An introduction to the Hirota bilinear method
José de Jesús Martı́nez
Department of Mathematics, Kobe University,
Rokko, Kobe 657-8501, Japan
Contents
1 Introduction
1
2 Wave equations
2.1 Linear nondispersive waves . .
2.2 Linear dispersive waves . . . .
2.3 Nonlinear nondispersive waves .
2.4 Nonlinear dispersive waves . . .
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2
2
2
2
3
3 A new binary operator
3.1 Exponential Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Exponential Identity properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 The Exchange Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
5
5
6
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4 Bilinearization
4.1 Rational transformation . . . . . . . . . . .
4.1.1 The KdV equation . . . . . . . . . .
4.1.2 The modified KdV equation . . . . .
4.1.3 The nonlinear Schrödinger equation
4.2 Logarithmic transformation . . . . . . . . .
4.2.1 The KdV equation . . . . . . . . . .
4.3 Bi-logarithmic transformation . . . . . . .
4.3.1 The mKdV equation . . . . . . . . .
4.3.2 The sine-Gordon equation . . . . . .
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7
7
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10
10
5 Solutions to Bilinear equations
5.1 One-soliton solution . . . . .
5.2 Two-soliton solution . . . . .
5.3 Three-soliton solution . . . .
5.4 N-soliton solution . . . . . . .
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11
12
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6 Visualization
6.1 One-soliton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Two-soliton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Three-soliton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
15
16
18
1
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Introduction
In this report I give an introduction to the Hirota direct method to solve soliton equations. Soliton theory is
a subject which began to attract the attention of many scientist with the study of nonlinearity in dynamical
systems. There are two main methods to solve such integrable nonlinear evolution equations with soliton
properties: The inverse scattering transform (IST) and the Hirota direct method.
The main advantage of the IST is that it may be employed to solve initial-value problems (IVP). However,
this is an analytic approach which could become difficult to work with. On the other hand, the Hirota direct
method can be applied to many more equations and it is algebraic. The major advantage over the IST is that
with the direct method one can construct multisoliton solutions (N-soliton solutions). This method quickly
proved to be useful in producing solutions to equations such as the KdV, modified-KdV, sine-Gordon, and
nonlinear Schrodinger equations.
1
So what are solitons? A soliton is a sort of solitary wave that is not destroyed when it collides with another
similar wave. Nonetheless, numerical simulations seems to show that solitons are not conserved. However, to
be sure one must find exact solutions.
2
Wave equations
wave equations
solitary waves
solitons
One can categorize waves to be either linear or nonlinear and dispersive or nondispersive. A wave equation
with soliton solutions is both nonlinear and dispersive. Let us first study these four types of waves in order to
understand how solitons can come into being.
2.1
Linear nondispersive waves
Consider the most basic wave equation,
∂
∂
+ v0
∂t
∂x
f (x, t) = 0,
(1)
where v0 represents the speed of the wave. Assuming that this wave is periodic, the most basic solution is given
by
f (x, t) = exp[i(ωt − kx)],
(2)
where ω is the angular frequency and k represents the number of waves. By substituting this solution into
equation (1) we find that the relation between ω and k is given by
ω = v0 k.
The above is called the dispersion relation. Since it is linear, the equation depends on the relation, making it
a nondispersive wave. Hence, it does not conserve its original shape.
2.2
Linear dispersive waves
Consider the following equation,
∂
∂
∂3
+ v0
+δ 3
∂t
∂x
∂x
f (x, t) = 0.
This is once again a linear equation and it has a solution similar to (2). By substituting the solution into the
equation above we find the following dispersion relation,
ω = v0 k − δk 3
which is nonlinear with respect to k. The above is called a phase velocity. But to obtain a group velocity we
differentiate ω with respect to k,
∂ω
= v0 − 3δk 3
∂k
Notice that even though this shows that the wave is dispersive, the velocity of each phase varies with k which
means that the wave ”spreads” out. Therefore, the wave does not preserve it’s original shape.
