Analytical solution with tanh-method and fractional sub-equation method for non-linear partial differential equations and corresponding fractional differential equation composed with Jumarie fractional derivative
Uttam Ghosh (1), Srijan Sengupta (2a), Susmita Sarkar (2b), Shantanu Das (3)
(1): Department of Mathematics, Nabadwip Vidyasagar College, Nabadwip, Nadia, West Bengal, India; Email: [email protected] (2): Department of Applied Mathematics, University of Calcutta, Kolkata, India Email (2b): [email protected], (3)Scientist H+, RCSDS, Reactor Control Division BARC Mumbai India Senior Research Professor, Dept. of Physics, Jadavpur University Kolkata Adjunct Professor. DIAT-Pune UGC-Visiting Fellow. Dept of Appl. Mathematics; Univ. of Calcutta, Kolkata Email (3): [email protected] Abstract The solution of non-linear differential equation, non-linear partial differential equation and non- linear fractional differential equation is current research in Applied Science. Here tanh-method and Fractional Sub-Equation methods are used to solve three non-linear differential equations and the corresponding fractional differential equation. The fractional differential equations here are composed with Jumarie fractional derivative. Both the solution is obtained in analytical traveling wave solution form. We have not come across solutions of these equations reported anywhere earlier.
Keywords: Tanh-method, Fractional Sub-Equation Method, Boussinesq equation, Non-linear Evolution equation, Fractional Differential Equation, Jumarie Fractional Derivative
1.0 Introduction
The advanced analytical methods to get solution to non-linear partial differential equation is current research field in different branches of applied science such as Applied Mathematics, Physics, Biology and Engineering applications. On the other hand fractional calculus and consequently solutions of fractional differential equations is also another growing field of science especially in Mathematics, Physics, Engineering and other scientific fields like controls etc [1-5, 25]. The exact solutions of non-linear partial differential equation and the corresponding fractional differential equations are physically very important [25-27]. To solve such type non- linear differential equations different methods are used, such as Adomain Decomposition Method (ADM) [6-7, 25], the Variational Iterative Method (VIM) [8-10], the Homotopy Perturbation Method (HPM) [11-13], Differential Transform Method (DTM) [14]. Another 1
important method to find the traveling wave solution of the non-linear partial differential equation is tanh- method was first introduced by Huibin and Kelin [15]. This method was used directly into a higher-order KdV equation, to get its solutions. Wazwaz [16, 17] used tanh- method to find traveling wave solutions of some non-linear equations generalized Fisher’s equation, generalized KdV equation, Zhiber-Shaba equation and other related equations. Recently Zhang and Zhang [18] introduce Fractional Sub-Equation Method to find travelling wave solution of the non-linear fractional differential equation. The principle of this method is expressing the solution of the non-linear differential equation as polynomials, where the variables of which is one of the solution of simple and solvable ordinary and partial differential equation. The method is also based on Homogeneous Balance Principle [19, 20] and applied on Jumarie’s modified Riemann-Liouvelli derivative [21, 22]. The Fractional Sub-Equation Method used by many authors [19, 21, 24] to find the analytic solution of non-linear fractional differential equations. Rest of the paper is as follows-in section 2.0 we describe tanh-method, in section 3.0 we describe Fractional Sub-Equation method, in section 7.0-12.0 we describe the solution of three non-linear equations in both integer order partial differential and fractional order partial differential equation types. In this paper we find the solution of three non-linear equation using tanh-method and the corresponding non-linear fractional differential equation using Fractional Sub-Equation method. In all the cases analytical solutions obtained in travelling J α ()α wave solution form. In this paper the symbols for fractional differential operator used 0 Dx , f means Jumarie fractional derivative.
2.0 Tanh-method
Let the non-linear partial differential equation is of the form
Huu( ,txxx , u , u ....)= 0...... (1)
Where uuxt= (,)is function of x, t .
