Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a polygon $S$ in the plane. Then, we perform a triangulation of this polygon. Consider the space $Q$ the space of functions such that the restriction of any function in the space $Q$ to any triangle in the polygon $S$ is a linear polynomial and continuous at the center of the triangle edges. Can you provide a counter-example of a function in $Q$ which is not globally continuous on $S$?

share|improve this question
    
What makes you think such a counter-example exists? –  nbubis Dec 8 '12 at 14:30
    
Any boundary conditions? (At the outer edges.) –  WimC Dec 8 '12 at 14:32

1 Answer 1

up vote 3 down vote accepted

This is a standard homework problem for a class on analysis of finite element methods; what you are describing is known as the Crouzeix-Raviart element.

It's not hard to construct a counterexample if you consider a specific pair of triangles (one counterexample will do). Let, say, $T_1$ with the vertices $(0,0),(1,0),(1,1)$ and $T_2$ with the vertices $(1,1),(0,1),(0,0)$. The midpoint of the common edge $e$ is then $(1/2,1/2)$. If you prescribe the value $0$ for the interpolant $v$ there, you can pick the remaining values such that $v|_{T_1}$ and $v|_{T_2}$ have opposite sign on $e$. One example is $v|_{T_1} = 1-x-y$ and $v|_{T_2} = x+y-1$, for which $$v|_{T_1}(1/2,1/2) = 0 = v|_{T_2}(1/2,1/2),$$ but $$v|_{T_1}(0,0) = 1 \quad\text{and}\quad v|_{T_2}(0,0) = -1.$$ Hence, $v$ is not continuous on $e$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.