Collision of wave packets in the sine-Gordon equation

(1)

Geerten Koers

Collision of wave packets in the sine-Gordon equation

Bachelor’s thesis

Supervisor: Dr. M. Chirilus-Bruckner

Date Bachelor Exam: July 19, 2016

Mathematical Institute, Leiden University

(2)

(3)

This equation arises in numerous physical systems. A classic example is given by classical mechanics [9]. We consider an infinitely long spring, with pendula attached to it at even intervals. These pendula can pivot around their attachment point with the spring. Let u(x, t) be the angle that a pendulum at position x makes with the equilibrium position, at time t. When a pendulum moves, it will twist the spring. The spring will then exert a force on the neighbouring pendula, which in turn will start to move. If we repeatedly take smaller intervals and lighter pendula, the limit case will be described by the sG equation. It also arises in quantum mechanics, astronomy and various other physical systems [5]. This thesis is motivated by an application in the field of photonics, namely the understanding of the collision of two pulses in the sG equation[3, 10].

Motivated by these applications, we look for travelling wave solutions for the sG equation.

It allows for three different kind of solutions, namely solitons, periodic solutions and unbounded solutions. The solitons, or kinks, are the most important ones.

For linear differential equations, every linear combination of two solutions is again a solution, which is the so called superposition principle. The sG equation is non linear, so we cannot apply this principle to construct new solutions. However, it is still possible to use the earlier found solutions to construct new solutions. We do this by using a Bäcklund transformation, which transforms a solution of equation (1.1) into another solution of equation (1.1). This transformation decouples the sG equation into a system of first order differential equations. By solving this system, you get a new solution for the sG equation. For general cases, these differential equations are hard to solve. Surprisingly, applying multiple Bäcklund transformations is vastly easier than performing one. This is due to Bianchi’s Commutativity Theorem. This theorem gives us an expression for the Bäcklund transformation of a Bäcklund transformation. Hence there is no need to solve differential equations directly the second time. This solution can be expressed in terms of the previously found solutions. This means that once you

(5)

can do one B¨acklund transformation, you can do an arbitrary amount of transformations.

The Commutativity Theorem provides a ”non linear superposition principle”. Using this technique, we derive an equation for kink solutions, which is a travelling wave. We also derive a formula for breather solutions, which behave as wave packets.

By applying Bianchi’s Commutativity Theorem multiple times, one can construct solutions that behave like the sum of a kink and a breather as t → ±∞. They interact nonlinearly when they collide, since the sG equation is not linear. It turns out that after they have collided, the trajectories of the solutions will have shifted. We give an exact formula for this shift in terms of the parameters of the solution.

By choosing the right parameters, we can also create a solution to the sG equation that behaves approximately as the sum of two breathers. If the breathers have different speeds, they will collide and their trajectories will get shifted. We compute this shift numerically.

Another way to construct new solutions is by applying a so-called Lorentz boost to a known solution. One can do this for a standing breather, which results in the ’moving breather’. It is given by the following formula:

u(x, t) = 4 arctan m ω

sin ω˜t cosh m˜x

!

, (1.2)

for any m, w ∈ R such that m²+ ω² = 1 and m, ω 6= 0, with ˜x = ^√^x−ct

1−c² and ˜t = ^√^t−cx

1−c², and for any c ∈ R such that |c| < 1. It is also possible to find this solution by applying two B¨acklund transformations.

The B¨acklund transformation is a shortcut for a larger, and more difficult solution scheme, namely the Inverse Scattering Transform, which associates an eigenvalue problem to solutions of the sG equation (see [7] for an introduction). We introduce parts of this technique and how it relates to the B¨acklund transformation.

(6)

2 Constructing solutions

2.1 Travelling wave solutions

In this section, we study travelling wave solutions of the sG equation. We derive the equations for the two-dimensional first order system, and find an explicit solution for a heteroclinic orbit.

