Simulating Solitons of the Sine-GordonEquation using Variational Approximations

and Hamiltonian Principles

ByEvan Foley

A SENIOR RESEARCH PAPER PRESENTED TO THE DEPARTMENT OF MATHEMATICSAND COMPUTER SCIENCE OF STETSON UNIVERSITY IN PARTIAL FULFILLMENT OF

THE REQUIREMENTS FOR THE DEGREE OF BACHELOR OF SCIENCE

STETSON UNIVERSITY

1

0.1 Acknowledgements

I would like to thank Dr. Vogel for his patience, cooperation, and guidance with the research.

I would like to thank the Stetson Undergraduate Research Experience (SURE) program for

providing me the opportunity to work with Dr. Vogel in an one on one environment over the

summer of 2012.

2

Contents

0.1 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

0.2 Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1 Introduction 5

2 Approximation Method on Korteweg-de Vries Equations 6

2.1 Original . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Modified . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Approximation Method on the Sine-Gordon Equation 9

3.1 Gaussian Wave Trial Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.1 Traveling Wave Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.2 Dynamic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.2 Odd Oriented Gaussian Wave Trial Function . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 Traveling Wave Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.2 Dynamic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Future Research 19

5 References 20

6 Appendices 20

6.1 Mathematica Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6.2 List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3

0.2 Abstract

Simulating Solitons of the Sine-Gordon Equation using Variational Approximations andHamiltonian Principles

ByEvan FoleyMay 2013

Advisor: Dr. Thomas VogelDepartment: Mathematics and Computer Science

This project examines the underlying phenomena of solitons in partial differential equations

using Hamilton’s Principal and Variational Approximations. William Hamilton formulated the idea

that physical models can be described in terms of energies. Using the concept of functionals, we

can describe these energies in the form of the Lagrangian to find approximate solutions. Equations

such as the Korteweg-de Vries (KdV), Modified KdV, and the Sine-Gordon will be examined to

find localized structure (i.e. soliton) solutions. Interestingly enough, soltions behave differently

than typical waves, due to their unique characteristics which will be addressed later. The solutions

to these equations shall yield two results, prove the existence of solitons in the solution and provide

an approximate solution for the PDEs that are being analyzed. These solutions will be compared

to the exact solutions, to determine how accurate the approximation works on those particular

equations.

4

1 Introduction

In 1834, a Scottish naval engineer, named John Scott Russell, was conducting an experiment at

the Union Canal in Scotland. He tried to establish a conversion factor between steampower and

horsepower by tying two horses on opposite sides of the canal to a steamboat, while he observed

on a horse on one side. While the horses were pulling the stationary steamboat, the ropes snapped

and something unexpected happened. The boat was shaken around and a wave popped out in

front of the boat, but the wave was not an ordinary wave. Russell described the wave as ”a

rounded, smooth, and well-defined heap of water.” He followed it for a couple of miles until the

canal prevented him from moving onward. Russell tried to recreate the wave that he saw but had

not been successful. Around that time, Newton and Bernoulli had formulated linear equations for

explaining hydrodynamics, so Russells observation was rejected from physicists. By 1895, Diederik

Korteweg and Gustav de Vries had derived a non-linear, partial differential equation, called the

KdV equation, that would model the behavior of what will be known as a soliton wave observed

by John Scott Russell [1].

A soliton can simply be described as a solitary wave with a localized structure that does not

diminish over time. The soliton wave is also known as the ’Wave of Translation’. A soliton wave is

said to have 4 distinct properties:

1. Stable and travels over long distances.

2. Speed depends on height and width of wave on the depth of water.

3. Multiple waves never merge, rather, they pass through each other.

4. If the wave is too big for depth of water, it will split into two.

William Hamilton developed the concept of energies describing physical systems and formalized

Hamilton’s Principle. Hamilton’s Principle can be explained as follows: Of all possible paths along

which a dynamical system may evolve within a specified time interval, the actual path followed is

that which minimizes the time integral of the difference between the kinetic and potential energies

which is called the action, S.

S =

∫ t2

t1

(T − U)dt (1)

5

where

L = T − U (2)

where L=f[x,y(x),y’(x)] and is called the Lagrangian. In our case, we will integrate the action over

R in the spatial domain.

To minimize the action, we use the formula:

δS

δy= Sy −

d

dxSy = 0 (3)

where y is a parameter of the action [1]. One thing to note is that the action is not considered

a function, but a functional. In Variational Calculus, a functional is an operator that outputs

numbers where the input is a function. For instance, a definite integral is considered a functional.

Keeping the action in mind, there is a necessary condition that the lagrangian must satisfy called

the Euler-Lagrange equation [2, p. 125]:

Ly −d

dxLy′ = 0 (4)

where y=f(x). By applying the correct Lagrangian to this condition, the original partial differnetial

equation should arise.

