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Some new remarks on MHD Jeffery-Hamel fluid flow problem

  • Remus-Daniel Ene EMAIL logo and Camelia Pop
Published/Copyright: December 29, 2017

Abstract

A Hamilton-Poisson realization of the MHD Jeffery-Hamel fluid flow problem is proposed. Tthe nonlinear stability of the equilibrium states is discussed. A comparison between the analytic solutions obtained using the OHAM method and the exact solutions provided by the Hamilton-Poisson realization are presented.

1 Introduction

The well known Jeffery-Hamel problem deals with the flow of an incompressible viscous fluid between the non-parallel walls. This flow situation was initially formulated by Jeffery [1] and Hamel [2]. Jeffery-Hamel flows are exact similarity solutions of the Navier-Stokes equations in the special case of two-dimensional flow through a channel with inclined plane walls meeting at a vertex, and with a source or sink at the vertex. A lot of papers propose different methods to solve the nonlinear magnetohydrodynamics (MHD) Jeffery-Hamel blood flow problem: numerical solutions [3], analytical solutions [4, 5, 6, 7, 8], or solutions obtained via stochastic numerical methods based on computational intelligence techniques [9].

Recently, several techniques have been used for solving different nonlinear differential equations, such as: stochastic numerical methods [10], spectral analysis based on continuous wavelet transform [11], wavelet analysis [12], and the fractional derivative technique [13].

The challenge of this paper is to find some new properties of the MHD Jeffery-Hamel fluid flow problem which can gives us a different point of view from the classical ones. The main goals of our work are to find a Hamilton-Poisson realization (see [14]) of the MHD Jeffery-Hamel fluid flow problem and to point out some of its geometrical and dynamical properties from a mechanical geometry point of view. In addition, once the Casimir functions of the Hamilton-Poisson structure are found, the exact solution of the equation is the intersection between the surfaces H = const. and C = const. As a consequence, we can sketch a comparison with the analytic solutions proposed in [15].

The structure of this paper is as follows. In the second section of this work we prepare the framework of our study by writing the nonlinear differential equation as a Hamilton-Poisson one. The Poisson structure of the system, the corresponding Casimirs and the phase portrait are presented here.

The spectral stability and the nonlinear stability of the equilibrium states are the subjects of the third section. In the last section a comparison of the exact solution provided by the Hamilton-Poisson realization and the analytic solution given in [15] is proposed.

For the beginning, let us recall very briefly the definitions of general Poisson manifolds and the Hamilton-Poisson systems.

Definition

Let M be a smooth manifold and let C(M) denote the set of the smooth real functions on M. A Poisson bracket onM is a bilinear map from C(MC(M) into C(M), denoted as:

(F,G)F,GC(M),F,GC(M)

which verifies the following properties:

  1. skew-symmetry:

    F,G=G,F;
  2. Jacobi identity:

    F,G,H+G,H,F+H,F,G=0;
  3. Leibniz rule:

    F,GH=F,GH+GF,H.

Proposition

Let {⋅, ⋅} a Poisson structure on Rn. Then for any f, g ∈ C(Rn,R) the following relation holds:

f,g=i,j=1nxi,xjfxigxj.

Let the matrix given by:

Π=xi,xj.

Proposition

Any Poisson structure {⋅,⋅} on Rn is completely determined by the matrix Π via the relation:

f,g=(f)tΠ(g).

Definition

A Hamilton-Poisson system on Rn is the triple (Rn, {⋅,⋅}, H), where {⋅,⋅} is a Poisson bracket on Rn and HC(Rn,R) is the energy (Hamiltonian). Its dynamics is described by the following differential equations system:

x.=ΠH

where x = (x1, x2,…xn)t.

Definition

Let {⋅,⋅} a Poisson structure on Rn. A Casimir of the configuration (Rn, {⋅,⋅} ) is a smooth function CC(Rn,R) which satisfy:

f,C=0,fC(Rn,R).

