(Galerkin) Finite element approximations - TU/ehulsen/cr/slides2.pdf · (Galerkin) Finite element...

(Galerkin) Finite element approximations

The finite element method (FEM): special choice for the shape functions φ˜

.

x = a x = b

Ω4

Ne = 5

Ω2 Ω3 Ω5Ω1

Subdivide Ω into elements Ωe:

Ω =Ne⋃e=1

Ωe

Ωe1 ∩ Ωe2 = ∅

Approximate u on each element separately by a polynomial of some degree p, forexample by Lagrangian interpolation (using p+ 1 nodal points per element). Theend points of an element must be nodal points.

Example: linear elements

Global shape functions:

x3 x4 x5x1 = a x6 = bx2

Ω4

Ne = 5

Ω2 Ω3 Ω5Ω1

φ3(x)1 uh(x) =

n∑i=1

uiφi(x) = φ˜

T (x)u˜

n: number of global nodal points.

Local element shape functions:

xe1 xe

2 xe1 xe

2

1

φ2(x)φ1(x)

1

ueh(x) = ue1φ1(x)+ue2φ2(x) = φ˜

T (x)u˜e

with φ˜

T = [φ1(x), φ2(x)] and

φ1(x) =x− xe2xe1 − xe2

, φ2(x) =x− xe1xe2 − xe1

Example: quadratic elements

Local element shape functions:

φ1(x) φ2(x) φ3(x)

1

xe1 xe

3xe2 xe

1 xe3 xe

1 xe2 xe

3

1 1

xe2

ueh(x) = ue1φ1(x) + ue2φ2(x) = φ˜

T (x)u˜e

with φ˜

T = [φ1(x), φ2(x), φ3(x)] and

φ1(x) =(x− xe2)(x− xe3)

(xe1 − xe2)(xe1 − xe3), φ2(x) =

(x− xe1)(x− xe3)(xe2 − xe1)(xe2 − xe3)

,

φ3(x) =(x− xe1)(x− xe2)

(xe3 − xe1)(xe3 − xe2)

Example: higher-order elements

General polynomials of order p:

ueh(x) =p+1∑i=1

ueiφi(x) = φ˜

T (x)u˜e

with φ˜

T = [φ1, φ2, . . . , φp+1].

Various expansions possible:

B Gauss-Lobatto integration points (includes end points) and Lagrangianinterpolation (Spectral elements).

B Hierarchical base functions: end points are nodes but internal shape functionshave no nodes (similar to Legrende polynomials). (hp-FEM).

B Legrendre polynomials in discontinuous Galerkin methods.

Global numbering

Ω1 Ω2 Ω3

x3x2x1 = a x4 = b

u1

u3

u4

u2

Ne = 3 uh(x)

Ω1

1

Ω2 Ω3

φ3(x)Ne = 3

x3x2x1 = a x4 = b

uh(x) =4∑i=1

uiφi(x) = φ˜

T (x)u˜

u˜

=

u1

u2

u3

u4

, φ˜(x) =

φ1(x)φ2(x)φ3(x)φ4(x)

Local numbering in elements

Weak form: Find uh in Sh such that

(dvhdx

,Aduhdx

) + vh(b)hb = (vh, f) for all vh ∈ Vh

Split:Ne∑e=1

(dvhdx

,Aduhdx

) + vh(b)hb =Ne∑e=1

(vh, f) for all vh ∈ Vh

Write in each element e:

ueh(x) =2∑i=1

ueiφi(x) = φ˜

T (x)u˜e, veh(x) =

2∑i=1

veiφi(x) = φ˜

T (x)v˜e

where u˜Te = (ue1, u

e2) and φ

˜

T (x) = (φ1(x), φ2(x)).

Element matrix and vector

We get:Ne∑e=1

(v˜TeK

¯eu˜e) + vnhb =

Ne∑e=1

v˜Te f

˜e

where

K¯e =

(dφ˜dx,Adφ

˜

T

dx

)e

=∫

Ωe

dφ˜dxAdφ

˜

T

dxdx

f˜e = (φ

˜, f)e =

∫Ωe

φ˜f dx

are the element matrix K¯e and element vector f

˜e.

