international journal of pure and applied mathematics ...gntgli/publication/ijpam2011.pdf168 a....

International Journal of Pure and Applied Mathematics————————————————————————–Volume 73 No. 2 2011, 165-188

THREE-DIMENSIONAL MHD STAGNATION POINT-FLOW

OF A NEWTONIAN AND A MICROPOLAR FLUID

Alessandra Borrelli1 §, Giulia Giantesio2, Maria Cristina Patria3

1,2,3Department of MathematicsVia Machiavelli, 35, Ferrara, ITALY

Abstract: The steady three-dimensional stagnation-point flow of an electri-cally conducting Newtonian or micropolar fluid in the presence of a uniformexternal magnetic field H0 is analysed and some physical situations are exam-ined.

In particular, we prove that, if we impress an external magnetic field H0,and we neglect the induced magnetic field, then the steady three-dimensionalMHD stagnation-point flow is possible if, and only if, H0 has the direction ofone of the coordinate axes.

In all cases it is shown that the governing nonlinear partial differential equa-tions admit similarity solutions. We find that the flow has to satisfy an ordinarydifferential problem whose solution depends on H0 through the Hartmann num-ber M2.

Finally, the skin-friction components along the axes are computed.

AMS Subject Classification: 76W05, 76D10Key Words: Newtonian fluids, Micropolar fluids, MHD flow, three-dimensionalstagnation-point flow

1. Introduction

An important example of flow past a body, where the three velocity componentsappear, is the three-dimensional stagnation-point flow. Such a flow occurs whena jet of fluid impinges on a rigid wall.

Received: July 21, 2011 c© 2011 Academic Publications, Ltd.§Correspondence author

166 A. Borrelli, G. Giantesio, M.C. Patria

The steady three-dimensional stagnation-point flow of a Newtonian fluid hasbeen studied by Homman ([9]), Howarth ([10]), Davey and Schoffield ([1], [14]).In the similarity transformations which reduce the Navier-Stokes equations toan ODE nonlinear system two constant parameters a, b appear. As it is known,if both a and b are positive then there is a nodal point of attachment and if oneand only one of a, b is negative, such that a+b > 0, then there is a saddle pointof attachment ([10], [1], [3], [14]). Further suitable boundary conditions have

to be assigned. The ODE system obtained depends upon a parameter c =b

awhich is a measure of three-dimensionality.

Guram and Anwar Kamal ([6]) studied the steady three-dimensional stag-nation-point flow of a micropolar fluid. We recall that the micropolar fluidsintroduced by Eringen ([4]) physically represent fluids consisting of rigid ran-domly oriented particles suspended in a viscous medium which have an intrinsicrotational micromotion (for example biological fluids in thin vessels, polymericsuspensions, slurries, colloidal fluids). Extensive reviews of the theory and itsapplications can be found in [5] and [11].

The aim of this paper is to study how the steady three-dimensional stagnation-point flow of an electrically conducting Newtonian and micropolar fluid is in-fluenced by a uniform external electromagnetic field (E0,H0).

As it is customary in the literature, we assume that at infinity, the velocityv and the pressure p of the fluid under consideration approach the flow of aninviscid fluid for which the stagnation-point is shifted from the origin ([2], [12]and [13]). Moreover for the micropolar fluid we suppose that at infinity themicrorotation w is given by w = 1

2∇× v, i.e. the micropolar fluid behaves like

a classical fluid far from the wall.

For this reason, first of all, we study the steady three-dimensional stagnation-point flow of an inviscid fluid in the presence of a uniform external magneticfield H0. As it is customary in MHD, we assume that the magnetic Reynoldsnumber is very small, so that the induced magnetic field is negligible in com-parison with the imposed field. Under this assumption we prove that H0 hasto be parallel to one of the coordinate axes.

Then we consider the same problems for a Newtonian and a micropolarfluid. Taking into account the results obtained for an inviscid fluid, we findthat the flow has to satisfy an ordinary differential problem depending on twoparameters: c and M2 (M2 is the Hartmann number).

Moreover we analyse the skin-friction components τ1, τ3 along x1 and x3axes, which are physically interesting.

The paper is organized in this way:

THREE-DIMENSIONAL MHD STAGNATION POINT-FLOW... 167

In Section 2, we study the three-dimensional stagnation-point flow of aninviscid fluid in the presence of a uniform external electromagnetic field. Weprove in Theorem 2 that, if we impress an external magnetic field H0, and weneglect the induced magnetic field, then the steady three-dimensional MHDstagnation-point flow of such a fluid is possible for any value of c > −1 if, andonly if, H0 has the direction parallel to one of the axes. Further if c = 1, orc = −1

2, we prove that the magnetic field can be parallel also to the plane

Ox1x3, or Ox2x3 respectively.In Section 3, we determine the nonlinear ODE problems for a Newtonian

fluid corresponding to the cases treated in Theorem 2. Moreover we prove thatif c = 1 or c = −1

2, then the three-dimensional MHD stagnation point flow

is not possible when H0 is parallel to the plane Ox1x3 or Ox2x3, unlike whatoccurs to the inviscid fluid.

