Galerkin Method:

Choose $w_i(x)=\phi_i(x)$, the basis of the trial space. If we used a different basis we have Petrov-Galerkin. So

$$\int_{\Omega} r(x)\phi_i(x0 \mathrm{d}x = 0$$

ie we require that $r(x)$ be orthogonal to $\phi_i(x)$.

Example:

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

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

Again the nodes are $x=0$, $x=0.5$ and $x=1$.

$$v(x)=\sum_{j=1}^3 v_j \phi_j(x)$$

$v_1=1$ and $v_3=0$ by boundary conditions. So we only have to work out things for node 2:

$$\int_0^1 \left( \frac{ \mathrm{d}^2 v}{ \mathrm{d}x^2} + k^2v \right) \phi_2(x) \mathrm{d}x =0$$

or

$$\sum_{j=1}^3 v_j \int_0^1 \left( \frac{ \mathrm{d}^2 \phi_j}{ \mathrm{d}x^2} + k^2\phi_j \right) \phi_2(x) \mathrm{d}x =0$$

However we can rewrite using integration by parts

$$\int_0^1 \left( \frac{ \mathrm{d}^2 v}{ \mathrm{d}x^2} + k^2v \... ...mathrm{d}x + \left.\frac{ \mathrm{d}v}{ \mathrm{d}x} \phi_2 \right\vert _0^1$$

but $\phi_2=0$ at $x=0$ and $x=1$ so the last term is zero.

$$\begin{multline*} \int_0^1\left( -\frac{ \mathrm{d}v}{ \mathrm{d}x}\frac{ \m... ...+ v_3\left[\frac{1}{\Delta x} + \frac{k^2\Delta x}{6} \right] =0 \end{multline*}$$

$\Box$

Another Comparative Example

Solve the

$$BVP\left\{\begin{array}{ll} Y''=6t & 0\le t\le 1\\ Y(0)=0 & \mbox{}\\ Y(1)=1 & \mbox{} \end{array}\right.$$

The exact solution is clearly $Y(t)=t^3$. First, let's solve this problem using:

A) Collocation Technique

seek a function $u(t)$ that satisfies BVP at a discrete set of mesh points in interval. We choose $u(t)$ as a simple polynomial, capable of satisfying the boundary conditions and with the regularity suggested by the BVP.

For illustration $\rightarrow$ only 1 point $t=0.5$ and $y=0$, at $t=1$.

$$\begin{array}{ll} \mbox {Pick } & u(t)=x_0+x_1t+x_2t^2\\ \m... ...o } & u'(t)=x_1+2x_2t\\ \mbox {and } & u''(t)=2x^2 \end{array}$$

So for

$$\left\{\begin{array}{ll} Y''=f(t,Y,Y') & x_0\le t=b\\ Y(a)=\alpha &\\ Y(b)=\beta \end{array}\right.$$

we require that $u=Y$ at 3 points requires 3 equations:

$$\begin{array}{ll} x_0+x_1a+x_2a^2=\alpha \end{array}$$ (86)

$$x_0+x_1b+x_0b^2=\beta$$ (87)

equations (86) and (87) lead to

$$x_0=0$$

$$x_1+x_2=1$$

and for some $t\in(a,b)\quad u''(t)=f(t,u(t),u'(t)).$

For us $u''(0.5)=f(0.5, u(0.5), u'(0.5))$

Thus

$$2x_2=6(0.5)=3$$ (88)

So solving (86), (87), (88) get $x_0=0$, $x_1=-5$, $x_2=1.5$ thus the approximate solution is $u=-0.5 t+1.5 t^2$. A comparison to the exact solution appears in Figure (15)

Figure 15: Comparison of exact and approximate solution via collocation

$$u(0.5) = -0.5(0.5) + 1.5(0.5)^2=(0.5)^2(-1+1.5)=(0.5)^3$$

$$Y(0.5) = (0.5)^3$$

B) FEM/Galerkin Method:

Same BVP and use same 3 mesh points, which now become ``knots'' in the piecewise polynomial approximation. Take ``hat'' basis or elements, which are shown in Figure (16).

Figure 16: Hat functions, on the unit interval.

So $Y(t)\approx u(t)=x_1\phi(t)+x_2\phi_2 (t)+x_3\psi(t)$.

Applying the boundary conditions,

$$u(0) = 0=x_1\phi_1(0)+x_2\phi_2(0)+x_3\phi_3(0)\Rightarrow x_1=0$$

$$u(1) = 1=x_1\phi_1(1)+x_2\phi_2(1)+x_3\phi_3(1)\Rightarrow x_3=1$$

Galerkin condition applied at $t=0.5\Rightarrow$ residual must be orthogonal to space spanned by the basis functions and hence to each basis individually

$$\mbox {Orthogonality condition:} \begin{array}{l} \\ \displaysty... ...displaystyle \int^1_0 u''\phi_2(t)dt-6\int^1_0 6\phi_2(t)dt\equiv 0 \end{array}$$

integrate by parts:

$$=-\int^1_0 u\overbrace{\phi'_2(t)dt+u'(t)\phi_2(t)}\vert^1_0-\frac32=0$$

$$\phi_2(0)=\phi_2(1)=0\therefore 2^{nd } ,$$

$$=+\int^1_0u'\phi'_2(t)dt+\frac32=0.$$

$$\int^1_0\left(\sum^3_{i=1}x_i\phi'_i(t)\right)\phi'_2(t)+\frac32=0$$

$$\sum^3_{i=\alpha}x_i\int^1_0\phi'_i(t)\phi'_2(t)dt+\frac32=0\Righ... ...ht)+x_2\left(\frac{2}{h}\right)+x_3\left(-\frac{1}{h}\right)+\frac32}_{(\$)}=0,$$

where $h=1/2$. Substituting $x_1$ and $x_3$ gives $x_2=\displaystyle \frac18$ in ($). We conclude that the approximation is

$$u(t)=0.125\phi_2(t)+\phi_3(t)$$

A comparison of the exact and approximated solution appears in Figure (17).

Figure 17: Comparison of exact and FEM approximate solution

Remark: One particularly nice feature of Galerkin/FEM and collocation methods is that the approximation $u(t)$ of the solution $Y(t)$ is defined over all of the range of $t$ prescribed in the problem statement. This is not true for the finite difference solution, which only gives you an approximation $y$ of $Y$, at specified locations $t_i$ defined by the grid.

We'll consider more BVP issues in the context of PDE's, which is in the next part of the courseII.

$\Box$