Local → global (assembling)

The local vectors u˜e and v

˜e are part of the global vectors:

u˜e = P

¯eu˜, v

˜e = P

¯ev˜,

So we get

Ne∑e=1

v˜TeK

¯eu˜e =

Ne∑e=1

v˜T P

¯TeK

¯eP

¯e︸︷︷︸

K¯e

u˜

= v˜T( Ne∑e=1

K¯e

)u˜

Ne∑e=1

v˜Te f

˜e =

Ne∑e=1

v˜T P

¯Te f

˜e︸︷︷︸

f˜e

= v˜T( Ne∑e=1

f˜e

)

Weak form

Substitution into the weak form:

v˜TK

¯u˜

= v˜Tf

˜for all v

˜

orK¯u˜

= f˜with

K¯

=Ne∑e=1

K¯e

f˜

=Ne∑e=1

f˜e

+

00...0−hb

Example (1)

For example:

u˜

2 =(u2

u3

)=(

0 1 0 00 0 1 0

)︸︷︷︸

P¯

2

u1

u2

u3

u4

= P¯

2u˜

K¯

2 = P¯T2K

¯2P

¯2 =

0 01 00 10 0

(K211 K2

12

K221 K2

22

)(0 1 0 00 0 1 0

)

=

0 0 0 00 K2

11 K212 0

0 K221 K2

22 00 0 0 0

Example (2)

Assembly of K¯

:

K¯

=

K1

11 K112 0 0

K121 K1

22 0 00 0 0 00 0 0 0

+

0 0 0 00 K2

11 K212 0

0 K221 K2

22 00 0 0 0

+

0 0 0 00 0 0 00 0 K3

11 K312

0 0 K321 K3

22

=

K1

11 K112 0 0

K121 K1

22 +K211 K2

12 00 K2

21 K222 +K3

11 K312

0 0 K321 K3

22

Similar for assembly of f

˜.

Exercise 5

The ‘bandwidth’ of K¯

for linear shape functions is 3. How large is the bandwidthfor K

¯when using polynomial shape functions of order k, k ≥ 1?

Dirichlet conditions

Renumber and split vector and matrix (‘u’ unknown, ‘p’ prescribed)

u˜

=(u˜u

u˜p

), v

˜=(v˜u

v˜p

), K

¯=(K¯uu K

¯up

K¯pu K

¯pp

), f

˜=(f˜u

f˜p

)Note: u

˜p prescribed values, v

˜p = 0

˜. Weak form

v˜TK

¯u˜

= v˜Tf

˜for all v

˜

with v˜p = 0

˜leads to

v˜Tu (K

¯uuu

˜u +K

¯upu

˜p) = v

˜Tuf

˜u for all v

˜u

orK¯uuu

˜u = f

˜u −K

¯upu

˜p

1D convection-diffusion-reaction equation

1D convection-diffusion-reaction Eq.: find u(x) such that for x ∈ (a, b)

∂u

∂t+ a

∂u

∂x− ∂

∂x(A∂u

∂x) + bu = f

and

u = ua(t), at x = a, t > 0 (ΓD)

−Adudx

= hb(t) at x = b, t > 0 (ΓN)

Notes:

B Strong form; Classical (strong) solution u(x, t)B f(x, t) ∈ C0(a, b) (continuous) then u ∈ C2(a, b) (twice continuously

differentiable)

Oldroyd-B/UCM viscoelastic model

λ5τ +τ = 2ηD

where5τ= τ −L · τ − τ ·LT

or∂τ

∂t︸︷︷︸∂u∂t

+ ~u · ∇τ︸︷︷︸a∂u∂x

−L · τ − τ ·LT +τ

λ︸︷︷︸bu

= 2η

λD︸︷︷︸f

Diffusion is missing (A = 0).

Dimensionless form (1)

Scaling:

t = tct∗ tc : characteristic time

u = Uu∗ U : characteristic value solution

x = Lx∗ L : characteristic length scale

Dimensionless variables: O(1).

U

tc

∂u∗

∂t∗+aU

L

∂u∗

∂x∗− AUL2

∂2u∗

∂x∗2+ bUu∗ = f

Relative to convection:

L

atc

∂u∗

∂t∗+∂u∗

∂x∗− 1

Pe∂2u∗

∂x∗2+bL

au∗ =

L

aUf

Pe =aL

A: Peclet number, convection/diffusion.

Dimensionless form (2)

Time scales:

B convection:L

a

B diffusion:L2

A

B source:1b

(’relaxation time’)

B time scales in b.c.

