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16.920J/SMA 5212 Numerical Methods for PDEs

Thanks to Franklin Tan

Finite Differences: Parabolic Problems

B. C. Khoo

Lecture 5

SMA-HPC 2002 NUS

Outline

Governing Equation Stability Analysis 3 Examples Relationship between and h Implicit Time-Marching Scheme Summary

2

SMA-HPC 2002 NUS

[ 2 2 0,u x t =

GoverningEquation

3

Consider the Parabolic PDE in 1-D

0subject to at 0, atu u x u u x = = =

If viscosity Diffusion Equation If thermal conductivity Heat Conduction Equation

0x = x =

0u

( ), u x t =

] u x

= u

?

SMA-HPC 2002 NUS

Stability Analysis Discretization

4

Keeping time continuous, we carry out a spatial discretization of the RHS of

There is a total of 1 grid points such that , 0,1, 2,....,

jN j x j

+ = =

0x = x = 0x 1x 3x 1Nx Nx

2

2

u t

=

x N

u x

SMA-HPC 2002 NUS


5

2 1 2

2

2 ( j j

j

u uu O x x

+ =

2

2Use the Central Difference Scheme for u

x

which is second-order accurate.

Schemes of other orders of accuracy may be constructed.

1 2 )j u

x + +

SMA-HPC 2002 NUS 6


We obtain at

0

0

Note that we need not evaluate at and since and are given as boundary conditions.

N

N

u x x x u

= =

1 1 22: 2 )o

du x u udt x

= + 2

2 2 32: 2 )du x u udt x

= +

1 2: 2 )j

j j j

du x u u

dt x

+ = +

1 1 12: 2 )

N N N N

du x u udt x

= +

x u

1 ( u

1 ( u

1( j u

2 ( N u

SMA-HPC 2002 NUS 7

Stability Analysis Matrix Formulation

Assembling the system of equations, we obtain 1

1 2

2

2

2

1 1 2

2 1

01 1

0

1 1

1 0

1

o

jj

NN N

du u udt x du

udt

udu x dt

udu u xdt

=

0

0

A

2

2

2

+

SMA-HPC 2002 NUS 8

Stability Analysis PDE to Coupled ODEs

Or in compact form

du Au bdt

= G G G

We have reduced the 1-D PDE to a set of Coupled ODEs!

[ ]1 1where T Nu u u = G

2 0 0 T

o u b x

= G

+ 2 u

20 Nu

x

SMA-HPC 2002 NUS 9

Stability Analysis Eigenvalue and

Eigenvector of Matrix A

If A is a nonsingular matrix, as in this case, it is then possible to find a set of eigenvalues { 1 1, ,...., ,....,j = For each eigenvalue , we can evaluate the eigenvector consisting of a set of mesh point values , i.e.

j j

j i

V v

1 1 Tj j j

NV v v =

( from det 0.A }2 N

2 j v

)I =

SMA-HPC 2002 NUS 10

Stability Analysis Eigenvalue and

Eigenvector of Matrix A

The ( 1) ( 1) matrix formed by the ( 1) columns diagonalizes the matrix byj

N E N V

1E AE = 1

2

1

where

N

= 0

0

N A

SMA-HPC 2002 NUS 11

Stability Analysis Coupled ODEs toUncoupled ODEs

Starting from du Au bdt

= + G G G

1Premultiplication by yieldsE

1 1duE E Au E bdt

= G G G

( 1 1 1duE E A EE u E bdt = G G G

I

( 1 1 1duE E AE E u E bdt = G G G

1 +

)1 + )1 +

SMA-HPC 2002 NUS 12

Stability Analysis Coupled ODEs to Uncoupled ODEs

1 Let and , we haveU E u F E b = GG G d U Fdt

= + JG JGG

which is a set of Uncoupled ODEs!

1 1duE u E bdt

= + G G G

Continuing from

1= G

U

1 E

SMA-HPC 2002 NUS 13

Stability Analysis Coupled ODEs to Uncoupled ODEs

Expanding yields

Since the equations are independent of one another, they can be solved separately.

