Partial Differential Equations in Applied Mathematics 11 (2024) 100813
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Partial Differential Equations in Applied Mathematics
journal homepage: www.elsevier.com/locate/padiff
Integration of the Kaup–Boussinesq system via inverse scattering method
B.A. Babajanov a,b ,∗, Sh.O. Sadullaev a , M.M. Ruzmetov a
a
b
Urgench State University, 14, H.Alimdjan st., Urgench, 220100, Uzbekistan
Khorezm branch of Uzbekistan Academy of Sciences V.I.Romanovskiy Institute of Mathematics, Urgench, 220100, Uzbekistan
ARTICLE
INFO
Keywords:
Nonlinear soliton equation
Kaup–Boussinesq system
Inverse scattering method
Quadratic pencil of one-dimensional
Schrodinger operators
Time evolution of scattering data
ABSTRACT
In this paper, we investigate the Cauchy problem for the KB system in the class of rapidly decreasing functions
and give an algorithm of the constructing a solution to the Cauchy problem of the KB system in the class of
rapidly decreasing functions via IST method when the scattering coefficient 𝑎(𝑘) of the problem (6) has multiple
zeros. In this case, the class of initial functions will be expanded by removing the assumption about simplicity
of the scattering coefficient 𝑎(𝑘) of the problem (6). We also present an efficient method to obtain the time
evolution of scattering data. The advantage of this method is its reliability and the possibility of using other
soliton equations to obtain the time evolution of scattering data. The resulting equalities completely determine
the scattering data at any 𝑡, which allows applying the IST method to solve the Cauchy problem for the KB
system. The results obtained will be useful in carrying out further analysis in the context of shallow water
waves that arises in the context of oceanography.
1. Introduction
One of the most well-known topics in nonlinear science is the
study of soliton equations. Numerous scientific disciplines, including
fluid mechanics, solid-state physics, plasma physics, biology, optics,
earthquakes, and surface water waves—all exhibit nonlinear behaviour.
The use of the inverse scattering theory (IST) method, which was
originally presented by Gardner, Greene, Kruskal, and Miura in 1967,1
in nonlinear science has become the main achievement of mathematical
physics of the twentieth century, but there are still many unsolved
problems in this area. In particular, active research is underway to
extend the inverse problem method to new classes of equations.
The Kaup–Boussinesq(KB) system
{
𝜂𝜏 = 𝛷𝑥𝑥 + 𝛽 2 𝛷𝑥𝑥𝑥𝑥 − 𝜀 ⋅ (𝛷𝑥 𝜂)𝑥
(1)
𝜂 = 𝛷𝜏 + 12 𝜀 ⋅ 𝛷𝑥2 ,
describing wave propagation in shallow water was first derived by
Boussinesq.2 In Ref. 3, D.J. Kaup, using IST method, proved that this
system is completely integrable. Many efforts have been devoted to
seeking solitary and periodic wave solutions to KB system. For instance,
the periodic solutions of the KB system was investigated using Riemann
problem.4 In Ref. 5, a coupled trial equation method is used for a
obtaining the cosine function solution of KB system which shows an
important periodic motion. In Ref. 6, the simplified form of Hirota’s
direct method is used to determine multiple soliton solutions and
multiple singular soliton solutions of the KB system. In Ref. 7, the
authors studied the dual equation of the KB system by employing the
bifurcation theory of planar dynamical systems to investigate the travelling wave solutions. Other aspects on integration of soliton equations
were presented in Refs. 8–28.
After the following transformation (see Ref. 9)
𝜂=
4𝛽 2
4𝛽 2
4𝛽
1
1
(𝑣 + 𝑢2 ) + , 𝛷𝜏 =
(𝑣 + 3𝑢2 ) + , 𝛷𝑥 = − 𝑢, 𝑡 = 𝑖𝛽𝜏,
𝜀
𝜀
𝜀
𝜀
𝑖𝜀
the KB system (1) takes a simpler form
{
𝑣𝑡 = 𝑢𝑥𝑥𝑥 − 4𝑣𝑢𝑥 − 2𝑢𝑣𝑥 ,
𝑢𝑡 = −6𝑢𝑢𝑥 − 𝑣𝑥 .
