# Why is this bilinear form elliptic on $H^1_0$?

I have the following bilinear form:

$$a(v,w) = \int_\Omega \nabla v(x) \cdot \nabla w(x)dx + \int_\Omega \left( \sum_{i=1}^n \beta_i v_{x_i}(x) \right) w(x) dx$$

where $\Omega \subset \mathbb{R}^n$ is a bounded and open set, $\beta \in \mathbb{R}^n$, and $v,w \in H^1_0(\Omega)$.

I have to prove that it is elliptic, i.e. $\exists \gamma >0$ with $$\gamma ||v||^2_{H^1_0} \leq a(v,v)$$ for all $v\in H^1_0$, where

$$||v||^2_{H^1_0} = (v,v)_{L^2} + \sum_{i=1}^n (v_{x_i},v_{x_i})_{L^2}.$$

My idea was to prove it first for $v\in C^\infty_0(\Omega) \subset H^1_0$ and use a density argument later.

However, I do not see where to start now. Even worse, I think the statement has to be wrong. Because for $\Omega = (a,b)$ and $v\in C^\infty_0(\Omega)$ I obtain

$$\int_a^b v'(x)v(x)dx = [v(x)v(x)]_a^b - \int_a^b v'(x)v(x)dx$$ which implies directly $$\int_a^b \beta_1 v'(x)v(x)dx = 0 .$$

Am I doing anything wrong?

Edit: I think I just made it too complicated, it's simply given by:

$$||v ||^2_{H^1_0} = (v,v)_{L^2} + (\nabla v, \nabla v)_{L^2} \leq (1 + \alpha) (\nabla v, \nabla v)_{L^2}$$ for $v\in H^1_0$ due to the Poincaré Inequality, and therefore $a$ is elliptic with $\gamma = (1 + \alpha)^{-1}$. It only remains to show $$\int_\Omega \left( \sum_{i=1}^n \beta_i v_{x_i}(x) \right) v(x) dx = 0$$ for $v \in H^1_{0}$.

Edit 2: Using the formula: $$\int_\Omega \frac{\partial u(x)}{\partial x_i} v(x) dx = \int_{\partial\Omega} u(x)v(x) \nu_i(x) d\sigma - \int_\Omega \frac{\partial v(x)}{\partial x_i} u(x) dx \$$ where $\nu$ is the normal of $\partial \Omega$, everything works analogously.

• You are starting the right way (1-D example, the right calculation etc.). Now also use the term $\int_a^b (v'(x))^2dx$ which is the first (leading) part of the bilinear form. You'll notice that you are "almost" there, with a term $\int_a^b v^2(x) dx$ missing from the identity that you produce. To get that, recall Poincare's inequality. Note that $v'(x) v(x) = \frac{d}{dx} \frac{1}{2} v^2(x)$, with an obvious generalization to higher dimensions. – Hans Engler Jun 18 '14 at 13:22
• @HansEngler if the second term vanishs then elliptic is clear from Poincare's ineqaulity for $v \in C^\infty_0$ right? Why would I need $v'(x)v(x) = (0.5 v^2(x))'$ ?I am only left with using a density argument, right? – Adam Jun 18 '14 at 13:32
• right, after using Poincare you have the estimate for $C_0^\infty$ functions and then a density argument does the rest. Now generalize to $\Omega \subset \mathbb{R}^n$. – Hans Engler Jun 18 '14 at 15:16
• @HansEngler Seems difficult to proof it for $\mathbb{R}^n$ Green's Identities do not really help here.. any cloue? – Adam Jun 18 '14 at 16:57
• looks like you're done! – Hans Engler Jun 18 '14 at 19:38