metro metallurgical training on-linefluid.ippt.pan.pl/metro/cdrom-pl/kursy/metro-pdf-en/... ·...

METRO MEtallurgical TRaining On-line

Education and Culture

Spatial discretization techniques

Arkadiusz NagórkaCzUT

2METRO – MEtallurgical TRaining On-line Copyright © 2005 Arkadiusz Nagórka - CzUT

Introduction

• Many physical phenomena are described by partial differential equations (PDEs).

• Analytical solutions are impossible to obtain except for linear equations on simple geometries.

• Since the computer memory is limited, discretization of the problem is necessary. Numerical methods can give approximate solutions of PDEs.

• Spatial discretization - obtain the solution in a set of points rather that in the entire domain.


Discretization of a simple problem

Simple one-dimensional Poisson equation

with Dirichlet boundary conditions

will be solved numerically using– the finite difference method (FDM),– the finite element method (FEM),– the finite volume method (FVM).

10 ,222


Solution of a simple problem using the finite difference method

• Introduce a grid

0xa = 1x 2x 3x 4x 5x 1−Nx bxN =

h

1,1,0 ,1 −==−= + Nihxxh iii K

• Approximate the differential equation at each grid point• Solve the resulting system of algebraic equations


Solution of a simple problem using FDM, cont.

hxuhxu

dxdu

h

)()(lim0

−+=

→• By definition the derivative

ix• Finite difference approximation at point

huu

hxuhxu

dxdu iiii

i

−=

−+≈⎟

⎠⎞

⎜⎝⎛ +1)()( forward difference

huu

dxdu ii

i

1−−≈⎟⎠⎞

⎜⎝⎛

huu

dxdu ii

i 211 −+ −≈⎟

⎠⎞

⎜⎝⎛

backward difference central difference



• Forward, backward and central differences

• Central difference is the most accurate1−ix ix 1+ix

exact derivative

backwardforward

central

x

u



• Approximation of second derivative

hxdxduhxdxdu

dxdu

dxd

dxud ))(/())(/(2

2 −+≈⎟

⎠⎞

⎜⎝⎛=

• Use central differences for dxdu /

21111

2

2 21h

uuuhuu

huu

hdxud iiiiiii +−−+ +−=⎥⎦

⎤⎢⎣⎡ −−

−≈



• In our problem assume three grid points

0.5h ,1 ,5.0 ,0 210 ==== xxx

22 2210 =

+−h

uuu• Difference equation for the point 1x

25.0

0202

1 =+− u

or

• One equation is needed for one unknown

25.01 −=u• The solution is


Solution of a simple problem using the finite element method

Subdivide the domain into finite elements (e.g. line segments in 1D, triangles or quadrilaterals in 2D, tetrahedra in 3D)

)( 0e )( 1−Ne

h

ax = bx =

0x 1−Nx Nx)( 1e1x )( 2e2x


Solution of a simple problem using FEM, cont.

xxu h 21)0( )( αα +=

• Assume approximating function in each element, e.g. a linear polynomial

for the element 0

• Parameters jα can be determined easily

021)0( 0)0( uu h =+= αα

121)0( )( uhhu h =+= αα , hence ,01 u=α h

uu 012

−=α

110010)0( )()(1)( uxNuxNu

hxu

hxxu h +=+⎟⎠⎞

⎜⎝⎛ −=

• Substituting yields



The function is a trial function. It has the property)( xN j

⎩⎨⎧

=elsewhere

at 01

)( jjx

xNj

N

jjh uxNxu ∑

=

=0

)()(Then

2N 3N

0x Nx

1

1x 2x 3x 1−Nx



0x 1x 2x 3x 1−Nx Nx

3N2N

Trial function is called shape function when restricted to a finite element

)( 2e

2)2(

0 NN =3

)2(1 NN =

2x 3x x

trial functions

shape functions

1



Obtain an integral form of the problem, for example using the weighted residual method

2)( 22

−=dxudxR hMake the residual “vanish”, e.g.

