faculteit Wiskunde en

Natuurwetenschappen

Topological and

nontopological solitons

Bacheloronderzoek wiskunde en natuurkunde

Juli 2012

Student: Rik van Breukelen

Begeleider: Prof. Dr. D. Boer

Begeleider: Dr. A. Kiselev

Contents

1 Introduction 2

2 Solitary waves in wave equations 4

3 KdV-equation 73.1 properties of the KdV equation . . . . . . . . . . . . . . . . . . . . . . 73.2 the Lax form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Inverse scattering method 13

5 KdV Hierarchy 195.1 pseudo differential operator . . . . . . . . . . . . . . . . . . . . . . . . 195.2 symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6 Hirota method 24

7 Derrick’s Theorem 30

8 Q-ball 32

9 Topological conservation laws 40

10 CPN model 45

11 Conclusion 48

A Matlab script, transmission coefficient 49

B Derivation of the Marchenko equation 50

1 Introduction

In this thesis the phenomena called solitons will be reviewed, furthermore some methodsin obtaining them are discussed. In 1834 John Scott Russell first observed a solitarywave [1]. “I believe I shall best introduce this phenomenon by describing the circum-stances of my own first acquaintance with it. I was observing the motion of a boatwhich was rapidly drawn along a narrow channel by a pair of horses, when the boatsuddenly stopped - not so the mass of water in the channel which it had put in motion ;it accumulated round the prow of the vessel in a state of violent agitation, then suddenlyleaving it behind, rolled forward with great velocity, assuming the form of a large solitaryelevation, a rounded, smooth and well-defined heap of water, which continued its coursealong the channel apparently without change of form or diminution of speed. I followedit on horseback, and overtook it still rolling on at a rate of some eight or nine miles an

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hour, preserving its original figure some thirty feet long and a foot to a foot and a halfin height.”

Russell thought that these waves were important and should be studied more. How-ever many of his contemporaries, including George Stokes, thought that the wave thatRussell observed was impossible. They thought the wave would disperse. This problemwas solved in 1895 when Diederik Korteweg and Gustav de Vries derived the shallowwater equation; we call this equation the Korteweg-de Vries (KdV) equation. Of thisequation the wave profile observed by Russell is an exact solution.

Because solitons were observed before there was a mathematical description of soli-tons, there is no universally accepted definition of solitons. Only a short list of qualitiesdefines what we see as solitons:

• Wave profile

• Finite energy

• Localized phenomenon

• Stable in time

• Stable under interaction processes

That the soliton is a localized phenomena means that the soliton goes towards a vacuumstate at least exponentially fast. There are more phenomena that are named solitonsthan we will discuss in this thesis. Examples of these are phenomena with, periodicboundary conditions, discontinuous derivative and many more. However in this thesiswe will discus only the most basic examples.

In section 2 we will describe dispersive effects on a solitary wave by a higher orderderivative term in the wave equation. Furthermore the steepening effects of adding anon-linear term will be discussed. In the case of the KdV-equation these effects cancelout, therefore stable solitary waves can be obtained.

In 1965 Norman Zabusky and Martin Kruskal showed by numerical calculation thatwhen two of the solitary waves (obtained from the KdV-equation) interacted, they didnot merge or combine but the faster wave simply overtaking the slower one. Furthermorethe shapes of the waves were unchanged, incurring only a phase difference. Because ofthis particle-like interaction, Zabusky and Kruskal named these waves solitons.

In section 3 we study more details of the KdV-equation, as this equation is centralin the historical development of the subject. Moreover many of the qualities of theKdV-equation are shared with other soliton equations 1. In section 6 we will introducethe Hirota method which is used to obtain soliton solution in many equations. We willuse the KdV-equation as an example and re-derive the single soliton solution and alsoderive an explicit form for the 2-soliton solution.

A more powerful and general method of finding soliton solutions is the inverse scat-tering transform. A full description of this method is beyond the scope of this thesisbut we will explain how this works in case of the KdV-equation in section 4.

1soliton equations are equations that have solitons as solutions

3

Not only in this mathematical context are soliton solutions interesting, in field theo-ries soliton solution can also occur. In these theories, when there is enough non-linearity,stable bound states can exist. However in section 7 Derricks theorem provides restric-tions on these theories.

To circumvent Derricks theorem we look at two types of solitons in local field the-ories, non-topological and topological solitons. In section 8 we will discuss Q-balls, anon-topological soliton. Finally in sections 9 and 10 we will discuss topological solitons.

2 Solitary waves in wave equations

In this section we will attempt to obtain the solitary waves observed by Russell. Al-though waves, that resemble the wave observed by Russell, can be obtained in the mostbasic wave equation, they will periodic and therefore not solitary. Instead we study theeffect of dispersion caused by an higher order derivative term and the effect of makingthe equation nonlinear. We try to obtain the solitary wave in each case, however we thewave will not be stable unless these two effect cancel. At the end of the section someresults are shown where these effect are balanced [2].

We start our discussion with the most basic wave equation:(∂2

∂t2− v2

∂2

∂x2

)f(x, t) = 0 (1)

we assume that the velocity v is constant in time. The wave equation can be rewrittenas: (

∂

∂t+ v

∂

∂x

)(∂

∂t− v

∂

∂x

)f(x, t) = 0(

∂

∂t− v

∂

∂x

)(∂

∂t+ v

∂

∂x

)f(x, t) = 0 (2)

This splits the left moving part and the right moving part of the wave. If we restrictourselves to right moving waves, we remain with:(

∂

∂t+ v

∂

∂x

)f(x, t) = 0 (3)

Any solution of equation (3) still satisfies equation (1). Because f(x, t) is a right movingwave, we can rewrite its argument:

f(x, t) = f(x− vt) (4)

Solving this for a periodic wave gives the plane wave result:

f(x, t) = aei(ωt−kx) (5)

Definition 2.1. The relation between the angular frequency ω and the wave numberk is called the dispersion relation.

4

In this case the dispersion relation is ω = vk which is linear. Because the dispersionrelation is linear, the phase velocity vp =

ωk

and the group velocity vg =∂ω∂k

are equal.Therefore in any superposition of waves, all the waves move at the same speed andthe composed wave stays together. Waves with a linear dispersion relation are callednon-dispersive.

By adding a third order derivative term we make the dispersion relation non-linear:(∂

∂t+ v

∂

∂x+ δ

∂3

∂x3

)f(x, t) = 0 (6)

Using the plane wave ansatz (5), we obtain the non-linear dispersion relation:

ω = vk− δk3 (7)

The phase velocity is given by:

vp =ω

k= v− δk2 (8)

While the group velocity is given by:

vg =∂ω

∂k= v− 3δk2 (9)

The group velocity and phase velocity are different for δ 6= 0 and therefore a wavecomposed by a superposition of multiple waves with different k values will spread out,as some waves move faster than others.

Now, instead of adding a higher order derivative, we add a non-linear term to ourequation. Replace v as a constant with v(f) = v0+αf(x), where v0andα are constants.The wave equation now looks like:(

∂

∂t+ v(f)

∂

∂x

)f(x, t) = 0 (10)

With the following solution:f(x, t) = f(x− v(f)t) (11)

We are interested in the case where the wave is a solitary wave. In that case the wavehas a maximum. The speed v(f) is increasing with the amplitude of the wave, thereforethe apex of the wave is moving faster than the rest of the wave, causing the wave totopple over.

One can obtain to obtain an equation which supports stable solitary waves by com-bining the dispersive properties of the higher order derivative and the non-linear term.The resulting wave equation is:(

∂

∂t+ v(f)

∂

∂x+ δ

∂3

∂x3

)f(x, t) = 0 (12)

5

Example 2.1. As an example we take the Korteweg-deVries (KdV) equation. Bywriting v(f) explicitly we obtain:(

∂

∂t+ (v0 + αf(x, t))

∂

∂x+ δ

∂3

∂x3

)f(x, t) = 0 (13)

To obtain the KdV equation, we set the parameters δ = 1 and α = 6 and shift thefunction: u ≡ f− v0/6.

ut + 6uux + uxxx = 0 (14)

Figure 1: 2-soliton interaction [3]

This equation has solitary wave solutions, in sections 4 and 6 we will obtain thesesolution. In figure 1 two of these waves are shown. On the left side of the figurethe two waves (one large and one small) are shown on three different times. On theother side of figure the gray lines represent the sum of the solution, however as the

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equation is nonlinear the superposition principle does not apply. The black line showsthe time evolution of the initial conditions. When the two waves meet, they do notmerge or destroy each other but pass through only incurring a delay, or phase shift, inthe interaction.

This behavior was first observed by Kruskal and Zabusky. This conservation underinteraction is also the reason why they called these waves solitons, as the -on is usuallyused to name particles (electron, proton and muon for example). As said, the superpo-sition principle does not apply tot nonlinear equation, for solitons however there doesappear to be some use for a superposition. The initial condition can be superimposedas long as they are separated. This can be seen by the first of the time steps, there isno difference between sum of the waves or the proper evolution (note: this is not theinitial condition of the simulation).

The KdV equation was the first equation of which these soliton solution where dis-covered, many other equation also have soliton solution, which have the same qualitiesas the solitons in the KdV equation. However in this thesis the KdV equation willbe our favored example, therefore we will discus some of the properties of the KdVequation in the next section.

3 KdV-equation

3.1 properties of the KdV equation

Maybe the most famous non-linear dispersive wave equation is the Korteweg-deVries(KdV) equation:

ut + 6uux + uxxx = 0 (15)

First introduced (although not in the form above) by Korteweg and De Vries in 1895[4] to describe the solitary wave observed by Russell, this equation is also known as theshallow water equation.

Remark 3.1. Although we have taken the KdV equation (15) with these parameters,any equation of the form

ut + auux + buxxx = 0 (16)

is a KdV equation. This can be shown by the following scaling

x→ b−1/3x u→ a

6b1/3u (17)

This transforms (15) into (16). Although all equations of the form (16) are KdV-equations, we will often use the KdV-equation in the form of (15) .

Remark 3.2. [5] In the same way as we have shown that the KdV-equation can berewritten we can show that it is scale invariant:

x→ cx t→ c3t u→ c−2u (18)

7

This leads to:

ut + 6uux + uxxx = 0→c−5ut + c−56uux + c−5uxxx=c−5(ut + 6uux + uxxx) = 0 (19)

The KdV-equation is also translational invariant for both the x coordinate as the tcoordinate:

x→ x+ c1 t→ t+ c2 (20)

Both do not change the equation. Next to scale invariant and transformational invari-ant, the KdV-equation is also Galilean invariant:

x→x+ vtu→u−

v

6(21)

ut + 6uux + uxxx→ut + vux + 6(u−v

6

)ux + uxxx

=ut + 6uux + uxxx (22)

Remark 3.3. If the initial conditions are given the soliton solution is unique [5]. Let vand u be solutions of the KdV-equation, which both satisfy the same initial conditions.We assume that that the initial conditions are at least continuously differentiable.

ut + 6uux + uxxx = 0

vt + 6vvx + vxxx = 0 (23)

Subtracting v from u and setting w = u− v leads to:

∂(u− v)

∂t=∂w

∂t= −6u

∂u

∂x+ 6v

∂v

∂x−∂3(u− v)

∂x3

= −6(w+ v)∂(w+ v)

∂x+ 6v

∂v

∂x−∂3w

∂x3

= −6u∂w

∂x− 6w

∂v

∂x−∂3w

∂x3(24)

We multiply equation (24) with w and integrate over x:∫∞−∞

∂( 12w2)

∂tdx =

∫∞−∞ −6wu

∂w

∂x− 6w2

∂v

∂x−w

∂3(w)

∂x3dx (25)

The soliton solutions are localized, therefore we assume that u, v, w and their deriva-tives go to zero sufficiently fast at the boundary. The last part of the previous equation

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then becomes: ∫∞−∞w

∂3w

∂x3dx = wwxx|

∞−∞ −

∫∞−∞

1

2

∂

∂xw2xdx

= 0−1

2w2x|

∞−∞ = 0 (26)

We rewrite the left hand side of equation (25) as follows:∫∞−∞

∂( 12w2)

∂tdx =

∂

∂t

∫∞−∞

1

2w2dx =

∂E

∂t(27)

where E =

∫∞−∞

1

2w2dx

The remaining part of the right hand side is:∫∞−∞ 6wu

∂w

∂x+ 6w2

∂v

∂x

=

∫∞−∞ 6w

2

(vx −

1

2ux

)+ 3

∂(uw2)

∂xdx

The last part is again zero because of the boundary conditions.

