General Magneto-Static Model
Xing-Bin Pan
School of Mathematics,
East China Normal University, and NYU-ECNU Institute of Mathematical Sciences at NYU Shanghai,
Shanghai 200062, P.R. China. [email protected]
Abstract.
In this paper we study a nonlinear magneto-static model on a general domain which is multiply-connected and has m holes, and under a nonlinear relation between magnetic induction B and magnetic field H. The equation contains a Neumann field h1∈H1(Ω) and a Dirichlet field h2∈H2(Ω), which represent the effects of domain topology. For a general electric current, the equation contains an unknown gradient ∇p(x), which represents the electric field. Existence results of solutions of boundary value problems of this model under various types of boundary conditions are proved, which exhibit the effects of domain topology.
Key words and phrases:
magneto-static model; quasilinear curl system; domain topology; weak solution; existence
2010 Mathematics Subject Classification:
35J61; 35J62; 35Q60; 35Q61; 78A25
1. Introduction
In this paper we study a nonlinear magneto-static model on a general domain in R3, which is multiply-connected and has m holes (namely the boundary of ∂Ω consists of m+1 connected components). We are interested in the effects of domain topology on solvability of the boundary value problems.
In the recent years quasilinear magneto-static models in electromagnetism have been studied by many authors. A typical equation is the quasilinear Maxwell type equation of the following form:
[TABLE]
where J is a given vector field. This system is a time-independent version of the eddy current model for a magnetic and anisotropic material where the relation between magnetic induction field B and the magnetic field H is given by the BH-curve H=H(B) which is usually nonlinear,
see For instance, Milani and Picard [MP1, MP2], R. Picard [Pi], Jochmann [Jo], F. Bachinger, U. Langer and J. Schöberl [BLS], Jiang and Zheng [JZ1, JZ2], L. Yousept [Yo], Pan [P4], and the references therein.
It was shown in [P4] that for a system similar to (1.1) with J replaced by a nonlinear vector-valued function f(x,u), it is in general necessary to introduce a potential term into the equation and one needs to consider a Maxwell-Stokes type system
[TABLE]
It was also shown in [P4] that the type of boundary condition for p should be determined according to the topology of Ω. If Ω has no holes, then it is natural to consider Dirichlet boundary condition for p. If Ω has holes, then one has to consider Neumann type boundary condition for p. In this paper we will explain the physical significance of the term ∇p.
In this paper we assume that the relation between magnetic induction B and magnetic field intensity H is nonlinear:
[TABLE]
To avoid the complexity of electric part, we assume that the relation between electric displacement D and electric field intensity E is linear: D=εE, where ε is scalar or matric valued function. Finally, we shall assume the electric current density j satisfies the Ohm’s law. We shall introduce a vector field u and a scalar function p, such that the electric field E and magnetic induction B can be represented by
[TABLE]
where h1∈H1(Ω), h2∈H2(Ω), which describe the topology of the domain Ω. Then we can derive the magneto-static model. If the electric current is given, then the equation is
[TABLE]
and if the electric current depends on the magnetic induction, then the equation is
[TABLE]
We shall examine solvability of (1.4) subjected to a boundary condition on u. For (1.5) we need also a boundary condition for p. Following the observation in [P4], we shall consider Neumann type boundary condition for p. Regularity of weak solutions of BVPs of (1.4) and (1.5) can be obtained by the methods in [P4] and will not be discussed in this paper.
We mention that the gradient term ∇p in (1.5) has a clear physical meaning. In fact, from (1.3) we see that
[TABLE]
so ∇p represents both electric field E and domain topology.
We also mention that both (1.4) and (1.5) contain the topological terms h1,h2, and we will see that solvability of the boundary value problems (BVPs for short) for (1.4) and (1.5) depends on the choice of h1,h2, which is an effect of domain topology on these equations.
In this paper we shall use various results on vector fields, including the divergence-curl-gradient inequalities, and regularity of div-curl system and of time-independent Maxwell’s system, which can be fund in literature, see for instance [DaL, GR, MMT, Sc, W, BW, NW, AS, AuA, CDN, KY] and the references therein.
We mention that time-independent semilinear Maxwell system has also been studied by many authors, see for instance [Y, BF, JS, P3] and the references therein. We also mention that the Meissner states of type II superconductors can be described by a quasilinear system of the form similar to (1.1), see [Ch, Mon, BaP, P1, LP].
This paper is organized as follows. In section 2 we list notation for various spaces of vector fields, and collect some facts on these spaces that we will need. In section 3 we derive the magneto-static equations. Sections 4, 5, and 6 are devoted to study of existence of weak solutions to BVPs of (1.4) under various type of boundary conditions on u. Solvability of BVPs for (1.5) is studied in section 7.
2. Preliminaries
2.1. Notation
Most of materials in this subsection can be fund in [DaL, GR].
Let Ω be a bounded domain in R3 and ν be the unit outer normal vector of the boundary ∂Ω.
We assume that Ω has the following properties ([DaL, p.217]):
Ω is a bounded domain in R3 with a Cr boundary
∂Ω, r≥1; Ω is locally situated on one side of
∂Ω; ∂Ω has a finite number of connected components
Γ1,⋯,Γm+1, where m≥0 and
Γm+1 denoting the boundary of the unbounded connected
component of the set R3∖Ωˉ.
There exist N manifolds of dimension 2 and of class Cr
denoted by Σ1,⋯,ΣN, N≥0, such that
Σi∩Σj=∅ for i=j and non-tangential
to ∂Ω, such that Ω˙=Ω∖(∑j=1NΣj) is simply-connected and Lipschitz.
2.1.1. The spaces of vector fields
As in [BaP, P1, P4], we denote the Sobolev spaces of scalar functions by Wk,p(Ω), Hk(Ω) and Hs(∂Ω) etc, and denote the Sobolev spaces of vector fields by Wk,p(Ω,R3), Hk(Ω,R3), Hs(∂Ω,R3) etc.
We use same notation for the norm of scalar functions and for vector fields. For instance, the norms in Hk(Ω) and in Hk(Ω,R3) are both denoted by ∥⋅∥Hk(Ω).
We denote
[TABLE]
It is well-known that dimH1(Ω)=N and dimH2(Ω)=m, where m and N are given in (O1) and (O2). We denote by Hj(Ω)L2(Ω)⊥ the orthogonal complementary of Hj(Ω) in L2(Ω,R3). When ∂Ω is of C2, using regularity of div-curl system we have
[TABLE]
If X(Ω) denotes a space of vector fields on Ω, we write
[TABLE]
For instance,
[TABLE]
If Y(Ω) is a space of scalar functions on Ω, we denote
[TABLE]
Denote
[TABLE]
We also need spaces of tangential vector fields:
[TABLE]
We denote
[TABLE]
2.1.2. Decomposition of L2(Ω,R3)
Let us recall that
(see [DaL, p.225-226]):
[TABLE]
and recall image of operator curl (see [DaL, p.222, Proposition 3; p.226, Remark 5]):
[TABLE]
where
[TABLE]
These yield the following decompositions of the kernel of the operators curl and div :
[TABLE]
It follows that H1(Ω)⊂HΓ(Ω,div0) and H2(Ω)⊥L2(Ω)H1(Ω).
From the first equality of (2.3) and the first equality of (2.5) we have
[TABLE]
From the second equality of (2.3) and the second equality of (2.5) we have
[TABLE]
2.2. Assumptions on H(x,z)
When Ω is bounded, we say H∈Clock,α(Ωˉ×R3,R3) if H∈Clock,α(Ωˉ×K,R3) for any bounded set K⊂R3.
Now we list the conditions on H(x,z) that we may need.
H∈Cloc0(Ωˉ×R3,R3), and there exist positive constants c1,c2, and functions g1∈L1(Ω), g2∈L2(Ω) such that, for all x∈Ωˉ and z∈R3,
[TABLE]
H∈Cloc1(Ωˉ×R3,R3), and there exist positive constants μ,M1,M2 such that for all x∈Ωˉ and z,ξ∈R3,
[TABLE]
H(x,z) has an inverse B(x,w), namely,
z=B(x,w) if and only if w=H(x,z).
Remark 2.1**.**
If H satisfies (H2) then it is Lipschitz in z, and has the following property:
There exists a constant μ>0 such that
[TABLE]
We can show that (H1),(H2),(H3) imply the following properties of B:
B∈Cloc0(Ωˉ×R3,R3), and there exist positive constants c3,c4, and functions g3∈L1(Ω), g4∈L2(Ω) such that, for all x∈Ωˉ and w∈R3,
[TABLE]
B∈Cloc1(Ωˉ×R3), and there exist positive constants λ0,N1,N2 such that for all x∈Ωˉ and w,η∈R3,
[TABLE]
Remark 2.2**.**
If H satisfies (H1), (H3), then B satisfies (B1), with
[TABLE]
If H satisfies (H2), (H3), then B satisfies (B2), where λ0 and N2 depend on μ and M2, N1 depends on μ,M1,M2. In fact,
[TABLE]
where C0 is an absolute constant.
Remark 2.3**.**
If (B2) holds, then for any w1,w2∈T3 and x∈Ωˉ,
[TABLE]
2.3. Assumption on f(x,z)
f∈Cloc1(Ωˉ×R3,R3), and there exist positive constants K0,K1,K2 and f0∈L2(Ω) such that, for all x∈Ωˉ and z∈R3,
[TABLE]
2.4. The Projections Pν and Pn
Definition 2.4**.**
We introduce the projection Pν as follows. If w∈L2(Ω,R3), then Pν[w]=w+∇ϕw, where ϕw∈H˙1(Ω) is the unique weak solution of
[TABLE]
namely, for any ϕ∈H1(Ω) and
[TABLE]
Remark 2.5**.**
Assume Ω is abounded domain in R3 with a C2 boundary. Then
Pν:L2(Ω,R3)↦HΓ(Ω,div0)
is a continuous projection.