(2)
The system (2) can be viewed as the compatibility condition
(see Ref. 9)
𝑦𝑥𝑥𝑡 − 𝑦𝑡𝑥𝑥 ≡ [(𝑣𝑡 − 𝑢𝑥𝑥𝑥 + 4𝑣𝑢𝑥 + 2𝑢𝑣𝑥 ) + 2𝑘(𝑢𝑡 + 6𝑢𝑢𝑥 + 𝑣𝑥 )]𝑦 = 0
for the system of linear equations
{
−𝑦𝑥𝑥 + 𝑣𝑦 + 2𝑘𝑢𝑦 − 𝑘2 𝑦 = 0,
𝑦𝑡 + 2𝑘𝑦𝑥 + 2𝑢𝑦𝑥 − 𝑢𝑥 𝑦 = 0.
The first of these equations is called the quadratic pencil of onedimensional Schrodinger operators. The IST method for the problem
(6) was investigated in the works.29–32
In this paper, IST method of the problem (6) is used to obtain
a representation of the solution of the Cauchy problem for the KB
∗ Corresponding author.
E-mail addresses: a.murod@mail.ru (B.A. Babajanov), serzodsadullaev27@gmail.com (Sh.O. Sadullaev), rmurod2002@gmail.com (M.M. Ruzmetov).
https://doi.org/10.1016/j.padiff.2024.100813
Received 2 June 2024; Received in revised form 5 July 2024; Accepted 8 July 2024
Available online 11 July 2024
2666-8181/© 2024 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/bync-nd/4.0/).
Partial Differential Equations in Applied Mathematics 11 (2024) 100813
B.A. Babajanov et al.
system in the class of rapidly decreasing functions. In works,33–37 the
Cauchy problem for the KB system was studied for both physical and
mathematical reasons, and solution of the problem was constructed by
the IST method under the assumption that the scattering coefficient 𝑎(𝑘)
of the problem (6) has only simple zeros. If this condition is removed,
then a new soliton solution of the Cauchy problem for the KB system
can be obtained via IST method.
Motivated by this reports, we investigate the Cauchy problem for
the KB system in the class of rapidly decreasing functions and give an
algorithm of the constructing a solution to the Cauchy problem of the
KB system via IST method when the scattering coefficient 𝑎(𝑘) of the
problem (6) has multiple zeros. In this case, the class of initial functions
will be expanded by removing the assumption about a simplicity of
the scattering coefficient 𝑎(𝑘) of the problem (6). We also present an
efficient method to obtain the time evolution of scattering data. The
advantage of this method is its reliability and the possibility of using
other soliton equations to obtain the time evolution of scattering data.
The resulting equalities completely determine the scattering data at any
𝑡, which allows applying the IST method to solve the Cauchy problem
for the KB system. The results obtained will be useful in carrying out
further analysis in the context of shallow water waves that arises in the
context of oceanography.
Moreover, the function 𝑎(𝑘, 𝑡) admits an analytic continuation to the
half-plane 𝐼𝑚𝑘 > 0 and can have at most a finite number of zeros
𝑘1 (𝑡), 𝑘2 (𝑡), … , 𝑘𝑁 (𝑡).
2. Formulation of the problem
𝑗 = 0, 1, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁.
The set of the quantities
{
We denote by 𝑚𝑛 (𝑡) the multiplicity of the root 𝑘𝑛 (𝑡) of the equation
𝑎(𝑘, 𝑡) = 0. There exist chains of numbers
+
+
+
+
{𝜒𝑛,0
(𝑡), 𝜒𝑛,1
(𝑡), 𝜒𝑛,2
(𝑡), … , 𝜒𝑛,𝑚
(𝑡)}
−1
𝑛
and
−
−
−
−
{𝜒𝑛,0
(𝑡), 𝜒𝑛,1
(𝑡), 𝜒𝑛,2
(𝑡), … , 𝜒𝑛,𝑚
(𝑡)},
−1
𝑛
satisfying inequality
𝑗
∑
|
|
1 𝑑𝑗
1 𝑑𝜈
±
|
𝑓
(𝑥,
𝑡,
𝑘)
=
𝑓 (𝑥, 𝑡, 𝑘)||
𝜒𝑛,𝑗−𝜈
(𝑡)
,
∓
|
𝜈 ±
𝑗! 𝑑𝑘𝑗
𝜈!
𝑑𝑘
|𝑘=𝑘𝑛 𝜈=0
|𝑘=𝑘𝑛 (𝑡)
+
− (𝑡) = 1,
wherein 𝜒𝑛,0
(𝑡)𝜒𝑛,0
| 𝜒 ± (𝑡)
| 𝑛,1
|
| 𝜒 ± (𝑡)
𝑗
| 𝑛,2
(−1)
∓
|
𝜒𝑛,𝑗 (𝑡) = ±
𝑗+1
(𝜒𝑛,0 ) (𝑡) ||
...
|
| ±
| 𝜒𝑛,𝑗 (𝑡)
|
𝑢𝑡 = −6𝑢𝑢𝑥 − 𝑣𝑥 , 𝑥 ∈ R, 𝑡 > 0,
∫−∞
...