2 Approximation Method on Korteweg-de Vries Equations

2.1 Original

The first equation that will be analyzed is the Korteweg-de Vries (KdV) Equation which is as

follows [3]:

ut + uux + uxxx = 0 (5)

This equation was originally designed to model the motion of a wave in shallow water.

To start off, assume the soliton is in a traveling wave solution, meaning the only parameter

6

changing with respect with to time is it’s position which can be described as

ξ = x− ct (6)

To start the approximation, the trial function needs to be defined. Based on the geometry of a

soliton wave, the gaussian wave will be used as the trial function:

u(ξ) = Ae− ξ2

ρ2 (7)

Applying the necessary condition (3), the following Lagrangian can be found:

L = −1

2cu2 +

1

6u3 − 1

2(u′)2 (8)

The resulting action is:

S =A2√π(2

√3Aρ2 − 9

√2(cρ2 + 1))

36ρ(9)

When minimizing the action, two Euler Lagrange equations show up since there are two parameters

that we are interested in.

∂S

∂ρ=

A2√π(2√3Aρ2 − 9

√2(cρ2 − 1))

36ρ2= 0 (10)

∂S

∂A=

A√π(√3Aρ2 − 3

√2(cρ2 + 1))

6ρ= 0 (11)

Figure 1 shows the amplitude (red) and speed (blue) of the wave vary with respect to the wave

width. Since the exact solution for the KdV equation is known, namely

u(ξ) = 3c sech2(

√cξ

2) (12)

the approximations (blue) can be compared to the exact (red) solution in Figures 2 and 3. Looking

at the graphs, the solutions obtained from the appoximation method very closely resemble the

exact solutions, this shows how accurate this method is.

7

2.2 Modified

Aside from the regular KdV equation, many modified versions have been created and tested, how-

ever, this research paper will only examine one [3]:

ut + 6u2ux + uxxx = 0 (13)

There will still be a the assumption that the soliton is in a stationary state from earlier, with the

same trial function:

u(ξ) = Ae− ξ2

ρ2 (14)

Applying the necessary condition (3), and the following Lagrangian is obtained

L = −1

2cu2 +

1

2u4 − 1

2(u′)2 (15)

The action becomes

S =A2√π(A2ρ2 −

√2(cρ2 + 1))

4ρ(16)

The two Euler Lagrange equations that spawn from minimizing the action are:

∂S

∂ρ=

A2√π(A2ρ2 +√2(1− cρ2))

4ρ2= 0 (17)

∂S

∂A=

A√π(2A2ρ2 −

√2(cρ2 + 1))

2ρ= 0 (18)

Figure 4 displays Amplitude (red) and speed (blue) vs. width of the soliton wave. Since the exact

solution to this Modified KdV Equation is known, namely

u(ξ) =√c sech

√cξ (19)

then a comparison is in order to determine how accurate the approximations (blue) are compared

to exact (red) in Figures 5 and 6. Based on the graphs, the method used is producing very accurate

results.

8

3 Approximation Method on the Sine-Gordon Equation

Now that the method is known to produce reasonable results, it is time to finally test the method

on the Sine-Gordon equation. Since there are multiple solutions for the Sine-Gordon Equation, we

will examine a few of them. We will also use different trial functions since some of the solutions

resemble the trial functions. The Sine Gordon equation is

utt − uxx + sinu = 0 (20)

3.1 Gaussian Wave Trial Function

The first trial funtion we will use is the same as the trial function used in the Korteweg-de Vries

equations.

3.1.1 Traveling Wave Solution

We first examine the Traveling Wave Solution (TWS), meaning the wave just moves along the

x-axis and does not change shape or form. Recall the trial function being:

u(ξ) = Ae−ξ2

ρ2 (21)

By applying the necessary condition (3) as well as the power series representation of sin(x), we get

the following Lagrangian:

L =1

2

(1− c2

)(u′)2 +

u2

2!− u4

4!+

u6

6!− u8

8!(22)

This results in the following action:

S =

√πA2

((−3

√2A6 + 112

√6A4 − 10080A2 + 120960

√2)ρ2 − 120960

√2c2 + 120960

√2)

483840ρ(23)

9

The two following Euler-Lagrangian equations spawn when the action is minimized:

(24)

∂S

∂A=

√πA

((−3

√2A6 + 112

√6A4 − 10080A2 + 120960

√2)ρ2 − 120960

√2c2 + 120960

√2)

241920ρ

+

√π(−18

√2A5 + 448

√6A3 − 20160A

)A2ρ

483840

∂S

∂ρ=

√πA2

(−3

√2A6 + 112

√6A4 − 10080A2 + 120960

√2)

241920

−√πA2

((−3

√2A6 + 112

√6A4 − 10080A2 + 120960

√2)ρ2 − 120960

√2c2 + 120960

√2)

483840ρ2

(25)

Using numerics, the amplitude and the wave width can be approximated and solved for at varying

speeds, which is shown in Figure 7.

