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Classical global estimate for $H^1$ error

Computational Science Asked on July 12, 2021

I’m having lots of troubles in understanding the proof the estimation of the classical $H^1$ error using finite elements of degree $r$.

$$||u-u_h||_{H^1(Omega)} leq frac{M}{alpha} C h^r |u|_{H^{r+1}(Omega)}$$
where $M$ and $alpha$ are the operator norm and the coercivity constant, respectively and $C$ is a constand independent on $h$.

My book (Quarteroni – Numerical methods for differential problems) shows this by using the Cèa’s lemma and the estimate on the seminorm of the interpolation error. More precisely, it states:

$$||u-u_h||_{H^1(Omega)} leq frac{M}{alpha} inf_{v_h in V_h} ||u – v_h||_{H^1(Omega)} leq frac{M}{alpha} ||u-Pi^ru||_{H^1(Omega)}$$
Now use the estimate $$|u-Pi^r u|_{H^m(Omega)} leq Ch^{r+1-m} |u|_{H^{m}(Omega)}$$ (valid for $m=0,1$ and $rgeq 1$.
) with $m=1$ to conclude

The very last step is what I cannot do: I can’t see how to pass from the $H^1$ seminorm to the full $H^1$ norm. I mean, if I need to bound $||u-Pi^ru||_{H^1(Omega)}$, how can I use the bound on the seminorm?

One Answer

I think $u - Pi u$ is zero on a part of the boundary. Then you can use Poincare inequality to bound $| u - Pi u |_0 leq C| u - Pi u|_1$.

It could also be possible to look at the case $m = 0$ and argue the $L^2$ part is of higher order and, hence, smaller in the asymptotic limit.

Answered by knl on July 12, 2021

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