*3.2. Model 2*

The same contrast as described for Model 1 along with solutions are summarized as:

• System of Order Zero

$$K\_{0,2} = \frac{\partial^2 u\_{0,2}}{\partial y^2} \tag{33}$$

• System of Order One

$$K\_{1,2} = \frac{\partial^2 u\_{1,2}}{\partial y^2} - \frac{\alpha}{2} \frac{\partial}{\partial y} \left( \sigma\_0 \left( \frac{\partial u\_{0,2}}{\partial y} \right)^3 \right) \tag{34}$$

• Zeroth Order Solution

$$u\_{0,2} = \frac{1}{2} \left( -2 + h\_1 h\_2 K\_{0,2} - h\_1 K\_{0,2} y - h\_2 K\_{0,2} y + K\_{0,2} y^2 \right) \tag{35}$$

• First Order Solution

The solution of above equation is examined directly and is prescribed as:

$$\begin{array}{ll} \mu\_{1,2} = & \frac{1}{160(h\_1 - h\_2)}(h\_1 - y)(h\_2 - y)(-80h\_2K\_{1,2} + 9h\_1^3K\_{0,2}^3\alpha - h\_2^3K\_{0,2}^3\alpha - \\ & 6h\_2^2K\_{0,2}^3y\alpha + 14h\_2K\_{0,2}^3y^2\alpha - 16K\_{0,2}^3y^3\alpha - h\_1^2K\_{0,2}^3(h\_2 + 26y)\alpha + \\ & h\_1\left(80K\_{1,2} + K\_{0,2}^3\left(9h\_2^2 - 16h\_2y + 34y^2\right)\alpha\right) \end{array} \tag{36}$$

here,

$$\mathbb{K}\_{0,2} = -\frac{12(-1 - d + h\_1 - h\_2 + \mathcal{Q})}{\left(h\_1 - h\_2\right)^3} \tag{37}$$

$$\begin{array}{lcr} \mathbf{K}\_{1,2} = & -\frac{1}{5(h\_1 - h\_2)} 12(-5h\_1^4(1+d-Q) - 5h\_2^4(1+d-Q) + h\_1^3(20h\_2(1+d-Q)) \\ & -27\alpha) + 27h\_2^3\alpha + 81h\_2^2(1+d-Q)\alpha + 81h\_2(1+d-Q)^2\alpha + 27(1+d-Q)^3 \\ & \alpha + h\_1^2(-30h\_2^2(1+d-Q) + 81h\_2\alpha + 81(1+d-Q)\alpha) + h\_1(20h\_2^3(1+d-Q)) \\ & -81h\_2^2\alpha - 162h\_2(1+d-Q)\alpha - 81(1+d-Q)^2\alpha) \end{array} \tag{38}$$

Pressure rise Δ*p* over one wavelength in dimensionless format is obtained by

$$
\Delta p = \int\_0^1 \text{Kdx} \tag{39}
$$

The integral in Equation (41) is evaluated numerically using software package Mathematica 7.0.
