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Physics
Letters
A
www.elsevier.com/locate/pla
Nonlocal
KdV
equations
Metin Gürses
a,
1,
Aslı Pekcan
b,
∗
,
1aDepartmentofMathematics,FacultyofScience,BilkentUniversity,06800Ankara,Turkey bDepartmentofMathematics,FacultyofScience,HacettepeUniversity,06800Ankara,Turkey
a
r
t
i
c
l
e
i
n
f
o
a
b
s
t
r
a
c
t
Articlehistory:
Received23April2020
Receivedinrevisedform10August2020 Accepted14September2020
Availableonline18September2020 CommunicatedbyB.Malomed
Keywords:
Hirota-Satsumasystem LocalandnonlocalKdVequations Ablowitz-Musslimanireductions Hirotamethod
WritingtheHirota-Satsuma(HS)systemofequationsinasymmetricalformwefinditslocalandnew nonlocal reductions. It turns out that all reductions of the HS system are Korteweg-de Vries (KdV), complexKdV,andnewnonlocalKdVequations.Weobtainone-solitonsolutionsoftheseKdVequations byusingthemethodofHirotabilinearization.
©2020ElsevierB.V.Allrightsreserved.
1. Introduction
After Ablowitz and Musslimani introduced nonlocal type of reductions [1–3] we witnessed several works about finding new integrable nonlocal nonlinear partial differential equations and obtaining different kinds of solutions of these local and nonlocal equations. In partic-ular, most of the works focused on nonlocal nonlinear Schrödinger equations (NLS) [1–14], nonlocal modified Korteweg-de Vries (mKdV) equations [2–4], [8], [15–18], nonlocal sine-Gordon (SG) equations [2–4], [19], and so on [20–33].
Recently, it has been shown that systems admitting nonlocal reductions have discrete symmetry transformations leaving the systems invariant. In [34] we showed that a special case of discrete symmetry transformations are actually the nonlocal reductions of the same systems. The connection between local and nonlocal reductions was given in [35], [36]. Among all the nonlocal equations presented so far in the literature, nonlocal KdV equation is missing. It is not possible to obtain it from the nonlocal reductions of the AKNS system. For this purpose, we use the Hirota-Satsuma (HS) system and its reductions.
The original integrable HS system of equations [37–46] is given by apt
=
1
2pxxx
+
3ppx−
6qqx,
(1)aqt
= −
qxxx−
3pqx.
(2)This system has a recursion operator [39], [40], [46], N-soliton
solutions, and infinitely many conserved quantities [
52]. Several methods were applied to find solutions of (1) and (2) such as Hirota bilinear method [37], [43], [47], dressing operator technique [48], Darboux transformation method [49], [50], formally variable separation approach [51], and so on. The above form of the HS system is not appro-priate for consistent local and nonlocal reductions. It is more useful to map it to a more symmetrical version. By lettingp
=
12
(
u+
v),
q=
γ
(
u−
v)
(3)with γ2
=
14, we obtain the symmetrical integrable HS system [53–57]
*
Correspondingauthor.E-mailaddresses:[email protected](M. Gürses),[email protected](A. Pekcan).
1 Theseauthorscontributedequallytothiswork. https://doi.org/10.1016/j.physleta.2020.126894
aut
= −
uxxx+
3vxxx−
6uux+
6vux+
12vxu,
(4)avt
= −
vxxx+
3uxxx−
6v vx+
6uvx+
12vux,
(5)where a is
a constant. We can also obtain the recursion operator of the symmetrical HS system from the original one. By the use of this
recursion operator it is possible to obtain(
2+
1)
-dimensional negative HS system.It is possible to obtain soliton solutions of the above symmetrical HS system by the Hirota bilinearization method. The bilinearizing transformation and the Hirota bilinear form of the original system (1) and (2) were given in [37]. By using the result of [37] with the transformation (3) and introducing two additional constants u0 and v0 and by letting u
=
u0+
2(
ln f)
xx+
gf and v=
v0+
2(
ln f)
xx−
gf, we obtain the Hirota bilinear form of the system (4) and (5) as(
aDt+
4D3x+ (
18u0−
6v0)
Dx)
{
g·
f} =
0,
(6)(
aDxDt−
2D4x−
6(
u0+
v0)
Dx)
{
f·
f} = −12g
2.
