optimally blended finite/spectral element scheme for wave … · 2011-09-23 · optimally blended...

Optimally Blended Finite/Spectral Element Scheme

for Wave Propagation

Mark Ainsworth(joint work with Hafiz Abdul Wajid, COMSATS, Pakistan)

Mathematics and Statistics, Strathclyde University,Glasgow, Scotland.

Modern Techniques for Numerical Solution of PDEs, ACMAC, Crete, September 2011 – p. 1/54

Outline

(Very) Quick Overview of Spectral Element Method

Typical Practical Application

Analysis and Properties

Optimal Blending


Model Problem

Seek u : (−1, 1)× (0, T ) 7→ C:

utt − uxx = f(x, t) in (−1, 1)× (0, T ),

subject to u(x, 0) = ut(x, 0) = 0, x ∈ (−1, 1) and

u(−1, t) = g(t), ux(1, t) + ut(1, t) = 0 for t ∈ (0, T )


Variational Formulation

Seek u ∈ H1(−1, 1) : u(−1, t) = g(t)

d2

dt2(u, v) +

d

dtuv|x=1 + (ux, vx) = (f, v),

for all v ∈ H1(−1, 1) : v(−1) = 0, where (u, v) =∫ 1

−1uv dx.


Variational Formulation

Seek u ∈ H1(−1, 1) : u(−1, t) = g(t)

d2

dt2(u, v) +

d

dtuv|x=1 + (ux, vx) = (f, v),

for all v ∈ H1(−1, 1) : v(−1) = 0, where (u, v) =∫ 1

−1uv dx.

Discretise by introducing a finite dimensional subspace X ⊂ H1(−1, 1).


Semi-Discretisation in Space

Seek U ∈ X : U(−1, t) = g(t)

d2

dt2(U, v) +

d

dtUv|x=1 + (Ux, vx) = (f, v),

for all v ∈ X : v(−1) = 0.


Semi-Discretisation in Space

Seek U ∈ X : U(−1, t) = g(t)

d2

dt2(U, v) +

d

dtUv|x=1 + (Ux, vx) = (f, v),

for all v ∈ X : v(−1) = 0.

Introduce basis φiNi=0 for X , and write

U(x, t) =N∑

i=0

αi(t)φi(x).


Semi-Discrete Scheme

Find αiNi=0 such that

N∑

i=0

((φi, φj)

d2αi

dt2+ φi(1)φj(1)

dαi

dt+ (φi,x, φj,x)αi

)= (f, φj)

for j = 1, . . . , N , with

αi(0) =dαi

dt(0) = 0

with α0(t) = g(t), t ∈ (0, T ).



Find ~α such that

Md2~α

dt2+C

d~α

dt+K~α = ~r

with

~α(0) =d~α

dt(0) = ~0



Find ~α such that

Md2~α

dt2+C

d~α

dt+K~α = ~r

with

~α(0) =d~α

dt(0) = ~0

and

M ij =

∫ 1

−1

φi(x)φj(x) dx ‘Mass Matrix’

Kij =

∫ 1

−1

φ′

i(x)φ′

j(x) dx ‘Stiffness’


Fully Discrete Scheme

Simple leapfrog scheme in time

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn

where ~α0 = ~α1 = ~0, and

M ij =

∫ 1

−1

φi(x)φj(x) dx Mass

Kij =

∫ 1

−1

φ′

i(x)φ′

j(x) dx Stiffness


Fully Discrete Scheme

Simple leapfrog scheme in time

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn


M ij =

∫ 1

−1


Kij =

∫ 1

−1

φ′

i(x)φ′

j(x) dx Stiffness

... requires inversion of mass matrix M at each time step.


Spectral Element Method: General Idea

Spectral element spatial discretisation gives

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+ K~αn = ~rn


M ij ≈

∫ 1

−1


Kij ≈

∫ 1

−1

φ′

i(x)φ′

j(x) dx Stiffness


Spectral Element Method: General Idea

Spectral element spatial discretisation gives

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+ K~αn = ~rn


M ij ≈

∫ 1

−1


Kij ≈

∫ 1

−1

φ′

i(x)φ′

j(x) dx Stiffness

Spectral element method characterised by particular combinations of

quadrature rule for integrals

nature and form of basis functions.


