A conservative FE-discretisation of the Navier-Stokes equation

A conservative FE-discretisation of the Navier-Stokes equation

JASS 2005, St. PetersburgThomas Satzger

2

Overview

• Navier-Stokes-Equation– Interpretation– Laws of conservation

• Basic Ideas of FD, FE, FV

• Conservative FE-discretisation of Navier-Stokes-Equation

3

Navier-Stokes-Equation

• The Navier-Stokes-Equation is mostly used for the numerical simulation of fluids.

• Some examples are- Flow in pipes- Flow in rivers- Aerodynamics- Hydrodynamics

4


The Navier-Stokes-Equation writes:

gupuuut

1

Equation of momentum

0 u Continuity equation

with u Velocity field

Pressure fieldp

Density

Dynamic viscosity

5


The interpretation of these terms are:

gupuuut

1

Derivative of velocity field

Convection

Pressure gradientDiffusion

Outer Forces

6


The corresponding for the components is:

1122

2

121

2

11

221

111

1 gux

ux

px

ux

uux

uut

2222

2

221

2

22

222

112

1 gux

ux

px

ux

uux

uut

022

11

u

xu

x

for the momentum equation, and

for the continuity equation.

7


With the Einstein summation jiji

m

jjiji xayxay

1

and the abbreviation we get:tt

i

i x

iijjiijjii gupuuu

1

)2,1( i

0 jju

for the momentum equation, and

for the continuity equation.

8


Now take a short look to the dimensions:

iijjiijjii gupuuu

1

2sm

2

1sm

ssm

2

1sm

msm

sm

22

13 s

mmsm

kgm

mkg

22

13 s

mms

m

mkgsm

kg

9

Navier-Stokes-Equation - Interpretation

We see that the momentum equations handles with accelerations. If we rewrite the equation, we get:

iijjiijjii gupuuu

1

This means:

Total acceleration is the sum of the partial accelerations.

10


Interpretation of the Convection

1x

2x

fluid particle

iiu pi

1ijj u

igijj uu

Transport of kinetic energy by moving the fluid particle

11


Interpretation of the pressure Gradient

1x

2x

fluid particle

iiu ijj uu ijj u

igpi

1

Acceleration of the fluid particle by pressure forces

12


Interpretation of the Diffusion

iiu ijj uu pi

1igijj u

1x

2x

fluid particle

Distributing of kinetic Energy by friction

13


Interpretation of the continuity equation 0 jju

• Conservation of mass in arbitrary domain

h

h

11u 12u

21u

22u

this means: influx = out flux

huhu 11 21 huhu

22 21

01212 2211

huu

huu

for 0h we get

02211 uu

14

Navier-Stokes-Equation - Laws of conservation

Conservation of kinetic energy:

We must know that the kinetic energy doesn't increase, this means:

Proof:

duuduE iikin 212

21

duu

duuuuduuduudtdE

dtd

iti

iititiiitiikin 21

21

21

0kinEdtd

15


With the momentum equation it holds

Using the relations (proof with the continuity equation)

and

0Re1

puuuu iijjijjii

puuuuuu

puuuuduuEdtd

iiijjiijji

iijjijjiitikin

Re1

Re1

ijjiji uuuu

ijjiijijijji

ijijijii

product

ruleijii

Gauß

ii

uuuuuuuuu

uuuuuuuuuuuu

2

0

16


Additionally it holds

Therefore we get

Due to Greens identity we have

00

Re1

puuuuuuE

dtd

iiijjiijjikin

jj

continuity

equationjjjj

product

rulejj

Gauß

upupuppuup0

0

0 ijij

Green

ijji uuuu

17


This means in total

We have also seen that the continuity equation is very important for energy conservation.

0kinEdtd

18

Basic Ideas of FD, FE, FV

We can solve the Navier-Stokes-Equations only numerically.Therefore we must discretise our domain. This means, we regard our Problem only at finite many points.There are several methods to do it:

•Finite Difference (FD)One replace the differential operator with the difference operator, this mean you approximate by

or an similar expression.

f

h

xfhxfxf

19


• Finite Volume (FV)- You divide the domain in disjoint subdomains- Rewrite the PDE by Gauß theorem- Couple the subdomains by the flux over the

boundary• Finite Elements (FE)

- You divide the domain in disjoint subdomains- Rewrite the PDE in an equivalent variational

problem- The solution of the PDE is the solution of the

variational problem

20


Comparison of FD, FE and FV

Finite Difference

Finite Element

Finite Volume

21


Advantages and DisadvantagesFinite Difference:

+ easy to programme- no local mesh refinement- only for simple geometries

Finite Volume:+ local mesh refinement+ also suitable for difficult geometries

Finite Element:+ local mesh refinement+ good for all geometries

BUT:Conservation laws aren't always complied by the discretisation. This can lead to problems in stability of the solution.