For instant
∂∂2uu − u = 0 ∂∂x2 t
is of such an equation as Huuu(,txx , )= 0, with H a linear /non-linear operator. Using travelling wave transformation ξ = + ctkx , with k constant called wave-number (or propagation constant) and c a constant; as velocity of propagating wave, equation (1) reduce to an ordinary differential equation in the form
Huu( ,ξξξ , u ....)= 0...... (2)
2
Whereuu= ()ξ , is function of one variableξ . In summary we are making several variables (,x yz ,,)t to a single variableξ , with travelling wave transformation such as
ξ =+kxxy ky++kzz ct, with corresponding constants kx , kkyz,, c as described above, and ∂ ξ is in units of distance. Thus the operator ∂t , in terms of ξ with ξ = kx + ct is following
∂ ∂∂ξ d ≡=c ∂∂∂ttξ dξ and similarly ∂2 is ∂x2
2 2 ∂∂∂∂∂∂∂∂∂⎛⎞ ξξ⎛⎞⎛⎞2 d 2 ≡=⎜⎟ ⎜⎟⎜⎟ =kkk =2 ∂∂∂∂∂∂∂∂∂xxx⎝⎠ ξ x⎝⎠⎝⎠ξξξ x dξ
The transformed equation ∂∂2uu−=u 0 , is thus, ∂x2 ∂t
dd2uu kcu22−=0or kucuu −=0 ddξξ2 ξξ ξ
which is Huu(,ξξξ , u )= 0.
Again, letφ = φ ξ)( be such type of function which is a solution of Riccati equation ′ )( += φσξφ 2 . Whereσ is a real constant. That is dφ = σ +φ 2 , dξ = dφ dξ ∫ ∫ σ +φ 2
Then
⎧−−σσξtanh( − )⎪⎫ ⎪ ⎬ forσ < 0 ⎪−−σσξcoth( − )⎭⎪ ⎪ ⎪ σσξtan( ) ⎪⎫ φξ()=>⎨⎬forσ 0 ⎪− σσξcot( ) ⎭⎪ ⎪ 1 ⎪−=forσ 0 ⎪ ξ ⎩
The above expression is detailed below briefly.
Case-I For σ < 0 then the Riccati equation can be written equation, considering integral constant as zero
3
⎧ 1 − tanh−1 φ ()−σ dφ ⎪ −σ dξξ==⎨ ∫∫σφ+ 2 1 ⎪− coth −1 φ ()−σ ⎩⎪ −σ Therefore, by inverting above we get the following ⎧ ⎪−−σ tanh ( −σξ ) φ = ⎨ ⎪−−σ coth −σξ ⎩ () Case-II For σ = 0 then the Riccati equation can be written equation, ignoring integration constant d1φ 1 dξξφ==−=− ∫∫φ 2 φξ Case-III For σ > 0 then the Riccati equation can be written equation, considering integral constant as zero ⎧ 1 tan−1 φ ()σ dφ ⎪ σ dξξ==⎨ ∫∫σφ+ 2 1 ⎪− cot −1 φ ()σ ⎩⎪ σ Therefore, inverting above we get the following ⎧ ⎪ σ tan ( σξ ) φ = ⎨ ⎪− σ cot σξ ⎩ () The solution of the equation (2) will be expressed in-terms of φ in the form
n i uSa()ξφ==∑ i ...... (3) i=0
Where n is obtained balancing the highest order derivative term and the non-linear terms in equation (2). Then put the value of u()ξ from (3) to the equation (2) and comparing the coefficient of φ i the values of coefficients can be obtained.
Consider the equation ∂2u= u ∂u. Using the travelling wave transformation ξ = + ctkx the ∂x2 ∂t considered equation reduces to
2 kuξξ= cuuξ ……...... (4)
4
Then using the transformation as defined in (3) in-terms of φ and using the chain rule of derivative we obtain
dddφ d d ==sec h22ξφ =− (1 ) dddξξφ dφ dφ
22 dddd⎛⎞22 ddφ 2 d 2d 22==−⎜⎟(1)φφφ =−−− (1)2(1)φ ddddξ ξξ φ⎝⎠ dd φ ξ d φ dφ
Putting this transformation in equation (4) we get
2 222⎡⎤ddSS2 2dS k⎢⎥(1−−−=−φφφφ )2 2 (1 )cS (1 ) ……………………………(5) ⎣⎦ddφ φφd
In the highest order derivative term the highest order ofφ is n + 2 and in the non-linear term it is 21n + . Equating them we get n = 1 .We have (3) as following
uSaa()ξ = =+01φ
Therefore
ddSS2 ==a and 0 ddφφ1 2
Putting the above value in equation (5) we get the following
22⎡⎤ 2 kacaa⎣⎦2(1φφ−=+− )101( φ)(1 φ )a1
Comparing the like powers of φ from both sides of above expression we get
2k 2 aa==0and − 01c
Hence the corresponding solution is
22kk22 uS()ξ ==−φξ =− tanh() cc
that is
2k 2 uxt( , )=− tanh(kxct + ) c
5
3.0 Fractional sub-equation method
The space and time fractional partial differential equation is of the form