3.1.2 Dynamic Solution

Previously, the Sine-Gordon was examined and analyzed for a traveling wave solution, however,

this assumes the wave’s shape and speed do not change over time. It is more accurate to allow for

”breather” soliton solutions to form. This is accomplished by allowing the wave’s amplitude and

width to vary with respect to time. Leaving behind the trial function:

u(x, t) = A(t)e−x2

ρ(t)2 (26)

By taking this into effect, there is no transformation we can use that all of the previous examples

used. Since the equation is not going to be transformed into an ODE, a new necessary condition

for the lagrangian needs to be met. The new Euler-Lagrange equation becomes [2, p. 139]

Lu − ∂

∂xLux − ∂

∂tLut = 0 (27)

Using the new necessary condition, the following Lagrangian can be obtained:

L =−1

2u2t +

1

2u2x +

u2

2!− u4

4!+

u6

6!− u8

8!(28)

10

This results in the following action

S = −√π

483840ρ(t)(120960

√2A(t)ρ(t)A′(t)ρ′(t) + 120960

√2ρ(t)2A′(t)2

−30240√2A(t)2(−3ρ′(t)2+4ρ(t)2+4)+3

√2A(t)8ρ(t)2−112

√6A(t)6ρ(t)2+10080A(t)4ρ(t)2)

(29)

with the two Euler-Lagrangian equations

SA(t) −d

dtSA(t) = −

√πρ′(t)

483840ρ(t)2(120960

√2ρ(t)A′(t)ρ′(t)− 60480

√2A(t)(−3ρ′(t)2 + 4ρ(t)2 + 4)

+ 24√2A(t)7ρ(t)2 − 672

√6A(t)5ρ(t)2 + 40320A(t)3ρ(t)2)

−√π

483840ρ(t)(120960

√2ρ(t)A′(t)ρ′(t)− 60480

√2A(t)(−3ρ′(t)2 + 4ρ(t)2 + 4)

+ 24√2A(t)7ρ(t)2 − 672

√6A(t)5ρ(t)2 + 40320A(t)3ρ(t)2)

+

√π

483840ρ(t)(120960

√2ρ(t)A′′(t)ρ′(t) + 120960

√2ρ(t)A′(t)ρ′′(t)

+ 120960√2A′(t)ρ′(t)2 − 60480

√2A′(t)(−3ρ′(t)2 + 4ρ(t)2 + 4)

+ 168√2A(t)6ρ(t)2A′(t)− 3360

√6A(t)4ρ(t)2A′(t) + 120960A(t)2ρ(t)2A′(t)

+ 48√2A(t)7ρ(t)ρ′(t)− 1344

√6A(t)5ρ(t)ρ′(t) + 80640A(t)3ρ(t)ρ′(t)

− 60480√2A(t)(8ρ(t)ρ′(t)− 6ρ′(t)ρ′′(t)))

(30)

11

Sρ(t) −d

dtSρ(t) = − 1

483840ρ(t)2√πρ′(t)(6

√2ρ(t)A(t)8 − 224

√6ρ(t)A(t)6 + 20160ρ(t)A(t)4

− 241920√2ρ(t)A(t)2 + 120960

√2A′(t)ρ′(t)A(t) + 241920

√2ρ(t)A′(t)2)

−√π

483840ρ(t)(6√2ρ(t)A(t)8 − 224

√6ρ(t)A(t)6 + 20160ρ(t)A(t)4

− 241920√2ρ(t)A(t)2 + 120960

√2A′(t)ρ′(t)A(t) + 241920

√2ρ(t)A′(t)2)

+

√πρ′(t)

241920ρ(t)3(3√2ρ(t)2A(t)8 − 112

√6ρ(t)2A(t)6 + 10080ρ(t)2A(t)4

− 30240√2(4ρ(t)2 − 3ρ′(t)2 + 4)A(t)2 + 120960

√2ρ(t)A′(t)ρ′(t)A(t)

+ 120960√2ρ(t)2A′(t)2) +

√π

483840ρ(t)2(3√2ρ(t)2A(t)8 − 112

√6ρ(t)2A(t)6

+ 10080ρ(t)2A(t)4 − 30240√2(4ρ(t)2 − 3ρ′(t)2 + 4)A(t)2

+ 120960√2ρ(t)A′(t)ρ′(t)A(t) + 120960

√2ρ(t)2A′(t)2)

