1 

 

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: uttam_math@yahoo.co.in

(2): Department of Applied Mathematics, University of Calcutta, Kolkata, India Email (2b): susmita62@yahoo.co.in,

(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): shantanu@barc.gov.in 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

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 wave solution form. In this paper the symbols for fractional differential operator used 0

JxDα , ( )f α

means Jumarie fractional derivative.

2.0 Tanh-method

Let the non-linear partial differential equation is of the form

( , , , ....) 0.................................................................(1)t x xxH u u u u =

Where is function of( , )u u x t= ,x t .

For instant

2

2 0u uux t∂ ∂

− =∂ ∂

is of such an equation as ( , , ) 0t xxH u u u = , with H a linear /non-linear operator. Using travelling wave transformation ctkx +=ξ , with constant called wave-number (or propagation constant) and a constant; as velocity of propagating wave, equation (1) reduce to an ordinary differential equation in the form

kc

( , , ....) 0..................................................................(2)H u u uξ ξξ =

2 

 

Where ( )u u ξ= , is function of one variableξ . In summary we are making several variables ( , , , )x y z t to a single variableξ , with travelling wave transformation such as

x yk x k y zk z ctξ = + with corresponding constants ,xk described above, and + + , c as, ,y zk k

ξ is in units of distance. Thus the operator t∂∂ , in terms of ξ with kx ctξ = + is following

dd

ct t

ξξ ξ

∂ ∂ ∂≡ =

∂ ∂ ∂

and similarly 2

2x∂∂

is

2 22

2 2

dd

k k kx x x x x

ξ ξξ ξ ξ ξ

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞≡ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ξ

The transformed equation 2

2 0u utx

u∂ ∂∂∂

− = , is thus,

22 2

2

d d 0 ord d

u uk cu k u cuuξξ ξξ ξ− = − 0=

which is . ( , , ) 0H u u uξ ξξ =

Again, let )(ξφφ = such type of function which is a solution of Riccati equation 2)( φσξφ +=′

. Where

be

σ is a real constant. That is d 2dφξ σ φ= + , 2

dd φσ φ

ξ+

=∫ ∫

Then

tanh( )for 0

coth( )

tan( )( ) for 0

cot( )1 for 0

σ σξσ

σ σξ

σ σξφ ξ σ

σ σξ

σξ

⎧ ⎫− − − ⎪⎪ <⎬⎪− − − ⎪⎭⎪

⎫⎪ ⎪= >⎨ ⎬− ⎪⎪ ⎭⎪⎪− =⎪⎩

The above expression is detailed below briefly.

Case-I For 0σ < then the Riccati equation can be written equation, considering integral constant as zero

3 

 

( )

( )

1

21

1 tanhdd

1 coth

φσ

φσ

φ σξ ξσ φ

σ

−−

−−

⎧−⎪ −⎪= = ⎨+ ⎪−

⎪ −⎩

∫ ∫

Therefore, by inverting above we get the following

( )( )

tanh

coth

σ σξφ

σ σξ

⎧− − −⎪= ⎨− − −⎪⎩

Case-II For 0σ = then the Riccati equation can be written equation, ignoring integration constant

2d 1d 1φξ ξ φφ φ ξ

= = −∫ ∫ = −

Case-III For 0σ > then the Riccati equation can be written equation, considering integral constant as zero

( )

( )

1

21

1 tandd

1 cot

φσ

φσ

φ σξ ξσ φ

σ

−

−

⎧⎪⎪= = ⎨+ ⎪−⎪⎩

∫ ∫

Therefore, inverting above we get the following

( )( )

tan

cot

σ σξφ

σ σξ

⎧⎪= ⎨−⎪⎩

The solution of the equation (2) will be expressed in-terms of φ in the form

0( ) ......................................................................(3)

ni

ii

u S aξ φ=

= =∑

Where is obtained balancing the highest order derivative term and the non-linear terms in equation (2). Then put the value of

n( )u ξ from (3) to the equation (2) and comparing the

coefficient of the values of coefficients can be obtained. iφ

Consider the equation 2

2u

txu∂

∂∂= u∂ . Using the travelling wave transformation ctkx +=ξ the

considered equation reduces to

2k u cuuξξ = ξ ……..................................................................(4)

4 

 

Then using the transformation as defined in (3) in-terms of φ and using the chain rule of derivative we obtain