The idea then is to solve for and determineU EU= G GG

1 1 1 1

dU U dt

= + 2

2 2 dU U dt

= +

j j j

dU U

dt = +

1 1 1

N N N

dU U dt

=

u

F

2 F

j F

1 N F +

SMA-HPC 2002 NUS 14

Considering the case of independent of time, for the general equation,th

b j

G Stability Analysis

Coupled ODEs to Uncoupled ODEs

1jt j j

j

U e F = is the solution for j = 1,2,.,N1.

Evaluating, ( ) 1 tu EU E ce E E b = JJJJGG G Complementary

(transient) solution Particular (steady-state)

solution

( 11 1 1where j N Tt tt t j ce c e c e c e c e = JJJJG

j c

1= G

) 2 2 t N

SMA-HPC 2002 NUS

We can think of the solution to the semi-discretized problem

15

Stability Analysis Stability Criterion

( ) 1 tu E ce E E b = JJJJGG G

This is the criterion for stability of the space discretization (of a parabolic PDE) keeping time continuous.

Since the transient solution must decay with time,

for all j( )Real 0j

Each mode contributes a (transient) time behaviour of the form to the time-dependent part of the solution.j t

j e

as a superposition of eigenmodes of the matrix operator A.

1

SMA-HPC 2002 NUS 16

Stability Analysis Use of Modal (Scalar)

Equation

It may be noted that since the solution is expressed as a contribution from all the modes of the initial solution, which have propagated or (and) diffused with the eigenvalue

, and a contribution frj

u

G

om the source term , all the properties of the time integration (and their stability properties) can be analysed separately for each mode with the scalar equation

jb

j

dU U Fdt

= +

SMA-HPC 2002 NUS 17

Stability Analysis Use of Modal (Scalar)

Equation

The spatial operator A is replaced by an eigenvalue , and the above modal equation will serve as the basic equation for analysis of the stability of a time-integration scheme (yet to be introduced) as a function of the eigenvalues of the space-discretization operators.

This analysis provides a general technique for the determination of time integration methods which lead to stable algorithms for a given space discretization.

SMA-HPC 2002 NUS 18

1 11 1 12 2

2 21 1 22 2

du a u a udt du a u a udt

=

=

1 1 12

2 1 22

Let , u a

u u a =

G du Audt

= G G

Consider a set of coupled ODEs (2 equations only):

Example 1 Continuous Time

Operator

+

+

1

2

a A

a =

SMA-HPC 2002 NUS 19

Proceeding as before, or otherwise (solving the ODEs directly), we can obtain the solution

1

1

1 11 2 12

2 21 2 22

t

t

u e c e u e c e

= =

11 21 1

21 22

1

where and are eigenvalues of and and are

eigenvectors pertaining to and respectively.

A

( )j As the transient solution must decay with time, it is imperative that Real 0 for 1, 2.j

Example 1 Continuous Time

Operator

2

2

1

1

t

t

c c

+ +

2

2

=

SMA-HPC 2002 NUS 20

Suppose we have somehow discretized the time operator on the LHS to obtain

1 1 1 1 12 2

1 2 1 1 22 2

n n

n n

u u a u

u u a u

= =

where the superscript n stands for the nth time level, then

1 11 12 1

21 22

where and Tn n n a u u u u u A

a = =

G G

Since A is independent of time, 1 0

.... n n nu u AAu A u

= = = G G G

Example 1 Discrete Time Operator

1 1

1 2

n

n

a

a

+ +

2

n n a Aa

= G

2 n A

= G

SMA-HPC 2002 NUS 21

Example 1 Discrete Time Operator

1 0 1

2

0where =

0

n n n

n u E u

= JJGG

' 1 1 11 1 2 12 2

' 2 1 21 1 2 22 2

n n

n n

u c

u c

= =

1 1 0

2

' where are constants.

' c

E u c

= JJG

1

1 1 0

,

.... n

A E E

u E E E E E u

= =

JJGG As

A A A

n E

'

'

n

n

c

c

+ +

1 E

SMA-HPC 2002 NUS 22

Example 1 Comparison

Comparing the solution of the semi-discretized problem where time is kept continuous

[ 1 2

1 1 12 1

2 1 22

t

t

u e c

u e

= to the solution where time is discretized

[ 1 1 12 1 1 2 1 22 2

' n n

n

u c

u

= coTh nte di inuofference equation where time is hus exponential

solution as

. te

The difference equation where time is hasdiscretized power solution . n

] 12 2

c

] 12 2

' c

SMA-HPC 2002 NUS 23

Example 1 Comparison

In equivalence, the transient solution of the difference equation must decay with time, i.e.

for this particular form of time discretization.

