# An Analytic Study of the Reversal of Hartmann Flows by Rotating Magnetic Fields

Authors: Manuel Núñez
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

The effects of a background uniform rotating magnetic field acting in a conducting fluid with a parallel flow are studied analytically. The stationary version with a transversal magnetic field is well known as generating Hartmann boundary layers. The Lorentz force includes now one term depending on the rotation speed and the distance to the boundary wall. As one intuitively expects, the rotation of magnetic field lines pushes backwards or forwards the flow. One consequence is that near the wall the flow will eventually reverse its direction, provided the rate of rotation and/or the magnetic field are large enough. The configuration could also describe a fixed magnetic field and a rotating flow.


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## Body

## 1. Introduction

In [1], a generalization of the Hartmann flow was announced. The Hartmann flow is a parallel one influenced by a strong transversal magnetic field [2, 3] and has been studied intensively for its role in many aspects of plasma physics, such as liquid metal pumping [4, 5], plasma convection [6, 7], and geophysics [8]. We now will allow the background magnetic field to rotate while keeping its spatial uniformity. Our aim in this paper is to analyze in depth the action of this field upon the flow through the Lorentz force, emphasizing the possibility of flow reversal at some time near the walls confining the fluid. Given the Galilean invariance of the relevant equations, we could change to a rotating reference frame and interpret the problem as consisting in a fixed magnetic field and a rotating flow; this configuration could perhaps be more easily constructed and have wider applications.Let us begin by recalling briefly the MHD equations appropriate for our configuration. The flow will be assumed two dimensional, and the background magnetic field will be uniform in space but rotating in time at a rate given by the angleθ(t):
(1.1)B(t)=B(cos⁡θ(t),sin⁡θ(t)).Although this field will create a secondary oneb, it is assumed that all the terms in b may be neglected when an analogous one for B is present; thus, |u×b|≪|u×B|, |b˙|≪|B˙|. This does not hold for the current density, as J=∇×B=0, j=∇×b≠0. The approximate induction equation becomes
(1.2)∂B∂t=∇×(-ηj+u×B),
where u is the flow velocity and η the resistivity. By substituting B by its value in (1.1), uncurling (1.2) to find j and taking it to the Lorents force j×B, we may write the momentum equation as
(1.3)∂u∂t+u⋅∇u=νΔu+σ(u×B)×B-σB2θ˙(y,0)+∇f,
where σ=η-1 is the conductivity, ν the viscosity, and f is the sum of the pressure and the gauge obtained by uncurling (1.2). Detailed calculations may be found in [1]. If in analogy with the classical Hartmann flow we assume a horizontal flow u(x,y)=(u(y),0), (1.3) reduces to
(1.4)∂u∂t=ν∂2u∂y2-σB2usin⁡2θ-σB2θ˙y+C,
where C includes all the conservative forces upon the fluid, such as the pressure and a possible electrostatic field. The interpretation is rather easy; as the magnetic field lines rotate at angular velocity θ, their linear velocity grows linearly with the radius y. This has the undesirable effect of providing infinite energy to the flow, but this should not deter us from studying the model. Other classical flows, such as Karman's, are the object of a vast literature in relation with swirling flows [9, 10] and also possess infinite energy. Their local properties are generalizable to many realistic situations.We assume that the fluid is bounded either by one no-slip wall or two. In the first case, initial and boundary conditions for (1.4) are
(1.5)u(0,t)=0,u(y,0)=ϕ(y),
whereas for the second one, if the walls are separated by a distance L,
(1.6)u(0,t)=0,u(L,t)=0,u(y,0)=ϕ(y).
Obviously ϕ must satisfy at the walls the same conditions as u.


