computational science equilibrium analysis by heider jeffer

7/28/2019 Computational Science Equilibrium Analysis by Heider Jeffer

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The dancer's equilibrium is unstable. If she didn't constantly make tiny adjustments, she would tip over.

Topics in Informatics (Fall 2009)

instructor: Mehdi Jazayeri

assistant: Sasa Nesic

class date:speaker:

topic: Computational Science

Equilibrium Analysis

Hayder ALMusawi

Heider Jeffer


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Hayder ALMusawi Computational Science

MechanismA body may be in one of five states of equilibrium Stable , Unstable , Neutral ,

Metastable and Metastable with multiple stable states(globally stable).

Stability

the ability to maintain normal characteristics.

Stable Equilibrium

a body is in stable equilibrium if it returns to its equilibrium position after it has been

displaced slightly.

Unstable Equilibrium

a body is in unstable equilibrium if it does not return to its equilibrium position anddoes not remain in the displaced position after it has been displaced slightly.

Neutral Equilibrium

a body is in neutral equilibrium if it stays in the displaced position after if has been

displaced slightly.

Metastable Equilibrium

the body returns towards equilibrium as long as disturbances are not too large. The

equilibrium is not globally stable because once the ball has crossed the unstableequilibrium (the peak), it is unlikely to return to the stable point.

Metastable Equilibrium with multiple stable states

the body returns towards one or the other equilibrium depending on which side of the

unstable equilibrium it is. The system is globally stable.

Stableandunstableequilibria.

Equilibrium Analysis By Heider Jeffer________________________________________________________


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A:Stable Equilibrium

B:Unstable Equilibrium

C:Neutral Equilibrium

D:Metastable Equilibrium

E:Metastable Equilibrium with multiple stable states

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AttractorsandTheirTypes

Whenthedynamicsofapopulationmodelistracedoveralargenumberofgenerations,thenthemodelexhibitsanasymptoticbehaviorwhichisalmost

independentfrominitialconditions(e.g.,initialpopulationdensity).Asymptotic

trajectoryiscalled"attractor"becausemodeltrajectoriesconvergetothis

asymptotictrajectory.There may be several attractors in a model. In this case each attractor has a domain of attraction.

Model trajectory converges to that attractor in which domain initial conditions were located.

In this example, there are two attractors: a limit cycle (at the left) and a stable equilibrium (at the right).

Domains of attraction are colored blue, they never overlap. For different starting places (initial

conditions), trajectories converge to different attractors.

Types of attractors:

1. Stable equilibrium (=steady state)2. Limit cycle3. Chaos

Examples:

This is a chaotic attractor (Lorenz attractor)

Equilibrium Analysis By Heider Jeffer___________________________________________________________


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Example

A chemical plant requires as input a chemical substance that goes through two processing stages in acascade of reactors. Each reactor is modelled as a reservoir: the variation of mass over time is equalto the net inflow. The mass contained in each reactor is labelled as xi. The outflow from each reactoris proportional to the mass of substance: axi.

The outflow from the last reactor is perturbed by a disturbance d, which is coming from another planthat is only occasionally active: z(t) = y(t) + d(t).

x1

x2

a*x1

u

y=a*x2d

Chemical Plant

z=y+d

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Solution

Equations

The system has two state variables

x1x2

, and its output is z:

x1 = ax1 + ux2 = ax2 + ax1z = ax2 + d

The state-space representation of the system is the following:x1x2

=

a 0a a

x1x2

+

10

u

z =

0 a x1

x2

+ d

For the equilibrium:x1 = 0x2 = 0

ax1 + u = 0ax2 + ax1 = 0

x1 =

u

a

x2 = x1

x1 =

u

a

x2 =u

a

The equilibrium point is (ua, ua

)

For the system stability, we calculate its eigenvalues from the characteristic equation:

|A I| = 0

a 0a a = 0 (a )2 = 0 = a

The system has a double eigenvalue that is a negative real. Therefore it is stable and not oscil-lating.

Equilibrium Analysis By Heider Jeffer__________________________________________________________


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Example

Solution

Given the system:x1 = x1x2 + x

22

x2 = x1x2 5x2 + u

An equilibrium state is a situation where the system doesnt change its behaviour, so the deriva-tives are set to 0:

x1x2 + x22 = 0

x1x2 5x2 + u = 0from the first equation we obtain

x1x2 + x22 = 0 x2(x1 + x2) = 0This equation has two solutions: The first is

x1 = x2And the second is:

x2 = 0

For the first solution:

We substitute x1 = x2 into the second equation:x22 5x2 + u = 0 x22 + 5x2 u = 0

The polynomial has two solutions:

x2 =525 + 4u

2

and, since x1 = x2, we obtain two solutions that produce two equilibria state:

Equil. 1 : (x1, x2)a =5

25+4u

2,5+

25+4u

2

Equil. 2 : (x1, x2)b =5+25+4u

2,5

25+4u

2

For the second solution:

When x2 = 0 the second equation reduces to u = 0. If this is the case (i.e. u = 0) there is anequilibrium for any values ofx1, that satisfy x2 = 0. This leads to an infinite number of solutions(the line x2 = 2). In case u = 0, we end up to an impossible situation, i.e., there are no equilibriafrom this solution.

