$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\halfi}{{1/2}}
\newcommand{\xpoint}{\boldsymbol{x}}
\newcommand{\normalvec}{\boldsymbol{n}}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\It}{\mathcal{I}_t}
\newcommand{\setb}[1]{#1^0} % set begin
\newcommand{\sete}[1]{#1^{-1}} % set end
\newcommand{\setl}[1]{#1^-}
\newcommand{\setr}[1]{#1^+}
\newcommand{\seti}[1]{#1^i}
\newcommand{\Real}{\mathbb{R}}
$$
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 \( \uex(x_i,t_n) \).