In Section 4, we obtain the nonlinear ODE problems for a micropolar fluidcorresponding to the cases treated in Theorem 2. Also for this fluid the three-dimensional MHD stagnation point flow does not exist if c = 1 and H0 isparallel to the plane Ox1x3 or c = −1

2and H0 is parallel to the plane Ox2x3.

2. Preliminaries

Consider the steady three-dimensional MHD flow of an electrically conductinghomogeneous incompressible inviscid fluid near a stagnation point filling thehalf-space S, given by

S = {(x1, x2, x3) ∈ R3 : (x1, x3) ∈ R

2, x2 > 0}. (1)

The boundary of S, i.e. x2 = 0, is a rigid, fixed, non-electrically conductingwall.

As it is well known in the three-dimensional stagnation-point flow the ve-locity field is given by

v1 = ax1, v2 = −(a+ b)x2, v3 = bx3, (x1, x2, x3) ∈ S, (2)

where a, b are constants.

We suppose a > 0, b 6= 0, and c =b

a. We exclude the case c ≤ −1 because

we impose the condition v2 < 0, so that the fluid moves towards the wall x2 = 0.

Remark 1. If c = 1, the velocity is axial symmetric with respect to x2axis:

v1 = ax1, v2 = −2ax2, v3 = ax3.


x1

x3

x2

O

Figure 1: Flow description.

The equations governing such a flow in the absence of external mechanicalbody forces are:

ρv · ∇v = −∇p+ µe(∇×H)×H,

∇ · v = 0,

∇×H = σe(E + µev ×H),

∇×E = 0, ∇ ·E = 0, ∇ ·H = 0 in S (3)

where v is the velocity field, p is the pressure, E and H are the electric andmagnetic fields, respectively, ρ is the mass density (constant > 0), µe is themagnetic permeability, σe is the electrical conductivity (µe, σe = constants > 0).

We suppose that a uniform external magnetic field H0 is impressed andthat the electric field is absent. As it is customary in the literature, we assumethat the magnetic Reynolds number is very small, so that the induced magneticfield is negligible in comparison with the imposed field. Then

(∇×H)×H ≃ σeµe(v ×H0)×H0. (4)

Now our aim is to prove the following:

Theorem 2. Let a homogeneous, incompressible, electrically conductinginviscid fluid occupy the half-space S. If we impress an external magneticfield H0, and we neglect the induced magnetic field, then the steady three-dimensional MHD stagnation-point flow of such a fluid is possible for all c > −1if, and only if, H0 is parallel to one of the coordinate axes.

Proof. For brevity sake, we will denote by H the external magnetic field:

H = H1e1 +H2e2 +H3e3, (5)


where (e1, e2, e3) is the canonical base of R3.On substituting the approximation (4) in (3)1, taking into account that the

velocity field v is given by (2), we get:

∂p

∂x1= −ρa2x1 + σea[cB1B3x3 − (B2

2 +B23)x1 − (1 + c)B1B2x2],

∂p

∂x2= −ρa2(1 + c)2x2 + σea[B1B2x1 + (1 + c)(B2

1 +B23)x2 + cB2B3x3],

∂p

∂x3= −ρa2c2x3 + σea[−(1 + c)B2B3x2 − c(B2

1 +B22)x3 +B1B3x1], (6)

where B = µeH.

It is possible to find a function p = p(x1, x2, x3) satisfying (6) if and only if

∂2p

∂xi∂xj=

∂2p

∂xj∂xi, i, j = 1, 2, 3, i 6= j. (7)

On the other hand we have

∂2p

∂x1∂x2= −σea(1 + c)B1B2,

∂2p

∂x2∂x1= σeaB1B2, (8)

∂2p

∂x1∂x3= σeacB1B3,

∂2p

∂x3∂x1= σeaB1B3, (9)

∂2p

∂x2∂x3= σeacB2B3,

∂2p

∂x3∂x2= −σea(1 + c)B2B3. (10)

Therefore, since c is arbitrary (> −1), conditions (8), (9), (10) are satisfied ifand only if B = Be1 or B = Be2 or B = Be3.

Finally if B = Be1 then we deduce

p = −ρv2(x1, x2, x3)

2+

a

2σeB

2[(1 + c)x22 − cx23] + p0; (11)

if B = Be2 then we deduce

p = −ρv2(x1, x2, x3)

2− a

2σeB

2[x21 + cx23] + p0; (12)

if B = Be3 then we deduce

p = −ρv2(x1, x2, x3)

2− a

2σeB

2[x21 − (1 + c)x22] + p0. (13)


Remark 3. The results obtained in Theorem 2 hold for any c > −1. Weremark that if c = 1, then it is possible to consider also the magnetic fieldparallel to the plane Ox1x3 as one can see from (9). In this case the pressurebecomes

p = −ρv2(x1, x2, x3)

2+ aσe(B

21 +B2

3)x22 −

a

2σe(B3x1 −B1x3)

2 + p0.

