Computer Physics Communications Volume 180 issue 9 2009 [doi 10.1016_j.cpc.2009.04.009] M.M. Rashidi; E. Erfani -- New analytical method for solving Burgers' and nonlinear heat transfer

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Computer Physics Communications 180 (2009) 1539–1544

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Computer Physics Communications

www.elsevier.com/locate/cpc

New analytical method for solving Burgers’ and nonlinear heat transfer equations

and comparison with HAM

M.M. Rashidi ∗, E. Erfani

Engineering Faculty of Bu-Ali Sina University, PO Box 65175-4161 Hamedan, Iran

a r t i c l e i n f o a b s t r a c t

Article history:

Received 22 January 2009Accepted 15 April 2009

Available online 17 April 2009

Keywords:

Differential transform method (DTM)

Burgers’ equation

Nonlinear differential equation

Homotopy analysis method (HAM)

Fin

In this study, we present a numerical comparison between the differential transform method (DTM) and

the homotopy analysis method (HAM) for solving Burgers’ and nonlinear heat transfer problems. The

first differential equation is the Burgers’ equation serves as a useful model for many interesting problems

in applied mathematics. The second one is the modeling equation of a straight fin with a temperature

dependent thermal conductivity. In order to show the effectiveness of the DTM, the results obtained

from the DTM is compared with available solutions obtained using the HAM [M.M. Rashidi, G. Domairry,

S. Dinarvand, Commun. Nonlinear Sci. Numer. Simul. 14 (2009) 708–717; G. Domairry, M. Fazeli, Commun.

Nonlinear Sci. Numer. Simul. 14 (2009) 489–499] and whit exact solutions. The method can easily be

applied to many linear and nonlinear problems. It illustrates the validity and the great potential of the

differential transform method in solving nonlinear partial differential equations. The obtained results

reveal that the technique introduced here is very effective and convenient for solving nonlinear partial

differential equations and nonlinear ordinary differential equations that we are found to be in good

agreement with the exact solutions.

© 2009 Elsevier B.V. All rights reserved.

1. Introduction

Most phenomena in real world are described through nonlinear

equations. Nonlinear phenomena play important roles in applied

mathematics, physics and in engineering problems in which each

parameter varies depending on different factors. The importance

of obtaining the exact or approximate solutions of nonlinear par-

tial differential equations (NLPDEs) in physics and mathematics, it

is still a hot spot to seek new methods to obtain new exact or

approximate solutions. Large class of nonlinear equations does not

have a precise analytic solution, so numerical methods have largely

been used to handle these equations. There are also some ana-

lytic techniques for nonlinear equations. Some of the classic ana-

lytic methods are the Lyapunov’s artificial small parameter method,perturbation techniques and δ-expansion method. In the recent

years, many authors mainly had paid attention to study solutions

of nonlinear partial differential equations by using various meth-

ods. Among these are the Adomian decomposition method (ADM),

tanh method, homotopy perturbation method (HPM), sinh–cosh

method, HAM, DTM and variational iteration method (VIM).

In 1992, Liao [1] employed the basic ideas of homotopy in

topology to propose a general analytic method for nonlinear prob-

lems, namely the HAM. Based on homotopy of topology, the va-

* Corresponding author. Tel.: +98 811 8257409; fax: +98 811 8257400.

E-mail address: [email protected] (M.M. Rashidi).

lidity of the HAM is independent of whether or not there exist

small parameters in the considered equation. Therefore, the HAM

can overcome the foregoing restrictions of perturbation methods.

In recent years, the HAM has been successfully employed to solve

many types of nonlinear problems, see [2–6] and the references

therein.

The concept of differential transform method method was first

introduced by Zhou [7] in 1986 and it was used to solve both

linear and nonlinear initial value problems in electric circuit anal-

ysis. The main advantage of this method is that it can be applied

directly to NLPDEs without requiring linearization, discretization,

or perturbation. It is a semi analytical–numerical technique that

formulizes Taylor series in a very different manner. This method

constructs, for differential equations, an analytical solution in theform of a polynomial. Not like the traditional high order Taylor se-

ries method that requires symbolic computation, the DTM is an

iterative procedure for obtaining Taylor series solutions. Another

important advantage is that this method reducing the size of com-

putational work while the Taylor series method is computationally

taken long time for large orders. This method is well addressed in

[23–30].

