galileo.dm.uniba.it · Math. Z. 232, 73–102 (1999) c Springer-Verlag 1999 Solitons and the electromagnetic ﬁeld ? V. Benci1, D. Fortunato2, A. Masiello2, L. Pisani2 1 Dip

Math. Z. 232, 73–102 (1999)

c© Springer-Verlag 1999

Solitons and the electromagnetic field?

V. Benci1, D. Fortunato2, A. Masiello2, L. Pisani2

1 Dip. di Matematica Applicata “U. Dini”, Universita degli Studi di Pisa, Via Bonanno 25/b,I-56126 Pisa, Italy

2 Dip. Interuniversitario di Matematica, Universita e Politecnico di Bari, Via Orabona 4,I-70125 Bari, Italia

Received November 11, 1997; in final form July 3, 1998

1. Introduction

In a recent paper [4], it has been introduced a Lorentz invariant equation inthree space dimensions, having soliton like solutions. We recall that, roughlyspeaking, a soliton is a solution whose energy travels as a localized packetand which preserves this form of localization under small perturbations (see[6], [15], [13], [10]).

The equation introduced in [4] is the Euler Lagrange equation of anaction functional

S1(ψ) =∫ t1

t0

∫R3

L1dxdt

whereL1 = L1(ψ,∇ψ,ψt) is a suitable Lagrangian density (see Sub-sect. 1.1) for the 4-dimensional vector field

ψ = (ψ1, ψ2, ψ3, ψ4)

defined in the space-timeR4. Here∇ψ (resp.ψt) denotes the derivatives ofψ with respect to the space variablex ∈ R3 (resp. the time variablet).

One of the main features of these soliton solutions is that they behave asrelativistic particles. In fact, by using the Noether theorem, we can introducethe energyE(ψ) and the massm(ψ) and it can be proved (see [4]) that thecelebrated Einstein relation

E(ψ) = m(ψ) c2

? Sponsored by M.U.R.S.T. (40% and 60% funds); the authors from Bari were sponsoredalso by E.E.C., Program Human Capital Mobility (Contract ERCBCHRXCT 940494).

74 V. Benci et al.

holds true. Moreover a topological invariant is associated to these solitons.If we interpret this invariant as the electric charge, it is natural to analyze

the interaction between the solitonψ and the electromagnetic field and totry to construct a simple Lorentz invariant model for the electromagnetictheory namely a model describing particle-like matter interacting with theelectromagnetic field through (deterministic) differential equations definedin a Newtonian space-time.

In the following,(A, φ) will denote the gauge potentials associated tothe electromagnetic field(E,H) by the relations

E = −(At + ∇φ)(1.1)

H = ∇ × A(1.2)

In order to carry out this analysis, we need to define the Lagrangian den-sityL2 of the electromagnetic field and the Lagrangian densityL3 describingthe interaction betweenψ and the electromagnetic field.

Now L2 andL3 can be defined in the standard way

L2 =18π

(|E|2 − |H|2)

=18π

(|At + ∇φ|2 − |∇ × A|2

)

L3 = (J(ψ,∇ψ,ψt) | A) − ρ(ψ,∇ψ)φ

where the dependence of electric currentJ(ψ,∇ψ,ψt) and the electric den-sityρ(ψ,∇ψ) onψ and its derivatives will be defined later (see Subsect. 1.2).

The total action will be

S = S(ψ,A, φ)= S1(ψ) + S2(A, φ) + S3(ψ,A, φ)

with

Si =∫ t1

t0

∫R3

Li dxdt

The model we introduce permits to describe the interaction of a relativis-tic particle with an electromagnetic field by using only concepts of classicalfield theory (for more details on this point see [3]).

In this paper we confine ourselves to analyze some mathematical ques-tions related to the existence of solutions for this model. More preciselywe prove the existence of static solutions (with non trivial charge) of theEuler-Lagrange equations

dS = 0(1.3)

Solitons and the electromagnetic field 75

namely solutions(u,A, φ) (u = (u1, ..., u4), A = (A1, A2, A3)) whichdo not depend ont ∈ R. Let us point out that these solutions give riseto travelling solutions(ψ,Av, φv) (Av = (Av,1, Av,2, Av,3)) with velocity(v, 0, 0) where

ψ(x, t) = u(x1 − vt√1 − v2/c2

, x2, x3)

Av,1(x, t) =A1( x1−vt√

1−v2/c2 , x2, x3) − vcφ( x1−vt√

1−v2/c2 , x2, x3)√1 − v2/c2

Av,2(x, t) = A2(x1 − vt√1 − v2/c2

, x2, x3)

Av,3(x, t) = A3(x1 − vt√1 − v2/c2

, x2, x3)

φv(x, t) =φ( x1−vt√

1−v2/c2 , x2, x3) − vcA1( x1−vt√

1−v2/c2 , x2, x3)√1 − v2/c2

ψ is a travelling soliton ”surrounded” by the electromagnetic field(Av, φv).

In the next two subsections, first we shall recall some heuristic argumentswhich have suggested the Lagrangian densityL1 related toψ, then weshall analyze the LagrangianL3 describing the interaction ofψ with theelectromagnetic field.

1.1. The Lagrangian for the fieldψ

In order to get the Lorentz invariance, we assume thatL1 has the form

L1 = L1(ψ, σ) = −12α(σ) − V (ψ)(1.4)

where

σ = (σ1, σ2, σ3, σ4)

σj = c2∣∣∇ψj∣∣2 −

(ψjt

)2

and the potential functionV is defined onΣ, an open subset of the targetspaceR4. We refer to the target spaceR4 as an internal parameter spacewith coordinates(ξ1, ξ2, ξ3, ξ4).

If we consider the action functional

S1(ψ) =∫ t1

t0

∫R3

(−1

2α(σ) − V (ψ)

)dxdt,

76 V. Benci et al.

the Euler-Lagrange equations relative toS1 are

∂

∂t

(∂α

∂ξj(σ)ψjt

)− c2∇

(∂α

∂ξj(σ)∇ψj

)+∂V

∂ξj(ψ) = 0,(1.5)

(1 6 j 6 4).

