Mixed Finite Elements Method

Transcription

1 Mixed Finite Elements Method A. Ratnani 34, E. Sonnendrücker 34 3 Max-Planck Institut für Plasmaphysik, Garching, Germany 4 Technische Universität München, Garching, Germany Contents Introduction 2. Notations Poisson equation 2 3 Stokes equation 4 3. First weak formulation Second weak formulation Time Harmonic Maxwell problem 5 5 Time Domain Maxwell problem 6 5. First formulation Second formulation Exercises 7

2 Introduction. Notations Scalar-valued test functions will be denoted by ϕ, while ϕ i will denote a scalar basis function, after reordering all the basis functions of a given discrete space. Vector-valued functions will be written in bold, like u, v. Ψ will denote a vector-valued test function, whileψ i will denote a vector-valued basis function (after reordering the basis functions). Even if most of what follows holds for both the 2D and 3D cases, we will restrict our studies to the 2D one. We recall that in 2D, there are two curl operators, one acting on scalars φ = ( y φ, x φ ) and one acting on vectors Ψ = x Ψ y y Ψ x. Differential operators that return a vector (grad, curl) will be written in bold (, ). We also recall the Green formula for the divergence and curl/rotational operators ( F) G = F G + (F n) G, F H(div, ), G H () () ( G) F = G F (G n) F, F H(curl, ), G H () (2) If R d, u = (u, u 2,..., u d ) and Ψ = (Ψ, Ψ 2,..., Ψ d ), we recall the notation 2 Poisson equation u : Ψ = d i= u i Ψ i (3) Let s start with the Poisson problem { 2 φ = f in, φ = 0 on, (4) As we know, the associated variational formulation can be written as a minimization problem over H (). Let us now, introduce an auxiliary variable u = φ. The Poisson problem is then equivalent to the following system u = f in, u = φ in, φ = 0 on, In order to have a weak formulation, we need to multiply the first equation by a scalar-valued test function φ, while the second one will be multiplied by a vector-valued test function Ψ. Therefor, ( u) ϕ = f ϕ, u Ψ = (6) φ Ψ, (5) 2

3 Now let us use the Green s formula for the divergence. Note that the integration by parts will only be done on the second equation. This leads to the following weak formulation ( u) ϕ + f ϕ = 0, u Ψ + φ Ψ = 0, (7) Now, let us consider the bilinear form B (( ) ( )) ( ) u Ψ, = ( u) ϕ + u Ψ + φ Ψ φ ϕ ( ) u Where we consider the trial vector-valued function and the test vector-valued function φ For the sake of clarity, we rewrite the blinear form as the following B (( ) ( )) ( ) u Ψ, = u Ψ + ( u ) ϕ + φ Ψ φ ϕ (8) ( ) Ψ. ϕ Therefor, it is easy to see that our blinear form is symmetric. Up to now, we didn t mention what are the correct spaces for this formulation. Because, there is no differential operator over the unknown φ and the scalar-valued test function ϕ, these functions can be placed in L 2 (). On the other hand, both u, Ψ, u and Ψ must be in L 2 (). Therefor, we need the Hilbert space H(div, ) defined as (9) H(div, ) = { Ψ L 2 (), Ψ L 2 () } (0) Therefor, the mixed weak formulation for the Poisson equation is Find (u, φ) H(div, ) L 2 () such that (Ψ, ϕ) H(div, ) L 2 () u Ψ + φ Ψ = 0, ( u) ϕ = f ϕ, () Now let us introduce the bilinear form a(u, Ψ) = u Ψ, u, Ψ H(div, ) and b(φ, Ψ) = φ Ψ, (φ, Ψ) L2 () H(div, ). The later problem can be written as Find (u, φ) H(div, ) L 2 () such that (Ψ, ϕ) H(div, ) L 2 () a(u, Ψ) + b(φ, Ψ) = r (Ψ), b(ϕ, u) = r 2 (ϕ), (2) 3

