solving daes - lecture 9

8/11/2019 Solving DAEs - Lecture 9

1/23

Systems Realization Laboratory

Solving

Differential Algebraic Equations(based on Continuous System Simulation by Cellier and Kofman)

Chris Paredis

2009, Chris Paredis

. .

Systems Realization Laboratory

Product and Systems Lifecycle Management CenterGeorgia Institute of Technology

Lecture Overview

Converting implicit DAE into semi-explicit DAE Graph-based algorithm

Tearing algorithm

Challenge 2: Structural singularities What is it? How to deal with it?

Higher-index problems

Challenge 3: Handling Discrete Events

Systems Realization Laboratory 2009, Chris Paredis

Zero crossings

Overview of Dymola solvers and their properties

For more info refer to: Continuous System Simulation by Cellier and Kofman


2/23

DAE problems in Dymola

Object-oriented model results in implicit DAE system:

0 ( , , , )F y y z t = &

Before solving, Dymola converts this into a semi-explicit DAE:

0

( , , )

( , , ) 0

(0)

y f y z t

g y z t

y y

=

= =

&

0 ( , , , )F y y z t = &


ro ems a nee o e a resse :

Selection of state variables,y

Causality assignment: For each equation determine which variable is thedependent variable

Order in which equations should be evaluated

Simple System Convert to Explicit ODE

0

1 1 1

2 2 2

( )u f t

u R i

u R i

=

=

=

0 1

LL

CC

C

diu L

dt

dui C

dt

u u u

=

=

= +


1 2

2

0 1

1 2

L

C

L

C

u u

i i i

i i i

==

= +

= +


3/23

Causality Assignment and State Selection

State variables? Variables for which derivatives appear

State variables can be considered as known

0

1 1 1

2 2 2

( )u f t

u R i

u R i

=

=

=

Which are the unknowns?

Causality assignment: If equation has only 1 unknown

0 1

LL

CC

C

diu L

dt

dui C

dt

u u u

=

=

= +


use t to so ve or un nown

If unknown appears only in 1 equation use it to solve for unknown

1 2

2

0 1

1 2

L

C

L

C

u u u

u u

i i i

i i i

=

=

= +

= +

Causality Assignment and State Selection

State variables? Variables for which derivatives appear

State variables can be considered as known

0

1 1 1

2 2 2

( )u f t

u R i

u R i

=

=

=

Which are the unknowns?

Causality assignment: If equation has only 1 unknown

0 1 2 0 1 2CL

L C

dudiu u u u i i i i

dt dt

0 1

LL

CC

C

diu L

dt

dui C

dt

u u u

=

=

= +


use t to so ve or un nown

If unknown appears only in 1 equation use it to solve for unknown

1 2

2

0 1

1 2

L

C

L

C

u u

i i i

i i i

==

= +

= +


4/23

Graph-Based Approach for Causality Assignment:

Tarjan Algorithm

0 0

1 1 1 1

( )u f t u

u R i u

=

=

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= += +

Step 0: Connect Equations and Unknowns

0 0

1 1 1 1

( )u f t u

u R i u

=

=

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= +

= +


5/23

Step 1: Identify Equations with Only 1 Unknown

Color Scheme: Acausal equation = only black and blue lines

Causal equation = one red line

=

Known variable = one red line

Algorithm step 1:

f or al l acausal equations doi f equation has only one black line t hen


col or l i ne red

f ol l ow l i ne t o var i abl e

col or al l ot her l i nes f or vari abl e bl ue

Step 2: Identify Variables in Only 1 Equation

Color Scheme: Acausal equation = only black and blue lines

Causal equation = one red line

=

Known variable = one red line

Algorithm step 1:

f or al l unknown variables doi f variable has only one black line t hen


col or l i ne red

f ol l ow l i ne t o equat i on

col or al l ot her l i nes f or equat i on bl ue


6/23

Final Result

(1)

(5)

0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

(4)

(7)

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)

(8)

(6)

2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= += +

Final Result Sorted & Causality Assigned

0 0

2 2

: ( )

:C

u f t u

u u u

=

=

(1)

(2)

2 2 2 2

1 0 1

11 1 1

1 2

1 2

:

: /

:

:

C

CC

LL

u u u u

ii u R

ii i i

uu u u

=

=

=

= +

(4)

(5)

(6)

(7)


00 1:

// : /

// : /

L

CC C

LL L

ii i i

du dt du dt i C

di dt di dt u L

= +

=

=

(8)

