EXACT COMPLEX SOLUTIONS FOR SOME NONLINEAR PARTIAL
DIFFERANTIAL EQUATIONS
MÜNEVVER TUZ
Fırat Universty, Science and Art Faculty
Department of Mathematics,
23119, Elazığ, Turkey
munevveraydin23@hotmail.com
Tel: 00.90. 424. 237 00 00/3509, Fax: 00.90. 424. 233 00 62
ABSTRACT
In this paper, we obtain complex solutions of the seventh degree Lax KDV
equation, (2+1) dimensional Konopelchenka-Dubravsky (KD) equation by using direct
algebraic method which is obtained by Zhang [1].
Key Words: Seventh degree Lax KDV equation, (2+1) dimensional KonopelchenkaDubravsky (KD) equation, direct algebraic method, complex solutions
BAZI NONLİNEER KISMİ DİFERANSİYEL DENKLEMLER İÇİN TAM
KARMAŞIK ÇÖZÜMLER
ÖZET
Bu makalede Zhang [1]daki direk cebirsel metodlar kullanılarak (2+1) boyutlu
Konopelchenka-Dubravsky (KD) denklemi elde edildi,aynı zamanda yedinci mertebeden Lax
KDV dalga hareket denkleminin çözümü elde edildi.
Anahtar Kelimeler: Yedinci mertebeden Lax KDV denklemi, (2+1) boyutlu
Konopelchenka-Dubravsky (KD) denklemi, direk cebirsel metod,kompleks çözümler.
1
1.INTRODUCTION(GİRİŞ)
Solution of nonlinear differential equations have important role in mathematic and
physics. Especially, traveling wave solutions of nonlinear equations have an crucial role in
mathematical physics and solution theory [2]. Recently, it has been stuied traveling wave
solutions of nonlinear differential equations through using symbolical computer programs
such as Maple, Matlab, etc. [3-8]. There are many methods which find exact solutions of
nonlinear partial differential equations. [9-14].
2. AN ANALYSIS OF THE METHOD AND APPLICATIONS(YÖNTEM VE
UYGULAMALARIN BİR ANALİZİ)
Before starting to give a direct algebraic method, we will give a simple description of
the direct algebraic method. For doing this, one can consider in a two variables general form
of nonlinear PDE
Qu, ut , u x , u xx ,  0 ,
(2.1)
and transform Eq.(2.1) with ux, t   u  ,   ik x  ct  ,where k, c are real constants. After
transformation, we get a nonlinear ODE for u 


Q' u,ikcu , iku ,k 2 u ,  0.
(2.2)
we are looking for solution of the equation (2.2) as following
n
U     a m F m  ,
(2.3)
m 0
where   ik x  ct .
i)
n is a positive integer that can be determined by balancing the highest order
derivate and with the highest nonlinear terms in equation.
ii)
a m and 
can be determined. Substituting solution (2.3) into Eq. (2.2) yields a
set of algebraic equations for F m and m  0,1,2,.
iii)
All coefficients of F m have to vanish. After this separated algebraic equation, we
could found coefficients a 0 , a m and  .
In this work, we obtain complex solutions of the Lax seventh-order KdV equation
equation and (2+1) dimensional Konopelchenko-Dubrovsky (KD) equation by using the
direct algebraic method which is introduced by Zhang [1].
F    b  F 2  
(2.4)
2
where F  
dF
and b is a constant. Some of the solutions of F    b  F 2   are given
d
in [1].
3.EXAMPLES(ÖRNEKLER)
Example 3.1. Consider lax seventh-order KdV equation,
u t  140u 3u x  70u x3  280uu x u xx  70u 2 u xxx  70u xx u xxx  42u x u xxxx  14uu xxxxx  u xxxxxxx  0,
(3.1.1)
We
can
use
transformation
with
Eq.
(2.1)
then
Eq.
(3.1.1)
become
cu  140u 3u   70k 2  u   280k 2uu u   70k 2u 2u   70k 4u u   42k 4u u    14k 4uu   
3
4
5
-k 6u  7   0,
(3.1.2)
3
When balancing u 3 u , u  , uu u , u 2 u , u u , u u 4  and uu 5  with u 7  then gives n=2. We
may choose
u  a0  a1 F  a 2 F 2 .
(3.1.3)
Substituting (3.1.3) into Eq. (3.1.2) yields a set of algebraic equations for a0 , a1 , a 2 , b . These
systems are finding as
140a03 a1b  a1bc  140a02 a1b 2 k 2  70a13b3k 2  560a0 a1a2b3k 2  224a0 a1b3k 4  952a1a2b 4 k 4 
272a1b 4 k 6  0,
420a02 a12b  280a03a2b  2a2bc  840a0 a12b 2 k 2  1120a02 a2b 2 k 2  980a12 a2b3k 2  1120a0 a22b3k 2 
1176a12b3k 4  3808a0 a2b3k 4  3584a22b 4 k 4  7936a2b 4 k 6  0,
140a03a1  420a0 a13b  1260a02 a1a2b  a1c  560a02 a1bk 2  910a13b 2 k 2  6440a0 a1a2b 2 k 2 
2520a1a22b3k 2  1904a0 a1b 2 k 4  16240a1a2b3k 4  3968a1b3k 6  0,
420a02 a12  280a03a2  140a14b  1680a0 a12 a2b  840a02 a22b  2a2c  2240a0 a12bk 2  2800a02 a2bk 2 
7140a12 a2b 2 k 2  7840a0 a22b 2 k 2  1680a23b3k 2  5656a12b 2 k 4  17248a0 a2b 2 k 4 
31136a22b3k 4  56320a2b3k 6  0,
420a0 a13  1260a02 a1a2  700a13a2b  2100a0 a1a22b  420a02 a1k 2  1890a13bk 2  12880a0 a1a2bk 2 
14420a1a22b 2 k 2  3360a0 a1bk 4  53648a1a2b 2 k 4  12096a1b 2 k 6  0,
140a14  1680a0 a12 a2  840a02 a22  1260a12 a22b  840a0 a23b  1400a0 a12 k 2  1680a02 a2 k 2 
12460a12 a2bk 2  13440a0 a22bk 2  8400a23b 2k 2  8008a12bk 4  23520a0 a2bk 4  81312a22b 2 k 4 
3
129024a2b 2 k 6  0,
700a13a2  2100a0 a1a22  980a1a23b  1050a13k 2  7000a0 a1a2 k 2  22680a1a22bk 2  1680a0 a1k 4 
63056a1a2bk 4  13440a1bk 6  0,
1260a12 a22  840a0 a23  280a24b  6300a12 a2 k 2  6720a0 a22 k 2  12320a23bk 2  3528a12 k 4 
10080a0 a2 k 4  84000a22bk 4  120960a2bk 6  0,
980a1a23  10780a1a22 k 2  24696a1a2 k 4  5040a1k 6  0,
280a24  5600a23 k 2  30240a22 k 4  40320a2 k 6  0.
(3.1.4)
From the solutions of the system, we can found
Case 1.
4bk 2
a0 