±
𝜒𝑛,0
(𝑡)
...
...
...
...
±
𝜒𝑛,𝑗−1
(𝑡)
±
𝜒𝑛,𝑗−2
(𝑡)
...
|
|
|
|
0
|
|,
|
|
...
|
|
±
𝜒𝑛,1 (𝑡) ||
0
(13)
and
{
(4)
𝑟+ (𝑡, 𝑘) = −
in which 𝑣0 (𝑥) and 𝑢0 (𝑥) are real functions satisfying the inequalities
∞
0
±
𝜒𝑛,1
(𝑡)
(3)
under the initial condition
𝑣(𝑥, 𝑡)|𝑡=0 = 𝑣0 (𝑥), 𝑢(𝑥, 𝑡)|𝑡=0 = 𝑢0 (𝑥), 𝑥 ∈ R,
±
𝜒𝑛,0
(𝑡)
}
{
𝑏(𝑡, 𝑘)
−
𝑟− (𝑡, 𝑘) =
, 𝑘 ∈ 𝑅, 𝑘𝑛 (𝑡), 𝜒𝑛,𝑗
(𝑡), 𝑗 = 0, 1, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁
𝑎(𝑡, 𝑘)
We consider the simpler form KB system
𝑣𝑡 = 𝑢𝑥𝑥𝑥 − 4𝑣𝑢𝑥 − 2𝑢𝑣𝑥 ,
(12)
̄ 𝑘)
𝑏(𝑡,
+
, 𝑘 ∈ 𝑅, 𝑘𝑛 (𝑡), 𝜒𝑛,𝑗
(𝑡), 𝑗 = 0, 1, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁
𝑎(𝑡, 𝑘)
(14)
∞
|𝑢0 (𝑥)|𝑑𝑥 < ∞,
∫−∞
(1 + |𝑥|)[|𝑣0 (𝑥)| + |𝑢′0 (𝑥)|]𝑑𝑥 < ∞.
}
(5)
The main aim of this work is to derive representations for the
solution {𝑣(𝑥, 𝑡), 𝑢(𝑥, 𝑡)} of the problem (3)–(5) with the use of the
IST method of the quadratic pencil of one-dimensional Schrodinger
operators 𝐿 = 𝐿(𝑡, 𝑘):
are called the left and right scattering data of Eq. (6). Functions
𝑟− (𝑡, 𝑘) and 𝑟+ (𝑡, 𝑘) are called the left and right reflection coefficients,
respectively.
The main equation of the IST for the problem (6) has the form
(see Ref. 29):
𝐿𝑦 ≡ −𝑦𝑥𝑥 + 𝑣(𝑥, 𝑡)𝑦 + 2𝑘𝑢(𝑥, 𝑡)𝑦 − 𝑘2 𝑦 = 0, 𝑥 ∈ R, 𝑡 ≥ 0.
𝑒𝑖𝛼+ (𝑥,𝑡) 𝐹+ (𝑥 + 𝑦, 𝑡) + 𝐾+ (𝑥, 𝑦, 𝑡) +
∞
(6)
∫𝑥
𝐾+ (𝑥, 𝜏, 𝑡)𝐹+ (𝜏 + 𝑦, 𝑡)𝑑𝜏 = 0, 𝑥 ≤ 𝑦 < ∞,
(15)
3. Preliminaries
In this section, we give main facts about the IST for problem (6)
(see Ref. 29).
Let us denote by 𝑓+ (𝑥, 𝑡, 𝑘) and 𝑓− (𝑥, 𝑡, 𝑘) the Jost solutions of
problem (6) satisfying the asymptotics
𝑒𝑖𝛼− (𝑥,𝑡) 𝐹− (𝑥 + 𝑦, 𝑡) + 𝐾− (𝑥, 𝑦, 𝑡) +
𝑓+ (𝑥, 𝑡, 𝑘) = 𝑒𝑖𝑘𝑥 [1 + 𝑜(1)], 𝑥 → +∞,
(7)
where
𝑓− (𝑥, 𝑡, 𝑘) = 𝑒−𝑖𝑘𝑥 [1 + 𝑜(1)], 𝑥 → −∞.