## 2. Flow with a Single Wall

We will solve (1.4), (1.5) for t≥0, y≥0. To simplify calculations and highlight the contributions of the different terms, we will consider separately the following three problems:
(2.1)∂u1∂t=ν∂2u1∂y2-σB2u1sin⁡2θ+C,u1(0,t)=0,u1(y,0)=0,∂u2∂t=ν∂2u2∂y2-σB2u2sin⁡2θ-σB2θ˙y,u2(0,t)=0,u2(y,0)=0,∂u3∂t=ν∂2u3∂y2-σB2u3sin⁡2θ,u3(0,t)=0,u3(y,0)=ϕ(y).
Then, u=u1+u2+u3 is the desired solution. From now on, we will denote
(2.2)F(t)=σB2∫0t‍sin⁡2θ(s)ds,wj(y,t)=eF(t)uj(y,t).
Notice that wj satisfies the same equation as uj, except that the term in σB2ujsin2θ disappears, and the forcing is multiplied by eF(t). Although the general analysis is valid for any function θ(t), in order to obtain explicit results we will consider a constant rotation θ(t)=λt; the rotation is counterclockwise if λ>0, clockwise otherwise. Then,
(2.3)F(t)=σB2(t2-sin⁡2λt4),
so that for large t(2.4)eF(t)~eσB2t/2,e-F(t)~e-σB2t/2.


### 2.1. Effect of the Constant Forcing

Leterf denote the error function,
(2.5)erf⁡(x)=2π∫0xe-t2dt.
Then the solution of
(2.6)∂w1∂t=ν∂2w1∂y2+CeF(t)w1(0,t)=0w1(y,0)=0
is
(2.7)w1(y,t)=∫0terf⁡(y4ν(t-s))CeF(s)ds,
so that
(2.8)u1(y,t)=∫0terf⁡(y4ν(t-s))Ce-F(t)+F(s)ds.
This implies
(2.9)∂u1∂y=2Cπ∫0t‍14ν(t-s)exp⁡(-y24ν(t-s)-F(t)+F(s))ds.Let us study the limit of this function whent→∞ in the case of constant rotation, θ(t)=λt, λ>0. It is an easy consequence of Lebesgue's theorem that, for any fixed t0>0,
(2.10)lim⁡t→∞∫0t014ν(t-s)exp⁡(-y24ν(t-s)-F(t)+F(s))ds=0.Taket0 large enough that we may substitute F(t) by σB2t/2. It is enough to consider
(2.11)2Cπ∫t0t14ν(t-s)exp⁡(-y24ν(t-s)-σB2(t-s)2)ds=2Cπ∫0t-t014νsexp⁡(-y24νs-σB2s2)ds,
whose limit when t→∞ is
(2.12)2Cπ∫0∞14νsexp⁡(-y24νs-σB2s2)ds,
which is obviously a finite integral. It reaches its maximum at y=0, where its value is
(2.13)Cπν∫0∞s-1/2e-σB2s/2ds=Cπν2σB2∫0∞r-1/2e-rdr=Cπν2σB2Γ(12)=Cπν2σB2π=C2νσB2=C2BPm,
where Pm is the magnetic Prandtl number. Therefore, u1 also has a limit when t→∞ for every y>0, given by the integral of (2.12) with respect to y.


### 2.2. Effect of the Lorentz Force

w
2 satisfies
(2.14)∂w2∂t=ν∂2w2∂y2-σB2θ˙y,w2(0,t)=0,w2(y,0)=0.
Although there exists an appropriate Green function for this problem, in this particular case it is simpler to assume that the solution is linear in y (which is correct). Setting first
(2.15)w2(y,t)=h(t)y,h(0)=0,
and recovering u2 from it, one finds
(2.16)u2(y,t)=-yσB2∫0texp⁡(-F(t)+F(s))θ˙(s)ds.
Therefore,
(2.17)∂u2∂y(y,t)=-σB2∫0texp⁡(-F(t)+F(s))θ˙(s)ds.
This function also has a limit when t→∞. Its value may be found by an argument similar to the previous one, although simpler; in this case
(2.18)lim⁡t→∞-λσB2∫t0texp⁡(-σB2(t-s)2)ds=-2λ.
Hence, u2(y,t) tends to -2λy when t→∞. Notice how the Lorentz force pushes the fluid backwards, the more rapidly the higher the distance from the wall.


