**3. Solution Procedure**

The analytical solutions of Equations (15)–(17) have been determined by regular perturbation method. To solve the problem under consideration, we presented the flow quantities wherein the velocity, concentration, and pressure interns of small Weissenberg number (We) have the following form:

$$u(x,y) = u\_{0,i} + (\text{We})^2 u\_{1,i} + (\text{We})^4 u\_{2,i} + \dots, \; i = 1, 2. \tag{18}$$

$$
\sigma = \sigma\_0 + (\text{We})^2 \sigma\_1 + (\text{We})^4 \sigma\_2 + \dots \tag{19}
$$

$$K = (K)\_{0,i} + (\text{We})^2 (K)\_{1,i} + (\text{We})^4 (K)\_{2,i} + \dots, \quad i = 1, 2, \dots \tag{20}$$

in which *K* = <sup>d</sup>*<sup>p</sup>* <sup>d</sup>*<sup>x</sup>* . After the implementation of above expressions in Equations (15)–(17) and equating the exponents of We, one obtains the following systems of equations along with associated boundary conditions.

### *3.1. Model 1*

• System of Order Zero

$$K\_{0,1} = \frac{1}{2} \left( \alpha \frac{\partial \sigma\_0}{\partial y} \frac{\partial u\_{0,1}}{\partial y} + (1 + \alpha \sigma\_0) \frac{\partial^2 u\_{0,1}}{\partial y^2} \right) \tag{21}$$

$$\frac{1}{S\_{\varepsilon}} \frac{\partial^2 \sigma\_0}{\partial y^2} = 0 \tag{22}$$

Along with the boundary conditions:

$$
\mu\_{0,1}(h\_1) = -1, \ u\_{0,1}(h\_2) = -1, \ \sigma\_0(h\_1) = 0, \ \sigma\_0(h\_2) = 1 \tag{23}
$$

• System of Order One

$$\begin{array}{ll} K\_{1,1} = & \frac{1}{2} \left( \mathfrak{a} \frac{\partial \sigma\_{0}}{\partial y} \frac{\partial u\_{1,1}}{\partial y} + (1 + \mathfrak{a}\sigma\_{0}) \frac{\partial^{2} u\_{1,1}}{\partial y^{2}} \right) + \frac{\mathfrak{a}\mathfrak{a}}{2} \frac{\partial \sigma\_{0}}{\partial y} \left( \frac{\partial u\_{0,1}}{\partial y} \right)^{3} \\ & + \frac{3\mathfrak{a}}{2} (1 + \mathfrak{a}\sigma\_{0}) \left( \frac{\partial u\_{0,1}}{\partial y} \right)^{2} \frac{\partial^{2} u\_{0,1}}{\partial y^{2}} \end{array} \tag{24}$$

$$\frac{1}{S\_{\text{c}}} \frac{\partial^2 \sigma\_1}{\partial y^2} = 0 \tag{25}$$

and the boundary conditions:

$$
\mu\_{1,1}(h\_1) = 0, \; \mu\_{1,1}(h\_2) = 0, \; \sigma\_1(h\_1) = 0, \; \sigma\_1(h\_2) = 0. \tag{26}
$$

Obtaining the solutions of the above sets of equations by making use of mathematical software Mathematica 7.0, we have the following results:

*Coatings* **2018**, *8*, 407

• Zeroth Order Solution

$$u\_{0,1} = \frac{1}{a(\ln(h\_2 - h\_1) - a\ln(h\_2 - h\_1)(1+a))} \begin{bmatrix} (2(h\_1 - h\_2)\,\, \mathcal{K}\_{0,1}(h\_2 - y) - a)\,\ln(h\_2 - h\_1) \\ -(2(-h\_2 + h\_1)\,\, \mathcal{K}\_{0,1}(h\_1 - y) - a)(a+1)\,\ln(h\_2 - h\_1) \\ +2(h\_1 - h\_2)^2 \,\mathcal{K}\_{0,1}\,\ln(h\_2 + ya - h\_1(1+a)) \end{bmatrix} \tag{27}$$
 
$$\sigma\_0 = \frac{1 - y + a\cos(2\pi x)}{1 + d + a\cos(2\pi x) + b\cos(2\pi x + \varphi)}\tag{28}$$
