Show that $b(\ ., \ .)$ it is not coercive.

Question

Let $b(\ .,\ .)$ a bilinear operator so that

$$b(u, \phi)= \int_0^T((u(t), \ \phi(t)))dt- \int_0^T(u(t), \ \phi'(t))dt,$$

where $((\ ., \ .))$ it is the inner product of $H_0^1(\Omega)$ (Sobolev Space) and $(\ .,\ .)$ of $L^2(\Omega)$. Show that $b(\ .,\ .)$ is not coercive in $W$, where

$$W=\{\phi \in L^2(0,T;H_0^1(\Omega));\ \phi' \in L^2(0,T;L^2(\Omega)) \ \mbox{and} \ \phi(T)=0\}.$$

Answer 1

First note that you have \begin{align*}b(u, u) &= \int_0^T((u(t), \ u(t)))dt- \int_0^T(u(t), \ u'(t))dt \\ &= \int_0^T((u(t), \ u(t)))dt- \frac12 \, \int_0^T \frac{d}{dt}(u(t), \ u(t))dt \\ &= \int_0^T((u(t), \ u(t)))dt- \frac12 \, (u(T), \ u(T)) + \frac12 \, (u(0), \ u(0)) \\ &= \int_0^T((u(t), \ u(t)))dt+ \frac12 \, (u(0), \ u(0)). \end{align*} From here, you can find easy examples showing that $b$ is not coercive. Try $u(t,x) = u_1(t) \, u_2(x)$ with arbitrary $u_2$ and suitable $u_1$.