(E\wedge
B)dx\right)\cdot\int x\wedge(e\wedge B)dx=0$ (4.77)

and

$\int B\cdot b\,dx-\frac{1}{I}\left(M-\int x\wedge(E\wedge
B)dx\right)\cdot\int x\wedge(E\wedge b)dx=0.$ (4.78)

Proof Note that for the soliton we have $\frac{\displaystyle 1}{\displaystyle
I}(M_{\omega}-\int x\wedge(E_{\omega}\wedge B_{\omega})dx)=\omega$. Substitute
to the LHS of (4.77) and obtain

$\int E_{\omega}\cdot e\,dx-\omega\cdot\int x\wedge(e\wedge B_{\omega})dx.$

First note that

$\int E_{\omega}\cdot
e\,dx=\int\hat{e}\cdot\overline{-i\frac{k\hat{\rho}(k)}{k^{2}}}=0,$

since $\hat{e}$ is orthogonal to $k$ by (4.75). Further,

$\omega\cdot\int x\wedge(e\wedge B_{\omega})dx=\omega\cdot\int e(x\cdot
B_{\omega})-(x\cdot e)B_{\omega}dx=$
$\omega\cdot\int\left[\hat{e}\,\,\overline{(-i\nabla_{k}\cdot\hat{B}_{\omega})}-(-i\nabla_{k}\cdot\hat{e})\overline{\hat{B}_{\omega}}\right]dk=i\omega\cdot\int\left[\hat{e}(\nabla_{k}\cdot\overline{\hat{B}_{\omega})}+(\nabla_{k}\cdot\hat{e})\overline{\hat{B}_{\omega}}\right]dk=$
$i\omega\cdot\int\left[\hat{e}(\nabla_{k}\cdot\overline{\hat{B}_{\omega})}-(\hat{e}\cdot\nabla_{k})\overline{\hat{B}_{\omega}}\right]dk$
(4.79)

by partial integration.

Let us compute the first summand of the integrand in (4.79). By rotational
invariance of $\rho(x)$ one has $\hat{\rho}(k)=\rho_{r}(r)$, $r:=|k|$. Denote
$\partial_{j}:=\partial_{k_{j}}$. Then

$\partial_{j}r=\frac{k_{j}}{r},\,\,\,\nabla_{k}\hat{\rho}(k)=\rho^{\prime}_{r}(r)\frac{k}{r}\,\,\,\,\,{\rm
and}\,\,\,\,\,\hat{B}_{\omega}(k)=-\frac{k\wedge(\omega\wedge\nabla_{k}\hat{\rho}(k))}{r^{2}}=$
$-\frac{\omega(k\cdot\nabla_{k}\rho_{r}(r))-\nabla_{k}\rho_{r}(r)(k\cdot\omega)}{r^{2}}=\frac{\rho^{\prime}_{r}(r)((k\omega)k-k^{2}\omega)}{r^{3}}.$

Note that $\hat{B}_{\omega}(k)$ is a real-valued function by (1.24), thus,
$\overline{\hat{B}_{\omega}}(k)=\hat{B}_{\omega}(k)$. For $j=1,2,3$ put
$\alpha_{j}=(k\omega)k_{j}-k^{2}\omega_{j}$, then
$\partial_{j}\alpha_{j}=k\omega-k_{j}\omega_{j}$. One has

$\partial_{j}\frac{\rho_{r}^{\prime}\alpha_{j}}{r^{3}}=\frac{\rho_{r}^{\prime\prime}r^{2}k_{j}\alpha_{j}+\rho_{r}^{\prime}r^{3}(k\omega-
k_{j}\omega_{j})-3\rho_{r}^{\prime}rk_{j}\alpha_{j}}{r^{6}}.$

Note that $\sum_{j}k_{j}\alpha_{j}=0$. Then

$\nabla_{k}\cdot{\hat{B}_{\omega}}=\frac{\rho_{r}^{\prime\prime}r^{2}\sum_{j}k_{j}\alpha_{j}-3\rho_{r}^{\prime}r\sum_{j}k_{j}\alpha_{j}+2\rho_{r}^{\prime}r^{3}k\omega}{r^{6}}=\frac{2\rho_{r}^{\prime}r^{3}k\omega}{r^{6}}=\frac{2\rho_{r}^{\prime}k\omega}{r^{3}}.$

