On 10/20/2016 02:55 PM, JAEKWANG KIM wrote:
Hi, all.
I got a question on the symmetry of system matrices.
I am looking for ways to solve Non-linear PDE using deal.ii and Dr.
Bangerth's video lecture from 31.5~31.7 is greatly helpful to understand
basic of this part.
In many elementary codes in tutorial lists, we have used Conjugate
Gradient Method to solve the matrix.
CG Method basically assumes that matrix A is symmetric.
I also checked that deal.ii offers other solver method, GMRES solver for
non-symmetric system matrix A.
Question 1. My question is where will the non-symmetry arise?
To be more specific, let's take a example of "linearized" non-linear
minimal surface equation with Picard iteration (as Dr. Bangerth's
lecture 31.65)
<https://lh3.googleusercontent.com/-TYzXkQCHcTA/WAksZOt6tjI/AAAAAAAAA7E/kX0e1wXWcfMcKc-dDYeq2rOOmXi-YtHjwCLcB/s1600/Screen%2BShot%2B2016-10-20%2Bat%2B3.42.54%2BPM.png>
where u is scalar function.
Because we use previous solution u_k , the term "1/sqrt(1+(nabla_u)^2)"
will work as a scalar coefficient that varies over the domain.
Does this fact will make our system matrix non-symmetric?
No. As you mention correctly, the additional factor is just a scalar
coefficient. It depends on the *previous* solution, but for the purposes
of symmetry, this is not important.
Question 2. Can we expect our numerical error will also behave ~O(h^p)
for non-linear system that includes iterative method?
Not only mesh spacing and shape function approximation degree, but also
many other parameters might influence error.
(e.g. mapping degree, and degree of gauss quadrature...)
For Non-linear problem with iterative method, the tolerance between
previous solution and solution might also affect error
However, if I use enough small tolerance between abs(u_{k}-u_{k+1}),
will the error will also show ~O(h^p)?
Correct, the error introduced by the nonlinear iteration is a component
of the overall error. But you can control it if you run the Newton
iteration with a small enough tolerance. In order to achieve O(h^p), you
will of course have to ensure that the iteration error also becomes
smaller and smaller as you refine the mesh, i.e., you have to reduce the
Newton tolerance with every mesh refinement.
Best
W.
--
------------------------------------------------------------------------
Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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