Moreover, if c = −1

2then it is possible to consider the magnetic field parallel

to the plane Ox2x3, as one can see from (10). The corresponding pressure is

p = −ρv2(x1, x2, x3)

2− a

2σe(B

22 +B2

3)x21 +

a

4σe(B3x2 −B2x3)

2 + p0.

Remark 4. As it is well known, in the absence of the external magneticfield the stagnation point is the point where the pressure assumes its maximum.We notice that, from (11), (12), (13), the pressure along the wall x2 = 0 takesits maximum in the stagnation-point.

Remark 5. In order to study the three-dimensional stagnation-point flowfor other fluids, it is convenient to consider a more general flow. More precisely,we suppose the fluid impinging on the flat plane x2 = C and

v1 = ax1, v2 = −(a+ b)(x2 − C), v3 = bx3, (x1, x3) ∈ R2, x2 ≥ C, (14)

with C some constant.In this way, the stagnation point is not the origin but the point (0, C, 0).As it is easy to verify, in the cases of Theorem 2 and Remark 3 the pressure

must be modified by replacing x2 with x2 − C.

3. Newtonian Fluids

Consider the steady three-dimensional stagnation-point flow of an electricallyconducting homogeneous incompressible Newtonian fluid towards a flat surfacecoinciding with the plane x2 = 0, the flow being confined to the half-space S.In the absence of external mechanical body forces, the MHD equations for sucha fluid are

v · ∇v = −1

ρ∇p+ ν△v+

µe

ρ(∇×H)×H,

∇ · v = 0,


∇×H = σe(E+ µev×H),

∇×E = 0, ∇ ·E = 0, ∇ ·H = 0 in S (15)

where ν is the kinematic viscosity.As far as the boundary condition for v is concerned, we prescribe the ad-

herence condition, i.e.

v|x2=0 = 0. (16)

We search v in the following form:

∀(x1, x2, x3) ∈ S

v1 = ax1f′(x2), v2 = −[af(x2) + bg(x2)], v3 = bx3g

′(x2), (17)

where f, g are sufficiently regular unknown functions. As for an inviscid fluid, wesuppose a > 0, b 6= 0, so that if b > 0 then we have a flow in the neighborhood

of a nodal point of attachment, while if b < 0 and c =b

a> −1, then we have a

flow in the neighborhood of a saddle point of attachment.The condition (16) supplies

f(0) = 0, f ′(0) = 0, g(0) = 0, g′(0) = 0. (18)

Moreover, as is customary when studying the stagnation-point flow for vis-cous fluids, we assume that at infinity, the flow approaches the flow of an inviscidfluid, whose velocity is given by (14).

Therefore, to (18) we also must append the following conditions

limx2→+∞

f ′(x2) = 1, limx2→+∞

g′(x2) = 1. (19)

The constant C in (14) is related to the behaviour of f and g at infinity.Actually, if

limx2→+∞

[f(x2)− x2] = −A, limx2→+∞

[g(x2)− x2] = −B (20)

with A,B some constants, then

limx2→+∞

[f(x2) + cg(x2)− (1 + c)x2] = −(1 + c)C, (21)

where

C =A+ cB

1 + c, c > −1.


The constants A,B,C are not assigned a priori, but their values can be foundby solving the problem.

In order to study the influence of a uniform external electromagnetic field,we continue to use the approximation (4), where v is given by (17). As a resultof the Theorem 2, we consider the following three cases.

3.1. Case I

H0 = H0e1.

(∇×H)×H ≃ σeµea{[H20 (f + cg)]e2 − cH2

0g′x3e3}. (22)

We substitute (17), and (22) in (15)1 to obtain

ax1

[

νf ′′′ + af ′′(f + cg)− af ′2]

=1

ρ

∂p

∂x1,

− νa(f ′′ + cg′′)− a2(f ′ + cg′)(f + cg) +σea

ρB2

0(f + cg) =1

ρ

∂p

∂x2,

acx3

[

νg′′′ + ag′′(f + cg)− acg′2 − σe

ρB2

0g′

]

=1

ρ

∂p

∂x3. (23)

Then, by integrating (23)2, we find

p =− 1

2ρa2[f(x2) + cg(x2)]

2 − ρaν[f ′(x2) + cg′(x2)]

+ σeaB20

∫ x2

0

[f(s) + cg(s)]ds + P (x1, x3),

where the function P (x1, x3) is determined supposing that, far from the wall,the pressure p has the same behaviour as for an inviscid fluid, whose velocityis given by (14) and the pressure is given by (11) replacing x2 by x2 − C.

