Three weeks special project

EPTMElectrohydrodynamical Perturbation Theory in Microfluidics

Laust Tophøj & Søren Møllers021645 s021816

Supervisor: Henrik Bruus

MIC – Department of Micro and NanotechnologyTechnical University of Denmark

January 21st 2005

Contents

1 Hydrodynamics:Shape Perturbation 31.1 Poiseuille flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Coordinate Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Expanding the velocity field . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Finding the velocity terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Test of the found solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Electrostatics:Perturbation of the Potential 102.1 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Expanding the field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Solving the ODE’s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Test of the Found Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Electrohydrodynamics 183.1 The reduced system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 First order perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 First Order Outlook . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4 Second order perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Conclusion 30

A Tedious calculation 32

1

Introduction

In the present report we are going to study and apply the techniques of perturbation the-ory on an electrohydrodynamic problem stemming from microfluidics. We will only dealwith exact solutions.

Perturbation theory can used when an exact solution to a problem involving differen-tial equations, linear or nonlinear is unobtainable or difficult to find. One starts out witha problem that has a known exact solution. Hereafter a slight disturbance (perturbation)is introduced and described by some parameter. All fields and operators entering thegoverning equations and boundary conditions are then expanded in this parameter, whichare thereafter solved, one order at a time.

The partial sum containing only the first few orders of perturbation will then provideadequate approximation for sufficiently small parameters, while the area of practical ap-plicability increases, as more and more terms are included.

To illustrate this, and to get acquainted with perturbation theory ourselves, we will warmup with two manageable examples taken from the periphery of the microfluidic subject.These will present the techniques in use in a relatively simple context. When the basicsare all in place, we will attack the full electrohydrodynamic problem consisting of thesolution of 6 coupled nonlinear partial differential equations (PDE’s).

On notation

Throughout the text, we will refer to equation no. X by (X).

2

Chapter 1

Hydrodynamics:Shape Perturbation

As a first illustration of the perturbation techniques, we will investigate a Poiseuille flowproblem, in which the channel cross-section is perturbed slightly from a familiar example.

1.1 Poiseuille flow

Given a steady flow uniform in the flow direction, that is, with velocity v = ezv(r, φ),(ez is a unit vector in the direction of the fixed z-axis of standard cylindrical (r, φ, z) co-ordinates ), the nonlinear term ρ(v · ∇)v in the Navier-Stokes Equation and the timedependence term ρ∂v

∂t drop out. Neglecting gravity and other external forces, we are leftwith 0 = −∇p+η∆v; p = p(r, φ, z) being the pressure field and ∆ the Laplacian operator.

Since v and thus ∆v only have components in the z-direction, so does ∇p, and wehave p = p(z). We are left with the scalar equation ∆v(r, φ) = 1

ηdpdz (z).

The two sides are independent, and we must have

1η

dp

dz(z) = constant ≡ P. (1.1)

We have thus derived the equation:

∆v(r, φ) = P, (1.2)

which, together with the no-slip boundary condition (BC), formulates a standard Poiseuilleflow problem.

1.2 Coordinate Transformation

Our starting point will be a circular cross-section of radius a, where the problem hasa known analytical solution. The boundary is perturbed in a small parameter ε, to be

3

described by the curve r = a(1 + ε sin kφ) in ordinary polar coordinates (r, φ). The wavenumber k is a positive integer. Practically any simple closed curve can be written asr = a(1 + ε

∑k ck sin kφ), and since the governing PDE is linear, the solution to a more

general Poiseuille flow problem could be found by superposition, once the problem at handis solved.

We start by introducing a transformation, which maps a circular disc in abstract (ρ, θ)-space onto the flow domain in physical (r, φ)-space:

φ(ρ, θ) = θ, 0 ≤ θ < 2π, (1.3a)r(ρ, θ) = aρ(1 + ε sin kθ), 0 ≤ ρ ≤ 1. (1.3b)

The problem is tackled in abstract space, where the boundary curve takes the simple formρ = 1.

The Laplacian must be expressed as an operator on the abstract space via eqs. (1.3).To this end, the inverse mapping is found to be

θ(r, φ) = φ, (1.4a)

ρ(r, φ) =r

a(1 + ε sin kφ)≡ r

aA(φ). (1.4b)

The Laplacian on physical space is ∆ = ∂2r + 1

r∂r + 1r2 ∂2

φ. The individual operators aretransformed into operators on abstract space via the chain rule e.g. ∂r = ∂rρ∂ρ + ∂rφ∂φ,and the functions ρ(r, φ) and θ(r, φ) are inserted in order to eliminate the physical coor-dinates.

The Laplacian is found, using ∂rθ = 0, ∂φθ = 1:

∆ = (∂rρ)2∂2ρ +

1ρaA

(∂rρ)∂ρ +1

(ρaA)2{(∂φρ)2∂2

ρ + (∂2φρ)∂ρ + 2(∂φρ)∂θ∂ρ + ∂2

θ

}, (1.5)

where A = A[φ(θ)] is to be inserted. The other derivatives are found: ∂rρ = 1aA and

∂φρ = −ρkεA cos kθ.

The operator is expanded in ε,

∆ = ∆0 + ε∆1 + ε2∆2 + · · · . (1.6)

And we find the first couple of expansion coefficients, which are shown dimensionless form:

a2∆0 =∂2ρ +

1ρ∂ρ +

1ρ2

∂2θ , (1.7a)

a2∆1 =− 2 sin(kθ)∂2ρ +

1ρ(k2 − 2) sin(kθ)∂ρ − 1

ρ2k cos(kθ)∂ρ∂θ − 1

ρ22 sin(kθ)∂2

θ , (1.7b)

a2∆2 =12

[k2 + 3 + (k2 − 3) cos(2kθ)

]∂2

ρ +12ρ

[3− k2 + (5k2 − 3) cos(2kθ)

]∂ρ

+1ρ3k sin(2kθ)∂ρ∂θ +

1ρ2

3 sin2(kθ)∂2θ . (1.7c)

4

1.3 Expanding the velocity field

The velocity field in the perturbed system is described as a function of the abstract coor-dinates and ε, that is: v = v(ρ, θ; ε) . The dependence of physical coordinates can easilybe found via the inverse transformations eqs. (1.4). We now expand this function in ε:

v = v0 + εv1 + ε2v2 + · · · . (1.8)

We assume convergence of these series, at least for small ε:Inserting the expanded Laplacian and velocity field in the Navier-Stokes equation yields

P = ∆v(x, y) = (∆0 + ε∆1 + ε2∆2 + . . . )(v0 + εv1 + ε2v2 + . . . ) (1.9)

and, sorting in powers of ε,

0 = (∆0v0 − P ) + ε(∆1v0 + ∆0v1) + ε2(∆2v0 + ∆1v1 + ∆0v2) +O(ε3), (1.10)

where O is the error function.Each coefficient must be zero because of the linear independence of the ε-powers. This

yields a number of equations. The first, corresponding to ε0, can be directly solved. Withv0 thus obtained, the second equation (ε1) can be directly solved; with these two solutionsthe third equation (ε2) can be solved, and so on. This is the perturbation method.

1.4 Finding the velocity terms

v0

The PDE on v0 is from (1.10):

Pa2 =[∂2

ρ +1ρ∂ρ +

1ρ2

∂2θ

]v0. (1.11)

This field, corresponding to zeroth order perturbation, exhibits total rotational symmetry,thus v0 = v0(ρ), and we are left with the ordinary differential equation (ODE)

∂ρ2v0 +1ρ∂ρv0 = Pa2,

{v0(1) = 0|v0(0)| < +∞.

(1.12)

For a particular solution vP0 to this inhomogeneous ODE we make a guess vP

0 ≡ Aρ2, Aconstant. Insertion in the ODE yields 4A = Pa2, which is solved for A.

The corresponding homogeneous equation is an example of an Euler differential equa-tion and has a constant solution. We choose the constant such that the homogeneous(no-slip) BC is fulfilled, and obtain, by superposition, from the two solutions:

v0 = −Pa2

4(1− ρ2). (1.13)

It is known, that a solution to the Poisson equation with sufficient BC’s is uniquelydetermined. Therefore this is the only solution.

5

v1

The first order perturbation field v1 is determined from the second term of (1.10):

∆0v1 = −∆1v0. (1.14)

Because of the rotational symmetry of v0, the θ-derivatives in ∆1 drop out. We find:

∆1v0 =P

2sin(kθ)(k2 − 4). (1.15)

As before, we guess a particular solution. It is seen that every term in the operator ∆0

serves to bring down the degree of a polynomial in ρ by two. So how about the guessvP1 ≡ Aρ2 sin(kθ), with A constant? Insertion in the PDE yields 2A sin(kθ) = Pa2 sin(kθ),

which is then solved for A.We now need a solution vH

1 , to the corresponding homogeneous equation, in order tofulfill the BC. We look for a solution going as sin(kθ) on the boundary. In the followingthis will relieve us from finding the full solution, by means of the uniqueness of solutionsto the Poisson equation.

The corresponding homogeneous equation is solved using the technique of separation ofvariables. As an ansatz we insert the product solution vH

0 = R(ρ)Θ(θ). We immediatelymultiply by ρ2/RΘ and obtain:

[ρ2 R′′

R+ ρ

R′

R

]+

Θ′′

Θ= 0. (1.16)

The two first and the last terms depend only on ρ and θ, respectively. These are indepen-dent, and the equation can only be fulfilled on the open domain of the PDE if each termis constant. Thus

Θ′′ + λΘ = 0, (1.17a)

ρ2R′′ + ρR′ − λR = 0, (1.17b)

where λ is a constant of separation to be determined. For λ = k2, (1.17a) has a solutionΘ = sin(kθ). With this separation constant, (1.17b) has a solution R = ρk, which takesthe value one on the boundary. We can thus construct a solution

v1 =Pa2

2(ρ2 − ρk) sin(kθ), (1.18)

which meets our requirements.

v2

From the third term of (1.10) we now find a PDE on v2:

∆0v2 = −∆2v0 + ∆1v1

=P

4

{[(k2 − 2)− 3k3ρk−2

]cos(2kθ) + 2− k3ρk−2

}. (1.19)

6

Again a particular solution can be guessed. Inspired by the preceding section we insertv2 = Aρ2 cos(2kθ) + Bρk cos(2kθ) + Cρ2 + Dρk, and determine the constants.

