A MHD problem on unbounded domains - Coupling of FEM and BEM

A MHD problem on unbounded domains - Coupling of FEM and BEM Wiebke Lemster and Gert Lube Abstract We consider the MHD problem on R 3 = Ω Ω E, where Ω is a bounded, conducting Lipschitz domain and Ω E is an insulating region. After semidiscretization in time, we apply a finite element approach in Ω. A boundary element approach is used in Ω E. We present results on the well-posedness of the continuous problem and for the semidiscrete coupled problems arising within each time step. Finally we show the quasi-optimality of a regularised FE discretisation within each time step. 1 Introduction Magnetohydrodynamics MHD is the study of the flow of electrically conducting fluids in the presence of magnetic fields. In the so-called direct problem, the magnetic induction B and the electric field E are unknown. Some efforts have been made to treat this problem with Lagrange finite elements for bounded domains cf. [2, 6]. Our aim is to extend this to R 3. In the bounded, simply connected conducting domain the model is time-dependent and nonlinear. For this reason we apply finite elements there. We regularise the system with a pressure stabilisation like method to get a saddle point problem. In the unbounded part, the model reduces to a Laplace equation. Therefore, we apply a boundary element technique for this domain. We use a symmetric coupling technique cf. [15, 4] to link both methods. For more informations of symmetric coupling of FEM and BEM, see [4, 7, 9]. Properties of the used spaces and other helpful results can be found in [1, 5, 11, 12, 16]. The paper is organised as follows: In Section 2 and 3 we derive a variational formulation and Wiebke Lemster Department of Mathematics and Computer Science, University Göttingen, Germany, e-mail: lemster@math.uni-goettingen.de Gert Lube Department of Mathematics and Computer Science, University Göttingen, Germany, e-mail: lube@math.uni-goettingen.de 1

2 Wiebke Lemster and Gert Lube show the well-posedness. Section 4 deals with the time-discretised problem. Section 5 is related to the finite element approach. In Section 6 we give some final remarks. 2 Variational Formulation of the MHD Model In this section we state the problem to find a vector potential for the interior solution and a scalar potential in the exterior domain. By we denote the L 2 -norm in the conducting domain Ω which is induced by the L 2 scalar product,. We seek the interior solution in the space H 1 0,T ;H := { B L 2 0,T ;H B L 2 0,T ;H }, H := { B L 2 Ω B L 2 Ω }. Ω is divided into two disjoint parts, Ω 1 and Ω 2. We require Ω 2 = /0. We assume the known velocity w and the function f to be zero outside of Ω 2. The electric field is denoted by E the, the magnetic induction by B, the magnetic permeability by 0 < µ 1 µ µ 2 and the electric conductivity by 0 σ 1 σ σ 2. We can state the following problem: B = E, 1 t 1 µ B = σ E + w B + R f 1 + s B 2 B, 2 B = 0, 3 B = o x 1 for x. 4 We want to replace B by a potential ansatz. For the interior domain Ω we set B = u. From 1 we get E = u t + φ c. Since φ c is only unique up to a constant we choose it such that Ω φ c dx = 0. In the non-conducting region Ω E it holds that σ = 0 and µ 0 := µ is constant. Therefore 2 simplifies to B = 0. We set B = Φ in Ω E. Hence equation 3 reduces to a Laplace equation. The fundamental solution for this equation is denoted by Ux,y := 4π 1 1 x y and its normal derivative at the boundary by T x,y. We have the representation for Φ by the single-layer- and the double-layer-potential Φx = Ux,y n Φy dsy + T x,yφy dsy, x Ω E. 5 To normalise the vector potential u we use the spaces H 1 Ω := { q H 1 Ω q,1 = 0 }, H := { v H v, q = 0 q H 1 Ω }.

