Mathematical and numerical analysis of a steady

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Mathematical and numerical analysis of a steady

magnetohydrodynamic problem

M. Discacciati

RICAM-Report 2006-38

Mathematical and numerical analysis of a steady magnetohydrodynamic problem

Marco Discacciati

Johann Radon Institute for Computational and Applied Mathematics (RICAM), Altengerberstraße 69, A-4040 Linz, Austria. [email protected]

Abstract. In this paper we consider the steady flow of a conductive incompressible fluid con- fined in a bounded region and subject to the Lorentz force exerted by the interaction of electric currents and magnetic fields. We study a nonlinear coupled problem where all the relevant phys- ical unknowns appear, i.e., magnetic field, velocity and pressure of the fluid, electric current and potential. We prove the well-posedness of the coupled problem and we study its conforming fi- nite element approximation. Finally, we propose and analyze an iterative decoupled method to compute its solution, and we present some numerical examples.

Keywords. Magnetohydrodynamics, Maxwell’s equations, Navier-Stokes equations, Conforming finite elements.

1 Introduction and problem setting

Magnetohydrodynamics (MHD) concerns the interaction of electrically conductive fluids and elec- tromagnetic fields. Applications are manifold, including astronomy and geophysics, as well as many industrial processes, especially concerning the production of metals (see, e.g., [8, 28] for a general introduction to this subject). Indeed, the interaction of magnetic fields and electric currents in a conductive fluid gives rise to a Lorentz force which permits to influence the motion of the fluid itself in a contactless way, which is clearly very useful in metallurgical industry.

The mathematical modeling of the processes taking place in such industrial plants is very involved.

Indeed, it requires to take into account many phenomena, e.g., multi-phase and free-surface flows, magnetic fields, electric currents, temperature effects, chemical reactions. However, the core model describing the interaction between the fluid and magnetic fields is a nonlinear system formed by Navier-Stokes’ and Maxwell’s equations coupled by Ohm’s law and Lorentz’s force.

The literature concerning both the mathematical analysis and the finite element approximation of this coupled problem is broad (see, e.g., [11, 14, 16, 15, 17, 19, 20, 21]).

Recently, optimal control methods have also been applied to the MHD system [13] providing a mathematical tool for the optimization of magnetic fields, in order to drive the fluid flow in a desired state.

In this paper, we consider a steady MHD problem in a formulation where the physical quantities:

magnetic field, velocity and pressure of the fluid, electric currents and potential appear, so that we have a nonlinear coupled system in five unknowns. After discussing the well-posedness of this problem under suitable smallness assumptions on the physical data (Sect. 2), we propose and analyze an iterative method to compute its solution (Sect. 3). In particular, we set up an operator- splitting based method which allows us to compute the global solution by independently solving the magnetic-field, the fluid-flow and the electric-currents subproblems. Concerning the finite element approximation, we consider a conforming discretization based on the classical N´ed´elec, Taylor-Hood and Raviart-Thomas elements (Sect. 4). Finally, we discuss the algebraic formulation of the iterative schemes that we have proposed, and we present some preliminary numerical results (Sect. 5).

The formulation of the problem that we consider is as follows.

Let Ωf⊂R³be a bounded Lipschitz domain, filled by an electrically conductive fluid. An external conductor Ωs is attached to a part of the boundary Γs ⊂ ∂Ωf, so that an electric current can be injected into the fluid domain. Finally, let Ωe be an additional external device, separated by the fluid domain, which possibly generates a magnetic fieldB^†_e in the whole space R³, that also influences the motion of the fluid in Ωf. A schematic representation of the domain is shown in fig. 1.

Ωf

Ωs

Ωe

Γs

Γs

Figure 1: Schematic representation of the setting.

In the conductive device Ωs, we assign an electric currentJssuch that divJs= 0 in Ωs,Js·n= 0 on∂Ωs\Γs, andJs·n=json Γs, where the (known) functionjsfulfills the compatibility condition R

Γsjs= 0. ndenotes the unit normal vector directed outward of ∂Ωf. The currentJsoriginates a magnetic fieldB^†_sin R³, which may be represented by the Biot-Savart law:

B^†_s(x) =− µ 4π

Z

Ωs

x−y

|x−y|³ ×Js(y)dy , x∈R³,

µ >0 being the magnetic permeability, that we assume to be constant in the whole space.

We suppose that the contact interface Γsbetween Ωf and Ωs is perfectly conductive, i.e. it holds Js·n=js=Jf·n on Γs. (1) Thus, in the fluid domain Ωf we have a currentJf which generates a magnetic field B^†_f:

B^†_f(x) =−µ 4π

Z

Ωf

x−y

|x−y|³ ×Jf(y)dy , x∈R³.

In conclusion, we have a global magnetic fieldB^† in the spaceR³due to the superposition of three components: B^†(x) =B^†_e(x) +B^†_s(x) +B^†_f(x).

The motion of the incompressible conductive fluid in Ωf is described by the steady Navier-Stokes’

equations: find the velocityuand the pressurepsuch that

−η△u+ρ(u· ∇)u+∇p−Jf×B^†_|Ω

f = 0 in Ωf,

divu = 0 in Ωf, (2)

with the Dirichlet boundary conditionu=gon∂Ωf,gbeing an assigned velocity field such that R

∂Ωfg·n= 0. Jf×B^†_|Ω

f is the Lorentz force exerted on the fluid by the interaction of the magnetic fieldB^† and the electric currentJf, whileη, ρ >0 are the fluid viscosity and density, respectively,

Finally, the electric currentJf satisfies

σ^ư1Jf+∇φưu×B^†_|Ω

f = 0 in Ωf,

divJf = 0 in Ωf, (3)

whereφis the electric potential andσ >0 is the electric conductivity of the fluid. We impose the boundary condition (1) on Γs, while we setJf·n= 0 on ∂Ωf\Γs, i.e. we assume that this part of the boundary is perfectly insulated.

