elliptic pdes and associated bilinear forms for Lax-Milgram I have a simple question on elliptic pdes, actually I can not understand clearly from definitions. Thats why I want to try think on an example.
Let us have an elliptic pde $$-A \Delta u(x)+Bu(x)=F(x)$$ where $A,B$ are constants and $x\in\mathbb{R}^3$. Thus, I can define an elliptic operator $L$, $$Lu=(-A \Delta+B)u=F(x).$$ Associated bilinear form to $L$ is given $b:H \times H\rightarrow\mathbb{R}$  for $u,v\in H$$$b\left(u,v\right)=\int \limits_{\mathbb{R}^3}A\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx+B\int \limits_{\mathbb{R}^3}u(x)v(x)dx.$$ Then I am trying to check Lax-Milgram's Lemma,$$\begin{eqnarray}
\left|b\left(u,v\right)\right|&=&\left|\int \limits_{\mathbb{R}^3}A\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx+B\int \limits_{\mathbb{R}^3}u(x)v(x)dx\right|\\&\leq&\int \limits_{\mathbb{R}^3}A\left|\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx\right|+B\int \limits_{\mathbb{R}^3}\left|u(x)v(x)dx\right|\end{eqnarray}$$ As you see, I am failed in the first step and can not see even how to apply Cauchy-Schwarz here. Coerciness is then much worse then continuity. Can you please help me on this example?
 A: We can assume $A=1$ so that your equation is 
$$
-\Delta u +bu= f
$$
with $b=A^{-1}B$ and $f=A^{-1}F$. Your bilinear form, which I'll call $B$, then satisfies
$$
|B(u,v)|=\left| \int_{\mathbb{R}^3} \nabla u \cdot \nabla v dx +b \int_{\mathbb{R}^3} uvdx \right| \leq \| \nabla u\| _{L^2} \|\nabla v \|_{L^2} +|b| \| u\|_{L^2}\| v\|_{L^2}\\ \leq \| u\|_{H^1}\| v\| _{H^1} +|b| \| u\|_{H^1}\| v\|_{H^1}
$$
where the first inequality is Cauchy-Schwarz applied to each summand, and the second follows from the definition of the $H^1$ norm.
For coercivity, I can only prove it when $b> 0$ (when it is obvious since in this case $B(u,u)=\| \nabla u\|_{L^2}^2 +b\| u\|_{L^2}^2$). 
Edit: A variant of $f_k(x)=|x|^{-\frac{3}{2}-\frac{1}{k}}$ shows that when $b\leq0$ the bilinear form is not coercive.
A: I assume you want $H = H^1(\Bbb R^3)$ here.
Note that
\begin{align}
|b(u,v)| & \leq \max \{|A|, |B|\} \left| \int_{\Bbb R^3} \sum_{i = 1}^3 \frac{\partial u}{\partial x_i} \frac{\partial v}{\partial x_i} ~dx + \int_{\Bbb R^3} uv ~dx \right| \\
& \leq \max \{|A|, |B|\} |(u,v)_{H^1(\Bbb R^3)}| \\
& \leq \max \{|A|, |B|\} \|u\|_{H^1(\Bbb R^3)} \|v\|_{H^1(\Bbb R^3)}.
\end{align}
On the other hand,
\begin{align}
b(u,u) & = A \int_{\Bbb R^3} |\nabla u|^2 ~dx + B \int_{\Bbb R^3} u^2 ~dx \\
& \geq \min\{A, B\} \|u\|_{H^1(\Bbb R^3)}^2.
\end{align}
Note that we need that $B/A > 0$ in order for the above to prove coercivity of $b(u,v)$. (Jose27's answer takes a nicer approach by dividing through by $A$ at the start)
From the above we see that $b(u,v)$ satisfies the hypotheses of the Lax-Milgram lemma.
