We illustrate the accuracy of OAFM by comparing obtained approximate solutions with the numerical integration results obtained by means of a fourth-order Runge-Kutta method in combination with the shooting method. In all cases, the unknown parameters are optimally identified via Galerkin method. For this, we use the following eleven weighted functions *f*_{i} given by [41]:

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}{f}_{1}(\eta )=\gamma {e}^{-3K\eta}+{\eta}^{2}{e}^{-2K\eta}+\delta \eta {e}^{-K\eta},}\\ {\displaystyle \text{\hspace{0.17em}}{f}_{2}(\eta )=\eta {e}^{-K\eta},\phantom{\rule{1em}{0ex}}{f}_{3}(\eta )={\eta}^{2}{e}^{-K\eta},}\\ {\displaystyle \text{\hspace{0.17em}}{f}_{4}(\eta )=1+\gamma \eta {e}^{-K\eta},\phantom{\rule{1em}{0ex}}{f}_{5}(\eta )={e}^{-2K\eta},}\\ {\displaystyle \text{\hspace{0.17em}}{f}_{6}(\eta )={e}^{-K\eta},\phantom{\rule{1em}{0ex}}{f}_{7}(\eta )=\eta {e}^{-4K\eta},\phantom{\rule{1em}{0ex}}{f}_{8}(\eta )={e}^{-4K\eta},}\\ {\displaystyle \text{\hspace{0.17em}}{f}_{9}(\eta )=\delta \eta {e}^{-2K\eta}+{\eta}^{3}{e}^{-2K\eta},}\\ {\displaystyle {f}_{10}(\eta )={e}^{-K\eta}+{K}_{1}{e}^{-2K\eta}+{K}_{3}\eta {e}^{-3K\eta},}\\ {\displaystyle {f}_{11}(\eta )=\eta {e}^{-K\eta}+{K}_{2}{e}^{-3K\eta}+{K}_{4}{e}^{-4K\eta},}\end{array}$$(44)

where *γ*, *δ*, *K*_{1}, *K*_{2}, *K*_{3} and *K*_{4} are unknown parameters.

The parameters *K*, *K*_{1}, *K*_{2}, *K*_{3}, *K*_{4}, *γ*, *δ*, *C*_{1}, …, *C*_{6} are determined from equations (Galerkin method):

$$\begin{array}{}{\displaystyle {J}_{j}=\underset{0}{\overset{\mathrm{\infty}}{\int}}R(m,\lambda ,\alpha ,\eta ){f}_{j}(\eta )\phantom{\rule{thinmathspace}{0ex}}d\eta =0,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,...,6,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j=1,\phantom{\rule{thinmathspace}{0ex}}...,11,}\end{array}$$(45)

where the residual *R*(*m*, *α*, *λ*, *η*) is given by Eq. (23):

$$\begin{array}{}{\displaystyle R(m,\lambda ,\alpha ,\eta )={\overline{\mathit{\Phi}}}^{\u2034}(\eta )+m\overline{\mathit{\Phi}}(\eta ){\overline{\mathit{\Phi}}}^{\u2033}(\eta )-{\overline{\mathit{\Phi}}}^{\prime}(\eta {)}^{2}}\end{array}$$(46)

and *Φ*(*η*) is given by Eq. (15) with the initial / boundary conditions (10)-(12).

#### Example 5.1

Consider planar stretching case with impermeable sheet, *m* = 1, *α* = 0 and *λ* = 1. In this case, from Eqs. (45) and (46) we obtain

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=0.4631238249,\phantom{\rule{1em}{0ex}}\gamma =0.6858214854,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-0.1052314539,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=2.13\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.21\cdot {10}^{-6},}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}{C}_{1}=-0.0119658357,\phantom{\rule{1em}{0ex}}{C}_{2}=0.0007396024,}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}{C}_{3}=0.0642359221,\phantom{\rule{1em}{0ex}}{C}_{4}=0.0326512044,}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}{C}_{5}=0.0530417312,\phantom{\rule{1em}{0ex}}{C}_{6}=\mathrm{0.0102396495.}}\end{array}$$