The value of this solution on the boundary determines a BC on the correspondinghomogeneous problem, which is solved in complete analogy with the preceding section,only this time we look for solutions constant on the boundary and solutions going ascos(2kθ) on the boundary.We find

v2 =P

4

{[− 1

2ρ2 + kρk − (k − 1

2)ρ2k]cos(2kθ) + (k − 1

2) +12ρ2 − kρk

}. (1.20)

1.5 Test of the found solution

We define vPn as the nth partial sum vP

n ≡ ∑ni=0 εiφi.

The found solution is tested as follows: We let the ordinary Laplacian operate onthe velocity field as a function of physical coordinates, vP

2 [ρ(r, φ), θ(r, φ)] and find, usingMaple, appendix A.

∆vP0 = P +O(ε), (1.21)

∆vP1 = P +O(ε2), (1.22)

∆vP2 = P +O(ε3). (1.23)

This confirms the validity of our solutions.

1.5.1 Flux

In search of a single scalar quantity characterizing the nth partial sum, we find the totalnormalized flux Qn trough a normalized cross-section.

For fixed wave number k this is defined as the total fluid flux through a cross-sectiondivided by the area of this surface. A designates the unit disc of abstract space.

Qn(ε) ≡ 1N(ε)

∫

AvPn (ρ, θ) |J(ρ, θ)| dρdθ, (1.24)

where the area N is found

N(ε) =∫

A|J(ρ, θ)| dρdθ. (1.25)

Here |J(ρ, θ)| is the Jacobian of the mapping from abstract space all the way into carte-sian coordinates of physical space. This mapping is described by (1.3) having Jaco-bian |J1(ρ, θ)| followed by the standard mapping from polar coordinates into cartesian(x, y)(r, φ) = (r cosφ, r sinφ), having Jacobian |J2(ρ, θ)| = r(ρ, θ).|J(ρ, θ)| ≡ J1(ρ, θ)J2(ρ, θ) is then the Jacobian of the composite function:

|J1(ρ, θ)| =∣∣∣∣

∂ρr(ρ, θ) ∂θr(ρ, θ)∂ρφ(ρ, θ) ∂θφ(ρ, θ)

∣∣∣∣ = aA[φ(ρ, θ)] , |J2(ρ, θ)| = r = aA[φ(ρ, θ)].

The computation is done in Maple.

7

Figure 1.1: The first three perturbation fields, corresponding to zeroth, first and second orderrespectively. Bright areas correspond to strong flow in the direction of the driving force −∇p.Note that the fields are actually independent of ε, which only distorts the shape and is included forclarity. Here ε = 0.15.While the first order perturbation serves to correct the obvious distortedness of v0, the second isgenerally inhibiting the flow. Note that for clarity the amplitude of v2 has been scaled down by afactor 5 compared to the two others.What would v3 look like?

Figure 1.2: Different flow profiles for ε = 0.15. From left the first and second order approxima-tions. The zeroth order approximation is seen in figure 1.1, far left. Notice that the velocity at thecentral axis is diminished in vP

2 compared to vP1 by what corresponds to one level curve.

1.6 Conclusion

We have used this first example to get our hands on the perturbation techniques, and wehave succeeded in solving the Poiseuille flow problem in the perturbed circular cylinderto the second order. We have used the total fluid flux through a normalized cross-sectionas a scalar measure for the effect of the perturbation, enabling us to plot this flux as afunction of the perturbation parameter ε.

A natural continuation of this would, in the absence of known analytical solutions tothe problem, be to use a numerical Finite-Element Method, and thereby obtain realisticsolutions for comparison. The flux plots could then have been extended with actual errorflux plots, characterizing the error made in the approximation we offer.

8

1

0,8

0,6

0,4

0,2

00,20,150,10,050

ε

Q0

Q1

Q2

1

0,8

0,6

0,4

0,2

00,20,150,10,050

ε

Q0

Q1

Q2

1

0,8

0,6

0,4

0,2

00,20,150,10,050

ε

Q0

Q1

Q2

Figure 1.3: The normalized flux through a normalized cross section for the different partial sums.It is seen that Q0 is constant. From left the values of k are [3, 5, 7]. It is seen that the effect ofQ2 is to slow down the flux corresponding to the greater hydraulic resistance of the more perturbedshape, as was seen in figure 1.1.

9

Chapter 2

Electrostatics:Perturbation of the Potential

In this chapter, we shall use the perturbation techniques to examine the usefulness ofthe Debye-Huckel approximation to the Poisson-Boltzmann equation for the behaviour ofbinary electrolytes in the presence of electric fields. Again we look at a static case, wherewe include the electrostatic body force ρelE due to an external electrical field E and thecharge density ρel.

Using the electrostatic Maxwell equations on the electrical field we obtain

∇×E = 0 ⇒ E = −∇φ∇ · (εelE) = ρel

}∇2φ(r) = − 1

εelρel(r), (2.1)

where we have introduced the electrostatic potential φ.Over the interface between an electrolyte and a solid, chemical phenomena cause charge

to be transferred across the interface, giving rise to a potential difference ζ. The systemas a hole is considered neutral, so the exterior boundary conditions on the potential arehomogeneous.

Electrical forces make the ions in the liquid migrate, effectively shielding the bulk liquidfrom the electrical field. Thermally induced diffusion opposes the concentration of ionsonto the interface, giving the disturbed layer a finite thickness. This layer is called theDebye layer and will be the subject of this section.

2.0.1 The Poisson-Boltzmann equation

A binary solute, while altogether neutral, has ion concentrations c+ and c− and oppositevalencies ±Z. Neutrality implies that both concentrations must approach the same con-stant value c0 far from the interface, where the potential is zero. The chemical potentialsat position r are then given by

µ±(r) = µ0 + kBT log(

c±c0

)± Zeφ(r), (2.2)

where c0 is the concentration of the neutral liquid, kB is the Boltzmann constant, log isthe natural logarithm and µ0 a constant. The second and third terms describe the thermal

10

energy from statistical mechanics and the electrostatic energy, respectively. The chemicalpotential is defined as ∂F

∂N , i.e. the differential free energy F per molecule. Thus it mustbe constant throughout a system in equilibrium. Otherwise energy could be gained by theexchange of molecules between different subsystems. In other words ∇µ = 0 everywhere,leading to:

kBT∇ log(

c±c0

)± Ze∇φ(r) = 0. (2.3)

The two scalar fields can thus only differ by a constant. Far from the interface, wherec± = c0 and φ = 0, the fields are equal zero and the constant too must be zero. Thus

kBT log(

c±c0

)± Zeφ(r) = 0, (2.4)

leading to

c± = c0 exp(∓ Ze

kBTφ(r)

). (2.5)

The two functions are added giving the charge density ρel(r) = Ze[c+− c−]This is insertedinto the Poisson equation (2.1), yielding the Poisson-Boltzmann equation, where all fieldsbut the electrostatic potential φ are eliminated:

∆φ(r) = 2Zec0

εelsinh

(Ze

kBTφ(r)

). (2.6)

2.0.2 The Debye length, λD

As a first examination we use the Debye-Huckel approximation, where the nonlinear termis replaced by its first expansion coefficient sinhx ≈ x, corresponding to the limit Zeζ

kBT ¿ 1:

∆φ = 2(Ze)2c0

εelkBTφ(r). (2.7)

This is solved in the 1-D planar case, φ = φ(x), where 0 < x < ∞ is the distanceto the interface, with boundary conditions φ(0) = ζ, φ(∞) = 0. This yields φ(x) =ζ exp [−x/λD], where we introduce the very important Debye length

λD ≡√

εelkBT

2(Ze)2c0, (2.8)

which depends on the electrolyte and its temperature and has order of magnitude λD ∼ 10 nm.

11

2.1 Scaling

We look at the planar case φ = φ(x) with boundary conditions

φ(0) = ζ, φ(∞) = 0. (2.9)

In (2.6) we wish to introduce dimensionless variables (marked by a tilde) by known con-stants:

x =1

λDx ⇒ ∆ = λ2

D∆, φ =1ζφ. (2.10)

Substituting and introducing the dimensionless energy ratio (thermal to electrical):

ε ≡ Zeζ

kBT, (2.11)

we obtain

∆φ = 2λ2

D

ζεelZec0 sinh(εφ), (2.12)

wherein the Debye length is inserted:

∆(εφ) = sinh(εφ). (2.13)

From now on we discard the tildes working only in dimensionless space.

∆(εφ) = sinh(εφ). (2.14)

2.2 Expanding the field

ε describes the perturbation from zeta potential zero corresponding to trivial potential.We expand the function φ(x; ε) in ε, tacitly assuming convergence.

φ = φ0 + εφ1 + ε2φ2 + · · · . (2.15)

The sum must fulfill the boundary conditions given by (2.9) implying:

φ(0) = 1, φ(∞) = 0. (2.16)

We let the zero order term fulfill these and enforce homogenous boundary conditions onall higher order terms.

The hyperbolic sine function has the (everywhere convergent) expansion around argumentzero:

sinhx = x +13!

x3 +15!

x5 + · · · . (2.17)

12

In (2.14) the expansions are inserted:

∆[εφ0 + ε2φ1 + ε3φ2 + · · · ] =

[εφ0 + ε2φ1 + ε3φ2 + · · · ]+ 1

3![εφ0 + ε2φ1 + ε3φ2 + · · · ]3+· · · .

