My question comes from the topic of Finite Elements (and domain-composition-algorithms).

Consider the following setting: We have some domain $\Omega = (0,1)^2$ which is decomposed into rectangular subdomains. Each subdomain $\Omega_i$ of it with $diam(\Omega) = H$ and $\tau^{h}$ ($h$ the step size between two nodes) a regular triangulation of $\Omega$ by $P_1$-elements (i.e. polynoms $ax + by + c$ on each triangle). Let $\theta_\nu$ be some nodal basis function for $\nu$ a vertex of $\Omega_i$, i.e. $\theta_\nu(x) = 1$ for $x = \nu$ and $\theta_\nu(x) = 0$ for $x$ being any other node of $\Omega$.

Now to the question: Why does $$|\theta_\nu|_{H^1(\Omega_i)} \leq C$$ and $$ ||\theta_\nu||_{L^2(\Omega_i)} \leq C \cdot h^2 $$ independent of $H$ and $h$ hold?

Idea for the first inequality:

Picture of the triangulation

W.l.o.g. triangulate the domain $\Omega$ as done in the picture. On $\Omega_i$ the vertex basis function $\theta_{\nu}$ is per definition only nonzero on the triangle $T$ and there it is a $P_1$ element which is bounded by 1, i.e. ${\theta_{\nu}}_{|T} = ax + by + c \leq 1$ for some reel coefficients $a,b,c \in \mathbb{R}$. In especially we have that $a^2 + b^2 \leq 1$ and thus $$|\theta_\nu|_{H^1(\Omega_i)} = \int_{\Omega_i}|\nabla \Omega_i|^2 d(x,y) = \int_{T}|\nabla(ax + by + c)|^2 d(x,y)$$ $$ = \int_{T} a^2 + b^2 d(x,y) = (a^2 + b^2) \cdot \lambda(T)$$ where $\lambda(T)$ is meant to be the Lebesgue-measure of the triangle T which has finite measure because the triangulation is regular! Are these ideas correct? What do you mean?

Thanks a lot for any approach!!!



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.