Sitnikov Map

7/27/2019 Sitnikov Map

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sitnikovMap.m: The program simulates the Sitnikov problem (actually the

continuation of the Sitnikov problem into the three body problem, but with

very small perturbation parameter). It is used to produce the results in the

introduction set of notes that pertain to the Sitnikov problem. The output is a

couple of Poincare maps and plots of the height of the third body and the

configuration of the entire system in physical space (R^3). The program lets

the user define both the number of initial conditions to be included in the

map, and the number of intersections of each trajectory with the Poincare

section.

clearformat long

%--------------------------------------------------------------------------%----------------------NOTES-----------------------------------------------

%--------------------------------------------------------------------------

%Dependencies:%This program calls 'nBodyWpar.m', and 'sitnikowVewton.m'

%USE:%Program simulates the Sitnikov Problem.%This program has many userdefined parameters. Several define theshape of%the orbits of the primaries, which are ellipses. Several more haveto do%with computing the poincare map of the motion of the third body.

They are%marked clearly below but are located in TWO seprate sections.

%Once the user sets these (or leaves them as they are now)%then save and type 'sitnikovMap' at the MatLab command prompt withthe%present program and it's dependencies in the search path.

%OUTPUT:%The output is several plots.

%One plot shows the orbits of the three bodies in physical space.

%One shows the height of the third body abov or below the xy plane%as a function of time.

http://www.math.utexas.edu/users/jjames/MatLabCode/sitnikovMap.m





%Then there are two Poincare maps. One in angular coordinates.%and the other lifted to the square.

%PURPOSE:%The purpose of the program is to study the Poincare mappingassociated

%with the Sitnikov problem. The program was used to generate theresults%about this problem at the end of the first set on notes where theSitnikov%proplem is discussed in some detail. The program lets you decide howmany%initial conditions to simulate, and how many intersections with thePoincare%section to compute for each initial condition.

%------------------------DATA-----------------------------------------

-----

N=3; %number of bodies%gravatational constantG=1;%Massesm1=5.0;m2=5.0;m3=0.00000001;Mass=[m1 m2 m3];

%--------------------------------------------------------------------------%----------------------SHAPE PARAMETERS------------------------------------%---------------------------FOR THE----------------------------------------%-----------------------PRIMARY BODIES--------------------------------------%--------------------------------------------------------------------------

%--------Compute the appropriate initial conditions for the primaries------

e=0.65 %eccentricity: Should be between zero and one (butnot 1).

% For e=0 the problem is completely% integrable. e can be considered the% perturbation parameter for the

problem.d=5 %max distanceM=m1+m2

%---------------Compute the needed two body initial conditions-------------

%Kepler constants

Ec2=(G*M)^2*(e^2-1)/2c2=d*G*M*(1-(1+(2*Ec2)/(G*M)^2)^(1/2))



c=sqrt(c2)thetaDotZero=c/d^2v_0=d*thetaDotZero

p=c2/(G*M)a=p/(1-e^2)

%conversion to two body coordinatesx2_0=d/(1+m2/m1)x1_0=-(m2/m1)*x2_0

x2Dot_0=v_0/(1+m2/m1)x1Dot_0=-(m2/m1)*x2Dot_0

T=2*pi*a^(3/2)/sqrt(G*M)

%---------------------------------------------------------------------

-----

%----------------------CONTROL PARAMETERS----------------------------------

%NOTE: The first 4 variables are set by the user

%This variable decides how many initial conditions will be used to%generate the Poincare map, or how many points in the domain of

the%mapping to consider.

k=300

%This variable decides how many crossings of the xy-plane to keepfor

%every initial condition. This is equivalent to specifying howmany

%itterates of the poincare map to keep for every point.iterates=1000

%NOTE: To get really good looking graphics one should use k=100-300 and

%poincareItterates=1000. However the program will run for a verylong

%time with these values (days even). As they are set now the

program%runs in a few minutes and finds an invariant torus or two.

%NOTE: zPrime=0 is a fixed point of the map. Setting thevariable

%below is deciding how far from the fixed point you want to map.%The boundary between orbits which escape to infinity and those%which are bounded is near z_0=3.643815. After this orbits escape

to%infinity.

%initial z velocityzPrime_0=0.5



%This determines the length of interval of initial velocitieswhich

%will be usedzPrime_max=1.5

%NOTE: The following parameters should not need to be changed.