Definition 2.6**.**
Assume ν×H0 satisfies (6.8). We introduce the Neumann projection Pn associated with ν⋅curlHT0 as follows. If w∈L2(Ω,R3), then Pn[w]=w+∇ψw, where ψw∈H˙1(Ω) is the unique weak solution of
[TABLE]
namely, for all η∈H1(Ω) it holds that
[TABLE]
Lemma 2.7**.**
Assume Ω is a bounded domain in R3 with a C2 boundary, and ν×H0 satisfies (6.8). Then
Pn:L2(Ω,R3)→HΓ(Ω,div0)
is a continuous projection.
Proof.
Denote by H0 the divergence-free and tangential component preserving extension on HT0. For any w∈L2(Ω,R3), Pn[w]=w+∇ψw∈H(Ω,div0). Hence ν⋅Pn[w]∈H−1/2(∂Ω). Denote the connected components of ∂Ω by Γj, j=1,⋯,m+1, and let ηj denote the harmonic function in Ω such that ηj=δij on Γi. Then ν×∇ηj=0 on ∂Ω. So
[TABLE]
Hence Pn[w]∈HΓ(Ω,div0). Taking ψw as a test function we have
[TABLE]
So
[TABLE]
Using the trace theorem and applying the Poincaré inequality to ψ∈H˙1(Ω), we get
[TABLE]
∎
3. Derivation of the Magneto-static Equations
3.1. Time-dependent problem: quasi-static magnetic fields
3.1.1. The Maxwell equations and our assumptions
We start with the general case of magnetic and anisotropic materials, occupying a bounded C2 domain in R3, which is multiply-connected and has holes. The operators div , curl , ∇ and Δ act on the space variables x.
The Maxwell equations are of the following form
[TABLE]
where E is the electric field, B is the magnetic induction, H is the magnetic field, j is the current density, ρ is the charge density, and ε is the permittivity.111From the first and third equations in (3.1) we get
∂tρ+divj=0,
which is the conservation law of electric charge.
Assumption 3.1**.**
(Assumption on the H-B relation.) There exists a nonlinear vector-valued function B(x,z) such that
magnetic induction B can be represented by magnetic field H:
[TABLE]
B(x,⋅)* has an inverse H(x,⋅), which means that*
[TABLE]
Assumption 3.2**.**
(Assumption on the magnetic induction B.)
The magnetic induction B(t,x) is such that
[TABLE]
Assumption 3.3**.**
(Assumption on the electric field E.)
The electric field E(t,x) is such that
[TABLE]
Later on we shall assume also the Ohm’s law holds (see for instance [LL, p.199-120]):
Assumption 3.4**.**
(Assumption on the current density j.) The current density j satisfies the Ohm’s law
[TABLE]
where σ≥0 is the electric conductivity, and ja is the applied current density.
3.1.2. Rewrite (3.1) in term of E and a vector potential A
Step 1. General case.
From (3.1)(b) we have divB=0.
From the first line of (2.4) and second line of (2.5) we see that there exist vector fields w(t,x) and h2(t,x), with w(t,⋅)∈Hn01(Ω,div0) and h2(t,⋅)∈H2(Ω), such that
[TABLE]
Since H2(Ω) is of finite dimensions, and curlw(t,⋅) is orthogonal to H2(Ω), it follows that
[TABLE]
From (3.1)(d) and (3.7) we have
[TABLE]
From this, (3.5) and (3.8) we have, for any t>0,
[TABLE]
From this and the last equality in (3.8) we have
∂th2=0, so curl(E+∂tw)=−∂th2=0. By the first line of (2.5), there exists a scalar function ξ(t,x) and a vector field h~1(t,x), with ξ(t,⋅)∈H1(Ω), h~1(t,⋅)∈H1(Ω), such that
[TABLE]
Using this, (3.5) and (3.8), and since H1(Ω) is of finite dimensions and ∇ξ(t,⋅) is orthogonal to H1(Ω), we can show that
[TABLE]
Introduce a vector field
[TABLE]
From the properties of w and ξ we see that
[TABLE]
Now we can represent E,B,H by A (see (3.9), (3.7), (3.3)):
[TABLE]
Plugging the relation H=H(x,B) into (3.1) (c)
we get
[TABLE]
Finally from (3.1)(a) and (3.11) we have
[TABLE]
Step 2. Assume the Ohm’s law (3.6). Then from (3.1) (c) and (3.11) we have
[TABLE]
Plugging the relation H=H(x,B) into the above equality
we get
[TABLE]
3.1.3. Quasi-static approximation
Next we consider a quasi-static approximation of the Maxwell equations (3.1) (in the form of (3.12)) by neglecting the displacement current ∂t(εE).
Step 1. General case. From (3.12) with the displacement current neglected we have:
[TABLE]
We introduce a function Φ such that for t≥0,
[TABLE]
Set
[TABLE]
From (3.10) we have
[TABLE]
Now Eq. (3.15) is transferred to
[TABLE]
here u and j may depend on t. Eq. (3.13) takes the form
[TABLE]
Step 2.
Assume the Ohm’s law (3.6). Then a quasi-static approximation of the Maxwell equations (3.1) (in the form of (3.14)) gives
[TABLE]
After introducing Φ(t,x) by (3.16), and introducing u,u0,q by (3.17), we have (3.19), and
(3.20) is transferred to
[TABLE]
In the special case when σ and ε are positive constants, from (3.19) and (3.21) we get
[TABLE]
3.2. Time-independent problems: magneto-static fields
Step 1. Assume that
[TABLE]
Then the steady state problem of (3.18) takes the following form:
[TABLE]
This equation is different to (3.18) in the sense that in (3.24) u and j are independent of t.
(3.24) should be coupled with (3.19), which takes the form
[TABLE]
Step 2. Assume the Ohm’s law (3.6), and assume
[TABLE]
Let us introduce p,h1 by
[TABLE]
Since ∂tΦ=−q(x)=−p(x)/σ, there exists a function Φ0(x) such that
[TABLE]
Now the relations between A,E,B,H,q and u,h1,h2,Φ0,p are the following:
[TABLE]
The steady state problem of (3.21) takes the following form:
[TABLE]
Now the Gauss’s law of electricity (3.19) is reduced to (3.22).
[TABLE]
and (3.1) is reduced to (3.29)-(3.30). ρ should not be arbitrarily given, instead, it is a function determined by h1 and p.
So we conclude that:
Lemma 3.5**.**
Assume the assumptions 3.1, 3.2, 3.3.
After neglecting the displacement current ∂t(εE) from the Maxwell equations (3.1), and introducing the magnetic potential u, and assume u and j are independent of t, then (3.1) is reduced to (3.24)-(3.25).
Assume furthermore the Ohm’s law (3.6) holds, σ is a positive constant, and assume u,h~1,js,q,ε,ρ are independent of t.
Solution (E,B,H) of (3.1) is represented by (3.28), and (3.1) is reduced to (3.29)-(3.30).
If ja and ρ are independent of u, then ρ is determined by ja by (3.22), and the magneto-static problem is reduced to a single system (3.29).
Remark 3.6**.**
Assume the Ohm’s law (3.6) and assume (3.26).
Equation (3.29) indicates the relation between magnetic field H, electric field E, and electric current j=σE+ja:
[TABLE]
(About the topological effects on electromagnetism.) (3.28) and (3.29) show that the electromagnetic fields depend on h1,h2 which represents topology of the domain. (3.28) indicates that E contains h1 and B contains h2 explicitly, which suggests that E is more directly linked to simply-connectedness of Ω, and H is more directly linked to connectedness of ∂Ω.
(About the physical meaning of ∇p in (3.29).) Mathematically we may say that, the gradient term ∇p is necessarily introduced into the equation in (3.29) to balance the equality, which is necessary for solvability of the boundary value problem. On the other hand, from (3.28) and (3.6) we see that
[TABLE]
so ∇p represents both electric field E and domain topology.
Remark 3.7**.**
In [P4] we considered the following system
[TABLE]
This is a particular case of (3.29) without the topological effect terms h1,h2, and without the potential term ∇p, where ja is either given or ja=f(x,u).
It was shown in [P4] that (3.31) under some Dirichlet boundary condition may not be well-posed, and a potential term should be introduced to balance the equation:
[TABLE]
The boundary condition for p considered in [P4] is Dirichlet type if Ω has no holes, and Neumann type if Ω has holes. Now we can say that ∇p gives the electric field E, hence a Dirichlet boundary condition on p is to prescribe the tangential component of E, and a Neumann boundary condition on p is to prescribe the normal component of E.
(3.29) is a better model than (3.32) in the following sense. (3.32) says that magnetic field (described by u) and electric field (described by ∇p) depend on each other, (3.29) says that magnetic field (described by u) and electric field (described by ∇p and h1) depend on each other, and both depend on domain topology (described by h1,h2).
3.3. BVPs where the current is given
Assume ja(x) is given and independent of u,h1,h2.
As mentioned in Lemma 3.5, we should assume ρ is determined by p,h1 by (3.30).
Then (3.1) is reduced to a single system (3.29).
If we write ja(x)=J(x),
then the first equation in (3.29) can be written as follows:
[TABLE]
Under a suitable boundary value condition, the unknown p(x) can be solved independent of u. In this case, replacing J+∇p by J, the equation is in the form of (1.4).
Then we may pose one of the following boundary conditions for u on ∂Ω:
Dirichlet boundary condition uT=uT0;
Tangential curl boundary condition ν×curlu=ν×B0;
Normal curl boundary condition ν⋅curlu+ν⋅h2(x)=Bn0;
Natural boundary condition ν×H(x,curlu+h2(x))=ν×H0;
Co-normal boundary condition ν⋅H(x,curlu+h2(x))=Hn0;
3.4. BVPs where the current depends on magnetic induction
Assume ja depends on the magnetic induction B=curlu+h2.