2 The MHD Jeffery-Hamel fluid flow problem

The MHD Jeffery-Hamel fluid flow problem can be written as [15]:

F(η)+2αReyF(η)F(η)+4Haα2F(η)=0,(1)

where η > 0, Rey is the Reynolds number, Ha ≥ 0 is the Hartmann number, 0 < α ≪ 1 is the flow angle and prime denotes derivative with respect to η. Also, the physical model is presented in [15].

Using the notations:

F(η)=f1(η),F(η)=f2(η),F(η)=f3(η),

the nonlinear equation Eq. (1) becomes:

f1(η)=f2(η)f2(η)=f3(η),η>0,f3(η)=2αReyf1(η)f2(η)4Haα2f1(η)(2)

Proposition

The system Eq. (2) has the Hamilton-Poisson realization

(R3,Π,H),

where

Π=010102αReyf1++(4Ha)α202αReyf10(4Ha)α2

is the minus Lie-Poisson structure and

H(f1,f2,f3)=12f2223αReyf134Ha2α2f12f1f3f3αReyf12(4Ha)α2f1

is the Hamiltonian.

Proof: Indeed, we have:

ΠH=f1f2f3

and the matrix Π is a Poisson matrix, see [14].

The next step is to find the Casimirs of the configuration described by the above Proposition. Since the Poisson structure is degenerate, there exist Casimir functions. The defining equations for the Casimir functions, denoted by C, are

ΠijjC=0.

It is easy to see that there exists only one functionally independent Casimir of our Poisson configuration, given by C : R3R,

C(f1,f2,f3)=f3+αReyf12+(4Ha)α2f1.

Consequently, the phase curves of the dynamics Eq. (2) are the intersections of the surfaces H(f1,f2,f3) = const. and C(f1,f2,f3) = const. see the Figures.

3 Nonlinear stability problem

The concept of stability is an important issue for any differential equation. The nonlinear stability of the equilibrium point of a dynamical system can be studied using the tools of mechanical geometry, so this is another good reason to find a Hamilton -Poisson realization. For more details, see [14]. We start this section with a short review of the most important notions.

Definition

An equilibrium state xe is said to be nonlinear stable if for each neighborhood U of xe in D there is a neighborhood V of xe in U such that trajectory x(t) initially in V never leaves U.

This definition supposes well-defined dynamics and a specified topology. In terms of a norm ‖ ‖, nonlinear stability means that for each ε > 0 there is δ > 0 such that if

x(0)xe<δ

then

x(t)xe<ε,()t>0.

It is clear that nonlinear stability implies spectral stability; the converse is not always true.

The equilibrium states of the dynamics Eq. (2) are

e1M=(M,0,0),e2=(4Ha)α2Rey,0,0,,MR.

Proposition 1

The equilibrium states e1M = (M,0,0) are nonlinearly stable for any MR*.

Proof

We will use energy-Casimir method, see [14] for details. Let

Fφ(f1,f2,f3)=H(f1,f2,f3)+φ[C(f1,f2,f3)]==12f2223αReyf134Ha2α2f12f1f3f3αReyf12(4Ha)α2f1+φ(f3+αReyf12+(4Ha)α2f1)

be the energy-Casimir function, where φ : RR is a smooth real valued function.

Now, the first variation of Fφ is given by

δFφ(f1,f2,f3)=f2δf2f3δf1+[f11++φ˙(f3+αReyf12+(4Ha)α2f1)]××(2αReyf1δf1+(4Ha)α2δf1+δf3)

so we obtain

δFφ(e1M)=M1+φ˙(αReyM2+(4Ha)α2M)××2αReyMδf1+(4Ha)α2δf1+δf3

that is equals zero for any MR* if and only if

φ˙αReyM2+(4Ha)α2M=M+1.(3)

The second variation of Fφ at the equilibrium of interest is given by

δ2Fφ(e1M)=(2αReyM+(4Ha)α2)2φ¨(αReyM2++(4Ha)α2M)(2αReyM+(4Ha)α2)(δf1)2++(δf2)221(2αReyM+(4Ha)α2)××φ¨(αReyM2+(4Ha)α2M)δf1δf3++φ¨(αReyM2+(4Ha)α2M)(δf3)2.