Spectral Element Method: Quadrature Rule

Gauss-Lobatto quadrature rule:

∫ 1

−1

f(ζ) dζ ≈ Qp(f) =

p∑

ℓ=2

wℓf(ζℓ) +2

p(p+ 1)[f(−1) + f(1)],

where

Nodes: ζℓpℓ=2 zeros of L′

p

Weights: wℓ = 2/p(p− 1)[Lp−1(ζℓ)]2 > 0

with Lp Legendre polynomial of degree p. Rule exact for f ∈ P2p−1.


Spectral Element Method: Basis Functions

Lagrange basis functions φℓp+1ℓ=1 :

φℓ ∈ Pp : φℓ(ζm) =

1, ℓ = m

0, ℓ 6= m




−1 −0.5 0 0.5 1−0.5

0

0.5

1

1.5Quadratic GL−Shape Functions




−1 −0.5 0 0.5 1−0.5

0

0.5

1

1.5Cubic GL−Shape Functions




−1 −0.5 0 0.5 1−0.5

0

0.5

1

1.5Order six GL−Shape Functions


... so what’s the big deal?

Stiffness matrix K:

Kℓm = Qp(φ′

ℓφ′

m),

recall φ′

ℓ ∈ Pp−1 and quadrature rule exact for P2p−1, so

Kℓm = (φ′

ℓ, φ′

m) = Kℓm

Hence, stiffness matrix unchanged.



Stiffness matrix K:

Kℓm = Qp(φ′

ℓφ′

m),

recall φ′

ℓ ∈ Pp−1 and quadrature rule exact for P2p−1, so

Kℓm = (φ′

ℓ, φ′

m) = Kℓm

Hence, stiffness matrix unchanged.... so what is the big deal?



Mass matrix M :

M ℓm = Qp(φℓφm) =

p+1∑

k=1

wkφℓ(ζk)φm(ζk).

(recall φℓ(ζk) = δℓk)

M ℓm =

p+1∑

k=1

wkδℓkδmk = wℓδℓm.



Mass matrix M is diagonal

M =

w1 0 0 · · · 0

0 w2 0 · · · 0

0 0 w3 · · · 0...

......

. . . 0

0 0 0 0 wp+1

... finite elements with ‘Gauss-point mass lumping’Ref: G. Cohen, Higher-Order Numerical Methods for Transient WaveEquations, Springer 2002.


Spectral Element Method

Spectral Element/Gauss-point Mass Lumped Finite Element Scheme

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+ K~αn = ~rn

where ~α0 = ~α1 = ~0, with

M = Diag(w1, . . . , wp+1)




1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+ K~αn = ~rn


M = Diag(w1, . . . , wp+1)

Same structure and approach applicable to

multiple spatial dimensions

acoustics, electromagnetics, structures, ...




1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+ K~αn = ~rn


M = Diag(w1, . . . , wp+1)

Same structure and approach applicable to

multiple spatial dimensions

acoustics, electromagnetics, structures, ...

Advantages:

retains geometric flexibility of finite elements

obtain extremely efficient time-stepping.


Application: Seismic Simulation


Application: Seismic Simulation

Earthquake: 9/9/2001 Hollywood CA MW = 4.3.

Surface topography of Southern California viewed from Southeast.