Then we may write ja=f(x,curlu+h2),
and (3.29) takes the form of (1.5).
We may pose one of the above mentioned boundary conditions for u, and either Dirichlet or Neumann condition for p. Other type boundary conditions are also possible. However, following the discussions in [P4], we have the following observations.
If Ω has no holes, then BVP of (1.5) with Dirichlet boundary condition for p (and with a suitable boundary condition on u) is well-posed.
If Ω has holes, then BVP of (1.5) with Neumann boundary condition for p (and with a suitable boundary condition on u) is well-posed, but the problem with Dirichlet boundary condition for p is not well-posed.
4. Dirichlet BVP of the Maxwell System with a Given Current
In this section we consider the Dirichlet BVP of the Maxwell system with a given current J:
[TABLE]
In (4.1) there are no boundary conditions for h1 and h2, and we may view h1,h2 as parameters. If u is a solution of (4.1), then for any h∈H2(Ω), u+h is also a solution of (4.1). To avoid non-uniqueness due to H2(Ω), we may require the solution u⊥L2(Ω)H2(Ω).
As in [P3, Lemma 2.1] we can show that the condition J∈HΓ(Ω,div0) is necessary for (4.1) to have a weak solution.
Throughout this section we assume that
[TABLE]
and
[TABLE]
(4.3) implies that there exists a divergence-free and curl-minimizing extension U of uT0, see [P2]. Set b=curlU and v=u−U. Then (4.1) can be written as follows:
[TABLE]
In the following we derive solvability of (4.1) by variational methods, monotone operator methods, and the reduction methods. See Proposition 5.7 in [P4]) for solvability by compact operator methods.
4.1. Solvability of (4.1) by variational methods
Proposition 4.1**.**
Let Ω,J,h1,h2,uT0 satisfy (4.2) and (4.3), H(x,z)=∇zP(x,z) for a function P satisfying
P∈Cloc1,α(Ωˉ×R3), P(x,z) is strictly convex in z, and there exist positive constants c1,c2,c3, functions p1∈L1(Ω) and p2∈L2(Ω) such that
[TABLE]
Then (4.1) has a solution u∈H1(Ω,div0).
Proof.
Condition (P1) here is slightly more general than the condition (P) in [P4].
By the argument in the proof of [P4, Proposition 3.1] we know that the energy functional has a minimizer u in Ht01(Ω,div0)∩H2(Ω)⊥.
Hence for all w∈Ht01(Ω,div0)∩H2(Ω)⊥ it holds that
[TABLE]
This equality remains true for all w∈Ht01(Ω,div0) because J+h1 is perpendicular to H2(Ω).
So we can use a variant form of the De Rham lemma for functionals vanishing on Ht01(Ω,div0) (see [P4, Lemma 2.2] and [P5, Lemma 4.2]) to conclude that there exists p∈L2,−1/2(Ω) such that the equality
[TABLE]
holds as elements in Ht01,∗(Ω,R3), and γ(p)=0 on ∂Ω. This means that
[TABLE]
Let ζ be such that
[TABLE]
Then w=∇ζ∈Ht01(Ω,R3). Plugging it into the above equality we find
[TABLE]
because div(J+h1)=0 in Ω and ζ=0 on ∂Ω. Hence p=0.
So v is a weak solution of (4.4).
∎
Corollary 4.2**.**
Let Ω,J,h1,h2,uT0 satisfy (4.2) and (4.3), H(x,z)=a(x,∣z∣2)z, where a(s,x) satisfies the following condition:
a∈Cloc1(Ωˉ×R3), ∇xa∈Clocα(Ωˉ×R3,R3) with 0<α<1, and there exist positive constants λ,Λ and δ∈(0,1) such that
[TABLE]
Then (4.1) has a weak solution u∈H1(Ω,div0).
Proof.
The proof is similar to the proof of Corollary 3.2 in [P4].
Let
c(t,x)=∫0∣t∣a(s,x)ds, P(x,z)=c(x,∣z∣2).
From (a) we see that P(x,z) satisfies (P1). In particular,
[TABLE]
So P(x,z) is strictly convex in z.
Hence the conclusion follows from Proposition 4.1.
∎
4.2. Solvability of (4.1) by monotone operator methods
Theorem 4.3**.**
Assume Ω,J,h1,h2,uT0 satisfy (4.2), (4.3), and H satisfies (H1),(H2). Then (4.1) has a unique solution u∈H1(Ω,div0)∩H2(Ω)L2(Ω)⊥. Moreover, the solution map (h1,J)↦u is continuous from H1(Ω)×HΓ(Ω,div0) to H1(Ω,div0).
Proof.
Step 1.
Define
[TABLE]
Endowed with the norm in H1(Ω,R3), X is a separable Hilbert space.
By the div-curl-gradient inequalities we know that
[TABLE]
Hence both ∥w∥H1(Ω) and ∥curlw∥L2(Ω) are the equivalent norms in X. Denote by X∗ the dual space of X, and denote by ⟨⋅,⋅⟩X∗,X the paring between X∗ and X.
For any given h2∈H2(Ω), w∈X, we define a functional Ah2,w on X by
[TABLE]
Using the condition (H1) we see that Ah2,w is a bounded and linear functional on X and
[TABLE]
Thus there is an element in X∗, which is denoted by Ah2(w), such that
[TABLE]
The mapping w↦Ah2(w) defines an operator
Ah2:X→X∗, and we have
[TABLE]
From (H2) we know that H(x,z) is Lipschitz in z, hence Ah2 is continuous. On the other hand, (H2) implies (H4), hence
[TABLE]
Hence Ah2 is strongly monotone from X to X∗. By the Browder-Minty theorem (see for instance [Z2, p.557, Theorem 26.A]) we know that Ah2 is surjective.
Step 2. Given any u∈L2(Ω,R3) we can define a functional L[u] on X by
[TABLE]
Then L[u] is a bounded and linear functional on X, and there is an element in X∗, which is denoted by L(u), such that
[TABLE]
Since Ah2 is surjective and L(J+h1)∈X∗, there exists w∈X such that
Ah2(w)=L(J+h1), that is,
[TABLE]
From (4.8) and since HΓ(Ω,div0)⊂H2(Ω)L2(Ω)⊥ we have
[TABLE]
From these and (4.9) we have
[TABLE]
Hence by the De Rham lemma on Ht01(Ω,div0) (see [P4, Lemma 2.2] and [P5, Lemma 4.2]) we know that there exists p∈L02,−1/2(Ω) such that, in the sense of functionals in Ht01(Ω,div0) we have
[TABLE]
Using the argument in the proof of Proposition 4.1 we can show that p=0. Thus w∈Ht01(Ω,div0) is a weak solution of (4.4).
By the strongly monotonicity of Ah2 we know that for any J+h1, the weak solution w is unique, and the map J+h1↦w is continuous.
∎
4.3. Solvability of (4.1) by the reduction method
A description of the reduction method can be found in [P3, P4]. Since J+h1∈HΓ(Ω,div0), there exists j0∈H1(Ω,R3) such that
[TABLE]
such a j0 is unique. The following lemma is obvious.
Lemma 4.4**.**
Assume Ω,J,h1,h2 satisfy (4.2) and (4.3), and j0 is the solution of (4.11).
System (4.4) has a weak solution v∈Ht01(Ω,div0) if and only if there exists ϕ∈H1(Ω) and h1′∈H1(Ω) such that the following system has a solution v:
[TABLE]
If furthermore H satisfies (H3), then (4.12) is equivalent to the following
[TABLE]
Lemma 4.5**.**
Given h1′∈H1(Ω), system (4.13) has a solution v∈Ht01(Ω,div0) if and only if the following equation
[TABLE]
has a solution ϕ∈H1(Ω) which satisfies the following orthogonality condition
[TABLE]
Proof.
Recall that (4.13) has a solution if and only if
[TABLE]
On the other hand, ϕ satisfies (4.14) if and only if
[TABLE]
Here we have used the equality
ν⋅curlU=ν⋅curlUT=ν⋅curluT0.
Hence (4.16) holds if and only if ϕ satisfies (4.14) and
[TABLE]
Since h2∈H2(Ω)⊂H1(Ω)L2(Ω)⊥, the above orthogonality is equivalent to (4.15).
∎
Lemma 4.6**.**
Assume H satisfies (H3) and (H4). Then for any given j0, U and h2, there exists at most one (ϕ,h1′)∈H˙1(Ω)×H1(Ω) such that (4.13) is solvable.
Proof.
Suppose there exist ϕj∈H˙1(Ω) and h1,j′∈H1(Ω), j=1,2, such that (4.13) with ϕ=ϕj and h1′=h1,j′ has a solution vj. We may assume vj∈H2(Ω)L2(Ω)⊥.
Write uj=vj+U. Then
[TABLE]
Multiplying it by curl(u2−u1) and integrating, and using condition (H4) we have
[TABLE]
In the last line we used the fact that ν×u2T=ν×u1T=ν×uT0. So
curl(u2−u1)=0, hence u2−u1∈H2(Ω).
This together with the fact u2−u1=v2−v1∈H2(Ω)L2(Ω)⊥ implies that
u2−u1=0, so v2−v1=0.
Finally, since u2=u1,
[TABLE]
Since h1,2′−h1,1′∈H1(Ω), and ∇(ϕ2−ϕ1)∈∇H1(Ω) which is orthogonal to H1(Ω), we conclude that h1,2′−h1,1′=0 and ∇(ϕ2−ϕ1)=0.
∎
4.3.1. Existence of solutions of (4.14) by the monotone operator method
(4.14) is a quasilinear co-normal problem. Existence and regularity of weak solutions to such type problems have been well studied, see for instance [LU, Chapter 10], [Li, Theorem 11.20].