(7)If the local or nonlocal reductions of integrable systems are done consistently the reduced equations must also be integrable. In this work we find the integrable reductions of the HS system (4) and (5). All the reductions of the HS system gives a kind of KdV equation. We find the standard KdV equation
aut
=
2uxxx+
12uux (8)from the real reduction, complex KdV equation
aut
= −
uxxx+
3u¯
xxx−
6uux+
6uu¯
x+
12uu¯
x (9)from the complex reduction, nonlocal KdV equation
aut
(
x,
t)
= −
uxxx(
x,
t)
+
3uxxx(
−
x,
−
t)
−
6u(
x,
t)
ux(
x,
t)
+
6u(
−
x,
−
t)
ux(
x,
t)
+
12ux(
−
x,
−
t)
u(
x,
t)
(10) from the real nonlocal reduction, and complex nonlocal KdV equationsaut
(
x,
t)
= −
uxxx(
x,
t)
+
3uxxx(
ε
1x,
ε
2t)
−
6u(
x,
t)
ux(
x,
t)
+
6u(
ε
1x,
ε
2t)
ux(
x,
t)
+
12ux(
ε
1x,
ε
2t)
u(
x,
t)
(11) from the nonlocal complex reductions, where ε21
=
ε
22=
1. We give the derivation of all these equations in Section 3. In Section 4 we obtain the one-soliton solutions of the above local and nonlocal KdV equations by using the one-soliton solutions of the HS system obtained in Section2and the reduction constraints.2. One-solitonsolutionsoftheHSsystem
To obtain the soliton solutions of the reduced equations we need the soliton solutions of the HS system. For this purpose, we use the Hirota bilinear equations given in (6) and (7).
Let g
=
ε
g1 and f=
1+
ε
2f2 where g1=
ceθ1 forθ
1=
k1x+
ω
1t+ δ
1and c isa constant, in the Hirota bilinear equations (
6) and (7). From the coefficient of ε, we get the dispersion relationω
1=
−
4k31
+ (
6v0−
18u0)
k1a (12)
if c
=
0. The coefficient of ε2 givesaf2,xt
−
2 f2,xxxx−
6(
u0+
v0)
f2,xx= −
6c2e2k1x+2ω1+2δ1 (13) yielding f2=
A1ek2x+ω2t+δ2+
A2e2k1x− 8k31 a t+2δ1,
(14) whereω
2=
2k32+
6(
u0+
v0)
k2 a,
(15) and A2=
c2 8k21(
k21+
2u0)
,
(16)and A1is an arbitrary constant. For c
=
0, A1=
0, the coefficients of ε3 and ε4vanish ifk2
=
k1±
−
k21−
4u0.
(17)u
(
x,
t)
=
u0+
c(
eθ1+
A 2e3θ1+
A1eθ1+θ2)
+
8k21A2e2θ1+
2 A1k22eθ2+
2 A1A2(
2k1−
k2)
2eθ2+2θ1(
1+
A1eθ2+
A2e2θ1)
2,
(18) v(
x,
t)
=
v0+
−
c(
eθ1+
A 2e3θ1+
A1eθ1+θ2)
+
8k21A2e2θ1+
2 A1k22eθ2+
2 A1A2(
2k1−
k2)
2eθ2+2θ1(
1+
A1eθ2+
A2e2θ1)
2,
(19)where
θ
1=
k1x+
ω
1t+ δ
1,θ
2=
k2x+
ω
2t+ δ
2 with the dispersion relations (12) and (15), and A2 is given by (16). Here A1,
a,
c,
k1,
δ
1,
δ
2 are arbitrary constants and k2=
k1±
−
k21
−
4u0. For certain choices of the parameters the functions u and v candevelop singularities
for finite time. These singularities will be transferred to the local and nonlocal reductions of the HS system.3. ReductionsoftheHSsystem
For the HS system (4) and (5) we have two local and two nonlocal reductions. (a)Localreductions:
(i) Real
reduction:
v(
x,
t)
=
ku(
x,
t)
, k isa real constant.
The system (4) and (5) consistently reduces to the well-known KdV equation
aut
=
2uxxx+
12uux.
(20)(ii) Complex
reduction:
v(
x,
t)
=
ku¯
(
x,
t)
, k isa real constant.
Under this reduction the system (4) and (5) reduces to the complex KdV equation
aut
= −
uxxx+
3u¯
xxx−
6uux+
6uu¯
x+
12uu¯
x,
(21)consistently, if a
= ¯
a.(b)Nonlocalreductions:
(i) Real
nonlocal reductions:
v(
x,
t)
=
ku(
ε
1x,
ε
2t)
, ε21=
ε
22=
1, k isa real constant.
