Lecture1 Cornell

8/13/2019 Lecture1 Cornell

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N. Zabaras 1

INTRODUCTION TO THE FINITE ELEMENT METHOD

FOR ELLIPTIC PROBLEMS

Variational weak formulation for a 1D model problem

FEM for the model problem using piecewise linear functions

Error estimation for the 1D model problem

FEM for the 2D Poisson equation

Introduction to Hilbert spaces L2(),H1() andH10()

The best error approximation: Geometric interpretation

The Neumann problem: Natural and essential boundary conditions

FEM Programming

Lecture 1: Introduction to FEM for el liptic problems Materials Process Design and Control Laboratory


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INTRODUCTION TO FEM FOR ELLIPTIC PROBLEMS

Variational formulation of a one-dimensional model problem

u(x) =f(x), 0< x


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SOME EXAMPLE PROBLEMS

An elastic bar

=Eu (Hookes law) =f, (Equilibrium equ.)u(0) =u(1) = 0 (Boundary conds.)

An elastic cord

Heat conduction

q=Ku

(Fouriers law)q =f, (Conservation of energy)u(0) =u(1) = 0 (Temperature boundary conds.)



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THE MINIMIZATION PROBLEM (M)

Define the linear functionalF :V as:

F(v) = 12(v, v) (f, v)

Findu V such thatF(u) F(v) vV

where:

V ={v:vis a continuous function on [0, 1],v is piecewise continuous and

bounded on [0, 1], andv(0) =v(1) = 0}

(v, w) =10v(x)w(x)dx, for real valued piecewise continuous functions.

(M) corresponds to thePrinciple of Minimum Potential Energy in Me-

chanics.



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THE VARIATIONAL (WEAK) PROBLEM (V)

FinduV such that (u, v) = (f, v) vV

where:

v is the so called test function

(V) corresponds to the Principle of Virtual Work in Mechanics.



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The solution of (D) is also a solution of (V)

We multiply the equation u = f by an arbitraryfunctionvV

Integrate over (0, 1) which gives:

(u, v) = (f, v)

We now integrate the left-hand side by parts using thefact thatv(0) =v(1) = 0 to get:

(u

, v) =u

(1)v(1) +u

(0)v(0) + (u

, v

) = (u

, v

)We conclude that: (u, v) = (f, v) vV, i.e. uis a

solution of (V).



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The solution of(V) is also a solution of(M)

Let u be a solution to (V), let v V and set w = v u so thatv=u+w,wV.

We can write: F(v) =F(u+w) = 12(u +w, u +w) (f, u+w)

= 12(u, u) (f, u) + (u, w) (f, w) + 12(w

, w)

=F(u) + 0 + 12(w, w)

F(u)

Thus,uis the minimizer ofF(v) (uis a solution of (M)).



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The solution of(M) is also a solution of (V)

Let u be a solution to (M), then for any real number and anyvVwe have thatu+v V, which implies:

F(u) F(u+v)

Thus the differentiable functiong() =F(u+v) has a minimumat= 0 and hence g(0) = 0.

We have:

g() =

1

2(u

, u

) +(u

, v

) +

2

2 (v

, v

) (f, u) (f, v) andg() = (u, v) +(v, v) (f, v).

Usingg(0) = 0 andg(0) = (u, v) (f, v), results in:

(u, v) = (f, v), i.e. uis a solution of (V).



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Problem (V) has a unique solution

Let (V) admit two solutions u1, u2V.

(u1, v) = (f, v) vV and

(u2, v) = (f, v) vV

Subtracting these equations leads to the following:10 (u

1 u

2)v

dx= 0 vV.

Choosingv=u1 u2V results to:10 (u1 u2)2dx= 0, which shows that:

u1(x) u2(x) = 0 x[0, 1].

It follows thatu1(x) u2(x) =constanton [0, 1], which together

with the boundary conditionu1(0) =u2(0) = 0 gives that:u1(x) =u2(x) x[0, 1], and the uniqueness is proved.



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Ifu is a solution of(V) then u also satisfies (D)

LetuVbe a solution of (V), then:10u

vdx 10fvdx= 0 vV ()

We assume thatu exists and is continuous (regularity condition).

