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# Homotopy Perturbation Method to Solve Heat Conduction Equation

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Submitted: 23 November 2011 Published: 24 October 2012

DOI: 10.5772/51509

From the Edited Volume

## Heat Transfer Phenomena and Applications

Edited by Salim N. Kazi

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## 1. Introduction

Fins are extensively used to enhance the heat transfer between a solid surface and its convective, radiative, or convective radiative surface. Finned surfaces are widely used, for instance, for cooling electric transformers, the cylinders of air-craft engines, and other heat transfer equipment. In many applications various heat transfer modes, such as convection, nucleate boiling, transition boiling, and film boiling, the heat transfer coefficient is no longer uniform. A fin with an insulated end has been studied by many investigators [Sen, S. Trinh(1986)]; and [Unal (1987)]. Most of them are immersed in the investigation of single boiling mode on an extended surface. Under these circumstances very recently, [Chang (2005)] applied standard Adomian decomposition method for all possible types of heat transfer modes to investigate a straight fin governed by a power-law-type temperature dependent heat transfer coefficient using 13 terms. [Liu (1995)] found that Adomian method could not always satisfy all its boundaries conditions leading to boundaries errors.

The governing equations for the temperature distribution along the surfaces are nonlinear. In consequence, exact analytic solutions of such nonlinear problems are not available in general and scientists use some approximation techniques to approximate the solutions of nonlinear equations as a series solution such as perturbation method; see [Van Dyke M. (1975)], and Nayfeh A.H. (1973)], and homotopy perturbation method; see [He J. H. (1999, (2000), and(2003)].

In this chapter, we applied HPM to solve the linear and nonlinear equations of heat transfer by conduction in one-dimensional in two slabs of different material and thickness L.

## 2. The perturbation method

Many physics and engineering problems can be modelled by differential equations. However, it is difficult to obtain closed-form solutions for them, especially for nonlinear ones. In most cases, only approximate solutions (either analytical ones or numerical ones) can be expected. Perturbation method is one of the well-known methods for solving nonlinear problems analytically.

In general, the perturbation method is valid only for weakly nonlinear problems[Nayfeh (2000)]. For example, consider the following heat transfer problem governed by the nonlinear ordinary differential equation, see [Abbasbandy (2006)]:

(1+εu)u,+u=0,u(0)=1E1

where ε > 0 is a physical parameter, the prime denotes differentiation with respect to the time t. Although the closed-form solution of u(t) is unknown, it is easy to get the exact result u,(0)=1/(1+ε), as mentioned by [Abbasbandy (2006)]. Regard that ε as a perturbation quantity, one can write u(t) into such a perturbation series

u(t)=u0(t)+εu1(t)+ε2u2(t)+ε3u3(t)+....................E2

Substituting the above expression into (1) and equating the coefficients of the like powers of ε, to get the following linear differential equations

u0,+u0=0,u0(0)=1,E3
u1,+u1=u0u0,,u1(0)=0,E4
u2,+u2=(u0u1,+u1u0,),u2(0)=0,E5
u3,+u3=(u0u2,+u1u1,+u2u0,),u3(0)=0,E6

Solving the above equations one by one, one has

u0(t)=et

u1(t)=ete2t

u2(t)=12et2e2t+32e3tE9

Thus, we obtain u(t)as a perturbation series

u(t)=et+ε(ete2t)+ε2(12et2e2t+32e3t)+.............E10

which gives at t = 0 the derivative

u,(0)=1+εε2+ε3ε4+ε5ε6+ε7ε8+ε9ε10+.............E11

Obviously, the above series is divergent for ε1, as shown in Fig. 1. This typical example illustrates that perturbation approximations are valid only for weakly nonlinear problems in general. In view of the work by [Abbasbandy (2006)], the HAM extends a series approximation beyond its initial radius of convergence. Figure 1.Comparison of the exact and approximate solutions of (1). Solid line: exact solution u,(0)=−1/(1+ε); Dashed-line: 31th-order perturbation approximation; Hollow symbols: 15th-order approximation given by the HPM; Filled symbols: 15th-order approximation given by the HAM when h=−1/(1+2ε).

