Consider the linear BVP, $$-\frac{d}{dx}\left(p(x) \frac{dy}{dx}\right)+q(x)y=f(x),\quad 0<x<1, y(0)=y(1)=0$$
We assume that $p(x)\geq\delta>0$ for some constant $\delta$, and $q(x)\geq0$ on $[0,1]$. If we multiply both sides of the equation by a test function $u(x)$, and integrate over the domain we get, $$-\int_0^1 \frac{d}{dx}\left(p(x) \frac{dy}{dx}\right)dx+\int_0^1 u(x)q(x)y(x)dx=\int_0^1 u(x)f(x)dx$$
Let $C^2[0, 1]$ be the space of all functions with two continuous derivatives on [0, 1], and let $C^2_0[0, 1]$ be the space of all functions in $C^2[0, 1]$ that are equal to zero at the endpoints $x=0$ and $x=1$. If we require that our test function $u(x)$ belongs to $C^2_0[0, 1]$, then $u(0) = u(1) = 0$ then, $$\int_0^1 \left[p(x)\frac{du}{dx}\frac{dy}{dx}+q(x)u(x)y(x)\right]dx=\int_0^1 u(x)f(x) dx$$ Furthermore, $y$ is a solution to the BVP if and only if it is the unique function in $C^2_0[0,1]$ that minimizes the functional,
$$I[u]=\frac12 \int_0^1 p(x)\left(\frac{du}{dx}\right)^2+q(x)\left(u(x)\right)^2dx-\int_0^1 u(x)f(x) dx$$
Question: I didn't understand how minimizing the functional related with the solution of the BVP? And I couldn't manage to get the $\frac12$ before the integrand.
There is a close connection between this functional, and the functional on $\mathbb R$,
$$\phi(\mathbb x)=\frac12 \mathbf{x^T A x}-\mathbf{x^Tb}$$
The matrix $A$ corresponds to the operator $Lu=-\frac{d}{dx}\left(p(x)\frac{du}{dx}\right)+q(x)u$. To find an approximation of the minimizer of $I[u]$, we restrict ourselves to a subspace of $C_0^2[0, 1]$ by requiring that, $$y(x)=\sum_{j=1}^n c_j\phi_j(x)$$ where $\phi_j$ form a basis of trial functions. Substituting this form yields, $$I[y]=\frac12 \int_0^1 p(x)\left[\sum_{j=1}^n c_j\phi_j(x)\right]^2+q(x)\left[\sum_{j=1}^n c_j\phi_j(x)\right]^2 dx-\int_0^1 f(x)\left[\sum_{j=1}^n c_j\phi_j(x)\right]dx$$ Computing the gradient of $I[y]$ with respect to $c_1,c_2\cdots$ and requiring it to vanish yields the system of equations, $$\sum_{j=1}^n \left[ p(x)\phi_i'(x)\phi_j'(x)+q(x)\phi_i(x)\phi_j(x) dx\right]c_j=\int_0^1 \phi_i(x)f(x)dx,\quad i=1,2,3,\cdots,n$$
Question: How they compute the gradient? Isn't there should be double summation for $i$ and $j$ indices?