Thus, the first summand of the integrand in (4.79) equals
$\hat{e}\frac{\displaystyle 2\rho_{r}^{\prime}k\omega}{\displaystyle r^{3}}$.
Now, the whole integral (4.79) equals

$i\,\int\left[\omega\cdot\hat{e}\frac{2\rho_{r}^{\prime}k\omega}{r^{3}}-\omega\cdot(\hat{e}\cdot\nabla_{k}){\hat{B}_{\omega}}\right]dk.$
(4.80)

In the integrand, the component $\hat{e}_{1}$ is multiplied by the expression

$\frac{2\omega_{1}\rho_{r}^{\prime}k\omega}{r^{3}}-\omega\cdot(\partial_{1}{\hat{B}_{\omega}}).$
(4.81)

One has

$\partial_{1}({\hat{B}_{\omega}})_{1}=\frac{\rho_{r}^{\prime\prime}r^{2}k_{1}\alpha_{1}+\rho_{r}^{\prime}r^{3}(k\omega-
k_{1}\omega_{1})-3\rho_{r}^{\prime}rk_{1}\alpha_{1}}{r^{6}},$

and

$\partial_{1}({\hat{B}_{\omega}})_{j}=\frac{\rho_{r}^{\prime\prime}r^{2}k_{1}\alpha_{j}+\rho_{r}^{\prime}r^{3}(\omega_{1}k_{j}-2k_{1}\omega_{j})-3\rho_{r}^{\prime}rk_{1}\alpha_{j}}{r^{6}}$

for $j=2,3$. Then in (4.80) $\hat{e}_{1}$ is multiplied by

$k_{1}\cdot\beta(k),\,\,\,\,\beta(k):=\frac{\rho_{r}^{\prime\prime}r^{2}\sum_{j}\omega_{j}\alpha_{j}-3\rho_{r}^{\prime}r\sum_{j}\omega_{j}\alpha_{j}-2\rho_{r}^{\prime}r^{3}\omega^{2}}{r^{6}}.$
(4.82)

Note that $\beta(k)$ does not depend on the number of the coordinate.
Similarly, in (4.80) $\hat{e}_{j}$ is multiplied by $k_{j}\cdot\beta(k)$,
$j=2,3$. Finally, in (4.80) $\hat{e}$ is multiplied by the vector
$k\cdot\beta(k)$ which is proportional to $k$. Hence, the scalar product
equals zero. We have proved that the soliton satisfies the first variational
equation (4.77).

Let us proceed to the equation (4.78). Insert $E=E_{\omega}$, $B=B_{\omega}$
and obtain

$\int b\cdot B_{\omega}dx-\omega\cdot\int x\wedge(E_{\omega}\wedge b)dx=\int
b\cdot B_{\omega}dx-\omega\cdot\int\left(E_{\omega}(x\cdot b)-b(x\cdot
E_{\omega})\right)dx=$
$\int\hat{b}\cdot\hat{B}_{\omega}dk-\omega\cdot\int\left((i\nabla_{k}\cdot\hat{b})\overline{\hat{E}}_{\omega}-\hat{b}(\overline{i\nabla_{k}\cdot\hat{E}_{\omega}})\right)dk.$
(4.83)

Here the first integral equals

$\int\hat{b}\cdot\frac{\rho_{r}^{\prime}((k\omega)k-r^{2}\omega)}{r^{3}}dk=-\int\omega\cdot\hat{b}\frac{\rho_{r}^{\prime}}{r}dk,$

since $\hat{b}$ is orthogonal to $k$. Insert
$\hat{E}_{\omega}=-\frac{\textstyle ik\rho_{r}}{r^{2}}$ and obtain, first,

$\int(i\nabla_{k}\cdot\hat{b})\overline{\hat{E}}_{\omega}dk=\int(i\nabla_{k}\cdot\hat{b})\frac{ik\rho_{r}}{r^{2}}=\int(\hat{b}\cdot\nabla_{k})\frac{k\rho_{r}}{r^{2}}$

by partial integration. Within this integral, $\hat{b}_{1}$ is multiplied by
$\partial_{1}\frac{\textstyle k\rho_{r}}{\textstyle r^{2}}$. Note that

$\partial_{1}\frac{k_{1}\rho_{r}}{r^{2}}=\frac{\rho_{r}(k^{2}-2k_{1}^{2})+\rho_{r}^{\prime}rk_{1}^{2}}{r^{4}},\,\,\,\,\partial_{1}\frac{k_{j}\rho_{r}}{r^{2}}=\frac{\rho_{r}^{\prime}rk_{1}k_{j}-\rho_{r}2k_{1}k_{j}}{r^{4}},\,\,\,\,j=2,3.$

