[PPT] - Spacetime finite element methods in biomedical applications Olaf PowerPoint Presentation

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Space–time finite element methods in biomedical applications

Olaf Steinbach

Institut f¨ ur Angewandte Mathematik, TU Graz http://www.applied.math.tugraz.at SFB Mathematical Optimization and Applications in Biomedical Sciences LEAD Project Mechanics, Modeling and Simulation in Aortic Disscetion

Subproject Parallel Space–Time Finite Element Methods

O. Steinbach

Magdeburg, 24.10.2018 1 / 23

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FETI Domain Decomposition Methods for the Simulation of Biological Tissues [C. Augustin 2014; C. Augustin, G. Holzapfel, OS 2014]

O. Steinbach

Magdeburg, 24.10.2018 2 / 23

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Cardiac Electromechanics [E. Karabelas 2015] Active stress Ta

O. Steinbach

Magdeburg, 24.10.2018 3 / 23

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Optimal control of Blood Flow in an Artery with Aneurysm [L. John 2014] Wall shear stress distribution [L. John, P. Pustejovska, OS 2017]

O. Steinbach

Magdeburg, 24.10.2018 4 / 23

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Lobe Pumpe [M. Neum¨

uller 2013]

O. Steinbach

Magdeburg, 24.10.2018 5 / 23

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Space–Time Finite Element Methods

◮ Formulation in Bochner spaces ◮ Formulation in anisotropic Sobolev spaces ◮ standard lowest order (piecewise linear) finite elements ◮ Heat equation [with M. Zank] ◮ Bidomain system [with H. Yang] ◮ (Navier–) Stokes [with D. Pacheco] ◮ Wave equation [with M. Zank]

The aim is to provide an adaptive simulation, simultaneously in space and time, which allows an accurate resolution of wave type solutions, and which allows, i.e. requires, the use of parallel solvers.

O. Steinbach

Magdeburg, 24.10.2018 6 / 23

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Dirichlet boundary value problem for the heat equation α∂tu(x, t) − ∆xu(x, t) = f (x, t) for (x, t) ∈ Q := Ω × (0, T), u(x, t) = for (x, t) ∈ Σ := Γ × (0, T), u(x, 0) = for x ∈ Ω.

O. Steinbach

Magdeburg, 24.10.2018 7 / 23

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Dirichlet boundary value problem for the heat equation α∂tu(x, t) − ∆xu(x, t) = f (x, t) for (x, t) ∈ Q := Ω × (0, T), u(x, t) = for (x, t) ∈ Σ := Γ × (0, T), u(x, 0) = for x ∈ Ω. Find u ∈ L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω)) such that

T

Ω
α∂tu(x, t)v(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ L2(0, T; H1

0(Ω)).

O. Steinbach

Magdeburg, 24.10.2018 7 / 23

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Dirichlet boundary value problem for the heat equation α∂tu(x, t) − ∆xu(x, t) = f (x, t) for (x, t) ∈ Q := Ω × (0, T), u(x, t) = for (x, t) ∈ Σ := Γ × (0, T), u(x, 0) = for x ∈ Ω. Find u ∈ L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω)) such that

T

Ω
α∂tu(x, t)v(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ L2(0, T; H1

0(Ω)).

Bilinear form a(u, v) = T

Ω
α∂tu(x, t)v(x, t) + ∇xu(x, t) · ∇xv(x, t)
dxdt

= T

Ω
α∂tu(x, t) − ∆xu(x, t)
v(x, t) dxdt
O. Steinbach

Magdeburg, 24.10.2018 7 / 23

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Quasi–static Dirichlet problem for u ∈ L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω))

−∆xw(x, t) = α∂tu(x, t) − ∆xu(x, t) for (x, t) ∈ Q, w(x, t) = for (x, t) ∈ Σ.

O. Steinbach

Magdeburg, 24.10.2018 8 / 23

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Quasi–static Dirichlet problem for u ∈ L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω))

−∆xw(x, t) = α∂tu(x, t) − ∆xu(x, t) for (x, t) ∈ Q, w(x, t) = for (x, t) ∈ Σ. Find w ∈ L2(0, T; H1

0(Ω)) such that

T

Ω

∇xw(x, t)·∇xv(x, t) dxdt = T

Ω
α∂tu(x, t)−∆xu(x, t)
v(x, t) dxdt

is satisfied for all v ∈ L2(0, T; H1

0(Ω)).

