I know that the piecewise linear finite element approximation $u_h$ of $$ \Delta u(x)=f(x)\quad\text{in }U\\ u(x)=0\quad\text{on }\partial U $$ satisfies $$ \|u-u_h\|_{H^1_0(U)}\leq Ch\|f\|_{L^2(U)} $$ provided that $U$ is smooth enough and $f\in L^2(U)$.

Question: If $f\in H^{-1}(U)\setminus L^2(U)$, do we have the following analogous estimate, in which one derivative is taken away on both sides: $$ \|u-u_{h}\|_{L^2(U)}\leq Ch\|f\|_{H^{-1}(U)}\quad? $$

Can you provide references?

Thoughts: Since we still have $u\in H^1_0(U)$, it should be possible to obtain convergence in $L^2(U)$. Intuitively, this should even be possible with piecewise constant functions.

  • $\begingroup$ I think that you get $\|u-u_h\|_0 \leq C h \|u-u_h\|_1$ from the standard Nitsche trick even for $u \in H^1$. You can find this e.g. in Braess - Finite elements. $\endgroup$ – knl Feb 3 '17 at 14:53

Yes, this is the standard Aubin-Nitsche (or duality) trick. The idea is to use the fact that $L^2$ is its own dual space to write the $L^2$-norm as an operator norm $$\|u\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u,\phi)}{\|\phi\|_{L^2}}.$$ We thus have to estimate $(u-u_h,\phi)$ for arbitrary $\phi\in L^2$. To do that, we "lift" $u-u_h$ to $H^1_0$ by considering first for arbitrary $\phi\in L^2$ the solution $w_\phi\in H^1_0$ of the dual problem $$\label{eq1}\tag{1} (\nabla w_\phi,\nabla v) =(\phi,v) \qquad\text{for all }v\in H^1_0. $$ Using the standard regularity of the Poisson equation, we know that $$ \|w_\phi\|_{H^2}\leq C \|\phi\|_{L^2}.$$

Inserting $v=u-u_h\in H^1_0$ in \eqref{eq1} and using Galerkin orthogonality for any finite element (in your case, piecewise linear) function $w_h$ yields the estimate $$ \begin{aligned} (\phi,u-u_h) &= (\nabla w_\phi,\nabla (u-u_h))\\ &= (\nabla w_\phi-\nabla w_h,\nabla (u-u_h))\\ &\leq C \|u-u_h\|_{H^1}\|w_\phi-w_h\|_{H^1}. \end{aligned} $$ Since this holds for all $w_h$, the inequality is still true if we take the infimum over all piecewise linear $w_h$. We therefore obtain $$\label{eq2}\tag{2} \|u-u_h\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u-u_h,\phi)}{\|\phi\|_{L^2}} \leq C \|u-u_h\|_{H^1} \sup_{\phi\in L^2\setminus\{0\}} \frac{\inf_{w_h}\|w_\phi-w_h\|_{H^1}}{\|\phi\|_{L^2}}. $$ This is the Aubin-Nitsche-Lemma.

The next step is now to use standard error estimates for the best finite element approximation of solutions to the Poisson equation. Since $u$ is only in $H^1$, we don't get a better estimate than $$\label{eq3}\tag{3} \|u-u_h\|_{H^1} \leq \inf_{v_h}\|u-v_h\|_{H^1} \leq c \|u\|_{H^1} \leq C \|f\|_{H^{-1}}. $$ But fortunately, we can use the fact that $w_\phi$ has higher regularity since the right-hand side $\phi\in L^2$ instead of $H^{-1}$. In this case, we have $$\label{eq4}\tag{4} \inf_{w_h}\|w_\phi-w_h\|_{H^1} \leq c h \|w_\phi\|_{H^2} \leq C h \|\phi\|_{L^2} $$ Inserting \eqref{eq3} and \eqref{eq4} into \eqref{eq2} now yields the desired estimate.

(Note that the standard estimates require that the polynomial degree $k$ of the finite element approximation and the Sobolev exponent $m$ of the true solution satisfy $m<k+1$, so this argument doesn't work for piecewise constant ($k=0$) approximation. We also have used that $u-u_h\in H^1_0$ -- i.e., that we have a conforming approximation -- which is not true for piecewise constants.)

Since you asked for a reference: You can find a statement (even for negative Sobolev spaces $H^{-s}$ instead of $L^2$) in Theorem 5.8.3 (together with Theorem 5.4.8) in

Susanne C. Brenner and L. Ridgway Scott, MR 2373954 The mathematical theory of finite element methods, Texts in Applied Mathematics ISBN: 978-0-387-75933-3.

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    $\begingroup$ And I get to make use of our shiny new citation feature :) $\endgroup$ – Christian Clason Feb 3 '17 at 12:55
  • $\begingroup$ Thanks for your answer, but continuous functions aren't embedded into $ H^1_0$ are they? $\endgroup$ – Bananach Feb 3 '17 at 14:12
  • $\begingroup$ Yes, sorry, I stroked off there -- they're dense, but not embedded. The duality argument works the same, though (just work with $H^1_0$ and $H^{-1}$ directly). I'll edit my answer accordingly. $\endgroup$ – Christian Clason Feb 3 '17 at 14:18
  • $\begingroup$ Thanks for the extensive update. And for finding another shiny citation $\endgroup$ – Bananach Feb 3 '17 at 16:12
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    $\begingroup$ @Praveen I don't think you need any theory here. Simple choose $v_h$ to be constant zero. $\endgroup$ – Bananach Feb 4 '17 at 17:51

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