Let $\omega \in (\pi, 2\pi)$ and $\Omega=\{(r\cos\phi, r\sin \phi); 0<r<1, 0<\phi<\omega\}$. Furthermore, let $u$ be the solution of \begin{eqnarray} -\Delta u=0;&~ \textrm{in}~ \Omega\\ u=g;&~ \textrm{on} ~\partial\Omega \end{eqnarray}

where $g(r,\phi)=r^{\pi/\phi}\sin(\frac{\pi}{\omega}\phi)$.

Consider a uniform triangulation of $\Omega$ and the finite element space $V_h^1(\Omega)$ of continuous and piecewise linear finite elements. Prove the estimate $$\|u-I_hu\|_{W^{1,2}(\Omega)}\leq C(u,\omega)h^{\pi/\omega}$$

where $I_h$ is interpolation operator on $V_h^1(\Omega)$.

I know basic theorems about interpolation error estimate where one gets bound with integer power of $h$ and seminorm of $u$. Here one can apply this estimate on some $\Omega-B_{\alpha}(0)$ where $u \in W^{2,2}(\Omega-B_{\alpha}(0))$ but then it stays to give a good estimate on $W^{1,2}$ norm on $B_{\alpha}$.

Any ideas how to deal with this problem?


1 Answer 1


There exist interpolation estimates that are more general and take into account the reduced regularity. The following inequality is directly from Ciarlet's book The Finite Element Methods for Elliptic Problems (p. 123): $$ |v-\Pi_h v|_{m,q,K} \leq C (\text{meas}(K))^{1/q-1/p} \frac{h_K^{k+1}}{\rho_K^m} |v|_{k+1,p,K}, $$ with $v \in W^{k+1,p}(K)$. After you figure out for which $z$ you have $g \in W^{z,2}(K)$ then you can apply this theorem to get a result like that.

I cheated a bit, though. You cannot have a well-defined $\Pi_h$ in the sense of Ciarlet (point evaluations) since your regularity is probably so low that point values are not well-defined. Thus, you would need to use a more general interpolation operator, like Clement or Scott$-$Zhang interpolant. The result, however, that you'd get is the same (modulo a constant) with a very high probability.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .