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Pseudo arc-length continuation

  1. Jun 8, 2006 #1
    Hi...

    I'm searching for info on pseudo arc-length continuation algorithm, but i can't find something that can guide me from the ground up.

    I found a presentation which describes an iterative algorithm for calculating dx and dl, but it doesn't seem to work. I don't know how to calculate the last element in the right hand side vector. The equation from the presentation (i don't have the link any more, but i can search it again if you want) is:

    (lamda_k - lamda_prev)^2 + |x_k - x(lamda_prev)|^2 - arc^2

    How should i calculate arc?

    If anybody has a comment / paper / tutorial / anything that describes the method in detail, i'd very glad to hear it.

    Thanks in advance and forgive my english.

    HellRaiZer
     
  2. jcsd
  3. Jun 8, 2006 #2

    J77

    User Avatar

  4. Jun 8, 2006 #3
    Thanks for the link, but unfortunatelly i've already seen it and it is a bit confusing. I'm looking for something more in the form of pseudo code. Not a complete library. I already have a Newton solver for calculating the function before the turning point (where the Newton coverges) and i'm looking for a way for turn it into pseudo arc-length solver, in case to overcome the turning point singularity.

    My problem is that i don't know which values must be kept fixed, which of them correspond the previous iteration and which are for the next! I've seen a lot of different ways to formulate the problem, and i'm a bit confused.

    Can somebody explain how exactly should i construct the augmented matrixs and the right hand side vector?

    Thanks again.

    HellRaiZer
     
  5. Jun 8, 2006 #4

    J77

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    I believe...

    Assuming steady state continuation...

    The system to solve should be just the righthand side evaluated at the steady state (fixed state [tex]x^*[/tex] and parameters [tex]\eta[/tex]) - [tex]f(x^*,\eta)=0[/tex].

    For simple bifurcations, eg. fold and Hopf, you need the characteristic evaluated at the appropriate eigenvalues, using the eigenvector (not the determinant), and a condition on this eigenvector...

    If you're near the turning point and don't converge, you have to adjust the length of the projection step or search back through the points you've already found and eg. take a bisection of the last two pairs of points, computed the new projection from these points.
     
    Last edited: Jun 8, 2006
  6. May 21, 2009 #5
    Hi, have you find the answer of your question?
     
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