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Pseudo-arclength continuation implementation

  1. Nov 13, 2011 #1
    Hi there,

    Apologies if this is the wrong section for this question; I'm not a regular here. I assume a mod will move it if it's not the right place.

    I have a quick question regarding pseudo-arclength continuation. As some background, I am a chemical engineer, not a mathematician, applied or otherwise, so my knowledge of numerical methods is limited to the "standard" engineering stuff, which, unfortunately, does not include this.

    I've been reading up as extensively as I can on pseudo-arclength continuation, but unfortunately, it's all from second-hand sources; I don't have access to Keller's original paper (not that I'm sure that would help me). Here's what I understand at this point:

    We want to solve a problem [itex]F(x,\lambda)=0[/itex]. We assume that the solution is known at [itex]x^0[/itex] and [itex]\lambda^0[/itex]. To avoid the singularity of the Jacobian, and therefore the breakdown of Newton's method, at turning points, [itex]x[/itex] and [itex]\lambda[/itex] both become parameterised by arclength ([itex]s[/itex]), and we end up with an augmented system of equations to solve:

    [itex]\left(u-u^{0}\right)\mathrm{d}u^{0}/\mathrm{d}s+\left(\lambda-\lambda^{0}\right)\mathrm{d}\lambda^{0}/\mathrm{d}s-\Delta S=0[/itex]

    While this seems simple enough, how does one obtain the derivatives w.r.t [itex]s[/itex]? Not a single text seems to mention this. Ideas that spring to mind are forward differences using, say, cubic splines; however, that seems horrendously inefficient to me. There must be a better way!

  2. jcsd
  3. Nov 13, 2011 #2
    I suppose the question boils down to: is there a straightforward way of expressing [itex]x[/itex] and [itex]\lambda[/itex] as functions of [itex]s[/itex]?
  4. Nov 22, 2011 #3
    I found the answer: a procedure is described in Kubicek's Algorithm 502 in ACM Trans. Math. Software. dl.acm.org/citation.cfm?id=355675
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