When we apply this reduction to the system (4) and (5), it yields that to have a consistent reduction we must have k
=
1 and ε1ε
2=
1 i.e. ε1=
ε
2= −
1. Hence we get the reduced nonlocal ST-reversal KdV equationaut
(
x,
t)
= −
uxxx(
x,
t)
+
3uxxx(
−
x,
−
t)
−
6u(
x,
t)
ux(
x,
t)
+
6u(
−
x,
−
t)
ux(
x,
t)
+
12ux(
−
x,
−
t)
u(
x,
t).
(22) (ii) Complexnonlocal reductions:
v(
x,
t)
=
ku¯
(
ε
1x,
ε
2t)
, ε21=
ε
22=
1, k isa real constant.
This reduction reduces the system (4) and (5) consistently if k
=
1 and a= ¯
aε
1ε
2. Therefore we have the following nonlocal complex KdV equations:aut
(
x,
t)
= −
uxxx(
x,
t)
+
3u¯
xxx(
ε
1x,
ε
2t)
−
6u(
x,
t)
ux(
x,
t)
+
6u¯
(
ε
1x,
ε
2t)
ux(
x,
t)
+
12u¯
x(
ε
1x,
ε
2t)
u(
x,
t).
(23) Explicitly, here we have three nonlocal complex KdV equations.(a)S-reversalnonlocalcomplexKdVequation:
aut
(
x,
t)
= −
uxxx(
x,
t)
+
3u¯
xxx(
−
x,
t)
−
6u(
x,
t)
ux(
x,
t)
+
6u¯
(
−
x,
t)
ux(
x,
t)
+
12¯ux(
−
x,
t)
u(
x,
t),
(24) where a isa pure imaginary number.
(b)T-reversalnonlocalcomplexKdVequation:
aut
(
x,
t)
= −
uxxx(
x,
t)
+
3u¯
xxx(
x,
−
t)
−
6u(
x,
t)
ux(
x,
t)
+
6u¯
(
x,
−
t)
ux(
x,
t)
+
12¯ux(
x,
−
t)
u(
x,
t),
(25) where a isa pure imaginary number.
(c)ST-reversalnonlocalcomplexKdVequation:
aut
(
x,
t)
= −
uxxx(
x,
t)
+
3u¯
xxx(
−
x,
−
t)
−
6u(
x,
t)
ux(
x,
t)
+
6u¯
(
−
x,
−
t)
ux(
x,
t)
+
12¯ux(
−
x,
−
t)
u(
x,
t),
(26) where a isa real number.
4. One-solitonsolutionsofthereducedKdVequations
Using the soliton solutions of the HS system obtained in Section 2and the reduction formulas we obtain one-soliton solutions of the reduced equations.
Fig. 1. One-soliton solution for (21) with the parameters k1=12,u0= −1,α=a=A1=1, δ1= δ2=0.
4.1. One-solitonsolutionsofthereducedlocalKdVequations
(i) v
(
x,
t)
=
u(
x,
t)
(KdV equation).This reduction relation holds if u0
=
v0 and c=
0 yielding A2=
0 and g(
x,
t)
=
0. Hence we obtain the one-soliton solution of the equation (20) as u(
x,
t)
=
u0+
2 A1k22eθ2(
1+
A1eθ2)
2,
(27) whereθ
2=
k2x+
(2k3 2+12u0k2)a t
+ δ
2and A1=
eδ. Here k2,
a,
u0,
δ
2,
δ
are some arbitrary constants. Let us take all the parameters real. Then the above solution is nonsingular and bounded soliton solution for A1>
0. If A1<
0, the real-valued solution is singular.(ii) v
(
x,
t)
= ¯
u(
x,
t)
, a= ¯
a (ComplexKdV equation).
Let−
k21
−
4u0≥
0 i.e. u0≤ −
k2 1
4, k1
,
u0∈ R
yielding k2∈ R
. If we use the Type 1 approach [7], [8], [15] we get the following constraints on the parameters of the one-soliton solution of the reduced complex local KdV equation (21):1
)
v0=
u0,
c= −¯
c,
eδ1=
e¯δ1,
eδ2=
e¯δ2,
A1= ¯
A1,
(28)2
)
v0=
u0,
c= ¯
c,
eδ1= −
e¯δ1,
eδ2=
e¯δ2,
A1= ¯
A1,
(29)3
)
v0=
u0,
c= ¯
c,
eδ1= −
e¯δ1,
eδ2= −
e¯δ2,
A1= − ¯
A1.