2
2.3
Nonlinear nondispersive waves
Consider the following equation
∂
∂
+ v(f )
∂t
∂x
f (x, t) = 0,
where v(f ) = v0 + αf m . We analyze the formal solution to this equation
f (x, t) = f x − v(f )t
If v(f ) is an increasing function, then that would imply that the wave travels faster as the amplitude
increases. In other words, the wave steepens and breaks. Consequently, the top of the wave travels faster than
the base. Once again, it eventually disappears.
2.4
Nonlinear dispersive waves
We consider the next equation,
∂
∂
∂3
∂
+ v0
+ αf m
+δ 3
∂t
∂x
∂x
∂x
f (x, t) = 0,
(3)
This wave has a pulse-like wave solution. To verify whether this wave can exist, we check that the velocity at
the top and base of wave have the same value.
First let us introduce a new space coordinate variable η such that
η = px − Ωt,
where v = Ω/p and p is a free parameter. Note that as p increases the pulse becomes sharper. If t = 0 then
η = px, which means that η is proportional to x. If t 6= 0 then η = p(x − vt), which means that η is proportional
to x − vt. Thus, η allows the wave to travel at a constant speed v regardless of t.
Let A be the maximum wave amplitude,
f ∼ A(1 − constant × η 2 )
Note that in the neighborhood around the top of the wave, f satisfies the equation
∂
∂
+ [v0 + αf (x, t)m ] + f (x, t) ∼ 0,
∂t
∂x
which is an equation for nonlinear nondispersive waves. We saw in the previous section that v(t) = v0 + αf m .
Therefore,
v(t) = v0 + αAm
For the base, we cannot consider a linear equation. Note that f m → 0 since the amplitude is 0 at the base.
Instead we choose an exponentially decaying wave solution of the form,
f ∼ exp [±(px − Ωt)]
3
By substituting into (3) we obtain the following relation,
v=
Ω
= v0 + δp2
p
If δp2 = αAm then the velocities at the top and base would be the same and therefore the wave would be
able to travel without changing its shape.
Note When finding solutions to nonlinear wave equations we should choose exponentially decaying solutions
to be able to expand f using taylor series.
3
A new binary operator
In the next section we define the new operator and give some properties and identities.
Definition
b(x),
We define a new binary differential operator, D-operator, acting on a pair of functions a(x) and
Dtm Dxn a(t, x) · b(t, x)
∂m ∂n
≡ m n a(t + s, x + y)b(t − s, x − y) |s=0,y=0
∂s ∂y
n
∂
Dxn ≡ n a(x + y)b(x − y) |y=0
∂y
where m, n = 0, 1, 2, 3, . . . .
Let ∂xn = ∂ n /∂xn . Now the normal derivative can also be defined using the same notation,
∂tm ∂xn a(t, x) · b(t, x)
≡ ∂sm ∂yn a(t + s, x + y)b(t + s, x + y) |s=0,y=0
We notice that the main difference between the D-operator and the normal derivative is that when the
second function has an odd-number of derivatives there is a minus sign. Here are some examples,
∂x a · b = ax b + bx a
Dx a · b = ax b − bx a
..
.
∂x2 a · b = axx b + 2bx ax + abxx
Dx a · b = axx b − 2bx ax + abxx
..
.
We can use Mathematica to write a module to calculate the nth derivative
It is convenient to think of this new operator in terms of normal derivatives. We also let Dz = (Dt + εDx )
and ∂z = (∂t + ε∂x ).
4
Here are some useful properties which can be proven using the definition,
Dtm Dxn a · b = Dxn Dtm a · b = Dxn−1 Dt Dx a · b
t
Dtm Dxn a · 1 = ∂xn ∂m
a
Dz = (Dt + εDx ) , ∂z = (∂t + ε∂x )
Dzn b · a = (−1)n Dzn a · b
Dz (Dz a · b) · c + Dz (Dz b · c) · a + Dz (Dz c · a) · b = 0
Note
From the last property we note that when Dz a · b = [a, b] we can write the Jacobi Identity:
[[a, b], c] + [[b, c], a] + [[c, a], b] = 0
Note
We see from Dzn b · a = (−1)n Dzn a · b that when n is odd,
Dzn a · a = 0,
This is an important property since it shows that D-operators often yield zero as a result. Hence Hirota
equations hold many more types of solutions.
3.1
Exponential Identity
We can also define the D-operator using exponential functions,
exp(δDz )a(z) · b(z) = a(z + δ)b(z − δ)
(4)
where δ is a parameter.
Proof First we expand f using the parameter δ,
f (x + δ) = f + δf 0 +
δ 2 00 δ 3 000
f + f + ...
2!
3!
We see that if a(z) and b(z) are continuously differentiable, then by Taylor series expansions in δ we have
that,
δ2
exp(δDz )a · b = 1 + δDz + Dz2 + . . . a · b
2!
δ2
=ab + δ(az b − abz ) + (azz b − 2bz az + abzz ) + ...
2! δ2
= a + δaz + azz + . . .
2!
(−δ)2
× b + (−δ)bz +
bzz + . . .
2!
=a(z + δ)b(z − δ)
3.2
Exponential Identity properties
For two products ab and cd we have,
exp(δDz )ab · cd = [exp(δDz )a · c][exp(δDz )b · d]
= [exp(δDz )a · d][exp(δDz )b · c]
5
Proof Using the definition and the commutativity of multiplication,
exp(δDz )a(z)b(z) · c(z)d(z)
= a(z + δ)b(z − δ)c(z + δ)d(z − δ)
= a(z + δ)c(z − δ)c(z + δ)b(z − δ)
= [exp(δDz )a · c][exp(δDz )b · d]
By expanding and collecting terms with respect to the powers of δ, we obtain the following properties,
Dz ab · c = az bc + aDz b · c
Dz ab · b = az b2
Dz2 ab · c = azz bc + 2az Dz b · c + aDz2 b · c
Dz2 ab · b = azz b2 + aDz2 b · b
Dz2 ab · cd = (Dz2 a · c)bd + 2(Dz a · c)(Dz b · d) + ac(Dz2 b · d)
Dz2 ac · bc = (Dz3 a · b)c2 + 3(Dz a · b)(Dz2 c · c)
Dzn exp(pz)a(z) · exp(pz)b(z) = exp(2pz)Dzm a(z) · b(z)
Dzn a · b = Dzn−1 Dz a · b
Dzn a · a = 2Dzn−1 az · a
Consider the following relation which can be proved using the definition,
Dzn exp(p1 z) · exp(p2 z) = (p1 − p2 )n exp[(p1 + p2 )z]
Then we compute the same expression using normal derivatives,
∂zn exp(p1 z) · exp(p2 z) = (p1 + p2 )n exp[(p1 + p2 )z]
By solving for exp[(p1 + p2 )z] on the second equation, we see that the following expression holds,
Dzn exp(p1 z) · exp(p2 z)
1
n
n
= (p1 − p2 )
∂ exp(p1 z) · exp(p2 z)
(p1 + p2 )n z
n
h
i
p1 − p2 )
=
∂zn exp (p1 + p2 )z
p1 + p2
This latter will prove useful when solving for the two-soliton solution of the KdV equation.
3.3
The Exchange Formulae
Now we study the Exchange formulae. This formula will be beneficial when deriving Backlund transformations.
We consider the most basic form,
exp(αDz )[exp(βDz )a · b] · [exp(γDz )c · d]
β+γ
β−γ
Dz exp α +
Dz a · d
= exp
2
2
β+γ
· exp −α +
Dz c · b
2
The proof is done using the definition and commutativity of the operator.
We make note of a couple of identities that hold for the D-operator.
i) For arbitrary functions a, b, c and d, the following formula holds,
[sinh(δDz )a · b][exp(δDz )c · d]
+ [sinh(δDz )b · c][exp(δDz )a · d]
+ [sinh(δDz )c · a][exp(δDz )b · d] = 0
6
When expanding with respect to δ, the coefficient for δ yields the following formula,
c Dz a · b + a Dz b · c + b Dz c · a = 0
ii) The following formula also holds,
sinh(δDz )[sinh(δDz )a · b][cosh(δDz )c · d]
+ sinh(δDz )[sinh(δDz )b · c][exp(δDz )a · d]
+ sinh(δDz )[sinh(δDz )c · a][exp(δDz )b · d] = 0
When expanding with respect to δ, the coefficient for δ 2 yields the Jacobi Identity,
4
Bilinearization
In this section we discuss the main ideas behind the bilinearization of nonlinear equations. One neat feature
of bilinear forms is that they appear as polynomials of simple exponentials. Fortunately, there are several
transformations and fundamental formulas that can be used. We present three fundamental dependent variable
transformations:
1. Rational transformation
2. Logarithmic transformation
3. Bi-logarithmic transformation
4.1
Rational transformation
a
b
When applying this transformation we obtain some homogeneous expression with respect to a and b.