B externally or internally generated frequencies (’von Karman vortex’)

We choose tc = L/a (convection) and get

∂u∗

∂t∗+∂u∗

∂x∗− 1

Pe∂2u∗

∂x∗2+ b∗u∗ = f∗

with b∗ =bL

aand f∗ =

L

aUf . Pe > 1 : convection dominated

Exercise 6

Assume b = 0 and f = 0. We take for the typical time scale tc =L2

A(diffusion

time scale). Show that we now have the non-dimensional form

∂u∗

∂t∗+ Pe

∂u∗

∂x∗− ∂

2u∗

∂x∗2= 0

When will this non-dimensional form be preferable over the one on the previousslide?

Steady state 1D convection-diffusion-reaction equation

steady state: ∂u/∂t = 0

L(u) = adu

dx− d

dx(Adu

dx) + bu = f

L: linear operator, with b.c.

u = ua, at x = a (ΓD)

−Adudx

= hb at x = b (ΓN)

Weak form

Multiply with test function v and integrate:

(v,Lu− f) = 0 for all v

Partial integration of the diffusion term and inserting b.c. we get: Find u ∈ Ssuch that

(dv

dx,Adu

dx) + (v, a

du

dx) + (v, bu) + v(b)hb = (v, f) for all v ∈ V

where S and V are appropriate spaces.

Galerkin FEM approximations (1)

Approximation spaces Sh and Vh:

uh(x) =n∑i=1

uiφi(x) = φ˜

T (x)u˜

vh(x) =n∑i=1

viφi(x) = φ˜

T (x)v˜

where φ˜(x) are global shape functions. Substituting this into the weak form gives:

Find uh ∈ Sh such that

(dvhdx

,Aduhdx

) + (vh, aduhdx

) + (vh, buh) + vh(b)hb = (vh, f) for all vh ∈ Vh

orv˜TK

¯u˜

= v˜Tf

˜for all v

˜


and thusK¯u˜

= f˜

with

K¯

= (dφ

˜dx,Adφ

˜

T

dx) or Kij = (

dφidx

,Adφjdx

)

+ (φ˜, adφ

˜

T

dx) + (φi, a

dφjdx

)

+ (φ˜, bφ

˜

T ) + (φi, bφj)

f˜

= (φ˜, f)− hbφ

˜(b) fi = (φi, f)− hbφi(b)


Build from element matrices. For example with constant coefficients and linearelement shape functions:

(dφ

˜dx,dφ

˜

T

dx)e =

1h

(1 −1−1 1

)“stiffness matrix”

(φ˜,dφ

˜

T

dx)e =

12

(−1 1−1 1

)“convection matrix”

(φ˜, φ

˜

T )e =h

6

(2 11 2

)“mass matrix”

where h is the element length.

Global stiffness matrix

Global stiffness matrix (uniform element size):

1h

1 −1 0 . . . . . . . . . . . . . 0−1 2 −1 0 . . . . . . . . 00 −1 2 −1 0 . . . 0... ...... ...... ...0 . . . 0 −1 2 −1 00 . . . . . . . . 0 −1 2 −10 . . . . . . . . . . . . . 0 −1 1

Finite difference scheme: (divide by h)

−ui−1 + 2ui − ui+1

h2= −d

2u

dx2(xi) +O(h2)

Global convection matrix

Global convection matrix (uniform element size):

12

−1 1 0 . . . . . . . . . . . . . 0−1 0 1 0 . . . . . . . . 00 −1 0 1 0 . . . 0... ...... ...... ...0 . . . 0 −1 0 1 00 . . . . . . 0 −1 0 10 . . . . . . . . . . . 0 −1 1


−ui−1 + ui+1

2h=du

dx(xi) +O(h2)

Global mass matrix

Global mass matrix (uniform element size):

h

6

2 1 0 . . . . . . . . . 01 4 1 0 . . . . . . 00 1 4 1 0 . . . 0... ...... ...... ...0 . . . 0 1 4 1 00 . . . . . . 0 1 4 10 . . . . . . . . . 0 1 2


ui−1 + 4ui + ui+1

6= u(xi) +O(h2)

Global right-hand side

Sometimes the following approximation is applied:

f(x) ≈∑k

fkφk(x) fk : nodal point values of f

In that case:fi = (φi, f) =

∑k

(φi, φk)fk

with (φi, φk) the mass matrix.Finite difference scheme: (divide by h)

fi−1 + 4fi + fi+1

6= f(xi) +O(h2)

Note that the usual approach is to use numerical integration.