+

√π

483840ρ(t)(6√2ρ′(t)A(t)8 + 48

√2ρ(t)A′(t)A(t)7 − 224

√6ρ′(t)A(t)6

− 1344√6ρ(t)A′(t)A(t)5 + 20160ρ′(t)A(t)4 + 80640ρ(t)A′(t)A(t)3

− 241920√2ρ′(t)A(t)2 − 483840

√2ρ(t)A′(t)A(t) + 120960

√2ρ′(t)A′′(t)A(t)

+ 120960√2A′(t)ρ′′(t)A(t) + 362880

√2A′(t)2ρ′(t) + 483840

√2ρ(t)A′(t)A′′(t))

−√π

483840ρ(t)2(6√2ρ(t)ρ′(t)A(t)8 + 24

√2ρ(t)2A′(t)A(t)7 − 224

√6ρ(t)ρ′(t)A(t)6

− 672√6ρ(t)2A′(t)A(t)5 + 20160ρ(t)ρ′(t)A(t)4 + 40320ρ(t)2A′(t)A(t)3

− 30240√2(8ρ(t)ρ′(t)− 6ρ′(t)ρ′′(t))A(t)2 + 120960

√2A′(t)ρ′(t)2A(t)

− 60480√2A′(t)(4ρ(t)2 − 3ρ′(t)2 + 4)A(t) + 120960

√2ρ(t)ρ′(t)A′′(t)A(t)

+ 120960√2ρ(t)A′(t)ρ′′(t)A(t) + 362880

√2ρ(t)A′(t)2ρ′(t)

+ 241920√2ρ(t)2A′(t)A′′(t))

(31)

In order to numerically solve for the amplitude and width of the wave using the two ODEs that

were just calculated, four initial conditions must be known. It is somewhat displeasing to just plug

in random numbers and see where the solution is heading towards. It is more beneficial to use

initial conditions that had some sort of relevance to one each other. This can be accomplished by

taking an ordered triplet of values from the TWS section. By taking a particular speed, amplitude,

and width trio, we can use these values as initial conditions. However, we still need two more

conditions.

In general, it can be observed that a wave’s amplitude tends to change more often and rapidly

than it’s width. With this idea, we can assume the initial change in the wave width is roughly zero.

With that in mind, now we need one more condition. We would like the last initial condition to be

12

how fast the wave’s amplitude is changing initially. This can be acquired by observation and some

simple algebra. Let us examine the derivatives of u.

ut(x, t) = A′(t)e−x2

ρ(t)2 +2x2e

−x2

ρ2 A(t)ρ′(t)

ρ(t)3(32)

ux(x, t) = −2xe−x2

ρ(t)2 A(t)

ρ(t)2(33)

Thus,

ut(x, t) = A′(t)e−x2

ρ(t)2 − xρ′(t)

ρ(t)ux(x, t) (34)

Observe that

x = 0 (35)

⇒ ux(x, t) = 0 (36)

ut(x, 0) ≈ c (37)

(37) holds true since ut represents how fast the wave is moving with respect to time, which is the

same as the wave speed.

Plugging in x = t = 0 into (34) yields

A′(0) = c (38)

The final four intital conditions to the problem are

A(0) = A (39)

ρ(0) = ρ (40)

A′(0) = c (41)

ρ′(0) = 0 (42)

where A, ρ, c are chosen values from the TWS trial earlier. With these values, A(t) and ρ(t) can

be numerically calculated. Unfortunately, according to the calculations, the wave immediately dies

13

off after a fraction of a second. Possible solutions to this problem will be addressed later.

3.2 Odd Oriented Gaussian Wave Trial Function

There is a set of solutions that do not resemble the trial that was previously being used. This other

trial function can be thought of as a gaussian wave oriented as an odd function, hence the title.

3.2.1 Traveling Wave Solution

With a slight modification, and the Lorentzian transformation, the new trial function that will be

used is

u(ξ) = −Aξe−ξ2

ρ2 (43)

Once again we have the Lagrangian

L =1

2(1− c2)(u′)2 +

u2

2!− u4

4!+

u6

6!− u8

8!(44)

Hence the action is

S =A2√πρ(1528823808

√2(1− c2) + 509607936

√2ρ2 − 7962624A2ρ4 + 16384

√6A4ρ6 − 81

√2A6ρ8)

8153726976

(45)

The two Euler Lagrange equations become

(46)∂S

∂A=

A2√πρ(−15925248Aρ4 + 65536√6A3ρ6 − 486

√2A5ρ8)

8153726976

+A√πρ(1528823808

√2(1− c2) + 509607936

√2ρ2 − 7962624A2ρ4 + 16384

√6A4ρ6 − 81

√2A6ρ8)

4076863488

∂S

∂ρ=

A2√πρ(1019215872√2ρ− 31850496A2ρ2 + 98304

√6A4ρ5 − 648

√2A6ρ7)

8153726976

+A2√π(1528823808

√2(1− c2) + 509607936

√2ρ2 − 7962624A2ρ4 + 16384

√6A4ρ5 − 81

√2A6ρ8)

8153726976(47)

Once again, letting the wave speed vary, we can solve for the amplitude and width, represented in

Figure 8.