(8)
𝛼+ (𝑥, 𝑡) =
(9)
𝐹± (𝑥, 𝑡) =
𝑥
∫−∞
𝐾− (𝑥, 𝜏, 𝑡)𝐹− (𝜏 + 𝑦, 𝑡)𝑑𝜏 = 0, −∞ < 𝑦 ≤ 𝑥,
(16)
∞
∫𝑥
𝑥
𝑢(𝜏, 𝑡)𝑑𝜏,
𝛼− (𝑥, 𝑡) =
∫−∞
𝑢(𝜏, 𝑡)𝑑𝜏,
For real 𝑘 ≠ 0, the following relations hold
𝑓+ (𝑥, 𝑡, 𝑘) = 𝑏(𝑡, 𝑘)𝑓− (𝑥, 𝑡, 𝑘) + 𝑎(𝑘, 𝑡)𝑓̄− (𝑥, 𝑡, 𝑘),
𝑁
∑
∞
𝑃𝑛± (𝑥, 𝑡)𝑒±𝑖𝑘𝑛 (𝑡)𝑥 +
𝑛=1
̄ 𝑡)𝑓+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑘, 𝑡)𝑓̄+ (𝑥, 𝑡, 𝑘).
𝑓− (𝑥, 𝑡, 𝑘) = −𝑏(𝑘,
1
𝑟 (𝑡, 𝑘)𝑒±𝑖𝑘𝑥 𝑑𝑘,
2𝜋 ∫−∞ ±
(17)
here 𝑃𝑛± (𝑥, 𝑡) is a polynomial in 𝑥 of degree 𝑚𝑛 − 1, having the form
(10)
The bar over the function here and below denotes complex conjugation.
It is easy to see that
{
}
1
𝑊 𝑓+ (𝑥, 𝑡, 𝑘), 𝑓− (𝑥, 𝑡, 𝑘) ,
𝑎(𝑘, 𝑡) = −
2𝑖𝑘
{
}
1
𝑏(𝑘, 𝑡) =
𝑊 𝑓+ (𝑥, 𝑡, 𝑘), 𝑓̄− (𝑥, 𝑡, 𝑘) , 𝑘 ∈ 𝑅∗ = (−∞, ∞)∖{0}.
(11)
2𝑖𝑘
𝑚𝑛 −1
𝑃𝑛± (𝑥, 𝑡) = −𝑖𝑒∓𝑖𝑘𝑛 (𝑡)𝑥
∑
𝑠=0
±
𝜒𝑛,𝑚
(𝑡)
−1−𝑠
𝑛
1
𝑠!
{
𝑚
𝑑 𝑠 (𝑘 − 𝑘𝑛 (𝑡)) 𝑛 𝑒
𝑑𝑘𝑠
𝑎(𝑡, 𝑘)
±𝑖𝑘𝑥 } |
|
.
|
|𝑘=𝑘𝑛 (𝑡)
Eqs. (15) and (16) are an analogue of the Gelfand–Levitan–Marchenko
integral equation for problem (6). To ensure that Eqs. (15) and (16)
have a unique solution for any 𝑡 ≥ 0, we add the condition
2
Partial Differential Equations in Applied Mathematics 11 (2024) 100813
B.A. Babajanov et al.
{
∞
𝜔+ (𝑦, 𝑡) = ∫𝑥 𝐹+ (𝑦 + 𝜏, 𝑡)𝜔+ (𝜏, 𝑡)𝑑𝜏, 𝑥 ≤ 𝑦 < ∞,
𝑥
𝜔− (𝑦, 𝑡) = ∫−∞ 𝐹− (𝑦 + 𝜏, 𝑡)𝜔− (𝜏, 𝑡)𝑑𝜏, −∞ < 𝑦 ≤ 𝑥,
(
⇒
𝜔+ (𝑦, 𝑡)
𝜔− (𝑦, 𝑡)
)
(
=
0
Here and hereafter, the dot over a letter denotes the derivative with
respect to 𝑡 and the prime denotes the derivative with respect to 𝑥.
⇒
)
Proof. Differentiating (6) with respect to 𝑡, we obtain
.
0
̇
𝐿𝑦̇ = −𝐿𝑦,
That is, we require that a homogeneous system of integral equations
have only a zero solution. If this condition is met, then the problem is
uniquely solvable. For example, this condition is automatically satisfied
in the following two cases:
(i) the discrete spectrum is absent;
(ii) the spectrum of the pencil consists only of the discrete spectrum,
i.e., 𝑟+ (𝑡, 𝑘) = 0.