,0)()(∫Ω

=ΩdxRxW )( xW is a test or weighting function

is a possible choice. Then)()( xNxW =

,N,,idxRxN i K10 ,0)()( ==Ω∫Ω

Galerkin finite element method



Assumed solution is piecewise linear and doesnot have second-order derivative. Integrate by parts

∫∫ +⎟⎠⎞

⎜⎝⎛ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

b

a

b

a

hii

hib

a

hi dx

duNdxNdxdu

dxdNdx

dxudN 222

2

Now only the first derivative is necessary

)1()( and 1)( −⋅⋅dxadu

dxbdu hh are normal fluxes nq

imposed in Neumann boudary conditions



• The weak form of the problem reads

( ) ( ) ( )∫ ∫ ++−=b

anini

b

ai

hi bqNaqNdxNdxdxdu

dxdN 2

,N,,i K10=

• Derivative of the solution is also approximated

j

N

j

jh udxdN

dxdu ∑

=

=0

j

N

jjh uNu ∑

=

=0



Ni ,,1,0 K=Substituting and shifting sum gives for

( ) ( ) ( )bqNaqNdxNudxdxdN

dxdN

nini

b

aij

N

j

b

a

ji ++−=⎟⎟⎠

⎞⎜⎜⎝

⎛∫∑ ∫

=

20

which is a system of linear algebraic equations

∑=

=+=N

jiijij Nigfuk

0,,1,0 , K or gfKu +=

- stiffness matrix- source vector

K ug

- unknown nodal values- flux vectorf



• Integrals can be evaluated by summing contributions from individual elements

( ) ( )∑ ∫∫ΩΩ

Ω⋅=Ω⋅elements

e

dd

• Integrals involving are nonzero only in elements containing the node , e.g.

iNi

( ) ( ) ( )∫∫∫+Ω

+

ΩΩ

Ω−+Ω−=Ω−1

)1(0

)(1 222

ii

dNdNdN iii



This leads to a local system of equations for each element e

)()()()( eeee gfuK +=

∫Ω

==e

jidxdxdN

dxdNk

ej

eie

ij 1,0, ,)()(

)(

( ) dxNfe

ei

ei ∫

Ω

−= )()( 2

nei

ei qNg

)()( = (only in elements with Neumann boundary condition)



• The element integrals can be evaluated easily

⎥⎦

⎤⎢⎣

⎡−

−=

11111)(

heK ⎥

⎦

⎤⎢⎣

⎡−=

11)( hef

• Assuming the mesh with 3 nodes and 2 elements, the local systems are

⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−11

11111 ,

11

11111

)1(1

)1(0

)0(1

)0(0 h

uu

hh

uu

h



Using global indices and augmenting gives

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

011

000011011

1

2

1

0

huuu

hfor element (0)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

110

110110

0001

2

1

0

huuu

hfor element (1)



• Assemble the global system by summing up local systems

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−

−

111

1

1101111

0111

2

1

0

huuu

h

• Local contributions are overlapped (superimposed)



• In our example 020 == uu

• Equation for the node 1

5.025.0

2 210 ⋅−=−+− uuu hence 25.01 −=u

• The solution is identical to that of FDM (for this simple problem and geometry)


Solution of a simple problem using the finite volume method

• Partial differential equations that we solve express conservation of a quantity (energy, mass,etc.)

• Conservation equations can be written in integral form, e.g. for the Poisson equation

∫∫ =1

0

1

02

2

2dxdxdxud

• Change the volume integral to a surface integral

Balance: flux of a quantity entering and leaving the domain equals the amount produced internally by the source∫=

1

0

1

0

2dxdxdu


Solution of a simple problem using FVM, cont.

The method utilises control volumes (finite volumes) and control surfaces

ax =0 bxN =1x 2x

CV0 CV1 CV2

h

CS0 CS1 CS2

CV1 around node 1 has control surfaces CS0 and CS1. Control surfaces are located at line midpoints.



Balance equation is valid for each control volume

∫=0

0

2CV

CS

a

dxdxdu for control volume 0

∫=1

1

0

2CV

CS

CS

dxdxdu

for control volume 1

∫=2

2

1

2CV

CS

CS

dxdxdu

for control volume 2



• Flux is evaluated at control surfaces using e.g. finite differences

• By summing up equations for control volumes we obtain the global equation

b

a

b

CS

CS

CS

CS

CS

CS

a dxdu

dxdu

dxdu

dxdu

dxdu

N

=+++−1

2

1

1

0

0

K

∫∫ ∫∫Ω

⋅=⋅++⋅+⋅ dxdxdxdxVC VCVC N

)()()()(10

K

(since fluxes through control surfaces cancel out)



• Flux is conserved between CV0 and CV1 through CS0 and between CV1 and CV2 through CS1 etc. provided approximation of du/dx is the same on both sides

Automatic conservation of a physical quantity is a distinct feature of FVM

• Flux at the control surface can be approximated in many ways, e.g.

huu

dxdu ii

CSx i

−≈ +

=

1



• Assume three nodes and three control volumes for our problem. Balance equation for the volume 1 is