14

3.2.2 Dynamic Solution

Earlier, the Sine-Gordon was examined and analyzed for a traveling wave solution, however, this

assumes that the wave’s shape and speed do not change over time, only position would change. It

is better to include the extra possibilities such as a change in amplitude, width, and speed. I.e.:

u(x, t) = −A(t)xe−(x−ζ(t))2

ρ(t)2 (48)

The Lagrangian for the dynamically changing Sine-Gordon Wave is

L =−1

2u2t +

1

2u2x +

u2

2!− u4

4!+

u6

6!− u8

8!(49)

The following action can be calculated

S = −√π

285380444160ρ(t)(17836277760

√2A(t)ρ(t)A′(t)(8ζ(t)ρ(t)ζ ′(t) + 4ζ(t)2ρ′(t) + 3ρ(t)2ρ′(t))

+ 17836277760√2ρ(t)2A′(t)2(4ζ(t)2 + ρ(t)2)

− 4459069440√2A(t)2(4ζ(t)2(−4ζ ′(t)2 − 3ρ′(t)2 + 4ρ(t)2 + 4)− 48ζ(t)ρ(t)ζ ′(t)ρ′(t)

+ ρ(t)2(−12ζ ′(t)2 − 15ρ′(t)2 + 4ρ(t)2 + 12))

+27√2A(t)8ρ(t)2(114688ζ(t)6ρ(t)2+53760ζ(t)4ρ(t)4+6720ζ(t)2ρ(t)6+65536ζ(t)8+105ρ(t)8)

− 114688√6A(t)6ρ(t)2(720ζ(t)4ρ(t)2 + 180ζ(t)2ρ(t)4 + 576ζ(t)6 + 5ρ(t)6)

+ 92897280A(t)4(48ζ(t)2ρ(t)4 + 64ζ(t)4ρ(t)2 + 3ρ(t)6))

(50)

15

which leaves the three Euler-Lagrangian equations:

SA(t) −d

dtSA(t)

= −√πρ′(t)

285380444160ρ(t)2(216

√2ρ(t)2(65536ζ(t)8 + 114688ρ(t)2ζ(t)6 + 53760ρ(t)4ζ(t)4

+ 6720ρ(t)6ζ(t)2 + 105ρ(t)8)A(t)7

− 688128√6ρ(t)2(576ζ(t)6 + 720ρ(t)2ζ(t)4 + 180ρ(t)4ζ(t)2 + 5ρ(t)6)A(t)5

+ 371589120(3ρ(t)6 + 48ζ(t)2ρ(t)4 + 64ζ(t)4ρ(t)2)A(t)3

− 8918138880√2(4(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)ζ(t)2 − 48ρ(t)ζ ′(t)ρ′(t)ζ(t)

+ ρ(t)2(4ρ(t)2 − 12ζ ′(t)2 − 15ρ′(t)2 + 12))A(t)

+ 17836277760√2ρ(t)A′(t)(4ρ′(t)ζ(t)2 + 8ρ(t)ζ ′(t)ζ(t) + 3ρ(t)2ρ′(t)))

−√π

285380444160ρ(t)(216

√2ρ(t)2(65536ζ(t)8 + 114688ρ(t)2ζ(t)6 + 53760ρ(t)4ζ(t)4

+ 6720ρ(t)6ζ(t)2 + 105ρ(t)8)A(t)7

− 688128√6ρ(t)2(576ζ(t)6 + 720ρ(t)2ζ(t)4 + 180ρ(t)4ζ(t)2 + 5ρ(t)6)A(t)5

+ 371589120(3ρ(t)6 + 48ζ(t)2ρ(t)4 + 64ζ(t)4ρ(t)2)A(t)3

− 8918138880√2(4(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)ζ(t)2 − 48ρ(t)ζ ′(t)ρ′(t)ζ(t)

+ ρ(t)2(4ρ(t)2 − 12ζ ′(t)2 − 15ρ′(t)2 + 12))A(t)

+ 17836277760√2ρ(t)A′(t)(4ρ′(t)ζ(t)2 + 8ρ(t)ζ ′(t)ζ(t) + 3ρ(t)2ρ′(t)))

+

√π

285380444160ρ(t)(432

√2ρ(t)(65536ζ(t)8 + 114688ρ(t)2ζ(t)6 + 53760ρ(t)4ζ(t)4

+ 6720ρ(t)6ζ(t)2 + 105ρ(t)8)ρ′(t)A(t)7 + 216√2ρ(t)2(524288ζ ′(t)ζ(t)7 + 229376ρ(t)ρ′(t)ζ(t)6