αα2 α Luuuu( ,txx , , .....)=<< 0 0α 1...... (6) where uuxt= (,)and L is functions (linear or non-linear operator) of the u and its derivatives. α is order of fractional derivative of Jumarie type, defined as follows
x ⎧ 1 −−α 1 ⎪ ()xf−<ξξξα ()d, 0 Γ−()α ∫ ⎪ 0 ⎪ 1dx J Dfxαα()== f() () x (x −ξξ )−α f () − f (0)d, ξ 0 < α< 1 0 x [] ⎨ ∫ () ⎪Γ−(1α ) dx 0 ⎪ ()α −n ()n ⎪()fx() , n≤<+α n1, n ≥ 1 ⎩⎪
We consider that at x < 0 the function fx()= 0and also fx()− f (0)0= for x < 0 . The fractional derivative considered here in the fractional differential equation are using Jumarie [22] modified Riemann-Liouville (RL) derivative which is defined as above. The first expression above is fractional integration of Jumarie type. The modification by Jumarie is to carry RL fractional integration or RL fractional derivation is by forming a new function by offsetting the original function by subtraction the function value at the start point; and then operate the RL definition. The above defined fractional derivative has the following properties
Γ+(1γ ) J Dxαγ⎡⎤=>xγα− ,0,γ 0 x ⎣⎦Γ+−(1γα ) JJJααα 000DfxgxgxDfxxxx[]()()=+ ()()() () fxDgx () () JJα ααJα 00Dxgxg[] f(()) gx== f′′() gx ()() D[] gx ()()0 D[] f (())( gx gx )
The equation (7) is example of the fractional differential equation
∂∂2ααuu = u …………………………………………………….(7) ∂∂x2α tα
Using the travelling wave transformationξ = + ctkx , the equation (6) reduce to
αα2 Luu( ,ξξξ , u ,.....)=<< 0 0α 1...... (8)
Where,uu= ()ξ is function of one variable ξ
6
Using the same transformation, the equation (7) reduce to dd2αu= u αu, as described below ddξ 2α ξα
Thus the operator ∂α , in terms of ξ with ξ = kx+ ct is ∂tα
ααα α ∂∂∂ξ α d J α ααα≡==cc α()Dξ ∂∂∂ttξξd
and similarly ∂2α is ∂x2α
2ααα ααααα α 2α ∂∂∂∂∂∂∂⎛⎞ξξ ⎛ ⎞ddα ⎛αα ⎞22 d αJ 2α 2ααα≡=⎜⎟ ααααα ⎜ ⎟ =kkkkD ⎜ α ⎟ ==2α ()ξ ∂∂∂∂∂∂∂xxx⎝⎠ξξ x ⎝ x ⎠dd ξξ ⎝ ⎠ d ξ
The transformed equation is thus, kcu2αα∂∂2ααuu− = 0 that is Luu(,αα , u2 )= 0. ∂∂ξξ2αα ξξξ
The solution of the equation (8) is taken in the following series form
n i ub(ξ )=Φ∑ i ...... (9) i=0
Where Φ ξ )( satisfy the fractional Riccati equation
Φ=+Φα ()ξσ 2 , 0 <<α 1
dΦ 2 Here σ is a real constant. Say for α = 1 we obtain the Riccati equation as dξ =+Φσ . From this we write dΦ = dξ , integrating both sides once we get dΦ = ξ . In terms of anti-derivative σ +Φ2 ∫ σ +Φ2 symbol, we write as ξ = D−1 ⎡1 ⎤. In similar way we write in terms of fractional integration Φ ⎣σ +Φ2 ⎦ (Jumarie type) the following
−α ⎡ 1 ⎤ ξ = DΦ ⎣⎢σ + Φ2 ⎦⎥
Similar to the previous method the coefficients can be obtained. Zhang et al [29] develop the method finding the solution fractional Riccati equation in the following form using generalized hyperbolic and trigonometric function
7
⎧−−σσξtanh − ⎫ α ( )⎪ ⎪ forσ < 0 ⎪ ⎬ −−σσξcoth − ⎪ ⎪ α ()⎭ ⎪ ⎪ σσξtan ⎫ ⎪⎪α () Φ=()ξσ⎨⎬for> 0 ⎪⎪− σσξcotα () ⎪ ⎭ ⎪ Γ+(1α ) −=forσ 0 ⎪ ξωα + ⎪ ⎩⎪
Where, the defined higher order transcendental functions are as follows
sinh (xx ) cosh ( ) tanh (xx ) ==ααcoth ( ) ααcosh (xx ) sinh ( ) αα Ex()α −+ Ex ()ααEx() Ex ()α sinh (xx ) ==αα cosh ( ) αα αα22
sin (xx ) cos ( ) tan (xx ) ==ααcot ( ) ααcos (xx ) sin ( ) αα Eix()α −+ Eix ()ααEix() Eix ()α sin (xx ) ==αα cos ( ) αα αα22
∞ α Where Ez()= z is the one parameter Mittag-Leffler function. α ∑ k=0 Γ+(1kα )
4.0 Nonlinear evolution equations (NEEs) of shallow water waves (SWW) partial and fractional differential equations
The nonlinear evolution equation of the shallow water wave equation is
uuxt+= xx u xxxy + puuquu x xt + t xx ...... (10)