Let c ∈ R such that |c| 6= 1. Since we are looking for travelling wave solutions, we begin with the ansatz u(x, t) = f (x − ct). Plugging this into the sG equation and rearranging the terms gives us

f⁰⁰(z) = sin f (z)

1 − c² , (2.1)

with z = x − ct. Using g(z) = f⁰(z), we get the system ( f⁰(z) = g(z),

g⁰(z) = ^{sin f (z)}_1−c₂ . (2.2)

This system has its critical points at (f (z), g(z)) = (kπ, 0) for all k ∈ Z. Figure 2.1 suggests that there are multiple heteroclinic orbits. We now derive an exact solution and show that it is indeed a heteroclinic orbit between (0, 0) and (2π, 0).

If we multiply equation (2.1) by f⁰(z) and integrate, we get 1

2(f⁰(z))²= −cos f (z) 1 − c² + C,

for some constant C ∈ R. Because we are looking for a heteroclinic orbit, we want f (z) → 2π and f⁰(z) → 0 as z → ∞, hence C = _1−c¹₂. By using 1 − cos f (z) = 2 sin(¹₂f (z))², we see

f⁰(z)² = 4

1 − c² sin(¹₂f (z))². Assuming that |c| < 1, we end up with the following equation:

|f⁰(z)| =

√ 2

1 − c²sin(¹₂f (z)) .

(7)

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

−3

−2

−1 0 1 2 3

f g

Figure 2.1: The phase portrait of equation (2.2) with c = ³₅.¹

By looking at the phase portrait, we can assume both terms inside the absolute bars to be positive. Rearranging the terms now gives us

1 sin(¹₂f (z))

1

2f⁰(z) = 1

√ 1 − c².

This is a separable equation, which means that we can solve it using standard techniques.

From this last equation is follows that log

1

sin¹₂f (z) −cos¹₂f (z) sin¹₂f (z)

= z

√

1 − c² + C⁰ for some constant C⁰ ∈ R. From here it is easy to see that

1 − cos¹₂f (z) = sin(¹₂f (z)) exp

z

√

1 − c² + C⁰

.

Again by using 1 − cos(¹₂f (z)) = 2 sin(¹₄f (z))² and sin(¹₂f (z)) = 2 sin(¹₄f (z)) cos(¹₄f (z)) we see

f (z) = 4 arctan exp

z

√

1 − c² + C⁰

. (2.3)

1The data for the black and the green orbits were generated by using the Forward Euler method, with ∆z = 10⁻⁴. The blue orbits were plotted using equation (2.3).

(8)

We now show that this is indeed a heteroclinic orbit. Since C⁰ is just a shift of the solution, we can take C⁰= 0. Note that lim_z→−∞f (z) = 0 and limz→∞f (z) = 2π, so f connects these two critical points.

We now verify that the behavior of f as z → ±∞ corresponds to the behavior of the (un)stable manifolds of (0, 0) and (2π, 0). In order to see the behavior of f near these critical points, we need to know the limit of ^dg_df as z → ±∞. Hence we need to look at

dg df =

sin x 1−c²

y = f⁰⁰(z) f⁰(z). The first and second derivative of f are

f⁰(z) = 4

√

1 − c² arctan

exp z

√ 1 − c²

· 1

exp^√^2z

1−c²

+ 1,

f⁰⁰(z) = 4 1 − c² ·

exp^√^2z

1−c²

+ 1· exp^√^z

1−c²

− 2 exp^√^z

1−c²

· exp^√^2z

1−c²

exp^√^2z

1−c²

+ 1²

.

Therefore we have

f⁰⁰(z)

f⁰(z) = 1

√

1 − c² ·

1 − exp^√^2z

1−c²

1 + exp^√^2z

1−c²

. We conclude that

z→−∞lim f⁰⁰(z)

f⁰(z) = 1

√ 1 − c² and

z→∞lim f⁰⁰(z)

f⁰(z) = − 1

√ 1 − c².

In order to verify that these limits correspond to the (un)stable manifolds of (0, 0) and (2π, 0), we can look at the Jacobian matrix. It is equal to

J (f, g) = 0 1

cos f 1−c² 0

! . Therefore

J (0, 0) = J (2π, 0) = 0 1

1 1−c² 0

! . The eigenvalues of this matrix are equal to λ± := ±^√¹

1−c², which means that (0, 0) and (2π, 0) are saddle points of equation (2.2). The eigenvectors of this matrix are v1 := (√

1 − c², 1)^| and v₂ := (−√

1 − c², 1)^|, belonging to the eigenvectors λ₊ and λ−, respectively. Therefore v₁ corresponds to the unstable manifold of (0, 0) and v2 corresponds to the stable manifold of (2π, 0). This is what we expected, since lim_z→−∞f (z) = 0 and limz→∞ = 2π. We also see that the eigenvectors are also in correspondence with the limits of ^dg_df as z → ±∞, so f is indeed the heteroclinic orbit.

