Introduction to FEM

Kengo NakajimaInformation Technology Center

FDM and FEM

• Numerical Method for solving PDE’s– Space is discretized into small pieces (elements, meshes)

• Finite Difference Method (FDM)– Differential derivatives are directly approximated using

Taylor Series Expansion.

• Finite Element Method（FEM）– Solving “weak form” derived from integral equations.

• “Weak solutions” are obtained.

– Method of Weighted Residual (MWR), Variational Method – Suitable for Complicated Geometries

• Although FDM can handle complicated geometries ...

2FEM-intro

Finite Difference Method (FDM)Taylor Series Expansion

∆x ∆x

φi-1

φi

φi+1

( ) ( )K

iiiii x

x

x

x

xx

∂∂∆+

∂∂∆+

∂∂∆+=+ 3

33

2

22

1 !3!2

φφφφφ

( ) ( )K

iiiii x

x

x

x

xx

∂∂∆−

∂∂∆+

∂∂∆−=− 3

33

2

22

1 !3!2

φφφφφ

2nd-Order Central Difference

( )K

ii

ii

x

x

xx

∂∂∆×+

∂∂=

∆− −+

3

3211

!3

2

2

φφφφ

3FEM-intro

FEM1D 4

2nd –Order Differentiation in FDM• Approximate Derivative at×(center of i and i+1)

∆x ∆x

φi-1

φi

φi+1

× xdx

d ii

i ∆−≈

+

+

φφφ 12/1

∆x→0: Real Derivative

• 2nd-Order Diff. at i

211

11

2/12/12

2 2

xxxx

x

dx

d

dx

d

dx

d iiiiiii

ii

i∆

+−=∆

∆−−

∆−

=∆

−

≈

−+−+

−+ φφφφφφφφφ

φ

i+1/2

∆x ∆x

φi-1

φi

φi+1

×i+1/2

×i-1/2

1D Heat Conduction

• Linear Equation at Each Grid Point

211

11

2/12/12

2 2

xxxx

x

dx

d

dx

d

dx

d iiiiiii

ii

i∆

+−=∆

∆−−

∆−

=∆

−

≈

−+−+

−+ φφφφφφφφφ

φ

• 2nd-Order Central Difference

222

11

1)(,

2)(,

1)(

)1()()()()(

xiA

xiA

xiA

NiiBFiAiAiA

RDL

iRiDiL

∆=

∆−=

∆=

≤≤=×+×+× +− φφφ

02

2

=+ BFdx

d φ

)1(0)(121

)1(0)(2

12212

211

NiiBFxxx

NiiBFx

iii

iii

≤≤=+∆

+∆

−∆

≤≤=+∆

+−

−+

−+

φφφ

φφφ

5FEM-intro

FDM can handle complicated geometries: BFC

Handbook of Grid Generation

6FEM-intro

History of FEM• In 1950’s, FEM was originally developed as a method

for structure analysis of wings of airplanes under collaboration between Boeing and University of Washington (M.J. Turner, H.C. Martin etc.).– “Beam Theory” cannot be applied to sweptback wings for

airplanes with jet engines.

• Extended to Various Applications– Non-Linear: T.J.Oden– Non-Structure Mechanics: O.C.Zienkiewicz

• Commercial Package– NASTRAN

• Originally developed by NASA• Commercial Version by MSC• PC version is widely used in industries

7FEM-intro

Recent Research Topics

• Non-Linear Problems– Crash, Contact, Non-Linear Material– Discontinuous Approach

• X-FEM

• Parallel Computing– also in commercial codes

• Adaptive Mesh Refinement (AMR)– Shock Wave, Separation– Stress Concentration– Dynamic Load Balancing (DLB) at Parallel Computing

• Mesh Generation– Large-Scale Parallel Mesh Generation

8FEM-intro

FEM-intro 9

• Numerical Method for PDE (Method of Weighted Residual)

• Gauss-Green’s Theorem• Numerical Method for PDE (Variational Method)

FEM-intro 10

Approximation Method for PDEPartial Differential Equations: 偏微分方程式

• Consider solving the following differential equation (boundary value problem), domain V, boundary S :

fuL =)(

• u (solution of the equation) can be approximated by function uM (linear combination)

i

M

i

iM au Ψ=∑=1

iΨ

ia

Trial/Test Function （試行関数）(known function of position, defined in domain and at boundary. “Basis” in linear algebra.