Remark that instead of (1), one could prescribe a difference of potential on Γs to generate the electric current Jf in the fluid domain, i.e. one would impose the natural boundary condition φ=φson Γs, for a suitable given functionφs.

2 Weak formulation and well-posedness analysis

In view of the numerical solution of the MHD problem, we restrict ourselves to a bounded domain Ω⊂R³, which includes the fluid domain Ωf ⊂Ω, and is assumed to be of class C^1,1 (see, e.g., [1]).

We define the following functional spaces:

H¹₀(Ωf) =

v∈H¹(Ωf)|v=0on∂Ωf , withH¹(Ωf) =

v∈(H¹(Ωf))³ , (4) H¹_0,div(Ωf) =

v∈H¹₀(Ωf)|divv= 0 in Ωf , (5)

H(div; Ωf) =

v∈(L²(Ωf))³|divv∈L²(Ωf) , (6) H₀(div; Ωf) ={v∈H(div; Ωf)|v·n= 0 on∂Ωf}, (7) H⁰₀(div; Ωf) ={v∈H₀(div; Ωf)|divv= 0 in Ωf}, (8) Q0=

(

q∈L²(Ωf) Z

Ωf

q= 0 )

, (9)

H(curl; Ω) =

v∈(L²(Ω))³|curl v∈(L²(Ω))³ , (10) H₀(curl; Ω) ={v∈H(curl; Ω)|v×n=0on∂Ω}. (11) We denote by k · kH¹(Ωf), k · kdiv,Ωf, k · kL²(Ωf) and k · k^curl,Ω the usual norms in the spaces H¹(Ωf), H(div; Ωf),L²(Ωf) and H(curl; Ω), respectively. Remark that kvkL²(Ωf) =kvkdiv,Ωf

for allv∈H⁰₀(div; Ωf).

Finally, let us recall the Poincar´e inequality (see, e.g., [18]):

∃Cp>0 : kvkL²(Ωf)≤Cpk∇vkL²(Ωf) ∀v∈H¹₀(Ωf), (12) and the following inequality:

∃C3>0 : Z

Ωf

(f ×g)·h≤C3kfkL³(Ωf)kgkL⁶(Ωf)khkL²(Ωf), (13) where the functionsf,g,hare assumed to be regular enough.

We introduce a continuous extension operator Ef : (H^1/2(∂Ωf))³ → H¹(Ωf), such that, taken g∈(H^1/2(∂Ωf))³ withR

∂Ωfg·n= 0,Efg∈H¹(Ωf) is the divergence-free extension ofg, such that Efg =g on∂Ωf. More precisely, letE^′_fg ∈H¹(Ωf) be a continuous extension of g such thatE^′_fg=gon∂Ωf. Then, we can construct a functionE^′′_fg∈H¹₀(Ωf) such that

ư Z

Ωf

q div (E^′′fg) =˜ Z

Ωf

q div (E^′fg)˜ ∀q∈L²(Ωf). (14)

The solvability of (14) is guaranteed by the inf-sup condition (see, e.g., [26] p. 157): there exists a positive constantβ^∗>0 such that

∀q∈L²(Ωf) ∃v∈H¹₀(Ωf), v6=0: − Z

Ωf

qdivv≥β^∗kvkH¹(Ωf)kqkL²(Ωf).

Finally, we indicate byEfg=E^′_fg+E^′′_fgthe divergence-free extension of the datumg. Remark that by constructionEfg=gon∂Ωf, and that, thanks to (14), it holds

Z

Ωf

qdiv (Efg) = 0 ∀q∈L²(Ωf). Then, we splitu=u0+Efg, withu0∈H¹₀(Ωf).

We introduce a continuous extension operator Es:L²(Γs)→H(div; Ωf), Es :ϕ→Esϕ, for all ϕ∈L²(Γs) withR

Γsϕ= 0. More precisely, we consider the auxiliary Neumann problem

−△χ= 0 in Ωf,

∂χ

∂n =ϕ on Γs,

∂χ

∂n = 0 on∂Ωf\Γs,

which has a unique solution in H¹(Ωf)/R. Then, we takeEsϕ =∇χ ∈ L²(Ωf). Remark that Esϕ·n=ϕ on Γs, Esϕ·n= 0 on ∂Ωf \Γs, and div (Esϕ) = 0 in Ωf. Thus, we can consider the divergence-free extensionEsjs of the boundary datumjs such thatEsjs·n= 0 on∂Ωf\Γs, Esjs·n = js on Γs, div (Esjs) = 0 in Ωf, and we decompose Jf = Jf,0+Esjs, with Jf,0 ∈ H⁰₀(div; Ωf).

Let us now consider the electric current:

Jf,0=





Jf,0 in Ωf, 0 in Ωs,

0 in Ω\(Ωf∪Ωs),

which satisfies divJf,0= 0 in Ω andJf,0·n= 0 on∂Ω. We can represent the magnetic field, say B0,f, generated byJf,0 as the solution of the problem

curl(µ⁻¹Bf,0) = Jf,0 in Ω, divBf,0 = 0 in Ω, Bf,0·n = 0 on∂Ω,

(15)

where we have introduced the fictitious essential boundary conditionBf,0·n= 0 on∂Ω.

Bf,0 can be equivalently expressed as the extension B^Ω(Jf,0) of Jf,0 by the linear continuous operatorB^Ω:X_BΩ →X_T(Ω), where

X_BΩ={J∈H(div; Ω)|J·n= 0 on∂Ω, divJ= 0 in Ω},

XT(Ω) ={B∈H(curl; Ω)∩H(div; Ω)|B·n= 0 on∂Ω, divB= 0 in Ω}. B^Ωapproximates in Ω the bounded linear Biot-Savart operatorB(see, e.g., [20]):

B: (L²(Ω))³→W¹(R³), B(Jf,0) =−µ 4π

Z

Ω

x−y

|x−y|³ ×Jf,0(y)dy, ∀x∈R³. With a slight abuse in notation we will often writeB^Ω(Jf,0) instead ofB^Ω(Jf,0).

Under the assumption that ∂Ω is at least of class C^1,1, thanks to Theorem 2.9 of [2] and via Sobolev immersion, we haveX_T(Ω)֒→(H¹(Ω))³ ֒→(L⁶(Ω))³ ֒→(L³(Ω))³. Then, in particular, the following inequalities hold:

kBΩ(Jf,0)kL³(Ωf) ≤ C_s¹kBΩ(Jf,0)kL⁶(Ωf)

≤ C_s¹kBΩ(Jf,0)kL⁶(Ω)≤C_s²kBΩ(Jf,0)kH¹(Ω)≤C_s³kBΩ(Jf,0)k^XT, (16) with positive constantsC_s¹, C_s², C_s³>0, and there holds

kBΩ(Jf,0)k^XT ≤ C kcurlBΩ(Jf,0)kL²(Ω)+kdivBΩ(Jf,0)kL²(Ω)

= CµkJf,0kL²(Ω)≤CµkJf,0k^div,Ωf, for C >0. Therefore, there exists a positive constantC >0 such that

kBΩ(Jf,0)kL³(Ωf)≤CµkJf,0kdiv,Ωf. (17) In analogous way, we can consider the currentJs:

Js=





Esjs in Ωf, Js in Ωs,

0 in Ω\(Ωf∪Ωs), which generates a magnetic fieldBs such that

curl(µ⁻¹Bs) = Js in Ω, divBs = 0 in Ω, Bs·n = 0 on∂Ω. In this case, there holds:

kBskL³(Ω_f)≤CµkJskdiv,Ω≤Cµ kEsjskdiv,Ωf +kJskdiv,Ωs

. (18)

Thus, we can approximate the magnetic fieldB^† in the bounded domain Ω by B =Be+Bs+ BΩ(Jf,0), where Beis the restriction ofB^†_e to Ω. BΩ(Jf,0) is the only unknown component ofB, since it depends on the unknown currentJf,0.

The representation of the magnetic fieldBf,0 in terms of the linear operatorBΩwill be useful for the analysis of the MHD problem, as we will see in the next section.

2.1 Weak form of the MHD problem

Using the notations introduced in the previous section, we can write the weak form of the MHD problem: findu0∈H¹₀(Ωf),p∈Q0,Jf,0∈H₀(div; Ωf),φ∈Q0 such that

Z

Ωf

η∇u0· ∇v+ Z

Ωf

ρ[((u0+Efg)· ∇)(u0+Efg)]·v− Z

Ωf

pdivv

− Z

Ωf

[(Jf,0+Esjs)×(Be+Bs+BΩ(Jf,0))]·v=− Z

Ωf

η ∇(Efg)· ∇v (19) Z

Ωf

qdivu0= 0 (20)

Z

Ω_f

σ⁻¹Jf,0·K− Z

Ω_f

φdivK+ Z

Ω_f

[K×(Be+Bs+BΩ(Jf,0))]·(u0+Efg)

=− Z

Ωf

σ⁻¹Esjs·K (21) Z

Ωf

ϕdivJf,0= 0 (22)

for allv∈H¹₀(Ωf),q∈Q0,K∈H₀(div; Ωf),ϕ∈Q0. Remark that the trilinear forms

− Z

Ωf

[Jf ×(Be+Bs+B^Ω(Jf,0))]·v and Z

Ωf

[K×(Be+Bs+B^Ω(Jf,0))]·u (23) realize the coupling between the fluid, the electric and the magnetic-field problems in Ωf. Notice that they make the MHD problem nonlinear, despite the fact, e.g., to replace the Navier-Stokes’

equations by the Stokes ones, that is to consider, instead of (19), the momentum equation Z

Ωf

η ∇u0· ∇v− Z

Ωf

pdivv− Z

Ωf

[Jf,0×(Be+Bs+B^Ω(Jf,0))]·v

− Z

Ωf

[Esjs× B^Ω(Jf,0)]·v=− Z

Ωf

η ∇(Efg)· ∇v+ Z

Ωf

[Esjs×(Be+Bs)]·v,(24) where the nonlinear convective term (u· ∇)uhas been omitted.

In the next sections, we consider (24) instead of (19), and we prove the well-posedness of the MHD problem (24), (20)-(22). To this aim, we will use some classical results that, for the sake of clarity, we anticipate (in a fairly abstract form) in the following section.

2.1.1 General existence and uniqueness results

Let us recall some existence and uniqueness results for nonlinear saddle-point problems, referring the reader to, e.g., [4, 5, 6, 7, 12] for a rigorous study.

