Initial-boundary value problems for the equations of the global atmospheric electric circuit

J. Math. Anal. Appl. 450 (2017) 112–136

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Initial-boundary value problems for the equations of the global atmospheric electric circuit Alexey V. Kalinin a,b , Nikolay N. Slyunyaev a,∗ a

Institute of Applied Physics, Russian Academy of Sciences, 46 Ulyanova St., 603950 Nizhny Novgorod, Russia b Lobachevsky State University of Nizhny Novgorod—National Research University, 23 Gagarina Ave., 603950 Nizhny Novgorod, Russia

a r t i c l e

i n f o

Article history: Received 27 May 2016 Available online 16 January 2017 Submitted by B. Kaltenbacher Keywords: Maxwell’s equations Evolution problem Weak solution Well-posed problem Global electric circuit

a b s t r a c t Time-dependent and steady-state problems for the equations of the global atmospheric electric circuit are studied in the quasi-stationary approximation. Maxwell’s equations in an anisotropic medium are written in terms of the electric potential; several types of boundary conditions (motivated by applications) are considered, including the analogues of the classical Dirichlet, Neumann and mixed conditions, as well as a non-standard condition relating the potential and normal current density at symmetric points on one of the boundary components. In each case the variational formulation of the problem is derived and is shown to be well-posed. Also, certain stabilisation theorems are proved for the solution of timedependent problem in case the main parameters do not depend on time. © 2017 Elsevier Inc. All rights reserved.

1. Introduction The concept of the global electric circuit (GEC) is fundamentally important for the theoretical understanding of atmospheric electricity (see, e.g., [25,31,32]). The term ‘global electric circuit’ refers to the electric current distribution in the Earth’s atmosphere; this distribution includes, for instance, lightning currents, precipitation currents and corona discharge currents, but its most important constituent is the so-called quasi-stationary current, which flows continuously and, according to the hypothesis originally proposed by Wilson [33,34], is maintained by permanent charge separation in thunderstorms and other electrified clouds. Roughly speaking, this current flows upwards in thunderstorm regions and flows downwards in fair-weather regions, the circuit being completed by the highly conductive Earth’s surface and lower ionosphere. The GEC modelling has been given much attention over the past few decades [4,13–15,19]. The majority of the existing GEC models are aimed at finding the distributions of the quasi-stationary current density * Corresponding author. E-mail addresses: [email protected] (A.V. Kalinin), [email protected] (N.N. Slyunyaev). http://dx.doi.org/10.1016/j.jmaa.2017.01.025 0022-247X/© 2017 Elsevier Inc. All rights reserved.

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and the electric field in the atmosphere, given the GEC generators, which correspond to electrified clouds. Usually these generators are regarded as a certain distribution of the so-called source current density, which enters into Ohm’s law along with the conduction current density. In order to study the GEC from a mathematical perspective and to further develop the existing GEC models, one must deal with Maxwell’s equations. The mathematical foundations of the electromagnetic theory have been established in the last few decades; the well-posedness of various steady-state and timedependent problems for Maxwell’s equations has been studied in detail, allowing for inhomogeneous media, inhomogeneous boundary conditions and domains with complicated topology (e.g., [1,2,5,7,11,24,26]). However, particular applications may lead to new problems, which, owing to certain peculiarities, cannot be reduced to the problems described in the literature and which have yet to be considered in detail. In this article we consider the time evolution of the GEC, described by Maxwell’s equations within the quasi-stationary approximation; in this approximation the electric field may be described in terms of a scalar potential. The most important aspects of this problem include a broad variety of possible boundary conditions (corresponding to different physical problems), the non-trivial topology of the Earth’s atmosphere and the fact that the main equation for the electric potential is not resolved with respect to the time derivative. The variational problems described in this article provide the basis for novel numerical models of the GEC and GEC-related physical problems. 2. Motivation We shall start with a preliminary discussion of the equations describing the atmospheric GEC. In this discussion we do not concern ourselves with rigorous formulations, as it is intended only to explain our motivation and the origin of the problem. Therefore we assume, for the moment, that all the functions, domains and boundary surfaces that appear in our reasoning are sufficiently smooth. We suppose that the atmosphere occupies the region Ω, whose boundary consists of the Earth’s surface Γ1 and a surface Γ2 representing the lower limit of the ionosphere, the latter encompassing the former (see Fig. 1a). We also suppose that both the dielectric permittivity and the magnetic permeability of the atmosphere are equal to 1, in which case non-stationary Maxwell’s equations read as follows1 : 1 ∂E 4π + j, c ∂t c 1 ∂H curl E = − , c ∂t div H = 0,

curl H =

div E = 4πρ,

(2.1) (2.2) (2.3) (2.4)

where E(t, x) is the electric field, H(t, x) is the magnetic field, j(t, x) is the current density, ρ(t, x) is the charge density, x denotes the spatial coordinates, t stands for time and c stands for the speed of light. These equations must be supplemented with Ohm’s law j = σE + js ,

(2.5)

where σ(t, x) is the conductivity and js (t, x) represents the source current density, as well as with suitable initial and boundary conditions. We regard thunderstorms as distributed current sources and suppose that the source current density is non-zero only within thunderclouds and other electrified clouds. In other words, at every moment in time the positions of thunderstorms correspond to the spatial distribution of js (see Fig. 1a). 1

Hereafter we use the Gaussian unit system.

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Fig. 1. (a) A schematic illustrating the geometry of the problem: the domain Ω ⊂ R3 bounded by the surfaces Γ1 and Γ2 . Inside Ω the distribution of the source current density is described by the function js (which is usually considered non-zero only inside electrified clouds). (b) A schematic illustrating the geometry of the geomagnetic field lines; x and x∗ are a pair of magnetic conjugate points.

Assuming that both surfaces Γ1 and Γ2 are perfect conductors, we have the boundary conditions E × n|Γ1 = 0,

E × n|Γ2 = 0,

(2.6)

where ‘×’ means the cross product in R3 and n denotes the outward unit normal vector to the boundary of Ω (in other words, these conditions demand that the tangential component of E at the boundary be equal to zero). The initial condition may be written in the form E|t=0 = E0 ,

(2.7)

E0 (x) being a stationary electric field satisfying (2.6). In modelling the GEC, it is common to analyse the problem in the quasi-stationary approximation—that is to say, to neglect the variation of H with time in (2.2). This yields the equation curl E = 0,

(2.8)

which means that we are primarily concerned with quasi-stationary currents and neglect fast processes; see, e.g., [6] for a more detailed discussion of this issue. Substituting relationship (2.5) into (2.1) gives curl H =

1 ∂E 4π + (σE + js ) . c ∂t c

(2.9)

Given σ and js , equations (2.6)–(2.9) form a system of equations in E and curl H. What we want now is to eliminate the magnetic field from equation (2.9). For brevity, let us denote the right-hand side of (2.9) by X. Let us observe that for a bounded domain Ω ⊂ R3 whose boundary consists of two connected components Γ1 and Γ2 , the following statements are equivalent2 : (i) There exists a vector field H such that X = curl H. (ii) div X is equal to zero, as is the flux of X through Γ1 . From this general observation we find that equation (2.9) is equivalent to the pair of equations ∂ (div E) + 4π div (σE) = −4π div js , ∂t 2

Of course, in (ii) we could have used Γ2 instead of Γ1 .

(2.10)

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  ∂E + 4πσE · n ds = −4π js · n ds. ∂t

Γ1

115

(2.11)

Γ1

If we have found E satisfying (2.10) and (2.11), we can find H by solving equations (2.9), (2.3) with suitable boundary conditions, which we do not specify here. Therefore in order to solve the original system of equations (2.1)–(2.4) within the quasi-stationary approximation, it is sufficient to find E. Since Ω is simply-connected, equation (2.8) makes it possible to introduce the electric potential, the function φ(t, x) such that E = − grad φ, and to reformulate other equations in terms of this function. Conditions (2.6) mean that φ does not vary over each of the two boundary surfaces, and thus if it is set equal to zero at Γ1 , its value at Γ2 is a certain function of time U (t). Condition (2.7) requires that at the moment t = 0 φ be equal to φ0 (x), the potential of E0 satisfying φ0 |Γ1 = 0 and φ0 |Γ2 = const. Thus we arrive at the system of equations ∂ Δφ + 4π div (σ grad φ) = 4π div js , ∂t     ∂ grad φ + 4πσ grad φ · n ds = 4π js · n ds, ∂t Γ1

(2.12) (2.13)

Γ1

∃U (·) : φ|Γ2 = U (·), 3

φ|Γ1 = 0,

φ|t=0 = φ0 ,

(2.14) (2.15)

where U (t) is an unknown function of time alone, satisfying U (0) = φ0 |Γ2 . A similar system for the corresponding steady-state problem (when nothing depends on t) may be written as follows: 

div (σ grad φ) = div js ,  (σ grad φ) · n ds = js · n ds,

Γ1

(2.16) (2.17)

Γ1

φ|Γ1 = 0,

∃U : φ|Γ2 = U,

(2.18)

where U is an unknown constant. In studies of the GEC, the potential difference U between the Earth’s surface and the lower ionosphere is termed the ionospheric potential. The fact that simultaneous measurements of this potential difference at remote locations yield very close results [20] provides important experimental evidence for the idea of the GEC. An important property of problems (2.12)–(2.15) and (2.16)–(2.18) is that U (which is a function of time in (2.14) and a constant in (2.18)) should not be specified explicitly and is actually considered as part of the solution, so that we can regard solutions of our problems as pairs (φ, U ). If (φ(1) (t, x), U (1) (t)) and (φ(2) (t, x), U (2) (t)) are two solutions to (2.12)–(2.15) with the same σ(t, x), js (t, x) and φ0 (x), then one can show that φ(1) = φ(2) and U (1) = U (2) (see [15,28]), and a similar result holds in the case of problem (2.16)–(2.18). In the sequel we shall prove that such problems are well-posed. The majority of GEC models [4,13–15], as well as many models describing the atmosphere–ionosphere interactions [10,16,23], are based on solving equation (2.12) or (2.16). Several GEC models are based precisely on problems (2.12)–(2.15) and (2.16)–(2.18), including the model developed by the authors of this paper (see [15,29]) and the model [4], which deals with equations equivalent to (2.16)–(2.18). Sometimes boundary conditions of other types are employed. For example, in the problem analysed in [23] an additional potential 3

Occasionally we write u(·) rather than u so as to emphasise that u depends on time or is regarded as a function of time.