+ 688128ρ(t)2ζ ′(t)ζ(t)5 + 215040ρ(t)3ρ′(t)ζ(t)4 + 215040ρ(t)4ζ ′(t)ζ(t)3 + 40320ρ(t)5ρ′(t)ζ(t)2

+ 13440ρ(t)6ζ ′(t)ζ(t) + 840ρ(t)7ρ′(t))A(t)7 + 1512√2ρ(t)2(65536ζ(t)8 + 114688ρ(t)2ζ(t)6

+ 53760ρ(t)4ζ(t)4 + 6720ρ(t)6ζ(t)2 + 105ρ(t)8)A′(t)A(t)6

− 1376256√6ρ(t)(576ζ(t)6 + 720ρ(t)2ζ(t)4 + 180ρ(t)4ζ(t)2 + 5ρ(t)6)ρ′(t)A(t)5

− 688128√6ρ(t)2(3456ζ ′(t)ζ(t)5 + 1440ρ(t)ρ′(t)ζ(t)4 + 2880ρ(t)2ζ ′(t)ζ(t)3

+ 720ρ(t)3ρ′(t)ζ(t)2 + 360ρ(t)4ζ ′(t)ζ(t) + 30ρ(t)5ρ′(t))A(t)5 − 3440640√6ρ(t)2(576ζ(t)6

+ 720ρ(t)2ζ(t)4 + 180ρ(t)4ζ(t)2 + 5ρ(t)6)A′(t)A(t)4 + 371589120(18ρ′(t)ρ(t)5

+ 96ζ(t)ζ ′(t)ρ(t)4 + 192ζ(t)2ρ′(t)ρ(t)3 + 256ζ(t)3ζ ′(t)ρ(t)2 + 128ζ(t)4ρ′(t)ρ(t))A(t)3

+ 1114767360(3ρ(t)6 + 48ζ(t)2ρ(t)4 + 64ζ(t)4ρ(t)2)A′(t)A(t)2 − 8918138880√2(4(8ρ(t)ρ′(t)

− 6ρ′′(t)ρ′(t)− 8ζ ′(t)ζ ′′(t))ζ(t)2 − 48ζ ′(t)ρ′(t)2ζ(t) + 8ζ ′(t)(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)ζ(t)

− 48ρ(t)ρ′(t)ζ ′′(t)ζ(t)− 48ρ(t)ζ ′(t)ρ′′(t)ζ(t)− 48ρ(t)ζ ′(t)2ρ′(t) + 2ρ(t)ρ′(t)(4ρ(t)2 − 12ζ ′(t)2

− 15ρ′(t)2 + 12) + ρ(t)2(8ρ(t)ρ′(t)− 30ρ′′(t)ρ′(t)− 24ζ ′(t)ζ ′′(t)))A(t)

+ 17836277760√2A′(t)ρ′(t)(4ρ′(t)ζ(t)2 + 8ρ(t)ζ ′(t)ζ(t) + 3ρ(t)2ρ′(t))

− 8918138880√2A′(t)(4(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)ζ(t)2 − 48ρ(t)ζ ′(t)ρ′(t)ζ(t)

+ ρ(t)2(4ρ(t)2 − 12ζ ′(t)2 − 15ρ′(t)2 + 12)) + 17836277760√2ρ(t)(4ρ′(t)ζ(t)2 + 8ρ(t)ζ ′(t)ζ(t)

+ 3ρ(t)2ρ′(t))A′′(t) + 17836277760√2ρ(t)A′(t)(4ρ′′(t)ζ(t)2 + 16ζ ′(t)ρ′(t)ζ(t) + 8ρ(t)ζ ′′(t)ζ(t)

+ 8ρ(t)ζ ′(t)2 + 6ρ(t)ρ′(t)2 + 3ρ(t)2ρ′′(t)))

(51)

16

Sρ(t) −d

dtSρ(t) = Refer to Appendix 1 for Mathematica Code (52)

Sζ(t) −d

dtSζ(t)

= −√πρ′(t)

285380444160ρ(t)2(27

√2ρ(t)2(524288ζ(t)7 + 688128ρ(t)2ζ(t)5 + 215040ρ(t)4ζ(t)3

+ 13440ρ(t)6ζ(t))A(t)8 − 114688√6ρ(t)2(3456ζ(t)5 + 2880ρ(t)2ζ(t)3 + 360ρ(t)4ζ(t))A(t)6

+ 92897280(96ζ(t)ρ(t)4 + 256ζ(t)3ρ(t)2)A(t)4

− 4459069440√2(8ζ(t)(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)− 48ρ(t)ζ ′(t)ρ′(t))A(t)2

