Scientific Computing I

Transcription

1 Scientific Computing I Module 8: An Introduction to Finite Element Methods Tobias Neckel Winter 2013/2014 Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 1

2 Part I: Introduction to Finite Element Methods The Model Problem FEM Main Ingredients Weak Forms and Weak Solutions FEM and the Residual Equation Test and Shape Functions Assembling the System of Equations Example Problem: 1D Poisson Stiffness Matrix for Linear Nodal Basis Functions A Road to Theory Time-Dependent Problems Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 2

3 The Model Problem: Poisson Equation 2D Poisson Equation on unit square: 2 2 u(x, y) + u(x, y) = f (x, y) in Ω = (0, 1)2 x 2 y 2 Dirichlet boundary conditions: u(x, y) = g(x, y) on Ω new for FEM: compute a function to approximate u(x, y) (instead of approximate values at certain grid points) Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 3

4 Finite Elements Main Ingredients 1. solve weak form of PDE to reduce regularity properties u = f v u dx = vf dx allows additional weak solutions 2. compute a function as numerical solution search in a function space W h : u h = j u j ϕ j (x), span{ϕ 1,..., ϕ J } = W h 3. find weak solutions of simple form: for example piecewise linear functions and choose basis functions with local support ( hat functions ) leads to system of linear equations Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 4

5 Weak Forms and Weak Solutions consider a PDE Lu = f (e.g. Lu = u or Lu = u ) transformation to the weak form: Lu, v = vlu dx = vf dx = f, v V a certain class of functions real solution u also solves the weak form however, additional, weak solutions are now allowed Lu, v a bilinear form; often written as: v V a (u, v) = f, v v V Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 5

6 Weak Form of the Poisson Equation Poisson equation with Dirichlet conditions: weak form: apply divergence theorem: v u dω = Ω u = f in Ω, u = 0 on δω v u dω = vf dω Ω Ω Ω v v u dω v u ds Ω choose functions v such that v = 0 on Ω: v u dω = vf dω Ω Ω v Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 6

7 Weak Form of the Poisson Equation (2) Poisson equation with Dirichlet conditions: u = f in Ω, u = 0 on δω transformed into weak form: v u dω = vf dω Ω Ω v weaker requirements for a solution u: twice differentiabale first derivative integrable Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 7

8 FEM and the Residual Equation for a general PDE Lu = f, the residual is defined as r(x) := f (x) Lu(x)! = 0 in the weak form (multiply with test function an integrate) we obtain v(x) r(x) = v(x) ( f (x) Lu(x) ) dx = 0 for all v remember that f (x)g(x) dx is the standard scalar product on functions hence: the residual of the weak solution has to be orthogonal to all test functions Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 8

9 Test and Shape Functions search for solution functions u h of the form u h = j u j ϕ j (x) the ϕ j (x) are typically called shape or ansatz functions the basis functions ϕ j (x) build a vector space (i.e., a function space) W h span{ϕ 1,..., ϕ J } = W h insert into weak formulation ( ) vl u j ϕ j (x) dx = j vf dx v V Module 8: An Introduction to Finite Element Methods, Winter 2013/2014 9

10 Test and Shape Functions (2) choose a basis {ψ i } of the test space V h then: if all basis functions ψ i satisfy ( ) ψ i L u j ϕ j (x) dx = j ψ i f dx ψ i then all v V h satisfy the equation the {ψ i } are therefore often called test functions we obtain a system of equations for unknowns u j : one equation for each test function ψ i V h is often chosen to be identical to W h (Ritz-Galerkin method) we then have as many equations as unknowns Module 8: An Introduction to Finite Element Methods, Winter 2013/

11 Example: Nodal Basis 1 h (x x i 1) x i 1 < x < x i 1 ϕ i (x) := h (x i+1 x) x i < x < x i+1 0 otherwise 1 0,8 0,6 0,4 0, ,2 0,4 0,6 0,8 x Module 8: An Introduction to Finite Element Methods, Winter 2013/

12 Assembling the System of Equations L linear system of linear equations ( ) ψ i Lϕ j (x) dx u j = j } {{ } =:A ij ψ i f dx ψ i aim: make matrix A sparse most A ij = 0 approach: local basis functions on a discretisation grid ψ j, ϕ j zero everywhere except in grid cells adjacent to grid point x j A ij = 0, if ψ i and ϕ j don t overlap Module 8: An Introduction to Finite Element Methods, Winter 2013/