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distribution was specified at the outer boundary so as to allow for various magnetospheric and ionospheric phenomena influencing the GEC. The zero-current condition at the upper boundary ( ∂φ/∂n|Γ2 = 0) was considered in [3]. Hays and Roble [13], in whose model the outer boundary was symmetric with respect to the geomagnetic equator (see Fig. 1b), demanded that the distribution of the electric potential on that boundary be symmetric and the distribution of the normal current density be antisymmetric (see also [22]); such a condition reflects the fact that in the ionosphere (i.e., at high altitudes) the geomagnetic field lines are nearly equipotential owing to essentially anisotropic conductivity (within this approach the difference between the normal component of the current density and its component parallel to the geomagnetic field is neglected). In this article we shall rigorously formulate and analyse time-dependent and steady-state problems related to the GEC with several different types of boundary conditions, including all of the aforementioned possibilities. Let us emphasise that whatever the boundary conditions may be, we must employ both equation (2.12) (or (2.16)) and integral condition (2.13) (or (2.17)) in order for the solution φ to be consistent with Maxwell’s equations. Equations of the form (2.12) have been studied, e.g., by Showalter and Ting [27] under the assumption of continuous differentiability of the solution with respect to time. Such equations are often called pseudoparabolic; in the past few decades the theory of equations of this type has been extensively developed. However, the problems that we consider here originate from Maxwell’s equations and differ from those commonly found in the literature, originating from fluid dynamics and seepage theory. Other distinctive features of our study are the necessity to consider equation (2.12) together with condition (2.13) and a broader variety of possible boundary conditions, dictated by the natural structure of quasi-stationary Maxwell’s equations. 3. Function spaces and trace operators Hereafter we suppose that Ω ⊂ R3 is a bounded domain with Lipschitz boundary Γ = Γ1 ∪ Γ2 , consisting of two connected components Γ1 and Γ2 , of which the latter encompasses the former (see Fig. 1a). The assumption of Lipschitz boundary guarantees that the outward unit normal vector n exists almost everywhere on Γ [21, Ch. 2, Lemma 4.2]. Let us note that although we confine ourselves to this geometry, our analysis can be readily extended to the case where Ω is still simply connected but has more boundary components. Let us briefly discuss the function spaces which we shall use. We will frequently employ the space L2(Ω) of square integrable functions on Ω, its analogue for R3 -valued functions L2 (Ω) = {L2 (Ω)}3 , the space L∞ (Ω) of essentially bounded functions on Ω and the Sobolev space H 1 (Ω). All these spaces are equipped with standard norms and (in the cases of Hilbert spaces L2 (Ω), L2 (Ω) and H 1 (Ω)) with standard inner products. We will also use the space H(div, Ω) = {u ∈ L2 (Ω) : div u ∈ L2 (Ω)} with the norm  1/2 uH(div, Ω) = u2L2 (Ω) +  div u2L2 (Ω) , the trace space H 1/2 (Γ) (see, e.g., [21, Ch. 2] for its definition and properties) and its dual H −1/2 (Γ). As usual, we define D (Ω) as the space of infinitely differentiable real-valued functions with compact support in Ω, and we define D (Ω) as the space of restrictions to Ω of functions from D (R3 ). There exists a continuous trace operator γ from H 1 (Ω) to (actually, onto) the trace space H 1/2 (Γ) extending the mapping u → u|Γ defined on D (Ω) [21, Ch. 2, Theorem 5.5], which possesses a continuous right inverse μ from H 1/2 (Γ) to H 1 (Ω) [21, Ch. 2, Theorem 5.7]. As a consequence, there exists a trace operator γn from H(div, Ω) to H −1/2 (Γ) extending u → u · n|Γ defined on {D (Ω)}3 [12, Ch. I, Theorem 2.5]. We shall use the notation

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u|Γ = γu and u · n|Γ = γn u for any u ∈ H 1 (Ω) and any u ∈ H(div, Ω). As Γ consists of two connected components Γ1 and Γ2 , one can show that H 1/2 (Γ) = H 1/2 (Γ1 ) × H 1/2 (Γ2 ), H −1/2 (Γ) = H −1/2 (Γ1 ) × H −1/2 (Γ2 ),

(3.1) (3.2)

and we have well-defined restrictions u|Γ1 ∈ H 1/2 (Γ1 ), u|Γ2 ∈ H 1/2 (Γ2 ) for u ∈ H 1 (Ω) and u · n|Γ1 ∈ H −1/2 (Γ1 ), u · n|Γ2 ∈ H −1/2 (Γ2 ) for u ∈ H(div, Ω). There are several possible approaches to time-dependent problems generalising (2.12)–(2.15). Here we make use of the variational approach described, e.g., in [8, Ch. XVIII]; in particular, this approach is convenient for numerical modelling with the help of the Galerkin approximation. It is common to study time-dependent problems on finite time intervals, and without loss of generality we may suppose that we study the GEC evolution for t ∈ [0, T ] with some T > 0. We shall often use various spaces of Bochner integrable functions, namely, L2 (0, T ; X), L∞ (0, T ; X) with usual norms (see, e.g., [8, Ch. XVIII, § 1]) and H 1 (0, T ; X) = {u ∈ L2 (0, T ; X) :

∂u ∈ L2 (0, T ; X)} ∂t

with the norm  uH 1 (0, T ; X) =

2 uL2 (0, T ; X)

1/2  2  ∂u    +  , ∂t L2 (0, T ; X)

where X is a certain Banach space and ∂u/∂t denotes the distributional derivative on (0, T ) of an X-valued function u ∈ L2 (0, T ; X). Let us denote the space of continuous functions from [0, T ] to X by C([0, T ]; X); we have an embedding H 1 (0, T ; X) → C([0, T ]; X)

(3.3)

(e.g., [8, Ch. XVIII, § 1, Theorem 1]), which makes it possible to define the traces of functions from H 1 (0, T ; X) in X at t = 0; for u ∈ H 1 (0, T ; X) we denote such trace by u|t=0 or simply by u(0). For arbitrary Banach spaces X and Y let L (X, Y ) be the space of continuous linear maps from X to Y . As we have noted above, there exist the trace operator γ ∈ L (H 1 (Ω), H 1/2 (Γ)) and its right inverse μ ∈ L (H 1/2 (Γ), H 1 (Ω)). Let us demonstrate that these maps can be ‘extended’ to spaces of time-dependent functions. Lemma 3.1. The mapping u → u|Γ defined on D (Ω × [0, T ]) can be extended to operators γH ∈ L (H 1 (0, T ; H 1 (Ω)), H 1 (0, T ; H 1/2 (Γ))) and γC ∈ L (C([0, T ]; H 1 (Ω)), C([0, T ]; H 1/2 (Γ))), which possess continuous right inverses μH and μC respectively. Moreover, μH and μC can be chosen such that the diagram

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H 1 (0, T ; H 1 (Ω))

C([0, T ]; H 1 (Ω))

γH

γC

H 1 (0, T ; H 1/2 (Γ))

C([0, T ]; H 1/2 (Γ))

t→0

t→0 γ

H 1 (Ω)

H 1/2 (Γ)

and its counterpart for μH , μC and μ are commutative. Proof. First of all, we define γH from L2 (0, T ; H 1 (Ω)) to L2 (0, T ; H 1/2 (Γ)) by sending u(·) to γu(·): u(·) ∈ L2 (0, T ; H 1 (Ω)) means that u(·) is strongly measurable on [0, T ] and square integrable in the Bochner sense, from which it follows that so is γu(·). Now let u ∈ H 1 (0, T ; H 1 (Ω)); in this case we immediately observe that γH u ∈ L2 (0, T ; H 1/2 (Γ)) and γH (∂u/∂t) ∈ L2 (0, T ; H 1/2 (Γ)). To show that γH u ∈ H 1 (0, T ; H 1/2 (Γ)), let us verify that it has a distributional derivative in L2 (0, T ; H 1/2 (Γ)). Indeed, for any φ ∈ D (0, T )

T 

 