To overcome the restrictions of perturbation techniques, some non-perturbation techniques are proposed, such as the Lyapunov's artificial small parameter method [Lyapunov A.M. (1992)], the δ-expansion method [Karmishin et al(1990)], the homotopy perturbation method [He H., J.,(1998)], and the variational iteration method (VIM), [He H., J.,(1999)],. Using these non-perturbation methods, one can indeed obtain approximations even if there are no small/large physical parameters. However, the convergence of solution series is not guaranteed. For example, by means of the HPM, we obtain the same and exact approximation of Eq.(1), as the perturbation result in Eq.(9), that is divergent for ε > 1, as shown in Fig.1. ; For details, see [Abbasbandy (2006)]. This example shows the importance of the convergence of solution series for all possible physical parameters. From physical points of view, the convergence of solution series is much more important than whether or not the used analytic method itself is independent of small/large physical parameters. If one does not keep this in mind, some useless results might be obtained. For example, let us consider the following linear differential equation [Ganji et al (2007)]:

ut+ux=2uxxt,xR,t>0E12
u(x,0)=exE13

uexact(x,t)=extE14

By means of the homotopy perturbation method, [Ganji et al (2007)] wrote the original equation in the following form:

(1p)φ(x,t:p)t+p[φ(x,t:p)t+φ(x,t:p)x3φ(x,t:p)x2t]=0E15

subject to the initial condition

φ(x,0:p)=exE16

where p[0;1]is an embedding parameter. Then, regarding p as a small parameter, [Ganji et al (2007)] expanded φ(x,t:p) in a power series

φ(x,t:p)=u0(x,t)+m=1+um(x,t).pmE17

which gives the solution. For p = 1, and substitute (15) into the original equation (13) and initial condition in (14), then equating the coefficients of the like powers of p, one can get governing equations and the initial conditions forum(x,t). In this way, [Ganji et al (2007)] obtained the mth-order approximation

u(x,t)u0(x,t)+k=1muk(x,t)E18

uHPM(x,t)ex720[t6+66t5+1470t4+13320t3+47440t2+45360t+720]E19

However, for any givenx0, the above approximation enlarges monotonously to the positive infinity as the time t increases, as shown in Fig.2. Unfortunately, the exact solution monotonously decreases to zero! Let

δ(t)=|uexactuHAMuexact|E20

where δ(t) denotes the relative error of the HPM approximation (17). As shown in Fig. 2, the relative error δ(t) monotonously increases very quickly:

In fact, it is easy to find that the HPM series solution (16) is divergent for all x and t except t = 0 which however corresponds to the given initial condition in (11). In other words, the convergence radius of the HPM solution series (17) is zero. It should be emphasized that, the variational iteration method (VIM) obtained exactly the same result as (17) by the 6th iteration see; [He H. J., (1999)], and [Ganji et al (2007)]. This example illustrates that both of the HPM and the VIM might give divergent approximations. Thus, it is very important to ensure the convergence of solution series obtained. Figure 2.Approximations of (10) given by the homotopy perturbation method. Dashed-line: exact solution(12); Solid line: the 5th-order HPM approximation(17); Dash-dotted line: the relative error δ(t) defined by(18).

## 3. Outline of Homotopy Perturbation Method (HPM)

The homotopy analysis method (HAM) has been proposed by Liao in his PhD dissertation in [Liao (1992)]. Liao introduced the so-called auxiliary parameter in [Liao (1997a)] to construct the following two-parameter family of equation:

(1p)L(uu0)=hpN(u)E21

where u0is an initial guess. [Liao (1997a)] pointed out that the convergence of the solution series given by the HAM is determined by h, and thus one can always get a convergent series solution by means of choosing a proper value of h. Using the definition of Taylor series with respect to the embedding parameter p (which is a power series of p ), [Liao (1997b)] gave general equations for high-order approximations.

[He J. H. (1999)] followed Dr. Liao’s early idea of Homotopy Perturbation Method (HPM) when he constructed the one-parameter family of equation:

(1p)L(u)+pN(u)=0E22

where Eq.(20) represented special case of Eq.(19) for convergent solution of (HAM) at h=1. To illustrate the basic ideas of this method, consider the following general nonlinear differential equation [see Ghasemi et al (2010)].

A(u)f(r)=0,rΩE23

With boundary conditions

B(u,un)=0,rΓE24

where A is a general differential operator, B is a boundary operator, f(r) is a known analytic function, and is the boundary of the domain.

