# An Optimal Homotopy Asymptotic Approach Applied to Nonlinear MHD Jeffery-Hamel Flow

Authors: Vasile Marinca; Nicolae Herişanu
Journal: Mathematical Problems in Engineering
 (2011)

---

## Abstract

A simple and effective procedure is employed to propose a new analytic approximate solution for nonlinear MHD Jeffery-Hamel flow. This technique called the Optimal Homotopy Asymptotic Method (OHAM) does not depend upon any small/large parameters and provides us with a convenient way to control the convergence of the solution. The examples given in this paper lead to the conclusion that the accuracy of the obtained results is growing along with increasing the number of constants in the auxiliary function, which are determined using a computer technique. The results obtained through the proposed method are in very good agreement with the numerical results.


---

## Body

## 1. Introduction

In various fields of science and engineering, nonlinear evolution equations, as well as their analytic and numerical solutions, are fundamentally important. The problem of an incompressible, viscous fluid between nonparallel walls with a sink or source at the vertex was pioneered by Jeffery [1] and Hamel [2]. Hamel mentioned an example of an exact nonsteady solution of the Navier-Stokes equations which describes the process of decay of a vortex through the action of the viscosity and considered the distribution of the tangential velocity component with respect to the radial distance and time and a particular case of the flow through a divergent channel was discussed and exactly solved. Jeffery-Hamel flows are exact similarity solution as the Navier-Stokes equations in the special case of two-dimensional flow through a channel with inclined plane walls meeting at a vertex with a source or sink at the vertex and have been studied by several authors and discussed in many text books and articles [3–5]. Sadri [6] denoted that Jeffery-Hamel flow used an asymptotic boundary condition to examine steady two-dimensional flow of a viscous fluid in a channel by means of certain symmetric solution of the flow although asymmetric solution are both possible and of physical interest [7].The classical Jeffery-Hamel problem was extended in [8] to include the effects of external magnetic field in conducted fluid. The magnetic field acts as a control parameter, along with the flow, Reynolds number, and the angle of the walls.Most scientific problems such as Jeffery-Hamel flows and other fluid mechanics problems are inherently nonlinear. Excepting a limited number of these problems, most do not have analytical solutions. Therefore, these nonlinear equations should be solved using other methods [9].The aim of the present work is to propose an accurate approach to the Jeffery-Hamel flow problem using an analytical technique, namely, OHAM [10, 11].The efficiency of our procedure, which does not require a small parameter in the equation, is based on the construction and determination of the auxiliary functions combined with a convenient way to optimally control the convergence of the solution.


## 2. Problem Statement and Governing Equation

We consider a system of cylindrical polar coordinates(r,θ,z) with a steady two-dimensional flow of an incompressible conducting viscous fluid from a source or sink at channel walls lying in planes, with angle 2α, as shown in Figure 1.Figure 1
Geometry of the Jeffery-Hamel flow problem.Assuming that the velocity is only along the radial direction and depends onr and θ, V(u(r,θ),0) [3–5], using the continuity Navier-Stokes equations in polar coordinates, the governing equations are(2.1)ρr∂∂r(ru(r,θ))=0,(2.2)u(r,θ)∂u(r,θ)∂r=-1ρ∂p∂r+v[∂2u(r,θ)∂r2+1r∂u(r,θ)∂r+1r2∂2u(r,θ)∂θ2-u(r,θ)r2],(2.3)-1ρr∂p∂ρ+2vr2∂u(r,θ)∂θ=0,
where ρ is the fluid density, p is the pressure, and v is the kinematic viscosity. From (2.1) and using dimensionless parameters we get(2.4)f(θ)=ru(r,θ),(2.5)F(x)=f(θ)fmax⁡,x=θα.Substituting (2.5) into (2.2) and (2.3) and eliminating the pressure, we obtain an ordinary differential equation for the normalized function profile F(x):(2.6)F′′′(x)+2αReF(x)F′(x)+4α2F′(x)=0,
where prime denotes derivative with respect to x and the Reynolds number is(2.7)Re=αfmax⁡v=umax⁡v(divergentchannel:α>0,umax⁡>0convergentchannel:α<0,umax⁡<0)
and umax is the maximum velocity at the centre of the channel.The boundary conditions for (2.6) are(2.8)F(0)=1,F′(0)=0,F(1)=0.