Second,

$\int\hat{b}(\overline{i\nabla_{k}\cdot\hat{E}_{\omega}})dk=\int\hat{b}(-i\nabla_{k}\cdot\frac{ik\rho_{r}}{r^{2}})dk=\int\hat{b}\cdot\frac{\rho_{r}+\rho_{r}^{\prime}r}{r^{2}}dk,$

since

$\nabla_{k}\cdot\frac{k\rho_{r}}{k^{2}}=\frac{\rho_{r}r^{2}+\rho_{r}^{\prime}r^{3}}{r^{4}},$

Finally, in (4.83) $\hat{b}_{1}$ is multiplied by

$-\omega_{1}\frac{\rho_{r}^{\prime}}{r}-\omega_{1}\frac{\rho_{r}(r^{2}-2k_{1}^{2})+\rho_{r}^{\prime}rk_{1}^{2}}{r^{4}}-\omega_{2}\frac{\rho_{r}^{\prime}rk_{1}k_{2}-2\rho_{r}k_{1}k_{2}}{r^{4}}-$
$\omega_{3}\frac{\rho_{r}^{\prime}rk_{1}k_{3}-2\rho_{r}k_{1}k_{3}}{r^{4}}+\omega_{1}\frac{\rho_{r}+\rho_{r}^{\prime}r}{r^{2}}=k_{1}\cdot\gamma(k),\,\,\,\gamma(k):=\frac{2\rho_{r}k\omega-\rho_{r}^{\prime}rk\omega}{r^{4}}.$

Here $\gamma(k)$ does not depend on the number of coordinate. Similarly,
$\hat{b}_{j}$ is multiplied by $k_{j}\cdot\gamma(k)$, $j=2,3$. As the result,
in (4.83) $\hat{b}$ is multiplied by the vector $k\cdot\gamma(k)$ which is
proportional to $k$. This equals zero and the variational equation (4.78) is
satisfied. $\Box$

###  4.2 Instability of the soliton on the surface of constant angular
momentum

####  4.2.1 Energy increment at the soliton

Let us compute

$H_{M}(E_{\omega}+e,B_{\omega}+b)-H_{M}(E_{\omega},B_{\omega})=\frac{1}{2I}(M-\int
x\wedge(E_{\omega}\wedge B_{\omega})dx-m)^{2}-\frac{1}{2I}(M-\int
x\wedge(E_{\omega}\wedge B_{\omega})dx)^{2}+$
$\frac{1}{2}\int(|E_{\omega}+e|^{2}+|B_{\omega}+b|^{2})dx-\frac{1}{2}\int(|E_{\omega}|^{2}+|B_{\omega}|^{2})dx,$

where we put

$m=m_{1}+m_{2}+m_{3},\,\,\,\,m_{1}=\int x\wedge(E_{\omega}\wedge
b)dx,\,\,\,m_{2}=\int x\wedge(e\wedge B_{\omega})dx,\,\,\,m_{3}=\int
x\wedge(e\wedge b)dx.$

Note that $M-\int x\wedge(E_{\omega}\wedge B_{\omega})dx=I\omega$. Thus, we
obtain $H_{M}(E_{\omega}+e,B_{\omega}+b)-H_{M}(E_{\omega},B_{\omega})=$

$\frac{1}{2I}(I\omega-m)^{2}-\frac{1}{2I}(I\omega)^{2}+\int E_{\omega}\cdot
e\,dx+\int B_{\omega}\cdot b\,dx+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx=$
$\frac{1}{2I}m^{2}-\omega\cdot m+\int E_{\omega}\cdot e\,dx+\int
B_{\omega}\cdot b\,dx+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx=$
$\frac{1}{2I}m^{2}-\omega\cdot m_{1}-\omega\cdot m_{2}-\omega\cdot m_{3}+\int
E_{\omega}\cdot e\,dx+\int B_{\omega}\cdot
b\,dx+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx.$

We have $\int E_{\omega}\cdot e\,dx=0$; $\omega\cdot m_{2}=0$ by the
variational equation (4.77), $-\omega\cdot m_{1}+\int B_{\omega}\cdot b\,dx=0$
by the variational equation (4.78). Finally, we obtain

$H_{M}(E_{\omega}+e,B_{\omega}+b)-H_{M}(E_{\omega},B_{\omega})=\frac{1}{2I}m^{2}-\omega\cdot
m_{3}+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx.$ (4.84)

Note that for $\omega=0$ the increment is nonnegative which corresponds to
stability of the zero soliton.