When

α(σ1, σ2, σ3, σ4) = σ1 + σ2 + σ3 + σ4,(1.6)

(1.5) reduces to a classical system of nonlinear wave equations

ψjtt − c2∆ψj +∂V

∂ξj(ψ) = 0,(1.7)

(1 6 j 6 4).

Now it can be shown (Derrick theorem) that, ifV (ξ) ≥ 0, any static solutionwith finite energy (see [6])

ψ(x, t) = u(x)

of (1.7) is trivial, i.e. it takes a constant value which is a minimum point ofV (see e.g. [6]).

In [4] it has been considered the “simplest” correction of (1.6), namely

α(σ1, σ2, σ3, σ4) =4∑i=1

(σi +

ε

3(σi)3)

(1.8)

beingε > 0. Then (1.5) can be written

∂

∂t

((1 + ε(σj)2

)ψjt

)− c2∇ ((1 + ε(σj)2

)∇ψj)+∂V

∂ξj(ψ) = 0,

(1 6 j 6 4)

or, as in [4],

2ψj + ε26ψj +

∂V

∂ξj(ψ) = 0.

where

26ψj =

∂

∂t

((σj)2ψjt

)− c2∇ ((σj)2∇ψj)

and

σj = c2∣∣∇ψj∣∣2 −

(ψjt

)2.

On the functionV , we make the following assumptions:


– V ∈ C2(Σ,R) where

Σ = R4 \ ξ– V (ξ) > V (0) = 0 for everyξ ∈ Σ \ 0 ; and0 is a non degenerate

minimum;– there existc, r > 0 such that∣∣ξ − ξ

∣∣ < r ⇒ V (ξ) >c∣∣ξ − ξ∣∣6(1.9)

Assume thatψ(x, t) is continuous and that

lim|x|→∞

ψ(t, x) = 0 for everyt ∈ R;

then a topological invariant,ch(ψ(·, t)), can be associated toψ(·, t); thisinvariant is called topological charge and it is an integer number (see [4]).Since the topological charge is an homotopic invariant, then the function

t 7→ ch(ψ(·, t)) ∈ Z

is constant.The definition of the charge given in [4] is recalled in Sect. 3. We notice

here that the topological charge can be also characterized by means of the3-form inΣ which is closed but non exact. Clearly, this form is unique upto a multiplicative constant; it is given by

η =∑

1≤a<b<c≤4

ηabc(ξ)dξa ∧ dξb ∧ dξc

where

ηabc(ξ) =1

|S3|(−1)k+1ξk∣∣ξ − ξ

∣∣4 ;(1.10)

here∣∣S3∣∣ denotes the measure of the unit sphereS3 andk is the unique

index in1, ..., 4 different froma, b, c. For every fixedt ∈ R, we considerthe function

x ∈ R3 7→ ψ(t, x)

and the pull-back ofη with respect to this function

ψ(·, t)∗(η) =∑

1≤a<b<c≤4

ηabc(ψ(x, t)) det∂(ψa, ψb, ψc)∂(x1, x2, x3)

dx1 ∧ dx2 ∧ dx3;

78 V. Benci et al.

we have that

ch(ψ(·, t)) =∫ψ(·, t)∗(η).

If the target spaceΣ has a more complicated topology, assumee.g.Σ = R4\ξ1, ξ2, . . . , ξk, then to each mapψwe can associate a topologicalinvariant(m1,m2, . . . ,mk) ∈ Zk (see [5]).

Other choices of Lagrangian densities are possible if we do not assumethat the Lagrangian splits as in (1.4), as it is in the Skyrme model [14].We point out that in these models (see [8], [9] and therein references), thetopological invariant follows from the fact that the fieldU takes value in asuitable manifold.

1.2. The Lagrangian for the interaction

We consider the 1-form in the space time

ω = A1dx1 +A2dx

2 +A3dx3 − φdt

and the 3-formψ∗(η) in R3 × R represented by the pullback ofη by asmooth functionψ : Ω × R → Σ

ψ∗(η) =∑

1≤a<b<c≤41≤l<m<n≤4

ηabc(ψ(x, t)) det∂(ψa, ψb, ψc)∂(xl, xm, xn)

dxl ∧ dxm ∧ dxn

Then we describe the interaction between the electromagnetic field and thesoliton by the following 4-form in the space timeR3 × R

S3 =∫ψ∗(η) ∧ ω

It is easy to check that we can write the interaction in the following usualform

S3 =∫ t1

t0

∫R3

L3dxdt =∫ t1

t0

∫R3

(J1A1 + J2A2 + J3A3 − φρ) dxdt

whenever we set fori = 1, 2, 3

Ji(ψ,∇ψ,ψt) = (−1)i∑

1≤a<b<c≤41≤l<m<n≤4

ηabc(ψ(x, t)) det∂(ψa, ψb, ψc)∂(xl, xm, xn)


being the sum on the right hand side extended on those indicesl,m, ndifferent fromi, and

ρ(ψ,∇ψ) =∑

1≤a<b<c≤4

ηabc(ψ(x, t)) det∂(ψa, ψb, ψc)∂(x1, x2, x3)

Now, sinceα is closed, alsoψ∗(η) is closed and then

dψ∗(η) = 0;

this implies that the 4-vector(J, ρ) associated to the formψ∗(η) satisfiesthe continuity equation:

∇J +∂ρ

∂t= 0;(1.11)

Moreover straightforward computations show that

∫R3ρ dx =

∫ψ(·, t)∗(η) = ch(ψ)(1.12)

and hence

∫R3ρ dx ∈ Z.(1.13)

The (1.11) allows to interpretJ andρ as electric current and charge den-sity. Moreover the (1.12) identifies the topological charge with the electriccharge.