(9)

2.2 The B¨ acklund transformation

In this section, we treat the B¨acklund transformation. This gives us a mechanism to construct new solutions. We also state the Commutativity Theorem by Bianchi, which enables us to perform multiple B¨acklund transformations without solving more than one differential equation. This is primarily based on [6].

Using the transformation X = ^x+t₂ and T = ^x−t₂ , we can write the sG equation as

u˜_XT = sin ˜u, (2.4)

where u(x, t) = ˜u(X, T ). We drop the ∼ in the notation for the calculations and results, but it should be clear from context if laboratory (with x and t) or characteristic coordinates (with X and T ) are used.

We now state the B¨acklund transformation and then prove that solving this systems yields a new solution.

Definition 2.1. Let u(X, T ) be a solution of equation (2.4) and a ∈ C a non-zero complex number. Then the B¨acklund transformation of u is defined as the solution U (X, T ) of the system











∂

∂T U − u

2 = a sin

U + u 2

,

∂

∂X U + u

2 = 1

asin

U − u 2

.

(2.5) (2.6) It turns out that U is another solution of the sG equation.

Theorem 2.1. Let u be a solution of equation (2.4) and let U be the B¨acklund trans- formation of u. Then U is also a solution of equation (2.4).

Proof. Rewrite equation equation (2.6) as

∂U

∂X = −∂u

∂X +2 asin

U − u 2

. Taking the derivative with respect to T results in

∂²U

∂T ∂X = − ∂²u

∂T ∂X + ∂

∂T

2 asin

U − u 2

= − sin u + 2 acos

U − u 2

· ∂

∂T

U − u 2

= − sin u + 2 acos

U − u 2

· a sin

U + u 2

= − sin u + 2 cos

U − u 2

· sin

U + u 2

.

(10)

By using the formula

2 cos(b) sin(a) = sin(a + b) + sin(a − b), we see that

∂²U

∂X∂T = − sin u + sin

U + u

2 +U − u 2

+ sin

U + u

2 −U − u 2

= − sin u + sin U + sin u

= sin U,

and thus U is a solution of the sG equation.

We now use this to construct a non-trivial solution. We take the solution u ≡ 0 and define V = ^U₂. Then equation (2.5) reduces to _∂T^∂ V = a sin V . This is a separable equation, which means that it is readily solved. Rearranging this equation and integrating, we see that^R _{sin V}¹ dV =^R a dT = aT + C(X) for some function C only depending on X.

Because

Z 1

sin V dV = log

1

sin V −cos V sin V , we get

1

sin V −cos V

sin V = exp(aT + C(X)).

Then

1 − cos V = exp(aT + C(X)) sin V.

Because 1 − cos V = 2 sin^V₂² and sin(V ) = 2 sin^V₂cos^V₂, this reduces to sin

V 2

= exp(aT + C(X)) cos

V 2

. Solving this for V yields

V = 2 arctan(exp(aT + C(X))) and therefore

U = 4 arctan(exp(aT + C(X))).

Using the similarities between equation (2.5) and equation (2.6) for u ≡ 0, we see that C(X) = ¹_aX + C⁰, for some constant C⁰. Hence U equals

U (X, T ) = 4 arctan

exp

aT +1

aX + C⁰

, (2.7)

or written in laboratory coordinates:

U (x, t) = 4 arctan exp a²+ 1

2a x +1 − a² 2a t + C⁰

!!

(2.8)

(11)

−100 −5 0 5 10 2

4 6

x u(·, 0)

−10 −5 0 5 10

−20 0 20

x t

Figure 2.2: Left: the profile of a kink at t = 0 for a = 2, given by equation (2.8). On the right, the contour plot for the same kink. The position of the red dot on the left is indicated by the red line in on the right.

u

₁

u

₂

U

a1

a2

a₂

a1

Figure 2.3: A diagram illustrating Bianchi’s Commutativity Theorem.