Now we introduce a fundamental formula to obtain the homogenous expression,
u=
exp(δ∂z )
a
exp(δDz )a · b
=
b
cosh(δDz )b · b
(5)
Proof We can easily prove this expression using equation(4),
exp(δ∂z )
a(z + δ)
a
=
b
b(z + δ)
a(z + δ)b(z − δ)
= 1
2b(z
+
δ)b(z
−
δ)
2
=
=
a(z + δ)b(z − δ)
1
2
exp(δDz )b · b + exp(−δDz )b · b
exp(δDz )a · b
cosh(δDz )b · b
By expanding both sides with respect to δ and collecting the terms for the powers of δ we obtain the
following set of formulas,
a
Dz a · b
=
b
b2
2
a
D a · b a Dz b · b
∂z2 = z 2 −
b
b
b b2
3
a
D a·b
D a · b Dz2 b · b
∂z3 = z 2 − 3 z 2
b
b
b
b2
..
.
∂z
Now we can proceed to work on an example.
7
(6)
4.1.1
The KdV equation
ut + 6uux + uxxx = 0
First we apply the rational transformation u = G/F to obtain,
∂t
G G
G
G
+ 6 ∂x + ∂x3 = 0
F
F F
F
by using formulae (6), and simplifying,
[(Dt + Dx3 )G · F ]F 2 = 3[Dx G · F ][Dx2 F · F − 2GF ]
We introduce an arbitrary function λ to decouple the equation,
(
(Dt + Dx3 )G · F = 3λDx G · F
Dx2 F · F − 2GF = λF 2
The latter being the bilinear form for the KdV equation.
4.1.2
The modified KdV equation
wt + 6w2 wx + wxxx = 0
Again we make use of the rational transformation w = G/F ,
G
∂t + 6
F
G
F
2
∂x
G
G
+ ∂x3 = 0
F
F
Using formulae (6) and simplifying,
[(Dt + Dx3 )G · F ]F 2 = 3[Dx G · F ][Dx2 F · F − 2G2 ]
for which we introduce an arbitrary function λ to decouple the equation,
(
(Dt + Dx3 )G · F = 3λDx G · F
Dx2 F · F − 2G2 = λF 2
The latter being the bilinear form for the mKdV equation.
4.1.3
The nonlinear Schrödinger equation
iΨt + Ψxx + 2c|Ψ|2 Ψ = 0
where c = ±1.
Once more, we use the rational transformation Ψ = G/F and formulae (6),
[(iDt + Dx2 )G · F ]F 2 = [GF ][Dx2 F · F − 2c|G|2 ]
Using an arbitrary function λ, it turns into the bilinear form for the Schrodinger equation,
(
(iDt + Dx2 )G · F = λGF
Dx2 F · F − 2c|G|2 = λF 2
4.2
Logarithmic transformation
u = 2(log f )xx
(7)
Subsequently we discuss in detail the logarithmic transformation. It is good to keep in mind that this
transformation only has one dependent variable f , whereas the rational transformation used two.
Just as with the rational transformation, we make use of a fundamental formulae which will facilitate the
procedure,
2 cosh(δ ∂z ) log f (z) = log[cosh(δ Dz )f (z) · f (z)]
8
We can prove the equation above essentially in the same manner as equation (5).
Now we can construct a formulae using taylor expansions and collecting the coefficients for the powers of δ.
Dt2 f · f
f2
Dx Dt f · f
2∂x ∂t log f =
f2
D2 f · f
2∂x2 log f = x 2
f
2∂t2 log f =
2∂x4
D4 f · f
log f = x 2 − 3
f
..
.
