= Finite element solution of the one-dimensional Poisson equation
{title2=$-u''=1$, $u(0)=u(1)=0$}
Consider the <Poisson equation> on an <interval> with homogeneous <Dirichlet boundary conditions>:
$$
-u''(x)=1\quad(0<x<1),\qquad u(0)=u(1)=0.
$$
Two <integrations> and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a <test function> $v$ that vanishes at both endpoints. <Integration by parts> gives
$$
\int_0^1(-u'')v\,dx
=\bigl[-u'v\bigr]_0^1+\int_0^1u'v'\,dx
=\int_0^1v\,dx.
$$
The boundary term is zero. Thus the <weak formulation> is to find $u\in V=H_0^1(0,1)$ such that
$$
a(u,v):=\int_0^1u'v'\,dx=\ell(v):=\int_0^1v\,dx\qquad(v\in V).
$$
The <zero-trace Sobolev space> $H_0^1(0,1)$ consists of square-integrable functions with square-integrable <weak derivatives> and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous <piecewise linear functions> are admissible despite corners. The <Poincaré inequality> makes $a$ a <coercive bilinear form> on $V$. Strang derives this one-dimensional <weak formulation>, the <piecewise-linear hat functions>, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.https://ocw.mit.edu/courses/18-085-computational-science-and-engineering-i-fall-2008/676ba0a00eef45a6cb3823c4da61ee11_18-085F08-L18.pdf[Strang, MIT 18.085, Lecture 18, transcript pp. 2–8: weak form, hat basis, and assembly]{ref}
Take the uniform <finite element mesh> $x_j=jh$, $j=0,\ldots,n$, with $h=1/n$. The continuous $P_1$ trial space is
$$
V_h=\{v_h\in C([0,1]):v_h(0)=v_h(1)=0,\quad v_h|_{[x_j,x_{j+1}]}\text{ is affine for every }j\}.
$$
Here $P_1$ means <polynomials> of degree at most one on each <interval>. This is a <conforming finite element space> in $V$, with $n-1$ interior <piecewise-linear hat functions> $\phi_i$ as a <basis>. Their endpoint values enforce the <Dirichlet boundary conditions> automatically. Set $u_h=\sum_{i=1}^{n-1}c_i\phi_i$ and test the <weak formulation> against each $\phi_i$:
$$
K_{ij}=\int_0^1\phi_j'\phi_i'\,dx,\qquad
F_i=\int_0^1\phi_i\,dx=h,\qquad Kc=F.
$$
Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared <interval>, giving $-1/h$. All other derivative products vanish. The <stiffness matrix> is the <tridiagonal matrix>
$$
K_{ii}=\frac2h,\qquad K_{i,i+1}=K_{i+1,i}=-\frac1h,\qquad K_{ij}=0\quad(|i-j|>1).
$$
The load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The <piecewise-linear hat function> article explains how the same entries arise by adding local two-node contributions.
With four <mesh> intervals, $h=1/4$, there are three unknowns:
$$
K=\begin{pmatrix}8&-4&0\\-4&8&-4\\0&-4&8\end{pmatrix},\qquad
F=\begin{pmatrix}1/4\\1/4\\1/4\end{pmatrix},\qquad
c=\begin{pmatrix}3/32\\1/8\\3/32\end{pmatrix}.
$$
For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The <coefficients> are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, <linear interpolation> gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.
In this model the <Galerkin method> solution equals the nodal approximation $I_hu$ obtained by <linear interpolation> on any <finite element mesh>. Indeed, on each <interval> $[a,b]$, $(I_hu)'$ is the average of $u'$ and $v_h'$ is constant. Therefore $\int_a^b(u'-(I_hu)')v_h'\,dx=0$. Summing proves <Galerkin orthogonality> for $I_hu$, and uniqueness gives $u_h=I_hu$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for <finite element methods> with other operators or integration rules.
On a uniform <mesh>, direct subtraction of the chord from the <parabola> gives, on every element $[a,b]$ of length $h$,
$$
e(x):=u(x)-u_h(x)=\frac{(x-a)(b-x)}2,\qquad
e'(x)=\frac{a+b}2-x.
$$
The largest value error is at an element midpoint. Its size and the <energy norm> error are exactly
$$
\|e\|_{\infty}=\frac{h^2}{8},\qquad
\|e\|_a^2=\int_0^1|e'|^2\,dx
=n\int_{-h/2}^{h/2}t^2\,dt=\frac{h^2}{12},\qquad
\|e\|_a=\frac{h}{\sqrt{12}}.
$$
Thus $h=1/4$ gives $\|e\|_\infty=1/128=0.0078125$ and $\|e\|_a=1/(4\sqrt{12})\approx0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx0.0360844$. Halving the <mesh> spacing quarters the maximum value error but only halves the <energy norm> error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.
\Image[/finite-element-poisson-refinement.png]
{title=Exact Poisson parabola and continuous piecewise-linear finite element solutions on four and eight intervals, with midpoint errors shown separately}
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