Subdomain Method

We minimize the residual $r(x)$ by forcing it to zero in a weighted average sense as follows: minimize $r(x)$ over $\Omega$ by forcing the arithmetic average of $r(x)$ taken over discrete intervals of $\Omega$ to be zero.

Choose weights

$$w_i(x)= \begin{cases} 1 \qquad x\in\Omega_i \\ 0 \qquad x \neq \Omega_i \end{cases}$$

where $\{\Omega_i\}_{i=1}^M$ are monimtersecting subregions within $\Omega$ whose union covers $\Omega$. For piecewise linear on nodes $\{x_i\}_{i=1}^N$ a good choice of $\Omega_i$ is

$$(x_i-\frac{1}{2}(x_i-x_{i-1}))<x<(x_i+\frac{1}{2}(x_{i+1}-x_i))$$

. In this case

$$w_i(x)= \begin{cases} 1 \qquad x_i - \frac{\nabla x_i}{2}<x<... ...\frac{\Delta x_i}{2} \\ 0 \qquad \mbox{otherwise} \end{cases}$$

where $\nabla x_i=x_i-x_{i-1}$ and $\Delta x_i= x_{i+1}-x_i$. When $x_i$ is near a boundary node the $\Omega_i$ is taken as only the region residing within $\Omega$.

Example:

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

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

$k$ is a given real number. The exact solution to the problem is
$u(x)=\cos(kx)-\cot(k)\sin(kx)$.

Let us use piecewise linear functions and the subdomain method. Choose $x_1=0$, $x_2=0.5$ and $x_3=1.0$, here $\Delta x = 0.5$.

We want to have

$$\int_0^1 r(x)w_1 \mathrm{d}x = 0,$$

$$\int_0^1 r(x)w_2 \mathrm{d}x = 0,$$

$$\int_0^1 r(x)w_3 \mathrm{d}x = 0.$$

Since $v_1=v(0)=1$ and $v_3=v(1)=0$ and $f=0$. So we have only to ensure that

$$\int_0^1 r(x)w_2 \mathrm{d}x = \int_0^1 \left(\frac{ \mathrm{d}^2 v}{ \mathrm{d}x^2}+k^2 v \right) w_2(x) \mathrm{d}x = 0$$

or

$$\sum_{j=1}^3v_j \int_{0.25}^{0.75}\left(\frac{ \mathrm{d}^2 \phi_j}{ \mathrm{d}x^2} + k^2 \phi_j\right) \mathrm{d}x = 0$$

Let's write

$$\frac{ \mathrm{d}\phi_2}{ \mathrm{d}x} = \frac{1}{\Delta x} H(x-0) - \frac{2}{\Delta x} H(x-0.5) + \frac{1}{\Delta x} H(x-1)$$

where $H(x-x_p)$ is the Heaviside function

$$H(x-x_p)= \begin{cases} 1 \qquad x\geq x_p\\ 0 \qquad x <x_p\\ \end{cases}$$

and

$$\frac{ \mathrm{d}}{ \mathrm{d}x}H(x-x_p)=\delta(x-x_p)$$

where $\delta(x-x_p)$ is the Dirac delta function. This function is zero for $x\neq x_p$ and unbounded at $x=x_p$ such that we have

$$\int_{-\infty}^\infty \delta(x-x_p) \mathrm{d}x = 1$$

$$\int_{-\infty}^\infty \delta(x-x_p) F(x) \mathrm{d}x = F(x_p)$$

where $F(x)$ is a well-defined function. It follows that

$$\frac{ \mathrm{d}^2 \phi_j}{ \mathrm{d}x^2} = \frac{1}{\Delta x... ...elta(x-0) - \frac{2}{\Delta x} \delta(x-0.5) + \frac{1}{\Delta x} \delta(x-1)$$

and

$$\begin{multline*} v_1\int_{0.25}^{0.75} \left[\frac{1}{\Delta x} \delta(x-0.5)+... ...{1}{\Delta x} \delta(x-0.5)+k^2\phi_3(x) \right] \mathrm{d}x=0 \end{multline*}$$

Since $\delta x = 0.5$

$$v_1\left[2+\frac{k^2}{16}\right]+v_2\left[-4+\frac{3k^2}{8}\right] +v_3\left[2+\frac{k^2}{16}\right]=0$$

Since $v_1=1$ and $v_3=0$ we find the following system of equations

$$\left[ \begin{array}{ccc} 1 & 0 & 0 \\ \frac{32+k^2}{16} & ... ...ight] = \left[ \begin{array}{c} 1 0 0 \end{array}\right]$$

Solving gives $(v_1,v_2,v_3)=(1,\frac{32+k^2}{64+6k^2},0)$.

Compare this to $(u_1,u_2,u_3)=(1,\cos(0.5k)-\cot(k)\sin(0.5k),0)$.