Article View: sci.math.num-analysis
Article #1097Re: FEM with SYMMETRIC but NON-POSITIVE stiffness
From: W. Donald Rolph
Date: Tue, 26 Jan 1993 21:45
Date: Tue, 26 Jan 1993 21:45
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4021 bytes
In article <ROGER.93Jan27083432@kea.grace.cri.nz>, roger@maths.grace.cri.nz (Roger Young) writes:
|>
|> I have been following the correspondence in the num-analysis
|> newsgroup concerning FE methods as applied to symmetric but
|> non-positive definite matrices.
|>
|> I appear to have a similar problem with the poro-elastic
|> equations. These equations are: (1) the elastic equilibrium
|> equations with the pressure gradient as a source term (or,
|> in fact, it can be convenient to work with the variation
|> of these equations)
|>
|> grad(j)d(stress)(ij) = grad(i)d(pressure)
|>
|> (2) the pressure diffusion equation with an additional time
|> derivative de/dt of the elastic dilatation e =
|> grad(i)(displacement)(i)
|>
|> S.d(pressure)/dt + de/dt = -grad(i)q(i)
|>
|> where the storage S can be taken as constant, and q is the
|> Darcy velocity which is proportional to pressure gradient
|> q(i) prop grad(i)(pressure).
|>
|> In the displacement-pressure formulation the field vector
|> is chosen as z = (displacement, pressure)^T. Then according
|> to Lewis and Roberts "The Finite Element Method in Porous
|> Media Flow" (in Fundamentals of Transport Phenomena in Porous
|> Media, eds Bear and Corapcioglu), the finite element equations
|> for the coupled problem can be written in the form
|>
|> K(dz/dt) + Hz = f
|>
|> where K is SYMMETRIC.
|>
|> If elasticity is decoupled from pressure, then it appears that
|> the reduced stiffness matrix is positive definite for displacement
|> and pressure individually. However, for the coupled problem
|> numerical analysis appears to show that K has both +ve and -ve
|> eigenvalues, ie is not positive definite.
|>
|> My question(s) are: is this really true? How does one set about
|> proving (or disproving) positive-definiteness? what is the
|> physical significance of positive-definiteness?
|>
|> Any help from readers would be appreciated...
This at first glance appears to akin to the problem that crops up with Lagrange
multipliers (which can masquerade as pressures under many conditions). If this
is true, then solving for the all elasticity terms connected with a given node
before you solve for any pressure terms for a given node you should find typical
factorization algorithms to work (dont try cholesky decomposition however).
Postive definitenes is in my experience as muchg a religious phenomena as
anything else (opps here go the battles). Consider however that if I multply
both sides of a positive definite system by -1 I get a negative definite system,
which is still very well behaved. In my experience and from my religious
framework positive definite means that I have a positive energy functional which
can be minimized (negative definiteness means to me that I have a negative enrgy
functional which needs to be maximized!).
Problems seem to occur for the following conditions:
indefinite - zero eigenvalues
semidefinite - some are postiive, some are negative
which seems to be your case. In some cases a semidefinite system can be
organized as a positive definite system followed by a negative definite system.
Hope these philosophical ramblings more or less answered your question.
--
Regards.
Don Rolph a722756@pan.mc.ti.com WD3 MS10-13 (508)-699-1263
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Newsgroups: sci.math.num-analysis Path: arc1-ca.pugleaf.net!sparky!uunet!math.fu-berlin.de!ira.uka.de!scsing.switch.ch!news.univie.ac.at!paladin.american.edu!howland.reston.ans.net!usc!cs.utexas.edu!csc.ti.com!tilde.csc.ti.com!pan.mc.ti.com!a722756 From: a722756@roper.mc.ti.com (W. Donald Rolph) Subject: Re: FEM with SYMMETRIC but NON-POSITIVE stiffness Message-ID: <1993Jan26.164528@roper.mc.ti.com> Originator: a722756@roper.mc.ti.com Sender: usenet@pan.mc.ti.com (USENET News System) Organization: Texas Instruments / Attleboro Mass / USA References: <ROGER.93Jan27083432@kea.grace.cri.nz> Distribution: inet Date: Tue, 26 Jan 1993 21:45:28 GMT Lines: 77