A MHD problem on unbounded domains - Coupling of FEM and BEM 3 The transmission conditions [B n] = 0 and [H n] = 0 imply the interface conditions 1 1 µ u n = Φ n and u n = Φ on. µ 0 n This leads to the following problem: Problem 1 Find u H 1 0,T ;H such that for all v H and almost all t 0,T 0 = σ u 1 1 t,v µ u, v µ u n,v σ f u R 1 + s u 2,v σw u,v. 6 In order to replace the boundary term in 6 by a formulation of the exterior problem, we use the Stokes formula on the boundary to get [ ] 1 µ u n v ds = 1µ0 ΦT n v ds with T n v := v n. If we apply this result to 5 and take the boundary values, we derive the following Calderòn equations 2V T n ux = Φx + 2KΦx, 2DΦx = T n ux 2K T n ux, where the integral operators are given by cf. [3] V ψx := K ψ x := Ux,yψy dsy, By defining the bilinear forms Dψx := n x n x Ux,yψy dsy, Kψx := T x, yψy dsy, T x, yψy dsy. A u,v := µ 1 u, v σw u,v, A t u,v := σ u t,v, K φ,v := 1 2 Id + Kφ,T nv, Dφ,ψ := Dφ,ψ + φ,1 ψ,1, V u,v := V T n u,t n v and the forms A V u,φ,v := 1 µ 0 V T n u 1 2 φ Kφ,T nv, A D u,φ,ψ := DΦ + 1 2 T nu + K T n u,ψ + Φ,1 ψ,1, f A nl u,v := R σ 1 + s u 2 u,v,

4 Wiebke Lemster and Gert Lube we get the variational problem: Problem 2 Find u,φ H 1 0,T ;H L 2 0,T ;H 1 2 such that for all v,ψ H H 1 2 and almost all t 0,T A t u + A u + A nl u + 1 µ 0 V u 1 µ 0 K Φ = 0, 7 DΦ + K u = 0. 8 3 Continuous Problem We sketch the proof of the existence and uniqueness of a solution for the continuous problem. All operators are bounded, D is invertible cf. e.g. [9] and A + A nl fullfills a Garding inequality cf. [2, 10]. Hence, the second equation can be transformed in such a way that we get from 7 an equation which only depends on u. Problem 3 Find u H 1 0,T ;H such that for almost all t 0,T Su + A t u + A nl u := A + 1 V + 1 K D 1 K u + A t u + A nl u = 0. 9 µ 0 µ 0 Therefore, we need an existence theorem for non-linear evolution problems cf. [17] Theorem 30.A.. Theorem 3.1 Let V X V be a Gelfand triple with dim V =. Assume that the operators A := S + A nl : V V fullfill the following conditions for p, q 1, with 1 p + 1 q = 1 and 0 < T < : a At is coercive for all t 0,T, i.e., there exist constants M > 0 and Λ 0 such that Atv,v M v p V Λ v V t 0,T. b At : V V is monotone and hemicontinuous for all t 0,T. c There exist a nonnegative function K 1 L q 0,T and a constant K 2 > 0 such that for all v V and t 0,T Atv V K 1 t + K 2 v p 1 V. d The function t At is weakly measurable, i.e., the function t Atu,v V is measurable for t 0,T and all u, v V. Then there exists a unique solution u W 1,p 0,T ;X for u 0 X of u t + Atut = 0, u0 = u 0. We have to modify Problem 3 in order to satisfy a coercivity condition. We multiply 9 with the function e κt, where κ is the constant for the L 2 -term in the Garding

A MHD problem on unbounded domains - Coupling of FEM and BEM 5 inequality which S+A nl satisfies. We denote by ũ := e κt u the new scaled potential. S is linear, so we can replace e κt Su by Sũ. To get an equivalent problem, we have to scale the constant s in A nl by e 2κt. In addition, we have e κt A t u = A t ũ + σκũ. We set u := ũ and S := S + σκid. Proof. We set p = q = 2, V = H r and X = L 2 Ω. a To show the first condition, we first note V + K D 1 K v,v 0 v H. The coercitivity follows with η = κ and c k 1 µ 2 from A nl tv,v + Stv,v 1 v 2 σ 2 u µ L Ω v v 2 σ 2 R f L Ω v v + σ 1η v 2 1 c k v 2 κc k v 2 + σ 1 η v 2 µ 2 2 { 1 min,σ 1 η κc } k v 2 H 2µ 2 2. b The monotonicity can be shown similarly. A nl is Lipschitz continuous, f Φ 1 1 + s Φ 1 2 f Φ 2 1 + s Φ 2 2,v 3 f L Ω Φ 1 Φ 2 v Φ 1,Φ 2,v H, and S is linear and bounded. Therefore both operators are hemicontinuous cf. [17] Figure 27.1. c+d The third condition follows from the boundedness of the two operators S and A nl. The operator A does not depend explicitly on time. By Gronwall inequality one can derive with 2 := σ2 2 w 2 L Ω + R2 f 2 L Ω, Lt := µ 2 t σ 1 0 2 s ds the following a-priori estimate for t [0, T ] ut u0 e Lt, u 2µ 2 t u0 e Lt. 4 Semidiscrete Problem We discretise Problem 3 in time by the implicit Euler scheme. An existence result is given. The proof relies on the main theorem of strongly monotone operators.