22

Conservative FE-Elements

We use a partially staggered grid for our discretisation.

h

h

u

v

u

v

u

v

u

vp

We write: N for the number of grid pointsiu

ivfor the horizontal velocity in the i-th grid pointfor the vertical velocity in the i-th grid point

23


The FE-approximation is an element of an finite-dimensional function space with the basis

The approximation has the representation

Nff 21,..., 2: IRfi whereby

N

iNiiiih fvfuu

1

24


If we use a Nodal basis, this means

we can rewrite the approximation

01

point grid th-iif

0

0points gridother allNif

1

0point grid th-iNif

01

points gridother allif

and

and

N

i vNi

uNii

vi

uii

h

h

ff

vff

uvu

1 ,

,

,

,

25


Every approximation should have the following properties:continuousconservative

In the continuous case the continuity equation was very important for the conservation of mass and energy.

If the approximation complies the continuity pointwise in the whole area, e.g. , then the approximation preserves energy.

hu

0 hu

26


Now we search for a conservative interpolation for the velocities in a box.

We also assume that the velocities complies the discrete continuity equation.

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

27




1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

hvvhuu

22

4331

28




1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

hvvhuu

22

4331 hvvhuu

22

2142

29



We also assume that the velocities complies the discrete continuity equation:

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

hvvhuu

22

4331 hvvhuu

22

2142

043314321 vvvvuuuu (1)

30


The bilinear interpolation isn't conservative1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

1 2 3 4

1 2 3 4

, 1 1 1 1

, 1 1 1 1

h

h

x y x y x y x yu x y u u u u

h h h h h h h hx y x y x y x y

v x y v v v vh h h h h h h h

31



3 3, u v 4 4, u v

h

h

1 2 3 4

1 2 3 4

, 1 1 1 1

, 1 1 1 1

h

h

x y x y x y x yu x y u u u uh h h h h h h hx y x y x y x y

v x y v v v vh h h h h h h h

It is easy to show that

0general in

hh vy

ux

32



3 3, u v 4 4, u v

h

h

, , ,,1, 3, 4,2,

,2 4,,1 4,

2 3 4

2

3

1

1 3

, 1 1 1 1

, 1 1 1

f x y f x y f x yf x yu u uu

f x yvxv

h

y

h

f f

x y x y x y x yu x y u u u uh h h h h h h hx y x y x yv x y v v vh h h h h h

4

, ,4, 4 4,

1

x y f x yv v

x yvh h

Basis on the box

33


These basis function for the bilinear interpolation are calledPagoden.

The picture shows the function on the whole support.

, i iu v

h

h

34


Now we are searching a interpolation of the velocities which complies the continuity equation on the box.

How can we construct such an interpolation?

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

35




1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

Divide the box in four triangles.

36




1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

Divide the box in four triangles.Make on every triangle an linear interpolation.

37


What's the right velocity in the middle?

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

38



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

We must have at every point in the box the following relations:

, , 0 , ,

, ,

u x y v x y u x y v x yx y x y

v x y u x yy x

39




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

40




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

41




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux 4

ux

42




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux 4

ux

3vy

43




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

44




, , 0 , ,

, ,


v x y u x yy x

1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

45



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

5 33u u

46



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 35 33 2

u uhu uh

47



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 33 2 2

u u v vh hu uh h

48



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33

1 12 2 2 2u u v vh hu u u u u v v u u v vh h

49



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33


5 42u u

50



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33


2 45 42 2

h u uu u

h

51



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33


2 4 2 45 42 2 2

h u u h v vu u

h h

52



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33


2 4 2 45 4 4 2 4 2 4 2 4 2 42

1 12 2 2 2h u u h v v

u u u u u v v u u v vh h

53



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

1 3 1 35 3 3 1 3 1 3 1 3 1 33


2 4 2 45 4 4 2 4 2 4 2 4 2 42

1 12 2 2 2h u u h v v

u u u u u v v u u v vh h

free are and

41 55 uu

54



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

4 3 4 35 3 3 4 3 4 3 3 4 3 44

1 12 2 2 2v v u uh hv v v v v u u u u v vh h

55



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

51u52u53u

54u

51v52v53v

54v

1ux

1ux

4ux

4ux

3vy

3

vy

2

vy

2

vy

4 3 4 35 3 3 4 3 4 3 3 4 3 44

1 12 2 2 2v v u uh hv v v v v u u u u v vh h

2 1 2 15 1 1 2 1 2 1 1 2 1 21

1 12 2 2 2h v v h u u

v v v v v u u u u v vh h

free are and

23 55 uv

56


Till now we have: 5 1 3 1 33

12

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112

v u u v v free are and 23 55 uv

free are and 41 55 uu

57


Till now we have:

With the discrete continuity equation

we get

5 1 3 1 3312

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112



1 2 3 4 1 2 3 4 0 (1)u u u u v v v v

5 53 2u u 5 54 1v v

58


Till now we have:

With the discrete continuity equation

we get

Therefore we choose

5 1 3 1 3312

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112



1 2 3 4 1 2 3 4 0 (1)u u u u v v v v

5 53 2u u 5 54 1v v

5 5 5 5 5 1 2 3 4 1 2 3 41 2 3 4

5 5 5 5 5 1 2 3 4 1 2 3 41 2 3 4

1: : : : :4

1: : : : :

4

u u u u u u u u u v v v v

v v v v v u u u u v v v v

59



1u 2u

3u 4u

h

h

1v 2v

3v 4v

h

h

1 2

3 4

1 2

3 4

5u 5v

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

60


Now we calculate the basis., ,

, ,1

Ni u i N uh

i ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

5 5, u v

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

61


Now we calculate the basis.

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

5 5, u v

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

, ,

, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1i i

62



, ,1i

Ni u i N uh

ii i N vih v

f fu

uv

vf f

1 1, u v 2 2, u v

3 3, u v 4 4, u v

h

h

5 5, u v

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

h

h

h

h

,i uf ,i vf

0 0 0

0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

63


Now we calculate the basis.

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

, ,

, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

3 3, u v 44, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

64



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

3 3, u v 44, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

45 1 2 3 1 2 3 4

5 1 2 3 4 1 2 3 4

:

1:4

14

u u u u v v v v

v

u

u u u u v v v v

65



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

3 3, u v 44, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

45 1 2 3 1 2 3 4

5 1 2 3 4 1 2 3 4

:

1:4

14

u u u u v v v v

v

u

u u u u v v v v

14

66



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

33, u v 4 4, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

14

67



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

33, u v 4 4, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

14

5 1 2 4 1 2 3 4

5 1 2 3 4 1 2

3

3 4

:

1

1

:

4

4

u u u u v v v v

v u u u u

u

v v v v

68



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

1 1, u v 2 2, u v

33, u v 4 4, u v

h

h

5 5, u v

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

14

5 1 2 4 1 2 3 4

5 1 2 3 4 1 2

3

3 4

:

1

1

:

4

4

u u u u v v v v

v u u u u

u

v v v v

14

69



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

5 1 3 4 1 2 3 4

5 1 2 3 4 1 4

2

2 3

:

1

1

:

4

4

u u u u v v v v

v u u

u

u u v v v v

1 1, u v 22, u v

3 3, u v 4 4, u v

h

h

5 5, u v0

0

h

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

14

14

14

70



, ,1

hi

Ni u i N u

ii v i N vh i

f fuu v

f fv

5 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:

1

1

:

4

4

u u u u v v v v

v u u u v

u

u v v v

11, u v 2 2, u v

3 3, u v 4 4, u v

h

h

5 5, u v

0

h

h

h

h

h

,i uf ,i vf

0 0 0

0 0

0 0 0

1

0 0 0

0 0

0 0 0

0i i

14

14

14

14

71



, ,1i

Ni u i N uh

ii i N vih v

f fu

uv

vf f

5 1 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

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72



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73



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79


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Linear interpolation providesthe basis.

80


View on conservative elements in 3D

81


View on conservative elements in 3D

Partially staggered grid in 3D

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

p

hh

h

82


We also search for a conservative interpolation of the velocities.

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

p

hh

h

83


We also search for a conservative interpolation of the velocities.

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

, , u v w

hh

h

Divide every box into 24 tetrahedrons, on which you make a linear interpolation

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A conservative FE-discretisation of the Navier-Stokes equation