The operator A can be generally divided into linear and nonlinear parts, say L and N. Therefore (21) can be

written as

L(u)+N(u)f(r)=0E25

[He (1999)] constructed a homotopy v(r,p):Ω×[0,1]Rwhich satisfies:

H(v, p) = (1 -p)[L(v) -L(v0)] + p[A(v) - f(r)] = 0E26

where rΩ, p[0,1]that is called homotopy parameter, and v0 is an initial approximation of (19). Hence, it is obvious that:

H(v, 0) = L(v) -L(v0) = 0E27
and
H(v, 1) = [A(v)- f(r)]= 0E28

In topology, L(v) -L(v0) is called deformation, and [A(v)- f(r)]is called homotopic. The embedding parameter p monotonically increases from zero to unit as the trivial problem H(v, 0)=0 in (25) is continuously deforms the original problem in (26), H(v, 1)=0. The embedding parameter p[0,1] can be considered as an expanding parameter. [Nayfeh A.H. (1985)] Apply the perturbation technique due to the fact that 0p1, can be considered as a small parameter, the solution of (21) or (23) can be assumed as a series in p, as follows:

v=v0+pv1+p2v2+p3v3+....................E29

when p1, the approximate solution, i.e.,

u=limp1v=v0+v1+v2+v3+....................E30

The series (28) is convergent for most cases, and the rate of convergence depends on A(v), [He, L. (1999)].

## 4. Application of Homotopy Perturbation Method HPM

An analytic method for strongly nonlinear problems, namely the homotopy analysis method (HAM) was proposed by Liao in 1992, six years earlier than the homotopy perturbation method by [He H., J.,(1998)], and the variational iteration method by [He H., J.,(1999)]. Different from perturbation techniques, the HAM is valid if a nonlinear problem contains small/large physical parameters.

More importantly, unlike all other analytic techniques, the HAM provides us with a simple way to adjust and control the convergence radius of solution series. Thus, one can always get accurate approximations by means of the HAM. In the next section, HPM is applied to solve the linear and nonlinear equations of heat transfer by conduction in one-dimensional in a slab of thickness (L). [Anwar (2010)] solved the linear and non-linear heat transfer equations by means of HPM.

### 4.1. Non-Linear Heat transfer equation

Consider the heat transfer equation by conduction in one-dimensional in a slab of thickness L. The governing equation describing the temperature distribution is:

ddx(kdTdx)=0,x[0,L]E31

Where the two faces are maintained at uniform temperatures T1 and T2 with T1>T2the slab make of a material with temperature dependent thermal conductivity k=k(T); see [Rajabi A.(2007)]. The thermal conductivity k is assumed to vary linearly with temperature, that is:

k=k2[1+εTT2T1T2]T(0)=T1,T(L)=T2E32

where ε is a constant and k2is the thermal conductivity at temperature T2. Introducing the dimensionless quantities

θ=TT2T1T2, ε=k1k2k2 X=xL
X[0,1]E33

where k1 is the thermal conductivity at temperature T1, then (29) reduces to

d2θdX2+ε(dθdX)2(1+εθ)=0,X[0,1]θ(0)=1,θ(1)=0E34

The problem is formulated by using (19) as:

(1p)L[θ(X,p)θ0(X)]+pN[θ(X,p)]=0E35

Where the Linear operator:

L(θ)=θ,,E36
and,
θ0XX=0E37

from Eq.(31), the initial guess is:

θ0(X)=1XE38

and the linear operator:

L[C1X+C2]=0E39

and the nonlinear operator of θ(X,p)is:

N[θ(X,p)]=θ(X,p)XX+ε[θ(X,p).θ(X,p)XX+(θ(X,p)X)2]=0E40
and:
θ(0,p)=1,θ(1,p)=0E41

wherep[0,1] is an embedding parameter. For p = 0 and 1, we have

θ(X,0)=θ0(X)θ(X,1)=θ(X)E42
θ0(X)tends to θ(X) as p varies from 0 to 1. Due to Taylor’s series expansion:
θ(X,p)=θo(X)+s=11s!sθ(X,p)ps|p=0E43

and the convergence of series (40) is convergent at p=1. Then by using (35) and (36) one obtains

θ(X)=θ0(X)+s=0θs(X)E44

For the s-th- order problems, if we first differentiate Eq.(32) s times with respect to p then divide by s! and setting p = 0 we obtain:

L[θs(X)usθs1(X)]+[θs1,,(X)+εn=0s1(θs1n,.θn,+θs1nθn,,)]=0E45

Where:

θs(0)=θs(1)=0E46
us={0,s11,s>1E47

The general solutions of (42) can be written as:

θs(X)=θ0(X)+θs*(X)E48
(45)

where θs*(X)is the particular solution.

