Telescoping Decomposition Method for Solving Second Order Nonlinear Differential Equations

DOI : 10.17577/IJERTV3IS20067

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Telescoping Decomposition Method for Solving Second Order Nonlinear Differential Equations

1Manjak Nibron H, 2A. K. Mishra and 3Kabala D.

1Department of Mathematical Science, ATBU, Bauchi3, Nigeria, 2Department of Mathematics, Gombe State University1, Nigeria 3College of Education, Waka- Biu, Borno State, Nigeria.

Abstract – In this paper we present a reliable algorithm for solving first and second order nonlinear differential equations. The Telescoping Decomposition Method (TDM) is a new iterative method to obtain numerical and analytical solutions for first order nonlinear differential equations. We aim to extend the works of Mohammed Al-Refaiet al (2008) and make progress beyond the achievements made so far in this regard. We have modified the function f (t,u,ut ) to solve second order nonlinear differential equations. The method is a modified form of Adomian Decomposition Method (ADM) where the computation of the Adomian Polynomial is a difficult task in most cases. The TDM is easier to apply when compared to ADM and offers better convergent to the exact solution while it is not the case in ADM.

Keywords: Telescoping Decomposition Method, Adomian Decomposition Method, Adomian Polynomial and Nonlinear Operators.

  1. INTRODUCTION

    Differential equations appear in various application problems in the physical sciences and engineering. Most of the differential equations coming from real life application are nonlinear and the seek of analytical solutions post a real challenge to mathematics. The Adomian decomposition methodG. Adomian (1994), G.Adomian (1988) and G. Adomian (1984) has been applied to a wide class of stochastic and deterministic problems in physics, biology and chemical reactions. For nonlinear models, the method has shown reliable results in supplying analytical approximation that converges very rapidly. It is well known by many authors that the decomposition method decomposes the linear term u(x,t) into an infinite sum of components un(x,t) defined by

    Where An are the so called Adomian polynomials G. Adomian (1994), G. Adomian (1988) and G. Adomian (1984) formally introduced formulas that can generate Adomianpolynomials for all forms of nonlinearity.

    Recently, a great deal of work has been done by Abdul- Majid Wazwaz (1999,2000), G. Adomian (1986) and V. Seng et al (1996) among others to develop a practical method for the calculation of Adomian polynomials An. The concern was to develop and calculate Adomian polynomials in a practical way without any need for formulas introduced by Adomian, G. Adomian (1986) and

    V. Seng, et al (1996) require a huge size of calculations and employ several formulas identical in spirit to that used by Adomian. It is worth nothing that calculating Adomian polynomials is difficult for large n, and the

    formular given by Adomian,G.Adomian(1986), G. Adomian(1992) and manjak(2008 ) cannot be applied if f is a function of several variables, such as

    f (u,u,u )

    Also, the ADM is shown to be divergent for certain problems Hosseini and Nasabzadeh (2006). The objective of this paper is to introduce the Telescoping decomposition method (TDM) for solving first and second nonlinear initial value problem. We will use the idea of the ADM for the linear part and introduce a new way of computing the nonlinear part by avoiding the calculation of the Adomian polynomials using the formulas discussed in Mohammed et al (2008). In section 2, we present the expansion procedure of the TDM.

  2. TELESCOPING DECOMPOSITION METHOD EXPANSION PROCEDURE

    We consider the initial value problem of the form

    u(x, t) un (x, t) (1.1)

    n0

    And identifies the nonlinear term F(u) by the

    ut f (t,u,ut ),t … (2.1)

    decomposition series

    F (u) An (x, t) (1.2)

    u(0) u0 (2.2)

    Where [0,T ] is compact subset of R . By

    n0

    integrating the equation we have

    t

    u(t) u(0) f ( , u( ),u ( ))d

    0

    We consider a solution of the form

    u(x, t) u n (t).

    n0

    … (2.3)

  3. THE TELECOPING DECOMPOSITION METHOD APPLIED TO FIRST-ORDER LINEAR AND NONLINEAR DIFFERENTIAL EQUATIONS.

    Test problem 1

    Consider the first order linear differential equation

    u 2tu 0, with,u(0) 1chama(2007) (3.1)