The profile of this solution is depicted in Figure 2.2. Note that we can shift this travelling wave solution on the x-axis by changing C⁰. This solution is the same as the solution we found by using the ansatz u(x, t) = f (x − ct). The following theorem makes it possible to create new solutions without solving more differential equations.

Theorem 2.2 (Bianchi’s Commutativity Theorem). Let a₁, a₂ be distinct, non-zero complex numbers and let u be a solution of equation (2.4). Furthermore, define u₁ and u₂ as the B¨acklund transformations of u with parameter a₁ and a₂ respectively.

Denote with U₁ the B¨acklund transformation of u₁ with parameter a₂, and let U₂ be the B¨acklund transformation of u₂ with parameters a₁. Then U₁ = U₂, and they satisfy

tan

U₁− u 4

= a₁+ a₂ a1− a₂tan

u₁− u₂ 4

. (2.9)

(12)

Proof. Let u be a solution of equation (2.4). Suppose u₁ and u₂ are the B¨acklund transformation of u with parameter a₁ and a₂, respectively. By solving equation (2.9) for U₁, we see that

U₁ = 4 arctan

a₁+ a₂ a₁− a₂tan

u₁− u₂ 4

+ u + 4kπ

for some k ∈ Z. We now show that U1 satisfies equation (2.5). Taking the derivative of

U1−u₁

2 with respect to T gives

∂

∂T

U₁− u₁ 2

= ∂

∂T

2 arctan

a₁+ a₂ a1− a₂ tan

u₁− u₂ 4

+ ∂

∂T

u − u₁ 2

=

a1+a2

a1−a₂

a1+a2

a1−a₂tan ^u¹^−u₄ ²²+ 1 1 cos ^u¹^−u₄ ²²

∂

∂T

u1− u₂ 4

− a₁sin

u1+ u 2

= (a²₁− a²₂)(a₁sin û¹₂^+u− a₂sin û²₂^+u) − a₁sin û¹₂^+ua²₁− 2a₁a2cos û¹^−u₂ ²+ a²₂ a²₁− 2a₁a₂cos û¹^−u₂ ²+ a²₂

= −a²₁a₂sin û²₂^+u− 2a²₂a₁sin û¹₂^+u+ a₂³sin û²₂^+u+ 2a²₁a₂sin û¹₂^+ucos û¹^−u₄ ² a²₁− 2a₁a2cos û¹^−u₂ ²+ a²₂ . Expanding the right-hand side of equation (2.5), we see that

a2sin

U1+ u₁ 2

= a₂sin

2 arctan

a1+ a₂ a₁− a₂ tan

u1− u₂ 4

+u + u1

2 + 2kπ

= a₂sin

2 arctan

a1+ a₂ a₁− a₂ tan

u1− u₂ 4

cos

u + u1

2

+ a₂cos

2 arctan

a₁+ a₂ a1− a₂ tan

u₁− u₂ 4

sin

u + u₁ 2

.

By using sin(arctan(x)) = _1+x^2x2 and cos(arctan(x)) = ^1−x_1+x²2, we see that a2sin

U + u1

2

= a₂

2(a1+a2)

a1−a2 tan ^u¹^−u₄ ²

a1+a2

a1−a2 tan ^u¹^−u₄ ²²+ 1 cos

u + u₁ 2

+ a₂sin

u + u₁ 2

1 −^a_a¹^+a²

1−a2tan ^u¹^−u₄ ²² 1 +^a_a¹^+a²

1−a2tan ^u¹^−u₄ ²²

= a2a²₁ sin û¹^−u₂ ²cos û¹₂^+u+ sin û¹₂^+ucos û¹^−u₂ ²− 2a₁a²₂sin û¹₂^+u+ a³₂sin û+u₂ ² a²₁− 2a₁a2cos û¹^−u₂ ²+ a²₂ .

