expansion method - İstanbul Ticaret Üniversitesi

Transkript

expansion method - İstanbul Ticaret Üniversitesi
İstanbul Ticaret Üniversitesi
Fen Bilimleri Dergisi Yıl: 14 Sayı: 28 Güz 2015 s. 85-92
Research Article
TRAVELING WAVE SOLUTIONS OF SOME NONLINEAR
PARTIAL DIFFERENTIAL EQUATIONS BY USING
EXTENDED (G ′ G ) - EXPANSION METHOD
Doğan KAYA 23
Asıf YOKUŞ 1
ABSTRACT
In this study, we use extended
(G ′ G ) -expansion method which is suggested in (Wang et al., 2008).
By
this method, we find traveling wave solutions of nonlinear Burgers and KdV equations. It is obtained
graphics of these solutions by using Mathematica.
Keywords: Burgers equation, KdV equation, extended
(G ′ G ) -expansion
method, traveling wave
solutions.
Araştırma Makalesi
GENİŞLETİLMİŞ (G ′ G ) -AÇILIM METODU KULANILARAK BAZI LİNEER
OLMAYAN KISMİ DİFERANSİYEL DENKLEMLERİN YÜRÜYEN DALGA
ÇÖZÜMLERİ
ÖZ
Bu çalışmada bir genişletilmiş
(G ′ G ) –açılım
metodu inşa edildi. Biz bu metodu lineer olmayan
denklemi, Burgers ve KdV denkleminin yürüyen dalga çözümünü bulmak için bu metodu uyguladık. Biz
ayrıca bu metodun diğer lineer olmayan denklemlere uygulanabilir olduğunu gösterdik.
Anahtar kelimeler: Burgers denklemi; KdV denklemi;
Makale Gönderim Tarihi: 03.03.2016
(G ′ G ) -açılım metodu; yürüyen dalga çözümleri.
Kabul Tarihi : 06.04.2016
Corresponding Author/Sorumlu Yazar: Fırat Üniversitesi, Fen Fakültesi, Aktüerya Bilimleri Bölümü,
e-mail/e-ileti: [email protected]
İstanbul Ticaret Üniversitesi, Fen Edebiyat Fakültesi, Matematik Bölümü, [email protected]
1
2
Asıf YOKUŞ, Doğan KAYA
1. INTRODUCTION
In this paper, we used the (G ′ G ) -expansion method. Firstly, this method is
introduced in (Wang et al., 2008). Then, a new extended version of this method was
presented in (Guo and Zhou, 2010; Guo and Zhou, 2010) have suggested a new
expansion to obtain traveling wave solution of nonlinear partial differential equation
(in short, PDE) in his paper. We find the traveling wave solutions of the Burgers
equation and KdV equation by using Guo and Zhou’s recommendations.
To find exact solutions of nonlinear partial differential equations always impressed
scientists. It is too important to find exact solutions of nonlinear partial differential
equations. These equations are mathematical models of complex physical
occurrences that arise in engineering, chemistry, biology, mechanics and physics.
Solution of a nonlinear partial differential equation gives much knowledge to the
scientists about the nature of these events. Therefore, various effective methods
have been developed to understand the mechanisms of these physical models, to
help physicists and engineers and to ensure knowledge for physical problems and its
applications. There are many methods to obtain traveling wave solutions of
nonlinear partial differential equation in literature (Fan, 2000; Clarkson, 1989;
Wazwaz, 2005; Parkes et al., 1996; Fan, 2000; Elwakil et al., 2002; Zheng et al.,
2003; He et al., 2006; Inan, 2010).
2. AN ANALYSIS OF THE METHOD AND APPLICATIONS
Before starting to give the (G ′ G ) - expansion method, we will give a simple
description of the (G ′ G ) -expansion method. For doing this, one can consider in a
two-variables general form of nonlinear ordinary differential equations
(1)
Q(u , u t , u x , u xx , ) = 0 ,
and transform Eq. (1) with u (x, t ) = u (ξ ) , ξ = x − Vt , where V is constant. After
transformation, we get nonlinear ordinary differential equations for u (ξ )
Q' (u ′, u ′′, u ′′′,) = 0.
(2)
The solution of the Eq. (2) we are looking for is expressed as