=

∫∞−∞ 6w

2

(vx −

1

2ux

)dx

6∫∞−∞ 6w

2mdx = 6mE (28)

where m = sup

(∣∣∣∣vx − 1

2ux

∣∣∣∣)This supremum exists because the initial conditions are continuously differentiable andtherefore so are the time evolution of those conditions. It would take an infinite amountof energy to do otherwise. Instead of equation (25), we have the inequality:

∂E

∂t6 6mE

⇒ E(t) 6 E(0)e6mt (29)

From w(x, 0) = u(x, 0) − v(x, 0) = 0, we conclude:

E(0) =

∫∞−∞

1

2w(0)2dx =

∫∞−∞ 0dx = 0

E(t) =

∫∞−∞

1

2w(t)2dx > 0

⇒ 0 6 E(t) 6 0⇒ E(t) = 0⇒ w(x, t) = 0⇒ u(x, t) = v(x, t) (30)

Therefore given the initial conditions the solution is unique.

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3.2 the Lax form

Remark 3.4. Instead of introducing the KdV-equation as a model for waves in shallowwater, we introduce the equation as follows: given the Schrodinger equation with aparameter a dependent potential [6]:

d2y(x)

dx2+ [λ− u(x, a)]y(x) = 0 (31)

We rewrite equation (31) as follows:

Ly = λy (32)

where L = D2 − u(x, a) and D = ddx

and ∂x∂a

= xa.

(Ly)a = Lya + Lay = λay+ λya

(Ly)a = ((D2 − u)y)a = yxxa − uya − uay = Lya − uay⇒ La = −ua (33)

If we assume that the a dependence of y can be expressed as a differential operator(ya = By), we obtain:

(Ly)a = Lya + Lay = λay+ λya

LBy− uay = λay+ λBy

LBy− uay = λay+ BLy

(−ua + LB− BL)y = λay (34)

We want to find the shape of the potential that leaves the eigenvalue λ invariant:

λa = 0 (35)

Therefore we obtain:−ua + [L, B] = 0 (36)

We will interpret a as a time parameter and write t instead of a.

Example 3.1. Let B1 be a first order differential operator:

B1 = aD (37)

where a is constant (ut − (D2 − u)(aD) + (aD)(D2 − u)

)y = 0 (38)⇒ (ut − aux)y = 0 (39)

This means that the potential satisfies ut − aux = 0. Therefore the potential is anyfunction of x+ at.

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Example 3.2. Let B3 be a third order differential operator:

B3 = bD3 + fD+ g (40)

then [L, B3]y = (2fx + 3bux)D2y+ (fxx + 2gx + 3buxx)Dy+ (gxx + buxxx + fux)y

(41)

We require that the terms D2y and Dy vanish, therefore:

2fx + 3bux = 0 (42)

⇒ f = −3

2bu+ c1 (43)

fxx + 2gx + 3buxx = 0 (44)

⇒ g = −3

4bux + c2 (45)

⇒ [L, B3]y = (gxx + buxxx + fux)y =1

4b(uxxx − 6uux) + c1ux)y (46)

This leads to (after a transformation x→ x + c1t and b = −4) an equation which thepotential must satisfy, namely:

ut − 6uux + uxxx = 0 (47)

Definition 3.1. The KdV-equation can be written using the compatibility in the sys-tem of the linear differential equations:

L = D2 + u (48)

B3 = bD3 + fD+ g (49)

where:

b = −4 (50)

f = −3

2bu+ c1 (51)

g = −3

4bux + c2 (52)

(53)

and c1 and c2 are arbitrary constants.

Lt = [B3, L] (54)

This is called the KdV-equation in Lax form.

By considering higher order derivatives higher order KdV equations can be obtained,this will be discussed in section 5

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Example 3.3. We can use the line of thought of example 3.2 to obtain a explicitsolution of the KdV-equation. Assume that there exists a potential for the Schrodingerequation

y ′′ + (k2 − u)y = 0 (55)

such that the solution can be written as:

y = eikxf(k, x) (56)

where f(k, x) is polynomial in k. The simplest form is the zeroth order, y0 = eikxa(x).

We use this as an ansats for the Schrodinger equation:

−k2eikxa+ 2ikeikxax + eikxaxx + (k2 − u)eikxa = 0 (57)⇒ 2ikax + axx − ua = 0 (58)

a(x) is independent of the parameter k, therefore we can separate the part containingk and the part not containing k.

k1 : 2iax = 0⇒ ax = 0, axx = 0 (59)

k0 : axx − ua = 0⇒ ua = 0⇒ u = o∨ a = 0 (60)

This leads to either the trivial solution, which tells us nothing about the shape ofthe potential, or leads to the trivial potential, which leads us back to the plane wavesolution because a(x) is a constant in this case. Using a first order polynomial inequation (56) gives us more interesting results. The constants are chosen with an eyeon future convenience.

y1 = eikx(2k+ ia(x)) (61)

This leads to the following solution:

−k2eikx(2k+ ia) − 2keikxax + ieikxaxx + (k2 − u)eikx(2k+ ia(x)) = 0 (62)⇒ −2kax + iaxx − 2ku− iua = 0 (63)

Again separating k0 and k1 we obtain:

k1 : ax = −u

k0 : axx = ua (64)

⇒axx = −aax = −1

2(a2)x

⇒ax + 1

2a2 = c1 (65)

When we substitute a = 2wxw

we obtain a linear equation.

⇒ 2wwx − 2w2x

w2+1

2

4w2xw2

= c1 (66)

⇒ 2wwx − 2w2x + 2w

2x = c1w

2 (67)

⇒ wx −1

2c1w = 0 (68)

⇒ w = αec2x + βe−c2x (69)

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where c1 is a constant of integration and 2c22 = c1. We have a solution for w andtherefore we also have a solution for u:

a =2wx

w= 2(ln(w))x (70)

u =− ax = −2(ln(w))xx

=− 2(ln(αec2x + βe−c2x))xx

=− 2c2

(αe2c2x − β

αe2c2x + β

)x

=− 2c22c2(αe

2c2x + β)e2c2x − 2c2(αe2c2x − β)e2c2x

(αe2c2x + β)2

=− 2c22

(2ec2x−1/2 ln(β/α)

e2(c2x−1/2 ln(β/α)) + 1

)2=− 2c22(sech

(c2x−

1

2ln(β/α)

)2=− 2c22sech2(c2x− c3) (71)

where c3 =12

ln(β/α). In the solution of w the α and β are independent of x but mayby a function of t. Therefore c3 is a function of time. We use the KdV-equation todetermine this time dependence. Put this into the KdV-equation:

0 = uτ − 6uux + uxxx

0 = −4c22c3sech(c2x− c3)2 tanh(c2x− c3) (72)

+ 6 · 2c22sech(c2x− c3)2 · 4c32sech(c2x− c3)

2 tanh(c2x− c3)

+ 16c52(sech(c2x− c3)

2 tanh(c2x− c3)3 − 2sech(c2x− c3)

4 tanh(c2x− c3))

0 = −c3 + 12c32sech(c2x− c3)

2 + 4c32(tanh(c2x− c3)2 − 2sech(c2x− c3)

2)

c3 = 4c32(tanh(c2x− c3)

2 + sech(c2x− c3)2)

c3 = 4c32t+ η0

u = −2c22sech2(c2x− 4c32t+ η0) (73)

this is the one-soliton solution of the KdV-equation.

4 Inverse scattering method

In 1967 a method to obtain solutions to the KdV-equation was found by Gardner,Green, Kruskal and Miura [7]. When a plane wave in the Schrodinger equation (31)interacts with the potential, the wave will be partly reflected and partially transmitted,of which the amplitude can be calculated when the potential is known. Furthermorewhen the potential is known possible bound states can also be calculated, togetherwith the reflection and transmission this is called the scattering data. Gardner, Green,

13

Kruskal and Miura obtained a method to construct the shape of the potential whenthe scattering data is known, this is known as the inverse scattering transform. In thecase of solitons the reflection coefficient will be zero, making this a powerful method toobtain these solutions.

In the Schrodinger equation when the potential is attractive and finite there will befinite many bound eigenstates. As we are looking for soliton solutions, which are finiteand localized, these bound states will be possible. When the energy eigenvalue λ < 0,a bound state will form, let λ1, λ2, . . . , λN denote the eigenvalues of these bound states.As shown in quantum mechanics the eigenstates yn corresponding to these negativeeigenvalues λn are square integrable, therefore |yn|→ 0 as |x|→∞.

Let us write λn = (iκn)2, with κn > 0. When |x| → ∞ then u → 0, in that region

the Schrodinger equation isyxx − κ

2y = 0 (74)

which can be solved, and by requiring that |yn|→ 0 as x→∞ we obtain:

yn ≈ cn(t)e−κx (75)

where cn(t) is normalization coefficient. In section 3 we derived that in the Schrodingerequation

yxx + [λ− u(x, a)]y = 0 (76)

where the potential satisfies the KdV-equation, the time evolution of y is given by:

yt = B3y = (−4D3 + 6uD+ 3ux)y = −4yxxx + 6uyx + 3uxy (77)

which reduces to:yt = −4yxxx (78)

when |x| → ∞ because u is localized. With this we obtain the time evolution of thenormalization coefficient.

dcn

dt= 4κ3cn (79)

In the case where λ > 0, there is a continuum of unbound states. These states willbehave like the plane wave solution when |x| → ∞. When a plane wave incident fromthe left interacts with the potential, part of the wave will reflect and a part will betransmitted. Let λ = k2 where k > 0.

y− = eikx + R(k)e−ikx x→ −∞y+ = T(k)eikx x→∞ (80)

where R(k) is the reflection coefficient and T(k) is the transmission coefficient. Whenthe wave is perfectly transmitted |T | = 1 and |R| = 0,therefore this is a reflectionlesspotential. Again using the time evolution of y for x→∞ we obtain:

dT

dt= 4ik3T (81)

Thus if |T(k, 0)| = 1 then T(k, t)| = 1, ∀t.

