An Efficient Approach for Fractional Harry Dym Equation by Using Homotopy Analysis Method

An Efficient Approach for Fractional Harry Dym Equation by Using Homotopy Analysis Method

Bhuvaneshwari Shunmugarajan

VIT University,Cennai.

Tamil Nadu,India.

Abstract – In this article, a structured approach based on Homotopy analysis method is implemented to resolve nonlinear fractional Harry Dym equation. The fractional derivative is described in Caputo sense. The ultimate goal of this paper is to find error analysis which shows our final

In case of =1, the equation(fractional Harry Dym equation) simplify into classical nonlinear Harry Dym equation. The exact solution of the Harry Dym equation is given by[15]

concluded result converge to the exact and approximate solution. The solution which was derived from convergent series form proves the effectiveness of the method in resolving different type of fractional differential equation. Henceforth

(, ) = ( 3 ( + ))23

2

Where a and b are suitable constant.

——1.3

this technique is efficient and less time consuming when compare to other methods of finding approximate and exact solutions for nonlinear partial differential equations.

Keywords: Harry Dym Equation, Caputo derivative, Homotopy analysis method, approximate solution

1. INTRODUCTION

   Fractional calculus is a separate study of investigating the characteristics of integral and derivatives of non integers order. This fractional calculus is specifically popular for solving differential equation involving fractional derivatives of unknown function which is also called fractional differential equation. The number of problems in science and engineering involving in fractional derivatives is growing continuously and perhaps the fractional calculus of the twenty five century[1-6].The idea appeared in a letter of Leibniz to LHospital in 1965.The analytical results related to fractional differential equation have been studied by many authors[7,8].During the last decades, multiple number of numerical and analytical methods have been suggested to resolve fractional differential equations. The most accepted methods are fractional difference method[9,10],Adomin decomposition method[11],variational iteration method[12,13].Recently the Homotopy perturbation method and Lagrange multiplier method were used extensively to solve multiorder fractional differential equation[14] .

   In this paper, we refer the below nonlinear time fractional Harry Dym equation of the form

   (, ) = 3(, )3(, ), 0 < 1 —–1.1

   This Harry Dym Dynamical equation is used to identify applications in several physical systems. It is also appeared(Harry Dym equation) in Kruskal and Moser[16] and is attributed in an unpublished paper by Harry Dym in 1973-1974.Harry Dym is a completely integrable nonlinear evolution equation. Meanwhile it obeys an infinite number of conversion laws; but it doesnt possess the Painleve property. The primary development of fractional calculus was achieved with the help of book by Oldhamnand Spanier[17].The essential results related with the solution of fractional differential equation may be found in book[18-21].The Homotopy analysis method was developed by liao[22-31].It made an analytical study on fractional Kdv, K(2,2),Burgers, BBM-Burgers, Cubic Boussineq, Boussinesq-like B(m,n) equation and coubled Kdv.

   In the present paper, we implemented the Homotopy analysis method to determine the approximate solution of nonlinear fractional differential equation. Our aim of this paper is to extend the application of Homotopy analysis method to obtain analytic and approximate solution to the nonlinear time fractional Harry Dym equation. HAM provides the solutions in terms of convergent series with easily computable components in a direct way which is not achievable using linearization, perturbation. The error analysis will be made by comparing the approximate and the exact solution. The results are represented in the form of graph.

2. BASIC DEFINITIONS OF FRACTIONAL CALCULUS

In this part, we give some definitions and properties of the

With the initial condition

(, 0) = ( 3 )23

——-1.2

fractional calculus[32]

Definition 2.1: A real function f(t), t > 0, is said to be in the space , ,if there exists a real number > ,

such that () = () where () (0, ) and it is

2 1

1

()

Where denotes the order of the fractional derivative and

(, ) refers to a function of x and t.The fractional derivative is present in the Caputo sense.

said to be in the space , if and only if

,

Definition 2.2: The Riemann-Liouville fractional integral operator ()of order 0, of a function , 1 if defined as [32]

X(t) = Xo(t) + 1 ()

=

=

X (t)= 1 () |

(3.3)

(3.4)

n !

=0

() = () =

Differentiating the Zeroth-Order deformation equation n-

1

0

0

()

1( )1(), ( > 0)

—-2.1

times with respect to q at q = 0 and then dividing it by n,we have the following nth-order deformation equation

0() = () ———-2.2

L[Xn(t)- nXn-1(t)] = Rn(1 (t)); (3.5)

Where () is the well known Gamma function.Some of

R (

(t)) = 1

1(()) |

(3.6)

1, > 1

1, > 1

the properties of the operator (), according to our requirement are given below.

n 1

(1)!