Exercise 7

a. Verify the element stiffness, convection and mass matrix by performing theintegrations.

b. Assume a non-equidistant grid and denote the element sizes by h1, h2,. . . ,hNe.How does the global stiffness matrix now look like? Has the global convectionmatrix still a zero on the diagonal?

Example: 1D steady convection-diffusion (1)

Equation:du

dx− 1

Ped2u

dx2= 0 on (0, 1)

with b.c.u(0) = 0, u(1) = 1

Exact solution:

u(x) =1− exp(Pex)1− exp(Pe)


Ne = 10; Pe = 1, 10, 100.

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)

x

Pe=1 exactPe=1 GFEMPe=10 exactPe=10 GFEMPe=100 exactPe=100 GFEM


Ne = 10, 20; Pe = 25

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)

x

Pe=25 exactPe=25 Ne=10Pe=25 Ne=20

Observations (1)

B Boundary layer of width1

Pe.

B Global upstream “wiggles” appear when the boundary layer is not resolved bythe mesh.

B Governed by the mesh Peclet number Peh:

Peh =ah

2A=

12h

LPe

where wiggles appear when

Peh > 1 orh

L>

2Pe

or wiggle-free whenh

L<

2Pe

Observations (2)

The problem is twofold:

1. Rapid change in the solution over small lengths (large gradients)

2. Mesh size h is too large to resolve the length scale

As long as the length scale is resolved GFEM works well for convection-diffusionproblems.

Example: 1D steady convection with source term (1)

Equation (Pe→∞):du

dx= f on (0, 1)

with

u(0) = 0, f =1√πσ

exp[−(x− 12)2

σ2]

Exact solution:

u(x) =12

[erf(

x− 12

σ) + erf(

12σ

)]

Notes:

B length scale σ

B for σ → 0u(x) = H(x− 1

2) (Heaviside)

du

dx= f(x) = δ(x− 1

2)


Ne = 20; σ = 0.45, 0.15, 0.05.

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)

x

σ=0.45 exactσ=0.45 GFEMσ=0.15 exactσ=0.15 GFEMσ=0.05 exactσ=0.05 GFEM


Ne = 11, 20, 21; σ = 0.05

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)

x

σ=0.05 exactσ=0.05 Ne=11σ=0.05 Ne=20σ=0.05 Ne=21

Observations

B Inaccurate solutions and upstream wiggles when layer with small length scaleσ is not resolved by the mesh.

B By experimentation we find that

h

σ<

12

to get good smooth solutions with GFEM.

B Strong mesh dependence when solution is underresolved.

Problem is again lack of resolution for rapidly changing solutions (large gradients).

Summary of the examples

B GFEM works well for smooth solutions and the mesh resolves rapidly changingsolutions.

B For non-smooth solutions and/or a too coarse mesh wiggles can easily appear.

B The GFEM is not robust and local errors can easily be amplified throughoutthe domain.

Some might consider the latter a good thing (“Don’t suppress wiggles they tellyou something”). However:

B In many practical problems resolving all large gradients is not an option andwould require too much computer resources.

B In viscoelastic flow problems the GFEM method often performs very poorly.

⇒ Stabilization necessary

Analysis of the discrete equations

Equidistant mesh (mesh size h), inner node i:

aui+1 − ui−1

2h−Aui+1 − 2ui + ui−1

h2=fi−1 + 4fi + fi+1

6

Notes:

B central differencing

B consistent averaging of source term

B for f = 0:

ui+1 − ui−1 − 1Peh

(ui+1 − 2ui + ui−1) = 0or

ui+1 − ui =1 + Peh1− Peh

(ui − ui−1)or

sign(ui+1 − ui) = − sign(ui − ui−1) for Peh > 1 Wiggles!

B for Peh →∞ (no diffusion): Simpson rule for integration, which is O(h4).

B for Peh →∞ (no diffusion), matrix:

a

2h

−1 1 0 . . . . . . . . . . . . . 0−1 0 1 0 . . . . . . . . 00 −1 0 1 0 . . . 0... ...... ...... ...0 . . . 0 −1 0 1 00 . . . . . . 0 −1 0 10 . . . . . . . . . . . 0 −1 1

= . . .

zeros on the diagonal, “leap frogging”.

