Integral Transform Method to Solve the Problem of Porous Slider without Velocity Slip

Abstract

This study is about the lubrication of a long porous slider in which the fluid is injected into the porous bottom. The similarity transformation reduces the Navier-Stokes equations to couple nonlinear, ordinary differential equations, which are solved by a new algorithm. The proposed technique is based on integral transformation. Apparently, there is great symmetry between proposed method and variation iteration method, Adomian decomposition method but in integral transform method all the boundary conditions are applied, then a recursive scheme is used for the analytical solutions, which is unlike the Variational Iteration Method, Adomian Decomposition Method, and other existing analytical methods. Solutions are obtained for much larger Reynolds numbers, and they are compared with analytical and numerical methods. Effects of Reynolds number on velocity components are presented.

1. Introduction

The flow between two plates is an important subject in mathematical physics [1]. Among the various problems, the porous slider is of great importance [2]. A large amount of literature is available related to long porous sliders (LPS). Naeem studied the effects of Reynolds number on a circular porous slider [3]. Rao investigated the effect of permeability and different slip velocities at the porous interface using couple-stress fluids on the slider bearing load carrying mechanism [4]. Kumar analyzed a porous elliptical slider with a semi-analytical technique [5]. Zhu studied multi-scale soft porous lubrication [6]. Lang investigated theoretical and experimental study of transient squeezing flow in a highly porous film [7]. The fluid dynamics in a slider bearing were discussed in a previous study [8]. A porous slider was numerically studied by Wang for large Reynolds numbers [9]. There are many numerical techniques to deal with the highly nonlinear problems related to porous sliders [10]. Numerical techniques are time consuming and need powerful computers, and it is not easy to deal with the stability criteria [11].

To overcome this difficulty, many authors have solved problem of LPS by using different analytical techniques. Khan et. al. used Homotopy Perturbation Method (HPM) to solve the problem [2]. Vishwanath et. al. applied the Homotopy Analysis Method (HAM) [12]. Khan et. al. tackled the same problem using Adomian Decomposition Method (ADM) and presented improved results [13]. These are similar analytical techniques with slight differences. According to some authors Homotopy perturbation is a sub-case of the homotopy analysis method [14]. Comparison between ADM and Variation Iteration Method (VIM) was done in a previous study [15]. VIM is better than ADM according to some authors, and the opposite is true according to others. Researchers are divided between Variational Iteration-I and Variational Iteration-II [16].

Although these methods have their own merits, in cases of highly nonlinear boundary value problems, they require immense calculations to obtain good approximate solutions. Solving two-point BVPs of the form (6) to (9) using ADM, VIM, HAM, or HPM requires a transcendental equation with unknown coefficients, or a sequence of highly nonlinear algebraic equations. Moreover, the unknown coefficients may not be uniquely determined for some cases. This paper proposes an integral transform method that comprises both variational iteration algorithm-II and Adomian decomposition to reduce the computational work. Apparently, the final formulation of the proposed method have great symmetry with the existing methods such as VIM and ADM but this method does not need to calculate the Lagrange multiplier separately and gives direct formulation of VIM-II, which avoids the unnecessary calculations. The ultimate goal of our study is to introduce a new method based on integral transformation to cover the shortcomings of the VIM-I, VIM-II, and ADM for solving nonlinear boundary value problems (6), (7), and (8). The LPS solved by Khan [2] and Khan [13] in 2011 involves unnecessary and repeated calculations.

In this article, a brief history about the solutions given by different authors in different geometries is given. The second section is problem formulation. Formulation and implementation of the integral transform method is presented in Section 3. Results and discussion are given in Section 4. Finally, we conclude our work in Section 5.

2. Problem Formulation

Consider a two dimensional LPS of dimensions $L_1$ and $L_2$ (Figure 1). The fluid is injected through the porous bottom to create a width $\phi$ with velocity $b$ and slider velocities $-U$ (lateral) and $-V$ (longitudinal). The Navier-Stokes equations are:

$$\nabla ·\mathrm{V}=0,$$

(1)

$$\rho \left(\mathrm{V}·\nabla \right)\mathrm{V}=-\nabla p+div\mathrm{V}$$

(2)

where $\rho$ denotes the density of the fluid and $p$ is the unknown part of the pressure, $\mathrm{V}=\left(u,v,w\right)$, in which $u,v$ and $w$ are the velocity components of the fluid along $x,y$ and $z$ directions, respectively, and the del operator is defined as $\nabla =\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\right)$.