    Where un(t) has to determine sequentially upon the following

    From (2.3) we have that

    t t

    u(t) u(0) (2u )d 1 (2u )d

    0

    u 0 u

    t

    , u1 (t)

    0 0

    f ( , u 0 ( ),u 0 ( ))d … (2.4)

    t

    0

    t 1 1 t

    u(t) 1 (2u )d

    0

    u 2 (t) f ( , u k ( ), u k ( ))d

    f ( , u 0 ( ),u 0 ( ))d .

    u 0 1

    0 k 0

    t 2

    k 0

    2

    0

    t 1 1

    Take

    t t

    1 0 2

    u3 (t)

    f ( , u k ( ), u k ( ))d

    f ( , u k ( ), u k ( ))d .

    u (t) (2u )d (2t)d t

    0 k 0

    k 0

    0 k 0

    k 0 0 0

    t t t 4

    t 3 3

    t 2 2

    u 2 (t) 2 (u 0 u1)d (2u 0 )d 2 (1 2 )d (t 2 )

    u 4 (t)

    f ( , u k ( ), u k ( ))d

    f ( , u k ( ), u k ( ))d. 0 0 2

    0 k 0

    k 0

    0 k 0

    k 0

    t t 4

    t 4 t 6

    u3 (t) 2 (u0 u1 u 2 )d 2 (u 0 u1 )d 2 (1 2

    )d (t 2 )

    . . . . . 0 0

    .

    t t

    4 6

    2 2 6

    t 4 t6 t8

    . . . . .

    u4 (t) 2 (u0 u1 u2 u3)d 2 (u0 u1 u2 )d 2 (1 2

    )d (t 2

    )

    . 0 0

    2 6 2 6

    24.

    t n1

    n1

    t n2

    n2

    . . . .

    u n (t) f ( , u k ( ), u k ( ))d

    f ( , u k ( ), u k ( ))d .

    0 k 0

    k 0

    0 k 0 k 0

    (2.5)

    . . .

    . . . . .

    Adding these equations (2.4) through ( 2.5), we obtain

    . . .

    n t n1

    n1

    u n (t) (1)

    n t 2n

    u k (t)

    f ( , u k ( ), u k ( ))d , n 1

    (2.6) n!

    k 0

    0 k 0

    k 0

    o

    Therefore,

    4 6 8

    We remind here that the choice of u

    in (2.4) is not unique,

    u(t) u n (t) 1 t 2 t t t ….. et 2

    we can chose it to be any function of t and this depends on the nature of the problem in question as will be seen in our

    n0

    2 3! 4!

    subsequent test problems. Also if the problem is a simple linear case, then the ADM and TDM will coincide and give the exact solution of the problem. However, the decomposition series (2.5) requires simple calculation and yields a series solution with excellent accelerated convergence of the linear differential equation.

    Which is the exact solution of the problem in it closed form and in this case the ADM and the TDM coincides as earlier discussed.

    Test Problem 2

    Consider the following first order nonlinear autonomous differential equation

    u 2u u 2 , with,u(0) 1

    Applying TDM, we have that

    u(t) u(0) (2u u 2 )dt 1 (2u u 2 )dt

    2 2 2 5t3 11t 4 29t5 68t6 311t7 3t8 497t9 19t10 167t11 t12 t13 2

    u [(1 t )(1 t t ) ]dt

    Take

    3 6 15 45 315 5 1620 150 3300 60 325

    u 0 1

    (t

    2 7t 3 7t 4 32t 5 68t 6 311t 7 3t8 497t 9 19t10 167t11 t12 t13

    2t

    )

    u1 (2u0 u0 2 )dt t

    u 2 [2(u0 u1 ) (u0 u1 )2 ]dt (2u0 u0 2 )dt

    3 3 15 45 315 5 1620 150 3300 60 325

    3 4 5 6 7 8 9 10

    3

    [2(1 t) (1 t)2 ]dt dt t 2

    u 2 t 2 7t

    3

    19t

    6

    53t

    15

    31t

    10

    601t

    315

    211t

    420

    2031t

    2835

    90727t

    56700

    ….