### 2.3. Effect of the Initial Condition

w
3 satisfies
(2.19)∂w3∂t=∂2w3∂y2w3(0,t)=0w3(y,0)=ϕ(y).ϕ must satisfy ϕ(0)=0. The solution is found by the method of images: ϕ is extended to (-∞,∞) as an odd function. Then,
(2.20)w3(y,t)=14πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ(z)dz,
so that
(2.21)u3(y,t)=e-F(t)4πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ(z)dz.
Therefore, provided ϕ is differentiable,
(2.22)∂u3∂y(y,t)=e-F(t)4πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ′(z)dz.
All this supposes that ϕ and ϕ′ satisfy certain reasonable conditions: for this example. if both are bounded. In that case, given that
(2.23)14πνt∫-∞∞exp⁡(-(y-z)24νt)dz=1,
both u3 and ∂u3/∂y are bounded by a function of the form Mexp(-F(t)), which means that they tend to zero as t→∞.Thus, the limit whent→∞ of u(0,t) is
(2.24)k=C2BPm-2λ.Ifk<0, we may be certain that the fluid inverts its direction and flows backwards from some point on. The factors contributing to this are a small (or negative) potential force C, which includes pressure and electric fields; a large background magnetic field; a large magnetic Prandtl number, which means that the viscosity dominates the resistivity; finally and most obviously, a large rate of rotation of the magnetic field.


## 2.1. Effect of the Constant Forcing

Leterf denote the error function,
(2.5)erf⁡(x)=2π∫0xe-t2dt.
Then the solution of
(2.6)∂w1∂t=ν∂2w1∂y2+CeF(t)w1(0,t)=0w1(y,0)=0
is
(2.7)w1(y,t)=∫0terf⁡(y4ν(t-s))CeF(s)ds,
so that
(2.8)u1(y,t)=∫0terf⁡(y4ν(t-s))Ce-F(t)+F(s)ds.
This implies
(2.9)∂u1∂y=2Cπ∫0t‍14ν(t-s)exp⁡(-y24ν(t-s)-F(t)+F(s))ds.Let us study the limit of this function whent→∞ in the case of constant rotation, θ(t)=λt, λ>0. It is an easy consequence of Lebesgue's theorem that, for any fixed t0>0,
(2.10)lim⁡t→∞∫0t014ν(t-s)exp⁡(-y24ν(t-s)-F(t)+F(s))ds=0.Taket0 large enough that we may substitute F(t) by σB2t/2. It is enough to consider
(2.11)2Cπ∫t0t14ν(t-s)exp⁡(-y24ν(t-s)-σB2(t-s)2)ds=2Cπ∫0t-t014νsexp⁡(-y24νs-σB2s2)ds,
whose limit when t→∞ is
(2.12)2Cπ∫0∞14νsexp⁡(-y24νs-σB2s2)ds,
which is obviously a finite integral. It reaches its maximum at y=0, where its value is
(2.13)Cπν∫0∞s-1/2e-σB2s/2ds=Cπν2σB2∫0∞r-1/2e-rdr=Cπν2σB2Γ(12)=Cπν2σB2π=C2νσB2=C2BPm,
where Pm is the magnetic Prandtl number. Therefore, u1 also has a limit when t→∞ for every y>0, given by the integral of (2.12) with respect to y.


## 2.2. Effect of the Lorentz Force

w
2 satisfies
(2.14)∂w2∂t=ν∂2w2∂y2-σB2θ˙y,w2(0,t)=0,w2(y,0)=0.
Although there exists an appropriate Green function for this problem, in this particular case it is simpler to assume that the solution is linear in y (which is correct). Setting first
(2.15)w2(y,t)=h(t)y,h(0)=0,
and recovering u2 from it, one finds
(2.16)u2(y,t)=-yσB2∫0texp⁡(-F(t)+F(s))θ˙(s)ds.
Therefore,
(2.17)∂u2∂y(y,t)=-σB2∫0texp⁡(-F(t)+F(s))θ˙(s)ds.
This function also has a limit when t→∞. Its value may be found by an argument similar to the previous one, although simpler; in this case
(2.18)lim⁡t→∞-λσB2∫t0texp⁡(-σB2(t-s)2)ds=-2λ.
Hence, u2(y,t) tends to -2λy when t→∞. Notice how the Lorentz force pushes the fluid backwards, the more rapidly the higher the distance from the wall.


