The optimality of IPG methods for odd degrees of polynomial - - PowerPoint PPT Presentation

The optimality of IPG methods for odd degrees of polynomial approximation

Oto Havle Vít Dolejší

Charles University Prague

Workshop Dresden-Prague on Numerical Analysis 2010

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Contents

1

Introduction

2

Main result and numerical evidence

3

Sketch of the proof

4

Conclusion

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Contents

1

Introduction

2

Main result and numerical evidence

3

Sketch of the proof

4

Conclusion

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DGFEM for the elliptic model problem

Interior Penalty Discontinuous Finite Element Galerkin discretization. u ∈ H2(Ω) : −∆u = f in Ω + boundary conditions ↓ ↓ uh ∈ Shp : Bh(uh, vh) = Lh(vh) for all vh ∈ Shp Test and trial space Shp of piecewise polynomial functions

• f degree at most p ≥ 1.

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IIPG, NIPG, SIPG; Interior penalty

Bh(u, v) =

• K∈Th
• K

∇u · ∇v dx −

• Γ∈Fh
• Γ
• u [v] + θ v [u]
• dS

+

• Γ∈Fh

cW HΓ

• Γ

[u] [v] dS where [v] denotes the jump and v the average value of v on the mesh face Γ, and θ =      1, SIPG (symmetric variant) 0, IIPG (incomplete variant) −1, NIPG (nonsymetric variant)

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IIPG, NIPG, SIPG; Interior penalty

Bh(u, v) =

• K∈Th
• K

∇u · ∇v dx −

• Γ∈Fh
• Γ
• u [v] + θ v [u]
• dS

+

• Γ∈Fh

cW HΓ

• Γ

[u] [v] dS where [v] denotes the jump and v the average value of v on the mesh face Γ, and HΓ = hL + hR 2 , max(hL, hR), diam(Γ) etc. cW > 0, large enough

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What is the optimal convergence behavior?

A priori error estimate in a given norm · u − uh = O(hk) Best possible approximation in the trial space Shp inf

vh∈Shp

u − vh = O(hℓ) The error estimate is optimal if k = ℓ. For example: · H1(Ω,Th) · L2(Ω) DGFEM - SIPG p p + 1 DGFEM - NIPG, IIPG p p classical FEM p p + 1 L2-projection p p + 1

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Facts about L2-convergence of nonsymetric DGFEM

Numerical experiments Optimal O(hp+1) for odd p, suboptimal O(hp) for even p. Suboptimal behavior possible for odd p on special nonuniform meshes [Guzmán and Riviére, 2009]. General theoretical results Suboptimal estimate O(hp) follows from discrete Poincaré-Friedrichs inequality. Optimal estimates can be proven for overpenalized DGFEM.

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Facts about L2-convergence of nonsymetric DGFEM (contd.)

The optimal L2 convergence has been proved for: One-dimensional NIPG/IIPG with odd p and uniform

• meshes. [Larson, Niklasson, 2004]

2D NIPG, linear triangles, special assumptions on the

• mesh. [Burmann, Stamm, 2008]

2D/3D IIPG/NIPG, bilinear/trilinear elements on uniform Cartesian meshes. [Wang, Wang, Sun, Wheeler 2009] Our results: Optimal error estimate for one-dimensional IIPG with odd p, nonuniform meshes, special assumption on the penalty parameter.

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Contents

1

Introduction

2

Main result and numerical evidence

3

Sketch of the proof

4

Conclusion

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1D model problem

Continuous problem −u′′ = f in (0, 1), u′(0) = gN u(1) = uD locally quasi-uniform mesh hk = xk+1 − xk, 1 CH ≤ hk hk+1 ≤ CH. IIPG discretization Bh(uh, vh) =

N−1

• k=0

xk+1

xk

u′

hv′ h dx − N

• k=1
• u′

h

• k [vh]k

+

N

• k=1

cW Hk [uh]k [vh]k Lh(vh) = 1 fvh dx − gNvh(0) + cW HN uDvh(1)

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Assumptions

The interior and boundary penalization term Jh(uh, vh) =

N

• k=1

cW Hk [uh]k [vh]k depends on parameters{cW, Hk : k = 1, . . . , N}. Hk is given by Hk = H(hk−1, hk) for k < N, HN = H(hN−1, hN−1) H(·, ·) is a continuous function and H(a, b) > 0, H(a, b) = H(b, a), H(κa, κb) = κH(a, b) cW ≥ c⋆

