**5. Applications of Homotopy perturbation method and new Homotopy perturbation method**

For understanding the application of HPM and NHPM, we will solve the one-dimensional heat equation given by

$$\frac{\partial U}{\partial \theta} = \beta \frac{\partial^2 U}{\partial z^2} \tag{18}$$

with boundary conditions

$$U(\mathbf{0}, \theta) = \mathbf{0}, U(\mathbf{1}, \theta) = \mathbf{0},\tag{19}$$

and initial condition

$$U(z,0) = \sin\frac{2\pi z}{L}, 0 \le z \le L. \tag{20}$$

The homotopy for the diffusion equation given by (18) is obtained as follows [2].

$$
\left(\frac{\partial v}{\partial \theta} - \frac{\partial u\_0}{\partial \theta}\right) + \mathfrak{f}\left(\frac{\partial u\_0}{\partial \theta} - \beta \frac{\partial^2 v}{\partial \mathbf{z}^2}\right) = \mathbf{0} \tag{21}
$$

Let *<sup>u</sup>*<sup>0</sup> <sup>¼</sup> *sin* <sup>2</sup>*π<sup>z</sup> <sup>L</sup> cos π*<sup>2</sup>*θ* be the initial approximation, which satisfies boundary conditions given by (19).

Let solution of (18) has the following form

$$
\nu = \nu\_0 + \mathfrak{p}v\_1 + \mathfrak{p}^2 v\_2 + \mathfrak{p}^3 v\_3 + \mathfrak{p}^4 v\_4 + \dots \tag{22}
$$

On substituting the value of *v* in Eq. (21) and comparing the coefficients of like powers of ƥ we obtain

$$\mathfrak{h}^0 : \frac{\partial v\_0}{\partial \theta} = \frac{\partial u\_0}{\partial \theta}.$$

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$$\begin{aligned} \text{f}^1: \frac{\partial v\_1}{\partial \theta} &= \beta \frac{\partial^2 v\_0}{\partial z^2}, \upsilon\_1(0, \theta) = 0 = \upsilon\_1(L, \theta) \\ \text{f}^2: \frac{\partial v\_2}{\partial \theta} &= \beta \frac{\partial^2 \upsilon\_1}{\partial z^2}, \upsilon\_2(0, \theta) = 0 = \upsilon\_2(L, \theta) \end{aligned}$$

$$\begin{aligned} \text{f}^3: \frac{\partial \upsilon\_3}{\partial \theta} &= \beta \frac{\partial^2 \upsilon\_2}{\partial z^2}, \upsilon\_3(0, \theta) = 0 = \upsilon\_3(L, \theta) \end{aligned}$$

$$\begin{aligned} \text{f}^n: \frac{\partial \upsilon\_n}{\partial \theta} &= \beta \frac{\partial^2 \upsilon\_{n-1}}{\partial z^2}, \upsilon\_n(0, \theta) = 0 = \upsilon\_n(L, \theta) \end{aligned} \tag{23}$$

On solving the system of Eq. (23) using Mathematica 5.2

$$\nu\_0 = u\_0 = \sin\frac{2\pi x}{L}\cos\pi^2\theta$$

$$\frac{\partial v\_1}{\partial \theta} = -\frac{4\beta\pi^2}{L^2}\nu\_0 \Rightarrow \nu\_1 = -\frac{\beta\sin\left[\frac{2\pi x}{L}\right]\sin\left[\pi^2\theta\right]}{L^2} + \sin\left[\frac{\pi x}{L}\right]$$

$$\frac{\partial v\_2}{\partial \theta} = -\frac{4\beta\pi^2}{L^2}\nu\_1 \Rightarrow \nu\_2 = -\frac{\beta\left(L^2\pi^2\theta + \arccos\left[\pi^2\theta\right]\right)\sin\left[\frac{2\pi x}{L}\right]}{L^4} + \frac{L^4\sin\left[\frac{2\pi x}{L}\right] + \beta^2\sin\left[\frac{2\pi x}{L}\right]}{L^4}$$

$$\frac{\partial v\_3}{\partial \theta} = -\frac{4\beta\pi^2}{L^2}\nu\_2 \Rightarrow \nu\_3 = \frac{\beta\left(\pi^2\theta\left(-2L^4 + L^2\pi^2\theta\theta - 2\theta^2\right) + 2\beta^2\sin\left[\pi^2\theta\right]\right)\sin\left[\frac{2\pi x}{L}\right]}{2L^6} + \sin\left[\frac{2\pi x}{L}\right]$$

$$\frac{\partial v\_4}{\partial \theta} = -\frac{4\beta\pi^2}{L^2}\nu\_3 \Rightarrow \nu\_4$$

$$\begin{array}{ll} \partial\theta & \frac{\pi}{2} \\ &= \frac{1}{6L^8} \left( \beta \left( \pi^2 \theta \left( -6L^6 + 3L^4 \pi^2 \beta \theta - L^2 \pi^4 \theta^2 \beta^2 + 3 \rho^3 \pi^2 \theta \right) 6 \rho^3 \cos\left[ \pi^2 \theta \right] \right) \sin\left[ \frac{2\pi x}{L} \right] \right) \\ &+ \frac{L^8 \sin\left[ \frac{2\pi x}{L} \right] - \rho^4 \sin\left[ \frac{2\pi x}{L} \right]}{L^8} \end{array}$$