The Burgers’ equation is a nonlinear partial differential equation

of second order. Burgers’ equation was first introduced by Bateman

[8] and then treated by Burgers’ [9,10] as a mathematical model

for turbulence. This equation has a large variety of applications

in modeling of water in unsaturated soil, dynamic of soil water,

statistics of flow problems, mixing and turbulent diffusion, cosmol-

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1540 M.M. Rashidi, E. Erfani / Computer Physics Communications 180 (2009) 1539–1544

ogy and seismology [11–13]. The Burgers’ equation is a nonlinear

equation, very similar to the Navier–Stokes equation and there is

analogy between the Burgers’ equation and Navier–Stokes equation

due to the form of nonlinear terms. This single equation has a con-

vection term, a diffusion term, and a time-dependent term.

In Burgers’ equation, discontinuities may appear in finite time,

even if the initial condition is smooth. They give rise to the phe-

nomenon of shock waves with important applications in physics[14]. These properties make Burgers’ equation a proper model for

testing numerical algorithms in flows where severe gradients or

shocks are anticipated. Several numerical methods to solve this

system have been given such as algorithms based on cubic spline

function technique [15], the explicit–implicit method [16], Adomi-

an’s decomposition method [17]. The variational iteration method

was used to solve the 1D Burgers’ and coupled Burgers’ equations

[18]. Rashidi and et al. [19] used the HAM to solve Burgers’ equa-

tion.

Fins or extended surfaces are frequently used to enhance the

rate of heat transfer from the primary surface. The rectangular fin

is widely used, probably, due to simplicity of its design and it is

less difficult in manufacturing process. However, it is well-known

fact that the rate of heat transmission from a fin base diminishesalong its length. Kern and Kraus [20] made an extensive review on

this issue. Aziz and Hug [21] used the regular perturbation method

and a numerical solution method to compute a closed form so-

lution for a straight convective fin with temperature-dependent

thermal conductivity. The HAM was used by Domairry and Fazeli

to solve Rectangular purely convective fin with temperature depen-

dent thermal conductivity [22].

In this paper, we extend the application of the differential

transform method to construct analytical approximate solutions

of the Burgers’ equation (5) and the modeling equation of a

straight fin with a temperature-dependent thermal conductivity

equation (15). Then we compare our results with the previously

obtained results by using the HAM in [19,22] and exact solutions.

With this technique, it is possible to obtain highly accurate resultsor exact solutions for differential equations.

2. Basic idea of the differential transform method

The basic definitions and fundamental operations of the two-

dimensional differential transform are defined in [23–30]. Consider

a function of two variable w( x, y) be analytic in the domain Ω and

let ( x, y) = ( x0, y0) in this domain. The function w( x, y) is then

represented by one series whose centre at located at w( x0, y0).

The differential transform of the function is the form

W (k, h) =1

k!h!

∂k+h w( x, y)

∂k x∂h y

( x0, y0)

, (1)

where w( x, y) is the original function and W (k, h) is the trans-

formed function. The transformation is called T-function and the

lower case and upper case letters represent the original and trans-

formed functions respectively. Then its inverse transform is defined

as

w( x, y) =

∞

k=0

∞

h=0

W (k, h)( x− x0)k( y − y0)h. (2)

The relations (1) and (2) imply that

w( x, y) =

∞k=0

∞h=0

1

k!h!

∂k+h w( x, y)

∂k x∂h y

( x0, y0)

( x− x0)k( y − y0)h. (3)

In a real application and when ( x0, y0) are taken as (0, 0), then

the function w( x, y) is expressed by a finite series and Eq. (2) can

be written as

w( x, y) ∼=

mk=0

nh=0

W (k, h) xk yh, (4)

in addition, Eq. (4) implies that ∞

k=m+1∞

h=n+1 W (k, h) xk yh is

negligibly small. Usually, the values of m and n are decided byconvergences of the series coefficients.