O. Steinbach

Magdeburg, 24.10.2018 8 / 23

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Quasi–static Dirichlet problem for u ∈ L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω))

−∆xw(x, t) = α∂tu(x, t) − ∆xu(x, t) for (x, t) ∈ Q, w(x, t) = for (x, t) ∈ Σ. Find w ∈ L2(0, T; H1

0(Ω)) such that

T

Ω

∇xw(x, t)·∇xv(x, t) dxdt = T

Ω
α∂tu(x, t)−∆xu(x, t)
v(x, t) dxdt

is satisfied for all v ∈ L2(0, T; H1

0(Ω)).

Lemma α∂tu − ∆xu2

L2(0,T;H−1(Ω)) = w2 L2(0,T;H1

0(Ω)) = a(u, w)

Corollary α∂tu − ∆xuL2(0,T;H−1(Ω)) ≤ sup

0=v∈L2(0,T;H1

0(Ω))

a(u, v) vL2(0,T;H1

0(Ω))

O. Steinbach

Magdeburg, 24.10.2018 8 / 23

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Norm equivalence 1 2

α∂tu2

L2(0,T;H−1(Ω)) + u2 L2(0,T;H1

0(Ω))

≤

α∂tu − ∆xu2

L2(0,T;H−1(Ω))

≤ 2

α∂tu2

L2(0,T;H−1(Ω)) + u2 L2(0,T;H1

0(Ω))

Lemma

1 √ 2

α∂tu2

L2(0,T;H−1(Ω)) + u2 L2(0,T;H1

0(Ω)) ≤

sup

0=v∈L2(0,T;H1

0(Ω))

a(u, v) vL2(0,T;H1

0(Ω))

[Schwab, Stevenson 2009; Urban, Patera 2014; Andreev 2013; Mollet 2014; OS 2015; . . . ]

O. Steinbach

Magdeburg, 24.10.2018 9 / 23

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Adjoint variational formulation to find u ∈ L2(0, T; H1

0(Ω)) such that

T

Ω
−u(x, t)α∂tv(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ L2(0, T; H1

0(Ω)) ∩ H1 ,0(0, T; H−1(Ω)).

O. Steinbach

Magdeburg, 24.10.2018 10 / 23

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Adjoint variational formulation to find u ∈ L2(0, T; H1

0(Ω)) such that

T

Ω
−u(x, t)α∂tv(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ L2(0, T; H1

0(Ω)) ∩ H1 ,0(0, T; H−1(Ω)).

Operator adjoint formulation L : L2(0, T; H1

0(Ω)) → [L2(0, T; H1 0(Ω)) ∩ H1 ,0(0, T; H−1(Ω))]′

Operator primal formulation L : L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω)) → [L2(0, T; H1 0(Ω))]′

O. Steinbach

Magdeburg, 24.10.2018 10 / 23

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Adjoint variational formulation to find u ∈ L2(0, T; H1

0(Ω)) such that

T

Ω
−u(x, t)α∂tv(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ L2(0, T; H1

0(Ω)) ∩ H1 ,0(0, T; H−1(Ω)).

Operator adjoint formulation L : L2(0, T; H1

0(Ω)) → [L2(0, T; H1 0(Ω)) ∩ H1 ,0(0, T; H−1(Ω))]′

Operator primal formulation L : L2(0, T; H1

0(Ω)) ∩ H1 0,(0, T; H−1(Ω)) → [L2(0, T; H1 0(Ω))]′

Can we define L : L2(0, T; H1

0(Ω))∩H1/2 0, (0, T; L2(Ω)) → [L2(0, T; H1 0(Ω))∩H1/2 ,0 (0, T; L2(Ω))]′

O. Steinbach

Magdeburg, 24.10.2018 10 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Primal variational formulation: Find u ∈ H1

0,(0, T) such that

a(u, v) = T u′(t)v(t) dt = T f (t)v(t) dt for all v ∈ L2(0, T) Solution operator B1 : H1

