Weak formulation with Dirichlet boundary conditions Let's consider the one-dimensional ODE: $$ u_{,xx}(x)+1=0,\quad \forall x\in ]0,\ell[$$
together with the two Dirichlet boundary conditions $u(0)=0$ and $u(\ell)=0$. 
The corresponding weak formulation is: Find $u\in H_0^1(0,\ell)$ such that $\forall v\in H_0^1(0,\ell)$: $$\int_0^\ell v_{,x}(x)u_{,x}(x)\,\mathrm{d}x=\int_0^\ell v(x)\,\mathrm{d}x$$
My question is on the fact that the test function $v$ has to belong to $H_0^1(0,\ell)$. I am aware that if $v\in H^1(0,\ell)$, then the weak formulation involves boundary terms in $vu_{,x}$ that are unknown, hence we specify $v\in H_0^1(0,\ell)$ to somehow get rid of them but:


*

*How do we know that, by doing so, we are not modifying the equivalence between the strong and weak formulations?

*If we rewrite the weak formulation as follows: Find $u\in H_0^1(0,\ell)$ such that $\forall v\in H^1(0,\ell)$: $$\int_0^\ell v_{,x}(x)u_{,x}(x)\,\mathrm{d}x=\int_0^\ell v(x)\,\mathrm{d}x+v(x)u_{,x}(x)\Big|_0^\ell$$
it cannot be solved, but why exactly? Is there some kind of operator that cannot be inverted (this is what happens in the discretized version)

 A: The strong formulation implies the weak formulation; this follows after integrating by parts. To show that the weak formulation actually implies the strong formulation for functions $u$ with two derivatives that have sufficient regularity, note that the weak formulation implies that $$\int_0^lu_{xx}v=-\int_0^lv_xu_x=-\int_0^lv,$$ which implies that $$\int_0^l(u_{xx}+1)v=0$$ for all $v\in H_0^1(0,l)$. If again $u_{xx}$ is sufficiently regular, there exists a sequence $(v_n)$ in $C_c^{\infty}(0,l)$ that approximates $u_{xx}+1$ in the $L^2$ norm; this will imply that $$\int_0^l(u_{xx}+1)^2=\lim_{n\to\infty}\int_0^l(u_{xx}+1)v_n=0.$$ Therefore, $u_{xx}+1=0$, which is the strong formulation.
For your second question, usually what happens here is that, to obtain coercivity, we want to control the $H_0^1$ norm of a function $v$ just by the $L^2$ norm of $u_x$; this follows from Poincare's inequality. But, Poincare's inequality does not hold for general $H^1$ functions.
A: To answer your second question: If $u$ belongs merely to $H_0^1(0,\ell)$, you have no chance to define $u_{,x}(0)$ and $u_{,x}(\ell)$ in a reasonable manner. So your weak definition is not well-defined.
To give an example: It is easy to verify that $u$ defined by
$$u(x) = x^{2/3} \, (1-x)$$
belongs to $H_0^1(0,\ell)$ and its weak derivative is
$$u_{,x}(x) = \frac23 \, x^{-1/3} \, (1-x) - x^{2/3}.$$
This weak derivative, however, has a singularity at $x = 0$ (and it would be easy to add another one at $x = \ell$).