3. Application

3.1. The Burgers’ equation

Consider the Burgers’ equation [31]

ut + uu x − u xx = 0, x ∈ R, (5)

with the exact solution [32]

u( x, t ) =1

2 −

1

2 tanh

1

4

x−

1

2t

, (6)

and with the initial condition

u( x, 0) =1

2 −

1

2 tanh

x

4

. (7)

Taking the two-dimensional transform of Eq. (5) by using the re-

lated definitions in Table 1, we have

(h + 1)U (k, h + 1)+

kr =0

hs=0

(k − r + 1)U (r , h − s)U (k − r + 1, s)

− (k + 1)(k + 2)U (k + 2, h) = 0, (8)

by applying the differential transform into Eq. (7), the initial trans-

formation coefficients are thus determined by

Table 1

The operations for the two-dimensional differential transform method.

Original function Transformed function

w( x, y) = u( x, y)± v( x, y) W (k, h) = U (k, h)± V (k, h)

w( x, y) = λu( x, y) W (k, h) = λU (k, h), λ is a constant

w( x, y) = ∂u( x, y)

∂ x W (k, h) = (k+ 1)U (k+ 1, h)

w( x, y) = ∂u( x, y)

∂ y W (k, h) = (h + 1)U (k, h + 1)

w( x, y) = ∂ r +s u( x, y)

∂ xr ∂ ys W (k, h) = (k+ 1)(k + 2) . . . (k+ r )(h + 1)(h + 2) . . . (h + s)U (k+ r , h+ s)

w( x, y) = u( x, y)v( x, y) W (k, h) =k

r =0

hs=0 U (r , h − s)V (k − r , s)

w( x, y) = ∂u( x, y)

∂ x∂ v( x, y)

∂ x W (k, h) =k

r =0

hs=0(r + 1)(k − r + 1)U (r + 1, h − s)V (k− r + 1, s)

w( x, y) = ∂u( x, y)

∂ x∂ v( x, y)

∂ y W (k, h) =

kr =0

hs=0(k− r + 1)(h − s + 1)U (k − r , s)V (r , h− s + 1)

w( x, y) = u( x, y)∂ v( x, y)

∂ x W (k, h) =kr =0 h

s=0(k− r + 1)U (r , h − s)V (k − r + 1, s)

w( x, y) = u( x, y)∂ 2 v( x, y)

∂ x W (k, h) =k

r =0

hs=0(k− r + 2)(k − r + 1)U (r , h− s)V (k− r + 2, s)



M.M. Rashidi, E. Erfani / Computer Physics Communications 180 (2009) 1539–1544 1541

∞k=0

U (k, 0) xk=

1

2 −

x

8 +

x3

384 −

x5

15360 +

17 x7

10321920

−31 x9

743178240 +

691 x11

653996851200 + · · · . (9)

Hence from Eq. (9)

U (k, 0) = 0, if k = 2, 4, 6, . . . ,

U (1, 0) =−1

8, U (3, 0) =

1

384,

U (5, 0) =−1

15360, . . . . (10)

Substituting Eq. (10) in Eq. (8), and by recursive method we

can calculating another values of U (k, h), some results are listed as

follows in Table 2. Hence, substituting all U (k, h) into Eq. (4) we

have series solution as below

Table 2

Some values of U (k, h) of Example 3.1.

U (1, 1) = 0 U (2, 3) = 13!×211 U (5, 6) = 691

3!3×224×52 U (1, 8) =− 31

3!2×224×5×7

U (2, 2) = 0 U (3, 2) =− 1

3!×210 U (6, 5) =− 691

3!3×223×52 U (8, 1) = 31

3!2×217×5×7

U (3, 3) = 0 U (3, 4) = 173!×217 U (2, 5) =−

173!×219×5

U (3, 6) =− 31

3!3×219×5

u( x, t ) ∼=

mk=0

nh=0

U (k, h) xkt h

= U (0, 0)+ U (1, 0) x+ U (0, 1)t

+ U (1, 1) xt +· · · + U (m, n) xmt n

=1

2

+t

16

−t 3

3072

+t 5

491520

− x

8

+t 2 x

512

−t 4 x

49152

−tx2

256

+t 3 x2

12888 −

17t 5 x2

15728640 +

x3

384 −

t 2 x3

6144 +

17t 4 x3

4718592 +

tx4

6144

−17t 3 x4

2359296 +

31t 5 x4

188743680 −

x5

15360

+17t 2 x5

1966080 −

31t 4 x5

94371840 + · · · . (11)

Our approximation has one more interesting property, if we ex-

pand exact solution (6) using Taylor’s expansion about (0, 0), we

have the series same as the our approximation (11).