The first-order approximate solution given by Eq. (15) can be written in the form

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.7549045180+(-0.0011830813{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0280632559\eta -0.1942637382){e}^{-0.4631238249\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-(0.0821765397\eta +0.5165462911){e}^{-0.9262476499\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0440894886{e}^{-1.3893714748\eta}.}\end{array}$$(47)

#### Example 5.2

In the second case for the same planar stretching case with impermeable sheet *m* = 1, *α* = 0 but *λ* = 5, from Eqs. (45) the values of the parameters are

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=0.5252370049,\phantom{\rule{1em}{0ex}}\gamma =4.8796805607,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-1.2890550305,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=1.24\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=1.02\cdot {10}^{-6},}\\ {\displaystyle {C}_{1}\text{\hspace{0.17em}}=-0.0279185716,\phantom{\rule{1em}{0ex}}{C}_{2}=-1.9918945423\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=6.2493581461\cdot {10}^{-12},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{4}=3.90211197551\cdot {10}^{-12},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-9.621041312\cdot {10}^{-12},\phantom{\rule{1em}{0ex}}{C}_{6}=0.}\end{array}$$

Approximate solution (15) becomes:

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.5251657155}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-1.3352013705\cdot {10}^{-9}{\eta}^{2}-0.0000374387\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.5251657155){e}^{-0.5252370049\eta}}\\ \phantom{\rule{thinmathspace}{0ex}}{\displaystyle +(6.7324778915\cdot {10}^{-12}\eta +1.4853702952\cdot {10}^{-11})}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\times {e}^{-1.0504740099\eta}+(9.1284670601\cdot {10}^{-13}\eta}\\ \phantom{\rule{thinmathspace}{0ex}}{\displaystyle +5.7165228355\cdot {10}^{-12}){e}^{-1.5757110148\eta}.}\end{array}$$(48)

#### Example 5.3

In the third case for the planar stretching case with impermeable sheet *m* = 1, *α* = 0 and with sleep parameter *λ* = 10, from Eqs. (45) we obtain

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=0.4331821853,\phantom{\rule{1em}{0ex}}\gamma =3.0639340595,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-0.5722383481,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=1.03\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=2.11\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=-0.0125089730,\phantom{\rule{1em}{0ex}}{C}_{2}=-9.6275057188\cdot {10}^{-7},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=6.6010897191\cdot {10}^{-12},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{4}=-3.360346796\cdot {10}^{-12},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-1.0095931161\cdot {10}^{-11},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{6}=-2.0561054881\cdot {10}^{-12}.}\end{array}$$

The first-order approximate solution (15) is

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.4331053011}\\ {\displaystyle +(-1.2809679965\cdot {10}^{-9}{\eta}^{2}-0.0000332989\eta}\\ {\displaystyle -0.4331053011){e}^{-0.433182185\eta}}\\ {\displaystyle +(1.0336892501\cdot {10}^{-11}\eta +2.7423124117\cdot {10}^{-11})}\\ {\displaystyle \times {e}^{-0.8663643706\eta}+(+0.932245039\cdot {10}^{-12}\eta}\\ {\displaystyle +1.065716162\cdot {10}^{-11}){e}^{-1.2995465559\eta}.}\end{array}$$(49)

The values of *Φ*″(0) and *Φ*(∞) are given in for the same planar stretching case with impermeable sheet, calculated by means OAFM, and by numerical integration.