13 Example Problem: 1D Poisson in 1D: u (x) = f (x) on Ω = (0, 1), hom. Dirichlet boundary cond.: u(0) = u(1) = 0 weak form: 1 v (x) u (x) dx = v(x)f (x) dx v computational grid: x i = ih, (for i = 1,..., n 1); mesh size h = 1/n V h = W h : piecewise linear functions (on intervals [x i, x i+1 ]) Module 8: An Introduction to Finite Element Methods, Winter 2013/

14 Nodal Basis 1 h (x x i 1) x i 1 < x < x i 1 ϕ i (x) := h (x i+1 x) x i < x < x i+1 0 otherwise 1 0,8 0,6 0,4 0, ,2 0,4 0,6 0,8 x Module 8: An Introduction to Finite Element Methods, Winter 2013/

15 Nodal Basis Derivatives and Integrals Compute derivative of basis functions: 1 h ϕ i (x) := x i 1 < x < x i 1 h x i < x < x i+1 0 otherwise Compute integrals of weak form: xi ϕ i (x)ϕ i (x) dx = 1 x i 1 h 1 xi+1 ( h dx + 1 ) ( 1 ) dx = 2 x i h h h xi+1 ( ϕ i (x)ϕ i+1 (x) dx = 1 ) 1 x i h h dx = 1 h ϕ i (x)ϕ i 1 (x) dx = 1 (same computation) h ϕ i (x)ϕ j (x) dx = 0 for all j i, i ± 1 (supports of basis functions do not overlap!) Module 8: An Introduction to Finite Element Methods, Winter 2013/

16 Nodal Basis System of Equations stiffness matrix: 1 h right hand sides (assume f (x) = α R): 1 ϕ i (x)f (x) dx = ϕ i (x)α dx = αh system of equations very similar to finite differences exercise: compute right-hand-side vector, if f (x) := f j φ j (x) Module 8: An Introduction to Finite Element Methods, Winter 2013/

17 Outlook: A Road to Theory weak formulation is equivalent to variational approach: solution u minimises an energy functional best approximation property: u uh FE inf u u h a a u h V in terms of the norm induced by the bilinear form a (energy norm) thus: error bounded by interpolation error (in energy norm) (details in lecture Numerical Programming II... ) Module 8: An Introduction to Finite Element Methods, Winter 2013/

18 Time-Dependent Problems Example: 1D Heat Equation u t = u xx + f on domain Ω = [0, 1] for t [0, t end ] spatial discretisation: weak form vu t dx = ) vu dx = ( t spatial discretisation finite elements: vu xx dx + vu xx dx + t (M hu h ) = A h u h + f h M h : mass matrix, A h : stiffness matrix, u h = u h (t) vf dx vf dx Module 8: An Introduction to Finite Element Methods, Winter 2013/

19 Time-Dependent Problems (2) Solve a system of ordinary differential equations: after spatial discretisation (M h constant): t M h (u h ) = A h u h + f h u h a vector of time-dependent functions: u h = (u 1 (t),..., u i (t),..., u n (t)) T usually: approximate M h by a simpler matrix (diagonal matrix, e.g.) mass lumping Module 8: An Introduction to Finite Element Methods, Winter 2013/

20 Part II: Implementation of Finite Element Methods Element Stiffness Matrices Element-Oriented Computation of Stiffness Matrices Example: 1D Poisson Example: 2D Poisson Typical Workflow Reference Elements and Stiffness Matrices Element-Oriented Computation on Unstructured Meshes Accumulation of Global Stiffness Matrix Simple Example: 1D Poisson Outlook: Extension to 2D and 3D Outlook: Further Components and Aspects of FEM Module 8: An Introduction to Finite Element Methods, Winter 2013/

21 Element Stiffness Matrices Finite Element methods are often used for complicated meshes: Questions: how to set up the system of equations efficiently? is it possible to use stencil notation?? Switch to element stiffness matrices! Module 8: An Introduction to Finite Element Methods, Winter 2013/

22 Element Stiffness Matrices (2) domain Ω subdivided into finite elements Ω (k) : Ω = Ω (1) Ω (2) Ω (n) observation: basis functions are defined element-wisely use: b a f (x) dx = c a f (x) dx + b f (x) dx c element-wise evaluation of the integrals: v u dx = Ω k Ω vf dx = k v u dx Ω (k) vf dx Ω (k) Module 8: An Introduction to Finite Element Methods, Winter 2013/

