Dyons of Unit Topological Charges in Gauged Skyrme Model

WU Zhong-lin,LI Dong-ya

(School of Mathematics and Statistics,Huanghuai University,Zhumadian,Henan,463000,China)

Abstract:Dyons are an important family of topological solitons carrying both electric and magnetic charges and the presence of a nontrivial temporal component of the gauge field essential for the existence of electricity often makes the analysis of the underlying nonlinear equations rather challenging since the governing action functional assumes an indefinite form.In this work,developing a constrained variational technique,We establish an existence theorem for the dyon solitons in a Skyrme model coupled with SO(3)-gauge fields,formulated by Brihaye,Kleihaus,and Tchrakian.These solutions carry unit monopole and Skyrme charges.

Key words:Skyrme model;gauge fields;dyons;calculus of variations for indefinite action functionals;weak convergence

§1.Introduction

Physical Background and Origins.Dyon is a hypothetical particle carrying both electric and magnetic charges.In contemporary physics,dyons and monopoles are relevant theoretical constructs for an interpretation of quark confinement[20,23].As early as in 1894,P.Curie[13]first formulated the concept of a magnetic monopole,a particle with only one magnetic pole,whose existence was widely suspected.In 1931,Dirac[15]explored the electromagnetic duality in the Maxwell equations and obtained a mathematical formalism of magnetic monopoles.It is sophisticated and highly challenging to obtain the existence of monopole and dyons in the Yang—Mills theory by mathematical methods.

As is well known,the Skyrme model is important for baryon physics[1,9,21]and gauge fields are often introduced to investigate inter-particle forces[2,6-7,11,14,16].Motivated by these applications,here we are interested in the combined system consisting of the Skyrme model coupled with SO(3)gauge fields originally presented by Brihaye,Kleihaus,and Tchrakian[6].We may mention that dyons of the Georgi—Glashow model[22]and the SO(3)gauged Skyrme model[7,18]have been obtained recently.Combining these two models,we encounter a new system.The numerical study of the Skyrme dyon solution conducted in[6]provides us evidence of existence of the dyons.The purpose of this paper is to give an analytic proof for the existence of such solutions.The difficulty of the indefinite action functional still appears as stated.To overcome this,we need to obtain suitable uniform estimates for a minimizing sequence at singular boundary points and achieve strong convergence for the sequence of the negative terms as seen in Gao and Yang[18].

Mathematical Problem.Let a,f,g,h be real-valued functions of variable x∈ [0,∞)and satisfying the two-point boundary condition

where θ0and q are some parameters.Consider the energy density functions E1and E2given by

where λ,κ ≥ 0 and ξ＞ 0 are constants.We shall aim at obtaining a critical point of the indefinite action functional

subject to the boundary condition(1.1)and the finite-energy condition

Such a critical point is a spherically symmetric particle-like solution,of the equations of motion of the classical Skyrme model coupled with gauge fields taking values in the Lie algebra of the orthogonal group SO(3),carrying unit topological charge and coined as‘dyon’in quantum field theory.

The rest of the paper is organized as follows.In Section 2,we review the SO(3)gauged Skyrme model of Brihaye—Kleihaus—Tchrakian[6].In the subsequent three sections,we establish our main existence theorem.Specifically,in Section 3,we prove the existence of a finiteenergy critical point of the indefinite action functional by formulating and solving a constrained minimization problem as in[18].In Section 4,we show that the critical point obtained in the previous section for the constrained minimization problem solves the original equations of motion by proving that the constraint does not give rise to an unwanted Lagrange multiplier problem.In Section 5,we study the properties of the solutions.In particular,we obtain some uniform decay estimates which allow us to describe the dependence of the(’t Hooft)electric charge on the asymptotic value of the electric potential function at infinity.