Therefore, by virtue of (19), and (20), we get

P (x1, x3) = −ρa2

2(x1

2 + c2x23)−a

2σeB

20cx

23 + p∗0,

where p∗0 is a suitable constant. Finally, the pressure field assumes the form

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23} − ρaν[f ′(x2) + cg′(x2)]


+ σeaB20

{∫ x2

0

[f(s) + cg(s)]ds − c

2x23

}

+ p0, (24)

where the constant p0 is the pressure at the origin.In consideration of (24), we obtain the ordinary differential system

ν

af ′′′ + (f + cg)f ′′ − f ′2 + 1 = 0,

ν

ag′′′ + (f + cg)g′′ − cg′

2+ c+M2(1− g′) = 0, (25)

where

M2 =σeB

20

ρa

is the Hartmann number. To these equations we append the boundary condi-tions (18), and (19).

As far as the other two cases are concerned, if we proceed as previously weget

3.2. Case II

H0 = H0e2.

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23} − ρaν[f ′(x2) + cg′(x2)]

− σeaB20(x

21 + cx23) + p0, (26)

ν

af ′′′ + (f + cg)f ′′ − f ′2 + 1 +M2(1− f ′) = 0,

ν

ag′′′ + (f + cg)g′′ − cg′

2+ c+M2(1− g′) = 0. (27)

3.3. Case III

H0 = H0e3.

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23} − ρaν[f ′(x2) + cg′(x2)]


+ σeaB20

{∫ x2

0

[f(s) + cg(s)]ds − x212

}

+ p0, (28)

ν

af ′′′ + (f + cg)f ′′ − f ′2 + 1 +M2(1− f ′) = 0,

ν

ag′′′ + (f + cg)g′′ − cg′

2+ c = 0. (29)

We have the following:

Theorem 6. Let a homogeneous, incompressible, electrically conductingNewtonian fluid occupy the half-space S. If we impress the external magneticfield H0 parallel to one of the coordinate axes and if we neglect the inducedmagnetic field, then the steady three-dimensional MHD stagnation-point flowof such a fluid has the form


′(x2), E = 0,

and

1. if H0 = H0e1, then the pressure field is given by (24) and (f, g) satisfiesproblem (25), (18), and (19);

2. if H0 = H0e2, then the pressure field is given by (26) and (f, g) satisfiesproblem (27), (18), and (19);

3. if H0 = H0e3, then the pressure field is given by (28) and (f, g) satisfiesproblem (29), (18), and (19).

Remark 7. We remark that from (24), (26), (28) the pressure along thewall x2 = 0 takes its maximum again in the stagnation-point.

We now analyse the cases considered in Remark 3.

Proposition 8. Let a homogeneous, incompressible, electrically conduct-ing Newtonian fluid occupy the half-space S. If we neglect the induced magneticfield and we suppose either

i) c = 1, H0 parallel to the plane Ox1x3,

or

ii) c = −1

2, H0 parallel to the plane Ox2x3,


then there is no solution to the problem of the steady three-dimensional MHDstagnation-point flow.

Proof. i) If c = 1 and the external magnetic induction field is B = B1e1 +B3e3 (B1, B3 6= 0), then from equations (15), (17), proceeding as in Case I,after some calculations, we deduce:

νf ′′′ + a(f + g)f ′′ − af ′2 +B23

σe

ρ(1− f ′) + a = 0,

σea

ρB1B3[g

′ − 1] = 0,

νg′′′ + a(f + g)g′′ − ag′2+B2

1

σe

ρ(1− g′) + a = 0,

σea

ρB1B3[f

′ − 1] = 0. (30)

From (30)2, (30)4 one has f ′ = g′ = 1 for all x2 ≥ 0 which contradicts theboundary conditions (18).

ii) Then we examine the case c = −1

2and B = B2e2 +B3e3 (B2, B3 6= 0).

By proceeding as above, we obtain

νf ′′′ + a(f − g

2)f ′′ − af ′2 + (B2

2 +B23)σe

ρ(1− f ′) + a = 0,

νg′′′ + a(f − g

2)g′′ +

a

2g′

2+B2

2

σe

ρ(1− g′)− a

2= 0,

σea

ρB2B3(f − g +A−B) = 0. (31)

From (31)3 evaluated at x2 = 0 and (18), we obtain f = g. So (31)1, (31)2become

νf ′′′ + af

2f ′′ − af ′2 + (B2

2 +B23)σe

ρ(1− f ′) + a = 0,

νf ′′′ + af

2f ′′ +

a

2f ′2 +B2

2

σe

ρ(1− f ′)− a

2= 0. (32)

By subtracting (32)1 from (32)2, we arrive at

(f ′ − 1)

[

3

2(f ′ + 1) +B2

3

σe

aρ

]

= 0. (33)

At x2 = 0, (33) gives the absurdum

3

2+

σe

aρB2

3 = 0.


Remark 9. If c = 1, f = g, H0 = H0e2, the axial symmetric case isobtained.