Notice that the Lagrangian describing the interaction betweenψ and theelectromagnetic field

S3(ψ,A, φ) =∫ t1

t0

∫R3

L3 dxdt

is invariant under the gauge transformation

A′ = A + ∇hφ′ = φ− ∂h

∂t

80 V. Benci et al.

whereh = h(x, t) is an arbitrary smooth function. In fact, we have

S3(ψ,A′, φ′) =∫ t1

t0

∫R3

[(J | A′)− ρφ′] dxdt

=∫ t1

t0

∫R3

[(J | A + ∇h) − ρ

(φ− ∂h

∂t

)]dxdt

= S3(ψ,A, φ) +∫ t1

t0

∫R3

[(J | ∇h) + ρ

∂h

∂t

]dxdt

= S3(ψ,A, φ) +∫ t1

t0

∫R3

(∇J +

∂ρ

∂t| h)dxdt

= S3(ψ,A, φ)

The last equality is a consequence of the continuity equation (1.11).

1.3. Existence of static solutions

The aim of this paper is to prove the existence of static solutions of theEuler-Lagrange equations relative to the action functional

S = S(ψ,A, φ)

=∫ t1

t0

∫R3

(−1

2α(σ) − V (ψ)

)dxdt

+18π

∫ t1

t0

∫R3

(|At + ∇φ|2 − |∇ × A|2

)dxdt

+∫ t1

t0

∫R3

((J(ψ,∇ψ,ψt) | A) − ρ(ψ,∇ψ)φ) dxdt

First we take the variation with respect toA

dS[δA] =∫ t1

t0

∫R3

14π

(At + ∇φ | (δA)t)

− 14π

(∇ × A | ∇ × δA) + (J(ψ,∇ψ,ψt) | δA) dx dt

=∫ t1

t0

∫R3

(− 1

4π∂

∂t(At + ∇φ)

− 14π

∇ × ∇ × A + J(ψ,∇ψ,ψt) | δA)dx dt

Therefore we get

dS[δA] = 0


if and only if

∇ × (∇ × A) = 4πJ(ψ,∇ψ,ψt) − ∂

∂t(At + ∇φ)(1.14)

Consider now

dS[δφ] =∫ t1

t0

∫R3

14π

(At + ∇φ | ∇ (δφ)) − (ρ(ψ,∇ψ) | δφ) dx dt

=∫ t1

t0

∫R3

(− 1

4π∇ (At + ∇φ) − ρ(ψ,∇ψ) | δφ

)dx dt

Therefore we get

dS[δφ] = 0

if and only if

−∇ (At + ∇φ) = 4π ρ(ψ,∇ψ)(1.15)

By (1.1) and (1.2), we get

∇ × H = 4πJ(ψ,∇ψ,ψt) + Et(1.16)

∇ · E = 4πρ(ψ,∇ψ)(1.17)

which complete the Maxwell equations (1.1) and (1.2).Now, if we want to take the variation with respect to thej-th component

of ψ, we notice that it has a complicated form. Anyway we can write theequation

dS[δψ] = 0

in the following form:

2ψj + ε26ψj +

∂V

∂ξj(ψ) = Fj(1.18)

where the left hand side derives from the variation of the actionS1 describingthe matter fieldψ. The right hand sideFj of (1.18), which derives from theinteraction termS3, depends onψ (and its first and second derivatives) andonA andφ (and their first derivatives). Observe that, whenFj = 0, (1.18)reduces to the equation studied in [4].

We confine ourselves to search static solutions, namely fieldsψ, A, φwhich do not depend ont. Then, since each componentJi contains a factor∂ψa

∂t (see Subsect. 1.2), we get immediately

J(ψ,∇ψ,ψt) = 0

82 V. Benci et al.

then, (1.14), (1.15) and (1.18) give respectively

−∆φ = 4π ρ(ψ,∇ψ)(1.19)

∇ × (∇ × A) = 0(1.20)

−∆ψj − ε∆6ψj +

∂V

∂ξj(ψ) = Gj(1.21)

whereGj depends onψ (and its first and second derivatives) andφ (andits first derivatives).

ClearlyA = 0 (as well asA = ∇h) solves (1.20), so the unknowns ofour problem are(ψ, φ). In particular, since our fieldψ does not depend ont, from now on, we rename itu.

Finally we can state our main result.

Theorem 1.1. There exist two fields

u : R3 → R4

φ : R3 → R

such thatch(u) 6= 0 and (u, 0, φ) is a (weak) static solution of the Euler-Lagrange equation (1.3).

2. The functional framework

In order to prove our result, we need to recall the following lemma

Lemma 2.1. LetL = L(Ψ,∇Ψ, Ψt) be a smooth Lagrangian density rela-tive to the field

Ψ : R3 × R → Rn

Ψ =(Ψ1, . . . , Ψn

)and denote byS(Ψ) its action

S(Ψ) =∫ t1

t0

∫R3

L dxdt(2.1)

LetE = E(Ψ) be the energy functional related to (2.1), i.e.

E(Ψ) =∫R3

(n∑i=1

∂L∂Ψ it

∂Ψ it∂t

− L)dx


LetF = E|M , whereM is the subset of the fields

U : R3 → Rn

which are constant int, that is

F(U) = −∫R3

L(U,∇U, 0) dx.

ThenU ∈ M solves the Euler-Lagrange ofS if and only if it solves theEuler Lagrange equations ofF .

Proof. The Euler-Lagrange equation ofS are

∂

∂t

∂L∂Ψ it

+ ∇ ∂L∂Ψ ix

− ∂L∂Ψ i

= 0(2.2)

(1 ≤ i ≤ n).

SoU =(U1(x), . . . , Un(x)

)is a solution of (2.2) if and only if it solves

∇ ∂L∂Ψ ix

(U,∇U, 0) − ∂L∂Ψ i

(U,∇U, 0) = 0

(1 ≤ i ≤ n).

which are the Euler Lagrange equations ofF . utBy the above lemma and by the fact thatA = 0, the solutions of our

problem can be obtained as critical points of the functional

E(u, φ) =∫R3

(12

|∇u|2 +ε

6|∇u|6 + V (u)

)dx+

−12

∫R3

|∇φ|2 dx+∫R3φρ(u,∇u)dx.(2.3)

Now we choose suitable function spaces for the pair(u, φ) in which itis convenient to study the problem.