(30)The above sets of the constraints give similar solutions. Let us consider the first set of the constraints. Let c
=
iα
, α∈ R
. Hence one-soliton solution of (21) is u(
x,
t)
=
u0+
iα
[
eθ1+
A 1eθ1+θ2−
α 2 8k21(k21+2u0)e 3θ1] +
2 A 1k22eθ2−
α 2 k21+2u0e 2θ1−
A1α2 4k21(k21+2u0)e θ2+2θ1 1+
A1eθ2−
α 2 8k21(k21+2u0)e 2θ1 2,
(31) whereθ
1=
k1x+
−4k 3 1−12u0k1 a t+ δ
1,θ
2=
k2x+
2k32+12u0k2 a t+ δ
2 with k2=
k1±
−
k2 1−
4u0. For u0<
−
k2 1 2≤ −
k2 14 with A1
>
0, the above solution is nonsingular and bounded. When these conditions are relaxed the solution (31) may develop singularity in a finite time. Example1.Let us take the parameters as
k1=
12,
u0= −
1,
α
=
a=
A1=
1,
δ
1= δ
2=
0. Hence one-soliton solution of (21) becomes|
u(
x,
t)
|
2=
H1(
x,
t)
(
1+
e(1+ √ 15 2 )x+4t+
2 7ex+11t)
4,
(32) where H1(
x,
t)
=
1+
ex+11t+
36 49e 2x+22t+
4 49e 3x+33t+
16 2401e 4x+44t+
8 49e (3+√15)x+(33+3√15)t+
2e(3+ √ 15 2 )x+(33−3 √ 15 2 )t+
e(2+ √ 15)x+(22−3√15)t− (
2√
15+
12)
e(3+3 √ 15 2 )x+(33−9 √ 15 2 )t− (
2√
15+
12)
e(1+ √ 15 2 )x+(11−2 √ 15 2 )t+ (
12√
15+
53)
e(1+ √ 15)x+(11−3√15)t− (
8√
15+
20 49)
e (5+ √ 15 2 )x+(55+3 √ 15 2 )t.
(33)Fig. 2. One-soliton solution for (22) with the parameters k2=2,u0= −1,a=8.
4.2. One-solitonsolutionsofthereducednonlocalKdVequations
(i) v
(
x,
t)
=
u(
ε
1x,
ε
2t)
, ε21=
ε
22=
1 (Nonlocal KdV equation).Recall that the system (4) and (5) reduces to (22) consistently by this reduction if ε1
=
ε
2= −
1. When we consider the reduction with the solutions u(
x,
t)
and v(
x,
t)
with Type 1 approach, we obtain that k1=
k2=
0 giving the trivial solution u(
x,
t)
=
u0. Therefore we apply Type 2 [8], [15]. In this case we have two possibilities:1
)
A1=
0,
u0=
v0,
e2δ1A2= −1
,
(34)2
)
c=
0,
u0=
v0,
A1eδ2= ±1
.
(35)Take also all the parameters as real numbers and consider the following particular cases.
(i).1) From
the first set of the constraints we obtain the one-soliton solution of the reduced nonlocal equation (
22) as u(
x,
t)
=
u0+
c(
ek1x+ω1t+δ1−
e3k1x+3ω1t+δ1)
−
8k2 1e2k1x+2ω1t(
1−
e2k1x+2ω1t)
2,
(36) where e2δ1=
−8k 2 1(k21+2u0)c2 . This solution is a singular solution on the line k1x
+
ω
1t=
0.(i).2) In
the second case we have
A2=
0 and the one-soliton solution of the equation (22) is obtained asu
(
x,
t)
=
u0+
σ
12k22ek2x+ω2t
(
1+
σ
1ek2x+ω2t)
2,
σ
1= ±
1.
(37)Let σ1
=
1, k2,
u0,
a∈ R
. Then the above solution can be rewritten asu
(
x,
t)
=
u0+
k22 2sech
2
(
k2x
+
ω
2t).
(38)This real-valued solution is nonsingular and bounded. The solution (37) is singular if σ1
= −
1.Example2.
Let
us choose the parameters as k1=
2,
u0= −
1 giving k2=
2, A1=
1,
δ
2=
0 satisfying A1eδ2=
1, and a=
8. Hence a nonsingular and bounded one-soliton solution of (22) isu
(
x,
t)
= −
1+
2sech2(
2x−
t).
(39)The graph of the above solution is given in Fig.2.