2 2d d d d dsec h (1 )d d d d d

φ ξ φξ ξ φ φ φ= = = −

2 22 2 2

2 2

d d d d d d d(1 ) (1 ) 2 (1 )d d d d d d d d

φφ φ φ 2 dφξ ξ ξ φ φ ξ φ φ

⎛ ⎞= = − = − − −⎜ ⎟

⎝ ⎠

Putting this transformation in equation (4) we get

22 2 2 2 2

2

d d(1 ) 2 (1 ) (1 )d d

S Sk φ φ φ φ dd

ScSφ φ φ

⎡ ⎤− − − = −⎢ ⎥

⎣ ⎦ ……………………………(5)

In the highest order derivative term the highest order ofφ is n 2+ and in the non-linear term it is2 1n Equating them we get n We have (3) as following + . .1=

0 1( )u S a aξ φ= = +

Therefore

2

1 2

d dand 0d d

S Saφ φ= =

Putting the above value in equation (5) we get the following

( )2 21 0 12 (1 ) (1 )k a c a aφ φ φ φ⎡ ⎤− = + −⎣ ⎦

21a

Comparing the like powers of φ from both sides of above expression we get

2

0 120 and ka ac

= = −

Hence the corresponding solution is

2 22 2( ) tanh( )k ku Sc c

ξ φ ξ= = − = −

that is

22( , ) tanh( )ku x t kx ctc

= − +

5 

 

3.0 Fractional sub-equation method

The space and time fractional partial differential equation is of the form

2( , , , .....) 0 0 1..............................................(6)t x xL u u u uα α α α= < <

where and is functions (linear or non-linear operator) of the u and its derivatives. ( , )u u x t= Lα is order of fractional derivative of Jumarie type, defined as follows

[ ] ( )

( )

1

0

( )0

0( )( )

1 ( ) ( )d , 0( )

1 d( ) ( ) ( ) ( ) (0) d , 0 1(1 ) d

( ) , 1, 1

x

xJ

x

nn

x f

D f x f x x f fx

f x n n n

α

α α α

α

ξ ξ ξ αα

ξ ξ ξ αα

α

− −

−

−

⎧− <⎪Γ −⎪

⎪⎪= = − − <⎨Γ −⎪⎪

≤ < + ≥⎪⎪⎩

∫

∫ <

We consider that at the function0x < ( ) 0f x = and also ( ) (0) 0f x f− = for . 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

0x <

[ ] ( ) ( )[ ] ( ) [ ]( ) [ ]( )

0

0 0 0

0 0 0

(1 ) , 0,(1 )

( ) ( ) ( ) ( ) ( ) ( )

( ( )) ( ) ( ) ( ( )) ( )

Jx

J J Jx x x

J J Jx g x g

D x x

D f x g x g x D f x f x D g x

D f g x f g x D g x D f g x g

α γ γ α

α α α

xα α α α

γ γγ α

−Γ +⎡ ⎤ = >⎣ ⎦ Γ + −

= +

′ ′= =

The equation (7) is example of the fractional differential equation

2

2

u u ux t

α α

α α

∂ ∂=

∂ ∂ …………………………………………………….(7)

Using the travelling wave transformation ctkx +=ξ , the equation (6) reduce to

2( , , ,.....) 0 0 1..........................................(8)L u u uα αξ ξξ α= < <

Where, ( )u u ξ= is function of one variable ξ

6 

 

Using the same transformation, the equation (7) reduce to 2

2d dd d

u uα uα

α αξ ξ= , as described below

Thus the operatort

α

α∂∂

, in terms of ξ with kx ctξ = + is

( )dd

Jc ct t

α α α α

Dα αξα α α α

ξξ ξ

∂ ∂ ∂≡ = =

∂ ∂ ∂

and similarly 2

2x

α

α∂∂

is

( )2 2

2 22 2

d d dd d d

Jk k k k Dx x x x x

α α α α α α α α α α2α α α αξα α α α α α α α α α

ξ ξξ ξ ξ ξ ξ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂≡ = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

α

The transformed equation is thus, 2

22 0u uk c uα α

α αα α

ξ ξ∂ ∂∂ ∂

− = that is . 2( , , ) 0L u u uα αξ ξξ =

The solution of the equation (8) is taken in the following series form

0( ) ........................................................................(9)

ni

ii

u bξ=

= Φ∑

Where )(ξΦ satisfy the fractional Riccati equation

2( ) , 0 1α ξ σ αΦ = +Φ < <

Here σ is a real constant. Say for 1α = we obtain the Riccati equation as 2ddξ σΦ = +Φ . From this

we write 2d d

σξΦ

+Φ= , integrating both sides once we get 2

dσ

ξΦ+Φ

=∫ . In terms of anti-derivative

symbol, we write as 2 ⎤⎦1 1D

σξ −

Φ +Φ⎡= ⎣ . In similar way we write in terms of fractional integration

(Jumarie type) the following

2

1D αξσ

−Φ

⎡ ⎤= ⎢ ⎥+Φ⎣ ⎦

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 

 