Seismic Simulation: Hollywood 9/9/2001 MW = 4.3

• Hollywood (Full)• Hollywood (Zoom)


Seismic Simulation: Computational Details

average distance between grid points at the surface roughly 335m

mesh contains 672,768 elements

time step size of 9 msec and 20,000 timesteps in total.

polynomial degree p = 4

each element contains 3(p+ 1)3 = 375 degrees of freedom

136 million degrees of freedom in total

144 processor Beowulf PC cluster using MPI

6.5 hours needed to compute seismograms with a duration of 3 min


Seismic Simulation: Computational Details

average distance between grid points at the surface roughly 335m

mesh contains 672,768 elements

time step size of 9 msec and 20,000 timesteps in total.

polynomial degree p = 4

each element contains 3(p+ 1)3 = 375 degrees of freedom

136 million degrees of freedom in total

144 processor Beowulf PC cluster using MPI

6.5 hours needed to compute seismograms with a duration of 3 min

... state of the art back in 2004. Same technology SPECFEM3D beingused for global seismic simulation–see cover of Science, May 2005.


Seismic Simulation: Cross-Section of Portion of Mesh

Note: Highly structured translation invariant mesh in bulk of the domain.


Basic Issues in Computational Wave Propagation

A numerical scheme should, as far as possible:

propagate waves at correct speed i.e. control numerical dispersion;

preserve conserved quantities (energy, momentum, ...) i.e. controlnumerical dissipation;






... but there is no silver bullet.






Every scheme needs at least two degrees of freedom per wavelength justto resolve the wave, unless we know something about structure of thesolution and take it into account.






Every scheme needs at least two degrees of freedom per wavelength justto resolve the wave, unless we know something about structure of thesolution and take it into account.Fairly poor understanding of properties and behaviour of high orderspectral element schemes for approximation of waves

how high should the order be?

how much numerical dispersion? ... numerical dissipation?

how does behaviour compare with (consistent) finite elementscheme?

how compare to the ‘two degrees of freedom per wavelength’?Modern Techniques for Numerical Solution of PDEs, ACMAC, Crete, September 2011 – p. 20/54

Numerical Observations

Consider the model problem with time-harmonic excitation:

utt − uxx = 0

subject to

u(−1, t) = exp(iωt), ux(1, t) + ut(1, t) = 0, t ∈ (0, T ).

Solution has form u(x, t) = U(x) exp(−iωt) where U satisfies U(−1) = 1,U ′(1)− iωU(1) = 0 and

−U ′′(x)− ω2U(x) = 0, x ∈ (−1, 1).

True solution U(x) = exp(iω(x+ 1)).


Piecewise Linear Approximation

Finite element method (exact quadrature) on uniform mesh of size h:

(K − κ2

M + iκC)~U = ~r

where κ = ωh/2, C = ~eN~etN ,

K =1

h

1 −1 0 · · · 0

−1 2 −1 · · · 0...

......

. . . −1

0 0 0 −1 1

; M =

h

6

2 1 0 · · · 0

1 4 1 · · · 0...

......

. . . 1

0 0 0 1 2

.

Easy to invert M of course, but becomes more problematic onunstructured grids in higher dimensions.



Spectral element method (GLL quadrature) on uniform mesh of size h:

(K − κ2

M − iκC)~U = ~r

where κ = kh/2, C = ~eN~etN ,

K =1

h

1 −1 0 · · · 0

−1 2 −1 · · · 0...

......

. . . −1

0 0 0 −1 1

; M =

h

2

1 0 0 · · · 0

0 2 0 · · · 0...

......

. . . 0

0 0 0 0 1

.

Mass matrix M diagonal.



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5

−1

−0.5

0

0.5

1

1.5

2

U

Numerical real wave obtained using the spectral element schemeNumerical real wave obtained using the finite element schemeNumerical real wave obtained using the optimal schemeExact real wave



Finite elements exhibit phase lead. Spectral elements phase lag.

Magnitude is same for both schemes.


Piecewise Quadratic Approximation

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5

−1

−0.5

0

0.5

1

1.5

2

U



Piecewise Quadratic Approximation


Magnitude is smaller for spectral elements.


Piecewise Cubic Approximation

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5

−1

−0.5

0

0.5

1

1.5

2

U



Piecewise Cubic Approximation


Magnitude is even smaller for spectral elements.


How to measure this behaviour?