0,(0, T) → L2(0, T)

O. Steinbach

Magdeburg, 24.10.2018 11 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Primal variational formulation: Find u ∈ H1

0,(0, T) such that

a(u, v) = T u′(t)v(t) dt = T f (t)v(t) dt for all v ∈ L2(0, T) Solution operator B1 : H1

0,(0, T) → L2(0, T)

Dual variational formulation: Find u ∈ L2(0, T) such that T u(t)v ′(t) dt = − T f (t)v(t) dt for all v ∈ H1

,0(0, T)

Solution operator B0 : L2(0, T) → [H1

,0(0, T)]′

O. Steinbach

Magdeburg, 24.10.2018 11 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Primal variational formulation: Find u ∈ H1

0,(0, T) such that

a(u, v) = T u′(t)v(t) dt = T f (t)v(t) dt for all v ∈ L2(0, T) Solution operator B1 : H1

0,(0, T) → L2(0, T)

Dual variational formulation: Find u ∈ L2(0, T) such that T u(t)v ′(t) dt = − T f (t)v(t) dt for all v ∈ H1

,0(0, T)

Solution operator B0 : L2(0, T) → [H1

,0(0, T)]′

Question Bs : [L2(0, T), H1

0,(0, T)]s → [[H1 ,0(0, T)]′, L2(0, T)]s

for s ∈ (0, 1), s = 1 2 Related work: [Fontes 1996; Langer, Wolfmayr 2013; Larsson, Schwab 2015]

O. Steinbach

Magdeburg, 24.10.2018 11 / 23

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Solution operator B1 : H1

0,(0, T) → L2(0, T)

Adjoint operator B′

1 : L2(0, T) → [H1 0,(0, T)]′

u, B′

1vL2(0,T) = B1u, vL2(0,T)

for all u ∈ H1

0,(0, T), v ∈ L2(0, T)

O. Steinbach

Magdeburg, 24.10.2018 12 / 23

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Solution operator B1 : H1

0,(0, T) → L2(0, T)

Adjoint operator B′

1 : L2(0, T) → [H1 0,(0, T)]′

u, B′

1vL2(0,T) = B1u, vL2(0,T)

for all u ∈ H1

0,(0, T), v ∈ L2(0, T)

Define A := B′

1B1 : H1 0,(0, T) → [H1 0,(0, T)]′

Eigenvalue problem Au = λu

O. Steinbach

Magdeburg, 24.10.2018 12 / 23

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Solution operator B1 : H1

0,(0, T) → L2(0, T)

Adjoint operator B′

1 : L2(0, T) → [H1 0,(0, T)]′

u, B′

1vL2(0,T) = B1u, vL2(0,T)

for all u ∈ H1

0,(0, T), v ∈ L2(0, T)

Define A := B′

1B1 : H1 0,(0, T) → [H1 0,(0, T)]′

Eigenvalue problem Au = λu Au, vL2(0,T) = B1u, B1vL2(0,T) = T ∂tu(t)∂tv(t) dt = λ T u(t)v(t) dt −u′′(t) = λu(t) for t ∈ (0, T), u(0) = 0, u′(T) = 0 vk(t) = sin π 2 + kπ t T , λk = 1 T 2 π 2 + kπ 2 , k = 0, 1, 2, 3, . . .

O. Steinbach

Magdeburg, 24.10.2018 12 / 23

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u ∈ H1

0,(0, T)

u(t) =

∞

k=0

uk sin π 2 + kπ t T , uk = 2 T T u(t) sin π 2 + kπ t T dt

O. Steinbach

Magdeburg, 24.10.2018 13 / 23

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u ∈ H1

0,(0, T)

u(t) =

∞

k=0

uk sin π 2 + kπ t T , uk = 2 T T u(t) sin π 2 + kπ t T dt Time derivative (B1u)(t) = u′(t) = 1 T

∞

k=0

uk π 2 + kπ

cos

π 2 + kπ t T

O. Steinbach

Magdeburg, 24.10.2018 13 / 23

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u ∈ H1

0,(0, T)

u(t) =

∞

k=0

uk sin π 2 + kπ t T , uk = 2 T T u(t) sin π 2 + kπ t T dt Time derivative (B1u)(t) = u′(t) = 1 T