%sets the tolerence for the map. The poincare map looks forcrossings

%of the xy plane. epsilon determines how close a trajectory pointmust

%be to the plane to be considered "on" the plan. Standard valueis

%ten to the minus six.epsilon=0.000001

%The program integrates for a time, then checks for crossings. Itrepeats

%process untill it has found the desired number of crossings. Thevariable

%below defines how long to integrate before looking for crossings.Thsi is

%somewhat arbitrary.

timeStep=1.0;

%The poincare map is no a bounded map on all of R^2. There is%z_0' above which trajectories escape to infinity. Then the

domain of%the map is a topological disk whos boundary we do not know a

priori.%This has to be garded against. The while loop will stop

executing if%the z component of some trajectory gets too big. If this happens

we will%conclude that we are out of domain and the program stops.

outOfDomain=0;escape=8; %We consider the trajectory to have

escaped if%the z or z' component exceeds this

tolerence.boundDomain=0; %We will also keep track of the last

trajectory%that was in the domain. This gives an%estimate as to the location of the

boundary

%--------------------------------------------------------------------------

%---------------Derived Parameters-----------------------------------------%-----------(Should not need to touch)-------------------------------------



%period of the primaries

P=T;

%NOTE: The following parameters are necessary but depend on the above

%if k=1 no steps are necessare and the following variable makes%no sense.if k>1

%determine the step size between successive initial conditonszStep=(zPrime_max-zPrime_0)/(k-1);

else%if k=1 just set step to zero so it is at least defined laterzStep=0;

end %end seeing if steps are necessary

%INITIALIZE:

%This is the number of iterates of the current initial conditions%found this far through the while loopz_nIterates=0;%This is the total number of points in the map so fartotalIterates=0;%Local name of initial conditions:initial_n=0;%keep track of how long an initial condition is integratedtrajectoryTime=0;

%number of times through the while loopcheckWhile=0;%number o times the various loops are calledinterpTimes=0;ReIntegrate=0;checkTotal=0;calledNewton=0;

%--------------------------------------------------------------------------%----------------------COMPUTATION OF POINCARE MAP-------------------------%--------------------------------------------------------------------------

for n=1:k%I like to see what loop I'm in at command line, just to see how

fast%the program is running and where it is. Comment this out or

delete it%if you prefer. Same goes for any 'check' variablecheck_n=ncheckTotal=totalIterates%First the positioninitial_n = [x1_0; 0.0; 0.0;

x2_0; 0.0; 0.0;0.0; 0.0; 0.000000001;0.0 ; x1Dot_0; 0.0;0.0; x2Dot_0; 0.0;0.0; 0.0; zPrime_0+(n-1)*zStep]';

%THE MAIN WHILE LOOP: This computes the number of times the



%trajectory with initial condition 'initial_n' crosses thepoincare

%section with positive y velocity. The desired number of suchcrossings

%is given by the integer 'iterates'. The count of how many havebeen

%found so far is in the raviable:x_nIterates=0; %This should be set back to zero before%every run through the while loop.

%set trajectory time back to zero for each new initial conditiontrajectoryTime=0;

%Begining of while loopwhile (x_nIterates <=iterates) & (outOfDomain ~= 1)

checkWhile=checkWhile+1;

%Intigrate this for time 'timeLength'.tspan=[0 timeStep];options=odeset('RelTol',1e-12,'AbsTol',1e-12);[t,trajectory_n] =

ode113('nBodyWpar',tspan,initial_n,options,flag,N,G,Mass);temp=size(trajectory_n);numSteps=temp(1,1);

if ((abs(trajectory_n(numSteps,9))>=2*escape) |(abs(trajectory_n(numSteps,18))>=escape))

outOfDomain=1;boundDomain=trajectory_n(1,18);

end %end of outOfdomain conditional

%Now check for pairs of points along the trajectory betweenwhich

%the sign of y changes and the sign of ydot is positivefor i=2:numSteps

if (outOfDomain == 1)break

end

%Check for sign change of y and correct sign of ydot.

if (sign(trajectory_n(i,9)) ~= sign(trajectory_n(i-1,9)))%note the approximate time of the crossingroughTime=t(i-1);%It could happen that one of these two points is

already a%crossing to the desired tolerence. We check for

this:if (abs(trajectory_n(i-1,9))<epsilon)

intersection=trajectory_n(i-1,:);intersectionTime=t(i-1);

elseif (abs(trajectory_n(i,9))<epsilon)intersection=trajectory_n(i,:);intersectionTime=t(i);

else%If (as is most likely the case) neither of the points



%for which the sign change occurs are in tolerence%(to be considered as crossings),%then a Newton method determines the%crossing to the desired accuracy.

