I have arrived at the weak form for a PDE which has the following form.
Let $V=\{ v\in H^1(0,1):v(1)=0 \}$, $F$ is a bounded linear functional on V.
$$\left\{\begin{array}{ll} \textrm{Given } F\in V^* \textrm{, find } u\in V \textrm{ such that}\\ \int_{0}^{1}{u'(x)v'(x) dx} =F(v) \\ \forall v\in V \end{array} \right.$$
I understand that the Poincare-Friedrichs inequality can be used to show that when $V=H_0^1 (0,1)$ the coercitivity of the Lax-Milgram holds, but what about for when I have only one Dirichlet boundary condition and I don't have completely defined Dirichlet conditions?
Does the Poincare-Friedrichs inequality apply to my space $V$, like so
$$\int_0^1 u(x)^2 dx \le \int_0^1 u'(x)^2 dx \quad \textrm{?}$$
If $u \in V$ then $$u(x) = u(x) - u(1) = - \int_x^1 u'(t) \, dt$$ so that $$|u(x)| \le \int_x^1 |u'(t)| \, dt \le \left( \int_0^1 |u'(t)|^2 \right)^{1/2}$$ and consequently $$\int_0^1 |u(x)|^2 \, dx \le \int_0^1 \int_0^1 |u'(t)|^2 \, dt \, dx = \int_0^1 |u'(t)|^2 \, dt.$$