## 3. Fundamentals of the OHAM

We consider the following nonlinear differential equation [10, 11]:(3.1)L(F(x))+f(x)+N(F(x))=0
subject to a boundary condition(3.2)B(F)=0,
where L is a linear operator, f(x) is a known analytical function, N is a nonlinear operator, and B is a boundary operator. By means of the OHAM one constructs a family of equations:(3.3)(1-p)[L(ϕ(x,p))+f(x)]=h(x,p)[L(ϕ(x,p))+f(x)+N(ϕ(x,p))],
and the boundary condition is(3.4)B(ϕ(x,p))=0.In (3.3), ϕ(x,p) is an unknown function, p∈[0,1] is an embedding parameter, and h(x,p) is an auxiliary function such that h(x,0) = 0 and h(x,p)≠0 for p≠0. When p increases from 0 to 1, the solution ϕ(x,p), changes from the initial approximation F0(x) to the solution F(x). Obviously, when p=0 and p=1 it holds that(3.5)ϕ(x,0)=F0(x),ϕ(x,1)=F(x).Expandingϕ(x,p) in series with respect to the parameter p, one has(3.6)ϕ(x,p)=F0(x)+pF1(x)+p2F2(x)+….If the initial approximationF0(x) and the auxiliary function h(x,p) are properly chosen so that the series (3.6) converges at p=1, one has(3.7)F(x)=F0(x)+F1(x)+F2(x)+….Notice that the series (3.6) contains the auxiliary function h(x,p) which determines their convergence regions. The results of the mth-order approximations are given by(3.8)F̅(x)≈F0(x)+F1(x)+F2(x)+⋯+Fm(x).We propose an auxiliary functionh(x,p) of the form(3.9)h(x,p)=pK1(x)+p2K2(x)+⋯+pmKm(x),
where Ki(x), i=1,2,…,m can be functions on the variable x.Substituting (3.6) into (3.1) we obtain(3.10)L(F(x))+f(x)+N(F(x))=N0(F0(x))+pN1(F0(x),F1(x))+p2N2(F0(x),F1(x),F2(x))+….If we substitute (3.9) and (3.10) into (3.3) and we equate to zero the coefficients of various powers of p, we obtain the following linear equations:(3.11)L(F0(x))+f(x)=0,B(F0(x))=0,(3.12)L(Fi)-L(Fi-1)-∑j=1iKjNi-j(F0,F1,…,Fi-j)=0,B(Fi)=0,i=1,2,…,m-1(3.13)L(Fm)-L(Fm-1)-∑j=1m-1KjNm-1-j-KmN0=0,B(Fm)=0.
At this moment, the mth-order approximate solution given by (3.8) depends on the functions K1,K2,…,Km. The constants C1, C2, …, Cq which appear in the expression of Ki(x) can be identified via various methodologies such as the least square method, the Galerkin method, and the collocation method.The constantsC1, C2, …, Cq could be determined, for example, if we substitute (3.8) into (3.1) resulting in the following residual:(3.14)R(x,Ci)=L(F̅(x,Ci))+f(x)+N(F̅(x,Ci)),i=1,2,….Forxi∈ (a,b) where a and b are two values depending on the given problem and we substitute xi into (3.14), we obtain the system of equations(3.15)R(x1,Ci)=R(x2,Ci)=⋯=R(xq,Ci),i=1,2,…,q,
where q is the number of constants Ci which appear in the expression of the functions K1(x), K2(x),…,Km(x).One can observe that our procedure contains the auxiliary functionh(x,p) which provides us with a simple but rigorous way to adjust and control the convergence of the solution. It must be underlined that it is very important to properly choose the functions K1,…,Km(x) which appear in the approximation (3.8).