    2

    u3 [2(u0 u1 u2 ) (u0 u1 u2 )2 ]dt [2(u0 u1) (u0 u1)2 ]dt

    u(t)

    1 t

    2t 3 4t 4 8t 5 143t 6 58t 7 41t8 11603t 9 23221t10

    ….

    u3 [2(1 t

    t ) (1 t

    3

    3

    t )2 ]dt [2(1 t) (1 t)2 ]dt 3

    3 3 5 90

    63 420

    11340

    8100

    2t 5 t 7

    15 63

    T

    Exact solution

    Adomian Method

    Telescoping Method

    Adomian Absolute Error

    Telescoping

    Absolute Error

    0.0

    1.000000000

    1.000000000

    1.000000000

    0.000000000

    0.000000000

    0.1

    1.100000000

    1.101273788

    1.100817677

    0.001273788

    0.000817677

    0.2

    1.200000000

    1.209530299

    1.208091029

    0.009530299

    0.008091029

    0.3

    1.300000000

    1.359283717

    1.333942014

    0.059283717

    0.033942014

    0.4

    1.400000000

    1.590982619

    1.500438418

    0.190982619

    0.100438418

    0.5

    1.500000000

    2.019639740

    1.741629316

    0.519639740

    0.241629316

    0.6

    1.600000000

    2.882724292

    2.082235597

    1.282724292

    0.482235597

    0.7

    1.700000000

    4.683061501

    2.520426644

    2.983061501

    0.820426644

    0.8

    1.800000000

    8.479851566

    2.031286526

    6.679851566

    0.231286526

    0.9

    1.900000000

    16.495189280

    2.12062087

    14.59518928

    1.77937913

    u4 [2(u0 u1 u2 u3 ) (u0 u1 u2 u3 )2 ]dt [2(u0 u1 u2 ) (u0 u1 u2 )2 ]dt

    Is the analytic solution of the equation.

    To showcase the accuracy of the TDM, we have computed the numerical values for some values of t compares to ADM

    3

    and the results depicted in Table 1 below.

    3

    5

    u4 [2(1 t t t

    3 15

    • t 7

      63

      ) (1 t t

      3

      3

      2t5

      15

    • t 7

      63

      )2 ]dt [2(1 t

      t ) (1 t

      3

      3

      t )2 ]dt 3

      Table1: The Solution of u(t) for different values of t.

      22t 7 11t 9

      134t11

      4t13

      t15

      ………

      315 2835

      Therefore

      51975

      12285

      59535

      3

      u(t) 1 t t

      3

      2t 5

      15

      17t 7

      315

      11t 9

      2835

      134t11

      51975

      4t13

      12285

      t15

      59535

      ……… 1 tanh(t)

      This is the exact Taylor series of the analytical solution.

      Test Problem 3

      Consider the nonlinear initial value problem

      u (1 t 2 )u 2 t 4 2t 3 2t 1

      with u(0)=1, Hosseini and Nasdzadeh (2006). Applying theTDM, We have that

      u(t) u(0) (1 2t 2t 3 t 4 )dt (1 t 2 )u 2 dt u(t) 1 (1 2t 2t 3 t 4 )dt (1 t 2 )u 2 dt

      take

      u 0 1 2t t 2

      2t 3

      3

      • t 4

        2

      • t 5

        5

        u1 [(1 t 2 )(u0 )2 ]dt

        3 4 5

        [(1 t 2 )(1 2t t 2 2t t t )2 ]dt

        t 2t 2

        7t 3

        3

        7t 4

        3

        32t 5

        15

        3

        68t 6

        45

        2

        311t 7

        315

        5

        3t8

        5

        497t 9

        1620

        19t10

        150

        167t11

        3300

        t12

        60

        t13

        325

        u 2 [(1 t 2 )(u0 u1)2 ]dt [(1 t 2 )(u0 )dt

        t n1

        n1

        n1

        t n2

        n2

        n2

        Test Problem 4

        un (x,t)

        f ( , uk (x, ), uk (x, ),uk (x, )d f ( , uk (x, ),uk (x, ),uk (x, ))d.