Integration of the first term in Equ. (*), and using v(0) = v(1) = 0,leads to: 1

0 (u +f)vdx = 0 vV

By the assumption that (u

+f) is continuous, the above relation canonly hold if:u(x) +f(x) = 0,x [0, 1]

Souis a solution of (D).



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Continuous piecewise linear basis functionsj(x),j = 1, 2, . . . , M

A functionvVh has the representation:

v(x) =Mi=1 i i(x),x[0, 1], where:

i=v(xi) and

j(xi) =ij

1 if i=j0 if i=j

i, j = 1, . . . , M .

The spaceVh is a linear space of dimension Mwith basis{i}Mi=1.



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Finite dimensional problems (Mh) and (Vh)

(Mh): FinduhVh such thatF(uh) F(v) vVh.

(Vh): FinduhVh such that (uh, v

) = (f, v) vVh.

where:uh(x) =M

j=1 jj(x), with

j =uh(xj) (nodal values ofuh(x)).



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Linear system of equations A=b

(Vh): FinduhVhsuch that (uh,

i) = (f, i),i= 1, 2, . . . , M .

We finally obtain the following linear system of equations:M

j=1 j (j,

i) = (f, i), i= 1, 2, . . . , M

where

A= (aij) is theM Mstiffness matrix withaij = (

j,

i)

b= (bi) is the force vector with: bi= (f, i) and

= (i) is the solution vector with: i=uh(xi), i= 1, 2, . . . , M .



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Properties of the stiffness matrix A

A is symmetric,Aij =Aji, (j, i) = (i, j), i, j = 1, 2, . . . , M .

Ais sparse (i.e. only a few elements ofAare nonzero)

(j, j) =xj

xj11

h2jdx+xj+1

xj1

h2j+1dx

= 1hj + 1hj+1, j = 1, 2, . . . , M .

(j, j1) = (

j1,,

j) =xjxj1

1h2j

dx

= 1hj ,j = 2, . . . , M .

(i, j) = 0 if |i j|>1.

A is positive definite. Indeed for M we obtain:

A=Mi=1

Mj=1 iAijj =

Mi=1

Mj=1 i(

i,

j)j

= (M

i=1 ii,M

j=1 jj) = (v, v)0, v=M

i=1 ii(x).

Also, A= 0 only ifj = 0,j = 1, . . . , M .



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Properties of the stiffness matrix A

SinceAis a positive definite matrix, we conclude thatAis non-singular.

It follows that the system A=bhas a unique solution.

For the particular case ofhj =h= 1M+1, the systemA=bbecomes:

1h

2 1 0 0 0 01 2 1 0 0 0

0 1 2 1 0 0 0 0 1 2 1 0 0 0 0 0 1 2 10 0 0 0 1 2

1

M

=

b1

bM



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An error estimate for FEM for the model problem

Let uh be the solution of the finite element problem(Vh), i.e. (u

h, v

) = (f, v), v Vh ()

Letube the solution to problem (D), it then satisfies:

(u

, v

) = (f, v), v V.SinceVhV, we can also write:

(u, v) = (f, v), v Vh ()

Subtracting the equations (*) above results in:

(u uh, v) = 0, v Vh



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Theorem: For any vVh we have (u uh) (u v)

Letv Vh be arbitrary and setw=uh v.

Thenw Vh and using the error condition

(u uh, v) = 0, v Vh, forv=w, we obtain:(u uh, w

) = 0.

Using the Cauchy inequality we also have:

(u uh)

2

=(u uh)

, (u uh)

+(u uh)

, w

=(u uh)

, (u uh+w)=(u uh)

, (u v)

(u uh) (u v)

Dividing by (u uh) , we obtain the required theorem.

We conclude that (u uh) is as small as possible.



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The finite element interpolant

Based on the previous theorem, (u uh) (u v) , v Vh, we

can obtain estimates of the error (u uh) .

