## 2 Formulation
### \"Crocco Variables\" and the Resulting Singularities  
We introduce the following set of variables:  
\[z=f^{\prime}(\eta),\quad h(z)=(f^{\prime\prime}(\eta))^{n}. \tag{9}\]  
These are similar to the classical Crocco variables which are given with new variables defined by \(f(\eta)\) and \(f^{\prime}(\eta)\). The following transformed problem is obtained from (7)-(9)  
\[h^{\prime\prime}(z)=\left(\alpha z-m(z^{2}-1)\frac{h^{\prime}(z)}{h(z)}\right) h^{-1/n}(z),\quad 0<z<1, \tag{10}\]  
subject to  
\[h(1)=0,\quad h(0)=(f^{\prime\prime}(0))^{n},\quad h^{\prime}(0)=-mn/f^{\prime \prime}(0), \tag{11}\]  
and where  
\[\alpha=2 mn-nk=\frac{(3m-1)n}{n+1}. \tag{12}\]  
This transformation proves to be useful in the mathematical analysis of the original problem (7) and (8) and its parameters. It also provides an _alternative means_ to obtain numerical results. This is in contrast to the technique that was used in [6], where the authors discuss the need for some careful treatments in determining the correct convergence for the boundary condition on the open end (\(+\infty\)) for \(n<1\), whereas careful treatment was also needed when \(n>1\), as the solutions reach \(f^{\prime\prime}=0\) identically at finite \(\eta\) (this observation was also proved at the analytical level for a similar problem in [12]). The advantage of using (10) and (11) in obtaining numerical solutions is that we deal with a finite domain and do not have to consider where or when to \"stop\". However, there is a problem that arises here and has to be addressed. The problem is dealing with the singularities of the equation. It is remarked that the singularity at the right end can be dealt with easily, since, as was illustrated in [26], solutions approach the point (1,0) linearly for \(n>1\). On the other hand, the solutions are integrable at \(z=1\) for \(0<n<1\), despite the fact that gradients become infinite. In particular, the criterion for determining the values of \(m\) remains very simple as Runge-Kutta methods (used by Matlab) are effective at that end. On the other hand, however, dealing with the singularity at \(z=0\) could lead to a serious problem on regular computing machines. This becomes more obvious when attempting to solve the ODE for \(f^{\prime\prime}(0)=0\), which is not mathematically possible in (11) above. **Remark 2**. : _Solving the ODE for \(f^{\prime\prime}(0)=0\) is crucial for us, as it corresponds to the cut-off values of \(m\), as was observed in [6; 26], and as is discussed in the introduction. The cut-off values of \(m\) are of much and particular interest in this fluid flow problem. We note that this is also a singularity of the original ODE (7)._  
To circumvent this problem, it is necessary to take values of \(f^{\prime\prime}(0)\) approaching (very close to) zero but the question arises of how small the values can be. Computing machines could then give erroneous and contradictory results when solving ODE (10) and seeking to determine the relevant parameter values that are of interest. These are typical and expected machine errors as smaller values of \(f^{\prime\prime}(0)\) lead to extremely large derivatives, but also, and more importantly, they lead to extremely small values of \(h(0)\), causing a problem that may be more serious for larger values of \(n\) (this may not occur due to the numerical method itself but rather due to a problem of the computing machines themselves). This problem, however, can be overcome as follows: Integrating Equation (10) from \(0\) to \(z>0\) leads to  
\[h^{\prime}(z)=-mn(1-z^{2})h^{-1/n}+\int_{0}^{z}(\alpha-2 mn)\mu h^{-1/n}(\mu)d\mu \tag{13}\]  
Observe that \(h^{\prime}(0)=-mn/f^{\prime\prime}(0)\) so that the initial conditions on both sides of the equation cancel out. Now, let \(\epsilon(z)=\int_{0}^{z}(\alpha-2 mn)\mu h^{-1/n}(\mu)d\mu\) represent the error in the computation of \(h^{\prime}(z)\) in (13) above.