§2.Dyons in The Combined Skyrme Model

The field-theoretical model under consideration is an elegant combination of the classical Georgi—Glashow model and of the semi-locally gauged Skyrme model investigated previously in[3,8].In the model,the matter field is an R3-valued Higgs field,Φα(α =1,2,3)over the(3+1)-dimensional Minkowski spacetime R3,1of signature(+---),of coordinates denoted by xμ(μ =0,1,2,3).Thus,usingto denote the real-valued gauge fields in a fixed representation of the gauge group SO(3).We can express the induced gauge-covariant derivatives and gauge field strength tensors as

where the late Greek letters μ,ν,...=0,1,2,3 label the Minkowski spacetime indices,while the early Greek letters α,β,···=1,2,3 label the indices of the algebra of the gauge group SO(3).With the above notation,the Lagrangian density of the Georgi—Glashow model is given by

The finite-energy condition implies that Φ =(Φα)maps the 2-sphere at the infinity of the space R3into the 2-sphere|Φ|2= η2(η ＞ 0)which naturally associated Φ with an integer class in π2(S2)=Z.Such an integer is called the monopole number of the model and is denoted by QM,which is also referred to as the monopole charge.

In the classical Skyrme model,one is concerned with a map φ =(φa)where a=(α,4)from the Minkowski spacetime into the standard 3-sphere so that(a=1,2,3,4)may also be expressed as(φα,φ4)(α =1,2,3).Our gauged Skyrme model is so gauged that there is the usual global O(4)symmetry but there is a local SO(3)symmetry imposed on the φαpart.Therefore the gauge-covariant derivatives are semi-local and defined by

As a consequence,the semi-local gauged Skyrme model is governed by the Lagrangian density

Following[3,8],we shall consider also the equations of motion coming from the coupled model resulting from(2.2)and(2.4)given by the Lagrangian density

Interestingly,the non-Abelian Higgs field Φ and the Skyrme scalar field φ are now elegantly coupled through a single Lie algebra so(3)-valued gauge field,Aμ=(Aαμ).

Restricting to spherically symmetric static field configurations,we have

which should not be confused with that denoting a point in R3before.

Substituting the Ansatze(2.6)—(2.8)into the static version of the equations of motion of the Lagrangian density(2.5),we arrive at following one-dimensional(radial)Lagrangian density,which is still denoted by the same letter L,defined by

where

and′denotes the differentiation,ξ＞ 0,λ ≥ 0 and κ ≥ 0,such that the associated Hamiltonian(energy)density is given by

It should be noted that,although there are six free coupling constants, λ0,λ1,λ2,and κ0,κ1,κ3in the original Lagrangian action density(2.5),the radially symmetrically reduced action density(2.10)as seen in(2.11)and(2.12)contains only three free coupling constants,ξ,λ,κ,after some appropriate rescaling formulation.See[3,8]for details.

The equations of motion of the original Lagrangian density(2.10)now become the variational equation

of the static action

which is indefinite.Explicitly,the equation(2.14)may be expressed in terms of the unknowns a,f,h,g as

We are to solve these equations under suitable boundary conditions.First we observe in view of the ansatz(2.6)—(2.8)that the regularity of the fields φ and Aμimposes at x=0 the boundary condition

Furthermore,the finite-energy condition

the non-triviality of the g-sector lead us to the boundary condition at x=∞,given as

where 0<θ0<,q＞0(say)is a parameter,to be specified later,which defined the asymptotic value of the electric potential at infinity.

Note that QMis a topological charge,in fact a homotopy invariant defined from the Higgs scalar,which should not be confused with the magnetic charge Qm,to be addressed below,arising from the underlying electromagnetism.

It can be directly checked that the above radially symmetric ansatz leads to the unit monopole charge,QM=1.There is another topological quantity,however,called the Skyrme charge,that needs to be elaborated upon in some detail.Recall that φ maps R3into S3.In the classical situation without local symmetry,φ has a limiting position at infinity of R3which enables us to compactify R3into S3so that φ defines a map from S3into itself and is thus characterized by an integer called the Skyrme charge,QSin the homotopy group π3(S3).In the semi-locally gauged situation,however,the gauge coupling complicates the asymptotic behavior at infinity in such a way that the aforementioned compactification breaks down due to a spontaneous symmetry breaking effect caused by gauge fields which makes the topological characterization of the field configuration much more sophisticated.Nevertheless,formally,the same integral formula that defines the Skyrme charge has the same expression[18]:

In our case,QSis no longer an integer.In fact,applying the boundary condition(2.20)and(2.22)in(2.23),we obtain

which is strictly decreasing for θ0∈ [0,π/2].