In Fig. 1, we study the diagrams of the results obtained by the

DTM for m = 50, n = 50 in comparison with the HAM [19] and

exact solution (6).

Note that the solution series obtained by the HAM contains theauxiliary parameter h, which provides us with a simple way to ad-

just and control the convergence of the solution series. As pointed

Fig. 1. The results obtained by the DTM (m = 50, n = 50) Eq. (11) and the HAM [19] by 9th-order approximate solution, in comparison with the exact solution (6), when

5 t 12, (a) x = 0.25; (b) x = 0.75; (c) x = 1; (d) x = 1.5.




Table 3

Comparison of the exact solution (6) whit the DTM (m = 50, n = 50) and the HAM (h =−0.6) by 9th-order approximate solution in case of x = 0.5, x = 1.5 and 0 t 12.

t x = 0.5 x = 1.5

Exact HAM (h =−0.6) DTM Exact HAM (h =−0.6) DTM

1 0.500000000 0.499967668 0.500000000 0.377540668 0.377496222 0.377540668

2 0.562176500 0.562186654 0.562176500 0.437823499 0.437785782 0.437823499

3 0.622459331 0.622618884 0.622459331 0.500000000 0.500133672 0.499999999

4 0.679178699 0.679494353 0.679178699 0.562176500 0.562684851 0.5621765005 0.731058578 0.731289069 0.731058578 0.622459331 0.623385326 0.622459331

6 0.777299861 0.776880106 0.777299861 0.679178699 0.680114881 0.679178699

7 0.817574476 0.815694315 0.817574476 0.731058578 0.730852749 0.731058578

8 0.851952801 0.847843276 0.851952801 0.777299861 0.773875815 0.777299861

9 0.880797077 0.874236662 0.880797060 0.817574476 0.807988448 0.817574479

10 0.904650535 0.896665720 0.904646851 0.851952801 0.832782589 0.851953323

11 0.924141819 0.917848146 0.923677794 0.880797077 0.848926292 0.880854270

12 0.939913349 0.941425185 0.901792839 0.904650535 0.858478394 0.908702365

Fig. 2. The comparison of the errors in answers results by the DTM ( m = 50, n = 50)

and the HAM (h =−0.6) by 9th-order approximate solution at t = 10.

in [19], the valid region of h in this case is −1.6 < h < −0.4. Whenh =−1, the solution by the HAM is as the same solution series ob-

tained by HPM, which proposed in 1998 by Dr. He [33]. Therefore,

the HAM logically contains the HPM. We can find that the best

value of h in this case is −0.6.

In Table 3, we present a numerical comparison between the

HAM (h =−0.6) by 9th-order approximate solution, the DTM (m =

50, n = 50), and exact solution (6) in this case of x = 0.5, x = 1.5

and 0 t 12.

Table 3 indicates that the results obtained by the DTM for the

case of 0 t 8 have nine digits precision whit the exact solu-

tions.

In Figs. 2 and 3, we present the comparison of the errors

in answers results by the DTM (m = 50, n = 50) and the HAM

(h =−0.6) by 9th-order approximate solution at t = 10 and t = 11respectively for the case of −1 x 1. Considering these two fig-

ures, we find out that errors of the DTM are very less than those

of the HAM even for large t .

3.2. Rectangular purely convective fin with temperature dependent

thermal conductivity

Consider a straight fin with a temperature-dependent thermal

conductivity, arbitrary constant cross-sectional area Ac , perimeter

P , and length b (see Fig. 4). The fin is attached to a base surface of

temperature T b , extends into a fluid of temperature T a and its tip

is insulated. The one-dimensional energy balance equation is given

Ac d

dx

k(T )

dT

dx

− P h(T b − T a) = 0. (12)

Fig. 3. The comparison of the errors in answers results by the DTM ( m = 50, n = 50)

and the HAM (h =−0.6) by 9th-order approximate solution at t = 11.