1 =0

For , 1 , , 0 1:

and n = {0, 1

(3.7)

(1) () = +()

(2) () = ()

Here 1 = [0(), 1(), 2(), 3() . . 1()]

(3) = (+1)

(++1)

+

If the series (3) converges at q = 1 we have X(t) = Xo(t) + 1 ()

(3.8)

Definition 2.3: The fractional derivative of () in

Caputo sense defined as

=

1. HAM SOLUTION OF HARRY DYM EQUATION:

   Consider the Harry Dym equation:

   () = 1

   ()()

   ,

   () 0 ()+1

   (, ) = 3(, )3 (, ), 0 < 1 —— 4.1

   1

   1

   where 1 < , , > 0,

   With the initial condition

   The two basic properties of the Caputos fractional derivative are given below

   (, 0) = ( 3 )23

   2

   ——-4.2

   Lemma 2.1: If 1 < , ,

   , 1 then

   Where denotes the order of the fractional derivative and

   (, ) refers to a function of x and t.

   ()() = (),

   1

   Solution:

   ()() = () (0+)

   Step 1: Let us consider the initial approximation from the

   =0

   !

   initial condition (2)

   1. BASIC HOMOTOPY ANALYSIS METHOD

      (, ) = ( 3 )23

      ——–4.3

      The Homotopy analysis method (HAM) has been proposed 0 2

      by Liao in [22]. The application of HAM is use to calculate the exact solution and the approximate solutions of linear and nonlinear Differential equations. Consider the non-

      Step 2: Let us consider the nonlinear operator

      ((, ; )) = (, ; ) 3(, ; )3(, ; )-

      linear differential equation

      N[X(t)]=0; (3.1)

      Where N is a nonlinear auxiliary operator and X(t) is an

      Step 3: Let us consider the Linear operator

      ——-44

      unknown function. The boundary and initial conditions are ignored, which can be treated in the similar way. The zero- order deformation equation is constructed as

      (1-q)L[Xn(t)-x0(t)]=q N[X(t),A], (3.2)

      Where q [0,1] the embedding parameter,

      0 a nonzero auxiliary parameter,

      L an auxiliary linear operator,

      X(t) an unknown function.

      Also we can choose whatever auxiliary unknown in HAM because of its properties. By Taylors theorem, X(t) can be expanded with respect to the embedding parameter q as

      ((, ; )) = (, ; ) ——–4.5

      Step 4: Using the above definition, we construct the zeroth order deformation equation

      (1 )[(, : ) 0(, )] = [(, ; )] –4.6 Obviously when q=0 and q=1, we can write from

      equation (4.6)

      (; 0) = 0() = (0) and

      (; 1) = ()

      Step 5: According to equation 3.4 and 3.7 we obtain mth order deformation equation

      [(, ) 1()] = [1() ]

      Step 8: Taking = 1, We get from equation 4.3,4.7,4.8 and 4.9 after some rigorous calculation,

      (, ) = ( 3 )23

      1 1[(;)]

      ——–4.7

      0

      (, ) =

      2

      32( 3 )13

      Where [1() ] = (1)! [

      = 1

      1 ]=0

      [ 1 {(, ; )

      1 (+1) 2

      3(, ; )3(, ; )}]

      (1)!

      1

      (, ) = 2

      3 ( 3 )43

      =0

      =[(1

      2 (2+1) 2 2

      3

      3

      (, ) =

      92(

      =1 =1 =1 ] —

      3 (3+1)

      3 )73[ 15 (2+1)

      16]

      ———–4.8

      Step 6: From equation 4.7 and 4.8 we can write

      2 2 2((+1)2)

      and so on.

      [(, ) 1()]

      = [( 1

      Step 9: By using equation 3.3, we obtain

      () = () + () , for q=1

      0 =1

      3 ]

      =1 =1

      =1

      1. The second order approximation of

         () is given

         (, ) = 0() + 1 =1 ()

         =

         () + 1[ ]

         1

         1

         =0() + 1()

         1[ 3]

         =1 =1

         =1

         =( 3 )23 32(

         2

         = 1() + [1]

         3 )13

         2

         (+1)

         [ 3]

      2. The third order approximation of ()

         is given

         =1 =1

         =1

         ——————-4.8a

         (, ) = 0() + 2 =1 ()

         Step 7: For m=1, equation 4.8a becomes

         3 3

         =0() + 1() + 2()

         =( 3 )23 32(

         1 = [0 0]

         3 )13

         2

         3 ( 3 )43

         2

         For m=2, equation 4.8a becomes

         2 (+1) 2 2

         (2+1)

         2 = 1 + [1] [331 +

      3. The third order approximation of ()

      32130]

      0

      is given

      0

      For m=3, equation 4.8a becomes

      3 = 2 + [2] [332 +

      (, ) = 0() + 3 =1 ()

