Homotopy analysis method for the solution of lubrication of a long porous slider

If q ∈ [0,1] then the zeroth order deformation problem can be constructed as

(1−q)Lh[h(η,q)−h0(η)]=qℏhℵh[h(η,q),f(η,q),g(η,q)],(1−q)Lf[f(η,q)−f0(η)]=qℏfℵf[h(η,q),f(η,q),g(η,q)],(1−q)Lg[g(η,q)−g0(η)]=qℏgℵg[h(η,q),f(η,q),g(η,q)].}$$\begin{array}{} \displaystyle \label{eq10} \left.\begin{matrix} (1-q)L_{h}[h(\eta,q)-h_{0}(\eta)] & = & q\hbar_{h}\aleph_{h}[h(\eta,q),f(\eta,q),g(\eta,q)],\\ (1-q)L_{f}[f(\eta,q)-f_{0}(\eta)] & = & q\hbar_{f}\aleph_{f}[h(\eta,q),f(\eta,q),g(\eta,q)],\\ (1-q)L_{g}[g(\eta,q)-g_{0}(\eta)] & = & q\hbar_{g}\aleph_{g}[h(\eta,q),f(\eta,q),g(\eta,q)]. \end{matrix}\right\} \end{array}$$(7)

The boundary conditions becomes

h(0,q)=0,h(1,q)=1,h′(0,q)=0,h′(1,q)=0,f(0,q)=0,f(1,q)=0,g(0,q)=1,g(1,q)=0,}$$\begin{array}{} \displaystyle \left.\begin{matrix} h(0,q)=0, h(1,q)=1, & {h}'(0,q)=0, & {h}'(1,q)=0,\\ f(0,q)=0, f(1,q)=0, & g(0,q)=1, & g(1,q)=0, \end{matrix}\right\} \end{array}$$

where q ∈ [0,1] is an embedding parameter. Here ħ_h,ħ_f and ħ_g are non-zero auxiliary parameters. Further ℵ is the non-linear differential operator and is given by

ℵh[h(η,q)]=∂4h(η,q)∂η4−R(∂h(η,q)∂η∂2h(η,q)∂η2−h(η,q)∂3h(η,q)∂η3),ℵf[h(η,q),f(η,q)]=∂2f(η,q)∂η2−R(f(η,q)∂h(η,q)∂η−h(η,q)∂f(η,q)∂η),ℵg[h(η,q),g(η,q)]=∂2g(η,q)∂η2+Rh(η,q)∂g(η,q)∂η.}$$\begin{array}{} \displaystyle \left.\begin{matrix} \aleph_{h}[h(\eta,q)] & = & \frac{\partial^{4}h(\eta,q)}{\partial \eta^{4}}-R(\frac{\partial h(\eta,q)}{\partial \eta}\frac{\partial^{2} h(\eta,q)}{\partial \eta^{2}}-h(\eta,q)\frac{\partial^{3} h(\eta,q)}{\partial \eta^{3}}),\\ \aleph_{f}[h(\eta,q),f(\eta,q)] & = & \frac{\partial^{2}f(\eta,q)}{\partial \eta^{2}}-R(f(\eta,q)\frac{\partial h(\eta,q)}{\partial \eta}-h(\eta,q)\frac{\partial f(\eta,q)}{\partial \eta}),\\ \aleph_{g}[h(\eta,q),g(\eta,q)] & = & \frac{\partial^{2}g(\eta,q)}{\partial \eta^{2}}+Rh(\eta,q)\frac{\partial g(\eta,q)}{\partial \eta}. \end{matrix}\right\} \end{array}$$

For q = 0 and q = 1, Eqn. (7) have the solutions

h(η,0)=h0(η)h(η,1)=h(η),f(η,0)=f0(η)f(η,1)=f(η),g(η,0)=g0(η)g(η,1)=g(η).}$$\begin{array}{} \displaystyle \label{eq13} \left.\begin{matrix} h(\eta,0)=h_{0}(\eta) & \quad \quad & h(\eta,1)=h(\eta),\\ f(\eta,0)=f_{0}(\eta) & \quad \quad & f(\eta,1)=f(\eta),\\ g(\eta,0)=g_{0}(\eta) & \quad & g(\eta,1)=g(\eta). \end{matrix}\right\} \end{array}$$(8)