The boundary conditions are:

$$u=U,v=V,w=0 \mathrm{at} \eta =0$$

(3)

$$w=-b,u=v=0 \mathrm{at} \eta =\phi .$$

(4)

We introduce the similarity transforms:

$$u=Uf\left(\eta \right)+b{h}^{/}(\eta ),v=Vg\left(\eta \right),w=-bh\left(\eta \right).$$

(5)

Assuming that $L_2\ge L_1\ge \phi$, the end effects can be neglected. After substituting (5) in (4), (3), and (2), we obtain

$$h^{iv}=R\left(h^{/}{h}^{//}-h{h}^{///}\right)$$

(6)

$$f^{//}=R\left(f{h}^{/}-h{f}^{/}\right)$$

(7)

$$g^{//}=-R\left(h{g}^{/}\right)$$

(8)

$R=b\xi /\phi$ is the Reynolds number (R); $\eta =\xi /\phi$ is for transformation of the above equations into ordinary differential equations with the following boundary conditions:

$$\begin{array}{l}h\left(0\right)=h^{/}\left(0\right)=0,g\left(0\right)=f\left(0\right)=1,\\ h\left(1\right)=1,h^{/}\left(1\right)=g\left(1\right)=f\left(1\right)=0\end{array}$$

(9)

3. Integral Transform Method

To describe the formulation of Integral Transform Method (ITM), we consider the general nonlinear second order differential equation:

$${\psi }^{//}\left(\eta \right)=f\left(\eta ,\psi \left(\eta \right),{\psi }^{/}\left(\eta \right)\right)$$

(10)

Along with the following boundary conditions:

$$\begin{array}{l}\eta ={\alpha }_0:\psi \left({\alpha }_0\right)=\alpha \\ \eta ={\beta }_0:\psi \left({\beta }_0\right)=\beta \end{array}$$

(11)

Integrating Equation (10) with respect to $\eta$ from ${\alpha }_0$ to $\eta$ twice yields

$$\psi \left(\eta \right)=\psi \left({\alpha }_0\right)+\left(\eta -{\alpha }_0\right){\psi }^{/}\left({\alpha }_0\right)+{\displaystyle \underset{{\alpha }_0}{\overset{\eta }{\int }}\left(\eta -\zeta \right)f\left(t,\psi \left(\zeta \right),{\psi }^{/}\left(\eta \right)\right)d\zeta }$$

(12)

$$\psi \left(\eta \right)=\alpha +\left(\eta -{\alpha }_0\right){\psi }^{/}\left({\alpha }_0\right)+{\displaystyle \underset{{\alpha }_0}{\overset{\eta }{\int }}\left(\eta -\zeta \right)f\left(\zeta ,\psi \left(\zeta \right),{\psi }^{/}\left(\zeta \right)\right)d\zeta }$$

(13)

By using second boundary condition we can evaluate ${\psi }^{/}\left({\alpha }_0\right)$, which is

$$\psi \left({\beta }_0\right)=\beta =\alpha +\left({\beta }_0-{\alpha }_0\right){\psi }^{/}\left({\alpha }_0\right)+{\displaystyle \underset{{\alpha }_0}{\overset{{\beta }_0}{\int }}\left({\beta }_0-\zeta \right)f\left(\zeta ,\psi (\zeta ),{\psi }^{/}\left(\zeta \right)\right)d\zeta }$$

(14)

$$\frac{\left(\beta -\alpha \right)}{\left({\beta }_0-{\alpha }_0\right)}-\frac{1}{\left({\beta }_0-{\alpha }_0\right)}{\displaystyle \underset{{\alpha }_0}{\overset{{\beta }_0}{\int }}\left({\beta }_0-\zeta \right)f\left(\zeta ,\psi (\zeta ),{\psi }^{/}\left(\zeta \right)\right)d\zeta }={\psi }^{/}\left({\alpha }_0\right)$$

(15)

Substituting Equation (15) into (14) results in

$$\psi \left(\eta \right)=\alpha +\left(\eta -{\alpha }_0\right)\left[\frac{\left(\beta -\alpha \right)}{\left({\beta }_0-{\alpha }_0\right)}-\frac{1}{\left({\beta }_0-{\alpha }_0\right)}{\displaystyle \underset{{\alpha }_0}{\overset{{\beta }_0}{\int }}\left({\beta }_0-\zeta \right)f\left(t,\psi (\zeta )\right)dt}\right]+{\displaystyle \underset{{\alpha }_0}{\overset{\eta }{\int }}\left(\eta -\zeta \right)f\left(\zeta ,\psi \left(\zeta \right),{\psi }^{/}\left(\zeta \right)\right)d\zeta }$$

(16)

$$\begin{array}{ll}\psi \left(\eta \right)& ={\psi }_0\left(\eta \right)-{\displaystyle \underset{{\alpha }_0}{\overset{{\beta }_0}{\int }}K\left(\eta ,\zeta \right)f\left(\zeta ,\psi (\zeta ),{\psi }^{/}\left(\zeta \right)\right)d\zeta }\\ {\psi }_0\left(\eta \right)& =\alpha +\frac{\left(\eta -{\alpha }_0\right)\left(\beta -\alpha \right)}{\left({\beta }_0-{\alpha }_0\right)}\\ K(\eta ,\zeta )& =\frac{\left(\eta -\zeta \right)\left({\beta }_0-{\alpha }_0\right)-\left(\eta -{\alpha }_0\right)\left({\beta }_0-\zeta \right)}{\left({\beta }_0-{\alpha }_0\right)},{\alpha }_0<\zeta <\eta \\ & =\frac{\left(\eta -{\alpha }_0\right)\left({\beta }_0-\zeta \right)}{\left({\beta }_0-{\alpha }_0\right)},\eta <\zeta <{\beta }_0\end{array}$$

(17)

Equation (17) is the standard form of variational iteration method-II. After applying this method, Equations (6) to (9) are expressed as follows

$$\begin{array}{lll}h\left(\eta \right)& =3{\eta }^2-2{\eta }^3+{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)\left[R\left(h^{/}{h}^{//}-h{h}^{///}\right)\right]d\zeta }& \\ K\left(\eta ,\zeta \right)& =\frac{{\eta }^2}{2}\left[{\left(1-\zeta \right)}^3-{\left(1-\zeta \right)}^2\right]-\frac{{\eta }^3}{2}\left[{\left(1-\zeta \right)}^2-\frac{2}{3}{\left(1-\zeta \right)}^3\right]-\frac{{\left(\eta -\zeta \right)}^3}{3!},& 0<\zeta <\eta \\ & =\frac{{\eta }^2}{2}\left[{\left(1-\zeta \right)}^3-{\left(1-\zeta \right)}^2\right]+\frac{{\eta }^3}{2}\left[{\left(1-\zeta \right)}^2-\frac{2}{3}{\left(1-\zeta \right)}^3\right],& \eta <\zeta <1\end{array}$$

(18)

$$\begin{array}{ll}f\left(\eta \right)& =1-\eta +{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)R\left[f{h}^{/}-h{f}^{/}\right]d\zeta }\\ K(\eta ,\zeta )& =\zeta \left(\eta -1\right),0<\zeta <\eta \\ & =\eta \left(1-\zeta \right),\eta <\zeta <1\end{array}$$

(19)

$$\begin{array}{ll}g\left(\eta \right)& =1-\eta -{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)R\left[h{g}^{/}\right]d\zeta }\\ K(\eta ,\zeta )& =\zeta \left(\eta -1\right),0<\zeta <\eta \\ & =\eta \left(1-\zeta \right),\eta <\zeta <1\end{array}$$

(20)