Coefficients (unknown)

FEM-intro 11

Method of Weighted Residual MWR: 重み付き残差法

• uM is exact solution of u if R (residual：残差)= 0:

fuLR M −= )(

• In MWR, consider the condition where the following integration of R multiplied by w (weight/weighting function：重み関数) over entire domain is 0

0)( =∫V

M dVuRw

• MWR provides “smoothed” approximate solution, which satisfies R=0 in the domain V

FEM-intro 12

Variational Method (Ritz) (1/2)変分法

• It is widely known that exact solution u provides extreme values (max/min) of “functional：汎関数” I(u) – Euler equation: differential equation satisfied by u, if

functional has extreme values（極値）– Euler equation is satisfied, if u provides extreme values of

I(u).

– provide extreme values：停留させる（or stationarize）

• For example, functional, which corresponds to governing equations of linear elasticity (principle of virtual work, equilibrium equations), is “principle of minimum potential energy (principle of minimum strain energy)（エネルギー最小，歪みエネルギー最小）” .

FEM-intro 13

Variational Method (Ritz) (2/2)変分法

i

M

i

iM au Ψ=∑=1

• Substitute the following approx. solution into I(u), and calculate coefficients ai under the condition where IM=I(uM) provides extreme values, then uMis obtained:

• Variational method is theoretical method, and can be only applied to differential equations, which has equivalent variational problem.– In this class, we mainly use MWR– Brief overview of Ritz method will given later today.

FEM-intro 14

Finite Element Method (FEM)有限要素法

• Entire region is discretized into fine elements（要素）, and the following approximation is applied to each element:

i

M

i

iM au Ψ=∑=1

• MWR or Variational Method is applied to each element

• Each element matrix is accumulated to global matrix, and solution of obtained linear equations provides approx. solution of PDE.

• Details of FEM will be provided after next class

FEM-intro 15

Example of MWR (1/3)

• Thermal Equation

02

2

2

2

=+

∂∂+

∂∂

Qy

T

x

Tλ

0=T at boundary S

in V

S

V

• Approximate Solution

j

n

j

jaT Ψ=∑=1

λ：Conductivity，Q：Heat Gen./Volume

• Residual

Qyx

ayxaR jjn

j

jj +

∂Ψ∂

+∂

Ψ∂= ∑

=2

2

2

2

1

),,( λ

FEM-intro 16

Example of MWR (2/3)

• Multiply weighting function wi, and apply integration over V:

0=∫ dVRwV

i

• If a set of weighting function wi is a set of ndifferent functions, the above integration provides a set of n linear equations:

• # trial/test functions = # weighting functions

),...,1(1

2

2

2

2

nidVQwdVyx

wan

j V

ijj

V

ij =−=

∂Ψ∂

+∂

Ψ∂∑ ∫∫

=

λ

FEM-intro 17

Example of MWR (3/3)

• Matrix form of the equations is described as follows:

[ ]{ } { }QaB =

dVQwQdVyx

wBV

iijj

V

iij ∫∫ −=

∂Ψ∂

+∂

Ψ∂= ,

2

2

2

2

λ

Actual approach is slightly different from this(more detailed discussions after next week)

FEM-intro 18

Various types of MWR’s

• Various types of weighting functions

• Collocation Method 選点法• Least Square Method 最小自乗法• Galerkin Method ガラーキン法

FEM-intro 19

Collocation Method

• Weighting function: Dirac’s Delta Function δ

( )ixx −= δiw x：location

• In collocation method, R (residual) is set to 0 at ncollocation points by feature of Dirac’s Delta Fn. δ :

( )ixxi

xx ==−∫ |RdVRV

δ

( )( ) ( )∫

∞+

∞−=≠=

=∞=

1,00

0

dzzzifz

zifz

δδ

δ

• If n increases, R approaches to 0 over entire domain.

