runge-kutta differential equations

8/8/2019 Runge-Kutta Differential Equations

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RungeKuttaNumericalMethodsusedinEvaluationof

DifferentialEquations

DanielCook

Abstract

Periodicmotionisoneofthemostcommonengineeringissuesfaced. Thegoaloftheseproblemsisto

modelthisperiodicmotionasafunctionoftime,resultinginadifferentialequation. Tosolvethese

differentialequationsseveralmethodsarecommonlyemployed. Themethodsthatwillbedescribedin

thispaperareallvariationsofRungeKuttamethods,specificallytheEulermethod,Heunsmethod,

Midpointmethod,andtheclassicfourthorderRungeKuttamethod.

Introduction

Theneedtosolveperiodicmotionisseeninmanyaspectsofengineering.Wavesandpendulumsare

twoexamplesofperiodicmotionfoundineverydayproblems.Pendulumsspecificallyhavefoundmany

importantuses. Consideredtobeoneofthemostexactformsoftimekeepinguntiltheinventionofthe

quartzclock,pendulumsplayedavitalroleinearlytimemanagement.

Periodicmotionisalsoseeninmolecularvibration.Molecularvibrationiswhenamoleculeasawhole

hasconstanttranslationalandrotationalmotion.Theinformationgainedfromknowingtheismolecular

vibrationisveryimportantingaininginsightintotheSchrodingerwaveequation,whichlaysthe

foundationformodernquantummechanics[1].

Tomodeltheseproblems,positionandvelocityoftheobjectmustbeknownasafunctionoftime. The

functionsthatmodelthismotionarecommonlyseenasdifferentialequations,meaningthatthevalues

ofthefunctionarebasedoffofthefunctionitselfanditsderivatives. Differentialequationscanbecome

verycomplex,andnumericalmethodsarecommonlyemployedtosolvetheseequations. Fourmethods

willbeusedinthispapertosolvethemotionofapendulum.ThesemethodsaretheEulermethod,

Heunsmethod,theMidpointmethod,andfinallytheFourthorderRungeKuttamethod.

PhysicalAnalysis

Asstatedabove,thispaperwillfocusontheharmonicmotionofapendulumsystem.Thissystemis

showninfigure1.


2/25

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yticalsolutio

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(3)

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3/25

(4)

And

sin (5)

Thenumericalmethodslistedabovecanthenbeusedtosolvethiscoupledsystem.

Additionally,bysubstitutingequation(3)intoequation(4)weobtainafirstorderdifferentialequation

thatcanbeusedtopredictthevelocity[2]:

sin cos (6)

Whichcanbesolvedforbyintegration.

NumericalAnalysis

AllfourofthenumericalmethodsmentionedarevariationsoftheRungeKuttamethods. TheRunge

Kuttamethodsallrelyonthesamegeneralideaoffindinganewvaluebasedonapreviousvaluebeing

modifiedbytheslopeorderivative,andanincreaseintheindependentvariable(astep).

Inequationformthiscanberepresentedas:

Orinamoremathematicalview[3]:

(7)

Where istheslopeorderivative,andhisthestepsize.

EulersMethod(FirstOrderMethod)

Eulersmethodusesthefirstderivativeofthefunctionattheoriginalxvaluetoextrapolateanewy

value. Thismeansthatforthismethod =f(xi,yi)=y,whichisthederivativeattheinitialxvalue. The

generalformulaforthismethodis[3]:

,


4/25

Graphical

Theitera

findanot

HeunsM

Heunsm

pointswh

valueare

interval.

