*3.1. Zeroth Order Deformation Problems*

Introducing the nonlinear operator ℵ as

$$\begin{split} \left[ \mathcal{N}\_{f} \left[ f(\zeta, j), g(\zeta, j) \right] \right] &= B\_{1} \frac{\partial^{4} f(\zeta, j)}{\partial \zeta^{4}} + \text{Re} \left[ 2f(\zeta, j) \frac{\partial^{3} f(\zeta, j)}{\partial \zeta^{3}} + 2g(\zeta, j) \frac{\partial g(\zeta, j)}{\partial \zeta} - \text{MB}\_{2} \frac{\partial^{2} f(\zeta, j)}{\partial \zeta^{2}} \right] - \\ &\quad k\_{2} \text{Re} B\_{1} \frac{\partial^{2} f(\zeta, j)}{\partial \zeta^{2}} - 2k\_{3} \text{Re} \frac{1}{\rho\_{\text{Im}f}} \frac{\partial f(\zeta, j)}{\partial \zeta} \frac{\partial^{2} f(\zeta, j)}{\partial \zeta^{2}}, \end{split} \tag{44}$$

$$\mathcal{N}\_{\xi}[f(\zeta\_{\prime}j),g(\zeta\_{\prime}j)] = B\_{1}\frac{\partial^{2}g(\zeta\_{\prime}j)}{\partial\zeta^{2}} + \text{Re}\left[2f(\zeta\_{\prime}j)\frac{\partial g(\zeta\_{\prime}j)}{\partial\zeta^{2}} - \text{MB}\_{2}\frac{\partial g(\zeta\_{\prime}j)}{\partial\zeta^{2}}\right] - k\_{2}\text{B}\_{1}g(\zeta\_{\prime}j) - k\_{3}\frac{1}{\rho\_{\text{lnf}}}\left[g(\zeta\_{\prime}j)\right]^{2},\tag{45}$$

$$\mathcal{R}\_{\theta}[f(\zeta,j),g(\zeta,j),\theta(\zeta,j)] = B\_3 \frac{k\_{\text{hrf}}}{k\_f} \frac{\partial^2 \theta(\zeta,j)}{\partial \zeta^2} + \frac{1}{Rd} Pr \text{Re} \left[ 2f(\zeta,j) \frac{\partial \theta(\zeta,j)}{\partial \zeta} + M B\_4 \text{Ec} \left( \frac{\partial f(\zeta,j)}{\partial \zeta} \right)^2 + \left( g(\zeta,j) \right)^2 \right],\tag{46}$$

$$\mathcal{N}\_{\boldsymbol{\theta}}[f(\boldsymbol{\zeta},\boldsymbol{j}),\boldsymbol{\varrho}(\boldsymbol{\zeta},\boldsymbol{j})] = \frac{\partial^{2}\boldsymbol{\varrho}(\boldsymbol{\zeta},\boldsymbol{j})}{\partial\boldsymbol{\zeta}^{2}} + \mathrm{Re}\mathrm{Sc}\left[2f(\boldsymbol{\zeta},\boldsymbol{j})\frac{\partial\boldsymbol{\varrho}(\boldsymbol{\zeta},\boldsymbol{j})}{\partial\boldsymbol{\zeta}} + k\_{4}\boldsymbol{\varrho}(\boldsymbol{\zeta},\boldsymbol{j})(1-\boldsymbol{\varrho}(\boldsymbol{\zeta},\boldsymbol{j}))^{2}\right],\tag{47}$$

where *j* is the homotopy parameter such that *j* ∈ [0, 1].

Moreover

$$[(1-j)\mathbf{L}\_f[f(\zeta,j)-f\_0(\zeta)] = j\hbar\_f \aleph\_f[f(\zeta,j),\mathbf{g}(\zeta,j)],\tag{48}$$

$$(1 - j)\mathcal{L}\_{\mathcal{S}}[\mathbf{g}(\boldsymbol{\zeta}, j) - \mathbf{g}\_0(\boldsymbol{\zeta})] = j\hbar\_{\mathcal{S}}\aleph\_{\mathcal{S}}[f(\boldsymbol{\zeta}, j), \mathbf{g}(\boldsymbol{\zeta}, j)],\tag{49}$$

$$\mathbf{h}\left(\mathbf{1}-j\right)\mathbf{L}\_{\theta}\left[\theta(\zeta\_{\prime}j)-\theta\_{0}(\zeta)\right]=j\hbar\_{\theta}\aleph\_{\theta}\left[f(\zeta\_{\prime}j),\mathbf{g}\left(\zeta\_{\prime}j\right),\theta(\zeta\_{\prime}j)\right],\tag{50}$$

$$\frac{1}{2}(1-j)\mathcal{L}\_{\varphi}[\varphi(\zeta,j)-\varphi\_{0}(\zeta)]=j\hbar\_{\varphi}\aleph\_{\varphi}[f(\zeta,j),\varphi(\zeta,j)],\tag{51}$$

where ¯*hϕ*, ¯*h<sup>f</sup>* , ¯*h<sup>θ</sup>* and ¯*h<sup>g</sup>* are the convergence control parameters.