## 4. Application of the Jeffery-Hamel Flow Problem

We introduce the basic ideas of the proposed method by considering (2.6) and (2.8). We choose f(x)=0 and the linear operator(4.1)L(ϕ(x,p))=∂3ϕ(x,p)∂x3.The nonlinear operator is(4.2)N(ϕ(x,p))=2αReϕ(x,p)∂ϕ(x,p)∂x+4α2∂ϕ(x,p)∂x,
and the boundary conditions are(4.3)ϕ(0,p)=1,∂ϕ(0,p)∂x=0,ϕ(1,p)=0.
Equation (3.11) becomes(4.4)F0′′′(x)=0,F0(0)=1,F0′(0)=0,F0(1)=0.
It is obtained that(4.5)F0(x)=1-x2.
From (4.2) and (3.10), we obtain the following expression:(4.6)N0(x)=F0′′′(x)+2αReF0(x)F0′(x)+4α2F0′(x).
If we substitute (4.5) into (4.6), we obtain(4.7)N0(x)=4αRex3-4(αRe+2α2)x.There are many possibilities to choose the functionsKi, i=1,2,…. The convergence of the solutions Fi, i=1,2,…m and consequently the convergence of the approximate solution F̅(x) given by (3.8) depend on the auxiliary functions Ki. Basically, the shape of Ki(x) should follow the terms appearing in (4.7), (3.12), and (3.13) which are polynomial functions. We consider the following cases (m=2).Case 1.
IfK1(x) is of the form
(4.8)K1(x)=C1,
where C1 is an unknown constant at this moment, then (3.12) for i=1 becomes
(4.9)F1′′′(x)-F0′′′(x)-K1N0(F0)=0.
Substituting (4.5), (4.7), and (4.8) into (4.9), we obtain the equation in F1:
(4.10)F1′′′(x)-4C1αRex3-4C1(αRe+2α2)x=0,F1(0)=F1′(0)=F1(1)=0.
The solution of (4.10) is given by
(4.11)F1(x)=C1αRe30x6-C1(αRe+2α2)6x4+2αRe+5α215C1x2.
Equation (3.13) for m=2 can be written in the form
(4.12)F2′′′(x)-F1′′′(x)-C1N1(F0,F1)-K2(x)N0(F0)=0,
where N1 is obtained from (3.10):
(4.13)N1(F0,F1)=F1′′′+2αRe(F0F1′+F0′F1)+4α2F1′.
If we consider
(4.14)K2(x)=C2,
where C2 is an unknown constant, then from (4.5), (4.11), (4.12), (4.13), and (4.14) we obtain the following equation in F2:
(4.15)F2′′′+8α2Re2C1215x7-12α2Re2+24α2Re5C12x5-[4αRe(2C1+C2)+60αRe+36α2Re2+40α3Re-16α415C12]x3+[4(αRe+2α2)(C1+C2)+60αRe+120α2-8α2Re2-36α3Re-40α415C12]x=0.
So, the solution of (4.15) is given by
(4.16)F2(x)=-α2Re2C121350x10+α2Re2+2α2Re140C12x8+[αRe(2C1+C2)30+15αRe+9α2Re2+10α3Re-4α4450C12]x6+[-(αRe+2α2)(C1+C2)6+2α2Re2+9α3Re+10α4-15αRe-3α290C12]x4+[2αRe+5α215(C1+C2)+2520αRe-1932α4+360α2-919α2Re2-2580α3Re18900]x2.
