Lecture3 Cornell

8/13/2019 Lecture3 Cornell

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N. Zabaras 1

EXAMPLES OF FINITE ELEMENTS AND FINITE

ELEMENT SPACES

Piecewise polynomial finite element spaces on a triangulation Th={K}of a bounded domain d, d= 1, 2, 3,into elementsK.

The 3-noded triangular finite element basis functions for P1(K).The 6-noded triangular finite element basis functions for P2(K).

TheP3(K) andP5(K) finite element spaces.

Rectangular finite elements in

2

.Pr(K) spaces defined with tetrahedrons.

Lecture 3: Examples of finite elements and finite element spaces Materials Process Design and Control Laboratory


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FINITE ELEMENT SPACES: REGULARITY

REQUIREMENTS

The finite element spaces examined here consist ofpiecewisepolynomialfunctions on a triangulation Th = {K} of a bounded domain d, d= 1, 2, 3,into finite elementsK.

Ford= 1, the elementsKwill be intervals,

Ford= 2, triangles or quadrilaterals, and

Ford= 3, tetrahedrons.

VhH1() VhC0(), for 2nd order boundary value problems

VhH2() VhC

1(), for 4th order boundary value problems

where = and

C0

() =v:vis a continuous function defined on

C1() =vC0() :DvC0(), ||= 1



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VhH1()VhC0() andVhH2()VhC1()

Vh H1()the functions vVh are continuous, i.e. Vh C

0().This equivalence arises in piecewise polynomial approximations as fol-lows:

A functionvVh is a polynomial on each finite elementK

vis taken to be continuous across adjoining element boundaries, thusthe derivativesDv, ||= 1, exist and are piecewise continuous, e.g.vH1().

If v is not continuous across a certain inter-element boundary S,i.e., v / C0(), then the derivatives Dv, || = 1, would be -functions supported bySwhich are not functions inL2() and thusv /H1()

Similarly, Vh H2() the functions v Vh andtheir first deriva-

tives are continuous.



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DEFINING A FINITE ELEMENT SPACE

To define a finite element spaceVh, one needs to specify:

1. The triangulationTh={K}in

2. The nature (type of polynomials) of the functions v in Vh on each

elementK:linear,

quadratic,

cubic, etc.

3. The parameters to be used to describe the functions in Vh:

the values ofv at the nodes ofTh,

the values ofv at points on the sides ofTh (e.g. midpoints),

the values ofv at the center ofTh, etc.



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Formal definition of a finite element

Define a finite element to mean a triple (K, PK,) where

Kis a geometric object, for example a triangle,

PKis a finite-dimensional linear space of functions defined on K,

is a set of degrees of freedom,

such that a function v PK is uniquely determined by the degrees offreedom.

For example for the triangle with degrees of freedom the nodal values, wehave that (K, PK,) is a finite element, where

K is a triangle,

PK=P1(K),

are the values at the vertices ofK



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FINITE ELEMENT SPACE EXAMPLES IN 2 WITH

POLYGONAL BOUNDARY

LetTh={K}be a given triangulation of into triangles K.

For eachr= 0, 1, 2, . . . ,let us define

Pr(K) ={v:vis a polynomial of degree ronK}

For example,P1(K) is the space of linear functions v defined on K, i.e.

v(x) =a00+ a10x1+ a01x2, x K, aij

The basis ofP1(K) is (dimP1(K) = 3): 1(x) = 1, 2(x) =x1, 3(x) =x2

P2(K) is the space of quadratic functions v onK, i.e. v(x)P2(K) is ofthe form:

v(x) =a00+ a10x1+ a01x2+ a20x21+ a11x1x2+ a02x22, xK, aij R

The basis ofP2(K) is (dimP1(K) = 6): {1, x1, x2, x21, x1x2, x

22}



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FINITE ELEMENT SPACE EXAMPLES IN 2 WITH

POLYGONAL BOUNDARY

For eachr= 0, 1, 2, . . . ,we defined:

Pr(K) ={v:vis a polynomial of degree ronK}

Pr(K) =

v : v(x) =

0i+jr

aijxi1x

j2forxK, whereaij

dimPr(K) =(r+ 1) (r+ 2)2



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FINITE ELEMENT SPACE EXAMPLE:

Vh={vC0

() :v|KP1(K), KTh}

Vh is the space of continuous piecewise linear functions discussed earlier.

As global degrees of freedom to describe the functions in Vh, we choose

the values at the node points ofTh (including the node points on ).EahKTh is a triangle with verticesa

i, i= 1, 2, 3.

