- #1
TokenMonkey
- 5
- 0
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 F(x,\lambda)=0. We assume that the solution is known at x^0 and \lambda^0. To avoid the singularity of the Jacobian, and therefore the breakdown of Newton's method, at turning points, x and \lambda both become parameterised by arclength (s), and we end up with an augmented system of equations to solve:
F(x,\lambda)=0
\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
While this seems simple enough, how does one obtain the derivatives w.r.t s? 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!
Thanks,
TM
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 F(x,\lambda)=0. We assume that the solution is known at x^0 and \lambda^0. To avoid the singularity of the Jacobian, and therefore the breakdown of Newton's method, at turning points, x and \lambda both become parameterised by arclength (s), and we end up with an augmented system of equations to solve:
F(x,\lambda)=0
\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
While this seems simple enough, how does one obtain the derivatives w.r.t s? 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!
Thanks,
TM