(2.18)In order to expand the powers of the parenthesis, some combinatorics is necessary. Becauseof the linear independence of the powers, we obtain the following equations for the firstfew terms of the series:

ε1 : ∆φ0 − φ0 = 0 , (2.19a)

ε2 : ∆φ1 − φ1 = 0 , (2.19b)

ε3 : ∆φ2 − φ2 =13!

φ30, (2.19c)

ε4 : ∆φ3 − φ3 =13!

3φ20φ1, (2.19d)

ε5 : ∆φ4 − φ4 =13!

[3φ0φ

21 + 3φ2

0φ2

]+

15!

φ50, (2.19e)

ε6 : ∆φ5 − φ5 =13!

[φ3

1 + 3!φ0φ1φ2 + 3φ20φ3

]+

15!

5φ40φ1, (2.19f)

ε7 : ∆φ6 − φ6 =13!

[3φ0φ

22 + 3φ2

1φ2 + 3!φ0φ1φ3 + 3φ20φ4

]+

15!

[5φ4

0φ2 + 5·42 φ3

0φ21

]+

17!

φ70.

(2.19g)

2.3 Solving the ODE’s

We define a differential operator L ≡ ∆− 1.

φ0

(2.19a) with BC’s is written explicitly:

Lφ0 = 0, 0 < x < ∞, φ0(0) = 1, φ0(∞) = 0. (2.20)

This ODE has the general solution φ0 = A exp(x) + B exp(−x), where the BC’s yieldA = 0, B = 1. We thus have

φ0(x) = e−x. (2.21)

φ1

(2.19b) with homogenous BC’s has only the trivial solution φ1 = 0.

We are now going to show that all odd expansion coefficients are trivial. We will dothis by showing that any equation in (2.18) yielding φn, where n is odd, contains an odd-numbered coefficient as a factor in each term on its right hand side. Then, by recursion,

13

given φ1 = 0, any such equation is homogenous and, in analogy with (2.19b), has onlytrivial solutions.

General: On equations (2.19) the right hand side terms are all of the form:φp1φp2 · · ·φpm , where m is odd, and pi ∈ {0, 1, 2, . . .} for i ∈ {0, 1, 2, . . . , m}.If such a term is to be part of the nth equation in (2.19), it must be associated with afactor εn in (2.18). Evidently n = m + p1 + p2 + . . . + pm

Odd coefficients: Consider any one of the considered equations, where the corre-sponding order n of ε is even. Since m is always odd, p1 + p2 + . . . + pm is also odd. Butany odd sum of integers must contain at least one odd term. This completes the proof.

φ2

In (2.19c) we insert φ0 and find

Lφ2 =16e−3x.

We immediately guess a particular solution φP2 = A exp(−3x). Insertion yields LφP

2 =8A exp(−3x), and thus A = 1/48 by comparison.

This function takes the value A at x = 0, and we therefore have for the homogenoussolution φH

2 , the BC’s: φH2 (0) = −A, φH

2 (∞) = 0. This is easily solved in analogy with(2.20) and we obtain the full solution

φ2(x) =148

[e−3x − e−x

]. (2.22)

φ4

In (2.19e) we insert φ0, φ1 and φ2 and find

Lφ4 =196

e−3x +3

160e−5x.

We guess a particular solution φP4 = A exp(−3x) + B exp(−5x). Insertion yields LφP

2 =8A exp(−3x) + 24B exp(−5x), and thus A = −1/768 and B = 1/1280 by comparison.

Just as with φ2 the homogeneous solution is found as exp(−x) with the opposite coef-ficients, and we find:

φ4(x) =1

3840

[− 5

(e−3x − e−x

)+ 3

(e−5x − e−x

)](2.23)

=1

3840

[2e−x − 5e−3x + 3e−5x

]. (2.24)

φ6

In (2.19g) we insert φ0, φ2, φ3 and φ4 and find

Lφ6 =11

23040e−3x − 1

512e−5x +

31792

e−7x.

14

We immediately guess a particular solution φP6 = A exp(−3x)+B exp(−5x)+C exp(−7x).

Insertion yields LφP6 = 8A exp(−3x) + 24B exp(−5x) + 48C exp(−7x), and thus A =

11/184320, B = −1/12288 and C = 1/28672 by comparison.The full solution is:

φ6(x) =1

1290240

[77

(e−3x − e−x

)− 105(e−5x − e−x

)+ 45

(e−7x − e−x

)](2.25)

=1

1290240

[− 17e−x + 77e−3x − 105e−5x

]. (2.26)

We tried to write down the φn term and thereby find the general series for our perturbedsolution. Then maybe we would have been able to find a closed expression by summingthis. Even though we tried hard, we were not able to find this generalization. We foundsome sort of pattern, but it was disturbed when going to the 7th order in ε. Even whentrying to use the expansion of the known exact expression, we failed to find any system.

2.4 Test of the Found Solution

We now want to compare our found solution to the known exact solution (Chapman-Gouy)

ψ(x) =4kBT

Zearctanh

[tanh

(Zeζ

4kBT

)e−xλD

], (2.27)

where arctanh denotes the inverse hyperbolic tangent. In order to compare the expressions,we choose to scale ψ with the same parameters as we did with φ:

ψ(x) =4kBT

Zeζarctanh

[tanh

(Zeζ

4kBT

)e−xλD

], (2.28)

ψ(x) =4ε

arctanh[tanh

(ε

4

)e−x

]. (2.29)

Again the tildes will be left out for convenience. We now define the nth partial sum ofour expanded potential

φPn (x; ε) ≡

n∑

i=0

εiφi(x), (2.30)

and the associated approximation error

Fn(x; ε) ≡ φPn (x; ε)− ψ(x). (2.31)

To confirm our work we Taylor expand F7(x, ε). We obtain:

F7(x; ε) = ψ(x)− φP7 (x) = f(x)O(ε8), (2.32)

where f is a function of x alone. Thus we have now found an expression, which is correctto the 8th order in ε.

15

2.4.1 Scalar error measure

To investigate the size of the error as a function of ε, we find the l2 norm of the error givenby:

‖Fn‖2(ε) ≡∫ ∞

0

[Fn(x; ε)

]2dx. (2.33)

The norm is found for n ∈ {0, 2, 4, 6} and plotted in Maple as seen in the left hand side offigure 2.2.

As a second approach to the problem of finding a scalar representation of the error asa function of epsilon, we look at the l∞ norm:

‖Fn‖∞(ε) ≡ supx{|Fn(x; ε)|}. (2.34)

The norm is calculated in Matlab and plotted in the right hand side of figure 2.2. Wefound that the maximal error always occurred at x = 1, which corresponds to the Debyelength.

2.5 Conclusion

Surprisingly the two norms are almost identical. For low values of ε all partial sumsprovide fair approximations to the exact solution. The higher order of the partial sum thehigher accuracy. As ε grows the different errors converge and intersect at ε ≈ 8. For evenhigher ε all the approximations diverge, most violently the higher order. partial sums. Acritical point is from the exact solution (2.29) seen to be ε = 4.

The fact that the approximations collapse at roughly the same time (figure 2.2), in-dicates that the perturbation approach is not very fruitful in this case, where we haveexceeded the Debye-Huckel approximation only for small ε.

16

032

1

510 4

0,8

0,2

0,6

0,4

x/λD

Figure 2.1: The exact solution (fat red) compared to the partial sums for ε = 6. The large value ischosen to make the deviation visible. 0th, 2nd, 4th and 6th partial sum is marked by green, black,blue, purple, respectively.

1e-116420

.1e2

1.

.1

.1e-1

.1e-2

.1e-3

1e-05

1e-06

1e-07

10

1e-08

1e-09

8

1e-10

ε

‖Fn(ε)‖2

F0

F2

F4

F6

0 2 4 6 8 1010

−6

10−5

10−4

10−3

10−2

10−1

100

ε

‖Fn(ε)‖∞

F0

F2

F4

F6

Figure 2.2: An overview of the errors of the different partial sums of φ(x) compared to the knownexact solution. To the left, the l2-norm is displayed and to the right the l∞-norm. The two plotsare almost identical, but only the one to the right can be directly interpreted. It is seen (as expected)that for low values of ε the highest partial sums provide the best approximations. Also the higherorder partial sums diverge more strongly for large ε.Perhaps more surprising is that all of our approximations collapse at about the same value of ε,indicating that the improvement obtained by going from Debye-Huckel approximation (F0) to sixthorder perturbation, while considerable for ε ¿ 4, is quickly lost as ε increases.

17

Chapter 3

Electrohydrodynamics

We now feel well prepared to take on the full electrohydrodynamical problem, which is themajor subject of our work. The problem was previously described by the group of Bruusin [Mortensen].

A binary electrolyte is like in chapter 2 confined to the halfspace x ≥ 0. The systemis translation invariant in one direction and the problem can thus be treated in 2 dimen-sions with the other Cartesian coordinate y describing the direction along the interfaceplane.

In the electrolyte the following physical fields are necessary for our description: Thehydrodynamic velocity and pressure fields v(x, y, t) = vx(x, y, t)ex + vy(x, y, t)ey andp = p(x, y, t) , the electrostatic potential φ(x, y, t) and the concentrations of positiveand negative ions c+(x, y, t) and c−(x, y, t).

An electric potential harmonic in time and y is applied on the interface x = 0 by electrodesembedded in the wall.

An insulating layer asserts that no charge penetrates the interface. In microsystemsapplications, this layer could be an oxide layer with thickness comparable to λD. Themodulation in y has wavelength 2π/q, which in practice must be much larger than λD.