1n <

SMA-HPC 2002 NUS 24

Consider a typical modal equation of the form

t

j

du u aedt

= where is the eigenvalue of the associated matrix .j A (For simplicity, we shall henceforth drop the subscript j). We shall apply the leapfrog time discretization scheme given as

Substituting into the modal equation yields

( 1 2 n

t t nh

u u aeh

+

= =

n nu e=

1

where 2

n du u u h dt h

+ =

Example 2 Leapfrog Time Discretization

+

)1 n u + ha+

1 n

t

=

SMA-HPC 2002 NUS

Solution of u consists of the complementary solution cn, and the particular solution pn, i.e.

un = cn + pn

There are several ways of solving for the complementary and particular solutions. shift operator S and characteristic polynomial.

The time shift operator S operates on cn such that

Scn = cn+1 S2cn = S(Scn) = Scn+1 = cn+2

25

( 1 1 2 2 2 n

n n n n n hnu u e u h u u ha eh

+

+ = =

Example 2 Leapfrog Time Discretization Time Shift Operator

One way is through use of the

)1 1n hu a +

SMA-HPC 2002 NUS 26

The complementary solution cn satisfies the homogenous equation

1

2

2

2

2

1( ) 0

( 1) 0

n n

n n

n n

n

c c c cSc h c S

S c h Sc c S

cS S S

+ = =

=

=

2( ) ( 2 1) 0p S S h S= = characteristic polynomial


1 0

0

2

2

n

n

n

h

h

SMA-HPC 2002 NUS 27

The complementary solution to the modal equation would then be

1 1 2 2 n nc =

The particular solution to the modal equation is 2 2

2

hn h n

h

ahe e p e e

=

Combining the two components of the solution together,

( ) ( )n nu p= ( ) ( )2 2 2 2 1 2 21 2 hn hn h ahe eh h h e e = + + + +

The solution to the characteristic polynomial is 2 2( ) 1h h h = = + 1 and 2 are the two roots


n+

1 h h

n c +

2 1 1

n

h h h

+

S

SMA-HPC 2002 NUS 28

For the solution to be stable, the transient (complementary) solution must not be allowed to grow indefinitely with time, thus implying that

is the stability criterion for the leapfrog time discretization scheme used above.

( ) ( )

2 2 1

2 2 2

1

1

h

h

= + <

= <

Example 2 Leapfrog Time Discretization Stability Criterion

1

1

h

h

+

+

SMA-HPC 2002 NUS 29

Im( )

Re( )-1 1

Region of Stability

The stability diagram for the leapfrog (or any general) time discretization scheme in the -plane is

Example 2 Leapfrog Time Discretization Stability Diagram

SMA-HPC 2002 NUS 30

In particular, by applying to the 1-D Parabolic PDE 2

2

u t

= the central difference scheme for spatial discretization, we obtain

which is the tridiagonal matrix.


2

2 1 1 1

1 1

A x

=

0

0

u x

2

2

SMA-HPC 2002 NUS 31


According to analysis of a general triadiagonal matrix B(a,b,c), the eigenvalues of the B are

2

2 cos , 1,..., 1

2 2 cos

j

j

jb c j NN

j N

= + = +

The most dangerous mode is that associated with the eigenvalue of largest magnitude

max 2

4 x =

( (

2 max1 ax max

2 max2 ax max

1

1

h h

h h

= + = +

i.e.

which can be plotted in the absolute stability diagram.

a

x

=

) )

2 m

2 m

h

h

+

SMA-HPC 2002 NUS 32

Example 2 Leapfrog Time Discretization Absolute Stability Diagram for

As applied to the 1-D Parabolic PDE, the absolute stability diagram for is

Region of stability

Unit circle

Region of instability

2 at h = t = 0

2 with h increasing

1with h increasing 1 at h = t = 0

Re()

Im()

SMA-HPC 2002 NUS 33

Stability Analysis Some Important

Characteristics Deduced

A few features worth considering:

1. Stability analysis of time discretization scheme can be carried out for all the different modes .

2. If the stability criterion for the time discretization scheme is

j valid for

all modes, then the overall solution is stable (since it is a linear combination of all the modes).