Proposition 4.7**.**
Assume that Ω,H,h2,uT0 satisfy (4.2) and j0∈H1(Ω,R3).
For any h1′∈H1(Ω), (4.14) has a unique solution ϕ∈H˙1(Ω).
There exists h1′∈H1(Ω) such that (4.14) has a solution ϕ∈H˙1(Ω) which satisfies (4.15).
Proof.
(i). As in the proof of Theorem 4.3, we fix h1′∈H1(Ω) and define a map T such that
[TABLE]
Using condition (B1), (B2) we can show that T:H˙1(Ω)↦H˙1(Ω)∗ is strongly monotone. By the Browder-Minty theorem T is surjective, hence there exists a unique ϕ∈H˙1(Ω) such that, for any η∈H˙1(Ω),
[TABLE]
Then ϕ is the weak solution of (4.14).
(ii). Choose an orthonormal basis {e1,⋯,eN} of H1(Ω), and write
ξ=(ξ1,⋯,ξN)t, h1′=h1,ξ′=∑i=1Nξiei,
and write the unique solution of (4.14) associated with h1,ξ′ by ϕ=ϕξ.
Step 2.1. We show that the map ξ↦ϕξ
is continuous from RN into H˙1(Ω). To prove, let ξ0∈RN and assume ξj→ξ0. Since H1(Ω) is of finite dimensions, and ξj→ξ0, so vj=h1,ξj′−h1,ξ0′→0 in H1(Ω,R3).
Let us denote
[TABLE]
Set ψj=ϕξj−ϕξ0, vj=h1,ξj′−h1,ξ0′, w0=h1,ξ0′+∇ϕξ0.
We have
[TABLE]
Hence ∥∇ψj∥L2(Ω)≤λ0N∥vj∥L2(Ω)→0.
Step 2.2. We can further show that there exists a constant C(Ω), which depends also on the constants in (B1),(B2), such that
[TABLE]
In fact, using the definition of weak solutions to (4.14) we have
[TABLE]
Using (B1) and (B2) we get from the above inequality that
[TABLE]
From this and (B1) we get (4.18).
Step 2.3. We claim that
[TABLE]
Since h1,ξ′∈H1(Ω), so h1,ξ′ and ∇ϕξ are orthogonal to each other with respect to the L2 inner product. So
[TABLE]
So (4.19) is true.
Step 2.4. Using the basis {e1,⋯,eN} of H1(Ω), (4.15) can be written as
[TABLE]
Let us write
[TABLE]
Then (4.20) can be written as
[TABLE]
Since the map ξ↦ϕξ is continuous from RN to H˙1(Ω), and the map
w↦B(x,w(x)) is continuous from L2(Ω,R3) to L2(Ω,R3), so the map
ξ↦B1(h1,ξ′+∇ϕξ)
is continuous from RN to L2(Ω,R3). Thus f(ξ) is continuous from RN to RN.
Step 2.5.
For any ξ∈RN we have
[TABLE]
where B0=B(x,j0), C1=∥B0∥L2(Ω) and C2=∥ν⋅(h2+curluT0)∥H−1/2(∂Ω). From this, (4.18), (4.19) and the Cauchy inequality we get
[TABLE]
Using (4.19) and (4.22) we conclude that there exists R0>0 such that
[TABLE]
It follows that there exists R>0 such that the following acute angle condition is satisfied:
[TABLE]
Here we denote SR={ξ∈RN:∣ξ∣=R} and BR={ξ∈RN:∣ξ∣≤R}. It then follows that there exists ξ0∈BR such that f(ξ0)−c=0 (see Proposition 2.8 of [Z1], and Theorem
1.11 of [G]). This ξ0 solves equation (4.21).
For this ξ0, let ϕξ0 be the solution of (4.14) associated with h1,ξ0′. Then (h1,ξ0′,ϕξ0) solves problem (4.14)-(4.15).
∎
Summarizing the above discussions we get the following
Theorem 4.8**.**
Assume (4.2) and (4.3).
Then (4.1) has a weak solution u∈H1(Ω,div0).
Proof.
Let j0 be determined by J and h1 by (4.11). By Proposition 4.7 (ii) we know that we can choose h1′ (depending on h1,h2) such that (4.14) has a solution ϕ=ϕh1′∈H˙1(Ω) which satisfies the orthogonality condition (4.15). Then by Lemmas 4.4 and 4.5 we know that for the given h2 and for the above h1′, (4.13) has a solution v∈Ht01(Ω,div0), hence (4.12) has a solution u∈H1(Ω,div0), and u is actually a solution of (4.1).
∎
5. Tangential Curl BVP of the Maxwell System with a Given Current
In this section we consider the tangential curl BVP of the Maxwell system with a given current J:
[TABLE]
We assume condition (4.2) and
[TABLE]
We shall look for weak solutions u∈H(Ω,curl,div0).
If u0 is a solution, then the general solution can be written as
u=u0+h1′+∇ϕ for any h1′∈H1(Ω) and ϕ∈H1(Ω)∩ker(Δ).
To avoid non-uniqueness we may look for a solution in Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥.
Denote by γτ the tangential trace map u↦ν×u. It yields a tangential trace map
[TABLE]
A necessary condition for (5.1) to have a weak solution is J∈HΓ(Ω,div0) and
[TABLE]
In fact, if (5.1) has a weak solution u∈H1(Ω,R3), then curlu∈HΓ(Ω,div0), hence
ν×B0=ν×curlu=T1(curlu)∈T1[HΓ(Ω,div0)].
Remark 5.1**.**
(i)* If ν×B0 has a divergence-free L2 extension to Ω, namely if there exists B∈H(Ω,div0) such that ν×B=ν×B0 on ∂Ω, then (5.4) is satisfied.*
(ii)* If Ω is a bounded domain in R3 with a C2 boundary and ν×B0∈TH1/2(∂Ω,R3), then (5.4) is satisfied.*
Proof.
(i) If ν×B0 has an L2 extension B∈H(Ω,div0), we can write
B=B1+h with B1∈HΓ(Ω,div0) and h∈H2(Ω).
Then ν×B0=ν×B=ν×B1∈T1[HΓ(Ω,div0)].
(ii) If ∂Ω∈C2 and ν×B0∈TH1/2(∂Ω,R3), then BT0=(ν×B0)×ν∈TH1/2(∂Ω,R3). So there exists a tangential component-preserving and divergence-free extension B∈H(Ω,div0) such that BT=BT0, see [P2]. Then ν×B=ν×BT=ν×BT0=ν×B0. Then from (i) we see that ν×B0 satisfies (5.4).
∎
Assume ν×B0 satisfies (5.4). Then there exists b∈HΓ(Ω,div0) such that
ν×b=ν×B0, and we can find U∈Hn01(Ω,div0) such that curlU=b.
Let w=u−U. Then (5.1) is reduced to the following system
[TABLE]
5.1. Reduce (5.1) to a nonlinear first order div-curl system (5.6)-(5.7)
Lemma 5.2**.**
Let Ω,J,h1,h2 satisfy (4.2) and b∈HΓ(Ω,div0).
If w∈H1(Ω,div0) is a weak solution of (5.5), and let Z=curlw, then
Z is a weak solution of
[TABLE]
and
[TABLE]
If (5.6) has a solution Z with
[TABLE]
then there exist h^2∈H2(Ω) and w∈Hn01(Ω,div0), such that curlw=Z−h^2, and w is a weak solution of (5.5) with h2 replaced by h2+h^2.
If furthermore Z satisfies (5.7), then h^2=0, hence w is a solution of (5.5) with the given h2.
Proof.
(i) is obvious. We prove (ii).
If (5.6) has a solution Z satisfying (5.8),
then since divZ=0, using the second equality of (2.5) we can write
Z=g+h^2 with g∈HΓ(Ω,div0) and h^2∈H2(Ω).
Take w∈Hn01(Ω,div0) such that curlw=g. Then w satisfies
[TABLE]
where h2′=h2+h^2∈H2(Ω). Thus, (5.1) with the given h2 replaced by h2′ has a solution u=w+U.
If Z satisfies (5.7), then h^2=0, and curlw=Z. Hence u=w+U is a solution of (5.5) for the given h2.
∎
Lemma 5.3**.**
Assume Ω,J,h1,h2 satisfy (4.2), H(x,z) is of C1 and satisfies (H3), and
b∈C1(Ωˉ,R3)∩HΓ(Ω,div0) satisfying ν×b=ν×B0.
If Z is a C1 solution of (5.6) and set Y(x)=H(x,Z(x)+b(x)+h2(x)),
then Y(x) is a C1 solution of
[TABLE]
If Y(x) is a C1 solution of (5.9) and set
[TABLE]
then Z is a C1 solution of (5.6). Moreover, Z∈HΓ(Ω,div0) if and only if
[TABLE]
Proof.
(i) is obvious and we prove (ii). If Y is a solution of (5.9) and set Z by (5.10), then Z is of C1 and
H(x,Z(x)+b(x)+h2(x))=Y. Using this and (5.9) we see that Z is a solution of (5.6).
Since Z∈H(Ω,div0), from the second equality of (2.5) we see that, Z∈HΓ(Ω,div0) if and only if (5.11) holds.
∎
5.2. Reduce (5.6) to a second order scalar equation (5.14)
Lemma 5.4**.**
Assume Ω,J,h1,h2 satisfy (4.2), H(x,z) satisfies (H3) and b∈C1(Ωˉ,R3)∩HΓ(Ω,div0) satisfying ν×b=ν×B0. Then we can find Y0 satisfying
[TABLE]
If (5.9) has a solution Y∈H(Ω,curl), then Y can be written as
[TABLE]
and ϕ is a weak solution of the following problem
[TABLE]
If there exists h10∈H1(Ω) such that (5.14) has a solution ϕ∈H1(Ω), and set Y by (5.13), then Y is a solution of (5.9).