Boundary conditions of Equation (48) are

$$f(0,j) = 0, \quad f'(0,j) = k\_{6'} \quad f(1,j) = 0, \quad f'(1,j) = k\_8. \tag{52}$$

Boundary conditions of Equation (49) are

$$g(0,j) = 1, \quad g(1,j) = \Omega. \tag{53}$$

Boundary conditions of Equation (50) are

$$
\theta(0, j) = 1, \qquad \theta(1, j) = 0. \tag{54}
$$

Boundary conditions of Equation (51) are

$$
\varphi'(0,j) = k\_7 \varphi(0,j), \qquad \varphi(1,j) = 1. \tag{55}
$$

Characterizing *j* = 0 and *j* = 1, the calculations obtained as

$$j = 0 \Rightarrow f(\zeta, 0) = f\_0(\zeta) \quad \text{and} \quad j = 1 \Rightarrow f(\zeta, 1) = f(\zeta), \tag{56}$$

$$j = 0 \Rightarrow \mathbf{g}(\zeta, \mathbf{0}) = \mathbf{g}\_0(\zeta) \quad \text{and} \quad j = 1 \Rightarrow \mathbf{g}(\zeta, 1) = \mathbf{g}(\zeta), \tag{57}$$

$$j = 0 \Rightarrow \theta(\zeta, 0) = \theta\_0(\zeta) \qquad \text{and} \qquad j = 1 \Rightarrow \theta(\zeta, 1) = \theta(\zeta), \tag{58}$$

$$j = 0 \Rightarrow \varphi(\zeta, 0) = \varphi\_0(\zeta) \quad \text{and} \quad j = 1 \Rightarrow \varphi(\zeta, 1) = \varphi(\zeta). \tag{59}$$

*f*(*ζ*, *j*) becomes *f*0(*ζ*) and *f*(*ζ*) as *j* assumes the values zero and one. *g*(*ζ*, *j*) becomes *g*0(*ζ*) and *g*(*ζ*) as *j* assumes the values zero and one. *θ*(*ζ*, *j*) becomes *θ*0(*ζ*) and *θ*(*ζ*) as *j* assumes the values zero and one. Finally, *ϕ*(*ζ*, *j*) becomes *ϕ*0(*ζ*) and *ϕ*(*ζ*) as *j* assumes the values zero and one.

Applying Taylor series expansion on the Equations (56)–(59), the results are obtained as

$$f(\zeta, j) = f\_0(\zeta) + \sum\_{m=1}^{\infty} f\_m(\zeta) j^m, \quad f\_m(\zeta) = \frac{1}{m!} \frac{\partial^m f(\zeta, j)}{\partial j^m} \mid\_{j=0} \tag{60}$$

$$g(\zeta, j) = g\_0(\zeta) + \sum\_{m=1}^{\infty} g\_m(\zeta) j^m, \quad g\_m(\zeta) = \frac{1}{m!} \frac{\partial^m g(\zeta, j)}{\partial j^m} \mid\_{j=0} \tag{61}$$

$$\theta(\zeta, j) = \theta\_0(\zeta) + \sum\_{m=1}^{\infty} \theta\_m(\zeta) j^m, \quad \theta\_m(\zeta) = \frac{1}{m!} \frac{\partial^m \theta(\zeta, j)}{\partial j^m} \mid\_{j=0} \tag{62}$$

$$\varphi(\zeta,j) = \varphi\_0(\zeta) + \sum\_{m=1}^{\infty} \varrho\_m(\zeta) j^m, \qquad \wp\_m(\zeta) = \frac{1}{m!} \frac{\partial^m \varphi(\zeta,j)}{\partial j^m} \mid\_{j=0} \tag{63}$$

*h*¯ *<sup>ϕ</sup>*, *h*¯ *<sup>f</sup>* , *h*¯ *<sup>θ</sup>* and *h*¯ *<sup>g</sup>* are adjusted to obtain the convergence for the series in Equations (60)–(63) at *j* = 1, so Equations (60)–(63) transform to

$$f(\zeta) = f\_0(\zeta) + \sum\_{m=1}^{\infty} f\_m(\zeta),\tag{64}$$

$$\mathbf{g}(\boldsymbol{\zeta}) = \mathbf{g}\_0(\boldsymbol{\zeta}) + \sum\_{m=1}^{\infty} \mathbf{g}\_m(\boldsymbol{\zeta}),\tag{65}$$

$$\theta(\zeta) = \theta\_0(\zeta) + \sum\_{m=1}^{\infty} \theta\_m(\zeta)\_{\prime} \tag{66}$$

$$
\varphi(\zeta) = \varphi\_0(\zeta) + \sum\_{m=1}^{\infty} \varphi\_m(\zeta). \tag{67}
$$