Table 1 Values of *Φ*″(0) and *Φ*(∞) in the case of planar stretching case with impermeable sheet *m* = 1, *α* = 0

#### Example 5.4

In this case, we consider planar stretching case but suction sheet, *m* = 1, *α* = 3 and with slip parameter *λ* = 1. The parameters obtained by means of Eqs. (45) are:

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.0265261151,\phantom{\rule{1em}{0ex}}\gamma =3.011278740,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-0.9721507438,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=1.002\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.52\cdot {10}^{-6},}\\ {\displaystyle {C}_{1}=2.6781920761\cdot {10}^{-11},}\\ {\displaystyle {C}_{2}=-4.2525109956\cdot {10}^{-12},}\\ {\displaystyle {C}_{3}=-4.655928821\cdot {10}^{-9},}\\ {\displaystyle {C}_{4}=1.6964107211\cdot {10}^{-9},}\\ {\displaystyle {C}_{5}=-6.9684341491\cdot {10}^{-10},}\\ {\displaystyle {C}_{6}=0.0001204043}\end{array}$$

and the first-order approximate solution (15) becomes

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0795956234}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-2.3525169361\cdot {10}^{-12}{\eta}^{2}+2.0134806302\cdot {10}^{-11}\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-4.9211079655\cdot {10}^{-11})\times {e}^{-1.0265261151\eta}}\\ +(8.7180902556\cdot {10}^{-11}\eta \\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-6.9432001142\cdot {10}^{-11}){e}^{-2.0530522302\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(1.3753249143\cdot {10}^{-6}\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0795955933){e}^{-3.0795783454\eta .}}\end{array}$$(50)

#### Example 5.5

For planar stretching case and suction sheet, *m* = 1, *α* = 3 and slip parameter *λ* = 5, we have

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.0016112565,\phantom{\rule{1em}{0ex}}\gamma =2.9079023283,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-0.9038557714,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=0.55\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.62\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=1.3658235079\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{2}=-2.1653508962\cdot {10}^{-7},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=-0.0005801616156,\phantom{\rule{1em}{0ex}}{C}_{4}=0.0002087511,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.9806268698,\phantom{\rule{1em}{0ex}}{C}_{6}=0.0158334825}\end{array}$$

and therefore, the first-order approximate solution (15) has the form

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0205594824}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-5.4255801019\cdot {10}^{-7}{\eta}^{2}+4.6777709040\cdot {10}^{-6}\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0000118225){e}^{-1.0016112565\eta}+(0.0000200454\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0000156837){e}^{-2.0032225131\eta}+(0.0003400695\eta}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0205318961){e}^{-3.0048337696\eta}.}\end{array}$$(51)

#### Example 5.6

The planar stretching case and suction sheet, *m* = 1, *α* = 3 but slip parameter *λ* = 10 we obtain from Eqs. (45):

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.1395297367,\phantom{\rule{1em}{0ex}}\gamma =2.9650196892,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-0.9588570845,\phantom{\rule{1em}{0ex}}{K}_{1}=0,\phantom{\rule{1em}{0ex}}{K}_{2}=0,}\\ {\displaystyle {K}_{3}=1.002\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=1.32\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=0.0000646091,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0000113762,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=-0.0863263871,\phantom{\rule{1em}{0ex}}{C}_{4}=0.02944192315,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.2391591505,\phantom{\rule{1em}{0ex}}{C}_{6}=-0.06102791273}\end{array}$$

and the first-order approximate solution (15) can be written as

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0106792212+(-0.0000661253{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0005189796\eta -0.0011722736){e}^{-1.1395297367\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0027526798\eta -0.0032398505){e}^{-2.2790594735\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-0.0012678992\eta -0.0062671170){e}^{-3.4185892103\eta}.}\end{array}$$(52)

In , we present the values of *Φ*″(0) and *Φ*(∞) in the case of planar stretching case with suction sheet.