Proof.
We only need to prove (i). Assume (5.9) has a solution Y∈H(Ω,curl). Then
[TABLE]
So there exists a unique Y0 satisfying (5.12).
Since curl(Y−Y0)=0, from the first equality in (2.5) we get (5.13). Since div(b+h2)=0 in Ω and ν×(b+h2)=ν×B0 on ∂Ω, from (5.9) we see that ϕ satisfies (5.14).
∎
5.3. Reduce (5.14) to Dirichlet problem (5.15)
We assume that H(x,z) satisfies (H3) and the following condition:
For any x∈∂Ω and y,z∈R3, the equality zT=BT(x,y) holds if and only if yT=HT(x,zT).
Here we use the notation
[TABLE]
We shall show that, BVP (5.14) is equivalent to the following Dirichlet problem
[TABLE]
where ϕ0 is any function on ∂Ω satisfying
[TABLE]
For a function ϕ defined on ∂Ω, we denote by ∇Tϕ the tangential gradient of ϕ. If ϕ is extended over Ωˉ, then
[TABLE]
Recall (see [NW, Lemma 2.5]) that, given a vector field w∈TC1(∂Ω,R3), a necessary and sufficient condition for existence of a function ϕ satisfying ∇Tϕ=w is
[TABLE]
Proposition 5.5**.**
Assume Ω,J,h1 satisfy (4.2), ν×B0 satisfies (5.2), H satisfies (H3) and (H3T).
Given Y0 satisfying (5.12) and h10∈H1(Ω), a necessary condition for solvability of (5.14) is
[TABLE]
If BT0 satisfies (5.18) and h10∈H1(Ω), then there exists a function ϕ0 on ∂Ω, which depends on h10, such that (5.16) holds, and BVP (5.14) is reduced to the Dirichlet BVP (5.15).
In addition to the assumption in (ii), if furthermore H(x,z) satisfies (H1) with g2∈L2(Ω)∩L2(∂Ω), then
we can choose ϕ0∈H1(∂Ω) such that
[TABLE]
Moreover, if ∂Ω is of C2,α, J∈Cα(Ωˉ,R3), and if the right side of (5.16) is of C1,α, then we can choose ϕ0∈C2,α(∂Ω) such that
[TABLE]
Proof.
(i). By (H3T), boundary condition in (5.14) is equivalent to
[TABLE]
(5.19) implies that the vector field HT(x,BT0)−YT0−h1,T0 is a tangential component of a gradient, which is possible if and only if (see (5.17))
[TABLE]
This condition is equivalent to (5.18) because
[TABLE]
(ii). Assume BT0 satisfies (5.18). Then we can find ϕ0∈H1(Ω) satisfying (5.16), such that
(5.19) can be written as (∇ϕ)T=(∇ϕ0)T on ∂Ω.
So we can reduce (5.14) to (5.15).
(iii) From (H1) we have
[TABLE]
This implies that ϕ0∈H1(∂Ω). From (5.12) we have
[TABLE]
Using this and (2.1) we have
[TABLE]
Next assume (5.18) holds, ∂Ω is of C2,α, and the right side of (5.16) is of C1,α. Then (∇ϕ0)T is of C1,α, so ϕ0∈C2,α(∂Ω). Since Y0∈Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥ and curlY0=J+h1∈Cα(Ωˉ,R3), by the Schauder regularity of the div-curl system we have Y0∈C1,α(Ωˉ,R3), and
[TABLE]
Here we used the fact that h1,h10∈H1(Ω), hence
[TABLE]
Finally from (5.19) we have
[TABLE]
∎
5.4. Solvability of (5.15)
The Dirichlet problem (5.15) has been very well studied.
We list the following existence result for (5.15), see for instance [LU, Chpater 4, Theorem 8.3] and [GT, Theorem 15.11].
Lemma 5.6**.**
Assume 0<α<1, Ω is a bounded domain in R3 with a C2,α boundary, B∈Cloc1,α(Ωˉ×R3,R3) satisfies (B1),(B2), Y0,h10∈C1,α(Ωˉ,R3) and ϕ0∈C2,α(∂Ω). Then (5.15) has a unique solution ϕ∈C2,α(Ωˉ).
Proof.
Given ϕ0∈C2,α(∂Ω), we can extend ϕ0 to a C2,α function on Ωˉ and set ψ=ϕ−ϕ0. Then ϕ solves (5.15) if and only if ψ solves
[TABLE]
As in the proof of Proposition 4.7 (i), we define a map T such that
[TABLE]
Using conditions (B1) and (B2) we can show that T:H01(Ω)→H−1(Ω) is hemi-continuous and strongly monotone, hence it is surjective. So there exists a unique ψ∈H01(Ω) such that
T(ψ)=0,
Thus ψ is the unique solution of (5.20) in H01(Ω).
Then using the conditions on B and by the Schauder regularity theory of elliptic equations we see that ψ∈C2,α(Ωˉ).
∎
5.5. Solvability of (5.1)
Using Lemma 5.6 we have existence for (5.1) under the following assumption
[TABLE]
Theorem 5.7**.**
Assume Ω,H,ν×B0,J,h1 satisfy (5.21), Y0 is determined by J and h1 by (5.12).
For any h10∈H1(Ω) such that (5.18) holds, there exists an h20∈H2(Ω) which depends on h10, such that (5.1) with h2=h20 has a solution u∈C2,α(Ωˉ,R3).
In particular, if Ω has no holes, and if there exists h10∈H1(Ω) such that (5.18) holds, then (5.1) has a solution.
Proof.
Since ν×B0∈TC1,α(∂Ω,R3), from Remark 5.1 (ii) we know that ν×B0∈T1[C1,α(Ωˉ,R3)∩HΓ(Ω,div0)]. Hence there exists b∈C1,α(Ωˉ,R3)∩HΓ(Ω,div0) such that ν×b=ν×B0 on ∂Ω. Then b=curlU for some U∈Hn01(Ω,div0)∩C2,α(Ωˉ,R3), and we reduce (5.1) to (5.5).
Since ∂Ω is of C2,α, so h1,h10∈H1(Ω)⊂C1,α(Ωˉ,R3)∩HΓ(Ω,div0).
Since J∈Cα(Ωˉ,R3)∩HΓ(Ω,div0), then J+h1∈C1,α(Ωˉ,R3)∩HΓ(Ω,div0). So we can find a unique Y0 satisfying (5.12), and by regularity of div-curl system we have Y0∈C1,α(Ωˉ,R3).
Since BT0,Y0,h10 satisfy (5.18) and H(x,z) satisfies (H3T), from Proposition 5.5 we can find a function ϕ0 defined on ∂Ω such that (5.16) holds. Since H is of C1,α, BT0 and h10 are of C1,α(∂Ω,R3), so the right side of (5.16) is of C1,α, hence ϕ0 can be chosen such that ϕ0∈C2,α(∂Ω).
By Lemma 5.6 the Dirichlet problem (5.15) has a solution ϕh10∈C2,α(Ωˉ), and by Proposition 5.5, ϕh10 is a solution of (5.14).
Set
[TABLE]
By Lemma 5.4 Yh10 is a C1,α solution of (5.9).
Now we show that, for the above given h10, we can find a unique h20∈H2(Ω) (and h20 depends on h10), such that
B(x,Yh10(x))−b(x)−h20(x)
satisfies (5.11).
To prove, let us write the orthonormal basis of H2(Ω) as {y1,⋯,ym}, and write the orthogonality condition (5.11) for B(x,Yh10) as follows:
[TABLE]
Obviously, for the given h10∈H1(Ω) we can find a unique h20∈H2(Ω) such that the above equalities hold. In fact we can choose
[TABLE]
Then (5.11) holds.
For the above choice of h20, let
[TABLE]
From Lemmas 5.2, 5.3, Z=Zh10,h2 is a C1,α solution of (5.6) and satisfies (5.7). Hence Zh10,h20=curlw for some w∈Hn01(Ω,div0). Then w is a weak solution of (5.5) with h2=h20. Since Zh10,h20∈C1,α(Ωˉ,R3), we see that w∈C2,α(Ωˉ,R3). Let u=U+w. Then u∈C2,α(Ωˉ,R3) is a solution of (5.1) with h2=h20.
∎
6. Other BVPs of the Maxwell System with a Given Current
6.1. The normal curl BVP of the Maxwell System
In this subsection we consider the following problem
[TABLE]
We assume (4.2) and
[TABLE]
It is natural to consider weak solutions of (6.1) in H(Ω,curl,div0).
A necessary condition for (6.1) to have a weak solution is ∫∂ΩBn0dS=0 and J∈HΓ(Ω,div0).
Lemma 6.1**.**
Assume J,h1 satisfy (4.2) and Bn0 satisfies (6.2). Then (6.1) is solvable in H(Ω,curl) for some h2∈H2(Ω) if and only if the following system has a solution Z∈H(Ω,div0):
[TABLE]
Proof.
If u∈H(Ω,curl) is a solution of (6.1) and let Z=curlu+h2, then Z∈H(Ω,div0) is a solution of (6.3).
On the other hand, if (6.3) has a solution Z∈H(Ω,div0), then we can always take h2∈H2(Ω) (which depends on Z) such that Z−h2⊥L2(Ω)H2(Ω). Then Z−h2∈HΓ(Ω,div0), so there exists u∈Hn01(Ω,div0) such that curlu=Z−h2. Then u is a solution of (6.1) corresponding to this h2.
∎
Lemma 6.2**.**
Assume that H satisfies (H1),(H3), J,h1 satisfy (4.2) and Bn0 satisfies (6.2). Let Y0 be determined by J and h1 by (5.12).