Finally, it is convenient to write the boundary value problems in Theorem6 in dimensionless form in order to reduce the number of the parameters. Tothis end we put

η =

√

a

νx2, ϕ(η) =

√

a

νf

(√

ν

aη

)

, γ(η) =

√

a

νg

(√

ν

aη

)

. (34)

So we can rewrite equations (25) as:

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c+M2(1− γ′) = 0; (35)

equations (27) as

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 +M2(1− ϕ′) = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c+M2(1− γ′) = 0; (36)

equations (29) as

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 +M2(1− ϕ′) = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c = 0. (37)

Of course we obtain three different ordinary differential problems by adjoiningthe boundary conditions in dimensionless form:

ϕ(0) = 0, ϕ′(0) = 0,

γ(0) = 0, γ′(0) = 0,

limη→+∞

ϕ′(η) = 1, limη→+∞

γ′(η) = 1. (38)

Remark 10. If M = 0, then equations (35), (36), (37) reduce to

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c = 0 (39)

which are the dimensionless equations governing the flow of a Newtonian fluidin the absence of H0. We recall that in [1] it is proved that the problem (39),(38) does not admit solution for c < −1. As far as existence of solutions isconcerned, we refer to [7], [8].


Remark 11. It is physically interesting to determine the skin-frictioncomponents τ1, τ3 along x1 and x3 axes :

τ1 = µ

(

∂v1

∂x2

)

x2=0

= ρ(ν)1/2a3/2x1ϕ′′(0),

τ3 = µ

(

∂v3

∂x2

)

x2=0

= cρ(ν)1/2a3/2x3γ′′(0). (40)

Formally τ1, τ3 have the same expression as in the absence of H0, but of courseϕ′′(0), γ′′(0) depend on H0 through the Hartmann number M2.

Remark 12. It is easy to verify that regular solutions to problem (36),(38) are invariant under the following transformation

ϕ

(

η,1

c,M2

c

)

=√c γ

(

η√c, c,M2

)

,

γ

(

η,1

c,M2

c

)

=√cϕ

(

η√c, c,M2

)

, c > 0.

Moreover we notice that by means of the previous transformation the regularsolutions to problem (35), (or (37)), (38) are transformed in the solutions toproblem (37), (or (35)), (38).

4. Micropolar Fluids

Consider now the steady three-dimensional stagnation-point flow of an electri-cally conducting homogeneous incompressible micropolar fluid towards a flatsurface coinciding with the plane x2 = 0, the flow being confined to the half-space S. In the absence of external mechanical body forces and body couples,the MHD equations for such a fluid are ([11])

v · ∇v = −1

ρ∇p+ (ν + νr)△v + 2νr(∇×w) +

µe

ρ(∇×H)×H,

∇ · v = 0,

Iv · ∇w = λ△w + λ0∇(∇ ·w)− 4νrw + 2νr(∇× v),

∇×H = σe(E+ µev×H),

∇×E = 0, ∇ · E = 0, ∇ ·H = 0 in S (41)

where w is the microrotation field, ν is the kinematic newtonian viscosity co-efficient, νr is the microrotation viscosity coefficient, λ, λ0 (positive constants)


are material parameters related to the coefficient of angular viscosity and I isthe microinertia coefficient.

We notice that in [4], [5], eqs. (41)1,3 are slightly different, as they arededuced as a special case of much more general model of microfluids. For thedetails, we refer to [11], p.23.

As far as the boundary conditions are concerned, we prescribe the appro-priate boundary condition for the velocity v and the microrotation w, i.e.

v|x2=0 = 0, w|x2=0 = 0 (strict adherence condition). (42)

Other boundary conditions are possible. We refer to Eringen ([4], p.17-18) fora complete discussion. However in our studies we will always assume the strictadherence condition.

We search v, w in the following form


′(x2),

w1 = −cx3F (x2), w2 = 0, w3 = x1G(x2), (x1, x2, x3) ∈ S, (43)

where f, g, F,G are sufficiently regular unknown functions and c =b

a.

The conditions (42) supply

f(0) = 0, f ′(0) = 0, g(0) = 0, g′(0) = 0,

F (0) = 0, G(0) = 0. (44)

Moreover, we assume that at infinity, the flow approaches the flow of aninviscid fluid, whose velocity is given by (14). Therefore, to (41) we mustappend also the following conditions

limx2→+∞

f ′(x2) = 1, limx2→+∞

g′(x2) = 1,

limx2→+∞

F (x2) = 0, limx2→+∞

G(x2) = 0. (45)

Conditions (45)3,4 mean that at infinity, w =1

2∇× v = 0, i.e. the micropolar

fluid behaves like an ideal fluid.

The constant C is related to the asymptotic behaviour of f and g at infinityas in the Newtonian case. So equation (21) continues to hold.

In order to study the influence of a uniform external electromagnetic field,we continue to use the approximation (4), where v is given by (43)1,2,3. As aresult of the Theorem 2, we consider the following three cases.


4.1. Case I

H0 = H0e1.