∞

k=0

uk π 2 + kπ

cos

π 2 + kπ t T B1u, vL2(0,T) = 1 T T

∞

k=0

uk π 2 + kπ

cos

π 2 + kπ t T v(t) dt = 1 T

∞

k=0

∞

ℓ=0

ukuℓ π 2 + kπ T cos π 2 + kπ t T cos π 2 + ℓπ t T dt = 1 2

∞

k=0

u2

k

π 2 + kπ

= u2

H1/2

0, (0,T)

if we choose v(t) =

∞

ℓ=0

uℓ cos π 2 + ℓπ t T ,

O. Steinbach

Magdeburg, 24.10.2018 13 / 23

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Transformation operator (HTu)(t) =

∞

k=0

uk cos π 2 + kπ t T , u(t) =

∞

k=0

uk sin π 2 + kπ t T ,

O. Steinbach

Magdeburg, 24.10.2018 14 / 23

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Transformation operator (HTu)(t) =

∞

k=0

uk cos π 2 + kπ t T , u(t) =

∞

k=0

uk sin π 2 + kπ t T , Lemma HT : Hs

0,(0, T) → Hs ,0(0, T),

HTuHs

,0(0,T) = uHs 0,(0,T)

O. Steinbach

Magdeburg, 24.10.2018 14 / 23

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Transformation operator (HTu)(t) =

∞

k=0

uk cos π 2 + kπ t T , u(t) =

∞

k=0

uk sin π 2 + kπ t T , Lemma HT : Hs

0,(0, T) → Hs ,0(0, T),

HTuHs

,0(0,T) = uHs 0,(0,T)

Inverse transformation operator (H−1

T v)(t) = ∞

k=0

vk sin π 2 + kπ t T , v(t) =

∞

k=0

vk cos π 2 + kπ t T

O. Steinbach

Magdeburg, 24.10.2018 14 / 23

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Transformation operator (HTu)(t) =

∞

k=0

uk cos π 2 + kπ t T , u(t) =

∞

k=0

uk sin π 2 + kπ t T , Lemma HT : Hs

0,(0, T) → Hs ,0(0, T),

HTuHs

,0(0,T) = uHs 0,(0,T)

Inverse transformation operator (H−1

T v)(t) = ∞

k=0

vk sin π 2 + kπ t T , v(t) =

∞

k=0

vk cos π 2 + kπ t T Lemma ∂tHTu = −H−1

T ∂tu

O. Steinbach

Magdeburg, 24.10.2018 14 / 23

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Transformation operator (HTu)(t) =

∞

k=0

uk cos π 2 + kπ t T , u(t) =

∞

k=0

uk sin π 2 + kπ t T , Lemma HT : Hs

0,(0, T) → Hs ,0(0, T),

HTuHs

,0(0,T) = uHs 0,(0,T)

Inverse transformation operator (H−1

T v)(t) = ∞

k=0

vk sin π 2 + kπ t T , v(t) =

∞

k=0

vk cos π 2 + kπ t T Lemma ∂tHTu = −H−1

T ∂tu

Lemma: For u ∈ H1/2

0, (0, T) and w ∈ H1/2 ,0 (0, T) we have

HTu, wL2(0,T) = u, H−1

T wL2(0,T)

O. Steinbach

Magdeburg, 24.10.2018 14 / 23

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Corollary ∂tu, HTv(0,T) = HTu, ∂tv(0,T)

O. Steinbach

Magdeburg, 24.10.2018 15 / 23

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Corollary ∂tu, HTv(0,T) = HTu, ∂tv(0,T) Lemma v, HTvL2(0,T) > 0 for all 0 = v ∈ H1/2

0, (0, T)

O. Steinbach

Magdeburg, 24.10.2018 15 / 23

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Corollary ∂tu, HTv(0,T) = HTu, ∂tv(0,T) Lemma v, HTvL2(0,T) > 0 for all 0 = v ∈ H1/2

0, (0, T)

