en forma matricial $$ \begin{equation*} u = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} = \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} \end{equation*} $$
reemplazando \( x_{1} = 0 \), \( x_{2} = \frac{L}{2} \) y \( x_{3} = L \) $$ \begin{align*} \alpha_{0} + \alpha_{1}(0) + \alpha_{2}(0)^{2} &= u_{1} \\ \alpha_{0} + \alpha_{1} \bigg( \frac{L}{2} \bigg) + \alpha_{2} \bigg( \frac{L}{2} \bigg)^{2} &= u_{2} \\ \alpha_{0} + \alpha_{1}(L) + \alpha_{2}(L)^{2} &= u_{3} \end{align*} $$
simplificando $$ \begin{align*} \alpha_{0} &= u_{1} \\ \alpha_{0} + \frac{L}{2} \alpha_{1} + \frac{L^{2}}{4} \alpha_{2} &= u_{2}\\ \alpha_{0} + L \alpha_{1} + L^{2} \alpha_{1} &= u_{3} \end{align*} $$
en forma matricial $$ \begin{align*} \begin{bmatrix} 1 & 0 & 0 \\ 1 & \frac{L}{2} & \frac{L^{2}}{4} \\ 1 & L & L^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{align*} $$
resolviendo $$ \begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -\frac{3}{L} & \frac{4}{L} & -\frac{1}{L} \\ \frac{2}{L^{2}} & -\frac{4}{L^{2}} & \frac{2}{L^{2}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*} $$
reemplazando las incógnitas $$ \begin{align*} u &= \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} \\ &= \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ -\frac{3}{L} & \frac{4}{L} & -\frac{1}{L} \\ \frac{2}{L^{2}} & -\frac{4}{L^{2}} & \frac{2}{L^{2}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{3}{L} x + \frac{2}{L^{2}} x^{2} & \frac{4}{L} x - \frac{4}{L^{2}} x^{2} & -\frac{1}{L} x + \frac{2}{L^{2}} x^{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} & N_{2} & N_{3} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{align*} $$
Reescribiendo \( u \) $$ \begin{equation*} u = \bigg( 1 - \frac{3}{L} x + \frac{2}{L^{2}} x^{2} \bigg) u_{1} + \bigg( \frac{4}{L} x - \frac{4}{L^{2}} x^{2} \bigg) u_{2} + \bigg( -\frac{1}{L} x + \frac{2}{L^{2}} x^{2} \bigg) u_{3} \end{equation*} $$