the finite element method defined
DESCRIPTION
The Finite Element Method Defined. The Finite Element Method (FEM) is a weighted residual method that uses compactly-supported basis functions. Brief Comparison with Other Methods. Finite Difference (FD) Method: - PowerPoint PPT PresentationTRANSCRIPT
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The Finite Element Method Defined
The Finite Element Method (FEM) is a weighted residual method that uses compactly-supported basis functions.
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Brief Comparison with Other Methods
Finite Difference (FD) Method:
FD approximates an operator (e.g., the derivative) and solves a problem on a set of points (the grid)
Finite Element (FE) Method:
FE uses exact operators but approximates the solution basis functions. Also, FE solves a problem on the interiors of grid cells (and optionally on the gridpoints as well).
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Brief Comparison with Other Methods
Spectral Methods:
Spectral methods use global basis functions to approximate a solution across the entire domain.
Finite Element (FE) Method:
FE methods use compact basis functions to approximate a solution on individual elements.
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MGWS
Overview of the Finite Element Method
Strong
form
Weak
form
Galerkin
approx.
Matrix
form
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Axial deformation of a bar subjected to a uniform load
(1-D Poisson equation)
Sample Problem
0p=xp
0
00
2
=dx
duEA
=u
p=dx
udEA
Lx
02
L
L=x 0,u = axial displacement
E=Young’s modulus = 1
A=Cross-sectional area = 1
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Strong Form
The set of governing PDE’s, with boundary conditions, is called the “strong form” of the problem.
Hence, our strong form is (Poisson equation in 1-D):
0
00
2
=dx
du
=u
p=dx
ud
Lx
02
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We now reformulate the problem into the weak form.
The weak form is a variational statement of the problem in which we integrate against a test function. The choice of test function is up to us.
This has the effect of relaxing the problem; instead of finding an exact solution everywhere, we are finding a solution that satisfies the strong form on average over the domain.
Weak Form
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Weak Form
0
0
0
0
2
0
2
2
=vdxpdx
ud
=pdx
ud
p=dx
ud
L
2
2
02
Strong Form
Residual R=0
Weak Form
v is our test function
We will choose the test function later.
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Why is it “weak”?
It is a weaker statement of the problem.
A solution of the strong form will also satisfy the weak form, but not vice versa.
Analogous to “weak” and “strong” convergence:
Weak Form
fxfxf
xx
n
n
lim :weak
lim :strong
n
n
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Weak Form
Choosing the test function:
We can choose any v we want, so let's choose v such that it satisfies homogeneous boundary conditions wherever the actual solution satisfies Dirichlet boundary conditions. We’ll see why this helps us, and later will do it with more mathematical rigor.
So in our example, u(0)=0 so let v(0)=0.
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Returning to the weak form:
LL
2
L
2
vdxp=vdxdx
ud
=vdxpdx
ud
0 0
0
2
0
0
2
0
Weak Form
IntegrateIntegrate LHS by parts:
00
00
0
)(
xLx
L
Lx
x
L
dx
duv
dx
duLvdx
dx
dv
dx
du
dx
duxvdx
dx
dv
dx
du
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Weak FormRecall the boundary conditions on u and v:
0)0(
0
00
v
=dx
du
=u
Lx
LL
xLx
L
vdxpdxdx
dv
dx
du
dx
duv
dx
duLvdx
dx
dv
dx
du
0 0
0
00
0
HHence,The weak form satisfies Neumann conditions automatically!
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Weak Form
functionallinear a ,functionalbilinear a
)(,such that Find
:statement lVariationa10
1
0 0
0
FB
HvvFvuBHu
vdxp=dxdx
dv
dx
du LL
Why is it “variational”?
u and v are functions from an infinite-dimensional function space H
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We still haven’t done the “finite element method” yet, we have just restated the problem in the weak formulation.
So what makes it “finite elements”?