Thismeth

averaged

corrector

Predictor:

Corrector

Thiscan

lythiscanbe

iveprocess

ernewpoin

ethod(Seco

ethodiskno

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averagedto

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withtheslo

formulasare

:

edepicted

g

seenas:

fextrapolati

tcontinuesf

dOrderMe

nasasecon

inganewval

obtainaslop

findingapr

eattheiniti

asfollows[3

raphicallyas:

Figure2:

ganewpoi

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hod)

dordermet

ue. Thederi

ethatisabe

dictor,whic

lvaluetoob

]:

Euler'sMetho

tbasedofft

nofthefun

hodbecause

ativeatthe

tterrepresen

istheslope

tainthecorr

, ,

[4]

hederivativ

tioninterval

ittakesthef

initialvalue

tationofthe

attheendp

ectorequati

,

,thenusing

.

unctionsderi

ndthederiv

actualslope

oint. Thispr

n. Thepred

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vativeattw

tiveatthee

acrossthe

dictoristhe

ictorand

ntto

nd


5/25

Midpoint

Themidp

tofindne

midpoint

methodt

[3]:

Graphical

Method(Se

ointmethod

wvalues. Th

methodtake

akestheave

lytheMidpo

ondOrder

isverysimila

emaindiffer

sthederivati

ageofthed

intmethodc

Figure3:

ethod)

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encebetwee

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rivativesatt

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Heun'sMetho

ethodintha

nthemidpoi

pointbetwe

heinitialand

.,

das:

[5]

titusesderi

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.

ativesattw

ndHeunsm

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separatep

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e,whileHeu

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ints

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6/25

FourthO

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Figure4:

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7/25

Gauss

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ulas

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aMethod[7]

tosolvethei

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validwithin

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at.

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]:

)


8/25

Thetwo

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hesevaluesi

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Figure6:Analy

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aussLegendre

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Ev

X

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XCoefficients[3]

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=.86113612

=.33998104

=.33998104

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solution.

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generalesti

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9/25

Thisconfi

lookatth

Eachmet

Eulerme

Eulersm

solution:

Andwith

solution:

rmsthatthe

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hodisevalua

hod

thodwitha

astepsizeo

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stepsizeofh

h=.05,Euler

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evariousnu

stepsizes,h

=.01yieldst

Figure

smethodyi

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efollowing

7:Euler.01Ste

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10/25

HeunsM

Thismeththetrues

results:

ethod

od,asstatedlopeofthef

above,takenctionmore

Figure

thederivaticlosely.Heu

8:Euler.05Ste

veattwopoinsmethod

p

ntstomakeithaninter

theextrapolalofh=.01yi

tedslopefoleldsthefollo

lowwing


11/25

Withast

psizeofh=.05,Heunsm

Figure

ethodyields

Figure

9:Heun.01Ste

thefollowin

10:Heun.05St

p

results:

p


12/25

MidpointMethod

Withastepsizeofh=.01,theMidpointmethodreturnsthefollowingresults:

Figure11:Midpoint.01StepAndwithastepofh=.05,theMidpointresultsare:

0 0.5 1 1.5 2-4

-3

-2

-1

0

1

2

3

4Midpoint Angular Velocity Step Size=0.01

Time (s)

AngularVelocity(rad/sec

0 0.5 1 1.5 2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8Midpoint Angular Displacement Step Size=0.01

Time (s)

AngularDisplac

ement(rad)


13/25

Figure12:Midpoint.05Step

FourthOrderRangeKutta

Forastepsizeofh=.01,thefourthorderRangeKuttamethodreturnsthefollowingresults:

Figure13:RK.01Step

0 0.5 1 1.5 2-4

-3

-2

-1

0

1

2

3

4Midpoint Angular Velocity Step Size=0.05

Time (s)


0 0.5 1 1.5 2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8Midpoint Angular Displacement Step Size=0.05

Time (s)

AngularDisplacement(rad)

0 0.5 1 1.5 2-4

-3

-2

-1

0

1

2

3

44th Order RK Angular Velocity Step Size=0.01

Time (s)


0 0.5 1 1.5 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

14th Order RK Solution Angular Displacment Step Size=0.01

Time (s)



14/25

Andwithastepofh=.05

Figure14:RK.05Step

Finallywecanexaminethevaluesofeachofthemethodsatt=2,andcomparethesevalueswiththe

integrationvaluesofequation(6)usingthetwoandfourpointGaussLegendreMethods:

Method

StepSize

Velocity

at

t=2

Analytical N/A 3.1054Euler 0.01 3.470153

0.05 4.454886

Heun 0.01 3.0405760.05 3.074375

Midpoint 0.01 3.0407060.05 3.076673

4thOrder

RK

0.01 3.1812530.05 3.726941

TwoPointGauss N/A 9.800047FourPointGauss N/A 3.285373

Table2:ComparisonofResultsatt=2

0 0.5 1 1.5 2-4

-3

-2

-1

0

1

2

3

44th Order RK Angular Velocity Step Size=0.05

Time (s)


0 0.5 1 1.5 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

14th Order RK Solution Angular Displacment Step Size=0.05

Time (s)



15/25

Discuss

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]

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16/25

TaketheEulermethodintheabovegraphasanexample. Becausethenextyvaluewillbebasedupon

thatline,thecloserthatlineistothenexttruepoint,thebetterthenewyvaluewillbe. Itcanalsobe

seenthatasthestepgrowslarger,theslopebecomesfartherfromtheexactsolution,socuttingstep

sizecangreatlyimproveaccuracy. Byreducingthestepsizebyhalf,theglobalerrorforEulersmethod

isalsohalved.

Itisworthnotingthelocalandglobaltruncationerrorforeachofthenumericalmethods:

Method LocalError GlobalErrorEuler O(h2) O(h)

Heun O(h3) O(h2)

Midpoint O(h3) O(h2)

4thOrderRK O(h5) O(h4)Table3:ComparisonofError[3]

WhatthistableessentiallyshowsisthatadecreaseinstepsizewillcausetheHeun/Midpoint/fourth

ordermethodstodecreaseinerroratafasterratethanEulersmethod

Conclusion

Ascanbeseenfromtheaboveresults,therearemanymethodsthatcanbeusedtosolveacoupled

systemofequations. Allofthesemethodshavetheirtradeoffswhencomparedtoeachother. Eulers

methodisrelativelyinaccurate,butitisnotintensiveoncomputerresourcesanditiseasiertoprogram.

ThefourthorderRungeKuttaisthemostaccurateofthemethodsexamined,butitcanbetaxingon

systemresourcesandismorelaborintensivetoprogram. Thisiswhyonemethodcannotbefavored

overanother,anditisimportanttoassesswhatisneededfromtheoutput;accuracyorlowtime/money

investment.

References

[1]"Molecularvibration Wikipedia,thefreeencyclopedia."Wikipedia,thefreeencyclopedia.N.p.,n.d.Web.15Nov.2010..

[2]Drapaca,Corina.Project3:NumericalAnalysisoftheSwingingSimplePendulum.PennsylvaniaStateUniversity,2010.Print.

[3]Chapra,StevenC.,andRaymondP.Canale.Numericalmethodsforengineers.6thed.Boston:McgrawHillHigherEducation,2010.Print.

[4]"Euler'sMethodforFirstOrderDifferentialEquations."Swarthmore.N.p.,n.d.Web.15Nov.2010.

.


17/25

[5]"NumericalTutorials:Euler'sMethod."ComputerEngineering,ChulalongkornUniversity.N.p.,n.d.Web.15Nov.2010..

[6]"MidpointMethod."Wikipedia,thefreeencyclopedia.N.p.,n.d.Web.15Nov.2010..

[7]"ODE2:RungeKutta."BostonCollegePhysics.N.p.,n.d.Web.15Nov.2010..

[8]"Runge Kutta."DrexelUniversity:DepartmentofPhysics:Home.N.p.,n.d.Web.15Nov.2010..

Appendix

Variables

ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;

Eulers

%Set the step size and number of calculations

h=.01;

nc=(tf-ti)/h+1;

%Euler's algorithm

for i=2:nc

theta(i) = theta(i-1) +v(i-1)*h;

v(i) = v(i-1) + - g/L *sin(theta(i-1))*h;

end

fprintf('\nThe velocity at t=2 by the Euler Method with step-size 0.01 is %f

\n\n', v(i))

%graphing results

t=0:0.01:2;

figure

hold on, subplot(1,2,2), plot(t,v,'r');

title('Eulers Angular Velocity Step Size=0.01')

xlabel('Time (s)')

ylabel('Angular Velocity (rad/sec')

grid on

hold on, subplot(1,2,1), plot(t,theta,'r');

title('Eulers Angular Displacement Step Size=0.01')

xlabel('Time (s)')

ylabel('Angular Displacement (rad)')