3. When there is more than one root , then one of them is the principal root which represents

( 0

an approximation to the physical behaviour. The principal root is recognized by the fact that it tends towards one as 0, i.e. lim 1. (The other roots are spurious, which affect the stability

h h

but not the accuracy of the scheme.) )h =

SMA-HPC 2002 NUS 34

Stability Analysis Some Important

Characteristics Deduced

1

4. By comparing the power series solution of the principal root to , one can determine the order of accuracy of the time discretization scheme. In this example of leapfrog time discretization,

1

he

h

= ( ( 1 2 2 2 2 4 42 2 2

1

2 2

1 .1 2 1 2 !

1 ...2

and compared to

1 ..2!

is identical up to the second order of . Hence, the above scheme is said to be second-order accurate.

h

h h h

hh

h e

h

+ + + +

= + + +

= + + +

+ ) )1 2 .

2

.

h

h

=

SMA-HPC 2002 NUS 35

Analyze the stability of the explicit Euler-forward time discretization 1n du u u

dt t

+ = as applied to the modal equation

du udt

= 1

1

Substituting where

into the modal equation, we obtain (1 ) 0

n

n

du u h h tdt

u u +

+

= = +

Euler-Forward Time Discretization Stability Analysis

Example 3

n

n

n

u

h

+ =

SMA-HPC 2002 NUS 36

Making use of the shift operator S 1 (1 ) (1 ) [ (1 )] 0n n n nc c Sc h c S h c + + = + = + =

Therefore ( ) 1 and n

h c

= +

=

characteristic polynomial

The Euler-forward time discretization scheme is stable if

Euler-Forward Time Discretization Stability Analysis

Example 3

1 h + or bounded by 1 s.t. 1 in the -plane.h = <

n h

n

h

1 < h

SMA-HPC 2002 NUS 37

Euler-Forward Time Discretization Stability Diagram

Example 3

Im(h)

0-1-2

Unit Circle

Region of Stability

Re(h)

The stability diagram for the Euler-forward time discretization in the h-plane is

SMA-HPC 2002 NUS 38

Euler-Forward Time Discretization Absolute Stability Diagram

Example 3

max 2

4As applied to the 1-D Parabolic PDE, x =

max

2The stability limit for largest h =

1-1

leaves the unit circle at h = 2

at h (=t) = 0

Re()

Im()

with h increasing

=

t

SMA-HPC 2002 NUS 39

Relationshipbetween and h

= (h)

Thus far, we have obtained the stability criterion of the time discretization scheme using a typical modal equation. generalize the relationship between and h as follows: Starting from the set of coupled ODEs

du Au bdt

= + G G G

Apply a specific time discretization scheme like the leapfrog time discretization as in Example 2

1

2

n du u u dt h

+ =

We can

1 n

SMA-HPC 2002 NUS 40


= (h)

The above set of ODEs becomes 1

2

n nnu Au h

+ = + G GG

Introducing the time shift operator S

1

2

2

n nn

nn

uSu hAu hbS

S SA I u bh

= +

G GG

GG

1

1 Premultiplying on the LHS and RHS and introducing

operating on n E

I EE u

=

i G 1

1 1 1

2 nS SE AE E E E u E b

h

GG

1 n u bG

2 n + =

G

1 =

SMA-HPC 2002 NUS 41


= (h)

Putting 1 , nn nU u F E b = GG G

1

2j j S S U

h

=

we obtain 1

1

2 n S SE U F

h

G G

1

2 S S

h

i.e. 1

2 n S S U

h

= G G

which is a set of uncoupled equations.

Hence, for each j, j = 1,2,.,N-1,

1 n E = G

j F

nE =

nF

SMA-HPC 2002 NUS 42


= (h)

Note that the analysis performed above is identical to the analysis carried out using the modal equation

j

dU U Fdt

= All the analysis carried out earlier for a single modal equation is applicable to the matrix after the appropriate manipulation to obtain an uncoupled set of ODEs.