T 

T ∂u ∂u ∂u (τ )φ(τ ) dτ = γ (τ )φ(τ ) dτ γ (τ ) φ(τ ) dτ = γ ∂t ∂t ∂t

0

0

T = −γ 0

dφ u(τ ) (τ ) dτ = − dt

0

T 0



dφ γ u(τ ) (τ ) dt



T dτ = −

(γu(τ ))

dφ (τ ) dτ, dt

0

which means that ∂ (γH u) /∂t = γH (∂u/∂t) ∈ L2 (0, T ; H 1/2 (Γ)). The operator γC is constructed pointwise in similar fashion; the continuity of γu(·) follows from that of u(·). To demonstrate that the diagram is commutative, it is sufficient to observe that each equivalence class of functions in H 1 (0, T ; X) has exactly one representative in C([0, T ]; X), given by the embedding (3.3). We define μH and μC using μ in the same manner as we did for γH and γC using γ. Our assertions concerning these operators can be proved by using exactly the same reasoning as above. From γ ◦ μ = id we infer that γH ◦ μH = id and γC ◦ μC = id. 2 Likewise, the following lemma ‘extends’ the operator γn ∈ L (H(div, Ω), H −1/2 (Γ)). The proof is similar. Lemma 3.2. The mapping u → u · n|Γ defined on {D (Ω × [0, T ])}3 can be extended to an operator γn, L ∈ L (L2 (0, T ; H(div, Ω)), L2 (0, T ; H −1/2 (Γ))). Henceforth we shall write u|Γ instead of γH u and γC u and we shall write u · n|Γ instead of γn, L u. Note that owing to the commutativity of the diagram in Lemma 3.1, the traces u|Γ, t=0 are well defined. Also, let us observe that relationships (3.1) and (3.2) yield similar decompositions for H 1 (0, T ; H 1/2 (Γ)) and L2 (0, T ; H −1/2 (Γ)), which make u|Γ1 , u|Γ2 , u · n|Γ1 and u · n|Γ2 well-defined for all u ∈ H 1 (0, T ; H 1 (Ω)) and u ∈ L2 (0, T ; H(div, Ω)). 4. Conductivity A typical GEC problem can be formulated as follows: given the conductivity, source current density, initial and boundary conditions, find the electric potential. It is natural to describe the source current density by

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a certain R3 -valued function js . We shall assume that js ∈ L2 (0, T ; H(div, Ω)) or js ∈ L2 (0, T ; L2 (Ω)) in the case of the time-dependent problem and js ∈ H(div, Ω) or js ∈ L2 (Ω) in the case of the steady-state problem. From now on we shall denote by (·, ·) the usual inner product in L2 (Ω) or L2 (Ω). Concerning the conductivity, in the time-dependent case we assume that for almost all t ∈ [0, T ] the conductivity distribution is described by an operator σ(t) ∈ L (L2 (Ω), L2 (Ω)),

(4.1)

by sending u(·) to σ(·)u(·) these operators induce the operator σ ∈ L (L2 (0, T ; L2 (Ω)), L2 (0, T ; L2 (Ω))),

(4.2)

and the inequalities σ(t)uL2 (Ω) ≤ const · uL2 (Ω) , 2

|(σ(t)u, u)| ≥ const · uL2 (Ω)

(4.3) (4.4)

hold for almost all t ∈ [0, T ] with both constants independent of t and the latter of them strictly positive. Likewise, in the steady-state case we assume that the conductivity distribution is described by an operator σ ∈ L (L2 (Ω), L2 (Ω))

(4.5)

2

(4.6)

and the inequality |(σ(t)u, u)| ≥ const · uL2 (Ω)

holds with a strictly positive constant. Let us ascertain that realistic distributions of the atmospheric conductivity satisfy these requirements. Considering the fact that the conductivity is essentially anisotropic at high altitudes, we suppose that it is described by a tensor σ, which can be represented in Cartesian coordinates as  σ=

σ 0 0

0 σP σH

0 −σH σP

(4.7)

if the first coordinate axis is chosen parallel to the Earth’s magnetic field H0 (σP and σH are termed Pedersen conductivity and Hall conductivity respectively; see, e.g., [6,30]); thus we can rewrite (2.5) as j = σ (E · h0 ) h0 + σP (E − (E · h0 ) h0 ) − σH E × h0 + js , where h0 = H0 / |H0 |. Therefore the conductivity distribution can be described by three scalar functions σ , σP and σH . The case of isotropic conductivity (e.g., when high altitudes are not considered) can be covered by setting σ = σP , σH = 0. In order for our requirements to be satisfied, in the time-dependent case it is sufficient to suppose that σ , σP , σH ∈ L∞ (0, T ; L∞ (Ω)),    ess inf ess inf σ > 0, ess inf ess inf σP > 0 

[0, T ]

Ω

[0, T ]

Ω

(4.8) (4.9)

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and at any x ∈ Ω there exists a Cartesian coordinate system continuously depending on x such that for almost all t ∈ [0, T ] almost everywhere in Ω the conductivity operator σ(t, x) is defined and described in this coordinate system by a matrix σ ˜ ∈ L (R3 , R3 ) of the form (4.7);

(4.10)

similarly, in the steady-state case it is sufficient to suppose that σ , σP , σH ∈ L∞ (Ω), ess inf σ > 0, Ω

ess inf σP > 0 Ω

(4.11) (4.12)

and at any x ∈ Ω there exists a Cartesian coordinate system continuously depending on x such that almost everywhere in Ω the conductivity operator σ(x) is defined and described in this coordinate system by a matrix σ ˜ ∈ L (R3 , R3 ) of the form (4.7).

(4.13)

Indeed, it is easy to see that assumptions (4.8) and (4.10) imply (4.1), (4.2) and (4.3), and condition (4.9) leads to (4.4). Similarly, (4.5) and (4.6) follow from (4.11)–(4.13). Therefore requirements (4.1)–(4.4) and (4.5), (4.6) generalise requirements (4.8)–(4.10) and (4.11)–(4.13). 5. Dirichlet problem In this section we shall rigorously formulate and analyse the Dirichlet problem for the GEC equations, a straightforward generalisation of problems (2.12)–(2.15) and (2.16)–(2.18). Roughly speaking, we assume that the potential is specified at one connected component of the boundary and is specified up to a constant or function of time at another boundary component. Such an approach is motivated by various models of atmosphere–ionosphere interactions, which deal with the so-called ‘downward mapping’ of the potential pattern from the upper boundary to the lower atmosphere (see, e.g., [23]). We begin with the time-dependent case. We suppose that the conductivity σ satisfies requirements (4.1)–(4.4). Let us assume for the time being that js ∈ L2 (0, T ; H(div, Ω)), and let φ1 ∈ H 1 (0, T ; H 1/2 (Γ1 )), φ2 ∈ H 1 (0, T ; H 1/2 (Γ2 )) and φ0 ∈ H 1 (Ω) be such that φ0 |Γ1 = φ1 |t=0 ,

φ0 |Γ2 = φ2 |t=0 .

(5.1)

We consider the following problem (a generalisation of (2.12)–(2.15)) for φ ∈ H 1 (0, T ; H 1 (Ω)):   ∂φ + 4πσ grad φ = 4π div js , div grad ∂t

    ∂φ + 4πσ grad φ · n , 1 = 4π js · n|Γ1 , 1 Γ , grad 1 ∂t Γ1

(5.2) (5.3)

Γ1

φ|Γ1 = φ1 ,

∃U ∈ H 1 (0, T ) : φ|Γ2 = φ2 + U ⊗ 1, 4 φ|t=0 = φ0 ,

(5.4) (5.5)

where U ∈ H 1 (0, T ) is an unknown function and ·, ·Γ1 denotes the duality pairing between H −1/2 (Γ1 ) and H 1/2 (Γ1 ). First of all, let us make sure that equations (5.2)–(5.5) make sense. Indeed, now that js ∈ 4

Here and below u(·) ⊗ v denotes the function equal to u(t) · v(x) at (t, x).

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L2 (0, T ; H(div, Ω)), the right-hand side of (5.2) lies in L2 (Ω) for almost all t ∈ [0, T ], hence its left-hand side does so as well. Since grad ∂φ/∂t + 4πσ grad φ obviously lies in L2 (Ω) for almost all t ∈ [0, T ], we find out that almost everywhere on [0, T ] both js and grad ∂φ/∂t + 4πσ grad φ lie in H(div, Ω). Therefore the traces of their normal components on Γ1 are defined and (5.3) makes sense for almost all t ∈ [0, T ]. Note that here we regard (5.2) and (5.3) as equations (in L2 (Ω) and R respectively) valid almost everywhere on [0, T ], but it is easy to see that we can also regard them as equalities in the sense of distributions. Indeed, using the fact that js ∈ L2 (0, T ; H(div, Ω)) and Lemma 3.2, one can show that div js ∈ L2 (0, T ; L2 (Ω)) and  s  j · n|Γ1 , 1 Γ ∈ L2 (0, T ), hence actually (5.2) holds in L2 (0, T ; L2 (Ω)) and (5.3) holds in L2 (0, T ). 1 Let us recall that by (·, ·) we denote the usual inner product in L2 (Ω) or L2 (Ω). For all v ∈ H(div, Ω) and ψ ∈ H 1 (Ω) we have the following Green’s formula:     (v, grad ψ) + (div v, ψ) = v · n|Γ1 , ψ|Γ1 Γ + v · n|Γ2 , ψ|Γ2 Γ 1

2

(5.6)

(see, e.g., [12, Ch. I, equation (2.17)]). Setting ψ = 1 in (5.6), we derive from (5.2) and (5.3) that

 ∂φ + 4πσ grad φ · n , 1 grad ∂t Γ2



  = 4π js · n|Γ2 , 1 Γ . 2

(5.7)