( )( )

( )( )

tanhfor 0

coth

tan( ) for 0

cot

(1 ) for 0

α

α

α

α

α

σ σξσ

σ σξ

σ σξξ σ

σ σξ

α σξ ω

⎧ ⎫− − − ⎪⎪ <⎬⎪− − − ⎪⎪ ⎭⎪

⎫⎪⎪ ⎪Φ = >⎨ ⎬−⎪ ⎪⎭⎪Γ +⎪− =⎪ +⎪

⎪⎩

Where, the defined higher order transcendental functions are as follows

sinh ( ) cosh ( )tanh ( ) coth ( )cosh ( ) sinh ( )

( ) ( ) ( ) ( )sinh ( ) cosh ( )2 2

x xx xx x

E x E x E x E xx x

α αα α

α α

α α αα α α α

α α

= =

− += =

α

sin ( ) cos ( )tan ( ) cot ( )cos ( ) sin ( )

( ) ( ) ( ) ( )sin ( ) cos ( )2 2

x xx xx x

E ix E ix E ix E ixx x

α αα α

α α

α α αα α α α

α α

= =

− += =

α

Where (1 )0( ) z

kkE z α

α α∞

Γ +==∑ is the one parameter Mittag-Leffler function.

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

................................................(10)xt xx xxxy x xt t xxu u u pu u qu u+ = + +

Where . The Corresponding space and time fractional differential equation is ( , , )u u x y t=

2 2 4 2 2 ........................................(11)xt xx xxxy x xt t xxu u u pu u qu uα α α α α α α+ = + +

Where for 0 1α< < , we have the following

2 42 4 2

3 2xt xxxy xxu uu u u

t x x y t

α αα α α

2 uαα α α α

∂ ∂= =∂ ∂ ∂ ∂ ∂ α

∂=

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

( 6 ) ............................................................(12)yy t x xxx xu u uu u= + +

Where . The Corresponding space and time fractional differential equation is ( , , )u u x y t=

2 3( 6 ) ......................................................(13)y t x x xu u uu uα α α α α= + +

Where

2 32 3

2 3 , 0y xu uu u

y t

α αα α

α α α∂ ∂= =∂ ∂

1< <

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

23( ) ...................................................(14)tt xx xx xxxxu u u u= + +

Where .The Corresponding space and time fractional equation is ( , )u u x t=

2 2 2 2 43( ) .....................................................(15)t x x xu u u uα α α α= + +

Where

2 42 4

2 4 , 0t xu uu u

t t

α αα α

α α α∂ ∂= =∂ ∂

1< <

9 

 

7.0 Solution of Nonlinear evolution equations (NEEs) of shallow water waves (SWW) using tanh method

Equation (10) that is xt xx xxxy x xt t xxu u pu u qu u+ = + +u is non-linear equation in

u Using travelling wave transformation in the form

( , , )u x y t= . mykxct ++=ξ and following substitution

2

3 2

2

xt xx

xxxy x xt

t xx

u cku u k u

u k mu u u k cu u

u u k cu u

ξξ ξξ

ξξξξ ξ ξξ

ξ ξξ

= =

= =

=

in equation (10), is reduced to the ordinary differential equation form

2 ( ) ( ) 0............................................(16)k mu ck p q u u c k uξξξξ ξ ξξ ξξ+ + + + =

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

2 22 ( ) 2( ) 0............................................(17)k mu ck p q u c k uξξξ ξ ξ+ + + + =

Here we consider the solution in the form 0 2

0 1 2( ) ... .................................................(18)nnu S a a a aξ φ φ φ φ= = + + + +

Where )tanh()( ξξφ = Then

2d d(1 )d d

φξ φ= −

2 22 2 2

2 2

d d(1 ) 2 (1 )d d

φ φ φ ddξ φ φ

= − − −

3 3 22 3 2 2 2 2

3 3 2

d d d(1 ) 6 (1 ) 2 (1 )(1 3 )d d d

φ φ φ φ φ φ ddξ φ φ

= − − − − − −φ

10 

 

4 4 32 4 2 3

4 4 3

22 2 2 2 2

2

d d d(1 ) 12 (1 )d d d

d d(36 8)(1 ) 2 (1 )(8 12 )d d

φ φ φξ φ φ

φ φ φ φ φφ φ

= − − − −

− − + − −

Putting the value of u and using the transformation define above equation (17) reduce to the form