4 3 Stokes equation Let u be the velocity of a fluid and p the pressure in the fluid. We denote by f the body forces. The Stokes problem writes 2 u + p = f in, u = g in, u = 0 on, When g = 0, the fluid is said to be incompressible. Because u = u n = 0, the source term g must fullfill the compatitbility condition g = 0. Therefor, the function g will be in the space L0 2 (), defined as } L0 {ϕ 2 () = L 2 (); ϕ = 0 (4) Solving the Stokes equation consists of looking for u H () and p L 2 0 (), where f H () and g L 2 0 (). 3. First weak formulation The first formulation consists on putting the constraint u = 0 directly into the continuous (and discrete) space. We start by introducing the space Find u V such that Ψ H 0 (div, ) (3) H 0 (div, ) = { Ψ H 0 (), Ψ = 0} (5) u : Ψ = f Ψ (6) This bilinear form can be shown to be coercive and one can directly apply the Lax-Milgram theorem. However, this formulation has a major drawback from the practical point of view: the pressure is not solved directly. Moreover, on the discrete level, we need to construct an increasing sequence (V h ) h of subspaces of V. This can be done thanks to the discrete DeRham sequence. 3.2 Second weak formulation Now let us keep the space for u as big as possible (which means the space H0 ()) and then add the constraint u = 0, as a Lagrange multiplier. In this case, the problem writes Find (u, p) H0 () L2 0 () such that (Ψ, ϕ) H 0 () L2 0 () u : Ψ p Ψ = f Ψ (7) ϕ u = 0 4

5 Now let us introduce the bilinear form a(u, Ψ) = u : Ψ, u, Ψ H(div, ) and b(p, Ψ) = p Ψ, (p, Ψ) L2 () H(div, ). The later problem can be written as Find (u, p) H 0 () L2 0 () such that (Ψ, ϕ) H 0 () L2 0 () a(u, Ψ) + b(p, Ψ) = r (Ψ), b(ϕ, u) = r 2 (ϕ), (8) 4 Time Harmonic Maxwell problem The Time Harmonic Maxwell problem writes Find E 0 such that { E = ω 2 E in, E n = 0 on, (9) Two cases occure: either ω is provided and then we only look for E. The other case is known as an eigenvalue problem, and the problem must be solved for both E and ω. If we multiply by a vector-valued function Ψ, and integrating by parts using the curl/rotational Green formula, we get E Ψ = ω 2 E Ψ (20) Again, the space where E lives was not mentioned; let us find what this space can be. We need both E, Ψ, E and Ψ to be in L 2 (), while satisfying the boundary condition E n = 0 on the boundary. Therefor, we need the Hilbert space H 0 (curl, ) defined as where H 0 (curl, ) = {Ψ H(curl, ), Ψ n = 0 on } (2) H(curl, ) = { Ψ L 2 (), Ψ L 2 () } (22) Therefor, in weak form, the problem writes Find E H 0 (curl, ), E 0 such that, Ψ H 0 (curl, ) E Ψ = ω 2 E Ψ (23) 5

6 5 Time Domain Maxwell problem The Maxwell equations write D = ρ B = 0 t B = E t D = H J In the case of linear, isotropic and non-dispersive materials, we have two additional relations: B = µh and D = ɛe. Furthermore, we assume that J = σe. Pluging these relations in (Eq. 24), we get (ɛe) = ρ (µh) = 0 (25) µ t H = E ɛ t E = H J Now, let us assume that both ɛ and µ are constants and equal to. We also, restrict our study to the 2D case and we consider the Transverse Electric (TE) mode. Therefor, the problem is only involving the variables E x, E y and H z. In this case, the Time Domain Maxwell problem writes t H = E t E = H J (26) E = ρ where we consider here that E = (E x, E y ) and H = H z. 5. First formulation The first variational formulation, in the case of perfectly conductin boundary conditions, can be derived using the curl/rotational Green formula in Ampere s law. Therefor, the TE Maxwell problem reads Find (E, H) H 0 (curl, ) L 2 () such that (Ψ, ϕ) H 0 (curl, ) L 2 () d dt E Ψ H Ψ = J Ψ d dt Hϕ + (27) ( E)ϕ = 0, 5.2 Second formulation Using the divergence Green formula in Faraday s law yields the second variational formulation. Find (E, H) H(div, ) H () such that (Ψ, ϕ) H(div, ) H () d dt E Ψ ( H) Ψ = J Ψ d dt Hϕ + (28) E ( ϕ) = 0 (24) 6