[timeCorrection, intersection]=sitnikovNewton(t(i-1),trajectory_n(i-1,:),t(i),trajectory_n(i,:),epsilon,N,G,Mass);

calledNewton=calledNewton+1; %(Print to scren)end %end of the intersection finding conditional.

%Keep track of how many iterates of the currentinitial

%condition have been foundx_nIterates=x_nIterates+1;

%Keep track of how many total intersections have beenfound

totalIterates=totalIterates+1 %(print to screne)%Stor the intersection information for plotting later

%The actual time of the crossing ismoserTime=trajectoryTime+roughTime+timeCorrection;

%store the poincarePointmoser(totalIterates,1)=mod(2*pi*moserTime/P,2*pi);moser(totalIterates,2)=abs(intersection(18));

end %end of pair checking if statementend %end of pair checking loop (i=2:numSteps index)

%make sure variables are ready for next while loop: unless allthe

%iterates have been found we will go fhrought the while loopagain.

%In that case the new initial condition is the final conditionof

%trajectory_n

%How long has this initial condition been integrated

initial_n=trajectory_n(numSteps,:);trajectoryTime=trajectoryTime+timeStep;

end %end of main while loop (x_nIterates <= iterates)

%The set up for the next time through the 'n' loop is done at the%begining of the loop instead of here at the end.%So just end and go back.

%if the trajectories are blowing up then breakif (outOfDomain == 1)

breakend %end break conditional

end %end of outmost for loop (n=1:k index)

%--------------------------------------------------------------------------%--------------------ANALYSIS OF PERFORMANCE-------------------------------%--------------------------------------------------------------------------

%We want to know how many meaning full intersections have beencomputed.



%This can be found by subtracting from the number of points in%'PoincareMap' the number of times the point [-mu, 0, 0, 0, 0, 0]occurs.%This is described in the comments of the program'CRTBPpoincareNewton'.%Then count these now:

%'PoincareMap' will contain 'k'*'iterates' intersections.computedIntersections = k*iterates%check this:Temp=size(moser)checkPMsize=Temp(1,1)checkTotal = totalIterates

%All three of the above should be the same number. If so then the%variable sizesGood will be one.if ((computedIntersections+1)==checkPMsize) &

(checkPMsize==checkTotal)sizeGood=1

elsesizeGood=0end

%initialize the counterfalsePositives=0

for p=1:computedIntersectionsif moser(p,:)==[0 0]

falsePositives=falsePositives+1;end %end of falsePositives conditional loop

end %end of computedIntersections for loop

%output the resultscheckFalsePositives=falsePositivestrueIntersections = computedIntersections - falsePositives

%How many times was the interpolator called?checkInterpTimes=interpTimes

%How many times is the ReIntegrate loop being called?checkReIntegrate=ReIntegrate

%bound domain: If this is zero then the boundary was not found%If it is nonzero then it is the z' coordinate of the last orbit thatdid

%not escape.checkBound=boundDomain

save sitnikovData1 moser

%--------------------------------------------------------------------------



%If running on UNIX in Background the following should be commentedout

%---------------------------------------------------------------------

-----%-------------------------PLOTTING-----------------------------------------%--------------------------------------------------------------------------

%--------Integrate the primarise and have a look at their orbits----------

%integrate%numSteps = 4000;%tf=moserTime;

%tspan = linspace(0, tf, numSteps);%initial positions and velocities%options=odeset('RelTol',1e-12,'AbsTol',1e-12);%[t, y] = ode113('nBodyWpar',tspan,y0,options,flag,N,G,Mass);

%hold on%plot3(y(:,1), y(:,2), y(:,3),'b', y(:,4), y(:,5), y(:,6), 'g',y(:,7), y(:,8), y(:,9), 'r')%title('physical configuration of the Sitnikov problem')

%figure%hold on%plot(t(:),y(:,9),'b')%title('height of the third body vs. time')

%--------------------------------------------------------------------------

%------------------------Moser Mapping-------------------------------------figureplot(moser(:,1),moser(:,2),'b.')title('lift of the Poincare map to the square')

figurepolar(moser(:,1),moser(:,2),'b.')title('Poincare section of the Sitnikov problem')%------------------------PROGRAM OVER--------------------------------------

Sitnikov Map

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