Numerical P.D.E. An introduction to Finite Elements Methods Nicolas Fourrier University of Virginia
[email protected]
November 9, 2010
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Overview
1
Introduction
2
Solving a problem with F.E.M.
3
F.E.M Process
4
Results
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Introduction
Definition
Introduction
Computational technique to approximate solutions to PDE. Solution of the PDE is approximated by a simple polynomial function on each element. Process: 1 2 3 4
Scheme Physical System Mathematical Model Discrete Method Solution
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PDE Seminar
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Introduction
Context
Context
Initially, it was uniquely used for elliptic problem. Nowadays, it is also used to solve parabolic and hyperbolic problems. Anticipate the behavior of a structure, optimize its shape are some of the most common application of F.E.M.
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PDE Seminar
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Introduction
Model Problem
Model Problem General equation φ
δu ~ = uˆf + div (−D∇u) + div (Au) δt
φ δu δt : the accumulation term div (−D∇u): the diffusion term ~ div (Au): the advection term uˆf : the source ~A: Darcy velocity ~ vanishes or is sufficiently small, we neglect the advection If the velocity A term and obtain a diffusion equation. Focus on the heat equation: div (−D∇u) = f . Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Solving a problem with F.E.M.
Overview
1
Introduction
2
Solving a problem with F.E.M. Presentation of the problem Example
3
F.E.M Process
4
Results
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Solving a problem with F.E.M.
Presentation of the problem
Problem
Domain 2 Consider Ω ⊂ R2 , an open S set of R . Consider ∂Ω = Γ = ΓN ΓD the border of Ω.
Nicolas Fourrier (U.V.A.)
PDE Seminar
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Solving a problem with F.E.M.
Presentation of the problem
Domain
Γ3 In this picture: 1
Ω: Plate
2
Γi : Border
3
Lx : Length
4
Ly : Width
Ly
Ω
Γ4
Γ2
Γ1 Lx
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Solving a problem with F.E.M.
Presentation of the problem
Problem definition
Problem definition −∆u = f (P) : u|Γd = u¯ ∂u|ΓN = g
(Dirichlet’s conditions) (Neumann’s conditions)
Function’s domains 2 f ∈ L (Ω) 1 u¯ ∈ H 2 (Γd ) g ∈ L2 (Γn )
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Solving a problem with F.E.M.
Presentation of the problem
Find u ∈ V u¯ V u¯ = {v ∈ V with v |γN = u¯ }. After multiplying by a test function v and integrating both members of the equality of the strong fomulation, we get the variational formulation. Variational formulation Find u ∈ V u¯ R R − Ω ∆u.v dΩ = Ω fv dΩ (P) ⇔ ∂n u|ΓN = g u|Γd = u¯
Nicolas Fourrier (U.V.A.)
PDE Seminar
(Neumann’s conditions) (Dirichlet’s conditions)
November 9, 2010
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Solving a problem with F.E.M.
Presentation of the problem
Integration by parts
u ¯ Find R u∈ V R ∇u ∇v dΩ − Γ δn uv dΓ = Ω fvdΩ Ω (P) ⇔ g (Neumann’s conditions) ∂n u|ΓN = (Dirichlet’s conditions) u|Γd = u¯
R
System Au = f After making an integration by parts, we make appear the different components of the system Au = f .
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
11 / 35
Solving a problem with F.E.M.
Presentation of the problem
Well-Posed
Lax-Milgram lemma Let V be an Hilbert space and a : V × V → R. If a is a continuous, bilinear and coercive form, then it exists one and only one u ∈ V such that a(u, v ) = f (v ) ∀v ∈ V . And the problem is well-posed. Well-posed Problem Lax-Milgram R lemma allows to say thatR this problem is well-posed by letting a(u, v ) = Ω ∇u ∇v dΩ and f (v ) = Ω fv dΩ.
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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Solving a problem with F.E.M.
Example
Two problems to solve Problem 1 −∆u = 0 u |Γ1 = x (P1 ) ∂ u | = ∂n u |Γ3 = 1 n Γ2 u |Γ4 = y Problem 2 −∆u = −4 u |Γ1 = x 2 ∂n u |Γ2 = 40 (P2 ) ∂n u |Γ3 = 20 u |Γ4 = y 2 Nicolas Fourrier (U.V.A.)
u1 (x, y ) = x + y
u2 (x, y ) = x 2 + y 2
PDE Seminar
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F.E.M Process
Overview 1
Introduction
2
Solving a problem with F.E.M.