FEM-intro 20

Least Square Method• Weighting function:

ii a

Rw

∂∂=

• Minimize the following integration according to ai(unknowns):

[ ] dVaRaIV

ii ∫=2),()( x

[ ] 0),(),(2)( =

∂∂=

∂∂

∫ dVaaR

aRaIa

V i

iii

i

xx

0),(

),( =

∂∂

∫ dVaaR

aRV i

ii

xx

FEM-intro 21

Galerkin Method

• Weighting Function = Test/Trial Function:

iiw Ψ=

• Galerkin, Boris Grigorievich – 1871-1945– Engineer and Mathematician of Russia– He got a hint for Galerkin Method while

he was imprisoned because of anti-czarism (1906-1907).

FEM-intro 22

Example (1/2)

• Governing Equation

)10(02

2

≤≤=++ xxudx

ud

• Boundary Conditions: Dirichlet0@0 == xu1@0 == xu

• Exact Solution

xx

u −=1sin

sin

FEM-intro 23

Exact Solution xx

u −=1sin

sin

0

0.02

0.04

0.06

0.08

0.00 0.25 0.50 0.75 1.00

x

u

FEM-intro 24

Example (2/2)• Assume the following approx. solution:

221122

121 )1()1())(1( Ψ+Ψ=−+−=+−= aaaxxaxxxaaxxu

• Residual is as follows:

)1(),1( 221 xxxx −=Ψ−=Ψ

232

12

21 )62()2(),,( axxxaxxxxaaR −+−+−+−+=

• Let’s apply various types of MWR to this equation– We have two unknowns (a1, a2), therefore we need two

independent weighting functions.

Test/trial function satisfies u=0@x=0,1

FEM-intro 25

Collocation Method

• n=2，x=1/4，x=1/2 for collocation points:

• Solution:

0)2

1,,(,0)

4

1,,( 2121 == aaRaaR

=

−2/1

4/1

8/74/7

64/3516/29

2

1

a

a

217

40,

31

621 == aa

)4042(217

)1(x

xxu +−=

232

12

21 )62()2(),,( axxxaxxxxaaR −+−+−+−+=

FEM-intro 26

Least Square Method• Weighting functions, Residual:

• Solution:

=

399

55

1572707

101202

2

1

a

a

32

22

2

11 62,2 xxxa

Rwxx

a

Rw −+−=

∂∂=−+−=

∂∂=

0)62(),,(),,(

0)2(),,(),,(

321

021

2

1

021

21

021

1

1

021

=−+−=∂∂

=−+−=∂∂

∫∫

∫∫

dxxxxxaaRdxa

RxaaR

dxxxxaaRdxa

RxaaR

246137

41713,

246137

4616121 == aa

)4171346161(246137

)1(x

xxu +−=

232

12

21 )62()2(),,( axxxaxxxxaaR −+−+−+−+=

FEM-intro 27

Galerkin Method• Weighting functions, Residual:

• Results:

=

20/1

12/1

105/1320/3

20/310/3

2

1

a

a

41

7,

369

7121 == aa

)1(),1( 22211 xxwxxw −=Ψ=−=Ψ=

0)(),,(),,(

0)(),,(),,(

321

0212

1

021

21

0211

1

021

=−=Ψ

=−=Ψ

∫∫

∫∫dxxxxaaRdxxaaR

dxxxxaaRdxxaaR

)6371(369

)1(x

xxu +−=

232

12

21 )62()2(),,( axxxaxxxxaaR −+−+−+−+=

FEM-intro 28

Results

• Galerkin Method provides the most accurate solution– If functional exists, solutions of variational method and

Galerkin method agree.• A kind of analytical solution (later of this material)

• Many commercial FEM codes use Galerkin method.• In this class, Galerkin method is used.• Least-square may provide robust solution in Navier-

Stokes solvers for high Re.

X AnalyticalCollocation

0.25-0.50Collocation

0.33-0.67Least-Square

Galerkin

0.25 0.04401 0.04493 0.04462 0.04311 0.04408

0.50 0.06975 0.07143 0.07031 0.06807 0.06944

0.75 0.06006 0.06221 0.06084 0.05900 0.06009

FEM-intro 29

Homework (1/2)• Apply the following two method is the next page to

the same equations:– Method of Moment– Sub-Domain Method– Results at x=0.25, 0.50, 0.75

• Compare the results of “collocation method” on “non-collocaion points” with exact solution– Explain the behavior – Try different collocation points

FEM-intro 30

Homework (2/2)

• Method of Moment（モーメント法）

)1(1 ≥= − iw ii x– Weighting functions ?