Let (X,k · kX) and (Y,k · kY) be two real Hilbert spaces. Consider a continuous bilinear form b(·,·) :X ×Y →R, (v, q)→b(v, q), and a trilinear form a(·;·,·) :X×X×X →R, (w, u, v)→ a(w;u, v), where, for w ∈ X, the mapping (u, v) → a(w;u, v) is a continuous bilinear form on X×X.

Then, consider the following problem: givenl∈X^′, find (u, p)∈X×Y satisfying a(u;u, v) + b(v, p) = hl, vi ∀v∈X

b(u, q) = 0 ∀q∈Y. (25)

Introducing the linear operatorsA(w)∈ L(X;X^′) forw∈X, andB ∈ L(X;Y^′):

hA(w)u, vi= a(w;u, v), hBv, qi= b(v, q), ∀u, v∈X, ∀q∈Y , problem (25) becomes: find (u, p)∈X×Y such that

A(u)u+B^Tp = l in X^′, Bu = 0 in Y^′. TakingV = Ker(B), we associate (25) to:

findu∈V : a(u;u, v) =hl, vi ∀v∈V , (26) or, equivalently: find u ∈ V such that ΠA(u)u = Πl in V^′, where the linear operator Π ∈ L(X^′;V^′) is defined byhΠl, vi=hl, vi,∀v∈V.

If (u, p) is a solution of (25), thenusolves (26). The converse may be proved as well provided an inf-sup condition holds. More generally, the following results hold.

Theorem 2.1 (Existence and uniqueness) Suppose that:

1. the bilinear forma(w;·,·) is uniformly elliptic in the Hilbert space V with respect to w, i.e.

there exists a constant α >0such that

a(w;v, v)≥αkvk²X ∀v, w∈V ;

2. the mappingw→ΠA(w)is locally Lipschitz-continuous inV, i.e. there exists a continuous and monotonically increasing function L:R⁺→R⁺ such that, for all m >0,

|a(w1;u, v)−a(w2;u, v)| ≤L(m)kukXkvkXkw1−w2kX, ∀u, v∈V, w1, w2∈Sm, with Sm={w∈V|kwkX ≤m};

3. it holds

kΠlkV^′

α² L

kΠlkV^′

α

<1. Then (26) has a unique solutionu∈V.

Theorem 2.2 Assume that the bilinear form b(·,·) satisfies the inf-sup condition: ∃β >0

q∈Yinf sup

v∈X

b(v, q) kvk^Xkqk^Y ≥β .

Then, for each solutionuof (26), there exists a uniquep∈Y such that(u, p)is a solution of (25).

2.2 Analysis of the Stokes-MHD problem

We introduce the product spacesX=H¹₀(Ωf)×H₀(div; Ωf),Q₀=Q0×Q0, and the subspace X0⊂X withX0=H¹_0,div(Ωf)×H⁰₀(div; Ωf). Moreover, we define the norms

k(v,K)k^X =

kvk²H¹(Ωf)+kKk²div,Ωf

1/2

∀(v,K)∈X, k(q, ϕ)k^Q=

kqk²L²(Ωf)+kϕk²L²(Ωf)

1/2

∀(q, ϕ)∈Q₀.

We introduce: the trilinear form a(·;·,·) : X ×X ×X → R, such that for all (vi,Ki) ∈ X (i= 1,2,3),

a((v1,K1); (v2,K2),(v3,K3)) = Z

Ωf

η∇v2· ∇v3− Z

Ωf

[K2×(Be+Bs+B^Ω(K1))]·v3− Z

Ωf

[Esjs× B^Ω(K2)]·v3

+ Z

Ωf

σ⁻¹K2·K3+ Z

Ωf

[K3×(Be+Bs+B^Ω(K1))]·v2+ Z

Ωf

[K3× B^Ω(K2)]·Efg, the bilinear formb(·,·) :X×Q₀→R:

b((v,K),(q, ϕ)) =− Z

Ωf

q divv− Z

Ωf

ϕdivK ∀(v,K)∈X, (q, ϕ)∈Q₀, and the linear functionalF :X →R:

F(v,K) = − Z

Ωf

η∇(Efg)· ∇v− Z

Ωf

σ⁻¹Esjs·K

+ Z

Ωf

[Esjs×(Be+Bs)]·v− Z

Ωf

[K×(Be+Bs)]·Efg, ∀(v,K)∈X.

With these notations, the coupled problem (24), (20)-(22) becomes: find (u0,Jf,0)∈X, (p, φ)∈ Q₀ such that

a((u0,Jf,0); (u0,Jf,0),(v,K)) +b((v,K),(p, φ)) = F(v,K) ∀(v,K)∈X,

b((u0,Jf,0),(q, ϕ)) = 0 ∀(q, ϕ)∈Q₀, (27) and it can be reformulated on the kernel ofb as:

find (u0,Jf,0)∈X₀: a((u0,Jf,0); (u0,Jf,0),(v,K)) =F(v,K) ∀(v,K)∈X₀. (28) Then, we can state the following result.

Proposition 2.1 Assume that Be|Ωf ∈ (L³(Ωf))³. If the boundary data g ∈ (H^1/2(∂Ωf))³, js∈L²(Γs)and the assigned magnetic fieldBeare small enough, and the physical parametersµ⁻¹, η, σ⁻¹ are sufficiently large, then the MHD problem (27) has a unique solution (u0,Jf,0)∈ X, (p, φ)∈Q₀.

Proof. The proof is made of several steps and it will be based on Theorems 2.1 and 2.2.