## 2.3. Effect of the Initial Condition

w
3 satisfies
(2.19)∂w3∂t=∂2w3∂y2w3(0,t)=0w3(y,0)=ϕ(y).ϕ must satisfy ϕ(0)=0. The solution is found by the method of images: ϕ is extended to (-∞,∞) as an odd function. Then,
(2.20)w3(y,t)=14πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ(z)dz,
so that
(2.21)u3(y,t)=e-F(t)4πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ(z)dz.
Therefore, provided ϕ is differentiable,
(2.22)∂u3∂y(y,t)=e-F(t)4πνt∫-∞∞exp⁡(-(y-z)24νt)ϕ′(z)dz.
All this supposes that ϕ and ϕ′ satisfy certain reasonable conditions: for this example. if both are bounded. In that case, given that
(2.23)14πνt∫-∞∞exp⁡(-(y-z)24νt)dz=1,
both u3 and ∂u3/∂y are bounded by a function of the form Mexp(-F(t)), which means that they tend to zero as t→∞.Thus, the limit whent→∞ of u(0,t) is
(2.24)k=C2BPm-2λ.Ifk<0, we may be certain that the fluid inverts its direction and flows backwards from some point on. The factors contributing to this are a small (or negative) potential force C, which includes pressure and electric fields; a large background magnetic field; a large magnetic Prandtl number, which means that the viscosity dominates the resistivity; finally and most obviously, a large rate of rotation of the magnetic field.


## 3. Flow between Two Walls

When the fluid is stationary along two walls, the Lorentz force does not have so clear an effect as in the previous case, as it is stopped by the upper no-slip wall. We may again decompose the solution in the three components given by (2.1) but with the boundary and initial conditions of (1.6). The solution to these problems may be found again by the method of images, but it is equivalent and somewhat simpler to postulate a Fourier series expression of the form
(3.1)∑k=1∞ck(t)sin⁡πkLy,
and find the coefficients ck.