Where uuxyt= (, ,). The Corresponding space and time fractional differential equation is
22αα 4 α αααα 2 2 uuuxt+= xx xxxy + puuquu x xt + t xx ...... (11)
Where for 0<<α 1 , we have the following
∂∂24ααuu∂2αu uu24αα== u 2 α= xt ∂∂txα ααxxxy ∂ x32 ∂ yαxx ∂ tα
8
5.0 Kadomtsev-Petviashvili (KP) partial and fractional differential equation
The KP equation was first written in 1970 by Soviet physicists Boris B. Kadomtsev and Vladimir I. Petviashvili. It came as a natural generalization of the KdV equation (derived by Korteweg and De Vries in 1895). Whereas in the KdV equation waves are strictly one-dimensional, in the KP equation this restriction is relaxed. The KP equation usually written as
uuuuuyy= ( t++ 6 x xxx ) x ...... (12)
Where uuxyt= (, ,). The Corresponding space and time fractional differential equation is
23αα ααα uuuuuyt=+( 6 xxx + ) ...... (13)
Where
∂∂23ααuu uu23αα==,0<<α 1 yx∂∂yt23αα
6.0 Forth order Boussinesq equations partial and fractional differential equation
In fluid dynamics, the Boussinesq approximation for water waves is an approximation valid for weakly non-linear and fairly long waves. The approximation is named after Joseph Boussinesq, who first derived them in response to the observation by John Scott Russell of the wave of translation (also known as solitary wave or soliton). The 1872 paper of Boussinesq introduces the equations now known as the Boussinesq equations. The forth order Boussinesq equation can be written in the following form. Solitons are solitary waves with huge amplitude locally, and zero amplitude at far away, with very high travelling velocity
2 uutt=+ xx 3( u )xx + u xxxx ...... (14)
Where uuxt= (,) .The Corresponding space and time fractional equation is
22αα 224 αα uutx=+3( u ) xx + u ...... (15)
Where
∂∂24ααuu uu24αα==,0<<α 1 tx∂∂tt24αα
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7.0 Solution of Nonlinear evolution equations (NEEs) of shallow water waves (SWW) using tanh method
Equation (10) that is uxt+=u xx u xxxy + puuquu x xt + t xx is non-linear equation in u= uxyt(, ,). Using travelling wave transformation in the form ξ = + + mykxct and following substitution
2 uckuukuxt ==ξξ xx ξξ 32 ukmuuukcuuxxxy ==ξξξξ x xt ξ ξξ 2 uutxx= k cuuξξξ in equation (10), is reduced to the ordinary differential equation form
2 kmuξξξξ ++ ckp( quu )ξ ξξ ++= ( c ku )ξξ 0...... (16)
Since the solitary waves are localized in space therefore the solution and its derivatives at large distance from the pulse are known to be extremely small and they vanishes ξ →±∞ [28]. Integrating once w.r.t ξ the above equation and by assuming the condition the derivative terms tend to zero as ξ ∞→ we get
22 2kmuξξξ ++++= ckp ( qu )ξ 2( c ku )ξ 0...... (17)
Here we consider the solution in the form 02n uSaaa()ξφφφφ==012 + + ++... an ...... (18) Where φ ξ = ξ)tanh()( Then
dd =−(1φ 2 ) ddξ φ
dd22d =−(1φφφ22 ) − 2 (1 −2 ) ddξ 22φφd
dd33 d 2 d =−(1φφφφφφ23 ) − 6 (1 −22 ) − 2 (1 −2 )(1 − 3 2 ) ddξ 33φφ d 2 dφ
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dd44 d 3 =−(1φφφ24 ) − 12 (1 −23 ) − ddξφ44 d φ 3
dd2 (36φφ22−− 8)(1 )2 + 2 φφφ (1 − 22 )(8 − 12 ) ddφ 2 φ
Putting the value of u and using the transformation define above equation (17) reduce to the form dd32SS 2(1)6(1)km223−−−φφφ22 ddφφ32 2 22ddSS⎛⎞2 −−2(1φφ )(1 − 3 ) +ck ( p + q )⎜⎟ (1 −φ ) ddφφ⎝⎠ dS ++2(ck )(1 −φ 2 ) = 0...... (19) dφ In the highest order derivative term the highest order ofφ is n + 3 and in the non-linear term it is 2n + 2. Equating them we get n = 1 .