6 Wiebke Lemster and Gert Lube Theorem 4.1 Suppose X is a real Hilbert space and the operator A : X X is strongly monotone and Lipschitz continuous on X. Then, for each b X, the operator equation has a unique solution u X. Au = b For simplicity we use an equidistant partition of the time interval [0,T ] into M parts with time step size τ = M T. Hence we obtain the following semidiscrete problem: Problem 4 Find u n,φ n H H 1 2 such that for all v,ψ H H 1 2 0 = σ un u n 1 τ 1,v + µ un, v R σ σw n u n,v + A V u n,φ n,v, 0 = A D u n,φ n,ψ. f n u n 1 + s u n 2,v For an estimate of the semidiscrete solution we need the discrete Gronwall lemma. Remark 4.2 Let {z n } Nτ n=1 z n C 1 + τc 2 be a sequence of nonnegative real numbers which fullfill z i i=0,...,n 1 for n = k,...,nτ with C 1 and C 2 independent of τ. Let z i z/k for i = 1,...,k 1. Then we obtain z n τc 2 z +C 1 1 + τc 2 n k for n = k,...,nτ. Hence, we get the main theorem of this section: Theorem 4.3 Problem 4 has a unique solution, if τ is chosen such that for c k < 1 µ 2 it holds c 1 τ := 1/τ κ n /2 8R 2 f n 2 L Ω /2c k > 0. Furthermore we obtain for constant σ the following estimate: max 1 n M un 2 L 2 Ω + τ n=1,...,m u n 2 L 2 Ω C u 0 2 L 2 Ω. Proof. The strong monotonicity and the boundedness of S n + A nl, defined by S n v, v := 1 1 1 τ v,ṽ + µ v, ṽ + V + K D 1 K v,ṽ µ 0 σw n v,ṽ, R f Anl n n u,v := 1 + s u 2 u, v, can be proved by similar arguments as presented in the previous section. Therewith we can apply Theorem 4.1 to prove existence and uniqueness.

A MHD problem on unbounded domains - Coupling of FEM and BEM 7 5 Finite Element and Boundary Element Approach By introducing a Lagrange multiplier p n 4 can be regularised with an pressure stabilisation like term ε p n, q. Problem 5 Find ũ n, p n, Φ n H H 1 Ω H 1 2 such that for all test functions v,q,ψ H H 1 Ω H 1 2 σ u n 1,v = τ σ 1,v τ ũn + µ ũn, v + A nlũn n,v σw n ũ n,v + A V ũ n, Φ n,v + p n,v, ũ n, q = ε p n, q + p n,1q,1, 0 = A D ũ n, Φ n,ψ. One can also introduce the Lagrange multiplier p n in the original semi-discrete problem. For time steps like in Theorem 4.3 we get the error for the regularisation c 1 τ u n ũ n 2 + u n ũ n 2 + εµ 2 p n p n 2 εµ 2 p n 2. Consider the Galerkin discretisation of the coupled Problem 5. Let X h H be the lowest order edge element space see [13] and M h H 1 Ω the space of piecewise linear elements on a shape-regular tetrahedral mesh of Ω with mesh size h. The space W h H 1 2 is defined by the traces on of piecewise linear nodal elements. Problem 6 Find ũ n h, pn h, Φ n h X h M h W h such that for all test functions v,q,ψ X h M h W h σ u n 1 τ h,v = σ 1 τ ũn h,v + µ ũn h, v σw n ũ n h,v + A n nlũn h,v + A V ũ n h, Φ n h,v + pn h,v, ũ n h, q = ε pn h, q + pn h,1q,1, 0 = A D ũ n h, Φ n h,ψ. Lemma 5.1 The error between the regularised semi-discrete problem and its Galerkin formulation can be estimated as follows ũ n ũ n h H inf v h X h u n v h H + inf q h M h p q h H 1 Ω + inf ψ h W h Φ ψ h H 1 2. Proof. Defining an operator S n like S n for the new problem leads to the additional term B C 1 Bv,ṽ. The ellipticity of this operator and some other transformations lead to the statement cf. [9, 10].

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