Collocation Method:

Minimize residual by forcing it to pass through zero at a finite number of discrete points within $\Omega$. So here

$$w_i(x)=\delta(x-x_i^c),\qquad x_i^c\in\Omega$$

where $x_i^c$ is the $i$-th collocation point. The choice of collocation points is an important consideration in the method. We aim to have

$$\int_\Omega r(x)w_i(x) \mathrm{d}x = \int_\Omega r(x)\delta(x-x_i^c) \mathrm{d}x = r(x_i^c)=0$$

In other words, at each collocation point the trial functions are required to satisfy the differential equation exactly. The number of collocation points is related to the number of $c_j$.

In general for an $M$-th order equation: choose polynomials of degree $(2M-1)$ with $C^{M-1}(\Omega)$ continuity. $m$ collocation points are chosen within each element located at the roots of the $M$-th degree Legendre polynomial over each element.

Example: Solve

$$\mathcal{L}u = \left(\frac{ \mathrm{d}^2}{ \mathrm{d}x^2} + k^2\right) u = 0 \qquad 0<x<1$$

$$u(0)=1 \qquad u(1)=0$$

Because $r(x)$ involves second order derivatives of trial functions, the trial space must be at least $C^1(\Omega)$ for $r(x)$ to remain bounded.

We choose piecewise cubic Hermite polynomials

$$v(x)=\sum_{j=1}^N v_j\phi{0j}(x)+\frac{ \mathrm{d}v_j}{ \mathrm{d} x}\phi_{1j}(x)$$

Take $N=2$. Nodes are located at $x=0$ and $x=1$. For second order differential equations with piecewise cubic Hermitian trial space, two collocation points should be chosen per element. Coupled with two boundary conditions we get $2N$ algebraic equations for $2N$ unknown nodal values. Further, if we choose the Gauss-Legendre quadrature points we obtain $O(\Delta x^4)$ accuracy. So the collocation points will be $x_1^c=(3-\sqrt{3})/6$ and $x_2^c=(3+\sqrt{3})/6$.

$$\begin{split}r(x_1^c)&= \sum_{j=1}^2 v_j \left. \frac{ \math... ...left. \phi_{1j}\right\vert _{x=x_2^c} \right] = 0 \end{split}$$

So the system to solve is

$$\left[ \begin{array}{cccc} 1 & 0 & 0 & 0\\ -5.657 + 0.998 k... ... = \left[ \begin{array}{c} 1 0 0 0 \end{array}\right]$$