We consider

u ∈ Hdef= W 1,2(R3,R4) ∩W 1,6(R3,R4)

φ ∈ Ddef= D1,2(R3,R)

We recall thatW 1,p(R3,R4) is the closure ofC∞0 (R3,R4) with respect

to the norm

‖u‖p =(∫

R3(|∇u|p + |u|p) dx

)1/p

.

84 V. Benci et al.

H is a Banach space and we have the continuous imbedding

H →u ∈ C(R3,R4) | lim

|x|→∞u(x) = 0

.(2.4)

The spaceD is the closure ofC∞0 (R3,R) with respect to the norm

‖φ‖D =(∫

R3|∇φ|2 dx

)1/2

.

D is a Hilbert space and, by well known Sobolev inequality, it is continuouslyembedded inL6(R3,R).

Since we are intersted in functionsu(x) 6= ξ, we set

Λ =u ∈ H | u(x) 6= ξ

.

By (2.4) we have thatΛ is well defined and it is an open subset ofH.Now we are going to show that, for every(u, φ) ∈ Λ×D,

E(u, φ) < +∞.

Using (2.4), sinceV isC2and∫R3

|u(x)|2dx < +∞

it is easy to see that ∫R3V (u) dx < +∞.

So, it remains to study the summability of the termφρ(u,∇u).We recall that, for everyu ∈ Λ,

ρ(u,∇u) =∑

16a<b<c64

ηabc(u) det∂(ua, ub, uc)∂(x1, x2, x3)

Soρ(u,∇u) is the sum of many terms; each of them has the form

ηabc(u)∂ua

∂xl

∂ub

∂xm

∂uc

∂xn

We notice thatηabc(u) ∈ L∞(R3,R). On the other hand we have

∂ua

∂xl∈ L2(R3,R) ∩ L6(R3,R)

so we have also

∂ua

∂xl∈ Lp(R3,R), ∀p ∈ [2, 6].


In particular we have

∂ua

∂xl∈ L18/5(R3,R)

so that

∂ua

∂xl

∂ub

∂xm

∂uc

∂xn∈ L6/5(R3,R).

We conclude

ρ(u,∇u) ∈ L6/5(R3,R)

and, by the Holder inequality, sinceφ ∈ D1,2(R3,R) ⊂ L6(R3,R),wehave that ∫

R3| φρ(u,∇u) | dx < +∞

We show that the functional

E : Λ×D → R

defined in (2.3) isC2.

Proposition 2.2. The functionalE isC2 in Λ×D. More precisely:

a) the functional

u 7→∫R3

(12

|∇u|2 +ε

6|∇u|6 + V (u)

)dx

isC2 in Λ;b) the functional

φ 7→ 12

∫R3

|∇φ|2 dx

isC∞ in D;c) the map

u 7→ ρ(u,∇u) ∈ L6/5(R3,R)

isC∞ in Λ;d) the functional

(u, φ) 7→∫R3φρ(u,∇u)dx

isC∞ in Λ×D.

86 V. Benci et al.

Proof. Statements a) and b) are obvious, we prove c) and d).We recall again thatρ(u,∇u) is the sum of many terms having the form

ηabc(u)∂ua

∂xl

∂ub

∂xm

∂uc

∂xn

We notice thatηabc (which has been defined in (1.10)) isC∞ in Σ, then themap

u ∈ Λ 7→ ηabc(u) ∈ L∞(R3,R)

isC∞. On the other hand, the other factors∂ua

∂xltake values inL18/5 and are

linear inu, so that

u ∈ Λ 7→ ∂ua

∂xl

∂ub

∂xm

∂uc

∂xn∈ L6/5(R3,R)

is aC∞ map. Therefore the map

u ∈ Λ 7→ ρ(u,∇u) ∈ L6/5(R3,R)

isC∞.Moreover the map

(u, φ) 7→∫R3φρ(u,∇u)dx

is linear in the variableφ ∈ L6(R3,R). The proof is thereby complete.utIt is easy to check that the functional is strongly indefinite,i.e. it is

unbounded both from above and from below, even modulo compact pertur-bations; thus it is difficult to deal with it directly. So, it is convenient to usethe following variational principle in order to deal with a functional whichis bounded from below.

Theorem 2.3. LetH1,H2 be two Hilbert spaces,A1 ⊂ H1,A2 ⊂ H2 twoopen sets,

f : A1 ×A2 → R

aC1 functional. Set

Z2 = (x1, x2) ∈ A1 ×A2 | fx2(x1, x2) = 0wherefx2(x1, x2) is the “partial derivative” with respect tox2 ∈ H2 andassume thatZ2 is the graph of aC1 map

Φ : A1 → A2.


Define the “reduced” functional

g : A1 → R

as follows

g(x1) = f(x1, Φ [x1]).

Then the following propositions are equivalent:

a) x = (x1, x2) is a stationary point forf , i.e.f ′(x) = 0;b) x1 is a stationary point forg, i.e. g′(x1) = 0, andx2 = Φ [x1].

The proof of this result can be straightaway obtained.In order to apply this result to our functionalE(u, φ), we need the fol-

lowing lemma.

Lemma 2.4. For every datumρ ∈ L1 ∩Lp, with 6/5 < p ≤ 2, there existsa uniqueφ ∈ D solution of the equation

∆φ = ρ(2.5)

such that

‖φ‖2D 6 C

(‖ρ‖2

L1 + ‖ρ‖2Lp

)(2.6)

So there exists a linear continuous map

ρ ∈ L1 ∩ Lp 7→ φ = ∆−1(ρ) ∈ D

Proof. We consider the functional

F : D → R

F (φ) =∫R3

(12

|∇φ|2 + ρφ

)dx

whose critical points give the weak solutions of (2.5). SinceF is coerciveand strictly convex, it has only one critical point, so we get existence anduniqueness of a solution inD.

