= Solution
Assume $p\in L^\infty(-1,1)$ with $p(x)\geq p_0>0$, $q\in L^\infty(-1,1)$ with $q\geq0$, and $f\in L^2(-1,1)$. The natural energy space is the <Sobolev space> $V=H_0^1(-1,1)$. Multiplying
$$
-(py')'+qy=f
$$
by $v\in V$ and using <integration by parts> gives the <weak formulation>
$$
\boxed{a(y,v)=\ell(v)\quad(v\in V),}
$$
where
$$
a(u,v)=\int_{-1}^1(pu'v'+quv)\,dx,
\qquad
\ell(v)=\int_{-1}^1fv\,dx.
$$
The form is bounded, and the <Poincare inequality> together with $p\geq p_0$ makes it coercive:
$$
a(v,v)\geq p_0\|v'\|_{L^2}^2\geq c\|v\|_{H^1}^2.
$$
The functional $\ell$ is bounded by the <Cauchy-Schwarz inequality>. The <Lax-Milgram theorem> therefore gives a unique weak solution.
Define the energy
$$
J(v)=\frac12a(v,v)-\ell(v).
$$
If $y$ is the weak solution, then for every $w\in V$,
$$
J(y+w)-J(y)
=a(y,w)-\ell(w)+\frac12a(w,w)
=\frac12a(w,w)\geq0.
$$
Thus $y$ is the unique minimizer of $J$. Conversely, differentiating $J(y+tw)$ at $t=0$ recovers $a(y,w)=\ell(w)$, so the minimization and weak problems are equivalent.
The <Ritz method> chooses a finite-dimensional conforming space $V_h\subset V$ and minimizes $J$ over $V_h$. Its minimizer $y_h$ satisfies
$$
a(y_h,v_h)=\ell(v_h)qquad(v_h\in V_h).
$$
For a basis $\{\phi_j\}$ this becomes $Kc=F$, with
$$
K_{ij}=\int_{-1}^1(p\phi_j'\phi_i'+q\phi_j\phi_i)\,dx,
\qquad
F_i=\int_{-1}^1f\phi_i\,dx.
$$
Coercivity makes the stiffness matrix $K$ <positive-definite matrix>[symmetric positive definite], so the discrete minimizer exists and is unique.
Subtracting the continuous and discrete equations gives <Galerkin orthogonality>,
$$
a(y-y_h,v_h)=0\qquad(v_h\in V_h).
$$
In the <energy norm> $\|v\|_a=\sqrt{a(v,v)}$, for any $v_h\in V_h$,
$$
\begin{aligned}
\|y-y_h\|_a^2
&=a(y-y_h,y-v_h)+a(y-y_h,v_h-y_h)\\
&=a(y-y_h,y-v_h)\\
&\leq\|y-y_h\|_a\|y-v_h\|_a.
\end{aligned}
$$
After cancellation and minimization over $v_h$, this proves the sharp symmetric form of the <Céa lemma>:
$$
\boxed{\|y-y_h\|_a\leq\inf_{v_h\in V_h}\|y-v_h\|_a.}
$$
For a mesh $-1=x_0<x_1<\cdots<x_N=1$, take $V_h$ to be the continuous functions that are affine on each interval and vanish at the endpoints. The interior <piecewise-linear hat functions> form a basis, and each overlaps only its two neighbours, so $K$ is symmetric tridiagonal. In the illustrative case $p=1$ and $q=0$ on a uniform mesh, its diagonal and neighbouring entries are
$$
K_{ii}=\frac2h,
\qquad
K_{i,i+1}=K_{i+1,i}=-\frac1h.
$$
If $y\in H^2(-1,1)$, its nodal interpolant $I_hy\in V_h$ satisfies $\|y-I_hy\|_{H^1}=O(h)$. The best-approximation estimate therefore gives $\|y-y_h\|_{H^1}=O(h)$, and a standard duality argument improves the $L^2$ error to $O(h^2)$ under the corresponding elliptic regularity. This completes the route from the boundary-value problem through its variational principle to a sparse, stable finite-element approximation.
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