As q vary from 0 to 1, h(η,q), f(η,q),g(η,q) also vary from the initial guesses h₀(η), f₀(η),g₀(η) to the final solutions h(η), f (η),g(η). With the help of Taylor’s theorem, Eqn. (8) can be written as

h(η,q)=h0(η)+∑m=1∞hm(η)qm,f(η,q)=f0(η)+∑m=1∞fm(η)qm,g(η,q)=g0(η)+∑m=1∞gm(η)qm,}$$\begin{array}{} \left. {\begin{array}{*{20}{c}} {h(\eta ,q)}& = &{{h_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{h_m}} (\eta ){q^m},}\\ {f(\eta ,q)}& = &{{f_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{f_m}} (\eta ){q^m},}\\ {g(\eta ,q)}& = &{{g_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{g_m}} (\eta ){q^m},} \end{array}} \right\} \end{array}$$(9)

where hm(η)=1m!∂mh(η,q)∂qm|q=0$\begin{array}{} \displaystyle {h_m}(\eta ) = \frac{1}{{m!}}\frac{{{\partial ^m}h(\eta ,q)}}{{\partial {q^m}}}{|_{q = 0}} \end{array}$,fm(η)=1m!∂mf(η,q)∂qm|q=0$\begin{array}{} \displaystyle {f_m}(\eta ) = \frac{1}{{m!}}\frac{{{\partial ^m}f(\eta ,q)}}{{\partial {q^m}}}{|_{q = 0}} \end{array}$, gm(η)=1m!∂mg(η,q)∂qm|q=0$\begin{array}{} \displaystyle {g_m}(\eta ) = \frac{1}{{m!}}\frac{{{\partial ^m}g(\eta ,q)}}{{\partial {q^m}}}{|_{q = 0}} \end{array}$. The convergence of the above series (14) depends on the auxiliary parameters ħ_h, ħ_f and ħ_g. In order to select the values of ħ_h, ħ_f and ħ_g in such a way that the series (14) is convergent at q = 1, we have

h(η,q)=h0(η)+∑m=1∞hm(η),f(η,q)=f0(η)+∑m=1∞fm(η),g(η,q)=g0(η)+∑m=1∞gm(η).}$$\begin{array}{} \displaystyle \left. {\begin{array}{*{20}{c}} {h(\eta ,q)}& = &{{h_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{h_m}} (\eta ),}\\ {f(\eta ,q)}& = &{{f_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{f_m}} (\eta ),}\\ {g(\eta ,q)}& = &{{g_0}(\eta ) + \sum\nolimits_{m = 1}^\infty {{g_m}} (\eta ).} \end{array}} \right\} \end{array}$$(10)

Differentiating the zeroth order deformation problem (7) m times with respect to the embedding parameter q and then dividing by m!, finally setting q = 0. The resulting mth-order deformation problem becomes

Lh[hm(η)−χmhm−1(η)]=ℏhℜmh(η),Lf[fm(η)−χmfm−1(η)]=ℏfℜmf(η),Lg[gm(η)−χmgm−1(η)]=ℏgℜmg(η).}$$\begin{array}{} \displaystyle \label{eq16} \left.\begin{matrix} L_{h}[h_{m}(\eta)-\chi_{m}h_{m-1}(\eta)] & = & \hbar_{h}\Re_{m}^{h}(\eta),\\ L_{f}[f_{m}(\eta)-\chi_{m}f_{m-1}(\eta)] & = & \hbar_{f}\Re_{m}^{f}(\eta),\\ L_{g}[g_{m}(\eta)-\chi_{m}g_{m-1}(\eta)] & = & \hbar_{g}\Re_{m}^{g}(\eta). \end{matrix}\right\} \end{array}$$(11)

The homogeneous boundary conditions are

hm(0,q)=0,hm(1,q)=0,hm′(0,q)=0,hm′(1,q)=0,fm(0,q)=0,fm(1,q)=0,gm(0,q)=0,gm(1,q)=0,}$$\begin{array}{} \displaystyle \left.\begin{matrix} h_{m}(0,q)=0, h_{m}(1,q)=0, & h_{m}^{'}(0,q)=0, & h_{m}^{'}(1,q)=0,\\ f_{m}(0,q)=0, f_{m}(1,q)=0, & g_{m}(0,q)=0, & g_{m}(1,q)=0, \end{matrix}\right\} \end{array}$$(12)

where

ℜmh(η)=hm−1⁗+R∑n=0m−1[hnhm−1−n‴−hn′hm−1−n″],ℜmf(η)=fm−1″+R∑n=0m−1[hnfm−1−n′−fnhm−1−n′],ℜmg(η)=gm−1″+R∑n=0m−1[hngm−1−n′],$$\begin{array}{} \left.\begin{matrix} \Re_{m}^{h}(\eta) & = & h_{m-1}^{''''}+R\sum\nolimits_{n=0}^{m-1}[h_{n}h_{m-1-n}^{'''}-h_{n}^{'}h_{m-1-n}^{''}],\\ \Re_{m}^{f}(\eta) & = & f_{m-1}^{''}+R\sum\nolimits_{n=0}^{m-1}[h_{n}f_{m-1-n}^{'}-f_{n}h_{m-1-n}^{'}],\\ \Re_{m}^{g}(\eta) & = & g_{m-1}^{''}+R\sum\nolimits_{n=0}^{m-1}[h_{n}g_{m-1-n}^{'}], \end{matrix}\right\} \end{array}$$

and

χm={0,m≤1;1,m>1;$$\begin{array}{} \displaystyle {\chi _m} = \left\{ \begin{array}{*{20}{c}} {0,\; {\rm{ }}m \le 1;}\\ {1,\; {\rm{ }}m > 1;} \end{array}\right. \end{array}$$