If there exists h2∈H2(Ω) such that (6.1) has a solution u∈H(Ω,curl), then there exists h10∈H1(Ω) (depending on h2) such that the following problem
[TABLE]
has a solution ϕ∈H1(Ω) which satisfies the orthogonality condition
[TABLE]
If there exists h10∈H1(Ω) such that (6.4) has a solution ϕ, then there exists h2∈H2(Ω) (depending on h10) such that (ϕ,h2) satisfies (6.5), and hence (6.1) has a solution u∈Hn01(Ω,div0).
Proof.
Z is a solution of (6.3) if and only if Y(x)=H(x,Z(x)) is a solution of
[TABLE]
Similar to the proof of Lemma 5.4 we can show that (6.6) is equivalent to (6.4).
∎
Theorem 6.3**.**
Assume Ω,H(x,z),J,h1 satisfy (4.2) and Bn0 satisfies (6.2). For any h10∈H1(Ω), there exists
h2∈H2(Ω) (depending h1,h10) such that (6.1) has a solution.
Proof.
Let Y0 be determined by J and h1 by (5.12), and let h10∈H1(Ω).
As in the proof of Proposition 4.7 (i), we define a map T by
[TABLE]
Conditions (B1), (B2) imply that T is hemi-continuous and strongly monotone from H˙1(Ω) to H˙1(Ω)∗, hence T is surjective. So there exists a unique ϕ∈H˙1(Ω) such that
[TABLE]
ϕ is the unique weak solution of (6.4) in H˙1(Ω). Note that ϕ depends on h1,h10. By choosing h2∈H2(Ω) depending on h1,h10 such that (6.5) holds, and using Lemma 6.2 (ii), we conclude that Eq. (6.1) with this choice of h2 has a weak solution u.
∎
6.2. The natural BVP of the Maxwell System
In this section we consider the natural BVP of the Maxwell system with a given current222Let us mention that in [MP1], the electro-static problem is reduced to a problem of the form (6.7).:
[TABLE]
We assume (4.2) and
[TABLE]
Let us denote by πτ the tangential trace map w↦wT from H(Ω,curl) to H−1/2(∂Ω,R3), where wT=(ν×w)×ν. In particular HT0=(ν×H0)×ν.
Definition 6.4**.**
Assume Ω,J,h1,h2 satisfy (4.2) and ν×H0 satisfies (6.8).
u is called a weak solution of (6.7) if u∈H(Ω,curl,div0),
H(x,curlu(x)+h2(x))∈Lt2,−1/2(Ω,R3), and for any w∈H(Ω,curl) it holds that
[TABLE]
u* is called an H1-weak solution of (6.7) if u is a weak solution and u∈H1(Ω,R3).*
If u is a solution of (6.7), then for any v∈H(Ω,curl0,div0), u+v is also a weak solution of (6.7).
Therefore solvability of (6.1) in H(Ω,curl,div0) is equivalent to solvability of H1-weak solutions in Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥.
Lemma 6.5**.**
Assume Ω,J,h1,h2 satisfy (4.2) and ν×H0 satisfies (6.8).
If (6.7) has a weak solution u∈H(Ω,curl,div0), then
[TABLE]
Proof.
Assume (6.7) has a weak solution u∈H1(Ω,R3). Then
[TABLE]
Taking w=h∈H1(Ω) in (6.9) we get the last equality in (6.10).
∎
Remark 6.6**.**
If J∈L2(Ω,R3), ν×H0∈TH1/2(∂Ω,R3) and they satisfy the first equality in (6.10), then
[TABLE]
Proof.
Let H0 be a divergence-free and tangential component preserving extension of HT0 on to Ω. From the first equality in (6.10), for any ϕ∈H1(Ω) we have
[TABLE]
∎
We have the following two existence results.
Proposition 6.7**.**
Assume Ω,J,h1,h2,ν×H0 satisfy (4.2), (6.8) and (6.10), and
H(x,z)=∇zP(x,z)
for a function P satisfying (P1). Then (6.7) has an H1-weak solution u∈Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥.
Proof.
Set P2(x,z)=P(x,z+h2(x)).
Then
∇zP2(x,z)=H(x,z+h2(x)), and
P2(x,z) also satisfies the condition (P1) (with different constants and functions g1,g2).
Define
[TABLE]
and
[TABLE]
By the div-curl-gradient inequalities we have
[TABLE]
As in the proof of Proposition 3.1 in [P4] we can show that E is coercive in Y, and E has a minimizer w0∈Y.
From (6.10) and (6.11) we have
[TABLE]
Hence
E[w+h+∇ϕ]=E[w] for all w∈Hn01(Ω,div0), h∈H1(Ω) and ϕ∈H1(Ω).
Recall that H(Ω,curl)=Hn01(Ω,div0)⊕gradH1(Ω). Therefore the minimizer w0 of E on Y is also a minimizer of E on H(Ω,curl).
∎
Theorem 6.8**.**
Assume Ω,J,h1,h2,ν×H0 satisfy (4.2), (6.8) and (6.10), and H satisfies (H1),(H2). Then (6.7) has an H1-weak solution u, and u is unique in Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥. The solution map (h1,J)↦u is continuous from H1(Ω)×HΓ(Ω,div0) to Hn01(Ω,div0).
Proof.
The proof is similar to the proof of Theorem 4.3.
Define Y as in (6.12).
For any given h2∈H2(Ω), we can define a map Ah2:Y→Y∗ such that
[TABLE]
We can show that Ah2:Y→Y∗ is hemi-continuous and strongly monotone, hence surjective. We can also define an operator L:L2(Ω,R3)→Y∗, such that
[TABLE]
Since Ah2 is surjective, there exists u∈Y such that
Ah2(u)=L(J+h1),
that is,
[TABLE]
For all h∈H1(Ω) and ϕ∈H1(Ω) we have
[TABLE]
On the other hand, from (6.10) we have
[TABLE]
From these and (6.13) we see that (6.9) holds for all w=v+h+∇ϕ, where v∈Y, h∈H1(Ω) and ϕ∈H1(Ω). So (6.9) holds for all w∈H(Ω,curl).
Hence u∈Y is a weak solution of (6.7).
By the strongly monotonicity of Ah2 we know that the weak solution u is unique in Y, and the map J+h1↦u is continuous from Y∗ to Y.
∎
6.3. The co-normal BVP of the Maxwell System
In this section we consider the following
[TABLE]
Definition 6.9**.**
Assume Ω,J,h1,h2 satisfy (4.2) and Hn0∈H−1/2(∂Ω). We say u is a weak solution of (6.14) if
u∈H(Ω,curl,div0), H(x,curlu(x)+h2(x))∈Lν2,−1/2(Ω,R3);
the equality ν⋅H(x,curlu(x)+h2(x))=Hn0 holds in the H−1/2(∂Ω) sense;
for all v∈Ht01(Ω,R3) it holds that
[TABLE]
To prove existence of solutions to (6.14), in addition to (4.2), we assume the following conditions:
For any x∈∂Ω and z,w∈R3, the equality ν⋅z=ν⋅B(x,w) holds if and only if ν⋅w=ν⋅H(x,z).
Hn0∈H1/2(∂Ω), and there exists H0∈H1/2(∂Ω,R3) such that ν⋅H0=Hn0.
Lemma 6.10**.**
Assume that Ω,J,h1 satisfy (4.2), H satisfies (H1),(H3),(H3N), Hn0 satisfies (H0). Let Y0 be determined by J and h1 by (5.12).
Given h2∈H2(Ω), if (6.14) has a weak solution u∈H(Ω,curl), then there exists h10∈H1(Ω) (depending on h1,h2) such that the following problem
[TABLE]
has a weak solution ϕ, and ϕ satisfies orthogonality condition (6.5).
If there exists h10∈H1(Ω) such that (6.15) has a weak solution ϕ∈H1(Ω), and let h2∈H2(Ω) (depending on h1,h10) be such that (6.5) holds, then (6.14) has a weak solution u∈Hn01(Ω,div0).
Proof.
Step 1. If u∈H(Ω,curl) is a weak solution of (6.14), let Z=curlu. From the first equation in (6.14) we see that there exist h10∈H1(Ω) and ϕ∈H1(Ω) such that
H(x,Z+h2)=Y0+h10+∇ϕ.
By (H3) we have
Z=B(x,Y0+h10+∇ϕ)−h2,
and from the second and third equalities in (6.14) we have
[TABLE]
Since Z∈curlH(Ω,curl,0)=HΓ(Ω,div0), it satisfies the orthogonality condition (6.5).
From (H3N) and (H0) we see that the boundary condition in (6.16) is equivalent to the boundary condition in (6.15).
Step 2. Assume (h10,ϕ) solves (6.15) and there exists h2 such that (6.5) holds. Then
[TABLE]
Hence there exists u∈Hn01(Ω,div0) such that curlu=Z. Then
[TABLE]
By (H3), (H3N), (H0) and (6.15) we see that u is a weak solution of (6.14) for this h2.
∎
Theorem 6.11**.**
Assume Ω,J,h1,H(x,z) satisfy (4.2), (H3N), and Hn0 satisfies (H0). For any h10∈H1(Ω), there exists
h2∈H2(Ω) (depending h1,h10) such that (6.14) has a weak solution u∈Hn01(Ω,div0).
Proof.
Given any h1, let Y0 be determined by J and h1 by (5.12). Then Y0∈H1(Ω,R3).
By using the monotone operator method (see the proofs of Proposition 4.7 (i), Lemma 5.6), we can show that for any h10∈H1(Ω), (6.15) has a unique solution ϕ∈H˙1(Ω)
(and ϕ depends on h1,h10). We choose h2∈H2(Ω) which depends on h1,h10 such that (6.5) holds. Then from Lemma 6.10, for this choice of h2, (6.14) has a weak solution u.
