= Dirichlet Poisson solution by tridiagonal elimination
{c}
{title2=$u(x)=1+2x-x^2$}
Consider the original small <Poisson equation> example
$$
-u''(x)=2\quad(0<x<1),\qquad u(0)=1,\qquad u(1)=2.
$$
Its exact solution is the <quadratic polynomial> $u(x)=1+2x-x^2$. Use a uniform <mesh> $x_i=ih$ with $h=1/4$ and three interior unknowns $U_1,U_2,U_3$. The <finite difference method> replaces the negative <second derivative> by
$$
\frac{-U_{i-1}+2U_i-U_{i+1}}{h^2}=2.
$$
Multiplying by $h^2$ and moving the prescribed <Dirichlet boundary conditions> to the right gives
$$
\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}
\begin{pmatrix}U_1\\U_2\\U_3\end{pmatrix}
=\begin{pmatrix}1+2h^2\\2h^2\\2+2h^2\end{pmatrix}
=\begin{pmatrix}9/8\\1/8\\17/8\end{pmatrix}.
$$
The endpoint values contribute $1$ and $2$ to the first and last equations; they are not additional unknowns. The <Thomas algorithm> gives
$$
\begin{array}{c|c|c|c}
i&m_i&p_i&g_i\\\hline
1&\text{unused}&2&9/8\\
2&-1/2&3/2&11/16\\
3&-2/3&4/3&31/12
\end{array}
$$
and <back substitution> yields
$$
U_3=\frac{31}{16},\qquad U_2=\frac74,\qquad U_1=\frac{23}{16}.
$$
All three values equal $u(x_i)$ exactly: the centered <finite difference> for the <second derivative> is exact on a <quadratic polynomial>. This special nodal exactness does not assert exactness for arbitrary forcing or for the piecewise-linear curve joining the nodes.
\Image[/dirichlet-poisson-tridiagonal-solution.png]
{title=Exact quadratic Dirichlet Poisson solution and the three interior values recovered by tridiagonal elimination}
For any $n$ interior points with $h=1/(n+1)$, the scaled <tridiagonal matrix> has $b_i=2$ and $a_i=c_i=-1$. Its <pivot elements> are $p_i=(i+1)/i$: start at $p_1=2$ and use $p_i=2-1/p_{i-1}$. All are positive. Interior rows have equality in <diagonal dominance>, so <strict diagonal dominance> is not the reason this example works. Instead it is a <symmetric positive-definite matrix>, since for a real <vector> $v\ne0$ with $v_0=v_{n+1}=0$,
$$
v^TAv=\sum_{i=0}^{n}(v_{i+1}-v_i)^2>0.
$$
Zero would force every neighbouring value equal and hence all values zero. This connects the boundary constraints directly to the <safe pivots for tridiagonal elimination>.
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