Each equation can be solved independently for and the 's can then be coupled through .

th

n n n j

j U u EU= GG

+

n j U

SMA-HPC 2002 NUS 43


= (h)

1. Uncoupling the set, 2. Integrating each equation in the uncoupled set,

3. Re-coupling the results to form the final solution.

These 3 steps are commonly referred to as the

ISOLATION THEOREM

Hence, applying any consistent numerical technique to each equation in the set of coupled linear ODEs is mathematically equivalent to

SMA-HPC 2002 NUS 44

Implicit Time-Marching Scheme

Thus far, we have presented examples of explicit time-marching methods and these may be used to integrate weakly stiff equations.

Implicit methods are usually employed to integrate very stiff ODEs efficiently. solution of a set of simultaneous algebraic equations at each time-step (i.e. matrix inversion), whilst updating the variables at the same time.

Implicit schemes applied to ODEs that are inherently stable will be unconditionally stable or A-stable.

However, use of implicit schemes requires

SMA-HPC 2002 NUS 45


Euler-Backward

Consider the Euler-backward scheme for time discretization 1 1n n du u u

dt h

+ + =

tdu u aedt

=

(

( ( 1

11

111

n n n

n n

u u eh h u u ahe

+ ++

++

= =

Applying the above to the modal equation for Parabolic PDE

yields

n

+

)

) ) n

h

hn

u a+

SMA-HPC 2002 NUS 46


Euler-Backward

Applying the S operator, ( ) ( 11 n nh S u ahe + =

the characteristic polynomial becomes

( ) ( ) ( )1 0S S = = The principal root is therefore

2 21 which, upon comparison with 1 .... , is only2

first-order accurate.

he h = + + +

2 21 1 .... 1

h h

= = + + +

The solution is (

( 11

1 1

n u n

h

aheU h e

+ =

)1 h

1 h =

h h

)

) 1 h

h +

SMA-HPC 2002 NUS 47


Euler-Backward

For the Parabolic PDE, is always real and < 0. Therefore, the transient component will always tend towards zero for large n irregardless of h ( t). The time-marching scheme is always numerically stable.

In this way, the implicit Euler/Euler-backward time discretization scheme will allow us to resolve different time-scaled events with the use of different time-step sizes. scaled events, and then a large time-step size used for the longer time-scaled events. hmax.

A small time-step size is used for the short time-

There is no constraint on

SMA-HPC 2002 NUS 48


Euler-Backward

However, numerical solution of u requires the solution of a set of simultaneous algebraic equations or matrix inversion, which is computationally much more intensive/expensive compared to the multiplication/ addition operations of explicit schemes.

SMA-HPC 2002 NUS

Summary

49

Stability Analysis of Parabolic PDE Uncoupling the set. Integrating each equation in the uncoupled set

modal equation.

Re-coupling the results to form final solution. Use of modal equation to analyze the stability |(h)| < 1.

Explicit time discretization versus Implicit time discretization.

16.920J/SMA 5212

Numerical Methods for Partial Differential Equations

Lecture 5

Finite Differences: Parabolic Problems

B. C. Khoo

Thanks to Franklin Tan

19 February 2003


2

OUTLINE

Governing Equation Stability Analysis 3 Examples Relationship between and h Implicit Time-Marching Scheme Summary

Slide 2

GOVERNING EQUATION

Consider the Parabolic PDE in 1-D

If viscosity Diffusion Equation If thermal conductivity Heat Conduction Equation

Slide 3

STABILITY ANALYSIS Discretization

Keeping time continuous, we carry out a spatial discretization of the RHS of

[ ]2

2 0,u u

xt x

pi

=

0subject to at 0, at u u x u u xpi pi= = = =

0x = x pi=

0u upi

( ), ?u x t =

2

2u u

t x

=

0x = x pi=

0x 1x 2x 1Nx Nx


3

Slide 4


which is second-order accurate.

Schemes of other orders of accuracy may be constructed.

Slide 5

Construction of Spatial Difference Scheme of Any Order p

The idea of constructing a spatial difference operator is to represent the spatial differential operator at a location by the neighboring nodal points, each with its own weightage.