23 Element Stiffness Matrices (3) leads to local stiffness matrices for each element: φ i φ j dx Ω (k) } {{ } =:A (k) ij and respective element systems: A (k) x = b (k) accumulate to obtain global system: A (k) x = k k }{{} =:A b (k) Module 8: An Introduction to Finite Element Methods, Winter 2013/

24 Element Stiffness Matrices (4) Some comments on notation: assume: 1D problem, n elements (i.e. intervals) in each element only two basis functions are non-zero! hence, almost all A (k) ij are zero: A (k) ij = φ i φ j dx Ω (k) only 2 2 elements of A (k) are non-zero therefore convention to omit zero columns/rows leaves only unknowns that are in Ω (k) notation: write this matrix as Â(k) Module 8: An Introduction to Finite Element Methods, Winter 2013/

25 Example: 1D Poisson Ω = [0, 1] splitted into Ω (k) = [x k 1, x k ] nodal basis; leads to element stiffness matrix: ( ) Â (k) 1 1 = 1 1 consider case with only three unknowns (i.e., 3 basis functions): A (1) + A (2) = in stencil notation: [ 1 1 ] + [ 1 1 ] [ ] Module 8: An Introduction to Finite Element Methods, Winter 2013/

26 Example: 2D Poisson u = f on domain Ω = [0, 1] 2 splitted into Ω (i,j) = [x i 1, x i ] [x j 1, x j ] bilinear basis functions pagoda functions ϕ ij (x, y) = ϕ i (x)ϕ j (y) Module 8: An Introduction to Finite Element Methods, Winter 2013/

27 Example: 2D Poisson (2) leads to element stiffness matrix: Â (k) = accumulation leads to 9-point stencil exercise at home: compute Â(k) and assemble global matrix (and compare with stencil) Module 8: An Introduction to Finite Element Methods, Winter 2013/

28 Typical Workflow 1. choose elements: quadratic or cubic cells triangles (structured, unstructured) tetrahedra, etc. 2. set up basis functions for each element Ω (k) ; for example, at all nodes x i Ω (k) ϕ i (x i ) = 1 ϕ i (x j ) = 0 for all j i 3. for element stiffness matrix, compute all  (k) ij = ϕ i Lϕ j dω Ω (k) 4. accumulate global stiffness matrix A (and rhs) Module 8: An Introduction to Finite Element Methods, Winter 2013/

29 Element-Oriented Computation on Unstructured Meshes Recall: FEM often used for complicated meshes: (Re-)computation of element matrices required for each element? left grid: no (up to scaling with mesh size) right grid: yes? requires a closer look at the computation of element matrices Module 8: An Introduction to Finite Element Methods, Winter 2013/

30 Element Stiffness Matrices Notation Revisited recall notation for local stiffness matrices A (k) ij : A (k) ij = φ i (x) φ j (x) dx Ω (k) which basis functions φ i, φ j belong to element k? switch to element-local numbering of basis functions: Â (k) (k) µν = φµ (x) φ (k) ν (x) dx Ω (k) where φ (k) µ (x) = φ i (x) and φ (k) ν = φ j (x) in Ω (k) requires a mapping φ (k) µ φ i (and φ (k) ν φ j ), i.e., a function ι (k) with i = ι (k) (µ) and j = ι (k) (ν) Module 8: An Introduction to Finite Element Methods, Winter 2013/

31 Accumulation of Global Stiffness Matrix note the different dimensions of A (k) and Â(k) :  (k) omits zero rows and columns of A (k) formulated via a projection matrix: A (k) = ( P (k))t  (k) P (k) with P (k) µ,i := recall computation of global stiffness matrix: { 1 if i = ι (k) (µ) 0 otherwise A := k A (k) = k ( P (k) )T  (k) P (k) hence, for each element: 1. compute element stiffness matrix Â(k) 2. determine position of matrix elements Â(k) in A (k) based on mapping ι (k) (µ) 3. add non-zero elements of A (k) = ( P (k))t  (k) P (k) to global matrix A Module 8: An Introduction to Finite Element Methods, Winter 2013/

32 Element Stiffness Matrices and Meshes Recall our initial question: (Re-)computation of Â(k) required for each element? left grid: no (up to scaling with mesh size) right grid: yes? efficient computation of Â(k) via reference elements Module 8: An Introduction to Finite Element Methods, Winter 2013/