### 3.1. Effect of the Constant Forcing

We have(3.2)∂w1∂t=ν∂2w1∂y2+CeF(t),w1(0,t)=w1(L,t)=0,w1(y,0)=0.
First we extend the forcing CeF(t) as an odd function of y in the interval [-L,L]. Its Fourier series is
(3.3)CeF(t)∑k=0∞‍4π(2k+1)sin⁡π(2k+1)Ly,
so that if we set
(3.4)w1(t)=∑n=1∞w^1(n,t)sin⁡πnLy,
we have
(3.5)∂w^1∂t(2k,t)=-νπ2(2k)2L2w^1(2k,t),w^1(2k,0)=0,∂w^1∂t(2k+1,t)=-νπ2(2k+1)2L2w^1(2k+1,t)+4π(2k+1)CeF(t),w^1(2k+1,0)=0.
Hence, all the even terms w^1(2k,t) vanish and
(3.6)w^1(2k+1,t)=4π(2k+1)exp⁡(-νπ(2k+1)2tL2)×(∫0t‍exp⁡(νπ(2k+1)2sL2)Cexp⁡(F(s))ds).
If we define
(3.7)Fm(t)=σB2∫0tsin⁡2θ(s)ds+νπ2m2tL2,
we obtain
(3.8)u^1(2k+1,t)=4Cπ(2k+1)exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds.
Thus,
(3.9)u1(y,t)=∑k=0∞4Cπ(2k+1)(exp⁡(-F2k+1(t))∫0texp⁡(F2k+1(s))ds)×sin⁡π(2k+1)Ly,∂u1∂t(y,t)=∑k=0∞4CL(exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds)×cos⁡π(2k+1)Ly.
In particular
(3.10)∂u1∂t(0,t)=∑k=0∞4CL(exp⁡(-F2k+1(t))∫0texp⁡(F2k+1(s))ds),(3.11)∂u1∂t(L,t)=-∑k=0∞4CL(exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds).
To evaluate the limit of these expressions when t→∞, let us assume as before that θ(t)=λt, λ>0. Analogously to (2.4), we have now
(3.12)Fk(t)~(σB22+νπ2k2L2)t,-Fk(t)~-(σB22+νπ2k2L2)t,
for t large. Thus, for t0 fixed,
(3.13)lim⁡t→∞exp⁡(-Fk(t))∫0t0exp⁡(Fk(s))ds=0.
By using an argument similar to the one in (2.10) and glossing over the technicalities concerning the infinite terms in the sums of (3.10) and (3.11) (which hold anyway), we obtain that if we denote
(3.14)αk=lim⁡t→∞exp⁡(-Fk(t))∫0texp⁡(Fk(s))ds,
then
(3.15)αk=(σB22+π2νk2L2)-1;
moreover
(3.16)∂u1∂t(0,t)=4CL∑k=0∞α2k+1,(3.17)∂u1∂t(L,t)=-4CL∑k=0∞α2k+1.
To find these sums, we denote
(3.18)γ=BLπσ2ν.
Then,
(3.19)αk=2σB2γ2γ2+k2.
It may be found by elementary means (e.g,  the residue theorem) that
(3.20)∑k=1∞γ2γ2+k2=-12+πγ2cotanhπγ,Therefore(3.21)∑k=1∞α2k=2σB2(γ2)2∑k=1∞1(γ/2)2+k2=1σB2(-1+πγ2cotanhπγ2),∑k=0∞α2k+1=1σB2(πγcotanhπγ-πγ2cotanhπγ2).This expression may be written in another form, which is more convenient if we takeL and ν as fixed and consider that the variation of γ comes from σB2. This is
(3.22)∑k=0∞α2k+1=L22πνγ(cotanhπγ-12cotanhπγ2).Given the behavior of the hyperbolic cotangent, it is easy to see that this sum tends to∞ when γ→0 and to 0 when γ→∞. Combining this with (3.16) and (3.17), we obtain
(3.23)lim⁡t→∞∂u1∂t(0,t)=2CLπνγ(cotanhπγ-12cotanhπγ2),(3.24)lim⁡t→∞∂u1∂t(L,t)=-2CLπνγ(cotanhπγ-12cotanhπγ2).


### 3.2. Effect of the Lorentz Force

The Fourier series of the functiony→-σB2θ˙y is
(3.25)σB2θ˙∑k=1∞2L(-1)kπksin⁡πkLy.
Hence, w^2 satisfies
(3.26)∂w^2∂t(k,t)=-π2νk2L2w^2(k,t)+2L(-1)kπkσB2θ˙(t)exp⁡(F(t))w^2(k,0)=0,
whose solution is
(3.27)w^2(k,t)=2L(-1)kπkσB2∫0texp⁡(Fk(s))θ˙(s)ds.
Therefore,
(3.28)u^2(k,t)=2L(-1)kπkσB2∫0texp⁡(-Fk(t)+Fk(s))θ˙(s)ds,u2(y,t)=∑k=1∞2L(-1)kπkσB2(∫0texp⁡(-Fk(t)+Fk(s))θ˙(s)ds)sin⁡πkLy,
which implies
(3.29)∂u2∂t(y,t)=2∑k=1∞(-1)kσB2(∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds)cos⁡πkLy.
In particular
(3.30)∂u2∂t(0,t)=2∑k=1∞(-1)kσB2∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds,∂u2∂t(L,t)=2∑k=1∞σB2∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds.To find the limit of this expression whent→∞ and θ(t)=λt, we use the same argument as in the previous paragraph. We find
(3.31)lim⁡t→∞∂u2∂t(0,t)=2σB2λ∑k=1∞(-1)kαk,lim⁡t→∞∂u2∂t(L,t)=2σB2λ∑k=1∞αk.
In terms of γ,
(3.32)lim⁡t→∞∂u2∂t(0,t)=4λ∑k=1∞(-1)kγ2γ2+k2,(3.33)lim⁡t→∞∂u2∂t(L,t)=4λ∑k=1∞‍γ2γ2+k2.
We know already (3.20) the value of the sum in (3.33). The one in (3.32) sums
(3.34)∑k=1∞(-1)kγ2γ2+k2=-12+πγ2cosechπγ.
This sum is a decreasing function of γ, tending to 0 when γ→0 and to -1/2 when γ→∞.