Furthermore, we can write Equations (18) to (20) in an iteration form as follows

$$\begin{array}{l}{h}_{n+1}=h_0+{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)\left[R\left(h_n^{/}{h}_n^{//}-h_n{h}_n^{///}\right)\right]d\zeta }\\ where\\ h_0=3{\eta }^2-2{\eta }^3\\ K\left(\eta ,\zeta \right)=\frac{{\eta }^2}{2}\left[{\left(1-\zeta \right)}^3-{\left(1-\zeta \right)}^2\right]-\frac{{\eta }^3}{2}\left[{\left(1-\zeta \right)}^2-\frac{2}{3}{\left(1-\zeta \right)}^3\right]-\frac{{\left(\eta -\zeta \right)}^3}{3!},0<\zeta <\eta \\ =\frac{{\eta }^2}{2}\left[{\left(1-\zeta \right)}^3-{\left(1-\zeta \right)}^2\right]+\frac{{\eta }^3}{2}\left[{\left(1-\zeta \right)}^2-\frac{2}{3}{\left(1-\zeta \right)}^3\right],\eta <\zeta <1\end{array}$$

(21)

$$\begin{array}{l}{f}_{n+1}=f_0+{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)\left[f_n{h}_n^{/}-h_n{f}_n^{/}\right]d\zeta }\\ where\\ f_0=1-\eta \\ K(\eta ,\zeta )=\zeta \left(\eta -1\right),0<\zeta <\eta \\ =\eta \left(1-\zeta \right),\eta <\zeta <1\end{array}$$

(22)

$$\begin{array}{l}{g}_{n+1}=g_0-{\displaystyle \underset{0}{\overset{1}{\int }}K\left(\eta ,\zeta \right)\left[h_n{g}_n^{/}\right]d\zeta }\\ where\\ g_0=1-\eta \\ K(\eta ,\zeta )=\zeta \left(\eta -1\right),0<\zeta <\eta \\ =\eta \left(1-\zeta \right),\eta <\zeta <1\end{array}$$

(23)

By using the software MATHEMATICA, we calculated the series solution of the above mentioned problem. For reference, first few iterations of the problem are given in Appendix A.

4. Graphs and Discussion

The graphical and numerical trend of $h,f,$ and $g$ are portrayed for 10th order approximation, which are calculated with Mathematica.

Table 1. Comparison with homotopy Padé approximation method [12] for larger Reynolds numbers.

	HAM [12]	Present	Error
R	$-h^{///}(0)$
0.2	12.4653341665	12.465334166	−5E-10
1	14.365346654	14.365346654	−0E+0
2	16.8235673555	16.8235673555	−0E+0
3	19.3565458558	19.3565458558	−0E+0
4	21.9473510772	21.9473510772	−0E+0
5	24.5830707891	24.55234070789	−5E-10
6	27.2533109825	27.2533109818	−7E-10
7	29.9476951163	29.9476951136	−2.7E-9
8	32.6611632416	32.6611632329	−8.7E-9
9	35.3924710097	35.3924709848	−2.49E-8
10	38.1449125507	38.1449125256	−2.51E-8
51.6	149.6714726105	149.6714725836	−2.69E-8
70	197.9636925814	197.9636925523	−2.91E-8
100	275.9321654987	275.9321651777	−3.21E-7
300	788.4398765432	788.4398762142	−3.29E-7
500	1291.1159263487	1289.1236547891	−1.99
1000	2526.4	2398.1243	−128.27

,

Table 2. Comparison with homotopy Padé approximation method [12] for larger Reynolds numbers.

	HAM [12]	Present	Error
R	$-f^{/}(0)$
0.2	1.0880258369	1.0880258369	-0E-0
1	1.4053692581	1.4053692581	-0E+0
2	1.7445321654	1.7445321654	-0E+0
3	2.0361159263	2.0361159263	-0E+0
4	2.2866753421	2.2866753421	-0E+0
5	2.5287418529	2.5287418529	-0E-0
6	2.9136985214	2.9136985214	-0E-0
7	3.172839654	3.172839654	-0E-0
8	3.6543219871	3.6543219866	-5E-10
9	3.8293711234	3.8293711226	-8E-10
10	4.1125836917	4.1125836759	-1.58E-8
51.6	7.5531472583	7.5531472314	-2.69E-8
70	8.7573572419	8.7573572108	-3.11E-8
100	10.40147311	10.401472583	-5.27E-7
300	17.81537007	17.815369258	-8.12E-7
500	22.670	22.661	-0.009
1000	30.432	30.429	-3.00

and

Table 3. Comparison with homotopy Padé approximation method [12] for larger Reynolds numbers.