(∇×H)×H ≃ σeµea{[H20 (f + cg)]e2 − cH2

0g′x3e3}. (46)

We substitute (43) in (41)1,3 to obtain

ax1

[

(ν + νr)f′′′ + af ′′(f + cg) − af ′2 +

2νra

G′

]

=1

ρ

∂p

∂x1,

− (ν + νr)a(f′′ + cg′′)− a2(f ′ + cg′)(f + cg) − 2νr(cF +G)

+σea

ρB2

0(f + cg) =1

ρ

∂p

∂x2,

cx3

[

(ν + νr)g′′′ + ag′′(f + cg) − cg′

2+

2νra

F ′ − σe

ρB2

0g′

]

=1

ρ

∂p

∂x3,

c{λF ′′ + Ia[F ′(f + cg)− cFg′]− 2νr(2F + ag′′)} = 0,

λG′′ + Ia[G′(f + cg)−Gf ′]− 2νr(2G+ af ′′) = 0. (47)

Since we are interested in three-dimensional flow, we assume c 6= 0 and soequation (47)4 can be replaced by

λF ′′ + Ia[F ′(f + cg)− cFg′]− 2νr(2F + ag′′) = 0. (48)

Then, by integrating (47)2, we find

p =− 1

2ρa2[f(x2) + cg(x2)]

2 − ρa(ν + νr)[f′(x2) + cg′(x2)]

− 2νrρ

∫ x2

0

[cF (s) +G(s)]ds + σeaB20

∫ x2

0

[f(s) + cg(s)]ds

+ P (x1, x3),

where the function P (x1, x3) is determined supposing that, far from the wall,the pressure p has the same behaviour as for an inviscid fluid, whose velocityis given by (14) and the pressure is given by (11) replacing x2 by x2 − C.

Therefore, under the assumption F,G ∈ L1([0,+∞)), by virtue of (45),(20), we get

P (x1, x3) = −ρa2

2(x1

2 + c2x23)−a

2σeB

20cx

23 + p∗0,


where p∗0 is a suitable constant. Finally, the pressure field assumes the form

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23}

− ρa(ν + νr)[f′(x2) + cg′(x2)]− 2νrρ

∫ x2

0

[cF (s) +G(s)]ds

+ σeaB20

{∫ x2

0

[f(s) + cg(s)]ds − c

2x23

}

+ p0, (49)

where the constant p0 is the pressure at the origin.

In consideration of (49), we obtain the ordinary differential system

ν + νr

af ′′′ + (f + cg)f ′′ − f ′2 + 1 +

2νra2

G′ = 0,

ν + νr

ag′′′ + (f + cg)g′′ − cg′

2+ c+

2νra2

F ′ +M2(1− g′) = 0, (50)

where M2 =σeB

20

ρais the Hartmann number. To these equations we append

equations (47)5, and (48) and the boundary conditions (44), and (45).

As far as the other two cases are concerned, if we proceed as previously weget

4.2. Case II

H0 = H0e2.

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23}

− ρa(ν + νr)[f′(x2) + cg′(x2)]

− 2νrρ

∫ x2

0

[cF (s) +G(s)]ds − σeaB20(x

21 + cx23) + p0, (51)

ν + νr

af ′′′ + (f + cg)f ′′ − f ′2 + 1 +

2νra2

G′ +M2(1− f ′) = 0,

ν + νr

ag′′′ + (f + cg)g′′ − cg′

2+ c+

2νra2

F ′ +M2(1− g′) = 0. (52)


4.3. Case III

H0 = H0e3.

p =− ρa2

2{x21 + [f(x2) + cg(x2)]

2 + c2x23}

− ρa(ν + νr)[f′(x2) + cg′(x2)]− 2νrρ

∫ x2

0

[cF (s) +G(s)]ds

+ σeaB20

{

∫ x2

0

[f(s) + cg(s)]ds − x212

}

+ p0, (53)

ν + νr

af ′′′ + (f + cg)f ′′ − f ′2 + 1 +

2νra2

G′ +M2(1− f ′) = 0,

ν + νr

ag′′′ + (f + cg)g′′ − cg′

2+ c+

2νra2

F ′ = 0. (54)

Thus, we have the following:

Theorem 13. Let a homogeneous, incompressible, electrically conductingmicropolar fluid occupy the half-space S. If we impress the external magneticfield H0 parallel to one of the coordinate axes and if we neglect the inducedmagnetic field, then the steady three-dimensional MHD stagnation-point flowof such a fluid has the form


′(x2),

w1 = −cx3F (x2), w2 = 0, w3 = x1G(x2), E = 0,

and

1. if H0 = H0e1, then the pressure field is given by (49) and (f, g, F,G) satis-fies problem (50), (47)5, (48), (44), and (45), provided F,G ∈ L1([0,+∞));

2. if H0 = H0e2, then the pressure field is given by (51) and (f, g, F,G) satis-fies problem (52), (47)5, (48), (44), and (45), provided F,G ∈ L1([0,+∞));

3. if H0 = H0e3, then the pressure field is given by (53) and (f, g, F,G) satis-fies problem (54), (47)5, (48), (44), and (45), provided F,G ∈ L1([0,+∞)).

Remark 14. We remark that, from (49), (51), (53), the pressure alongthe wall x2 = 0 takes its maximum again in the stagnation-point.