Solving the problem locally on elements
Finite-dimensional approximation to an infinite-dimensional space → Galerkin’s Method
Galerkin’s Method
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L N
j
L N
jjj
N
i
ii
jj
L L
j
N
jjj
j
N
jjj
N
iii
dxxbpdxxdx
dbx
dx
dc
vdxpdxdx
dv
dx
du
bxbxv
cxcxu
01
01
01
0 0 0
1
1
:formour weak into seInsert the
choseny arbitraril ,
for solve tounkowns ,
Then,
basis finite Choose
Galerkin’s Method
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dxpdxdx
d
dx
dc
dxpbdxdx
d
dx
dcb
dxxbpdxxdx
dbx
dx
dc
i
LN
j
Lij
j
i
LN
ii
N
i
N
j
Lij
ji
L N
j
L N
jjj
N
i
ii
jj
0 01
0
0 011 1
0
01
01
01
: Cancelling
:gRearrangin
Galerkin’s Method
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effect. without , einterchang
can wesince symmetric be will seealready can We
and
,
unknowns, of vector a is
where, problemmatrix a have now We
0 0
0
0 01
0
ji
K
dxpF
dxdx
d
dx
dK
c
dxpdxdx
d
dx
dc
ij
i
L
i
Lij
ij
j
i
LN
j
Lij
j
FKc
Galerkin’s Method
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Galerkin’s Method
So what have we done so far?
1) Reformulated the problem in the weak form.
2) Chosen a finite-dimensional approximation to the solution.
Recall weak form written in terms of residual:
00 00 02
2
L
i
L
ii
Ldxbvdxvdxp
dx
ud RR
This is an L2 inner-product. Therefore, the residual is orthogonal to our space of basis functions. “Orthogonality Condition”
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Orthogonality Condition
00 00 02
2
L
i
L
ii
Ldxbvdxvdxp
dx
ud RR
The residual is orthogonal to our space of basis functions:
u
uh
H
Hh
φi
Therefore, given some space of approximate functions Hh, we are finding uh that is closest (as measured by the L2 inner product) to the actual solution u.
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Discretization and Basis Functions
Let’s continue with our sample problem. Now we discretize our domain. For this example, we will discretize x=[0, L] into 2 “elements”.
0 h 2h=L
1Ω 2Ω
In 1-D, elements are segments. In 2-D, they are triangles, tetrads, etc. In 3-D, they are solids, such as tetrahedra. We will solve the Galerkin problem on each element.
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Discretization and Basis Functions
For a set of basis functions, we can choose anything. For simplicity here, we choose piecewise linear “hat functions”.
Our solution will be a linear combination of these functions.
x1=0 x2=L/2 x3=L
unity. ofpartition esatisfy th they Also, ory.interpolat
be illsolution wOur :satisfy functions Basis ijji x
φ1 φ2φ3
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Discretization and Basis Functions
To save time, we can throw out φ1 a priori because, since in this example u(0)=0, we know that the coefficent c1 must be 0.
x1=0 x2=L/2 x3=L
φ2φ3
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Basis Functions
otherwise 0
, if 12
otherwise 0
, if 2
2
,0 if 2
23
2
2
2
LxL
x
LxL
x
xL
x
L
L
L
x1=0 x2=L/2 x3=L
φ2 φ3
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Matrix Formulation
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exact isit since ,quadratureGaussian use toFEMin standard isIt
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and known, are functions basis thesince advance,in done
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and slide previous on thechosen insert thecan We
: problemmatrix our Given
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21
0
0 01
0
K
FK
FKc
FKc
FKc
L
p
L
dxpdxdx
d
dx
dc
i
i
LN
j
Lij
j
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Solution
2
:is problem for thissolution analyticalexact The
,n whe)(
,0n whe
:solution numerical finalour
gives functions basisby multiplieden which wh,
: tscoefficienour obtain we
slide, previous on the problemn eliminatioGaussian theSolving
20
0
22
041
2043
2
83
20
20
xpLxpxu
LxLxLp
xLxpx
c
L
L
iLp
Lp
i
c
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Solution
0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.2 0.4 0.6 0.8 1
x
u(x) Exact
Approx
Notice the numerical solution is “interpolatory”, or nodally exact.
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Concluding Remarks
•Because basis functions are compact, matrix K is typically tridiagonal or otherwise sparse, which allows for fast solvers that take advantage of the structure (regular Gaussian elimination is O(N3), where N is number of elements!). Memory requirements are also reduced.
•Continuity between elements not required. “Discontinuous Galerkin” Method