We use Mathematica to solve the linear system of equations (11) with the appropriate homogeneous boundary conditions (12) up to first few orders of approximations for the series h, f and g

h1=ℏhR(−835η2+2770η3−310η5+15η6−235η7),h2=1646800ℏhR(−1+η)2(−9240(16+5η−6η2+4η3)+ℏh(−9240(16+5η−6η2+4η3)+R(−761−7380η−13999η3−2905η4−504η5−1568η6+448η7))),f1=ℏfR(920η−η3+34η4−15η5,f2=125200ℏfRη(−6(−1+η)(210(9+9η−11η2+4η3)+ℏhR(−52−52η+268η2−217η3−55η4+155η5−85η6+20η7))+ℏf(−1260(−9+20η2−15η3+4η4)+R(1592−2835η3+2268η4−2520η5−2700η6−1485η7+280η8))),g1=ℏgR(320η−14η4+110η5,g2=−125200ℏgRη(−2(630(3−5η3+2η4)+ℏhR(−52+240η3−243η4+90η6−45η7+10η8))+ℏg(−1260(3−5η3+2η4)+R(−8−945η3+378η4+1800η6−1575η7+350η8))).$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{h_1}{\rm{ }}} \hfill & = \hfill & {{\hbar _h}R( - \frac{8}{{35}}{\eta ^2} + \frac{{27}}{{70}}{\eta ^3} - \frac{3}{{10}}{\eta ^5} + \frac{1}{5}{\eta ^6} - \frac{2}{{35}}{\eta ^7}),{\rm{ }}} \hfill \\ {{h_2}{\rm{ }}} \hfill & = \hfill & {\frac{1}{{646800}}{\hbar _h}R{{( - 1 + \eta )}^2}( - 9240(16 + 5\eta - 6{\eta ^2} + 4{\eta ^3}) + {\hbar _h}( - 9240(16 + 5\eta - 6{\eta ^2} + 4{\eta ^3}) + R( - 761} \hfill \\ {} \hfill & - \hfill & {7380\eta - 13999{\eta ^3} - 2905{\eta ^4} - 504{\eta ^5} - 1568{\eta ^6} + 448{\eta ^7}))),} \hfill \\ {{f_1}{\rm{ }}} \hfill & = \hfill & {{\hbar _f}R(\frac{9}{{20}}\eta - {\eta ^3} + \frac{3}{4}{\eta ^4} - \frac{1}{5}{\eta ^5},} \hfill \\ {{f_2}{\rm{ }}} \hfill & = \hfill & {\frac{1}{{25200}}{\hbar _f}R\eta ( - 6( - 1 + \eta )(210(9 + 9\eta - 11{\eta ^2} + 4{\eta ^3}) + {\hbar _h}R( - 52 - 52\eta + 268{\eta ^2} - 217{\eta ^3} - 55{\eta ^4}} \hfill \\ {} \hfill & + \hfill & {155{\eta ^5} - 85{\eta ^6} + 20{\eta ^7})) + {\hbar _f}( - 1260( - 9 + 20{\eta ^2} - 15{\eta ^3} + 4{\eta ^4}) + R(1592 - 2835{\eta ^3} + 2268{\eta ^4}} \hfill \\ {} \hfill & - \hfill & {2520{\eta ^5} - 2700{\eta ^6} - 1485{\eta ^7} + 280{\eta ^8}))),} \hfill \\ {{g_1}} \hfill & = \hfill & {{\hbar _g}R(\frac{3}{{20}}\eta - \frac{1}{4}{\eta ^4} + \frac{1}{{10}}{\eta ^5},} \hfill \\ {{g_2}} \hfill & = \hfill & { - \frac{1}{{25200}}{\hbar _g}R\eta ( - 2(630(3 - 5{\eta ^3} + 2{\eta ^4}) + {\hbar _h}R( - 52 + 240{\eta ^3} - 243{\eta ^4} + 90{\eta ^6} - 45{\eta ^7} + 10{\eta ^8}))} \hfill \\ {} \hfill & + \hfill & {{\hbar _g}( - 1260(3 - 5{\eta ^3} + 2{\eta ^4}) + R( - 8 - 945{\eta ^3} + 378{\eta ^4} + 1800{\eta ^6} - 1575{\eta ^7} + 350{\eta ^8}))).} \hfill \\ \end{array} \end{array}$$(13)