If we choose now φ such that the relation (3) is valid and

(2αReyM+(4Ha)α2)φ¨(αReyM2+(4Ha)α2M)1>0,

then the second variation of Fφ at the equilibrium of interest is positive defined and so our equilibrium states e1M are nonlinearly stable.

Proposition 2

The equilibrium state e2 = (4Ha)α2Rey,0,0, is nonlinearly stable for 0 ≤ Ha < 4.

Proof

We will use energy-Casimir method, see [14] for details. Let

Fφ(f1,f2,f3)=H(f1,f2,f3)+φ[C(f1,f2,f3)]==12f2223αReyf134Ha2α2f12f1f3f3αReyf12(4Ha)α2f1+φ(f3+αReyf12+(4Ha)α2f1)

be the energy-Casimir function, where φ : RR is a smooth real valued function.

Now, the first variation of Fφ is given by

δFφ(f1,f2,f3)=f2δf2f3δf1+f11++φ˙(f3+αReyf12+(4Ha)α2f1)××2αReyf1δf1+(4Ha)α2δf1+δf3

so we obtain

δFφ(e2)=(4Ha)α22Rey1+φ˙(4Ha)2α42Rey××(2α)(1α)α2(4Ha)δf1+δf3

that is equals zero if and only if

φ˙(4Ha)2α42Rey(2α)=1(4Ha)α22Rey.(4)

The second variation of Fφ at the equilibrium of interest is given by

δ2Fφ(e2)=(1α)2α4(4Ha)2φ¨(4Ha)2α42Rey××(2α)+(α1)α2(4Ha)(δf1)2+(δf2)2++21(α1)α2(4Ha)××φ¨(4Ha)2α42Rey(2α)δf1δf3++φ¨(4Ha)2α42Rey(2α)(δf3)2.

If we choose now φ such that the relation (4) is valid and

(1α)α2(4Ha)φ¨((4Ha)2α42Rey(2α))1<0

for 4 − Ha > 0, then the second variation of Fφ at the equilibrium of interest is positive defined and so our equilibrium state e2 is nonlinearly stable. □

4 Comparison of the exact solution and analytical solution

Consequently we have derived the following result:

Remark

The phase curves of the dynamics (2) are the intersections of the surfaces H(f1,f2,f3) = const. and C(f1,f2,f3) = const.

H(f1,f2,f3)=12f2223αReyf134Ha2α2f12f1f3f3αReyf12(4Ha)α2f1=Cst1C(f1,f2,f3)=f3+αReyf12+(4Ha)α2f1==Cst2,(5)

where

Cst1=12f22(0)23αReyf13(0)4Ha2α2f12(0)f1(0)f3(0)f3(0)αReyf12(0)(4Ha)α2f1(0)==12F(0)223αReyF(0)34Ha2α2F(0)2F(0)F(0)F(0)αReyF(0)2(4Ha)α2F(0),Cst2=f3(0)+αReyf12(0)+(4Ha)α2f1(0)==F(0)+αReyF(0)2+(4Ha)α2F(0).

Using the physical conditions as [15]:

f1(0)=1,f2(0)=0,f1(1)=0,(6)

the numerical values of the second-order derivative f3(0) = F″(0) were obtained via Optimal Homotopy Perturbation Method [15] for some values of the physical parameters, i.e. Rey = 50, (α = π/24, Ha = 250), (α = π/24, Ha = 500), (α = π/24, Ha = 1000), (α = π/36, Ha = 250), (α = π/36, Ha = 500) and (α = π/36, Ha = 1000), respectively. These values are presented in Table 1.

Table 1

Some numerical values of the second order derivative f3(0) = F(0) for Rey = 50 and different values of the parameters α and Ha respectively

αHaF(0)
π/24250-3.276689211655049
π/24500-2.344963905917293
π/241000-1.2436274085401224
π/36250-3.0222270064039733
π/36500-2.588448071307733
π/361000-1.916352763377657

Finally, we compare the phase curves given by Eq. (5) of the dynamics (2) with the corresponding approximate analytic solutions from [15] for the physical values presented in Table 1.