In our model we therefore consider only the limit, where this layer is of vanishing thick-ness.

All fields are assumed to be undisturbed from static equilibrium at x = ∞. This doesnot conflict with the real physical problem, as the typical width of a micro channel is∼ 100µm.

3.0.1 Governing equations

The Nernst-Planck equation describes the ionic current densities i± caused by convectionand by diffusion, thermal and forced.

18

The ionic concentrations are denoted c±.

i± = −D∇c± + vc± ∓ µc±∇φ, (3.1a)

where the diffusion constant D is related to the ionic mobility µ via the Einstein relationD = µ kBT/Ze and typically of the order D ∼ 10−9 m2/s. The ionic continuity equations:

∂tc± = −div i±. (3.1b)

The Poisson equation:

∆φ = − 1εel

ρel, (3.1c)

where the charge density ρel is given by the ionic densities, ρel = Ze(c+ − c−). Navier-Stokes equation in Stokes flow regime, where the nonlinear term (v ·∇)v cancels out:

ρm∂tv = −∇p + η∆v − ρel∇φ. (3.1d)

Finally the hydrodynamic continuity equation for incompressible flow:

divv = 0. (3.1e)

We start with a little manipulation:(3.1a) is inserted in (3.1b), using , div(ab) = b ·∇a + adivb for fields a and b, and (3.1e):

∂tc± = D∆c± − v ·∇c± ± µ div(c±∇φ). (3.2)

For later convenience we introduce a change of variables in the concentrations:

ν ≡ c+ − c−, Σ ≡ c+ + c−, (3.3)

which serves to isolate the field relevant to the charge density ν(x, y, t), and the essentiallynon-electrical field Σ(x, y, t).

3.1 The reduced system

In terms of the fields ν, Σ, φ, p and v we have the governing equations.

∂tν = D∆ν − v ·∇ν + µ∇Σ ·∇φ + µΣ∆φ, (3.4a)∂tΣ = D∆Σ− v ·∇Σ + µ∇ν ·∇φ + µν∆φ, (3.4b)

∆φ = − 1εel

Zeν, (3.4c)

ρm∂tv = −∇p + η∆v − Zeν∇φ, (3.4d)divv = 0. (3.4e)

19

3.1.1 Boundary Conditions

In order to solve the equations we need BC’s. At x = 0 no ions can penetrate the surface.Therefore coupled BC’s are obtained by setting the flux densities, n · i±, equal to zero,where n is a vector normal to the surface, and i± is found from (3.1a).

0 = D∂xc± ± µc±∂xφ. (3.5)

The ionic densities are as before assumed to approach the same constant value, c0, at theboundary x = ∞.

In terms of the new fields, (3.3), this leads to:

(D∂xν + µΣ∂xφ)x=0 = 0, ν|x=∞ = 0, (3.6a)(D∂xΣ + µν∂xφ)x=0 = 0, Σ|x=∞ = 2c0. (3.6b)

The potential is subject to the BC:

φ|x=0 = Vext(y, t) ≡ εV0 cos(qy)eiωt, φ|x=∞ = 0, (3.6c)

where ε is a small dimensionless parameter, which will be our perturbation parameter.The complex notation is used for convenience only, the physical quantities are taken asthe real part of expressions. Note that in the nonlinear terms the product must be takenof the real parts of the factors.

The velocity is subject to a no-slip BC which, together with the assumption that theapplied potential only causes a spatially limited disturbance of the flow, gives

v|x=0 = 0, v|x=∞ = 0. (3.6d)

The pressure p must be zero far from the interface. For x = 0 we find from the Navier-Stokes equation (3.8d) using the no-slip BC on v:

p|x=∞ = 0. (3.6e)

Note that one BC for p is actually enough, since p is really determined through ∇p. Thusit is in principle only a first order PDE on p requiring just one BC.

Note that the periodicity of Vext implies that all fields must be periodic in y with longestperiod 2π/q.

3.1.2 Expanding the fields

For zero perturbation, ε = 0, corresponding to the liquid in equilibrium with a zero poten-tial, the solution to (3.4) and (3.6) is obvious. All fields but Σ trivial and Σ = Σ0 ≡ 2c0

is a solution. By the uniqueness properties of the system of linear PDE’s, this is the only

20

solution.

All five fields are expanded in ε. Due to the argument above only the Σ field has azero order term, while all other fields are expanded from first order:

ν = εν1 + ε2ν2 + · · · , (3.7a)

Σ = Σ0 + εΣ1 + ε2Σ2 + · · · , (3.7b)

v = εv1 + ε2v2 + · · · , (3.7c)

p = εp1 + ε2p2 + · · · . (3.7d)

This is inserted in eqs. 3.4. It is assumed that the sums are convergent for ε in an openinterval around 0.

The sums must satisfy the BC’s (3.6), which must be fulfilled for every order of ε.

3.2 First order perturbation

We collect the coefficients to ε1 in eqs. 3.4, using Σ0 constant:

∂tν1 = D∆ν1 + µΣ0∆φ1, (3.8a)∂tΣ1 = D∆Σ1, (3.8b)

∆φ1 = − 1εel

Zeν1, (3.8c)

ρm∂tv1 = −∇p1 + η∆v1, (3.8d)divv1 = 0. (3.8e)

Inserting the expansions yield BC’s:

(D∂xν1 + µΣ0∂xφ1)x=0 = 0, ν1|x=∞ = 0, (3.9a)∂xΣ1|x=0 = 0, Σ1|x=∞ = 0, (3.9b)

φ1|x=0 = Vext(y, t), φ1|x=∞ = 0, (3.9c)v1|x=0 = 0, v1|x=∞ = 0, (3.9d)

p1|x=∞ = 0. (3.9e)

We note that the electrodynamics and the hydrodynamics are completely decoupled inthis perturbation order. Also the Σ1 field is decoupled from all other fields.

Starting with the simplest, trivial v1, p1 and Σ1 are seen to be admissible solutions tothe relevant homogeneous boundary value problems and thus represent the correct fields;

v1 = 0, (3.10)p1 = 0, (3.11)Σ1 = 0. (3.12)

21

3.2.1 Solving for ν1

(3.8c) is inserted in (3.8a), yielding

∂tν1 = D∆ν1 − µΣ0Ze

εelν1 =

(D∆− 1

λ2D

D

)ν1, (3.13)

by the Einstein relation. This is solved by separation of variables, inserting ν1 = T (t)X(x)Y (y).We obtain

T ′

T= D

(X ′′

X+

Y ′′

Y

)− 1

λ2D

D (3.14)

The two sides are independent and each is set equal to a constant of separation. In orderto obtain time-periodic solutions of angular frequency ω in T , we set the constant equalto iω. We are thus left with

X ′′

X+

Y ′′

Y− 1

λ2D

(1 + i

ω

Dλ2

D

)= 0. (3.15)

Again each term must be constant. Looking for periodic solutions Y = cos(qy), we get

X ′′

X= κ2, κ ≡ 1

λD

√(qλD)2 + 1 + i

ω

Dλ2

D, (3.16)

yielding two exponential solutions, where the one increasing in x is discarded due toboundedness. We thus have the total solution

ν1 = A1e−κx cos(qy)eiωt, (3.17)

where the constant A1 is to be determined.

3.2.2 Solving for φ1

Insertion of ν1 in (3.8c) gives

∆φ1 = − 1εel

ZeA1e−κx cos(qy)eiωt. (3.18)

A good guess for a particular solution is φP1 = B1e

−κx cos(qy)eiωt, where B1 is a constant.We find

∆φP1 = (κ2 − q2)φP

1 =1

λ2D

(1 + i

ω

Dλ2

D

)B1e

−κx cos(qy)eiωt, (3.19)

and thereforeA1 = − εel

Zeλ2D

(1 + i

ω

Dλ2

D

)B1 ≡ γB1. (3.20)

Now for the solution φH1 to the homogeneous equation: The total solution must fit Vext

at x = 0. Thus φH1 |x=0 = Vext − φP

1 |x=0. By separation of variables we obtain the mostgeneral (bounded) solution:

φH1 = C(t) cos(qy)e−qx, (3.21)

22

where C(t) = C1eiωt, due to the temporal variation. The found BC then implies C1 + B1 = V0.

To first order the BC (3.6a) yields

D∂xν1 + µΣ0∂xφ1 = 0. (3.22)

The found solutions are inserted:

DκA1 + 2µc0

[κB1 + q(V0 −B1)

]= 0. (3.23)

Here (3.20) is inserted and we solve for B1:

B1 = − 2µc0q

Dκγ + 2µc0(κ− q)V0 ≡ −C1V0. (3.24)

And we are ready to present:

3.2.3 The Full solution

The full solution, for all first order perturbation fields, is:

ν1 = −γC1 V0e−κx cos(qy)eiωt, (3.25a)

Σ1 = 0, (3.25b)

φ1 =[− C1e

−κx + (1 + C1)e−qx]

V0 cos(qy)eiωt, (3.25c)v1 = 0, (3.25d)p1 = 0, (3.25e)

where B is given by (3.24).We have tested the solution manually by direct insertion in the governing equations

and BC’s.

Surprisingly we have found that the liquid velocity is independent of the perturbationto first order.

3.3 First Order Outlook

We have now found the first order perturbation field. We found that in this order theonly non-trivial fields were the potential and the density ν1. We therefore will move onto second order perturbation, where we hope to determine the full solution. Here thepotential and the density, ν2, become trivial, but now the velocity and pressure fieldsbecome interesting. If you dare to read on, you will see that we are almost there. Wehave determined the pressure and velocity fields down to a constant, P0, which we hopeto determine from the last BC.