Dispersion relation for wave equation

utt − uxx = 0

obtained by seeking non-trivial solution of form u(x, t) = U(x)e−iωt, whereU has Bloch wave property

U(x+ h) = eikhU(x).

Gives k = ±ω.




utt − uxx = 0


U(x+ h) = eikhU(x).

Gives k = ±ω.Seek non-trivial discrete Bloch wave uh(x, t) = Uh(x)e

−iωt such that

Uh(x+ h) = eikhUh(x)

with k to be determined.




utt − uxx = 0


U(x+ h) = eikhU(x).

Gives k = ±ω.Seek non-trivial discrete Bloch wave uh(x, t) = Uh(x)e

−iωt such that

Uh(x+ h) = eikhUh(x)

with k to be determined.

k − k measures phase error


Example: First Order SEM

Spectral elements (p = 1):

1

h(Uj−1 + 2Uj − Uj+1)− ω2hUj = 0




1

h(Uj−1 + 2Uj − Uj+1)− ω2hUj = 0

Seek non-trivial discrete U of form Uj = Ceikjh, such that

1

h

(e−ikh − 2 + eikh

)− ω2h = 0.

Hence,

cos kh = 1−Ω2

2, Ω = ωh




1

h(Uj−1 + 2Uj − Uj+1)− ω2hUj = 0

Seek non-trivial discrete U of form Uj = Ceikjh, such that

1

h

(e−ikh − 2 + eikh

)− ω2h = 0.

Hence,

cos kh = 1−Ω2

2, Ω = ωh

For Ω ≪ 1, obtain

kh− ωh =Ω3

24+O(Ω)5.


Discrete Dispersion Relation (Ω = ωh)

Spectral Element Scheme

Order p Rp(Ω) cos−1Rp(Ω)− Ω

1 1− Ω2

2Ω3

24

2 Ω4−22Ω2+482(Ω2+24)

Ω5

2880

3 −Ω6+92Ω4−1680Ω2+3600

2(Ω4+60Ω2+1800)Ω7

604800

......

...


Discrete Dispersion Relation (Ω = ωh)

Spectral Element Scheme


1 1− Ω2

2Ω3

24

2 Ω4−22Ω2+482(Ω2+24)

Ω5

2880

3 −Ω6+92Ω4−1680Ω2+3600

2(Ω4+60Ω2+1800)Ω7

604800

......

...

Finite Element Scheme (Ainsworth, SINUM 2004)


1 −2Ω2+6Ω2+6 −Ω3

24

2 3Ω4−104Ω2+240

Ω4+16Ω2+240 − Ω5

1440

3 −4Ω6+540Ω4−11520Ω2+25200

Ω6+30Ω4+1080Ω2+25200 − Ω7

201600...

......


General Form of Dispersion Relation for Finite Elements

Discrete dispersion relation for order p elements:

cos(kh) = Rp(hω)

where Rp is the rational function

Rp(2κ) =[2No/2No − 2]κ cot κ − [2Ne + 2/2Ne]κ tanκ

[2No/2No − 2]κ cot κ + [2Ne + 2/2Ne]κ tanκ.

and Ne = ⌊p/2⌋ and No = ⌊(p+ 1)/2⌋.

Simple form in terms of Padé approximants ...

(Ainsworth, SINUM 2004)


General Form of Dispersion Relation Spectral Elements

Define sequences ap∞

p=0 and bp∞

p=0 recursively by the rule

ap+1 = − 2p+1κ bp + ap−1,

bp+1 = 2p+1κ ap + bp−1,

for p ∈ N with a0 = 1, a1 = 1, b0 = 0 and b1 = 1/κ. Then, the discretedispersion relation for spectral element method is given by

Rp(2κ) = cos(kh) = (−1)p[ap (κbp−1 + pap) + bp (κap−1 − pbp)

ap (κbp−1 + pap)− bp (κap−1 − pbp)

].

(Ainsworth and Wajid, SINUM 2008)

Not related to Padé approximants! (c.f. finite element case)


Phase Error for ωh ≪ 1

Phase error for finite element method satisfies

kh− ωh = −1

2

[p!