We have (18) asuSaa()ξ ==01 +φξ(). Putting the value of S in (19) we get
2 22 2 2 2 −−−++−++−=4km (1φφ )(1 3 )a111 ckp ( q )( (1φ ) a) 2( c k )(1φ ) a 0...... (20)
Comparing the powers of φ from both side the following relations obtained for a1 ≠ 0 02 φ :2(−++ck ) ackpq1 ( += ) 4 km 12 φ :2(ck+− ) 2 ackpq1 ( +=− ) 16 km 32 φαβ:(ack1 += )12 k m 12km 2 Solving the above we get a1 = cpq()+ and =+ 4)( mkkc .
Therefore the solution is 12km uxyt(, ,)=+ a tanh()ctkxmy++ 0 cp()+ q for all 0 ∈ Ra .
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Solution of Nonlinear evolution equations
400
200
0
-200
-400 4 2 4 0 2 0 -2 -2 t-axis -4 -4 x-axis
Figure 1: Graphical presentation of u for fixed y and considering p = q= m = k =1, c = 3.
8.0 Solution of Nonlinear fractional evolution equations (NEEs) of shallow water waves (SWW) using Fractional Sub-equation method
Using the same travelling wave transformation as define in section-7.0 and with following substitutions
22αααα uckuxt = ξ 222ααα ukuxx = ξ 43α αα4 α ukmuxxxy = ξ ααα ukux = ξ ααα ucut = ξ
22α αα 4 ααα 2 2α in equation (16) , that is uuuxtxxxxxyxxttxx+= + puuquu + reduce to the form
24αα α αα 2αα α α2 α kmuξξ++ kcpquu( )ξξ =+ ( k cu ) ...... (21)
Consider the solution in the form (22)
2 n uaaa()ξ =+Φ+Φ++Φ01 2 ... an ...... (22)
Where Φ is define in section 3.0 Comparing the highest powersΦ from the highest order derivative term and the non-linear derivative term from (21) and using u from (22) we get
for nn+=43+ 3or n = 1. Therefore, = + aau 10 Φ ξ )( . Putting the value of u from the
12
above in (21) and comparing the like powers of Φ the following relations is obtained in
simplest form and using the relation a1 ≠ 0 12αα αα 2αα2 Φ+=++:2(ck)σ ack1 (p q)16σσ km 32αα αα αα Φ+=++:2(ck) 4 ack1 (p q )40σ km 52αα αα Φ=:24km2a1 ck(p+ q) Solving the above relation we get a =− 12kmαα and ()4ckαβ+ +σ km2 αβ=0 . 1 cα ()α +β Therefore the solution is following
⎧ 12kmαα ⎫ ⎫ ac+−−++σσtanh (tkxmy ) ⎪ 0 α α ()⎪ ⎪ ⎪ cpq()+ ⎪ ⎪ ⎬ forσ < 0 ⎪ 12kmαα ⎪ ac+−−+σσcoth (t+kxmy ) ⎪ ⎪ ⎪ 0 α α () ⎪ cpq()+ ⎭⎪ ⎪ ⎪ αα ⎪ ⎪⎪12km Γ+(1α ) uxyt(,,) =+⎨⎬ a0 ααforσω=0, − Cons tan t...... (23) ⎪⎪cpqctkxmy()(++++ )ω ⎪⎪12kmαα ⎫ ⎪⎪ac−+σσtan ( tkx+my) 0 α α ()⎪ ⎪⎪cpq()+ ⎪ for σ > 0 ⎪⎪αα ⎬ 12km ⎪⎪a + σσcot (ct++ kx my) ⎪ 0 α α ()⎪ ⎩⎪⎪cpq()+ ⎭ ⎭
For all 0 ∈ Ra .
Solution of Nonlinear evolution equations for alpha=0.4 Solution of Nonlinear evolution equations for alpha=0.6 Solution of Nonlinear evolution equations for alpha=0.8
60 40 20
20 40 10
0 20 0 -20
0 -10 -40
-20 -60 -20 4 4 4 2 4 2 4 2 4 0 2 0 2 0 2 0 0 0 -2 -2 -2 -2 -2 -2 -4 -4 -4 -4 -4 -4 t-axis x-axis t-axis x-axis t-axis x-axis
Figure 2: Graphical presentation of u for fixed y and considering p = q = m = k = 1, c = 3, forσ =−1.