23

3.4 Second order perturbation

In second order we get from (3.4):

∂tν2 = D∆ν2 + µΣ0∆φ2, (3.26a)∂tΣ2 = D∆Σ2 + µ∇ν1 ·∇φ1 + µν1∆φ1, (3.26b)

∆φ2 = − 1εel

Zeν2, (3.26c)

ρm∂tv2 = −∇p2 + η∆v2 − Zeν1∇φ1, (3.26d)divv2 = 0. (3.26e)

And from (3.6) the BC’s:

(D∂xν2 + µΣ0∂xφ2)x=0 = 0, ν2|x=∞ = 0, (3.27a)(D∂xΣ2 + µν1∂xφ1)x=0 = 0, Σ2|x=∞ = 0, (3.27b)

φ2|x=0 = 0, φ2|x=∞ = 0, (3.27c)v2|x=0 = 0, v2|x=∞ = 0, (3.27d)

p2|x=∞ = 0. (3.27e)

3.4.1 Solving for ν2

It is seen that ν2 and φ2 in eqs. 3.26 and eqs. 3.27 are decoupled from the other fieldsand they can thus be separately determined. A trivial solution does the trick.

The Σ1 field is independent of the other second order fields and is therefore of limitedinterest. If, however, the perturbation analysis is to be carried on to a higher order, thisfield must be determined.

We are now left with a forced Navier-Stokes equation with body force −Zeν1∇φ1.

3.4.2 Futile analytical attempt

The body force becomes quite complicated when the real parts of ν1 and ∇φ1 are takenand multiplied. In the x direction it consists of 20 independent terms, and in the y direc-tion 10 independent terms.

In our attempt to guess a solution we started with the y component of the Navier-Stokesequation. We naturally tried inserting one of the 10 terms e.g.

e−κ1x cos(2κ2x) sin(qy)ei2ωt, (3.28)

where we have expanded κ in real and imaginary parts κ ≡ κ1 + iκ2, κ1, κ2 ∈ R.Applying the operator, A ≡ (ρm∂t − η∆), we find however, that the result is a linear

combination of several of the functions involved the infamous 10 terms and one extraindependent function. This invites us to look for a guess in the form of a linear combination

24

of the 11 functions, which we call fi(x, y, t). We will use matrix algebra consideringfi as basis for a vector space. The linear combination is described by the vector a =(a1, . . . , a11). In terms of the function base A can be defined as a matrix operator

Aai = Aijaj , (3.29)

where the Einstein summation convention implies sum over j on the right hand side.The matrix A must be determined by finding the coefficient Aij to fi in Afj . We haveperformed this lengthy calculation one long Friday afternoon.

Once the matrix A is known, a suitable guess can be found as follows. A general vectorof unknown coefficients is inserted, here a. The operator acts on the corresponding linearcombination, yielding another, which is set equal to b, where bi is the coefficient to fi inthe body force term, and b11 = 0. This corresponds to the matrix equation

Aa = b, (3.30)

where tilde denotes the transposed vector. The unknown vector a is then determined bystandard Gauss elimination. The equation has a solution as long as the rank of A is noless then the rank of the augmented matrix Ab, and the method relies on this.

We have done this in Maple Appendix C, and in principle confirmed the approach. Theexpressions, however, become gruesome and we were unable to make Maple reduce themproperly. This appears to be a major obstacle to the method, since we have no manageableway to treat or even test our solution. Thus we are not pursuing this idea any further.

3.4.3 A fantastically successful analytical attempt

It turned out to be much much better to keep the complex notation when calculating thebody force Re(ν1)Re(∇φ1) = 1

4(ν1 +ν∗1)(∇φ1 +∇φ∗1) where * denotes complex conjugate.In the x direction we get the body force Fx with 14 independent terms:

Fx ≡ −Zeν1∇xφ1 = Bxi fx

i (x, y, t). (3.31a)

Throughout this section we relegate lengthy results to section A. Expressions for Bxi and

fxi (x, y, t) can be found there.

In the y direction we get a body force Fy with 7 terms.

Fy ≡ −Zeν1∇yφ1 = Byi fy

i (x, y, t). (3.31b)

Byi and fy

i are found in section A.The operator A is a scalar operator on each of these functions corresponding to a

diagonal matrix operator. We thus have Afxi = Ax

i fxi (no sum) for i ∈ {1, . . . 14}; Afy

i =Ay

i fyi (no sum) for each i ∈ {1, . . . 7} , where the coefficients Ax

i and Ayi are found in

section A.The guess vF (read: Force) for the velocity:

vF ≡ vFx ex + vF

y ey, (3.32)

25

where

vFx ≡

14∑

i=1

1Ax

i

Bxi fx

i (x, y, t), (3.33)

vFy ≡

7∑

i=1

1Ay

i

Byi fy

i (x, y, t), (3.34)

thus takes care of the driving force. Note in section A that some of the functions fxi and

fyi are constant in time implying an biased flow. More disturbing is the fact that some

the function fxi in vx, i.e. the normal direction, are independent of y. This is impossible

for continuity reasons - Clearly we must make sure that the final velocity has zero fluxthrough every plane x =constant.

We therefore go to work with the non-zero divergence of this field, looking for a guessedparticular solution,

vP2 ≡ vF + vD, (3.35)

to (3.26d) and the continuity equation, where the correction term vD is supposed to killthe divergence. We thus demand

divvF = −divvD, (3.36)

The inhomogeneous Navier-Stokes equation, (3.26d), requires

AvD = −∇p. (3.37)

Taking the divergence and inserting eq. (3.36) we get an equation for the pressure

The pressure

∆p = A(divvF ). (3.38)

The pressure is from this Poisson equation indeterminate down to any harmonic function.However, appropriate BC’s will be employed at a later stage, and the pressure will beuniquely determined.

The divergence operator on the right hand side is examined. The fxi and fy

i functionsare all mapped into the space of fx

i functions. The operator acts as a scalar on fxi in vx,

∂xfxi = Dif

xi , (3.39)

where the coefficients Di are defined in Maple Appendix D.In vy the operator ∂y shifts the sine function in fy

i to a cosine and multiplies by 2q.We have

∂yfyi = 2qfx

ki, (3.40)

26

where ki picks the function fxki

, that enters the image of fyi under ∂y, and is defined

properly in (A.5), section A. In (3.38), A thus operates on a linear combination of fxi s,

where it is known from the solution of vF that Afxi = Ax

i fxi (no sum). The right hand

side of (3.38) is thus known, and we are ready to solve the Poisson equation.The Laplacian is found to be a scalar operator on fx

i , so that ∆fxi = Lif

xi , where Li is

defined in Maple Appendix D. We thus have a (particular) solution p:

p ≡ Eifxi , (3.41)

where

Ei ≡ 1Li

[BiDi + 2q

7∑

n=1

Byn

AynAx

knδi,kn

]no sum over i. (3.42)

δ is the Kronecker delta, the sum in the second term only serves to put in the one term(possibly zero) from fy

i that was transferred by the divergence operator.

Looking for a homogeneous solution we use separation of variables. We find the mostgeneral admissible harmonic solution pH :

pH = (P+ei2ωt + P−e−i2ωt) cos(2qy)e−2qx ≡ pH+ + pH

− , (3.43)

where the final definition is obvious.

Killing the divergence

With the pressure p ≡ pP +pH known, vD is determined by (3.37), in the same way as wefound vF from (3.31). First the gradient must be calculated. ∇x acts in the same way asthe divergence on the x-component of a vector field, and ∇y acts similar to the divergenceon the y component, only here the derivative of the cosine is −2q sin(2qy).

As for the pH , we find ∇xpH± = −2qpH± , and ∇ypH± = −2qpH

±S , where pH±S is pH± with

cos(2qy) replaced by sin(2qy), and pHS is defined accordingly.

∇xp =14∑

i=1

DiEifxi − 2qpH , (3.44)

∇yp = −2q14∑

i=1

Eify

k−1i

− 2qpHS = −2q

7∑

i=1

Ekify

i − 2qpHS . (3.45)

Just as in (3.31), the solution vD to (3.37) is found by dividing fi by the correspondingscalar operator Ai and noting that ApH± = ±i2ωρpH± , ApH

±S = ±i2ωρpH±S .

vD = vDx ex + vD

y ey, (3.46)

vDx = −

14∑

i=1

Di

Axi

Eifxi −

i

2ωρ(pH

+ − pH− ), (3.47)

vDy =

7∑

i=1

2q

Ayi

Ekifyi −

i

2ωρ(pH

+S − pH−S). (3.48)

27

We now have a divergence free particular solution vP2 ≡ vF + vD, where we are very

relieved to see that all the terms uniform in y cancel in the sum (coefficients tofx

i , i ∈ {3, 4, 7, 8, 10, 13, 14}).

The homogeneous solution vH will be utilized to make the total flow perturbation v2 ≡vP +vH meet the homogeneous velocity BC’s (3.27d). The general admissible vH is foundthrough separation of variables .

vHx = ax

+ cos(2qy)e−q

4q2−i 2ωρη

x cos(2qy)ei2ωt (3.49)

+ ax− cos(2qy)e−

q4q2+i 2ωρ

ηx cos(2qy)ei2ωt (3.50)

+ ax0 cos(2qy)e−2qx cos(2qy), (3.51)

vHy = ay

+ sin(2qy)e−q

4q2−i 2ωρη

x cos(2qy)ei2ωt (3.52)

+ ay− sin(2qy)e−

q4q2+i 2ωρ

ηx cos(2qy)ei2ωt (3.53)

+ ay0 sin(2qy)e−2qx cos(2qy). (3.54)

where continuity requires that the constants in x and y be equal: ax± = ay± ≡ a± and

ax0 = ay

0 ≡ a0.