        Consider the nonlinear inhomogeneous advection partial

        0 k 0

        x

        k 0

        xx

        k 0

        0 k 0

        x

        k 0

        xx

        k 0

        differential equation

        2

        ut uux x xt ,u(x,0) 0

        Abdul Majid (1999)

        and we are solving for utt then a two-fold integration is then considered.

        Test problem 5

        Applying TDM (2.3), we have that

        u(x,t) u(x,0) xdt xt 2 dt uux dt

        We consider the following nonlinear reaction- diffusion equation.

        With

        ut uxx u u ,u(x,t) u e (4.3)

        2 2 x

        x

        0

        u0 xt

        Applying TDM (4.1-4.2), we have

        xt 3

        xt 3

        xt 3

        ut f (u,ux ,uxx )

        u1

        xt 2 dt 0 3 3 3

        u(x,t) u(x,0)

        f (u,ux

        ,uxx

        )dt

        u(x,t) xt

        This gives the exact solution with only a step, unlike the

        Take

        0

        u u0 ex

        ADM which until the noise terms cancels each other in

        u1

        f (u0 ,u0 ,u0 )dt

        (u0 (u0 )2 (u0 )2 )dt

        (ex (ex )2 (ex )2 )dt te x .

        x xx xx x

        1 1 1

        u0and u2, u2 and u3 and so on before arriving at the exact

        u 2 f (u k (x, t), u k (x, t), u k (x, t))dt f (u 0 , u 0 , u 0 )dt

        k 0

        x xx

        o 0

        x xx

        solution.

  4. THE TDM FOR SOLVING SECOND ORDER LINEAR AND NONLINEAR DIFFERENTIAL

f [(ex te x ), (ex te x ),(ex te x )]dt te x

[(ex te x ) (ex te x )2 (ex te x )2 ]dt te x

(e x te x )dt te x t 2 e x

2

t 2 2 2 t 1 1 1

u3 (x,t)

f (uk (x, ), uk (x, ), uk (x, )d f (uk (x, ), uk (x, ), uk (x, ))d.

EQUATIONS.

0 k 0

x

k 0

xx

k 0

0 k 0

x

k 0

xx

k 0

In this section, we have modified equation (2.1) to read

[(ex te x t 2ex ) (ex te x t 2ex )2 (ex te x t 2ex )2 ]dt (te x t 2ex )

as

ut=f(t,x,u,ux,uxx) or utt=f(u,ux,uxx) 4.1

If we are solving for ut then write

t

[(e x

:

:

2

  • te x

  • t 2 e x

    2

    2

    )]dt (te x

    2

    )

  • t 2 e x

    2

    2

    t 3e x

    6

    x xx

    u1 f ( , u 0 , u 0 ,u 0 )dt

    0

    Thus,

    n

    x x t 2 e x t 3e x t 4 e x

    1 1 1

    u(x,t) u

    (x, t) e

    te

    ……

    u 2

    f ( , u k (x, t), u k (x,t), u k (x, t))dt

    f ( , u 0 , u 0 , u 0 )dt

    n0

    2 6 24

    k 0

    x xx

    o 0

    x xx

    n x t 2

    t 3 t 4 t 5

    t 2 2 2

    t 1 1 1

    ie,u(x, t) u

    n0

    (x,t) e

    (t ……) 2! 3! 4! 5!

    u3 (x,t)

    f ( ,uk (x, ),uk (x, ),uk (x, )d

    f (,uk (x, ),uk (x, ),uk (x, ))d.

    And in closed form, coincide with the exact solution given

    0 k 0

    t 3

    x

    k 0

    3

    xx

    k 0

    3

    0 k 0

    t 2

    x

    k 0

    2

    xx

    k 0

    2

    by u(x,t)=e

    x+t

    u4 (x,t)

    f ( ,uk (x, ),uk (x, ),uk (x, )d

    f ( ,uk (x, ),uk (x, ),uk (x, ))d.

    0 k 0

    x

    k 0

    xx

    k 0

    0 k 0

    x

    k 0

    xx

    k 0

    Problem 6

    . . . .

    Consider a nonhomogenous wave equation

    . . . 4.2 .

    uxx

    • uutt

      2 2(t 2 x2 ), Kaya, (1999)

      . . . .