*∂v*5 *<sup>∂</sup><sup>θ</sup>* ¼ � <sup>4</sup>*βπ*<sup>2</sup> *<sup>L</sup>*<sup>2</sup> *<sup>v</sup>*<sup>4</sup> ) *<sup>v</sup>*<sup>5</sup> <sup>¼</sup> �<sup>1</sup> <sup>24</sup>*L*<sup>10</sup> *β π*<sup>2</sup> *<sup>θ</sup>* <sup>24</sup>*L*<sup>8</sup> � <sup>12</sup>*L*<sup>6</sup>*π*<sup>2</sup> *βθ* <sup>þ</sup> <sup>4</sup>*L*<sup>4</sup>*π*<sup>4</sup>*θ*<sup>2</sup> *<sup>β</sup>*<sup>2</sup> � � � *<sup>L</sup>*<sup>2</sup> *π*<sup>6</sup>*θ*<sup>3</sup> *<sup>β</sup>*<sup>3</sup> <sup>þ</sup> <sup>4</sup> �<sup>6</sup> <sup>þ</sup> *<sup>π</sup>*<sup>4</sup>*θ*<sup>2</sup> � �*β*<sup>4</sup> � � <sup>þ</sup> <sup>24</sup>*β*<sup>4</sup> sin *<sup>π</sup>*<sup>2</sup> *<sup>θ</sup>* � �Þsin <sup>2</sup>*π<sup>z</sup> L* � �! <sup>þ</sup> sin <sup>2</sup>*π<sup>z</sup> L* � � *∂v*<sup>6</sup> *<sup>∂</sup><sup>θ</sup>* ¼ � *βπ*<sup>2</sup> *<sup>L</sup>*<sup>2</sup> *<sup>v</sup>*<sup>5</sup> ) *<sup>v</sup>*<sup>6</sup> <sup>¼</sup> �<sup>1</sup> <sup>120</sup>*L*<sup>12</sup> *β π*<sup>2</sup> *<sup>θ</sup>* <sup>120</sup>*L*<sup>10</sup> � <sup>60</sup>*L*<sup>8</sup>*π*<sup>2</sup> *βθ* <sup>þ</sup> <sup>20</sup>*L*<sup>6</sup>*π*<sup>4</sup>*θ*<sup>2</sup> *β*<sup>2</sup> � � � �5*L*<sup>4</sup>*π*<sup>6</sup>*θ*<sup>3</sup> *<sup>β</sup>*<sup>3</sup> <sup>þ</sup> *<sup>L</sup>*<sup>2</sup> *<sup>π</sup>*<sup>8</sup>*θ*<sup>4</sup>*β*<sup>4</sup> � <sup>5</sup>*π*<sup>2</sup> *<sup>θ</sup>* �<sup>12</sup> <sup>þ</sup> *<sup>π</sup>*<sup>4</sup>*θ*<sup>2</sup> � �*β*<sup>5</sup> Þ <sup>þ</sup>120*β*<sup>5</sup> cos *<sup>π</sup>*<sup>2</sup> *<sup>θ</sup>* � �Þsin <sup>2</sup>*π<sup>z</sup> L* � �Þ þ *L*<sup>12</sup> sin <sup>2</sup>*π<sup>z</sup> L* � � <sup>þ</sup> *<sup>β</sup>*<sup>6</sup> sin <sup>2</sup>*π<sup>z</sup> L* � � *L*<sup>12</sup> and so on … (24)

The approximate solution of (1) by setting ƥ ¼ 1 in (23) is given by

$$\mu = \lim\_{p \to 1} v = v\_0 + v\_1 + v\_2 + v\_3 + v\_3 + \dots \tag{25}$$

On substituting values of *vi* 0 *s* in Eq. (25), solution is obtained in terms of a summation of infinite series which gives results near to the exact solution.

Now we will solve the Eq. (18) using NHPM. First of all, following homotopy is constructed for solving heat conduction equation using NHPM

$$p(1-p)\left(\frac{\partial T}{\partial \theta} - U\_0\right) + p\left(\frac{\partial T}{\partial \theta} - \beta \frac{\partial^2 T}{\partial \mathbf{z}^2}\right) = \mathbf{0} \tag{26}$$

Taking *<sup>L</sup>*�<sup>1</sup> <sup>¼</sup> <sup>Ð</sup> *<sup>θ</sup> θ*0 ð Þ*: dθ* i.e. inverse operator on Eq. (26), then

$$T(\mathbf{z}, \theta) = \int\_0^{\theta} U\_0(\mathbf{z}, \theta) d\theta - p \int\_0^{\theta} \left( U\_0 - \beta \frac{\partial^2 T}{\partial \mathbf{z}^2} \right) d\theta + T(\mathbf{z}, \mathbf{0}). \tag{27}$$

Let the solution of the (27) is

$$T = T\_0 + pT\_1 + p^2T\_2 + p^3T\_3 + \dots,\tag{28}$$

where, *T*0, *T*1, *T*2, … are to be determined.