The final velocity field

We now have

v2 =vF + vD + vH (3.55)

= ex

14∑

i=1

(1

Axi

Bxi −

Di

Axi

Ei

)fx

i −i

2ωρ(pH

+ − pH− ) + vH

x (3.56)

+ ey

7∑

i=1

(1

Ayi

Byi +

2q

Ayi

Eki

)fy

i −i

2ωρ(pH

+S − pH−S) + vH

y (3.57)

The constants should determined by the homogeneous BC in v2, (3.27d):

v2|x=0 = 0. (3.58)

In each case there are three independent functions in time.

For each one the corresponding coefficient a+, a− and a0, together with P+ and P−,is determined.

3.4.4 Where it all goes terribly wrong

It turns out that the system of linear equations in a± and P± has no solution! The corre-sponding functions are linearly dependent and thus cannot solve in both the y directionand the x direction.

28

The reason for this to happen is, that the velocity field stemming from the introductionof the homogeneous pressure is per construction divergence free and thus subject to thesame kind of constraint as the homogeneous velocity field.

Anyway it seems that the homogeneous pressure, which seemed to offer our last degreeof freedom, contributes in just the same way to the velocity field, and thus it gives usnothing.

We are very annoyed to stop at this stage, but we are running out of time. At thismoment we are not aware of the flaw in our treatment. We hope to come back and fix itin the near future.

3.4.5 Continuing to third order..

To third order we note that the hydrodynamic equations, (3.4d) and (3.4e), give

ρm∂tv3 = −∇p3 + η∆v3, (3.59)divv3 = 0, (3.60)

because the nonlinear terms in the Navier-Stokes equations vanish. The BC’s are homo-geneous in v3. In p3 we get from (3.6e)

−∂xp3 + η∂2xv3 x = 0. (3.61)

Thus trivial hydrodynamic fields satisfy the system. The found velocity field is thus exactup to third order:

v = ε2v2 +O(ε4). (3.62)

3.4.6 conclusion

This is where we got to. We have almost completed the investigation of lowest orderfields of interest (excluding Σ). In the process we have found that the electrodynamic andhydrodynamic fields come in to play separately changing place for every order.

This seems quite intuitively correct, as the effect of the perturbation in the appliedpotential has an indirect influence on the fluid motion.

We got very close, but got no cigar.

29

Chapter 4

Conclusion

In this project we have introduced ourselves to perturbation theory and applied it to threeproblems all relevant to microfluidics.

The first two problems, Poiseuille flow through a perturbed circular cylinder and thesolution of the Poisson-Boltzmann equation in the planar case, have been solved com-pletely to 2nd and 7th order, respectively. By these problem, we have gotten a feel forthe perturbation techniques and prepared ourselves to the third and more difficult problem.

The third problem was treating a complete electrohydrodynamic model of a microflu-idic pump or mixer. We have completed the calculations for zeroth and first order, andhave carried on well into second order. Here we experienced some problems, and due tothe limited time at our disposal, we were not able to complete the calculations.

The perturbation approach seems to have been very effective and offers a nice way ofsolving the system of PDE’s in a systematic way. We think the full result to second orderis close at hand, and we are eager to come back and finish our work.

30

Bibliography

[Bruus] Henrik Bruus. Theoretical microfluidics. MIC - Department of Micro and Nan-otechnology, Technical University of Denmark, first edition, 2004.

[Mortensen] Niels Asger Mortensen et al. Electro-hydrodynamics of binary electrolytesdriven by modulated surface potentials. MIC - Department of Micro and Nan-otechnology.

[Lautrup] Benny Lautrup. Physics of Continuous Matter. The Institute of Physics, NBI,first edition, 2004.

31

Appendix A

Tedious calculation

x direction

Fx ≡ −Zeν1∇yφ1 = Bxi fx

i (x, y, t), (A.1a)

with coefficients:

Bx1 ≡ σxγq(1 + C1), fx

1 ≡ e−(κ+q)x cos(2qy)ei2ωt, (A.1b)

Bx2 ≡ −σxγκC1, fx

2 ≡ e−2κx cos(2qy)ei2ωt, (A.1c)

Bx3 ≡ Bx

1 , fx3 ≡ e−(κ+q)x ei2ωt, (A.1d)

Bx4 ≡ Bx

2 , fx4 ≡ e−2κx ei2ωt, (A.1e)

Bx5 ≡ σxγ∗q(1 + C1), fx

5 ≡ e−(κ∗+q)x cos(2qy), (A.1f)

Bx6 ≡ −σx2Re(γκC1), fx

6 ≡ e−2Re(κ)x cos(2qy), (A.1g)

Bx7 ≡ Bx

5 , fx7 ≡ e−(κ∗+q)x, (A.1h)

Bx8 ≡ Bx

6 , fx8 ≡ e−2Re(κ)x, (A.1i)

Bx9 ≡ σxγq(1 + C∗

1 ), fx9 ≡ e−(κ+q)x cos(2qy), (A.1j)

Bx10 ≡ B9, fx

10 ≡ e−(κ+q)x, (A.1k)

Bx11 ≡ σxγ∗q(1 + C∗

1 ), fx11 ≡ e−(κ∗+q)x cos(2qy)e−i2ωt, (A.1l)

Bx12 ≡ −σxγ∗κ∗C∗

1 , fx12 ≡ e−2κ∗x cos(2qy)e−i2ωt, (A.1m)

Bx13 ≡ Bx

11, fx13 ≡ e−(κ∗+q)x e−i2ωt, (A.1n)

Bx14 ≡ B12, fx

14 ≡ e−2κ∗x e−i2ωt, (A.1o)

where the common real factor is given by σx ≡ 14ZeV 2

0 .

32

We have Afxi = Ax

i fxi (no sum over i) for each i ∈ {1, . . . 14}, where the constants Ax

i

are found:

Ax1 = 2iρω + η(3q2 − 2κq − κ2), (A.2a)

Ax2 = 2iρω + 4η(q2 − κ2), (A.2b)

Ax3 = 2iρω − η(q + κ)2, (A.2c)

Ax4 = 2iρω − 4ηκ2, (A.2d)

Ax5 = η(3q2 − 2κ∗q − κ∗2), (A.2e)

Ax6 = 4η[q2 − Re(κ)2], (A.2f)

Ax7 = − η(q + κ∗)2, (A.2g)

Ax8 = − 4ηRe(κ)2, (A.2h)

Ax9 = η(3q2 − 2κq − κ2) = Ax∗

5 , (A.2i)

Ax10 = − η(q + κ)2 = Ax∗

7 , (A.2j)

Ax11 = −2iρω + η(3q2 − 2κ∗q − κ∗2) = Ax∗

1 , (A.2k)

Ax12 = −2iρω + 4η(q2 − κ∗2) = Ax∗

2 , (A.2l)

Ax13 = −2iρω − η(q + κ∗)2 = Ax∗

3 , (A.2m)

Ax14 = −2iρω − 4ηκ∗2 = Ax∗

4 . (A.2n)

y direction

Fy ≡ −Zeν1∇yφ1 = Byi fy

i (x, y, t), (A.3a)

with coefficients:

By1 ≡ σyγC2

1 , fy1 ≡ e−2κx sin(2qy)ei2ωt, (A.3b)

By2 ≡ −σyγC1(1 + C1), fy

2 ≡ e−(κ+q)x sin(2qy)ei2ωt, (A.3c)

By3 ≡ σy2Re(γ)|C1|2, fy

3 ≡ e−2Re(κ)x sin(2qy), (A.3d)

By4 ≡ −σyγ

∗C∗1 (1 + C1), fy

4 ≡ e−(κ∗+q)x sin(2qy), (A.3e)

By5 ≡ −σyγC1(1 + C∗

1 ), fy5 ≡ e−(κ+q)x sin(2qy), (A.3f)

By6 ≡ σyγ

∗|C1|2, fy6 ≡ e−2κ∗x sin(2qy)e−i2ωt, (A.3g)

By7 ≡ −σyγ

∗C∗1 (1 + C∗

1 ), fy7 ≡ e−(κ∗+q)x sin(2qy)e−i2ωt, (A.3h)

where the common real factor is given by σy ≡ 12ZeV 2

0 q.We have Afy

i = Ayi f

yi (no sum over i) for each i ∈ {1, . . . 7}, where the constants Ay

i

33

are found:

Ay1 = 2iρω + 4η(q2 − κ2), (A.4a)

Ay2 = 2iρω + η(3q2 − 2κq − κ2), (A.4b)

Ay3 = 4η[q2 − Re(κ)2], (A.4c)

Ay4 = η(3q2 − 2κ∗q − κ∗2), (A.4d)

Ay5 = η(3q2 − 2κq − κ2) = Ay∗

4 , (A.4e)

Ay6 = −2iρω + 4η(q2 − κ∗2) = Ay∗

1 , (A.4f)

Ay7 = −2iρω + η(3q2 − 2κ∗q − κ∗2) = Ay∗

2 . (A.4g)

We have now gotten enough of this typing. The rest of the coefficients (Di, Li and Ei)are listed in Maple Appendix C.

