Justification of the Two-Dimensional Model of
Electroconductivity for the Earth’s Ionosphere
Valery V. Denisenko
1,21Institute of Computational Modelling, Krasnoyarsk, 660036, Russia
2Siberian Federal University, 79 Svobodny Prospect, Krasnoyarsk 660041
∗Corresponding Author: [email protected]
Copyright c⃝2013 Horizon Research Publishing All rights reserved.
Abstract
Conventional two dimensional model for electric fields in the Earth’s ionosphere is analyzed to estimate its error. The main difficulties arise due to asymmetry of the conductivity tensor. We use the energy method and small parameter expansion. To make it possible in spite of asymmetry of the tensor coefficients the problem is reduced to the problem of minimum of proper quadratic energy functional. The variational principle is stated and proved for the 3-D boundary value problem. The error of the 2-D approximation is analyzed in the case, when conductor occupies a flat layer 0< z < z0 and is homogeneous in z direction, and the vector of magnetic field has onlyz component. The results of numerical simulation of the electric field penetration from ground to the Earth’s ionosphere with reduction of the 3-D model of the ionospheric conductor to the 2-D model are presented. Precision of such an approach is demonstrated.Keywords
elliptical equation, hyrotropic medium, energy method, electric field, atmosphere, ionosphere1
Introduction
The tensor of electric conductivity in the Earth’s iono-sphere is asymmetric, and conductivity in the direction of magnetic field is much larger than other components. So, mathematical models of large scale electric fields and currents in the Earth’s ionosphere are usually based on the assumption of infinite conductivity in the direction of magnetic field. Then magnetic field lines are equipo-tentials, and the model of electric potential distribution is two-dimensional.
A correction to such two-dimensional potential can be found as a solution for the equation that is stated in [12]. Typical values of the terms of this equation are estimated there to demonstrate that the correction is small. Electric fields in low latitude ionosphere are calculated in [18] in frames of 3−D and 2−D models, and it is shown that the results are close to each other as far as their 2−D features are concerned.
This paper is devoted to justification of the two-dimensional approximation with estimation of the error. A simple case is under analysis. The conductor occupies a flat layer 0< z < z0, it is homogeneous inzdirection, vector of magnetic field has onlyzcomponent.
We use the energy method [15] and expand the solu-tion as a power series of a small parameterεthat is the ratio of transversal and field-aligned conductivities. It works for the problems with asymmetric tensor coeffi-cients because problems of this kind are reduced to the problems of minimum of proper quadratic energy func-tionals in [2, 3]. Since the 3-D boundary value problem under analysis is a specific one, the variational principle is stated and proved in the section 4.
The 2-D model is used for numerical simulation of the electric field penetration from ground to the Earth’s Ionosphere in the last section. Its accuracy is shown for real height distributions of the coefficients of conductiv-ity σ which are not constant in contrast with such a restriction in our mathematical proofs.
2
Statement of the 3-D boundary
value problem
In a three-dimensional domain Ω occupied by a con-ductor, electric field strength e and current density j
satisfy the equations
divj=q curle=g
j=σe. (1)
The components of the conductivity tensor σ and right-hand sidesq,gare given functions of Cartesian co-ordinatesx, y, z.
Here we analyze the domain Ω that is a flat layer 0 < z < z0. We suppose periodicity in x, y directions with periods a, b
jx| x=a
x=0= 0, ey| x=a
x=0= 0, ez|
x=a x=0= 0, jy|
y=b
y=0= 0, ex|
y=b
y=0= 0, ez| y=b
When the upper boundaryz=z0and the bottom one z= 0 are isolators
jz|z=0,z0 = 0. (3)
For the problem (1-3) to be solvable, right-hand sides in (1) must satisfy the following constraints
divg= 0,
∫ ∫
gz(x, y,0)dxdy= 0, ∫
q dΩ = 0. (4) We do not write the limits for integrals when integrate over the whole domain or the whole interval.
Introduce an auxiliary function ˜e
˜ ex=
∫ z
0
gy dz′, ˜ey=− ∫ z
0
gx dz′, ˜ez= 0.
For the difference e−˜e the right-hand side in the second equation (1) becomes (0,0,˜gz), where
˜
gz=gz+
∂ ∂x
∫ z
0
gxdz′+
∂ ∂y
∫ z
0
gy dz′.
Interchange differentiation and integration and use the Newton-Leibnitz formula for the functiongz(x, y, z):
˜
gz(x, y, z) =gz(x, y,0) + ∫ z
0
(
∂gx
∂x + ∂gy
∂y + ∂gz
∂z
)
dz′.
As the result of the first constraint (4) the integrand equals zero. Hence the new right-hand side
˜
gz(x, y, z) =gz(x, y,0)
does not depend on z. The right-hand side in the first equation (1) changes also. To avoid new designations, we suppose without loss of generality that it is done beforehand and so from the very beginning
gx=gy = 0, gz=gz(x, y). (5)
To get a generalized solution for the problem (1-3), suppose that the right-hand sides in (1) are bounded in the norm of the spaceL2(Ω)
∫
q2dΩ< q20,
∫
|g|2 dΩ< g02. (6)
These constraints may be replaced with the less rigid ones that are so called divergency constraints [15]. How-ever, we use more rigid constraints to construct an ap-proximate solution. The function q and its derivatives with respect tox, yare assumed to be bounded
∂kq
∂xi ∂yj
≤ c3
xk
0
, k= 0,1,2, i+j =k. (7)
Herec3, x0are positive constants.
Since magnetic field is vertical, and the medium with-out magnetic field is isotropic, the tensorσ is invariant with respect to rotation aroundzaxis. Hence it may be written as
σ=
σP −σH 0
σH σP 0
0 0 σ∥
. (8)
Here σP, σH, σ∥, σC = (σ2 P +σ
2
H)/σP are referred as Pedersen, Hall, field-aligned and Cowling conductivities. The Hall parameter β =σH/σP is a measure of asym-metry.
In the main conducting layer of the Earth’s ionosphere σ∥ ≫σP,σ∥ ≫σH.
Let us introduce a small parameter εso that ˜
σ∥ =εσ∥ (9) is of the same order of magnitude as σP. Unbounded increase ofσ∥ is equivalent toε→0.
Denote the matrixσwith ˜σ∥ instead ofσ∥ as ˜σ. It is independent of ε. The coefficients of the matrix ˜σ are assumed to be uniformly bounded and the symmetrical part of ˜σ is assumed to be uniformly positive definite. These properties of ˜σmay be written as follows:
c1≤σP, (σ2P+σ2H)/σP ≤1/c1, (10) c1≤σ˜∥ ≤1/c1, (11) where c1 is some positive constant.