(13)

Hence

∂

∂T

U₁− u₁ 2

− a₂sin

U₁+ u₁ 2

= a²₁a₂− sin û²₂^+u+ sin û¹₂^+ucos û¹^−u₂ ²− sin û¹^−u₂ ²cos û¹₂^+u a²₁− 2a₁a2cos û¹^−u₂ ²+ a²₂

= a²₁a₂ − sin û²₂^+u+ sin û+u₂ ² a²₁− 2a₁a₂cos û¹^−u₂ ²+ a²₂

= 0.

Therefore _∂T^∂ Û¹^−u₂ ¹= a₂sinÛ¹^+u₂ ¹. Showing that _∂X^∂ Û¹^+u₂ ¹= _a¹

2 sin^U¹^−u₂ ¹ is similar. So U₁ is the B¨acklund transformation of u₁ with parameter a₂. By switching a1, a₂, u₁ and u₂ in equation (2.9) we see that two B¨acklund transformations indeed commute.

2.3 Lorentz boost

In this section, we define the Lorentz boost and show that it indeed yields a new solution.

Definition 2.2. Let u be a solution to equation (1.1) and let c ∈ R such that |c| < 1.

Define ˜x = ^√^x−ct

1−c² and ˜t = ^√^t−cx

1−c². Then the Lorentz boost of u is the function ˜u defined as ˜u(x, t) = u(˜x, ˜t).

Theorem 2.3. Let u be a solution of equation (1.1). Then the Lorentz boost ˜u of u is also a solution of equation (1.1).

Proof. This is a straight forward calculation:

d

dxu(x, t) =˜ d

dxu(˜x, ˜t)

= 1

√

1 − c²u₁ x, ˜˜ t− c

√

1 − c²u₂ x, ˜˜ t. From here, we see that

d²

dx²u(x, t) =˜ d dx

√ 1 1 − c²

u1 x, ˜˜ t− c · u₂ x, ˜˜ t

= 1

√1 − c²

1

√1 − c²u₁₁ x, ˜˜ t− c

√1 − c²u₁₂ x, ˜˜ t

−c ·

1

√

1 − c²u21 x, ˜˜ t− c

√

1 − c²u22 x, ˜˜ t

= 1

1 − c² h

u11(˜x, ˜t) − 2c · u12(˜x, ˜t) + c²u22(˜x, ˜t)ⁱ.

(14)

In a similar fashion, we get d²

dt²u(x, t) =˜ 1 1 − c²

hc²u11(˜x, ˜t) − 2c · u12(˜x, ˜t) + u22(˜x, ˜t)ⁱ. Therefore

d²

dt²u(x, t) −˜ d²

dx²u(x, t) =˜ 1 1 − c²

hc²u₁₁(˜x, ˜t) − 2c · u₁₂(˜x, ˜t) + u₂₂(˜x, ˜t)ⁱ

− 1

1 − c² h

u11(˜x, ˜t) − 2c · u12(˜x, ˜t) + c²u22(˜x, ˜t)ⁱ

= 1

1 − c²

h(u₁₁(˜x, ˜t) − u₂₂(˜x, ˜t)) + c²(u₂₂(˜x, ˜t) − u₁₁(˜x, ˜t))ⁱ

= 1

1 − c² h

sin(u(˜x, ˜t)) − c²sin(u(˜x, ˜t))ⁱ

= sin(˜u(x, t)), so ˜u satisfies equation (1.1).

2.4 The moving breather

We now show how to derive the breather solution, given by u(x, t) = 4 arctan

m ω

sin ωt cosh mx

. (2.10)

Let m, ω ∈ R such that m²+ ω²= 1 and m, ω 6= 0. We apply the B¨acklund transforma- tion to u ≡ 0 with the parameters a₁ = m + iω and a₂ = m − iω. We use equation (2.8).

Note that

a²₁+ 1

2a₁ = (m + iω)²+ 1 2(m + iω) = m.

The other coefficients are found in a similar way. We get the solutions u1(x, t) = 4 arctan(e^mx+iωt)

and

u₂(x, t) = 4 arctan(e^mx−iωt).

Define z := mx + iωt. From equation (2.9) we conclude that U (x, t) = 4 arctan m + iω + m − iω

m + iω − m + iωtan 4 arctan(e^z) − 4 arctan(e^z) 4

!!