The order of accuracy, p of a spatial difference scheme is represented as ( )pO x . Generally, to represent the spatial operator to a higher order of accuracy, more nodal points must be used.

Consider the following procedure of determining the spatial operator j

dudx

up to the

order of accuracy ( )2O x :

There is a total of 1 grid points such that ,0,1,2,....,

jN x j xj N

+ = =

2

2Use the Central Difference Scheme for u

x

21 1 2

2 2

2 ( )j j jj

u u uu O xx x

+ +

= +

j2

j1

j

j+1

j+2

j

dudx


4

1. Let j

dudx

be represented by u at the nodes j1, j, and j+1 with 1 , 0 and

1 being the coefficients to be determined, i.e.

( )1 1 0 1 1 pj j jj

duu u u O x

dx

+

+ + + =

2. Seek Taylor Expansions for 1ju , ju and 1ju + about ju and present them in a table as shown below.

(Note that p is not known a priori but is determined at the end of the analysis when the s are made known.)

uj uj uj uj

ju

0 1 0 0

1 1ju 1 1x 2

112

x

3 116

x

0 ju 0 0 0 0

1 1ju + 1 1x 2

112

x 3 116

x

1

1

k

j k j kk

u u=

+=

+ 1S 2S 3S 4S

( 1 )

This column consists of all the terms on the LHS of (1).

Each cell in this row comprises the sum of its corresponding column.


5

where

1

1 2 3 41

....

k

j k j kk

u u S S S S=

+=

+ = + + + +

3. Make as many iS s as possible vanish by choosing appropriate k s.

In this instance, since we have three unknowns 1 , 0 and 1 , we can therefore set:

1

2

3

000

SSS

=

=

=

(Note that in the Taylor Series expansion, one starts off with the lower-order terms and progressively obtain the higher-order terms. We have deliberately set the iS pertaining to the lower-order terms to zero, thereafter followed by increasingly higher-order terms.)

Hence,

1

0

1

01 1 111 0 1

1 0 1 0x

=

Solving the system of equations, we obtain

1

0

1

12

01

2

x

x

=

=

=

( )( )

1 1 0 1

2 1 1

2 23 1 1

3 34 1 1

1

1 12 2

1 16 6

j

j

j

j

S u

S x x u

S x x u

S x x u

= + +

= +

= +

= +


6

4. Substituting the k s into 1

1 2 3 41

....

k

j k j kk

u u S S S S=

+=

+ = + + + +

yields

( ) 21 11 12 6j j j ju u u x ux + = + higher-order terms

In other words,

( )1 1 2 ....2j jj ju udu

u O xdx x

+

= = + +

i.e. the above representation is accurate up to ( )2O x .

Some useful points to note:

1. These 4 steps are the general procedure used to obtain the representation of the spatial operator up to the order of accuracy ( )pO x .

2. For other spatial operators, say 2

2j

d udx

, we simply replace j

dudx

in (1) with

the said spatial operator.

3. For one-sided representations, one can choose nodal points , 0j ku k+ . This may be important especially for representations on a boundary. For example

( )0 1 1 2 2 .... pj j jj

duu u u O x

dx + +

+ + + + =

One possibility is

( )1 2 23 42j j jju u udu O x

dx x+ + +

+ =

which is also second-order accurate.

(We can also use a similar procedure to construct the finite difference scheme of Hermitian type for a spatial operator. This is not covered here).

( ) , 2pO x p =


7


We obtain at 11 1 22: ( 2 )odu

x u u udt x

= +

22 1 2 32: ( 2 )

dux u u u

dt x

= +

1 12: ( 2 )jj j j jdu

x u u udt x

+= +

11 2 12: ( 2 )NN N N N

dux u u u

dt x

= +

0

0

Note that we need not evaluate at and since and are given as boundary conditions.

N

N

u x x x x

u u

= =

Slide 6

STABILITY ANALYSIS Matrix Formulation

Assembling the system of equations, we obtain

Slide 7

11 2

2

2

2

1 1 2

2 1

01 2 1

0

1 2 1

1 0

1 2

o

jj

NN N

du uudt x

duudt

udu xdt

udu uxdt

= +

0

0

A


8

STABILITY ANALYSIS PDE to Coupled ODEs

Or in compact form

We have reduced the 1-D PDE to a set of Coupled ODEs!