(9)

(10)


7/23

Lecture Overview


Tearing algorithm





Zero crossings


System 2 Algebraic Loop

0

1 1 1

( )u f t

u R i

=

=

= 2 2 2

3 3 3

0 1 3

LL

diu L

dt

u R i

u u u

u u u

=

=

= +

= +


2 3

0 1

1 2 3

L

u u

i i i

i i i

=

= +

= +


8/23

0 0

1 1 1 1

( )u f t u

u R i u

=

=


2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= += +

0 0

1 1 1 1

( )u f t u

u R i u

=

=

Step 1: Identify Equations

(1)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= +

= +


9/23

0 0

1 1 1 1

( )u f t u

u R i u

=

=

Step 2: Identify Variables

(1)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +

(10)


3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= += +

(9)

0 0

1 1 1 1

( )u f t u

u R i u

=

=

Step 1: Identify Equations None Available

(1)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +

(10)


3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= +

= +

(9)


10/23

0 0

1 1 1 1

( )u f t u

u R i u

=

=

Step 2: Identify Variables

(1)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +

(10)

(8)


3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= += +

(9)

STUCK!! Algebraic Loop

(1) 0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(8)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


(9)

3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= +

= +


11/23

Algebraic Loop Tearing Algorithm

6 equations in 6 unknowns Simplify problem by assuming one of the unknown

variable to be known: e. .i If we knew i3 , we could compute i3newGuess i3 , and iterate until the residual is zero:

i3new -i3 = 0 (residual equation)

Also works for nonlinear equations

With symbolic differentiation,

3 3 3

2 3

2 2 2

1 0 3

1 1

:

:

: /

:

: /i

u R i

u u

i u R

u u u

i u R

=

=

=

=

=


era on s very as

Multiple tearing variables may be needed

Tricky part: How to choose the tearing variable? can only be determined using heuristics

3 1 2:newi i i=

Tearing Algorithm

(1) 0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(8)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


(9)

tear3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= +

= +


12/23

Tearing Algorithm

(1)0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(8)

(2*)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


(9)

tear3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= += +

Tearing Algorithm

(1) 0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(8)

(2*)

(3*)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


(9)

tear(4*) 3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= +

= +


13/23

Tearing Algorithm

(1)

(5*)

*

0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(8)

(2*)

(3*)

2 2 2 2

3 3 3 0

0 1 3 1

1 2 2

( / )

L L L

L

u L di dt u

u R i i

u u u i

u u u i

=

=

= +

= +


(9)

tear(4*)

residual

3 2 3

0 1

31 2 3

/L L

u u i

i i i di dt

ui i i

=

= += +

Algebraic Loop Example Summary

( , , )

( , , ) 0

y f y z t

g y z t

=

= =

&

0 ( , , , )F y y z t = &

0

: /LL

diu L=

3 3 3

2 3

2 2 2

:

:

: /

u R i

u u

i u R

=

=

=0 : ( )u f t=

1 2:

:

Lu u u

i i i

= +

= +

Algebraic Equations

Differential Equation


1 0 3

1 1

3 1 2

: /

:

i

new

i u R

i i i

=

=

=

Solve Iteratively


14/23

Lecture Overview


Tearing algorithm





Zero crossings


System 3 Structural Singularity

0

1 1 1

2 2 2

( )u f t

u R i

u R i

=

=

=

0 1

LL

CC

L

diu L

dt

dui C

dt

u u u

=

=

= +


1 2

2

0 1

1 2

C

L

C

L

u u

i i i

i i i

==

= +

= +


15/23

0 0

1 1 1 1

( )u f t u

u R i u

=

=


2 2 2 2

0

0 1 1

21 2

( / )

( / )

L L L

C C

L

C

u L di dt u

i C du dt i

u u u i

iu u u

=

=

= +

= +


2

0 1

1 2

/

/

CL

LC

CL

iu u

di dt i i i

du dt i i i

=

= += +

0 0

1 1 1 1

( )u f t u

u R i u

=

=

Apply Both Rules

(1)

2 2 2 2

0

0 1 1

21 2

( / )

( / )

L L L

C C

L

C

u L di dt u

i C du dt i

u u u i

iu u u

=

=

= +

= +

(10)

(9)


2

0 1

1 2

/

/

CL

LC

CL

iu u

di dt i i i

du dt i i i

=

= +

= +

(8)


16/23

STUCK! Structural Singularity

(1)0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

2 2 2 2

0

0 1 1

21 2

( / )

( / )

L L L

C C

L

C

u L di dt u

i C du dt i

u u u i

iu u u

=

=

= +

= +


(8)

2

0 1

1 2

/

/

CL

LC

CL

iu u

di dt i i i

du dt i i i

=

= += +

What is the problem?

