# Simple question about find the element indicator

It is necessary for me to find the element indicator for poisson equation by linear basis in FEM. Therefore I have to find the following:

$$\eta_k=h_k||f+\Delta u_h||_{L^2(k)}+\frac{1}{2}h_k^{1/2}||n.\nabla u_h||,$$

So, I should approximate $\Delta u_h$ numerically, and I need to second order derivative of the shape functions. But they are linear and this means $\Delta u_h=0$. I can not understand what happen here, and what does $||f+\Delta u_h||_{L^2(k)}$ mean? I am so confuesed by this term. Is there anybody can help me? Am i wrong?

• Yes for linear elements $\Delta u_h|_K = 0$ and the indicator still works. Have you have seen the proof that this is an upper bound for the $H^1$ error? Do you want some intuitive explanation or what? I do not fully understand the question. Care to elaborate? – knl Jan 24 '16 at 15:47
• Thanks a lot, i just could not understand how it works when $\Delta u=0$. – Rosa Jan 24 '16 at 16:37

It is correct that the error estimator

$$\eta_k=h_k||f+\Delta u_h||_{L^2(k)}+\frac{1}{2}h_k^{1/2}||n.\nabla u_h||,$$

simplifies to

$$\eta_k=h_k||f||_{L^2(k)}+\frac{1}{2}h_k^{1/2}||n.\nabla u_h||,$$

if you have linear elements. However, the term then is suboptimal. (It does have the right convergence order, but its value is too large.) A finer analysis shows that you can replace it by

$$\eta_k=h_k||f-f_h||_{L^2(k)}+\frac{1}{2}h_k^{1/2}||n.\nabla u_h||,$$

where $f_h$ is a piecewise constant interpolant of the function $f$ on each cell. This term is called "data oscillation".

• Interesting. Do you get a smaller global error if you construct mesh families using the modified estimator to refine elements with e.g. 10% of total error? – knl Jan 27 '16 at 9:08
• @knl: I wouldn't know, to be honest. Most "reasonable" error indicators produce very similar meshes that basically don't differ in any substantial way with regard to their approximation quality. My best guess is that you won't get better meshes using the modification, but you get a better estimate of the error. – Wolfgang Bangerth Jan 27 '16 at 13:29

The proof goes roughly as follows: Let $u$ be exact solution and $u_h$ discrete solution. Moreover, let $e=u-u_h$ and $\pi_h$ be the interpolation operator from $H^1$ to the discrete finite element space. Facts used in proving the a posteriori error estimate are the following:

1. Coercivity of the bilinear form.
2. Galerkin orthogonality.
3. Problem statement and elementwise integration by parts.
4. Cauchy-Schwarz for inner products and for sums.
5. Interpolation estimates $\|e-\pi_h e\|_{0,K} \lesssim h_K |e|_{1,K}$ and $\|e-\pi_h e\|_{0,E} \lesssim h_E^{1/2} |e|_{1,K}$ and the definition of $\|\cdot\|_1$.

\begin{align*} \|u-u_h\|_1^2&\lesssim (\nabla(u-u_h),\nabla(u-u_h))\\ &=(\nabla(u-u_h),\nabla(e-\pi_h e))\\ &=(f,e-\pi_h e)+\sum_{K}(\Delta u_h, e-\pi_h e)_K-\sum_{E}([[\nabla u_h\cdot n]],e-\pi_h e)_E\\ &\lesssim \big(\sum_{K} h_K^2 \|\Delta u_h+f\|_{0,K}^2\big)^{1/2} \big(\sum_{K} h_K^{-2}\|e-\pi_h e\|_{0,K}^2\big)^{1/2}\\ &\phantom{=}+\big(\sum_{E} h_E \|[[\nabla u_h\cdot n]]\|_{0,E}^2\big)^{1/2} \big(\sum_{E} h_E^{-1}\|e-\pi_h e\|_{0,E}^2\big)^{1/2}\\ &\lesssim \Big(\big(\sum_{K} h_K^2 \|\Delta u_h+f\|_{0,K}^2\big)^{1/2}+\big(\sum_{E} h_E \|[[\nabla u_h\cdot n]]\|_{0,E}^2\big)^{1/2} \Big)\|u-u_h\|_1 \end{align*}

Finally you divide by $\|u-u_h\|_1$ to get the bound.

In particular, note that none of the steps are invalid for $u_h|_K \in P^1(K)$. Thus, $\Delta u_h|_K=0$ should work just fine.