W

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The main result

Then two following assertions are equivalent. (A) There exists a constant CE > 0 such that uh − uL2(0,1) ≤ CEh p+1 where u is the weak solution. (B) The degree of approximation p is an odd number and the function H is a multiple of Hp(a, b) =          a p+1 − b p+1 a p − b p , a = b

p+1 p a,

a = b

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Numerical experiments

Continuous problem: −u′′(x) = x10 for x ∈ (0, 1), u(0) = u(1) = 0 Computational meshes: uniform, hk = h = 1/N non-uniform, N is a multiple of 3 h3i = 3 16N , h3i+1 = 5 · 3 16N , h3i+2 = 10 · 3 16N various choices of H(·, ·)

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Results on uniform meshes

p= 1 NIPG IIPG SIPG N ehL2(Ω) EOC ehL2(Ω) EOC ehL2(Ω) EOC 12288 1.087E-09 2.000 2.207E-10 2.000 1.319E-10 1.999 24576 2.717E-10 2.000 5.518E-11 2.000 3.297E-11 2.000 49152 6.792E-11 2.000 1.380E-11 2.000 8.244E-12 2.000 p= 2 12288 3.407E-10 2.000 1.278E-11 2.001 5.863E-15 3.000 24576 8.516E-11 2.000 3.194E-12 2.000 7.329E-16 3.000 49152 2.129E-11 2.000 7.985E-13 2.000 9.161E-17 3.000 p= 3 12288 1.554E-18 4.001 4.372E-19 4.000 3.178E-19 4.000 24576 9.711E-20 4.000 2.732E-20 4.000 1.987E-20 4.000 49152 6.069E-21 4.000 1.708E-21 4.000 1.242E-21 4.000 p= 4 12288 3.169E-19 4.000 1.288E-20 4.001 9.840E-24 5.000 24576 1.981E-20 4.000 8.048E-22 4.000 3.075E-25 5.000 49152 1.238E-21 4.000 5.030E-23 4.000 9.610E-27 5.000

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Results on nonuniform meshes, H(a, b) = max(a, b)

p= 1 NIPG IIPG SIPG N ehL2(Ω) EOC ehL2(Ω) EOC ehL2(Ω) EOC 12288 2.082E-07 0.974 2.076E-08 1.020 3.360E-10 1.999 24576 1.050E-07 0.987 1.031E-08 1.010 8.403E-11 2.000 49152 5.275E-08 0.994 5.136E-09 1.005 2.101E-11 2.000 p= 2 12288 3.283E-10 2.003 1.680E-11 2.002 3.501E-14 3.000 24576 8.198E-11 2.002 4.196E-12 2.001 4.375E-15 3.000 49152 2.048E-11 2.001 1.049E-12 2.001 5.469E-16 3.000 p= 3 12288 6.677E-15 2.993 3.040E-17 3.088 3.023E-18 4.000 24576 8.366E-16 2.997 3.694E-18 3.040 1.889E-19 4.000 49152 1.047E-16 2.998 4.557E-19 3.019 1.181E-20 4.000 p= 4 12288 9.695E-19 4.004 5.214E-20 4.002 1.903E-22 5.000 24576 6.052E-20 4.002 3.256E-21 4.001 5.946E-24 5.000 49152 3.780E-21 4.001 2.034E-22 4.001 1.858E-25 5.000

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Results on nonuniform meshes, H(a, b) = Hp(a, b)

p= 1 NIPG IIPG SIPG N ehL2(Ω) EOC ehL2(Ω) EOC ehL2(Ω) EOC 12288 3.178E-07 0.977 9.167E-10 1.999 3.274E-10 1.999 24576 1.602E-07 0.989 2.292E-10 2.000 8.189E-11 1.999 49152 8.040E-08 0.994 5.732E-11 2.000 2.048E-11 2.000 p= 2 12288 3.380E-10 2.004 1.837E-11 2.004 3.479E-14 3.000 24576 8.440E-11 2.002 4.587E-12 2.002 4.349E-15 3.000 49152 2.109E-11 2.001 1.146E-12 2.001 5.437E-16 3.000 p= 3 12288 6.755E-15 2.993 3.993E-18 4.000 3.020E-18 4.000 24576 8.465E-16 2.997 2.496E-19 4.000 1.887E-19 4.000 49152 1.059E-16 2.998 1.560E-20 4.000 1.180E-20 4.000 p= 4 12288 9.714E-19 4.004 5.306E-20 4.003 1.902E-22 5.000 24576 6.063E-20 4.002 3.313E-21 4.001 5.944E-24 5.000 49152 3.787E-21 4.001 2.069E-22 4.001 1.857E-25 5.000