7 6 Exercises Exercise 6.. Let us consider the mixed formulation (Eq. 29) of the 2D Poisson problem. Let V h H(div, ) and W h L 2 () be finite dimensional subspaces. The Galerkin method applied to (Eq. 29) leads to the following discrete problem Find (u h, φ h ) V h W h such that (Ψ, ϕ) V h W h u h Ψ + φ h Ψ = 0, We assume that the computational domain is the unit square. ( u h) ϕ = f ϕ, (29) Figure : The computational domain and its triangulation.. Let us assume that V h = span {Ψ, Ψ 2,..., Ψ n } and W h = span {ϕ, ϕ 2,..., ϕ m }. Write the global matrices and the corresponding right hand sides. 2. The latter linear system can be written as ( A B T B 0 ) ( ) u = φ ( ) f g (30) (a) We assume that A is invertible. Show that the system 30 has a unique solution if the matrix BA B T is invertible. (This method is called the Schur Complement) (b) If A is positive definite, show that BA B T is also positive definite. (c) Use the symmetry of A to show that BA B T is also a symmetric matrix. 7

8 3. We consider a triangulation T h constructed as the following: we take a uniform mesh of n n subsquares, where every subsquare is devided into two triangles as in the figure (Fig. ). We consider V h to be the space of piecewise linear vector fields and W h the space of continuous linear scalar fields, (a) For T T h, a given triangle, construct a function v W h, such that T v = 0. Show how to construct a piecewise linear function that is orthogonal to all piecewise constants. (b) Show that v ϕ = 0, ϕ V h, (c) Show that the stiffness matrix is singular. (Hint: Show that the function (0, v) belongs to the kernel of the stiffness matrix), Exercise 6.2 (The Checkerboard instability). In this exercise, we study the Stokes equation with the second weak formulation, when the inf-sup condition is not fullfilled. The compuational domain will be the unit square, with a uniform mesh of n n subsquares Q h = n i, j= Q i, j, as described in figure (Fig. 2). Let us recall that the variational problem can be written as Figure 2: The computational domain and the its mesh. Find (u, p) H 0 () L2 0 () such that (Ψ, ϕ) H 0 () L2 0 () u : Ψ p Ψ = f Ψ ϕ u = 0 (3) We consider V h to be the space of continuous linear vector fields and W h the space of piecewise 8

9 constants scalar fields, i.e. V h = { u h [C 0 ( )] 2 ; T T h, u h [P ] 2, u h = 0 } { } W h = φ h L 2 () C 0 ( ); T T h, φ h P, φ h = 0 For better clarity, we define the function ϕ W h, which is a piecewise constant, by its value in the center of our meshes, i.e. ϕ := ϕ i+ 2, j+, while for any function Ψ V h is defined by its 2 values on the vertices of our mesh, i.e. Ψ i, j = ( Ψi, x j, Ψy i, j). The aim of this exercise is to construct a non zero scalar-valued function q such that q Ψ = 0, Ψ V h.. Show that for all (Ψ, ϕ) V h W h, we have ϕ Ψ = ( Ψ x ( n 2 i, j δ x [ϕ] ) i, j + ( Ψy i, j δ y [ϕ] ) i, j) 2 i, j n (32) where the operators δ x and δ y are defined as ( δ x [ϕ] ) i, j := n { } ϕ 2 i+ 2, j+ ϕ 2 i 2, j+ + ϕ 2 i+ 2, j ϕ 2 i 2, j 2 ( δ y [ϕ] ) i, j := n { } ϕ 2 i+ 2, j+ + ϕ 2 i 2, j+ ϕ 2 i+ 2, j ϕ 2 i 2, j 2 and Ψ = (Ψ x, Ψ y ) 2. Prove that ϕ Ψ = 0, Ψ V h if and only if for any i, j n we have ϕ i+ 2, j+ 2 ϕ i 2, j+ 2 = ϕ i 2, j 2 = ϕ i+ 2, j 2 3. Construct such a function by using the values and +. Such oscillatory function is known as a spurious mode. 9