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9.0 Solution of Kadomtsev-Petviashvili (KP) equation using tanh method
Using the same travelling wave transformation as define in section-7.0 and 23 umuukuuukuuyy ===ξξξ,,xxx ξξξx & ucut =ξequation (12), that is
uuuuuyy=+(6 t x + xxx ) x reduces to
23 muξξ =++ kcu(ξ 6 kuuξ ku ξξξ) ξ ...... (24)
Integrating both side two time w.r.t ξ and then using the condition the solitary waves are localized in space that is ,, uuu ξξξ ..all tends to zero when ζ→∞ [28]. We get after first 23 integration muξ =+ kcu(6ξξkuu+ kuξξξ)and after second integration we get
2224 (m−= kcu ) 3 ku + kuξξ ...... (25)
We consider the solution as defined in equation (18), that is uSaaa()ξ == +φφ +2 ++... a φn 01 2 n . Then putting the values of u from (18) in (25) we get
2 22242⎛⎞2ddSS 2 (mkcSkSk−= ) 3 +⎜⎟ (1 −φφφ )2 −− 2 (1 ) ...... (26) ⎝⎠ddφφ
Comparing the highest order derivative and the non-linear term we get for n = 2 . Therefore, 2 uSaaa()ξ ==01 +φξ() + 2 φ () ξ , Putting the value of u in (26) and comparing the like powers of φ the following relations are obtained
02 224 φ :(mkcakaak−+ )3002 + 2 = 0 12 2 4 φ :(mkcakaaak−+ )61011 − 2 = 0 22 22 4 φ :(mkcaka−= )3(2)82102 + aaak −2 324 φ :6aak12+= 4 ak 1 0 4224 φ :3ak22+= 6 ak 0
Solving the above we get a = 8kmkc42+−, a = 0 and a=−2k2 with the relation += 289 kka 422 0 6k 2 1 2 0
Hence the solution is
8kmkc42+− uxyt(, ,)=−+2k22 tanh() ctkxmy+ 6k 2
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Solution of Kadomtsev-Petviashvili equation
8 x 10
5
0
-5
-10
-15
-20 4 2 4 0 2 0 -2 -2 -4 -4 t-axis x-axis
Figure 3: Graphical presentation of u for fixed y and considering m = k = 1, c = 3.68.
10.0 Solution of Kadomtsev-Petviashvili (KP) equation using Fractional Sub- equation method
2α ααα3α The equation (13), that is uuuuuy =+(6txx + )x can be written in the following form
22242αα ααα uuuuuutx+++=6( x ) 6 xx xxxx yy ...... (27)
Using the same travelling wave transformation ξ = ct++ kx my and with following substitutions
22αα utx = kcuξξ αα22 2 ()ukux = ()ξ
444ααα ukuxxxx = ξ 222αα umuyy = ξ equation (27) reduce to
αα22 α α 2 α α 2 2 α 2 α 44 α α (ck−+ m ) uξξ 6 k ( ux ) + 6 k uu += k uξ0...... (28)
We want to find the solution in the form as define in equation (22), that is 2 n u(ξ) =+Φ+Φ++Φaaa01 2 ... an , and then comparing the highest order derivative term and the 2 non-linear term we get for n = 2 . Therefore ξ)( = au 10 aa 2Φ+Φ+ ()( ξξ ) . Putting in equation (27) and comparing the coefficients of powers of Φ ξ )( we get
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02αα α 22α 2224α 3 Φ−+++=:2(ck m ) a20σσσσ 6(2 k aa212 a )160 k a 12αα α 22α 24α 2 Φ−+++=:2(ck m ) a10σσσσ 6(26 k aa11 aa21 )160 k a 2222αα α α 24α 2 Φ−++++: 8(ck m ) a20σ 6 k (8 aa2122 σσσ 4 a 2 a ) 136 k a σ= 0 322αα α α 4α Φ−+++:2(ck m )26(218)40 a10σσ k aa11 aa2 k a1 = 0 42αα α 22α 24α Φ−++++=:6(ck m )6(6163)2400 a20 k aa2212 aσσ a k a 52αα 4 Φ+=:72240kaa12 ak 1 62αα24 Φ+=: 60ka22 120 ak 0
Solving we get a = mkkc24α −−8σ ααα, a = 0 and ak=−2 2α . Hence the solution is following 0 6k 2 1 2