	HAM [12]	Present	Error
R	$-g^{/}(0)$
0.2	1.0301258369	1.0301258369	-0E-0
1	1.1531951623	1.1531951623	-0E+0
2	1.3096314785	1.3096314785	-0E+0
3	1.4658951623	1.4658951623	-0E+0
4	1.6187258369	1.6187258369	-0E+0
5	1.766159741	1.766159741	-0E-0
6	1.9086456321	1.9086456321	-0E-0
7	2.0464158231	2.0464158231	-0E-0
8	2.1850474125	2.1850474125	-0E-0
9	2.3368115987	2.3368115987	-0E-0
10	2.5264314755	2.5264314755	-0E-0
51.6	5.3011463295	5.30114632825	-1.25E-9
70	6.1449236681	6.14492366789	2.1E-8
100	7.2881583541	7.2881580341	-3.27E-7
300	12.5161258	12.5160446	-8.12E-5
500	15.368124	15.258154	-010997
1000	20.062	19.1583	-0.9037

present the direct comparison of the suggested method with existing methods. Through error analysis, we can see that ITM is showing close agreement with HAM (see Table 1, Table 2 and Table 3). Figure 2, Figure 3 and Figure 4 are plotted for different values of Reynolds Number, ranging from 1 to 4. Increasing Reynolds numbers result in increasing $h$ and decreasing $f$ (Figure 2 and Figure 3). According to Figure 3, $g$ first increases and then decreases for increasing values of Reynolds numbers.

Table 4. Comparison between HPM, Numerical (Num.), and the presented technique for different Reynolds numbers.

	HAM [12]	Num. [9]	Present	HPM [12]	Num. [9]	Present	HPM [12]	Num. [9]	Present
R	$-h^{///}(0)$			$-f^{/}(0)$			$-g^{/}(0)$
0.2	12.465	12.465	12.465	1.0880	1.088	1.088	1.0301	1.0301	1.030
1	14.365	14.365	14.365	1.405	1.405	1.405	1.1531	1.153	1.153
2	-	-	16.8235	-	-	1.7445	-	-	1.30963
3	-	-	19.356	-	-	2.0361	-	-	1.4658
4	-	-	21.9473	-	-	2.2866	-	-	1.6187
5	24.586	24.584	24.584	2.4417	2.528	2.528	-	-	1.766
6	-	-	27.253	-	-	-	-	-	1.9086
7	-	-	29.9476	-	-	-	-	-	2.04641
8	-	-	32.6611	-	-	-	-	-	2.18504
9	-	-	35.3924	-	-	-	-	-	2.33681
10	-	-	38.1449	-	-	-	-	-	2.52643
11	-	-	40.9267	-	-	-	-	-	2.79796
12	-	-	43.7471	-	-	-	-	-	3.2218
13.8	48.9043	48.484	48.484	4.0229	4.022	4.022	-	-	4.70844
15	-	-	52.2461	-	-	-	-	-	6.53822
16	-	-	54.6612	-	-	-	-	-	8.72019
51.6	149.67	149.67	149.67	7.553	7.553	7.553	5.301	5.301	5.301

is consistent with the results obtained in the graphical description.

5. Conclusions

This paper presented a method for solving two-point nonlinear BVPs utilizing a recursive algorithm. This algorithm has the advantage of fewer calculations for obtaining the model parameters compared with VIM, HPM, ADM, and HAM. With this algorithm, a direct recursion scheme is obtained to solve the problem. This method is simple and does not have the difficulties that occur in numerical techniques, including perturbation, linearization, and discretizing the differential equations. The superiority of the presented method is demonstrated via comparison of error estimates with existing methods.