(2p)!

]2(ωh)

2p+1

2p+ 1+O (ωh)2p+3





kh− ωh = −1

2

[p!

(2p)!

]2(ωh)

2p+1

2p+ 1+O (ωh)2p+3


Phase error for spectral element method satisfies

kh− ωh =1

2p

[p!

(2p)!

]2(ωh)2p+1

2p+ 1+O (ωh)

2p+3.





kh− ωh = −1

2

[p!

(2p)!

]2(ωh)

2p+1

2p+ 1+O (ωh)2p+3


Phase error for spectral element method satisfies

kh− ωh =1

2p

[p!

(2p)!

]2(ωh)2p+1

2p+ 1+O (ωh)

2p+3.


Finite elements exhibit phase lead and spectral elements exhibitphase lag.

Spectral elements a factor 1/p times more accurate than finiteelements.


Numerical Dispersion for Large Order and Large Wavenumber

0 10 20 30 40 50 6010

−14

10−12

10−10

10−8

10−6

10−4

10−2

100

102

104

p

|Rp−

cos(

ω h

)|

Case: ω h = 80 ω h >> 1

Error in the discrete dispersion relation with full integrationError in the discrete dispersion relation with reduced integration

Note: Very sharp transition between garbage and essentially exactModern Techniques for Numerical Solution of PDEs, ACMAC, Crete, September 2011 – p. 33/54

Numerical Dispersion for Large Order and Large Wavenumber

Suppose that ωh ≫ 1. Then the error Ep = cos kh− cosωh in the discretedispersion relation passes through distinct phases as the order p ∈ N isincreased:

Unstable Phase: For p = O(1), Ep ≈ (−1)p (ωh)2

2 .

Oscillatory Phase: For 1 ≪ 2p+ 1 < ωh− o(ωh)1/3, Ep oscillates anddecays to O(1) as p is increased.

Transition Zone: For ωh− o(ωh)1/3 < 2p+ 1 < ωh+ o(ωh)1/3, theerror Ep oscillates without further decrease.

Super-Exponential Decay: For 2p+ 1 > ωh+ o(ωh)1/3, Ep decreasesat a super-exponential rate.



One Consequence of the Analysis

What is significance of condition 2p+ 1 ≈ ωh which appears in analysis?




Degree p approximation corresponds to p+ 1 modes per element (0, h).





=⇒ π(2p+ 2)/ωh modes per wavelength λ = 2π/ω






Wave essentially resolved when 2p+ 1 = ωh+ o(ωh)1/3







Hence,

π modes per wavelength are required to resolve wave.

Provides rigorous justification of ‘well-known’ rule of thumb governing useof spectral element methods.







Hence,



... spectral element method not bad at all, in view of fact that no generalscheme can suffice with fewer than two degrees of freedom perwavelength.







Hence,



... spectral element method not bad at all, in view of fact that no generalscheme can suffice with fewer than two degrees of freedom perwavelength.

Ref: Ainsworth and Wajid, SINUM 2008


Blended Finite Element/Spectral Element Schemes


Blended FEM/SEM Schemes

Idea: Can we obtain better scheme by averaging (blending) finiteelements and spectral elements?



Finite Element Scheme:

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn




1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn

Spectral Element Scheme:

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn




1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn

Spectral Element Scheme:

1

(∆t)2M

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn

Blended FEM/SEM Scheme:

1

(∆t)2M τ

(~αn+1 − 2~αn + ~αn−1

)+

1

∆tC

(~αn − ~αn−1

)+K~αn = ~rn

where, for τ ∈ [0, 1] (to be determined)

M τ = (1− τ)M + τM .




Magnitude is same for both schemes. Suggests a symmetricweighting τ = 1/2.



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5

−1

−0.5

0

0.5

1

1.5

2

U




Dispersion analysis reveals for blended scheme (τ ∈ [0, 1]):

kh− ωh =1

24(ωh)3(2τ − 1) +O(ωh)5.



