1
$\begingroup$

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?

1 Answers 1

1

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.

  • 0
    @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.2012-06-19