(ii) v
(
x,
t)
= ¯
u(
ε
1x,
ε
2t)
, a= ¯
aε
1ε
2, ε21=
ε
22=
1 (Nonlocal complex KdV equations).If we use Type 1, we obtain the below constraints that must be satisfied by the parameters of the one-soliton solution of the nonlocal reduced complex KdV equations (23),
k1
=
ε
1k¯
1,
v0= ¯
u0=
u0,
a= ¯
aε
1ε
2 (40)with the following possibilities:
1
)
c= −¯
c,
A1= ¯
A1,
eδ1=
e¯δ1,
eδ2=
e¯δ2,
(41)2
)
c= ¯
c,
A1= ¯
A1,
eδ1= −
e¯δ1,
eδ2=
e¯δ2,
(42)yielding A2
= ¯
A2, ω1=
ε
2ω
¯
1, ω2=
ε
2ω
¯
2, and k2=
ε
1k¯
2 foru0
≤ −
k21
4
.
(44)Here for finding one-soliton solutions of (23) we pick the constraints (40) with (41). We have three types of nonlocal reduced complex equations and we will consider one-soliton solutions of them separately. Note that we will mainly focus on some particular cases giving nonsingular and bounded solutions.
(a)S-reversalnonlocalcomplexKdVequation:
In this case, according to the constraints (40) and (41) with a
= ¯
aε
1ε
2 and (44), since(
ε
1,
ε
2)
= (−
1,
1)
we have k1= −¯
k1and a= −¯
a. Let k1=
iα
1, k2=
iα
1(
1±
−
1+
4u0α2 1
)
=
iα
2, c=
iα
3, a=
iα
4 for αj,
j=
1,
2,
3,
4∈ R
. Hence we obtain the one-soliton solution of the S-reversal nonlocal complex equation (24) asu
(
x,
t)
=
u0+
B1(
x,
t)
(
B2(
x,
t))
2,
(45) where B1(
x,
t)
=
iα
3(
eiα1x+ω1t+δ1+
A1ei(α1+α2)x+(ω1+ω2)t+δ1+δ2+
A2e3iα1x+3ω1t+3δ1)
−
2α
22A1eiα2x+ω2t+δ2−
8α
21A2e2iα1x+2ω1t+2δ1−
2 A1A2(
2α
1−
α
2)
2ei(2α1+α2)x+(2ω1+ω2)t+2δ1+δ2 (46) and B2(
x,
t)
=
1+
A1eiα2x+ω2t+δ2+
A2e2iα1x+2ω1t+2δ1,
(47) whereω
1=
4α
3 1−
12u0α
1α
4,
ω
2=
−2
α
3 2+
12u0α
2α
4,
A2=
α
2 3 8α
12(
2u0−
α
1)
.
(48)If we consider the real-valued solution we get
|
u(
x,
t)
|
2=
H1(
x,
t)
(
H2(
x,
t))
2,
(49) where H2(
x,
t)
=
1+
2 A1eω2t+δ2cos(
α
2x)
+
2 A2e2ω1t+2δ1cos(
2α
1x)
+
A21e2ω2t+2δ2+
A22e4ω1t+4δ1+
2 A1A2e(2ω1+ω2)t+2δ1+δ2cos((
2α
1−
α
2)
x).
(50)Here H1
(
x,
t)
is a huge expression so we are not giving it here explicitly. We will consider some particular cases.(a).1) Let us consider the case when A1
=
0, u0=
α2 1
3
≥
α2 1
4, and α3
=
0. The function H2(
x,
t)
in the denominator of the solution (49) becomesH2
(
x,
t)
= (
A2e2ω1t+2δ1+
cos(
2α
1x))
2+
sin2(
2α
1x).
(51) Therefore the solution (49) corresponding to this case is singular for x=
2nαπ1 and A2e
2ω1t+2δ1
+ (−
1)
n=
0, where n isan integer.
Note that in addition to A1=
0 if we take u0=
α2 1 3 yielding ω1
=
0, we have|
u(
x,
t)
|
2=
H1(
x,
t)
4 A22e4δ1(
B+
cos(
2α
1x))
2,
B=
1+
A2 2e4δ1 2 A2e2δ1,
(52) where H1(
x,
t)
=
1 9α
4 1(
A42e8δ1+
1)
+
α
32e2δ1(
A22e4δ1+
1)
+
484 9α
4 1A22e4δ1+
2 9α
4 1A22e4δ1cos(
4α
1x)
+
2 A2e2δ1(
α
32e2δ1−
22 9α
4 1(
A22e4δ1+
1))
cos(
2α
1x)
+
2 3α
2 1α
3A2e3δ1(
A2e2δ1−
1)
sin(
3α
1x)
+ (
2 3α
2 1α
3eδ1(
A23e6δ1−
1)
+
44 3α
2 1α
3A2e3δ1(
A2e2δ1−
1))
sin(
α
1x).