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Remark 4.1. We have stated that the soliton solution is the reflectionless potential. Infigure 2 the transmission coefficient is shown for2:

yxx + (k2 − Vsech2(x))y = 0 (82)

This is calculated using the Matlab [11] script 3, shown in appendix A. Figure 2 shows

Figure 2: A reflectionless potential is obtained for V = 0, 2, 6, 12, 20

that a reflectionless potential is obtained for V = 0, 2, 6, 12, 20. Furthermore the valueof V for which the potential is reflectionless is k independent. However as k increases|T | will be closer to one everywhere which will cause rounding errors making the graphunreadable.

These results are the same to the results obtained by exact calculations done byLamb [6]. The potential is reflectionless when V = n(n+ 1) for n = 0, 1, 2, 3, . . . .

The scattering of waves on a potential is most easily described when scatteringpulses. Therefore we return to the most basic wave equation:

yxx −1

c2ytt = 0 (83)

Any function of the form f(x± ct) is a solution of this equation, for example:

y = δ(t± x/c) (84)

This is a pulse wave. When we include a potential term to the wave equation,

yxx −1

c2ytt − u(x)y = 0 (85)

the pulse wave will pass through while leaving behind some disturbance. A left incidentwave, leaves behind a disturbance described by K(x, ct)

y = δ(t− x/c) + cθ(t− x/c)K(x, ct) (86)

2In appendix A figure 8 shows the points at which the potential is reflectionless till V = 1003In the initial conditions the reflection coefficient is omitted, this will lead to an error, however the

reflection is relatively small and when |T | = 1 the reflection coefficient vanishes, therefore we assumethat the error vanishes as |T |→ 1.

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This K(x, ct) has the information we need to determine the shape of the potential.Substituting (86) into the wave equation (85), we obtain:

δ(t− x/c)

(2∂K(x, ct)

∂x+2

c

∂K(x, ct)

∂t+ u(x)

)−cθ(t− x/c)

(∂2K(x, ct)

∂x2−1

c2∂2K(x, ct)

∂t2+ u(x)K(x, ct)

)= 0 (87)

Integrating over time from x/c− ε to x/c+ ε we obtain:

−2∂K(x, ct)

∂x

∣∣∣∣ct=x

−2

c

∂K(x, ct)

∂t

∣∣∣∣ct=x

= u(x) (88)

−2d

dxK(x, x) = u(x) (89)

We still need to determine this disturbance, for which we will use the Fourier trans-form of y:

Yl(x,ω) =

∫∞−∞ dte

iωty(x, t)

= eikx +

∫∞x

dx ′K(x, x ′)eikx′

(90)

where x ′ = ct and ω/c = k. When multiplied with u(x) and using equation (85) foruy, we obtain (after integration by parts) the Schrodinger equation.

Yxx +(k2 − u(x)

)Y = 0 (91)

A wave incident from the right also leaves behind a disturbance, described by L(x, ct),and in similar manner as before we obtain:

Yr(x,ω) = e−ikx +

∫∞x

dx ′L(x, x ′)e−ikx′

(92)

These two solutions are not independent, they are related by the transmission andreflection coefficient [6]:

T(k)Yr(x, k) = R(k)Yl(x, k) + Yl(x,−k) (93)

again taking the Fourier transform, we obtain the Marchenko equation (see appendixB for the derivation):

K(x, y; t) + B(x+ y; t) +

∫∞x

K(x, z; t)B(y+ z; t)dz = 0 (94)

where B(ξ; t) =1

2π

∫∞−∞ R(k, t)e

ikξdk+∑n

cn(t)2e−κnξ (95)

16

To solve the inverse scattering, we need to solve the Marchenko equation. Becausethe reflection is zero B(ξ; t) simplifies to:

B(ξ; t) =

N∑n

cn(t)2e−κnξ =

N∑n

cn(0)2e−8ik

3t−κnξ (96)

Therefore B(x+ y) can be separated:

B(x+ y) =

N∑n

Fn(x)Gn(y) (97)

where Fn(x) = cn(t)2e−κnx (98)

and Gn(y) = e−κny (99)

It follows that K(x, y) can also be separated.

K(x, y) =

N∑n

Hn(x)Gn(y) (100)

Instead of giving Hn explicitly, we eliminate it from our equation:

N∑n

Hn(x)Gn(y) +

N∑n

Fn(x)Gn(y) +

∫∞x

N∑m

Hm(x)Gm(z)

N∑n

Fn(z)Gn(y)dz = 0 (101)

N∑n

(Hn(x)Gn(y) + Fn(x)Gn(y) +

N∑m

∫∞x

Gm(z)Fn(z)dzHm(x)Gn(y)

)= 0 (102)

Because the left term is only depending on x and the right term is only depending ony, this equation spits into N equations:

N∑n

(Hn(x) + Fn(x) +

N∑m

Anm(x)Hm(x)

)Gn(y) = 0

where Anm(x) =

∫∞x

Gm(z)Fn(z)dz

δnmHm(x) + Fn(x) +

N∑m

Anm(x)Hm(x) = 0

Fn(x) +

N∑m

Anm(x)Hm(x) = 0

where Anm(x) = δnm +Anm(x)

Hm(x) = −

N∑n

A−1mn(x)Fn(x) (103)

⇒ K(x, x) = −

N∑n

N∑m

Gn(x)A−1nm(x)Fm(x) (104)

17

With this we can obtain an expression for the N-soliton solution

Anm(x) = δnm +

∫∞x

Gm(z)Fn(z)dz

= δnm + c2n(t)

∫∞x

e−(κn+κm)ydy

= δnm + c2n(t)e−(κn+κm)x

κn + κm(105)

Therefore we obtain an expression for the disturbance:

K(x, x) = −∑n

∑m

Gn(x)A−1nm(x)Fm(x)

=

N∑n

N∑m

e−κnxA−1nm(−c

2m(t)e

−κmx)

=

N∑n

N∑m

A−1nm

d

dxAmn

= Tr

(A−1 d

dxA

)=

1

det(A)

d

dxdet(A)

=d

dxlog det A (106)

Now we obtain the potential:

u(x, t) = −2d

dxK(x, x; t) = −2

d2

dx2log det A(x; t)

= −2d2

dx2log det

(δnm + c2n(t)

e−(κn+κm)x

κn + κm

)(107)

Example 4.1. With this we solve the KdV equation for the single soliton solution.There is only a single eigenvalue k, therefore

u(x, t) = −2d2

dx2log

(1+ c2n(0)

e−8k3t−kx

k

)

= −2d2

dx2log

(1+ e−8k

3t−kx+η0)

= −2∂

∂x

ke−8k3t−kx+η0

1+ e−8k3t−kx+η0

= −2(1+ e−8k

3t−kx+η0)k2e−8k3t−kx+η0 − k2e2(−8k

3t−kx+η0)

(1+ e−8k3t−kx+η0)2

= −2k2e−8k

3t−kx+η0

(1+ e−8k3t−kx+η0)2= −

k2

2sech2

(−8k3t− kx+ η0

2

)(108)

18

When we scale x → 2kx, we obtain the the potential for which we have confirmed that

it is reflectionless.

Although we have focused on the KdV equation this approach can be generalized,moreover the inverse scattering transform can be used to obtain solutions when thereflection is not zero (this however makes solving the Marchenko equation very difficult).The inverse scattering transform is a powerful tool, however there is no general methodto obtain the time dependence of the scattering data.

5 KdV Hierarchy

5.1 pseudo differential operator

Before we continue with our discussion we first introduce an pseudodifferential operator.Which will be useful in the future. Let ∂ denote differentiation to x then we can write:

∂ ◦ f = (∂f) + f∂ (109)

Therefore we can apply ∂f to a function g:

∂ ◦ f(g) = ((∂f) + f∂)g = (∂f)g+ f(∂g) (110)

which is the Leibniz rule. We generalize this for higher orders:

∂n ◦ f =n∑j=0

(n

j

)(∂jf) · ∂n−j (111)

The binomial coefficient (n

j

)=n(n− 1) · · · (n− j+ 1)

j(j− 1) · · · 1(112)

is well defined when j is a natural number, furthermore it is set to zero when j > n.This means we can extend the summation of equation (111) without any problems:

∂n ◦ f =∞∑j=0

(n

j

)(∂jf) · ∂n−j (113)

We obtain differential operators with negative powers, these are pseudodifferential op-erators. From equation (113) we obtain a expression for ∂−1:

∂−1 ◦ f = f∂−1 − (∂f)∂−2 + (∂2f)∂−3 + . . . (114)

We know ∂n · ∂m = ∂n+m, therefore we require ∂−1 · ∂ = 1 = ∂ · ∂−1. We will show thisfor ∂ · ∂−1.

∂ · ∂−1f = ∂(f∂−1 − (∂f)∂−2 + (∂2f)∂−3 + . . . ) (115)

∂ · ∂−1f = f∂∂−1 + (∂f)∂−1 − (∂f)∂∂−2 − (∂2f)∂−2 + (∂2f)∂∂−3 + (∂3f)∂−3 + . . . (116)

∂ · ∂−1f = f (117)

19

Definition 5.1. In general the expression:

P =

∞∑j=0

gj∂α−j (118)

is called a pseudodifferential operator of order 6 α.

Example 5.1. Using the pseudodifferential operator we calculate the square root ofthe operator L = ∂2 +u, which we have used in the lax-form of the KdV-equation (54)(although we replace u→ −u).

X = ∂+

∞∑n=1

fn∂−n (119)

L = X2 (120)

X2 = ∂2 + 2

∞∑n=1

fn∂1−n +

∞∑n=1

(∂fn)∂−n +

∑n.m>1,l>0

(−n

l

)fn(∂

lfm)∂−m−n−l (121)

X = ∂+1

2u∂−1 −

1

4ux∂

−2 +1

8(uxx − u

2)∂−3 + . . . (122)

We split the pseudo-differential operator into two parts, the part containing thenonnegative powers of ∂ and the rest:

P+ =

α∑j=0

gj∂α−j (123)

P− = P − P+ (124)

using this we can reformulate the lax-form of the KdV-equation. When change theparameters of the KdV-equation (to absorb the constants of integration), we obtain:

L = ∂2 + u (125)

B3 = ∂3 +

3

2u∂+

3

4ux (126)

We already know L = X2. Moreover we can express B3 in terms of X:

(X3)+ = ((∂+1

2u∂−1 −

1

4ux∂

−2 +1

8(uxx − u

2)∂−3 + (. . . )∂−4 . . . )(∂2 + u))+

= (∂3 +1

2u∂−

1

4ux +

1

8(uxx − u

2)∂−1 + (. . . )∂−2 + ux + u∂+ . . . )+

= ∂3 +3

2u∂+

3

4ux = B3 (127)

Furthermore we know the following:

[X2, Xl] = X2+l − Xl+2 = 0

[X2, Xl] = [X2, (Xl)− + (Xl)+]⇒ [X2, (Xl)+] = −[X2, (Xl)−] (128)

20

We reformulate the Lax-form of the KdV-equation:

Lt = [L, (L3/2)+] (129)

We have stated before (see section 3.2) that including higher order derivatives that weobtain so called higher order KdV-equations. This is done by choosing l 6 3 in thefollowing equation:

Lt = [L, (Ll/2)+] = [(Ll/2)−, L] (130)

5.2 symmetries

Definition 5.2. An evolutionary equation

∂u

∂t= K(u) (131)

is said to have a symmetry of the form:

∂u

∂s= K(u) (132)

when we let u depend on two independent time variables s and t, and solving first fors and then for t or solving first for t and then for s give the same result, see figure 3:

Figure 3: Solving first for s and then for t or the other way around [8].