Γ2

Let us introduce the space5 VD (Ω) = {u ∈ H 1 (Ω) : u|Γ1 = 0, ∃c ∈ R : u|Γ2 = c}, which is obviously a closed subspace of H 1 (Ω); if we take ψ ∈ VD (Ω) in (5.6), equations (5.2), (5.3) and (5.7) readily yield that 

∂φ grad + 4πσ grad φ, grad ψ ∂t

 = 4π (js , grad ψ)

∀ψ ∈ VD (Ω),

(5.8)

which again can be understood either as an equation valid for almost all t ∈ [0, T ] or as an equality in the sense of distributions on (0, T ). To proceed further, we need a simple lemma. Lemma 5.1. For any u ∈ H 1 (0, T ; H 1 (Ω)) and ψ ∈ H 1 (Ω) the following identity holds in the sense of distributions on (0, T ):  grad

∂u (·), grad ψ ∂t

 =

d (grad u(·), grad ψ) . dt

=

d (u(·), ψ)X dt

(5.9)

Proof. We use the fact that 

∂u (·), ψ ∂t

 X

(5.10)

if u ∈ H 1 (0, T ; X), ψ ∈ X and (·, ·)X denotes the inner product in a Hilbert space X [8, Ch. XVIII, § 1, Proposition 7]. Taking arbitrary u ∈ H 1 (0, T ; H 1 (Ω)) and ψ ∈ H 1 (Ω), we can apply to them (5.10) with X = H 1 (Ω) and (5.10) with X = L2 (Ω); subtracting the resulting equations from each other, we obtain (5.9). 2 5

Here ‘D’ stands for ‘Dirichlet problem’.

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Applying Lemma 5.1 in (5.8), we arrive at the following problem: d (grad φ, grad ψ) + 4π (σ grad φ, grad ψ) = 4π (js , grad ψ) ∀ψ ∈ VD (Ω), dt φ|Γ1 = φ1 , ∃U ∈ H 1 (0, T ) : φ|Γ2 = φ2 + U ⊗ 1,

(5.12)

φ|t=0 = φ0 ,

(5.13)

(5.11)

where it is convenient to understand equation (5.11) as an equality of distributions. Let us verify that for js ∈ L2 (0, T ; H(div, Ω)) this problem is equivalent to the problem from which we have started. Proposition 5.1. Suppose that js ∈ L2 (0, T ; H(div, Ω)) and conditions (4.1)–(4.4) are satisfied. Then for φ ∈ H 1 (0, T ; H 1 (Ω)) problem (5.2)–(5.5) is equivalent to problem (5.11)–(5.13). Proof. We have already shown that equations (5.2)–(5.5) imply (5.11)–(5.13). For the converse, we need to show that (5.11) entails (5.2) and (5.3). According to Lemma 5.1, identity (5.11) is equivalent to (5.8). Then, using Green’s formula (5.6) with ψ ∈ D (Ω), we immediately obtain (5.2). To derive (5.3), let us note that there exists a function ψ ∈ VD (Ω) such that ψ|Γ1 = 0 and ψ|Γ2 = 1 (to construct it, we can simply apply the operator μ). Substituting this ψ into Green’s formula, we obtain (5.7), which, together with (5.2), finally gives us (5.3), this time by Green’s formula with ψ = 1. 2 Problem (5.11)–(5.13) serves as the variational formulation of problem (5.2)–(5.5). As usual, we can regard the solutions to the variational problem as weak solutions to the original problem even if js ∈ / 2 s 2 2 L (0, T ; H(div, Ω)). Restricting ourselves to the case where j ∈ L (0, T ; L (Ω)), we will now show that problem (5.11)–(5.13) is well-posed. Proposition 5.2. Suppose that js ∈ L2 (0, T ; L2 (Ω)), conditions (4.1)–(4.4) are satisfied, and φ1 ∈ H 1 (0, T ; H 1/2 (Γ1 )), φ2 ∈ H 1 (0, T ; H 1/2 (Γ2 )) and φ0 ∈ H 1 (Ω) meet the compatibility conditions (5.1). Then there exists a unique solution φ ∈ H 1 (0, T ; H 1 (Ω)) to problem (5.11)–(5.13), satisfying  φL2 (0, T ; H 1 (Ω)) ≤ const · js L2 (0, T ; L2 (Ω)) + φ1 H 1 (0, T ; H 1/2 (Γ1 ))

 + φ2 H 1 (0, T ; H 1/2 (Γ2 )) + φ0 H 1 (Ω) ,

(5.14)

where the constant is independent of js , φ1 , φ2 and φ0 . Proof. Using the operator μH defined in Lemma 3.1, we immediately construct φΓ ∈ H 1 (0, T ; H 1 (Ω)) such that φΓ |Γ1 = φ1 ,

φΓ |Γ2 = φ2 .

Setting χ = φ − φΓ and using Lemma 5.1 to move one of the time derivatives inside the inner product, we obtain from (5.11)–(5.13) the following problem for χ ∈ H 1 (0, T ; VD (Ω)): d (grad χ, grad ψ) + 4π (σ grad χ, grad ψ) dt   ∂φΓ s = 4πj − grad − 4πσ grad φΓ , grad ψ ∂t

(5.15) ∀ψ ∈ VD (Ω),

χ|t=0 = φ0 − φΓ |t=0 , where, owing to (5.1), the right-hand side of the last equation lies in VD (Ω).

(5.16)

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To proceed with the proof, we need the following result (see [12, Ch. I, Lemma 3.1]), which, in a certain sense, generalises Poincaré’s inequality, valid for the space H01 (Ω). Lemma 5.2. Let Γ0 be a portion of Γ with strictly positive measure. Then the inner product u, v → (grad u, grad v) defines on the space {u ∈ H 1 (Ω) : u|Γ0 = 0} a norm equivalent to the norm inherited from H 1 (Ω) (here u|Γ0 means the restriction of u|Γ ∈ H 1/2 (Γ) to Γ0 ). Applying this lemma with Γ0 = Γ1 and passing to the subspace VD (Ω), we observe that u → grad uL2 (Ω) is a well-defined norm on VD (Ω) equivalent to u → uH 1 (Ω) .6 Let us equip VD (Ω) with this norm (which we hereafter denote by ·VD (Ω) ) and with the corresponding inner product (which we denote by (·, ·)VD (Ω) ), and let us identify VD (Ω) with its dual space (by means of the isometry given by the Riesz representation theorem). For almost all t ∈ [0, T ] we can introduce a bilinear form B(·) on VD (Ω) defined by B( · ; u, v) = 4π (σ(·) grad u, grad v)

(5.17)

and a linear functional f (·) on VD (Ω) defined by 

f (·), u =

4πjs (·) − grad

 ∂φΓ (·) − 4πσ(·) grad φΓ (·), grad u ; ∂t

(5.18)

this allows us to rewrite identity (5.15) as d (χ(·), ψ)VD (Ω) + B( · ; χ(·), ψ) = f (·), ψ dt

∀ψ ∈ VD (Ω).

(5.19)

Let us note that assumptions (4.3) and (4.4) imply that |B(t; u, v)| ≤ const · uVD (Ω) · vVD (Ω) 2

|B(t; u, u)| ≥ const · uVD (Ω)

∀u, v ∈ VD (Ω), ∀u ∈ VD (Ω)

(5.20) (5.21)

for almost all t ∈ [0, T ] with both constants independent of t and the latter of them strictly positive; in other words, the form B(·) is uniformly continuous and uniformly VD (Ω)-elliptic on a subset of full measure in [0, T ]. Also, one can easily show that B( · ; u, v) is measurable on [0, T ] for all u, v ∈ VD (Ω). Finally, it follows from (5.18) that7 f (·) is strongly measurable on [0, T ], f ∈ L2 (0, T ; VD (Ω)) and   f L2 (0, T ; VD (Ω)) ≤ const · js L2 (0, T ; L2 (Ω)) + φΓ H 1 (0, T ; H 1 (Ω)) .

(5.22)

Thus we have shown that all the conditions are satisfied for the solution of problem (5.15), (5.16) (or (5.16), (5.19)) to exist and be unique (see [8, Ch. XVIII, § 3, Theorems 1 and 2]). Furthermore, the corresponding continuity theorem [8, Ch. XVIII, §3, Theorem 3] states that   χL2 (0, T ; VD (Ω)) ≤ const · φ0 − φΓ |t=0 VD (Ω) + f L2 (0, T ; VD (Ω)) . 6 7

The equivalence is important, for the condition χ ∈ H 1 (0, T ; VD (Ω)) implicitly depends on the norm in VD (Ω). Note once again that we identify VD (Ω) and its dual.

(5.23)

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Recalling that φ = φΓ +χ, observing that φΓ = μH (φ1 , φ2 ) and using the continuity of the map φΓ → φΓ |t=0 , we deduce from (5.22) and (5.23) that (5.14) holds. 2 Note that from the relationship U ⊗ 1 = φ|Γ2 − φ2 and (5.14) one can obtain the inequality  U L2 (0, T ) ≤ const · js L2 (0, T ; L2 (Ω)) + φ1 H 1 (0, T ; H 1/2 (Γ1 ))

 + φ2 H 1 (0, T ; H 1/2 (Γ2 )) + φ0 H 1 (Ω) .

