- #1
nylonsmile
- 8
- 0
I'm struggling to implement a pseudo arclength continuation method for my system. Here is what I have so far.
I am trying solve the system of equations [tex]F(x, \lambda) = 0[/tex] but if I parameterise only by using lambda, I can't get around turning points, so I paramterise by "arclength" s and attempt to solve the system of equations:
[tex]F(x(s), \lambda (s)) = 0\\
N(x, \lambda, s) = 0[/tex]
Where N is the linearised normalisation equation (' denotes differentiation with respect to s):
[tex]N(x, \lambda, s) \equiv x' \delta x + \lambda ' \delta \lambda - \delta s = 0[/tex]
Where ds is a specified step length. I solve these using a Newton-type method:
[tex]
\begin{pmatrix}
F_{\textbf{x}} & F_{\lambda} \\
N_{\textbf{x}} & N_{\lambda}
\end{pmatrix}
\begin{pmatrix}
\delta x \\
\delta \lambda
\end{pmatrix}
= -\begin{pmatrix}
F_0 \\
N_0
\end{pmatrix}[/tex]
Where the subscripts denote differentiation. So far so normal, this is the scheme as it is usually presented. I'm having difficulty figuring out exactly what to do though, particularly with the bottom half of the right hand side vector. What should go in there? The unlinearised normalisation? [tex]||x'||^2 + \lambda '^2 - 1[/tex] But if I use that, then where does the step length come into it? I think I'm getting confused by the linearisation - usually with this type of thing I'd only have [tex]\delta[/tex] for variables I'm trying to solve for, but here we also have it for s. It's a constant, and is confusing me no end.
Any help would be greatly appreciated, even just pointers to books/papers. Feeling pretty stupid atm!
I am trying solve the system of equations [tex]F(x, \lambda) = 0[/tex] but if I parameterise only by using lambda, I can't get around turning points, so I paramterise by "arclength" s and attempt to solve the system of equations:
[tex]F(x(s), \lambda (s)) = 0\\
N(x, \lambda, s) = 0[/tex]
Where N is the linearised normalisation equation (' denotes differentiation with respect to s):
[tex]N(x, \lambda, s) \equiv x' \delta x + \lambda ' \delta \lambda - \delta s = 0[/tex]
Where ds is a specified step length. I solve these using a Newton-type method:
[tex]
\begin{pmatrix}
F_{\textbf{x}} & F_{\lambda} \\
N_{\textbf{x}} & N_{\lambda}
\end{pmatrix}
\begin{pmatrix}
\delta x \\
\delta \lambda
\end{pmatrix}
= -\begin{pmatrix}
F_0 \\
N_0
\end{pmatrix}[/tex]
Where the subscripts denote differentiation. So far so normal, this is the scheme as it is usually presented. I'm having difficulty figuring out exactly what to do though, particularly with the bottom half of the right hand side vector. What should go in there? The unlinearised normalisation? [tex]||x'||^2 + \lambda '^2 - 1[/tex] But if I use that, then where does the step length come into it? I think I'm getting confused by the linearisation - usually with this type of thing I'd only have [tex]\delta[/tex] for variables I'm trying to solve for, but here we also have it for s. It's a constant, and is confusing me no end.
Any help would be greatly appreciated, even just pointers to books/papers. Feeling pretty stupid atm!