3 22 2 3 2 2

3 2

22 2 2

2

d d2 (1 ) 6 (1 )d d

d d2(1 )(1 3 ) ( ) (1 )d d

d2( )(1 ) 0.............................................................................(19)d

S Sk m

S Sck p q

Sc k

φ φ φφ φ

φ φ φφ φ

φφ

− − −

⎛ ⎞− − − + + −⎜ ⎟

⎝ ⎠

+ + − =

In the highest order derivative term the highest order ofφ is 3n + and in the non-linear term it is . Equating them we get . 2n + 2 1n =

We have (18) as 0 1( ) ( )u S a aξ φ ξ= = + . Putting the value of in (19) we get S

( )22 2 2 2 21 1 14 (1 )(1 3 ) ( ) (1 ) 2( )(1 ) 0............(20)k m a ck p q a c k aφ φ φ φ− − − + + − + + − =

Comparing the powers of φ from both side the following relations obtained for 01 ≠a0 2

11 2

13 2

1

: 2( ) ( ) 4

: 2( ) 2 ( ) 16

: ( ) 12

c k a ck p q k m

c k a ck p q k m

a ck k m

φ

φ

φ α β

− + + + =

+ − + = −

+ =

Solving the above we get 121 ( )

kmc p qa += and . mkkc 24)( =+

Therefore the solution is

( )012( , , ) tanh( )

kmu x y t a ct kx myc p q

= + + ++

for all .0 Ra ∈

11 

 

12 

 

-4

-20

24

-4-2

0

24

-400

-200

0

200

400

x-axis

Solution of Nonlinear evolution equations

t-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

2 2

2 2 2

4 3

xt

xx

xxxy

x

t

u c k u

u k u

u k m u

u k u

u c u

α α α αξ

α α αξ

4α α α αξ

α α αξ

α α αξ

=

=

=

=

=

in equation (16) , that is 2 2 4 2 2xt xx xxxy x xt t xxu u u pu u qu uα α α α α α+ = + + α reduce to the form

2 4 2 2( ) ( ) .........................(21)k m u k c p q u u k c uα α α α α α α α α αξ ξ ξ ξ+ + = +

Consider the solution in the form (22)

20 1 2( ) ... ........................................................(22)n

nu a a a aξ = + Φ + Φ + + Φ

Where is define in section 3.0 Comparing the highest powersΦ Φ from the highest order derivative term and the non-linear derivative term from (21) and using from (22) we get for or . Therefore,

u4 3 3n n ++ = 1n = )(10 ξΦ+= aau . Putting the value of u from the

above in (21) and comparing the like powers of Φ the following relations is obtained in simplest form and using the relation 01 ≠a

1

)

)

2a

α α

α α

α α

1 2

13 2

15 2

: 2( ) 16

: 2( 4 ) 40

: 24 )

c k a c k q k m

c k a c k q k m

k m c k q

α α α

α α α

α α

2 2(

(

(

p

p

p

α

α

σ σ σ

σ

Φ + = + +

Φ + = + +

+

Φ =

13 

 

Solving the above relation we get 12cα1 ( )

k ma α α

α β+= − 0and 2( ) 4c k k mα β α βσ+ + = .

Therefore the solution is following

( )

( )

( )

0

0

0

0

0

12 tanh ( )( )

for 012 coth ( )

( )

12 (1 )( , , ) 0, Cons tan t.( ) ( )

12 tan (( )

12 cot (( )

k ma ct kx myc p q

k ma ct kx myc p q

k mu x y t ac p q ct kx my

k ma ct kx myc p q

k mac p q

α α

αα

α α

αα

α α

α α

α α

αα

α α

αα

σ σσ

σ σ

α σ ωω

σ σ

σ σ

⎫+ − − + + ⎪+ ⎪ <⎬

⎪+ − − + ⎪+ ⎭Γ +

= + = −+ + + +

− + ++

++

........(23)

for 0ct kx

σ

⎫⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ ⎪

⎨ ⎬⎪ ⎪⎪ ⎪⎫⎪ ⎪⎪⎪ ⎪⎪ >⎬⎪ ⎪

⎪⎪ ⎪+ + ⎪⎪ ⎪⎭⎩ ⎭

for

)

+

( ))my

For all .0 Ra ∈

-4

-20

24

-4-2

0

24

-20

0

20

40

60

x-axis

Solution of Nonlinear evolution equations for alpha=0.4

t-axis -4-2

Solution of Nonlinear evolution equations for alpha=0.6

-4-2

02

4

-4-2

0

24

-20

-10

0

10

20

x-axis

Solution of Nonlinear evolution equations for alpha=0.8

t-axis

Figure 2: Graphical presentation of u for fixed y and considering p = q = m = k = 1, c = 3, for 1σ =− .