3
1  i 3 



56 3 2 1  i 3 b 2 k 4
3 3 529200c  1254400b k  15420489728000b k   529200c  1254400b k
3
3 6
6 12


3 6 2
529200c  1254400b3k 6  15420489728000b6 k 12   529200c  1254400b3k 6 
2
840 3 2
a1  0, a2  2k 2 , k  0.
(3.1.5)
Case 2.
a0 
38bk 2
, a1  0 , a 2  2k 2 ,
5
c
867136b 3 k 6
, k  0.
25
(3.1.6)
Case 3.
a 0  0 , a1  0 , a 2  12k 2 , b  0 , c  0 , k  0 .
(3.1.7)
a0  0 , a1  0 , a 2  6k 2 , b  0 , c  736b 3 k 6 , k  0 .
(3.1.8)
Case 4.
Case 5.
a0  3
c
, a1  0 , a2  2k 2 , b  0 , k  0 .
140
(3.1.9)
With the aid of Mathematica substituting (3.1.5)-(3.1.6)-(3.1.7)-(3.1.8) and (3.1.9) into
(3.1.3), we have obtained the following complex solutions of equation (3.1.1). These solutions
are:
Family 1:
4

1  i 3 


56 3 2 1  i 3 b 2 k 4
4bk 2
u1 

3
3 529200c  1254400b k  15420489728000b k  529200c  1254400b k
3 6
3
3

529200c  1254400b3k 6  15420489728000b6 k 12   529200c  1254400b3k 6 
2

840 3 2



3 6 2
6 12
2
2k 2  bTanh  b  ikx  ikct  .
(3.1.10)
where b <0, k is an arbitrary real constant.
4bk 2
u2 

3
1  i 3 



56 3 2 1  i 3 b 2 k 4
3 529200c  1254400b k  15420489728000b k   529200c  1254400b k
3 6
3
3
6 12


3 6 2
529200c  1254400b3k 6  15420489728000b6 k 12   529200c  1254400b3k 6 
2

840 3 2


2
2k 2  bCoth  b  ikx  ikct  .
(3.1.11)
where b <0, k is an arbitrary real constant.
4bk 2
u3 

3
1  i 3 

2k 2



56 3 2 1  i 3 b 2 k 4
3 3 529200c  1254400b k  15420489728000b k   529200c  1254400b k
3 6
3
6 12

529200c  1254400b3k 6  15420489728000b6 k 12   529200c  1254400b3k 6 
2
840 3 2


3 6 2

2
bTan  b  ikx  ikct  .
(3.1.12)
where b >0, k is an arbitrary real constant.
4bk 2
u4 

3


56 3 2 1  i 3 b 2 k 4
3 529200c  1254400b k  15420489728000b k   529200c  1254400b k
3
3 6
6 12
5

3 6 2

1  i 3 


3
529200c  1254400b3k 6  15420489728000b6 k 12   529200c  1254400b3k 6 
840 3 2

2

2
2k 2  bCot  b  ikx  ikct  .
(3.1.13)
where b >0, k is an arbitrary real constant.
Family 2:
2


38bk 2
867136 3 7  

u5 
 2k 2    bTanh   b  ikx 
b k it   .
5
25

 


(3.1.14)
where b <0, k is an arbitrary real constant.
2


38bk 2
867136 3 7  

u6 
 2k 2    bCoth  b  ikx 
b k it   .
5
25

 


(3.1.15)
where b <0, k is an arbitrary real constant.
2

 
38bk 2
867136 3 7  
u7 
 2k 2  bTan  b  ikx 
b k it   .
5
25
 
 

(3.1.16)
where b >0, k is an arbitrary real constant.
2

 
38bk 2
867136 3 7  
u8 
 2k 2   bCot  b  ikx 
b k it   .
5
25

 


(3.1.17)
where b >0, k is an arbitrary real constant.
Family 3:
2
 1 
u 9  12k 2  
 .
 ikx 
(3.1.18)
where b  0, k is an arbitrary real constant.
Family 4:
2
1


u10  6k  
.
3 7 
 ikx  736b k it 
2
(3.1.19)
where b  0, k is an arbitrary real constant.
Family 5:
2
u11  3
c
1


 2k 2  
 .
140
 ikx  ikct 
6
(3.1.20)
where b  0, k is an arbitrary real constant.
Example 3.2. Consider (2+1) dimensional Konopelchenko-Dubrovsky (KD) equation,
3
u t  u xxx  6cuu x  d 2 u 2 u x  3v y  3du x v  0,
2
u y  v x  0.
(3.2.1)
where, c and d arbitrary real parameters,   ik x   y   t  . We can use transformation with
Eq. (2.1) then Eq. (3.2.1) become
3
  u   k 2 u   6cuu   d 2 u 2 u   3v   3dvu   0,
2
  u   v   0.
(3.2.2)
When balancing u 2 u  with u then gives n1  1 and u with v then gives n2  1 . We may
choose
u  a0  a1 F
v  b0  b1 F
.
(3.2.3)
Substituting (3.2.3) into Eq. (3.2.2) yields a set of algebraic equations for a0 , a1 , b0 , b1 . These
systems are finding as
 6a 0 a1bc  3a1bb0 d 
3 2
a 0 a1bd 2  2a1b 2 k 2  3bb1  a1b  0,
2
 6a12 bc  3a1bb1 d  3a 0 a12 bd 2  0,
 6a 0 a1c  3a1b0 d 
3 2
3
a 0 a1 d 2  a13 bd 2  8a1bk 2  3b1  a1   0,
2
2
(3.2.4)
 6a12 c  3a1b1 d  3a 0 a12 d 2  0,
3 3 2
a1 d  6a1 k 2  0,
2
 bb1  a1b  0,
 b1  a1  0.
From the solutions of the system, we can found
Case 1.
12c 2
 4bk 2  3 2  2
2
2c  d
2ik
2ik
a0 
, a1  
, b0  d
, b1 
, d  0, k  0 .
2
d
6d
d
d
Case 2.
7
(3.2.5)
12c 2
 3 2  2
2
2c  d
2ik
2ik
a0 
, a1 
, b0  d
, b1  
, d  0, k  0 .
2
d
6d
d
d
(3.2.6)
With the aid of Mathematica substituting (3.2.5) and (3.2.6) into (3.2.3), we have obtained the
following complex solutions of equation (3.2.1). These solutions are:
Family 1:
u1 

2c  d 2ik

 bTanh  b ik  x   y   t  
d2
d

12c 2
 4bk 2  3 2  2 
2
2ik
v1  d

 bTanh  b ik  x   y   t  
6d
d


(3.2.7)

(3.2.8)
where b <0, k is an arbitrary real constant.
u2 

2c  d 2ik

 bCoth  b ik  x   y   t  
d2
d

12c 2
 4bk 2  3 2  2
2
2ik
v2  d

 bCoth  b ik  x   y   t  
6d
d

where b <0, k is an arbitrary real constant.
u3 
2c  d 2ik

d2
d

bTan  b ik  x   y   t  
12c 2
 4bk 2  3 2  2
2
2ik
v3  d

6d
d


bTan  b ik  x   y   t  

(3.2.9)
where b >0, k is an arbitrary real constant.
u4 

2c  d 2ik

 bCot  b ik  x   y   t  
d2
d

12c 2
 4bk 2  3 2  2 
2
2ik
d
v4 

 bCot  b ik  x   y   t  
6d
d

where b >0, k is an arbitrary real constant.
Family 2:
u5 

2c  d 2ik 
1




d2
d  ik  x   y   t  
8

(3.2.10)
12c 2
 3 2  2
2

2ik 
1
v5  d

 

6d
d  ik  x   y   t  
(3.2.11)
where b  0, k is an arbitrary real constant.
3. CONCLISIONS(SONUÇLAR)
In this paper, it is obtained solutions for (2+1) dimensional (KD) equation and seventh
ordinary Lax KDV equations by using direct algebraic method. This method can be used to
many other nonlinear equations or couple ones. In addition, this model is also computerizable,
which allows us to perform complicated and tedious algebraic calculation on a computer.
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