Slide 8

STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A

If A is a nonsingular matrix, as in this case, it is then possible to find a set of eigenvalues

{ }1 2 1, ,...., ,....,j N =

( )from det 0.A I =

For each eigenvalue , we can evaluate the eigenvector consisting of a set of mesh point values , i.e.

jj

ji

Vv

Slide 9

STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A

The ( 1) ( 1) matrix formed by the ( 1) columns diagonalizes the matrix byj

N N E NV A

1E AE =

[ ]1 2 1where TNu u u u =

2 20 0 0T

o Nu ubx x

=

du Au bdt

= +

1 2 1 Tj j j j

NV v v v

=


9

Slide 10

STABILITY ANALYSIS Coupled ODEs to Uncoupled ODEs

Starting from du Au bdt

= +

1Premultiplication by yieldsE

1 1 1duE E Au E bdt

= +

Slide 11


Continuing from

1 1 1duE E u E bdt

= +

1 1Let and , we haveU E u F E b = =

1

2

1

where

N

=

0 0

( )1 1 1 1duE E A EE u E bdt = +

I

( )1 1 1 1duE E AE E u E bdt = +


10

d U U Fdt

= +

which is a set of Uncoupled ODEs!

Slide 12


Expanding yields

11 1 1

dU U Fdt

= +

22 2 2

dU U Fdt

= +

jj j j

dUU F

dt= +

11 1 1

NN N N

dU U Fdt

= +

Since the equations are independent of one another, they can be solved separately.

The idea then is to solve for and determine U u EU=

Slide 13


Considering the case of independent of time, for thegeneral equation,th

bj

1jtj j j

jU c e F =

is the solution for j = 1,2,.,N1.


11

Evaluating, ( ) 1 1tu EU E ce E E b = =

( ) 11 21 2 1where j N Tt tt tt j Nce c e c e c e c e =

The stability analysis of the space discretization, keeping time continuous, is based on the eigenvalue structure of A. The exact solution of the system of equations is determined by the eigenvalues and eigenvectors of A.

Slide 14


We can think of the solution to the semi-discretized problem

as a superposition of eigenmodes of the matrix operator A.

Each mode contributes a (transient) time behaviour of the form to the time-dependent part of the solution.jt

je

Since the transient solution must decay with time,

( )Real 0j for all j

This is the criterion for stability of the space discretization (of a parabolic PDE) keeping time continuous.

Slide 15

Complementary (transient) solution

Particular (steady-state) solution

( ) 1 1tu E ce E E b =


12

STABILITY ANALYSIS Use of Modal (Scalar) Equation

It may be noted that since the solution is expressed as acontribution from all the modes of the initial solution,which have propagated or (and) diffused with the eigenvalue

, and a contribution frj

u

om the source term , all theproperties of the time integration (and their stabilityproperties) can be analysed separately for each mode withthe scalar equation

jb

Slide 16

STABILITY ANALYSIS Use of Modal (Scalar) Equation

The spatial operator A is replaced by an eigenvalue , and the above modal equation will serve as the basic equation for analysis of the stability of a time-integration scheme (yet to be introduced) as a function of the eigenvalues of the space-discretization operators.

This analysis provides a general technique for the determination of time integration methods which lead to stable algorithms for a given space discretization.

Slide 17

EXAMPLE 1 Continuous Time Operator

Consider a set of coupled ODEs (2 equations only):

111 1 12 2

221 1 22 2

dua u a u

dtdu

a u a udt

= +

= +

1 11 12

2 21 22

Let , u a a du

u A Auu a a dt

= = =

Slide 18

j

dU U Fdt

= +


13

EXAMPLE 1 Continuous Time Operator

Proceeding as before, or otherwise (solving the ODEs directly), we can obtain the solution

1 2

1 2

1 1 11 2 12

2 1 21 2 22

t t

t t

u c e c e

u c e c e

= +

= +

11 211 2

21 22

1 2

where and are eigenvalues of and and are

eigenvectors pertaining to and respectively.

