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Choose a partition $0=x_0<x_1<\cdots <x_N=1$, with the usual nodal piecewise-linear hat functions ${\varphi }_i$, including the half hats at the endpoints. Write $u_h={\sum }_{i=0}^N{U}_i{\varphi }_i$ with $U_0=1$. Differentiating the Ritz method energy with respect to $U_i$, $1\le i\le N$, givesOn an element $[a,b]$ of length $\ell$, its two basis functions are $(b-x)/\ell$ and $(x-a)/\ell$. Direct integration yields the element matrixFor example the first weighted diagonal is ${\int }_a^{b}x(b-x{)}^2/{\ell }^{2}dx=\ell (3a+b)/12$; the off-diagonal is ${\int }_a^{b}x(b-x)(x-a)/{\ell }^{2}dx=\ell (a+b)/12$. This gives the affine-weighted hat mass matrix and explicit equations on any partition by assembling adjacent elements.

For the uniform choice $h=1/N$, $x_i=ih$, the assembled interior equations areThe last node contributes only one element, so its equation isIn the first interior equation the known $U_0$ term moves to the right, giving $1/h-h^2/12$. For $N=1$ only the endpoint equation is needed. The last row enforces the weak Neumann boundary condition, so no artificial value outside the interval is introduced. The matrix on $U_1,\dots ,U_N$ is a symmetric positive-definite matrix: its quadratic form is $a(v_h,v_h)>0$ for every nonzero trial variation $v_h$. The discrete minimizer is therefore unique.

Take a uniform finite element mesh $x_i=ih$, $0\le i\le J$, $h=1/J$, and interior piecewise-linear hat functions ${\varphi }_i$. Write $u_h={\sum }_{i=1}^{J-1}{U}_i{\varphi }_i$. The Ritz method gives $KU=F$ withThe derivative part has diagonal $2/h$, neighboring entries $-1/h$ and zero other entries. For the affine-weighted hat mass matrix, symmetry around $x_i$ givesOn $[x_i,x_{i+1}]$, the product of the two hats is symmetric about the midpoint, integrates to $h/6$, and therefore has weighted integralConsequently the explicit tridiagonal equations arefor $1\le i\le J-1$, with $U_0=U_J=0$. The reaction coefficient has been integrated exactly; replacing it by an unweighted constant mass matrix would give a different system. For a nonzero coefficient vector, the represented function is nonzero and $U^{T}KU=a(u_h,u_h)>0$, so the matrix is symmetric positive-definite and the algebraic solution is unique.

Uniform spacing is a choice, not a requirement of the method. On a general finite element mesh with $h_i=x_i-x_{i-1}$, the corresponding entries arewith symmetric lower entries. These reduce to the displayed uniform-grid system.

Put $h=1/(N+1)$ and $x_i=ih$. The interior piecewise-linear hat functions areThey are a basis of the continuous piecewise-affine functions vanishing at both endpoints. The support of a function for each hat only overlaps those of neighbouring hats, so the stiffness matrix is a tridiagonal matrix. Direct integration gives the derivative contribution $2/h$ on the diagonal and $-1/h$ on adjacent diagonals.

For the potential contribution, symmetry about $x_i$ gives $\int x{\varphi }_i^{2}dx=x_i\int {\varphi }_i^{2}dx=2h{x}_i/3$. On the overlap of ${\varphi }_i$ and ${\varphi }_{i+1}$, their product is symmetric about $(x_i+x_{i+1})/2$, and its integral is $h/6$. This is the affine-weighted hat mass matrix calculation. Consequently,and all other entries vanish. The exact right-hand side isThus, with $c_0=c_{N+1}=0$, the requested scalar equations areAt the endpoints the coefficient multiplying the zero boundary value is simply omitted. Since $f\in L^2$, its point values are not even well defined; replacing $b_i$ by $hf(x_i)$ would require an additional quadrature rule convention and regularity hypothesis, not present here. The exact Ritz method uses the displayed integrals. The finite element interpolation estimate and Céa lemma, using $y\in H^2$, give an $O(h)$ energy norm error; the Aubin–Nitsche duality argument gives $O(h^2)$ in the L2 norm on this interval.