>|> I need to solve a linear system equation. In fact, that's the stiffness
>|> matrix [K] from finite element method. The system is like :
>|> [K]*{U}={P}. where [K](nxn)
>|>
>|> [K] IS ALWAYS SYMETRIC, but not necessarily positive-definite. If it is
>|> singular, we think this system to be unstable. Because of the symetry, I
>|> always use Cholesky Decompsition to reduce the size of [K]. However, now
>|> I have some situation that [K] is negative-definite but I still have to
>|> solve the equation. I have not try Gauss method yet, however.
>|>
>|> Does anyone kow :
>|> 1) Is there any method I can use to reduce [K], and still solve the system
>|> when [K] is negative-definite ? (Note that [K] IS ALWAYS SYMETRIC)
>|> 2) If such method exists, what's the criteria for positive- or negative-
>|> definite?
>|> 3) If no such method exits,( We can not take advantage of symetry to
>|> reduce the size of [k], I mean.), does Gauss method work? What's
>|> the criteria for question (2)?
>|>
>|> NOTE : (1) In such system, we do not do pivotting among the rows of [K],
>|> if using Gauss method. If there is a zero on the diagonal,
>|> we consider it as unstable.
>|> (2) [ ] ====> matrix { } ====> column vector
>|>
>|> It will be greatly appreciated if anyone can give me some idea. If you
>|> do not want to type all that stuff to me, just tell me the theorem used.
>|> Of course, it will be better if you can tell me reference books. Any
>|> assistance will be greatly appreciated.
>|>
>|> Thank you very very very much !!!!!!
>|>
>
>
>I think I dont understand your problem. If K is the stiffness matrix from
>the finite element method, how can K be NEGATIVE DEFINITE?
>The qudratic form of K represents some kind of internal energy (eg. strain energy in solid mechanics), so K is always positive semi-definite, right???? It seems you ought not to worry about a negative definite K.
>
>I realize that I am not shedding any light on your problem, but would
>appreciate it if you could clarify this.
>In particular, let me know what physical problem gives rise to a negative definite K.
>Regards
>
>Makarand
>makarand@puma.larc.nasa.gov
Hi,
This is from Shrn-yeh Chen, for further discription of the problem.
This is hard to explain here. In fact, if a structure is under 'unloading'
condition, [k] will be negative-definite. That is, in general condition,
an incremental load will cause the total strain energy increase-if the
structure is 'intended' to be stable. Other wise, there is condition call
'snapping through ' happening.
For example, a loaded sturcture like the FIG1 will have the behavior