We calculate (u uh)

, where uh is the FEM interpolant ofu, i.e.

uh(xj) =u(xj),j = 0, 1, 2, . . . , M + 1



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Proof of |u (x) uh (x)| h2

8 max0y1 |u (y)| , x [0, 1]

1. Recall from the previous derivation that:u(x) u(x) = j+1m=j 12u(m)m(x)(xm x)2

2. With simplifications, we can write:u(x) u(x) = j+1m=j 12u(m)(x2m+x2 2xmx)m

3. Using mxm=xresults in:|u(x) u(x)| 12max0y1 |u

(y)| (j+1m=jx2mm x2) (*)

4. Let us defineE(x) =j+1

m=jx2mm x2 0. It is easy to calculate fromE(x) = 0that E(x) becomes maximum (E(x) = 2 < 0) at x =

hj2 +xj and that

E(x) = h2j4.

5. Using the above result, Equation (*) can be re-written as follows:

|u(x) u(x)| h2

8max0y1 |u (y)|

which completes the proof.



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Proof of (u uh) hmax0y1 |u

(y)|

1. From the main theorem we have that: (u uh) (u uh)

.

2. Using the results from the interpolation error calculations gives that:

(u uh) = (10(u(z) uh(z))2 dz)1/2

(u uh) = (10(u

(z) uh(z))2 dz)1/2

(10h2(max

0y1|u (y)|)2dz)1/2

=hmax0y1 |u (y)|

and the required result was obtained.



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THE MINIMIZATION PROBLEM (M)

Define the linear functionalF :V as:

F(v) = 12 a(v, v) (f, v)

Findu V such thatF(u) F(v) vV

where:

V ={v:v is a continuous function on , vx1 and vx2

are piecewise

continuous on , andv= 0 on }

a (u, v) =

u v dx=

ux1

vx1

+ ux2v

x2

dx

(f, v) =

fvdx

(M) corresponds to thePrinciple of Minimum Potential Energy in Me-chanics.


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THE VARIATIONAL (WEAK) PROBLEM (V)

FinduV such thata(u, v) = (f, v) vV

where:

V ={v:v is a continuous function on , vx1 and vx2

are piecewise

continuous on , andv= 0 on }

a (u, v) =

u v dx=

ux1

vx1

+ ux2v

x2

dx

(f, v) =

fvdx

v is the so called test function

(V) corresponds to the Principle of Virtual Work in Mechanics.


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FEM for the model problem with piecewise linear functions

Construct a finite-dimensional subspaceVh V as follows:

Approximate with a polygonal curve.

Make a triangulation of by subdividing into a set

Th={K1, K2, . . . , K m}of non-overlapping trianglesKi, such that

no vertex of one triangle lies on the edge of another triangle, =KThK=K1 K2 . . . Km

We introduce the mesh parameterhas follows: h= maxKThdiam (K),where, diam(K) = diameter ofK= longest side ofK.

Let us defineVhVto be the set of functions v such that:Vh={v: v is continuous on , v|K is linear forKTh, v= 0 on }

wherev|Kdenotes the restriction ofv toK(i.e. the function defined onKagreeing withv onK).


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A finite element triangulation


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Continuous piecewise linear basis functions j,j = 1, 2, . . . , M

To describe v Vh, we choose the values v(Ni) at the nodes Ni,i= 1, . . . , M , ofTh, but exclude the nodes on the boundary sincev= 0 on . v has then the representation:

v(x) =Mi=1 i i(x),x , where:

i=v(Ni) and

j(Ni) =ij 1 if i=j0 if i=j i, j = 1, . . . , M .

The spaceVh is a linear space of dimension Mwith basis{i}Mi=1.


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The basis function j


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Finite dimensional problems (Mh) and (Vh)

(Mh): FinduhVh such thatF(uh) F(v) vVh

(Vh): FinduhVh such thata(uh, v) = (f, v) vVh

where:

uh(x) =M

j=1 jj(x), with

j =uh(Nj) (nodal values ofuh(x)).