Both K(u) and K(u) are differential polynomials, meaning that they are polynomialin u and its derivatives (with respect to a certain variable, in this case x). This is calleda symmetry because it resembles an infinitesimal generator. That the order of solvingthe equation leaves the solution unchanged allows us to state the following:

∂K(u)

∂s=∂2u

∂s∂t=∂2u

∂t∂s=∂K(u)

∂t(133)

Example 5.2. The most simple symmetry for the KdV-equation is given by:

∂u

∂t= K(u) = 6uux − uxxx (134)

∂u

∂s= K(u) = ux (135)

∂K(u)

∂t= utx = (6uux − uxxx)x = 6(ux)

2 + 6uuxx − u4x (136)

∂K(u)

∂s= (6uux − uxxx)s = 6usux + 6uusx − uxxxs = 6(ux)

2 + 6uuxx − u4x (137)

21

These are the same. Therefore us = ux forms a symmetry of the KdV-equation. Thisis as we we expected because we have already showed (in remark 3.2) that the KdV-equation is invariant under translations and us = ux corresponds to u(x, t, s) = u(x+s, t) a translation.

Using equation (33) we can rewrite equation (130) as:

∂u

∂xl= [L, (Ll/2)−] = Kl(u) or

∂u

∂xl= −[L, (Ll/2)+] = Kl(u) (138)

These are the higher order KdV-equations. The operator L is of differential order 2,while the operator (Ll/2)− is of order −1∀l. Furthermore the commutator adds theorders of the terms and subtracts one from the order because the highest orders alwayscancel out. Therefore the operator [L, (Ll/2)−] is of order 0, because of the constructionof both L and (Ll/2)−, it is a differential polynomial in u with respect to x. We onlyconcern ourselves with equations of odd order as the even orders are all identically zero.Moreover these equation are symmetries of the KdV equation.

Example 5.3. The cases l = 1, 3 are not truly “higher order” KdV-equation. Howeverthey do give us insight on some of the infinitely many variables that we have introducedby the infinitely many symmetries.

∂u

∂x1= −[L, (L1/2)+] (139)

∂u

∂x1= −[∂2 + u, ∂] (140)

∂u

∂x1= ux (141)

x1 = x (142)

And for the l = 3 case we obtain:

∂u

∂x3= −[L, (L3/2)+] (143)

∂u

∂x3= −[L, B3] (144)

∂u

∂x3= 6uux − uxxx (145)

x3 = t (146)

Now we show that all these higher order equations are symmetries:

∂Kl

∂xj=∂Kj

∂xl(147)

We know:∂L

∂xl= −[L, (Ll/2)+] (148)

22

∂f(L)

∂xl= −[f(L), (Ll/2)+] (149)

and therefore∂(Lj/2)+∂xl

= (∂Lj/2

∂xl)+ = −([Lj/2, (Ll/2)+])+ (150)

With this we show that the equations are indeed symmetries.

∂Kj

∂xl= −

∂

∂xl[L, (Lj/2)+]

= −

[∂

∂xlL, (Lj/2)+

]−

[L,

∂

∂xl(Lj/2)+

]= [[L, (Ll/2)+], (L

j/2)+] + [L, ([Lj/2, (Ll/2)+])+]

= [[L, (Ll/2)+], (Lj/2)+] + [L, ([L

j/2+ + L

j/2− , (L

l/2)+])+]

= [[L, (Ll/2)+], (Lj/2)+] + [L, ([L

j/2+ , (L

l/2)+])+ + ([Lj/2− , (L

l/2)+])+]

= [[L, (Ll/2)+], (Lj/2)+] + [L, ([L

j/2+ , (L

l/2)+])+ − ([Lj/2+ , L

l/2])+]

= [[L, (Lj/2)+], (Ll/2)+] + [L, ([Ll/2, (Lj/2)+])+]

= −

[∂

∂xjL, (Ll/2)+] − [L,

∂

∂xj(Ll/2)+

]= −

∂

∂xj[L, (Ll/2)+] =

∂Kl

∂xj(151)

where we have used, in the fifth step of equation (151):

0 = ([Lj/2, Ll/2])+

= ([Lj/2+ , L

l/2])+ + ([Lj/2− , L

l/2+ ])+ + ([L

j/2− , L

l/2− ])+

= ([Lj/2+ , L

l/2])+ + ([Lj/2− , L

l/2+ ])+⇒ ([L

j/2− , L

l/2+ ])+ = −([L

j/2+ , L

l/2])+ (152)

In the last step of equation (152) we have used equation (128). Furthermore in thesixth step of equation (151) we have used the Jacobi identity:

[a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 (153)

Hence the higher order KdV-equation are symmetries of all other KdV-equations. Thisinfinite system of equations is called the KdV hierarchy. Many equations with solitonsolution show this structure, however there exists no general treatment and every equa-tion needs to be studied in its own way to determine whether it can be written in thisway. Instead of viewing the hierarchy as a system of infinitely many equations, it can beviewed as an infinite set of nonlinear differential equations in a function τ(x1, x2, x3, . . . )of infinitely many variables. For the KdV-equation this function is related to u as fol-lows:

u = 2∂2

∂x2log τ (154)

23

However the discussion of this involves a discussion of hierarchy of the Kadomtsev-Petviashvili (KP) equation, which is beyond the scope of this thesis. The τ function isneeded in the next section to obtain a powerful yet simple method for obtaining solitonsolutions for our equations.

6 Hirota method

When trying to find soliton solutions in the same way as solutions to ”ordinary” waveequations we hit a dead end. The basic approach to wave equations makes use of thesuperposition principle, when one solution is found, infinitely many (linearly-dependent)solutions are found and by combining solutions the boundary conditions and initialvalues can be satisfied. But in the case of soliton equations, the equations are non-linear. Therefore the superposition principle does not apply. Furthermore, becauseof this non-linearity, perturbation theory cannot be used. Perturbation theory wouldlinearize the equations and would negate the non-linearity which governs the solitonsolutions.

Definition 6.1. The Hirota method [8] brings the equations to a bilinear form, whichsimplifies the problem. First we need to introduce the Hirota derivative Dx(f · g) .Take two single variable functions f(x) and g(x) and make a Taylor expansion aroundy = 0 of the following (x is fixed):

h(y) = f(x+ y)g(x− y) (155)

Normally we would do the following:

h(y) =

∞∑j=0

yj

j!

djh(0)

dyj(156)

We would like to express this in terms of f(x) and g(x) and derivatives of those.

f(x+ y)g(x− y) = fg+ (fxg− fgx)y+ (fxxg− 2fxgx + fgxx)y2

2+ . . . (157)

This must be equal to (156) and we introduce the Hirota derivative in the followingway:

f(x+ y)g(x− y) =

∞∑j=0

yj

j!Djx(f · g) (158)

Comparing the powers of y in the expansion of f(x+y)g(x−y) we obtain the following:

Dx(f · g) = fxg− fgx (159)

D2x(f · g) = fxxg− 2fxgx + fgxx (160)

D3x(f · g) = fxxxg− 3fxxgx + 3fxgxx − fgxxx (161)

D4x(f · g) = fxxxxg− 4fxxxgx + 6fxxgxx − 4fxgxxx + fgxxxx (162)

24

In the equations (159)-(162) we observe a pattern, the Hirota derivative follows theLeibniz rule but includes a alternating minus sign. the Leibniz rule can rewritten as:

dn

dxnf · g =

(∂

∂x1+

∂

∂x2

)nf(x1)g(x2)|x2=x1=x (163)

We adept this to include the alternating minus sign and obtain the Hirota derivativefor a single variable [10]:

Dnx (f · g) =

(∂

∂x1−

∂

∂x2

)nf(x1)g(x2)|x2=x1=x (164)

This can also be done in a multi variable case in exactly the same way. (note: h(y) =∑∞j=0

yj

j!djh(0)dxj

= eyddxh(x)|x=0)

f(x1+y1, x2+y2, x3+y3, . . . )g(x1−y1, x2−y2, x3−y3, . . . ) = ey1Dx1+y2Dx2+y3Dx3+...f ·g

(165)Again comparing the powers of the variables defines the Hirota derivative, for example:

Dx1Dx2(f · g) = −fx1gx2 − fx2gx1 + fx1x2g+ fgx1x2 (166)

Putting g(x) = f(x) and x1 = t, x2 = x which we will need later on, we obtain:

DtDx(f · f) = −2ftfx + 2fxtf (167)

Other equations we will need are:

∂2

∂x∂tlog(f) =

∂

∂t

fx

f=ffxt − fxft

f2=

1

2f2(DtDx(f · f)) (168)

And:

∂4

∂x4log(f) =

∂3

∂x3fx

f=∂2

∂x2

(fxx

f−

(fx

f

)2)

=∂

∂x

(fxxx

f−fxxfx

f2− 2

fx

f

ffxx − fxfxf2

)=fxxxx

f− 4

fxxxfx

f2− 3

f2xxf2

+ 12ffxx − f

2x

f3− 6

f4xf4

=1

2f2D4x(f · f) − 6

(1

2f2(D2

x(f · f)))2

(169)

Example 6.1. To bring our equation to bilinear form we need to do a transformation.As an example we are going to bring the KdV-equation (15) to bilinear form using(170).

u = 2∂2

∂x2log τ (170)

25

ut + 6uux + uxxx = 0 (171)

⇒ 2∂3

∂t∂x2log τ+ 6

(2∂2

∂x2log τ

)(2∂3

∂x3log τ

)+ 2

∂5

∂x5log τ = 0 (172)

⇒ ∂

∂x

(2∂2

∂t∂xlog τ+ 3

(2∂2

∂x2log τ

)2+ 2

∂4

∂x4log τ

)= 0 (173)

Integrating once and using (168) and (169) we obtain:

2

(1

2τ2(DtDx(τ · τ))

)+12

(1

2τ2(D2

x(τ · τ)) + 21

2τ2(D4

x(τ · τ))−12

(1

2τ2(D2

x(τ · τ))

= c

(174)Which simplifies to:

[DtDx +D4x](τ · τ) = c (175)

We want τ to give us a solution of the KdV-equation. For example choosing τ = 1 leadsto u = 0, which is solution of the KdV-equation. We can conclude that c = 0.

[DtDx +D4x](τ · τ) = 0 (176)

This is the KdV-equation in bilinear form.

Remark 6.1. The equation (170) brought the KdV-equation (15) to bilinear form. How-ever (170) does not bring all equations to bilinear form. There is no general methodto bring an equation to bilinear form. Sometimes two equations are needed to bring aequation to bilinear form. For example the sine-Gordon equation:

uxt = sin(u) (177)

Is transformed by:

u = 4 arctan(F

G) (178)

With the bilinear system: {[DxDt − 1](F ·G) = 0D2x[(F · F) − (G ·G)] = 0 (179)

Moreover not all equation can be transformed into bilinear form. For example theMonge-Ampre equation

u2xy − uxxuyy = 0 (180)

cannot be brought to bilinear form [9].

