$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\sequencei}[1]{\left\{ {#1}_i \right\}_{i\in\If}}
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\psib}{\boldsymbol{\psi}}
\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\Xno}[1]{X_{(#1)}}
\newcommand{\xdno}[1]{\boldsymbol{x}_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
How to deal with the boundary conditions?
Since \( u(0)=0 \) and \( u(L)=0 \), we must force
$$ v(0)=v(L)=0,\quad \baspsi_i(0)=\baspsi_i(L)=0$$
Now we choose the finite element basis: \( \baspsi_i=\basphi_i \), \( i=0,\ldots,N_n \)
Problem: \( \basphi_0(0)\neq 0 \) and \( \basphi_{N_n}(L)\neq 0 \)
Solution: we just exclude \( \basphi_0 \) and \( \basphi_{N_n} \)
from the basis and work with
$$ \baspsi_i=\basphi_{i+1},\quad i=0,\ldots,N=N_n-2$$
Introduce index mapping \( \nu(i) \): \( \baspsi_i = \basphi_{\nu(i)} \)
$$ u = \sum_{j\in\If}c_j\basphi_{\nu(j)},\quad i=0,\ldots,N,\quad \nu(j) = j+1$$
Irregular numbering: more complicated \( \nu(j) \) table