Finally,the associated magnetic and electric charges of the solutions following the’t Hooft electromagnetism as calculated in[18]are given by Qm=1 and

respectively.

We can now state our main result regarding the existence of dyon solitons in the coupled George—Glashow and the gauged Skyrme model[6]as follows.

Theorem 2.1 For any parameters θ0and q satisfying

the equations of motion of the minimally gauged Skyrme model defined by the Lagrangian density(2.10),have a static finite-energy spherically symmetric solution described by the ansatz(2.6)—(2.8)so that(a,f,h,g)satisfies the boundary conditions(2.20)and(2.22).Furthermore,such a solution enjoys the property that f(x),g(x)are strictly increasing,and a(x)＞0,0

which may assume any value in the interval(-,0),a unit magnetic charge Qm=1,and an electric charge given by the integral

which depends on q and approaches zero as q→0.

The above theorem will be established in the subsequent sections.

§3.Constrained Minimization Problem

In this section,we will prove the theorem 2.1 by using a indefinite variational process.In functional(2.15),the difficulty arises from the negative terms and it will be overcomed by solving a constrained minimization problem.And we will show the solution solve the equations(2.16)—(2.19)in next section.

To proceed,we introduce the admissible space for our variational problem as follows

A={(a,f,h,g)|a,f,h,g are continuous functions over[0,∞)which are absolutely continuous on any compact subinterval of(0,∞),satisfy the boundary conditions a(0)=1,a(∞)=0,f(0)=0,f(∞)= θ0,h(∞)=1,g(∞)=q,and of finite-energy E(a,f,h,g)<∞}.

It is hard to find a critical point of the indefinite functional L directly in A.Then we can deal with the negative term with respect to g independently,and then consider a,f,h for fixed g.Following this,we need a further restriction:we assume that(a,f,h,g)satisfies the constraint

where G is an arbitrary test function satisfying G(∞)=0 and

The constrained class C is defined to be

In the rest of this section,we shall focus on the following constrained minimization problem

Here we emphasize that the constraint(3.1)partially freezes the temporal component of the gauge field,which,in view of the’t Hooft electromagnetic formalism(?)of non-Abelian gauge field theory,partially freezes the electric sector of the system of equations of motion.By doing so,we are able to tackle the negative component arising in the indefinite Lagrangian action functional.We then show that the minimizer of the full(indefinite)Lagrangian action(but not the positive definite energy functional)subject to the constraint(3.1)will be a finite-energy critical point of the Lagrangian action,which is a classical solution of the original equations of motion.

Lemma 3.1 Assume 0< θ0< π/2.For the problem(3.4),we may always restrict our attention to functions f satisfying 0≤f≤π/2.

Proof It is easy to see that L(a,f,h,g)=L(a,|f|,h,g).Hence we may assume f≥0 in the minimization problem.Noting f(∞)= θ0<,if there is some x0＞0 such that f(x0)＞,we see that there is an interval(x1,x2)with 0≤x1

Lemma 3.2 The constrained admissible class C defined in(3.3)is non-empty.Furthermore,if q＞0 and(a,f,h,g)∈C,we have 0

Proof To proceed,we rewrite the action functional as

Then any element(a,f,h,g)∈C may be obtained by first choosing suitable a,f,h such that F(a,f,h)< ∞ and then choosing a unique g such that g(∞)=q and g solves the problem(3.5).

In fact,the Schwartz inequality gives us the asymptotic estimate

which indicates that the limiting behavior G(∞)=q can be preserved for any minimizing sequence of the problem(3.5).Hence(3.5)is solvable.In fact,it has a unique solution,say g,for any given function a,in view of the functional J(a,·)is strictly convex.Since J(a,·)is even,we have g≥0.By the maximum principle in(2.19),we conclude with 0

Lemma 3.3 For any(a,f,h,g)∈C,g(x)is nondecreasing for x＞0 and g(0)=0.