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Contents

1

Introduction

2

Main result and numerical evidence

3

Sketch of the proof

4

Conclusion

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The proof of B = ⇒ A

p odd , H = Hp = ⇒ uh − uL2 ≤ Ch p+1 Step 1 The Aubin-Nitsche duality trick. −ψ′′ = uh − u, ψ′(0) = ψ(1) = 0 uh − u2

L2(0,1) = Bh(uh − u, ψ) − N

• k=1

ψ′(xk) [uh − u]k

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The proof of B = ⇒ A

p odd , H = Hp = ⇒ uh − uL2 ≤ Ch p+1 Step 1 The Aubin-Nitsche duality trick. Step 2 Identify the leading part of the jump [uh]k. cW Hk [uh]k = Bh(uh, w⋆

h,p,k) = Lh(w⋆ h,p,k)

= 1 f w⋆

h,p,k dx

(interior nodes) = Kp

• (−1) p−1h p

k−1 − h p k

• f (p−1)(xk) + O(hp+1

k

) = Kp

• h p

k−1 − h p k

• f (p−1)(xk) + O(hp+1

k

)

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The proof of B = ⇒ A

p odd , H = Hp = ⇒ uh − uL2 ≤ Ch p+1 Step 1 The Aubin-Nitsche duality trick. Step 2 Identify the leading part of the jump [uh]k. Step 3 Summation by parts.

N

• k=1

ψ′(xk) [uh − u]k ≈

N−1

• k=1

ψ′(xk)

• hp+1

k−1 − hp+1 k

• f (p−1)(xk)

≈

N−2

• k=1

hp+1

k

• ψ′f (p−1)xk+1

xk

Chp+1fHp(0,1)ψH2(0,1)

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The proof of A = ⇒ B

uh − uL2 ≤ Ch p+1 = ⇒ p odd , H = Hp Step 1 Perturbation of the penalization parameter cW. ucW1

h

− ucW2

h

L2 ≤ ucW1

h

− uL2 + ucW2

h

− uL2 ≤ Ch p+1 The difference ucW1

h

− ucW2

h

constant on each element ucW1

h

− ucW2

h

• (xℓ,xℓ+1) =
• 1 − cW1

cW2

• N
• k=ℓ+1
• uCW1

h

• k

We already have an expression for [uh]k. We can estimate ucW1

h

− ucW2

h

L2 for particular meshes and right hand side function f.

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The proof of A = ⇒ B

uh − uL2 ≤ Ch p+1 = ⇒ p odd , H = Hp Step 1 Perturbation of the penalization parameter cW. Step 2 Prove that p is odd. Use uniform mesh, and assume p even and f(x) = xp−1. Then ucW1

h

− ucW2

h

L2 ≥ Chp + O(hp+1) with C = 0.

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The proof of A = ⇒ B

uh − uL2 ≤ Ch p+1 = ⇒ p odd , H = Hp Step 1 Perturbation of the penalization parameter cW. Step 2 Prove that p is odd. Step 3 Use nonuniform mesh h3i+j = 3αj (α0 + α1 + α2)N , j = 0, 1, 2. to get an equation for H H(α0, α1)(αp

0 − αp 1) + H(α1, α2)(αp 1 − αp 2)

+ H(α2, α0)(αp

2 − αp 0) = 0.

Every solution H is a multiple of Hp.

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Contents

1

Introduction

2

Main result and numerical evidence

3

Sketch of the proof

4

Conclusion

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Conclusion

IIPG and NIPG are optimal for odd p on equidistant meshes. IIPG is optimal for odd p on non-equidistant meshes only if Hk is particularly chosen. NIPG is not optimal for odd p on non-equidistant meshes. Relation for optimal Hk is unknown (if any) Unable to prove the result in 2D or 3D.

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Conclusion

IIPG and NIPG are optimal for odd p on equidistant meshes. IIPG is optimal for odd p on non-equidistant meshes only if Hk is particularly chosen. NIPG is not optimal for odd p on non-equidistant meshes. Relation for optimal Hk is unknown (if any) Unable to prove the result in 2D or 3D.

Thank you for your attention.

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