In the general case, i.e., without an additional condition, we could
not solve the problem under consideration.
We now turn to the question of constructing 𝑢(𝑥, 𝑡) and 𝑣(𝑥, 𝑡) from
scattering data (14). To restore the coefficient functions 𝑢(𝑥, 𝑡) and
𝑣(𝑥, 𝑡) in Eq. (6) from the right reflection coefficient 𝑟+ (𝑡, 𝑘) we proceed
as follows (see Ref. 29):
(1) It is necessary to find the function 𝐹+ (𝑥, 𝑡) by formula (17) and
solve with respect to 𝐾+0 (𝑥, 𝑦, 𝑡) ∈ 𝐿1 (𝑥, ∞), 𝐾+1 (𝑥, 𝑦, 𝑡) ∈ 𝐿1 (𝑥, ∞)
integral equations
𝐹+ (𝑥 + 𝑦, 𝑡) + 𝐾+(0) (𝑥, 𝑦, 𝑡) +
∞
∫𝑥
where
𝐿̇ = 𝑣𝑡 + 2𝑘𝑢𝑡 .
Now we calculate 𝐿𝑧:
𝐿𝑧 = 𝐿(𝑦̇ + 2(𝑢 + 𝑘)𝑦′ − 𝑢𝑥 𝑦) = 𝐿𝑦̇ + 𝐿(2(𝑢 + 𝑘)𝑦′ − 𝑢𝑥 𝑦).
∞
∫𝑥
𝐿𝑧 = −(𝑞𝑡 − 𝑝𝑥𝑥𝑥 + 4𝑞𝑝𝑥 + 2𝑝𝑞𝑥 )𝑦 − 2𝑘(𝑝𝑡 + 6𝑝𝑝𝑥 + 𝑞𝑥 )𝑦 = 0.
Step 2. At this stage, a new equality is obtained for the time
derivative of the Jost solution to Eq. (6) using the asymptotics of
fundamental systems of solutions of Eq. (6).
Lemma 2. Let 𝑓− (𝑥, 𝑡, 𝑘) be Jost solution of Eq. (6). The following equality
is hold:
𝐾+(0) (𝑥, 𝜏, 𝑡)𝐹+ (𝜏 + 𝑦, 𝑡)𝑑𝜏 = 0, 𝑥 ≤ 𝑦 < ∞,
𝑓̇− (𝑥, 𝑡, 𝑘) = 𝐴𝑓− (𝑥, 𝑡, 𝑘) − 2𝑖𝑘2 𝑓− (𝑥, 𝑡, 𝑘),
𝐾+(1) (𝑥, 𝜏, 𝑡)𝐹+ (𝜏 + 𝑦, 𝑡)𝑑𝜏 = 0, 𝑥 ≤ 𝑦 < ∞.
Proof. For real 𝑘 ≠ 0, the pairs 𝑓− (𝑥, 𝑡, 𝑘), 𝑓̄− (𝑥, 𝑡, 𝑘) are fundamental
systems of solutions to Eq. (6). Then the solution 𝑧(𝑥, 𝑡, 𝑘) = 𝑓̇− (𝑥, 𝑡, 𝑘) −
𝐴𝑓− (𝑥, 𝑡, 𝑘) is linearly expressed in terms of 𝑓− (𝑥, 𝑡, 𝑘), 𝑓̄− (𝑥, 𝑡, 𝑘) as
follows:
(2) Next, the function 𝛼+ (𝑥, 𝑡) is defined as a solution to the nonlinear integral equation of the Volterra type
𝑓̇− (𝑥, 𝑡, 𝑘) − 𝐴𝑓− (𝑥, 𝑡, 𝑘) = 𝑐(𝑡, 𝑘)𝑓− (𝑥, 𝑡, 𝑘) + 𝑑(𝑡, 𝑘)𝑓̄− (𝑥, 𝑡, 𝑘).
∞
where
∫𝑥
(28)
where 𝐴𝑓− (𝑥, 𝑡, 𝑘) = 𝑢𝑥 𝑓− (𝑥, 𝑡, 𝑘) − 2(𝑢 + 𝑘)𝑓−′ (𝑥, 𝑡, 𝑘).