The function k : {1 . . . 7} 7→ U is defined to assign to the argument i a number givenby the table below. The function k−1 is defined as the extension of k’s inverse functiondone by defining fx

k−1i

≡ 0, whenever i /∈ U .

k ↓ 1 2 3 4 5 6 72 1 6 5 9 12 11

↑ k−1 (A.5)

34

A := 1 + ε sin k θ( )

Kr := 11 + ε sin k θ( )( ) a

Kphi := - r ε cos k θ( ) k

1 + ε sin k θ( )( )2 a

K2phi := 2 r ε 2 cos k θ( )2 k2

1 + ε sin k θ( )( )3 a +

r ε sin k θ( ) k2

1 + ε sin k θ( )( )2 a

ε

lapl0 := A2rho

a 2 + Arho

a 2 ρ + A2theta

ρ 2 a 2

lapl1 := - 2 sin k θ( ) A2rho

a 2 + - 2 sin k θ( )

a 2 ρ +

sin k θ( ) k2

ρ a 2

⎛⎜⎜⎝

⎞⎟⎟⎠

Arho - 2 cos k θ( ) k Arhotheta

ρ a 2 -

2 sin k θ( ) A2theta

ρ 2 a 2

lapl2 := 3 sin k θ( )2

a 2 +

cos k θ( )2 k2

a 2

⎛⎜⎜⎝

⎞⎟⎟⎠

A2rho + 3 sin k θ( )2

a 2 ρ +

2 ρ cos k θ( )2 k2 - sin k θ( )2 ρ k2

ρ 2 a 2 -

2 sin k θ( )2 k2

ρ a 2

⎛⎜⎜⎝

⎞⎟⎟⎠

Arho

+ 6 sin k θ( ) cos k θ( ) k Arhotheta

ρ a 2 +

3 sin k θ( )2 A2theta

ρ 2 a 2

ε

V0 := ρ → P a 2 ρ 2 - 1( )

4 η

lapl1V0 := sin k θ( ) P -4 + k2( )

2 η

V1 := rho , theta( ) → P a 2 ρ 2 - ρ k( ) sin k θ( )

2 η

lapl1V1 := -8 P ρ 2 + 8 P ρ 2 cos 2 k θ( ) - 8 P k2 ρ 2 cos 2 k θ( ) + P k3 ρ k + 3 P k3 ρ k cos 2 k θ( )

4 η ρ 2

lapl2V0 :=

32

P - 32

P cos 2 k θ( ) + 32

P k2 cos 2 k θ( )

η

RHS := P -2 ρ 2 + 2 k2 ρ 2 - 3 k3 ρ k( ) cos 2 k θ( )

4 η ρ 2 + P 2 ρ 2 - k3 ρ k( )

4 η ρ 2

lapl0V2 :=

∂ 2

∂ ρ 2 V2 ρ, θ( )

a 2 +

∂

∂ ρ V2 ρ, θ( )

a 2 ρ +

∂ 2

∂ θ2 V2 ρ, θ( )

ρ 2 a 2

ψ := rho , theta( ) →

P a 2 12

ρ 2 - 12

ρ 2 cos 2 k θ( ) - k ρ k + k ρ k cos 2 k θ( )⎛⎜⎝

⎞⎟⎠

4 η

V2 := rho , theta( ) → ψ ρ , θ( )

P 2 ρ 2 - 2 ρ 2 cos 2 k θ( ) + 2 k2 ρ 2 cos 2 k θ( ) - k3 ρ k - 3 k3 ρ k cos 2 k θ( )( )

4 η ρ 2

true

P a 2 12

- 12

cos 2 k θ( ) - k + cos 2 k θ( ) k⎛⎜⎝

⎞⎟⎠

4 η

BC1 = -

P a 2 - 12

+ k⎛⎜⎝

⎞⎟⎠

cos 2 k θ( )

4 η -

P a 2 12

- k⎛⎜⎝

⎞⎟⎠

4 η

χ ρ , θ( )

χ := rho , theta( ) → A0 ρ 2 k( ) cos 2 k θ( ) + B0

V2 := rho , theta( ) → χ ρ , θ( )

lapl0V2 = 0

A0 := -

P a 2 - 12

+ k⎛⎜⎝

⎞⎟⎠

4 η

B0 := -

P a 2 12

- k⎛⎜⎝

⎞⎟⎠

4 η

V2 := rho , theta( ) → ψ ρ, θ( ) + χ ρ , θ( )

true

V2 1, θ( ) = 0

V2 =

P a 2 12

ρ 2 - 12

ρ 2 cos 2 k θ( ) - k ρ k + k ρ k cos 2 k θ( )⎛⎜⎝

⎞⎟⎠

4 η -

P a 2 - 12

+ k⎛⎜⎝

⎞⎟⎠

ρ 2 k( ) cos 2 k θ( )

4 η -

P a 2 12

- k⎛⎜⎝

⎞⎟⎠

4 η

Vt := rho , theta( ) → V0 ρ( ) + ε V1 ρ , θ( ) + ε 2 V2 ρ, θ( )

V := r, theta( ) → subs ρ = ra A

, Vt ρ, θ( )⎛⎜⎝

⎞⎟⎠

∆ = Pη

+ O ε 3( )⎛⎜⎝

⎞⎟⎠

∆ = Pη

+ O ε 3( )⎛⎜⎝

⎞⎟⎠

Qnorm := -3.141592654 - 1.570796327 ε 2

Q0 := 1

Q1 := 1.256637061 ε

-3.141592654 - 1.570796327 ε 2

Q2 := 4.358959807 ε 2 + 9.738937226

-3.141592654 - 1.570796327 ε 2

φ0 := e-x( )

φ2 := 148

e-3 x( ) - 1

48 e

-x( )

φ4 := - 1768

e-3 x( ) + 1

1920 e

-x( ) + 11280

e-5 x( )

φ6 := 11184320

e-3 x( ) - 17

1290240 e

-x( ) - 112288

e-5 x( ) + 1

28672 e

-7 x( )

O ε 7( )

φ ε

renu1 := x, y, t( ) → e-κ1 x( ) cos q y( ) σ1 cos κ2 x( ) + σ2 sin κ2 x( )( ) cos ω t( ) + σ1 sin κ2 x( ) - σ2 cos κ2 x( )( ) sin ω t( )( )

regradphi1 := x, y, t( ) → cos q y( ) ( τ1 e-κ1 x( ) + τ2 e

-κ1 x( ) sin κ2 x( ) + π1 e-q x( )( ) cos ω t( )

+ τ1 e-κ1 x( ) sin κ2 x( ) - τ2 e

-κ1 x( ) cos κ2 x( ) - π2 e-q x( )( ) sin ω t( ))

nu1phi1 := x, y, t( ) → simplify expand Z e renu1 x, y, t( ) regradphi1 x, y, t( )( )( )

GAET := x, y, t( ) → f1 x, y, t( ) TOT1 + f2 x, y, t( ) TOT2 + f3 x, y, t( ) TOT3 + f4 x, y, t( ) TOT4 + f5 x, y, t( ) TOT5 + f6 x, y, t( ) TOT6 + f7 x, y, t( ) TOT7 + f8 x, y, t( ) TOT8 + f9 x, y, t( ) TOT9 + f10 x, y, t( ) TOT10 + f11 x, y, t( ) TOT1

b1, 1, b2, 1, b3, 1, b4, 1, b5, 1, b6, 1, b7, 1, b8, 1, b9, 1, b10, 1, 0[ ]

force0y = - 12

Z e γ1 C1 V0 2 e-κ1 x( ) e

I1 ω t( ) -C2 e-κ2 x( ) + 1 + C2( ) e

-q x( )( ) q eI2 ω t( )

BBy1 := 12

Z e γ1 C1 V0 2 C2 q

BBy2 := - 12

Z e γ1 C1 V0 2 1 + C2( ) q

By1 := 12

Z e γ2 C 2 V0 2 q

By2 := - 12

Z e γ2 C V0 2 1 + C( ) q

By3 := 12

Z e γ2 C V0 2 C q + 12

Z e γ2 C V0 2 C q

By4 := - 12

Z e γ2 C V0 2 1 + C( ) q

By5 := - 12

Z e γ2 C V0 2 1 + C( ) q

By6 := 12

Z e γ2 C2 V0 2 q

By7 := - 12

Z e γ2 C V0 2 1 + C( ) q

A := ρ D3 - η D1, 1 + D2, 2( )

Ay1 := 2 I ρ ω - 4 η κ 2 + 4 η q 2

Ay2 := 2 I ρ ω - η κ 2 - 2 η κ q + 3 η q 2

Ay3 := η -κ 2 - 2 κ κ - κ2 + 4 q 2( )

Ay4 := η -κ2 - 2 κ q + 3 q 2( )

Ay5 := η -κ 2 - 2 κ q + 3 q 2( )

Ay6 := -2 I ρ ω - 4 η κ2 + 4 η q 2

Ay7 := -2 I ρ ω - η κ2 - 2 η κ q + 3 η q 2

vyf := Z e γ2 C 2 V0 2 q e-2 κ x( ) sin 2 q y( ) e

2 I ω t( )

2 2 I ρ ω - 4 η κ 2 + 4 η q 2( ) - Z e γ2 C V0 2 1 + C( ) q e

- κ + q( ) x( ) sin 2 q y( ) e2 I ω t( )

2 2 I ρ ω - η κ 2 - 2 η κ q + 3 η q 2( )

+

12

Z e γ2 C V0 2 C q + 12

Z e γ2 C V0 2 C q⎛⎜⎝

⎞⎟⎠

e- κ + κ( ) x( ) sin 2 q y( )

η -κ 2 - 2 κ κ - κ2 + 4 q 2( )

- Z e γ2 C V0 2 1 + C( ) q e- κ + q( ) x( ) sin 2 q y( )

2 η -κ2 - 2 κ q + 3 q 2( )

- Z e γ2 C V0 2 1 + C( ) q e- κ + q( ) x( ) sin 2 q y( )

2 η -κ 2 - 2 κ q + 3 q 2( )

+ Z e γ2 C2 V0 2 q e

-2 x κ( ) sin 2 q y( ) e-2 I ω t( )

2 -2 I ρ ω - 4 η κ2 + 4 η q 2( )

- Z e γ2 C V0 2 1 + C( ) q e- κ + q( ) x( ) sin 2 q y( ) e

-2 I ω t( )

2 -2 I ρ ω - η κ2 - 2 η κ q + 3 η q 2( )

force0x = Z e γ1 C1 V0 2 e-κ1 x( ) cos q y( )2 e

I1 ω t( ) κ2 C2 e-κ2 x( ) - q 1 + C2( ) e

-q x( )( ) eI2 ω t( )