33 Simple Example: 1D Poisson Consider FEM for 1D Poisson on an adaptive grid: grid points given as x i, not necessarily equidistant; define elements Ω (k) = [x k, x k+1 ] piecewise linear basis functions ( hat functions ) φ i (x); φ i (x) non-zero only on two elements: Ω (i 1) and Ω (i) mapping local to global indices: ι (k) (1) := k and ι (k) (2) := k + 1 wanted: Â(k) µ,ν := Ω (k) x φ(k) µ (x) x φ(k) ν (x) dx for every element k Introduce reference element Ω ref with linear basis functions Φ µ : Φ 1 (ξ) := ξ Φ 2 (ξ) := 1 ξ ξ Ω ref := [0, 1] and a mapping x = χ (k) (ξ) for each Ω (k) defined as χ (k) (ξ) := x k + ξ(x k+1 x k ) such that φ (k) µ (x) = φ (k) ( µ χ (k) (ξ) ) := Φ µ (ξ) for x Ω (k). Module 8: An Introduction to Finite Element Methods, Winter 2013/

34 Simple Example: 1D Poisson (2) We need to compute  (k) µ,ν := x k+1 x k x φ(k) µ (x) x φ(k) ν (x) dx = Compare integration by substitution: g(b) g(a) χ (k) (1) χ (k) (0) x φ(k) µ (x) f (x) dx = x φ(k) ν (x) dx b f (g(t))g (t) dt 1  (k) ( µ,ν = x φ(k) µ χ (k) (ξ) ) ( x φ(k) ν χ (k) (ξ) ) (x k+1 x k ) dξ. 0 Note that: ξ χ k(ξ) = (x k+1 x k ) corresponds to the factor g (t) in the substitution rule we still have to deal with the partial derivatives in x Module 8: An Introduction to Finite Element Methods, Winter 2013/ a

35 Simple Example: 1D Poisson (3) We require the chain rule, f x = f g g x, to obtain: ( x φ(k) µ χ (k) (ξ) ) = x Φ µ(ξ) = ξ Φ µ(ξ) ξ x As x = χ (k) (ξ) := x k + ξ(x k+1 x k ), we have ξ = x x k and thus: x k+1 x k Finally, we get: ξ Φ µ(ξ) ξ x = ξ Φ 1 µ(ξ) x k+1 x k 1 Â (k) ( µ,ν = x φ(k) µ χ (k) (ξ) ) ( x φ(k) ν χ (k) (ξ) ) (x k+1 x k ) dξ = ξ Φ µ(ξ) ξ Φ 1 ν(ξ) (x k+1 x k ) 2 (x k+1 x k ) dξ. Module 8: An Introduction to Finite Element Methods, Winter 2013/

36 Simple Example: 1D Poisson (4) Altogether, element stiffness matrices are computed as: Â (k) µ,ν = 1 x k+1 x k 1 0 ξ Φ µ(ξ) ξ Φ ν(ξ) dξ With Φ 1 (ξ) = ξ and Φ 2 (ξ) = 1 ξ, we obtain that ( ) Â (k) = x k+1 x k 1 1 Observations: the integral values only depend on the reference element Ω ref = [0, 1] and the reference basis functions Φ µ, Φ ν all other element matrices are obtained by simple scaling we did not use that Φ µ, Φ ν are linear functions; same computation for more complicated basis functions! Module 8: An Introduction to Finite Element Methods, Winter 2013/

37 Outlook: Extension to 2D and 3D For 2D, 3D, etc., we require the multidimensional substitution rule: f (x, y, z) dω = f (x(ξ, η, ζ), y(ξ, η, ζ), z(ξ, η, ζ)) J dξ dη dζ Ω (k) Ω ref where J is the determinant of the Jacobian: x x x ξ η ζ J = y y y ξ η ζ z ξ z η and also the multidimensional chain rule. Result: similar, element-wise computation of stiffness matrices via reference elements Jacobian instead of factor 1 x k+1 x k ; more complicated, in general Module 8: An Introduction to Finite Element Methods, Winter 2013/ z ζ

38 Outlook: Further Components and Aspects of FEM Additional components and features of Finite Element Methods: mesh generation: structured/unstructured meshes consisting of triangles, tetrahedra, rectangles, cuboids, quadrilaterals, hexahedrals, prisms,... higher-order basis functions: polynomials of higher degree in various dimensions; nodal vs. spectral elements,... quadrature rules for integration: Gaussian quadrature, e.g., to compute integrals of weak forms; resp. choice of quadrature points for certain elements? solving the system of equations: e.g., set up a global system of equations explicitly, or work directly on element representation?... and many more... Module 8: An Introduction to Finite Element Methods, Winter 2013/