NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/sci/math/numanal/3909

Internet Message Format | 1993-01-21 | 5.8 KB

Path: sparky!uunet!noc.near.net!argo.hks.com!pegasus.hks.com!curran
From: [e-mail] (Bill Mills-Curran)
Newsgroups: sci.math.num-analysis
Subject: Re: HELP!!(Cholesky Decompsition for negative-definite matrix ???)
Date: 21 Jan 1993 13:11:08 GMT
Organization: Hibbitt, Karlsson & Sorensen, Inc.
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Distribution: world
Message-ID: <[e-mail]>
References: <[e-mail]> <[e-mail]> <[e-mail]>
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In article <[e-mail]>, [e-mail] (Chen, Shen Yeh) writes:
|> In article <[e-mail]>, [e-mail] (Makarand A. Kulkarni) writes...
|> >In article <[e-mail]>, [e-mail] (Chen, Shen Yeh) writes:
|> >|>
|> >|>    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
|> >[e-mail]
|> 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
|> described    above:
|>
|>  ###########################################################################
|>  ##   @                   ##   |      *  *                                ##
|>  ## A  \                  ##   |   *   |    *                      *      ##
|>  ##      \                ##   | *     |      *                   *       ##
|>  ##        \              ##  L|*      |        *  <DEFLECTION>  *        ##
|>  ##          \ <----------##  O|-------|------------------------*-----    ##
|>  ##          /    LOAD    ##  A|       |          *            *          ##
|>  ##        /              ##  D|       |            *         *           ##
|>  ##      /                ##   |       |              *     *             ##
|>  ## B  /                  ##   |       |                 *                ##
|>  ##   @                   ##           |C               D|                ##
|>  ########< FIG1 >##########################< FIG2 >#########################
|>
|>    Please see FIG2.    [K] will be sigular on C & D
|>                        [K] SHOULD be negative-definite from C to D
|>
|>   HOWEVER, SOMEONE TELL ME THAT,  IF [K] IS NEGATIVE, THEN  -[K] SHOULD BE
|>   POSITIVE!! SO THERE IS NO PROBLEM!!!!!!!
|>   IS THIS CORRECT ???   IF YES(OR NO), CAN ANYONE PROVE IT???
|>
|>   OR MAYBE [K] IS NOT NEGATIVE FROM C TO D. I HAVE PROVE IT TO BE NEGATIVE
|>   , ALTHOUGH IT MAY BE WRONG.
|>
|>
|>   THANK ALL OF YOU AGAIN !!
|>

The stiffness matrix for the physical problem you describe is not
negative definite, but is indefinite.  This problem cannot be solved
by traditional energy methods.  There are two approaches:

1.  Recognize the an unstable structure does not have a static
solution, so switch to a dynamic method.  The mass matrix then
provides numeric stability for implicit methods.  Explicit methods do
not require a positive definite stiffness matrix, so they can zip
right through the instability.

2.  Use the RIKS method (as in ABAQUS) to follow the load-deflection
path.

--
Bill Mills-Curran ([e-mail])
Hibbitt, Karlsson & Sorensen, Inc.    Tel: [phone]
1080 Main St., Pawtucket, RI 02860    Fax: [phone]