Proof Lemma 3.2 shows that 0

Assume otherwise,then there are ∈0,δ＞ 0,such that x2|g′(x)| ≥ ∈0for 0

Then(3.7)implies that there is a sequence{xk},such that xk→ 0 and x|g′(xk)|→ 0,as k→∞.Noting this fact and(2.19),we obtain

Therefore g′(x)≥ 0 and g(x)is nondecreasing.As a result,we see that there is number g0≥ 0 such that

It is necessary to prove g0=0.Otherwise,if g0＞ 0,we use the fact a(0)=1,x2g′(x)→ 0(x → 0)(see(3.7)),and L’Hopital’s rule to obtain

Then,there is a δ＞ 0,such that

Integrating(3.10),we obtain

which contradicts the existence of limit stated in(3.9).Then g0=0,and the lemma is proved.

Lemma 3.4 For any 0< θ0<π/2,0

where C1,C2＞0 are constants depending on q only.

Proof For any(a,f,h,g)∈ C,set g1=qh.Then g1satisfies g1(∞)=q.As a result,we have

and thus,

which implies the lower estimate(3.11).

Lemma 3.5 Under the conditions stated in Theorem 2.1,the constrained minimization problem(3.4)has a solution.

Proof Using Lemma 3.4,we see that

is well defined.Let{(an,fn,hn,gn)}be any minimizing sequence of(3.4).That is,(an,fn,hn,gn)∈C and L(an,fn,hn,gn)→η0as n→∞.Without loss of generality,we may assume L(an,fn,hn,gn)≤η0+1(say)for all n.Applying(3.11)and the Schwartz inequality,we obtain

where C＞0 is a constant independent of n.In particular,an(x)→1 uniformly as x→0,fn(x)→θ0and hn(x)→1 uniformly as x→∞.

For any(an,fn,hn,gn),the function Gn=qhnsatisfies Gn(∞)=q.Then,by the definition of gnand(3.11),we have

where C＞0 is a constant,which shows that J(an,gn)is bounded as well.

With the above preparation,we are now ready to study the limit of the sequence{(an,fn,hn,gn)}.We introduce the Hilbert space(X,(·,·)),where the functions in X are all continuously defined in x ≥ 0 and vanish at x=0 and the inner product(·,·)is defined by

Since{an-1}is bounded in(X,(·,·)),we may assume without loss of generality that{an}has a weak limit,say,a,in the same space,

as n→∞.

Similarly,we consider the Hilbert space(Y,(·,·))where the functions in Y are all continuously defined in x ＞ 0 and vanish at infinity with the inner product(·,·)defined by

Noting that{fn- θ0},hn-1 and{gn-q}are bounded in(Y,(·,·)),we may assume that there are functions f,h,g with f(∞)= θ0,h(∞)=1,g(∞)=q,and f- θ0,h-1,g-q ∈ (Y,(·,·)),such that

as n → ∞,for Wn=fn-θ0,ω =f-θ0,Wn=hn-1,ω =h-1,and Wn=gn-q,ω =g-q,respectively.

In the sequel,we need to prove that the weak limit(a,f,h,g)of the minimizing sequence{(an,fn,hn,gn)}obtained above actually lies in C.In other words,we need to show that(a,f,h,g)satisfies the boundary conditions and the constraint(3.1).Using the uniform estimates(3.6),(3.15)-(3.17),we conclude that a(0)=1,f(∞)= θ0,h(∞)=1,g(∞)=q.Moreover,by Lemma 3.4,we see that a ∈ W1,2(0,∞).Therefore a(∞)=0.To show f(0)=0,we use(3.15)to get a δ＞ 0 such that

Then,by(3.21),we obtain

Noting 0≤fn≤,we invert(3.22)to get the uniform estimate

where C＞0 is independent of n.Letting n→∞in(3.23),we obtain f(0)=0 as desired.Then the total boundary conditions are verified.

Then,the next thing is to verify(3.1).To this end,it is sufficient to establish the following results,

for any test function G satisfying(3.2)and G(∞)=0,as n→∞.