### 3.3. Effect of the Initial Condition

To deal with the problem(3.35)∂w3∂t=ν∂2w3∂y2,w3(0,t)=w3(L,t)=0,w3(y,0)=ϕ(y),
we extend ϕ to [-L,L] as an odd function, with Fourier series
(3.36)ϕ(y)=∑k=1∞ϕ^(k)sin⁡πkLy.
Then,
(3.37)∂w^3∂t(k,t)=-π2νk2L2w^3(k,t),w^3(k,0)=ϕ^(k),
whose solution is
(3.38)w^3(k,t)=exp⁡(-π2νk2L2t)ϕ^(k).
Therefore,
(3.39)u3(y,t)=∑k=1∞exp⁡(-Fk(t))ϕ^(k)sin⁡πkLy,∂u3∂y(y,t)=∑k=1∞πkLexp⁡(-Fk(t))ϕ^(k)cos⁡πkLy.
It is trivial to prove that both tend to zero when t→∞ for all y.When(3.40)∂ϕ∂y(0)>0,∂ϕ∂y(L)<0
the fluid flows initially in the positive sense. If we show that under certain conditions
(3.41)lim⁡t→∞∂u∂t(0,t)<0,
or
(3.42)lim⁡t→∞∂u∂t(L,t)>0,
the fluid reverses flow at least in one of the walls. Since u=u1+u2+u3 and we have found already these limits, we may combine first (3.23) and (3.34) to get
(3.43)lim⁡t→∞∂u∂t(0,t)=2CLπνγ(cotanhπγ-12cotanhπγ2)+4λ(-12+πγ2cosechπγ).
On the other hand, combining now (3.24), (3.20), and (3.33), we find
(3.44)lim⁡t→∞∂u∂t(L,t)=-2CLπνγ(cotanhπγ-12cotanhπγ2)+2λ(-1+πγcotanhπγ).The expression in (3.43) tends to -2λ<0 when γ→∞. Thus, for any C and L, when λ and/or σB2 are large enough, there exists a flow reversal at this wall.The limit of (3.44) when γ→∞ is ∞. Since this becomes positive for σB2 or λ large enough, the fluid will eventually flow backwards at y=L. Notice that given the respective size of the limits, this occurs much earlier than the flow reversal at y=0. This was to be expected, as the Lorentz force pushes backwards the flow the more rapidly the larger y is; magnetic field lines move faster the higher the radius.