∫∫ =−⇒=== 1011

1

0

22CVCSxCSxCV

CS

CS

dxdxdu

dxdudx

dxdu

• Approximate derivatives with finite differences

hh

uuuhhuu

huu 222 2100112 =+−⇒=−−−

• The solution is the same as in FDM and FEM (for this simple problem and geometry)


Imposing the Neumann condition

Consider again the Poisson equation

10 ,2)(22


Imposing the Neumann condition in FDM

• One approach is to add a fictitious node on the right

1−Nx2−Nx Nx 1+Nx

extra nodenqdxdu =/

),(2 211

NNNN xf

huuu

=+− +−

nNN q

huu

=− −+2

11

• By eliminating we obtain the equation for the node N

hqxf

huu n

NNN −=

−− )(21

21

1Nu +


Imposing the Neumann condition in FDM, cont.

• Us usual, assume 3 nodes in our problem

• Difference equation for the internal node 1

22 2210 =

+−h

uuu 25.0

202

21 =+− uuor

• Equation for the node on the Neumann boundary

5.01

22

5.0 221 −=

−uu

• The solution is 0 ,25.0 21 =−= uu


Imposing the Neumann condition in FEM

• Neumann condition apears “naturally” in FEM equations as the result of integration by parts

• Since u0 is known (Dirichlet b.c.) it can be eliminated from the system of equations

• In our two-element mesh only the last element (e1) contributes to the flux vector g

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−10

12

11121

2

1 huu

h

11)(2 =⋅= nn qqbN

• The solution is the same like in FDM (for this problem)


Imposing the Neumann condition in FVM

• Neumann data qn appears explicitly in the balance equation for the boundary volume

∫∫ =−⇒== 212

221 CVCSx

nCV

b

CS

dxdxduqdx

dxdu

• The system of equations is then

,20112 hhuu

huu

=−

−−

2212 h

huuqn =

−−

• The solution is the same as in FDM and FEM (for this problem)


FDM, FEM and FVM for a 1D equation - conclusions

• FDM is the easiest to understand.

• Derivation of FEM equations is tedious.

• Only FVM has inherent conservative property.

• Neumann conditions are approximated in FDM, but appear directly in the formulation in FEM and FVM.

• All the methods gave identical solution. However, this happens only for simple problems and geometries.


Two-dimensional problems

Consider the Poisson equation

Ωf(x,y)ukuk in yyxx

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

−⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

−

with boundary conditions

DDuu Ω= on (Dirichlet)

Nnyx qnyun

xuk

nuk Ω=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−=∂∂

− on (Neumann)

Ωf(x,y)uuk in yx 22

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−

For constant k

DΓ

NΓn

Ω


The finite difference method in 2D

Each grid point has two indices

xi hixx 0 +=

xh

ji, ji ,1+ji ,1−

1, −ji

1, +ji 1,1 ++ ji1,1 +− ji

1,1 −− ji 1,1 −+ ji

yj hjyy 0 +=yh

( )jiji yxuu ,, =


The finite difference method in 2D, cont.

Finite difference formulas used for 1D problems can be used to approximate the partial derivatives at i,j, e.g.

x

jiji

h huu

hyxuyhxu

xu ,,1

0

),(),(lim−

≈−+

=∂∂ +

→

y

jiji

h huu

hyxuhyxu

yu ,1,

0

),(),(lim−

≈−+

=∂∂ +

→

(forward differences)



Approximation of partial derivatives

x

jiji

ji h

ukukuk ,2/1,2/1

,

xxxx

−+

⎟⎠⎞

⎜⎝⎛

∂∂

−⎟⎠⎞

⎜⎝⎛

∂∂

≈⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

∂∂ (central difference)

x

x

jijiji

x

jijiji

ji hhuu

khuu

kuk

,1,,2/1

,,1,2/1

,xx

−−

++

−−

−

≈⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

(central differences)

⎟⎠⎞

⎜⎝⎛ +=+ jxiji yhxkk ,2

1,2/1 evaluated at the midpoint



• Approximation of the Poisson equation at i,j

Kx

x

jijiji

x

jijiji

hhuu

khuu

k ,1,,2/1,,1

,2/1−

−+

+

−−

−

−

jiy

y

jijiji

y

jijiji

fh

huu

khuu

k

,

1,,2/1,

,1,2/1,

=

−−

−

−

−−

++

for each grid point except on the boundary

ji,• Five unknowns per an equation



• FDM is the best suited for structured grids, where it is possible for the grid point i,j to

– compute its coordinates,– determine its neighbours.

• In a structured grid all internal the nodes – have the same number of neighbours,– have the same number of cells around them.

• A grid point is located at intersection of two lines, each belonging to one of the two families of lines.

