Weak formulation of boundary value PDE and its meaning

Energy functional
An energy functional is a mapping from a function space (often a Sobolev space) to the real numbers, which assigns a "total energy" value to each function in the space. The energy assigned typically depends on the function and its derivatives, reflecting physical or geometrical properties like potential energy, kinetic energy, or strain energy in various physical contexts.

$$J(u)=\frac{1}{2}{\int }_{-1}^1|u^{\prime }(x){|}^{2}dx-{\int }_{-1}^{1}f(x)u(x)dx$$
over $\mathring{H^1}(-1,1)$ where this Sobolev space is for functions that go to $0$ at boundary points for weak formulation.

Find $u\in \mathring{H^1}(-1,1)$such that
$${\int }_{-1}^1{u}^{\prime }(x)v^{\prime }(x)={\int }_{-1}^{1}f(x)v(x),v(x)\in \mathring{H^1}(-1,1)$$

Weak Formulation of a Boundary Value Problem

We consider a boundary value problem where we seek to find a function $u$ that satisfies the differential equation
$$-u^{″}(x)=f(x)\text{on}(-1,1),$$
with boundary conditions
$$u(-1)=u(1)=0.$$
Multiplying by a Test Function
To derive the weak form, multiply the differential equation by a test function $v(x)$, which is smooth and vanishes at the boundaries $(-1,1)$, hence $v(-1)=v(1)=0$. This test function $v(x)$ belongs to the space ${\mathring{H}}^1(-1,1)$, a subspace of $H^1(-1,1)$. We obtain:
$$-u^{″}(x)v(x)=f(x)v(x).$$
Integration Over the Domain
Integrate both sides over the interval $(-1,1)$:
$$-{\int }_{-1}^1{u}^{″}(x)v(x)dx={\int }_{-1}^{1}f(x)v(x)dx.$$
Integration by Parts
Use integration by parts on the left-hand side:
$$\begin{array}{rl}-{\int }_{-1}^1{u}^{″}(x)v(x)dx& ={[-u^{\prime }(x)v(x)]}_{-1}^1+{\int }_{-1}^1{u}^{\prime }(x)v^{\prime }(x)dx\\ & =0+{\int }_{-1}^1{u}^{\prime }(x)v^{\prime }(x)dx,\end{array}$$
                                                                                                        where the boundary terms vanish because $v(-1)=v(1)=0$.
Weak Formulation
Thus, we have the weak formulation of the boundary value problem:
$${\int }_{-1}^1{u}^{\prime }(x)v^{\prime }((x)dx={\int }_{-1}^{1}f(x)v(x)dx.$$
This equation must hold for all test functions $v(x)\in {\mathring{H}}^1(-1,1)$.
Meaning of the Weak Formulation
In this form, the differential equation $-u^{″}=f$ is translated into an integral equation that does not require the function $u$ to be twice differentiable. Instead, $u$ needs only to have its first derivative in $L^2(-1,1)$. This allows the inclusion of functions with less smoothness, accommodating more general solutions such as those exhibiting weak derivatives.