Coercivity of a bilinear form in one dimesnion

Question

Let $a$ a bilinear forme defined over $H_0^1(0,1) \times H_0^1(0,1)$ by: $$a(u,v) = \int\limits_0^1 {u'v' + u'v + uvdx} $$ the term who is making me a truble is ${u'v}$, is there any approach to deal with it ? Thanks.

Answer 1

Note that the norm of $H_0^1(0,1)$ is $$ \|u\|=\sqrt{\int\limits_0^1(|u'|^2 + |u|^2)dx}, u\in H_0^1(0,1).$$ So by using $2ab\ge-\frac12(a^2+b^2)$, one has, for $u\in H_0^1(0,1)$, \begin{eqnarray} a(u,u)&=& \int\limits_0^1(|u'|^2 + u'u + |u|^2)dx\\ &=&\|u\|^2+\int\limits_0^1u'udx\\ &\ge&\|u\|^2-\frac{1}{2}\int\limits_0^1(|u'|^2 + |u|^2)dx\\ &=&\frac12\|u\|^2, \end{eqnarray} namely $a(u,v)$ is coercive.