Artificial diffusion (1)

For Peh =ah

2A= 1 we have a diffusivity of:

A = A =ah

2

Adding −Ad2u

dx2to the equations, or

adu

dx− (A+ A)

d2u

dx2= f

removes wiggles, but we solve a modified equation. New mesh Peclet:

Peh =ah

2(A+ ah2 )

=Peh

1 + Peh< 1


Discretized equations with equidistant mesh (mesh size h), inner node i:

aui+1 − ui−1

2h− (A+

ah

2)ui+1 − 2ui + ui−1

h2=fi−1 + 4fi + fi+1

6

or

aui − ui−1

h︸︷︷︸upwind differencing

−Aui+1 − 2ui + ui−1

h2=fi−1 + 4fi + fi+1

6

h h

xi+1xi

ui−1

uiui+1

xi−1

ui − ui−1

h=du

dx(xi) +O(h)


In general artificial diffusivity:

A = βah

2with examples:

B β = 1, upwind differencingB Peh = 1 for Peh > 1:

β =

0 for Peh ≤ 11− 1

Pehfor Peh > 1

B Exact nodal point values (’optimal’ value) for

adu

dx−Ad

2u

dx2= constant

gives:

β = coth(Peh)− 1Peh

β(Peh)

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

coth(x)-1/x1-1/x

x/3

Exercise 8

Show that for β 6= 0 the first-order derivative can be interpreted as beingdiscretized by the following finite difference scheme

12(1− β)

ui+1 − uih

+ 12(1 + β)

ui − ui−1

h=du

dx(xi) +O(h)

i.e. a linear combination of ‘downwind’ and ‘upwind’ differencing.

Example: 1D steady convection-diffusion

Ne = 10; Pe = 25 (Peh = 1.25)

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)

x

Pe=25 exactGFEMβ=1β=0 for Peh≤1, 1-1/Peh for Peh>1β=coth(Peh)-1/Peh

Observations

B Upwind differencing rather inaccurate (first-order) near large gradients.

B Wiggles are gone. As expected the boundary layer is not resolved.

B ‘Optimal’ β(Peh) can reduce error significantly (weighted upwind/downwind).

B Away from the boundary layer the solution seems accurate.

Example: 1D steady convection with source term

Ne = 20, 40; σ = 0.05; β = 1

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

u(x)

x

σ=0.05 exactGFEM Ne=20GFEM Ne=40AD β=1 Ne=20AD β=1 Ne=40

Example: 1D steady convection with source term

Error in solution. Ne = 100 (Galerkin), 3000 (β = 1); σ = 0.05

-0.002

-0.0015

-0.001

-0.0005

0

0.0005

0.001

0.0015

0.002

0 0.2 0.4 0.6 0.8 1

max

k|u k

-uex

act|

x

AD β=1 Ne=3000GFEM Ne=100

Observations

B GFEM: global wiggles and inaccuracy for Ne = 20 but very accurate forNe = 40 (fast convergence)

B Upwind: no wiggles (robust) but convergence is very slow near large gradients.Seems to be accurate outside the layer with large gradients.

Question: Can we combine the accuracy (fast convergence) of GFEM with therobustness of upwind?

Set of equations (flow of a visco-elastic fluid)

Rewrite: Find (~u, p, τ ) such that,

ρ(∂~u

∂t+ ~u · ∇~u)−∇ · (2ηsD) +∇p−∇ · τ = ρ~b, in Ω

∇ · ~u = 0, in Ω

λ(∂τ

∂t+ ~u · ∇τ −L · τ − τ ·LT ) + τ = 2ηD, in Ω

Weak form of modified equation by AD

Artificial diffusion solves the modified equation:

adu

dx− d

dx

((A+ A)

du

dx

)+ bu = f, with A = β

ah

2

where we assume also the modified boundary condition:

u = ua, at x = a (ΓD)

−(A+ A)du

dx= hb at x = b (ΓN)

For the weak form we get: Find u ∈ S such that

(dv

dx, (A+ A)

du

dx) + (v, a

du


Streamline upwinding (SU) (1)

Compared to standard weak form we have added the term

(dv

dx, Adu

dx) = (

dv

dx, βah

2du

dx) = (

βh

2dv

dx, adu

dx)

This suggest another form of the modified weak form:

(dv

dx,Adu

dx) + (v +

βh

2dv

dx, adu


the convection term has been weighted with a modified test function

v = v +βh

2dv

dx

⇒ streamline upwinding (SU). For constant a, uniform size h, 1D→ equivalent toAD (artificial diffusion). Very inaccurate (overly diffusive) with time-dependent,reaction and source terms (similar to AD).