(19)
𝛼+ (𝑥, 𝑡) =
(27)
Substituting the expression (26) into (27) and after a long computation
we derive
(18)
𝑖𝐹+ (𝑥 + 𝑦, 𝑡) + 𝐾+(1) (𝑥, 𝑦, 𝑡) +
(26)
𝛷(𝑠, 𝛼+ (𝑠, 𝑡))𝑑𝑠, −∞ < 𝑥 < ∞,
(20)
(29)
From the following representation of Jost’s solutions
∞
𝑓+ (𝑥, 𝑘) = 𝑒𝑖𝛼+ (𝑥) ⋅ 𝑒𝑖𝑘𝑥 +
]
Φ (𝑠, 𝑧) = Re 𝐾+(0) (𝑠, 𝑠, 𝑡) − Im 𝐾+(1) (𝑠, 𝑠, 𝑡) sin 2𝑧+
[
]
]
[
[
+ 2 Re 𝐾+(1) (𝑠, 𝑠, 𝑡) sin2 𝑧 − 2 Im 𝐾+(0) (𝑠, 𝑠, 𝑡) cos2 𝑧
(21)
(22)
(3) Find the coefficients 𝑢(𝑥, 𝑡), 𝑣(𝑥, 𝑡) of Eq. (6) are from the formu𝜕𝛼+ (𝑥, 𝑡)
,
𝜕𝑥
𝑐(𝑡, 𝑘) = −2𝑖𝑘2 ;
(30)
Theorem 3. If the functions 𝑣 = 𝑣(𝑥, 𝑡), 𝑢 = 𝑢(𝑥, 𝑡) are solutions of the
problem (3)–(4), then the scattering data of the operator
(24)
4. Evolution of scattering data
𝐿𝑦 ≡ −𝑦𝑥𝑥 + 𝑣(𝑥, 𝑡)𝑦 + 2𝑘𝑢(𝑥, 𝑡)𝑦 − 𝑘2 𝑦 = 0, 𝑥 ∈ R, 𝑡 ≥ 0
In this section, we give an algorithm of the deriving time evolution
of the scattering data, which allows us to present a procedure for
solving the problem (3)–(5).
Step 1. At this stage, a new solution to Eq. (6) is obtained using
Eq. (3).
depend on 𝑡 as follows
Lemma 1. If the functions 𝑣 = 𝑣(𝑥, 𝑡), 𝑢 = 𝑢(𝑥, 𝑡) and 𝑦 = 𝑦(𝑥, 𝑡, 𝑘) are
solutions to the Eqs. (3) and (6) respectively, then the function
𝑧(𝑥, 𝑡, 𝑘) = 𝑦̇ + 2(𝑢 + 𝑘)𝑦 − 𝑢𝑥 𝑦
𝑑(𝑡, 𝑘) = 0.
Step 3. At this stage, the time evolution of the scattering data is
obtained using the results of steps 1 and 2.
2
′
𝐾− (𝑥, 𝜏)𝑒−𝑖𝑘𝜏 𝑑𝜏,
Setting (30) to (29) we deduce (28).
(23)
]
𝜕 {[
𝑣(𝑥, 𝑡) = −𝑢 (𝑥, 𝑡) − 2
Re 𝐾+ (𝑥, 𝑥, 𝑡) cos 𝛼+ (𝑥, 𝑡)
𝜕𝑥
[
]
}
+ Im 𝐾+ (𝑥, 𝑥, 𝑡) sin 𝛼+ (𝑥, 𝑡) .
∫−∞
one can easily derive the asymptotics: 𝑓̇− (𝑥, 𝑡, 𝑘) ∼ 0, 𝑓− (𝑥, 𝑡, 𝑘) ∼
𝑒−𝑖𝑘𝑥 , 𝑓−′ (𝑥, 𝑡, 𝑘) ∼ −𝑖𝑘𝑒−𝑖𝑘𝑥 , 𝑓̄− (𝑥, 𝑡, 𝑘) ∼ 𝑖𝑘𝑒𝑖𝑘𝑥 for 𝑥 → −∞. Proceeding
to the limit in equality (29) as 𝑥 → −∞, in view of these asymptotics
and condition (4), we get
las
𝑢(𝑥, 𝑡) = −
𝐾+ (𝑥, 𝜏)𝑒𝑖𝑘𝜏 𝑑𝜏,
𝑥
𝑓− (𝑥, 𝑘) = 𝑒𝑖𝛼− (𝑥) ⋅ 𝑒−𝑖𝑘𝑥 +
and
𝐾+ (𝑥, 𝑦, 𝑡) = 𝐾+(0) (𝑥, 𝑦, 𝑡) cos 𝛼+ (𝑥, 𝑡) + 𝐾+(1) (𝑥, 𝑦, 𝑡) sin 𝛼+ (𝑥, 𝑡) .