However we use more rigid conditions to construct an approximate solution. We additionally suppose σto be independent ofz
∂σ/∂z= 0, (12) and the second derivatives σP,σ˜∥ with respect to x, y are assumed to be bounded:
∂k ∂xi ∂yj
1 ˜ σ∥
≤
c2
xk0, k= 0,1,2, i+j=k,
∂σP
∂x
≤ c0
x0
, ∂σP ∂y
≤ c0
x0
, (13) where c2, c0, x0 are positive constants.
The new problem, which solution gives the solution to the problem (1-3), is stated below in the section 4. The original problem has a solution if the new problem has, but the uniqueness must be proved independently.
3
Uniqueness of the classical
so-lution
Suppose that there are two smooth solutions for the problem (1-3). Then their difference e, jsatisfies these equations with the zero right-hand sidesq= 0, g= 0.
We do not take into account a small parameter that means ε = 1 in (9) in this section. We do not use the additional assumptions (7, 12, 13) either.
In the proof we use the functionV:
e=−gradV. (14) Such a function exists, since the vector fieldeis irro-tational by the second equation (1) andg= 0.
By assumptione is not identically zero. Therefor we can find a numberζ >0 and a point such that|e|= 2ζ. In view of continuity of e we can choose a neighbor-hood of this point in wich |e| > ζ. Denote the volume of this neighborhood byx3
1.
Consider the following integral over the domain Ω: ω=
∫
Using the denotations (14), we can rewrite the integral as
ω=
∫
(gradV)T σT gradV dΩ. (15) We can replace the matrixσT in this integral with its symmetrical part, since we calculate the quadratic form
ω=
∫
(gradV)T σ+σ T
2 gradV dΩ.
By positive definiteness of (σ+σT)/2, (10, 11), the in-tegrand is nonnegative. Therefore, the integral over the domain Ω is not less than the integral over the selected neighborhood; hence,
ω ≥ x31 ζ2c1 > 0. (16) Now we turn to our integral written down in the shape (15). Transform identically the integrand:
ω=
∫
(−Vdiv (σgradV) + div (V σgradV))dΩ. The first term equals
Vdiv (σe) =V divj= 0,
since the first equation (1) holds withq= 0. Transform the remaining integral by using of the Gauss - Ostro-gradskii theorem:
ω=
∫
V (σgradV)n dΓ.
Returning to (14) we obtain ω=−
∫
V (σe)n dΓ.
In accordance with the last equation (1) it means ω=
∫
V (−jn)dΓ.
By the boundary condition (3) the integrand equals zero at the horizontal parts of the boundaryz= 0,z= z0. By the first periodicity condition (2) the integrals over the surfacesx= 0 andx=aare equal in magnitude and opposite in sign. Hence their sum equals zero. The same about the sum of integrals over the surfacesy= 0 and y = b. We obtain ω = 0 that contradicts (16). Hence, the assumption thateis different from identical zero is false. Thereby the uniqueness of the classical solution for the problem (1-3) is proved.
4
Variational principle
When the normal component of current density is given at the whole boundary, the equations of elec-troconductivity (1) are reduced to minimization of the quadratic energy functional in [3]. Here we give analo-gous statements and proofs for the problem (1-3).
We do not take into account a small parameter that meansε = 1 in (9) in this section. We do not use the additional assumptions (7, 12, 13) either.
One can obtain a new statement that has a symmet-rical operator by one of the next two heuristic ideas. It
is possible to introduce potentials similar to the usual scalar and vector potentials and to use them not sepa-rately but in pair. It is also possible to find the adjoint operator and multiply the original one by the adjoint operator from the right [3]. It is equivalent to intro-ducing of some potentials. Another way is to use the least square method that is equivalent to multiplication of the original operator by the adjoint operator from the left. In contrast with that method we do not dif-ferentiate the right-hand sides and the solutions for our problem are more smooth functions. Our way corre-sponds to the reducing of Cauchy - Rieman equations with nonzero right-hand sides to a pair of the Poisson equations in 2-D space. These equations are separate for the main boundary value problems and the solution of one of them is identical zero because of zero right-hand side. That is why only one potential is usually used, and the same is traditionally done in more general case, when a pair of potentials could be useful.
Omitting details of the presented heuristic ideas which are similar to those in [3], we consider some quadratic functional and prove that its minimization gives pair of potentials that permits to construct the solution for the original problem, which uniqueness is proved above.
4.1
The energy spaceWe consider pairs f,pof smooth functions satisfying the following conditions
f|xx==0a= 0, f|yy==0b = 0,
p|xx==0a = 0, p|yy==0b = 0, (17) px|z=0,z0 = 0, py|z=0,z
0 = 0, (18)
∫
f dΩ = 0,
∫
pzdΩ = 0. (19)
Consider the following symmetrical bilinear form [
(
u
v
)
,
(
f
p
)
] =
∫ (
gradu curlv
)T(
σ s σT −σ s
−s σT s
) (
gradf curlp
)
dΩ+
+
∫
divvdivp)dΩ. (20) Hereu,vandf,pare pairs of smooth functions which satisfy (17-19), and
s−1= (σ+σT)/2. (21) First, consider the auxiliary integral
∫
(gradf)TcurlpdΩ. (22) Transform the integrand:
∫
( div (fcurlp) − fdiv curlp)dΩ.
The second summand vanishes identically. Using the Gauss - Ostrogradskii formula, we can transform the rest integral, coming to
∫
The integrand equals zero at the boundaries z = 0, z =z0 by (18). By the first and the third periodicity conditions (17) the integrals over the surfaces x = 0 andx=aare equal in magnitude and opposite in sign. Hence their sum equals zero. The same about the sum of integrals over the surfacesy = 0 andy=b. Therefore the integrals (23) and (22) are equal to zero.
Consider the quadratic form to which the bilinear form (20) is reduced when (u,v) = (f,p). The ma-trix in (20) is degenerate since its upper blocks result from multiplying its lower blocks by−σ. After adding the duplicated integral (22), the matrix of the quadratic form under the integrating becomes
K=
(
σ s σT −σ s + I
−s σT + I s
)
,
where I is the identity matrix. Since the integral (22) equals zero, the value of the quadratic form is not changed.
The matrix K is symmetrical. Hence it has 6 real eigenvalues. Its specific shape can be obtain by substi-tution ofσ (8) ands(21). Without changing its eigen-values we can put the matrix in a block diagonal form by interchanging rows and columns. The blocks are
(
σC β β 1/σP
)
,
(
σC −β
−β 1/σP
)
, σ∥, 1/σ∥. (24) The eigenvalues of the first and the second blocks are equal. They can be obtained as the roots of the quadratic equation
λ2−(σC + 1/σP)λ+ 1 = 0.