Observation

If f(η) is the approximate analytic solution obtained via Optimal Homotopy Perturbation Method [15], then the corresponding residual functions are:

RH=12(f¯)223αReyf¯34Ha2α2f¯2f¯f¯f¯αRey(f¯)2(4Ha)α2f¯Cst1RC=f¯+αRey(f¯)2+(4Ha)α2f¯Cst2.(7)

Example 1

Rey = 50, α = π/24, Ha = 250. The phase curve and the corresponding residual are presented in Figs 1, 2 and 3 respectively.

Figure 1 The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 250
Figure 1

The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 250

Figure 2 The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 250
Figure 2

The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 250

Figure 3 The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha = 250
Figure 3

The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha = 250

Valuesoftheintegrals:01RH2(η)dη6.26536108;01RC2(η)dη5.62805108.

Example 2

Rey = 50, α = π/24, Ha = 500. The phase curve and the corresponding residual are presented in Figs 4, 5 and 6 respectively.

Figure 4 The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 500
Figure 4

The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 500

Figure 5 The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 500
Figure 5

The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 500

Figure 6 The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha=500
Figure 6

The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha=500

Valuesoftheintegrals:01RH2(η)dη3.37046106;01RC2(η)dη1.48946106.

Example 3

Rey = 50, α = π/24, Ha = 1000. The phase curve and the corresponding residual are presented in Figs 7, 8 and 9 respectively.

Figure 7 The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 1000
Figure 7

The phase curve given by Eq. (5) for Rey = 50, α = π/24, Ha = 1000

Figure 8 The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 1000
Figure 8

The residual RH given by Eq. (7) for Rey = 50, α = π/24, Ha = 1000

Figure 9 The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha = 1000
Figure 9

The residual RC given by Eq. (7) for Rey = 50, α = π/24, Ha = 1000

Valuesoftheintegrals:01RH2(η)dη1.27702106;01RC2(η)dη5.45545107.

Example 4

Rey = 50, α = π/36, Ha = 250. The phase curve and the corresponding residual are presented in Figs 10, 11 and 12 respectively.

Figure 10 The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 250
Figure 10

The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 250

Figure 11 The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 250
Figure 11

The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 250

Figure 12 The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 250
Figure 12

The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 250

Valuesoftheintegrals:01RH2(η)dη7.0543108;01RC2(η)dη6.16148108.

Example 5

Rey = 50, α = π/36, Ha = 500. The phase curve and the corresponding residual are presented in Figs 13, 14 and 15 respectively.

Figure 13 The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 500
Figure 13

The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 500

Figure 14 The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 500
Figure 14

The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 500

Figure 15 The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 500
Figure 15

The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 500

Valuesoftheintegrals:01RH2(η)dη5.09766107;01RC2(η)dη4.60622107.

Example 6

Rey = 50, α = π/36, Ha = 1000. The phase curve and the corresponding residual are presented in Figs 16, 17 and 18 respectively.

Figure 16 The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 1000
Figure 16

The phase curve given by Eq. (5) for Rey = 50, α = π/36, Ha = 1000

Figure 17 The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 1000
Figure 17

The residual RH given by Eq. (7) for Rey = 50, α = π/36, Ha = 1000

Figure 18 The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 1000
Figure 18

The residual RC given by Eq. (7) for Rey = 50, α = π/36, Ha = 1000

Valuesoftheintegrals:01RH2(η)dη2.86153106;01RC2(η)dη1.56449107.

5 Conclusions

The stability of a nonlinear differential problem governing the MHD Jeffery-Hamel fluid flow is investigated. Due to the existence of a Poisson formulation, the results were obtained using specific tools, such as the energy-Casimir method.

Finally, the analytical integration of the nonlinear system (obtained via the Optimal Homotopy Asymptotic Method and presented in [15]) is compared with the exact solution (obtained as intersections of the surfaces H(f1,f2,f3) = const. and C(f1,f2,f3) = const).

  1. Conflict of Interest: The authors declare that there are no conflicts of interest regarding the publication of this paper.

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Received: 2017-9-29
Accepted: 2017-11-2
Published Online: 2017-12-29

© 2017 Remus-Daniel Ene and Camelia Pop

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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