If we say a solution is weak/strong/classical/viscous, the following aspects are concerned (or more):
How we obtain the solution.
The regularity of the solution (how smooth this solution is, integrability, differentiability).
The solution satisfies the equation in what sense.
Weak solution:
- We can obtain the solution by Ritz-Galerkin formulation: find the minimizer of the following quadratic functional in an appropriate Hilbert space,
$$
\mathcal{F}(u) = \frac{1}{2}\int_{\Omega} |\nabla u|^2 - \int_{\Omega} fu.
$$
Smoothness depends on the right side data. If the $f\in H^{-1}$, then $u\in H^1$. If $f\in L^2$, then $u\in H^2_{loc}$. Moreover if $\Omega$ is $C^{1,1}$, we have an $H^2$-solution $u$ globally.
The solution satisfies the equation in distribution sense (see following explanation).
Why "weak":
The term "weak" normally refers to the 2 and 3: The solution $u$ is only in $H^1$ in the most general setting, this means that $u$ is the only differentiable once, notice $-\Delta$ has second partial derivative in it. The strong solution, however, indeed have twice differentiability, normally if we say $u$ is a strong solution, we mean that $u$ has $W^{2,p}$-regularity (Please refer to Gilbarg and Trudinger). The solution satisfies the equation only in the "weak" formulation
$$
\int_{\Omega} \nabla u \cdot \nabla v \, dx =
\int_{\Omega} fv \, dx
\quad \forall v \in V, \tag{1}
$$
where $V$ is certain Sobolev space.
Two ways to get this weak form: first is to write what condition the minimizer of $\mathcal{F}(u)$ must satisfy: if $u$ is a minimizer, then
$$
\lim_{\epsilon \to 0}\frac{d}{d\epsilon} \mathcal{F}(u+\epsilon v) =0
$$
and the weak form of Euler-Lagrange equation is (1).
Another is multiplying the original equation by a test function then integration by parts. The intuition behind this should be Riesz representation theorem (at least to me it makes sense), we have:
$$
\langle (-\Delta)u,v\rangle = l_u(v) = (u,v)_{V},
$$
from the differential operator $-\Delta$ $\to$ linear functional $l_u$ $\to$ representation using inner product $(\cdot ,\cdot)_V$. The inner product $(\cdot ,\cdot)_V$ on this Hilbert space $V$ is the left hand side of (1), if we make the test function space have zero boundary condition (We can use Poincaré inequality to prove the equivalence of the standard $H^1$-inner product). If you have taken any numerical PDE course in finite element, the professor would introduce Lax-Milgram theorem, and Lax-Milgram relies on Riesz.
Why weak form is useful in finite element method:
Short answer: Weak form is very handy in that it helps us formulate a linear equation system which can be solved by computer!
Long answer: The essential of Galerkin type approach is that we are exploiting the fact that the infinite dimensional Hilbert space has a set of basis $\{\phi_i\}_{i=1}^{\infty}$, if we can expand the $u$ in this basis:
$$
u = \sum_{i=1}^{\infty} u_i\phi_i,
$$
where $u_n$ is a number, pluggin back to (1), and let the test function $v$ run through all $\phi_j$ (same function, different subscript):
$$
\int_{\Omega} \nabla (\sum_{i=1}^{\infty} u_i\phi_i) \cdot \nabla \phi_j \, dx =\sum_{i=1}^{\infty} u_i \int_{\Omega} \nabla \phi_i \cdot \nabla \phi_j \, dx =
\int_{\Omega} f\phi_j \, dx
\quad \forall j =1,2,\ldots. \tag{2}
$$
We have an infinite dimensional linear equation system:
$$
AU = F,
$$
where $A_{ji} = \displaystyle\int_{\Omega} \nabla \phi_i \cdot \nabla \phi_j \, dx$, $U_i = u_i$, and $F_j = \displaystyle\int_{\Omega} f\phi_j\, dx$.
Finite element method essentially choose a finite dimensional subspace $V_h\subset V$ (may not be a subspace, please google Discontinuous Galerkin method), so that we approximate the solution in this finite dimensional subspace $V_h$! The summation in (2) does not have an infinite upper limit any more, instead there are finitely many $\phi_i$ and $v$ runs from $\phi_1$ to $\phi_N$, so that the linear system generated is still $AU = F$, but this time, it only has $N$ equations, and we can use computer to solve it.
The reason it's an inner product is that the inner product axioms are satisfied for $\phi(u,v)=\int_\Omega\nabla u\cdot\nabla v.$
The easy ones to show are symmetry and linearity, i.e., $\phi(u,v)=\phi(v,u)$ and $\phi(au,v)=a\phi(u,v)$ ($a$ constant). One has to also show that $\phi(u,u)\ge0$ for all $u$, and $\phi(u,u)=0$ if and only if $u=0$.
One has that $\phi(u,u)=\int_\Omega|\nabla u|^2\ge0$. Then, if $u=0$, then $\phi(u,u)=\int_\Omega0=0$, and if $\phi(u,u)=0$, then $\int_\Omega|\nabla u|^2=0$, and so $u$ must be constant, but also $u\in H^1_0(\Omega)$, so the constant must in fact be zero, so $u=0$.
Best Answer
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.