Implementation of the dg method

Transcription

1 Implementation of the dg method Matteo Cicuttin CERMICS - École Nationale des Ponts et Chaussées Marne-la-Vallée March 6, 2017

2 Outline Model problem, fix notation Representing polynomials Computing integrals Assembly, solve, postprocess Matlab code to solve 1D model problem

3 Model problem Let Ω R d with d {1, 2, 3} be an open, bounded and connected polytopal domain. We will consider the model problem u = f in Ω, u = 0 on Ω, with f L 2 (Ω). By setting V := H 1 0 (Ω), its weak form is Find u V such that ( u, v) Ω = (f, v) Ω for all v V.

4 Notation I: mesh and elements Let be a suitable subdivision of Ω in polytopal elements. We define the skeleton Γ := Moreover, we define: Γ int = Γ \ Ω + and generic elements sharing a face + n n + int e := + Γ int n + and n normals of + and on e

5 Notation II: jump and average Let q : Ω R and φ : R d Average: {q} e := 1 2 (q+ + q ) {φ} e := 1 2 (φ+ + φ ) Jump: [q ] e := q + n + + q n [φ] e := φ + n + + φ n If e belongs to the boundary of the domain (i.e. e Ω) we just drop the terms with : {φ} e := φ + and [q ] e := q + n +

6 Symmetric Interior Penalty dg SIP dg method is derived from the following equation: u v ({ u} [v ]+{ v} [u] η [u] [v ]) = Γ Ω fv = (f, v) In SIP dg we approximate the solution of our equation using piecewise continuous polynomials on the elements. S p h := { w h L 2 (Ω) : w h P k d( ), } SIP dg method will then be: Find u h S p h s.t. a(u h, v h ) = (f, v h ), v h S p h where a(u, v) : S p h Sp h R a(u, v) = u v ({ u} [v ] + { v} [u] η [u] [v ]) Γ

7 Representing polynomials We need to be able to represent d-variate polynomials of degree k on cells: p(x) P k d ( ). We introduce a basis of Pk d ( ): in 1D for example {1, x, x 2,...}. Once the basis is fixed, the coefficients p i fully determine the polynomial. N k d p(x) = p i φ i (x) i=1 where Nd k is the size of the basis for Pk d ( ): ( ) k + d Nd k = d he coefficients of the basis will be called degrees of freedom (DoFs).

8 Scaled monomial basis It is better, however, to use the so-called scaled monomial basis centered on the barycenter x of : { d } d P k d( ) = span x αi,i 1 i d 0 α i k. i=1 where x = (x x )/h and x,i is the i-th component of x. i= k=0 k=1 k=2 k=3 k=4 k=

9 Integrals and mass matrix We want to compute p(x)q(x), where p, q Pk d. As we discussed, we can express polynomials as linear combinations of basis functions: Introduce mass matrix: p(x)q(x) = Rewrite using mass matrix: N k d q i φ i (x) p j φ j (x) i=1 M ij = Let p = {p j } and q = {q i }: p(x)q(x) = N k d j=1 φ i (x)φ j (x) N k k d N d q i i=1 j=1 pq = q Mp. M ij p j.

10 Mass matrix he integral is now hidden inside the mass matrix M ij = φ i (x)φ j (x). How to compute it? We need to do numerical integration using quadrature rules.

11 Quadrature rules Quadrature Q = (Q w, Q p ): collection of Q points and associated weights. Definite integrals are computed as weighted sum of evaluations of the integrand on the points prescribed by the quadrature: 1 Q f(x) dx = w i f(x i ), 1 i=1 w i Q w, x i Q p A quadrature is given on a specific reference element. Because of that you need to map it on your physical element. In particular: Map points from the reference to physical (affine transform) Multiply weights by measure of physical element (Jacobian)

12 Quadratures in practice here are lots of different types of quadrature. Keywords for simplices: 1D: Gauss, Gauss-Lobatto,... 2D: Dunavant, Grundmann-Moeller,... 3D: Keast, ARBQ, Grundmann-Moeller,... On quads, we usually tensorize. Look here for code: jburkardt/. In Matlab code we use Golub-Welsch algorithm to compute Gauss quadrature.

13 Mass matrix and stiffness matrix We are now able to compute the mass matrix: M ij = φ i (x)φ j (x) = Q i=1 w i φ i ( x i )φ j ( x i ), where w i and x i are the quadrature weights and points after the transformations. It is possible to build the stiffness matrix in the same way: S ij = φ i (x) φ j (x) = hese matrices will have size N k d N k d. Q i=1 w i φ i ( x i ) φ j ( x i ).

14 A code he numerical solution of a PDE, in general, consists of three phases: Assembly: Compute the local contributions for every and put them in the global system matrix, Solve: Solve the linear system Au = f, Postprocess: Recover the values of the solution from the DoFs computed in the previous step.

15 Assembly - Cell contributions u h v h ({ u h } [v h ]+{ v h } [u h ] η [u h ] [v h ]) = (f, v h ) Γ Remember: v h can be any function in S p h ; we choose all the coefficients to be 1 for linearity, you can write one equation per basis function the coefficients u j are the unknowns hen, for the terms in red, locally we get for 1 n Nd k u 1 φ 1 φ u n φ n φ 1 = fφ 1 u 1. φ 1 φ n u n φ n φ n = fφ n We ve got Nd k local equations for each element in.

16 Assembly - Cell contributions We now put the equations we obtained in a global matrix card( ) N d k Local stiffness matrices card( ) N d k = Consider a 1D mesh composed on 5 elements (depicted in blue). Each element gets its own set of equations in the global matrix. he structure of the global matrix is related to the mesh. Knowing the mesh, it is easy to determine the size of the system. We haven t assembled the other terms yet. Note the decoupling.

17 Assembly - Face-related terms u h v h ({ u h } [v h ] + { v h } [u h ] η [u h ] [v h ]) = (f, v h ) Γ + We have three additional terms to assemble We expand them with the expressions for jump and average hey will couple adjacent elements

18 Assembly - Face-related terms [v ] = e{ u} 1 ( u + + u ) (v + n + + v n ) = 2 e = 1 [ ( u+ v + n + ) + ( u + v n ) + ( u v + n + ) + ( u v n ) ] 2 e he terms in red will be on the diagonal he terms in green will be off-diagonal A ++ 1 = 1 2 A + 1 = 1 2 e e u + v + n + A + 1 = 1 2 u v + n + A 1 = 1 2 e e u + v n u v n

19 Assembly - Face-related terms A ++ A + A + A Suppose + = 2 and = 3 You can see that the off-diagonal terms introduce a coupling between adjacent elements Remember that since in 1D faces are just points, integrating means that you need to just evaluate the functions there

20 Assembly - Face-related terms We have the two remaining terms, you handle them exactly as the previous one. e { v} [u] = 1 2 e ( v+ + v ) (u + n + + u n ) e η [u] [v ] = e η(u+ n + + u n ) (v + n + + v n ) Don t forget the two boundary faces!

21 Solve Once we have assembled the problem, we must solve it. In Matlab there are different ways: Use the backslash operator u = A\f Use one of the iterative solvers, pcg() is ok

22 Postprocess By solving, we computed the coefficients u i,n, 1 i Nd k for each element 1 n card( ). o recover the values of the solution at any point, we must evaluate them against the basis. We choose N p equispaced points on each element We evaluate there We plot the result N k d u n (x j ) = u i,n φ i (x i ), i=1 1 j N p