## 3.1. Effect of the Constant Forcing

We have(3.2)∂w1∂t=ν∂2w1∂y2+CeF(t),w1(0,t)=w1(L,t)=0,w1(y,0)=0.
First we extend the forcing CeF(t) as an odd function of y in the interval [-L,L]. Its Fourier series is
(3.3)CeF(t)∑k=0∞‍4π(2k+1)sin⁡π(2k+1)Ly,
so that if we set
(3.4)w1(t)=∑n=1∞w^1(n,t)sin⁡πnLy,
we have
(3.5)∂w^1∂t(2k,t)=-νπ2(2k)2L2w^1(2k,t),w^1(2k,0)=0,∂w^1∂t(2k+1,t)=-νπ2(2k+1)2L2w^1(2k+1,t)+4π(2k+1)CeF(t),w^1(2k+1,0)=0.
Hence, all the even terms w^1(2k,t) vanish and
(3.6)w^1(2k+1,t)=4π(2k+1)exp⁡(-νπ(2k+1)2tL2)×(∫0t‍exp⁡(νπ(2k+1)2sL2)Cexp⁡(F(s))ds).
If we define
(3.7)Fm(t)=σB2∫0tsin⁡2θ(s)ds+νπ2m2tL2,
we obtain
(3.8)u^1(2k+1,t)=4Cπ(2k+1)exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds.
Thus,
(3.9)u1(y,t)=∑k=0∞4Cπ(2k+1)(exp⁡(-F2k+1(t))∫0texp⁡(F2k+1(s))ds)×sin⁡π(2k+1)Ly,∂u1∂t(y,t)=∑k=0∞4CL(exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds)×cos⁡π(2k+1)Ly.
In particular
(3.10)∂u1∂t(0,t)=∑k=0∞4CL(exp⁡(-F2k+1(t))∫0texp⁡(F2k+1(s))ds),(3.11)∂u1∂t(L,t)=-∑k=0∞4CL(exp⁡(-F2k+1(t))∫0t‍exp⁡(F2k+1(s))ds).
To evaluate the limit of these expressions when t→∞, let us assume as before that θ(t)=λt, λ>0. Analogously to (2.4), we have now
(3.12)Fk(t)~(σB22+νπ2k2L2)t,-Fk(t)~-(σB22+νπ2k2L2)t,
for t large. Thus, for t0 fixed,
(3.13)lim⁡t→∞exp⁡(-Fk(t))∫0t0exp⁡(Fk(s))ds=0.
By using an argument similar to the one in (2.10) and glossing over the technicalities concerning the infinite terms in the sums of (3.10) and (3.11) (which hold anyway), we obtain that if we denote
(3.14)αk=lim⁡t→∞exp⁡(-Fk(t))∫0texp⁡(Fk(s))ds,
then
(3.15)αk=(σB22+π2νk2L2)-1;
moreover
(3.16)∂u1∂t(0,t)=4CL∑k=0∞α2k+1,(3.17)∂u1∂t(L,t)=-4CL∑k=0∞α2k+1.
To find these sums, we denote
(3.18)γ=BLπσ2ν.
Then,
(3.19)αk=2σB2γ2γ2+k2.
It may be found by elementary means (e.g,  the residue theorem) that
(3.20)∑k=1∞γ2γ2+k2=-12+πγ2cotanhπγ,Therefore(3.21)∑k=1∞α2k=2σB2(γ2)2∑k=1∞1(γ/2)2+k2=1σB2(-1+πγ2cotanhπγ2),∑k=0∞α2k+1=1σB2(πγcotanhπγ-πγ2cotanhπγ2).This expression may be written in another form, which is more convenient if we takeL and ν as fixed and consider that the variation of γ comes from σB2. This is
(3.22)∑k=0∞α2k+1=L22πνγ(cotanhπγ-12cotanhπγ2).Given the behavior of the hyperbolic cotangent, it is easy to see that this sum tends to∞ when γ→0 and to 0 when γ→∞. Combining this with (3.16) and (3.17), we obtain
(3.23)lim⁡t→∞∂u1∂t(0,t)=2CLπνγ(cotanhπγ-12cotanhπγ2),(3.24)lim⁡t→∞∂u1∂t(L,t)=-2CLπνγ(cotanhπγ-12cotanhπγ2).


## 3.2. Effect of the Lorentz Force

The Fourier series of the functiony→-σB2θ˙y is
(3.25)σB2θ˙∑k=1∞2L(-1)kπksin⁡πkLy.
Hence, w^2 satisfies
(3.26)∂w^2∂t(k,t)=-π2νk2L2w^2(k,t)+2L(-1)kπkσB2θ˙(t)exp⁡(F(t))w^2(k,0)=0,
whose solution is
(3.27)w^2(k,t)=2L(-1)kπkσB2∫0texp⁡(Fk(s))θ˙(s)ds.
Therefore,
(3.28)u^2(k,t)=2L(-1)kπkσB2∫0texp⁡(-Fk(t)+Fk(s))θ˙(s)ds,u2(y,t)=∑k=1∞2L(-1)kπkσB2(∫0texp⁡(-Fk(t)+Fk(s))θ˙(s)ds)sin⁡πkLy,
which implies
(3.29)∂u2∂t(y,t)=2∑k=1∞(-1)kσB2(∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds)cos⁡πkLy.
In particular
(3.30)∂u2∂t(0,t)=2∑k=1∞(-1)kσB2∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds,∂u2∂t(L,t)=2∑k=1∞σB2∫0t‍exp⁡(-Fk(t)+Fk(s))θ˙(s)ds.To find the limit of this expression whent→∞ and θ(t)=λt, we use the same argument as in the previous paragraph. We find
(3.31)lim⁡t→∞∂u2∂t(0,t)=2σB2λ∑k=1∞(-1)kαk,lim⁡t→∞∂u2∂t(L,t)=2σB2λ∑k=1∞αk.
In terms of γ,
(3.32)lim⁡t→∞∂u2∂t(0,t)=4λ∑k=1∞(-1)kγ2γ2+k2,(3.33)lim⁡t→∞∂u2∂t(L,t)=4λ∑k=1∞‍γ2γ2+k2.
We know already (3.20) the value of the sum in (3.33). The one in (3.32) sums
(3.34)∑k=1∞(-1)kγ2γ2+k2=-12+πγ2cosechπγ.
This sum is a decreasing function of γ, tending to 0 when γ→0 and to -1/2 when γ→∞.