24αααα ⎧mkkc−−8σ 22α ⎫ ⎪ +−+2tanh(kcσσα tkx+my ) 6k 2 ()⎪ ⎪ ⎬ forσ < 0 ⎪mkkc24αααα−−8σ +−+2coth(kc22ασσtkx+my )⎪ ⎪ 2 α ()⎪ ⎪ 6k ⎭ ⎪ 24αααα ⎛⎞2 ⎪mkkc−−8σ 2 ()Γ+(1α ) uk()ξσ=−⎨ 2 α ⎜⎟for= 0,ω− Cons tan t. 6k 2 ⎜⎟α 2 ⎪ ⎝⎠()()ct++ kx my +ω ⎪ ⎪mkkc24αααα−−8σ ⎫ − 2tan(kc22ασσtkxmy++ ) ⎪ 6k 2 α ()⎪ ⎪ ⎬ forσ > 0 ⎪mkkc24αααα−−8σ −+2cot(kc22ασσtkxm+y ) ⎪ ⎪ 2 α () ⎩ 6k ⎭⎪
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Solution of Kadomtsev-Petviashvili equation for alpha=0.4 Solution of Kadomtsev-Petviashvili equation for alpha=0.6 Solution of Kadomtsev-Petviashvili equation for alpha=0.8
24 x 10 24 21 x 10 x 10 4 2 3 3 1.5 2
2 1 1 0.5 1 0 0 0 -1 -0.5 -1 -2 4 -1 2 4 4 -3 0 2 2 4 4 0 0 2 2 4 -2 -2 0 2 t-axis -4 -4 -2 -2 0 x-axis 0 -4 -4 -2 t-axis x-axis -2 t-axis -4 -4 x-axis
Solution of Kadomtsev-Petviashvili equation for alpha=0.81 Solution of Kadomtsev-Petviashvili equation for alpha=0.85 Solution of Kadomtsev-Petviashvili equation for alpha=0.90
21 22 22 x 10 x 10 x 10 10 0.5 4 8 0 3 6 -0.5 4 2 -1 2 -1.5 1 0 -2 0 -2 4 -2.5 -1 2 4 4 4 0 2 2 4 0 2 4 -2 0 2 -2 0 0 2 -4 -4 -2 0 t-axis x-axis -2 -2 -2 t-axis -4 -4 x-axis -4 -4 t-axis x-axis
Figure 4: Graphical presentation of u for fixed y and considering m = k = 1, c = 3.68, forσ =−1.
11.0 Solution of Forth order Boussinesq equation using tanh Method
Equation (14), that is non-linear equation in uuxt= (,). Using travelling wave transformation in the form ξ += ctkx and with following substitution
2 ucutt = ξξ 2 ukuxx = ξξ
222 ()ukux = () ξ 4 ukuxxxx = ξξξξ in equation (14) the equation reduced to the form 22 22 4 (cku−= )ξξ 3 ku ( )ξξ + ku ξξξξ ...... (29)
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Here we consider the solution as defined in equation (18), that is 2 n u()ξ ==Saaa01 +φφ + 2 ++... an φ . Putting the values of u from (9) in equation (19) and using the transformation as define in section 7.0 we get 2 22⎡⎤ 22ddSS2 (ck−− )⎢⎥ (1φφφ )2 −− 2 (1 ) = ⎣⎦ddφφ
2 2 22⎡⎤2ddSS 2 22⎛⎞ d S 3{2(1)4(1)}2(1)kS⎢⎥−−−+−+φφφφ2 ⎜⎟ ⎣⎦ddφφ⎝⎠ d φ 43 ⎛⎞24ddSS23 ⎜⎟(1−−−φφφ ) 12 (1 ) ddφφ43 k 4 ⎜⎟ ⎜⎟2 22 2 ddSS2 2 ⎜⎟+−(1φφ ) (36 − 8)2 + 2 φφφ (1 − )(8 − 12 ) ⎝⎠ddφφ = 0...... (30)
Comparing the highest order derivative term and the non-linear term we obtain n = 2 . Therefore 2 ξ)( = Su 10 ++= aaa 2 ξφξφ )()( . Putting this value of S in equation (30) and comparing the powers of φ we get
0222 24 φ :2()3(42)16cka−=20 kaaa212 +− ak 1222 4 φ :2()3(412)16−−c k a10 =− k aa11 + aa2 + ak1 222224 φ :−− 8(cka )20 =− 3 k ( 16 aaa2 + 41 ) + 136 ak2 3222 4 φ :2()3(436)40c−= k a10 k aa11 − aa21 − ak 42222 4 φ : 6(cka−=−+ )210 3 k ( 26 a 12 aa2 ) − 240 ak2 5 42 φ :24a112kaak+=72 0 622 φ :60(2)0ak22 a+= k Solving the above set of equation we get a = 8kck422+− , a = 0 and a=−2k2 . Hence the solution 0 6k 4 1 2 is 8kck422+− uxyt(, ,)=−2k22 tanh() ctkx+ 6k 4
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Solution of Forth order Boussinesq equation
4 x 10
1
0
-1
-2
-3
-4
-5 4 2 4 0 2 0 -2 -2 -4 -4 t-axis x-axis
Figure 5: Graphical presentation of ufor fixed y and considering c = k = 1.