(5.24)

Now we shall briefly discuss the corresponding steady-state problem, which generalises problem (2.16)–(2.18). In this case we have an elliptic problem for the potential φ; considering that such problems have been already studied before, we confine ourselves to a concise summary of the main results. Supposing that the conductivity σ satisfies requirements (4.5) and (4.6) and assuming that js ∈ H(div, Ω), φ1 ∈ H 1/2 (Γ1 ) and φ2 ∈ H 1/2 (Γ2 ), we consider the following problem for φ ∈ H 1 (Ω): div (σ grad φ) = div js ,     (σ grad φ) · n|Γ1 , 1 Γ = js · n|Γ1 , 1 Γ , 1

φ|Γ1 = φ1 ,

1

∃U ∈ R : φ|Γ2 = φ2 + U,

(5.25) (5.26) (5.27)

where U is an unknown constant. As before, we observe that equations (5.25)–(5.27) make sense and all necessary traces are well defined. Problem (5.25)–(5.27) has much in common with typical electrostatic problems, where the dielectric permittivity plays the role played here by the conductivity and the space charge density plays the role of div js . Such electrostatic problems have been studied before (see, e.g., [11]), but here we have a slightly different situation in that we formulate the problem in terms of js instead of dealing with the function div js , the counterpart of the space charge density. Using the same technique as before, we find that the variational counterpart of problem (5.25)–(5.27) is the problem (σ grad φ, grad ψ) = (js , grad ψ) φ|Γ1 = φ1 ,

∀ψ ∈ VD (Ω),

∃U ∈ R : φ|Γ2 = φ2 + U.

(5.28) (5.29)

The analogues of Proposition 5.1 and Proposition 5.2 read as follows. Proposition 5.3. Suppose that js ∈ H(div, Ω) and conditions (4.5) and (4.6) are satisfied. Then for φ ∈ H 1 (Ω) problem (5.25)–(5.27) is equivalent to problem (5.28), (5.29). Proposition 5.4. Suppose that js ∈ L2 (Ω), conditions (4.5) and (4.6) are satisfied, φ1 ∈ H 1/2 (Γ1 ) and φ2 ∈ H 1/2 (Γ2 ). Then there exists a unique solution φ ∈ H 1 (Ω) to problem (5.28), (5.29), satisfying   φH 1 (Ω) ≤ const · js L2 (Ω) + φ1 H 1/2 (Γ1 ) + φ2 H 1/2 (Γ2 ) , where the constant is independent of js , φ1 and φ2 . The proofs of these results are completely similar to those of Proposition 5.1 and Proposition 5.2. In place of the theorems establishing that the solution of a variational evolution problem of the form (5.16), (5.19) exists, is unique and continuously depends on the right-hand side of (5.19) and the initial condition, here we invoke the Lax–Milgram theorem (see, e.g., [21, Ch. 1, Lemma 3.1]). Also, similarly to (5.24), we have

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  |U | ≤ const · js L2 (Ω) + φ1 H 1/2 (Γ1 ) + φ2 H 1/2 (Γ2 ) . To conclude our study of the Dirichlet problem, we will prove a stabilisation theorem establishing that the solution of the evolution problem (5.11)–(5.13) approaches the solution of the steady-state problem (5.28), (5.29) as t → ∞, provided that the conductivity, source current density and boundary conditions do not depend on t (cf. [18,27]). First of all, we remark that if φ is the solution to (5.11)–(5.13) set for [0, T ], then for any T  ∈ (0, T ) the restriction φ|[0, T  ] is the solution to the same problem restricted to [0, T  ]. This allows us to speak about the behaviour of the solution to (5.11)–(5.13) as t → ∞, once the conductivity, source current density and boundary conditions are defined for all t. Proposition 5.5. Let us suppose that js ∈ L2 (Ω), conditions (4.5) and (4.6) are satisfied, and φ1 ∈ H 1/2 (Γ1 ), φ2 ∈ H 1/2 (Γ2 ) and φ0 ∈ H 1 (Ω) are such that φ0 |Γ1 = φ1 ,

φ0 |Γ2 = φ2 .

Let φ(·) be a function from (0, ∞) to H 1 (Ω) such that for any T > 0 φ(·)|[0, T ] ∈ H 1 (0, T ; H 1 (Ω)) is the solution to the problem d (grad φ, grad ψ) + 4π (σ grad φ, grad ψ) = 4π (js , grad ψ) ∀ψ ∈ VD (Ω), dt ∃U ∈ H 1 (0, T ) : φ|Γ2 = 1 ⊗ φ2 + U ⊗ 1, φ|Γ1 = 1 ⊗ φ1 ,

(5.30) (5.31)

φ|t=0 = φ0 , and let φ ∈ H 1 (Ω) be the solution to the corresponding steady-state problem, that is, 

 ∀ψ ∈ VD (Ω), σ grad φ, grad ψ = (js , grad ψ) φ Γ = φ1 , ∃U ∈ R : φ Γ = φ2 + U . 1

2

(5.32) (5.33)

Then φ(t) tends to φ in H 1 (Ω) as t → ∞. Proof. For the moment, let us fix T > 0. Setting χ = φ − 1 ⊗ φ and subtracting (5.32) multiplied by 4π from (5.30) and (5.33) from (5.31), we arrive at the following problem for χ ∈ H 1 (0, T ; VD (Ω)): d (grad χ, grad ψ) + 4π (σ grad χ, grad ψ) = 0 dt

∀ψ ∈ VD (Ω),

(5.34)

χ|t=0 = φ0 − φ. Let us denote by A the operator associated with the bilinear form (5.17),8 defined by the relationship (Au, v)VD (Ω) = B(u, v) = 4π (σ grad u, grad v) for all u, v ∈ VD (Ω); such A exists as a result of the continuity of B and the Riesz representation theorem. The introduction of A along with (5.10) allows us to rewrite identity (5.34) as 

8

∂χ (·) + Aχ(·), ψ ∂t

 =0

∀ψ ∈ VD (Ω).

VD (Ω)

Now that the conductivity does not depend on t here, the form B does not depend on it either.

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For any ψ ∈ VD (Ω) this equation holds in the sense of real-valued distributions on (0, T ) and, by inference, almost everywhere on [0, T ]. Now that VD (Ω) is clearly separable, from this it follows that for almost all τ ∈ [0, T ] (∂χ/∂t) (τ ) + Aχ(τ ) = 0, whereupon for those values of τ we obtain that 



∂χ (τ ) + Aχ(τ ), χ(τ ) ∂t

= 0. VD (Ω)

Turning back to B, we find that 

∂χ (·), χ(·) ∂t

 = −4π (σ grad χ(·), grad χ(·)) VD (Ω)

almost everywhere on [0, T ]. For all u, v ∈ H 1 (0, T ; X) we have the formula (e.g., [8, Ch. XVIII, §1, Theorem 2])

T 

T 



∂u (τ ), v(τ ) ∂t

dτ + X

0

∂v (τ ), u(τ ) ∂t

0

 T

dτ = (u(τ ) · v(τ ))X |0 , X

where the right-hand side is well defined owing to embedding (3.3). Taking X = VD (Ω) and substituting χ(·) for both u(·) and v(·), we obtain that

T  2 χ(T )VD (Ω)

=

2 χ(0)VD (Ω)

+2



∂χ (τ ), χ(τ ) ∂t

0

dτ X

T =

2 χ(0)VD (Ω)

− 8π

(σ grad χ(τ ), grad χ(τ )) dτ, 0

if we require that χ(·) be continuous according to (3.3). Since increasing T does not change χ(τ ) for τ ≤ T , we can deduce that d 2 χ(·)VD (Ω) = −8π (σ grad χ(·), grad χ(·)) dt everywhere on (0, ∞); then from (5.20) one can easily infer that the right-hand side of this equation is continuous with respect to t, hence so is the derivative in the left-hand side. Finally, (5.21) implies that there exists a strictly positive constant α such that d 2 2 χ(·)VD (Ω) + α χ(·)VD (Ω) ≤ 0; dt multiplying this inequality by exp(αt) and integrating over [0, τ ], we find that 2

2

χ(τ )VD (Ω) ≤ χ(0)VD (Ω) · exp(−ατ ). Turning back to φ(·) and φ, we can rewrite (5.35) as   φ(t) − φ2

VD (Ω)

 2 ≤ φ0 − φV

which means that φ(t) → φ in H 1 (Ω) as t → ∞.