In overall:If

u

= 0 then we have two equilibria points ((x1, x2

)a

and (x1, x2

)b

indicated above).Ifu = 0 we have infinite equilibria points: The two (x1, x2)a and (x1, x2)b indicated above, andthe line x2 = 0

In the specific case u = 6, system has two equilibria points:

Equil. 1 : (x1, x2)a =525+462

,5+

25+462

= (1, 1)

Equil. 2 : (x1, x2)b =5+25+462

,5

25+462

= (6,6)

Equilibrium Analysis By Heider Jeffer____________________________________________________________


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ThreeGeneralTypesOfTheEquilibrium

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Example

Logisticmodel

Tofindequilibriawehavetosolvetheequation:dN/dt=0:

Thisequationhastworoots:N=0andN=K.Anequilibriummaybestableor

unstable.Forexample,theequilibriumofapencilstandingonitstipisunstable;

Theequilibriumofapictureonthewallis(usually)stable.

Anequilibriumisconsideredstable(forsimplicitywewillconsiderasymptotic

stabilityonly)ifthesystemalwaysreturnstoitaftersmalldisturbances.Ifthe

systemmovesawayfromtheequilibriumaftersmalldisturbances,thenthe

equilibriumisunstable.

Thenotionofstabilitycanbeappliedtoothertypesofattractors(limitcycle,chaos),however,thegeneraldefinitionismorecomplexthanforequilibria.

Stabilityisprobablythemostimportantnotioninsciencebecauseitrefersto

whatwecall"reality".Everythingshouldbestabletobeobservable.For

example,inquantummechanics,energylevelsarethosethatarestablebecause

unstablelevelscannotbeobserved.

Now,let'sexaminestabilityof2equilibriapointsinthelogisticmodel.

Inthisfigure,populationgrowthrate,dN/dt,isplottedversuspopulation

density,N.Thisisoftencalledaphase-plotofpopulationdynamics.If0


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increaseuntiltheyreachthestableequilibriumN=K.Afteranydeviationfrom

N=Kthepopulationreturnsbacktothisstableequilibrium.

Thedifferencebetweenstableandunstableequilibriaisintheslopeofthelineonthephaseplotneartheequilibriumpoint.Stableequilibriaarecharacterized

byanegativeslope(negativefeedback)whereasunstableequilibriaare

characterizedbyapositiveslope(positivefeedback).

Thesecondexampleisthebarkbeetlemodelwithtwostableandtwounstable

equilibria.Stableequilibriacorrespondtoendemicandepidemicpopulations.

Endemicpopulationsareregulatedbytheamountofsusceptibletreesinthe

forest.Epidemicpopulationsarelimitedbythetotalnumberoftreesbecause

massattackofbeetlefemalesmayovercometheresistanceofanytree.

Stabilityofmodelswithseveralvariables

Detectionofstabilityinthesemodelsisnotthatsimpleasinonevariable

models.Let'sconsiderapredatorpreymodelwithtwovariables:(1)densityof

preyand(2)densityofpredators.Dynamicsofthemodelisdescribedbythe

systemof2differentialequations:

Thisisthe2variablemodelinageneralform.Here,Histhedensityofprey,and

Pisthedensityofpredators.Thefirststepistofindequilibriumdensitiesofprey

(H*)andpredator(P*).Weneedtosolveasystemofequations:

Thesecondstepistolinearizethemodelattheequilibriumpoint(H=H*,P=P*)

byestimatingtheJacobianmatrix:

Equilibrium Analysis By Heider Jeffer________________________________________________________________


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Third,eigenvaluesofmatrixAshouldbeestimated.Thenumberofeigenvaluesis

equaltothenumberofstatevariables.Inourcasetherewillbe2eigenvalues.

Eigenvaluesaregenerallycomplexnumbers.Ifrealpartsofalleigenvaluesarenegative,thentheequilibriumisstable.Ifatleastoneeigenvaluehasapositive

realpart,thentheequilibriumisunstable.

Eigenvaluesareusedheretoreducea2dimensionalproblemtoacoupleof1

dimensionalproblemproblems.Eigenvalueshavethesamemeaningastheslope

ofalineinphaseplots.Negativerealpartsofeigenvaluesindicateanegative

feedback.ItisimportantthatALLeigenvalueshavenegativerealparts.Ifone

eigenvaluehasapositiverealpartthenthereisadirectionina2dimensional

spaceinwhichthesystemwillnottendtoreturnbacktotheequilibriumpoint.

Thereare2typesofstableequilibriainatwodimensionalspace:knotandfocus

Thereare3typesofunstableequilibriainatwodimensionalspace:knot,focus,

andsaddle

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Stability in discrete-time models

Consider a discrete-time model (a difference equation) with one state variable:

This model is stable if and only if :

wheredf/dNt istheslopeofathicklineingraphsbelow:

Equilibrium Analysis By Heider Jeffer_________________________________________________________________

computational science equilibrium analysis by heider jeffer

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