Remark 15. If c = 1, f = g, F = G, H0 = H0e2, the axial symmetriccase is obtained.

Now it is convenient to write the boundary value problems in Theorem 13 indimensionless form in order to reduce the number of the material parameters.To this end we put

η =

√

a

ν + νrx2, ϕ(η) =

√

a

ν + νrf

(

√

ν + νr

aη

)

,

γ(η) =

√

a

ν + νrg

(

√

ν + νr

aη

)

, Φ(η) =2νra2

√

a

ν + νrF

(

√

ν + νr

aη

)

,

Γ(η) =2νra2

√

a

ν + νrG

(

√

ν + νr

aη

)

. (55)

So system (50), (47)5, (48) can be written as

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 + Γ′ = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c+Φ′ +M2(1− γ′) = 0,

Φ′′ + c3Φ′(ϕ+ cγ)− Φ(c3cγ

′ + c2)− c1γ′′ = 0,

Γ′′ + c3Γ′(ϕ+ cγ)− Γ(c3ϕ

′ + c2)− c1ϕ′′ = 0, (56)

where

c1 =4ν2rλa

, c2 =4νr(ν + νr)

λa, c3 =

I

λ(ν + νr). (57)

The boundary conditions in dimensionless form become:

ϕ(0) = 0, ϕ′(0) = 0,

γ(0) = 0, γ′(0) = 0,

Φ(0) = 0, Γ(0) = 0,

limη→+∞

ϕ′(η) = 1, limη→+∞

γ′(η) = 1,

limη→+∞

Φ(η) = 0, limη→+∞

Γ(η) = 0. (58)

Moreover, equations (52) can be written as

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 + Γ′ +M2(1− ϕ′) = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c+Φ′ +M2(1− γ′) = 0; (59)


and equations (54) as

ϕ′′′ + (ϕ+ cγ)ϕ′′ − ϕ′2 + 1 + Γ′ +M2(1− ϕ′) = 0,

γ′′′ + (ϕ+ cγ)γ′′ − cγ′2+ c+Φ′ = 0. (60)

Of course we obtain other two different ordinary differential problems by ad-joining equations (56)3,4 and the boundary conditions (58).

We now analyse the cases considered in Remark 2. For sake of simplicitywe use the dimensionless equations.

Proposition 16. Let a homogeneous, incompressible, electrically con-ducting micropolar fluid occupy the half-space S. If we neglect the inducedmagnetic field and we suppose either

i) c = 1, H0 parallel to the plane Ox1x3,

or

ii) c = −1

2, H0 parallel to the plane Ox2x3,

then there is no solution to the problem of steady three-dimensional MHDstagnation-point flow.

Proof. i) If c = 1 and the external magnetic induction field is B = B1e1 +B3e3 (B1, B3 6= 0), then after some calculations we deduce:

ϕ′′′ + (ϕ+ γ)ϕ′′ − ϕ′2 +M23 (1− ϕ′) + 1 + Γ′ = 0,

M1M3(γ′ − 1) = 0,

γ′′′ + (ϕ+ γ)γ′′ − γ′2+M2

1 (1− γ′) + 1 + Φ′ = 0,

M1M3(ϕ′ − 1) = 0, (61)

where M2i =

σeB2i

ρa, i = 1, 3.

From (61)2, (61)4, we have ϕ′ = γ′ = 1, ∀η ≥ 0, which contradicts theboundary conditions (58)2,4.

ii) If c = −1

2and B = B2e2 +B3e3 (B2, B3 6= 0), then we arrive at:

ϕ′′′ +(

ϕ− γ

2

)

ϕ′′ − ϕ′2 + (M22 +M2

3 )(1 − ϕ′) + 1 + Γ′ = 0,

γ′′′ +(

ϕ− γ

2

)

γ′′ +γ′

2

2+M2

2 (1− γ′)− 1

2+ Φ′ = 0,


M2M3(ϕ− γ + α− β) = 0, (62)

where

α =

√

a

ν + νrA, β =

√

a

ν + νrB.

From (62)3 evaluated at η = 0 and using (58)1,3, we deduce ϕ = γ. Thereforewe have to solve the following overdetermined ODE system

ϕ′′′ +ϕ

2ϕ′′ − ϕ′2 + (M2

2 +M23 )(1− ϕ′) + 1 + Γ′ = 0,

ϕ′′′ +ϕ

2ϕ′′ +

ϕ′2

2+M2

2 (1− ϕ′)− 1

2+ Φ′ = 0,

Φ′′ + c3Φ′ϕ

2+ Φ

(c3

2ϕ′ − c2

)

− c1ϕ′′ = 0,

Γ′′ + c3Γ′ϕ

2− Γ(c3ϕ

′ + c2)− c1ϕ′′ = 0, (63)

together with the boundary conditions (58).Our aim is to prove that such a problem does not admit solution. To this

end, by subtracting (63)2 to (63)1, we obtain that (ϕ,Φ,Γ) solves the equation

3

2(ϕ′2 − 1) +M2

3 (ϕ′ − 1) + Φ′ − Γ′ = 0. (64)