The element degrees of freedomare the values i at the vertices ai, i =

1, 2, 3.

TheoremLetK Th be a triangle with vertices a

i =ai1, a

i2

, i= 1, 2, 3.. A function

v P1(K) is uniquely determined by the degrees of freedom, i.e., given the

valuesi, i= 1, 2, 3, there is a uniquely determined functionvP1(K) suchthat:

vai=i i= 1, 2, 3



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Theorem: LetKTh be a triangle with vertices ai = (ai1, ai2), i= 1, 2, 3.

A functionvP1(K) is uniquely determined by the degrees of freedom.

Need to show that given the values i, i = 1, 2, 3, there is a uniquelydetermined functionvP1(K) such that: v

ai=i, i= 1, 2, 3.

Since v(x) = c1x1+c2x2+c3 for some constants ci , we obtain thelinear system of equations below for the unknowns ci:

c1ai1+ c2a

i2+ c3=i, i= 1, 2, 3,

This system has a unique solution for giveni if and only if detB = 0:

B =

a11 a12 1

a21 a22 1

a31 a32 1

By basic linear algebra: detB/2 = area ofK, and thus detB= 0.

HenceB is non-singular, which proves the desired result.



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Theorem (another proof): LetKTh be a triangle with vertices

ai

, i= 1, 2, 3.A functionvP1(K) is uniquely determined by the degrees offreedom.

Recall that since v(x) = c1x1+c2x2+c3 for some constants ci R, weobtained the following linear system of equations for the unknownsci:

c1ai1+ c2ai2+ c3=i, i= 1, 2, 3, ()

We notice first that dimP1(K) = number of degrees of freedom (= 3)

i.e. (*) has the same number of unknowns as equations.

Solutions of (*) are unique if the only solution of (*) with i= 0, i= 1, 2, 3,is given byci= 0, i= 1, 2, 3,or formally,

IfvP1(K) andv(ai) = 0, i= 1, 2, 3, thenv0

This approach does not require knowing detB and can be generalized tohigher order polynomials in which case computing detB is complicated.



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Basis functions forP1(K)

We can now determine the basis functions for P1(K) associated with thenodal values, i.e. the functions iP1(K), i= 1, 2, 3,such that

i(aj) =ij =

1 ifi=j

0 ifi=j

i, j = 1, 2, 3.

v(x) = 3i=1

v(ai)i(x), xK, v(x) P1(K)

To determinei, we can solvec1ai1+ c2a

i2+ c3=i, i= 1, 2, 3,for three

special choices of the rhs, namely, (1, 0, 0), (0, 1, 0) and (0, 0, 1).



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Basis functions forP1(K)

Alternatively, to determine e.g. 1, letd1x1+d2x2+d3= 0 be the equationfor the straight line through the verticesa2 anda3. Then:

1(x) =(d1x1+ d2x2+ d3)

where the constantis chosen so that1(a1) = 1.

In the same way, we may determine2 and3.

If the triangleKhas vertices at (1, 0), (0, 1) and (0, 0), then:

1=x1, 2=x2, 3= 1 x1 x2



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Vh={v: v|KP1(K), KTh, andv is continuous at the nodes}

Vh={vC0

() :v|KP1(K), KTh}To show this equivalence, we need to prove that: vC0(). Indeed:

It suffices to show thatv is continuous across inter-element boundaries.

LetK1 andK2 be two triangles inTh having the common sideSwiththe end pointsN1andN2.

Let vi=v|KiP1(Ki), i= 1, 2, be the restrictions ofvto the elementsK1 andK2, respectively.

Then the functionw=v1 v2defined onSvanishes at the end pointsN1 andN2.

Sincewis linear onSit follows thatw vanishes onS.

Hence,v is continuous across Sand as suchvC0

().



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A finite element spaceVh using piecewise quadratic functionsv:

v|KP2(K)

Let K Th be a triangle with vertices ai, i = 1, 2, 3, and denote the

midpoints of the sides ofKbyaij, i < j, i, j = 1, 2, 3.

TheoremA functionv P2(K) is uniquely determined by the following degrees offreedom:

v(ai), i= 1, 2, 3,v(aij), i < j, i, j= 1, 2, 3



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Theorem: A functionvP2(K) is uniquely determined by the following

degrees of freedom: v(ai

), i= 1, 2, 3, andv(aij

), i < j, i, j = 1, 2, 3

Since dim P2(K) = 6 (=number of dof), it is sufficient to prove that if:vP2(K) andv(a

i) = 0, v(aij) = 0, i < j, i, j = 1, 2, 3, =v0.