(53)For B
>
1 or B<
−
1 the solution (52) is nonsingular and bounded solution. Here if−
1≤
B≤
1 the solution (52) is singular. Example3.For this case choose the parameters as α
1=
12,
α
3=
1,
δ
1=
0. This gives A2= −
6. Hence solution of (24) becomes|
u(
x,
t)
|
2=
24049+
8040 cos(
x)
+
72 cos(
2x)
+
16968 sin(
x
2
)
+
1008 sin(
3 2x)
144
(
37−
12 cos(
x))
2.
(54)Fig. 3. Periodic wave solution for (24) with the parametersα1=12,α3=1, δ1=0.
Fig. 4. Periodic wave solution for (24) with the parametersα1=12,A1=2, δ2=0.
(a).2) Let
us
now assume that α3=
0 i.e. c=
A2=
0, and u0=
α2 2
6, A1
=
0. In this case the function H2(
x,
t)
in the denominator of the solution (49) becomesH2
(
x,
t)
= (
A1eω2t+δ2+
cos(
α
2x))
2+
sin2(
α
2x).
(55) Therefore the solution (49) corresponding to this case is singular for x=
nαπ2 and A1eω
2t+δ2
+ (−
1)
n=
0, where n isan integer.
Note that in addition to α3=
0 if we also take u0=
α2 2
6
≥
α2 1
4 i.e. ω2
=
0, the solution (49) turns to be|
u(
x,
t)
|
2=
α
4 1(
7±
4√
3)
[
1+
A41e4δ2+
100 A2 1e2δ2− (
20 A31e3δ2+
20 A1eδ2)
cos(
α
1(
3±
√
3)
x)
+
2 A21e2δ2cos(
2α
1(
3±
√
3)
x)
]
[(
A1eδ2+
cos(
α
1(
3±
√
3)
x))
2+
sin2(
α
1(
3±
√
3)
x)
]
2.
(56) The above solution is nonsingular and bounded if A1eδ2<
−
1 or A1eδ2>
1.Example4.
Take α
1=
12,
α
2=
12(
3+
√
3
),
A1=
2,
δ
2=
0. Hence solution of (24) becomes|
u(
x,
t)
|
2=
(
7+
4√
3
)
[
417−
200 cos(
12(
3+
√
3)
x)
+
8 cos((
3+
√
3)
x)
]
16
[
5+
4 cos(
12(
3+
√
3)
x)
]
2.
(57)The above solution is a nonsingular and bounded periodic wave solution. The graph of this solution is given in Fig.4. (b)T-reversalnonlocalcomplexKdVequation:
According to the constraints (40) and (41) with a
= ¯
aε
1ε
2 and (44), since(
ε
1,
ε
2)
= (
1,
−
1)
here we have k1= ¯
k1 and a= −¯
a. Letc
=
iα
1 and a=
iα
2for αj,
j=
1,
2,
∈ R
yieldingω
1=
i(
4k31+
12u0k1)
α
2=
iα
3,
ω
2=
−
i(
2k31+
12u0k1)
α
2=
iα
4,
A2= −
α
12 8k21(
k21+
2u0)
.
(58)Therefore we obtain the one-soliton solution of the T-reversal nonlocal complex equation (25) as u
(
x,
t)
=
u0+
B1
(
x,
t)
(
B2(
x,
t))
2Fig. 5. One-soliton solution for (25) with the parameters k1=16,k2=16(3+ √ 3),A1=1, δ2=0. where B1
(
x,
t)
=
iα
1(
ek1x+iα3t+δ1+
A1e(k1+k2)x+i(α3+α4)t+δ1+δ2+
A2e3k1x+3iα3t+3δ1)
+
2k22A1ek2x+iα4t+δ2+
8k21A2e2k1x+2iα3t+2δ1+
2 A1A2(
2k1−
k2)
2e(2k1+k2)x+i(2α3+α4)t+2δ1+δ2 (60) and B2(
x,
t)
=
1+
A1ek2x+iα4t+δ2+
A2e2k1x+2iα3t+2δ1.
(61) Let us consider the real-valued solution|
u(
x,
t)
|
2. We have|
u(
x,
t)
|
2=
H1(
x,
t)
(
H2(
x,
t))
2,
(62) where H2(
x,
t)
=
1+
2 A2e2k1x+2δ1cos(
2α
3t)
+
2 A1ek2x+δ2cos(
α
4t)
+
A21e2k2x+2δ2+
A22e4k1x+4δ1+
2 A1A2e(2k1+k2)x+2δ1+δ2cos((
2α
3−
α
4)
t).