3
F.E.M Process Geometry & Meshing Definition of an element Stiffness Matrix Shape function Rewrite the problem at each node Computing the solution at each nodes Second Member Boundary Condition Resolution
4
Results Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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F.E.M Process
Geometry & Meshing
Any dimension Domain Ω can be in 1-D, 2-D or 3-D. In the case of the two problems, let Ω be a plate of length x and width y. Meshing Meshing the domain (here a plate) is one of the most difficult and decisive part in F.E.M.. Depending on the choices done at this step, one might get some very different results. Meshing need not to be completely regular. Around point of interests, it is common to increase the density of the meshing.
Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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F.E.M Process
Geometry & Meshing
Examples Different type of meshing: 1
2-D: Square, rectangles, triangles ...
2
3-D: Cube, polyhedral ...
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PDE Seminar
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F.E.M Process
Definition of an element
Element
Definition The plate is discretized as grid. Each point of the grid is a node (defined by its coordinates). If the elements are triangles, then a triple of nodes will form an element. Smaller the elements are, better the approximation is going to be. Difficulty One might would like to simply create an extremly dense meshing, but the storage and computation costs (both in time and power) is quickly becoming a big deal.
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PDE Seminar
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F.E.M Process
Stiffness Matrix
Recall F.E.M. leads to solve the system Au = f where u is the solution to the PDE. More precisely, u is the approximate solution at each node. Stiffness Matrix A is the stiffness matrix. Symmetric matrix. The stiffness matrix will be modified during the application of the boundary condictions (in the case of Dirichlet’s conditions).
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PDE Seminar
November 9, 2010
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F.E.M Process
Stiffness Matrix
Calculation of coefficients
Importance of Reference Element In order to generalize the process and make the computation easier, each real element is expressed in term of the reference element. All calculation are done on reference elements, then the reference element is tranformed into each real element.
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PDE Seminar
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F.E.M Process
Stiffness Matrix
Triangle Element η Real Element Consider a triangle element, generated by the 3 apexes: 1
(x1 , y1 )
2
(x2 , y2 )
3
(x3 , y3 )
Nicolas Fourrier (U.V.A.)
(x1 , y3 )
T (0, 1) (x1 , y1 ) (x1 , y2 ) Real Element ξ (0, 0) (1, 0) Reference Element
PDE Seminar
November 9, 2010
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F.E.M Process
Stiffness Matrix
Node vs. entry in the stiffness matrix
6 Coefficients per triangle element The stiffness matrix is a n × n matrix if there are n nodes in the system. The entry (i , j) corresponds to the node (i , j). Since the computation are done on reference element, it is enough to consider local stiffness matrices, and then sum up all these small matrices to get the global stiffness matrix. For instance, in the case of triangle element, the local stiffness matrix is defined as follows: 1
A′ is a 3 × 3 matrix.
2
A′ij = A′ji : Contribution of the node (i , j) to the current element.
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PDE Seminar
November 9, 2010
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F.E.M Process
Stiffness Matrix
Notation
Let m be the number of elements, and n the number of nodes. ¯ be the reference element. Let K be the real element and K Let T be the set of all elements. A is the stiffness matrix, n × n invertible matrix.
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F.E.M Process
Shape function
Shape function
To describe the elements (reference or real), it is important to describe mathematically an element. This is the purpose of the shape functions. Shape function describes how the nodal solutions (it can be displacements for example) are distributed throughout the element.
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PDE Seminar
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F.E.M Process
Shape function
Shape functions’ equation Triangle element - Reference In the case of triangle elements, the shape functions are: 1
φ1 = 1 − ξ − η
2
φ2 = η
3
φ3 = ξ
The index corresponds to one side of the triangle. Recall, we need to compute the gradiant these functions. Computation of the coefficients In order to compute each coefficient for each element, and to avoid the consideration of the geometry (shape and orientation of the triangle), it is possible to find a formula allowing to make the computation of each real element on the reference element. Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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F.E.M Process
Shape function
Recall
Problem 2 −∆u = −4 u |Γ1 = x 2 (P2 ) ∂n u |Γ2 = 40 ∂ u | = 20 n Γ3 2 u |Γ4 = y
u2 (x, y ) = x 2 + y 2
From Variational to F.E.M. R R R Recall: Ω ∇u ∇v dΩ − Γ δn uv dΓ = Ω fvdΩ Goal: discretize this variational form, and get a formula for each element. The shape functions represent the solution on each element.