Dispersion analysis reveals for blended scheme (τ ∈ [0, 1]):

kh− ωh =1

24(ωh)3(2τ − 1) +O(ωh)5.

... so choice τ = 1/2 is optimal and gives

kh− ωh =1

480(ωh)5 +O(ωh)7.

two additional orders of accuracy;

improved coefficient of leading order term.



A quote taken from the conclusion of the highly influential article of Marfurt(Geophysics, 49, 1984):

Frequency domain finite element solutions employing weighted averageof consistent and lumped masses yield the most accurate results,and they promise to be the most cost-effective method for [severalapplications in wave propagation]. Marfurt (1984).





Despite this bold statement, none of the citing articles or any other overthe past 20 years presents an optimal blended scheme for elements ofhigher than first order.





Despite this bold statement, none of the citing articles or any other overthe past 20 years presents an optimal blended scheme for elements ofhigher than first order.Is it possible to obtain a general result?



Let p ≥ 2 and τ ∈ [0, 1]. Then, error in discrete wavenumber for blendedscheme is given by

kh− ωh =1

2

[p!

(2p)!

]2(ωh)2p+1

2p+ 1(1 + 1/p)τ − 1+O(ωh)2p+3.




kh− ωh =1

2

[p!

(2p)!

]2(ωh)2p+1

2p+ 1(1 + 1/p)τ − 1+O(ωh)2p+3.

Optimal blending parameter given by τ = p/(p+ 1).




kh− ωh =1

2

[p!

(2p)!

]2(ωh)2p+1

2p+ 1(1 + 1/p)τ − 1+O(ωh)2p+3.

Optimal blending parameter given by τ = p/(p+ 1).

With this choice, we find that

kh− ωh =4

2p− 1

[(p+ 1)!

(2p+ 2)!

]2(ωh)2p+3

2p+ 3+O(ωh)2p+5.

two additional orders of accuracy;

coefficient of leading order term much smaller.

Ref: Ainsworth and Wajid, SINUM 2010.



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5

−1

−0.5

0

0.5

1

1.5

2

U



What about d > 1 dimensions?

Can we find τ such that blending mass matrices for FEM/SEM scheme in2D has higher accuracy?



Can we find τ such that blending mass matrices for FEM/SEM scheme in2D has higher accuracy?Bad news: there is no choice giving higher accuracy.



Can we find τ such that blending mass matrices for FEM/SEM scheme in2D has higher accuracy?Bad news: there is no choice giving higher accuracy.Don’t give up yet. Look in detail at structure of 2D equations on tensorproduct grid:

K ⊗M +M ⊗K − ω2M ⊗M (FEM)

and

K ⊗ M + M ⊗K − ω2M ⊗ M (SEM)



Can we find τ such that blending mass matrices for FEM/SEM scheme in2D has higher accuracy?Bad news: there is no choice giving higher accuracy.Don’t give up yet. Look in detail at structure of 2D equations on tensorproduct grid:

K ⊗M +M ⊗K − ω2M ⊗M (FEM)

and

K ⊗ M + M ⊗K − ω2M ⊗ M (SEM)

Suggests that in multi-dimensions should look for blending in the form

K ⊗Mτ +Mτ ⊗K − ω2Mτ ⊗Mτ (Blended)



Can we choose τ so that


has higher order accuracy?






Yes ... and for all dimensions d;

Optimal τ = p/(p+ 1) ... same as in 1D case;

Obtain two additional orders of accuracy and improved constant (asin 1D case).






Yes ... and for all dimensions d;

Optimal τ = p/(p+ 1) ... same as in 1D case;

Obtain two additional orders of accuracy and improved constant (asin 1D case).

but ...

nobody wants to assemble mass matrices for both SEM and FEM orto form Kronecker products.

what to do with non-affine elements?


New ‘Non-Standard’ Quadrature Rules

For τ ∈ [0, 1), define

∫ 1

−1

f(x)dx ≈ Q(p)τ (f) =

p∑

j=0

wjf(ξj)

where ξj zeros of Pp+1 − τPp−1 and

wj =2[p(1 + τ) + τ ]

p(p+ 1)Pp(ξj)[P ′

p+1(ξj)− τP ′

p−1(ξj)].