= 4 arctan

m

ωitan(arctan(e^z) − arctan(e^z))

.

(15)

Note that

sin(w) = e^iw − e^−iw 2i and

cos(w) = e^iw+ e^−iw

2 .

so

tan(w) =

_eiw−e^−iw 2i

_e_iw_+e_−iw

2

= i ·1 − e^2iw 1 + e^2iw

holds. Solving this last equation for w gives us an expression for arctan(w):

arctan(w) = 1 2i Log

1 − iw 1 + iw

. Then we have

tan(arctan(e^z) − arctan(e^z))

= i

1 − exp2i[arctan(e^z) − arctan(e^z)] 1 + exp(2i[arctan(e^z) − arctan(e^z)])

= i1 − exp(2i arctan(e^z)) · exp(−2i arctan(e^z)) 1 + exp(2i arctan(e^z)) · exp(−2i arctan(e^z)). Because for all w ∈ C

exp(2i arctan(e^w)) = exp

2i · 1

2i Log

1 − ie^w 1 + ie^w

= 1 + ie^w 1 − ie^w, we see that

tan(arctan e^z− arctan e^z) = i1 −^1+ie_1−ie^zz ·^1−ie^z

1+ie^z

1 +^1+ie_1−ie^zz ·^1−ie^z

1+ie^z

= i(1 − ie^z)(1 + ie^z) − (1 + ie^z)(1 − ie^z) (1 − ie^z)(1 + ie^z) + (1 + ie^z)(1 − ie^z)

= i2ie^z− 2ie^z 2 + 2e^z+z

= e^z− e^z 1 + e^z+z.

(16)

−5 0 5

−4

−2 0 2 4

x u(·,¹₂√

2π)

Figure 2.4: The profile of equation (2.10) with m = ω = ¹₂√

2 and t = ¹₂√ 2π.

We now substitute z = mx + iωt, to see that

tan(arctan(e^z) − arctan(e^z)) = e^mx+iωt− e^mx−iωt 1 + e^2mx

= e^iωt− e^−iωt e^−mx+ e^mx

= cos(ωt) + i sin(ωt) − cos(−ωt) + i sin(−ωt) e^−mx+ e^mx

= i sin ωt cosh mx. Therefore

U (x, t) = 4 arctan

m ω

sin ωt cosh mx

. The profile of this wave is depicted in Figure 2.4.

We now apply the Lorentz boost to the moving breather. This gives us a new solution:

u(x, t) = 4 arctan˜



 m

ω

sin ω^√^t−cx

1−c²

cosh m^√^x−ct

1−c²



. (2.11)

We can also find this solution by taking different parameters a₁ and a₂. Let R ∈ R \ {0}.

Let a₁ := R(m + iω) and a₂ := R(m − iω), again with m²+ ω² = 1 and m, ω 6= 0. If we use these parameters, the solution we find is

U (x, t) = 4 arctan



 m ω

sin^ω₂^R²_R⁻¹x −^R²_R⁺¹t

cosh^m₂ ^R²_R⁺¹x −^R_R²2⁻¹+1t



. (2.12)

(17)

−4 −2 0 2 4

−4

−2 0 2 4

x u(·, 0)

Figure 2.5: The profile of equation (2.12) with m = ¹₃, ω = q8

9 and t = 0. The dashed line indicates the envelope.

The profile of this solution is depicted in Figure 2.5. Inside the arctan we see that the moving breather is a product of two travelling waves. The _cosh¹ part can be described as the envelope of the breather. It moves with speed v₁ = ^R_R²₂⁻¹₊₁. So if |R| 6= 1 this is a moving breather. The oscillatory part stems from the sine. This is a travelling wave with speed _v¹

1. Since v₁ 6= _v¹

1 and the sine is periodic, there are infinitely many (x, t) such both the sine and _cosh¹ attain their maximum value of 1, so there are infinitely many (x, t) such that U (x, t) = 4 arctan(^m_ω).

2.5 Kink collisions

We now look at the collision of kink solutions. First we derive the general formula, and then look at how these behave for specific parameters.