D (Ω)

2

· exp(−αt),

(5.35)

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It is worth noting that if the conductivity distribution is described by a tensor of the form (4.7) and meets the conditions (4.11)–(4.13), then α in the preceding proof may be taken equal to   α = 8π min ess inf σ , ess inf σP . Ω

Ω

Also, it follows from the preceding proof that U (t) → U as t → ∞. Indeed, since χ|Γ2 = U ⊗ 1 − U , we have   U (t) − U ≤ U (t) − U 

H 1/2 (Γ2 )

≤ const · χ(t)H 1 (Ω) ≤ const · χ(t)VD (Ω) ,

and the desired result follows from (5.35). The results obtained in this section generalise the results obtained in [35] for the case where the conductivity is isotropic and is given by a function σ ∈ C([0, T ]; L∞ (Ω)), js ∈ C([0, T ]; L2 (Ω)), φ1 = 0, φ2 = 0, φ0 ∈ VD (Ω) and the potential φ is assumed to be continuously differentiable with respect to t (φ ∈ C 1 ([0, T ]; VD (Ω))). Here we have proved the corresponding results in a more general situation under weaker hypotheses. More general problems of this type have been discussed in [27], also under the assumption of continuous differentiability. 6. Neumann problem In this section we shall study the GEC equations with boundary conditions of another type: we shall assume that the normal component of the total current9 density, which includes the conduction current density −σ grad φ, source current density js and displacement current density −1/(4π) grad ∂φ/∂t, is specified on the boundary of Ω. In the case of the steady-state problem this means that the values of σ grad φ · n on Γ are known, and if the conductivity is isotropic, this condition reduces to the fact that ∂φ/∂n on Γ is set equal to a certain function. Therefore the problem which we shall study here generalises the Neumann problem. Note that in the time-dependent case we deliberately make use of the total current density (rather than the term σ grad φ alone) in the boundary conditions; this choice is motivated by the structure of the GEC equations and the fact that the total current plays an important role in the studies of the GEC (see, e.g., [17]).10 When we discuss the steady-state Neumann problem, it is convenient to use the quotient space H 1 (Ω)/R, where R is understood as the subspace of H 1 (Ω) consisting of constant functions. This space is equipped with the usual quotient norm: if u ˜ is an arbitrary representative of an equivalence class u ∈ H 1 (Ω)/R, then uH 1 (Ω)/R = inf ˜ u + cH 1 (Ω) . c∈R

(6.1)

For any u ∈ H 1 (Ω)/R we define grad u = grad u ˜ ∈ L2 (Ω) with u ˜ ∈ H 1 (Ω) any representative of u (it is clear that grad u does not depend on the choice of u ˜). Since in the steady-state Neumann problem only grad φ enters into the equations and boundary conditions, it is clear that the solution can be unique only up to addition of a constant, and it is natural to regard the potential φ as a function lying in H 1 (Ω)/R. As we will see below, in the case of the non-stationary Neumann problem the situation is quite similar, and it is convenient to set the problem in the quotient space H 1 (0, T ; H 1 (Ω))/H 1 (0, T ), where H 1 (0, T ) is embedded in H 1 (0, T ; H 1 (Ω)) as the subspace of functions of t alone; this quotient space is equipped with the 9

It is also called Maxwell current in the studies of atmospheric electricity. According to equation (2.9), such boundary condition (the normal component of the total current density specified on Γ) arises when we know the tangential component of the magnetic field H on the boundary. Similarly, we use the Dirichlet condition on Γ if we are given the tangential component of the electric field E. 10

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quotient norm similar to (6.1). One can readily check that for any u ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) we have well-defined ∂u/∂t ∈ L2 (0, T ; H 1 (Ω))/L2 (0, T ), grad u ∈ L2 (0, T ; L2 (Ω)) and grad ∂u/∂t ∈ L2 (0, T ; L2 (Ω)). As in the case of the Dirichlet problem, we primarily focus on the study of the time-dependent case. Let us suppose that the conductivity σ satisfies (4.1)–(4.4), js ∈ L2 (0, T ; H(div, Ω)), φ0 ∈ H 1 (Ω)/R, and let j1 ∈ L2 (0, T ; H −1/2 (Γ1 )) and j2 ∈ L2 (0, T ; H −1/2 (Γ2 )) be such that

j1 , 1Γ1 = 0,

j2 , 1Γ2 = 0.

(6.2)

We consider the following problem for φ ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ):   ∂φ + 4πσ grad φ = 4π div js , div grad ∂t 

   ∂φ grad + 4πσ grad φ · n , 1 = 4π js · n|Γ1 , 1 Γ , 1 ∂t Γ1 Γ1     1 ∂φ ∂φ 1 − grad − σ grad φ + js · n = j1 , grad − σ grad φ + js · n = j2 , − 4π ∂t 4π ∂t Γ1 Γ2 φ|t=0 = φ0 .

(6.3) (6.4)

(6.5) (6.6)

We need to verify that equations (6.3)–(6.6) make sense; it is convenient to understand (6.3)–(6.5) as equalities in the sense of distributions on (0, T ). First of all, as we noted above, grad φ and grad ∂φ/∂t are well defined for φ ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ). The condition js ∈ L2 (0, T ; H(div, Ω)) implies that grad ∂φ/∂t + 4πσ grad φ also lies in this space and conditions (6.4) and (6.5) make sense. Finally, the initial condition (6.6) makes sense owing to embedding (3.3), which implies that the functions from H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) possess well-defined traces in H 1 (Ω)/R. The former of the compatibility conditions (6.2) is necessary for (6.4) and (6.5) to be consistent, and Green’s formula (5.6) with ψ = 1 again yields (5.7), which makes necessary the latter of conditions (6.2). Note that if we had omitted equation (6.4), the only compatibility condition would read j1 , 1Γ1 + j2 , 1Γ2 = 0. Let us find the variational formulation of the problem (6.3)–(6.6). Using (6.3), (6.4) and (5.7), we derive from Green’s formula (5.6) that 

 ∂φ + 4πσ grad φ, grad ψ ∂t     = 4π (js , grad ψ) − 4π j1 , ψ|Γ1 Γ − 4π j2 , ψ|Γ2 Γ

grad

1

(6.7) ∀ψ ∈ H (Ω). 1

2

First of all, we observe that owing to conditions (6.2) this identity also holds with ψ ∈ H 1 (Ω)/R. Next, it is easy to see that Lemma 5.1 is also valid for u ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) and ψ ∈ H 1 (Ω)/R. Applying this lemma, we arrive at the problem d (grad φ, grad ψ) + 4π (σ grad φ, grad ψ) dt     = 4π (js , grad ψ) − 4π j1 , ψ|Γ1 Γ − 4π j2 , ψ|Γ2 Γ 1

φ|t=0 = φ0 ,

(6.8) 2

∀ψ ∈ H 1 (Ω)/R, (6.9)

where we understand (6.8) as an equality of distributions. We readily obtain the following proposition. Proposition 6.1. Suppose that js ∈ L2 (0, T ; H(div, Ω)) and conditions (4.1)–(4.4) and (6.2) are satisfied. Then for φ ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) problem (6.3)–(6.6) is equivalent to problem (6.8), (6.9).

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Proof. We slightly modify the proof of Proposition 5.1. We have already shown that (6.8), (6.9) follows from (6.3)–(6.6). To prove the converse, we observe that (6.8), in view of Lemma 5.1 and conditions (6.2), is equivalent to (6.7). Using Green’s formula (5.6) with ψ ∈ D (Ω), we obtain (6.3). For any u ∈ H 1/2 (Γ1 ) there exists a function ψ ∈ H 1 (Ω) such that ψ|Γ1 = u and ψ|Γ2 = 0 (it can be constructed with the help of the operator μ); since u is arbitrary, substituting this ψ into Green’s formula gives us the former of conditions (6.5), and similarly we get the latter of them. Finally, equation (6.4) follows automatically from (6.2) and (6.5). 2 Problem (6.8), (6.9) is the variational formulation of problem (6.3)–(6.6). Let us now suppose that js ∈ L2 (0, T ; L2 (Ω)); then the solutions to the variational problem can be considered weak solutions to the original problem. The following proposition is similar to Proposition 5.2 and demonstrates that problem (6.8), (6.9) is well-posed. Proposition 6.2. Suppose that js ∈ L2 (0, T ; L2 (Ω)), conditions (4.1)–(4.4) are satisfied, φ0 ∈ H 1 (Ω)/R, and j1 ∈ H 1 (0, T ; H −1/2 (Γ1 )) and j2 ∈ H 1 (0, T ; H −1/2 (Γ2 )) meet conditions (6.2). Then there exists a unique solution φ ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) to problem (6.8), (6.9), satisfying11  φL2 (0, T ; H 1 (Ω))/L2 (0, T ) ≤ const · js L2 (0, T ; L2 (Ω)) + j1 H 1 (0, T ; H −1/2 (Γ1 ))  + j2 H 1 (0, T ; H −1/2 (Γ2 )) + φ0 H 1 (Ω)/R ,

(6.10)

where the constant is independent of js , j1 , j2 and φ0 . Proof. Here we cannot simply apply the theorems establishing the existence, uniqueness and continuity of the solution to a variational evolution problem, as we did in the proof of Proposition 5.2; to employ these general results, we need to set our problem in H 1 (0, T ; H 1 (Ω)/R) rather than in H 1 (0, T ; H 1 (Ω))/H 1 (0, T ). Therefore together with (6.8), (6.9) we consider the problem    d  ˇ grad ψ + 4π σ grad φ, ˇ grad ψ grad φ, dt     = 4π (js , grad ψ) − 4π j1 , ψ|Γ1 Γ − 4π j2 , ψ|Γ2 Γ 1 2 = φ0 φˇ

(6.11) ∀ψ ∈ H 1 (Ω)/R,

t=0

(6.12)

for φˇ ∈ H 1 (0, T ; H 1 (Ω)/R) with the same σ, js , j1 , j2 and φ0 . Let us demonstrate that this problem is well-posed. We need the following lemma (see, e.g., [21, Ch. 1, Theorem 1.6]). Lemma 6.1. The inner product u, v → (grad u, grad v) defines on the space H 1 (Ω)/R a norm equivalent to the norm (6.1). Hereafter we denote by ·H 1 (Ω)/R the norm on H 1 (Ω)/R provided by this lemma, and we equip H 1 (Ω)/R with the corresponding inner product (which we denote by (·, ·)H 1 (Ω)/R ). Besides, we identify H 1 (Ω)/R with its dual space. Following the technique used in the proof of Proposition 5.2 and noting that for any u ∈ H 1 (Ω)/R and k = 1, 2 we have the inequality   jk (·), u|Γk Γ ≤ const · jk (·)H −1/2 (Γk ) · uH 1 (Ω)/R k

11

The space L2 (0, T ; H 1 (Ω))/L2 (0, T ) is equipped with the quotient norm.

