# Ritz projection error estimate for bilinear Lagrange element?

For second order elliptic problem \begin{align} -\Delta u=f,\quad in ~~\Omega\\ u=0,\quad on~~ \partial\Omega, \end{align} we have for the Ritz projection for $$P_1$$ conforming element \begin{align} \|u-R_hu\|_{W^{0,p}}+h\|u-R_hu\|_{W^{1,p}}\leq Ch^2\| u \|_{W^{2,p}}，\quad u\in W_0^{1,p}(\Omega)\cap W^{2,p}(\Omega). \end{align} Does this error estimate also hold for bilinear $$Q_1$$ element?

• You did not define the "Ritz projection". However, this looks like a superconvergence result, because typically for linear elements you get $O(h)$. What you have here, $O(h^2)$, is common for quadratic elements.
– knl
Jun 25 at 9:01
• For a quasi-uniform rectangular partition $\mathcal{T}_h$ for rectangular domain $\Omega$. The bilinear Lagrange finite element space is \begin{align*} V_h=\{v_h\in C(\bar{\Omega}),v_h|_{K}\in span\{1,x,y,xy\},v_h|_{\partial\Omega}=0\}. \end{align*} And the Ritz projection is \begin{align*} (\nabla(u-R_hu),\nabla v_h)=0,\quad \forall v_h\in V_h. \end{align*} Jun 25 at 10:03
• @knl no, that is the standard rate (note that $W^{0,p} = L^p$ -- for which you get $h^2$ using the Aubin--Nitsche trick -- and that the $W^{1,p}$ norm is multiplied by $h$, so by canceling you get the expected rate of $h$ only) Jun 25 at 13:32
• see Theorem 5.9 in arxiv.org/abs/1709.08618 (shameless plug). And, yes, the the same rate holds for bilinear elements, see Brenner and Scott, Chapter 4.6 (in the third edition). Jun 25 at 13:33
• I'm quite sure there wasn't $h$ in front of the second term when I read it first. Oh well, my mistake.
– knl
Jun 25 at 20:51