## 3.3. Effect of the Initial Condition

To deal with the problem(3.35)∂w3∂t=ν∂2w3∂y2,w3(0,t)=w3(L,t)=0,w3(y,0)=ϕ(y),
we extend ϕ to [-L,L] as an odd function, with Fourier series
(3.36)ϕ(y)=∑k=1∞ϕ^(k)sin⁡πkLy.
Then,
(3.37)∂w^3∂t(k,t)=-π2νk2L2w^3(k,t),w^3(k,0)=ϕ^(k),
whose solution is
(3.38)w^3(k,t)=exp⁡(-π2νk2L2t)ϕ^(k).
Therefore,
(3.39)u3(y,t)=∑k=1∞exp⁡(-Fk(t))ϕ^(k)sin⁡πkLy,∂u3∂y(y,t)=∑k=1∞πkLexp⁡(-Fk(t))ϕ^(k)cos⁡πkLy.
It is trivial to prove that both tend to zero when t→∞ for all y.When(3.40)∂ϕ∂y(0)>0,∂ϕ∂y(L)<0
the fluid flows initially in the positive sense. If we show that under certain conditions
(3.41)lim⁡t→∞∂u∂t(0,t)<0,
or
(3.42)lim⁡t→∞∂u∂t(L,t)>0,
the fluid reverses flow at least in one of the walls. Since u=u1+u2+u3 and we have found already these limits, we may combine first (3.23) and (3.34) to get
(3.43)lim⁡t→∞∂u∂t(0,t)=2CLπνγ(cotanhπγ-12cotanhπγ2)+4λ(-12+πγ2cosechπγ).
On the other hand, combining now (3.24), (3.20), and (3.33), we find
(3.44)lim⁡t→∞∂u∂t(L,t)=-2CLπνγ(cotanhπγ-12cotanhπγ2)+2λ(-1+πγcotanhπγ).The expression in (3.43) tends to -2λ<0 when γ→∞. Thus, for any C and L, when λ and/or σB2 are large enough, there exists a flow reversal at this wall.The limit of (3.44) when γ→∞ is ∞. Since this becomes positive for σB2 or λ large enough, the fluid will eventually flow backwards at y=L. Notice that given the respective size of the limits, this occurs much earlier than the flow reversal at y=0. This was to be expected, as the Lorentz force pushes backwards the flow the more rapidly the larger y is; magnetic field lines move faster the higher the radius.


## 4. Conclusions

A generalization of the classical Hartmann flow is obtained when we pose a background uniform magnetic field rotating in time. Since the Lorentz force varies both in time and space, the resulting flow cannot be stationary. If we add one or two walls with the no-slip property, the interaction between the magnetic field and the boundary layers creates an interesting phenomenology. We study in particular the phenomenon of flow reversal near the walls for a magnetic field rotating at a constant rate; it is found that for large enough values of this rate, the conductivity of the fluid, or the size of the magnetic field, there exists always flow reversal at some instant. Forces opposing this include the kinetic pressure and a possible electrostatic field. This reversal makes excellent sense, as the Lorentz force (visualized through the rotation of magnetic field lines) pushes backwards the fluid if the sense of rotation of the field is contrary to the original flow.



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*Source: 641738-2012-08-13.xml*