12.0: Solution of Forth order fractional Boussinesq equation using Fractional Sub-equation method Using the same type of travelling wave transformation ξ = ct+ kx and with following substitution
222ααα ucut = ξ , uku222ααα= , x ξ 22α 2αα 22 ()ukux = ()ξ 444ααα ukux = ξ
22α αα 224α the fractional Boussinesq equation (15) which is uutx=+3( u ) x + uxreduce to the following form
222ααα 22244 α ααα (cku−= )ξξξ 3 ku ( ) + ku ...... (31)
We want to find the solution in the form as define in equation (22), comparing the highest order 2 derivative term and the non-linear term we get n = 2 . Therefore, ξ)( 10 aaau 2Φ+Φ+= ξξ )()( . Comparing the like powers of Φ ξ)( we get the following
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022222243αα α α Φ−=++:2()cka20σσ 3 k (42)16 aaa21 ak2σ 1222αα α 224α Φ−=++:2(σσckakaaaa )12(1011 321σ )16 akσ 222222αα α 22α4 α2 Φ−=+++: 8σ (cka )20 24 kaaa (22122 )σσ 36 ak 136 akσ 3222αα α 2α 4 α Φ−=++: 2(cka )10 120 kaaaak11 1082σσ 40 ak 1 4 22αα 22 α 22α 4α Φ :6(cka−= )210 18 kaaaka ( ++ 222 ) 96σσ + 240 ak 2 542αα Φ+=:24720ak11 aak2 622αα Φ+=:60(2)0ak22 a k
Solving the above system of equation we get a = ckk242α −−8σ αα, a = 0 and ak=−2 2α . Hence the 0 6k 2 1 2 solution is 242ααα ⎧ckk−−8σ 22α ⎫ ⎪ +−+2tanh(kcσσα tkx ) 6k 2 ()⎪ ⎪ ⎬ forσ < 0 ⎪ckk242ααα−−8σ +−+2coth(kc22ασσtkx )⎪ ⎪ 2 α ()⎪ ⎪ 6k ⎭ ⎪ 242ααα ⎛⎞2 ⎪ckk−−8σ 2α ()Γ+(1α ) uxy(,) =−⎨ 2k ⎜⎟forσω=− 0, Cons tan t. 6k 2 ⎜⎟α 2 ⎪ ⎝⎠()()ct++ kx ω ⎪ ⎪ckk242ααα−−8σ ⎫ −+2tan(kc22ασσtkx ) ⎪ 6k 2 α ()⎪ ⎪ ⎬ forσ > 0 ⎪c2α −8σ kk42αα− −+2cot(kc22ασσtkx ) ⎪ ⎪ 2 α () ⎩ 6k ⎭⎪
Solution of Forth order Boussinesq equation for alpha=0.6 Solution of Forth order Boussinesq equation for alpha=0.8
Solution of Forth order Boussinesq equation for alpha=0.4
200
150 250 20 100 200 0 50 150 -20 0 -40 100 -50 -60 -100 50 -80 4 2 4 0 -100 0 2 4 0 -50 -2 2 4 -2 4 2 -4 -4 0 t-axis x-axis 0 2 4 -2 -2 0 2 -4 -4 t-axis x-axis 0 -2 -2 -4 -4 t-axis x-axis
Figure 6: Graphical presentation of u for fixed y and considering c = k= 1, forσ =−1.
From figure 1-6 it is clear that solutions of the fractional differential equations depend on order of fractional derivative.
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13.0 Conclusion
In all the three cases the solution obtained by tanh-method coincides with one solution of obtained by Fractional Sub-Equation method, when fractional order tends to one. Thus the solution of the non-linear partial differential equation and the non-linear fractional differential equation is obtained in analytic form. However, the solution obtained by fractional sub-equation method is more general compare to the tanh-method, as it encompasses the fractional generalization of the non-linear partial differential equations.
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