We apply the Commutativity Theorem to a₁, a2 ∈ R and u ≡ 0. Define ˜a1 := ^a_2a²¹⁺¹

1 ,

˜˜

a₂:= ^1−a

2 1

2a2 and similarly for a₂. First, we get

u1= 4 arctan(exp(˜a1x + ˜˜a1t))

and

u₂ = 4 arctan exp((˜a₂x + ˜a₂t)).

(18)

So we get

U = 4 arctan

a₁+ a₂

a1− a₂ tanu₁− u₂ 4

= 4 arctan

a₁+ a₂

a1− a₂ tanarctan exp ã₁x + ã˜₁t− arctan exp(ã₂x + ã₂t)

= 4 arctan a1+ a₂

a1− a₂ · exp ã1x + ã˜1t− exp(ã2x + ã2t) 1 + exp (ã₁+ ã₂)x + ã˜₁+ ˜ã₂t

!

. (2.13)

2.5.1 Kink anti-kink collision

In this section we consider the collision of a kink and an anti-kink. Since the sG equation is non linear, it is to be expected that the collision will affect the solitons nonlinearly. It turns out the trajectories of the solitons undergo a shift, for which we derive a formula.

The kink anti-kink collision is defined by parameters the 0 < a₁ < 1 and a₂ > 1, or

−1 < a₁< 0 and a2 < −1. The contour plot the former can be seen in Figure 2.6. We can see that the two kinks get shifted after they have collided.

−10 −5 0 5 10

−50 0 50

x t

Figure 2.6: The contour plot of equation (2.13) with a1= ³₅ and a2= ¹¹₁₀.

We assume that 0 < a₁< 1 and a2> 1. Note that ã1, ã˜1, ã2 > 0 and ˜ã2 < 0. We assume that ˜ã₁+ ˜ã₂ > 0. This last assumption also implies that ã₁− ã₂ > 0. If we take the

(19)

limits t → ∞ and x → ∞, we see

exp ˜a1x+˜˜a1t

−exp(ã2x+ã2t) 1+exp (ã1+ã2)x+ ˜ã1+˜ã2

t

exp(−ã₂x − ˜ã₂t) = exp(ã₁x + ˜ã₁t) − exp(ã₂x + ã˜₂t) exp(−ã₂x − ã˜₂t) + exp(ã₁x + ˜ã₁t)

∼ exp(ã1x + ˜ã1t) − exp(ã2x + ˜ã2t)

exp(˜a₁x + ˜˜a₁t) as x → ∞

= 1 − exp((ã2− ã1)x + (˜ã2− ˜ã1)t),

where we used that −ã₂ < 0 and ã₂− ã₁ < 0. This last expression tends to 1 as x → ∞.

Similarly, we can take the limit as t → ∞:

exp ˜a1x+˜˜a1t

−exp(ã2x+ã2t) 1+exp (ã1+ã2)x+ ˜ã1+ã˜2

t

exp(−ã₂x − ˜ã₂t) = exp(ã1x + ã˜1t) − exp(ã2x + ˜ã2t) exp(−ã₂x − ˜ã₂t) + exp(ã₁x + ˜ã₁t)

∼ exp(˜a₁x + ˜˜a₁t)

exp(−ã2x − ˜ã2t) + exp(ã1x + ˜ã1t) as t → ∞

= exp(˜a₁x)

exp(−ã2x − (˜ã1+ ˜ã2)t) + exp(ã1x), which also tends to 1 as t → ∞, since −(˜ã₁+ ã˜₂) < 0. So

exp ã1x + ˜ã1t− exp(ã2x + ã2t)

1 + exp (ã₁+ ã₂)x + ˜ã₁+ ã˜₂t ∼ exp(−ã2x − ˜ã2t) as x → ∞ and t → ∞. By manipulating this formula, we can write

4 arctan

a₁+ a₂

a1− a₂ · exp(−˜a₂x − ˜a˜₂t)

= −4 arctanexpa˜₂^h(x − δ) + ˜˜a₂· ˜a⁻¹₂ tⁱ

with

δ2 = 1

˜a₂ log

−a1+ a₂ a₁− a₂

. Let

δ₁ = 1

˜a₁ · log

−a₁+ a₂ a₁− a₂

.