The linear non-homogeneous (42) is solved for the order s = 1, 2, 3,..., for s=1, (42) becomes:

d2θ1dX2+ε(dθ0dXdθ0dX+θ0d2θ0dX2)=0θ1(0)=0,θ1(1)=0E49

Then

d2θ1dX2+ε=0E50

the solution of (47) gives :

θ1=ε2(XX2)E51

For s = 2, Eq.(42) becomes:

d2θ2dX2+ε(2dθ1dXdθ0dX+θ1d2θ0dX2+θ0d2θ1dX2)=0,θ2(0)=0,θ2(1)=0E52

Solution of (49) gives :

θ2=ε22[2X2X3X]E53

Then, solution of (31) is:

θ(X)=(1X)+ε2(XX2)+ε22[2X2X3X]E54

Results of θobtained for different values of εare presented in Table 1 and Fig. 3. Clearly, for small value for0ε1 then (51) is a good approximation to the solution. That means forε=0, thenk1=k2, forε=1, thenk1=2k2. However, as εincreases, (51) produces inaccurate divergent results. The results obtained via HPM are compared to those via General Approximation Method GAM obtained by Khan R. A.(2009). For this problem, it is found that HPM produces agreed results compared to GAM.

 X ε=.5 ε=.8 ε=1.0 ε=1.5 ε=2 0.1 0.912375 0.91008 0.9045 0.876375 0.828 0.2 0.824 0.82304 0.816 0.776 0.704 0.3 0.734125 0.73696 0.7315 0.692125 0.616 0.4 0.642 0.64992 0.648 0.618 0.552 0.5 0.546875 0.56 0.5625 0.546875 0.5 0.6 0.448 0.46528 0.472 0.472 0.448 0.7 0.3446249 0.36384 0.3735 0.386625 0.384 0.8 0.2359999 0.2537599 0.2639999 0.2839999 0.2959999 0.9 0.1213749 0.1331199 0.1404999 0.1573749 0.1719999

### Table 1.

Solutions θ via X for different values of ε.

Special computer program was used as special case, the temperature distribution along a road of length (L = 1 m) whenT1=100C0andT2=50C0, are presented in Table 2 and Fig.4.

 X ε=.5 ε=.8 ε=1.0 ε=1.5 ε=2 0.0 100 100 100 100 100 0.1 95.61875 95.504 95.225 93.81875 91.4 0.2 91.2 91.152 90.8 88.8 85.2 0.3 86.70625 86.848 86.575 84.60625 80.8 0.4 82.1 82.496 82.4 80.9 77.6 0.5 77.34375 78 78.125 77.34375 75 0.6 72.4 73.264 73.6 73.6 72.4 0.7 67.23125 68.192 68.675 69.33125 69.2 0.8 61.8 62.688 63.2 64.2 64.8 0.9 56.06874 56.656 57.025 57.86874 58.6 1.0 50 50 50 50 50

### Table 2.

Solutions T via X for different values of ε

### 4.2. Linear Heat transfer equation

In this section we consider the linear one-dimensional equation of heat transfer by conduction (diffusion equation) [Anderson (1984)]:

Ttα2Tx2=00x1,t>0E55

for initial condition Figure 3.Graphical results θ via X obtained by HPM for different values of ε. Figure 4.Graphical results of Temperature via X obtained by HPM for different values of ε.
T(x,0)=g(x)=sin(2π.x)E56

and boundary condition

T(0,t)=T(1,t)=0E57
αis thermal conductivity that is assumed constant with temperature. To solve the parabolic partial differential equation (52) using HPM, we consider a correction functional equation as:
(1p)[Ttu0t]+p[Ttα2Tx2]=0E58

Then:

Ttu0tpu0tαp2Tx2=0E59
Ttαp2Tx2=0E60
(T0+pT1+p2T2+p3T3+...)tαp2(T0+pT1+p2T2+p3T3+..)x2=0E61

For zeroth order of p:

T0t=0E62

Then

T0(x,t)=sin(2π.x)E63

For first order of p:

T1tα2T0x2=0E64
T1t+4π2αsin(2π.x)=0E65
T1(x,t)=sin(2π.x)4π2αsin(2π.x)tE66

For second order of p:

T2tα2T1x2=0E67
T2(x,t)=sin(2π.x)4π2αsin(2π.x)t+8π4α2sin(2π.x)t2E68

Using equation (56) for other orders of p, we can obtain the following results:

T(x,t)=sin(2π.x)[1(4π2α.t)+12(4π2α.t)2.....]E69

It is obvious that T(x,t) converge to the exact solution as increasing orders of p:

T(x,t)=sin(2π.x).exp(4π2α.t)E70

Fig.5 and Fig.6 represent the HPM solution T(x, t) for α=0.05, and α=0.1respectively for 0x1, 0t0.4.