The second-order approximate solution (m=2) is obtained from (3.8)
(4.17)F̅(x)=F0(x)+F1(x)+F2(x),
where F0, F1, and F2 are given by (4.5), (4.11), and (4.16), respectively.Case 2.
In this case we consider(4.18)K1(x)=C1,(4.19)K2(x)=C2x+C3,
where C1, C2, and C3 are unknown constants.
It is clear that the functionF1 is given by (4.11). Equations (3.13) or (4.12) becomes
(4.20)F2′′′+8α2Re2C1215x7-12α2Re2+24α2Re5C12x5-4αReC3x4-[4αRe(C1+C2)+60αRe-36α2Re2-120α3Re-80α415C12]x3+4(αRe+2α2)C3x2+[4(αRe+2α2)(C1+C2)+60αRe+12α2-8α2Re2-36α3Re-40α415C12]x=0,F2(0)=F2′(0)=F2(1)=0
and has the solution
(4.21)F2(x)=-α2Re2C121350x10+α2Re2+2α2Re140C12x8+2αRe105C3x7+[αRe(2C1+C2)30+15αRe+9α2Re2+10α3Re-4α4450C12]x6+αRe+2α215C3x5+[-(αRe+2α2)(C1+C2)6+2α2Re2+9α3Re+10α4-15αRe-3α290C12]x4+[2αRe+5α215(C1+C2)+5αRe+14α2105C3+2520αRe-1932α4+360α2-919α2Re-2580α3Re18900C12]x2.
The second-order approximate solution becomes(4.22)F̅(x)=F0(x)+F1(x)+F2(x),
where F0, F1, and F2 are given by (4.5), (4.11), and (4.21), respectively.Case 3.
In the third case we consider(4.23)K1(x)=C1+C2x,K2(x)=C3+C4x+C5x2.
Equation (3.12) for i=1 or (4.9) can be written as
(4.24)F1′′′-4αReC2x4-4αReC1x3+4(αRe+2α2)C2x2+4(αRe+2α2)C1x=0,F1(0)=F1′(0)=F1(1)=0.
From (4.24) we have
(4.25)F1(x)=2αReC2105x7+αReC130x6-(αRe+2α2)C215x5-(αRe+2α2)C16x4+[(αRe+2α2)(5C1+2C2)30-αRe(7C1+4C2)210]x2.
Equation (3.13) becomes
(4.26)F2′′′(x)+12α2Re235C22x9+92α2Re2C1C2105x8+(8α2Re2C1215-6α2Re2+12α3Re5C22)x7-18α2Re2+36α3Re5C1C2x6-[4αReC5+12α2Re2+24α3Re5C12+12αRe-2α2Re2-8α3Re-8α43C22]x5-[4αRe(C2+C4)+120αRe-46α2Re2-160α3Re-120α415C1C2-40α2Re2+112α3Re105C22]x4-[4αRe(C1+C3)-4(αRe+2α2)C3+60αRe-36α2Re2-120α3Re-80α415C12-40α2Re2+112α3Re105C1C2-4(αRe+2α2)C22]x3+[4(αRe+2α2)(C2+C4)+120αRe+240α2-8α2Re2-36α3Re-40α415C1C2+20α2Re2+96α3Re+112α4105C22]x2+[4(αRe+2α2)(C1+C3)+60αRe+120α2-8α2Re2-36α3Re-40α415C12-20α2Re2+96α3Re+112α4105C1C2]x,F2(0)=F2′(0)=F2(1)=0.
The solution of (4.26) is