Using the fact G∈Y and(3.20),we conclude that(**)is valid.

To get(3.24),we rewrite

for some positive constants 0< δ1< δ2< ∞,and we begin with

Noting(3.15)and(3.18),we see that there is a small δ＞ 0 such that gn∈ L2(0,δ)and there holds the uniform bound

for some constant K ＞ 0.Therefore,we may assume gn→ g weakly in L2(0,δ)as n→ ∞.In particular,g ∈ L2(0,δ)and ‖g‖L2(0,δ)≤ K.Besides,since in(3.2),the function a satisfies a(0)=1,we have G ∈ L2(0,δ)when δ＞ 0 is chosen suitable small.Then,using(3.15)and taking δ1≤ δ,we have

where C ＞ 0 is a constant independent of n.Therefore,for any ε＞ 0,we can take δ1＞ 0 sufficiently small to assure|I11|< ε.On the other hand,since gn→ g weakly in L2(0,δ)and G∈L2(0,δ),we have I12→0 as n→∞.

Since{an}and{gn}are bounded sequences in W1,2(δ1,δ2),using the compact embedding W1,2(δ1,δ2)C[δ1,δ2],we see that an→ a and gn→ g uniformly over[δ1,δ2]as n → ∞.Thus I2→0 as n→∞.

To estimate I3,we recall that{J(an,gn)}is bounded by(3.18),gn(x)→q uniformly as n→∞by(3.6),and G(x)=O()as x→∞by(3.2).In particular,since q＞0,we may choose x0＞ 0 sufficiently large so that

From the above,we arrive at

where x≥x0(cf.(3.30))and C＞0 is a constant.From(3.18),we see that for any ε＞0 we may choose δ2large enough to get|I3|< ε.

Summarizing the above discussion,we see that

which proves the desired conclusion(*).Thus,the claim(a,f,h,g)∈C follows.

To show that(a,f,h,g)is a solution of(3.4),we need to establish

It is difficult to get this fact due to the negative terms in the functional L.To overcome this,we may rewrite the Lagrange density(2.10)as

where

Therefore,to establish(3.33),we need to show that

We first prove(3.38).Since both(an,gn)and(a,g)satisfy(3.1),i.e.,

letting G=g-gnin the above equations and subtracting them,we obtain

where 0< δ1< δ2< ∞.

To estimate I1,we need to get some uniform estimate for the sequence{gn}as x→0.Using(3.15),we see that for any 0<γ<(say)there is a δ＞ 0 such that

We consider the comparison function of the form

Then

Since gnsolve(3.5)in Lemma 3.2,we have

Choose C ＞ 0 in(3.42)large enough so that Cδ1-γ≥ q.Noting gn

Which implies that the weak limit g of{gn}satisfies the same estimate.Thus,from the uniform estimates(3.15)and(3.45),we see that for any ε＞ 0 there is some δ1＞ 0(δ1< δ)such that|I1|<ε.

Moreover,from(3.6)and(3.30),we get

which may be made small than ε when δ2＞ 0 is large enough due to(3.18).

Furthermore,since an→ a and gn→ g in C[δ1,δ2],we see that I2→ 0 as n → ∞.

From the above results regarding I1,I2,I3in(3.40),we obtain the strong convergence

In particular,we obtain

We can also prove that

In fact,we have the fact that{(an,fn,hn,gn)}is bounded in(0,∞).Therefore the sequence is convergent in C[α,β]for any pair of numbers,0< α < β < ∞.Noting that an(x)→ 1 and gn(x)→ 0 as x → 0 uniformly,with respect to n=1,2,···,we see that an→ a and gn→ g uniformly over any interval[0,β](0< β< ∞).Hence,using this result and the uniform estimate(3.6),we get(3.49).

Then(3.38)follows from(3.48)and(3.49).

On the other hand,using the uniform estimate(3.17),we also have

Finally,using(3.50),and the condition q<1,i.e.,1-q2＞0,we get(3.37)and the proof of the lemma is complete.

