variational formulation of second order differential equation
The problem at hand can be reduced to a (somewhat more general) normed problem:
$$
\frac{d^2 T}{d\xi^2} - p^2 T(\xi) = F(\xi)
$$
The left hand side of this normed problem is handled with help of the following references:
- Understanding Galerkin method of weighted residuals
- Are there any two-dimensional quadrature that only uses the values at the vertices of triangles?
The second reference shows that vertex integration is the most stable one. If we employ this for the right hand side,
then the integral
$$
\int_0^1 F(\xi)f(\xi)\,d\xi
$$
results in a load vector $\vec{F}$ instead of $0$ . Giving for the system of equations as a whole (read the first reference):
$$
\begin{bmatrix} E_{0,0}^{(1)} & E_{0,1}^{(1)} & 0 & 0 & 0 & \cdots \\
E_{1,0}^{(1)} & E_{1,1}^{(1)}+E_{0,0}^{(2)} & E_{0,1}^{(2)} & 0 & 0 & \cdots \\
0 & E_{1,0}^{(2)} & E_{1,1}^{(2)}+E_{0,0}^{(3)} & E_{0,1}^{(3)} & 0 & \cdots \\
0 & 0 & E_{1,0}^{(3)} & E_{1,1}^{(3)}+E_{0,0}^{(4)} & E_{0,1}^{(4)} & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{bmatrix}
\begin{bmatrix} T_1 \\ T_2 \\ T_3 \\ T_4 \\ T_5 \\ \cdots \end{bmatrix} =
\begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ \cdots \end{bmatrix}
$$
with the boundary conditions properly imposed.
The original problem - with $x$ and $u$ instead of $\xi$ and $T$ - is recovered by employing the following transformations.
Herewith: $\xi_k \;\rightarrow\; x_k$ and $T_k \;\rightarrow\; u_k$ :
$$
x = (b-a)\xi+a \quad \Longrightarrow \quad
\begin{cases} x = a \;\leftrightarrow\; \xi = 0 \\ x = b \;\leftrightarrow\; \xi = 1 \end{cases}
\\
u = (\beta-\alpha)T+\alpha \quad \Longrightarrow \quad
\begin{cases} u = \alpha \;\leftrightarrow\; T = 0 \\ u = \beta \;\leftrightarrow\; T = 1 \end{cases}
$$
Note. Variational formulation and Galerkin method are the same in this case.