+ 17836277760√2ρ(t)A′(t)(8ρ(t)ζ ′(t) + 8ζ(t)ρ′(t))A(t) + 142690222080

√2ζ(t)ρ(t)2A′(t)2)

−√π

285380444160ρ(t)(27

√2ρ(t)2(524288ζ(t)7 + 688128ρ(t)2ζ(t)5 + 215040ρ(t)4ζ(t)3

+ 13440ρ(t)6ζ(t))A(t)8 − 114688√6ρ(t)2(3456ζ(t)5 + 2880ρ(t)2ζ(t)3 + 360ρ(t)4ζ(t))A(t)6

+ 92897280(96ζ(t)ρ(t)4 + 256ζ(t)3ρ(t)2)A(t)4

− 4459069440√2(8ζ(t)(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)− 48ρ(t)ζ ′(t)ρ′(t))A(t)2

+ 17836277760√2ρ(t)A′(t)(8ρ(t)ζ ′(t) + 8ζ(t)ρ′(t))A(t) + 142690222080

√2ζ(t)ρ(t)2A′(t)2)

+

√π

285380444160ρ(t)(54

√2ρ(t)(524288ζ(t)7 + 688128ρ(t)2ζ(t)5 + 215040ρ(t)4ζ(t)3

+ 13440ρ(t)6ζ(t))ρ′(t)A(t)8

+ 27√2ρ(t)2(3670016ζ ′(t)ζ(t)6 + 1376256ρ(t)ρ′(t)ζ(t)5 + 3440640ρ(t)2ζ ′(t)ζ(t)4

+ 860160ρ(t)3ρ′(t)ζ(t)3 + 645120ρ(t)4ζ ′(t)ζ(t)2 + 80640ρ(t)5ρ′(t)ζ(t) + 13440ρ(t)6ζ ′(t))A(t)8

+ 216√2ρ(t)2(524288ζ(t)7 + 688128ρ(t)2ζ(t)5 + 215040ρ(t)4ζ(t)3 + 13440ρ(t)6ζ(t))A′(t)A(t)7

− 229376√6ρ(t)(3456ζ(t)5 + 2880ρ(t)2ζ(t)3 + 360ρ(t)4ζ(t))ρ′(t)A(t)6

− 114688√6ρ(t)2(17280ζ ′(t)ζ(t)4+5760ρ(t)ρ′(t)ζ(t)3+8640ρ(t)2ζ ′(t)ζ(t)2+1440ρ(t)3ρ′(t)ζ(t)

+ 360ρ(t)4ζ ′(t))A(t)6 − 688128√6ρ(t)2(3456ζ(t)5 + 2880ρ(t)2ζ(t)3 + 360ρ(t)4ζ(t))A′(t)A(t)5

+ 92897280(96ζ ′(t)ρ(t)4 + 384ζ(t)ρ′(t)ρ(t)3 + 768ζ(t)2ζ ′(t)ρ(t)2 + 512ζ(t)3ρ′(t)ρ(t))A(t)4

+ 371589120(96ζ(t)ρ(t)4 + 256ζ(t)3ρ(t)2)A′(t)A(t)3

− 4459069440√2(−48ζ ′(t)ρ′(t)2 − 48ρ(t)ζ ′′(t)ρ′(t) + 8ζ ′(t)(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)

− 48ρ(t)ζ ′(t)ρ′′(t) + 8ζ(t)(8ρ(t)ρ′(t)− 6ρ′′(t)ρ′(t)− 8ζ ′(t)ζ ′′(t)))A(t)2

+ 17836277760√2A′(t)ρ′(t)(8ρ(t)ζ ′(t) + 8ζ(t)ρ′(t))A(t)

− 8918138880√2A′(t)(8ζ(t)(4ρ(t)2 − 4ζ ′(t)2 − 3ρ′(t)2 + 4)− 48ρ(t)ζ ′(t)ρ′(t))A(t)

+ 17836277760√2ρ(t)(8ρ(t)ζ ′(t) + 8ζ(t)ρ′(t))A′′(t)A(t)

+ 17836277760√2ρ(t)A′(t)(16ζ ′(t)ρ′(t) + 8ρ(t)ζ ′′(t) + 8ζ(t)ρ′′(t))A(t)

+ 142690222080√2ρ(t)2A′(t)2ζ ′(t) + 285380444160

√2ζ(t)ρ(t)A′(t)2ρ′(t)

+ 17836277760√2ρ(t)A′(t)2(8ρ(t)ζ ′(t) + 8ζ(t)ρ′(t)) + 285380444160

√2ζ(t)ρ(t)2A′(t)A′′(t))

(53)

In order to solve this system of ODEs, there needs to be initial conditions, which can be obtained

by examining the Traveling Wave Solution, similar to the Gaussian Wave Trial case. We need to

17

choose a trio of values for the amplitude and width relative to speed to use as the initial conditions.