§4.Ful fi llment of The Governing Equations

Let(a,f,h,g)be the solution of(3.4)obtained in the previous section.Due to the negative terms in L,it is not obvious say(a,f,h,g)satisfies the governing equations(2.16)—(2.19).Since we have solved a constrained minimization problem,we need to prove that the Lagrange multiplier problem does not arise as a result of the constraint,which would otherwise alter the original equations of motion.In fact,since the constraint(3.1)involves a and g only and(3.1)immediately gives rise to(2.19),we see that all we have to do is to verify the validity of(2.16)because(2.17)and(2.18)are the f-equation and h-equation respectively(3.1)does not involve f and h explicitly.

It is clear that the vanishing of I1implies(2.16)so that it suffices to prove that I2vanishes.To this end,from(3.1),we only need to show that~G satisfies the same conditions required of G in(3.1).

In(3.1),using the replacementswe have

Or,with gt=g+,we have

Applying the bounds 0≤g,gt≤q and the relation=gt-g in(4.5),we have

As a consequence,we get

Letting t→0 in(4.7)and(4.8),we obtain J(a,<∞and)=O()(for x large).In particular,)=0 andindeed satisfies all conditions required in(3.1)for G.Hence I2vanishes in(4.2).Therefore,the equation(2.16)has been fulfilled.

§5.Properties of The Solution Obtained

In this section,we investigate the properties of the solution(a,f,h,g).

Lemma 5.1 The solution(a,f,h,g)enjoys the properties a(x)＞0,h(x)＞0,0

Proof From Lemmas 3.1 and 3.2,we see that 0≤g≤q and 0≤f≤.Besides,it is obvious that a≥0 since both(2.11)and(2.12)are even in a.

If a(x0)=0 for some x0＞ 0,then x0is a minimizing point and a′(x0)=0.Using the uniqueness of the solution to the initial value problem consisting of(2.16)and a(x0)=a′(x0)=0,we get a≡0 which contradicts a(0)=1.Thus,a(x)＞0 for all x＞0.Similarly,we can prove that f(x)＞0,g(x)＞0 for all x＞0.Since(3.8)is valid,we see that g(x)is strictly increasing.In particular,g(x)

Lemma 3.1 already gives us f≤.We now strengthen it to f<θ0.First it is easy to see that f ≤ θ0.Otherwise there is a point x0＞ 0 such that f(x0)＞ θ0.Thus,we can find two points x1,x2,with 0≤x1

To see that f is non-decreasing,we assume otherwise that there are 0

The proof of the lemma is complete.

Lemma 5.2 For the solution(a,f,h,g),we have the asymptotic estimates

as x→∞,where ε∈(0,1)is arbitrarily small and

Proof To obtain the asymptotic estimate of a,we introduce a comparison function η,

From(2.16)and the obtained asymptotic behavior of a,f,h,g,we see that there is a suffi-ciently large xε＞ 0 so that

Take the coefficient C in(5.4)large enough to make a(xε)-b(xε)<0.Since a-b vanishes at infinity,applying the maximum principle in the above inequality,we derive the bound a

To get the estimate for g,from(3.8)we see that

which leads to

for x＞0 large,since a(x)vanishes exponentially fast at x=∞.

Similarly,we obtain the asymptotic estimate of f.

Set H=x(h-1),using the fact h(∞)=1 and the estimate of a in(decay),we have

Let c(x)=Ce-β(1-ε)xbe a comparison function.Then there is a sufficiently large xε＞ 0 so that

For fixed xε,we take the constant C in c large enough to make H(xε)+c(xε) ＞ 0.From the finite-energy condition we get that there exist a sequence{xj}:xj→∞(j→∞),such that H(xj)→0.Furthermore,we have H(xj)+c(xj)→0(j→∞).Applying the maximum principle in(5.7),we derive that H+c＞ 0,i.e.0

Lemma 5.3 For the solution(a,f,h,g)with fixed θ0∈ (0,π/2),the electric charge

enjoys the property Qe(q)→0 as q→0.

Proof For fixed θ0,since a vanishes exponentially fast at infinity uniformly with respect to q∈(0,1)and 0