∎
7. The Maxwell-Stokes System with Solution-dependent Current
In this section we consider the Maxwell-Stokes system where the current is of the form f(x,curlu+h2):
[TABLE]
Throughout this section we assume that
[TABLE]
It was observed in [P4] that, if (7.1) has a solution u, then
f(x,curlu+h2)+h1+∇p∈HΓ(Ω,div0),
which holds if and only if
[TABLE]
It suggests p satisfies Neumann type boundary condition. In this section, in addition to a boundary condition for u.333The case where u satisfies a Dirichlet boundary condition uT=uT0 can be treated by the methods in [P4], hence will not be discussed here. We also mention that most of analyzes in this section are valid for the nonlinear term of the form f(x,u+h2). we assume p satisfies the following boundary condition
[TABLE]
7.1. The normal curl BVP
In this subsection we consider the following problem
[TABLE]
We assume (7.2) and Bn0 satisfies (6.2).
Definition 7.1**.**
We say that (u,p) is a weak solution of (7.3) if u∈H(Ω,curl,div0) and p∈H1(Ω) such that
H(x,curlu(x)+h2(x))∈L2(Ω,R3), f(x,curlu(x)+h2(x))∈L2,−1/2(Ω,R3);
the equality ν⋅curlu(x)+ν⋅h2(x)=Bn0 holds in H−1/2(∂Ω);
[TABLE]
[TABLE]
We shall show the relation between (7.3) with the Z-system
[TABLE]
and we shall show the equivalence of Z-system (7.6) and the Y-system
[TABLE]
Here Pν is the projection operator onto HΓ(Ω,div0), see section 2.4.
Definition 7.2**.**
(i)* We say Z is a weak solution of (7.6) if Z∈H(Ω,div0) such that*
[TABLE]
(ii)* We say Y is a weak solution of (7.7) if Y∈L2(Ω,R3) such that*
[TABLE]
Lemma 7.3**.**
Assume (7.2) and (6.2).
If Y∈L2(Ω,R3) is a weak solution of (7.7), then Y∈H(Ω,curl), and the first equality in (7.7) holds a.e. in Ω.
If (u,p)∈H(Ω,curl,div0)×H1(Ω) is a weak solution of (7.3) and letting Z=curlu, then Z∈H(Ω,div0) and it is a weak solution of (7.6). On the other hand, if Z∈H(Ω,div0) is a weak solution of (7.6), then there exists h20∈H2(Ω) such that (7.3) with h2 replaced by h2+h20 has a weak solution (u,p)∈Hn01(Ω,div0)×H1(Ω), and curlu=Z−h20. If furthermore Z⊥L2(Ω)H2(Ω), then h20=0.
If Z∈H(Ω,div0) is a weak solution of (7.6) and letting
Y=H(x,Z(x)+h2(x)),
then Y∈H(Ω,curl) and it is a weak solution of (7.7). On the other hand, if Y∈L2(Ω,R3) is a weak solution of (7.7) and letting
Z=B(x,Y(x))−h2(x),
then Z∈H(Ω,div0) and it is a weak solution of (7.6).
Proof.
We only prove (i). Assume Y∈L2(Ω,R3) is a weak solution of (7.7). By Definition 7.2 (ii) and by (f1) we see that, for all v∈Ht01(Ω,R3) we have
[TABLE]
So curlY exists as a functional on Ht01(Ω,R3). For any w∈D(Ω,R3) we have
[TABLE]
Hence curlY∈L2(Ω,R3), and
curlY=Pν[f(x,B(x,Y))]+h1.
∎
To solve (7.7), we use (2.6) to write Y=w+h^1+∇ψ, with w∈H0Σ(Ω,div0), h^1∈H1(Ω), ψ∈H1(Ω).
If Y satisfies (7.7) then ψ satisfies
[TABLE]
Proposition 7.4**.**
Assume Ω,H(x,z),Bn0 satisfy (7.2) and (6.2), w∈H0Σ(Ω,div0) and h^1∈H1(Ω).
Then (7.8) has a unique solution ψ=ψw,h^1∈H˙1(Ω).
The map (w,h^1)↦∇ψw,h^1 is a locally Lipschitz and bounded map from H0Σ(Ω,div0)×H1(Ω) into gradH1(Ω).
Proof.
As in the proof of Proposition 4.7 (i), we can define a map T such that
[TABLE]
Using (B1) and (B2) we can show T:H˙1(Ω)↦H˙1(Ω)∗ is Lipschitz and strongly monotone, hence T is surjective. Hence there exists a unique ψ∈H˙1(Ω) such that
[TABLE]
From this and (B1) we have
[TABLE]
here we have used the Poincaré inequality for ψw,h^1∈H˙1(Ω).
Hence
[TABLE]
where C depends only on the constants and g3,g4 in (B1). Thus (w,h^1)↦∇ψw,h^1 is a bounded map with respect to the L2 norms.
Using (B2) and Remark 2.3 we have
[TABLE]
From this and using (B1) we have
[TABLE]
Here we have used (7.9). It follows that
[TABLE]
where the constant C is linearly increases in ∥w+h^1∥L2(Ω)+∥w′+h^1′∥L2(Ω). It follows that
the map (w,h^1)↦∇ψw,h^1 is locally Lipschitz in L2(Ω,R3).
∎
For any w∈H0Σ(Ω,div0), h^1∈H1(Ω), we denote
[TABLE]
We look for v∈H0Σ(Ω,div0) such that
[TABLE]
Lemma 7.5**.**
Assume (7.2) and (6.2). Let h^1∈H1(Ω).
For any w∈H0Σ(Ω,div0) and (7.10) has a unique solution v=Vh^1[w]∈H1(Ω,R3)∩H0Σ(Ω,div0).
[TABLE]
where C depends on Ω,H(x,z) and f(x,z). The map Vh^1 is a compact map in H0Σ(Ω,div0).
If in addition f(x,z) satisfies
[TABLE]
then Vh^1 has a fixed point wh^1.
Proof.
By the definition of Pν, we know that the right side of the first equation of (7.10) lies in HΓ(Ω,div0). Hence (7.10) has a unique solution v∈Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥=H1(Ω,R3)∩H0Σ(Ω,div0). We have
[TABLE]
Using (B1), (7.12), (7.9), and Lemma 2.7, we get (7.11).
Hence Vh^1 is continuous and bounded from H0Σ(Ω,div0) into H1(Ω,R3)∩H0Σ(Ω,div0). Then by the Sobolev imbedding theorem we know that Vh^1 is compact from H0Σ(Ω,div0) into itself.
If f(x,z) satisfies (7.12), then there exists R>0 such that Vh^1 maps the closed ball BR with center 0 and radius R in H0Σ(Ω,div0) into BR itself. By Schauder’s fixed point theorem we conclude that Vh^1 has a fixed point in BR.
∎
Theorem 7.6**.**
Assume (7.2), (6.2) and (7.12). Then for any h^1∈H1(Ω) there exists h2=h2(h1,h^1) such that (7.3) with h2=h2(h1,h^1) has a weak solution (u,p)∈H1(Ω,R3)×H1(Ω).
Proof.
Let wh^1 denote the fixed point of the map Vh^1 in H0Σ(Ω,div0), and let
[TABLE]
wh^1 satisfies
[TABLE]
Then Yh^1 is a solution of (7.7). By Lemma 7.3 (iii), Zh^1 is a solution of (7.6).
Then there exists h2=h2(h1,h^1) depending on h1 and h^1, such that Zh^1⊥L2(Ω)H2(Ω), namely
[TABLE]
Then Zh^1∈HΓ(Ω,div0). By Lemma 7.3 (ii), there exists (u,p)∈Hn01(Ω,div0)×H1(Ω) which solves (7.3) for the given h1.
∎
7.2. The natural-Neumann BVP
In this section we consider the natural BVP of the Maxwell-Stokes system
[TABLE]
We assume (7.2) and
[TABLE]
where HT0=(ν×H0)×ν.
Definition 7.7**.**
We say that (u,p) is a weak solution of (7.15) if u∈H(Ω,curl,div0), p∈H1(Ω), such that
H(x,curlu(x)+h2(x))∈Lt2,−1/2(Ω,R3),
f(x,curlu(x)+h2(x))∈L2(Ω,R3);
[TABLE]
[TABLE]
Regarding (7.15) we have the following Z-system
[TABLE]
and Y-system
[TABLE]
where Pn is the Neumann projection associated with ν⋅curlHT0, see section 2.4.
Definition 7.8**.**
(i)*
We say Z is a weak solution of (7.19) if Z∈H(Ω,div0) such that*
[TABLE]
and for all v∈H1(Ω,R3) it holds that
[TABLE]
(ii)* We say Y is a weak solution of (7.20) if Y∈Lt2,−1/2(Ω,R3) such that*
[TABLE]
and for all v∈H1(Ω,R3) it holds that
[TABLE]
Similar to Lemma 7.3 we have the following
Lemma 7.9**.**
Assume (7.2) and (7.16).
If Y∈Lt2,−1/2(Ω,R3) is a weak solution of (7.20), then Y∈H(Ω,curl), the first equality in (7.20) holds a.e. in Ω, and the equality ν×Y=ν×H0 holds in the sense of trace in H−1/2(∂Ω,R3).
If (u,p)∈H(Ω,curl,div0)×H1(Ω) is a weak solution of (7.15) and letting Z=curlu, then Z∈H(Ω,div0) and it is a weak solution of (7.19).
On the other hand, if Z∈H(Ω,div0) is a weak solution of (7.19), then there exists h20∈H2(Ω) such that (7.15) with h2 replaced by h2+h20 has a weak solution (u,p)∈Hn01(Ω,div0)×H1(Ω), and curlu=Z−h20. If furthermore Z⊥L2(Ω)H2(Ω), then h20=0.