Multiplying (2.5) byφ, one has∫R3

|∇φ|2 dx = −∫R3ρφ dx

and therefore, by Sobolev embedding, we get

‖φ‖2D ≤ ‖ρ‖L6/5 · ‖φ‖L6 ≤ c1 ‖ρ‖L6/5 · ‖φ‖D

Then

‖φ‖2D ≤ c2 ‖ρ‖2

L6/5(2.7)

88 V. Benci et al.

By Holder inequality we get

‖ρ‖2L6/5 ≤ ‖ρ‖2θ

L1 · ‖ρ‖2(1−θ)Lp(2.8)

where

θ =(5p− 6)6(p− 1)

.

So, by (2.7) and (2.8), we have

‖φ‖2D ≤ ‖ρ‖2θ

L1 · ‖ρ‖2(1−θ)Lp

which in turn yields (2.6) upon applying Young’s inequality to the right handside. ut

For every(u, φ) ∈ Λ×D, the “partial derivative” ofE with respect toφ is given by

Eφ(u, φ) : D → R

Eφ(u, φ)[v] = −∫R3

∇φ · ∇v dx+∫R3ρ(u,∇u)v dx

Let

Zφ = (u, φ) ∈ Λ×D | Eφ(u, φ) = 0 .We have thatZφ is the graph of aC∞ map

u ∈ Λ 7→ φ [u] ∈ D

Indeed,(u, φ) ∈ Zφ if and only if φ is a solution of the equation

−∆φ = ρ(u,∇u).(2.9)

Sinceρ(u,∇u) ∈ L1(R3,R) ∩ L2(R3,R), by the previous lemma, thesolution is unique and can write

φ = (−∆)−1 (ρ(u,∇u)) .So we have shown thatZφ is the graph of the map

u ∈ Λ 7→ φ[u] = (−∆)−1 (ρ(u,∇u)) ∈ D.(2.10)

Such a map isC∞; indeed we have already shown thatu 7→ ρ(u,∇u)isC∞, on the other hand,∆−1 is linear continuous, then it isC∞.

We notice also that the map (2.10) is bounded; in fact, by (2.6) appliedto the equation (2.9), we have

‖φ[u]‖2D ≤ C

(‖ρ(u,∇u)‖2

L1 + ‖ρ(u,∇u)‖2L2

)(2.11)


So we can consider the functional

J : Λ → RJ(u) = E(u, φ [u])

After multiplying (1.19) byφ [u] and integrating by parts, we obtain∫R3

|∇φ [u]|2 dx =∫R3ρ(u,∇u)φ [u] dx

Then the reduced functional takes the form

J(u) =∫R3

(12

|∇u|2 +ε

6|∇u|6 + V (u)

)dx+

12

∫R3

|∇φ [u]|2 dx

The proof of Theorem 1.1 is based on the following argument. We shallprove thatJ has a non trivial local minimum inΛ, then, by the variationalprinciple, we get a critical point ofE, i.e. a static solution for the Euler-Lagrange equation ofS.

More precisely, we are going to prove the existence of a minimum pointof J in the setΛ∗ of fieldsu with non trivial charge.

Let us remark that, since

E(u, φ[u]) = maxφ∈D

E(u, φ),

then we have the following characterization:

minu∈Λ∗ J(u) = min

u∈Λ∗ E(u, φ[u]) = minu∈Λ∗ max

φ∈DE(u, φ).

3. The existence argument

Since our fields are defined inR3, our minimization problem has lack ofcompactness. We shall overcome this difficulty by using a sort of concentra-tion-compactness argument.

First we study some properties of our functional.It is obvious thatJ is bounded from below; we need to show that it is

coercive.

Lemma 3.1. For every sequenceun ⊂ Λ, if ‖un‖H → ∞, then

limn→∞J(un) = +∞

90 V. Benci et al.

Proof. By the Sobolev inequality, the spaceH can be equipped with theequivalent norm

‖u‖H = ‖u‖L2 + ‖∇u‖L2 + ‖∇u‖L6 .

It is obvious that, if

‖∇un‖L2 + ‖∇un‖L6 → ∞then ∫

R3

(12

|∇un|2 +ε

6|∇un|6

)dx → ∞.

Assume now that

‖∇un‖L2 + ‖∇un‖L6 ≤ C1;(3.1)

‖un‖L2 → ∞.(3.2)

We shall prove that ∫R3V (un) dx → ∞.(3.3)

Since0 is a non degenerate minimum ofV , there existρ, λ > 0 suchthat

|ξ| < ρ ⇒ V (ξ) ≥ λ |ξ|2 .

For everyn ∈ N, we set

An =x ∈ R3 | |un(x)| ≤ ρ

.

From‖∇un‖L2 ≤ C1, we deduce

‖un‖L6 ≤ C2

so we have also

meas(CAn) ≤ C3.

Moreover, from (3.1), we deduce

‖un‖L∞ ≤ C4

which implies ∫CAn

|un|2 dx ≤ C5


Then we conclude∫R3V (un) dx ≥

∫An

V (un) dx ≥ λ

∫An

|un|2 dx

= λ

(‖un‖2

L2 −∫

CAn

|un|2 dx)

≥ λ(‖un‖2

L2 − C5

)which gives (3.3). ut

About the behavior ofJ(u), whenu approaches the boundary∂Λ of Λ,we need the same lemmas already used in [4] (Lemma 3.2 - Proposition3.4).

Lemma 3.2. For every sequenceun ⊂ Λ, if un u ∈ ∂Λ, then∫V (un) dx → +∞.

Lemma 3.3. For everyM > 0, there existsd > 0 such that, for everyu ∈ Λ ∫

R3V (un) dx < M ⇒ min

x∈R3

∣∣u(x) − ξ∣∣ ≥ d.

In order to prove thatJ is weakly lower semicontinuous on its sublevels,we need the following lemma.

Lemma 3.4. Let un ⊂ Λ be a sequence converging weakly tou in H;suppose that there existsd > 0 such that, for everyn ∈ N

minx∈R3

∣∣un(x) − ξ∣∣ ≥ d.(3.4)

Thenφ [un] φ [u] in D.

Proof. SinceD is a Hilbert space, we have to show that, for everyΦ ∈ D∫R3

∇φ [un] · ∇Φdx →∫R3

∇φ [u] · ∇Φdx(3.5)

The proof consists in two parts. We show that

i) the sequenceφ [un] is bounded inD;ii) the (3.5) holds true for everyΦ ∈ C∞

0 (R3,R).