Fig. 4. Geometry of a straight fin.

The thermal conductivity of the fin material is assumed a linear

function of temperature according to

k(T ) = ka

1+ λ(T − T a)

, (13)

where ka is the thermal conductivity at the ambient fluid temper-

ature of the fin and k is the parameter describing the variation of

the thermal conductivity.

Employing the following dimensionless parameters

θ =T − T a

T b − T a, ζ =

x

b, β = λ(T b − T a), ψ =

h P b2

ka Ac

1/2

.

(14)

The formulation of the problem reduces to

d2θ

dζ 2 + βθ

d2θ

dζ 2 + β

dθ

dζ

2−ψ 2θ = 0, (15a)



M.M. Rashidi, E. Erfani / Computer Physics Communications 180 (2009) 1539–1544 1543

Table 4

Comparison of the dimensionless temperature errors in answers results by the DTM (k = 15) and the HAM by 12th-order approximate solution for the case of constant

thermal conductivity, i.e. β = 0.

ζ ψ = 0.5 ψ = 1

DTM HAM (h =−1) HAM (h =−0.92) DTM HAM (h =−1) HAM (h =−0.791)

0 0.000000000 1.3600 × 10−12 4.8849 × 10−15 2.0095 × 10−14 1.7793×10−5 4.5732× 10−11

0.1 0.000000000 1.3433× 10−12 4.6629 × 10−15 2.0317× 10−14 1.7574× 10−5 4.6543× 10−11

0.2 0.000000000 1.2936× 10−

12 4.4408 × 10−

15 2.0539× 10−

14 1.6922×10−

5 4.7738× 10−

11

0.3 0.000000000 1.2121× 10−12 3.1086× 10−15 2.0983 × 10−14 1.5854×10−5 4.5748 × 10−11

0.4 1.1102×10−16 1.1013 × 10−12 2.9976× 10−15 2.1649 × 10−14 1.4395×10−5 3.5230× 10−11

0.5 0.000000000 9.6378 × 10−13 2.3314× 10−15 2.2648 × 10−14 1.2582×10−5 1.0373× 10−11

0.6 0.000000000 8.0147 × 10−13 5.2180× 10−15 2.3869× 10−14 1.0458×10−5 3.2824× 10−11

0.7 1.1102×10−16 6.2105× 10−13 6.7723 × 10−15 2.5091 × 10−14 8.0782×10−6 9.1056× 10−11

0.8 1.1102×10−16 4.2377× 10−13 1.2212 × 10−14 2.5979× 10−14 5.4985×10−6 1.4602× 10−10

0.9 1.1102×10−16 2.1505× 10−13 3.9968 × 10−15 2.3092× 10−14 2.7835×10−6 1.4952× 10−10

1. 0.000000000 6.9944 × 10−15 7.7715× 10−15 0.0000000000 0.000000000 5.3290× 10−15

dθ

dζ = 0 at ζ = 0, (15b)

θ = 1 at ζ = 1. (15c)

Taking the one-dimensional transform of Eq. (15a) by using the

related definitions, we have

(k + 1)(k + 2)Θ(k + 2)

+ β

kr =0

(k − r + 2)(k − r + 1)Θ(r )Θ(k − r + 2)

+ β

kr =0

(r + 1)(k − r + 1)Θ(r + 1)Θ(k − r + 1)

−ψ2Θ(k) = 0, (16)

where dimensionless temperature is given by

θ(ζ) =

∞

k=0

Θ (k)ζ k. (17)

Additionally, applying the DTM to Eq. (15b) the boundary condition

is given as follow

Θ(1) = 0, (18)

by assuming

Θ(0) = α, (19)

moreover, substituting Eqs. (18) and (19) into Eq. (16) and by re-

cursive method we can calculating another values of Θ(k), some

results are listed as follows

Θ(k) = 0, if k = 3, 5, 7, . . . ,

Θ(2) =αψ2

2!(1+αβ),

Θ(4) =α(1− 2αβ)ψ 4

4!(1+αβ)3 ,

Θ(6) =α(2αβ − 1)(14αβ − 1)ψ6

6!(1+ αβ)5 ,

Θ(8) =α(2αβ − 1)(1 + 4αβ(112αβ − 19))ψ8

8!(1+ αβ)7 ,

.