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grid on

for i = 1:201

position(i) = pi/4*cos(sqrt(g/L)*t(i));

velocity(i) = -sqrt(g/L)*(pi/4)*sin(sqrt(g/L)*t(i));

end

hold on, subplot(1,2,1), plot(t,position,'b');

hold on, subplot(1,2,2), plot(t,velocity,'b');

%Euler's Method (h=0.05)

clear all

%For Equations (5)n and (6)

%Initialize variables

ti=0;

tf=2;theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.05;

nc=(tf-ti)/h+1;

%Euler's algorithm

for i=2:nc

theta(i) = theta(i-1) +v(i-1)*h;

v(i) = v(i-1) + - g/L *sin(theta(i-1))*h;

end

fprintf('\nThe velocity at t=2 by the Euler Method with step-size 0.05 is %f

\n\n', v(i))

%graphing results

t=0:0.05:2;

figure


title('Eulers Angular Velocity Step Size=0.05')

xlabel('Time (s)')

ylabel('Angular Velocity (rad/sec')grid on


title('Eulers Angular Displacement Step Size=.05')

xlabel('Time (s)')


grid on

for i = 1:41



19/25


end



Heuns

%Initialize variablesti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.01;

nc=(tf-ti)/h+1;

%Heun's algorithm

for i=2:ncthetap(i) = theta(i-1) +v(i-1)*h;

vp(i) = v(i-1) + - g/L *sin(theta(i-1))*h;

theta(i) = theta(i-1) + (vp(i)+v(i-1))/2*h;

v(i) = v(i-1) + (-g/L*sin(theta(i-1))+(-g/L*sin(thetap(i))))/2*h;

end

fprintf('\nThe velocity at t=2 by the Heun Method with step-size 0.01 is %f

\n\n', v(i))

%graphing results

t=0:0.01:2;

figurehold on, subplot(1,2,2), plot(t,v,'r');

title('Heuns Angular Velocity Step Size =0.01)')

xlabel('Time (s)')


grid on


title('Heuns Angular Displacment Step Size=0.01')

xlabel('Time (s)')


grid on

for i = 1:201


velocity(i) = -sqrt(g/L)*(pi/4)*sin(sqrt(g/L)*t(i));end




20/25

%Heun's Method (h=0.05)

clear all

%For Equations (5) and (6)


ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.05;

nc=(tf-ti)/h+1;

%Heun's algorithm

for i=2:ncthetap(i) = theta(i-1) +v(i-1)*h;

vp(i) = v(i-1) + - g/L *sin(theta(i-1))*h;

theta(i) = theta(i-1) + (vp(i)+v(i-1))/2*h;

v(i) = v(i-1) + (-g/L*sin(theta(i-1))+(-g/L*sin(thetap(i))))/2*h;

end

fprintf('\nThe velocity at t=2 by the Heun Method with step-size 0.05 is %f

\n\n', v(i))

%graphing results

t=0:0.05:2;

figure


title('Heuns Angular Velocity Step Size=0.05)')

xlabel('Time (s)')


grid on


title('Heuns Angular Displacment Step Size=0.05)')

xlabel('Time (s)')


grid on

for i = 1:41

position(i) = pi/4*cos(sqrt(g/L)*t(i));velocity(i) = -sqrt(g/L)*(pi/4)*sin(sqrt(g/L)*t(i));

end




21/25

Midpoint

ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.01;

nc=(tf-ti)/h+1;

%Midpoint algorithm

for i=2:nc

thetap(i) = theta(i-1) +v(i-1)*h/2;

vp(i) = v(i-1) + - g/L *sin(theta(i-1))*h/2;

theta(i) = theta(i-1) + vp(i)*h;

v(i) = v(i-1) + (-g/L*sin(thetap(i)))*h;

end

fprintf('\nThe velocity at t=2 by the Midpoint Method with step-size 0.01 is

%f \n\n', v(i))

