finite element methods for flow simulations daniel...

Finite Element Methods for Flow Simulations

Daniel Arndt

Heidelberg University

Interdisciplinary Center for Scientific Computing

The Heidelberg Laureate Forum atMATCH

21. September 2016

Introduction Discretization Numerical Results Further Topics

Overview

1 Introduction

2 Discretization

3 Numerical Results

4 Further Topics

21. September 2016 FEM for Flow Simulations 2 Daniel Arndt 2 / 21


Incompressible Navier-Stokes Equations

Momentum Equation - Inertial Frame of Reference

∂t u − ν∆u + (u · ∇)u +∇p = fu∇ · u = 0

Motion of viscuous, incompressible fluids

u velocity

p velocity

ν kinematic viscosity

Parabolic PDE of second order

Non-linear

Challenge: ν → 0

C. Fukushima and J. Westerweel,Technical University of Delft(Wikimedia Commons)



Navier-Stokes Equations - Millenium Problem1

Momentum Equation - Inertial Frame of Reference

∂t u − ν∆u + (u · ∇)u +∇p = fu (1)

∇ · u = 0 (2)

Open question: Existence of a solution satisfying

(1), (2)

u, p ∈ C∞(R3 × [0,∞))

f u ≡ 0∫Rn |u(x, t)|2 dx ≤ C ∀t ≥ 0

for a given u(x, 0) ∈ C∞(R3), ∇ · u(x, 0) = 0.

1Charles L Fefferman. “Existence and smoothness of the Navier-Stokes equation”.In: The millennium prize problems (2006), pp. 57–67



Weak Formulation

Consider a bounded domain Ω and the function spaces

V := [W 1,20 (Ω)]d Q := L2

∗(Ω)

Find (u, p) : (t0, T )→ V × Q such that

(∂t u, v) + cu(u; u, v) + ν(∇u,∇v)− (p,∇ · v) = (fu, v),

(∇ · u, q) = 0

for all (v , q) ∈ V × Q and t ∈ (t0, T ) a.e.

cu(w ; u, v) :=12

[((w · ∇)u, v)− ((w · ∇)v , u)

]Existence X Uniqueness ?



Overview

1 Introduction

2 Discretization

3 Numerical Results

4 Further Topics



Weak Formulation - Conforming Methods

Finite dimensional spaces

Vh := [W 1,20 (Ω)]d Qh ⊂ Q := L2

∗(Ω)

Find (uh, ph) : (t0, T )→ V h × Qh such that

(∂t uh, vh) + cu(uh; uh, vh) + ν(∇uh,∇vh)− (ph,∇ · vh) = (fu, vh),

(∇ · uh, qh) = 0

for all (vh, qh) ∈ Vh × Qh and t ∈ (t0, T ) a.e.

→ Nonlinear finite dimensional problemLinearize−→ Linear algebra

Existence X Uniqueness X

Problem: ∇ · Vh 6⊂ Qh =⇒ ∇ · uh 6= 0



Problems

Dominant convection Shear and boundary layers

Poor mass conservation Large Coriolis forces(non-inertial frame of reference)



Stabilization

(∂t u, v) + cu(u; u, v) + ν(∇u,∇v)− (p,∇ · v) = (fu, v),

(∇ · u, q) = 0

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

=

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

+

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

Idea: Stabilize only small scales

Family of macro decompositions Mh

DM ⊂ [L∞(M)]d finite elemente ansatz space on M ∈Mh.

