Ondrej Certik wrote:

On Fri, Aug 1, 2008 at 4:54 PM, LUK ShunTim <shunt...@polyu.edu.hk> wrote:

Robert Cimrman wrote:

Ondrej Certik wrote:

On Fri, Aug 1, 2008 at 12:23 PM, Ondrej Certik <ond...@certik.cz> wrote:

> If you plugged the basis into the equation naively (e.g. directly), > you'd get unsymmetric matrices, so you first integrate the equation > per partes (or the green theorem, as you use in the paper). We do not care so much about symmetry (we use UMFPACK, right?), but BTW, we do actually care a lot, because in my case we are using an eigensolver and if the matrix is not symmetric, the eigenvalues are in general complex. Not mentioning that pysparse (imho) can only solve symmetric matrices. So using the per partes is actually crutial just to be able to solve it easily. Yes, in your case we do care.

The argument about linear elements is imho not that important, you can always use higher order elements. But as you said, it would not be that easy.

So from my point of view, you really want to rewrite the problem as symmetric just to move forward. Everything else is just a sideeffect, that can be fixed somehow. Symmetry is a win if one can achieve it, I agree with that. But believe me, weak formulations have other merits as well :)

r. Hi,

Let me chime in as a civil engineer. :-)

The 2nd order ODE for an axially loaded rod given by Ryan could also represent a tensioned string subject to lateral loads. For a point load, the deflected shape will be 2 straight line segments which has a discontinuity of slope at the location of the point load and hence not C^2. The weak formulation incorporates this very naturally. Further more, weak formulation is equivalent to the principle of virtual work in structural mechanics.

I don't know the equations in this case, but I guess it's the same as in electromagnetism, where you also get solutions that are discontinuous at the boundary (either tangent or perpedicular projection) and no one doubts that maxwells equations govern it. I mean I don't understand why you require C^2? The discontinuity in slope means that the derivative is not continuous, right? E.g. the laplace operator generates couple delta functions, correct? I have no problems with that and I don't need any weak formulation for that. C^2 is just something that mathematicians invented, so that they have some justifications to think that weak formulation is something more, but in fact everybody knows that the solution that you are looking for is not C^2 and you can get it from the original differential equation as easily, or is there something more that I don't see?

What is the definition of the (second) derivative you use? Once you use something with integrals and test functions (e.g. distributions), you have the weak formulation. You just do not like the term "weak", right? Well, it's just a definition of the exactly same thing you think about the solution.

Classical solution = C^2 Weak solution = H^1

It's a naming convention, and everybody uses it ;)

r.