1. Let (u,J),(v,K)∈X0. Then, using the inequalities (16), (17), (12) and (13), we have a((u,J); (v,K),(v,K))

=ηk∇vk²L²(Ωf)+σ⁻¹kKk²L²(Ωf)

− Z

Ω_f

[Esjs× B^Ω(K)]·v+ Z

Ω_f

[K× B^Ω(K)]·Efg

≥ η

1 +C_p²kvk²H¹(Ωf)+σ⁻¹kKk²L²(Ωf)

−Cµ˜ kEsjsk^div,ΩfkKk^div,ΩfkvkH¹(Ωf)−Cµ˜ kEfgkH¹(Ωf)kKk²div,Ω_f

≥ η

1 +C_p²kvk²H¹(Ωf)+σ⁻¹kKk²L²(Ωf)

−Cµ˜

2 kEsjsk^div,Ωf

kKk²div,Ωf +kvk²H¹(Ωf)

−Cµ˜ kEfgkH¹(Ωf)kKk²div,Ωf

≥ η

1 +C_p² −Cµ˜

2 kEsjsk^div,Ωf

!

kvk²H¹(Ωf)

+ σ⁻¹−Cµ˜

2 kEsjsk^div,Ωf −Cµ˜ kEfgkH¹(Ωf)

!

kKk²div,Ωf ≥αk(v,K)k²X, where we have denoted ˜C=CC3C_s²/C_s¹, and

α= min η

1 +C_p²−Cµ˜

2 kEsjsk^div,Ωf, σ⁻¹−Cµ˜

2 kEsjsk^div,Ωf −Cµ˜ kEfgkH¹(Ωf)

!

. (29) If

η >(1 +C_p²)Cµ˜

2 kEsjskdiv,Ωf and σ⁻¹>Cµ˜

kEsjskdiv,Ωf

2 +kEfgkH¹(Ω_f)

, (30)

then, the bilinear forma((u,J);·,·) is uniformly elliptic inX₀ with constantαgiven in (29).

2. Let (wi,Ji),(u,Y),(v,K)∈X₀, i= 1,2. Then,

|a((w1,J1); (u,Y),(v,K))−a((w2,J2); (u,Y),(v,K))|

= Z

Ωf

[Y× BΩ(J2−J1)]·v− Z

Ωf

[K× BΩ(J2−J1)]·u

≤Cµ˜ kJ2−J1k^div,Ωf kYk^div,ΩfkvkH¹(Ωf)+kKk^div,ΩfkukH¹(Ωf)

≤2√

2 ˜Cµk(w1,J1)−(w2,J2)k^Xk(u,Y)k^Xk(v,K)k^X. Thus, the bilinear formais Lipschitz continuous with constant 2√

2 ˜Cµ.

3. Let (v,K)∈X0and letk · k^X′ denote the norm of the dual space ofX. Then, we have:

F(v,K)≤ ηkEfgkH¹(Ωf)+C^′kEsjskdiv,Ωf(kBekL³(Ωf)+kBskL³(Ωf))

kvkH¹(Ωf)

+ σ⁻¹kEsjsk^div,Ωf +C^′kEfgkH¹(Ωf)(kBekL³(Ωf)+kBskL³(Ωf))

kKk^div,Ωf,

withC^′=C3C_s²/C_s¹. Thus,||F||^X′ ≤CF, where, owing to (18), CF =√

2 max ηkEfgkH¹(Ωf)+C^′kEsjskdiv,Ωf(kBekL³(Ωf)+CµkEsjskdiv,Ωf +CµkJskdiv,Ωs), σ⁻¹kEsjsk^div,Ωf +C^′kEfgkH¹(Ωf)(kBekL³(Ωf)+CµkEsjsk^div,Ωf+CµkJsk^div,Ωs)

.

4. If the data of the problem satisfy (30), and there holds 2√

2 ˜CCFµ < α², Theorem 2.1 guarantees that there exists a unique solution (u0,Jf,0)∈X₀ to (28).

5. The bilinear formb is continuous: indeed, for all (u,J)∈X, (q, ϕ)∈Q₀,

|b((u,J),(q, ϕ))| ≤ kqkL²(Ωf)kdivukL²(Ωf)+kϕkL²(Ωf)kdivJkL²(Ωf)≤ k(u,J)k^Xk(q, ϕ)k^Q. Moreover, it is well-known that inf-sup conditions hold between the spacesH¹₀(Ωf) andQ0(see, e.g., [26] p. 157), and betweenH₀(div; Ωf) andQ0(see, e.g., [25] p. 238), with positive constants, say,β1>0 andβ2>0, respectively:

∀q∈Q0 ∃u∈H¹₀(Ωf), u6=0: − Z

Ωf

qdivu≥β1kqkL²(Ω_f)kukH¹(Ω_f), (31)

∀ϕ∈Q0 ∃J∈H₀(div; Ωf), J6=0: − Z

Ωf

ϕdivJ≥β2kϕkL²(Ωf)kJkdiv,Ωf . (32) Then, for all (q, ϕ)∈Q₀, there exists (u,J)∈X, (u,J)6= (0,0), such that:

b((u,J),(q, ϕ)) ≥ β1kqkL²(Ω_f)kukH¹(Ω_f)+β2kϕkL²(Ω_f)kJkdiv,Ωf

≥ min(β1, β2)(kqkL²(Ωf)kukH¹(Ωf)+kϕkL²(Ωf)kJk^div,Ωf)