New ‘Non-Standard’ Quadrature Rules

For τ ∈ [0, 1), define

∫ 1

−1

f(x)dx ≈ Q(p)τ (f) =

p∑

j=0

wjf(ξj)

where ξj zeros of Pp+1 − τPp−1 and

wj =2[p(1 + τ) + τ ]

p(p+ 1)Pp(ξj)[P ′

p+1(ξj)− τP ′

p−1(ξj)].

Then

nodes are distinct and contained in (−1, 1);

weights well-defined and positive;

rule has precision 2p− 1 (under-integration).

... non-standard but perfectly reasonable quadrature rule.(Ainsworth-Wajid, SINUM 2010)


‘AW-Quad’ Quadrature Rules

ξ w

p = 1 ±0.816496580927726 1.000000000000000

p = 2 0.000000000000000 1.230769230769231

±0.930949336251263 0.384615384615385

p = 3 ±0.96433527587956 0.199826014447922

±0.429352058315787 0.800173985552078

p = 4 0.000000000000000 0.693766937669377

±0.978315678013417 0.121787277062268

±0.638731398345590 0.531329254103044

... non-standard but perfectly reasonable quadrature rules.



If you have FEM code on quads/hexahedra: Replace Gauss-Legendrequadrature rule with ‘AW-Quad’ rule with τ = p/(p+ 1), and solve.




OR




OR

If you have SEM code on quads/hexahedra: Replace Gauss-Lobattoquadrature rule with ‘AW-Quad’ rule with τ = p/(p+ 1) and solve (don’tchange basis functions).




OR


Resulting approximation is precisely the optimally blended FEM/SEMapproximation.




OR


Resulting approximation is precisely the optimally blended FEM/SEMapproximation.

... trivial implement the method at virtually no additional overhead.


Pekeris Waveguide (thanks to Jeff Zitelli, Univ. of Texas)

Planar case: harmonic excitation on line zS .Axisymmetric case: harmonic excitation at point zS .


Pekeris Waveguide - True Solution

Real part of pressure. Frequency 200Hz. Line source.


Pekeris Waveguide - Mesh

PML used to truncate domain.Polynomial order of elements (yellow = quartic, green = cubic).Mesh size: hk/(2p+ 1) ≈ 0.4 at 200Hz.


Pekeris Waveguide - FEM vs Optimal Blending

Magnitude of error for Standard FEM (top) and Optimal Blending (bottom).In this example, we take a 2D waveguide of 1km and frequency of 200Hz.Relative error in L2 norm:Standard FEM 18.1%; Optimally Blended 4.44%.


Pekeris Waveguide - FEM vs Optimal Blending

Comparison of errors in standard FEM and Optimally Blended Methodover a range of frequencies.


Pekeris Waveguide - Axisymmetric Case

Same behaviour observed in axi-symmetric case also!


Blended FEM/SEM Schemes: The story so far ...

general result for optimal blending and error analysis;

numerical performance is promising;

novel non-standard quadrature rules means easily implemented in anexisting code.


Blended FEM/SEM Schemes: The story so far ...

general result for optimal blending and error analysis;

numerical performance is promising;

novel non-standard quadrature rules means easily implemented in anexisting code.

References:

M. Ainsworth, “Discrete dispersion relation for hp-version finiteelement approximation at high wave number,” SINUM, vol. 42, no. 2,2004.

M. Ainsworth and H. A. Wajid, “Dispersive and dissipative behavior ofthe spectral element method,” SINUM, vol. 47, no. 5, 2009.

M. Ainsworth and H. A. Wajid, “Optimally blended spectral-finiteelement scheme for wave propagation and nonstandard reducedintegration,” SINUM, vol. 48, no. 1, 2010.


optimally blended finite/spectral element scheme for wave … · 2011-09-23 · optimally blended...

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