• Sub-Domain Method（部分領域法）– Domain V is divided into subdomains Vi (i=1-n), and

weighting functions wi are given as follows:

=0

1iw

for points in Vifor points out of Vi

– Two unknowns, two sub domains

FEM-intro 31

• Numerical Method for PDE (Method of Weighted Residual)

• Gauss-Green’s Theorem• Numerical Method for PDE (Variational Method)

FEM-intro 32

Gauss’s Theorem

( )dSWnVnUndVz

W

y

V

x

U

S

zyx

V∫∫ ++=

∂∂+

∂∂+

∂∂

• 3D （x,y,z）

• Domain V surrounded by smooth closed surface S

• 3 continuous functions defined in V :– U(x,y,z)，V(x,y,z)，W(x,y,z)

• Outward normal vector n on surface S: – nx，ny，nz: direction cosine

V

Sn

FEM-intro 33

Green’s Theorem (1/2)

z

BAW

y

BAV

x

BAU

∂∂=

∂∂=

∂∂= ,,

• Assume the following functions:

• Thus：

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂+

∂∂+

∂∂=

∂∂+

∂∂+

∂∂

z

B

z

A

y

B

y

A

x

B

x

A

z

B

y

B

x

BA

z

W

y

V

x

U2

2

2

2

2

2

• Apply Gauss’s theorem:

( ) dSnz

Bn

y

Bn

x

BAdSWnVnUn

dVz

B

z

A

y

B

y

A

x

B

x

AdV

z

B

y

B

x

BA

S

zyx

S

zyx

VV

∫∫

∫∫

∂∂+

∂∂+

∂∂=++=

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂+

∂∂+

∂∂

2

2

2

2

2

2

FEM-intro 34

Green’s Theorem (2/2)• (cont.)

• Finally:

∫∫ ∫

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂−

∂∂=

∂∂+

∂∂+

∂∂

VV S

dVz

B

z

A

y

B

y

A

x

B

x

AdS

n

BAdV

z

B

y

B

x

BA

2

2

2

2

2

2

dSn

BA

dSn

z

z

B

n

y

y

B

n

x

x

BAdSn

z

Bn

y

Bn

x

BA

S

SS

zyx

∫

∫∫

∂∂=

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂=

∂∂+

∂∂+

∂∂

n

B

∂∂

• Appears often after next week– From 2nd order differentiation to 1st order differentiation.

Gradient of B to the direction of normal vector

FEM-intro 35

In Vector Form

• Gauss’s Theorem

• Green’s Theorem

dSdVS

T

V

nww ∫∫ =⋅∇

( ) ( )( ) dVuvdSuvdVuvV

TT

SV

∇∇−∇=∆ ∫∫∫ n

FEM-intro 36

• Numerical Method for PDE (Method of Weighted Residual)

• Gauss-Green’s Theorem• Numerical Method for PDE (Variational Method)

FEM-intro 37

Variational Method (Ritz) (1/2)変分法

• It is widely known that exact solution u provides extreme values (max/min) of “functional：汎関数” I(u) – Euler equation: differential equation satisfied by u, if

functional has extreme values（極値）– Euler equation is satisfied, if u provides extreme values of

I(u).

– provide extreme values：停留させる（or stationarize）

• For example, functional, which corresponds to governing equations of linear elasticity (principle of virtual work, equilibrium equations), is “principle of minimum potential energy (principle of minimum strain energy)（エネルギー最小，歪みエネルギー最小）” .

FEM-intro 38

Variational Method (Ritz) (2/2)変分法

i

M

i

iM au Ψ=∑=1

• Substitute the following approx. solution into I(u), and calculate coefficients ai under the condition where IM=I(uM) provides extreme values, then uMis obtained:

• Variational method is theoretical method, and can be only applied to differential equations, which has equivalent variational problem.– In this class, we mainly use MWR– Brief overview of Ritz method will given later today.