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Linear system of equations A=b

(Vh): FinduhVh such thata(uh, i) = (f, i),i= 1, 2, . . . , M

We finally obtain the following linear system of equations:

Mj=1 j a(j, i) = (f, i), i= 1, 2, . . . , M

where

A= (aij) is theM Mstiffness matrix withaij =a(j, i)

b= (bi) is the force vector with: bi= (f, i) and

= (i) is the solution vector with: i=uh(Ni), i= 1, 2, . . . , M .


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Special case of uniform discretization

Consider to be a square with side length 1 and let Th be the uniform trian-gulation of with the indicated node enumeration (h= 1N+1).


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Special case of uniform discretization

1h

4 1 0 1 0 01 4 1 0 1 0

0 1 4 1 0 1 1 0 1 4 1 0 1 0

0 0 1 0 1 0 1 0 1 4 10 0 1 0 1 4

1

M

=

b1

bM

Note that the stiffness matrix is the same as that in the 5 point finite differencemethod. However, the force vector contains averages off around each nodeNi.


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Interpolation error and finite element method error

The best approximation theorem: u uh u v, v Vh,

v=a (v, v)12 =

|v|2 dx12

=

v

x1

2+

vx2

2 dx12

, v Vh

From the above theorem we have the following FEM error estimate:

u uh u uhwith uhthe FE interpolant, i.e. uhVhand uh (Ni) =u (Ni) , i= 1, . . . , M .

FEM error: IfK Th are not allowed to become too thin, we can show:u u

h Ch

u uh

(u uh)2 dx12

Ch2

whereCish-independent and depends only on the size of the second deriva-

tives ofuand the smallest angle of the triangles K in Th.The error and the gradient of the error u uhtend to zero in the norm

ashtends to zero (assuming that the exact solutionuis sufficiently regular).


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Some preliminary computational aspects

To calculate the stiffness matrix A, assemble over element contributions:

a (i, j) = KTh

aK(i, j) ,

where

aK(i, j) =K

i jdx

LetNi,Nj andNkbe the vertices of the triangle K. We define the elementstiffness matrix forKas follows:

aK(i, i) aK(i, j) aK(i, k)aK(j, j) aK(j, k)

sym aK(k, k)


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The basis function j associated with K.

To compute the element stiffness matrix of the triangle K, we work withthe restrictions of the basis functions j on K,j =j|K.

We calli,j andk the basis functions on the triangle K.

Ifw Vh, its representation on Kis the following:w(x) = w(Ni)i(x) + w(Nj)j(x) + w(Nk)k(x)


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The Hilbert spaces L2(), H1() and H10()

While working with variational forms of BVPsfor PDEs, it is natural and useful to work withfunction spaces V that are slightly larger thanthe spaces of continuous functions with piecewisecontinuous derivatives used previously.

We will endow the spaceVwith the scalar prod-

uct related to the BVP. The space V will be aHilbert space.


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Linear and bi-linear forms

Lis alinear formonV ifL: V , i. e.,L (v) forv V,andLis linear, i. e., for allv, w V and,

L (v+w) =L (v) +L (w)

We say thata (., .) is a bi-linear formonV V ifa:V V ,i. e.,a (v, w) forv, w V, and a is linear in each argument,i. e., for allu, v, w V and, , the following holds:

a (u,v+w) =a (u, v) +a (u, w)

a (u+v, w) =a (u, w) +a (v, w)The bilinear forma (., .) onV V is said to be symmetric if:

a (v, w) =a (w, v) , v, w V


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Inner product and norm related to the BVP

A symmetric bilinear form a (., .) onV V is said to be a scalarproductonV if:

a (v, v)>0 v V, v= 0

Thenorm a associated with a scalar producta (., .) is definedby

va= (a (v, v))12 , v V energy norm

Further, if< ., . > is a scalar product with corresponding norm , then we have the Cauchy-Schwarz inequality:

|< v, w >| v w


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Hilbert spaces

Let V be a linear space with a scalar product and correspondingnorm .

A sequencev1,v2,. . ., of elementsviin the spaceVwith norm is said to be a Cauchy sequence if for all > 0 there is a naturalnumberNsuch that

vi vj< i,j > N

viconverges tov ifv vi 0 asi

The space V is said to be a Hilbert space ifV is complete, i. e., ifeveryCauchy sequencewith respect to is convergent.