Definition 6.2. Equation (176) is called a Hirota equation. In general we write(D1, D2, . . . ) for the Hirota derivatives with respect to the variables (x1, x2, . . . ), thena Hirota equation is of the form:

P(D1, D2, . . . )(τ · τ) = 0 (181)

where P(D1, D2, . . . ) is a polynomial in (D1, D2, . . . ). Furthermore assume P(0, 0, . . . ) =0

26

Remark 6.2. Before we continue, we look at some qualities of the Hirota derivative.Using equation (164) we obtain the following:

Dx(f · g) = −Dx(g · f) (182)

Using this we can conclude:

P(Dx)(f · g) = P(−Dx)(g · f) (183)

Generalizing this to more variables:

P(D1, D2, . . . )(f · g) = P(−D1,−D2, . . . )(g · f) (184)

Other useful identities are:

Dnx (f · 1) =

(∂

∂x1−

∂

∂x2

)nf(x1)|x2=x1=x =

dnf

dxn(185)

⇒ P(D1, D2, . . . )(f · 1) = P(∂

∂x1,∂

∂x2, . . .

)f (186)

Dnx (e

kx · elx) =(∂

∂x1−

∂

∂x2

)nekx1 · elx2 |x2=x1=x = (k− l)ne(k+l)x (187)

⇒ P(Dx)(ekx · elx) = P(k− l)e(k+l)x (188)

Let us now restrict ourselves to the case where g(x) = f(x), which is the case for theequations we are interested in. We observe that:

P(D1, D2, . . . )(f · f) = P(−D1,−D2, . . . )(f · f) (189)

due to equation (184). Thus P must be an even function when applied to (f · f).We are going to construct the solutions, starting with the 0-soliton solution (an

n-soliton solution is the solution that has n solitons). We will show that these solutionare soliton solutions later. Take τ = 1, which will satisfy the Hirota equation (181):

P(D1, D2, . . . )(1 · 1) = P(0, 0, . . . )(1 · 1) = 0 (190)

We expand τ around this point:

τ = 1+ ετ1 + ε2τ2 + ε

3τ3 + . . . (191)

For the 1-soliton solution we only need to take the first order term τ1. BecauseP(0, 0, . . . ) = 0 the equations put homogeneous constraints on the other τ2, τ3, . . .of which τi = 0 are solutions and therefore these terms can be disregarded. We remainwith the following:

P(D1, D2, . . . )(1+ετ1 ·1+ετ1) = P(D1, D2, . . . )(1 ·1+1 ·ετ1+ετ1 ·1+ετ1 ·ετ1) (192)

27

We already know the following, using that P is an even function and P(0, 0, . . . ) = 0:

P(D1, D2, . . . )(1 · 1) = 0 (193)

P(D1, D2, . . . )(1 · τ1) = 0⇔ P(D1, D2, . . . )(τ1 · 1) = 0 (194)

We only need to look at:P(D1, D2, . . . )(τ1 · 1) = 0 (195)

According to equation (186) this is equivalent with:

P

(∂

∂x1,∂

∂x2, . . .

)τ1 = 0 (196)

This is a linear differential equation of which we know the solution. Choose k1, k2, · · · ∈C in such a way that P(k1, k2, . . . ) = 0 then:

τ1 = cek1x1+k2x2+... (197)

Here c ∈ C is a constant. This solves equation (196). We need to make sure that τ1also solves the following equation:

P(D1, D2, . . . )(τ1 · τ1) = 0 (198)

P(D1, D2, . . . )(τ1 · τ1) = P(D1, D2, . . . )(cek1x1+k2x2+... · cek1x1+k2x2+...

)(199)

Now we can use equation (188):

P(D1, D2, . . . )(cek1x1+k2x2+... · cek1x1+k2x2+...

)= P(k1 − k1, k2 − k2, . . . )e

2(k1x1+k2x2+... )

= P(0, 0, . . . )e2(k1x1+k2x2+... ) = 0 (200)

Therefore τ1 solves equation (192). The other τi are all chosen to be equal to zero.o Therefore our solution of the Hirota equation (181) is:

τ = 1+ cek1x1+k2x2+... (201)

We can absorb the c into the exponential by writing ε = eη0 , therefore our final solutionlooks like:

τ = 1+ eη0+k1x1+k2x2+... (202)

Example 6.2. Remember that in the case of the KdV-equation (15) the Hirota typeequation is (176): [

DtDx +D4x

](τ · τ) = 0

Therefore k1, k2 need to satisfy:

k1k2 + k42 = 0⇒ k1 = −k32 (203)

⇒ τ = 1+ ekx−k3t+η0 (204)

28

We transform back to u using equation (170).

u = 2∂2

∂x2log(τ) = 2

∂2

∂x2log(1+ ekx−k

3t+η0)

= 2∂

∂x

kekx−k3t+η0

1+ ekx−k3t+η0= 2

(1+ ekx−k3t+η0)k2ekx−k

3t+η0 − k2e2(kx−k3t+η0)

(1+ ekx−k3t+η0)2

=2k2ekx−k

3t+η0

(1+ ekx−k3t+η0)2=k2

2sech2

(kx− k3t+ η0

2

)(205)

The Hirota Method can also be used to obtain the N-soliton solution.

P(k1, k2, . . . ) = 0 (206)

defined the relation that k1, K− 2, . . . must satisfy to make

τ = 1+ eη0+k1x1+k2x2+... (207)

a solution of the equation. When there are multiple sets of solutions k(j)1 , k

(j)2 , . . . , we

can make a superposition of the solutions in the Hirota equation to obtain more thenone soliton. Let k

(1)1 , k

(1)2 , . . . and k

(1)1 , k

(1)2 , . . . (assume that these sets are different)

satisfy the Hirota equation. Then we can write:

P(D1, D2, . . . )(τ · τ) = 0 (208)

where τ = 1+ ε

2∑j=1

cjek(j)1 x1+k

(j)2 x2+... + ε2τ2 +O(ε3) (209)

⇒P(∂1, ∂2, . . . )τ2 + c1c2P(k(1)1 − k(2)1 , k

(1)2 − k

(2)2 , . . . )e

(k(1)1 +k

(2)1 )x1+(k

(1)2 +k

(2)2 )x2+... = 0

(210)

where we have used equation (188) on the cross terms between the exponentials of k(1)

and k(2). The other terms already satisfy the Hirota equation and thus disappear. Weobtain an expression for τ2

τ2 = −c1c2P(k

(1)1 − k

(2)1 , k

(1)2 − k

(2)2 , . . . )

P(k(1)1 + k

(2)1 , k

(1)2 + k

(2)2 , . . . )

e(k(1)1 +k

(2)1 )x1+(k

(1)2 +k

(2)2 )x2+... (211)

Substitute this back into τ together with the solutions corresponding to k(1) and k(2),then the 2-soliton solution is obtained. This can be generalized to obtain a N-solitonsolution. However it is not always possible to find N different sets of k

(j)1 , k

(j)2 , . . . for

a generic Hirota equation,however it turns out that as an empirical fact that the oneand two soliton solution can alway be obtain, however the n-soliton solution for n 6 3is equivalent to the integrability of the system [8].

The Hirota method is less calculation heavy than the inverse scattering transform.However where in the inverse scattering transform the difficulties are the calculation of

29

the time dependence of the scattering data and solving Marchenko integral equation,In the case of the Hirota method the problem is finding the transformation to bilinearform. This however is at the cost that the Hirota method only provides soliton solutions,while the inverse scattering transform can also be used to obtain non soliton solutions(in the case that the reflection coefficient is non zero).

7 Derrick’s Theorem

Remark 7.1. We will discuss the phenomena of solitons in field theories, remember thatthe list we of qualities of solitons was:

• Wave profile

• Finite energy

• Localized phenomenon

• Stable in time

• Stable under interaction processes

however the interpretation of this last point is not universal. This leads to solutionsthat interact very little but still being called solitons. A important notion is that thesoliton solution must be a stationary point of the energy, otherwise the solution woulddecay into a stationary point and would therefore be unstable. Often when studyingsolitons as solutions in field equation, we restrict ourselves to the stationary solutions.Moving solutions can be obtained by boosting the system.

Remark 7.2. We return to the KdV-equation (15), and look for a field equation withmirrors the behavior of that equation. As we have seen, the magnitude of the wave isrelated to the speed of the wave. A static wave solution was equal to zero everywhere.When a soliton moves with a certain speed, we can boost our system to follow thesoliton, which is not moving after the boost. We have already stated that a staticsolutions equals zero everywhere, therefore the soliton disappears after the boost. Thereis no relativistic KdV-soliton, which is no surprise as the KdV-equation is not Lorentz-invariant.

In a large class of field theories there exist bounds on whether soliton solution canexist. The most important of these is Derrick’s theorem [13].

Theorem 7.1. Derrick’s theorem. In theories with a Lagrangian of the form:

L(ϕ) =1

2∂µϕ∂

µϕ−U(ϕ) (212)

non-trivial time-independent scalar soliton solutions do not exist in more then one spa-tial dimension in either Euclidean or Minkowski space-time.

30

The proof of Derrick’s theorem is based on spatial rescaling of the energy functional,which is never zero for stationary solutions. Moreover these solutions which are minimaof the energy are found by a variational principle and should thus be stationary againstall variations of the field, including spatial rescaling. Therefore no stationary solutionexist except for the vacuum states. Lorentz boosts imply there are also no movingsolutions. Let us study the D-space dimensional energy functional corresponding tothe Lagrangian (212)[12]:

E[ϕ] =

∫dDx

(1

2~Oiϕ · ~Oiϕ+U(ϕ(x))

)≡ ES[ϕ] + EU[ϕ] (213)

where ES[ϕ] =

∫dDx

(1

2~Oiϕ · ~Oiϕ

)(214)

and EU[ϕ] =

∫dDx (U(ϕ(x))) (215)

Both ES[ϕ] and EU[ϕ] are non-negative. Now consider a time-independent solutionϕ(x) and apply a spatial rescaling: x → λx with λ > 0. Let ϕλ(x) be an parametricfamily of field configurations:

ϕλ(x) ≡ ϕ(λx) (216)

In order for the theory to be scale invariant, the energy functional transforms as follows:

E[ϕλ(x)] = ES[ϕλ(x)] + EU[ϕλ(x)] = λ2−DES[ϕ(x)] + λ

−DEU[ϕ(x)] (217)

Because ϕ(x) is a solution, it is a stationary point of the variation of the energy func-tional around λ = 1. Therefore we require the following:

dϕλ

dλ

∣∣∣∣λ=1

= 0 (218)

This leads to the following relation:

(2−D)ES[ϕ(x)] = DEU[ϕ(x)] (219)

Because both ES[ϕ] and EU[ϕ] are non-negative, this equation can only be solved inthe case D = 1, otherwise only the vacuum solution remains.

