$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\halfi}{{1/2}}
\newcommand{\xpoint}{\boldsymbol{x}}
\newcommand{\normalvec}{\boldsymbol{n}}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\It}{\mathcal{I}_t}
\newcommand{\setb}[1]{#1^0} % set begin
\newcommand{\sete}[1]{#1^{-1}} % set end
\newcommand{\setl}[1]{#1^-}
\newcommand{\setr}[1]{#1^+}
\newcommand{\seti}[1]{#1^i}
\newcommand{\Real}{\mathbb{R}}
$$
Analytical work with the discrete equations (1)
We want to show that \( \uex \) also solves the discrete equations!
Useful preliminary result:
$$
\begin{align}
\lbrack D_tD_t t^2\rbrack^n &= \frac{t_{n+1}^2 - 2t_n^2 + t_{n-1}^2}{\Delta t^2}
= (n+1)^2 -n^2 + (n-1)^2 = 2\\
\lbrack D_tD_t t\rbrack^n &= \frac{t_{n+1} - 2t_n + t_{n-1}}{\Delta t^2}
= \frac{((n+1) -n + (n-1))\Delta t}{\Delta t^2} = 0
\end{align}
$$
Hence,
$$ [D_tD_t \uex]^n_i = x_i(L-x_i)[D_tD_t (1+{\half}t)]^n =
x_i(L-x_i){\half}[D_tD_t t]^n = 0$$