Using a similar reasoning, we see that the solution behaves as follows:

U ∼ 4 arctanexp−ã₁^h(x − δ₁) + ã˜₁· ã⁻¹₁ tⁱ as t → −∞, x → ∞;

U ∼ −4 arctanexpã₁^h(x + δ₁) + ã˜₁· ã⁻¹₁ tⁱ as t → ∞, x → −∞;

U ∼ 4 arctanexpa˜₂^h(x + δ₂) + ˜˜a₂· ˜a⁻¹₂ tⁱ as t → −∞, x → −∞;

U ∼ −4 arctanexp−˜a2

h(x − δ₂) + ˜˜a2· ˜a⁻¹₂ tⁱ as t → ∞, x → ∞.

(20)

These are shifted (anti-)kinks with parameter ±a₁ and ±a₂. Writing these approximate functions in this form, we can see that the shift of the (anti-)kink with parameter a₁ is equal to

 = 2δ₁ = 2

˜a₁log

−a₁+ a₂ a₂− a₁

. The shift of the (anti-)kink with parameter a₂ is given by

 = 2

˜a2

log

−a₁+ a₂ a2− a₁

.

The cases with ˜ã1+ ã˜2 = 0 or ˜ã1+ ã˜2 < 0 are similar and yield the same result.

2.6 Moving breather and standing kink collision

We can concatenate multiple B¨acklund transformations by a clever use of Bianchi’s Commutativity Theorem, in order to construct new solutions. This is illustrated in Figure 2.7. By doing this, we get a formula for the solution consisting of a standing kink and a moving breather. The pointwise limits of this solution as t → ±∞ are shifted standing kinks. We derive a formula for this shift.

Let m, w ∈ R such that m²+ w² = 1 and with m, ω 6= 0. Let R ∈ R be non-zero. Define a₁ := R(m + iω), a₂ = 1 and a₃ = R(m − iω). Furthermore, define ˜a₁ = ^a_2a²¹⁺¹

1 and

˜˜

a1:= ^1−a_2a²¹

1 , and similarly for a₃. We take u ≡ 0. We now get u1 = 4 arctan(exp ˜a1x + ˜˜a1t);

u2 = 4 arctan(exp(x));

u₃ = 4 arctan(exp ˜a₃x + ˜˜a₃).

This gives us

u₄ = 4 arctan a₁+ a₂ a₁− a₂

exp(ã₁x + ã˜₁t) − exp(x) 1 + exp(ã1x + ã˜1t + x)

!

and

u₅= 4 arctan a₂+ a₃ a2− a₃

exp(x) − exp(ã₃x + ˜ã₃t) 1 + exp(x + ã₃x + ã˜₃)

! .

With the expressions for u₄ and u₅, one can find U by applying the Commutativity Theorem one more time.

(21)

u

₁

u

₂

u

₃

u

₄

u

₅

U

a₁

a2

a₃

a2

a₁

a₃ a2

a2

a₃

a₁

Figure 2.7: A ladder of B¨acklund transformations.

Proposition 2.1. Assume mR > 0. Let ˜˜˜ R = R + _R¹ and R = R −˜˜ _R¹. Define the followings functions:

Q(x, t) :=2Rw

2 exp

1

2m ˜Rx − 1 2mRt˜˜

cos

w 2

˜˜ Rx − w

2 Rt˜

· (1 − exp(2x))

− 2 exp(x) + 2 expm ˜R + 1x − mRt˜˜

+ 2(1 − R²) exp

1

2m ˜Rx − 1 2mRt˜˜

sin

w 2

˜˜ Rx − w

2 Rt˜

(1 + exp(2x))

and

W (x, t) :=2Rm

1 + 4 exp

x +m

2

Rx −˜ m 2

˜˜ Rt

cos

w 2

˜˜ Rx −w

2 Rt˜

+ exp2 + m ˜Rx − mRt˜˜ − exp(2x) − expm ˜Rx − mRt˜˜

− (1 + R²)^h1 + exp2 + m ˜Rx − mRt˜˜ + exp(2x) + expm ˜Rx − mRt˜˜ ⁱ.

Collision of wave packets in the sine-Gordon equation