However, this only satisifies three out of the six initial conditions necessary. The third condition

that is met is the ζ’(0) term which can be considered as the initial speed of the wave, in other

words

ζ ′(0) ≈ ut(x, 0) ≈ c (54)

Just like in the Gaussian trial case, we will assume ρ’(0) = 0 since the wave’s width does not change

much. For the ζ(0) case, we need to be careful about what value to start it at. It may not be

obvious, but choosing ζ(0) = 0 is dangerous because undefined answers appear during the solving

process, so we need to choose another value. It’s value must be something relatable to the other

conditions in some way, so we will choose a value to allow for simplicity in some sort of calculation.

Suppose

ζ(0) =ρ(0)√

2(55)

the reason behind this assumption may become apparent in the upcoming calculations. Using the

same argument and methods as in the Gaussian Trial case, meaning we solve for A’(0) by writing

ut in terms of ux, this leads to

A′(0) =−√2c

ρ(0)(1 +A(0)) (56)

This gives six initial conditions to use

A(0) = A (57)

A′(0) =−√2c

ρ(1 +A) (58)

ρ(0) = ρ (59)

ρ′(0) = 0 (60)

ζ(0) =ρ√2

(61)

ζ ′(0) = c (62)

By numerically solving the 3 Ordinary Differential Equations with the initial conditions above, we

come to the conclusion that the soliton once again dies quickly. A solution to the problem will be

addressed later.

18

4 Future Research

Negative amplitudes will be accounted for due to loss of generality. More values of speed will be

examined in the Traveling Wave Solution case to see if there are other solutions besides what is in

the interval between 0 and 1. There still needs to be a comparison to an exact solution to see how

accurate the results really are. There is the idea of examining multi-soliton solutions, once this

single soliton case is completed, however that may put off for future research in graduate school

and such.

To address the issues during the Dynamic solution cases, it seems the initial values are not

forming a practical system, it appears the system is stiff. More values of A, ρ, c, and the initial

conditions need to be tested in the system until a viable solution is obtained.

19

5 References

References

[1] D.J. Kaup, T.K. Vogel, Quantitative Measurement of Variational Approximations, Physics

Letters A PLA 362 (2007) 289-297

[2] J. David Logan Applied Mathematics. 2nd ed.

A Wiley-Interscience Publication. 1996.

[3] P.G. Drazin, R.S. Johnson Solitons: an introduction

Cambridge University Press. 1989.

6 Appendices

6.1 Mathematica Code

U[x,t]:=-A[t]*x*Eˆ((-(x-ζ[t])ˆ2)((ρ[t])ˆ2))

Ux=D[U[x,t],x]

Ut=D[U[x,t],t]

Lp=-12*(Ut)ˆ2+12*(Ux)ˆ2+1

2*(U[x,t])ˆ2-14!((U[x,t])ˆ4)+ 1

6!((U[x,t])ˆ6)- 18!((U[x,t])ˆ8)

S=Assuming[ρ[t]>0,Integrate[Lp,x,-Infinity,Infinity]]

el1=D[S,A[t]]-D[D[S,A[t]],t]

el2=D[S,[t]]-D[D[S,ρ[t]],t]

el3=D[S,[t]]-D[D[S,ζ[t]],t]

NDSolve[(el1==0,el2==0,el3==0,A[0]==3.485016648986174,ρ[0]==3.000380941991318,

ζ[0]==3.000380941991318(Sqrt[2]),ζ’[0]==”.5”,ρ’[0]==0,A’[0]==-1.2098800005826753),

(A[t],ρ[t],ζ[t]),(t,0,10)]

6.2 List of Figures

20

0 1 2 3 4 5

20

40

60

80

100

Figure 1: A plot of the wave’s amplitude and width vs. it’s speed.

100 200 300 400 500

200

400

600

800

1000

1200

1400

Figure 2: A plot of the wave’s approximate and exact amplitude vs. it’s speed.

21

0 10 20 30 40 50

1

2

3

4

5

Figure 3: A plot of the wave’s approximate and exact width vs. it’s speed.

0 1 2 3 4 5

20

40

60

80

100

Figure 4: A plot of the wave’s amplitude and speed vs. it’s width.

22

100 200 300 400 500

5

10

15

20

Figure 5: A plot of the wave’s approximate and exact amplitude vs. it’s speed.

0 10 20 30 40 50

1

2

3

4

5

Figure 6: A plot of the wave’s approximate and exact width vs. it’s speed.

23

0.0 0.2 0.4 0.6 0.8 1.00

5

10

15

20

Figure 7: The wave’s amplitude and width paired with varying values of speed.

0.0 0.2 0.4 0.6 0.8 1.00

5

10

15

20

Figure 8: The wave’s amplitude and width paired with varying values of speed.

24