The analytic expressions of h, f and g in terms of series are given in Eqn. (10) contains the auxiliary parameters and the convergence of the series strictly depends upon the value of parameters ħ_h, ħ_f and ħ_g which are called as convergence control parameters. These parameter plays a vital role in determining the convergence region and the rate of approximation for the HAM. For this purpose, we have drawn the line segment of the ħ curves parallel to η-axis. Fig. 2 shows the ħ curves for the series h‴(0), f′(0) and g′(0) for the 15th order of approximations. It is clearly indicates that the admissible ranges of ħ_h, ħ_f and ħ_g are −1.5 ≤ ħ_h ≤ −0.5, −1.5 ≤ ħ_f ≤ −0.25 and −2 ≤ ħ_g≤ −0.25 respectively for R = 1. Our calculation shows that the all the three series converges in the whole region of 0 ≤ η ≤ 1 when ħ_h = ħ_f = ħ_g = −1, other than this ħ values the results may diverge or converges slowly.

Fig. 2

The equation of motion for the lubrication of a long porous slider is governed by coupled nonlinear ODEs (5) together with the boundary conditions (6) is solved by HAM. This section discusses the effects of Reynolds number on velocity profiles, skin friction and lifts and drag coefficients. Table 1 shows the convergence of HAM solutions for different order of approximations; it is observed that the series solution converges at the thirty fifth orders of approximations.

Table 1

Convergence of HAM solutions for different order of approximations for R = 10 when ħ_h = ħ_f = ħ_g = −1.

Further, the homotopy-pade technique is used to accelerate the convergence of series solution, which is more efficient than the traditional Pade technique. The [m,m] homotopy-pade approximations of h‴(0), f′(0) and g′(0) are given in Table 2.

Table 2

The [m,m] homotopy-Pade approximations of h‴(0), f′(0) and g′(0) when R = 10.

The results for velocity profiles and lift and drag coefficients have been presented in the form of graphs and Tables. The effects of the cross flow Reynolds number R on the velocity components of h, f and g are shown in Figures 3-5.

Fig. 3

From the Fig. 3, it is observed that the velocity profiles of h increases with the increasing values of R. The influence of R on f is described in Fig. 4, it shows that velocity profiles are decreasing with increasing values of R. The effect of R on velocity components of g are shown in Fig. 5. Here, the velocity profile depicts the same behaviour as compared to f .

Fig. 4

Fig. 5

The lift and drag can be calculated by the following relations:

L=(ρW4L1L212γ2)(−h′′′(0)R3)LiftDx=ρUWL1L2(−f′(1)R)Draginx−directionDy=ρUWL1L2(−g′(1)R)Draginy−direction$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {\quad \quad \quad \quad \quad \quad L = (\frac{{\rho {W^4}{L_1}{L_2}}}{{12{\gamma ^2}}})(\frac{{ - h^{\prime \prime \prime} (0)}}{{{R^3}}})} & {{\rm{Lift}}} \\ {\quad \quad \quad \quad \quad \quad {D_x} = \rho UW{L_1}{L_2}(\frac{ - f^{\prime} (1)}R)} & {{\rm{Drag\; in}}\;x - {\rm{direction}}} \\ {\quad \quad \quad \quad \quad \quad {D_y} = \rho UW{L_1}{L_2}( \frac{- g^{\prime} (1)}R)} & {{\rm{Drag\; in}}\;y - {\rm{direction}}} \\ \end{array} \end{array}$$

Fig. 6 shows the normalized lift (L) and drag coefficients (D_x and D_y) as a function of R, it is seen that the magnitudes of the lift and drag were decreased with the increase in Reynolds number. In order to see the variations of h‴(0), f′(0) and g′(0) for different values of Reynolds number are given in Table 3.

Fig. 6

Table 3

Comparison of the Homotopy pade approximations (HPA) [40, 40] solution with the numerical solution [2] and long series (LS) [16] for different values of the Reynolds number R.

It is observed that the values of h‴(0), f′(0) and g′(0) decreases with increasing Reynolds number. From the Table 3, we have seen that the HAM results agree very well and have good accuracy as compared to previous literature results [2]- [16]. Also the HAM results are converges for much larger Reynolds number as compared to HPM, ADM and numerical results.

In this article, we describe the analysis of coupled boundary value problem for eighth order nonlinear ordinary differential equation over finite interval arising in three dimensional flow problems. The semi-analytical schemes described here offer advantages over solutions obtained by long series, HPM, ADM and other numerical methods etc. The convergence of the Homotopy analysis method is given. The results are presented in the form of Tables and graphs, also the effects of cross flow Reynolds number is discussed for much larger domain.