Bx1 := 14

Z e γ2 V0 2 q 1 + C( )

Bx2 := - 14

Z e γ2 V0 2 κ C

Bx3 := 14

Z e γ2 V0 2 q 1 + C( )

Bx4 := - 14

Z e γ2 V0 2 κ C

Bx5 := 14

Z e γ2 V0 2 q 1 + C( )

Bx6 := - 14

Z e γ2 V0 2 κ C - 14

Z e γ2 V0 2 κ C

Bx7 := 14

Z e γ2 V0 2 q 1 + C( )

Bx8 := - 14

Z e γ2 V0 2 κ C - 14

Z e γ2 V0 2 κ C

Bx9 := 14

Z e γ2 V0 2 q 1 + C( )

Bx10 := 14

Z e γ2 V0 2 q 1 + C( )

Bx11 := 14

Z e γ2 V0 2 q 1 + C( )

Bx12 := - 14

Z e γ2 V0 2 κ C

Bx13 := 14

Z e γ2 V0 2 q 1 + C( )

Bx14 := - 14

Z e γ2 V0 2 κ C

fx1 := x, y, t( ) → e- κ + q( ) x( ) e

2 I ω t( ) cos 2 q y( )

fx2 := x, y, t( ) → e-2 κ x( ) e

2 I ω t( ) cos 2 q y( )

fx3 := x, y, t( ) → e- κ + q( ) x( ) e

2 I ω t( )

fx4 := x, y, t( ) → e-2 κ x( ) e

2 I ω t( )

fx5 := x, y, t( ) → e- κ + q( ) x( ) cos 2 q y( )

fx6 := x, y, t( ) → e- κ + κ( ) x( ) cos 2 q y( )

fx7 := x, y, t( ) → e- κ + q( ) x( )

fx8 := x, y, t( ) → e- κ + κ( ) x( )

fx9 := x, y, t( ) → e- κ + q( ) x( ) cos 2 q y( )

fx10 := x, y, t( ) → e- κ + q( ) x( )

fx11 := x, y, t( ) → e- κ + q( ) x( ) e

-2 I ω t( ) cos 2 q y( )

fx12 := x, y, t( ) → e-2 κ x( ) e

-2 I ω t( ) cos 2 q y( )

fx13 := x, y, t( ) → e- κ + q( ) x( ) e

-2 I ω t( )

fx14 := x, y, t( ) → e-2 κ x( ) e

-2 I ω t( )

Ax1 := 2 I ρ ω - η κ 2 - 2 η κ q + 3 η q 2

Ax2 := 2 I ρ ω - 4 η κ 2 + 4 η q 2

Ax3 := 2 I ρ ω - η κ 2 - 2 η κ q - η q 2

Ax4 := 2 I ρ ω - 4 η κ 2

Ax5 := -η κ2 + 2 κ q - 3 q 2( )

Ax6 := -η κ 2 + 2 κ κ + κ2 - 4 q 2( )

Ax7 := -η κ + q( )2

Ax8 := -η κ + κ( )2

Ax9 := η -κ 2 - 2 κ q + 3 q 2( )

Ax10 := -η κ + q( )2

Ax11 := -2 I ρ ω - η κ2 - 2 η κ q + 3 η q 2

Ax12 := -2 I ρ ω - 4 η κ2 + 4 η q 2

Ax13 := -2 I ρ ω - η κ2 - 2 η κ q - η q 2

Ax14 := -2 I ρ ω - 4 η κ2

vxf := Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( ) e

2 I ω t( ) cos 2 q y( )

4 2 I ρ ω - η κ 2 - 2 η κ q + 3 η q 2( ) - Z e γ2 V0 2 κ C e

-2 κ x( ) e2 I ω t( ) cos 2 q y( )

4 2 I ρ ω - 4 η κ 2 + 4 η q 2( )

+ Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( ) e

2 I ω t( )

4 2 I ρ ω - η κ 2 - 2 η κ q - η q 2( ) - Z e γ2 V0 2 κ C e

-2 κ x( ) e2 I ω t( )

4 2 I ρ ω - 4 η κ 2( )

- Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( ) cos 2 q y( )

4 η κ2 + 2 κ q - 3 q 2( )

-

- 14

Z e γ2 V0 2 κ C - 14

Z e γ2 V0 2 κ C⎛⎜⎝

⎞⎟⎠

e- κ + κ( ) x( ) cos 2 q y( )

η κ 2 + 2 κ κ + κ2 - 4 q 2( )

- Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( )

4 η κ + q( )2

-

- 14

Z e γ2 V0 2 κ C - 14

Z e γ2 V0 2 κ C⎛⎜⎝

⎞⎟⎠

e- κ + κ( ) x( )

η κ + κ( )2

+ Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( ) cos 2 q y( )

4 η -κ 2 - 2 κ q + 3 q 2( )

- Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( )

4 η κ + q( )2 + Z e γ2 V0 2 q 1 + C( ) e

- κ + q( ) x( ) e-2 I ω t( ) cos 2 q y( )

4 -2 I ρ ω - η κ2 - 2 η κ q + 3 η q 2( )

- Z e γ2 V0 2 κ C e-2 x κ( ) e

-2 I ω t( ) cos 2 q y( )

4 -2 I ρ ω - 4 η κ2 + 4 η q 2( )

+ Z e γ2 V0 2 q 1 + C( ) e- κ + q( ) x( ) e

-2 I ω t( )

4 -2 I ρ ω - η κ2 - 2 η κ q - η q 2( )

- Z e γ2 V0 2 κ C e-2 x κ( ) e

-2 I ω t( )

4 -2 I ρ ω - 4 η κ2

( )

Dx1 := -κ - q

Dx2 := -2 κ

Dx3 := -κ - q

Dx4 := -2 κ

Dx5 := -κ - q

Dx6 := -2 ℜ κ( )

Dx7 := -κ - q

Dx8 := -2 ℜ κ( )

Dx9 := -κ - q

Dx10 := -κ - q

Dx11 := -κ - q

Dx12 := -2 κ

Dx13 := -κ - q

Dx14 := -2 κ

Dy1 := 2 cos 2 q y( ) qsin 2 q y( )

Dy2 := 2 cos 2 q y( ) qsin 2 q y( )

Dy3 := 2 cos 2 q y( ) qsin 2 q y( )

Dy4 := 2 cos 2 q y( ) qsin 2 q y( )

Dy5 := 2 cos 2 q y( ) qsin 2 q y( )

Dy6 := 2 cos 2 q y( ) qsin 2 q y( )

Dy7 := 2 cos 2 q y( ) qsin 2 q y( )

L := D1, 1 + D2, 2

L1 := κ 2 + 2 κ q - 3 q 2

L2 := 4 κ 2 - 4 q 2

L3 := κ + q( )2

L4 := 4 κ 2

L5 := κ2 + 2 κ q - 3 q 2

L6 := κ 2 + 2 κ κ + κ2 - 4 q 2

L7 := κ + q( )2

L8 := 4 ℜ κ( )2

L9 := κ 2 + 2 κ q - 3 q 2

L10 := κ + q( )2

L11 := κ2 + 2 κ q - 3 q 2

L12 := 4 κ2 - 4 q 2

L13 := κ + q( )2

L14 := 4 κ2

1 = 4 q 2

2 = 4 q 2

3 = 0

4 = 0

5 = 4 q 2

6 = 4 ℜ κ( )2 - κ 2 - 2 κ κ - κ

2 + 4 q 2

7 = 0

8 = 0

9 = 4 q 2

10 = 0

11 = 4 q 2

12 = 4 q 2

13 = 0

14 = 0

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ 2 + 2 κ q - 3 q 2( ) - Z e γ2 C V0 2 1 + C( ) q 2

κ 2 + 2 κ q - 3 q 2, Z e γ2 V0 2 κ 2 C

2 4 κ 2 - 4 q 2( ) + Z e γ2 C 2 V0 2 q 2

4 κ 2 - 4 q 2,

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ + q( )2, 1

8 Z e γ2 V0 2 C, Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ2 + 2 κ q - 3 q 2( )

+ Z e γ2 C V0 2 1 + C( ) q 2

-κ2 - 2 κ q + 3 q 2

,

-

2 - 14

Z e γ2 V0 2 κ C - 14

Z e γ2 V0 2 κ C⎛⎜⎝

⎞⎟⎠

ℜ κ( )

κ 2 + 2 κ κ + κ2 - 4 q 2

-

2 12

Z e γ2 C V0 2 C q + 12

Z e γ2 C V0 2 C q⎛⎜⎝

⎞⎟⎠

q

-κ 2 - 2 κ κ - κ2 + 4 q 2

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ + q( )2

, - - 1

4 Z e γ2 V0 2 κ C - 1

4 Z e γ2 V0 2 κ C

2 ℜ κ( ),

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ 2 + 2 κ q - 3 q 2( ) - Z e γ2 C V0 2 1 + C( ) q 2

κ 2 + 2 κ q - 3 q 2, Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ + q( )2,

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ2 + 2 κ q - 3 q 2( )

- Z e γ2 C V0 2 1 + C( ) q 2

κ2 + 2 κ q - 3 q 2

, Z e γ2 V0 2 κ2 C

2 4 κ2 - 4 q 2( )

+ Z e γ2 C2 V0 2 q 2

4 κ2 - 4 q 2

,

Z e γ2 V0 2 q 1 + C( ) -κ - q( )

4 κ + q( )2

, 18

Z e γ2 V0 2 C

C := 2 µ c0 qDi κ γ2 + 2 µ c0 κ - q( )

κ := q 2 λ 2

+ 1 + I ω λ 2

Diλ

γ2 := -

epsel 1 + I ω λ 2

Di

⎛⎜⎜⎝

⎞⎟⎟⎠

Z e λ 2