By Viete’s theoremλ1λ2= 1,λ1+λ2=σC + 1/σP. In view of (10) we obtain
c1/2≤λ1≤λ2≤2/c1. (25) In view of (11) more rigid restrictions are valid for the rest diagonal elements in (24). Hence all eigenvalues of the matrixK satisfy (25).
Since the integral (22) equals zero, the obtained esti-mations of the eigenvalues of the matrix K lead to the following estimations from below and above
[
(
f
p
)
,
(
f
p
)
]≤ 2
c1
∫
((gradf)2+ (curlp)2+ (divp)2)dΩ, (26)
[
(
f
p
)
,
(
f
p
)
]≥ c1
2
∫
((gradf)2+ (curlp)2+ (divp)2)dΩ. (27) The inequality c1 ≥ 1 that is a sequence of (10) or (11) was used.
To continue the estimations (27) from below and (26) from above, we consider the functions f and p sepa-rately.
Since the average value of the functionf equals zero (19), the Poincare’s inequality is valid for it. For the parallelepiped Ω it can be written in the shape [15]:
∫
f2 dΩ ≤ max(z02, a2, b2)
∫
(gradf)2 dΩ. (28)
For the functionpwe consider the identity (curlp)2+ (divp)2= (gradpx)2+ (gradpy)2+
(gradpz)2+ div (pdivp−(pgrad)p),
that can be easily verified in Cartesian coordinates. Let integrate it over Ω. In view of the Gauss - Ostrogradskii theorem the integral of the last term is equal to the flux of the following vector through the boundary of the domain
pdivp−(pgrad )p.
The normal component of the vector equals zero at the boundaries z = 0 and z = z0 since the vector p
has the only nonzero z− component by (18). The x− component equals
px
∂py
∂y +px ∂pz
∂z −py ∂px
∂y −pz ∂px
∂z .
Only tangential derivatives are involved. They satisfy periodicity conditions as a result of (17) as well as the vector p itself. So the values at the surfaces x = 0 and x=aare equal in magnitude and opposite in sign. Hence their sum equals zero. The same about the sum of the integrals over the surfaces y= 0 andy=b.
Thereby we obtain the identity
∫
((curlp)2+ (divp)2)dΩ =
∫ (
(gradpx)2+ (gradpy)2+ (gradpz)2 )
dΩ. (29) for the functionspwhich satisfy (17, 18).
It permits to transform the right-hand side of (27) [
(
f
p
)
,
(
f
p
)
]≥ c1 2
∫
((gradf)2+
(gradpx)2+ (gradpy)2+ (gradpz)2)dΩ. (30)
Since the average value of the functionpzequals zero
(19), it satisfies the Poincare’s inequality with the same constant as one for the function f (28):
∫
p2z dΩ ≤ max(z02, a2, b2)
∫
(gradpz)2dΩ. (31)
The function px equals zero atz= 0 (18). Hence
px(x, y, z) = ∫ z
0 ∂px
∂z dz.
To square the identity and use the Cauchy - Bun-yakovskii inequality
p2x≤z ∫ z
0
(
∂px
∂z
)2
dz≤z0
∫ (
∂px
∂z
)2 dz. By integration over the domain Ω we obtain
∫
p2x dΩ≤z02
∫ (
∂px
∂z
)2
dΩ≤z20
∫
(gradpx)2dΩ.
(32) Since the functionpyalso equals zero atz= 0 by (18)
it satisfies the same inequality
∫
p2y dΩ≤z20
∫
The inequalities (28, 31, 32, 33) permit to continue the estimation (30)
[
(
f
p
)
,
(
f
p
)
]≥ c1
2max(z20, a2, b2)
∫
(f2+|p|2)dΩ. (34) Such an inequality means the quadratic form (20) is positive definite in the spaceL2(Ω). In couple with (30) they mean positive definiteness in the spaceW2(1)(Ω).
Since the quadratic form that corresponds to the linear form (20) is positive definite, we can use the bi-linear form as the inner product in the set of pairsf,p
of smooth functions satisfying (17 - 19). It is referred to as the energy inner product.
By introducing some inner product we have con-structed some Hilbert space. We complete it by adjoin-ing limit elements. The resultant space is referred to as the energy space.
The inequality (26) in view of the identity (29) is the estimation of the quadratic form from above with the sum of norms of the functions f, px, py, pz as the
ele-ments of the spaceW2(1)(Ω). This estimation from above and the estimations from below (30, 34) imply equiva-lence of the energy norm and the norm of the space W2(1)(Ω).
4.2
The energy functionalWe define the energy functional W(f,p) = 1
2 [
(
f
p
)
,
(
f
p
)
] −
∫
(f q+pTg)dΩ. (35) The linear functionals can be estimated by the Cauchy - Bunyakovskii inequality
|
∫
f q dΩ| ≤
(∫
f2 dΩ
∫
q2 dΩ
)1/2
|
∫
pTgdΩ| ≤
(∫
pTpdΩ
∫
gTgdΩ
)1/2 . By inequalities (28, 30 – 33) the right-hand sides can be estimated from above by the energy norm with factors q0, g0 (6) which are independent of f,p. Hence linear functionals in (35) are bounded ones.
By the Riesz’s theorem each bounded linear functional is representable by means of the inner product into some unique element of the Hilbert space. Therefore we can write the energy functional (35) as
W(f,p) = 1 2 [
(
f
p
)
,
(
f
p
)
] − [
(
f0
p0
)
,
(
f
p
)
], wheref0,p0 is some element of the energy space.
We can transform the expression by distinguishing the square of the difference:
W(f,p) = 1 2[
(
f−f0
p−p0
)
,
(
f−f0
p−p0
)
]
− 1
2[
(
f0
p0
)
,
(
f0
p0
)
]. (36)
The second term is independent off,pand the first one is positive definite. Therefore the minimum of
W(f,p) is attained at f = f0,p = p0. Since the ele-ment f0,p0 belongs to the energy space and is defined uniquely, we have proved that the energy functional has a unique minimum in the energy space.
4.3
A weak solutionWe use the functionsf,pobtained as the result of the energy functional minimization to define the following vector functions
e=−sσTgradf +scurlp, j=σe. (37) Here we usee,jonly as a contraction of these expres-sions, but later we will demonstrate that it is the sought solution for the original problem (1-3).
The minimum condition for the energy functional can be written down, on use made of notations (37), as the following identity that must be hold for all smooth func-tionsu,vsatisfying conditions (17-19):
∫
(−(gradu)Tj+ (curlv)Te
+divvdivp−uq−vTg)dΩ = 0. (38) Takeu= 0 and definev by formula
v=−gradV,
where the functionV is obtained as the solution for the following boundary value problem for the Poisson equa-tion in the parallelepiped Ω:
−∆V =ξ V|z=0,z
0 = 0, V|
x=a
x=0 = 0, V|
y=b y=0= 0, ∂V
∂x
x=a
x=0
= 0, ∂V ∂y
y=b
y=0
= 0. (39)
The functionξ is chosen below.
