Galerkin and collocation solution of an ODE

Residual

With $u = a\,x(1-x)$: $u'' = -2a$. The trial function satisfies $u(0) = u(1) = 0$.

$$R(x) = u'' + u + x = -2a + a\,x(1-x) + x$$

(a) Galerkin method

The weight function equals the trial function shape, $W = x(1-x)$, and $\int_0^1 W R\,dx = 0$.

Needed integrals: $\int_0^1 x(1-x)dx = \dfrac{1}{6}$, $\int_0^1 x^2(1-x)^2dx = \dfrac{1}{30}$, $\int_0^1 x^2(1-x)dx = \dfrac{1}{12}$.

$$\begin{aligned} \int_0^1 x(1-x)\,R\,dx &= -2a\left(\frac{1}{6}\right) + a\left(\frac{1}{30}\right) + \frac{1}{12} \\ &= -\frac{3a}{10} + \frac{1}{12} = 0 \end{aligned}$$ $$a = \frac{10}{36} = \frac{5}{18} = 0.2778$$

$u(0.5) = 0.2778(0.5)(0.5) = 0.06944$.

(b) Point collocation at $x = 0.5$

$$R(0.5) = -2a + a(0.25) + 0.5 = -1.75a + 0.5 = 0 \Rightarrow a = \frac{2}{7} = 0.2857$$

$u(0.5) = 0.2857(0.25) = 0.07143$.

Comparison

Exact: $u(0.5) = \dfrac{\sin 0.5}{\sin 1} - 0.5 = \dfrac{0.47943}{0.84147} - 0.5 = 0.06975$.

Galerkin averages the residual over the whole domain, so it is more accurate than collocation, which forces zero residual only at one point.

Answer: Galerkin $u = 0.2778\,x(1-x)$, $u(0.5) = 0.0694$; collocation $u = 0.2857\,x(1-x)$, $u(0.5) = 0.0714$; exact $0.0698$.