− i +1
 G′ 




i
i
i
−
−
1
2
m 

 1  G′  
G
  G′ 
 G′ 
 G′ 



u (ξ ) = a0 + ∑ ai   + bi  
σ 1 +   + ci   + d i
,


2
G
G
i =1   G 
 1  G′   
 µG 


σ 1+   

 µ G  


 
(3)
where G = G (ξ ) satisfies the second order linear ordinary differential equation in the
form
(4)
G ′′ + µG = 0,
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where ai , , bi , , ci , , di , and µ are constants to be determined later, the positive
integer m can be determined by balancing the highest order derivative and with the
highest nonlinear terms into Eq. (2). Substituting solution (3) into Eq. (2) and using
Eq. (4) yields a set of algebraic equations for same order of (G ′ G ) ; then all
coefficients same order of (G ′ G ) have to vanish. After this separated algebraic
equation, we can find ai , , bi , , ci , , di , and µ constants. General solutions of
the Eq. (4) have been well known us, then substituting ai , , bi , , ci , , di , and the
general solutions of Eq. (4) into (3) we have more traveling wave solutions of Eq.
(1) (Guo et al., 2010; Yokus, 2011; Inan, 2010).
Example 1. Let’s consider classical Burgers equation (Gorguis, 2006)
(5)
u t + uu x + u xx = 0.
For doing this example, we can use transformation u ( x, t )= u (ξ ) , ξ= x − Vt then
Eq. (5) become
1
(6)
c − Vu + u 2 + δ u ′ = 0,
when balancing u 2 with
u′
2
then gives
m = 1 . Therefore, we may choose
−1
 1  G′  2 
 G′ 
 G′ 
u (ξ ) = a0 + a1   + b1 σ 1 +    + c1   + d1
 µG  
G
G


1
 1  G′  2 
σ 1 +   
 µ G  


,
(7)
substituting Eq. (7) into (6) yields a set of algebraic equations for
a0 , a1 , a2 , b1 , c1 , d1 , µ , σ and V . These systems are
a 02
b 2σ
+ c + a1c1 + b1 d1 − a 0V + c1δ − a1δµ + 1 = 0, d1δσ = 0,
2
2
2
c 2 + 2c1δµ
d12
a1 (a 0 − V ) = 0,
= 0, c1 (a 0 − V ) = 0,
= 0,
2
2σ
d1δσ
a12 µ + 2a1δµ − b12σ
b1c1 = 0,
= 0, a 0 d1 − d1V = 0,
= 0,
µ
2µ
b δσ
a1 d1 − b1δσ = 0, 1 = 0, a 0 b1 − b1V = 0, a1b1 = 0.
c1 d1 = 0,
µ
(8)
From the solutions system, we obtain the following with the aid of Mathematica.
Case 1:
a 0 = − 2 c − 8δ 2 µ ,
a1 = 2δ ,
b1 = 0,
c1 = −2δµ ,
d 1 = 0, V = − 2 c − 8δ 2 µ ,
(9)
87
Asıf YOKUŞ, Doğan KAYA
substituting Eq. (9) into (7) we have three types of traveling wave solutions of Eq.
(5):
(
( (
))
( (
 µ A Cos µ x + 2 c − 8δ 2 µ t − A Sin µ x + 2 c − 8δ 2 µ t
2
1

u1 (ξ ) =
− 2 c − 8δ 2 µ + 2δ 
2
2
 A1Cos µ x + 2 c − 8δ µ t + A2 Sin µ x + 2 c − 8δ µ t

(
))
( (
( (
( (
))
))
( (
( (
( (
 µ A Cos µ x + 2 c − 8δ 2 µ t − A Sin µ x + 2 c − 8δ 2 µ t
1
2
− 2δµ 
 A1Cos µ x + 2 c − 8δ 2 µ t + A2 Sin µ x + 2 c − 8δ 2 µ t
))
))
)))

)))  −



−1
.
Fig 1: Traveling wave solution of Eq. (5) for case 1 when
A1 = A2 = 1, c = 0.6, µ = 0.1 and δ = 0.2.
Case 2:
a 0 = − 2 c − 2δ 2 µ ,
a1 = 2δ ,
b1 = 0,
c1 = 0,
d 1 = 0, V = − 2 c − 2δ 2 µ ,
(10)
substituting Eq. (10) into (7) we have three types of traveling wave solutions of Eq.
(5) as following
(
( (
( (
))
( (
( (
 µ A Cos µ x + 2 c − 2δ 2 µ t − A Sin µ x + 2 c − 2δ 2 µ t
2
1
u 2 (ξ ) = − 2 c − 2δ 2 µ + 2δ 
 A1Cos µ x + 2 c − 2δ 2 µ t + A2 Sin µ x + 2 c − 2δ 2 µ t
))
Fig 2: Traveling wave solution of Eq. (5) for case 2 when
88
))
))).