It can be easily verified that the solution for the last problem exists and is unique. By the first boundary condition (39)
∫
∂V ∂z dΩ =
∫ ∫
V|z0
0 dxdy= 0.
Therefore the obtained pare of the functionsu,vsatisfy all the conditions (17 - 19) independently of the function ξ, which must be only enough smooth.
For such a pairu,vthe identity (38) takes the form
∫
ξdivpdΩ +
∫
(gradV)TgdΩ = 0. (40) Transform the second integral by using of the Gauss - Ostrogradskii theorem
∫
Vgn dΓ− ∫
VdivgdΩ.
the boundariesx= 0 andx=ais the integral over y, z of the difference
V(a, y, z)gx(a, y, z)−V(0, y, z)gx(0, y, z).
It equals zero because ofV and gxare periodic
func-tions. The sum of the integrals over the surfacesy = 0 andy=bequals zero by the same reason. Thereby from (40) we obtain the equality
∫
ξdivpdΩ = 0. (41) By using the arbitrariness of the function ξ, we can routinely deduce that divp= 0. Assuming the contrary, we take a point Ω at which divp̸= 0 and take a neigh-borhood of this point in which divppreserves sign. Take a functionξ that differs from identical zero, is nonneg-ative in this neighborhood and equals zero outside the neighborhood. Then the integral (41) is nonzero, that contradicts the identity.
Suppose additionally that all functions are smooth. Then the weak equality divp = 0, that is a result of the minimization of the energy functional, holds point-by-point:
divp = 0. (42)
After integration by parts using (42) the identity (38) takes shape
∫
u(divj−q)dΩ +
∫
vT(curle−g)dΩ−
−
∫ ∫
(ujz)|zz0=0dxdy− ∫ ∫
(ujx)|ax=0dydz
−
∫ ∫
(ujy)| b
y=0dxdz+ +
∫ ∫
[v×e]x| a x=0dydz +
∫ ∫
[v×e]y| b
y=0dydz= 0. (43) The integral
∫ ∫
[v×e]z| z0
z=0dxdy
is omitted because its integrand is identical zero by (18). Whenv= 0 anduequals zero at the boundaries the only first integral remains in the identity. By using arbitrari-ness of the functionuwe can routinely deduce
divj−q= 0. By the same way we obtain
curle=g.
Then only integrals over boundaries remain in the tity (43). Each of them produces the independent iden-tity.
Arbitrariness of uat the boundaries z = 0 and z = z0 gives jz = 0 there. Other identities give periodicity
conditions of the form (2).
In this way the pare of the functions f,p provid-ing a minimum to the energy functional enable us to construct by formulae (37) the solution to the original boundary value problem (1-3), if we additionally assume their smoothness. In general case we have got the weak
solution from the point of view of the identity (38) for arbitrary elementu,v of the energy space.
The conditions of a minimum to the energy functional for the functionsf,pthemselves can be obtained by sub-stitution (37) to the identity (38) that define the weak solutionf,p. Under assumption of smoothness substitu-tion of (37) to the equasubstitu-tions (1-3) produce the boundary value problem
div (−σsσTgradf + σscurlp) =q curl (−sσT gradf + scurlp) =g divp= 0 ∂f /∂z|z=0,z
0 = 0, ∂f /∂x|
x=a x=0 = 0, ∂pz/∂x|
x=a
x=0= 0, ∂py/∂x| x=a x=0 = 0, ∂f /∂y|yy==0b = 0, ∂pz/∂y|
y=b y=0= 0, ∂px/∂y|
y=b
y=0= 0, (44)
that is completed with the boundary conditions (17 -19).
The boundary conditions (2) for the functionsf,pcan be simplified taking into account the periodicity of the tangential derivatives that is a result of (17)
(σC∂f ∂x +β
∂pz
∂x)
x=a
x=0
= 0, (β∂f ∂x+
1 σP
∂pz
∂x)
x=a
x=0 = 0,
(1 σ∥
∂py
∂x)
x=a
x=0
= 0,(σC∂f ∂y +β
∂pz
∂y )
y=b
y=0 = 0,
(β∂f ∂y +
1 σP
∂pz
∂y )
y=b
y=0
= 0, (1 σ∥
∂px
∂y )
y=b
y=0 = 0. Since unit determinant of the matrix
(
σC β β 1/σP
)
,
the first pairs of these conditions can be resolved. It is already done in (44).
The boundary conditions (17 - 19) are referred to as the main ones because the functions, at which the min-imum is searching, must satisfy them. In contrast the boundary conditions in (44) are natural ones since they are satisfied as a result of minimization.
The converse assertion is also easily provable: the so-lution to the boundary value problem (44, 17-19) pro-vides minimum to the energy functional in the energy space.
In this way the existence and the uniqueness of the weak solution for the original boundary value problem (1-3) is proved and the minimum principle for the energy functional is justified.
Since the functionsf,pbelong to the energy space the functionf and the components of the vector functionp
belong to the spaceW2(1)(Ω). By formulae (37) it means that the vector functions e, j have finite norms in the spaceL2(Ω).
4.4
Orthogonal decomposition of the energy spaceConsider two sets of pairs of smooth functions. The first comprises the pairsf,p1 satisfying (17-19) and
The second comprises the pairs 0, V, where the func-tionsV are periodic inx, ydirections and equal zero at the boundariesz= 0 and z=z0
V|z=0,z
0 = 0, V|
a
x=0, V|
b
y=0. (45) Each functionsf,psatisfying (17 - 19) can be uniquely represented as a sum of elements of these sets. The function V is obtained as the solution for the Poisson equation
∆ V = divp,
with the boundary conditions (45), and the functionp1 is the differencep−gradV. And conversely the sum of arbitrary elements of these sets satisfies (17 - 19).
Substitution of the elementsf,p1and 0,gradV to the bilinear form (20) gives zero integrand because divp1= 0 and grad 0 = 0, curl gradV = 0. Hence the inner product equals zero.
We complete these two sets by adjoining limit ele-ments. The resultant spaces are two orthogonal sub-spaces of the energy space.
The functionsp1in the elementsf,p1of the first sub-space have zero divergence in weak sense. Hence
∫
( divp1 )2dΩ = 0.
The second subspace that contains only elements of the shape 0,gradV is in some sense excessive. The vec-tor functions e,j which are of interest in the original problem do not change if elements of this subspace are added. So it is possible to consider the first subspace as the whole energy space slightly changing the statement of the problem. This simplifies the proofs but creates dif-ficulties in constructing numerical algorithms. For this reason we avoid doing so.