(4.27)F2(x)=-α2Re2C223850x12-46α2Re2C1C251975x11+[-α2Re2C121350+α2Re2+2α3Re600C22]x10+α2Re2+2α3Re420C1C2x9+[αReC584+α2Re2+2α3Re140C12+6αRe-α2Re2-4α3Re-4α4504C22]x8+[2αRe(C2+C4)105+60Re-23α2Re2-80α3Re-60α41575C1C2-20α2Re2+56α3Re11025C222αRe(C2+C4)105()()]x7+[αRe(C1+C3)30-(αRe+2α2)C530+15αRe-9α2Re2-30α3Re-20α4450C12+5α2Re2+14α3Re1575C1C2-αRe+2α230C22αRe(C1+C3)30]x6-[(αRe+2α2)(C2+C4)15+30αRe+60α2-2α2Re2-9α3Re-10α4225C1C2-5α2Re2+24α3Re+28α41575C22(αRe+2α2)(C2+C4)15]x5-[(αRe+2α2)(C1+C3)6+15αRe+30α2-2α2Re2-9α3Re-10α490C12-5α2Re2+24α3Re+28α4630C1C2(αRe+2α2)(C1+C3)6]x4+[(2αRe+5α2)(C1+C3)15+(5αRe+14α2)(C2+C4)105+9αRe+28α2420C5+2520αRe-1260α4+6300α2-163α2Re2-900α3Re18900C12+19800αRe+55440α2-1433α2Re2-8514α3Re-10560α4207900C1C2-4774α4-32340α2-10395αRe+2695α3Re+380α2Re2485100C22]x2.
The second-order approximate solution is(4.28)F̅(x)=F0(x)+F1(x)+F2(x),
where F0, F1, and F2 are given by (4.5), (4.24), and (4.26), respectively.Case 4.
In the last case, we consider(4.29)K1(x)=C1+C2x,(4.30)K2(x)=C3+C4x+C5x2+C6x3+C7x4+C8x5+C9x6.The solution ofF1(x) is given by (4.24). On the other hand, (3.13) has the solution (4.31)F2(x)=(αRe330C9-α2Re2C223850)x12+(2αRe495C8-46α2Re251975C1C2)x11+[αReC7-(Re+2α2)C9180-α2Re21350C12+α2Re2+2α3Re600C22]x10+[αReC6-(αRe+2α2)C8126+α2Re2+2α3Re420C1C2]x9+[αReC5-(αRe+2α2)C784+α2Re2+2α3Re140C12+6αRe-α2Re2-4α3Re-4α4504C22αReC5-(αRe+2α2)C784]x8+[2αReC4-2(αRe+2α2)C6105+2αReC2105+60αRe-23α2Re2-80α3Re-60α41575C1C2-20α2Re2+56α3Re11025C222αReC4-2(αRe+2α2)C6105]x7+[αRe(C1+C3)-(αRe+2α2)C530+15αRe-9α2Re2-30α3Re-20α4450C12-5α2Re2+14α3Re1575C1C2-(αRe+2α2)C2230]x6+[(αRe+2α2)(C2+C4)152α2Re2+9α3Re+10α4-30αRe-60α2225C1C2-5α2Re2+24α3Re+28α41575C22-(αRe+2α2)(C2+C4)15]x5+[-(αRe+2α2)(C1+C3)62α2Re2+9α3Re+10α4-15αRe-30α290C12+5α2Re2+24α3Re+28α4630C1C2-(αRe+2α2)(C1+C3)6]x4+[5αRe+14α2105(C2+C4)+2αRe+5α215C1+2αRe+15α215C3+9αRe+28α2420C5+7αRe+24α2630C6+4αRe+15α2630C7+27αRe+110α26930C8+2520αRe-1260α4+6300α2-163α2Re2-900α3Re18900C12+5αRe+22α21980C9+10395αRe-380α2Re2+32340α2-2685α3Re-4774α4485100C22+19800αRe-1443α2Re2+55440α2-8514α3Re-10560α4207900C1C2]x2.The second-order approximate solution in this case is given by(4.32)F̅(x)=F0(x)+F1(x)+F2(x),
where F0, F1, and F2 are given by (4.5), (4.24), and (4.30), respectively.