Table 2 Values of *Φ*″(0) and *Φ*(∞) in the case of planar stretching case with suction sheet *m* = 1, *α* = 3

#### Example 5.7

Now, we consider the axisymmetric flow with impermeable sheet, *m* = 2, *α* = 0 and with slip parameter *λ* = 1. From Eqs. (45) we obtain the following set of parameters

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.2284861855,\phantom{\rule{1em}{0ex}}\gamma =-1.2107679860,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =5.9462560050,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=0.7335772195,}\\ {K}_{3}=1.11\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=1.06\cdot {10}^{-6},\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=-0.0272717347,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0098884303,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=-0.0130978383,\phantom{\rule{1em}{0ex}}{C}_{4}=-0.1950919615,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.06294893217,\phantom{\rule{1em}{0ex}}{C}_{6}=\mathrm{0.0018821549.}}\end{array}$$

In this case, the first-order approximate solution (15) is

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.5509446955+(-0.0077420177{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0679124177\eta -0.6421304231){e}^{-1.2284861855\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0469289624\eta +0.0831404603){e}^{-2.4569723711\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0041898146\eta +0.0075368933){e}^{-3.6854585567\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0000154947\eta +0.0005083699){e}^{-4.9139447423\eta}.}\end{array}$$(53)

#### Example 5.8

For the case of the axisymmetric flow with impermeable sheet, *m* = 2, *α* = 0 and with slip parameter *λ* = 5, from Eqs. (45) we have

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=0.8408529662,\phantom{\rule{1em}{0ex}}\gamma =-2.5936764722,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =12.6879512489,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=0.5953831575,}\\ {\displaystyle {K}_{3}=2.03\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=1.01\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=-0.0053548861,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0012448162,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=-0.00821412956,\phantom{\rule{1em}{0ex}}{C}_{4}=-0.03031382709,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.01091079627,\phantom{\rule{1em}{0ex}}{C}_{6}=0.00176755942}\end{array}$$

and therefore, the first-order approximate solution (15) becomes

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.3780173693+(-0.0023588678{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}-0.0315158198\eta -0.4396463038){e}^{-0.8408529662\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0208604015\eta +0.0573797361){e}^{-1.6817059324\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.001423055078\eta +0.0040581045){e}^{-2.5225588986\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-0.0000368811\eta +0.0001911237){e}^{-3.3634118648\eta}.}\end{array}$$(54)

#### Example 5.9

In this case we consider the axisymmetric flow with impermeable sheet, *m* = 2, *α* = 0 and with slip parameter *λ* = 10. From Eqs. (45) yields

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=0.5239724222,\phantom{\rule{1em}{0ex}}\gamma =1.6648535124,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =-1.8794890289,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=-3.4855129921,}\\ {\displaystyle {K}_{3}=0.43\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.31\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=0.0112870692,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0004699006,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=0.0366309932,\phantom{\rule{1em}{0ex}}{C}_{4}=-0.0059152635,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.0310033957,\phantom{\rule{1em}{0ex}}{C}_{6}=-\mathrm{0.00815982977.}}\end{array}$$

The first-order approximate solution can be written in the form

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=0.3106373548+(-0.0006102510{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0246579539\eta -0.3170175340){e}^{-0.5239724222\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-0.0247341994\eta +0.0442009306){e}^{-1.0479448445\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-0.0157948648\eta -0.0156201635){e}^{-1.5719172668\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(-0.0041191012\eta -0.0222006138){e}^{-2.095889689\eta}.}\end{array}$$(55)

In we present the values of *Φ*″(0) and *Φ*(∞) for the stretching flow with impermeable sheet.

Table 3 Values of *Φ*″(0) and *Φ*(∞) in the case of stretching flow with impermeable sheet *m* = 2, *α* = 0

#### Example 5.10

For the axisymmetric flow with suction sheet *m* = 2, *α* = 3 and slip parameter *λ* = 1 we obtain the following values of the parameters:

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.5574274146,\phantom{\rule{1em}{0ex}}\gamma =-0.7860056759,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =3.6690514687,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=-0.0507748232,}\\ {\displaystyle {K}_{3}=1.003\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.32\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=0.0107133300,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0024557002,}\\ {\displaystyle {C}_{3}=-9.2340545611,\phantom{\rule{1em}{0ex}}{C}_{4}=4.93780472001,}\\ {\displaystyle {C}_{5}=-2.72171778619,\phantom{\rule{1em}{0ex}}{C}_{6}=23.18498897742928,}\end{array}$$

with the first-order approximate solution obtained from (15):