Remark 7.3. When we study solitons in field equations for D > 1 we need to circumventthis restriction, this can be done in several ways [12]:

By adding an extra field for example gauge fields Aµ(x). This changes the energyfunctional in the following way:

E = ES + EU + EG (220)

ES[ϕ] =

∫dDxTn(Diϕ) (221)

EU[ϕ] =

∫dDxU(ϕ) (222)

EG[Aµ] =

∫dDxFijFij (223)

where Fµν = ∂µAν − ∂νAµ (224)

31

Here Tn is the n power of the derivatives of the scalar kinetic term (usually n = 2).The gauge field transforms as a 1-form, Aµλ ≡ λAµ(λx). The covariant derivativetransforms as followedDµ

λϕ(x) = (dµϕλ(x)+Aµϕλ(x)) ≡ λDµϕ(λx). The field strength

transforms as Fµνλ (x) ≡ λ2Fµν(x), then following the same procedure as before we obtainthe constrain:

(n−D)ES −DEU + (4−D)EG = 0 (225)

The following solitons use gauge fields to circumvent Derrick’s theorem:

• (2) space dimensions: CPN model [21]

• (3) space dimensions: ’t Hooft-Polyakov monopole [14]

• (4) space dimensions: Instanton [19]

Another way of circumventing Derrick’s theorem is to include a non-trivial time de-pendence ϕ(x) → ϕ(x, t). Q-balls show this structure by including a time dependentphase. Stability of the soliton is then governed by a Noether charge, which is relatedto the phase invariance of the Lagrangian. See section 8 for more about Q-balls. Byconsidering more complicated Lagrangians, for example higher powers of the derivative,one can also circumvent Derrick’s theorem and thus obtain solitons. Skyrme [15] andFaddeev-Niemi [16] models are examples of this kind.

Remark 7.4. We can split the solitons in field theories in two types, depending on themethod which stabilized them. If they are stable because of the topology of the theorythen they are called topological solitons. Otherwise they are called non-topologicalsolitons, usually a Noether charge stabilizes these solutions. However it is imaginablethat other more exotic theories also allow soliton solutions without a topological orNoether charge, which would also have to be called non-topological solitons.

8 Q-ball

In this section we will discuss the Q-ball [17], a non-topological soliton, which we studyusing the simple Lagrangian density of a complex scalar field ϕ in (1+1) dimensions:

L = ∂µϕ∂µϕ† −U(ϕ†ϕ) (226)

This Lagrangian density is phase invariant: if we substitute ϕ→ eiθϕ the Lagrangiandensity does not change. This invariance gives rise to a conserved charge by virtue ofNoether’s theorem. The Lagrangian density gives an equation of motion:

∂2ϕ−ϕdU(ϕ†ϕ)

d(ϕ†ϕ)= 0 (227)

and the conserved Noether current is:

jµ = −i(ϕ†∂µϕ−ϕ∂µϕ†) (228)

32

∂µjµ = −i∂µ(ϕ

†∂µϕ−ϕ∂µϕ†)

= −i(∂µϕ†∂µϕ+ϕ†∂2ϕ−ϕ∂2ϕ† − ∂µϕ†∂µϕ)

= −i(ϕ†∂2ϕ− ∂2ϕ†ϕ)

using the equation of motion

= −i

(ϕ†ϕ

dU(ϕ†ϕ)

d(ϕ†ϕ)−ϕϕ†

dU(ϕ†ϕ)

d(ϕ†ϕ)

)= −i(ϕ†ϕ−ϕ†ϕ)

dU(ϕ†ϕ)

d(ϕ†ϕ)

∂µjµ = 0 (229)

The conserved current can be used to obtain a conserved charge for the field.

Qdef

=

∫j0dx (230)

∂Q

∂t=

∫∂j0

∂tdx

using ∂µjµ = 0⇒ ∂j1

∂x−∂j0

∂t= 0

∂Q

∂t=

∫∂j1

∂xdx

∂Q

∂t= j1|boundary (231)

We are interested in solitons, which are localized. Therefore ϕ|boundary = 0, thenj1|boundary = 0. We can conclude that the charge is conserved in time:

∂Q

∂t= 0 (232)

In order to obtain soliton solutions we must put some restriction on the potential.The potential U must have a minimum at ϕ = 0, which we can choose to be zero byshifting the potential. When we have a soliton (or any solution except for the vacuum),Q 6= 0. This implies the following:

Q =

∫j0dx 6= 0 ⇒ j0 6= 0 (233)

j0 = −i(ϕ†∂0ϕ− ∂0ϕ†ϕ) 6= 0 ⇒ ∂0ϕ 6= 0 (234)

Our solution must be time dependent. In order to determine this time dependence, wewrite:

ϕ = (ϕR + iϕI)/√2 (235)

33

Where ϕR and ϕI are respectively the real and imaginary parts of the field ϕ. Werewrite the equation for the charge (230) as follows:

Q =

∫j0dx =

∫−i(ϕ†∂0ϕ− ∂0ϕ†ϕ)dx

=−i

2

∫(ϕR − iϕI)

∂(ϕR + iϕI)

∂t− (ϕR + iϕI)

∂(ϕR − iϕI)

∂tdx

=

∫ϕI∂ϕR

∂t−ϕR

∂ϕI

∂tdx (236)

The solution has minimal energy, satisfying the constraint of the conserved charge (ωis the Lagrange multiplier):

δ(E−ωQ) = 0 (237)

The equation for the energy can be obtained from the Lagrangian density (226) andthen be rewritten using (235):

E =

∫∂µϕ∂µϕ

† +U(ϕ†ϕ)dx

=

∫1

2

(ϕR − iϕI∂t

+ϕR − iϕI∂x

)(ϕR + iϕI∂t

−ϕR + iϕI∂x

)+U

(ϕ2R +ϕ

2I

2

)dx

=

∫1

2

∂ϕR

∂t

2

+1

2

∂ϕI

∂t

2

+1

2

∂ϕR

∂x

2

+1

2

∂ϕI

∂x

2

+U

(ϕ2R +ϕ

2I

2

)dx (238)

We solve equation (237) assuming that the spatial part already satisfies the minimalenergy requirement. Therefore the only variations in the field are of the time derivative.

∂ϕR

∂t→ ∂ϕR

∂t+ δR (239)

∂ϕI

∂t→ ∂ϕI

∂t+ δI (240)

Equation (237) then becomes:

δ(E−ωQ) =

∫1

2

∂ϕR

∂t

2

+∂ϕR

∂tδR +

1

2

∂ϕI

∂t

2

+∂ϕI

∂tδI +

1

2

∂ϕR

∂x

2

+1

2

∂ϕI

∂x

2

+U

(ϕ2R +ϕ

2I

2

)dx−ω

(∫ϕI

(∂ϕR

∂t+ δR

)−ϕR(

∂ϕI

∂t+ δI)

)dx

(241)

Collecting the terms δR and δI yield:

∂ϕR

∂t= ωϕI (242)

∂ϕI

∂t= −ωϕR (243)

34

The other parts of the equation already satisfy the minimal energy. Solving theseequations we obtain:

ϕ(x, t) = σ(x)e−iωt (244)

Where σ(x) is a real function. We can rewrite the charge (230) using this relation intoa more simple form:

Q =

∫j0dx = −i

∫(ϕ†∂0ϕ−ϕ∂0ϕ†)dx

= ω

∫σ(x)2dx (245)

To further study the Q-ball we take the potential [18]:

U(ϕ†ϕ) = m2|ϕ|2 −2

3a|ϕ|3 +

1

2b|ϕ|4 (246)

By rescaling we simplify the Lagrangian.

ϕ→ m2

aϕ (247)

x→ 1

mx (248)

The Lagrangian density then becomes:

L = ∂µϕ∂µϕ† − |ϕ|2 +

2

3|ϕ|3 −

1

2B|ϕ|4 (249)

where B =bm2

a2(250)

The equation of motion then becomes:

∂2ϕ−ϕ+ |ϕ|ϕ− B|ϕ|2ϕ = 0 (251)

Using equation (244) we obtain a differential equation for σ(x):

σ′′+ (ω2 − 1)σ+ σ2 − Bσ3 = 0 (252)

The soliton is localized and is stationary, therefore:

σ ′(0) = 0 (253)

σ(∞) = 0 (254)

This describes exactly the movement of a particle under the potential:

V =1

2(ω2 − 1)|σ|2 −

1

3|σ|3 +

B

4|σ|4 (255)

35

To obtain a soliton solution we require that the solution cannot be formed by a super-position of plane-waves. Therefore we require m2

eff = ω2 − 1 < 0. Furthermore for the

solution to be stable the potential must have a zero different from ϕ = 0. These twoconditions together give us the constrain:

1−2

9B< ω2 < 1 (256)

In order to have stability we require that the solution is a minimal energy configuration(δ2E > 0). Using the potential we can write equation (238) more explicitly:

E =

∫σ ′2 + σ2 −

2

3σ3 +

B

2σ4dx+

Q2∫σ2dx

(257)

Then:

δ2E =

∫δσOδσdx (258)

O = −d2

dx2+ (1− 2σ0 + 3Bσ

20) + 3ω

2 (259)

In order for δ2E > 0 it is enough to know that O has no negative eigenvalues. Todetermine whether there are negative eigenvalues a numerical study was done by M.Axenides, et al [18] shows the results shown in figure 4. When the parameters aresuitably chosen the Q-ball will be stable and will not decay into other Q-balls.

Figure 4: The stability of the Q-ball depending on ω2 vertical and B horizontal [18]

Furthermore their numerical analysis shows the interaction of Q-balls. The figures5(a), 5(b), 5(c) and 5(d) show the interaction between qballs. At low velocities 5(a) theQ-balls the Q-balls attract and scatter along the x-axis and again attract each other,creating breather type oscillations. At higher velocity 5(b) part of the charge breaksfree of the interaction traveling along the x-axis, the rest continues traveling in the y

36

direction. Each of these Q-balls escape towards infinity. At very high velocity 5(c) onlyforward scattering occurs and the Q-balls separate from each other, escaping towardsinfinity. The anti Q-ball , Q-ball interaction is only minimal, they exchange some chargeand then travel towards infinity.

Where the Solitons of the KdV equation did not change their shape under interac-tion, this is no longer the case for Q-balls. Under interaction Q-balls may decay intoother Q-balls depending on their relative velocity. However a single Q-ball is stable intime.

37

(a) Head-on collision of two Q-balls with thesame charge.t = 0, t = 41.9, t = 55.9. Initialvelocity of each Q-ball v = 0.2.[18]

(b) Head-on collision of two Q-balls with thesame charge.t = 0, t = 19.6, t = 69.8. Initialvelocity of each Q-ball v = 0.4. [18]

38

(c) Head-on collision of two Q-balls with thesame charge. t = 0, t = 12.6, t = 29.3. Initialvelocity of each Q-ball v = 0.8. [18]

(d) Head-on collision of two Q-balls with op-posite charge. Solid lines represent positivevalues and dashed-dotted lines negative val-ues. : t = 0, t = 28, t = 61.4. Initial velocityof each Q-ball v = 0.4. [18]

Figure 5:39

9 Topological conservation laws

Instead of a Noether’s charge, which provided stability for the Q-ball, we look at dif-ferent charges. These charges are not governed by a symmetry however the topologyof the theory provides a conserved quantity [19]. The solitons that a stable because ofthese charges are called topological solitons.