(63)Here again H1
(
x,
t)
is a huge expression so we are not expressing it here. Let us consider two particular cases.(b).1) Take
α
1=
0 yielding A2=
c=
0, and u0= −
k2 2
6 giving α4
=
0. Note that in this case k2= (
3±
√
3
)
k1. The solution (62) becomes|
u(
x,
t)
|
2=
k 4 1(
7±
4√
3)
[
1−
20 A1e(3± √ 3)k1x+δ2+
102 A2 1e2(3± √ 3)k1x+2δ2−
20 A3 1e3(3± √ 3)k1x+3δ2+
A4 1e4(3± √ 3)k1x+4δ2]
[
A1e(3± √ 3)k1x+δ2+
1]
4.
(64)The above solution is nonsingular and bounded for A1
>
0.Example5.
Take the parameters as
k1=
16,
k2=
16(
3+
√
3
),
A1=
1,
δ
2=
0. Hence one-soliton solution of (25) becomes|
u(
x,
t)
|
2=
(
7+
4√
3)
[
e23(3+√
3)x
−
20e12(3+√3)x+
102e13(3+√3)x−
20e61(3+√3)x]
1296(
e16(3+√
3)x
+
1)
4.
(65)The above solution is a nonsingular and bounded solution. The graph of this solution is given in Fig.5. (b).2) Choose
now α
3=
0 that is u0= −
k2 1
3 and α1
=
0 yielding A2=
c=
0. In this case k2= (
1±
1√
3
)
k1. Take the plus sign for k2. The solution (62) becomes|
u(
x,
t)
|
2=
H1(
x,
t)
9[(
A1e (1+√1 3)k1x+δ2+
cos(
α
4t))
2+
sin2(
α
4t)
]
2,
(66) where H1(
x,
t)
=
k41[
1+
A41e 4k1(1+√13)k1x+4δ2+
12 A21(
7+
4√
3)
e2k1(1+ 1 √ 3)k1x+2δ2+
2 A2 1e 2k1(1+√13)k1x+2δ2 cos(
2α
4t)
−
4 A1(
3+
2√
3)(
e3k1(1+√13)k1x+3δ2+
A2 1e k1(1+√13)k1x+δ2)
cos(
α
4t)
].
(67)Fig. 6. Periodic wave solution for (26) with the parametersα1=12, u0=2,δ2=0, A1= −4, a=4. Here α4
= −
8k3 1 √ 39α2 . The above solution is singular for A1e
k2x+δ2
+ (−
1)
n=
0 at t=
nπα4, where n is
an integer.
(c)ST-reversalnonlocalcomplexKdVequation:In this case according to the constraints (40) and (41) with a
= ¯
aε
1ε
2 and (44), since(
ε
1,
ε
2)
= (−
1,
−
1)
we have k1= −¯
k1 and a= ¯
a. Let k1=
iα
1, k2=
iα
1(
1±
−
1+
4u0 α2 1)
=
iα
2, c=
iα
3 for αj,
j=
1,
2,
3∈ R
yieldingω
1=
i(
4α
13−
12u0α
1)
a=
iα
4,
ω
2=
i(
−
2α
23+
12u0α
2)
a=
iα
5,
A2=
α
2 3 8α
2 1(
2u0−
α
12)
.
(68)Therefore we obtain the one-soliton solution of the ST-reversal nonlocal complex equation (26) as u
(
x,
t)
=
u0+
B1(
x,
t)
(
B2(
x,
t))
2,
(69) where B1(
x,
t)
=
iα
3(
eiα1x+iα4t+δ1+
A1ei(α1+α2)x+i(α4+α5)t+δ1+δ2+
A2e3iα1x+3iα4t+3δ1)
+
2 A1k22eiα2x+iα5t+δ2+
8k21A2e2iα1x+2iα4t+2δ1−
2 A1A2(
2α
1−
α
2)
2ei(2α1+α2)x+i(2α4+α5)t+2δ1+δ2 (70) and B2(
x,
t)
=
1+
A1eiα2x+iα5t+δ2+
A2e2iα1x+2iα4t+2δ1.
(71) When we consider the real-valued solution|
u(
x,
t)
|
2 we get|
u(
x,
t)
|
2=
H1(
x,
t)
(
H2(
x,
t))
2,
(72) where H2(
x,
t)
=
1+
2 A2e2δ1cos(
2α
1x+
2α
4t)
+
2 A1eδ2cos(
α
2x+
α
5t)
+
A21e2δ2+
A22e4δ1+
2 A1A2e2δ1+δ2cos((
2α
1−
α
2)
x+ (
2α
4−
α
5)
t).