Streamline upwinding (SU) (2)

For the Galerkin FEM:

vh = vh +βh

2dvhdx

= v˜T φ

˜with φ

˜= (φ

˜+βh

2dφ

˜dx)

where h may vary from element to element.

hi hi+1

β2

φi

φi

xi+1xixi−1

β2

Consistent weighting

Hughes & Brooks (1982): problem with SU in inconsistency : The exact solutiondoes not satisfy the weak form. Their solution is to modify the weighting of thedifferential equation from the standard form

(v,Lu− f) = 0 for all v

to

(v +βh

2dv

dx,Lu− f) = 0 for all v

where

L(u) = adu

dx− d

dx(Adu

dx) + bu

The space of the weight functions (not V !) is in effect different from the spaceof shape functions when applying “Vh = Uh” like in the Galerkin method (→Petrov-Galerkin).

Interpretation of second-derivatives

Interpretation of Hughes & Brooks (1982). Split:

(v +βh

2dv

dx,Lu− f) = (v,Lu− f)︸︷︷︸

I

+ (βh

2dv

dx,Lu− f)︸︷︷︸II

= 0 for all v

B Term I: partial integration of diffusion terms with fluxes on the boundary(standard procedure)

B Term II: interprete element wise:

(βh

2dv

dx,Lu− f) =

∑e

(βh

2dv

dx,Lu− f)e =

Streamline upwind Petrov-Galerkin (SUPG)

This leads to the SUPG method for the 1D steady convection-diffusion-reactionequation: Find u ∈ S such that

(dv

dx,Adu

dx) + (v, a

du

dx) + (v, bu) + v(b)hb

+∑e

( βh

2︸︷︷︸τa

dv

dx, adu

dx− d

dx(Adu

dx) + bu− f

)e

= (v, f) for all v ∈ V

Notes:

B A = 0 (no diffusion):∑e can be removed (viscoelastic CE)

B consistency

B usual definition: τa =βh

2with time scale τ

τ = βh

21a

= βah

21a2

= A1a2

B constant A, linear elements:d

dx(Adu

dx) = 0 in the interior of the element. If

reaction term bu and source term f are also absent (convection-diffusion only)SU=SUPG!

B optimal values of β(Peh) neccessary to improve accuracy of diffusion terms.


Ne = 20; σ = 0.05

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

u(x)

x

σ=0.05 exactGFEM Ne=20AD β=1 Ne=20SUPG β=1 Ne=20


σ = 0.05; maximum nodal error as a function of number of elements Ne.

0.001

0.01

0.1

1

10 100

max

k|u k

-uex

act|

Ne

ADSUPGGFEM


σ = 0.05; maximum nodal error as a function of number of elements Ne.

0.001

0.01

0.1

1

10 100

max

k|u k

-uex

act|

Ne

ADSUPGGFEM

7/x10/x2

Observations

B Compared to GFEM: SUPG greatly enhances the stability/robustness for coarsegrids in convection(-dominated) problems where solutions are non-smooth orunderresolved (no global wiggles).

B Compared to SU: SUPG is much more accurate because convergence is fasterthan linear.

Extension to 3D: convection-diffusion-reaction equation

Strong solution u(~x, t):

∂u

∂t+ ~a · ∇u−∇ · (A∇u) + bu︸︷︷︸

Lu= f on Ω

with boundary and initialconditions

u(~x, t = 0) = u0(~x) in Ω

u = uD on ΓD

−A∂u∂n

= −A~n · ∇u = hN on ΓN~n Ω

ΓD

~n: outside normal

ΓN

Weak form (1)

Multiply by testfunction v:

(v,∂u

∂t+ Lu− f) = 0 for all v

where the standard inner product on L2(Ω) is

(a, b) =∫

Ω

abdx

With∇ · (A∇u)v = ∇ · (A(∇u)v)

)−A∇u · ∇vand the divergence theorem (Gauss):∫

Ω

∇ · ~adx =∫

Γ

~n · ~adΓ

Weak form (2)

we get the weak form: Find u ∈ S such that

(v,∂u

∂t) + (v,~a · ∇u) + (∇v,A∇u) + (v, bu) + (v, hN)ΓN = (v, f)

for all v ∈ V . Notes:

B v = 0 on ΓD, u = uD on ΓDB we have silently introduced:

(~a,~b) =∫

Ω

~a ·~bdΩ

(a, b)Γ =∫

Γ

abdΓ

GFEM (1)

Ω = ∪eΩe, Ωe ∩ Ωf = ∅ for e 6= f

1 2

3

2

3

1

φ1(~x)