Lemma (HTu)(t) = 1 2T T

1

sin π

2 s−t T

+ 1 sin π

2 s+t T

u(s) ds
O. Steinbach

Magdeburg, 24.10.2018 15 / 23

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Corollary ∂tu, HTv(0,T) = HTu, ∂tv(0,T) Lemma v, HTvL2(0,T) > 0 for all 0 = v ∈ H1/2

0, (0, T)

Lemma (HTu)(t) = 1 2T T

1

sin π

2 s−t T

+ 1 sin π

2 s+t T

u(s) ds

Corollary: For T → ∞ (H∞u)(t) = 1 π ∞ u(s) s − t 2s s + t ds → Hilbert transformation → Construction of optimal test functions [Demkowicz, Gopalakrishnan 2011]

O. Steinbach

Magdeburg, 24.10.2018 15 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Variational formulation: Find u ∈ H1/2

0, (0, T) such that

∂tu, v(0,T) = f , v(0,T) for all v ∈ H1/2

,0 (0, T)

O. Steinbach

Magdeburg, 24.10.2018 16 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Variational formulation: Find u ∈ H1/2

0, (0, T) such that

∂tu, v(0,T) = f , v(0,T) for all v ∈ H1/2

,0 (0, T)

Variational formulation: Find u ∈ H1/2

0, (0, T) such that

∂tu, HTv(0,T) = f , HTv(0,T) for all v ∈ H1/2

0, (0, T)

O. Steinbach

Magdeburg, 24.10.2018 16 / 23

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Model problem u′(t) = f (t) for t ∈ (0, T), u(0) = 0 Variational formulation: Find u ∈ H1/2

0, (0, T) such that

∂tu, v(0,T) = f , v(0,T) for all v ∈ H1/2

,0 (0, T)

Variational formulation: Find u ∈ H1/2

0, (0, T) such that

∂tu, HTv(0,T) = f , HTv(0,T) for all v ∈ H1/2

0, (0, T)

Theorem ∂tu, HTu(0,T) = u2

H1/2

0, (0,T)

for all u ∈ H1/2

0, (0, T)

Corollary uH1/2

0, (0,T) ≤

sup

0=v∈H1/2

,0 (0,T)

∂tu, v(0,T) vH1/2

,0 (0,T)

for all u ∈ H1/2

0, (0, T)

O. Steinbach

Magdeburg, 24.10.2018 16 / 23

SLIDE 38

Galerkin FEM: Find uh ∈ Vh := S1

h(0, T) ∩ H1/2 0, (0, T) such that

∂tuh, HTvh(0,T) = f , HTvh(0,T) for all vh ∈ Vh Linear system Khu = f , Kh = K ⊤

h > 0

Numerical example: u(t) = 8t3 − 8t2 + 2t, t ∈ (0, 1) L N u − uhL2 eoc u′ − u′

hL2

eoc λmin λmax κ2 2 1.949 –1 2.196 0.417 0.960 2.30 1 4 4.366 –2 2.20 1.152 0.93 0.284 1.117 3.93 2 8 1.018 –2 2.10 5.790 –1 0.99 0.169 1.128 6.68 3 16 2.440 –3 2.10 2.893 –1 1.00 0.091 1.133 12.38 4 32 5.964 –4 2.00 1.445 –1 1.00 0.047 1.134 23.89 5 64 1.474 –4 2.00 7.222 –2 1.00 0.024 1.134 46.96

O. Steinbach

Magdeburg, 24.10.2018 17 / 23

SLIDE 39

Dirichlet boundary value problem for the heat equation α∂tu(x, t) − ∆xu(x, t) = f (x, t) for (x, t) ∈ Q := Ω × (0, T), u(x, t) = for (x, t) ∈ Σ := Γ × (0, T), u(x, 0) = for x ∈ Ω. Find u ∈ H1,1/2

0;0,

:= L2(0, T; H1

0(Ω)) × H1/2 0, (0, T; L2(Ω)) such that

T

Ω
α∂tu(x, t)v(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ H1,1/2

0;,0

:= L2(0, T; H1

0(Ω)) × H1/2 ,0 (0, T; L2(Ω)).