The energy functional can be rewritten as the sum of two functionals corresponding to the two orthogonal subspaces. To the second subspace there corresponds only
1 2
∫
(divp2)2dΩ,
since the linear part of the energy functional on the func-tions (f2,p2) = (0,gradV) is equal to the second inte-gral in (40) that equals zero as it was shown.
Thus we can independently minimize the energy func-tional in the two orthogonal subspaces constructed above.
4.5
ThermodynamicsIn view of notations (37) with s chosen in (21), we can write down the quadratic form that corresponds to the energy inner product (20) in terms of the original boundary value problem (1-3):
[
(
f
p
)
,
(
f
p
)
] =∫ ( eTj + (divp)2) dΩ. (46)
The producteTjis the density of the Joule dissipation, i.e., the thermal energy production that accompanies electric current.
As it was shown in the previous section, we can min-imize the energy functional in the subspace which ele-ments have zero divergence
∫
(divp)2 dΩ = 0.
Then the quadratic form (46) equals the total Joule dissipation in Ω, which justifies usage of the term ”en-ergy”.
Under the choice
s−1= (θσ+σTθ)/2, (47) instead of (21), we obtain another symmetrical state-ment for the original problem. Here θ ia an arbitrary symmetrical tensor that leaves the tensor s uniformly positive definite. In this event,
[
(
f
p
)
,
(
f
p
)
] =
∫ ∫
eTθjdΩ.
In particular, on taking θ to be the product of the identity tensor by the scalar function 1 / T, where T is the absolute temperature, we give this integral the meaning of the total entropy production in the domain Ω.
Thus, the energy norm makes sense from the stand-point of nonequilibrium thermodynamics.
Consider the energy functional in the shape (36). Since the second term (36) is independent of f,p, the minimization of the energy functional implies the mini-mization of the first term, that is the square of the en-ergy norm of the difference between an arbitrary element f,pand the exact solution. By (46) it is the minimiza-tion of the funcminimiza-tional
∫
δeT δjdΩ, (48) or the functional
∫
δeTθδjdΩ,
ifsis chosen by more general formula (47). Hereδstands for the difference from the exact solution. The quantity δemay be called as the electric field fluctuation.
In fact, it is impossible to calculate the integral (48), since the exact solution is unknown, but its minimiza-tion is equivalent to the minimizaminimiza-tion of the energy func-tional (35), in which all integrals can be calculated for arbitraryf,p.
Thus, from the viewpoint of thermodynamics the square of the energy norm is the entropy production, and the exact solution can be obtained by minimization of the energy of fluctuations.
5
Small parameter expansion
(1) and (8). Then the x− and y− components of the second equation (1) with zero right-hand sides (5) cause ex and ey independence of z. Since ez = 0 the whole
vectoreis independent ofz whenε= 0.
We fulfil the small parameter expansion of the so-lution under additional restrictions, which are due to the method of proof but not to the essence of the prob-lem. We are to suppose conductivity independece of the heightz (12). By the same reason from the very begin-ning we consider the flat conducting layer and suppose magnetic field to be vertical, when conductivity tensor has shape (8).
Represent the solution in the form
f =F(x, y) +εr, px=U(x, y) +εu,
py=V(x, y) +εv, pz=P(x, y) +εw. (49)
Denote the set (u, v, w) as the vectoru.
It will be shown that it is enough to put the conditions
∫
r dz= 0,
∫
w dz= 0 (50) to make this representation unique.
Use the main boundary conditions (17 - 19) separately for the two-dimensional functions and for the three di-mensional correction in the representation (49). Hence all functions are periodic over x, y, and
U(x, y) =V(x, y) = 0, (51)
F|xx==0a= 0, F|yy==0b = 0,
∫ ∫
F(x, y)dxdy= 0, P|xx==0a= 0, P|yy==0b= 0,
∫ ∫
P(x, y)dxdy = 0,(52)
u|xx==0a= 0, u|yy==0b = 0, u|z=0,z
0 = 0, v|z=0,z0 = 0.
(53) The conditions
∫
r dΩ = 0,
∫
w dΩ = 0, (54) are also satisfied as a result of (50).
Substitute the representation (49) to the energy func-tional (35). The integral that contains the products of the derivatives of two-dimensional functions F, P and those of three-dimensional corrections r,u is the inte-gral overx, yand overz of the following sum
(σC∂F ∂x −
∂P ∂y −β
∂P ∂x)
∂r ∂x+ (σC
∂F ∂y +
∂P ∂x −β
∂P ∂y)
∂r ∂y
+(−∂F ∂x −β
∂F ∂y +
1 σP
∂P ∂y)(
∂w ∂y −
∂v ∂z) +(−∂F
∂y +β ∂F
∂x − 1 σP
∂P ∂x)(
∂u ∂z −
∂w ∂x).
Every summand is a product of the functions indepen-dent ofzbecause of (12) and two-dimensional definition ofF, P (49) and the functions with zero integrals overz by (50) or (53). Hence the integral overzof every sum-mand equals zero and the whole integral can be omitted. All integrals of the shape (22) can be also omitted because they are equal zero by the same reasons as (22).
Therefore after some transformation using indepen-denceσofz (12) the energy functional takes the shape
W(f,p) =W0(F, P) +εW1(r)−εWq(r)+
+ε2W2(r) +ε2W3(r,u) +ε3W4(u, v), (55) where
W0(F, P) = z0 2
∫ ∫
{σC((∂F ∂x)
2+ (∂F ∂y)
2)
+ 1 σP((
∂P ∂x)
2+ (∂P ∂y)
2)− −2β(∂F ∂x
∂P ∂x +
∂F ∂y
∂P ∂y)
−2Fq(x, y)¯ −2P gz(x, y)}dxdy
W1(r) =1 2
∫
˜ σ∥(∂r
∂z) 2
dΩ
Wq(r) = ∫
r(q−q(x, y))¯ dΩ
W2(r) = 1 2
∫
σC((∂r ∂x)
2+ (∂r ∂y)
2)dΩ
W3(r,u) =1 2
∫
{ 1
σP( ∂w
∂y − ∂v ∂z)
2+ 1 σP(
∂u ∂z −
∂w ∂x)
2+
+2β∂r ∂x(
∂u ∂z −
∂w ∂x)−2β
∂r ∂y(
∂w ∂y −
∂v
∂z) + (divu) 2}dΩ
W4(u, v) = 1 2
∫
1 ˜ σ∥(
∂v ∂x−
∂u ∂y)
2 dΩ.
Here
¯
q(x, y) = 1 z0
∫
q dz. (56) By (4) and (5) the functions have zero average values
∫ ∫
¯
q(x, y)dxdy= 0,
∫ ∫
gz(x, y)dxdy= 0.