πM : [L2(M)]d → DM orthogonal L2 projection,κM = Id − πM fluctuation operator

(∇ · uh, q) −→ τu,gd,M(∇ · uh,∇ · vh)M

cu(uh; uh, vh) −→ τu,M(κM((uM · ∇)uh), κM(uM · ∇vh))M



Stabilization

(∂t u, u) + cu(u; u, u) + ν(∇u,∇u)− (p,∇ · u) = (fu, u),

(∇ · u, p) = 0

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

=

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

+

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5





(∇ · uh, q) −→ τu,gd,M(∇ · uh,∇ · vh)M




Stabilization

(∂t u, u) +cu(u; u, u) + ν(∇u,∇u)−(p,∇ · u) = (fu, u),

(∇ · u, p) = 0

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

=

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5

+

0 0.2 0.4 0.6 0.8 1

-1.5

-1

-0.5

0

0.5

1

1.5





(∇ · uh, q) −→ τu,gd,M(∇ · uh,∇ · vh)M




Analytical Results

U := (u, p), Uh := (uh, ph)

|||U |||2Stab := ν‖∇u‖20 +

∑M∈Mh

(γM‖∇ · u‖2

0,M + τM‖κM((uM · ∇)u)‖20,M

)Aims:

suitable choice of stabilization parameters

quasi-optimal error estimates

‖U − Uh‖2l∞(t0,T ;L2(Ω)) + |||U − Uh|||2l2(t0,T ;Stab)

≤ CeCG(T−t0)

(infWh

‖U −Wh‖2

l∞(t0,T ;L2(Ω)) + |||U −Wh|||2l2(t0,T ;Stab)

)semi-robust error estimates C(ν), CG(ν)



Overview

1 Introduction

2 Discretization

3 Numerical Results

4 Further Topics



Analytical Test Case - Setting

u(x, t) =

(− cos

(π2 x)· sin(π2 y) · sin(π · t)

sin(π2 x)· cos(π2 y) · sin(π · t)

)p(x, t) = −π · sin

(π2

x)· sin

(π2

y)· sin(π · t)

‖u‖k+1

‖p‖k= 1 Ω = [−1, 1]2

T = 4 · 10−1 ∆t = 4 · 10−5

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0



Analytical Test Case - Parameter Choice - ν = 10−4

10−4

10−2

100

102

104

10−4

10−3

10−2

10−1

100

101

102

Q2Q1

L2 velocity

H1 velocity

Linfty velocity

L2 div velocity

L2 pressure

H1 pressure

Linfty pressure



Analytical Test Case - Parameter Choice

h

5 ·10-2

10-1

2 ·10-1 5 ·10

-1

H1(u

) e

rro

r

10-6

10-4

10-2

100

102

104

h5 ·10

-210

-12 ·10

-1 5 ·10-1

L2(p

) e

rro

r

10-4

10-3

10-2

10-1

γ = 1 ν = 1

γ = 0 ν = 1

γ = 1 ν = 10−3

γ = 0 ν = 10−3

γ = 1 ν = 10−6

γ = 0 ν = 10−6

h2



Blasius Flow, Flow over Horizontal Plate

Prandtl’s Boundary Layer Equation

u = u∞f ′(η)

2f ′′′(η) + f (η)f ′′(η) = 0

f (0) = f ′(0) = 0, f ′(∞) = 1

where η = y√

u∞/(νx) is a non-dimensional variable.



Blasius Flow

η=y (u∞

/( ν x))1/2

0 2 4 6 8

u/u

∞

0

0.2

0.4

0.6

0.8

1

1.2

Blasius profilex=0.1x=0.5x=0.9

ν = 10−3, γ = 1, τM = 0, h = 2−5



Blasius Flow - ν = 10−3

Adaptively refined mesh, τM = 0 Globally refined mesh, τM = 1



Overview

1 Introduction

2 Discretization

3 Numerical Results

4 Further Topics



Further Topics

Not covered today

Time discretization

Linear algebra

Fast solverMultigridMatrix-Free methods

Implementation⇒ deal.II

Turbulence modeling

MHD, nonisothermal flow, . . .

k

100

101

102

E

10-4

10-3

10-2

10-1

100

τM = h/(2‖uh‖∞,M )τM = 1/(2‖uh‖∞,M )τM = 1

k−5/3


Thanks for your attention!

Rayleigh-Bénard Convection

Ra = 105 Ra = 107 Ra = 109

T = 1000, Pr = 0.786, N = 10 · 163, γM = 0.1

finite element methods for flow simulations daniel...

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