Computing (64), (63)3,4 at η = 0, gives

Φ′(0) − Γ′(0) =3

2+M2

3 , Φ′′(0) = Γ′′(0) = c1ϕ′′(0). (65)

If we differentiate (64), then we have

3ϕ′ϕ′′ +M23ϕ

′′ +Φ′′ − Γ′′ = 0. (66)

By means of (65)2, (66) we deduce

Φ′′(0) = Γ′′(0) = ϕ′′(0) = 0. (67)

After differentiating (63)3,4 we get

Γ′′′(0) = −c2(1 +M22 +M2

3 ) + (c1 − c2)ϕ′′′(0),

Φ′′′(0)− Γ′′′(0) = c2

(3

2+M2

3

)

, (68)

where we have used (65)1, and (63)1.


If we differentiate (66), then we obtain

3ϕ′′2 + 3ϕ′ϕ′′′ +M23ϕ

′′′ +Φ′′′ − Γ′′′ = 0 (69)

from which, taking account of (67)3, follows

ϕ′′′(0) = −Φ′′′(0) − Γ′′′(0)

M23

. (70)

Further, from (63)1,2 we get

Φ′(0) =1

2−M2

2 − ϕ′′′(0), Γ′(0) = −[1 +M22 +M2

3 + ϕ′′′(0)]. (71)

Differentiating of (69) furnishes:

9ϕ′′ϕ′′′ + 3ϕ′ϕIV +M23ϕ

IV +ΦIV − ΓIV = 0. (72)

Another differentiation of (63)3,4 gives

ΦIV (0) = ΓIV (0) = c1ϕIV (0),

so that (72) yields

ΦIV (0) = ΓIV (0) = ϕIV (0) = 0. (73)

If we differentiate (72) and evaluate the resulting equation at η = 0, we obtain

9ϕ′′′(0)2 +M23ϕ

V (0) + ΦV (0)− ΓV (0) = 0. (74)

On the other hand we can find ϕV (0) from (63)1 after two differentiations:

ϕV (0) = (M22 +M2

3 )ϕ′′′(0)− Γ′′′(0). (75)

Moreover, by means of another differentiation of (63)3,4 we arrive at

ΦV (0)− ΓV (0) = c2[Φ′′′(0)− Γ′′′(0)] − c3ϕ

′′′(0)[

2Φ′(0) +5

2Γ′(0)

]

. (76)

Finally on substituting (75), (76) into (74) and taking into account (68)1, (70),(71), we get

9

2(2 + c3)ϕ

′′′(0)2 +[

M23 (M

22 +M2

3 − c1 +5

2c3) +

c3

2(9M2

2 + 3)]

ϕ′′′(0)

+ c2M23 (1 +M2

2 +M23 ) = 0. (77)


The thesis follows easily because, as one can see from (68), (70), ϕ′′′(0) doesnot depend on c1 and by differentiating (77) with respect to c1, we obtain

M23ϕ

′′′(0) = 0, (78)

which gives the absurdum

3

2+

σe

aρB2

3 = 0,

as for a Newtonian fluid.

Remark 17. If M = 0, then equations (56)1,2 (or (59), or (60)), (56)3,4reduce to equations found by Guram and Anwar Kamal in [6].

Remark 18. The skin-friction components τ1, τ3 along x1 and x3 axesare given by (39). Actually ϕ′′(0), γ′′(0) depend on H0 through M2.

Remark 19. It is easy to verify that regular solutions of problem (59),(56)3,4, (58) are invariant under the following transformation

ϕ

(

η,1

c,c1

c,c2

c, c3,

M2

c

)

=√c γ

(

η√c, c, c1, c2, c3,M

2

)

,

γ

(

η,1

c,c1

c,c2

c, c3,

M2

c

)

=√c ϕ

(

η√c, c, c1, c2, c3,M

2

)

,

Φ

(

η,1

c,c1

c,c2

c, c3,

M2

c

)

=1√cΓ

(

η√c, c, c1, c2, c3,M

2

)

,

Γ

(

η,1

c,c1

c,c2

c, c3,

M2

c

)

=1√cΦ

(

η√c, c, c1, c2, c3,M

2

)

, c > 0.

Moreover we notice that by means of the previous transformation regular solu-tions of problem (56)1,2 (or (60)), (56)3,4, (58) are transformed in solutions ofproblem (60) (or (56)1,2), (56)3,4, (58).

The previous transformation continues to hold even if M = 0.

Remark 20. In all cases here considered the results continue to hold evenif there are external conservative body forces by modifying the pressure fieldappropriately.

Remark 21. The nonlinear boundary value problems here obtained arenot solvable by means of analytical functions, but only through numerical meth-ods. In a subsequent paper we will give the numerical integration of theseboundary value problems.


Acknowledgments

The Authors would like to express their gratitude to Professor Andrea Corlifor valuable suggestions and comments. This work was performed under theauspices of G.N.F.M. of INdAM and supported by Italian MIUR.

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