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(owing to the fact that our norm on H 1 (Ω)/R is equivalent to (6.1)), we find that there exists a unique solution φˇ ∈ H 1 (0, T ; H 1 (Ω)/R) to problem (6.11), (6.12), satisfying    ˇ φ

L2 (0, T ; H 1 (Ω)/R)

 ≤ const · js L2 (0, T ; L2 (Ω)) + j1 H 1 (0, T ; H −1/2 (Γ1 ))  + j2 H 1 (0, T ; H −1/2 (Γ2 )) + φ0 H 1 (Ω)/R .

(6.13)

It remains to link problems (6.8), (6.9) and (6.11), (6.12). Let p be the canonical projection from H 1 (Ω) to H 1 (Ω)/R, and let us denote by R⊥ the orthogonal complement of R in H 1 (Ω), R being regarded as the subspace of H 1 (Ω) consisting of constants. For the equivalent norm (6.1), p induces an isometry from R⊥ to H 1 (Ω)/R, hence p has a continuous right inverse q. Similarly to the construction of γH and μH in the proof of Lemma 3.1, we can introduce an operator pH ∈ L (H 1 (0, T ; H 1 (Ω)), H 1 (0, T ; H 1 (Ω)/R)) and its continuous right inverse qH , ‘extending’ p and q to spaces of time-dependent functions. It is easy to see that Ker pH = H 1 (0, T ). Indeed, suppose that u ∈ H 1 (0, T ; H 1 (Ω)) and pH u = 0 in H 1 (0, T ; H 1 (Ω)/R); then for almost all t ∈ [0, T ] we have pu(t) = 0, hence u(t) is equal to a constant in H 1 (Ω) for almost all t ∈ [0, T ], from which we can infer that u ∈ H 1 (0, T ). Now we can find the connection between the two problems; we state this result as a lemma. Lemma 6.2. The operator pH induces an isomorphism of the Banach spaces H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) and H 1 (0, T ; H 1 (Ω)/R), under which the solutions of problem (6.8), (6.9) precisely correspond to the solutions of problem (6.11), (6.12). ˇ = pH u ˜∈ Proof. Since Ker pH = H 1 (0, T ), for any u ∈ H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) we have well-defined u ˜ any representative of u. The inverse map sends u ˇ ∈ H 1 (0, T ; H 1 (Ω)/R) to the H 1 (0, T ; H 1 (Ω)/R) with u equivalence class of qH u ˇ in H 1 (0, T ; H 1 (Ω))/H 1 (0, T ): indeed, pH qH u ˇ=u ˇ and qH pH u ˜−u ˜ ∈ Ker pH , since pH (qH pH u ˜−u ˜ ) = pH u ˜ − pH u ˜ = 0. It is easy to see that both maps are continuous, thus we have an isomorphism of Banach spaces. Observing that grad (pH u) = grad u (in L2 (0, T ; L2 (Ω))) and pH u|t=0 = u|t=0 , we deduce that the constructed isomorphism provides a bijection between the solutions of problem (6.8), (6.9) and the solutions of problem (6.11), (6.12). 2 As we have shown before, problem (6.11), (6.12) admits a unique solution; in view of Lemma 6.2 this means that so does problem (6.8), (6.9). The estimate (6.10) easily follows from (6.13), the continuity of qH and the definition of the quotient norm on L2 (0, T ; H 1 (Ω))/L2 (0, T ). 2 Following the outline of the previous section, we can formulate the steady-state Neumann problem and its variational formulation and prove the stationary counterparts of Proposition 6.1 and Proposition 6.2. In this case we require that φ ∈ H 1 (Ω)/R. As Lemma 6.2 allows us to regard solutions of (6.8), (6.9) lying in H 1 (0, T ; H 1 (Ω))/H 1 (0, T ) as elements of the space H 1 (0, T ; H 1 (Ω)/R), we can ‘extend’ the time-dependent Neumann problem to the unbounded interval (0, ∞) and prove the corresponding stabilisation theorem, similar to Proposition 5.5. All the statements and proofs in this case are completely analogous to those corresponding to the Dirichlet problem. 7. Remarks concerning the mixed problem In some physical problems it seems natural to specify the total current on a certain part of the boundary of Ω (e.g., only under thunderstorms or only at the outer boundary); this leads us to the mixed problem

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for the GEC equations, in which Dirichlet and Neumann boundary conditions are established on disjoint parts of the boundary. Since this case can be analysed in the same manner as those discussed above, here we restrict ourselves to a few remarks, omitting the proofs and details. Supposing that the boundary Γ of Ω is partitioned into two distinct parts, a closed set Γ(1) and its (l) complement Γ(2) = Γ \ Γ(1) , we also introduce the sets Γk = Γk ∩ Γ(l) for k, l = 1, 2; we assume that each of them is either the empty set or a set with non-empty interior, and we also assume that the common boundary of Γ(1) and Γ(2) consists of a finite number of Lipschitz curves. Following the approach presented, e.g., in [9, Ch. VII, §2], we define the space H 1/2 (Γ(1) ) consisting of 1/2 the restrictions of functions from H 1/2 (Γ) to Γ(1) and the space H00 (Γ(2) ) of the functions in L2 (Γ(2) ) whose extensions by zero to the entire Γ lie in H 1/2 (Γ); these spaces are equipped with the norms uH 1/2 (Γ(1) ) =

inf

w∈H 1/2 (Γ) P w=u

wH 1/2 (Γ) ,

uH 1/2 (Γ(2) ) = ZuH 1/2 (Γ) , 00

where P w means the restriction of w to Γ(1) and Zu stands for the function equal to zero on Γ(1) and equal to u on Γ(2) . Generalising the theory of Section 3, we can define an operator δ from H 1 (Ω) to H 1/2 (Γ(1) ) by sending 1/2 ∗ u ∈ H 1 (Ω) to δu = P γu and an operator δn from H(div, Ω) to (H00 (Γ(2) )) by sending u ∈ H(div, Ω) to δn u = Z ∗ γn u (for u ∈ D (Ω) and u ∈ {D (Ω)}3 the operators δ and δn coincide with simple restrictions u → u|Γ(1) and u → u · n|Γ(2) ). Since δ is clearly continuous and surjective and considering that H 1 (Ω) is a Hilbert space, we can show that it has a continuous right inverse ν. After that we can prove the analogues of Lemma 3.1 and Lemma 3.2, ‘extending’ δ, ν and δn to spaces of time-dependent functions. In the case of the mixed problem we suppose that the potential is specified on Γ(1) and the normal component of the total current density is specified on Γ(2) . The boundary conditions and equivalent variational problems are slightly different depending on how the boundary components are partitioned. (1) (1) (2) (2) In the most complicated case, where all of Γ1 , Γ2 , Γ1 and Γ2 are non-empty, we consider the equations (5.2) and (5.3) for φ ∈ H 1 (0, T ; H 1 (Ω)) with boundary conditions12 φ|Γ(1) = φ1 , ∃U ∈ H 1 (0, T ) : φ|Γ(1) = φ2 + U ⊗ 1, (7.1) 1 2     1 ∂φ ∂φ 1 − grad − σ grad φ + js · n grad − σ grad φ + js · n = j1 , − = j2 , (2) (2) 4π ∂t 4π ∂t Γ Γ 1

1/2

(2)

2

∗

(1)

where jk ∈ L2 (0, T ; (H00 (Γk )) ) and φk ∈ H 1 (0, T ; H 1/2 (Γk )) for k = 1, 2, and with suitable initial condition. Proceeding as in the case of the Dirichlet problem, we can show that the equivalent variational problem consists of the identity d (grad φ, grad ψ) + 4π (σ grad φ, grad ψ) = 4π (js , grad ψ) dt     − 4π j1 , ψ|Γ(2) (2) − 4π j2 , ψ|Γ(2) − [ ψ|Γ(1) ] (2) 1

Γ1

2

2

Γ2

∀ψ ∈ VM (Ω)

together with conditions (7.1) and the initial condition; here we have introduced the space13 VM (Ω) = {u ∈ H 1 (Ω) : u|Γ(1) = 0, ∃c ∈ R : u|Γ(1) = c} 1

2

(1)

and have denoted by [ ψ|Γ(1) ] the constant to which ψ|Γ(1) is equal almost everywhere on Γ2 . 2

12 13

2

As usual, we write u|Γ(1) and u · n|Γ(2) instead of using δ, δn and their extensions. Here ‘M’ stands for ‘mixed problem’.