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Hilbert spaces for variational problems

IfI= (a, b), define the space of square integrable functions onIas:L2(I) =v: vis defined onIand

Iv2dx


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The Sobolev H1(I) space

We define the space H1(I) as follows:H1 (I) ={v: v andv L2(I)}

We equipH1(I) with the following scalar product:

(v, w)H1(I)=I (vw+v

w) dx

The corresponding norm is:

vH1(I)=

I

v2 + (v)2

dx

12


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The Sobolev H10(I) space

Recall our model problem: u(x) =f(x),x (0, 1),u(0) =u(1) = 0.

For this problem, we need to use the spaceH10(I) defined as:H10(I) =v H1 (I) :v (0) =v (1) = 0

with the same scalar product and norm as forH1(I).

The weak form for the model problem is then written as follows:Findu(x) H10(I) such that: (u

, v) = (f, v), v H10(I),

where (u, v) =I

u v dx.

The spaceH10(I) is the largest space for which the variational problemis meaningful.

The formulation in H10(I) is preferable to that in V in that the basic

error estimate for the FEM is an estimate in the H1 norm.


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The spaces L2(), H1() and H10(),

d, d= 2, 3

L2 () =v:vis defined on and

v2dx


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An example problem in d, d= 2, 3

Let us consider the following BVP:(D) Findu such: u=f in , u= 0 on .

The variational form is given as follows:

(V) Findu H1

0

() such that a (u, v) = (f, v) v H1

0

()

or equivalently

(M) Findu H10() such that F(u) F(v) v H10()

F(v) = 12a (v, v) (f, v),

a (u, v) =

u vdx

(f, v) =

fvdx

It may be difficult to prove the existence of classical solutions of (D).

However, the existence of solutions of (V) (weak solutions of (D)) maybe easier.


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1


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A finite element approximation in VhH10

u+u=f in ,

u= 0 on

Findu H10 such that:

< u, v >= (f, v) , v H10() ,where:

< u, v >=

[u v+uv] dxis the scalar product inH10().

Finduh VhH1

0() such that: < uh, v >= (f, v)v Vh


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A i i i f h FEM i H1


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A geometric interpretation of the FEM in H10

We can obtain: < u uh, v >= 0, v Vh, i.e.the erroru uh is orthogonal toVh with respect to< ., . >.

The finite element solution uonVh is the projectionwith respectto < ., . >of the exact solutionu in Vh, i.e. uh is the element in

Vh closest touwith respect to theH10() norm or in other words:

u uhH1() u vH1() v Vh


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Th N bl


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The Neumann problemNatural and essential boundary conditions

Consider the following Newmann problem (D):

u+u=f in u

n =g on

The variational formulation (V) becomes:

Findu H1 () such that,

a (u, v) = (f, v) +< g, v > v H

1

()wherea (u, v) =

[u v+uv] dx, (f, v) =

fvdx, < g,v >=

gvds

The Newmann condition un =g does not appear explicitly in (V).

The solutionu of (V) is only required to belong toH1() and is notexplicitly required to satisfy the Newmann condition.


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The Neumann problem


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The Neumann problem

LetTh be a triangulation of and letVh={v:vis continuous on ,v|Kis linear K Th} H

1()

As parameters to describe the functions inVh, we choose the valuesat the nodes including the nodes on the boundary .

The finite element method becomes: Find uh Vh such thata (uh, v) = (f, v) +< g, v > v Vh

The problem has a unique solution uh that can be determined by

solving a symmetric, positive definite linear system of equations.The following error estimate is obtained:u uhH1() u vH1() v Vh

and hence as shown earlieru uh

H

1

()

Ch, (if uis regular enough)

The solutionuhwill satisfy the Newmann condition approximately.


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Sh th t l ti H1() f (V ) i l l ti f (D)


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Show that a solution u H1() of (V) is also a solution of (D)

Using Greens theorem, we find that ifuis sufficiently regular

(f, v)+< g, v >=a(u, v) =

unvds+

(u+u)vdx

or rearranging terms:

(u+u f)vdx+

(un g)vds= 0,v H1()

Consider the above equation for the functionsvthat vanish on theboundary. We then conclude that: u+u f= 0, in .