≥ min(β1, β2) min(kukH¹(Ωf),kJkdiv,Ωf)(kqkL²(Ωf)+kϕkL²(Ωf))

≥ min(β1, β2)k(q, ϕ)k^Qmin(kukH¹(Ωf),kJk^div,Ωf)

·kukH¹(Ωf)+kJk^div,Ωf

kukH¹(Ω_f)+kJkdiv,Ωf

≥ min(β1, β2)k(q, ϕ)k^Qmin(kukH¹(Ω_f),kJkdiv,Ωf) kukH¹(Ωf)+kJkdiv,Ωf

k(u,J)k^X

≥ βk(q, ϕ)k^Qk(u,J)k^X. (33)

Thus, the bilinear form b satisfies an inf-sup condition with constantβ =Cβmin(β1, β2), where 0 < Cβ < 1/3. Remark that in order to guarantee the inf-sup condition (33), we need only to ensure that the “local” inf-sup conditions (31) and (32) hold for the velocity–pressure spaces, and for the electric currents–potential spaces, respectively. No additional compatibility condition is required.

6. Thanks to (33), Theorem 2.2 guarantees that there exists a unique solution (u0,Jf,0) ∈ X,

(p, φ)∈Q₀ of (27). 2

2.3 A simplified linear problem

In some applications like in the case of weakly conductive fluids (e.g., salty water), or, more generally, for low magnetic Reynolds numbers, the magnetic field BΩ(Jf,0) can be neglected in comparison to the assigned external magnetic fields, that we generically indicate byB^∗ (see, e.g., [8]).

In such cases the MHD system reduces to a linear saddle-point problem in Ωf. Indeed, the nonlinear coupling terms (23) become:

− Z

Ωf

[Jf×B^∗]·v and Z

Ωf

[K×B^∗]·u.

This saddle-point problem in Ωf may also be encountered as a step of an iterative scheme to solve the MHD system (27), when one linearizes the problem in Ωf considering the magnetic field, say B^(k), computed at the previous iterate. We will illustrate this issue in Sect. 3.

In this simplified case, the well-posedness of the linear MHD problem can be proved using the classical theory by Brezzi [3] without any assumption on the physical data. Let us briefly show it.

For all (u,J), (v,K)∈X, we define the bilinear formalin(·,·) :X×X →R, alin((u,J),(v,K)) =

Z

Ω_f

η∇u· ∇v+ Z

Ω_f

σ⁻¹J·K− Z

Ω_f

[J×B^∗]·v+ Z

Ω_f

[K×B^∗]·u, and the linear functional: F^lin:X→R

Flin(v,K) =− Z

Ωf

η∇(Efg)· ∇v− Z

Ωf

σ⁻¹Esjs·K+ Z

Ωf

[Esjs×B^∗]·v− Z

Ωf

[K×B^∗]·Efg. The weak form of the simplified linear MHD problem reads: find (u0,Jf,0)∈X, (p, φ)∈Q₀ such that

alin((u0,Jf,0),(v,K)) +b((v,K),(p, φ)) = F^lin(v,K) ∀(v,K)∈X,

b((u0,Jf,0),(q, ϕ)) = 0 ∀(q, ϕ)∈Q₀ . (34) We can prove the following result.

Lemma 2.1 Assuming that B^∗ ∈(L³(Ωf))³, alin(·,·) is continuous on X×X and coercive on X₀×X₀. Moreover, F^lin is a continuous linear functional on X.

Proof. Concerning the continuity ofalin, we have:

|alin((u,J),(v,K))| ≤ ηkukH¹(Ωf)kvkH¹(Ωf)+σ⁻¹kJk^div,ΩfkKk^div,Ωf

+C^′kB^∗kL³(Ωf)(kJkdiv,ΩfkvkH¹(Ωf)+kKkdiv,ΩfkukH¹(Ωf))

≤ γlink(u,J)k^Xk(v,K)k^X, whereγlin= 2 max(η, σ⁻¹, C^′kB^∗kL³(Ωf)). On the other hand,

alin((v,K),(v,K))≥ η

1 +C_p²kvk²H¹(Ωf)+σ⁻¹kKk²div,Ωf ≥αlink(v,K)k²X, ∀(v,K)∈X₀, withαlin= min(η/(1 +C_p²), σ⁻¹).

Finally, the functionalF^lin is continuous onX:

|Flin(v,K)| ≤ ηkEfgkH¹(Ωf)kvkH¹(Ωf)+σ⁻¹kEsjskdiv,ΩfkKkdiv,Ωf

+C^′kB^∗kL³(Ωf)(kEsjskdiv,ΩfkvkH¹(Ωf)+kKkdiv,ΩfkEfgkH¹(Ωf))

≤ CFlink(v,K)k^X, ∀(v,K)∈X, with constant

CFlin = √

2 max ηkEfgkH¹(Ωf)+C^′kB^∗kL³(Ωf)kEsjskdiv,Ωf, σ⁻¹kEsjsk^div,Ωf+C^′kB^∗kL³(Ωf)kEfgkH¹(Ωf)

.

2 Then, we can prove the well-posedness of (34).

Proposition 2.1 The linear coupled problem (34) admits a unique solution (u0,Jf,0) ∈ X, (p, φ)∈Q₀ which satisfies the a-priori estimates:

k(u0,Jf,0)k^X≤ CF_lin

αlin

, k(p, φ)k^Q ≤ 1 β

1 + γlin

αlin

CF_lin,

whereαlin, γlin, CFlin are defined in Lemma 2.1, and β is the inf-sup constant given in (33).