 X t=0 t = 0.1 t = 0.2 t = 0.3 t = 0.4 0 0 0 0 0 0 0.05 0.3091373 0.2082383 0.1402717 9.448861E-2 6.364859E-2 0.1 0.5879898 0.3960766 0.2668017 0.1797206 0.1210618 0.15 0.8092399 0.5451131 0.3671944 0.2473463 0.1666153 0.2 0.9512127 0.6407476 0.4316148 0.2907406 0.1958462 0.25 0.9999998 0.6736112 0.4537521 0.3056525 0.205891 0.3 0.9508218 0.6404843 0.4314375 0.2906211 0.1957657 0.35 0.8084964 0.5446123 0.366857 0.247119 0.1664622 0.4 0.5869664 0.3953872 0.2663373 0.1794078 0.1208511 0.45 0.3079342 0.207428 0.1397258 0.0941209 0.0634009 0.5 0 0 0 0 0 0.55 -0.3103399 -0.2090485 -0.1408174 -0.0948562 -6.389621E-2 0.60 -0.5890125 -0.3967655 -0.2672657 -0.1800332 -0.1212724 0.65 -0.8099824 -0.5456133 -0.3675313 -0.2475732 -0.1667681 0.70 -0.9516023 -0.64101 -0.4317916 -0.2908597 -0.1959264 0.75 -0.9999982 -0.6736101 -0.4537514 -0.3056521 -0.2058907 0.80 -0.9504291 -0.6402198 -0.4312593 -0.2905011 -0.1956849 0.85 -0.8077511 -0.5441103 -0.3665189 -0.2468912 -0.1663087 0.90 -0.5859417 -0.3946969 -0.2658723 -0.1790946 -0.1206401 0.95 -0.3067303 -0.206617 -0.1391795 -9.37529E-2 -6.315302E-2 1 0 0 0 0 0

### Table 3.

Solution T(x, t) for0x1, 0t0.4 atα=0.05.

 X t=0 t = 0.1 t = 0.2 t = 0.3 t = 0.4 0 0 0 0 0 0 0.05 0.3091373 0.2537208 0.2082383 0.1709092 0.1402717 0.1 0.5879898 0.4825858 0.3960766 0.3250752 0.2668017 0.15 0.8092399 0.6641742 0.5451131 0.4473952 0.3671944 0.2 0.9512127 0.7806966 0.6407476 0.5258861 0.4316148 0.25 0.9999998 0.8207381 0.6736112 0.5528585 0.4537521 0.3 0.9508218 0.7803758 0.6404843 0.52567 0.4314375 0.35 0.8084964 0.6635639 0.5446123 0.4469841 0.366857 0.4 0.5869664 0.4817458 0.3953872 0.3245094 0.2663373 0.45 0.3079342 0.2527334 0.207428 0.1702441 0.1397258 0.5 0 0 0 0 0 0.55 -0.3103399 -0.2547078 -0.2090485 -0.1715741 -0.1408174 0.60 -0.5890125 -0.4834251 -0.3967655 -0.3256406 -0.2672657 0.65 -0.8099824 -0.6647836 -0.5456133 -0.4478057 -0.3675313 0.70 -0.9516023 -0.7810164 -0.64101 -0.5261015 -0.4317916 0.75 -0.9999982 -0.8207368 -0.6736101 -0.5528576 -0.4537514 0.80 -0.9504291 -0.7800536 -0.6402198 -0.5254529 -0.4312593 0.85 -0.8077511 -0.6629522 -0.5441103 -0.4465721 -0.3665189 0.90 -0.5859417 -0.4809047 -0.3946969 -0.3239429 -0.2658723 0.95 -0.3067303 -0.2517453 -0.206617 -0.1695785 -0.1391795 1 0 0 0 0 0

### Table 4.

Solution T(x, t) for 0x1, 0t0.4at α=0.1.

## 5. Discussion

For example.1, clearly, for0ε1, (51) is a good approximation to the solution. That means for ε = 0, thenk1=k2, and for ε = 1, thenk1=2k2. However, as ε increases, (51) produces inaccurate divergent results. For example 2, (66) is a good approximation to the solution as α values decreased. That means as α increase, (66) produces inaccurate divergent results.