O. Steinbach

Magdeburg, 24.10.2018 18 / 23

SLIDE 40

Dirichlet boundary value problem for the heat equation α∂tu(x, t) − ∆xu(x, t) = f (x, t) for (x, t) ∈ Q := Ω × (0, T), u(x, t) = for (x, t) ∈ Σ := Γ × (0, T), u(x, 0) = for x ∈ Ω. Find u ∈ H1,1/2

0;0,

:= L2(0, T; H1

0(Ω)) × H1/2 0, (0, T; L2(Ω)) such that

T

Ω
α∂tu(x, t)v(x, t)+∇xu(x, t)·∇xv(x, t)
dxdt =

T

Ω

f (x, t)v(x, t)dxdt is satisfied for all v ∈ H1,1/2

0;,0

:= L2(0, T; H1

0(Ω)) × H1/2 ,0 (0, T; L2(Ω)).

Theorem 1 2 uH1,1/2

0;0,

(Q) ≤

sup

0=v∈H1,1/2

0;,0

(Q)

a(u, v) vH1,1/2

0;,0

(Q)

O. Steinbach

Magdeburg, 24.10.2018 18 / 23

SLIDE 41

Galerkin FEM: Find uh ∈ Vh := Q1

h(Q) ∩ H1,1/2 0;0, (Q) such that

∂tuh, HTvhQ + ∇xuh, ∇xHTvhL2(Q) = f , HTvhQ for all vh ∈ Vh Numerical example u(x, t) = sin 5π 4 t sin πx for (x, t) ∈ (0, 1) × (0, 2), α = 1 Nx Nt dof hx ht u − uhL2 eoc |u − uh|H1 eoc 2 2 2 0.50000 1.0000 0.91080 4.48436 4 4 12 0.25000 0.5000 0.15774 2.50 1.89083 1.20 8 8 56 0.12500 0.2500 0.02936 2.40 0.84239 1.20 16 16 240 0.06250 0.1250 0.00689 2.10 0.41495 1.00 32 32 992 0.03125 0.0625 0.00169 2.00 0.20679 1.00

O. Steinbach

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SLIDE 42

Heat equation with nonlinear reaction term e.g. Schl¨

gl model, Nagumo equation, Monodomain equation

∂tu − ∆xu + 1

ε(u3 − u) = 0,

Q = (17, 19) × (17, 19) × (0, 5), u(x1, x2, t) = 1 2

1 − tanh x1 − st

2 √ 2ε

, ε = 0.38, s = 3/(

√ 2ε) #Dofs u − uhL2(0,T;H1

0(Ω))

eoc 225 1.35 · 10−0 − 1377 8.23 · 10−1 0.71 9537 4.78 · 10−1 0.78 70785 2.48 · 10−1 0.95 545025 1.19 · 10−1 1.06

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Convergence history

Figure: uniform (− + −), octasection (− ◦ −) , bisection (− ∗ −), ideal (−)

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SLIDE 44

Adaptive refinement in space–time

Figure: Adaptive meshes: octasection (left) and bisection (right).

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SLIDE 45

References

1. M. Neum¨

uller, O. Steinbach: Refinement of flexible space–time finite element meshes and discontinuous Galerkin methods. Comput. Visual. Sci. 14 (2011) 189–205.

2. O. Steinbach: Space–time finite element methods for parabolic problems. Comput.
Meth. Appl. Math. 15 (2015) 551–566.
3. O. Steinbach, H. Yang: An algebraic multigrid method for an adaptive space–time

finite element discretization. Lecture Notes in Comput. Sci., 10665, Springer, Cham,

pp. 66–73, 2018.
4. O. Steinbach, H. Yang: Comparison of algebraic multigrid methods for an adaptive

space–time finite element discretization of the heat equation in 3D and 4D. Numer. Linear Algebra Appl. 25 (2018), no. 3, e2143, 17 pp.

5. O. Steinbach, H. Yang: A space–time finite element method for the linear bidomain

equations, submitted, 2018.

6. O. Steinbach, M. Zank: A stabilized space–time finite element method for the wave

equation, accepted, 2018.

7. O. Steinbach, H. Yang: Space–time finite element methods for parabolic evolution

equations: Discretization, a posteriori error estimation, adaptivity and solution, submitted, 2018.

8. O. Steinbach, M. Zank: Coercive space–time finite element methods for initial

boundary value problems, to be submitted.

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