Proof. It is a straightforward consequence of the theorem by Brezzi [3], whose hypotheses are satisfied thanks to Lemma 2.1, and to the fact that the bilinear form bis continuous and fulfills

an inf-sup condition as shown in Proposition 2.1. 2

3 Iterative solution methods

The proof of the well-posedness of (27), which is essentially based on Banach’s contraction theorem (see, e.g., [29]), suggests that the solution of the coupled problem might be computed by successive approximations. More precisely, starting from any (u⁽⁰⁾₀ ,J⁽⁰⁾_f,0) ∈ X, (p⁽⁰⁾, φ⁽⁰⁾) ∈ Q₀, one can construct the sequences (u^(k)₀ ,J^(k)_f,0)∈X, (p^(k), φ^(k))∈Q₀solving at each stepk≥0 the linearized problem

a((u^(k−1)₀ ,J^(k−1)_f,0 ); (u^(k)₀ ,J^(k)_f,0),(v,K)) +b((v,K),(p^(k), φ^(k))) = F(v,K) ∀(v,K)∈X b((u^(k)₀ ,J^(k)_f,0),(q, ϕ)) = 0 ∀(q, ϕ)∈Q₀. This fixed-point scheme converges under the same hypotheses of Proposition 2.1.

Each iteration of this method requires to compute the magnetic fieldBΩ(Jf,0), i.e. to solve (15).

This is usually done by introducing a vector potential A such that curl A = BΩ(Jf,0), and to rewrite (15) as:

curl(µ⁻¹curl A) = Jf,0 in Ω,

A×n = 0 on∂Ω. (35)

Remark that the boundary conditionA×n=0impliesBΩ(Jf,0)·n= 0 on∂Ω.

The problem (35) has not a unique solution, since adding any arbitrary gradient to A gives another solution. Notice that in (21) and (24) we always consider curl A, so that the presence of an arbitrary gradient does not influence the fluid/electric-currents problem in Ωf. However, uniqueness may be recovered either looking forAin the quotient space, say,H₀(curl; Ω)/∇ϕ, or adding a consistent penalization divergence term exploiting the fact that divA= 0.

Another possible approach is to regularize (35) by a perturbation term of orderO(ε):

curl(µ⁻¹curl A) +εA = Jf,0 in Ω,

A×n = 0 on∂Ω, (36)

0< ε≪1 being a suitably chosen regularization parameter.

The weak form of (36) reads: findA∈H0(curl; Ω) such that Z

Ω

µ⁻¹curl A·curl W+ε Z

Ω

A·W= Z

Ω

Jf,0·W ∀W∈H₀(curl; Ω). (37) Following the idea of successive approximations, we would like to set up an iterative method where the magnetic-field problem defined in Ω can be solved separately from the fluid/electric-currents problem in the fluid domain Ωf. Possibly, also a splitting of the fluid and the electric-currents subproblems might be envisaged, in order to set up a completely decoupled scheme. Thus, we propose the following algorithm.

GivenJ⁽⁰⁾_f,0∈H⁰₀(div; Ωf), let

J⁽⁰⁾_f,0=





J⁽⁰⁾_f,0 in Ωf, 0 in Ωs,

0 in Ω\(Ωf∪Ωs).

Then, fork≥0,

1. findA^(k)∈H₀(curl; Ω) such that Z

Ω

µ⁻¹curl A^(k)·curl W+ε Z

Ω

A^(k)·W= Z

Ω

J^(k)_f,0·W ∀W∈H0(curl; Ω) ; (38)

2. find (u^(k+1)₀ , p^(k+1))∈H¹₀(Ωf)×Q0, (J^(k+1)_f,0 , φ^(k+1))∈H₀(div; Ωf)×Q0 such that Z

Ωf

η ∇u^(k+1)₀ · ∇v− Z

Ωf

p^(k+1) divv− Z

Ωf

[J^(k+k_f,0 ^c)×(Be+Bs+curl A^(k))]·v

− Z

Ωf

[Esjs×curl A^(k)]·v=− Z

Ωf

η ∇(Efg)· ∇v+ Z

Ωf

[Esjs×(Be+Bs)]·v (39) Z

Ωf

qdivu^(k+1)₀ = 0 (40)

Z

Ωf

σ⁻¹J^(k+1)_f,0 ·K− Z

Ωf

φ^(k+1) divK+ Z

Ωf

[K×(Be+Bs+curl A^(k))]·u^(k+k₀ ^f) +

Z

Ωf

[K×curl A^(k)]·Efg=− Z

Ωf

σ⁻¹Esjs·K− Z

Ωf

[K×(Be+Bs)]·Efg(41) Z

Ωf

ϕdivJ^(k+1)_f,0 = 0 (42)

for all v ∈ H¹₀(Ωf), q ∈ Q0, K ∈ H₀(div; Ωf), ϕ ∈ Q0. kc and kf are suitably chosen iteration indeces.

Algorithm (38)-(42) solves the magnetic-field problem in Ω separately from the fluid/electric- currents problem in Ωf. Moreover, the latter can be dealt with in two ways according to the choice of the indeceskc andkf. Indeed, takingkc= 0 andkf = 1, the problem in Ωf is decoupled and we have to solve (39)-(40) before (41)-(42). On the other hand, ifkc=kf = 1, the problem in Ωf is coupled, and at each iteration we have to solve a linear saddle-point problem like (34), where nowB^∗=Be+Bs+curl A^(k).