.

.. (20)

By applying the boundary condition (15c), we can obtain α .

Hence, substituting all Θ(k) into (17) we have series solutionas below

Fig. 5. The comparison of the dimensionless temperature errors in answers results

by the DTM (k = 15) and the HAM by 12th-order approximate solution (ψ = 1, β =

0)

.

θ(ζ) ∼=

mk=0

Θ(k)ζ k = α +αψ2

2!(1+ αβ)ζ 2 +

α(1− 2αβ)ψ 4

4!(1+αβ)3 ζ 4

+α(2αβ − 1)(14αβ − 1)ψ6

6!(1+ αβ)5 ζ 6

+α(2αβ − 1)(1 + 4αβ(112αβ − 19))ψ8

8!(1+ αβ)7 ζ 8

+· · · . (21)

When the thermal conductivity is constant, i.e. β = 0 Eq. (15)

becomes a linear equation for which analytical solution is avail-

able. The analytical solution for dimensionless temperature distri-

bution θ(ζ) is

θ(ζ)analytical =eψζ + e−ψζ

eψ + e−ψ . (22)

In Table 4, numerical comparison of the dimensionless temper-

ature errors in answers results by the DTM and the HAM [22] were

compared for ψ = 0.5 and ψ = 1. Note that the solution series

obtained by the HAM contains the auxiliary parameter h, which

influence its convergence region and rate. We should therefore fo-

cus on the choice of h by plotting of errors in answers results by

the HAM for some values of h.

As pointed in [22], the valid region of h for the case of ψ = 0.5

and constant thermal conductivity is −1.3 < h < 0 and for the case

of ψ = 1 and constant thermal conductivity is −1.6 < h < −0.1.

In Fig. 5, we present the comparison of the errors in answersresults by the DTM (m = 15) and the HAM (h =−0.78, h =−0.785




Fig. 6. Comparison for dimensionless temperature variation for ψ = 1.

and h = −0.791) by 12th-order approximate solution for the case

of the constant thermal conductivity.From Table 4 and Fig. 5, it can be concluded that approximate

the DTM expression presented as a solution in this study gives

better results than the HAM approximation. We observe that the

best value of h is −0.791 for the case of (ψ = 1, β = 0) Fig. 5 and

−0.92 for the case of (ψ = 0.5, β = 0).

When conductivity varies whit temperature, Eq. (15) becomes

a nonlinear equation for which analytical solution is not avail-

able. Hence, in this paper a second analysis is also conducted via

the classical fourth-order Runge–Kutta for the purpose of testing

this method. In Fig. 6, dimensionless temperature distribution is

compared for ψ = 1 with various β values. From Fig. 6 it is seen

that, when the problem becomes nonlinear the obtained results by

those the DTM and the HAM agreement with numerical results.

4. Conclusions

In this paper, we presented a reliable algorithm based on

the DTM to solve some nonlinear equations. Some examples are

given to illustrate the validity and accuracy of this procedure. The

present method reduces the computational difficulties of the other

methods (same as the HAM, VIM, ADM and HPM) and all the

calculations can be made simple manipulations. The accuracy of

the method is very good. The method has been applied directly

without requiring linearization, discretization, or perturbation. The

obtained results demonstrate the reliability of the algorithm and

gives it a wider applicability to nonlinear differential equations.

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http://dx.doi.org/10.1016/j.chaos.2008.02.009

http://dx.doi.org/10.1016/j.chaos.2008.02.009

Documents

Computer Physics Communications Volume 180 issue 9 2009 [doi 10.1016_j.cpc.2009.04.009] M.M. Rashidi; E. Erfani -- New analytical method for solving Burgers' and nonlinear heat transfer