∫𝑥
𝑑𝑟+ (𝑡, 𝑘)
= −4𝑖𝑘2 𝑟+ (𝑡, 𝑘), 𝑘 ∈ R,
𝑑𝑡
(31)
𝑑𝑘𝑛 (𝑡)
= 0, 𝑛 = 1, 2, … , 𝑁,
𝑑𝑡
(32)
+
𝑑𝜒𝑛,0
(𝑡)
𝑑𝑡
(25)
+
𝑑𝜒𝑛,1
(𝑡)
is also a solution to Eq. (6).
𝑑𝑡
3
+
= −4𝑖𝑘2𝑛 𝜒𝑛,0
(𝑡),
(33)
+
+
= −4𝑖𝑘2𝑛 𝜒𝑛,1
(𝑡) − 8𝑖𝑘𝑛 𝜒𝑛,0
(𝑡),
(34)
Partial Differential Equations in Applied Mathematics 11 (2024) 100813
B.A. Babajanov et al.
+
𝑑𝜒𝑛,2
(𝑡)
𝑑𝑡
+
+
+
= −4𝑖𝑘2𝑛 𝜒𝑛,2
(𝑡) − 8𝑖𝑘𝑛 𝜒𝑛,1
(𝑡) − 4𝑖𝜒𝑛,0
(𝑡),
Using equality (42) and (43), taking into account the asymptotics of
Jost solution and its derivatives with respect to 𝑥 and 𝑡 for 𝑥 → +∞
and equating the coefficients for (𝑖𝑥)𝑙 𝑒𝑖𝑘𝑛 (𝑡)𝑥 , 𝑙 = 0, 1, … , 𝑚𝑛 − 1, we find
(33)–(36).
(35)
+ (𝑡)
𝑑𝜒𝑛,𝑝
+
+
+
= −4𝑖𝑘2𝑛 𝜒𝑛,𝑝
(𝑡) − 8𝑖𝑘𝑛 𝜒𝑛,𝑝−1
(𝑡) − 4𝑖𝜒𝑛,𝑝−2
(𝑡),
𝑑𝑡
𝑝 = 3, 4, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁.
(36)
Remark 1. The obtained results completely determine the time evolution of spectral data, which allows solving problem (3)–(4) by the
following algorithm. Assume that 𝑣0 (𝑥) and 𝑢0 (𝑥) are given.
1. For given functions 𝑣0 (𝑥) and 𝑢0 (𝑥), find the scattering data at
𝑡=0
{
}
+
𝑟+ (𝑘), 𝑘 ∈ 𝑅, 𝑘𝑛 , 𝜒𝑛,𝑗
, 𝑗 = 0, 1, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁
Proof. Differentiate (10) with respect to 𝑡 we obtain
̄̇ 𝑘)𝑓+ (𝑥, 𝑡, 𝑘) − 𝑏(𝑡,
̄ 𝑘)𝑓̇+ (𝑥, 𝑡, 𝑘)+
𝑓̇− (𝑥, 𝑡, 𝑘) = −𝑏(𝑡,
̇
̄
̄
+𝑎(𝑡,
̇ 𝑘)𝑓 (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓 (𝑥, 𝑡, 𝑘).
+
(37)
for 𝐿(0, 𝑘);
2. By Theorem 3, find the scattering data
{
}
+
𝑟+ (𝑡, 𝑘), 𝑘 ∈ 𝑅, 𝑘𝑛 (𝑡), 𝜒𝑛,𝑗
(𝑡), 𝑗 = 0, 1, … , 𝑚𝑛 − 1; 𝑛 = 1, 2, … , 𝑁
+
Substituting the expression (10) into (28) we derive
̄ 𝑘)𝑓+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘))′ +
𝑓̇− (𝑥, 𝑡, 𝑘) = −2(𝑢 + 𝑘)(−𝑏(𝑡,
𝑥
̄ 𝑘)𝑓+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘))−
+𝑢𝑥 (−𝑏(𝑡,
(38)
at 𝑡 > 0 for 𝐿(𝑡, 𝑘);
3. Now, the function 𝐹+ (𝑥, 𝑡) is determined from relation (17);
4. Substituting 𝐹+ (𝑥, 𝑡) into Eqs. (18), (19) and solving them, obtain
𝐾+(0) (𝑥, 𝑦, 𝑡) and 𝐾+(1) (𝑥, 𝑦, 𝑡);
5. Further, from (20)–(22), derive 𝐾+ (𝑥, 𝑦, 𝑡) and then find the
potentials 𝑣(𝑥, 𝑡) and 𝑢(𝑥, 𝑡) by formulas (23) and (24).