• The lines are not limited to horizontal/vertical.


Structured and unstructured grids

Unstructured grids are more flexible but connectivitymust be given explicitly

112

3

4

109

11

(10)(1)

(2)(3)

(4)

(5)(7)

(8) (9)(6)

connectivity6elem (1): 5 7 10 11elem (2): 11 6 5elem (3): 4 6 11

etc.

5

7

8 213

structured grid(can compute indices

of element nodes)

unstructured grid(no pattern in numbering)


The finite element method in 2D, cont.

• Finite element method shows its advantages on unstructured grids and complex geometries

• A great variety of shapes can be used

1D

2D

3D



• Derivation of the weak problem and finite element equations is similiar to that for 1D

• The weak form

NidsqNfdNdyu

yN

xu

xNk niiii ,,1,0 , K=−Ω=Ω⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

∂∂

∫∫∫ΓΩΩ

• Approximate solution in a triangular element (e) in terms of shape functions and nodal values

( )∑= ∂

∂=

∂∂ 2

0

)()()( ,

j

ej

ej

eh u

xyxN

xu( )∑

=

=2

0

)()()( ,j

ej

ej

eh uyxNu



• Finite element equations for the element (e)

∫∫∫Ω∂∩ΓΩΩ

−Ω=Ω⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

∂∂

+∂

∂∂

∂

eNee

dsqNfdNdyN

yN

xN

xNk n

ei

ei

ej

ei

ej

ei )()(

)()()()(

2,1,0, =ji

• It is a local system of equation. The size equals the number of nodes in the element (3 - triangle, 4 -quadrilateral, 6 - a second-order triangle)

• As usual, the global system of equations is assembled from such local systems



• Numerical integration is used for evaluation of volume and surface integrals

• It is much easier to integrate over regular triangles or squares. This involves mapping

y

1

2

x1

2

3η

(0,0)

eΩ (0,1)

3

ξ(1,0)

[ ] ξηηξηξηξξ

ddyxfdyxfe

),(),(),,(),(1

0

1

0

J∫ ∫∫−

Ω

=Ω


Shape functions of the linear triangular element

),( 00 yx

),( 11 yx

),( 22 yx

),( yx1A 0A

2A

• Area coordinates

AAyxLAAyxLAAyxL

22

11

00

),(),(),(

===

A - element area

⎩⎨⎧

=nodesother at 0

),( node at the1 iii

yxL

• Shape functions are the area coordinates2,1,0 , == iLN ii


Coordinate mapping for the linear triangular element

ξ

1

2

3x

12

3η

(0,0) (1,0)

(0,1)eΩy

ηξ

ηξ

====

−−==

22

11

00 1

LNLNLN

ηξηξηξηξ)()(),(

)()(),(

02010

02010

yyyyyyxxxxxx−+−+=−+−+=

const ,const =∂∂

=∂∂

yN

xN iiA2=J


The finite volume method in 2D

• Derivation of the balance equation is similiar to that for 1D

• Invoking the divergence theorem

( )dsqdsnunukdukuk nyx ∫∫∫ΓΓΩ

−=⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

=Ω⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

∂∂

xxyyxx

• Integral conservation equation for the Poisson equation is simply

∫∫ΩΓ

Ω= fddsqn


The finite volume method in 2D, cont.

Dual mesh is the mesh of finite volumes around nodes of a mesh of triangles or quadrilaterals

Voronoi region(midpoints and circumcenters)

Median dual element(midpoints and centers of gravity)



Balance equation for a finite volume around node i

i

CV 3CS

2CS

4CS

1CS

5CS

n

∫∫ΩΩ∂

Ω=ii

fddsqn

∫∑ ∫ Ω== CVi CS

n fddsqi,...2,1

∫∑ ∫ Ω=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−= CVi CS

yx fddsnyun

xuk

i,...2,1



• Integrals are evaluated numerically.

)Volume( ),( CSyxffd iiCV

⋅≈Ω∫

)(Area)middle( CSqdsq nCS

n ⋅≈∫

• Interpolation of partial derivatives in the middle of the control surface

– central finite differences– using finite element shape functions

Spatial discretization techniquesIntroductionDiscretization of a simple problemSolution of a simple problem using the finite difference methodSolution of a simple problem using FDM, cont.Solution of a simple problem using FDM, cont.Solution of a simple problem using FDM, cont.Solution of a simple problem using FDM, cont.Solution of a simple problem using the finite element methodSolution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.Solution of a simple problem using FEM, cont.

metro metallurgical training on-linefluid.ippt.pan.pl/metro/cdrom-pl/kursy/metro-pdf-en/... ·...

Documents