%graphing results

t=0:0.01:2;

figure


title('Midpoint Angular Velocity Step Size=0.01')

xlabel('Time (s)')


grid onhold on, subplot(1,2,1), plot(t,theta,'r');

title('Midpoint Angular Displacement Step Size=0.01')

xlabel('Time (s)')


grid on

for i = 1:201



end



%Midpoint Method (h=0.05)

clear all


22/25



ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.05;

nc=(tf-ti)/h+1;

%Midpoint algorithm

for i=2:nc

thetap(i) = theta(i-1) +v(i-1)*h/2;

vp(i) = v(i-1) + - g/L *sin(theta(i-1))*h/2;

theta(i) = theta(i-1) + vp(i)*h;

v(i) = v(i-1) + (-g/L*sin(thetap(i)))*h;

end

fprintf('\nThe velocity at t=2 by the Midpoint Method with step-size 0.05 is

%f \n\n', v(i))

%graphing results

t=0:0.05:2;

figure


title('Midpoint Angular Velocity Step Size=0.05')

xlabel('Time (s)')


grid on


title('Midpoint Angular Displacement Step Size=0.05')

xlabel('Time (s)')


grid on

for i = 1:41



end

hold on, subplot(1,2,1), plot(t,position,'b');hold on, subplot(1,2,2), plot(t,velocity,'b');


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FourthOrderRK

ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;L=2;


h=.01;

nc=(tf-ti)/h+1;

%4th Order RK algorithm

for i=2:nc

vp(i-1)=v(i-1)+2*(-g/L*sin(theta(i-1))*h/2);

thetap(i-1)=theta(i-1)+vp(i-1)*h/2;

vpp(i-1)=v(i-1)+(-g/L*sin(thetap(i-1))*h/2);thetapp(i-1)=theta(i-1)+vpp(i-1)*h;

vppp(i-1)=v(i-1)+(-g/L*sin(thetapp(i-1))*h);

theta(i)=theta(i-1)+1/6*h*(v(i-1)+2*vp(i-1)+2*vpp(i-1)+vppp(i-1));

v(i)=v(i-1)+(-g/L*sin(theta(i-1))*h);

end

fprintf('\nThe velocity at t=2 by the 4th order Runge-Kutta Method with step-

size 0.01 is %f \n\n', v(i))

%graphing results

t=0:0.01:2;

figure


title('4th Order RK Angular Velocity Step Size=0.01')

xlabel('Time (s)')


grid on


title('4th Order RK Solution Angular Displacment Step Size=0.01')

xlabel('Time (s)')


grid on

for i = 1:201position(i) = pi/4*cos(sqrt(g/L)*t(i));


end




24/25

%4th order RK Method (h=0.05)

clear all



ti=0;

tf=2;

theta(1)=pi/4;

v(1)=0;

g=32.2;

L=2;


h=.05;

nc=(tf-ti)/h+1;

%4th Order RK algorithm

for i=2:nc

vp(i-1)=v(i-1)+2*(-g/L*sin(theta(i-1))*h/2);thetap(i-1)=theta(i-1)+vp(i-1)*h/2;

vpp(i-1)=v(i-1)+(-g/L*sin(thetap(i-1))*h/2);

thetapp(i-1)=theta(i-1)+vpp(i-1)*h;

vppp(i-1)=v(i-1)+(-g/L*sin(thetapp(i-1))*h);

theta(i)=theta(i-1)+1/6*h*(v(i-1)+2*vp(i-1)+2*vpp(i-1)+vppp(i-1));

v(i)=v(i-1)+(-g/L*sin(theta(i-1))*h);

end

fprintf('\nThe velocity at t=2 by the 4th order Runge-Kutta Method with step-

size 0.05 is %f \n\n', v(i))

%graphing results

t=0:0.05:2;

figure


title('4th Order RK Angular Velocity Step Size=0.05')

xlabel('Time (s)')


grid on


title('4th Order RK Solution Angular Displacment Step Size=0.05')

xlabel('Time (s)')


grid on

for i = 1:41



end




25/25

runge-kutta differential equations

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