FEM-intro 39

Application of Variational Method (1/5)

( ) dVQuy

u

x

uuI

V

−

∂∂+

∂∂= ∫ 22

122

• Consider the following integration I(u) in 2D-domain V, where u(x,y) is unknown function of x and y:

S

V

• I(u) is “functional（汎関数）” of function u• u* is a twice continuously differentiable function and

minimizes I(u). η is an arbitrary function which satisfies η=0 at boundary S, and α is a parameter. Consider the following equation:

Q: known value

0=u at boundary S

( ) ( ) ( )yxyxuyxu ,,, * ηα ⋅+=

FEM-intro 40

Application of Variational Method (2/5)

( ) ( )*uIuI ≥• At this stage, the following condition is necessary:

( ) 00

* =⋅+∂∂

=α

ηαα

uI

• Assume that functional I(u*+αη) is a function of α. Functional I provides minimum value, if α=0. Therefore, the following equation is obtained:

• According to the definition of functional I(u), following equation is obtained

0**

=

−

∂∂

∂∂+

∂∂

∂∂

∫ dVQyyu

xx

u

V

ηηη

FEM-intro 41

( ) 00

* =⋅+∂∂

=α

ηαα

uI

0**

=

−

∂∂

∂∂+

∂∂

∂∂

∫ dVQyyu

xx

u

V

ηηη

( ) ( ) ηα

ηαα

Qu

QQu =∂

⋅+∂=∂∂ *

( ) dVQuy

u

x

uuI

V

−

∂∂+

∂∂= ∫ 22

122

( ) ( ) ( )yxyxuyxu ,,, * ηα ⋅+=

( )

yy

u

y

u

xx

u

x

u

xx

u

xx

u

x

u

x

u

x

u

x

u

x

u

∂∂

∂∂=

∂∂

∂∂

∂∂

∂∂=

∂∂

∂∂

⇒=∂∂=

∂∂

∂∂

∂∂+

∂∂=

∂⋅+∂=

∂∂

∂∂

∂∂⋅

∂∂=

∂∂

∂∂

ηα

ηα

αηα

ηαηααα

*2*2

**2

2

1,

2

10,

,2

1

FEM-intro 42

Application of Variational Method (3/5)• Apply Green’s theorem on 1st and 2nd term of LHS,

and apply integration by parts, then following equation is obtained:（A=η, B=u*）(next page) ：

• At boundary S, η=0:

0*

2

*2

2

*2

=∂∂+

+

∂∂+

∂∂− ∫∫ dSn

udVQ

y

u

x

u

SV

ηη

yx ny

un

x

u

n

uwhere

∂∂+

∂∂=

∂∂ *** Gradient of u* in the direction

of normal vector

02

*2

2

*2

=

+

∂∂+

∂∂− ∫ dVQy

u

x

u

V

η

• (A) is required, if the above is true for arbitrary η0

2

*2

2

*2

=+∂∂+

∂∂

Qy

u

x

u(A)

FEM-intro 43

Green’s Theorem• （A=η, B=u*）：

∫∫ ∫

∂∂

∂∂+

∂∂

∂∂−

∂∂=

∂∂+

∂∂

VV S

dVy

u

yx

u

xdS

n

udV

y

u

x

u ***

2

*2

2

*2 ηηηη

∫ ∫∫ ∂∂+

∂∂+

∂∂−=

∂∂

∂∂+

∂∂

∂∂

V SV

dSn

udV

y

u

x

udV

y

u

yx

u

x

*

2

*2

2

*2**

ηηηη

0**

=

−

∂∂

∂∂+

∂∂

∂∂

∫ dVQyyu

xx

u

V

ηηη

FEM-intro 44

Application of Variational Method (4/5)

• Equation (A) is called “Euler equation”– Necessary condition of u*, which minimizes functional I(u), is

that u* satisfies the Euler equation.

• Sufficient condition:– Assume that u* is solution of the Euler equation and αη=δu*

( ) ( )( ) ( )

dVy

u

x

udVuQ

y

u

x

u

uIuuI

VV

∂∂+

∂∂+

+

∂∂+

∂∂−

=−+

∫∫2*2*

*2

*2

2

*2

***

2

1 δδδ

δ

δI= 0First Variation第一変分

δI2≧0Second Variation第二変分

FEM-intro 45

Application of Variational Method (5/5)• It has been proved that u* (solution of Euler equation)

minimizes functional I(u).