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A1 = A2 = 1, c = 0.6, µ = 0.1 and δ = 0.2.
Case 3:
a0 =
2 c − 2δ 2 µ ,
a1 = 0,
c1 = −2δµ ,
b1 = 0,
d1 = 0,
(11)
2 c − 2δ 2 µ ,
V =
substituting Eq. (11) into (7) we obtain three types of traveling wave solutions of Eq.
(5) as following
(
( (
( (
))
( (
( (
 µ A Cos µ x − 2 c − 2δ 2 µ t − A Sin µ x − 2 c − 2δ 2 µ t
2
1
u 3 (ξ ) = 2 c − 2δ µ − 2δµ 
 A1Cos µ x − 2 c − 2δ 2 µ t + A2 Sin µ x − 2 c − 2δ 2 µ t
2
))
))
)))

−1
.
Example 2. Let’s consider KdV equation (Kaya D. 2002),
(12)
ut + uu x + δ u xxx = 0.
For doing this example, we can use transformation u ( x, t )= u (ξ ) , ξ= x − Vt then Eq.
(12) become
− Vu ′ + uu ′ + δu ′′′ = 0.
(13)
When balancing u ′u ,
u ′′′ then gives m = 2 . Therefore, we may choose
−1
 1  G′  2 
 G′ 
 G′ 
u (ξ ) = a0 + a1   + b1 σ 1 +    + c1   + d1
 µG 
G
G


1
 1  G′  2 
σ 1 +   
 µ G  


2
2
 G′ 
 G′   1  G′  
 G′ 
+ a2   + b2   σ 1 +    + c2   + d 2
G
 G   µ  G  
G
 G′ 
 
G
2
+
−1
 1  G′  2 
σ 1 +   
 µ G  


,
(14)
substituting Eq. (14) into (13) yields a set of algebraic equations for
a0 , a1 , a2 , b1 , b2 , c1 , c2 , d1 , d 2 , µ , σ and V. These systems are
a 02
b 2σ
+ c + a1c1 + a 2 c 2 + b1 d1 + b2 d 2 − a 0V + 2c 2δ + 2a 2δµ 2 + 1 = 0,
2
2
c 22
+ 6c 2δµ 2 = 0,
2
c1c 2 + 2c1δµ 2 = 0,
a0 c1 + a1c 2 + b1 d 2 − c1V + 2c1δµ = 0,
a1 a 2 + 2a1δ +
b1b2σ
µ
= 0,
2σ
b σ b σ
a
+
= 0,
+ a 0 a 2 + a 2V + 8a 2δµ +
2
2
2µ
(15)
d1 d 2
= 0,
σ
2
2
3d 2δσ
2
= 0,
2
1
3d1δσ 2 = 0,
µ2
= 0,
2
c12
d2
+ a 0 c 2 − c 2V + 8c 2δµ = 0, 3d 2δσ = 0, 1 = 0,
2
2
2
σ
µ
−
3d1δσ
a
b σ
+ 6a 2δ +
= 0,
2µ
2
6d 2δσ 2
µ
2
2
= 0,
2
2
6d1δσ 2
µ
−
2b1δσ 2
(σ ) 2 µ (σ ) 2 µ
3
2
a0 a1 + a 2 c1 + b2 d1 − a1V + 2a1δµ + b1b2σ = 0, d 2 = 0, c 2 d 2 + 2d 2δµ 2 = 0,
2
1
3d1δσ 2
3
= 0,
a 2 b2 + 2b2δ = 0,
c 2 d 1 + c1 d 2 = 0,
= 0, − d1δµσ = 0, d 2δµσ = 0,
3
3
(σ ) 2
(σ ) 2
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Asıf YOKUŞ, Doğan KAYA
−
2b2δσ 2
(σ ) µ
3
= 0, −
2
d 2δσ
(σ ) µ
3
2
−
b2δσ 2
(σ )
3
= 0, −
2
a0 d1 + a1 d 2 − d1V + b1δµσ = 0,
b1c2 = 0,
a 0 b1 + b2 c1 − b1V = 0,
b2δσ 2
(σ ) µ
3
2
8b2δσ = 0,
2
= 0,
−
2b1δσ 2
(σ )
3
2
µ
2
3b1δσ
= 0,
µ
a1d1 + a 2 d 2 + 3b2δµσ = 0,
= 0,
5b2δσ
µ
= 0,
a 2 d 1 + 4b1δσ = 0,
c1b1 + b2 c2 = 0, a1b1 + a0 b2 − b2V + 2b2δµ = 0,
a 2 b1 + a1b2 = 0,
From the solutions of the system, we obtain the following with the aid of
Mathematica.
Family 1:
a0 = −8δµ − 2 c + 128δ 2 µ 2 ,
a1 = 0,
a2 = −12δ
b1 = 0,
b2 = 0
c1 = 0,
c2 = −12δµ 2 ,
(16)
d1 = 0, d 2 = 0, V = − 2 c + 128δ 2 µ 2 .
Substituting Eq. (16) into (14) we have three types of traveling wave solutions of
Eq. (12) as following
u1 (ξ ) = −8δµ − 2 c + 128δ 2 µ 2
))
)))
( (
( (
))
))
( (
( (
µ ( A Cos ( µ ( x + 2 c + 128δ µ t ) ) − A Sin ( µ ( x + 2 c + 128δ µ t ) ) ) 