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0235364880+(0.0026339287{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0162169332\eta -0.0279371463){e}^{-1.5574274146\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0864947892\eta -0.0510827083){e}^{-3.1148548292\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0937682076\eta +0.0527720186){e}^{-4.6722822438\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0064922112\eta +0.0027118480){e}^{-6.2297096585\eta}.}\end{array}$$(56)

#### Example 5.11

For the case of the axisymmetric flow with suction sheet *m* = 2, *α* = 3 and slip parameter *λ* = 5, the parameters are

$$\begin{array}{}\text{\hspace{0.17em}}K=1.7780169491,\phantom{\rule{1em}{0ex}}\gamma =-0.5957717503,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =2.8222930255,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=-0.0131160498,\\ {K}_{3}=1.013\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=1.22\cdot {10}^{-6},\\ \text{\hspace{0.17em}}{C}_{1}=0.0000862400,\phantom{\rule{1em}{0ex}}{C}_{2}=-0.0000222658,\\ \text{\hspace{0.17em}}{C}_{3}=-0.3512786202,\phantom{\rule{1em}{0ex}}{C}_{4}=0.1994379874,\\ \text{\hspace{0.17em}}{C}_{5}=-1.6374819712,\phantom{\rule{1em}{0ex}}{C}_{6}=1.6614059805\end{array}$$

and therefore, the first-order approximate solution (15) has the form

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0053612258+(-0.0001065840{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0005858601\eta -0.0008994531){e}^{-1.7780169491\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0035222493\eta -0.0022439786){e}^{-3.55603389829\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0065444365\eta -0.0021759449){e}^{-5.3340508474\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0000797662\eta -0.0000397491){e}^{-7.1120677965\eta}.}\end{array}$$(57)

#### Example 5.12

In the last case we consider the axisymmetric flow with suction sheet *m* = 2, *α* = 3 and slip parameter *λ* = 10, such that the parameters are

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}K=1.8412120648,\phantom{\rule{1em}{0ex}}\gamma =-0.554362126,}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =2.6330245292,\phantom{\rule{1em}{0ex}}{K}_{1}={K}_{2}=-0.0066382243,}\\ {\displaystyle {K}_{3}=1.004\cdot {10}^{-6},\phantom{\rule{1em}{0ex}}{K}_{4}=0.52\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{1}=9.5831454966\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{2}=-2.5525898638\cdot {10}^{-6},}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{3}=-0.0821658733,\phantom{\rule{1em}{0ex}}{C}_{4}=0.04718459612,}\\ {\displaystyle \text{\hspace{0.17em}}{C}_{5}=-0.8534171847,\phantom{\rule{1em}{0ex}}{C}_{6}=\mathrm{0.5288479009.}}\end{array}$$

The first-order approximate solution (15) can be written as

$$\begin{array}{}{\displaystyle \overline{\mathit{\Phi}}(\eta )=3.0027282856+(-0.0000243991{\eta}^{2}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+0.0001301957\eta -0.0001939768){e}^{-1.8412120648\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0008131418\eta -0.0005329181){e}^{-3.6824241297\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0020278091\eta -0.0019878153){e}^{-5.5236361946\eta}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}+(0.0000116973\eta -0.0000135853){e}^{-7.3648482595\eta}.}\end{array}$$(58)

In are presented the values of *Φ*″(0) and *Φ*(∞) for stretching flow with suction sheet.

Table 4 Values of *Φ*′(0) and Φ(∞) in the case of stretching flow with suction sheet *m* = 2, *α* = 3

From - we deduced that there exist an excellent agreement between the numerical results and the results obtained by means of OAFM.