SinceC∞0 (R3,R) is dense inD, our claim will follows.

Proof of i)We need to show thatρ(un,∇un) is bounded inL1 ∩L2; then, from (2.11),it will follow that φ [un] is bounded inD.

92 V. Benci et al.

Now we recall the definition of

ρ(u,∇u) =∑

16a<b<c64

ηabc(u) det∂(ua, ub, uc)∂(x1, x2, x3)

We have to show that

– ηabc(un) is bounded inL∞(R3,R);– det ∂(ua

n,ubn,u

cn)

∂(x1,x2,x3) is bounded inL1 ∩ L2.

First of all we notice thatun is bounded inH, so it is bounded inL∞(R3,R) too.

Then boundedness ofηabc(un) in L∞(R3,R) follows from assumption(3.4) and the definition ofηabc (see (1.10)).

The boundedness of

det∂(uan, u

bn, u

cn)

∂(x1, x2, x3)

in L1 ∩ L2 follows from the fact that∂uan

∂xlis bounded inLp, for every

p ∈ [2, 6]. For p = 3 and forp = 6 we get the boundedness we haverequired.

Proof of ii)ChooseΦ ∈ C∞

0 (R3,R), and consider a ballB ⊂ R3 which contains thesupport ofΦ. We have to show that∫

B∇φ [un] · ∇Φdx →

∫B

∇φ [u] · ∇Φdx(3.6)

Sinceφ [u] is implicitly defined by (2.9), we have that (3.6) is equivalentto ∫

Bρ(un,∇un)Φdx →

∫Bρ(u,∇u)Φdx(3.7)

Indeed, by (2.9),

〈ρ(un,∇un), Φ〉 = 〈∆φ [un] , Φ〉 = − 〈∇φ [un] ,∇Φ〉= −

∫R3

∇φ [un] · ∇Φdx = −∫B

∇φ [un] · ∇Φdx

On the other hand

〈ρ(un,∇un), Φ〉 =∫R3ρ(un,∇un)Φdx =

∫Bρ(un,∇un)Φdx


So we have ∫B

∇φ [un] · ∇Φdx = −∫Bρ(un,∇un)Φdx

and, analogously,∫B

∇φ [u] · ∇Φdx = −∫Bρ(u,∇u)Φdx

If un u in H, sinceH is compactly imbedded inL∞loc(R

3,R4), wehave that

un(x) → u(x)

uniformly inB, so

ηabc(un) → ηabc(u)

in L∞(B,R).Moreover, by a result of Ball [1], fromun u in W 1,6(R3,R4), we

deduce

det∂(uan, u

bn, u

cn)

∂(x1, x2, x3) det

∂(ua, ub, uc)∂(x1, x2, x3)

in L2(R3,R).So we have that

ρ(un,∇un) ρ(u,∇u)in L2(B,R) and this implies (3.7). ut

Now we can prove thatJ is weakly lower semicontinuous on its sublevels

u ∈ Λ | J(u) ≤ MProposition 3.5. For every sequenceun ⊂ Λ, if un u ∈ Λ andJ(un)is bounded from above, then

lim infn→∞ J(un) ≥ J(u)(3.8)

Proof. Our functionalJ is the sum of two terms.The first one∫

R3

(12

|∇u|2 +ε

6|∇u|6 + V (u)

)dx

94 V. Benci et al.

is w.l.s.c., as it have been shown in [4] (Proposition 3.5). Then we have

lim infn→∞

∫R3

(12

|∇un|2 +ε

6|∇un|6 + V (un)

)dx ≥

≥∫R3

(12

|∇u|2 +ε

6|∇u|6 + V (u)

)dx(3.9)

Now we study the second term

∫R3

|∇φ [un]|2 dx

Using the previous lemma, we have that

φ[un] φ[u](3.10)

in D; indeedun u and (3.4) is satisfied sinceJ(un) is bounded.Now consider the functional

φ ∈ D 7→∫R3

|∇φ|2 dx

which is convex and continuous, then it is w.l.s.c. By using (3.10), we obtain

lim infn→∞

∫R3

|∇φ [un]|2 dx ≥∫R3

|∇φ [u]|2 dx(3.11)

From (3.9) and (3.11), we deduce (3.8).utLet

Λ∗ = u ∈ Λ | ch(u) 6= 0 .We are ready to prove a “splitting lemma” for sequencesun ⊂ Λ∗ suchthatJ(un) is bounded.

We note explicitly that, if we consider a static fieldu : R3 → R, thenthe definition of charge given in Subsect. 1.1. becomes

ch(u) =∫u∗(η) =

∫R3ρ(u,∇u) dx(3.12)

In order to prove a concentration- compactness result (the splitting lemma),we shall recall the definition of topological charge given in [4]. The equiva-lence of the next Definition 3.6 with (3.12) is a consequence of well knownfacts (seee.g.[11], [7]).

In the open setΣ = R4 \ ξ, we consider the 3-sphere centered atξ

S =ξ ∈ R4 | |ξ − ξ| = 1

.


OnΣ we consider the ”north pole”

ξN = 2ξ.

We consider also the projectionP : Σ → S defined by

P (ξ) = ξ +ξ − ξ∣∣ξ − ξ

∣∣Definition 3.6. For u ∈ Λ, we call support ofu the compact set

K(u) = x ∈ R3 | 1 < |u(x)|,Then the topological charge ofu coincides with the topological degree ofP u in the support ofϕ with respect to the north pole ofS, namely

ch(u) = deg(P u, int(K(u)), 2ξ).

Clearly the support ofu is the minimal subset where the degree ofP uwith respect to2ξ stabilizes. More precisely the following proposition holds.

