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