(8)
Dx2 f · f
f2
2
Note
Remember that Dzn = (Dt + ε Dx )n and ∂zn = (∂t + ε ∂x )n where ε is a parameter.
4.2.1
The KdV equation
In order exemplify the use of the logarithmic transformation, we study in detail the KdV equation.
ut + 6uux + uxxx = 0
First we represent (7) as a series of two transformation to facilitate the calculation,
(
u = wx ,
u = 2(log f )xx ∼
w = 2(log f )x .
First we use u = wx which is basically differentiating once with respect to x,
wxt + 6wx wxx + wxxxx = 0
By integrating this expression with respect to x we yield the following expression,
wt + 3wx2 + wxxx = c
Now we may use w = 2(log f )x ,
2
2∂x ∂t log f + 3 2∂x2 log f + 2∂x4 (log f ) = c
where c is the constant of integration. Next we employ the formulae (8) and substitute accordingly into the
equation above,
Dx (Dt + Dx3 )f · f = cf 2
The latter is the KdV bilinear equation. However, when searching for soliton equation it is necessary to let
c = 0. Therefore,
Dx (Dt + Dx3 )f · f = 0
4.3
Bi-logarithmic transformation
The bi-logarithmic transformation,
ϕ = log(a/b)
is usually seen along another dependent variable transformation,
ρ = log(a b).
The fundamental formulae used in this case are ,
2 sinh(δ∂x ) log(a/b)
= log[exp(δDx )a · b] − log[exp(−δDx )a · b]
2 cosh(δ∂x ) log(a/b)
= log[cosh(δDx )a · a] − log[cosh(δDx )b · b]
2 cosh(δ∂x ) log(a b)
= log[exp(δDx )a · b] + log[exp(−δDx )a · b]
9
Same as before, they can be easily proven using definition (4).
By expanding and collecting powers of δ we can produce the following formulae,
Dx a · b
ab
2
D
a · a D2 b · b
∂x2 log(a/b) = x 2 − x 2
2a
2b
2
2
D
a
·
b
D
xa · b
x
2
∂x log(a b) =
−
ab
ab
2
3
D a·b
D a·b
∂x3 log(a/b) = x
−3 x
ab
ab
..
.
∂x log(a/b) =
4.3.1
(9)
The mKdV equation
vt + 6v 2 vx + vxxx = 0
Let v = iϕx so that the above equation becomes,
ϕxt − 6(ϕx )2 ϕxx + ϕxxxx = 0
and integrate once with respect to x,
ϕt − 3ϕ2xx + ϕxxx = c.
Now we employ the bi-logarithmic transformation, φ = log(a/b),
∂t log(a/b) − 2(∂x log(a/b))3 + ∂x3 log(a/b) = c
By substituting the corresponding expression in formulae (9) into the equation above,
2
Dx a · b
Dx a · b
Dt a · b Dx3 a · b
+
−3
= c.
ab
ab
ab
ab
Note In the equation above we should choose c = 0 in order to decouple the equation.
By simplifying the latter equation we obtain,
a b (Dt + Dx3 )a · b − 3(Dx a · b)(Dx2 a · b) = 0
We introduce an arbitrary function λ to decouple the previous expression,
(Dt + Dx3 )a · b = 3λDx a · b
Dx2 a · b = λa b
The above is the bilinear form of the m-KdV equation.
4.3.2
The sine-Gordon equation
ϕxx − ϕtt = sin ϕ
We slightly modified the bi-logarithmic transformation,
ϕ = 2i log(f /f ∗ )
where f ∗ is the complex conjugate of f .
10
(10)
Note
By modifying equation (10) we see that,
iϕ = log (f ∗ /f )2
exp(iφ) = (f ∗ /f )2
Thus we can construct the following expressions,
1
exp(iϕ) − exp(−iϕ)
2i
1 f ∗2
f2 =
− ∗2
2
2i f
f
sin(ϕ) =
ϕxx =2i∂x2 log(f /f ∗ )
2
Dx2 f ∗ · f ∗
Dx f · f
−
=
f2
f ∗2
ϕtt =
Dt2 f · f
Dt2 f ∗ · f ∗
−
f2
f ∗2
By plugging this into the original equation and simplifying, we obtain,
!
