Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Omega_h$ denote to the domain that is bounded by a polygon, and $V_h$ denote to the space of all $v\in C^0(\Omega)$ such that $v_{|T}$ is linear on any (curved) triangle T and $v=0$ in the vertices of the boundary i.e. $V_h = \{v \in C^0(\overline{\Omega}): v|T \in P_1(T), v|_{\partial T}=0 \}$.

For a bilinear form $a$ that is elliptic and bounded on $V_h$, e.g. $\exists\ c_1, c_2 >0$ such that

$$a(v_h,v_h) \geq c_1 ||v_h||_{V_h}^2 \quad \text{and} \quad a(u,v_h) \leq c_2 ||u||_{V_h} ||v_h||_{V_h}$$ holds for all $v_h \in V_h$.

I have proved that

$$||u-v_h||\leq c(h||u||_{2,\Omega} + h^{\frac 3 2} ||u||_{2,\Omega}).\ \ \quad (*)$$

How can I show that $(*)$ remains true if I replace $a(u,v)$ by $\tilde{a}(u,v)=\int_{\Omega_h} a_0 u v dx + \int_{\Omega_h} \alpha \nabla u \nabla v dx$?

My first idea was to prove that $\tilde{a}$ is elliptic and continuous on $V_h$ as well but I failed to show this.

Who can help me?

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.