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PDE Seminar
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F.E.M Process
Rewrite the problem at each node
Rewrite the problem at each node
Recall, A is the stiffness matrix. Let K be a real element, and T the meshing. Z ∇φi ∇φj dΩ Aij = Ω XZ = ∇φi ∇φj dΩ K ∈T
=
Nicolas Fourrier (U.V.A.)
(2)
Ω
X Z
K ∈T (i ,j)∈K
(1)
∇φi ∇φj dΩ
(3)
Ω
PDE Seminar
November 9, 2010
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F.E.M Process
Computing the solution at each nodes
Computing the solution at each nodes We deduce the derivatives of the shape functions: −1 for node 1 −1 for node 1 dφi dφi for node 2 for node 2 dξ : 1 dη : 0 0 for node 3 1 for node 3 The coordinate x and y can be determined as followed: x(ξ, η) = (1 − ξ − η)X1 + ξX2 + ηX3 y (ξ, η) = (1 − ξ − η)Y1 + ξY2 + ηY3 And the determinant of Jacobian matrix J (the matrix which transforms real elements to reference elements) is: det(J) =
Nicolas Fourrier (U.V.A.)
∂x ∂y ∂y ∂x − ∂ξ ∂η ∂ξ ∂η
PDE Seminar
November 9, 2010
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F.E.M Process
Computing the solution at each nodes
Computing the solution at each nodes II Let C = ∇φi (x(ξ, η), y (ξ, η)).∇φj (x(ξ, η), y (ξ, η)) = constant. Z C .det(J)dξdη Aij = K Z 1 Z 1−η 1dξdη = C .det(J) 0
=
(4) (5)
0
C .det(J) 2
(6)
Extending this process from one element to whole system, we get: Aij =
X Cij .det(Jij ) 2
i ,j∈K
where the node (i , j) belongs to the element K. Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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F.E.M Process
Second Member
Second member Recall that we are solving a system of the form Au = f , so far we have defined the stiffness matrix. Let’s define the second member f. The construction is very similar to the one of A. For each element, there are 3 coefficients to calculate: F1 , F2 and F3 , where the index are given with respect to the reference element. Considering the whole system, the i th entry of the second member F corresponds to the i t h node. Again a node might be inside several elements. Thus, Fi is given by summing each contribution. In (P1 ), f = 0, we get that F1 = F2 = F3 = 0. In (P2 ), f = −4, we get that F1 = F2 = F3 = − 23 .det(J).
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PDE Seminar
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F.E.M Process
Boundary Condition
Boundary Condition Neumann It is essential to apply the Neumann conditions first, since they only affect the second member, whereas, the Dirichlet conditions will modify both the stiffness matrix and the second member. The Neumann conditions add a contribution to the second member at some nodes. The Neumann conditions will give us two coefficients per element. Indeed, we are not dealing anymore with triangular elements but with linear elements. Dirichlet The Dirichlet conditions give the solution at some nodes. In other words, the calculation done at the node where the conditions were applied are replaced. From a computational point of view, it is trivial to add the Dirichlet Condition. Nicolas Fourrier (U.V.A.)
PDE Seminar
November 9, 2010
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F.E.M Process
Boundary Condition
Application of Neumann
In general R The integral Γ ∂n uvdΓ corresponds to the Neumann conditions. And it is given by ∂uΓ = f (x, y ). Again, it is necessary to compute the contribution on a reference element to compute each contribution. In our case In =1 R (P1 ), we have∂n u |Γ2 = ∂Rn u |Γ31−ξ 1 1 Γ2 ∂n u.φ1 .det(J)dξ dη = Γ2 1. 2 .( 2 (y2 − y1 ))dξ dη = 2 (y2 − y1 ) Adding these contributions to the appropriate nodes of the second member, give the second member under Neumann Condition.
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F.E.M Process
Resolution
Algorithms
Resolution Finally, the problem to solve is Au = F where A is a symmetric, positive definite matrix. In addition, A contains a lot a zero. However, the matrix A is huge, so there is a matter of cost. Use Cholesky Algorithm.
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Results
Errors
Cea’s lemma If u ∈ H 2 (Ω) and triangulation is regular, then: ku − uh k ≤ C1 .ku − v k2,Ω ∀v ∈ A
Vh : finite dimensionnal supspace of H 2 (Ω) uh ∈ A: approximate solution u: analytic solution C1 : constant
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Results
Errors II
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Results
The end
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