Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In most of the books on numerical methods and finite difference methods the error is measured in discrete $L^2$ norm. I was wondering if people do the in Sobolev norm. I have never see that done and I want to know why no one uses that.

To be more specific look at the $$Au=f,$$ where assume $A_h$ is some approximation for $A$ and $U$ is the numerical solution for the system. Then if we plug the actual function $u$ into $A_hU=f$ and substruct we have $$A_h(u-U)=\tau$$ for $\tau$ being a local error. Thus I have an error equation $$e=A_h^{-1}\tau$$ What are the problems I am facing If I use discrete Sobolev norm?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

For one thing, it's a question of what norm measures how "accurate" the solution is. Which of the two error terms would you rather have: $0.1\sin(x)$ or $0.0001\sin(10000x)$? The first is smaller in the Sobolev norm, the second is smaller in the $L^2$ norm.

share|improve this answer
    
but if I choose to use Sobolev norm because I have control on not only how function behaves but also on the derivative, which might be of advantage, I have to estimate then $||A_h^{-1}||_{H^1}$ and $||\tau||_{H^1}$ is that right? I am not sure such a norm for matrix exist.... –  Medan Jun 19 '12 at 13:33
    
@Medan There is at least reasonable discrete Sobolev norm on vectors, namely $\sum_{k} |x_{k}-x_{k+1}|^2+(n^{-1}\sum_k x_k)^2$. I threw in the average of $x$ to make it a norm rather than a seminorm. For any vector norm $\|x\|$ there is a corresponding matrix norm, namely $\|A\|=\sup_{\|x\|\le 1}\|Ax\|$. This definition is not great from the computational viewpoint, though. Unfortunately, I don't know numerical analysis, so everything I say is just a layman's guess. Looking up "discrete Sobolev norm" is more likely to be helpful. –  user31373 Jun 19 '12 at 14:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.