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when applying a Finite Difference scheme for an IVP, two factors come to mind when considering stability:

One factor would be the condition number of the approximation operator. The other factor would be the CFL Condition. I think I understand both, but :

1) How come CFL condition only depends on the equation , while the Condition number is dependent on the approximation operator as well?

2) Are the two dependent? Can one be deduced from the other?

The problem in which I've encountered the dilemma is the one dimensional Biharmonic Linear Schrodinger equation, but I think the aforementioned is far more general then a specific problem.

$iu_t = u_{xxxx}$, $x$ is periodic in $[0, 2 \pi]$, $t>0$ and $u(x,0) = u_0 (x)\in H^2$

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closed as off-topic by Michael Renardy, Franz Lemmermeyer, Jan-Christoph Schlage-Puchta, Daniel Moskovich, Wolfgang Apr 22 '16 at 7:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Michael Renardy, Franz Lemmermeyer, Jan-Christoph Schlage-Puchta, Daniel Moskovich, Wolfgang
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is not a research-level question, and so is not appropriate for mathoverflow. You could ask on math.SE or scicomp.SE. But a lot of confusion is evident in the question; I recommend that you read an introductory text like This one to get some basic understanding. $\endgroup$ – David Ketcheson Nov 16 '14 at 11:31
  • $\begingroup$ Well , I am reading a different textbook (Gustafsson, Kreiss and Oliger), and I am still confused in something that interferes with my ongoing research. $\endgroup$ – Amir Sagiv Nov 16 '14 at 22:09
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How come CFL condition only depends on the equation , while the Condition number is dependent on the approximation operator as well?

The CFL condition is a restriction on the relation between the time and space steps in order for convergence to be possible. It does depend on what you call "the approximation operator", since the numerical domain of dependence of your method depends on the stencil width.

Can one be deduced from the other?

No.

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  • $\begingroup$ Could you explain why the two are not connected? $\endgroup$ – Amir Sagiv Apr 21 '16 at 18:47

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