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Without much difficulty we can prove the counterparts of all the propositions that we have proved for (1) (1) (2) (2) the Dirichlet (or Neumann) problem. Other cases, where one or two of Γ1 , Γ2 , Γ1 and Γ2 are empty, can be analysed in similar fashion. 8. The boundary conditions relating the potential and current at magnetic conjugate points Apart from specifying the potential or the normal component of the total current density at the boundary of the atmosphere, there is one more type of boundary conditions occasionally employed at the upper atmospheric boundary in GEC models [13,22]. It is a non-standard boundary condition, which reflects the fact that at high altitudes the conductivity tensor components (see (4.7)) satisfy σP , σH  σ , and thus it is reasonable to assume that the current flows along the geomagnetic field lines. If the outer boundary Γ2 of the model atmosphere is symmetric with respect to the geomagnetic equator, each pair of symmetric points on Γ2 lies on the same geomagnetic field line; such pairs are termed magnetic conjugate points (see Fig. 1b). A natural way to establish boundary conditions on Γ2 is to demand that each pair of magnetic conjugate points x, x∗ be at the same potential (as the magnetic field lines are nearly equipotential) and the current flowing outwards along the field line at x be equal to the current flowing inwards along the same field line at x∗ ; but since Γ2 is usually chosen such that at relatively high latitudes the geomagnetic field lines intersect it nearly at right angles, the latter of these requirements is replaced with the condition that at magnetic conjugate points the normal components of the current density be equal with opposite signs. This approximation works well except for the geomagnetic equatorial region, which is relatively small. Roughly speaking, the usual approach is to assume that Γ2 is symmetric and to require that the distribution of the electric potential on Γ2 be symmetric and the distribution of the normal component of the current density be antisymmetric. Here we shall study a slightly more general situation, assuming that at each pair of magnetic conjugate points we know the difference of the potentials and the sum of the normal current densities. As for the other boundary component Γ1 , we assume there a Dirichlet boundary condition; the case of a Neumann or mixed boundary condition on Γ1 can be analysed in similar manner. Until now we only assumed that Γ is Lipschitz and consists of two connected components Γ1 and Γ2 ; throughout this section we also assume that Γ2 is symmetric with respect to a certain plane intersecting Ω and there is a reflection ρ : R3 → R3 such that ρ(Γ2 ) = Γ2 . Obviously, ρ induces the map ρ∗ ∈ L (H 1/2 (Γ2 ), H 1/2 (Γ2 )) defined by ρ∗ u = u ◦ ρ, and its dual ρ∗ ∈ L (H −1/2 (Γ2 ), H −1/2 (Γ2 )) is given by ρ∗ f, uΓ2 = f, u ◦ ρΓ2 . As usual, we consider the time-dependent case. Let us suppose that the conductivity σ satisfies (4.1)–(4.4), js ∈ L2 (0, T ; H(div, Ω)), let φ2 ∈ H 1 (0, T ; H 1/2 (Γ1 )) and j2 ∈ L2 (0, T ; H −1/2 (Γ2 )) satisfy the conditions ρ∗ φ2 = −φ2 ,

ρ∗ j2 = j2 ,

j2 , 1Γ2 = 0,

(8.1)

(id − ρ∗ ) φ0 |Γ2 = φ2 |t=0 .

(8.2)

and let φ1 ∈ H 1 (0, T ; H 1/2 (Γ1 )) and φ0 ∈ H 1 (Ω) be such that φ0 |Γ1 = φ1 |t=0 ,

We consider the following problem for φ ∈ H 1 (0, T ; H 1 (Ω)):   ∂φ + 4πσ grad φ = 4π div js , div grad ∂t

    ∂φ + 4πσ grad φ · n , 1 = 4π js · n|Γ1 , 1 Γ , grad 1 ∂t Γ1

(8.3) (8.4)

Γ1

φ|Γ1 = φ1 ,

(id − ρ∗ ) φ|Γ2 = φ2 ,

(8.5)

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∗



(id + ρ )

∂φ 1 grad − σ grad φ + js − 4π ∂t



· n

133

= j2 ,

(8.6)

Γ2

φ|t=0 = φ0 .

(8.7)

From ρ2 = id it follows that ρ2∗ = id and (ρ∗ )2 = id as well; therefore the first two conditions in (8.1) are necessary owing to (8.5) and (8.6). The third condition in (8.1) makes (8.4) and (8.6) consistent, as Green’s formula, as usual, yields (5.7). Let us introduce the space14 VMC (Ω) = {u ∈ H 1 (Ω) : u|Γ1 = 0, ρ∗ u|Γ2 = u|Γ2 }, which is obviously a closed subspace of H 1 (Ω). With the help of Green’s formula (5.6), we obtain from (8.3) that   ∂φ grad + 4πσ grad φ, grad ψ = 4π (js , grad ψ) ∂t

  ∂φ s + 4πσ grad φ − 4πj · n , ψ|Γ2 + grad ∂t Γ2

(8.8) ∀ψ ∈ VMC (Ω).

Γ2

For brevity, we denote j=−

∂φ 1 grad − σ grad φ + js , 4π ∂t

and for any ψ ∈ VMC (Ω) we observe, by virtue of (8.6), that 

j · n|Γ2 , ψ|Γ2

 Γ2

    = j2 , ψ|Γ2 Γ − ρ∗ j · n|Γ2 , ψ|Γ2 Γ 2 2         = j2 , ψ|Γ2 Γ − j · n|Γ2 , ρ∗ ψ|Γ2 Γ = j2 , ψ|Γ2 Γ − j · n|Γ2 , ψ|Γ2 Γ , 2

2

2

2

whence 

j · n|Γ2 , ψ|Γ2

 Γ2

=

 1 j2 , ψ|Γ2 Γ . 2 2

(8.9)

Substituting it into (8.8) and using Lemma 5.1, we arrive at the following variational problem: d (grad φ, grad ψ) + 4π (σ grad φ, grad ψ) dt   = 4π (js , grad ψ) − 2π j2 , ψ|Γ2 Γ 2

φ|Γ1 = φ1 ,

(8.10) ∀ψ ∈ VMC (Ω),

(id − ρ∗ ) φ|Γ2 = φ2 , φ|t=0 = φ0 .

(8.11) (8.12)

Proposition 8.1. Suppose that js ∈ L2 (0, T ; H(div, Ω)) and conditions (4.1)–(4.4) and (8.1) are satisfied. Then for φ ∈ H 1 (0, T ; H 1 (Ω)) problem (8.3)–(8.7) is equivalent to problem (8.10)–(8.12). 14 Here ‘MC’ stands for ‘magnetic conjugate’, implying that the potential and current density at magnetic conjugate points are related through boundary conditions.

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Proof. As we have just derived (8.10)–(8.12) from (8.3)–(8.7), we only need to prove the converse. Suppose that (8.10) holds; with the help of Green’s formula we obtain (8.3) and (8.4) in the usual manner, so it remains to prove (8.6). Using Green’s formula once again, we observe that (8.9) holds for any ψ ∈ VMC (Ω); as for any u ∈ H 1/2 (Γ2 ) u + ρ∗ u is a trace of a certain function in VMC (Ω), we deduce from (8.9) that 

(id + ρ∗ ) j · n|Γ2 , u

 Γ2

  1 = j · n|Γ2 , (id + ρ∗ ) u Γ = j2 , (id + ρ∗ ) uΓ2 = j2 , uΓ2 2 2

owing to the second condition in (8.1). 2 Proposition 8.2. Suppose that js ∈ L2 (0, T ; L2 (Ω)), conditions (4.1)–(4.4) are satisfied, and φ1 ∈ L2 (0, T ; H 1/2 (Γ1 )), φ2 ∈ L2 (0, T ; H 1/2 (Γ2 )), j2 ∈ L2 (0, T ; H −1/2 (Γ2 )) and φ0 ∈ H 1 (Ω) meet conditions (8.1) and (8.2). Then there exists a unique solution φ ∈ H 1 (0, T ; H 1 (Ω)) to problem (8.10)–(8.12), satisfying  φL2 (0, T ; H 1 (Ω)) ≤ const · js L2 (0, T ; L2 (Ω)) + φ1 H 1 (0, T ; H 1/2 (Γ1 ))

 + φ2 H 1 (0, T ; H 1/2 (Γ2 )) + j2 H 1 (0, T ; H −1/2 (Γ2 )) + φ0 H 1 (Ω) ,

where the constant is independent of js , φ1 , φ2 , j2 and φ0 . Proof. We construct φΓ ∈ H 1 (0, T ; H 1 (Ω)) such that φΓ |Γ1 = φ1 ,

φΓ |Γ2 =

1 φ2 . 2

Setting χ = φ − φΓ , using Lemma 5.1 and observing that (id − ρ∗ )φ2 = 2φ2 because of the first condition in (8.1), we obtain from (8.10)–(8.12) the following problem for χ ∈ H 1 (0, T ; VMC (Ω)): d (grad χ, grad ψ) + 4π (σ grad χ, grad ψ) dt     ∂φΓ s = 4πj − grad − 4πσ grad φΓ , grad ψ − 2π j2 , ψ|Γ2 Γ 2 ∂t

∀ψ ∈ VMC (Ω),

χ|t=0 = φ0 − φΓ |t=0 , where, owing to (8.2), the right-hand side of the last equation lies in VMC (Ω). The rest of the proof is carried out similarly to that of Proposition 5.2 with the help of Lemma 5.2. 2 In the usual way we can prove the analogues of Proposition 8.1 and Proposition 8.2 for the corresponding steady-state problem, as well as the stabilisation theorem. Acknowledgments The authors are thankful to E.A. Mareev for valuable discussion regarding physical applications of this research. The work was supported by a grant from the Government of the Russian Federation (contract no. 14.B25.31.0023), by the Ministry of Education and Science of the Russian Federation (projects no. 2664 and no. 1727 of the state contract for 2014–2016 and a grant within the agreement no. 02.B.49.21.0003 of 27 August 2013), by the Federal Agency of Scientific Organisations (agreement no. 007-GZ/C3541/35 of 22 January 2016) and by RFBR grant no. 16-05-01086.

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