02

4

-4-2

0

24

-60

-40

-20

0

20

40

x-axist-axis

9.0 Solution of Kadomtsev-Petviashvili (KP) equation using tanh method

Using the same travelling wave transformation as define in section-7.0 and 2 3, , &yy xxx x tu m u u k u u u ku u u cuξξ ξξξ ξ= = =

( 6 )yy t x xxx xu u uu u= + +ξ= equation (12), that is

reduces to

2 3( 6 ) .......................................................(24)m u k cu kuu k uξξ ξ ξ ξξξ ξ= + +

Integrating both side two time w.r.t ξ and then using the condition the solitary waves are localized in space that is all tends to zero when ζ→∞ [28]. We get after first

integration

..,, ξξξ uuu

( 6kuu k2m u k cu 3 )uξ ξ ξ ξ= + ξξ+ and after second integration we get

2 2 2 4( ) 3 ....................................................................(25)m kc u k u k uξξ− = +

We consider the solution as defined in equation (18), that is 20 1 2( ) ... n

nu S a a a aξ φ φ φ= = + + + + .

Then putting the values of from (18) in (25) we get u

22 2 2 4 2 2 2

2

d d( ) 3 (1 ) 2 (1 ) ...............................(26)d d

S Sm kc S k S k φ φ φφ φ

⎛ ⎞− = + − − −⎜ ⎟

⎝ ⎠

Comparing the highest order derivative and the non-linear term we get for . Therefore, 2n =2

0 1 2( ) ( ) ( )u S a a aξ φ ξ φ ξ= = + + , Putting the value of in (26) and comparing the like powers of

uφ the following relations are obtained

0 2 2 2 40 0 2

1 2 2 41 0 1 1

2 2 2 22 1 0 2

3 2 41 2 1

4 2 2 42 2

: ( ) 3 2 0

: ( ) 6 2 0

: ( ) 3 ( 2 ) 8

: 6 4 0

: 3 6 0

m kc a k a a k

m kc a k a a a k

m kc a k a a a a k

a a k a k

a k a k

φ

φ

φ

φ

φ

− + + =

− + − =

− = + −

+ =

+ =

42

Solving the above we get 4 2

28

0 6k m kc

ka + −= , 1 0a = and with the relation 2

2 2a = − k 4220 289 kka +=

Hence the solution is

( )4 2

2 22

8( , , ) 2 tanh6

k m kcu x y t k ct kx myk

+ −= − + +

14 

 

-4-2

02

4

-4-2

0

24

-20

-15

-10

-5

0

5

x 108

x-axis

Solution of Kadomtsev-Petviashvili equation

t-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

The equation (13), that is 2 ( 6 )3y t x xu u uu u xα α α α= + + α can be written in the following form

2 2 2 4 26( ) 6 ..........................................................(27)tx x xx xxxx yyu u uu u uα α α α α+ + + =

Using the same travelling wave transformation ct kx myξ = + + and with following substitutions

2 2

2 2

4 4 4

2 2 2

( ) ( )tx

x

xxxx

yy

u kcu

u k u

u k u

u m u

α αξξ

α αξ

α α αξ

α αξ

=

=

=

=

2

,

Puttin

equation (27) reduce to

2 2 2 2 2 2 4 4( ) 6 ( ) 6 0.................... (28)xc k m u k u k uu k uα α α α α α α α α αξ ξ ξ− + + + =

We want to find the solution in the form as define in equation (22), that is(u and then comparing the highest order derivative term and the

non-linear term we get for 2n = . Therefore )(ξ = au g in equation (27) and comparing the coefficients of powers of )(

20 1 2) ... n

na a a aξ = + Φ + Φ + + Φ

)()( 2210 ξξ Φ+Φ+ aa .

ξΦ we get

15 

 

16 

 

2 =

0 2 2 2 2 2 2 4 32 0 2 1 2

1 2 2 2 2 4 21 0 1 1 2 1

2 2 2 2 2 42 0 2 1 2 2

3 2 2 41 0 1 1 2

: 2( ) 6 (2 ) 16 0

: 2( ) 6 (2 6 ) 16 0

: 8( ) 6 (8 4 2 ) 136 0

: 2( )2 6 (2 18 ) 40

c k m a k a a a k a

c k m a k a a a a k a

c k m a k a a a a k a

c k m a k a a a a k

α α α α α

α α α α α

α α α α α

α α α α α

σ σ σ σ

σ σ σ σ

σ σ σ σ σ

σ σ

Φ − + + + =

Φ − + + + =

Φ − + + + +

Φ − + + + 14 2 2 2 2 4

2 0 2 2 1 25 2 4

1 2 16 2 2 4

2 2

0

: 6( ) 6 (6 16 3 ) 240 0

: 72 24 0

: 60 120 0

a

c k m a k a a a a k a

k a a a k

k a a k

α α α α α

α α

α α

σ σ

=

Φ − + + + + =

Φ + =

Φ + =

Solving we get 2 4

28

0 6m k k c

ka α α α ασ− −= , 1 0a = and 2

2 2a k α= − . Hence the solution is following

( )