( ) ( )*** uIuuI ≥+δ• Therefore, boundary value problem by Euler equation

(A) with B.C. (u=0) is equivalent to variational problem.– Solving equivalent variational problem provides solution of

Euler equation (Poission equation in this case)– Functional must exist !

*u

FEM-intro 46

Approx. by Variational Method (1/4)

• Functional

( ) dxxuudx

duuI ∫

−−

=1

0

22

2

1

2

1

• Boundary Condition0@0 == xu1@0 == xu

• Obtain u, which “stationalizes” functional I(u) under this B.C.– Corresponding Euler equation is as follows (same as

equation in p.21):

)10(02

2

≤≤=++ xxudx

ud (B-1)

FEM-intro 47

Approx. by Variational Method (2/4)

• Assume the following test function with n-th order for function u, which is twice continuously differentiable:

( ) ( )123211 −++++⋅−⋅= nnn xaxaxaaxxu L• If we increase the order of test function, un is closer

to exact solution u. Therefore, functional I(u) can be approximated by I(un):– If I(un) stationarizes, I(u) also stationarizes.

• We need to obtain set of unknown coefficients ak, which satisfies the following stationary condition:

( ) ( )nka

uI

k

n ~10 ==∂

∂(B-3)

(B-2)

FEM-intro 48

Ritz Method

• Equation (B-3) is linear equations for a1-an.

• If this solutions is applied to equation (B-2), approximate solution, which satisfies Euler equation (B-1), is obtained.– Approximate solution, but satisfies Euler equation strictly （厳密解）

• This type of method using a set of coefficients a1-an is called “Ritz Method”.

FEM-intro 49

Approx. by Variational Method (3/4)

• Ritz Method, n=2

( ) ( ) ( ) ( ) 221212 111 axxaxxxaaxxu ⋅−⋅+⋅−⋅=+⋅−⋅=

( )⇒=

∂∂

01

2

a

uI ( )( )

( )( ) ( ){ } ( ) 0113221

311

1

0

22

1

0

232

1

1

0

22

=−+

−−−−+

+−−−

∫∫

∫

dxxxadxxxxxx

adxxxxx

( )⇒=

∂∂

02

2

a

uI ( )( ) ( ){ }

( )( ) ( ) 012232

13221

1

0

32

1

0

3232

1

1

0

232

=−+

−−+−+

−−−−

∫∫

∫

dxxxadxxxxxxx

adxxxxxx

FEM-intro 50

Supplementation for (3/4) (1/3)

• Ritz Method, n=2

( ) ( ) ( ) ( ) 221212 111 axxaxxxaaxxu ⋅−⋅+⋅−⋅=+⋅−⋅=

( ) dxxuudx

duuI ∫

−−

=1

0

22

2

1

2

1

( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]2312

2

22

1

2

22

1

22

11

112

13221

2

1

2

1

2

1

axxaxx

axxaxxaxxax

xuudx

du

⋅−⋅+⋅−⋅−

⋅−⋅+⋅−⋅−−+−

=−−

FEM-intro 51

Supplementation for (3/4) (2/3)

( ) ( ){ }

( )( ) ( ){ } ( ) 0113221

121

1

0

22

1

0

232

1

1

0

222

=−⋅−

−⋅−−−+

−⋅−−

∫∫

∫

dxxxadxxxxxx

adxxxx

( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]2312

2

22

1

2

22

1

22

11

112

13221

2

1

2

1

2

1

axxaxx

axxaxxaxxax

xuudx

du

⋅−⋅+⋅−⋅−

⋅−⋅+⋅−⋅−−+−

=−−

( )⇒=

∂∂

01

2

a

uI

FEM-intro 52

Supplementation for (3/4) (3/3)

( )( ) ( ){ }

( ) ( ){ } ( ) 01132

13221

1

0

32

1

0

2422

1

1

0

232

=−⋅−

−⋅−−+

−⋅−−−

∫∫

∫

dxxxadxxxx

adxxxxxx

( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]2312

2

22

1

2

22

1

22

11

112

13221

2

1

2

1

2

1

axxaxx

axxaxxaxxax

xuudx

du

⋅−⋅+⋅−⋅−

⋅−⋅+⋅−⋅−−+−

=−−

( )⇒=

∂∂

02

2

a

uI

FEM-intro 53

Approx. by Variational Method (4/4)

• Final linear equations are as follows:

=

20/1

12/1

105/1320/3

20/310/3

2

1

a

a

41

7,

369

7121 == aa

)6371(369

)1(x

xxu +−=

• This result is identical with that of Galerkin Method– NOT a coincidence !!