Consider the sidea2

a3

. Along this side, v has a quadratic variation andvanishes at the pointsa2, a23, anda3. Thusvvanishes identically ona23

and we can write:

v(x) =1(x)w1(x), xK

wherew1P1(K) andi, i= 1, 2, 3 are the basis functions forP1(K).

In the same way, we see that v also vanishes along the sidea1a3:

v(x) =1(x)2(x)w0, xK, w0= constant.

If we now finally takex=a12: 0 =v(a12) =w01(a12)2(a

12) =w01

2 12,

so thatw0= 0 and hencev0 and the proof is complete.



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A functionvP2(K)

ForvP2(K): v= 3i=1

v(ai)i (2i 1) + 3

i,j=1,i


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The finite element space: Vh=vC0

: v|KP2(K), KTh

We have seen that the global degrees of freedom of the functions v Vhcan be chosen as:

1. the values ofv at the nodes ofTh,

2. the values ofv at the mid points of all the sides of the triangles in Th.

Global basis functions



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The finite element spacev|KP3(K), KTh

LetKbe a triangle with vertices ai, i= 1, 2, 3,and define:

aiij = 1

3(2ai + aj), i , j = 1, 2, 3, i=j

a123 = 1

3(a1 + a2 + a3)



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Theorem: A functionvP3(K) is uniquely determined by the following

degrees of freedom: v(ai

), v(aiij

), i , j = 1, 2, 3, i=j andv(a123

)

Since dim P3(K) is equal to the number of dof (=10), it is sufficient to showthat ifv P3(K) andv(a

i) =v(aiij) =v(a123) = 0, i, j = 1, 2, 3, i=j,thenv0.

Observe that ifv has a cubic variation along the sidea2a3 thenv0 ona2a3. In the same way it follows that v vanishes on the sides a1a3 anda1a2 and hencev(x) =1(x)2(x)3(x), whereis a constant.

If we now choosex=a123

: 0 =v(a123

) =1

3 1

3 1

3 == 0 =v0.



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The finite element spaceVh=vC0

:v|KP3(K), KTh

Let vi = v|Ki P3(Ki), i = 1, 2, where K1 and K2 are twotriangles with common sideS

Suppose thatv1andv2take the same values at the end points and

the two pointsaiij

ofS.Sincev1 v2 varies cubically onS, it follows thatv1=v2onS.



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:v|KP3(K), KTh

We now introduce the finite element space Vh with the following degreesof freedom:

1. the values ofv at the nodes ofTh

2. the values ofv at the points aiij on the sides ofTh

3. the values ofv at the center of gravity for all KTh.



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AnotherP3(K)finite element space where Kis a triangle with vertices

a

i

, i= 1, 2, 3,and center of gravity a

123

.TheoremA functionv P3(K) is uniquely determined by the following dof: v(a

i), i=1, 2, 3, vxj

ai

, i= 1, 2, 3, j = 1, 2,andva123

Since dim P3(K) = # of dof, it suffices to prove that ifv P3(K) &vai= v

xj

ai=v

a123

= 0, i= 1, 2, 3, j = 1, 2, () =v0

From (*) vsai= vx1

ai

s1+ vx2

ai

s2= 0, i= 1, 2, 3.

We then have: vsa2

= vs

a3

= 0, where s is the direction from a2 to

a3. Usingv(a2) =v(a3) = 0 =vvanishes alonga2a3 sincevvaries as acubic polynomial along this side. Similarly,v vanishes ona1a2 anda1a3.

Hencev(x) =1(x)2(x)3(x), whereis a constant.

If we now choosex=a123: 0 =v(a123) =13

13

13

== 0 =v0.



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:v|KP3(K), KTh

We expressed every functionvP3(K) uniquely by the following degreesof freedom:

v(ai) i= 1, 2, 3,v

xj

ai

, i= 1, 2, 3, j = 1, 2,

va123

The corresponding finite element space VhC0

is given by

Vh={v: v|KP3(K), KTh,

v and vxi , i= 1, 2,are continuous at the nodes}

with the following degrees of freedom:

1. The values ofv and vxi, i= 1, 2,at the nodes ofTh,

2. The values ofv at the center of gravity of eachKTh.



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A finite element space satisfying VhC1()

We work with functions that are polynomials of degree 5 on eachK.Theorem

LetKbe a triangle with vertices ai, i = 1, 2, 3, and let aij be the midpointon the side aiaj, i , j = 1, 2, 3, i < j. A functionv P5(K) is uniquely

determined by the following dof:

Dv(ai), i= 1, 2, 3, || 2,

v

n

aij

, i , j = 1, 2, 3, i < j

where n denotes differentiation in the outward normal direction toS.