We thus have that: (

un g)vds= 0,v H1().

By varying v over H1() (which means that v will vary freely on), we finally get: un =g, on .


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Programming remarks


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Programming remarks

Consider the Newmann problem examined earlier:

u+u=f in u

n=g on

LetTh={K}be a triangulation of 2 with boundary and let

Vhbe the corresponding space of continuous piecewise linear functions.

Let Ni, i = 1, 2, . . . , M , denote the nodes ofTh and 1, . . . , M the

natural base forVh withi(Nj) =ij.We want to find the solution M of the linear system: A= b,

whereA= (aij) , b= (b1, . . . , bM) andaij = KTh

aKij , bi= KTh

bKi

aKij =K {i j+ij} dx, b

Ki =K f idx+ K gids


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Main tasks in finite element programming


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Main tasks in finite element programming

Input the dataf,g, and the coefficients of the equation.

Construct and represent the triangulationTh.

Compute the element stiffness matricesaK and element loadsbK.

Assemble the global matrixAand load vectorb.

Solve the system of equations A=b.

Present the results (post-processing, visualization, etc.).


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Construction of the triangulation Th


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Construction of the triangulation Th

Successive refinement by connecting the midpoints of each side (thisleads toquasi-uniform mesheswhere the triangles have essentiallythe same size in all parts of ):

In general, one needs smaller triangles in regions where the exact

solution varies quickly or where certain derivatives of the exactsolution are large (adaptive methods).


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Representation of the triangulation Th


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Representation of the triangulation Th

Let Ni, i = 1, 2 . . . , M and Kn, n = 1, . . . , N be enumerations ofthe nodes and triangles ofTh, respectively.

ThenTh is defined using two arrays:

Z(j, i),j = 1, 2 are the coordinates of node Ni.T(j, n),j= 1, 2, 3 are the number of the vertices of triangle Kn.

T =

1 1 2 3 4 5 3 3 8 7 7

3 4 4 8 8 8 6 7 7 6 104 2 5 4 5 9 7 8 9 10 9

When using the Gaussian elimination to solve A = b, the nodesneed to be enumerated such that the bandwidth ofAis minimal.


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Computation of the element stiffness matrices


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p

We know thataKij 0 only if bothNi andNj are nodes ofK.

LetKnTh.

T(, n),= 1, 2, 3 are the vertices ofKn, whereasZ(i, T(, n)),= 1, 2, 3,i= 1, 2 are thexi coordinates of these vertices.

The element stiffness matrix A(n) = (an), , = 1,2,3, for ele-mentKnis given as follows:

an= Kn

[ +] dx

whereis the linear function on Kn that takes the following values:

NT(,n)=

1 if=0 if= , = 1, 2, 3

We also compute:bn= Kn

fdx+ Kn

gdx, = 1, 2, 3


N. Zabaras 59

Assembly of global stiffness matrix


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y g

SetA (i, j) = 0, b (i) = 0, i, j = 1, . . . , M .

Forn= 1, . . . , N ,fetchA(n) =an

andb(n) = (bn) from scratchfile and for, = 1, 2, 3 set:

A (T(, n) , T(, n)) =A (T(, n) , T(, n)) +an,

b (T(, n)) =b (T(, n)) +bn


N. Zabaras 60

Solution of the linear system A=b


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y

If Gaussian elimination is to be used, A is stored as a band ma-trix with an arrangement of the node numbering that leads to aminimum bandwidth.

If an iterative method is used, only the nonzero elements ofAarestored.

Once the stiffness matrix A for the Newmann problem has beendetermined, we can easily derive the system A=bcorrespondingto any other type of boundary conditions.

For example, ifu = uo on 1, we can obtain the correspondingsystem A = b by deleting the rows of A corresponding to thenodes on 1 and by entering the values ofgiven by the Dirichletboundary condition.


Lecture1 Cornell

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Transcript of Lecture1 Cornell