####  4.2.2 Energy increment alternation as instability argument

For instability of a soliton of finite angular momentum we have to prove that

$\exists\varepsilon>0\,\,\forall\delta>0\,\,\exists(\Omega,e,b):\,\,|\Omega|+\|e\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm I\kern-1.09685ptR}^{3})}+\|b\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm I\kern-1.09685ptR}^{3})}<\delta\,\,\,{\rm and}$
$\exists t\in{\rm
I\kern-1.56693ptR}:\,\,|\omega(t)-\omega|+\|E(\cdot,t)-E_{\omega}\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm
I\kern-1.09685ptR}^{3})}+\|B(\cdot,t)-B_{\omega}\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm I\kern-1.09685ptR}^{3})}\geq\varepsilon.$ (4.85)

Clearly, it is sufficient to prove that

$\exists\varepsilon>0\,\,\forall\delta>0\,\,\exists(e,b):\,\,\|e\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm I\kern-1.09685ptR}^{3})}+\|b\|_{L^{2}({\rm
I\kern-1.09685ptR}^{3};{\rm I\kern-1.09685ptR}^{3})}<\delta\,\,\,{\rm
and}\,\,\,$
$|H_{M}(E_{\omega}+e,B_{\omega}+b)-H_{M}(E_{\omega},B_{\omega})|\geq\varepsilon.$
(4.86)

Indeed, consider the solution $(\omega(t),E(x,t),B(x,t))$ with the initial
data $(\omega,E_{\omega}+e,B_{\omega}+b)$. Then by the energy conservation
(2.28) one has $H(\omega(t),E(x,t),B(x,t))=H_{M}(E_{\omega}+e,B_{\omega}+b)$
and thus (4.86) implies (4.85).

To obtain (4.86) we construct special perturbation fields $(e,b)$.

The idea of the construction is rather simple. Let us concentrate ourselves on
the term $\omega\cdot m_{3}$ of (4.84). This term is linear while all the rest
terms are quadratic. Further, $m_{3}=\int x\wedge(e\wedge b)dx$. Suppose that
compact supports of $e$ and $b$ tend to infinity i.e. they are in the region
of large $|x|$. Then $m_{3}$ can be made sufficiently large compared to the
other terms such that $\omega\cdot m_{3}$ ($\omega\neq 0$) makes the increment
(4.84) alternating. We only have to provide that $e$ is odd, $b$ is even and
$\nabla\cdot e=0$, $\nabla\cdot b=0$. So, it is natural to start with
potentials of the fields $e$ and $b$. The exact construction is given below.

Consider a smooth non-increasing function $\alpha(r)$, $r\in{\rm
I\kern-1.56693ptR}$, $r\geq 0$ such that

$\alpha(r)\equiv d\,\,\,{\rm for}\,\,\,0\leq r\leq 1,\,\,\,\alpha(r)\equiv
0\,\,\,{\rm for}\,\,\,r\geq 2,$ (4.87)

where $d>0$ will be chosen later. Set

$\tilde{e}(x):=(\alpha(|x-a|)+\alpha(|x+a|),0,0),\,\,\,\,\tilde{b}(x):=(0,\alpha(|x-a|)-\alpha(|x+a|),0),$

where $a:=(0,c,0)$, $c>0$ is sufficiently large. Then $\tilde{e}(x)$ is even
in $x$, $\tilde{b}(x)$ is odd in $x$. Further, define the perturbation fields
as

$e(x)=\nabla\wedge\tilde{e}(x),\,\,\,\,b(x)=\nabla\wedge\tilde{b}(x).$

Then

$\nabla\cdot e=0,\,\,\,\,\nabla\cdot b=0,$

and the field $e$ is odd in $x$, the field $b$ is even in $x$. Now let us
compute $e$ and $b$ in detail:

$e=(0,\partial_{3}(\alpha(|x-a|)+\alpha(|x+a|),-\partial_{2}(\alpha(|x-a|)+\alpha(|x+a|))=$
$(0,\,\,f(|x-a|)\frac{x_{3}}{|x-a|}+f(|x+a|)\frac{x_{3}}{|x+a|},\,\,-f(|x-a|)\frac{x_{2}-c}{|x-a|}-f(|x+a|)\frac{x_{2}+c}{|x+a|}),$

where we put $f(r)=\alpha^{\prime}_{r}$. By the definition of $\alpha$, the
summands of the components of $e$ are non zero for either $|x-a|\leq 2$ or
$|x+a|\leq 2$, i.e. $|(x_{1},x_{2}-c,x_{3})|\leq 2$ or
$|(x_{1},x_{2}+c,x_{3})|\leq 2$. Hence, $\|e\|_{L^{2}}$ is bounded uniformly
in $c$. Further,

$b=(-\partial_{3}(\alpha(|x-a|)-\alpha(|x+a|),0,\partial_{1}(\alpha(|x-a|)-\alpha(|x+a|))=$
$(-f(|x-a|)\frac{x_{3}}{|x-a|}+f(|x+a|)\frac{x_{3}}{|x+a|},\,\,0,\,\,f(|x-a|)\frac{x_{1}}{|x-a|}-f(|x+a|)\frac{x_{1}}{|x+a|});$

$\|b\|_{L^{2}}$ is bounded uniformly in $c$.

Let us compute the term

$m_{1}=\int\,x\wedge(E_{\omega}\wedge b)\,dx=\int\,E_{\omega}(x\cdot
b)-b(x\cdot E_{\omega})\,dx.$

Since $x_{j}(E_{\omega})_{k}\in L^{2}({\rm I\kern-1.56693ptR},{\rm
I\kern-1.56693ptR})$ by (2.32), $|m_{1}|$ is bounded uniformly in $c$.

Similarly,

$m_{2}=\int\,x\wedge(e\wedge B_{\omega})\,dx=\int\,e(x\cdot
B_{\omega})-B_{\omega}(x\cdot e)\,dx.$

Since $x_{j}(B_{\omega})_{k}\in L^{2}({\rm I\kern-1.56693ptR},{\rm
I\kern-1.56693ptR})$ by (2.32), $|m_{2}|$ is bounded uniformly in $c$.

Let us compute the term

$m_{3}=\int\,x\wedge(e\wedge b)\,dx=\int\,e(x\cdot b)-b(x\cdot e)\,dx.$

We start with the first component $m_{3,1}$. The first component of $e(x\cdot
b)$ is clearly zero. Compute

$-\int\,b_{1}(x\cdot
e)\,dx=\int\,dx\left(f(|x-a|)\frac{x_{3}}{|x-a|}-f(|x+a|)\frac{x_{3}}{|x+a|}\right)(x\cdot
e).$ (4.88)

First,

$(x\cdot
e)=f(|x-a|)\frac{x_{2}x_{3}}{|x-a|}+f(|x+a|)\frac{x_{2}x_{3}}{|x+a|}-f(|x-a|)\frac{x_{3}(x_{2}-c)}{|x-a|}-f(|x+a|)\frac{x_{3}(x_{2}+c)}{|x+a|}=$
$cx_{3}\left(\frac{f(|x-a|)}{|x-a|}-\frac{f(|x+a|)}{|x+a|}\right).$

Now, the integral of (4.88) is the integral of four summands. Two of them are
of the kind $f(|x-a|)f(|x+a|)[...]=0$ for sufficiently large $c$, since the
supports of $f(|x-a|)$ and $f(|x+a|)$ do not intersect. Then

$m_{3,1}=-\int\,b_{1}(x\cdot
e)\,dx=c\int\,dx\,x_{3}^{2}\left(\frac{f^{2}(|x-a|)}{|x-a|^{2}}\right)+c\int\,dx\,x_{3}^{2}\left(\frac{f^{2}(|x+a|)}{|x+a|^{2}}\right).$

For the first integral let us change the variables as $y=x-a$, in detail,

$y_{1}=x_{1},\,\,\,\,y_{2}=x_{2}-c,\,\,\,\,y_{3}=x_{3}.$ (4.89)

Then we obtain

$c\int\,dx\,x_{3}^{2}\left(\frac{f^{2}(|x-a|)}{|x-a|^{2}}\right)=c\int\,dy\,y_{3}^{2}\left(\frac{f^{2}(|y|)}{|y|^{2}}\right).$

For the second integral we change the variables as $z=x+a$:

$z_{1}=x_{1},\,\,\,\,z_{2}=x_{2}+c,\,\,\,\,z_{3}=x_{3}.$ (4.90)