2

3

1

2

3

1

φ3(~x)φ2(~x)

GFEM (2)

Approximation spaces Sh and Vh:

uh(~x, t) =n∑i=1

ui(t)φi(~x) = φ˜

T (~x)u˜(t)

vh(~x, t) =n∑i=1

vi(t)φi(~x) = φ˜

T (~x)v˜(t)

where φ˜(~x) are global shape functions. Substituting into the weak form:

v˜TM

¯u˜

+ v˜TK

¯u˜

= v˜Tf

˜for all v

˜

orM¯u˜

+K¯u˜

= f˜

GFEM (2)

The ‘mass matrix’ M¯

is given by

M¯

= (φ˜, φ

˜

T ), Mij = (φi, φj)

and the ‘stiffnes matrix’ by

K¯

= (∇φ˜, A∇φ

˜

T ) or Kij = (∇φi, A∇φj)+ (φ

˜,~a · ∇φ

˜

T ) + (φi,~a · ∇φj)+ (φ

˜, bφ

˜

T ) + (φi, bφj)

f˜

= (φ˜, f)− (φ

˜, hN)ΓN fi = (φi, f)− (φi, hN)ΓN


u = 0

u = 0

|~a| = 1

ax∂u

∂x+ ay

∂u

∂y= f on (0, 1)× (0, 1)

with (ax, ay) = (12

√2, 1

2

√2) and

u(x, 0) = 0, u(0, y) = 0, f =1√πσ

exp[−(x+ y − 1)2

2σ2]

Exact solution goes from 0 for x + y < 1 to 1 for x + y > 1 over a length scaleof σ.


Ne = 15× 15; σ = 0.05; GFEM

−3.15E−01

−4.57E−02

2.23E−01

4.92E−01

7.61E−01

1.03E+00

max: 1.03E+00min:−3.15E−01


Ne = 30× 30; σ = 0.05; GFEM

−3.08E−02

1.79E−01

3.89E−01

5.99E−01

8.08E−01

1.02E+00

max: 1.02E+00min:−3.08E−02


Ne = 45× 45; σ = 0.05; GFEM

−1.61E−02

1.88E−01

3.91E−01

5.95E−01

7.99E−01

1.00E+00

max: 1.00E+00min:−1.61E−02


Ne = 100× 100; σ = 0.05

−3.47E−03

1.98E−01

3.99E−01

6.00E−01

8.01E−01

1.00E+00

max: 1.00E+00min:−3.47E−03

Example: 2D steady convection with discontinuous inflow (1)

u = 0

u = 1|~a| = 1

ax∂u

∂x+ ay

∂u

∂y= 0 on (0, 1)× (0, 1)

with (ax, ay) = (12

√2, 1

2

√2) and

u(x, 0) = 0, u(0, y) = 1

Exact solution is: 1 for y > x

0 for y < x


Ne = 15× 15; GFEM

−6.22E−02

1.63E−01

3.88E−01

6.13E−01

8.38E−01

1.06E+00

max: 1.06E+00min:−6.22E−02


Ne = 30× 30; GFEM

−6.22E−02

1.63E−01

3.88E−01

6.13E−01

8.38E−01

1.06E+00

max: 1.06E+00min:−6.22E−02

Observations for GFEM in 2D

B Inaccurate solutions and global upstream wiggles when layer with small lengthscale σ is not resolved by the mesh.

B Almost wiggle free, very localized cross-stream jumps possible even for coarsemeshes.

B Strong mesh dependence when solution is underresolved.


1D added diffusion:

−Ad2u

dx2with A = β

ah

2which stabilizes

adu

dx3D added diffusion:

−∇(A∇u) with A = β|a|h

2which stabilizes

~a · ∇uhowever, in all directions. This is too much added diffusion. Only diffusion in thedirection of ~a is needed since

~a · ∇u = |~a|~e · ∇u = |~a|∂u∂e

with ~e =~a

|~a|


Add anisotropic diffusion using a tensorial diffusion coefficient, with a componentin the direction of ~e (i.e. the “streamline” direction) only:

−∇(A~e~e · ∇u)

or

−∇(A · ∇u) with A = A~e~e = A~a~a

|~a|2In the weak form we get an extra term:

(∇v, A · ∇u)

SU (1)

Diffusion added in AD method:

(∇v, A · ∇u) = (∇v, A ~a~a|~a|2 · ∇u)