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Since dimP5(K) = 21 (# dof), need to show that if all dof = 0 v= 0.

Withsthe direction on the sidea2a3, we conclude that

v(ai) =v

s

(ai) =2v

2

s

(ai) = 0, i= 2, 3

Sincevis a polynomial on the sidea2a3 of degree at most 5 = vvanishesona2a3. Further, vn is a polynomial of degree at most 4 ona

2a3 and

v

na23

=

v

na

i=

sv

n

ai

= 0, i= 2, 3,

Thus bothv and vn vanish ona2a3.

We may factor (1(x))2 out ofv (x).

v (x) = (1 (x))2p3 (x) , xK, p3P3(K) =v=212223,

SincevP5(K) we conclude that = 0 so thatv= 0 onK.


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ConsiderK1 &K2 with a common side Sand letv1, v2P5(Ki).

Assume that: Dv1=Dv2, at the end points ofS, || 2,

v1

n =

v2

nat the midpoint ofS

Withw=v1 v2, we then conclude that:

w=w

s =

2w

2s = 0, at the end points ofS

w

n =

w

n =

w

s

w

n

= 0, at the midpoint ofS

It follows thatw= wn = 0 onS (*)

Fromw= 0 onS= ws = 0 onS(stangent toS) (*)

From (*) = v|Ki=viand its 1st derivatives vary continuously across S.


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Define the spaceVhC1

Vh = { v:v|KP5(K), KTh, Dvis continuous at the nodes for

|| 2 andv

nis continuous at the midpoints of each side }

Each function v|K P5(K) is determined with the following degrees offreedom:

Dv(ai), i= 1, 2, 3, || 2,

vn

aij

, i , j = 1, 2, 3, i < j


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A three-dimensional finite element: Tetrahedron

Consider that is the union ofTh = {K} of non-overlapping tetrahe-dronsKsuch that no vertex of one tetrahedron lies on a side of anothertetrahedron.

Forr= 1, 2, . . ., andKTh, we define:

Pr(K) = { v:vis a polynomial to degreer onK, i.e.,v has

the formv (x) =

i+j+mraijmx

i1x

j2x

m3, aijm }

For r = 1, a function v P1(K) is uniquely determined by the valuesv(ai), i= 1, 2, 3, 4, where theai are the vertices ofK.

We then introduce the space:

Vh={vC0() :v|KP1(K), KTh}

with global dof the values at the nodes ofTh.


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Rectangular finite elements in2: Bilinear functions

Letai, i= 1, 2, 3, 4 be the vertices of the rectangularKwith sides parallelto the axis in2.

Define: Q1(K) ={ v:vis bilinear onK, i.e.,v (x) =a00+ a10x1+ a01x2+ a11x1x2, xK, aij }

A function v Q1(K) is uniquely determined by the values v(ai), i =

1, 2, 3, 4.

IfK1, K2 have a common side S and v1, v2 Q1(Ki) agree at the end

points ofS, thenv1 v2= 0 onSsincev1 v2varies linearly onS.We defineVh={vC

0() :v|KQ1(K), KTh}where

Th = {K} is a subdivision of into non-overlapping rectangles suchthat no vertex of any rectangle lies on a side of another rectangle.

The values at the nodes are the global degrees of freedom.


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Rectangular finite elements in2: Biquadratic functions

We can also use polynomials of higher degree on each rectangle. Forexample, we may choose

Vh=vC0

:v|KQ2 (K) , KTh

whereQ2(K) is the set of biquadratic functions on K, i.e.,

Q2(K) =

v:v (x) =

2i,j=0

aijxi1x

j2, x K, aij

For this elements, we use as global degrees of freedom the following:1. the values at the nodes ofTh,

2. the values at the midpoints of the sides ofTh,

3. the values at the midpoint of each rectangle KTh.


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Finite element examples

In the elements shown, the various dof are denoted as follows: function values values of the first derivatives

values of the second derivatives

/ value of the normal derivative

value of the mixed derivative 2

vx1x2


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Local support of finite element basis functions

The support is always local.

If and are two basis functions associated with the nodes N1 andN2,then the supports of the functions and overlap only if N1 and N2belong to the same element.

The basis functions shown below correspond to a node, midpoint of a sideand a point in the interior.


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