Resolver \( u(0 < x < 1) \) para $$ \begin{align*} \frac{d^{2} u(x)}{d x^{2}} + x^{2} &= 0 \\ u(0) &= 0 \\ u(1) &= 0 \\ \end{align*} $$
Se utilizaran cuatro términos en la aproximación $$ \begin{equation*} \hat u(x) = a_{0} + a_{1} x + a_{2} x^{2} - a_{3} x^{3} \end{equation*} $$
reemplazando las condiciones de contorno $$ \begin{align*} \require{cancel} \hat u(0) &= a_{0} + a_{1} (0) + a_{2} (0)^{2} - a_{3} (0)^{3} = 0 \\ \hat u(1) &= \cancel{a_{0}} + a_{1} (1) + a_{2} (1)^{2} - a_{3} (1)^{3} = 0 \end{align*} $$
resolviendo el sistema $$ \begin{align*} a_{0} &= 0 \\ a_{2} &= a_{3} - a_{1} \end{align*} $$
reemplazando las constantes $$ \begin{equation*} \hat u(x) = a_{1} x + a_{3} x^{2} - a_{1} x^{2} - a_{3} x^{3} = a_{1} x (1- x) + a_{3} x^{2} (1 - x) \end{equation*} $$
\( \hat u_{x} \) es $$ \begin{equation*} \frac{d \hat u(x)}{d x} = a_{1} + 2 a_{3} x - 2 a_{1} x - 3 a_{3} x^{2} \end{equation*} $$
\( \hat u_{xx} \) es $$ \begin{equation*} \frac{d^{2} \hat u(x)}{d x^{2}} = 2 a_{3} - 2 a_{1} - 6 a_{3} x = 2(a_{3} - a_{1}) - 6 a_{3} x \end{equation*} $$
la función residual es $$ \begin{equation*} R(x) = \frac{d^{2} \hat u(x)}{d x^{2}} + x^{2} = 2(a_{3} - a_{1}) - 6 a_{3} x + x^{2} \end{equation*} $$
la función ponderada es $$ \begin{equation*} W(x) = \delta a_{1} x (1- x) + \delta a_{3} x^{2} (1 - x) \end{equation*} $$
la forma débil de la ecuación diferencial es $$ \begin{equation*} \int_{0}^{1} R(x) W(x) \ dx = \int_{0}^{1} [2(a_{3} - a_{1}) - 6 a_{3} x + x^{2}] [\delta a_{1} x (1- x) + \delta a_{3} x^{2} (1 - x)] \ dx = 0 \end{equation*} $$
puede escribirse como la suma de integrales $$ \begin{equation*} \int_{0}^{1} [2(a_{3} - a_{1}) - 6 a_{3} x + x^{2}] [\delta a_{1} x (1- x)] \ dx + \int_{0}^{1} [2(a_{3} - a_{1}) - 6 a_{3} x + x^{2}] [\delta a_{3} x^{2} (1 - x)] \ dx = 0 \end{equation*} $$
las constantes salen de la integral $$ \begin{align*} \delta a_{1} \int_{0}^{1} [2(a_{3} - a_{1}) - 6 a_{3} x + x^{2}] [x (1- x)] \ dx &= 0 \\ \delta a_{3} \int_{0}^{1} [2(a_{3} - a_{1}) - 6 a_{3} x + x^{2}] [x^{2} (1 - x)] \ dx &= 0 \end{align*} $$
multiplicando y ordenando $$ \begin{align*} \int_{0}^{1} - 2 a_{1} x + 2 a_{3} x + 2 a_{1} x^{2} - 8 a_{3} x^{2} + 6 a_{3} x^{3} + x^{3} - x^{4} \ dx &= 0 \\ \int_{0}^{1} - 2 a_{1} x^{2} + 2 a_{3} x^{2} + 2 a_{1} x^{3} - 8 a_{3} x^{3} + 6 a_{3} x^{4} + x^{4} - x^{5} \ dx &= 0 \end{align*} $$
integrando $$ \begin{align*} \bigg( - a_{1} x^{2} + a_{3} x^{2} + \frac{2}{3} a_{1} x^{3} - \frac{8}{3} a_{3} x^{3} + \frac{3}{2} a_{3} x^{4} + \frac{1}{4} x^{4} - \frac{1}{5} x^{5} \bigg) \bigg|_{0}^{1} &= 0 \\ \bigg( - \frac{2}{3} a_{1} x^{3} + \frac{2}{3} a_{3} x^{3} + \frac{1}{2} a_{1} x^{4} - 2 a_{3} x^{4} + \frac{6}{5} a_{3} x^{5} + \frac{1}{5} x^{5} - \frac{1}{6} x^{6} \bigg) \bigg|_{0}^{1} &= 0 \end{align*} $$
reemplazando límites de integración y simplificando $$ \begin{align*} - \frac{1}{3} a_{1} - \frac{1}{6} a_{3} &= - \frac{1}{20} \\ - \frac{1}{6} a_{1} - \frac{2}{15} a_{3} &= - \frac{1}{30} \end{align*} $$
resolviendo el sistema $$ \begin{align*} a_{1} &= \frac{1}{15} \\ a_{3} &= \frac{1}{6} \end{align*} $$
reemplazando en la solución aproximada $$ \begin{equation*} \hat u(x) = \frac{1}{15} x (1 - x) + \frac{1}{6} x^{2} (1 - x) = \frac{1}{15} x + \frac{1}{10} x^{2} - \frac{1}{6} x^{3} \end{equation*} $$