# Eigenvalue Solution Changes as Mesh changes

Originally I try to use adaptive mesh to solve some eigenvalue problems. But the results I get are far from the real values. Finally I test it with 1D harmonic oscilator and uniform mesh, and I find the calculated eigenvalue is proportional to the mesh size.

I also test the demo examples in dolfin/

When I run "demo_eigenvalu

If I add "mesh=refine(mesh)" and run it again, I get the result is about 6

Is the eigenvalue solution should be independent with the mesh size?

## Question information

- Language:
- English Edit question

- Status:
- Solved

- For:
- DOLFIN Edit question

- Assignee:
- No assignee Edit question

- Solved by:
- Houdong Hu

- Solved:
- 2013-03-20

- Last query:
- 2013-03-20

- Last reply:
- 2013-03-20

## This question was reopened

Jan Blechta (blechta) said : | #1 |

An assembled matrix of whatever operator is typpicaly dependent on mesh size. Hence are eigenvalues.

Jan Blechta (blechta) said : | #2 |

On Wed, 20 Mar 2013 12:21:02 -0000

Jan Blechta <email address hidden> wrote:

> Question #224702 on DOLFIN changed:

> https:/

>

> Status: Open => Answered

>

> Jan Blechta proposed the following answer:

> An assembled matrix of whatever operator is typpicaly dependent on

> mesh size. Hence are eigenvalues.

>

I was thinking and realized that this is not good explanation. Better

way to look at it is that if you halve mesh step then you enlarge

(finite-

And that can have very problem-specific conclusions. For example when

operator represents harmonic oscilator you can't wonder that finer mesh

supports shorter wave-lengths, can you?

Jan

Houdong Hu (vincehouhou) said : | #3 |

I see the problem,

Eigenvalue problems is Ax=lamda Mx, A is the coefficient matrix, and M is the basis normalization matrix. Originally I though I only need to set up A, and M will be set up automatically. Now it seems you have to set both A and M to solve the eigenvalue problem.

I think people are more interested in the real eigenvalue problem, not just the eigenvalue of A.

Thanks~