(73)Here also H1
(
x,
t)
is a huge expression so we are not giving it here explicitly. We will deal with some particular cases.(c).1) Consider
the case when α
3=
0 yielding A2=
0. Take also A1=
0. In this case the solution (72) of the nonlocal complex equation (26) becomes|
u(
x,
t)
|
2=
4u0A1(
u0−
α
2 2)
eδ2(
1+
A21e2δ2)
cos(
α
2x+
α
5t)
+
2u20A21e2δ2cos(
2α
2x+
2α
5t)
+
4 A21e2δ2(
u0−
α
22)
2[(
A1eδ2+
cos(
α
2x+
α
5t))
2+
sin2(
α
2x+
α
5t)
]
2.
(74)This solution is nonsingular and bounded if A1eδ2
>
1 or A1eδ2<
−
1. When these conditions are relaxed the solution (74) may develop singularity in a finite time.Example6.
Take the parameters of the solution (
74) as α1=
12, u0=
2,δ
2=
0, A1= −
4, and a=
4. Hence the solution (74) turns to be|
u(
x,
t)
|
2=
1284−
64(
1+
√
31
)
2+
4(
1+
√
31)
4+ (
136(
1+
√
31)
2−
1088)
cos(ψ )
+
128 cos(
2ψ )
[
17−
8 cos(ψ )
]
2,
(75)where
ψ
=
12(
1+
√
31)
x+ (
3(
1+
√
31)
−
161(
1+
√
31)
3)
t. This is a nonsingular and bounded periodic wave solution. The graph of the above solution is given in Fig.6.Fig. 7. Periodic wave solution for (26) with the parametersα1=12,u0=2, δ1=0,α3=5,a=4.
(c).2) Here
we again consider a particular case. Take
A1=
0 and A2=
0. Therefore the solution (72) of the nonlocal complex equation (26) becomes|
u(
x,
t)
|
2=
H1(
x,
t)
[(
A2e2δ1+
cos(
2α
1x+
2α
4t))
2+
sin2(
2α
1x+
2α
4t)
]
2,
(76) where H1(
x,
t)
=
2α
3eδ1[2u
0A2e2δ1−
u0−
2u0A22e4δ1+
u0A32e6δ1−
8 A2α
12e2δ1+
8 A22α
12e4δ1]sin
(
α
1x+
α
4t)
+
2e2δ1A 2[
2u20−
8u0α
21+
α
32e2δ1+
2u20A22e4δ1−
8u0A22α
21e4δ1]
cos(
2α
1x+
2α
4t)
+
2u0α
3A2e3δ1[
A2e2δ1−
1]
sin(
3α
1x+
3α
4t)
+
2u20A22e4δ1cos(
4α
1x+
4α
4t)
+
u20+
e2δ1[
α
2 3+
64 A22α
14e2δ1−
32u0A22α
12e2δ1+
4u20A22e2δ1+
α
23A22e4δ1+
u20A42e6δ1].
(77) This solution is nonsingular and bounded if A2e2δ1>
1 or A2e2δ1<
−
1. When these conditions are relaxed the solution (76) may develop singularity in a finite time.Example7.
Choose the parameters of the solution (
76) as α1=
12,
u0=
2,
δ
1=
0,
α
3=
5,
a=
4. Therefore the solution (76) becomes|
u(
x,
t)
|
2=
68449+
45780 sin(
1
2x
−
238t)
+
12600 sin(
32x−
698t)
+
39660 cos(
x−
234t)
+
7200 cos(
2x−
232t)
(
60 cos(
x−
234t)
+
109)
2.
(78)The above solution is a nonsingular and bounded periodic wave solution. The graph of this solution is given in Fig.7. 5. Conclusion
In this work, we have studied local and nonlocal integrable reductions of the symmetrical HS system. The local reductions yield KdV and complex KdV equations. The nonlocal reductions of the HS system give several type of new integrable nonlocal KdV equations. These are time (T)-, space (S)-, and space-time (ST)-reversal nonlocal complex KdV equations and space-time (ST)-reversal nonlocal KdV equation. By using the one-soliton solution of the HS system and the reduction formulas we obtained the one-soliton solutions of the reduced local and nonlocal KdV equations. Using our approach one can find also two- and three-soliton solutions of these KdV equations.
CRediTauthorshipcontributionstatement
Metin Gürses: Conceptualization, Methodology, Supervision, Conducting, Visualization, Writing-Review and Editing. Aslı Pekcan: Methodology, Validation, Formal Analysis, Writing-original draft, Writing-Review and Editing.
Funding
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors. Declarationofcompetinginterest
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.
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