Instead of studying the solutions, we focus on the initial value data. Because thesoliton solutions have a finite energy, the initial value date must also have finite en-ergy. Therefore we can conclude that the initial value data must be sufficiently smooth(continuously differentiable). Moreover the initial value data, at boundary

ϕ(±∞, t) def

= limx→±∞ϕ(x, t) (260)

is a zero of the potential energy. If the potential has a discrete set of zero’s, becausethe initial value data is smooth, this implies ):

∂0ϕ(±∞, t) = 0 (261)

This is a conservation law. When the potential has more then one zero, it is possiblethat at one end (x = +∞) the solution is connected to a different zero of the potentialthan at the other end (x = +∞). Between the two boundaries the solution connectsthem in a smooth manner, which is nonzero because the potential is at least partiallynonzero in between two zeros. This topological conserved charge (261) can be used toprove existence of nonzero and non-dissipating solutions.

Example 9.1. As an example, we take the sine-Gordon field-equation. The Lagrangiandensity is given by:

L =1

2∂µϕ∂µ +U(ϕ) (262)

where U(ϕ) = 1− cos(ϕ) (263)

where ϕ is a real scalar field. In figure 6 we have plotted the shape of the potential.

Figure 6: The shape of the potential of the sine-Gordon equation

The zeros of the potential form a discreet set. If a solution is at one zero (for example

40

Figure 7: The kink and anti-kink solutions of the sine-Gordon equation

ϕ = 0) at x = −∞ and when x =∞ at another zero (for example ϕ = 2π) somewhere inbetween a link between these two must form. In the case of the sine-Gordon equationthese look like figure 7. Because of the shape these solutions are called kinks (andanti-kinks). In the case of the sine-Gorden equation a kink and a anti-kink will forma breather pair, not dissimilar in behavior as the breather pair observed in section 8where two slow moving Q-balls formed a breather pair. However in other equations thathave kink and anti-kink solutions, the kink and anti-kink may annihilate each other.

As we have observed in section 7 gauge field can be used to circumvent Derrick’stheorem 7.1. We will review gauge fields and obtain some properties.

The theories which are of interest to us contain a set of n scalar fields. These wecombine into a n-vector, ϕ. Furthermore these theories contain a gauge group, G,which is a compact connected Lie group with a n-dimensional unitary representationD(g). We assume that G is a simple group, the procedure we follow will still be validin the case that G is not simple however this would add a notational complexity wewould like to avoid. We define a gauge transformation by:

g(x) : ϕ(x)→ D(g(x))ϕ(x) (264)

for any function g(x) from space-time to G. To simplify notation we identify the groupG with the representation D(g).

g(x) : ϕ(x)→ g(x)ϕ(x) (265)

Denote the generators of D(g) by Ta, these are n×n orthogonal Hermitian matricessatisfying the following commutation relations:

[Ta, Tb] = icabcT c (266)

Where cabc are the structure constants. To each of these generators Ta a gauge fieldAaµ isassociated. These are real co-vector fields and transform with the gauge transformationgiven by:

Aaµ(x)Ta → g(x)Aaµ(x)T

ag(x)−1 + ie−1(∂µg(x))g(x)−1 (267)

41

Where e is the gauge coupling constant, which is a real constant. Instead of this gaugetransformation it is more common to look at the infinitesimal gauge transformation:

g(x) = 1− iTbωb(x) (268)

δAaµ = cabcAbµδω

c + e−1∂µδωa (269)

We define the covariant derivative of ϕ as:

Dµϕ(x) =(∂µ + ieA

aµT

a)ϕ (270)

and we can define a field strength for the gauge fields:

Faµν = ∂µAaν − ∂νA

aµ − ec

abcAbµAcν (271)

these transform under gauge transformations as follows:

Dµϕ(x)→ g(x)Dµϕ(x) (272)

FaµνTa → g(x)FaµνT

ag(x)−1 (273)

We construct a Lagrangian density that defines our theory and is invariant under gaugetransformations:

L = −1

4FaµνF

aµν +Dµϕ†Dµϕ−U(ϕ) (274)

The potential U is a gauge invariant function of the scalar fields, and we can shift Uthereby ensuring that U becomes non-negative.

The ground states ϕ0 have zero energy, ϕ0 is constant and a zero of U. Furthermorethe gauge fields vanish in the ground state. Define H as the subgroup of G that leavesground states unchanged.

h ∈ H⇔ hϕ0 = ϕ0 (275)

As the potential is invariant under gauge transformations, if ϕ0 is a zero of U then sois gϕ0. Assume4 that all zeros of U can be written as gϕ0 for some g ∈ G. This meansthat we can identify the zeros of U by the coset space G/H.

Remark 9.1. In some cases it useful to rewrite the gauge fields in such a way that eitherthe time dependence or the radial dependence vanishes. Therefore we want to simplifythe gauge fields in such a way that the can be written as:

Aa0 = 0 (276)

To this end we need to look at the gauge fields in relation with paths in space-time.

4This neglects both accidental degeneracies and internal (non-gauge) symmetries. The accidentaldegeneracies will disappear when we include quantum theory. Furthermore the results we obtain canbe generalized to the case where internal symmetries are included. However both are beyond the scopeof this thesis [19].

42

Take a path P in space-time, traveling from x0 to x1. Map a parameter s ∈ [0, 1] tothis path such that P(0) = x0 and P(1) = x1. Consider a field ϕ such that the covariantderivative vanishes along P:

dxµ

dsDµϕ = 0 (277)

⇒ dϕ

ds= −ie

dxµ

dsAaµT

aϕ (278)

We want to write the final field ϕ(x1) in terms of the initial field ϕ(x0) and a transfor-mation g(P) ∈ G:

ϕ(x1) = g(P)ϕ(x0) (279)

This problem is similar to the time evolution given by a potential in quantum mechanicsand is solved using Dyson’s formula:

g(P) = Se−ie∫10dxµ

dsAaµT

ads (280)

Where S is an ordering based on s, in the same manner that in Dyson’s formula thereis a ordering based on t. These g(P) transform as [19]:

g(P)→ g(x1)g(P)g(x0)−1 (281)

Let P be the path from (x, 0) to x by a straight line, where x is any space-timepoint. Transform g(Px) with the gauge transformation g(x) = g(Px)

−1.

g(Px)→ g(Px)−1g(Px)g(P0) = g(P0) = 0 (282)

The last part holds because P0 is the path from and to the same point which changesnothing. After differentiation we obtain:

Aa0 = 0 (283)

This fixes the gauge in which we work, so we lose the freedom to choose any gauge ifwe set Aa0 = 0. A similar argument can be used to obtain Aar = 0, r > ε, the radialpart can be set to zero (everywhere but the origin as there the radial part is not definedthere).

Again studying stationary solutions, we can fix the time. We remove the timedependence from our formula. In the two dimensional case5 we look at the energy ofthe field:

E >∫∞ε

r

∫ 2π0

(∂rϕ

†∂rϕ+U(ϕ))dθdr (284)

Note, we have cut out the origin from the integral. Because the solution is smooth thisoffsets the energy only by a finite amount. We require that the total energy is finite

5By including an extra angle this procedure can be adopted for a three dimensional case.

43

therefore the radial derivative must vanish (for this we use the results of remark 9.1)and

φ(∞, θ) def

= limr→∞ϕ(r, θ) (285)

must be a zero of U. In order for the finite energy condition to hold we need to cancelthe energy contribution of the angular dependence of ϕ(∞, θ) with the gauge field:

limr→∞ rAaθTaϕ(∞, θ) = −ie−1

dϕ(∞, θ)dθ

(286)

We have obtained a mapping from the boundary (S1) to the zero’s of the potential(G/H) namely ϕ(∞, θ).Definition 9.1. Let T s(x) : X→ Y be a continuous map ∀s ∈ [0, 1], x ∈ X. If T s(x) isa continuous function of s, ∀x, then for s1, s2 ∈ [0, 1] the maps T s1(x) and T s2(x) arehomotopic to each other.

If S is homotopic to T we write S ∼ T . This is an equivalence relation, if S ∼ T

and T ∼ R then S ∼ R. This can easily be shown be combining the function T s(x) thatdeforms S into T with the function that deforms T into R and rescaling the parameters. These equivalence classes, denoted by T , together with a composition rule, form agroup. In the case where X = [0, 1] with the end points identified (effectively X = S1),it is given by (287)

(T1 · T2) ={

T1(2x) 0 > x > 1/2T2(2x− 1) 1/2 > x > 1

. (287)

The group that is formed in this manner is called the first homotopy group π1(Y). Thisgroup is the most interesting in this thesis, since we will mostly be working in twodimensions (where the border is S1). Therefore we can classify the mappings that weobtained by looking at the boundary (285), this classification is π1(G/H). Solutionscannot go from one class to another as this would mean that they would be deformeddiscontinuously, which costs an infinite amount of energy [20].

Example 9.2. G is U(1), and the fields consists of a single complex scalar field ϕ. Thepotential is given by:

U =λ

2(ϕ∗ϕ− 1)2 (288)

The zero’s of the potential U are:

|ϕ| = 1 (289)

ϕ = eiσ (290)

where σ is a real number. This forms a circle, thus G/H = S1. Therefore we look atπ1(S

1), this the question how can we map a circle on a circle. This characterized by theso called winding number, how often is the circle wrapped around the circle. Withoutgoing into the mathematical details, π1(S

1) = Z. This winding number is conserved andtherefore a solution with a nonzero winding number will not dissipate into the vacuumsolution.

44

Example 9.3. G = SO(3), and the fields form a three component real isovector fieldϕ. The potential is given by:

U =λ

2(ϕ2 − 1)2 (291)

The zero’s of the potential U are:

ϕ21 +ϕ22 +ϕ

23 = 1 (292)

This forms a sphere, thus G/H = S2. Therefore we look at π1(S2). When we place circle

on the sphere we can always pull the circle together into a point. Thus π1(S2) = 0,

therefore no conserved winding number exists and every solution may dissipate into thevacuum solution (unless something else in the theory stops this from happening).

However when considering this theory in three dimensions instead of two, we needto use the second homotopy group as the boundary now is a sphere. In this case weobtain π2(S

2) = Z therefore in three dimensions solitons are possible.

Topological conservation laws are a useful tool to construct non-dispersive solutions,however they give little information on the explicit form of the solution. Furthermorethe result of interactions can not be obtained from these conservation laws only. In thenext section we will give an example of a theory with a topological conservation law inwhich we obtain explicit solutions.

10 CPN model

A CPN field is set of N+ 1 complex field that are combined in CPN spinor [21]:

ϕσ =

ϕ1ϕ2. . .

. . .

ϕN+1

(293)

With the restriction that:N+1∑n=1

|ϕn|2 = 1 (294)

Furthermore it is invariant under the local gauge transformation:

ϕσ(x)→ ϕσ(x)eiΛ(x) (295)

Because of the presence of this gauge invariance, we introduce a covariant derivative.Otherwise the field equation (which are normally gradient dependent) would changeunder the gauge transformations.

~Dϕσ(x) = (~O+ i~A)ϕσ(x) (296)

~A = i(ϕσ)†~Oϕσ (297)

45

The most simple energy functional that is consistent with the two constrains is:

E =

∫dx(~Dϕσ(x)

†) · (~Dϕσ(x)) (298)

The equations of motion are easy to obtain, however we need to include a Lagrangemultiplier κ to take in account the first constraint (294).

~D · ~Dϕσ(x) + κϕσ(x) = 0 (299)

By multiplying with ϕ†σ we obtain:

ϕ†σ(x)~D · ~Dϕσ(x) + κϕ†σ(x)ϕσ(x) = 0⇒ ϕ†σ(x)

~D · ~Dϕσ(x) = −κ (300)

Which simplifies our equation of motion by removing the Lagrange multiplier.