FEM-intro 54

Galerkin Method• Weighting functions (which satisfy u=0@x=0,1),

Residual:

• Results:

=

20/1

12/1

105/1320/3

20/310/3

2

1

a

a

41

7,

369

7121 == aa

)1(),1( 22211 xxwxxw −=Ψ=−=Ψ=

0)(),,(),,(

0)(),,(),,(

321

0212

1

021

21

0211

1

021

=−=Ψ

=−=Ψ

∫∫

∫∫dxxxxaaRdxxaaR

dxxxxaaRdxxaaR

)6371(369

)1(x

xxu +−=

232

12

21 )62()2(),,( axxxaxxxxaaR −+−+−+−+=

FEM-intro 55

Ritz Method & Galerkin Method (1/4)( ) ( ) 2211212 1 wawaxaaxxu +=+⋅−⋅=

( )

[ ] 11

22

1

122111

22

22

1

122

11

2

1

2

2

2

1

2

1

2

1

wxa

uxxu

a

wwawaa

uuu

a

dx

dw

dx

dwa

dx

dwa

dx

du

adx

du

dx

du

a

⋅=∂∂⋅=

∂∂

⋅+=∂∂⋅=

∂∂

+=

∂∂⋅=

∂∂

( ) dxxuudx

duuI ∫

−−

=1

0

22

2

1

2

1

( )⇒=

∂∂

01

2

a

uI

( )⇒=

∂∂

02

2

a

uI

( ){ } 01

0

22111

1

0

221

1

21 =

++−

+

∫∫ dxxawawwdxadx

dw

dx

dwa

dx

dw

( ){ } 01

0

22112

1

0

2

2

21

21 =

++−

+ ∫∫ dxxawawwdxadxdw

adx

dw

dx

dw

FEM-intro 56

Ritz Method & Galerkin Method (2/4)( )

⇒=∂

∂0

1

2

a

uI

( ){ } 01

0

22111

1

0

221

1

21 =

++−

+

∫∫ dxxawawwdxadx

dw

dx

dwa

dx

dw

∫∫∫

−=

−

=

1

0

121

2

1

1

0

121

2

1

1

0

111

1

0

1

2

1 dxadx

wdwdxa

dx

wdw

dx

dwwadxa

dx

dw

∫∫∫

−=

−

=

1

0

222

2

1

1

0

222

2

1

1

0

212

1

0

221 dxa

dx

wdwdxa

dx

wdw

dx

dwwadxa

dx

dw

dx

dw

)1(

),1(2

22

11

xxw

xxw

−=Ψ=

−=Ψ= 21

2

1111

1 dx

wdw

dx

dw

dx

dw

dx

dww

x+=

∂∂

22

2

1212

1 dx

wdw

dx

dw

dx

dw

dx

dww

x+=

∂∂

FEM-intro 57

Ritz Method & Galerkin Method (3/4)( )

⇒=∂

∂0

1

2

a

uI

( ) 01

0

2211222

2

121

2

1 =

+++

+− ∫ dxxawawadx

wda

dx

wdw

01

0

222

2

1 =

++− ∫ dxxudx

udw

( )⇒=

∂∂

02

2

a

uI

( ) 01

0

2211222

2

121

2

2 =

+++

+− ∫ dxxawawadx

wda

dx

wdw

01

0

222

2

2 =

++− ∫ dxxudx

udw

Galerkin Method !!

02

2

=++ xudx

ud

2211 wawau +=

FEM-intro 58

Ritz Method & Galerkin Method (4/4)

• This example is a very special case. But, generally speaking, results of Galerkin method and Ritz method agree, if functional exists.

• Although Ritz method provides approx. solution, that satisfies Euler equation in strict sense. Therefore, solution of Ritz method is closer to exact solution.– This is the main reason that Galerkin method is accurate.

• Please just remember this.

• This relationship between Ritz and Galerkin is not correct if functional does not exist.– In these cases, Galerkin method is not necessarily the

best method from the viewpoint of accuracy and robustness.