The integral of r¯q equal zero by (50) is put into the expression ofWq(r) for convenience.
As it can be seen by (55), the functional is factor-ized, and the two-dimensional functionsF, P can be ob-tained separately. If they are smooth the minimization ofW0(F, P) is equivalent to the solution of the equations
Div (−σsσT GradF + σscurlP) = ¯q(x, y)
Curlz (−sσT GradF + sCurlP) =gz(x, y) (57)
with conditions (52). Here the operators are two-dimensional ones,
CurlP = (∂P/∂y,−∂P/∂x). By the formulae
E=−sσTGradF + sCurlP, J=σE (58) we obtain the periodical solution for the two-dimensional problem
DivJ= ¯q(x, y), Curlz E=gz(x, y), (59)
that simulate two-dimensional electric fields and cur-rents.
proof for the periodicity conditions because the differ-ences in the proofs are obvious. State the results only. The operator (57, 52) is symmetrical and positive defi-nite. There exists the unique pair of the functionsF, P that satisfies (52) and minimizes the value of the func-tionalW0(F, P). It is the weak solution for the problem (57, 52). By the formulae (58) it gives the periodical solution for the equations (59). The squared functions F, P are summable as well as their first derivatives. The squared vector functionsE, Jobtained by the formulae (58) are also summable.
The three-dimensional correction r,uto the solution of the two-dimensional problem ought minimize the rest terms of the sum (55).
It will be shown that the minimization ofW3(r,u) over
u results also W4(u, v) = 0, that means minimization of this nonnegative term. Since the function u is not involved to other terms of the sum (55), as a result of the minimization of W3(r,u) we obtain the expression ofuin terms of the functionrthat is fixed during this minimization overu.
6
Minimization
of
the
energy
functional over u
The functionris fixed in this section. First prove that divu= 0 as a result of minimization ofW3(r,u) overu. Take into account the set of the functions
˜
u=u−tgradV,
where the functionuminimizes the value ofW3(r,u),t is an arbitrary number,V is a solution for the problem (39) with an arbitrary functionξ. Then
W3(r,u˜) =W3(r,u)−t
∫
div gradV divudΩ+
+t 2 2
∫
(div gradV)2dΩ.
Since the functionuminimizes the value ofW3(r,u), the factor aftertequals zero. It means the identity
∫
div gradV divudΩ = 0
for an arbitraryV from the set under analysis. In view of (39) we obtain the identity with an arbitrary function ξ
∫
ξdivudΩ = 0.
Hence
divu= 0 (60)
and it is possible to reject the last term in the functional W3(r,u).
Now the minimum conditions forW3(r,u) overucan
be written as (60) and
− ∂
∂z( 1 σP
∂u ∂z) +
∂ ∂z(
1 σP
∂w ∂x) =
∂ ∂z(β ∂r ∂x) −∂ ∂z( 1 σP ∂v ∂z) +
∂ ∂z(
1 σP
∂w ∂y) =
∂ ∂z(β ∂r ∂y) ∂ ∂x( 1 σP ∂u ∂z) +
∂ ∂y(
1 σP
∂v ∂z)−
∂ ∂x( 1 σP ∂w ∂x) −∂ ∂y( 1 σP ∂w ∂y) =−
∂ ∂x(β
∂r ∂x)−
∂ ∂y(β
∂r
∂y) +c(x, y). (61) An arbitrary functionc(x, y) appears in the last equa-tion (61) because the restricequa-tion (50) for the funcequa-tionw. Show thatc(x, y) = 0. Integrate the third equation (61) over z between the limits 0 and z0 taking into account σP andβ independence ofzand (60)
∂ ∂x(
1 σP u|
z0
0 ) + ∂ ∂y(
1 σP v|
z0
0 )− ∂ ∂x( 1 σP ∂ ∂x ∫ w dz) − ∂ ∂y( 1 σP ∂ ∂y ∫ w dz)+ + ∂ ∂x(β ∂ ∂x ∫
rdz) + ∂ ∂y(β
∂ ∂y
∫
r dz) =z0c(x, y). All terms in the left-hand side equal zero by (53, 50). Hence
c(x, y) = 0.
The first and the second equations (61) are the sums of the derivatives overz. Therefore they can be integrated
− 1 σP ∂u ∂z + 1 σP ∂w ∂x =β
∂r
∂x+ξ(x, y)
− 1 σP ∂v ∂z + 1 σP ∂w ∂y =β
∂r
∂y+η(x, y). (62) Hereξ, η are arbitrary functions. Prove that they are zero. Integrate these equalities overzbetween the limits 0 andz0taking into accountσP andβ independence of z
− 1
σP u|
z0 0 + 1 σP ∂ ∂x ∫
w dz−β ∂ ∂x
∫
r dz=z0ξ(x, y)
− 1
σP v|
z0 0 + 1 σP ∂ ∂y ∫
w dz−β ∂ ∂y
∫
r dz=z0η(x, y). All terms in the left-hand side equal zero by (53, 50). Hence
ξ(x, y) =η(x, y) = 0. (63) Now deduce the equation for the functionwfrom (62, 63). Multiply both equations (62) by the factorσP. Dif-ferentiate the results with respect toxandyrespectively and sum them. Add to and subtract from the left-hand side∂w/∂z
− ∂2u
∂x∂z − ∂2v ∂y∂z −
∂2w ∂z∂z +
∂2w ∂x2 +
∂2w ∂y2 +
∂2w ∂z2 = ∂
∂x(σH ∂r ∂x) +
∂ ∂y(σH
∂r ∂y).
The sum of the first three terms equals to−∂divu/∂z that equals zero by (60). Therefore the equation for w takes shape
∂2w ∂x2 +
∂2w ∂y2 +
∂2w ∂z2 =
∂ ∂x(σH
∂r ∂x) +
∂ ∂y(σH
∂r
It defines the unique functionw taking into account the last condition (54) and
∂w/∂z|z=0,z
0 = 0, (65)
that is a result of (60)and (53).
However we need the function w that satisfy more strict condition (50). Prove that the obtained function satisfies it. Integrate the equation (64) overz.
( ∂ 2 ∂x2+
∂2 ∂y2)
∫
wdz+ ∂w ∂z
z0
0 =
∂ ∂x(σH
∂ ∂x
∫
rdz) + ∂ ∂y(σH
∂ ∂y
∫
rdz).
The right-hand side equals zero by the first condition (50). The second term in the left-hand side equals zero by (65). Hence it is two-dimensional Laplace equation for the function ∫w dz. Only a constant may be its periodical solution and it equals zero by (54). Hence the second condition (50) is satisfied.