## 5. Numerical Examples

In the following, using the algorithm described in Section3, with the help of a computer program which implement the procedure presented above, we will obtain the convergence-control constants Ci and we will show that the error of the solution decreases when the number of terms in the auxiliary function h(x,p) increases. Obviously, the computational effort increases along with increasing the number of convergence-control constants, but a significant improvement of the accuracy of results is observed.Example 5.1.
ForRe=50 and α=5 in Case 1 it is obtained two solutions for the constants C1 and C2:(a)
C1=0.017506079C2=-0.047286881,(b)
C1=-0.017506079C2=0.022737435but the second-order approximate solution (4.16) is the same in both cases:
(5.1)F̅(x)≈1-1.767845893x2+1.236877181x4-0.619019693x6+0.164176501x8-0.014188096x10Example 5.2.
ForRe=50 and α=5 in the Case 2, we obtain(a)
C1=0.007977212C2=-0.041469373C3=0.022761122,(b)
C1=-0.007977212C2=-0.009560525C3=0.022761122.
The second-order approximate solution (4.18) becomes
(5.2)F̅(x)≈1-1.769527092x2+1.40514047x4-0.45522244x5-0.319921792x6+0.108386295x7+0.03409066x8-0.002946101x10.Example 5.3.
ForRe=50 and α=5 in Case 3, the second-order approximate solution (4.27) can be written in the form
(5.3)F̅(x)≈1-1.769777647x2+1.295738478x4-0.010406134x5-0.871743272x6+0.178832164x7+0.299266794x8-0.090606944x9-0.040442155x10+0.00935599x11-0.0002178x12.Example 5.4.
ForRe=80 and α=-5 in Case 4, the second-order approximate solution (4.31) becomes
(5.4)F̅(x)≈1-0.399291819x2-0.461970063x4-0.014703786x5-0.12415397x6-0.07325724x7-0.08278982x8+0.45101379x9-0.648015234x10+0.47466473x11-0.121496588x12
The profile of theF(x) function is presented in Figure 2 for Re=50 and α=5.
It is easy to verify the accuracy of the obtained solutions if we compare these analytical solutions with the numerical ones or with results obtained by other procedures.
It can be seen from Tables1, 2, 3, and 4 that the analytical solutions of Jeffery-Hamel flows obtained by OHAM are very accurate.Table 1
The results of the second-order approximate solutions (5.1), (5.2), and (5.3) and numerical solution of F(x) for Re=50, α=5.
xF̅(x), (5.1)F̅(x), (5.2)F̅(x), (5.3)Numerical solution011110.10.982446110.9824403820.9824308420.982431240.20.9312259690.9313024690.9312259590.931225970.30.8504719970.8508107090.8506114450.850610630.40.7463793150.7470749960.7467907840.746790810.50.6262986260.6271920840.6269472530.626948180.60.4976659230.4983409840.4982350280.498234460.70.3669663450.3669663530.3669700880.366966350.80.2389520340.2381487820.2381423220.238123750.90.1163130190.1152603610.1152190250.1151519310000Table 2
Comparison between the OHAM and numerical solutions forRe=50 and α=5(error=|F(x)num-F(x)app|).
xError of the solution (5.1)Error of the solution (5.2)Error of the solution (5.3)00000.10.000014870.0000091420.0000003980.20.0000000010.0000764990.0000000110.30.0001386330.00020.0000008150.40.0004114950.0002841860.0000000260.50.0006495540.0002439040.0000009270.60.0005685370.0001065240.0000005680.70.0000000050.0000000030.0000037380.80.0008282840.0000249620.0000185720.90.00011610890.0001084310.0000670951000Table 3
Comparison between Differential Transformation Method (DTM) [4], Homotopy Perturbation Method (HPM) [4], Homotopy Analysis Method [4], and OHAM-(5.4) for Re=80, α=-5.
xF̅(x) (DTM)F̅(x) (HPM)F̅(x) (HAM)F̅(x) (OHAM)Numerical0111110.10.99596038870.99606718740.99959602420.9959606050.99596062780.20.98327454810.98369594240.98327552580.9832755480.98327553830.30.96017755510.96107587730.96017989110.9601799140.960179911390.40.92351707060.92492451560.92352157370.9235216430.92352158940.50.86845113490.87019976970.86845889970.8684589630.868458877720.60.78807854020.78983259370.78809101860.7880909230.788090920320.70.6732484480.67453349680.67314376900.6731436330.67314363460.80.51196440610.51283730950.51199099390.5119911070.51199108910.90.29152801220.29189369910.29155801780.2915587420.29155874261100−0.00000114900Table 4
Comparison between OHAM (5.4) and numerical solutions [4] for Re=80, α=-5.
xF̅(x),(5.4)NumericalError01100.10.9959606050.99596062780.0000000220.20.9832755480.98327553830.0000000090.30.9601799140.960179911390.0000000020.40.9235216430.92352158940.0000000530.50.8684589630.868458877720.0000000850.60.7880909230.788090920320.0000000020.70.6731436330.67314363460.0000000010.80.5119911070.51199108910.0000000170.90.2915587420.291558742610.0000000061000Figure 2
Sample profile of theF(x) function for Re=50 and α=5, given by (5.1).


## 6. Conclusions

In this paper the Optimal Homotopy Asymptotic Method (OHAM) is employed to propose a new analytic approximate solution for the nonlinear MHD Jeffery-Hamel flow problems. The proposed procedure is valid even if the nonlinear equation does not contain any small or large parameters.OHAM provides us with a simple and rigorous way to control and adjust the convergence of the solution through the auxiliary functionsh(x,p) involving several constants Ci which are optimally determined.From the results presented above, we can conclude that the following.(1)
Whenα>0 and steep of the channel is divergent, stream in value of Reynolds number is caused by decreasing in velocity.(2)
Whenα<0 and steep of the channel is convergent, the results are inverse. Increase in value of Reynolds number is caused by increasing in velocity.The examples related to the Jeffery-Hamel flow problem presented in this paper lead to the very important conclusion that the accuracy of the obtained results is growing along with increasing the number of constants in the auxiliary function. This paper confirmed that DTM, HPM, or HAM gives a good accuracy, but OHAM is by far the best method delivering faster convergence and better accuracy. In the proposed procedure, iterations are performed in a very simple manner by identifying some coefficients, and therefore very good approximations are obtained in few terms. Actually the capital strength of the proposed procedure is its fast convergence, since after only two iterations it converges to the exact solution, which proves that this method is very effective in practice. This version of the method proves to be very rapid and effective, and this is proved by comparing the analytic solutions obtained through the proposed method with the solutions obtained via numerical simulations or other known procedures.



---

*Source: 169056-2011-12-21.xml*