Suppose Eq. (25) is the solution of Eq. (24). Comparing the coefficients of powers of *p* and equating to zero and using Eq. (25) in Eq. (24), following are obtained:

$$p^0: T\_0(z, \theta) = \int\_0^{\theta} U\_0(z, \theta) d\theta + T(z, 0)$$

$$p^1: T\_1(z, \theta) = -\int\_0^{\theta} (U\_0(z, \theta) - \beta \frac{\partial^2 T\_0}{\partial z^2}) d\theta$$

$$p^2: T\_2(z, \theta) = \int\_0^{\theta} (\beta \frac{\partial^2 T\_1}{\partial z^2}) d\theta$$

$$p^3: T\_3(z, \theta) = \int\_0^{\theta} (\beta \frac{\partial^2 T\_2}{\partial z^2}) d\theta$$

$$\text{and so on.} \tag{29}$$

Consider initial approximation of Eq. (18) as

$$U\_0(\mathbf{z}, \boldsymbol{\theta}) = \sum\_{n=0}^{\infty} c\_n(\mathbf{z}) P\_n(\boldsymbol{\theta}), \\ T(\mathbf{z}, \mathbf{0}) = U(\mathbf{z}, \mathbf{0}), \\ P\_k(\boldsymbol{\theta}) = \boldsymbol{\theta}^k,\tag{30}$$

where, *P*1ð Þ*θ* , *P*2ð Þ*θ* , *P*3ð Þ*θ* , … and *c*0ð Þ*z* ,*c*1ð Þ*z* ,*c*2ð Þ*z* , … are specified functions and unknown coefficients respectively, depending on the problem.

Using Eq. (30) in (29), following are obtained:

$$T\_0(z, \theta) = \left(c\_0(z)\theta + c\_1(z)\frac{\theta^2}{2} + c\_2(z)\frac{\theta^3}{3} + c\_3(z)\frac{\theta^4}{4} + \dots\right) + \sin\frac{2\pi z}{L}$$

$$\begin{split} T\_1(z, \theta) &= \left(-c\_0(z) - \frac{4\beta\pi^2}{L}\sin\frac{2\pi z}{L}\right)\theta + \left(-\frac{1}{2}c\_1(z) + \frac{1}{2}\beta c\_0''(z)\right)\theta^2 \\ &+ \left(-\frac{1}{3}c\_2(z) + \frac{1}{3}c\_1''(z)\right)\theta^3 + \dots \end{split} \tag{27}$$

and so on … (31)

Now solving the above equations in such a manner that, *T*1ð Þ¼ *z*, *θ* 0. Therefore Eq. (31) reduces to

$$c\_0(\mathbf{z}) = -\frac{2^2 \beta \pi^2}{L} \sin \frac{2\pi z}{L}$$

$$c\_1(\mathbf{z}) = \frac{2^4 \beta^2 \pi^4}{L^2} \sin \frac{2\pi z}{L}$$

$$c\_2(\mathbf{z}) = -\frac{2^6 \beta^3 \pi^6}{L^3} \sin \frac{2\pi z}{L}$$

$$\begin{split} U(\mathbf{z}, \theta) = T\_0(\mathbf{z}, \theta) &= \sin \frac{2\pi z}{L} + c\_0(\mathbf{z})\theta + c\_1(\mathbf{z})\frac{\theta^2}{2} + c\_2(\mathbf{z})\frac{\theta^3}{3} + c\_3(\mathbf{z})\frac{\theta^4}{4} + \dots \\ &= \sin \frac{2\pi z}{L} \left[ 1 - \frac{2^2 \beta \pi^2}{L} \theta + \frac{2^4 \beta^2 \pi^4}{L^2} \frac{\theta^2}{2} - \frac{2^6 \beta^3 \pi^6}{L^3} \frac{\theta^3}{3} + \dots \right] \end{split}$$

which is same as the universally accepted exact solution for the problem which is shown in **Figure 1**.

(32)

The solution of one-dimensional heat conduction equation is solved using HPM and NHPM and then compared with the universally accepted exact solution obtained from method of separation of variable. **Figure 2** represents the comparison of solution of heat equation using HPM, NHPM and method of separation of variable. It is found that the solution obtained using HPM gives result near to the exact solution whereas solution using NHPM gives same results as the exact solution.

**Figure 1.** *Solution using NHPM.*

*Application of Perturbation Theory in Heat Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.95573*

**Figure 2.** *Comparison of HPM, NHPM and the exact solution.*