~D · ~Dϕσ(x) −(ϕ†τ(x)

~D · ~Dϕτ(x))ϕσ(x) = 0 (301)

We will look for solutions in the case where we work in two spatial dimensions.Furthermore we look for static solutions, using boosts to obtain the time dependence.The most basic solution is the zero energy solution:

ϕσ(~r) = cσeiΛr) (302)

~Dϕσ = ~DcσeiΛ(r)

= (~O+ i~A)cσeiΛ(r)

= ~OcσeiΛ(r) + i~Acσe

iΛ(r)

= icσeiΛ(r)∂Λ(r)

∂r− i(cτe

iΛ(r))†(~OcτeiΛ(r))cσe

iΛ(r)

= icσeiΛ(r)∂Λ(r)

∂r− icσe

iΛ(r)∂Λ(~r)

∂r= 0 (303)

where cσ is a constant and we have used that c†σ · cσ = 1. This solution has zero energyeverywhere. To obtain a non-zero but finite energy solution we need to include anangular dependence to the solution, however the energy contribution of this dependencemust vanish at the boundary, for which the gauge invariance can be used. Applying thegauge transformation ϕσ → ϕσe

−iΛ(θ) will put the energy contribution of the boundaryto zero. Furthermore the gauge invariance of the energy ensures us the boundary willvanish for all solutions of this shape.

limr→∞ϕσ(r, θ) = ϕσ(∞, θ) = bσeiΛ(θ) (304)

The eiΛ(θ) term maps a circle on the boundary, which is also a circle. This can beclassified with the first homotopy group applied to a circle π1[S1] = Z. For each n ∈ Z

46

there exists a solution. Spinors of the form,

ϕσ(z) = K(z)

1

ω2

...ωN+1

= K(z)ωσ (305)

where z = x+iy, K(z) normalizes the spinor andωi are holomorphic functions, are exactsolution of the equations of motion. This is easily shown because the differentiationof the holomorphic will drop out, therefore when we plug this spinor (305) into theequation of motion (301) we obtain:

(~D · ~DK(z))ωσ − ((K(z)ωτ)†(~D · ~DK(z))ωτ)K(z)ωσ

=(~D · ~DK(z))ωσ − ((K(z)ωτ)†K(z)ωτ)(~D · ~DK(z))ωσ

=(~D · ~DK(z))ωσ − (~D · ~DK(z))ωσ = 0 (306)

Which proves that (305) is an exact solution. Furthermore the complex conjugate of(305) can be proven to also be a solution.

A more explicit example of such a spinor is, where n > 0:

ϕσ(z) =1√

a2 +Nr2n

a

zn

zn

...zn

(307)

What is special about this solution is that it clearly shows the mapping of the circle atthe boundary.

ϕσ(z) =einθ√a2

r2n+N

a

rneinθ

1

1...1

(308)

ϕσ(θ)r→∞ =einθ√N

0

1

1...1

(309)

To obtain solutions with negative winding number, we must use the complex conjugate.Although we have constructed an explicit solution for the soliton. Under interactions

the solitons would need to add their winding number, however construction of a twosoliton solution of this model is difficult. A superposition of the initial conditions cannotbe made as this would violate the constraint (294). Therefore we cannot compare thesesolitons with for example the KdV solitons or Q-balls as we can only construct one.

47

11 Conclusion

A wide variety of phenomena are called solitons. In non-linear partial differential equa-tions solitons are solutions in the form of a solitary wave. These solutions can beobtained by the inverse scattering method and the Hirota method, however both meth-ods cannot be applied to every differential equation and no general treatment of allevolution equations for solitons solutions is known. In spatial dimensions higher thanone, Derrick’s theorem restricts which field-equation can have soliton solution. Thereare numerous ways to circumvent Derrick’s theorem, we discussed in this thesis Q-balls,a type of nontopological soliton, and topological conservation laws. Q-balls are stablein time both may decay in other solitons under interaction. Topological solitons canannihilate with opposing winding number and are therefore not stable under all in-teractions. Both of these solitons have different properties than the solitons obtainedin partial differential equations like the KdV equation. A more distinguishing namingscheme may be appropriate as the phenomena named solitons are very diverse.

References

[1] S. Russell, Report on Waves report fourteenth meeting british associa-tion advancement of science; held at york in september 1844 available at:http://www.archive.org/stream/reportofbritisha44brit/reportofbritisha44brit djvu.txt310-390 (1845).

[2] R. Hirota, The direct method in soliton theory Cambridge University Press (2004).

[3] D. A. Russell Acoustics and vibration animations available at:http://www.acs.psu.edu/drussell/Demos/Solitons/solitons.html (2004)

[4] D. J. Korteweg, G. de Vreis, On the Change of Form of Long Waves Advancing ina Rectangular Canal, and on a New Type of Long Stationary Waves PhilosophicalMagazine 39: 422443 (1895).

[5] A. Das, Integrable models World Scientific Publishing Co. (1989).

[6] G. L. Jr. Lamb Elements of soliton theory John Wiley & Sons, Inc (1980).

[7] C. S. Gardner, J. M. Greene, M. D. Kruskal, R. M. Miura method for solving thekorteweg-deVries equation physical review letters, V19 (1967)

[8] T. Miwa, M. Jimbo, E. Date, Solitons: Differential Equations, Symmetries, andinfinite Dimensional Algebras. Cambridge University Press (2000).

[9] P. P. Goldstein, Hints on the Hirota Bilinear Method Acta Physica Polonica A,V112 (2007).

[10] J. Hietarinta Introduction to the Hirota bilinear method eprint arXiv:solv-int/9708006v1 (1997).

48

[11] Matlab version 7.12.0 R2011a The MathWorks Inc. (2011)

[12] D. Rubiera Garca, Relativistic lagrangian non-linear field the-ories supporting non-topological soliton solutions available at:http://www.tesisenred.net/handle/10803/11131 (2008).

[13] G. H. Derrick, Comments on Nonlinear Wave Equations as Models for Elemen-taryParticles J. Math. Phys. 5, 1252 (1964).

[14] G. ’t Hooft Magnetic monopoles in unified gauge theories Nuclear Physics B79276-284 North-Holland Publishing Company (1974)

[15] S.M.H. Wong, What exactly is a skyrmion? eprint: arXiv:hep-ph/0202250v2(2002)

[16] L. Faddeev, Knotted solitons Proc. ICM2002, vol. 1, pp. 235244 (2002)

[17] T. D. Lee, Y. Pang Nontopological Solitons physics reports 221, 251-350 (1992).

[18] M. Axenides, S. Komineas, L. Perivolaropoulos, M. Floratos Dynamics of Non-topological Solitons - Q Balls Phys.Rev. D61 (2000).

[19] S. Coleman Aspects of symmetry Cambridge University Press (1985).

[20] B. Bakker, D. Boer Group Theory in Physics, Part III, Representations of Semi-Simple Lie algebras (2009)

[21] R. Rajaraman, Solitons in quantum Hall systems The European Physical JournalB, Volume 29 (2002)

A Matlab script, transmission coefficient

1 function [ Z ] = trans2 clear a l l3 close a l l4 global k a5

6 for t e l =1:2017 k=10;8 at =.1 ;9 a=at ∗( t e l −1) ;

10

11 t s t a r t = 0 ;12 tend = 60 ;13 n = 1000 ;14 tspan = linspace ( t s t a r t , tend , n) ;

49

15 x i n i t = [ 1 ; 1 i ∗k ] ;16

17 [ t , x ] = ode45 ( @ integ ra t ing funct i on , tspan , x i n i t ) ;18 y = x ( : , 1 ) ;19 B( t e l )=max(abs ( y ) ) ;20 a t e l ( t e l )=a ;21 end22

23 plot ( a t e l ,B, ’b ’ )24 end25

26 function [ dydt ] = i n t e g r a t i n g f u n c t i o n ( t , x )27 global k a28

29 dydt = zeros ( s ize ( x ) ) ;30 dydt (1 ) = x (2) ;31 dydt (2 ) = −(kˆ2+a∗sech ( t−30) ˆ2)∗x (1 ) ;32 end

Remark A.1. for information about ode45 see:http://www.mathworks.nl/help/techdoc/ref/ode45.html

Remark A.2.

Figure 8: A reflectionless potential is obtained for V = 0, 2, 6, 12, 20, 30, ..., n(n+ 1), ...

B Derivation of the Marchenko equation

In this appendix we preform the Fourier transform which leads to the Marchenko equa-tion.

T(k)Yr(x, k) = R(k)Yl(x, k) + Yl(x,−k) (310)

50

We do the Fourier transform:

R.H.S. =

∫∞−∞

dω

2πe−iωt

∫−∞ dze

−iωz/cr(z)

∫∞−∞ dt

′eiωt′y(x, t ′) + y(x,−t) (311)

where r(z) =

∫∞−∞

dk

2πeiktR(k) (312)

R.H.S. =

∫∞−∞ dz

∫∞−∞ dt

′r(z)y(x, t ′)δ(t ′ − t− z/c) + y(x,−t) (313)

= c

∫∞−∞ dt

′r(c(t ′ − t))y(x, t ′) + y(x,−t) (314)

= cr(x− ct) + c2∫∞x/c

dt ′r(c(t ′ − t))K(x, ct ′) + δ(t+ x/c) + cθ(−ct− x)K(x,−ct)

(315)

L.H.S. =

∫∞−∞

dw

2πe−iωtT(k)Yr(x, k) (316)

=

∫∞−∞

dw

2πe−iωtT(ω/c)

∫∞−∞ dt

′eiωt′(δ(t ′ + x/c) + cθ(t ′ + x/c)L(x,−ct ′))

(317)

=

∫∞−∞

dw

2πe−iω(t+x/c)(T(ω/c) − 1+ 1) (318)

+ c

∫∞−∞ dt

′∫∞−∞

dw

2πe−iω(t−t ′)(T(ω/c) − 1+ 1)θ(t ′ + x/c)L(x,−ct ′) (319)

= cΓ(t+ x/c) + δ(t+ x/c) + c2∫∞−x/c

dtΓ(t− t ′)L(x,−ct ′) + c

∫∞−x/c

dtδ(t− t ′)L(x,−ct ′)

(320)

= cΓ(t+ x/c) + δ(t+ x/c) + c2∫∞−x/c

dtΓ(t− t ′)L(x,−ct ′) + cθ(t+ x/c)L(x,−ct)

(321)

where Γ(z) =

∫∞−∞

dk

2πeikt(T(k) − 1) (322)

Because K(a, b) = 0 for a > b (it describes the wake behind the pulse) we consider thecase t + x/c < 0. The integral in Γ(t + x/c) can then be calculated as a contour plot,with finite many poles in the upper half plane, these poles correspond to the boundedstates [6].

Γ(t− x/c) = −∑n

cn(t)2e−κnξ (323)

51

equating the left and right hand side, we obtain the Marchenko equation:

K(x, y; t) + B(x+ y; t) +

∫∞x

K(x, z; t)B(y+ z; t)dz = 0 (324)

where B(ξ; t) =1

2π

∫∞−∞ R(k, t)e

ikξdk+∑n

cn(t)2e−κnξ (325)

52