      = () + () + () + ()

      0 0 1 2 3

      32131 + (302 + 322)30]

      0

      1 0

      =( 3 )23 32( 3 )13

      For 3 ,equation 4.8 becomes

      2

      3 ( 3 )43

      2

      2

      + 92(

      (+1)

      = + []

      2 2 (2+1)

      [

      3

      ]——-4.9

      3 )73[ 15 (2+1)

      16] 3

      =1

      =1

      =1

      2 2 2((+1)2)

      (3+1)

      TABLE 1: Comparison study between HPSTM,ADM,HAM and Exact solution, when =1 and for constant values of a=4 and b=1

      x	T	HPSTM[33]	ADM[33]	HAM	Exact solution
      0	1	1.843946953	1.843946953	1.843946953	1.843946953
      0.2	1	1.694117377	1.694117377	1.6918972844	1.691538112
      0.4	1	1.537581542	1.537581542	1.5365890571	1.534036644
      0.6	1	1.373028020	1.373028020	1.3737638298	1.367980757
      0.8	1	1.198654865	1.198654865	1.1949462024	1.191138425
      1	1	1.011880649	1.011880649	1.0113697992	1.000000000

      HAM solution U(x,t)

      HAM solution U(x,t)

      2.5

      2.5

      U(x,t)

      U(x,t)

      U(x,t)

      U(x,t)

      2 2

      1.5

      0

      0.2

      0.4

      0.6

      0.8

      t 1 0

      0.2

      0.4

      x

      0.6

      1

      0.8

      1.5

      0

      0.2

      0.4

      0.6

      0.8

      t 1 0

      0.2

      0.4

      x

      0.6

      1

      0.8

      Figure 1: HAM solution of U(x,t) when alpha=1 Figure 2: HAM solution of U(x,t) when alpha=2

      Absolute error

      Exact solution U(x,t)

      0

      E(x,t)

      E(x,t)

      -0.5

      -1

      -1.5

      1

      0.8

      0.6

      0.4

      x

      0.2

      0 0

      0.2

      0.4

      t

      0.6

      1

      0.8

      2.5

      U(x,t)

      U(x,t)

      2

      1.5

      1

      0

      0.2

      0.4

      0.6

      0.8

      t 1 0

      0.2

      0.4

      x

      0.6

      1

      0.8

      Figure 3: Absolute error E(x,t) Figure 4: Exact solution of U(x,t)

      2

      1.8

      1.6

      alpha=1 alpha=1/2 alpha=1/3

      2.5

      2

      alpha=1.2

      alpha=1.5 alpha=2

      U(x,t)

      U(x,t)

      U(x,t)

      U(x,t)

      1.4

      1.2

      1

      1.5

      0.8

      0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

      x

      1

      0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

      x

      Figure 5&6: Plot of U(x,t )versus x at t=1 for different value of alpha

      1. RESULTS & DISCUSSION

         In this article, detail study has been made on error analysis between exact and approximate solution which is specified by through figure 3 with high accuracy. The easiest and accuracy of the suggested method is depicted by calculating the absolute error 5(, ) = |(, )

         5(, )| at the constant value (a=4,b=1) where (, ) and

         5(, ) are the exact solution and approximate solution of equation (4.1) respectively. Figure 3 shows the error analysis between the exact and approximate solution which indicates significantly small convergence of the series solution very quickly. Hence during all numerical evolution we are going to take only third order of the series solution By introducing more terms of approximate solution, the

         accuracy of error analytical result will be improved. The characteristics of the approximate solution is specified in the figure 5 for 5(, ) for different value of = 1, ½ ,1/3 for standard Harry Dym equation. As per the Table 1 and

         graphs we inferred that HAM solution made a good agreement with the HPSTM and ADM solution. The figure shows that the accurate solution can only be improve introducing more terms of HAM solution. This is due to quick convergence of the HAM.

      2. CONCLUSION

In this paper HAM is efficiently applied to get approximate solution of time fractional Harry Dym equation. In HAM a Homotopy with embedding small parameter [0,1] which is helpful to implement full advantages of traditional perturbation method and Homotopy technique.HAM is different from other analtical method by providing the way to as to control the convergence region of solution series by with auxiliary parameter .This is the main useful features of HAM.HAM does not require any other methods or transformation techniques. The results of the compare solution (HAM third order solution, HPSTM, ADM) are specified in the table 1.From the table 1 it is observed that there is a good understanding between the HSPTM, ADM and exact solution. From the derived results it shows the reliability of the algorithm and it is greatly suitable for Non-linear fractional partial differential equation.

ACKNOWLDGEMENTS

The authors are grateful to the reviewers for interpretation this paper and for their valuable comments and suggestion for the improvement of the article.

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