CE 341/441 - Lecture 14 - Fall 2004

LECTURE

PRACTIC ATIONS

Derivatives

Consider

where i

Proceed i

Now eval

gp. 14.1

14

AL ISSUES IN APPLYING FINITE DIFFERENCE APPROXIM

of Variable Coefficients

the term

s a coefficient which is a function of

n steps. First let:

uate:

ddx------ g x( )

df x( )dx-------------

x

u g x( )dfdx------

dudx------ ui

1( )=

CE 341/441 - Lecture 14 - Fall 2004

Now evalp. 14.2

uate and

hi

i-1 ii-1/2 i+1/2 i+1

hi+1

x x

ui1( )

ui 12---+

ui 12---

12--- hi hi 1++( )------------------------------=

ui 12---+

ui 12---

ui 12---+

gi 12---+

fi 12---+

1( ) gi 12---+

f i 1+ f ihi 1+

----------------------

= =

ui 12---

gi 12---

fi 12---

1( ) gi 12---

f i f i 1hi

----------------------

= =

CE 341/441 - Lecture 14 - Fall 2004

Hence

where

When

Example

Fluids

Solids

h

h =p. 14.3

, then the approximation reduces to:

Applications with variable coefficients:

: spatially varying viscosities/diffusivities

: spatially varying elasticities

ui1( ) f i 1+

gi 1/2+havghi 1+----------------------- f i

gi 1/2+havghi 1+-----------------------

gi 1/2havghi----------------+ f i 1

gi 1/2havghi----------------+=

avghi hi 1++

2----------------------

hi 1+ hi=

ui1( ) 1

h2----- f i 1+ gi 12---+

f i gi 12---+g

i 12---+ f i 1 gi 12---+=

CE 341/441 - Lecture 14 - Fall 2004

Two Dimen

Define no

= sp

= sp

i

jp. 14.4

sional Finite Difference Approximations

des in a two-dimensional plane with

atial index in the -direction

atial index in the -direction

x

y

fi , j+2i, j+2 i+2, j+2

f

i, j+1

i, j-1

i, j-2

fi+2, jy x

i-2, j i-1, j i, j i+1, j i+2, jx

fi , j

fi+2 , j+2y

CE 341/441 - Lecture 14 - Fall 2004

When tak

and

Similarlyp. 14.5

ing partial derivatives w.r.t. , we hold constantx y

fx------ i j,

f i 1 j,+ f i 1 j,2x-------------------------------------=

2 fx2---------

i j,

f i 1 j,+ 2 f i j, f i 1 j,+x2

--------------------------------------------------------=

2 fy2---------

i j,

f i j 1+, 2 f i j, f i j 1,+y2

--------------------------------------------------------=

CE 341/441 - Lecture 14 - Fall 2004

Example 1 Consider

This mayp. 14.6

Laplaces equation and :

be viewed as follows in terms of a molecule:

x y h= =

f2 i j, 2 f

x2---------

2 fy2---------+=

2 f i j,1h2----- f i 1 j,+ f i 1 j, f i j 1+, f i j 1, 4 f i j,+ + +( )=

fi , j+1

fi-1, j fi, j fi+1, j

fi , j-1

1

1 1

1

-41h2

CE 341/441 - Lecture 14 - Fall 2004

Example 2 Lets cons

Draw a m

Obtain 4

x-----p. 14.7

ider:

olecule diagram for the 2nd order central difference formulation.

by finding

4 f 2 2 f( ) 4 f

x4--------- 2

4 fx2y2-----------------

4 fy4---------+ += =

f4----

2

x2--------

2 fx2---------

4 fx4---------

2

x2--------

2 fx2---------

=

4 fx4---------

2

x2--------

1x2--------- f i 1 j,+ 2 f i j, f i 1 j,+( ) =

CE 341/441 - Lecture 14 - Fall 2004p. 14.8

4 fx4---------

1x2---------

2

x2-------- f i 1 j,+( ) 2

2

x2-------- f i j,( )

2

x2-------- f i 1 j,( )+

=

4 fx4---------

1x2---------

1x2--------- f i 2 j,+ 2 f i 1+ j, f i j,+( )=

2x2--------- f i 1 j,+ 2 f i j, f i 1 j,+( )

1x2--------- f i j, 2 f i 1 j, f i 2 j,+( )+

4 fx4---------

1x4--------- f i 2 j, 4 f i 1 j, 6 f i j, 4 f i 1 j,+ f i 2 j,+++( )=

1 -4 6 1-41x4

CE 341/441 - Lecture 14 - Fall 2004

Similarlyp. 14.9

:

4 fdy4---------

1y4--------- f i j 2, 4 f i j 1, 6 f i j, 4 f i j 1+, f i j 2+,++( )=

1y4

1

-4

6

-4

1

CE 341/441 - Lecture 14 - Fall 2004

Evaluatiop. 14.10

n of the mixed derivative:

4 fx2y2-----------------

2

y2--------

2 fx2---------

=

4 fx2y2-----------------

2

y2--------

1x2--------- f i 1 j, 2 f i j, f i 1 j,++( ) =

4 fx2y2-----------------

1x2---------

1y2--------- f i 1 j 1, 2 f i 1 j, f i 1 j 1+,+( )=

2y2--------- f i j 1, 2 f i j, f i j 1+,+( )

1y2--------- f i 1 j 1,+ 2 f i 1 j,+ f i 1 j 1+,++( )+

CE 341/441 - Lecture 14 - Fall 2004

This thenp. 14.11

produces the following module

4 fx2y2-----------------

1x2y2------------------- f i 1 j 1, 2 f i 1 j, f i 1 j 1+,+[=

2 f i j 1, 4 f i j, 2 f i j 1+,+

f i 1 j 1,+ 2 f i 1 j,+ f i 1 j,+ ]++

1y2

1 -2 1

-2 4 -2

1 -2 1

x2

CE 341/441 - Lecture 14 - Fall 2004

We can al

Therefore

2

y22

x

1

-2

1p. 14.12

so obtain this module as follows:

2fx2

=

x21 -2 1

1 [ ]

2fy2

=

y21

1

-2

1

=

x21 1 -2 1y2

1

-2

1

f2 x x2

1y2

=

1 -2

-2 4

1 -2

CE 341/441 - Lecture 14 - Fall 2004

To obtain

If x =p. 14.13

, we must add up modules:

then:

4 f

4 f 4 f

x4--------- 2

4 fx2y2-----------------

4 fy4---------+ +=

y h=

-8

20

-8

1

2 2

-8 -81 1

1

2 2

1h4

LECTURE 14PRACTICAL ISSUES IN APPLYING FINITE DIFFERENCE APPROXIMATIONSDerivatives of Variable Coefficients Consider the termwhere is a coefficient which is a function of Proceed in steps. First let: Now evaluate: Now evaluate and Hence

where When , then the approximation reduces to: Example Applications with variable coefficients: Fluids: spatially varying viscosities/diffusivities Solids: spatially varying elasticities

Two Dimensional Finite Difference Approximations Define nodes in a two-dimensional plane with = spatial index in the -direction = spatial index in the -direction

When taking partial derivatives w.r.t. , we hold constantand Similarly

Example 1 Consider Laplaces equation and :

This may be viewed as follows in terms of a molecule:

Example 2 Lets consider: Draw a molecule diagram for the 2nd order central difference formulation. Obtain by finding

Similarly: Evaluation of the mixed derivative:

This then produces the following module We can also obtain this module as follows: Therefore To obtain , we must add up modules: If then: