**5.2 Application of the homotopy perturbation method to the fluid flow problem**

According to homotopy perturbation method (HPM), one can construct an homotopy for Eq. (36)–(39) as

$$\begin{aligned} H\_1(p,\eta) &= (\mathbf{1} - p) \left[ \left( \mathbf{1} + \frac{\mathbf{1}}{\beta} \right) f^{\dot{\boldsymbol{\nu}}} \right] + p \left[ \begin{pmatrix} \mathbf{1} + \frac{\mathbf{1}}{\beta} \boldsymbol{f}^{\dot{\boldsymbol{\nu}}} - \mathbf{S} \mathbf{1}\_1 (\mathbf{1} - \boldsymbol{\phi})^{2.5} \begin{pmatrix} \boldsymbol{\eta}f^{\boldsymbol{\nu}} + \mathbf{3}f'' \\ +f\boldsymbol{f}^{\boldsymbol{\cdot}} - \boldsymbol{f}'f'' \end{pmatrix} \\ &= \mathbf{0}, \end{aligned} \right] = \mathbf{0},$$

$$H\_2(p, \eta) = (1 - p) \left[ \left( 1 + \frac{4}{3} R \right) \theta' \right] + p \left[ \begin{aligned} & \left( 1 + \frac{4}{3} R \right) \theta' + \text{PrS} \left( \frac{A\_2}{A\_3} \right) \left( \theta' \xi - \eta \theta' \right) \\ & + \frac{\text{PrEc}}{A\_3 (1 - \phi)^{2.5}} \left( \left( f'' \right)^2 + 4 \delta^2 \left( f' \right)^2 \right) \end{aligned} \right] = \mathbf{0}, \tag{54}$$

Taking power series of velocity, temperature and concentration fields, gives

$$f = f\_0 + pf\_1 + p^2 f\_2 + p^3 f\_3 + \dots \tag{55}$$

(53)

and

$$
\theta = \theta\_0 + p\theta\_1 + p^2\theta\_2 + p^3\theta\_3 + \dots \tag{56}
$$

Substituting Eqs. (55) and (56) into Eq. (53) and (54) as well as the boundary conditions in Eq. (42), and grouping like terms based on the power of *p,* the fluid flow velocity equation is given as:

**Zeroth-order equations**

$$p^0: \qquad f\_0^{\dot{v}}(\eta) + \frac{1}{\beta} f\_0^{\dot{v}}(\eta) = \mathbf{0},\tag{57}$$

$$p^0: \qquad \left(\mathbf{1} + \frac{4}{3}\mathbf{R}\right)\theta\_0'' = \mathbf{0},\tag{58}$$

*Perturbation Methods to Analysis of Thermal, Fluid Flow and Dynamics Behaviors of… DOI: http://dx.doi.org/10.5772/intechopen.96059*

## **First-order equations**

$$\begin{aligned} \text{p}^1: \quad \frac{1}{\rho} f\_1^{iv}(\eta) + f\_1^{iv}(\eta) - \text{SA}\_1(1-\phi)^{2.5} \eta f\_0(\eta) - \frac{1}{Da} f\_0''(\eta) - Ha^2 f\_0''(\eta) \\ - \text{3SA}\_1(1-\phi)^{2.5} f\_0''(\eta) - \text{SA}\_1(1-\phi)^{2.5} f\_0'(\eta)f\_0''(\eta) + \text{SA}\_1(1-\phi)^{2.5} f\_0(\eta)f\_0''(\eta) = \text{0}, \\ \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \end{aligned} \tag{59}$$

$$p^1: \quad \left(1 + \frac{4}{3}R\right)\theta\_1'' + Pr\mathcal{S}\left(\frac{A\_2}{A\_3}\right)\left(\theta\_0'f\_0 - \eta\theta\_0'\right) + \frac{\text{Pr}\text{Ec}}{A\_3(1-\phi)^{2.5}}\left(\left(f\_0''\right)^2 + 4\delta^2\left(f\_0'\right)^2\right) = 0\tag{60}$$