2
∗2
(f
−
f
)
(Dx2 −Dt2 )f · f −
f ∗2
2
=
(Dx2
−
Dt2 )f ∗
!
(f ∗2 − f 2 )
·f −
f2
2
∗
By introducing an arbitrary function λ, we obtain the sine-Gordon bilinear equation along with its complex
conjugate,
(Dx2 − Dt2 )f · f − (f 2 − f ∗2 )/2 = λf 2 .
5
Solutions to Bilinear equations
To illustrate this, consider the bilinear KdV equation obtained by using u = 2(log f )xx as a logarithmic
transformation,
Dx (Dt + Dx3 )f · f = 0
(11)
Using the perturbation method on f with respect to the small parameter ε,
f = 1 + εf1 + ε2 f2 + ε3 f3 + . . .
(12)
Substituting the above into the bilinear equation (1) and collecting the coefficients for the powers of ε we
obtain the following,
ε : Dx (Dt + Dx3 )(f1 · 1 + 1 · f1 ) = 0
ε2 : Dx (Dt + Dx3 )(f2 · 1 + f1 · f1 + 1 · f2 ) = 0
ε3 : Dx (Dt + Dx3 )(f3 · 1 + f2 · f1 + f1 · f2 + 1 · f3 ) = 0
4 : Dx (Dt + Dx3 )(2f4 · 1 + 2f3 · f1 + f2 · f2 ) = 0
..
.
11
Note
We consider the following identities,
Dx Dt a · 1 = axt = Dx Dt 1 · a
Dx4 a · 1 = axxxx = Dx4 1 · a
Dxm Dtn exp η1 · exp η2 = (P1 − P2 )m (Ω1 − Ω2 )n
× exp(η1 + η2 )
where ηi = Pi x + Ωi t + c and Ωi + Pi = 0 for i = 1, 2, 3, . . .
The relation Ωi + Pi = 0 is called the nonlinear dispersion relation.
5.1
One-soliton solution
By simplifying the coefficient of order-ε we obtain,
∂x (∂t + ∂x3 )f1 = 0
(13)
for which we can choose the following to be a solution,
f1 = exp η1
Now we plug this solution into the coefficient of order-ε2 to obtain,
2∂x (∂t + ∂x3 )f2 = −Dx (Dt + Dx3 )f1 · f1 = 0
This means that we can choose the following solution to the equation above,
f2 = 0
We continue to plug these solutions into the terms for the powers of ε and see that the rest of them can
have solutions equal to 0. We substitute these solutions into equation (12) to find the one-soliton solution,
f = 1 + exp η1
5.2
Two-soliton solution
In order to find a two-soliton solution, we utilize the superposition principle. We may use this principle since
we are dealing with a bilinear equation and not a nonlinear one. Therefore, let
f1 = exp η1 + exp η2
be a solution to equation (13).
In the same manner as before, we substitute f1 into the coefficient for ε2 to obtain the following expression,
∂x (∂t + ∂x3 )f2 = −2(P1 − P2 )(Ω1 − Ω2 + (P1 − P2 )3 )
× exp(η1 + η2 )
A solution to the equation above would be of the form,
f2 = a12 exp(η1 + η2 )
where the constant a12 may be expressed as,
a12
h
i
2(P1 − P2 ) Ω1 − Ω2 + (P1 − P2 )3
h
i
=−
2(P1 + P2 ) Ω1 + Ω2 + (P1 + P2 )3
=
(P1 − P2 )2
(P1 + P2 )2
12
Note The constant a12 is chosen so as to represents the phase shift caused in the interaction between two
soliton waves.
Now we are able to substitute the solutions f1 and f2 into the coefficient for ε3 . The right hand side of the
equation will yield 0 and thus we obtain the following expression,
∂x (∂t + ∂x3 )f3 = 0
and for which we are able to choose f3 = 0 to be a solution. As we continue this method we see that we may
choose fr = 0 where r = 3, 4, 5, 6....