0

1 1 1

2 2 2

( )u f t

u R i

u R i

=

=

=

0 1

LL

CC

L

diu L

dt

dui C

dt

u u u

=

=

= +


1 2

2

0 1

1 2

C

L

C

L

u u

i i i

i i i

==

= +

= +


17/23

Structural Singularity

Main approach Pantelides Algorithm Use symbolic differentiation to remove singularity in equations

1. Determine constraint equation (could be hidden use tearing)

2. Differentiate constraint equation and assign dummy derivatives

3. Remove corresponding integration equations remain square

Details of algorithm are tricky

(For details see: Continuous System Simulation by Cellier and Kofman)


Algorithm is also called an index reduction algorithm

What is an index?

Characteristic of DAE: Perturbation Index

Loosely defined:

The number of differentiations needed

to turn a DAE into a pure ODE

Examples: Pure ODE is index-0


ODE with algebraic loop is index-1 Structurally singular system index-2 (or larger)


18/23

Lecture Overview


Tearing algorithm





Zero crossings


Mixed Systems: DAE + Discrete event

mg

At impact: v -v*k[]

Discrete Event


mg

when height


19/23

Mixed DAEs

Hybrid systems require additional solvercapabilities:1. Event Detection solving for zero-crossings

Height changes continuously as function of time step size needs to be selected such that height = radius

2. Event Handling Stop the integration algorithm

Find a new state consistent with the algebraic constraints

Restart the inte ration al orithm


Be careful!

Event handling can be slow Chattering

Summary of Method Characteristics

Appropriate for DAE or only ODE?

Capable of finding initial conditions?

Capable of solving Stiff systems? Stiff implies implicit method

Capable of handling overdetermined systems? More convenient than having to remove redundant constraints

manually


Order of method? Typically higher is better


20/23

Dymola Methods

Method Model

Type

Order Stiff Root

Find

Algorithm

DEABM ODE 1-12 No No Adams/Bashforth/Moulton; reliable, maybe slow

LSODE1 ODE 1-12 No No Adams/Bashforth/Moulton; faster than DEABM

LSODE2 ODE 1-5 Yes No Backward Difference Formulae; Gear method

LSODAR ODE 1-12, 1-5 Both Yes Adams/Bashforth/Moulton

DOPRI5 ODE 5 No No Runge-Kutta by Dormand and Prince

DOPRI8 ODE 8 No No Runge-Kutta by Dormand and Prince

GRK4T ODE 4 Yes No Linearly-implicit Rosenbrock


DASSL DAE 1-5 Yes Yes BDF; default choice in Dymola

ODASSL ODAE 1-5 Yes Yes Modified DASSL for overdetermined DAEs

MEXX ODAE 2-24 No No Special index-2 DAE; not for ODE!

Summary

Identify types and properties of ODE systems

Identify types and properties of ODE solvers Accuracy

Stability

Stiffness

Higher Order Methods and their properties

Causalization, tearing, index reduction

Complexities of mixed DAE problems



Solver implementation is for Professionals Only

Do not try this at home

Not-so-fine print:


21/23

Step 1: Identify Equations with Only 1 Unknown

(1)0 0

1 1 1 1

( )u f t u

u R i u

=

=

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= += +

Solution for Example 1

Step 2: Identify Variables in Only 1 Equation

(1) 0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)

(8)

2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= +

= +Solution for Example 1


22/23

Repeat Step 1: Identify Equations

(1)0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

(4)

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)

(8)

2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= += +


Repeat Step 2: Identify Variables

(1) 0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

(4)

(7)

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)

(8)

(6)

2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= +

= +Solution for Example 1


23/23

Repeat Step 1: Identify Equations

(1)

(5)

0 0

1 1 1 1

( )u f t u

u R i u

=

=

(10)

(9)

(4)

(7)

2 2 2 2

0

0 1 1

1 2 2

( / )

( / )

L L L

C C

C

L

u L di dt u

i C du dt i

u u u i

u u u i

=

=

= +

= +


(2)

(8)

(6)

2

0 1

1 2

/

/

C C

L L

CC

u u i

i i i di dt

du dt i i i

=

= += +


solving daes - lecture 9

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