A

( )jAs the transient solution must decay with time, it is imperative thatReal 0 for 1, 2.j =

Slide 19

EXAMPLE 1 Discrete Time Operator

Suppose we have somehow discretized the time operator on the LHS to obtain

1 11 11 1 12 2

1 12 21 1 22 2

n n n

n n n

u a u a u

u a u a u

= +

= +

where the subscript n stands for the nth time level, then

1 11 121 2

21 22

where and Tn n n n n a a

u Au u u u Aa a

= = =

Since A is independent of time,

1 2 0....

n n nn

u Au AAu A u

= = = =

In later examples, we shall apply specific time discretization schemes such as the leapfrog and Euler-forward time discretization schemes.

Slide 20


14

EXAMPLE 1 Discrete Time Operator

As

1 0 1

2

0 where =

0

nn

n n

nu E E u

=

' '

1 1 11 1 2 12 2' '

2 1 21 1 2 22 2

n n n

n n n

u c c

u c c

= +

= +

1 1 0

2

'

where are constants.'

cE u

c

=

Slide 21

Alternative View

Alternatively, one can view the solution as:

01 1

1 2 02 2

n

n n

n

U UU U

=

0 1where n nU U U E u= =

EXAMPLE 1 Comparison

Comparing the solution of the semi-discretized problem where time is kept continuous

[ ] 12

1 11 121 2

2 21 22

t

t

u ec c

u e

=

to the solution where time is discretized

[ ]1 11 12 11 22 21 22 2

' '

n n

n

uc c

u

!

! !

= " #" # " #

$% $ %

" #

$ %

1

1 1 1 0

,

....

n

A E E

u E E E E E E u

=

= &'& ((

A A A


15

difference equation where time is continuous has exponentialsolution The

.

te

The difference equation where time is discretized has powersolution .n

Slide 22

EXAMPLE 1 Comparison

In equivalence, the transient solution of the difference equation must decay with time, i.e.

1n <

for this particular form of time discretization.

Slide 23

EXAMPLE 2 Leapfrog Time Discretization

Consider a typical modal equation of the form

t

j

duu ae

dt

= +

where is the eigenvalue of the associated matrix .j A

(For simplicity, we shall henceforth drop the subscript j).

We shall apply the leapfrog time discretization scheme given as

1 1

where 2

n ndu u u h tdt h

+

= =

Substituting into the modal equation yields

1 1

2

n nu u

h

+

( )tt nh

u ae=

= +

n hnu ae= +

Slide 24


16

Reminder

Recall that we are considering a typical modal equation which had been obtained from the original equation

du Au bdt

= +

EXAMPLE 2 Leapfrog Time Discretization: Time Shift Operator

( )1 1 1 1 2 22n n

n hn n n n hnu u u ae u h u u ha eh

+

+ = + =

Solution of u consists of the complementary solution nc , and theparticular solution np , i.e.

n n nu c p= +

There are several ways of solving for the complementary andparticular solutions. One way is through use of the shift operator S and characteristic polynomial.

The time shift operator S operates on nc such that

1n nSc c +=

( )2 1 2n n n nS c S Sc Sc c+ += = =

Slide 25


The complementary solution nc satisfies the homogenous equation

1 12 0

2 0

n n n

nn n

c h c ccSc h cS

+ =

=


17

2

2

1( 2 ) 0

( 2 1) 0

n n n

n

S c h Sc cS

cS h SS

=

=

Slide 26


The solution to the characteristic polynomial is

2 2( ) 1h S h h = = +

The complementary solution to the modal equation would then be

1 1 2 2n nn

c = +

The particular solution to the modal equation is 22

2 1

hn hn

h hahe ep

e h e

= .

Combining the two components of the solution together,

nu ( ) ( )n nc p= + ( ) ( )2 2 2 21 2 2 21 1 2 1

hn hn n

h hahe eh h h h

e h e

= + + + + +

Slide 27

EXAMPLE 2 Leapfrog Time Discretization: Stability Criterion

For the solution to be stable, the transient (complementary) solution must not be allowed to grow indefinitely with time, thus implying that

( )( )

2 21

2 22

1 1

1 1

h h

h h

= + + E&~

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Documents

16920 spring 2001