A Cos ( µ ( x + 2 c + 128δ µ t ) ) + A Sin ( µ ( x + 2 c + 128δ µ t ) ) 

(
 µ A Cos µ x + 2 c + 128δ 2 µ 2 t − A Sin µ x + 2 c + 128δ 2 µ 2 t 
2
1

− 12δ 
2 2
2 2

 A1Cos µ x + 2 c + 128δ µ t + A2 Sin µ x + 2 c + 128δ µ t


− 12δµ 2 


2
−2
2
2
2
2
2
1
2
2
1
2
.
2
2
Family 2:
a 0 = −8δµ −
2 c + 8δ 2 µ 2 ,
a1 = 0,
a 2 = −12δ
b1 = 0,
b2 = 0
d1 = 0, d 2 = 0, V = − 2 c + 8δ µ .
2
2
c1 = 0,
c 2 = 0,
(17)
Substituting Eq. (17) into (14) we find three types of traveling wave solutions of Eq.
(12) as following
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u2 ( ξ ) =
−8δµ − 2 c + 8δ 2 µ 2
(
( (
))
( (
 µ A Cos µ x + 2 c + 8δ 2 µ 2 t − A Sin µ x + 2 c + 8δ 2 µ 2 t
2
1

− 12δ 
2 2
2 2
 A1Cos µ x + 2 c + 8δ µ t + A2 Sin µ x + 2 c + 8δ µ t

))
( (
))
( (
)))  .
2



Family 3:
a 0 = −8δµ −
2 c + 8δ 2 µ 2 ,
a1 = 0,
a 2 = 0,
b1 = 0,
b2 = 0
c1 = 0,
c 2 = −12δµ 2 ,
(18)
d1 = 0, d 2 = 0, V = − 2 c + 8δ 2 µ 2 .
Substituting Eq. (18) into (14) we write three types of traveling wave solutions of
Eq. (12) as following
−8δµ − 2 c + 8δ 2 µ 2
u3 (ξ ) =
(
( (
))
( (
 µ A Cos µ x + 2 c + 8δ 2 µ 2 t − A Sin µ x + 2 c + 8δ 2 µ 2 t
2
1

− 12δµ 2 
2 2
2 2
 A1Cos µ x + 2 c + 8δ µ t + A2 Sin µ x + 2 c + 8δ µ t

( (
))
( (
))
))) 
−2
 .


3. CONCLUSIONS
In this study, we consider to solve the traveling wave solutions of the
classical Burgers equation and KdV equation by using the extended (G ′ G ) -expansion
method. The method can be applied to many other nonlinear equations or coupled
ones. In addition, this method is also computerizable, which allows us to perform
complicated and tedious algebraic calculation on a computer.
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Asıf YOKUŞ, Doğan KAYA
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