If Z∈H(Ω,div0) is a weak solution of (7.19) and letting
Y=H(x,Z(x)+h2(x)),
then Y∈H(Ω,curl) and it is a weak solution of (7.20).
On the other hand, if Y∈Lt2,−1/2(Ω,R3) is a weak solution of (7.20) and letting
Z=B(x,Y(x))−h2(x),
then Z∈H(Ω,div0) and it is a weak solution of (7.19).
Theorem 7.10**.**
Assume (7.2), (7.16) and (7.12). Then for any given h2∈H2(Ω), there exists h1∈H1(Ω) which depends on h2 such that (7.15) has a weak solution (u,p)∈H1(Ω,R3)×H1(Ω).
Proof.
Step 1. To solve (7.20), we use (2.7) to write
Y=w+h^2+∇ψ, with w∈HΓ(Ω,div0), h^2∈H2(Ω) and ψ∈H01(Ω).
If Y satisfies (7.20) then ψ satisfies
[TABLE]
As in the proof of Proposition 7.4 (also see Proposition 4.7 (i)), we can show that the map T:H01(Ω)→H−1(Ω) defined by
[TABLE]
is hemi-continuous and strongly monotone. Then there exists a unique ψ=ψw,h^2∈H01(Ω) such that T[ψw,h^2]=0.
The map (w,h^2)↦∇ψw,h^2 is a locally Lipschitz and bounded map from HΓ(Ω,div0)×H2(Ω) into gradH01(Ω).
Step 2. Denote
[TABLE]
We look for v∈HΓ(Ω,div0) such that
[TABLE]
We shall show that, there exists a unique h1=h1,w,h^2∈H1(Ω) such that (7.22) has a unique solution v=Vh^2[w]∈H1(Ω,R3)∩HΓ(Ω,div0), and
[TABLE]
where C depends on Ω,H(x,z) and f(x,z).
To prove, we take a tangential component preserving extension of HT0, which is also denoted by H0, see [P2]. Then H0∈H1(Ω,div0). We can choose H0 such that
[TABLE]
Let z=v−H0.
Then (7.22) is reduced to
[TABLE]
Let us denote by g the right side of the first equation in (7.24). Then (7.24) is solvable in Ht01(Ω,div0) if and only if g∈H0Σ(Ω,div0).
By the definition of Pn we easily see that g∈H0(Ω,div0).
On the other hand, there exists a unique h1=h1,w,h^2 such that g⊥H1(Ω), namely
[TABLE]
Recalling that H0(Ω,div0)=H0Σ(Ω,div0)⊕L2(Ω)H1(Ω),
we see that for this choice of h1 we have g∈H0Σ(Ω,div0). Then (7.24) with h1=h1,w,h^2 has a unique solution z∈H1(Ω,div0)∩H2(Ω)L2(Ω)⊥=H1(Ω,R3)∩HΓ(Ω,div0).
To estimate h1,w,h^2, let {e1,⋯,eN} be a orthonormal basis of H1(Ω). Then
[TABLE]
Using the above equalities and (2.1) we have
[TABLE]
where C depends on Ω, H(x,z) and f(x,z).
For the solution z of (7.24) obtained above we have
[TABLE]
Hence (7.22) has a unique solution v=z+H0 in HΓ(Ω,div0). We denote this solution by v=Vh^2[w].
Then we have (7.23).
(7.23) shows that the map w↦Vh^2[w] is a bounded map from H(Ω,div0) into H1(Ω,R3)∩HΓ(Ω,div0). Using (H1),(H2),(H3) and (f1) we see that the maps
[TABLE]
are continuous in the L2 norm. Thus the map
w↦h1,w,h^2
is continuous in the L2 norm. By these facts and using the L2-div-curl-gradient inequalities we see that the map
w↦Vh^2[w]
is continuous from HΓ(Ω,div0)→H1(Ω,R3)∩HΓ(Ω,div0).
By the Sobolev imbedding theorem we know that Vh^2 is compact from HΓ(Ω,div0) into itself.
By the assumption (7.12), then there exists R>0 such that Vh^2 maps the closed ball BR with center 0 and radius R in HΓ(Ω,div0) into BR itself. By Schauder’s fixed point theorem we conclude that Vh^2 has a fixed point wh^2 in BR.
Step 3.
Let
[TABLE]
Since wh^2 satisfies
[TABLE]
we see that Yh^2 is a solution of (7.20), and by Lemma 7.9 (iii) Zh^2 is a solution of (7.19).
Using the method in the proof of Proposition 4.7 (ii), we can show that, for the given h2∈H2(Ω), there exists h^2∈H2(Ω) such that
[TABLE]
Namely, there exists h^2 such that Zh^2⊥L2(Ω)H2(Ω).
Then Zh^2∈HΓ(Ω,div0). By Lemma 7.9 (ii) we know that there exists (u,p)∈Hn01(Ω,div0)×H1(Ω) which solves (7.15) for the given h2.
∎
7.3. The Co-normal-Neumann BVP
In this section we consider the natural BVP of the Maxwell-Stokes system
[TABLE]
In the following we assume (7.2) and
Hn0∈H−1/2(∂Ω).
Definition 7.11**.**
We say that (u,p) is a weak solution of (7.27) if u∈H(Ω,curl,div0) and p∈H1(Ω) such that
H(x,curlu(x)+h2(x))∈Lν2,−1/2(Ω,R3), f(x,curlu(x)+h2(x))∈Lν2,−1/2(Ω,R3);
the equality ν⋅H(x,curlu(x)+h2(x))=Hn0 holds in the H−1/2(∂Ω) sense;
[TABLE]
[TABLE]
We shall show the relation between (7.27) with the Z-system
[TABLE]
and we shall show the equivalence of Z-system (7.28) and the Y-system
[TABLE]
Definition 7.12**.**
(i)*
We say Z is a weak solution of (7.28) if Z∈H(Ω,div0) such that*
[TABLE]
(ii)* We say Y is a weak solution of (7.20) if Y∈Lν2,−1/2(Ω,R3) such that*
[TABLE]
Similar to Lemma 7.3 we have
Lemma 7.13**.**
Assume (7.2) and Hn0∈H−1/2(∂Ω).
If Y∈Lν2,−1/2(Ω,R3) is a weak solution of (7.29), then Y∈H(Ω,curl), and the first equality in (7.29) holds a.e. in Ω.
If (u,p)∈H(Ω,curl,div0)×H1(Ω) is a weak solution of (7.27) and letting Z=curlu, then Z∈H(Ω,div0) and it is a weak solution of (7.28).
On the other hand, if Z∈H(Ω,div0) is a weak solution of (7.28), then there exists h20∈H2(Ω) such that (7.27) with h2 replaced by h2+h20 has a weak solution (u,p)∈Hn01(Ω,div0)×H1(Ω) with curlu=Z−h20. If furthermore Z⊥L2(Ω)H2(Ω), then h20=0.
If Z∈H(Ω,div0) is a weak solution of (7.28) and letting
Y=H(x,Z(x)+h2(x)),
then Y∈H(Ω,curl) and it is a weak solution of (7.29).
On the other hand, if Y∈Lν2,−1/2(Ω,R3) is a weak solution of (7.29) and letting
Z=B(x,Y(x))−h2(x),
then Z∈H(Ω,div0) and it is a weak solution of (7.28).
Theorem 7.14**.**
Assume (7.2), (H3N), (H0) and (7.12). Then for any given h1,h^1∈H1(Ω) and h2=h2(h^1)∈H2(Ω) which depends on h^1 such that (7.27) has a weak solution (u,p)∈H1(Ω,R3)×H1(Ω) for the given h1 and for h2=h2(h^1).
Proof.
Step 1. By Lemma 7.13, we first solve (7.29). Using (2.6), we write
[TABLE]
From the last two equalities in (7.29) and using (H0) we have
[TABLE]
Using (H3) and (H3N) we see that this BVP can be written as
[TABLE]
Hence (w,h^1,∇ψ) satisfies (7.30).
By the monotone operator methods we can show that, for any given w∈H0Σ(Ω,div0) and h^1∈H1(Ω), (7.30) has a unique solution ψ=ψw,h^1∈H˙1(Ω).
Step 2. Denote
[TABLE]
We look for v∈H0Σ(Ω,div0) such that
[TABLE]
By the definition of Pν we see that the right side of the first equation of (7.31) lies in HΓ(Ω,div0).
So (7.31) has a unique solution v=Vh^1[w]∈Hn01(Ω,div0)∩H1(Ω)L2(Ω)⊥=H1(Ω,R3)∩H0Σ(Ω,div0). Similar to Lemma 7.5 we can show that
[TABLE]
where C depends on Ω,H(x,z) and f(x,z).
Using the Sobolev imbedding theorem we know that Vh^1 is compact from H0Σ(Ω,div0) into itself.
Since f(x,z) satisfies (7.12), then there exists R>0 such that Vh^1 maps the closed ball BR with center 0 and radius R in H0Σ(Ω,div0) into BR itself. By Schauder’s fixed point theorem we conclude that Vh^1 has a fixed point wh^1 in BR.
Step 3.
Let
[TABLE]
wh^1 satisfies
[TABLE]
Then Yh^1 is a solution of (7.29), and by Lemma 7.13 (iii) we know that Zh^1,h2 is a solution of (7.28).
Now we choose h2=h2(h^1) such that
[TABLE]
Denote Zh^1=Zh^1,h2(h^1). Then Zh^1∈HΓ(Ω,div0). By Lemma 7.13 (ii), there exists (u,p)∈Hn01(Ω,div0)×H1(Ω) which solves (7.27) for the given h1 and for h2=h2(h^1).
∎
Acknowledgements
This work was partially supported
by the National Natural Science Foundation of China grants no. 11671143 and no. 11431005.