The functionris fixed during the construction of the functions u, v, w and w is already constructed. It only remains to integrate the equalities (62) over z taking into account (63) and the boundary conditions (53) at z= 0:
u(x, y, z) = ∂ ∂x
∫ z
0
(w−σHr)dz′,
v(x, y, z) = ∂ ∂y
∫ z
0
(w−σHr)dz′. (66) The conditions (53) at z =z0 are satisfied, because the integrals of the functions w, r between the limits 0 andz0 equal zero andσH is independent ofz.
By (62, 63) we obtain the expressions ofx−, y− com-ponents of the vector curlu, and by (66) itsz− compo-nent equals zero:
curlu=σH
−∂r/∂y∂r/∂x 0
(67)
This formula provides zero value forW4(u, v). The obtained expressions (67, 60) permit to express the minimal overuvalue ofW3(r,u) in terms ofr:
−1
2
∫ σ2 H σP((
∂r ∂x)
2+ (∂r ∂y)
2)dΩ.
In sum withε2W2(r) it equals
ε2Wxy(r) =
ε2 2
∫
σP{(∂r ∂x)
2+ (∂r ∂y)
2} dΩ.
After the performed minimization (55) overF, P and
uit remains to find the functionrthat satisfies (50) and minimizes the functional
εWr(r) =εW1(r)−εWq(r) +ε2Wxy(r). (68)
This way we obtain the minimum of the energy func-tionalW(f,p) (55) over all arguments.
7
Minimization
of
the
energy
functional over
r
The minimum conditions forWr(r) (68) over periodic
functionsrsatisfying the first condition (50) under addi-tional suggestion of smoothness is the following bound-ary value problem:
−∂
∂z(˜σ∥ ∂r ∂z)−ε{
∂ ∂x(σP
∂r ∂x) +
∂ ∂y(σP
∂r ∂y)}= q−q(x, y)¯ ∂r/∂z|z=0,z
0 = 0
∫
r dΩ = 0. (69) The last condition is added for uniqueness of the so-lution for the problem. Demonstrate that it can be used instead of the first condition (50). Integrate the first equation over z between the limits 0 and z0. Taking into account that σP is independent of z, ∂r/∂z= 0 at z = 0 andz = z0 and the definition of ¯q(x, y) (56) we obtain
−Div (σPGrad
∫
r dz) = 0, (70) where the common factor ε is omitted. The operators are two-dimensional ones. This is a two-dimensional el-liptical equation for the function ∫r dz. By (10) the coefficient σP is uniformly bounded and strictly posi-tive. The average value of the unknown function equals zero by the last condition (69). Therefore the unique periodical solution is zero
∫
r dz= 0. (71) Point out that the equation (70) for the function
∫
r dz is a particular case of the equation (57) with σH = 0. Then the system of the equations (57) is split into independent equations for F and P, the first of which differ from (70) only by nonzero right-hand sides. The obtained equality (71) means that the solution for the problem (69) also satisfies the first condition (50).
The solution for the problem (69) exists and is unique, since for an arbitrary positive ε this is the Neumann boundary value problem for an elliptical equation with uniformly bounded and uniformly positive definite sym-metrical coefficient tensor.
By the boundary condition (69) the functionrcan be formally expanded in Fourier series
r(x, y, z) = ∞
∑
n=1
rn(x, y) cos (nπz/z0). (72)
The Fourier series starts withn= 1 sincer0(x, y) = 0 by (71).
By this expansion the problem (69) is reduced to the set of two-dimensional problems with a small parameter ε:
−εDiv (σPGradrn(x, y)) +
n2π 2σ˜
∥
z2 0
rn(x, y) =qn(x, y). (73)
The shape of the right-hand side is simplified in com-paring with the first equation (69) because the only nonzero term of the Fourier series of the function ¯q(x, y) is ¯q0(x, y) = ¯q(x, y), and by (56)q0(x, y)−q0(x, y) = 0.¯ It corresponds tor0(x, y) = 0.
Integrate the equation (73) overx, y. Use the Gauss -Ostrogradskii formula for the first term. Taking into ac-count the periodicity of the involved functions it equals zero. There remains the equalities of the average values for each Fourier coefficient
∫ ∫
˜
σ∥(x, y)rn(x, y)dxdy =
z2 0 n2π2
∫ ∫
qn(x, y)dxdy.
The last condition (69) and the first condition (50) are satisfied in spite of nonzero average values of rn(x, y),
because of that the z− dependencies are of the shape cos(nπz/z0) andn≥1.
We construct the solution for each singular pertubed equation (73), following the Lusternik - Vishik method [19]. We expand the unknown function in a power series of the small parameterε. Since the boundary conditions for the equation (73) are the only periodicity overx, y, the assymptotic expansion does not contain boundary layers. Represent
rn(x, y) =ρn(x, y) +εδn(x, y), (74)
whereδn(x, y) is the remainder term that ought be
esti-mated.
The equations for the functions ρn(x, y), δn(x, y) we
derive from (73) by equating to zero the sum of the terms which do not containεand the sum of the rest terms:
n2π 2˜σ
∥ z2
0
ρn(x, y) =qn(x, y), (75)
n2π 2σ˜
∥ z2
0
δn(x, y)−εDiv (σPGradδn(x, y)) = Div (σPGradρn(x, y)). (76)
By (10) the coefficient σP is strictly positive and bounded. Hence at the point (x1, y1) where the func-tionδn(x, y) reaches its positive maximum
Div (σPGradδn(x, y))≤0.
By substitution this inequality to the equation (76) we obtain the estimation
δn(x1, y1)≤
z2 0 n2π2σ˜
∥(x1, y1)
Div (σPGradρn(x, y))|x1,y1.
Using the analogous estimation for the negative mini-mum of the functionδn(x, y) and the first condition (13)
we obtain
|δn(x, y)| ≤
z2 0c2
n2π2maxx,y |Div (σPGradρn(x, y))|. (77) The expansion in Fourier series corresponds to calcu-lation of the coefficients as the integrals overz
qn(x, y) =
2 z0
∫
q(x, y, z) cos (nπz/z0)dz, (78)
and the derivatives ofqn(x, y) overx, ycan be obtained
by differentiating ofq(x, y, z) under integrals. Hence the boundedness of the Fourier coefficients and their deriva-tives overx, yresult from (78) and (7)
∂kqn(x, y)
∂xi ∂yj ≤2c3
xk
0
, k= 0,1,2, i+j=k. (79) The function 1/˜σ∥ are presumed bounded by (13) as well as their first and second derivatives over x, y, and the estimations (79) are proved. Therefore the functions ρn(x, y) has the same property by virtue of (75)
∂kρn(x, y)
∂xi∂yj
≤ 2k+1z02c2c3 n2π2xk
0
, k= 0,1,2, i+j=k. (80) The obtained estimations of the Fourier coefficients give the estimations of the functions themselves
∂∂xkρ(x, y)i ∂yj
≤3.3z022kc2c3 π2xk
0
, k= 0,1,2, i+j=k. (81) The numerical factor is obtained taking into account that the sum of the numbers 1/n2 from n = 1 till ∞ equals ζ(2) and it can be seen by a table of Riemann’s ζ−function thatζ(2)<1.65.