=∫

Ω

∇v · A ~a~a|~a|2 · ∇u) dΩ

=∫

Ω

A

|~a|2(~a · ∇v)(~a · ∇u) dΩ

= (A

|~a|2~a · ∇v,~a · ∇u)

= (τ~a · ∇v,~a · ∇u)

with

τ =A

|~a|2 = βh

2|~a|

SU (2)

The weak form now becomes

· · ·+ (v + τ~a · ∇v,~a · ∇u)+ · · ·

i.e. the convection term is multiplied by a modified weighting function

v = v + τ~a · ∇v

SUPG

SU → SUPG: consistent weighting and interpret extra terms elementwise:

(v + τ~a · ∇v, ∂u

∂t+ Lu− f

)= 0

or (with elementwise interpretation):

(v,∂u

∂t+ Lu− f

)+∑e

(τ~a · ∇v, ∂u

∂t+ Lu− f

)Ωe

= 0

First term: partial integrate to standard weak form.

Notes on SUPG (1)

B τ =A

|~a|2, with A “the amount of streamline diffusion”. When the real diffusion

A 6= 0, we need to generalize A = β(Peh)ah

2to more dimensions. There are

various ways to do this. For example for a bilinear quadrilateral:

~e1

~e2

~a0~h1 = h1~e1

h1

h2

~h2 = h2~e2

A = β(Peh1)|~a · ~h1|

2+ β(Peh2)

|~a · ~h2|2

with

Peh1 =|~a · ~h1|

2A, Peh2 =

|~a · ~h2|2A

B For higher-order elements a constant τ is non-optimal.

Notes on SUPG (2)

B The choice of τ is somewhat arbitrary, in particular when diffusion is absent(A = 0). Note, that

τ =A

|~a|2 = βh

2|~a|In viscoelastic flow it is common to use:

τ =h

2U

where h is a characteristic element size and U is a characteristic velocity.There are various possible choices for h and U (see separate notes).

A bit of experimentation with the 1D “Gaussian distribution in the right-handside” example (source term, no diffusion) seems to suggest that the solutionis rather insensitive to the exact value of β in this case, as long as the scalingwith h is there.


u = 0

u = 0

|~a| = 1

ax∂u

∂x+ ay

∂u

∂y= f on (0, 1)× (0, 1)

with (ax, ay) = (12

√2, 1

2

√2) and

u(x, 0) = 0, u(0, y) = 0, f =1√πσ

exp[−(x+ y − 1)2

2σ2]

Exact solution goes from 0 for x + y < 1 to 1 for x + y > 1 over a length scaleof σ.


Ne = 15× 15; σ = 0.05; SUPG

−7.67E−03

1.98E−01

4.03E−01

6.08E−01

8.14E−01

1.02E+00

max: 1.02E+00min:−7.67E−03


Ne = 30× 30; σ = 0.05; SUPG

−4.83E−04

2.00E−01

4.01E−01

6.02E−01

8.03E−01

1.00E+00

max: 1.00E+00min:−4.83E−04


u = 0

u = 1|~a| = 1

ax∂u

∂x+ ay

∂u

∂y= 0 on (0, 1)× (0, 1)

with (ax, ay) = (12

√2, 1

2

√2) and

u(x, 0) = 0, u(0, y) = 1

Exact solution is: 1 for y > x

0 for y < x


Ne = 15× 15; SUPG

−4.47E−02

1.73E−01

3.91E−01

6.09E−01

8.27E−01

1.05E+00

max: 1.05E+00min:−4.47E−02


Ne = 30× 30; SUPG

−4.70E−02

1.72E−01

3.91E−01

6.11E−01

8.30E−01

1.05E+00

max: 1.05E+00min:−4.70E−02


Ne = 30× 30; AD method with isotropic diffusion of |a|h/2

0.00E+00

2.00E−01

4.00E−01

6.00E−01

8.00E−01

1.00E+00

max: 1.00E+00min: 0.00E+00


Ne = 100× 100; AD method with isotropic diffusion of |a|h/2

0.00E+00

2.00E−01

4.00E−01

6.00E−01

8.00E−01

1.00E+00

max: 1.00E+00min: 0.00E+00

Observations for SUPG in 2D

B In streamline direction: SUPG greatly enhances the stability/robustness forcoarse grids in convection(-dominated) problems where solutions are non-smooth or underresolved (no global wiggles).

B Normal to the streamline direction: SUPG shows a small ‘diffusion-like’smoothing that decreases with mesh size. The diffusion is much less than thediffusion in the AD method (when added isotropically).

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