On the other hand, considering the effect of slip parameter on the velocity *Φ*(*η*) in both flows, Figures 1-4 have been displayed. Figure 1 shows the variation of *Φ* for planar flow and impermeable sheet. In Figure 2 the variation of *Φ* for planar flow and suction sheet is plotted.

Figure 1 Variation of *Φ* by increasing the slip parameter *λ* for planar flow and impermeable sheet (*m* = 1, *α* = 0): — numerical solution; …… OAFM solution

Figure 2 Variation of *Φ* by increasing the slip parameter *λ* for planar flow and suction sheet (*m* = 1, *α* = 3): — numerical solution; …… OAFM solution

Figure 3 shows the variation of *Φ* for axisymmetric flow and impermeable sheet. Figure 4 shows the variation of *Φ* for axisymmetric flow and suction sheet. It is clear that the velocity components decrease with an increase in the slip parameter for all cases.

Figure 3 Variation of *Φ* by increasing the slip parameter *λ* for axisymmetric flow and impermeable sheet (*m* = 2, *α* = 0): — numerical solution; …… OAFM solution

Figure 4 Variation of *Φ* by increasing the slip parameter *λ* for axisymmetric flow and suction sheet (*m* = 2, *α* = 3): — numerical solution; …… OAFM solution

In Figures 5-7 the planar cases for every value of slip parameter *λ* have been plotted and in Figures 8-10 the stretching cases for different values of *λ* have been plotted. It is evident that the velocity is less for the axisymmetric flow when compared with the planar case.

Figure 5 Variation of *Φ* by increasing the coefficient *α* for planar flow (*m* = 1, *λ* = 1): — numerical solution; …… OAFM solution

Figure 6 Variation of *Φ* by increasing the coefficient *α* for planar flow (*m* = 1, *λ* = 5):
— numerical solution; …… OAFM solution

Figure 7 Variation of *Φ* by increasing the coefficient *α* for planar flow (*m* = 1, *λ* = 10): — numerical solution; …… OAFM solution

Figure 8 Variation of *Φ* by increasing the coefficient *α* for axisymmetric flow (*m* = 2, *λ* = 1): — numerical solution; …… OAFM solution

Figure 9 Variation of *Φ* by increasing the coefficient *α* for axisymmetric flow (*m* = 2, *λ* = 5): — numerical solution; …… OAFM solution

Figure 10 Variation of *Φ* by increasing the coefficient *α* for axisymmetric flow (*m* = 2, *λ* = 10):
— numerical solution; …… OAFM solution

Finally, the residual functions obtained for the approximate analytic solutions given by Eqs. (47)-(58) are plotted in Figures 11-22.

Figure 11 The residual *R*(1, 1, 0, *η*) for Eq. (47) obtained by OAFM

Figure 12 The residual *R*(1, 5, 0, *η*) for Eq. (48) obtained by OAFM

Figure 13
The residual *R*(1, 10, 0, *η*) for Eq. (49) obtained by OAFM

Figure 14 The residual *R*(1, 1, 3, *η*) for Eq. (50) obtained by OAFM

Figure 15 The residual *R*(1, 5, 3, *η*) for Eq. (51) obtained by OAFM

Figure 16 The residual *R*(1, 10, 3, *η*) for Eq. (52) obtained by OAFM

Figure 17 The residual *R*(2, 1, 0, *η*) for Eq. (53) obtained by OAFM

Figure 18 The residual *R*(2, 5, 0, *η*) for Eq. (54) obtained by OAFM

Figure 19 The residual *R*(2, 10, 0, *η*) for Eq. (55) obtained by OAFM

Figure 20 The residual *R*(2, 1, 3, *η*) for Eq. (56) obtained by OAFM

Figure 21 The residual *R*(2, 5, 3, *η*) for Eq. (57) obtained by OAFM

Figure 22 The residual *R*(2, 10, 3, *η*) for Eq. (58) obtained by OAFM

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