Therefore, we may write f as,
f = 1 + ε(exp η1 + exp η2 ) + ε2 a12 exp(η1 + η2 )
Because of the constant in ηi , the small parameter ε can be absorbed. Hence, the solution above can be written
as,
f = 1 + exp η1 + exp η2 + a12 exp(η1 + η2 )
(14)
Equation (14) is the two-soliton bilinear solution for the KdV equation. In terms of u it is written as,
u = 2(log f )xx
5.3
Three-soliton solution
In order to obtain the three-soliton solution we now choose,
f1 = exp η1 + exp η2 + exp η3
In the same manner, ultimately we find that,
f =1 + exp η1 + exp η2 ) + exp η3
+ a12 exp(η1 + η2 ) + a13 exp(η1 + η3 ) + a23 exp(η2 + η3 )
+ a123 exp(η1 + η2 + η3 )
where a123 = a12 a13 a23 is the three-soliton solution to the KdV equation in the bilinear form.
5.4
N-soliton solution
Before constructing the N-soliton solution, let us employ vector notation.
Consider the bilinear equation of the form,
F (Dt , Dx , Dy , . . . )f · f = 0
where F is a polynomial.
Now let D = (Dt , Dx , Dy , ...). Thus (15) becomes,
F (D)f · f = 0
We set a pair of conditions,
(i) F (−D) = F (D)
(ii) F (0, 0, 0, . . . ) = 0
13
(15)
Note Condition (i) is not truly necessary, but due to the antisymmetry of the D-operator, if F is an odd
function, it will automatically become 0. Condition (ii) is called the zero-soliton solution (or vacuum) and it is
essential. Otherwise, f is not a solution and F (0, 0, 0, ...)1 · 1 6= 0. Notice that it is necessary to pick f = 1 so
that u ≡ 0 since u = 2(log f )xx .
We may also make the following change in notation,
η i = Pi x + Ω i t + c
(i = 1, 2, 3, . . . )
⇒ ηi = Pi · xi + c
where Pi = (Pi , Ωi , Qi , . . . ) and xi = (x, t, y, . . . )
The nonlinear dispersive relation which is given by Ωi + Pi = 0 may now be written as,
F (Pi ) = 0
The phase shift can also be rewritten using the vector notation,
aij = −
F (Pi − Pj )
F (Pi + Pj )
(i, j = 1, 2, 3, . . . )
Let us recall the 1,2,3-soliton solutions,
f = 1 + exp η1
f = 1 + exp η1 + exp η2 + a12 exp(η1 + η2 )
f = 1 + exp η1 + exp η2 + exp η3
+ a12 exp(η1 + η2 ) + a13 exp(η1 + η3 )
+ a23 exp(η2 + η3 ) + a123 exp(η1 + η2 + η3 )
To construct some formula for any N -soliton, first let,
aij = exp Aij
(i, j = 1, 2, 3, . . . )
so that for example, the two-soliton solution may be written as,
f = 1 + exp η1 + exp η2 + exp(A12 + η1 + η2 )
We express f as,
f=
X


(N )
N
X
X
exp 
µi ηi +
Aij µi µj 
i=1
(16)
i<j
P
The first
represents
the sum of all possible combinations of µi = o, 1 where i = 1, 2, 3, . . . .
P
The third
represents the sum of all possible combinations of (i, j) from the set {N } with the condition
i < j. c Finally, for all KdV-type bilinear equations of the form,
F (D)f · f = 0
having N-soliton solutions, f is of the form (16).
Note
The polynomial F (D) is not arbitrary, there is a condition,
X
F
N
X
i=1
σi Pi
! (N )
Y
F (σi Pi − σj Pj )σi σj = 0
i<j
The condition above is called The Hirota Condition. The first
combinations of σN = 0, 1.
14
P
represents the summation of all possible
6
Visualization
In this section we present some examples of how one can plot soliton solutions using Mathematica.
6.1
One-soliton
The following is a mathematica code to produce a 3D plot of a one-soliton.
We may also write a code for a 2D plot,
15
6.2
Two-soliton
The following is a mathematica code to produce a 3D plot of a two-soliton.
16
We may also write a code for a 2D plot,
17
6.3
Three-soliton
The following is a mathematica code to produce a 3D plot of a three-soliton.
We may also write a code for a 2D plot,
18
19