# choice of the norm for the error of the numerical method

When I read books on finite differences they often end up using discrete $L^2$ norm for estimating the error as it naturally arises from weak formulation. I was wondering if people do that in Sobolev norm and when it is useful. I have not seen any at least in finite-differences book.

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? What would that be then, it should involve derivatives estimates, but can I do one for the local truncation error?

Thanks!

• How do you define Sobolev norm? The L2 norm is a Sobolev norm. Are you thinking of the H1 norm, for example? – Wolfgang Bangerth Jul 8 '12 at 19:21
• I think it has something to do with the associated scalar product, but I don't know further. Please answer this question. – vanCompute Jul 8 '12 at 21:18
• yes, I guess I am thinking of a discrete norm for H^1, so there should be some discretization of a continuous norm. – Kamil Jul 8 '12 at 23:56

Maybe a note on the motivation to use the $L^2$-error. It is natural to estimate the error in the input of a partial differential equation in the $L^2$-norm. The mathematical motivation of this is the Gauss-Markov theorem:
It states that, if your measurement satisfies intuitive probability laws, then a least square approximation based on measurements will provide the best approximation with high probability. This choice of norm propagates within partial differential equations, so you end with $L^2$-Sobolev norms.
The discrete $H^1$ norm of the error is simply $\sqrt{e^T A_h e}$, so you immediately get the equation $$e^T A_h e = e^T \tau = e^T A_h^{T/2} A_h^{-1/2} \tau$$ and thus $$e^T A_h e \le \sqrt{e^T A_h e } \sqrt{\tau^T A_h^{-1} \tau}$$ and $$\|e\|_{H^1 \textrm{discrete}} = \sqrt{e^T A_h e} \le \sqrt{\tau^T A_h^{-1} \tau}$$ You treat the term on the right as always with truncation errors.
• thanks, however, I'm not sure how to proceed. For example, in the case of $L^2$ discrete norm from above I'd have: $$||e||_{L^2,d}\leq ||A^{-1}_h||_{L^2,d}||\tau||_{L^2,d}$$ Then If I could show that $|\tau|<C_1h^2$ that would imply that $||\tau||_{L^2,d}<C_2h^2$ and then with a combination of boundedness of $||A^{-1}_h||_{L^2,d}$ we have the error estimates. Now that I see the discrete Sobolev norm, I'd have to estimate $||\tau||_{H^1discrete}$ which is a vector norm? What does it mean in the finite finite difference meaning? Can I use this norm to estimate the error of Crank-Nicolson? – Kamil Jul 9 '12 at 15:49