## 2 Materials and Methods
### Methodology
\right]=0 \tag{26}\]  
The corresponding boundary conditions are:  
\[\psi=0,\ \ \frac{\partial^{2}\psi}{\partial y^{2}}=0,\ \ \ \text{at}\ y=0\]  
\[\psi=q,\ \ \frac{\partial\psi}{\partial y}=u_{0},\ \ \ \text{at}\ y=h \tag{27}\]  
One decomposes the stream function \(\psi\) and the flux \(q\) in the following series:  
\[\psi=\psi_{0}+p_{1}\psi_{1}+\dots, \tag{28}\]  
\[q=q_{0}+p_{1}q_{1}+\dots, \tag{29}\]  
One chooses the initial guess:  
\[w_{0}=u_{0}+\frac{\left(y^{2}-h^{2}\right)}{2}\frac{dp_{0}}{dx} \tag{30}\]  
And, making use of Equations (28)-(30) in Equations (26) and (27) and the same powers of \(p_{1}\) on both sides, one obtains the following equations:  
\[p_{1}{}^{0}:\ \frac{\partial^{4}\psi_{0}}{\partial y^{4}}-\left(M^{2}+\frac{1}{K} \right)\frac{\partial^{2}\psi_{0}}{\partial y^{2}}=\frac{\partial^{4}w_{0}}{ \partial y^{4}}-\left(M^{2}+\frac{1}{K}\right)\frac{\partial^{2}w_{0}}{ \partial y^{2}} \tag{31}\]  
with the boundary conditions:  
\[\psi_{0}=0,\ \ \ \frac{\partial^{2}\psi_{0}}{\partial y^{2}}=0,\ \ \ \text{at}\ y=0\]\[\psi_{0}=q_{0},\ \ \ \frac{\partial\psi_{0}}{\partial y}=u_{0},\ \ \ \text{at}\ y=h \tag{32}\]  
\[p_{1}{}^{1}:\ \frac{\partial^{4}\psi_{1}}{\partial y^{4}}-\left(M^{2}+\frac{1}{k} \right)\frac{\partial^{2}\psi_{1}}{\partial y^{2}}=Re\beta\Big{(}\frac{ \partial\psi_{0}}{\partial y}\frac{\partial^{3}\psi_{0}}{\partial x\partial y^ {2}}-\frac{\partial\psi_{0}}{\partial x}\frac{\partial^{3}\psi_{0}}{\partial y ^{3}}\Big{)}-\lambda_{1}\beta\Big{(}\frac{\partial\psi_{0}}{\partial y}\frac{ \partial^{5}\psi_{0}}{\partial x\partial y^{4}}-\frac{\partial\psi_{0}}{ \partial x}\frac{\partial^{5}\psi_{0}}{\partial y^{5}}\Big{)}. \tag{33}\]  
with the boundary conditions:  
\[\psi_{1}=0,\ \ \ \frac{\partial^{2}\psi_{1}}{\partial y^{2}}=0,\ \ \ \text{at}\ y=0\]  
\[\psi=q,\ \ \frac{\partial\psi}{\partial y}=u_{0},\ \ \ \text{at}\ y=h \tag{27}\]0  
The solution of Equation (31) under the boundary conditions of Equation (32) can be easily found after using the value of \(\psi_{0}\) in Equation (33), and by solving the resulting equation under the boundary conditions of Equation (34) with the help of \"Mathematica\", one can obtains the solution in the following form:  
\[\psi=q,\ \ \frac{\partial\psi}{\partial y}=u_{0},\ \ \ \text{at}\ y=h \tag{27}\]1  
The series solution obtained by the HPM is convergent, as the coefficients of higher-order solution become very small. It is also noted that when \(\lambda_{1}\to 0\), then only a zero-order system gives a nontrivial solution, and the final result matches with the solution obtained by [PERSON] et al. (2015). Now, the pressure gradient can be found from the following expression:  
\[ \begin{array}{rl}\frac{\partial p}{\partial x}&=Re\beta\left(\frac{ \partial\psi}{\partial x}\frac{\partial^{2}\psi}{\partial y^{2}}-\frac{ \partial\psi}{\partial y}\frac{\partial^{2}\psi}{\partial x\partial y}\right) +\frac{\partial^{3}\psi}{\partial y^{3}}\\ &+\lambda_{1}\beta\Big{(}\frac{\partial\psi}{\partial y}\frac{\partial^{4} \psi}{\partial x\partial y^{3}}-\frac{\partial\psi}{\partial x}\frac{ \partial^{4}\psi}{\partial y^{4}}+\frac{\partial^{2}\psi}{\partial x\partial y }\frac{\partial^{3}\psi}{\partial y^{3}}+\frac{\partial^{3}\psi}{\partial x \partial y^{2}}\frac{\partial^{2}\psi}{\partial y^{2}}\Big{)}\\ &-\left(M^{2}+\frac{1}{K}\right)\left(\frac{\partial\psi}{\partial y}+1 \right)\\ \end{array}