( )( )

( )

2 42 2

2

2 42 2

2

22 42

22

2 42 2

2

8 2 tanh ( )6 for 0

8 2 coth ( )6

(1 )8( ) 2 for 0, Cons tan t.6 ( )

8 2 tan (6

m k k c k ct kx myk

m k k c k ct kx myk

m k k cu kk ct kx my

m k k c k ck

α α α αα

α

α α α αα

α

α α α αα

α

α α α αα

α

σ σ σσ

σ σ σ

ασξ σω

σ σ σ

⎫− −+ − + + ⎪⎪ <⎬

− − ⎪+ − + + ⎪⎭⎛ ⎞Γ +− − ⎜ ⎟= − = −⎜ ⎟+ + +⎝ ⎠

− −− ( )

ω

( )2 4

2 22

)for 0

8 2 cot ( )6

t kx my

m k k c k ct kx myk

α α α αα

α

σσ σ σ

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪

⎫⎪+ + ⎪⎪ ⎪⎪ >⎬

− −⎪ ⎪− + +⎪ ⎪⎭⎩

Solution of Kadomtsev-Petviashvili equation for alpha=0.6

-4-2

02

4

-4-2

0

24

-1

0

1

2

3

4

x 1024

x-axis

Solution of Kadomtsev-Petviashvili equation for alpha=0.4

t-axis

-4-2

02

4

-4-2

0

24

-3

-2

-1

0

1

2

3

x 1021

x-axis

Solution of Kadomtsev-Petviashvili equation for alpha=0.8

t-axis

2

x 1024

1.5

17 

 

Figure 4: Graphical presentation of u for fixed y and considering m = k = 1, c = 3.68, for 1σ =− .

11.0 Solution of Forth order Boussinesq equation using tanh Method

Equation (14), that is non-linear equation in ( , )u u x t= . Using travelling wave transformation in the form ctkx +=ξ and with following substitution

2

2

tt

xx

u c u

u k uξξ

ξξ

=

=

2 2 2

4

( ) ( )x

xxxx

u k u

u k uξ

ξξξξ

=

=

in equation (14) the equation reduced to the form 2 2 2 2 4( ) 3 ( ) ..............................(29)c k u k u k uξξ ξξ ξξξξ− = +

-4-2

02

4

-4-2

0

24

-1

-0.5

0

0.5

1

t-axis x-axis

-4-2

02

4

-4-2

0

24

-2.5

-2

-1.5

-1

-0.5

0

0.5

x 1022

x-axis

Solution of Kadomtsev-Petviashvili equation for alpha=0.85

t-axis

Solution of Kadomtsev-Petviashvili equation for alpha=0.90Solution of Kadomtsev-Petviashvili equation for alpha=0.81

-4-2

02

4

-4-2

0

24

-2

0

2

4

6

8

10

x 1021

4

x 1022

x-axis

3

t-axis

-4-2

02

4

-4-2

0

24

-1

0

1

2

t-axis x-axis

Here we consider the solution as defined in equation (18), that isu 2

0 1 2( ) ... nnS a a a aξ φ φ φ= = + + + + . Putting the values of u rom (9) in equation (19) and

using the transformation as define in section 7.0 we get f

22 2 2 2 2

2

222 2 2 2 2 2

2

4 32 4 2 3

4 34

22 2 2 2 2

2

d d( ) (1 ) 2 (1 )

d d d3 {2(1 ) 4 (1 ) } 2(1 )d d d

d d(1 ) 12 (1 )d d

d d(1 ) (36 8) 2 (1 )(8 12 )d d

0...........