̄ 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘))
−2𝑖𝑘2 (−𝑏(𝑡,
From (37) and (38) we get
̄̇ 𝑘)𝑓+ (𝑥, 𝑡, 𝑘) − 𝑏(𝑡,
̄ 𝑘)𝑓̇+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡,
− 𝑏(𝑡,
̇ 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘)+
(39)
′
̇
̄
̄
̄
+𝑎(𝑡, 𝑘)𝑓 (𝑥, 𝑡, 𝑘) = −2(𝑢 + 𝑘)(−𝑏(𝑡, 𝑘)𝑓 (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓 (𝑥, 𝑡, 𝑘)) +
+
+
+
𝑥
5. Conclusion
̄ 𝑘)𝑓+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘))−
+𝑢𝑥 (−𝑏(𝑡,
2
̄ 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘) + 𝑎(𝑡, 𝑘)𝑓̄+ (𝑥, 𝑡, 𝑘))
−2𝑖𝑘 (−𝑏(𝑡,
In this paper, we extended the class of initial functions of the Cauchy
problem for the KB system and presented an efficient method to obtain
the time evolution of scattering data, which allows applying the ITS
method to solve the Cauchy problem for the KB system in the class
of rapidly decreasing functions. Namely, we considered the case when
the scattering coefficient 𝑎(𝑡, 𝑘) of the problem (6) has multiple zeros.
This case includes a larger class of initial functions than previously
studied works. The results obtained will be useful in carrying out
further analysis in the context of shallow water waves that arises in
the context of oceanography.
Passing in the equality (39) to the limit 𝑥 → +∞ and using asymptotics
of the Jost solution and its derivatives with respect to 𝑥 and 𝑡, we obtain
̇ 𝑘) = 4𝑖𝑘2 𝑏(𝑡, 𝑘),
𝑏(𝑡,
(40)
𝑎(𝑡,
̇ 𝑘) = 0,
(41)
It follows from (41) that the zeros 𝑘𝑛 (𝑡), 𝑛 = 1, 2, … , 𝑁 of the function
𝑎(𝑡, 𝑘) are also independent of time, which means that relation (32) is
true. According to (40), (41) and the form of the function 𝑟+ (𝑡, 𝑘), we
obtain (31).
We now turn to find the evolution of the chains of numbers
Declaration of competing interest
+
+
+
{𝜒𝑛,0
, 𝜒𝑛,1
, … , 𝜒𝑛,𝑚
, 𝑛 = 1, 2, … , 𝑁}.
𝑛 −1
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to
influence the work reported in this paper.
To this end, we differentiate (28) 𝑚𝑛 −1 times with respect to 𝑘, putting
𝑘 = 𝑘𝑛 (𝑡), we get
Data availability
𝜕 𝑚𝑛 −1 𝑓−′ (𝑥, 𝑡, 𝑘𝑛 (𝑡))
𝜕 𝑚𝑛 −1 𝑓̇− (𝑥, 𝑡, 𝑘𝑛 (𝑡))
= −2(𝑢 + 𝑘𝑛 (𝑡))
+
𝜕𝑘𝑚𝑛 −1
𝜕𝑘𝑚𝑛 −1
+(𝑢𝑥 − 2𝑖𝑘2𝑛 (𝑡))
−2𝐶𝑚1 −1
𝜕𝑘𝑚𝑛 −1
𝜕 𝑚𝑛 −2 𝑓−′ (𝑥, 𝑡, 𝑘𝑛 (𝑡))
𝑛
− 4𝑖𝐶𝑚2 −1
𝑛
𝜕 𝑚𝑛 −1 𝑓− (𝑥, 𝑡, 𝑘𝑛 (𝑡))
𝜕𝑘𝑚𝑛 −2
𝜕𝑘𝑚𝑛 −3
References
−
− 4𝑖𝑘𝑛 (𝑡)𝐶𝑚1 −1
𝜕 𝑚𝑛 −3 𝑓− (𝑥, 𝑡, 𝑘𝑛 (𝑡))
No data was used for the research described in the article.
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𝜕𝑘𝑚𝑛 −2
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∑
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,
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𝜈=0
(43)
𝑗 = 0, 1, … , 𝑚𝑛 − 1.
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