The boundedness of the first and the second deriva-tives over x, y of the functions ρn(x, y) (80) and the
estimation (77) cause that the functions δn(x, y) are
bounded and decrease as 1/n4whennincreases:
|δn(x, y)| ≤
24z40c0c22c3 n4π4x2
0
. (82)
Therefore the Fourier expansion of the function δ(x, y) absolutely converges and
|δ(x, y)| ≤ 26z 4 0c0c22c3 π4x2
0
. (83)
The numerical factor is obtained taking into account that the sum of the numbers 1/n4 from n = 1 till ∞ equals ζ(4) andζ(4)<1.083.
The derivative of the functionδ(x, y, z) overzcan be calculated by differentiating of its Fourier series of the shape (72)
∂δ(x, y, z) ∂z =−
π z0
∞
∑
n=1
nδn(x, y) sin (nπz/z0).
By (82) the coefficients δn(x, y) decrease as 1/n4.
Therefore the Fourier expansion of its derivative over z absolutely converges and the derivative is bounded
∂δ∂z≤ 30z03c0c22c3 π3x2
0
. (84)
The numerical factor is obtained taking into account that the sum of the numbers 1/n3 from n = 1 till ∞ equals ζ(3) < 1.25. By the same reasons the second derivative overzis also bounded, but we do not use this property.
Since the Fourier expansions converge the summation of the Fourier coefficients (74) corresponds to the sum-mation of the functions themselves
Substitute the representation (85) of the function r(x, y, z) to the problem (69) with (71) instead of the last condition since they are equivalent to obtain the problem for the functionρ(x, y, z)
−∂2ρ
∂z2 = 1 ˜
σ∥(q−q(x, y))¯ ∂ρ/∂z|z=0,z
0= 0
∫
ρ dz= 0, (86) and for the rest termδ(x, y, z):
− ∂
∂z(˜σ∥ ∂δ ∂z)−ε{
∂ ∂x(σP
∂δ ∂x) +
∂ ∂y(σP
∂δ ∂y)}= ∂
∂x(σP ∂ρ ∂x) +
∂ ∂y(σP
∂ρ ∂y) ∂δ/∂z|z=0,z
0= 0
∫
δ dz= 0. (87) The problem (86) is the set of independent bound-ary value problems for the ordinbound-ary differential equation over z for each fixed pairx, y. The solution for such a problem exists, is unique and can be calculated by inte-gration:
ρ(x, y, z) = 1 ˜ σ∥
∫ z
0 (
∫ z′
0
(q(x, y, z′′)−q(x, y))¯ dz′′)dz′−ρ0(x, y), (88) where ρ0(x, y) is obtained from the last condition (86) as the average value overz of the first summand.
By (7, 13) we can independently of (81) derive from this formula the boundedness of the first and the second derivatives of the functionρ(x, y, z). In particular,
|ρ| ≤2z20c2c3,
∂ρ∂z≤2z0c2c3, (89)
Taking into account the representation (85) and proved boundedness of the functions ρ, δ and their derivatives overz(83, 84, 89) we obtain the boundedness ofrand∂r/∂z:
|r| ≤c5z0, ∂r ∂z
≤c5, (90)
wherec5= 2z0c2c3(1 +ε15z2
0c0c2/(π3x20).
8
The energy norms of the errors
Estimate the derivatives over x, y of the functions r and δin the energy norm. The derivatives of the func-tionρsatisfy the estimation (81).
Multiply the equation (69) byr and integrate it over the domain Ω. Transform the integral in the left-hand side by using of the Gauss - Ostrogradskii theorem. In view of the periodicity conditions over x, y and the boundary conditions (69) at z = 0, z0 the sum of the integrals over all boundaries equals zero. Therefore
∫
˜ σ∥(∂r
∂z)
2dΩ +ε
∫
σP((∂r ∂x)
2+ (∂r ∂y)
2)dΩ =
∫
r(q−q(x, y))¯ dΩ.
Use the Cauchy - Bunyakovskii inequality for the right-hand side and omit the nonnegative summand in the left-hand side
ε
∫
σP((∂r ∂x)
2+ (∂r ∂y)
2)}dΩ≤
√∫
r2dΩ
√∫
(q−q(x, y))¯ 2 dΩ.
By the definition of ¯q(x, y) (56) the last integral is not more than the integral of q2.
Using the estimationr(90), boundedness of the func-tionsq(7) and 1/σP (10) we obtain
∫
((∂r ∂x)
2+ (∂r ∂y)
2)dΩ≤c6
ε (91)
where c6=z20c3c5ab/c1.
To estimate the derivative of the functionδ overx, y we multiply the equation (87) byδand integrate it over the domain Ω. Transform the integrals in the left-hand side and in the right-hand side by using of the Gauss - Ostrogradskii theorem. We omit the integrals over the boundaries since their sum equals zero because of the periodicity conditions over x, y and the boundary conditions for δatz= 0, z0 (87).
∫
˜ σ∥(∂δ
∂z)
2 dΩ +ε
∫
σP((∂δ ∂x)
2+ (∂δ ∂y)
2)dΩ =
−
∫
σP∂δ ∂x
∂ρ ∂x dΩ−
∫
σP∂δ ∂y
∂ρ ∂y dΩ.
Omit the first integral in the left-hand side since it is nonnegative and use the Cauchy - Bunyakovskii inequal-ity for the integrals in the right-hand side
ε2 2
(∫
σP((∂δ ∂x)
2+ (∂δ ∂y)
2)dΩ
)2
≤
≤
∫
σP(∂δ ∂x)
2dΩ
∫
σP(∂ρ ∂x)
2dΩ+
∫
σP(∂δ ∂y)
2dΩ
∫
σP(∂ρ ∂y)
2dΩ≤
∫
σP((∂δ ∂x)
2+ (∂δ ∂y)
2)dΩ
∫
σP((∂ρ ∂x)
2+ (∂ρ ∂y)
2)dΩ.
Cancel the common nonzero factor
∫
σP((∂δ ∂x)
2+(∂δ ∂y)
2)dΩ≤ 2 ε2
∫
σP((∂ρ ∂x)
2+(∂ρ ∂y)
2)dΩ.
Using (81) and (10) we obtain the result estimation
∫
((∂δ ∂x)
2+ (∂δ ∂y)
2)dΩ≤ c7