Finally, an acceleration step might be considered:

J^(k+1)_f,0 ←θJ^(k+1)_f,0 + (1−θ)J^(k)_f,0 , with 0< θ≤1. (43)

3.1 Convergence of the iterative method

We consider now the issue of convergence of (38)-(42). We take the case kc = 0 and kf = 1 (completely decoupled algorithm) and we assumeε= 0. We can prove the following convergence result, which also proves the existence and uniqueness of the solution of the MHD problem in its formulation with the five unknowns: A,u,p,Jf and φ.

Proposition 3.1 Under the same hypotheses of Proposition 2.1, there exists a positive constant ρJ>0 such that ifJ^(k)_f,0∈BJ, withBJ={J∈H₀(div; Ωf)| kJkL²(Ωf)≤ρJ}, then the iterations (38)-(42) converge to the unique solution of (27).

Proof. The proof is based on Banach’s contraction theorem and it consists essentially of two parts.

1. There exists a positive valueρJ>0such that ifkJ^(k)_f,0kL²(Ωf)≤ρJ, thenkJ^(k+1)_f,0 kL²(Ωf)≤ρJ. Indeed, thanks to (17), there holds

kcurl A^(k)kL³(Ωf)≤CµkJ^(k)_f,0kL²(Ωf). (44) Setting v=u^(k+1)₀ in (39) and using (44), we have:

ku^(k+1)₀ kH¹(Ωf) ≤ C3(1 +C_p²)

η kBe+BskL³(Ωf)+CµkEsjskL²(Ωf)

kJ^(k)_f,0kL²(Ωf)

+C3C(1 +C_p²)µ

ηkJ^(k)_f,0k²L²(Ωf)+ (1 +C_p²)kEfgkH¹(Ωf). (45)

Then, taking K=J^(k+1)_f,0 in (41), we have σ⁻¹kJ^(k+1)_f,0 kL²(Ωf) ≤

kBe+BskL³(Ωf)+kcurl A^(k)kL³(Ωf)

·

ku^(k+1)₀ kH¹(Ωf)+kEfgkH¹(Ωf)

+σ⁻¹kEsjskL²(Ωf). Finally, (44) and (45) give

kJ^(k+1)_f,0 kL²(Ωf) ≤ kJ^(k)_f,0k³L²(Ωf)C3C²(1 +C_p²)µ²σ η +kJ^(k)_f,0k²L²(Ωf)C3C(1 +C_p²)µσ

η 2kBe+BskL³(Ωf)+CµkEsjskL²(Ωf)

+kJ^(k)_f,0kL²(Ωf)

C3(1 +C_p²)σ

ηkBe+BskL³(Ωf) kBe+BskL³(Ωf)

+CµkEsjskL²(Ωf)

+C(2 +C_p²)µσkEfgkH¹(Ωf)

+(2 +Cp²)σkEfgkH¹(Ωf)kBe+BskL³(Ωf). (46) ChoosingρJ = 2(2 +C_p²)σkEfgkH¹(Ωf)kBe+BskL³(Ωf), and assuming thatkJ^(k)_f,0kL²(Ωf)≤ ρJ (we may always take kJ⁽⁰⁾_f,0kL²(Ωf) ≤ρJ), then kJ^(k+1)_f,0 kL²(Ωf) ≤ ρJ, provided that, for example, σ is sufficiently small, or η is sufficiently large, or µ and kBe+BskL³(Ωf) are sufficiently small.

2. LetJ^(k)_i ∈H₀(div; Ωf) and letA^(k)_i ,u^(k+1)_i (i= 1,2) be the corresponding magnetic fields and fluid velocities given by (38) and (39), respectively. Then, it holds

kcurl(A^(k)₁ −A^(k)₂ )kL³(Ω_f)≤CµkJ^(k)₁ −J^(k)₂ kL²(Ω_f). (47) Moreover,u^(k+1)₁ −u^(k+1)₂ satisfies

Z

Ωf

η∇(u^(k+1)₁ −u^(k+1)₂ )· ∇v− Z

Ωf

(p^(k+1)₁ −p^(k+1)₂ )divv

− Z

Ωf

[(J^(k)₁ −J^(k)₂ )×(Be+Bs)]·v− Z

Ωf

[Esjs×curl(A^(k)₁ −A^(k)₂ )]·v

− Z

Ωf

[J^(k)₁ ×curl A^(k)₁ ]·v+ Z

Ωf

[J^(k)₂ ×curl A^(k)₂ ]·v= 0 ∀v∈H¹₀(Ωf). Using (47), we have

ku^(k+1)₁ −u^(k+1)₂ kH¹(Ωf) ≤ C3(1 +C_p²) η

kBe+BskL³(Ωf)+CµkEsjskL²(Ωf)

+Cµ(kJ^(k)₁ kL²(Ω_f)+kJ^(k)₂ kL²(Ω_f))

kJ^(k)₁ −J^(k)₂ kL²(Ω_f).(48)

Finally, from (41) it follows

kJ^(k+1)₁ −J^(k+1)₂ kL²(Ωf) ≤ C3σ

kBe+BskL³(Ωf)ku^(k+1)₁ −u^(k+1)₂ kH¹(Ωf)

+kcurl(A^(k)₁ −A^(k)₂ )kL³(Ω_f)kEfgkH¹(Ω_f)

+kcurl A^(k)₁ kL³(Ωf)ku^(k+1)₁ −u^(k+1)₂ kH¹(Ωf)

+kcurl(A^(k)₁ −A^(k)₂ )kL³(Ωf)ku^(k+1)₂ kH¹(Ωf)

.