## **Second-order equations**

$$p^2: \quad \frac{1}{\theta} f\_2^{\dot{w}}(\eta) + f\_2^{\dot{w}}(\eta) - \text{SA}\_1(1-\phi)^{2.5} \eta f\_1(\eta) - \frac{1}{Da} f\_1''(\eta) - Ha^2 f\_1''(\eta) - 3\text{SA}\_1(1-\phi)^{2.5} f\_1''(\eta)$$

$$-\text{SA}\_1(1-\phi)^{2.5} f\_1'(\eta)f\_0''(\eta) - \text{SA}\_1(1-\phi)^{2.5} f\_0'(\eta)f\_1''(\eta) + \text{SA}\_1(1-\phi)^{2.5} f\_1(\eta)f\_0'(\eta)$$

$$+\text{SA}\_1(1-\phi)^{2.5} f\_0(\eta)f\_1''(\eta) = 0,\tag{61}$$

$$\begin{aligned} \text{p}^2: \quad \left(1 + \frac{4}{3}R\right)\theta\_2'' + \text{PrS}\left(\frac{A\_2}{A\_3}\right)\left(\theta\_1'f\_0 + \theta\_0'f\_1 - \eta\theta\_1'\right) + \frac{2\text{PrEc}}{A\_3(1-\phi)^{2.5}}\left(f\_0''f\_1'' + 4\delta^2 f\_0'f\_1'\right) = 0. \end{aligned} \tag{62}$$

the boundary conditions are

$$f\_0 = f\_1 = f\_2 = \mathbf{0}, \quad f\_0'' = f\_1'' = f\_2'' = \mathbf{0}, \quad \theta\_0' = \theta\_1' = \theta\_2' = \mathbf{0}, \text{ when } \quad \eta = \mathbf{0}, \tag{6}$$

$$f\_0 = \mathbf{1}, \ f\_1 = f\_2 = \mathbf{0}, \quad f\_0' = f\_1' = f\_2' = \mathbf{0}, \ \theta\_0 = \mathbf{1}, \ \theta\_1 = \theta\_2 = \mathbf{0}, \text{ when } \quad \eta = \mathbf{1}, \tag{6}$$

In a similar way, the higher orders problems are obtained.

On solving Eqs. (57), (61) and (64) with their corresponding boundary conditions, we arrived at

$$f\_0(\eta) = \frac{1}{2} (3\eta - \eta^3) \tag{64}$$

$$f\_1(\eta) = -\frac{1}{6720(1+\beta)} \begin{pmatrix} \left(168\left(\frac{1}{Da}\right)\beta + 168Ha^2\beta + 419SA\_1(1-\phi)^{2.5}\beta\right)\eta \\\\ -\left(336\left(\frac{1}{Da}\right)\beta + 336Ha^2\beta + 873SA\_1(1-\phi)^{2.5}\beta\right)\eta^3 \\\\ +\left(168\left(\frac{1}{Da}\right)\beta + 168Ha^2\beta + 504SA\_1(1-\phi)^{2.5}\beta\right)\eta^5 \\\\ -28SA\_1(1-\phi)^{2.5}\beta\eta^6 - 24SA\_1(1-\phi)^{2.5}\beta\eta^7 \\\\ +2SA\_1(1-\phi)^{2.5}\beta\eta^8 \end{pmatrix} \eta^7 \tag{65}$$

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,

In the same manner, the energy equations are solved. Following the definition of the homotopy perturbation method as presented in Eq. (52), one could write the solution of the fluid flow equation as

$$\begin{aligned} f(\boldsymbol{u}) &= \frac{1}{2}(\boldsymbol{\delta}\_{\boldsymbol{u}} - \boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - \boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - \boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - \boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} + 3564\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} (1 - \boldsymbol{\delta}^{\prime})^{\beta}) - \left(364\left(\frac{1}{\Delta t}\right)^{2} + 3564\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} + 3752\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} (1 - \boldsymbol{\delta}^{\prime})^{\beta}\right) \boldsymbol{\delta}^{\beta} \\ &+ \left(486\left(\frac{1}{\Delta t}\right)^{2}\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} + 1862\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} + 9564\boldsymbol(\delta^{\prime}\_{\boldsymbol{u}} - \boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}}\right)\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - 2354\boldsymbol(1 - \boldsymbol{\delta}^{\prime})^{\beta}\boldsymbol{\delta}^{\prime} + 2454(1 - \boldsymbol{\delta}^{\prime})^{\beta}\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} \\ & \quad \left(\begin{aligned} \left(1 - 2564\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}}\right)^{2}\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - 2559\boldsymbol(1 - \left(\frac{1}{\Delta t}\right)^{\beta}\boldsymbol{\delta}^{\prime}\_{\boldsymbol{u}} - 15364\boldsymbol(\$$
