$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\uexd}[1]{{u_{\small\mbox{e}, #1}}}
\newcommand{\vex}{{v_{\small\mbox{e}}}}
\newcommand{\vexd}[1]{{v_{\small\mbox{e}, #1}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\halfi}{{1/2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Ddt}[1]{\frac{D #1}{dt}}
\newcommand{\E}[1]{\hbox{E}\lbrack #1 \rbrack}
\newcommand{\Var}[1]{\hbox{Var}\lbrack #1 \rbrack}
\newcommand{\Std}[1]{\hbox{Std}\lbrack #1 \rbrack}
\newcommand{\xpoint}{\boldsymbol{x}}
\newcommand{\normalvec}{\boldsymbol{n}}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\w}{\boldsymbol{w}}
\newcommand{\V}{\boldsymbol{V}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\F}{\boldsymbol{F}}
\newcommand{\stress}{\boldsymbol{\sigma}}
\newcommand{\strain}{\boldsymbol{\varepsilon}}
\newcommand{\stressc}{{\sigma}}
\newcommand{\strainc}{{\varepsilon}}
\newcommand{\I}{\boldsymbol{I}}
\newcommand{\T}{\boldsymbol{T}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\ii}{\boldsymbol{i}}
\newcommand{\jj}{\boldsymbol{j}}
\newcommand{\kk}{\boldsymbol{k}}
\newcommand{\ir}{\boldsymbol{i}_r}
\newcommand{\ith}{\boldsymbol{i}_{\theta}}
\newcommand{\iz}{\boldsymbol{i}_z}
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\It}{\mathcal{I}_t}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\setb}[1]{#1^0} % set begin
\newcommand{\sete}[1]{#1^{-1}} % set end
\newcommand{\setl}[1]{#1^-}
\newcommand{\setr}[1]{#1^+}
\newcommand{\seti}[1]{#1^i}
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\newcommand{\psib}{\boldsymbol{\psi}}
\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
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\newcommand{\Real}{\mathbb{R}}
\newcommand{\Integerp}{\mathbb{N}}
\newcommand{\Integer}{\mathbb{Z}}
$$
Analytical work with the discrete equations (1)
$$
\begin{align*}
\lbrack D_xD_x \uex\rbrack^n_i &=
(1+{\half}t_n)\lbrack D_xD_x (xL-x^2)\rbrack_i =
(1+{\half}t_n)\lbrack LD_xD_x x - D_xD_x x^2\rbrack_i \\
&= -2(1+{\half}t_n)
\end{align*}
$$
Now, \( f^n_i = 2(1+{\half}t_n)c^2 \) and we get
$$ [D_tD_t \uex - c^2D_xD_x\uex - f]^n_i = 0 - c^2(-1)2(1 + {\half}t_n
+ 2(1+{\half}t_n)c^2 = 0$$
Moreover, \( \uex(x_i,0)=I(x_i) \),
\( \partial \uex/\partial t = V(x_i) \) at \( t=0 \), and
\( \uex(x_0,t)=\uex(x_{N_x},0)=0 \). Also the modified scheme for the
first time step is fulfilled by \( \uex(x_i,t_n) \).