S Sc kd d

S S Sk S

S S

kS S

φ φ φφ φ

φ φ φ φφ φ φ

φ φ φφ φ

φ φ φ φ φφ φ

⎡ ⎤− − − − =⎢ ⎥

⎣ ⎦

⎡ ⎤ ⎛ ⎞− − − + − +⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦⎛ ⎞

− − −⎜ ⎟⎜ ⎟⎜ ⎟+ − − + − −⎜ ⎟⎝ ⎠

= .......................(30)

Comparing the highest order derivative term and the non-linear term we obtain 2n = . Therefore

)(ξ Su = this value of S in equation (30) and comparing the powers of

)()( 2210 ξφξφ aaa ++= . Putting

φ we get 0 2 2 2 2 4

2 0 2 1 2

1 2 2 21 0 1 1 2

2 2 2 2 22 0 2 1

3 2 2 2 41 0 1 1 2 1

4 2 2 2 22 1 0 2

5

: 2( ) 3 (4 2 ) 16

: 2( ) 3 ( 4 12 ) 16

: 8( ) 3 ( 16 4 ) 136

: 2( ) 3 (4 36 ) 40

: 6( ) 3 ( 26 12 ) 240

: 24

c k a k a a a a k

c k a k a a a a a k

c k a k a a a a k

c k a k a a a a a k

c k a k a a a a k

a

φ

φ

φ

φ

φ

φ

− = + −

− − = − + +

− − = − + +

− = − −

− = − + −4 2

1 1 26 2 2

2 2

72 0

: 60 ( 2 ) 0

k a a k

a k a kφ

+ =

+ =

41

42

42

Solving the above set of equation we get 4 2 2

48

0 6k c k

ka + −= , 1 0a = and . Hence the solution

is

22 2a = − k

( )4 2 2

2 24

8( , , ) 2 tanh6

k c ku x y t k ct kxk

+ −= − +

18 

 

-4-2

02

4

-4-2

0

24

-5

-4

-3

-2

-1

0

1

x 104

x-axis

Solution of Forth order Boussinesq equation

t-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 ct kxξ = + and with following substitution

2 2 2

2 2 2

2 2 2 2 2

4 4 4

,

,

( ) ( )

t

x

x

x

u c u

u k u

u k u

u k u

α α αξ

α α αξ

α α αξ

α α αξ

=

=

=

=

the fractional Boussinesq equation (15) which is 2 2 2 23( )t x xu u u u 4x

α α α= + + α reduce to the following form

2 2 2 2 2 2 4 4( ) 3 ( ) .............................(31)c k u k u k uα α α α α α αξ ξ ξ− = +

We want to find the solution in the form as define in equation (22), comparing the highest order derivative term and the non-linear term we get 2n = . Therefore, . Comparing the like powers of

)()()( 2210 ξξξ Φ+Φ+= aaau

)(ξΦ we get the following

19 

 

20 

 

4 2

0 2 2 2 2 2 2 4 32 0 2 1 2

1 2 2 2 2 2 41 0 1 1 2 1

2 2 2 2 2 2 2 22 0 2 1 2 2

3 2 2 2 2 41 0 1 1 2 1

4

: 2( ) 3 (4 2 ) 16

: 2 ( ) 12 ( 3 ) 16

: 8 ( ) 24 (2 ) 36 136

: 2( ) 120 108 40

: 6

c k a k a a a a k

c k a k a a a a a k

c k a k a a a a k a k

c k a k a a a a k a k

α α α α

α α α α

α α α α α

α α α α α

σ σ σ

σ σ σ σ

σ σ σ σ

σ σ

Φ − = + +

Φ − = + +

Φ − = + + +

Φ − = + +

Φ 2 2 2 2 2 2 42 1 0 2 2 2

5 4 21 1 2

6 2 22 2

( ) 18 ( 2 ) 96 240

: 24 72 0

: 60 ( 2 ) 0

c k a k a a a k a a k

a k a a k

a k a k

α α α α α

α α

α α

σ σ− = + + +

Φ + =

Φ + =

Solving the above system of equation we get 2 4 2

28

0 6c k k

ka α α ασ− −= , 1 0a = and 2

2 2a k α= − . Hence the

solution is

( )

( )( )

( )

( )

2 4 22 2

2

2 4 22 2

2

22 4 22

22

2 4 22 2

2

2

8 2 tanh ( )6 for 0

8 2 coth ( )6

(1 )8( , ) 2 for 0, Cons tan t.6 ( )

8 2 tan ( )6

8

c k k k ct kxk

c k k k ct kxk

c k ku x y kk ct kx

c k k k ct kxk

c

α α αα

α

α α αα

α

α α αα

α

α α αα

α

α

σ σ σσ

σ σ σ

ασ σ ωω

σ σ σ

σ

⎫− −+ − + ⎪⎪ <⎬

− − ⎪+ − + ⎪⎭⎛ ⎞Γ +− − ⎜ ⎟= − = −⎜ ⎟+ +⎝ ⎠

− −− +

− ( )4 2

2 22

for 02 cot ( )

6k k k ct kxk

α αα

α

σσ σ

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪

⎫⎪⎪⎪ ⎪⎪ >⎬

−⎪ ⎪− +⎪ ⎪⎭⎩

Solution of Forth order Boussinesq equation for alpha=0.6 Solution of Forth order Boussinesq equation for alpha=0.8

Figure 6: Graphical presentation of u for fixed y and considering c = k= 1, for 1σ =− .