Then

$c\int\,dx\,x_{3}^{2}\left(\frac{f^{2}(|x+a|)}{|x+a|^{2}}\right)=c\int\,dz\,z_{3}^{2}\left(\frac{f^{2}(|z|)}{|z|^{2}}\right).$

Finally, after redesignating the variables again by $x$, we obtain

$m_{3,1}=2c\int\,dx\,x_{3}^{2}\left(\frac{f^{2}(|x|)}{|x|^{2}}\right)=2c\cdot\frac{1}{3}\int\,dx\,|x|^{2}\cdot\frac{f^{2}(|x|)}{|x|^{2}}=c\cdot\frac{2}{3}\int\,dx\,f^{2}(|x|).$
(4.91)

Let us proceed to the second component

$m_{3,2}=\int\,e_{2}(x\cdot b)-b_{2}(x\cdot e)\,dx=\int\,e_{2}(x\cdot b)\,dx,$

since $b_{2}=0$. Note that

$x\cdot
b=-f(|x-a|)\frac{x_{1}x_{3}}{|x-a|}+f(|x+a|)\frac{x_{1}x_{3}}{|x+a|}+f(|x-a|)\frac{x_{1}x_{3}}{|x-a|}-f(|x+a|)\frac{x_{1}x_{3}}{|x+a|}=0.$

Thus, $m_{3,2}=0$.

For the third component we have

$m_{3,3}=\int\,e_{3}(x\cdot b)-b_{3}(x\cdot e)\,dx=-\int\,b_{3}(x\cdot
e)\,dx,$

since $x\cdot b=0$. Insert $b_{3}$ and $x\cdot e$ and obtain

$m_{3,3}=-c\int\,dx\left(\frac{f^{2}(|x-a|)x_{1}x_{3}}{|x-a|^{2}}\right)-c\int\,dx\left(\frac{f^{2}(|x+a|)x_{1}x_{3}}{|x+a|^{2}}\right).$

For the first integral we apply the change of variables (4.89) and obtain

$\int\,dx\left(\frac{f^{2}(|x-a|)x_{1}x_{3}}{|x-a|^{2}}\right)=\int\,dy\left(\frac{f^{2}(|y|)y_{1}y_{3}}{|y|^{2}}\right)=0,$

since the integrand is odd w.r.t $y_{1}$ or $y_{3}$. Similarly, the second
integral

$\int\,dx\left(\frac{f^{2}(|x+a|)x_{1}x_{3}}{|x+a|^{2}}\right)=0$

by the change of variables (4.90). As the result, $m_{3,3}=0$.

Further, we modify the perturbation fields as follows:

$\tilde{e}_{c}(x):=c^{-1/2}(\alpha(|x-a|)+\alpha(|x+a|),0,0),\,\,\,\,\tilde{b}_{c}(x):=c^{-1/2}(0,\alpha(|x-a|)-\alpha(|x+a|),0),$
$e_{c}(x)=\nabla\wedge\tilde{e}_{c}(x),\,\,\,\,b_{c}(x)=\nabla\wedge\tilde{b}_{c}(x).$

Then, as $c\to+\infty$,

$\|e_{c}\|_{L^{2}}={\cal O}(c^{-1/2}),\,\,\,\,\|b_{c}\|_{L^{2}}={\cal
O}(c^{-1/2});$ $m_{1,c}:=\int\,x\wedge(E_{\omega}\wedge b_{c})\,dx={\cal
O}(c^{-1/2}),\,\,\,\,m_{2,c}:=\int\,x\wedge(e_{c}\wedge B_{\omega})\,dx={\cal
O}(c^{-1/2}),$

while

$\omega\cdot m_{3,c}=\omega_{1}\cdot
m_{3,1,c}=\omega_{1}\cdot\frac{2}{3}\int\,dx\,f^{2}(|x|)$

by (4.91), thus, $\omega\cdot m_{3,1,c}$ does not depend on $c$.