Proposition 3.7. For everyu ∈ Λ and for everyR > 0 such thatK(u) ⊂BR(0),

ch(u) = deg(P u,BR(0), 2ξ).(3.13)

Now, using Definition 3.6, it is easy to check that

ch(u) 6= 0 ⇒ ‖u‖L∞ > 1.(3.14)

The last remark is concerned with a useful estimate. By well knownSobolev inequality, there exists∆∗ > 0 such that

‖u‖L∞ ≥ 1 =⇒ J(u) ≥ ∆∗(3.15)

Proposition 3.8 (Splitting Lemma).Letun ⊂ Λ∗ such that

J(un) 6 M.(3.16)

There existsl ∈ N,

1 6 l 6 M/∆∗,(3.17)

and there existu1, ....., ul ∈ Λ, x1n, ..., xln ⊂ R3, R1, ...., Rl > 0 such

that, up to subsequence,

un(· + xin) ui;(3.18)

‖ui‖∞ > 1;(3.19)

96 V. Benci et al.

|xin − xjn| → ∞ for i 6= j;(3.20)

l∑i=1

J(ui) 6 lim infn→∞ J(un);(3.21)

∀x ∈ C

(l⋃

i=1

BRi(xin)

): |un(x)| 6 1.(3.22)

Then we have also

ch(un) =l∑

i=1

ch(ui);(3.23)

lim supn→∞

∥∥∥∥∥un −l∑

i=1

ui(· − xin)

∥∥∥∥∥∞

6 1.(3.24)

Remark 1.We notice that, from (3.19), it follows

J(ui) > ∆∗.(3.25)

Proof. The proof is divided in two parts. In the first part, with an iterative pro-cedure, we prove the existence ofl ∈ N, u1, ....., ul ∈ Λ, x1

n, ..., xln ⊂R3, R1, ...., Rl > 0 such that (3.17-3.22) are satisfied; in the second partfrom these properties we shall easily deduce (3.23) and (3.24). For the sake ofsimplicity, whenever it is necessary, we shall tacitly consider a subsequenceof un.

First of all we arbitrarily chooseγ ∈]0, 1[.Letx1

n ∈ R3 be a maximum point for|un|; by (3.14) we have∣∣un (x1

n

)∣∣ >1. We set

u1n = un(· + x1

n)

and we obtain ∥∥u1n

∥∥∞ =

∣∣u1n(0)

∣∣ > 1.(3.26)

SinceJ(u1n) = J(un) and the functionalJ is coercive, then the sequence

u1n

is bounded inH and we have

u1n u1 ∈ H,(3.27)

From (3.26) it follows‖u1‖∞ > 1 .


Sinceu1n

⊂ Λ andJ(u1n) is bounded, by (3.27) and Lemma 3.2, we

getu1 ∈ Λ.SinceJ(un) ≤ M , we can apply Proposition 3.5 and we get

J(u1) 6 lim infn→∞ J(u1

n) = lim infn→∞ J(un)(3.28)

Now, using (2.4), we considerR1 > 0 such that

∀x ∈ CBR1(0) |u1(x)| 6 γ;(3.29)

for simplicity we set

B1n = BR1(x

1n).

Now we distinguish two cases: eitherA1) for n sufficiently large

∀x ∈ CB1n : |un(x)| 6 1;

orB1) eventually passing to a subsequence,

∃x ∈ CB1n s.t. |un(x)| > 1.

In the caseA1) the first part of Proposition is proved withl = 1; let usconsider the caseB1).

Letx2n be a maximum point for|un| inR3\B1

n; we have that∣∣un (x2

n

)∣∣ >1. We set

u2n = un(· + x2

n)

and we obtain ∥∥u2n

∥∥∞ =

∣∣u2n(0)

∣∣ > 1.

As foru1n

, we have that

u2n u2 ∈ Λ,(3.30)

with

‖u2‖∞ > 1.(3.31)

Now we have to show that∣∣x2n − x1

n

∣∣→ ∞.(3.32)

We set

yn = x2n − x1

n

98 V. Benci et al.

and, arguing by contradiction, we assume that the sequenceyn is boundedin R3; then, up to subsequence, we have that

yn → y.

Since|yn| =∣∣x2n − x1

n

∣∣ > R1, we have|y| > R1; then, using (3.29),

|u1(y)| 6 γ < 1.(3.33)

On the other hand we have

1 6∣∣un(x2

n)∣∣ = ∣∣un(yn + x1

n)∣∣ = ∣∣u1

n(yn)∣∣ ,

then, by (3.33),

0 < 1 − |u1(y)| 6∣∣u1n(yn)

∣∣− |u1(y)| 6∣∣u1n(yn) − u1(y)

∣∣ 66∣∣u1n(yn) − u1(yn)

∣∣+ |u1(yn) − u1(y)| 6

6(

sup|y−y|61

∣∣u1n(y) − u1(y)

∣∣)+ |u1(yn) − u1(y)| .

Taking the limit forn → ∞ we get a contradiction.Now we show that

J(u1) + J(u2) 6 J(un).(3.34)

Hereafter, for sake of simplicity, we set, for everyu ∈ Λ andA ⊂ R3

J|A(u) =∫A

(c2

2|∇u|2 + ε

c6

6|∇u|6 + V (u)

)dx+

12

∫A

|φ[u]|2 dx.

For a fixedη > 0, there existsρ > 0 such that

J|CBρ(0)(u1) < η/2 and J|CBρ(0)(u2) < η/2.

From (3.32) it follows that the spheresBρ(x1n) andBρ(x2

n) are disjoint forn sufficiently large, then we get:

lim infn→∞ J(un) > lim inf

n→∞(J|Bρ(x1

n)(un) + J|Bρ(x2n)(un)

)>

> lim infn→∞ J|Bρ(x1

n)(un) + lim infn→∞ J|Bρ(x2

n)(un)

= lim infn→∞ J|Bρ(0)(u

1n) + lim inf

n→∞ J|Bρ(0)(u2n) >

> J|Bρ(0)(u1) + J|Bρ(0)(u2) >> J(u1) + J(u2) − η.

From the arbitrariness ofη, we get (3.34).


Finally, as well as foru1, from (2.4) we getR2 > 0, such that

∀x ∈ CBR2(0) |u2(x)| 6 γ

and we set

B2n = BR2(x

2n)

Also in this second step we have an alternative: eitherA2) for n sufficiently large,

∀x ∈ C(B1n ∪B2

n) : |un(x)| 6 1;

orB2) up to a subsequence,

∃x ∈ C(B1n ∪B2

n) s.t. |un(x)| > 1.