From figure 1-6 it is clear that solutions of the fractional differential equations depend on order of fractional derivative.

-4-2

02

4

-4-2

0

24

-100

-80

-60

-40

-20

0

20

x-axis

Solution of Forth order Boussinesq equation for alpha=0.4

t-axis

-4-2

02

4

-4-2

0

24

-100

-50

0

50

100

150

200

x-axist-axis

-4-2

02

4

-4-2

0

24

-50

0

50

100

150

200

250

t-axis x-axis

21 

 

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. Reference

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[3] Samko SG, Kilbas AA, Marichev OI. Fractional Integrals and Derivatives. Gordon and Breach Science, Yverdon, Switzerland; 1993.

[4] Podlubny I. Fractional Differential Equations, Mathematics in Science and Engineering, Academic Press, San Diego, Calif, USA. 1999;198.

[5] Oldham KB, Spanier J. The Fractional Calculus, Academic Press, New York, NY, USA; 1974.

[6] El-Sayed AMA, Gaber M. The adomian decomposition method for solving partial differential equations of fractal order in finite domains. Phys. Lett. A. 2006;359(20):175-182. [7] A El-Sayed AMA, Behiry SH, Raslan WE. Adomian’s decomposition method for solving an intermediate fractional advection–dispersion equation. Comput. Math. Appl. 2010;59(5):1759-1765. [8] J.H. He, Commun. Nonlinear Sci. Numer. Simul. 2, 230-235 (1997).

[9] Wu GC, Lee EWM. Fractional variational iteration method and its application. Phys. Lett. A. 2010;374(25):2506-2509. [10] S. Guo and L. Mei, Phys. Lett. A 375, 309-313 (2011) 309.

[11] He J-H, Homotopy perturbation technique, Comput Methods. Appl. Mech. Eng.1999;178(3-4):257-262. [12] J.H. He, Inter. J. Non-Linear Mech. 35, 37-43 (2000).

[13] H. Jafari, S. Momani, Solving fractional diffusion and wave equations by modified homotopy perturbation method, Phys. Lett. 2007 A 370. 388–396.

[14] Z. Odibat and S. Momani, Appl. Math. Lett. 21, 194-199(2008).

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22 

 

[16] A. M. Wazwaz, 2004. The tanh method for travelling wave solutions of nonlinear equations. Applied Mathematics and Computation..154. 713-723. [17] A. M. Wazwaz, 2008. The tanh method for travelling wave solutions to the Zhiber-Shabat equation and other related equations. Communications in Nonlinear Science and Numerical Simulation. 13. 584-592. [18] S. Zhang and H. Q. Zhang, Fractional sub-equation method and its applications to nonlinear fractional PDEs, Phys. Lett. A, 375. 2011. 1069.

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[22]Jumarie G. Modified Riemann-Liouville derivative and fractional Taylor series of non differentiable functions further results. Comput. Math. Appl. 2006;51(9-10):1367-1376. [23] Jumarie G. Fractional partial differential equations and modified Riemann-Liouville derivative new methods for solution. J. Appl. Math. Comput. 2007;24(1-2):31-48. [24] S. Guo, L. Mei, Y. Li and Y. Sun, The improved fractional sub-equation method and its applications to the space-time fractional differential equations in fluid mechanics, Phys. Lett. A, 376. 2012. 407. [25] Das. S. Functional Fractional Calculus 2nd Edition, Springer-Verlag 2011. [26] S. Das, Fractional order boundary controller enhancing stability of partial differential wave equation systems with delayed feedback", International Journal of Mathematics & Computation IJMC . 19.( 2). 2013. 42-59. [27] S. Das, Mechanism of Wave Dissipation via Memory Integral vis-a-vis Fractional Derivative", International Journal of Mathematics & Computation . 19(2). 2013. 72-83. [28] D.G. Natsis. Solitary wave solutions of the one-dimensional Boussinesq equations. Springer. 2007. DOI 10.1007/s11075-007-9102-6. [29] S. Zhang, Q.A. Zong, D. Liu, Q. Gao, A generalized exp-function method for fractional Riccati differential equations , Commun. Fract. Calc. 1 (2010) 48-51