Let us compute in detail the energy increment (4.84) with $m_{j}=m_{j,c}$,
$j=1,2,3$, $e=e_{c}$, $b=b_{c}$; for simplicity of notation we omit the index
$c$:

$\frac{1}{2I}m^{2}-\omega\cdot m_{3}+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx=$
$\frac{1}{2I}(m_{1}^{2}+m_{2}^{2}+2m_{1}\cdot m_{2}+2m_{1}\cdot
m_{3}+2m_{2}\cdot m_{3})+\frac{1}{2}\int(|e|^{2}+|b|^{2})dx+$
$\frac{1}{2I}m_{3,1}^{2}-\omega_{1}\cdot m_{3,1},$

we take into account that $m_{3,2}=0$ and $m_{3,3}=0$. Here the second line
tends to zero as $c\to+\infty$ and can be ignored for sufficiently large $c$.
We have to analyze the third line. Put $\tilde{m}=m_{3,1}$. The quadratic
function $g$ of $\tilde{m}$,
$g=\frac{1}{2I}\tilde{m}^{2}-\omega_{1}\cdot\tilde{m}$ has the zeros at the
points $\tilde{m}_{1}=0$ and $\tilde{m}_{2}=2I\omega_{1}\neq 0$, the point of
minimum is $\tilde{m}_{0}=I\omega_{1}$, the minimal value equals
$-I\omega_{1}^{2}/2$. Fix a $\varepsilon>0$ such that
$-I\omega_{1}^{2}<-\varepsilon<0$, then the equation
$g(\tilde{m})=-\varepsilon$ has two roots $\tilde{m}_{1}^{\prime}$ and
$\tilde{m}_{2}^{\prime}$ between $\tilde{m}_{1}$ and $\tilde{m}_{2}$.

Finally, we choose a $d$ in (4.87) such that
$\tilde{m}\in(\tilde{m}_{1}^{\prime},\tilde{m}_{2}^{\prime})$. Then (4.86)
follows and hence the statement of Theorem 2.5.

Gratitudes. The author is grateful to Professor Herbert Spohn, Technical
University of Munich, Germany, for drawing the author’s attention to the
problem, and to Professor Alexander Komech, Moscow University, Russia and
University of Vienna, Austria, for fruitful discussions.

##  5 Appendix. Angular momentum of a soliton

Let us compute the angular momentum $M_{\omega}$ of a soliton
$(E_{\omega},B_{\omega},\omega)$ given by (2.35), in detail. We have

$\int x\wedge(E_{\omega}\wedge B_{\omega})dx=\int(E_{\omega}(x\cdot
B_{\omega})-B_{\omega}(x\cdot E_{\omega}))dx=$
$\int\left[(i\nabla_{k}\cdot\hat{B}_{\omega})\overline{\hat{E}_{\omega}}-\hat{B}_{\omega}\overline{(i\nabla_{k}\cdot\hat{E}_{\omega})}\right]dk=i\int\left[(\nabla_{k}\cdot\hat{B}_{\omega})\overline{\hat{E}_{\omega}}+\hat{B}_{\omega}(\nabla_{k}\cdot\overline{\hat{E}_{\omega}})\right]dk.$
(5.92)

Since

$\hat{B}_{\omega}=\frac{\rho_{r}^{\prime}((k\omega)k-k^{2}\omega)}{r^{3}},\,\,\,\,\overline{\hat{E}_{\omega}}=\frac{ik\overline{\tilde{\rho}}}{k^{2}},$

we obtain by the computation of Section 4.1 that

$\nabla_{k}\cdot\hat{B}_{\omega}=\frac{2\rho_{r}^{\prime}(k\omega)}{r^{3}},\,\,\,\,\nabla_{k}\cdot\overline{\hat{E}_{\omega}}=i\frac{\rho_{r}+\rho_{r}^{\prime}r}{r^{2}}.$

Then the expression in (5.92) equals

$-\int\left[\frac{2\rho_{r}^{\prime}\rho_{r}(k\omega)k}{r^{5}}+((k\omega)k-k^{2}\omega)\rho_{r}^{\prime}\frac{\rho_{r}+\rho_{r}^{\prime}r}{r^{5}}\right]dk.$

Since $\rho_{r}$ and $\rho_{r}^{\prime}$ are even in $k$, the last expression
simplifies to

$\frac{2}{3}\omega\int\frac{(\rho^{\prime}_{r})^{2}}{r^{2}}dk.$

Finally, the angular momentum of the soliton equals

$M_{\omega}=\omega\left(I+\frac{2}{3}\int\frac{(\rho_{r}^{\prime})^{2}}{r^{2}}dk\right).$
(5.93)

###### Remark 5.1

In $x$-representation (5.93) reads

$M_{\omega}=\omega\left(I+\frac{2}{3}\int\,x\rho\cdot\Delta^{-1}(x\rho)dx\right).$

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