      With initial conditions u(x,0)=x2, u(o,t)=t2, ux(0,t)=0.

      . . Applying TDM, we have that

      uxx f (x,t,u,utt )

      u(x,t) u(x,0) xux (x, o) f (x,t,u,utt )dxdx

      t 2 [2 2(t 2 x2 ) uutt ]dxdx

      x2

      polynomials for nonlinear terms. Finally, we conclude that,

      the idea of TDM can be developed to deal with higher order differential equations and various types of functional

      t 2 x2 x2t 2

      6

      Take

  • [uutt ]dxdx

    equations as well.

    REFERENCES

    u0 t 2 x2

    1. Abdul-Majid Wazwaz (1999). A Reliable modification of Adomian

      u1 x2t 2

      • x2

        6

        [2(t 2

  • x2

    )]dxdx

    decomposition method. Appl. Math. And Comput. Vol. 102, issue 1, pp77-80

    u1 x2t 2

    • x2

      6

  • x2t 2

    x2

    0

    6

    1. Abdul-Majid Wazwaz (2000). A new algorithm for calculating Adomian Polynomials for nonlinear equations, Elservier, Applied

      u(x, t) u n (x, t) t 2 x2 .

      n0

      Which is the exact solution of the problem, unlike the ADM, which can be easily observed that the self-canceling noise terms appear between various components as follows:

      Mathematics and Computation 111, 53-69.

    2. Ahmed H.M (2008). Adomian Decomposition Method: Convergence Analysis and Numerical Approximations. Master of Science Mathematics, McMaster University, Hamilton, Ontorio.

    3. Chama A. (2007). Numerical implementation of Adomian

      Decomposition Method for Voltera Integral equations of second kind

      u(x,t) t 2

  • x2

    • x2t 2

    • x4

      6

  • x2t 2

  • x4

    6

    • x4t 2

    3

    7×6

    90

    2x6t 2

    15

  • x8

    16

  • x4t 2

3

…….

with weakly singular Kernals. African Institute for Mathematical Sciences (AIMS)chama@aims.ac.za

Though the result will arrive to the exact solution u(x,t)=x2

+t2 by cancelling the third term in u0 and the first term in u1, the fourth term in u0 and the first term in u2 and so on, thus, consumed much time when compared to TDM.

CONCLUSION

We have presented and analyzed the Telescoping Decomposition Method (TDM) for solving linear and nonlinear initial value problem. Mohammmed Al-Refai et al (2008). We have applied the TDM tosolve some physical problems and obtained closed form as well as exact solutions for these problems. It has been shown in the forgoing discussions that once the initial value is identified, with proper use of TDM it is possible to obtain analytical solution to a class of linear and nonlinear first order differential equations. In our quest to find and discover if this method can be applicable for solving the second order differential equations, we have redefined the function f(x,u)given by Mohammed Al-Refai et al (2008) and with the few test examples used, we have successfully demonstrated the efficiency, accuracy and convergence of this method for second order linear and nonlinear equations. We have also confirmed that the use of TDM reduces the volume of computation by not requiring the Adomian

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    Kaya D. And Mustafa,Inc. (1999). The solution of the Wave equation by the Decomposition Method. Bull. Malysian Math. Soc. (2ndseries ) 22, 151-155.

  6. Mustafa Inc. (2004). On Numerical Solution of Partial Differential Equations by Decomposition Method, KragujevacJ.Math. 26, 153- 164.

  7. M.M Hosseini, H. Nasabzadeh (2006). On the convergence of Adomian decomposition method, Appl. Math. Comp. Issue 1, 536- 543.

  8. Manjak N.H N and Kwami A.M (2008). On Numerical solutions of partial differential equations by Decompositions Method. Science Forum: J. Pure and Appl. Sci. Vol 11 (1)

  9. Mohammed Al-Refai, Maysem Abu- Dalu and Ahmed Al-Rashdeh (2008). Telescoping Decomposition Method for solving first order Nonlinear Differencial Equations. Proceeding of the international

    Multi- Conference of engineers and Computer Scientists (IMECS), Hong kong. Vol 11, 19-21.

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