Hi Guys,
I know how to get access to the compiled data of continuation of equilibrium points within MatCont. The directory is the following
Systems/ToyModel/diagram/
Upon loading the interested data, the usual structure of the .matfile is as follow:
x, v, f, s, h, ....
Where x is the branch itself and f contains the associated eigenvalues at each point of the branch.
How to get the associated eigenvectors of the whole branch?
Cheers,
Many
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In my case the f is a 2-by-2 matrix with first entry with length the numbers of variables of my problem and second entry the number of points for the continuation. That is all.
Am I missing something into setting the Matcont Interface (windows) before running the computations for the continuation? Basically, I am applying something along the line of tutorial 2 in the documentation of MatCont.
Thanks
Many
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For equilibria, the variable f contains the eigenvaluesonce you've loaded the mat-file. In the 7px-versions, the Data Browser in the GUI will also indicate the eigenvalues (at the very bottom). Now, to come to your questions, in MatCont we do not compute or store the eigenvectors, this is too costly and typically not needed. If you want this for a specific point, you need to recompute the Jacobian (which we also do not store). Say you want to compute it for the 3rd point on the branch:
load mat file
global cds; cds=globals.cds; eds=globals.eds; %Getting some specific info
JAC=cjac(eds.odefile,eds.Jacobian,x(:,3),num2cell(eds.P0),eds.ActiveParams) [V,D]=eig(JAC); %now you have the eigenvectors in V
Perhaps, check the Starter Window and see if the box for computing eigenvalues is ticked.
Note a 2x2 matrix suggests that you had only two points, i.e. your run terminated quickly.
PS. edited once I realized you're asking about eigenvectors not eigenvalues.
Last edit: hilmeijer 2023-08-30
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My question is also about the data
x, v, f, s, h, ....
I understood that worksheet x refers to the data of the curve being calculated. The columns refer to each point on the curve. My question is about the lines. I work with 6 degrees of freedom, I noticed that for each column, or point on the curve, there are 161 values for each degree of freedom. I couldn't understand why it was 161 times.
Can you help me?
Please!
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161 is the hint:
If you have a periodic orbit discretized into ntst=40 meshpoints with ncol=4 on each interval + one final point, then that's 4*40+1=161 points.
So it seems you have been continuing limit cycles.
My question is also about the data
x, v, f, s, h, ....
I understood that worksheet x refers to the data of the curve being calculated. The columns refer to each point on the curve. My question is about the lines. I work with 6 degrees of freedom, I noticed that for each column, or point on the curve, there are 161 values for each degree of freedom. I couldn't understand why it was 161 times.
Can you help me?
Please!
[Getting all eigenvectors of a branch from Matcont's recordings]
Hi Guys,
I know how to get access to the compiled data of continuation of equilibrium points within MatCont. The directory is the following
Upon loading the interested data, the usual structure of the
.matfile is as follow:Where
xis the branch itself andfcontains the associated eigenvalues at each point of the branch.How to get the associated eigenvectors of the whole branch?
Cheers,
Many
Hello, you need to go to file f and scroll to the bottom. The last lines should correspond to your own eigenvectors.
In my case the
fis a 2-by-2 matrix with first entry with length the numbers of variables of my problem and second entry the number of points for the continuation. That is all.Am I missing something into setting the Matcont Interface (windows) before running the computations for the continuation? Basically, I am applying something along the line of tutorial 2 in the documentation of MatCont.
Thanks
Many
For equilibria, the variable f contains the eigenvaluesonce you've loaded the mat-file. In the 7px-versions, the Data Browser in the GUI will also indicate the eigenvalues (at the very bottom). Now, to come to your questions, in MatCont we do not compute or store the eigenvectors, this is too costly and typically not needed. If you want this for a specific point, you need to recompute the Jacobian (which we also do not store). Say you want to compute it for the 3rd point on the branch:
load mat file
global cds; cds=globals.cds; eds=globals.eds; %Getting some specific info
JAC=cjac(eds.odefile,eds.Jacobian,x(:,3),num2cell(eds.P0),eds.ActiveParams)
[V,D]=eig(JAC); %now you have the eigenvectors in V
Perhaps, check the Starter Window and see if the box for computing eigenvalues is ticked.
Note a 2x2 matrix suggests that you had only two points, i.e. your run terminated quickly.
PS. edited once I realized you're asking about eigenvectors not eigenvalues.
Last edit: hilmeijer 2023-08-30
Hi Guys,
My question is also about the data
x, v, f, s, h, ....
I understood that worksheet x refers to the data of the curve being calculated. The columns refer to each point on the curve. My question is about the lines. I work with 6 degrees of freedom, I noticed that for each column, or point on the curve, there are 161 values for each degree of freedom. I couldn't understand why it was 161 times.
Can you help me?
Please!
161 is the hint:
If you have a periodic orbit discretized into ntst=40 meshpoints with ncol=4 on each interval + one final point, then that's 4*40+1=161 points.
So it seems you have been continuing limit cycles.
Van: discussion@matcont.p.re.sourceforge.net discussion@matcont.p.re.sourceforge.net namens Jordana Lopes jordana-panel@users.sourceforge.net
Verzonden: zondag 10 september 2023 12:28
Aan: [matcont:discussion] 762214@discussion.matcont.p.re.sourceforge.net
Onderwerp: [matcont:discussion] Getting all eigenvectors of a branch from Matcont's recordings
Hi Guys,
My question is also about the data
x, v, f, s, h, ....
I understood that worksheet x refers to the data of the curve being calculated. The columns refer to each point on the curve. My question is about the lines. I work with 6 degrees of freedom, I noticed that for each column, or point on the curve, there are 161 values for each degree of freedom. I couldn't understand why it was 161 times.
Can you help me?
Please!
[Getting all eigenvectors of a branch from Matcont's recordings]