If caseA2) hold true, the first part of Proposition is proved withl = 2;in the caseB2) we consider a maximum point of|un| in C(B1

n ∪ B2n) and

we repeat the same argument used in the caseB1).This alternative process terminates in a finite number of steps. Indeed,

using (3.25), (3.21) and (3.16), we get (3.17); we notice that this estimate isindependent on the sequenceun.

Now we prove (3.23). We considern sufficiently large so that (3.22)holds and

Bin ∩Bj

n = ∅ for i 6= j.(3.35)

Then we have, by the additive property of the topological degree,

ch(un) = deg

(P un,

l⋃i=1

Bin, 2ξ

)=

l∑i=1

deg(P un, Bi

n, 2ξ)

=

=l∑

i=1

deg(P uin, BRi(0), 2ξ

).(3.36)

On the other hand, for everyi ∈ 1, ..., l, sinceuin converges uni-formly to ui onBRi(0) and

∀x ∈ CBRi(0) |ui(x)| 6 γ < 1,

we obtain, forn large enough,

deg(P uin, BRi(0), 2ξ

)= deg

(P ui, BRi(0), 2ξ

)= ch(ui).

Then, substituting in (3.36), we obtain (3.23).

100 V. Benci et al.

Finally, in order to prove (3.24), we assume that, for everyi ∈ 1, ..., l,∀x ∈ Bi

n :∣∣un(x) − ui(x− xin)

∣∣ < γ.(3.37)

We shall prove that, forn large enough,

∀x ∈ R3 :

∣∣∣∣∣un(x) −l∑

i=1

ui(x− xin)

∣∣∣∣∣ < 1 + lγ.(3.38)

Indeed, ifx ∈ ⋃li=1B

in, then, by (3.35), there exists a unique indexj ∈

1, ..., l such thatx ∈ Bjn, then∣∣∣∣∣un(x) −

l∑i=1

ui(x− xin)

∣∣∣∣∣ 6∣∣un(x) − uj(x− xjn)

∣∣+∑i6=j

∣∣ui(x− xin)∣∣

< γ + (l − 1)γ = lγ < 1 + lγ.(3.39)

On the other hand, ifx /∈ ⋃li=1B

in, then, by (3.22),∣∣∣∣∣un(x) −

l∑i=1

ui(x− xin)

∣∣∣∣∣ 6 |un(x)| +l∑

i=1

∣∣ui(x− xin)∣∣

6 1 + lγ.

Now fix η > 1; choosingγ sufficiently small we have

1 + lγ < η(3.40)

(taking in account (3.17), this kind of choice can be made a priori in theproof). Substituting (3.40) in (3.38), we get

∀x ∈ R3 :

∣∣∣∣∣un(x) −l∑

i=1

ui(x− xin)

∣∣∣∣∣ < η,

and, by the arbitrariness ofη > 1, we obtain (3.24). utRemark 2.Consider the function

∑li=1 ui(· − xin) which has been intro-

duced in (3.24); using (2.4) and (3.20), it is not difficult to show that, fornlarge enough,

l∑i=1

ui(· − xin) ∈ Λ,

ch

(l∑

i=1

ui(· − xin)

)=

l∑i=1

ch(ui).


Finally we can complete the proof of Theorem 1.1. We set

J∗ = inf J(Λ∗).

By (3.14) and (3.15), it follows

∆∗ 6 J∗.

Theorem 3.9. There existsu ∈ Λ∗ such thatJ(u) = J∗.

Proof. We consider a minimizing sequenceun ⊂ Λ∗. It has obviouslybounded energy; then we can apply Proposition 3.8. There existl ∈ N andu1, ....., ul ∈ Λ such that, up to a subsequence,

l∑i=1

J(ui) 6 lim infn→∞ J(un) = J∗;(3.41)

ch(un) =l∑

i=1

ch(ui).(3.42)

Sincech(un) 6= 0, from 3.42 we deduce that there existsi ∈ 1, ...., l,for sake of simplicityi = 1, such thatch(u1) 6= 0.

Then, by 3.41, we obtain

J∗ >l∑

i=1

J(ui) > J(u1) > J∗;

so we getJ(u1) = J∗. ut

Acknowledgements.The authors thank the referee for suggesting an easier proof of Lem-ma 2.4.

References

1. J.M. Ball, On the Calculus of Variations and Sequantially Weakly Continuous Maps, Or-dinary and Partial Differential Equations (Dundee 1976), Lecture Notes in Mathematics564, Springer-Verlag, Berlin Heidelberg New York.

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in dimension 3, Reviews in Mathematical Physics3 (1998), 315–3445. V. Benci, D. Fortunato, L. Pisani, Remarks on Topological Solitons, Topological Meth-

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6. R.K. Dodd, J.C. Eilbeck, J.D. Gibbon, H.C. Morris, Solitons and Nonlinear Wave Equa-tions, Academic Press, London, New York, 1982.

7. B. Doubrovine, S. Novikov, A. Fomenko, Geometrie et Topologie des Varietes, 2 partie,Editions Mir, Moscou (1985).

8. M.J. Esteban, P.L. Lions, Skyrmions and Symmetry, Asymptotic Anal.1 (1988), 187-192.

9. M.J. Esteban, S. Muller, Sobolev Maps with integer degree and applications to Skyrme’sproblem, Proc. R. Soc. Lond.436(1992) , 197-201.

10. S. Kichenassamy, Non linear wave equations, Marcel Dekker Inc., New York, Basel,Hong Kong (1996).

11. L. Nirenberg, Topics in Nonlinear Functional Analysis, Courant Institute Lecture Notes,New York University, New York (1974).

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Documents

galileo.dm.uniba.it · Math. Z. 232, 73–102 (1999) c Springer-Verlag 1999 Solitons and the electromagnetic ﬁeld ? V. Benci1, D. Fortunato2, A. Masiello2, L. Pisani2 1 Dip